Basic Principles and Calculations .n - UW Digital

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Basic Principles and Calculations • .n Chemical Engineering

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David M. Himmelblau -ith Special Applications to

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Professor of Chemical Engineering Unil'ersity of Texas

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'p'meering: Vol. 3; Process Afodeling,

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Introduction to Engineering Calculations

1

The allocation of resources in the United States and elsewhere is beginning a period of transition. Air and water, for example, at one time were the economist's favorite examples of "free goods." However, people are accepting the fact that air and water indeed are natural resources of ever-increasing value and are not simply infinite sources and sinks for man's uses in production and consumption. For you to learn how to appreciate and treat the problems that will arise in our modern technology, and especially in the technology of the future, it is necessary to learn certain basic principles and practice their application. This text describes the principles of making material and energy balances and illustrates their application ina wide variety of ways. We begin in this chapter with a review of certain background information. You have already encountered most of the concepts to be presented in basic chemistry and physics courses. Why, then, the need for a review? First, from experience we have found it necessary to restate these familiar basic concepts in a somewhat more general and clearer fashion; second. you will need practice to develop your ability to analyze and work engineering problems. To read and understand the principles discussed in this chapter is relatively easy; to apply them to different unfamiliar situations is not. An engineer becomes competent in his profession by mastering the techniques developed by his predecessorsthen, perhaps, he ca n pioneer new ones. 1

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Engineering Ca/cli/alions

Chap. J

This chapter begins with a discussion of units, dimensions, and conversion factors, and then goes on to review some terms you should already be acquainted with, such as (a) (b) (c) (d) (e)

Mole and mole fraction. Density and specific gravity. Measures of concentration. Temperature. Pressu reo

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It then provides some clues as to "how to solve problems" which should be of material aid in all the remaining portions of your work. Finally, the principles of stoichiometry are reviewed, and the technique of handling incomplete reactions is illustrated. Figure 1.0 shows the relation of the topics to be discussed to

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Fig. 1.0. Heirarchy of topics to be studied in this chapter (section numbers are in the upper left·hand corner of the boxes).

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other and to the ultimate goal of being able to solve problems involving material and energy balances. At the end of the chapter you will find a dlC(~ list of levels of skill you should possess after completing the chapter.

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1.1 Units and dimensions

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Chemical Equation and Stoichiometry

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..\t some time in every student's life comes the exasperating sensation of frustrat,nn in problem solving. Somehow. the answers or the calculations do not come ('lit as expected. Often this outcome arises because of inexperience in the hand'Ing of units. The use of units or dimensions along with the numbers in your , .• klliations requires more attention than you probably have been giving to \,)ur computations in the past, but it will help avoid such annoying experiences. '1 he proper use of dimensions in problem solving is not only sound from a logical viewpoint-it will also be helpful in guiding you along an appropriate path of analysis from what is at hand through what has to be done to the final solution. Dimensions are our basic concepts of measurement such as length, time, mass, temperatllre, etc; lin its are the means of expressing the dimensions, as/eet or centimeters for length. or hOllrs or seconds for time. Units are associated with some quantities you may have previously considered to be dimensionless. A good example is lIloleclilar weight, which is really the mass of one substance per mole of that substance. This method of attaching units to all numbers which are not fundamentally dimensionless has the following very practical benefits: (a) It diminishes the possibility of inadvertent inversion of any portion of the calculation. (b) It reduces the calculation in many cases to simple ratios, which can be easily manipulated on the slide rule. (c) It reduces the intermediate calculations and eliminates considerable time in problem solving. (d) It enables you to approach the problem logically rather than by remembering a formula and plugging numbers into the formula. (e) It demonstrates the physical meaning of the numbers,You use.

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The rule for handling units is essentially quite simple: treat the units as you would algebraic symbols. For example, you cannot add, subtract, multiply, or divide different units into each other and thus cancel them out-this may be done only for like units. You may be able to add pounds to pounds and calories to calories-subtract them, multiply them, or divide them-but you cannot divide 10 pounds by 5 calories and get 2 any more than you can change 2 apples into 2 bananas. The units contain a significant alllount of information content that cannot be ignored.

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Introduc/ioll 10 Engineering Calculations

EXAMPLE 1.1

Chap. J

Dimensions and units

Add the following: (a)

1 foot + 3 seconds

(b)

1 horsepower + 300 watts

Solution: The operation indicated by

Ift+3sec has no meaning since the dimensions of the two terms are not the same. One foc: has the dimensions of length, whereas 3 sec has the dimensions of time. In the case or I hp + 300 watts the dimensions are the same (energy per unit time) but the units are different. You must transform the two quantities into like units, such as horsepower, watts, or something else, before the addition can be carried out. Since 1 hp = 746 watts, 746 watts + 300 watts

=

1046 watts

EXAMPLE 1.2 Conversion of units If a plane travels at twice the speed of sound (assume that the speed of sound is 1100 ft/sec), how fast is it going in miles per hour?

Solution: 2

60 min I hr

1500 mi hr

or mi 60 _2+--1..c.1_00_ft+_h"r_ = 1500 mi sec 88~ hr sec

You will note in Example 1.2 the use of what is called the dimensional equation. It contains both units and numbers. The initial speed, 2200 ft!sec. is multiplied by a number of ratios (termed conversion/aclors) of equivalent val ue, of combinations of time, distance, etc., to arrive at the final desired answer The ratios used are simple well-known values and thus the conversion itself should present no great problem. Of course, it is possible to look up conversion ratios, which will enable the length of the calculation to be reduced; for instance. in Example 1.2 we could have used the conversion factor of 60 mi/hr equals 88 ft/sec. However, it usually takes less time to use values you know than te look up shortcut conversion factors in a handbook. Common conversion ratios are listed in Appendix A.

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Units and Dimensions 5

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The dimensional equation has vertical lines set up to separate each ratio, .",J these lines retain the same meaning as an / or multiplication sign placed

.:ore not the same. One foot :,ions of time. In the case of :ne units are different. You . as horsepower, watts, or "mee I hp = 746 watts,

',:~"ccn each ratio, The dimensional equation will be retained in this form ., r,'lIghout most of this text to enable you to keep clearly in mind the signifi.• ,"c of units in prohlem solving. It is recommended that units always be .. ,;lIen down next to the associated numerical value (unless the calculation is \en' simple) until you become quite familiar with the use of units and dimen'.\)~, and can carry them in your head. At any point in the dimensional equation you can determine the consoli,i.lled net units and see what conversions arc still required, This may be carried ,.,,1 r,mnally. as shown below by drawing slanted lines below the dimensional equation and writing the consolidated units on these lines, or it may be done by C)C. mentally canceling and accumulating the units.

2 x 1100 ft

60 sec

sec

60 min I hr

_'us

EXAMPLE 1.3 Use of units :!!

that the speed of sound is

Change 400 in,'jday to cm'jmin.

Solution: 400 in.' day 15(

I e,541 cm)' 1124dayhr I 601min hr In.

=

4 56 cm' . min

In this example note that not only lre the numbers raised to a power, but the units also are raised to the same power.

There shaH be one measure of wine throughout our kingdom, and one measure of ale, and one measure of grain ... and one breadth of cloth .... And of weights it shall be as of measures. is called the dimensional :uial speed, 2200 fll/sec, is :::tors) of equivalent values the final desired answer. :nus the conversion itself :ole to look up conversion ) be reduced; for instance, ractor of 60 mi/hr equals values you know than to Dmmon conversion ratios

So reads the standard measures clause of the Magna Carta (June 1215). The standards mentioned were not substantially revised until the nineteenth century. When the American Colonies separated from England, they retained, among other things, the weights and measures then in use. It is probable that at that time Ihese were the most firmly established and widely used weights and measures in the world. No such uniformity of weights and measures existed on the European continent. Weights and measures differed not only from country to country but even from town to town and from one trade to another. This lack of uniformity led the National Assembly of France during the French Revolution to enact a decree (May 8, 1790) that called upon the French Academy of Sciences to act in concert with the Royal Society of London to "deduce an invariable standard

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6

Introduction to Engineering Calclilations

Chap. J

for all of the measures and all weights." Having already an adequate system of weights and measures, the English did not participate in the French undertaking. The result of the French endeavor has evolved into what is known as the metric system. The metric system became preferred by scientists of the nineteenth century, partly because it was intended to be an international system of measurement, partly because the units of measurement were theoretically supposed to be independently reproducible, and partly because of the simplicity of its decimal nature. These scientists proceeded to derive new units for newly observed physical phenomena, basing the new units on elementary laws of physics and relating them to the units of mass and length of the metric system. The Americans and the British adapted their system of measurements to the requirements of new technology as time went on for both business and commerce as well as scientific uses despite the fact that the other countries, one after another, turned toward the metric system. Problems in the specification of the units for electricity and magnetism led to numerous international conferences to rectify inconsistencies and culminated in 1960 in the eleventh General Conference on Weights and Measures adopting the SI (Systeme International) system of units. As of this date (1973) the United States is the last large country not employing or engaged in transforming to some form of the S1 units. Table 1.1 shows the most common systems of units used by engineers in the last few decades. Note that the S1, the cgs, the fps (English absolute), and the British engineering systems all have three basically defined units and that the fourth unit is derived from these three defined units. Only the American engineering system has four basically defined units. Consequently, in the American engineering system you have to use a conversion factor, g" a constant whose numerical value is not unity, to make the units come out properly. We can use Newton's Jaw to see what the situation is with regard to conversion of units:

F= Cma where' F

=

(1.1)

force

m = mass a = acceleration C

=

a constant whose numerical value and units depend on those selected for F, m, ar
In the cgs system the unit of force is defined as the dyne; hence if C is selected to be C = I dynejCg)(cm)/sec 2 , then when I g is accelerated at I cm/sec 2 F

=

J dyne (g)(cm)

Ig

I c~ = I dyne sec·

"""SeC2 'A lisl of the nomenclature is included at 'he end of the book.

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an adequate system of Jate in the French underved into what is known as

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by -engineers in ; (English absolute), and the ,- defined units and that the :s. Only the American cngi-:lSequently, in the American 'actor, g,. a constant whose -:Je out properly. We can use ::ard to conversion of units:

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Chap. 1

Introduction to Engineering Calculations

Similarly, in the SI system in which the unit of force is defined to be the newton (N), if C = I N/(kg)(m)!sec', then when I kg is accelerated at I mlsec', F=

I N (kg)(m) sec 2

I kg

I miN sec'

However, in the American engineering system we ask that the numerical value of the force and the mass be essentially the same at the earth's surface. Hence', if a mass of I lb~ is accelerated at g ft/sec" where g is the acceleration of gravity (about 32.2 ftisec' depending on the location of the mass), we can make the force be I lb f by choosing the propcr numerical value and units for C:

F = C I Ibm

g ft = I Ib

sec '

,

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(1.2)

Observe that for Eq. (1.2) to hold, the units of C have to be

Co<>

Ib,

Ibm

(~) sec'

A numerical value of 1/32.174 has been chosen for the constant because g equals the numerical value of the average acceleration of gravity at sea level at 45° latitude when the latter is expressed in fUsec ' . The acceleration of gravity, you may recall, varies a few tenths of I percent rrom place to place on the surface of the earth and changes considerably as you rise up from the surface as in a rocket. With this selection of units and with the number 32.174 employed in the denominator of the conversion factor, the inverse of C is given the special symbol gc: (1.3) Division by gc achieves exactly the same result as multiplication by C. You can see, therefore, that in the American engineering system we have the convenience that the numerical value of a pound mass is also that of a pound force if the numerical value of the ratio glgc is equal to I, as it is approximately in most cases. Similarly, a l-Ib mass also usually weighs I Ib (weight is the force required to support the mass at rcst). However, you should be aware that these two quantities g and gc are not the same. Note also that the pound (mass) and the pound (force) are not same units in the American engineering system even though we speak of pounds to express force. weight. or mass. Furthermore, if a satellite with a mass of lIb (mass) "weighing" I Ib (force) on the earth's surface is shot up to a height of 50 mi, it will no longer "weigh" I Ib (force). although its mass will still be I lb (mass). In ordinary language most people. including scientists and engineers. omit the designation of "forcc" or "mass" associated with tbe pound or kilogram but pick up the meaning from the context of the statement. No one gets confused by the fact that a man is six feet tall but has only two feet. Similarly, onc should

'

• Chap. J

. defined to be the- newton c:-ated at I mjsec Z,

;'e ask that the numerical ':ne at the earth's surface. '~re g is the acceleration of :-.f the mass), we can make value and units for C:

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Units and Dimensions 9

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":arret the statement that a bottle "weighs" 5 kg as meaning the bottle has a ':.1>; of 5 kg and is attracted to the earth's surface with a force equal to a weight .·f 5 kg. Additional details concerning units and dimensions can be found in :~c articles by Silberberg and McKetta, Z and by Whitney, J in various books, 4. ',6 .lnd in the references at the end of this chapter. For systems used in the t. S. S. R., consult J'vleas. Tech. (USSR) (English TrailS.), no. 10 (April), 1964.

! ,"'IPLE 1.4

Use of g,

One hundred pounds of water are flowing through a pipe at the rate of 10.0 ft/sec. What is the kinetic energy of this water in (ft)(lb f )?

(1.2)

SO/II/;on: kinetic energy = K =

.'e to be

tri"

l:mv 2

Assume the 100 Ib of water means the mass of the water. (This would be numerically identical to the weight of the water if g = g,.) K = 1 100lbm 2

~

constant because g equals cravity at sea level at 45c:=celf'r···;')n of gravity, you .~ tc )n the surface of . fro," .. e surface as in a - oer 32.174 employed in the . is given the special symbol

C\,\:\lPLE 1.5 Use of g,

What is the potential energy in (ft)(Ibf ) ofa 100-lb drum hanging 10ft above the 'llrface of the earth with reference to the surface of the earth? SO/lIt;OIl: potential energy = P = mgh

(1.3) ::rltiplication by C. You can we have the convenience -:at of a pound force if the :"proximately in most cases. =nt is the force required to ware that these two quanti:mnd (mass) and the pound :-mg system even though we :rthermore, if a satellite with ,arth's surface is shot up to a J_ although its mass will still

Assume the 100lb means 100 Ib mass; g = acceleration of gravity = 32.2 ft!sec'.

=

:lentists and engineers. omit , the pound or kilogram but -:lent. No one gets confused [) feet. Similarly, olle should

p=IOOlbm

32.2 ft

sec'

10 ft

(ft)(lb) = 1000 (ft)(Ibf ) 32.174 (seC 2 )o'b ) f

~"lice

that in the ratio of gig" or 32.2 ftisec' divided by 32.174 (ft/sec 2 )(lb m /lb f ), ;he numerical values are almost equal. A good many people would solve the problem ~; saying 100lb x 10 ft = 1000 (ft)(Ib) without realizing that in effect they are canceling out the numbers in the gig, ratio. 'I. H. Silberberg and J. J. McKella, Jr., Petrolellm Refiner, v. 32, 179-183 (April 1953), Hi-150 (May 1953). 'H. Whitney, "The Mathematics of Physical Quantities," American Math. Monthly, v. 75, 115. 227 (1968). 4D. C. Ipsen. Units, Dimensions, and Dimensionless Numbers. McGraw-HilI, New York, 1%0. 'So J. Kline, Similitude and Approximation Theory, McGraw-Hili, New York, 1965. 'E. F. O'Day, Physical Quamities "lid Units (A Self-Instructiollal Programmed Mallua/), I'r~ntice-Hall, Englewood Clilfs, N. J., 1967.

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Introduction to Engineering Calculations

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EXAMPLE 1.6 Weight What is the difference in the weight in Newtons of a lOO-kg rocket at height of =. 9.76 m/sec 2 , as opposed to its weight o!l the surface of the earth, where g = 9.80 m/sec 2 ? 10 km above the surface of the earth, where g

Solution: The weight in Newtons can be computed in each case from Eq. (1.1) with a =!: if we ignore the tiny effect of centripetal acceleration resulting from the rotation of the earth (less than 0.3 percent): . h d'ff,

welg t

I

erence

=

100 kg

(9.80 - 9.76) m sec'

IN (kg)(m) = 4.00 N ~

Note that the concept of weight is not particularly useful in treating the dynamics of long-range ballistic missiles or of earth satellites because the earth is both round and rotating.

You should develop some facility in converting units from the S1 and cgs systems into the American engineering system and vice versa, since these are the three sets of units in this text. Certainly you are familiar with the common conversions in the American engineering system from elementary and high school work. Similarly, you have used the S1 and cgs system in physics, chem· istry, and mathematics. If you have forgotten, Table 1.2 lists a short selection of essential conversion factors from Appendix A. Memorize them. Common abbreviations also appear in this table. Unit abbreviations are written in lowercase letters, except for the first letter when the name of the unit is derived from a proper name. The distinction between uppercase and lowercase letters should be followed, even if the symbol appears in applications where the other lettering is in uppercase style. Unit abbreviations have the same form for both singular and plural, and they are not followed by a period (except in the case of inches). Other useful conversion factors will be discussed in subsequent sections of this book. One of the best features of the S1 system is that units and their multiples and submultiples are related by standard factors designated by the prefixes indicated in Table 1.3. Prefixes are not prefened for use in denominators (except for kg). The strict use of these prefixes leads to some amusing combinations of noneuphonious sounds, such as nanonewlOn, nembujoule. and so forth. Also. some confusion is certain to arise because the prefix M can be confused with m as well as with M ~ 1000 derived from the Roman numerical. When a COIllpound unit is formed by multiplication of two or more other units. its symbol consists of the symbols for the separate unnts joined by a centered dot (for example, N· m for newton meter). The dot may !be omiued in the case of familiar units such as watthour (symbol Wh) if no conflUsion will result. or if the symbols are separated by exponents as in N . m'kg-'. Hyphens should not be used in symbols

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12 Introduction to Engineering Calculations TABLE

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Chap. I

1.2

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1.3 51 PREFIXES

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Factor

Prefix

Symbol

10 12 1010' 10) 10 2

.Tera Giga Mega Kilo Hecto' Deka'

T G M k h da

10 1

Factor

Prefix

Symbol

10- 1

Deci' Centi' Milli Micro Nano Pico Femto Atto

d c m

10- 2 10-) 10-' 10--

10- 1l 10-1>

10- 18

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f a

• Avoid except for areas and volumes.

Dimension.11 ,,f terms or qua'll ,,'n\l~tent: i,~ .. CI

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for compound units. Positive and negative exponents may be used with the symbols for units. If a compound unit is formed by division of one unit by another, its symbol consists of the symbols for the separate units either separated by a solidus or multiplied by using negative powers (for example, mjs or m· S-1 for meters per second). We shall not use the center dot for multiplication in this text but shall use parentheses around the unit instead, a procedure less subject to misinterpretation in hand written material.

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EXAMPLE 1.7

.He

Application of dimensions

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'.In ~ •

11,,, C I he

A simplified equation for heat transfer from a pipe to air is

11= 0.026Go.6'Do .•

where h G D

= the heat transfer coefficient in Btu/(hr)(ftl)(°F) = mass rate of flow in Ib m /(hr)(ft 2 ) = outside diameter of the pipe in ft

If h is to be expressed in cal/(min)(cm')CC), what shf!>IUld the new constant in the equation be in place of 0.026? Solution: To convert to cal/(min)(cm 2 )(0C) we set up the din1el!1sional equation as follows:

7.2 The

mc~

1~l1rln,~ti{';'

... ~

',:~~t\

_,),1"', ,

! ".",

d :. f

I ft ) 1 12 in.

I(

I in \2 /1.8°F._: _4 GO.' cal \2.54cm/ 1°C - 2.l~ x 10 Do.4(min)(cm'jCC)

~

-, l ' i

., l.

..

(Note: Sec Sec. 1.4 for a detailed discussion of tel1lllperature conversion factors.) If G and D are to be used in units of G"

g . (min)(cm')

D':cm

~

I, ,

I

4ft

Chap. I

L.

...

t

a*s

hr.

. .......... -w.:...... aM

t-d'

.'..

2tt

.'$'

n

'Set



1ft 1'.

The Mole Unit

1.2

\r,

'''~

Prefix

tt t

t

ttra

13

an additional conversion is required:

Symbol

Deci-

d

CenliMilli Micro Nano Pieo Femlo Alto

e m

---:----,I,---~-'I-:-::---=--:-0.4 : D'(cm)

C.~ln~m) I C~ r~,)

=

_, (G')0.6cal 1.4 8 x 10 • (D')O 4(min)(em' WC)

p

n p

f

a

""1ts may be used with the ,Ivision of one unit by an,rate units either separated 70r example, m/s or m· S-I ,t for multiplication in this .j, a' '''dure less subject

Dimensional considerations can also be used to help identify the dimensions Equations must be dimensionally "onsistcnt; i.e., each term in an equation must have the same net dimensions and units as every other term to which it is added or subtracted. The use of dimen'I('nal consistency can be illustrated by an equation which represents gas ":havior and is known as van der Waals' equation, an equation which will be ,:Iscussed in more detail in Chap, 3:

<,r terms or quantities in terms in an equation.

(p + :,)(V -- b) = RT Inspection of the equation shows that the constant a must have the dimensions of [(pressure)(volume)'] in order for the expression in the first set of parentheses to be consistent throughout. I f the units of pressure are atm and those of volume are cm', then a will have the units specifically or [(atm)(cm)6]. Similarly, b must have the same units as V, or in this particular case the units of cm'.

-!e to air is "')

1.2 The mole unit

'mid the new constant in the

':1cnsional equation as follows:

GO .• cal

4 II x 10. =--.::'7-"=="",,,, Do.4(min)(em")('C)

':nperature conversion factors.)

What is a mole? The best answer is that a mole is a certain number of molecules atoms, electrons, or other specified types of particles.' In particular, the 1969 International Committee on Weights and Measures approved the mole (sometimes abbreviated as mol) in the SI system as being "the amount of a substance that contains as many elementary entities as there are atoms in 0.012 kg of carbon 12." Thus in the SI and cgs systems the mole contains a different number of molecules than it does in the American engineering system. In the SI and cgs systems a mole has about 6.02 x 10" molecules: we shall call this a gram mole (abbreviated g mole). In the American engineering system a pOllnd mole (abbreviated lb mole) has 6.02 >< 10" x 454 molecules. Thus the pound mole and the gram mole represent two different quantities. Here is another way to look at 'For a discussion of the requirements of the mole concept, refer to the series of articles inJ. Clrem. Edt/c., v. 38, 549-556 (1961).

I ! .)

I 1 J

I 1

14

Introduction to Engineering Calculations

Chap. I

the mole unit: the

g

,, ,

mole =

mass in g molecular weIght

(I.4)

=

mass in Ib molecular weight

(1.5)

the Ib mole or

mass in g = (mol. wt)(g mole) mass in Ib

= (mol.

wt)(lb mole)

(1.6)

(/.7)

The values of the molecular weights (relative molar masses) are built up from the tables of atomic weights based on an arbitrary scale of the relative masses of the elements. The terms atomic "weight" and molecular "weight" are universally used by chemists and engineers instead of the more accurate terms atomic "mass" or molecular "mass." Since weighing was the original method for determining the comparative atomic masses, as long as they were calculated in a common gravitational field, the relatilVe values obtained for the atomic "weights" were identical with those or the atomic "masses." There have always been and are today some questions in issue about the standard reference element and its atomic weight. Many of the old Silandards such as H (atomic hydrogen) = 1, 0 (atomic oxygen) = 100, 0 = land 0 = 16 have been discarded, and now I2C = 12 exactly is used as the reference point on the chemist's scale, i.e .. 12.000 g of carbon 12 contain I g mole or 6.023 x 1023 atoms (or 0.012 kg in the SI system). On this scale of atomic weights, hydrogen is 1.008, carbon is 12.01, etc. (In most of our calculations we shall round these off to I and 12, respectively.) In your calculations you may attach any unit of mass you desire to these atomic weights, as, for example, a gram atom, a pound atom, a ton atom, etc. Thus a gram atom of oxygen is 16.00 g, a pound atom of oxygen is 16.00 Ib, and a kilogram atom of hydrogen is 1.008 kg. A compound is composed of more than one atom, and the molecular weight of the compound is nothing more than the sum of the weights of the atoms of which it is composed. Thus H 2 0 consists of 2 hydrogen atoms and I oxygen atom, and the molecular weight of water is (2)(1.008) + 16.000 = 18.02. Thc;~e weights are all relative to the 12C atom as 12.0000. and again. any mass unit can be attached to the molecular weight of water, such as 18.02 gig mole, 18.02 Ibjlb mole, etc. You can compute average molecular weights for mixtures of constant composition even though they are not chemically bonded if their compositions arc known accurately. Thus later Oll in Example 1.12 we shall show how to calculate the average molecular weight of' air. Of course, for a material stich as fuel oil or coal whose composition may not be exactly known, you cannot determine 3n exact molecular weight, although you might estimate an approximate average molecular weight good enough for engineering calculations.

eo

Chap. 1



•• ai

V

¥

o.

1

roe.

..



*

t

t

Conventions in Methods of Anaiysis and Measurement

'ire 13

tM

,

UW

15

• '\.,"I'LE 1.8 Molecular weights

(1.4)

=

If 3 bucket holds 2.00 Ib of NaOH (mol. wi

40.0), how many

Pound moles of NaOH ~oes it contain? (b) Gram moles of NaOH does it contain?

(a)

(\.5)

Solution:

(\.6) (\.7) ~ses)

are built up :e of the relative :C:llar "weight" are -"Te accurate terms -Iginal method for were calculated in -j f( atomic '7hel ; always Teference elemen t 'atomic hydrogen) ~n discarded, and -nemist's scale, i.e., -ms (or 0.012 kg in

Basis: 2.00 1b NaOH

..

2.00 Ib NaOH II Ib mole NaOH 40.0 lb NaOH

\

2.00 Ib NaOH

=

0050 Ib

I Ib mole NaOH 454 g mole 40.0 lb NaOH ,lib mole

2.00 Ib NaOH 1 454 g 1 Ib

I N OH mo e a

.

11 g mole NaOH 40.0 g NaOH

=

= 227

. g

mole

I . g mo e

227

lXA:'I.tPLE 1.9 l'.loiecular weights

How many Ib of NaOH are in 7.50 g mole of NaOH?

'0

-:'Tbon is 12.01, etc. 12, respectively.) osire to these atomic -:1 atom, etc. Thus a is 16.00 lb, and a

Soilllion:

I

Basis: 7.50 g

7.50 g moles NaOH I Ib mole 7 54 g moles ----"'-----+--4

mol~

NaOH

40 Ib NaOH I Ib mole NaOH = 0.66 lb NaOH

-~d

'ne molecular weight

:!hts of the atoms of 2toms and I oxyg(~n .000 = 18.02. Thcse 'n. any 'mass unit can g!g mole, 18.02 lb/lb ',:res of constant com;eir compositions arc -now how to calculate -lal such as fuel oil or car etermllle an apt late average

ms.

i3 Conventions in methods of analysis and measurement There arc certain definitions and conventions which we should mention at this :ime since they will be constantly used throughout this book. If you memorize

:"cm now. you will immediately have a clearer perspective and saye considerit1le trouble latcr on .

1.3.1 Density. Density is the ratio of mass per unit volume, as, for "ample, gram/cm' or Ib'ft J It has both a numerical value and units. To deternunc the density of a substance, you must find both its volume and its mass or "~;eht. If the substance is a solid. a common method to determine its volume is :" dIsplace a measured quantity of inert liquid. For example. a known weight :' a nuterial can be placed into a container of liquid of known weight and ','Iume. and the final weight and volume of the combination measured. The ,~cnsity (or specific gravity) ofa liquid is commonly measured with a hydrometer IJ known weight and volume is dropped into the liquid and the depth to which it

1

I I

16 Introductioll to Engineering Calculations

Chap. I

penetrates into the liquid is noted) or a Westphal balance (the weight of a known slug is compared in the unknown liquid with that in water). Gas densities are quite difficult to measure; one device used is the Edwards balance. which com· pares the weight of a bulb filled with air to the same bulb when filled with the unknown gas. In most of your work using liquids and solids, density will not change very much with temperature or pressure, but for precise measurements for common substances you can always look up in handbook the variation of density with temperature. The change in density with temperature is illustrated in Fig. 1.1

.,

E

~ 0.7

r - -__ I"

I

I

Temperature, ·C Fig. 1.1. Densities of liquid H 2 0 and NH, as a function of temperature.

for liquid water and liquid ammonia. Figure 1.2 illustrates how density also varies with composition. In the winter you may put antifreeze in your car radiator. The service station attendant checks the concentration of antifreeze by measuring the specific gravity and, in elTect, the density of the radiator solution after it is mixed thoroughly. He has a little thermometer in his hydrometer kit in order to be able to read the density at the proper temperature.

1.3.2 Specific Gravity. Specific gravity is commonly thought of as a dimensionless ratio. Actually, it should be considered as the ratio of two densities-that of the substance of interest, A, to that of a reference substance. In symbols: _

'f]

sp gr - specl

lC

't _ (Ib/ft'), gravl y - (lbjj't ') ref

=

(g/cm'),.

(I g em ') ref

(kg/m'), g,ffi J) lef

= (k '

(1.8)

The reference substance for liquids and solids is normally water. Thus the specific gravity is the ratio of the density of the substance in question to the density of water. The specific gravity of gases frequently is referred to air, but may be referred to other gases, as will be discussed in more detail in Chap. 3.

.,

,

. ,j i

-

,.,.j

4

j't".'.... ·

Chap. I

tN-'.

3t

t

,e"t

••

tit

Conventions in Methods of Analysis and Measurement

17

:ance (the weight of a known m water). Gas densities are :1wards balance. which com;Je bulb when filkd with the density will not change very " measurements for common :ne variation of density with ::ure is illustrated in Fig. 1.1

, 10·C

j

400C

j

O.7501;---n'o----;f.----,&----,,';;;-------;'

°

Composilion, weight fraction alcohol

Fig. 1.2. Density of a mixture of ethyl alcohol and water as a function of composition. ----1.

~

=ction of temperature. ~

illustrates how density also -; put antifreeze in your car :::oncentration of antifreeze by =nsity of the radiator solution 'meter in his hydrometer kit in - lemperature. ~s commonly thought of as a :-lsidered as the ratio of two mat of a reference substance.

::m 3 )A' _

(kg/m 3 )A

::ro 3 )"r - (kg/m'),·,r

I

Liquid density can be considered to be nearly independent of pressure for most n)mmon calculations, but, as just mentioned, it varies somewhat with teml'erature; therefore, to be very precise when referring to specific gravity, state the temperature at which each density is chosen. Thus

80 ,>OJ 100

sp gr

=

1

"In be interpreted as follows: the specific gravity at 20°C referred to that of water

"t ~ Cis 0.73. Since the density of water at 4°C is very close to 1.0000 in the cgs ·\Stcm, the numerical values of the specific gravity and density in this system are essentially equal. Since densities in the American engineering system are expressed in Ib/ft 3 and the density of water is about 62.4lb/ft 3 , it can be seen that the '['ccific gravity and density values are not numerically equal in the American on~ineering system. I n the petroleum industry the specific gravity of petroleum products is usually reported in terms of a hydrometer scale called °API. The equation for the API scale is

(1.8)

is normally water. Thus the substance in question to the !.Quently is referred \0 air, but ~d in more detail in Chap. 3.

J

20°

0.73~

(1.9)

:5

J,

I

I

I !

or

60° 141.5 sp gr6O' = °API -I- 131.5

! (1.10)

lI

i >

,;;au,;: i4Q

I j

»

.v •• _ " ..... ' , · · · , - ..<.I~""

-

. ,»~,,,,, _ _ _ , __ ,

_ ••.• _~._"_. _ _ _ ~, ____ .... _ .... ~~

'ef'

18 Introduction to Engineering Calculations

Chap. i

The volume and therefore the density of petroleum products vary with temper.! ture, and the petroleum industry has established 60°F as the standard tcr.'. perature for volume and APl gravity, The National Bureau of Standards k published the National Standard Petroleum Oil Tables, NBS Circular C4i' (1936), which relates API, density, specific gravity, and temperature for :l_ petroleum oils. Thus you may convert any of the above properties at any ten> perature to any other temperature, There are many other systems of measuring density and specific gravit: which are somewhat specialized as, for example, the Baume (,Be) and the Twaddell (OTw) systems, Relationships among the various systems of densitl may be found in standard reference books.

1.3-3 Specific Volume. The specific volume of any compound is the inverse of the density, that is, the volume per unit mass or unit amount of material. Units of specific volume might be ft'/lb m , ftJ lIb mole, cm' /g, bbljlb", or similar ratios. 1.3-4 Mole Fraction and Weight Fraction. Mole fraction is simpl, the moles of a particular substance divided by the total number of moles present. This definition holds for gases, liquids. and solids. Similarly, the weight fraction is nothing more than the weight of the substance divided by the total weight o( all substances present. Mathematically these ideas can be expressed as mole fraction = mol~s of A tota mo les . . weIght fractIon

=

weight of A total weight

(1.11)

(1.121

Mole percent and weight percent are the respective fractions times 100.

1.3-5 Analyses. The analyses of gases such as air, combustion products. and the like are usually on a dry basis-i.e., water vapor is excluded from the analysis. Such an analysis is called an Otrsat analysis. In practically all cases the analysis of the gas is on a volume basis, which for the ideal gas is the same as cl mole basis. Consider a typical gas analysis, that of air, which is approximatel) 21 % o~ygcn 79% niiHogen 100 % tOlltal This means that any sample of air will lContain 21 percent oxygen by volume and also 21 mole percent oxygen. Pcrct:'nt. you should remember, is nothin, more than the fraction times 100; consC\ij}uently the mole fraction of oxygen " 0.21. Analyses of liquids and solids are uSlUally given by weight percent, but occa-

..

• Chap. J

vary with temperathe standard ternI of Standards has ms Circular C410 -emperature for all ':Jerties at any temRnd specific gravity ume eBe) and the systems of density

(1.11) (1.12) ons times 100.

i '\ -\\\PLE 1.10

19

Mole fraction and weight fraction

S,,/ution: Basis: 10.0 Ib of total solution weight fraction

Ib

component H 2O

5.00

i~ =

NaOH

5.00

io~

Total

10.00

mol. wt Ib moles

mole fraction

0.500

18.0

0.278

0.278 = 0699 0.403 .

= 0.500

40.0

0.125

0.125 = 0 311 0.403 .

0.403

1.00

1.000

n,C Ib moles are calculated as follows: 5.00 Ib H 2 0 II Ib mole H 2 0 18.0 Ib H 2 0

=

5.00 Ib NaOH lIb mole NaOH 40.0 Ib NaOH

0 278 Ib I H 0 . mo e 2

=

0 1251b I N OH . mo e a

Adding these quantities together gives the totallb moles. -1

EXAMPLE 1.11

i

Density and specific gravity

If dibromopentane has a specific gravity of 1.57, what is its density in Ib m /ft in g/cm', and in kg/m'?

j

3,

j

Solution: Our reference substance is water. 1.578 DBP I oog H 2 0 _--::c.cm-,-!);,.-t-_·__c_m_' = 1.57 g DBP cm' l.00 g H 2 0 cm' I 571bmDBP 62 41bmH20 . ftJ . ftJ 100lb m H 20

.

1.57 g DBP I em'

ft'

COOI mcrn)' I1000 I kg = g

=

97 91bmDBP . ft' 1.57

X

10' kg DBP m'

Note how the units of speCific gravity as used here clarify the calculation.

'eight percent, but occa-

t hie'

An industrial-strength drain cleaner contains 5.001b of water and 5.00 Ib of .. ,Oil. What is the weight fraction and mole fraction of each component in the bottle , dC.Jncr?

(b)

oxygen by volume remember, is nothing ~ fraction of oxygen is

nr.

, .rJlly by mole percent. In this text analyses of liquids and solids will always .., ",umed to be weight percent unless otherwise stated.

combustion products, - is excluded from the "ractically all cases the "al gas is the same as a chich is approximately

~~nt

.".

t

Conventions in Methods of Analysis and Measurement

\,;:,1.3

ny compound is the or unit amount of cnole, cm 3 fg, bblflb m ,

ole fraction is simply -ber of moles pres(:nt. ,'. H 'ht fraction ""y Ii "I weight of expressed as

»

I

Ii

j

J

!

_ _ _ ,_ _ ,.....,,__ ,_ _

20

__

~,-'--'.'t

Introduction to Engineering Calculations

Chap.

J

',,,

EXAMPLE 1.12 Compute the average molecular weight of air

/ I

(a) ,

Basis: 100 Ib mole of air

(h) 1

mol. wI moles = % Ib weight % 23.17 21.0 32 672 O2 79.0 28.2 76.83 2228 N2* Total 100 2900 100.00 'Includes Ar, C02, Kr, Ne, Xe, and is called atmospheric nitrogen. The molecula: weight is 28.2. Table 1.4 lists the detailed composition of air.

(e) \

component

--

The average molecular weight is 2900 Ib/lOO Ib mole

=

29.00.

lhcscqw -' numhcr • lor Ie \!.Iny till' hJ,is is I ~Ii('!lc

ana;

·'r percell ("ely. all' TABLE

1.4

COMPOSITION OF CLEAN. DRY AIR NEAR SEA LEVEL

Component Nitrogen Oxygen Argon Carbon dioxide Neon Helium Methane Krypton Nitrous oxide Hydrogen Xenon Ozone Ammonia Carbon monoxide Iodine Nitrogen dioxide Sulfur dioxide

Percent by volume = mole percent 78.084 20.9476 0.9,34 0.0314 0.001818 0.000524 0.0002 0.000114 0.00005 0.00005 0.0000087 Summer: 0-0.000007 Winter: 0-0.000002

O-trace O-trace

rXAMPL Arom .,od as mli it)

characrc

.hat is thl

Solwi If a h

I Consequel

0-0.000001 0-0.000002 0-0.0001

f:XAMPL

1.3-6 Basis.

Have you noted in the previous examples that the word basis has appeared at the top of the computations? This concept of basis i.· vitally important both to your understanding of how to solve a problem an, also to your solving it in the most expeditious manner. The basis is the reference chosen by you for the calculations you plan to make in any particular problen' and a proper choice of basis frequently makes the problem much easier to soh c The basis may be a period of time-for e.xample, hours, or a given weight o' material-such as 5 Ib of CO 2 or some otner convenient quantity. In selectinf a sound basis (which in many problems is predetermined for you but in some

Most .'lude ~()

:5iticaticl ~i.ticaIil'!

Gi\'cn

to eJ$

pcr . . . (.'

? SO/lIIi

Thc ( 'cOeclion .

• Chap. }

weight % ",72 23.17 :::228 76.83 -'100 100.00 "mheric nitrogen. lhe molecular /b

air.

::lJe

= 29.00.

S

"SS'

W&'5'

Canl'entians ill Me/hods of Analysis and Measuremellt 7 .•• nIcms

::iJt of air

z-wt'

.

21

is not so clear), you should ask yourself the following questions:

(a) What do I have to start with? (n) What do I want to find out? (e) What is the most convenient basis to use?

~ ~c~c questions and their answers will suggest suitable bases. Sometimes, when , -,;lIllbcr of bases seem appropriate, you may find it is best to use a unit basis , I or 100 of something, as, for example, pounds, hours, moles, cubic feet, etc. '~.In)' times for liquids and solids when a weight analysis is used, a convenient "'J'IS is I or 100 Ib or kg; similarly, 1 or 100 moles is often a good choice if a ,-,.,le analysis is used for a gas. The reason for these choices is that the fraction .; percent automatically equals the number of pounds, kg, or moles, respect.,c!y, and one step in the calculations is saved.

::Jt NEAR SEA LEVEL

= t by volume ~ mole percent 78.084 20.9' O. O.l

0.001818 0.000524 0.0002 0.000114 0.00005 0.00005 0.0000087 ~.000007

I:XA~IPLE

1.13

Aromatic hydrocarbons form 15 to 30 percent of the components of leaded fuels ,nJ as much as 40 percent of nonleaded gasoline. The carbon to hydrogen ratio helps :"characterize the fuel components. If a fuel is 80 percent C and 20 percent H by weight, ... hat is the CfH ratio in moles? Solution: If a basis of 100 Ib or kg of oil is selected, then the percent

= pounds or kg.

Basis: 100 Ib of oil (or 100 kg of oil) kg =

compollellt

C H

Ib =

% or %

80 20

mol. wt 12.0 1.008

kg moles or lb moles 6.67 19.84

100

~.000002

O-trace O-trace

Choosing a basis

Consequently the CjH ratio in moles is

CfH = 6.67/19.84 = 0.33

~.OOOOOI

O-n.000002 ~.ooi)]

EXAMPLE 1.14

',oous examples that the word =ns? This concept of basis is now to solve a problem and ::.ner. The basis is the reference .::..KC in any particular problem. :Jroblem much easier to solve. '" hours, or a given weight of -venient quantity. In selecting ::.~rmined for you but in some

Choosing a basis

Most processes for producing high energy content gas or gasoline from coal "'elude some type of gasifkation step to make hydrogen or synthesis gas. Pressure ".I
1 1

1

II

I

22 Introduction to Engineering Calculations

Chap. J

of this gas times pounds and expect the answer to mean anything. Thus the next step is to choose a "convenient basis" which is 100 Ib or kg moles of gas, and proceed as follows: Basis: 100 Ib moles of gas composition CO, CO CH. H,

% = Ibmoles 20.0 30.0 40.0 10.0 100.0

mol. wt 44.0 28.0 16.04 2.02

,... 1.4 At) ~ ;'\~dlnc

;rl rl\ a! \ p('fI'lJ~

Ib 880 840 642 20 2382

c

I (h) I

. 23821b average molecular weight = 100 lb moles = 23.8 Ib/lb mole

Ce) C

1 It is important that your basis be indicated at the beginning of the problem so that you will keep clearly in mind the real nature of your calculations and so that anyone checking your problem will be able to understand on what basis they are performed. If you change bases in the middle of the problem, a new basis should be indicated at that time. Many of the problems which we shall encounter will be solved on one basis and then at the end will be shifted to another basis to give the desired answer. The significance of this type of manipulation will become considerably clearer as you accumulate more experience. The ability to choose the basis that requires the fewest steps in solution can only come with practice. You can quickly accumulate the necessary experience if, as you look at each problem illustrated in this text, you determine first in your own mind what the basis should be and then compare your choice with the selected basis. By this procedure you will quickly obtain the knack of choosing a sound basis.

i,"" .;

,'I

r]

r~

1 -1

1.3-7 Concentrations. Concentration means the quantity of some solute per fixed amount of solvent or solution in a mixture of two or more components, as, for example, (a) Weight per unit volume (Ibm of solute/ft', g of solute/I, Ibm of solute/bbl, kg of solute/m'). (b) Moles per unit volume (Ib mole of solute/ft" g mole of solute/I, g mole of solute/cm'). (c) Parts per million-a method of expressing the concentration of ex· tremely dilute solutions. Ppm is equivalent to a weight fraction for solids and liquids because the total amount of material is of a much higher order of magnitude than the amount of solute; it is essentially a mole fraction for gases. (d) Other methods of expressing concentration with which you should be familiar are molarity (mole/I), normality (equivalents/I), and molality (mole/iOoo g of solvent).

.,'

-. ,'.,;1

, "

..

: \

.' .,

.. •

1, ~ ~

"'i

..._,,*_______....___

" " _.....~ __ 'O... ' -.. ' 11111_'.. ' ..•....._ '..C..i _ _ _" ' ,....~,_ _"'>tt.. · __,,,"_k.. W.. '",'_ _.... "'D"

"-"

Chap. }

Temperature 23

anything. Thus the next step moles of gas, and proceed a,

\ Iypical example of the use of these concentration measures is the set of • ,,:rlmcs by which the Environmental Protection Agency defined the extreme _"cis al which the five most common air pollutants could harm individuals over r"rlods of time.

Ib 880 840

(a) Sulfur dioxide: 2620 jlg/m', equal to I ppm averaged over a 24-hr period. (b) Particulate matter: IOOO jlg/m', or a "coefficient of haze" equal to 8, whichever is stricter, averaged over 24 hr. (Total absence of haze, if it could occur in the earth's atmosphere, would be equal to a zero haze coefficient.) (c) Carbon monoxide: 57.5 mg/m' (50 ppm) when averaged over an 8-hr period; 86.3 mg/m' (75 ppm) over a 4-hr average; 144 mg/m ' (125 ppm) when averaged over I hr. (d) Photochemical oxidants: 800 jlg/m' (0.4 ppm) when averaged over 4 hI'; 1200 jlg/m' (0.6 ppm) when averaged over 2 hI'; 1400 jlg/m' (0.7 ppm) when averaged over 1 hr. (e) Nitrogen dioxide: 3750 jlg/m' (2 ppm) averaged over 1 hI', 938 jlg/m' (0.5 ppm), 24-hr average.

642

20 2382

23.8Ib/lb mole

:2e beginning of the problem of your calculations and so ::derstand on what basis they ~_f the problem, a new basis 7llS which we shall encounter . De shifted to another basis ; tYT lanipulation will ~e ex, .ce. The ability to ::oiution can only come with .' experience if, as you look :lJne first in your own mind ;oice with the selected basis. ; of choosing a sound basis_ !ans the quantity of some a mixture of two or more of solute/I, Ibm of solute/bbl, :.:ft', g mole of solute/I, g " the concentration of exo a weight fraction for solids uteriai' is of a much higher :ute; it is essentially a mole . with which you should be equivalents/I}. and molality

It is important to remember that in an ideal solution, such as in gases or a simple mixture of hydrocarbon liquids or compounds of like chemical :-Jture. the volumes of the components may be added without great error to get : ::c total volume of the mixture. For the so-called non ideal mixtures this rule I"CS not hold, and the total volume of the mixture is bigger OJ!' smaller than the ".Im of the volumes of the pure components. ,11

1.4 Temperature Our concept of temperature probably originated with our physical sense of hot or cold. Attempts to be more specific and quantitative led to the idea of a temperature scale and the thermometer-a device to measure how hot or cold something is. We are all familiar with the thermometer used in !laboratories which holds mercury sealed inside a glass tube or the alcohol the-.rmometer used to measure outdoor temperatures. Although we do not have the space to discuss in detail the many methods of measuring temperature, we can point out some of the otmer more common !~(hniques with which you are probably already familiar: (a) The voltage produced by a junction of two dissimilar conductors changes with temperature and is used as a measure <<:If temperature (the thermocouple). (b) The property of changing electrical resistance wi lin temperature gives us a device known as the thermistor.

~¥tt'

24

Introduction to Engineering Calclilatiolls

Chap.

(c) Two thin strips of metal bonded together at one end expand at diffcw rates with change of temperature. These strips assist in the control, the flow of water in the radiator of an automobile and in the operatIc of air conditioners and heating systems. (d) High temperatures can be measured by devices called pyrometers whi, note the radiant energy leaving a hot body. Figure 1.3 illustrates the appropriate ranges for various lemperature-measurin; devices. As you also know, temperature is a measure of the thermal energy of tl".. random motion of the moleCUles of a substance at thermal equilibrium. Tcn~ perature is normally measured in degrees Fahrenheit or Celsius (Centigrade The common scientific scale is the Celsius scale,8 where 0° is the ice point 0 water and 100" is the normal boiling point of water. In the early 17005 Fahrenhci: a glassblower by trade, was able to build mercury thermometers that gave tem· perature measurements in reasonable aEreement with each other. The Fahrenhc;' scale is the one commonly used in everyday life in the United States. Its referenCe' points are of more mysterious origin. but it is reported that the fixed starlin: point, or 0° on Fahrenheit's scale, wa'S the temperature of an ice-salt mixture. and 96°, the temperature of the blood a healthy man, was selected as the UPF' point because it is easily divisible by 2,3,4,6, and 8, whereas 100° is not. In an:case, as now standardized, 32'F represents the ice point and 212 c F represent) the normal boiling point of water. In the SI system, temperature is measured if. kelvins, a unit named after the famous Lord Kelvin. Note that in the SI systerr the degree symbol is suppressed; i.e., the boiling point of water is 373 K. The Fahrenheit and Celsius scales are relali!'e scales; that is, their zer0 points were arbitrarily fixed by their inventors. Quite often it is necessary to use absolute temperatures instead of relative temperatures. Absolute temperature scales have their zero point at the lowest possible temperature which man believes can exist. As you may know, this lowest temperature is related both \c) the ideal gas laws and to the laws of thermodynamics. The absolute scale, which is based on degree units the size of those in the Celsius (Centigrade) scale. i, called the Kell'il1 scale; the absolute scale which corresponds to the FahrenhcII degree scale is called the Rankil1e scale in honor of a Scottish engineer. Th, relations between relative temperature and absolute temperature are illustratc(~ in Fig. 1.4. We shall round offabsolutc lero on the Rankine scale of -459.58'1' to -460 F; similarly, - 273.1S'C will be rounded 011' to -27rc. You should recognize that the unit degree, i.e., the unit temperature difference on the Kelvin-Celsius scale, is not the same size as that on the RankineFahrenheit scale. If we let d"F represent the unit temperature difference in the' Fahrenheit scale, d'R be the unit temperat\lre difference in the Rankine scale.

or

Q

'As originally devised by Celsius in 1742. the freezing point was designated as 100 Officially. ''C now slands for degrees Celsius.

..

b., ••

we

1f

1

t!

$

>;

t

'1".)'1

tiRe

rtH

t't

Chap. I

.dt Oflll elid expand at differen,

;;tfips assist in the control of RO

.3m5iJile ana in the operation

FO

CO

KO ~

~'Yi€i!s railed pyfometers which

:y. 5500 5000

:2ri{jll§ lempefatui'eomeasuring

~ 3000

Molybdenum melts

= of [email protected] lliefmal energy of the -=.1 thermal equilibrium. Tern:::oneil gf Celsius (tentigrade). .. wHere {)o is the ice point of _ j Ii Ihe early 1700s Fahrenheit. : ifieffficlfue!ers that gave tern"l,h eileli otlier, The Fahrenheir .c'le YIli!ed States. Its reference =:,erWI {hat the fixed starting :.:.iatw' "an ice'salt mixture. =nat ~Ieded as the upper b; W"_das 100" is not. In an\' :::.= p61tit ilnd 212'F represen 3, leffipefattire is measured in .:n, N6{e that in the SI system ?Gillf of water is 373 K. ,'e s€ales; that is, their zero ::...;le ef!en it is necessary to use :.:=jtifes, Absolute temperature :,:!lle femperature which man :~mt5efil!iite is related both to .:: ·=s, The absolute scale, which :;:e'lsliis (Centigrade) scale, is ::-:>ftesponds to the Fahrenheit . Of a Swttish engineer. The :.o::e temperature are illustrated c:: Rankine scale of --459.5S'F c- dff to - 273'C. :::. .. 1M unit temperature differ- size as that on the RankineitHtipetaiure diffaence in the .o:letei1ce in the Rankine scale,

ts

"rig po'ii!( ,vas dtSignated as IOtY

:~:j'um

melts

E

5000 4500

1).1.'00 boils

~

4500 4000

'" ....c ;;; '" c'" ~

:;;

2500

Q;

E

>-'"

E'

~

a.

'0

.."

0.

0

E

T:l:)nium melts

'" 'ii

2000

r.::',J:mum melts

0-

~~

3500 3000 Lead boils l~on melts N'del melts Sdcon melts

","

,g;

E-

",= co

~~

.... g

3000 2500

Eo "'u cO

zE

1500 Giass melts Gold melts

0-

a::~

>--

'-../

2500 2000

r'r--..,

:;;

. .

u c

u

.2

0

0

~ ~

:;;

Q;

1000

E 0 E

:;;

'g"

1500 1000

c

'"
E'

~

U

&' 1000

c

2

c

500

500

Q;

E 0 E

.

~ c 0

....uc

;;;

E 0 E

:,:.

Q;

-

1

E

0

§

c

;;; 0;;

~ ~

l!'

'0

E

~

u

'" 0 a:: c

E'

~

c

.~ . .

~

E

~

>--

f

'" c '" r--..,

l!'

-27318

O'rgen boils

;:Q;

'.." '.." '" 0;;

'0

212 _""500,,,,-_ 32 Waler freezes

~ ~

'0

=~

~

Q;

E 0 E

:;; !O "0

Q;

I--.~

rn

.

'-"

P,d'oqen boils

'i'''um bollS

E

.2

Sulphur boils Mercury boils

Q;

0

·45958

o

"Absolule zero"

Fig. 1.3. Temperature measuring instruments span lhe range from near absolute zero to beyond 3000 K. The chart indicates the preferred methods of thermal instrumentation for various temperature regions.

25

__ h--...,.....,.."',.........

_,...~

.,. ._...

"..".,>--

Chap.)

26 Introduction to Engineering Calculations Boiling point of water at 760 mm Hg

212 672

373 100

180

100 Freezing point of water

J32 492

01

273

255 -18

0460

0C = OF

-40 420

..

233 -40

'"

"

\.:

:t:

'" ...,'"c

c .;;

.c c

e!

c

~

0:

.c

~

? -.; ~

Q;

:.: u

0

Absolute zero

-4600

a -273

Fig. 1.4. Temperature scales.

and /loe and /loK be the analogous units in the other two scales, you should be aware that

/loF = /loR

(1.13)

/loe=/loK

(1.14)

Also, if you keep in mind that the /l °e is larger than the /l OF,

/loe /loF

=

1.8

or /loe = 1.8/loF

( US) (1.\6)

Unfortunately, the symbols /loe, /loF, /loK, and /loR are not in standard usage. and consequently the proper meaning of the symbols °e, OF, OK, and cR, as either the temperature or the temperature difference, mllst be illterpreted from the context of the equation or sentence being examined. You should learn how to conver! one temperature to another with ease. The relations between OR and OF and between OK and °e are. respectively, ( 1.17)

( 1.18)

' , ,'

------_.....

.

--..,

ttt

'US

ei'''' '4 ~~ttZ at

f

Chap. J

·'ctte'

. 7

'..

) rn

r

Temperature 27 f\~lU<'c the relative temperature scales do not have a common zero at the same !<",pcrature, as can be seen from Fig. 1.4, the relation between OF and °e is

T. p

:r

-

32

1.8 AOF)

=

T'e ( I

Aoe

(1.19)

Aflcr you have used Eqs. (1.17)-(1.19) a bit, they will become, so familiar that !crnpcrature conversion will become an automatic reflex. During your "learning" pcllOO," in case you forget them, just think of the appropriate scales side by side .~ In Fig. 1.4, and put down the values for the freezing and boiling points of ... atcr. From Fig. 1.4 you will notice that _40° is a common temperature on both the Centigrade and Fahrenheit scales. Therefore the foHowing relationships hold: (1.20) (1.21) In Fig. 1.4 all the values of the temperatures have been rounded off, but more figures can be used. ooe and its equivalents are known as standard ronditions of temperature.

~Ignificant

0-273

t:'XAMPLE 1.15 Temperature conversion :l

scales, you should be

Convert 100°C to (a) OK, (b) OF, (e) OR.

Soluriol1: (1.13)

(I)

(1.14)

( 100

+ 273)OC 1I AOK = AOC

3730K

or with suppression of the A symbol ( 100 (1.15)

10 another with ease. C are, respectively,

(1.17)

(Ll8)

0

(b)

(100°C) \~!

(e)

(212

(1.16) not in standard usage. :::, of, oK, and cR, as [S1 be interpreted from

+ 273)OC 1I"CK = 373°K + 32 = 212°F

+ 460)OF ::~ = 672°R

or (373°K) \~~

= 672°R

LXA:'.\PLE 1.16 Temperature conversion

The thermal conductivity of aluminum at 32°F is 117 Btu/(hr)(ft 2 )(OF/ft). Find the equivalent value at O°C in terms of Btu/(hrXft2)(OK/ft).

Solurion: Since 32°F is identical to oDe, the value is already at the proper temperature. The "OF" in the denominator of the thermal conductivity actually stands for A OF

28

Introduction to Engineering Calculations

Chap. J

so that the equivalent value is 117 (Btu)(ft) I t. DC ";'(h-'-,r)"'(f~t' '"')(-'.'t.''"F~) ' +-=7-Fr<-l·I-'it."'"'""K

= 211 (B t u) /(hr)(ft' )(' K/ ft)

or with the t. symbol suppressed 117 (Btu)(ft)

I DC

';'(h:.c.r)"';'(f;;:'t2<7)(20FFC),:...t-=~~'"'1ocir K

= 211

(Btu)/(hr)(ft2)(DK/ft)

1.5 Pressure Pressures, like temperatures, can also be expressed by either absolute or relative scales. Pressure is defined as "force per unit area." Figure 1.5 shows a column of mercury held in place by a sealing plate. Suppose that the column of mercury

Closure plate

Fig. 1.5. Pressure.

has an area of I em' and is 50 em high. From Table D. I we can find that the sp gr at 20 c e and hence the density, essentially, of the Hg is 13.55 g!cm'. ThllS the force exerted on the I em' section of that plate by the column of mercury is F= I3.55g 50cm em'

I em' 980cm I kg I mIN (sec)' 1000 g 100 em I (kg)(m) sec 2

= 6.64 N

The pressure on the section of the plate covered by the mercury is P =

6.64 N I em2

If we had started with units in the American engineering system, the pressure could be similarly computed as

trY'

Chap. J

(hr)(ft 2)('K/ft)

1

-

m'

*-

0";:

Mr'.

»W'1

-.1'

t

Pressure 29

"N.I.5

r

I in. 8461b m 50cm I ft' 2.54 em

=

I ft 12 in.

32.2 ft sec' 32. I 47(ft)(lb m l (sec)2(lb,)

1387 1b, ft2

=

:lr)(ft')CK/ft)

'ti

Whether relative or absolute pressure is measured in a pressure-measuring

,'t' ICC depends on the nature of the instrument used to make the measurements.

~v

either absolute or relative f"igure 1.5 shows a column Ihat the column of mercury

l.>r example, an open-end manometer (Fig. 1.6) would measure a relative rrc"urc, since the reference for the open end is the pressure of the atmosphere ,t the open end of the manometer. On the other hand, closing off the end of the m.lnomcter (Fig. 1.7) and creating a vacuum in the end results in a measurement >r.Hnst a complete vacuum, or against "no pressure." This measurement is called .. ··!,l/ute pressure. Since absolute pressure is based on a complete vacuum, a fixed ,,!crcncc point which is unchanged regardless of location or temperature or "c;lther or other factors, absolute pressure then establishes a precise, invariable 'llue which can be readily identified. Thus the zero point for an absolute presAir

~ t.h • 11.0 in. Kg

..

M : 40.10 in. Hg

Nz

11g. 1.6. Open-end manometer showing

I mIN i 100 cm I (kg)(m) sec'

lhe mercury is

10 4N :!eer •. _

L

1 1

Fig. 1.7. Absolute pressure manometer.

a pressure above atmospheric pressure.

·"e D.I we can find that the lhe Hg is 13.55 g(cm 3. Thus 7>y the column of mercury is ~

1

Vacuum

1--

~'ate

~ystem,

the' pressure

•1

sure scale corresponds to a perfect vacuum, whereas the zero point for a relative pressure scale usually corresponds to the pressure of the air which surrounds us Jt all times, and as you know, varies slightly. If the mercury reading is set up as in Fig. 1.8, the device is called a barometer .,nd the reading of atmospheric 'pressure is termed barometric pressure. An understanding of the principle upon which a manometer operates will aid you in recognizing the naVacuum ture of the pressure measurement taken from it. As shown In Fig. 1.6 for an open-end, U-tube manometer, if the pressure measured is greater than atmospheric, the liquid IS forced downward in the leg to which the pressure M ',)urce is connected and upward in the open kg. Eventuallya point of hydrostatic balance is reached in which the manometer fl~id stabilizes, and the difference in ~"Ight of the fluid in the open leg relative to that in the kg allached to the pressure source is exactly equal to Fig. 1.8. A barometer.

I

1

, ~

! j j

j

-- -

.-.......

............ .......

--~.~

.......-......... Chap. I

30 Introduction to Engineering Calculations

the difference between the atmospheric pressure and the applied pressure in the other leg. If vacuum were applied instead of pressure to the same leg of the manometer, the fluid column would rise on the vacuum side. Again, the difference in pressure between the pressure source in the tank and atmospheric pressure is balanced by the difference in the height of the two columns of fluid. Water and mercury are commonly used indicating fluids for manometers; the readings thus can be expressed in "inches of water" or "inches of mercury'" (In ordinary engine.ering calculations we ignore the vapor pressure of mercury and minor changes in the density of mercury due to temperature changes in making pressure measurements.) Bourdon Tube >.J

LI..---'___

Closed End

I

Ii'

J~-'--Link

Gear and

Pinion

"-

Coonectioo to Pressure Source

(0) "C' Bourdon

, • .F

\

Connection to Pressure Source (b) Spirol Bourdon

Corrugated Flexible Metal

Coonectioo to PressureSourCe

......._---+-

Motion Perpendicular to Flexible Tubing

(c) Convex Diophrogm Capsule Fig. 1.9. Bourdon and diaphragm pressure-rnneasuring devices.

1 , 1

¢. 1

_I

Mt

t

.,01

Chap. 1

',e applied pressure in the "e to the same leg of the :":11 side. Again, the differe tank and atmospheric :he two columns of fluid. ~:lds for manometers; the or "inches of ml:rcury." =nor pressure of mercury -, temperature changes in

Bourdon Tube

.1

'C 'tn',

W

Prt!ssure 31

Sir. 1.5

Another type of common measuring device is the visual Bourdon gauge d"lg. 1.9). which normally (but not always) reads zero pressure when open to 'h~ atmosphere. The pressure-sensing device in the Bourdon gauge is a thin ::lctai tube with an elliptical cross section closed at one end which has been bent ,nto an arc. As the pressure increases at the open end of the tube. it tries to ,:rJlghten out, and the movement of the tube is converted into a dial movement h' cears and levers. Figure 1.9 also illustrates a diaphragm capsule gauge. t ',g~re 1.10 indicates the pressure ranges for the various pressure-measuring devices. Pressure scales may be temporarily somewhat more confusing than temperature scales since the reference point or zero point for the relative pressure ~ales is not constant, whereas in the temperature scales the boiling point or the freezing point of water is always a fixed value. However, you will become ace uslOmed to this feature of the pressure scale with practice. The relationship between relative and absolute pressure is illustrated in Figs. J.11 and 1.12 and is also given by the following expression: gauge pressure

+ barometer pressure =

absolute pressure

(1.22)

Equation (1.22) can be used only with consistent units. Note that you must add the atmospheric pressure, i.e., the barometric pressure, to the gauge, or relative mmHg

Atm

.\

Connection to Pressure Source (bl Spiral Bourdon

.1

I

'" E

Q;

oc

j

o

E

,

,l

t =ing devices.

Fig. 1.10. Ranges of application for pressure-measuring devices.

j

I i

. . . . . .,.

r ..-...

t--,.......- -.. " - -...--~t·.w

32

------ . --.---- ..- ..~-

~--.~,,~~'-.

.

-

Introduction to Engineering Calculations

Chap. J

Pounds per SQuare inch

Inches

Newtons per

mercury

squore meter

93 5t 04,147

Standard pressure

2992,082

00 143

Barometric pressure

29.1 0

0.024 x 105 , 1.01 x 105

I -2.45 11.85

~

~

~

~

0 0

~

~

\!' """'~

~

\!' \!'

"-

" ~~"

~ ~

E

~ "- "-

u ""

'"0'" g;;'"

0

L>

'"



143 00

"

~ ~

'" g ~ '" -'" '-"

'" S?"

0 C>

I

0.17 x lei' 082 x 10 5

~

t" \!' "-

Perfect vacuum

! 098 x 10 5

000

24.1 -50 5.0

" ~"

~

00

.0


-0.98 x105 000

0-291 29.1

Fig. 1.11. Pressure comparisons when barometer reading is 29.1 in Hg.

'----

/ 5

Time

Fig. 1.12. Pressure terminology. The standard atmosphere is shown by the heavy horizontal line. The broken line illustrates the atmospheric (barometric) pressure which changes from time to time. I in the figure is a pressure of 19.3 psi referred to a complete vacuum or

5 psi referred to the barometric pressure. 2 is the complete vacuum. while 3 represents the standard atmosphere. 4 illustrates a negati,e relative pressure or a pressure less [han atmospheric. This type of measurement is described in the texl as a vacuum type of measurc~ ment. 5 also indicates a vacuum measurement. but one \\'hich is equivalent to an absolule pressure above lhe slandard almosphere.

Chap. J

\.t.

Pressure 33

J5

; .,,,ure (or manometer reading if open on one end), in order to get the absolute '(,,,urc.

Newtons per square mete' 034 x 105

1133 x105

"'I '" ·

0"" • 000

<0'

. 0.98 x 105

0.17 x 105 082 x 105

Another term applied in measuring pressure which is illustrated in Figs. 1 "od 1.12 is l'acuum. In effect, when you measure pressure as "inches of C!.ury \'acuum," you reverse the usual direction of measurement and measure '".11 the barometric pressure down to the vacuum, in which case a perfect '." UIIOl would be the nighest pressure that you could achieve. This procedure ",'uld be the same as evacuating the air at the top of a mercury barometer and ",Itching the mercury rise up in the barometer as the air is removed. The vacuum ",tern of measurement of pressure is commonly used in apparatus which operate "I pressures less than atmospheric-such as a vacuum evaporator or vacuum ':iter. A pressure which is only slightly below barometric pressure may somet,mes be expressed as a "draft" (which is identical to the vacuum system) in Inches of water, as, for example, in the air supply to a furnace or a water cooling • 1

tl\WCT.

-0.98 x 105 000

'lg is 29.1 in Hg.

As to the units of pressure, we have shown in Fig. 1.11 two common systems: pounds per square inch (psi) and inches of mercury (in. Hg). Pounds per 'quare inch absolute is normally abbreviated "psia," and "psig" stands for "pounds per square inch gauge." Figure 1.11 compares the relative and absolute pressure scales in terms of these two systems of measurement when the barometric pressure is 29.1 in. Hg (14.3 psia) with the SI system of units. Other ~)stems of expressing pressure exist; in fact you will discover there are as many dltrerent units of pressure as there are means of measuring pressure, Some of the other most frequently used systems are Millimeters of mercury Feet of water

(ft H,O)

Atmospheres

(atm)

Bars

(mm Hg)

(bar)

Newtons per square meter (N/m') To sum up our discussion of pressure and its measurement, you should now he acquainted with ,ere is shown by the :Ues the atmospheric me to time. I in the mmplete vacuum or ~e

complete vacuum,

Illustrates a negative cDherie. This type of lum type of meaSurcnt, but one which is :l.andard atmosphere.

(a) Atmospheric pressure-the pressure of the air and the atmosphere surrounding us which changes from day to day. (b) Barometric pressure-the same as atmospheric pressure, called "barometric pressure" because a barometer is used to measure atmospheric pressure. (e) Absolute pressure-a measure of pressure referred to a complete vacuum, or zero pressure. (d) Gauge pressure-pressure expressed as a quantity measured from (above) atmospheric pressure (or some other reference pressure).

f

1 J, '1

IJ

1 I

1

1

I

----~----.,-----

..

-..---

....-,,,#~.

Chap.]

34 Introduction to Engilleering Calculatiolls

,

\a

It

I

~

I

(e) Vacuum-a method of expressing pressure as a quantity below atmospheric pressure (or some other reference pressure). You definitely must not confuse the standard atmosphere with atmospheric pressure. The standard atmosphere is defined as the pressure (in a standard gravitational field) equivalent to 14.696 psi or 760 mm Hg at ooe or other equivalen, value, whereas atmospheric pressure is a variable and must be obtained from a barometer each time you need it. The standard atmosphere may not actually exist in any part of the world except perhaps at sea level on certain days, but it is extremely useful in converting from one system of pressure measurement to another (as well as being useful in several other ways to be considered later). Expressed in various units, the standard atmosphere is equal to 1.000 33.91 14.7 29.92 760.0 1.013 x 10'

atmospheres (atm) feet of water (ft H 2 0) (14.696, more exactly) pounds per square inch absolute (psia) (29.921, more exactly) inches of mercury (in. Hg) millimeters of mercury (mm Hg) newtons per square meter (N/m2)

1\.\\11

r, {

1\ \

\II

To convert from one set of pressure units to another, it is convenient to use the relationships among the standard pressures as shown in the examples below. If pressures are measured by means of the height of a column of liquid' other than mercury or water (for which the standard pressure is known), it is easy to convert from one liquid to another by means of the following expression: p = pgh (1.23) where p = density of the liquid g = acceleration of gravity By equating the same pressure yielded by two liquids with different densities, you will get the ratio between the heights of two columns of liquid. If one of these liquids is water, it is then easy to convert to any of the more commonly used liquids as follows: or

' . ,A

~-I!JM.

hll,o -

P

EXAMPLE 1.17 Pressure conversion

Convert 35 psia to in. Hg. Solutioll: It is desirable to use the ratio of 14.7 psia to 29.92 in. Hg, an identity, to carry

out this conversion. 'Sometimes these liquid columns are referred to as "heads" of liquid.

,.

t

Chap. 1

r

K t'

s't

t'

1$.'"

1

t

ret

$ ' .,

Pressure 35

Su.1.5

Basis: 35 psia

entity below atmos-

35 psia 29.92 in. Hg = 71 25' H 14.7 psia . lfl. g

-:)here with atmos':;5ure (in a standard :: at ooe or other .2.Dle and must bc '::'"'ldard atmospherc :ipS at sea level on :0 system of pressure other ways to be

'"--v---'

an identity

rX.\:-'IPLE 1.18 Pressure conversion The density of the atmosphere decreases with increasing altitude. When the rrcssure is 340 mm Hg, how many inches of water is it? How many kilonewtons per ..quare meter?

Solution:

Basis: 340 mm Hg

.;;qual to

340mm Hg \33,91 ft H,O \12 in. 760 mm Hg 1 ft

= 182' H 0

340 mm Hg 1.013 x 10' N/m' I kN 760.0 mm Hg 1000 N

::h absolute (psia) . Hg)

lfl.

2

= 45.4 kN/m2

EXAMPLE 1.19 Pressure conversion ~,nver

use the '! e). .; below., ::. coilliufi of liquid' 3sure is known), it the following ex.-

The pressure gauge on a tank of CO, used to fill soda water bottles reads 51.0 psi. At the same time the barometer reads 28.0 in. Hg. What is the absolute pressure In the tank in psia? Sec Fig. E1.19.

'0

(1.23)

different densities, ;)f liquid. If one of ::le more commonly ::l

Fig. E1.I9.

So/uriol1:

The pressure gauge is reading psig, not psia. From Eq. (1.22) the absolute pressure is the sum of the gauge pressure and the atmospheric (barometric) pressure expressed in the same units. Basis: Barometric pressure = 28.0 in. Hg . atmosp henc pressure

psia 1378· = 28.0 in. Hg 2914,7 92 111. Hg = . ps.a •

f

(Note: Atmospheric pressure does not equal 1 standard atm.) The absolute pressure in

.an identity, to carry

l

the tank is 51.0 + 13.78

=

I

64.78 psia

J

c:id.

f

I I j

- - - . -....

~~'~""~~-r-~-

.

"~"·"··."'I';-""""--''''J''P'---~·''''~·~~·'"'<-"'~'''--'-"''''''''''''''~'''''''"-''''''1!''''''':'''''''''''''''''''-'''''''--"""'.#..,....__'''',m!!''_~9'''....'''"',_

i

........... ,_w..,..,;,...._

i

'---.- ---'-'~------"-'- .'.-

-- ----.-.-----.- ...... ~-

36 Introduction to Engineering Calculations

Chap. 1

EXAMPLE 1.20 Pressure conversion Air is flowing through a duct under a draft of 4.0 in. H 2 0. The barometer indicates that the atmospheric pressure is 730 mm Hg. What is the absolute pressure of the gas in inches of Hg? See Fig. E1.20.

Air

~

, .r, Fig. E1.20.

Solution: Again we have to employ consistent units, and it appears in this case that the most convenient units are those of inches of mercury. Basis: 730 mm Hg of pressure . pressure atmosp henc

=

92 in. Hg 730 mm Hg 1 29 760. mm Hg

=

28 .9·In. H g

Basis: 4.0 in. H 2 0 draft (under atmospheric) 4 in. H 2 0

J ft 29.92 in. Hg 12 in. 33.91 ft H 2 0

029·

=.

In.

H

g

Since the reading is 4.0 in. H 2 0 draft (under atmospheric), the absolute reading in uniform units is 28.9 - 0.29 = 28.6 in. Hg absolute EXAMPLE 1.21

\

Vacuum Pressure Reading

Small animals such as mice can live at reduced air pressures down to 3.0 psia (although not comfortably). In a test a mercury manometer attached to a tank as shown in Fig. EI.21 reads 25.4 in. Hg and the barometer reads 14.79 psia. Will the mice survive?

,. f'

Fig El.21.

--~.-

. .-....

,..-.--~ ~"""'i"""'"

M

Chap.}

'N.

to

tit.

en

Ct#'

W! •

d' "

Physical and Chemical Properties of Compounds and Mixtures

1.6

- 1*'$'0'8'1*_,-'

37

SOllilion:

Basis: 25.4 in. Hg below atmospheric narometer indicates , pressure of the gas

We shaH assume that gig, ~ 1.00 and ignore any temperature corrections to the mercury to oce. Then, since the vacuum reading on the tank is 25.4 in. 1Ir. t>clow atmospheric, the absolute pressure in the tank is

,,'~\Cfl

1479' 25.4 in. Hg 14.7 psia . pSla 29.92 in. Hg

= 14,79 _ 12.49

=

2.3 psia

The mice probably will not survive.

1.6 Physical and chemical properties of compounds and mixtures m this case that the

28.9 '

the absolute reading

:Ires down to 3.0 psia ·ned to a tank as shown 'sia. Will the mice sur-

Publications of industrial associations provide a tremendous storehouse of data. For example, the American Petroleum Institute publishes both the Manual on Disposal of Refinery Wastes! 0 and the Technical Dala Book-Petroleum Refining! I. Tables 1.5 1.6 and 1.7 are typical of the tables that can be found in the former on p. 2-4, and p. 13-11. If you know the chemical formula for a pure compound, you can look up many of its physical properties in standard reference books, such as (a) Perry's Chemical Engineers' Handbook 12 (b) Handbook of Physics and Chemistry" (c) Lange's Handbook!' (d) The Properties of Gases and Liquids." Also, specialized texts in your library frequently list properties of compounds. For example, the book Fuel Flue Gases I 6 lists considerable data on gas mixtures which are primarily of concern in the gas utility field. Appendix D in this text tabulates the chemical formula, molecular weight, melting point, boiling point, ctc., for most, but not all, of the compounds involved in the problems you will tind at the end of the chapters. Many of the materials we talk about and use every day are not pure comIOAmerican Petroleum Institute, Div. of Refining, 1271 Ave. of the Americas, New York, N.Y., 1969. I I ibid., 2nd ed" 1970. 12). H, Perry, ed, Cltemical Engineers' Handbook, 4th ed., McGraw-Hill, New York, 1963. I l Handbook of Chemistry and Pity sics, Chemical Rubber Publishing Co., Cleveland, Ohio, annually, "N. A. Lange, Handbook of Cltemistry, McGraw-HilI, New York, annually, DR. C. Reid and T. K. Sherwood, The Properties of Gases and Liquids, 2nd ed., McGrawHill, New York, 1966, I6Fuel Flue Gases, American Gas Association, New York, 1941.

i

I

38

Chap. 1

Introduction to Engineering Calculations

pounds, and it is considerably more trouble to obtain information about the properties of these materials. Fortunately, for materials such as coal, coke, petroleum products, and natural gas-which are the main sources of energy in this country-tables and formulas are available in reference books and handbooks which list some of the specific gross properties of these mixtures. Some typical examples of natural gas analyses and ultimate analyses of petroleum and petroleum products as shown in Tables 1.5, 1.6, 1.7,1.8, and 1.9 point out the variety of materials that can be found. TABLE 1.5 DISSOLVED SOLIDS IN FRESH WATERS OF THE UNITED STATES Concentralion (mgjl)

Component

In 5% of the waters is less than

In 50% of the waters is less than

In 95% of the waterS is less than

72.0 40.0 11.0 3.0 0.2 15.0 3.5 6.0 0.1

169.0 90.0 32.0 9.0 0.9 28.0 7.0 10.0 0.3

400.0 180.0 90.0 170.0 4.2 52.0 14.0 85.0 0.7

Total dissolved solids Bicarbonate, HCO, Sulfate, SO. Chloride, CI Nitrate, NO, Calcium, Ca Magnesium, Mg

Sodium, Na, and Potassium, K Iron, Fe

SOURCE: Mallual on Disposal 0/ Refinery Wastes, Amer. Pet. Inst., Div. of Refining, pp. 2-4; New York, 1969. TABLE 1.6 CHARACTERISTICS OF WATERS WHICH SUPPORT A VARIED AND PROFUSE FISH FAUNA-A CRITERION OF QUALITY Value of Characteristic In 5% of the waters is less than

In 50% of the waters is less than

In 95% of the waters is less than

pH value

6.7

8.3

Dissolved oxygen, 02 Carbon dioxide, CO 2 Free Fixed Ammonia, NH,

5.0

7.6 Concentration (mg/I) 6.8

0.1 8.0 0.5

1.5 45.0 J.S Mho (25'C)

5.0 95.0 2.5

Specific conductivity

(50)(10- 6 )

(2iO)(I0-6)

(1100)(10- 6 )

Characteristic

9.8

SOURCE: Manllal on Disposal of Refinery Wasles, Amer. Pet. Inst., Div. of Refining, pp. 2-4; New York, 1969.

t

Chap. /

:mati on about the ::ch as coal, coke, .·lUrces of energy in books and hand',5e mixtures. Some .~ of petroleum and :: 1.9 point llut the

.~(C.

t

, ...

••

400.0 180.0 90.0 170.0 4.2 52.0 14.0 85.0 0.7

Meld

,.

,

TABLE 1.7 REFI"ERY BIOLOGICAL TREATMENT UNIT FEED CHARACTERISTICS

Ranges reported'

200-960 140--640 97-280 80-450 77-210 69-100 56-120 23-130 2(}-97 7.6-61 7.1-9.5 1.3-38 0.3-0.7

Chlorides, mg/I COD, mg/I. BOD" mgll Suspended solids, mgjl Alkalinity, mg/I as CaCO, Temperature, 'F Ammonia nitrogen, mg/I Oil, mg/I Phosphate, mg/I Phenolics, mg!1 pH Sulfides, mg/I Chromium, mg!1

:ng/I) In 95 % of the waters is less than

mer,

pte

Physical and Chemical Properties of Compounds alld Mixtures 39

/.6

.::> STATES

o

if

'j

SOURCE: Manllal on Disposal of Refinery Wastes, Amer. Pet. Inst., Diy. of Refining, pr. 2-4. New York, 1969. 'Values are the averages of the minima and maxima reported by 12 refineries treating total emuent. Individual plants have reponed data well outside many of these ranges.

TABLE 1.8 TYPICAL DRY GAS ANALYSES

Analysis (yol. %--excluding water vapor) Type

In 95% of the waters is less than

1

Natural Natural Natural :-;atural Butane

C02

gas gas, dry' gas, wet· gas, soud

O2

6.5 0.2 1.1 (H 2 S 6.4)

N2

CO

H2

CH.

77.5 99.2 87.0 58.7

0.6

C2H6 C,H. C.Hlo C,H I 2+

9.8 5.0 95.0 2.5 (1100)(10- 6 ) mst., Diy. of Refining,

1

4.1 16.5 2.0

2.6 9.9 3.5

2.0 5.0 75.4 18.1

3.4 3.5 n-butane isobutane

lIIuminants Reformed 2.3 refinery oil Coal gas, by-product 2.1 Producer gas 4.5 Blast furnace gas 5,4 Sewage gas 22.0

0.7

4.9

20.8

49.8

12.3

0.4 0.6 0.7

4.4 50.9 8.3 6.0

13.5 27.0 37.0

51.9 14.0 47.3 2.0

24.3 3.0 1.3 68.0

5.5

;4i.. ;:.~.-

>.

-~.~ '~~--.-~-'-""~""'-'" ,.-,~--~~-.,.

. . . . . . . . . ,. . ._

I

1\

1

3.4

I

*_,,,,".".,......

.. " . .'" ' . ".;.;. . . . .

j

3.7

SOURCE: Fliel FllIe Gases, American Gas Association. New York, 19-11. p. 20. 'Ory gas contains much ie'S propane te,H,) and higher hydrocarl::ons than wet gas does. t SOllr implies that the gas contains significant amounts of hydrogen sulfide.

·. . . .'-_1. . . . .

I, j

16.0

8.3 j)

~

j

,nst., Diy. of Refining,

VARIED

i 1,

Q...._

u"'*_. . . . .,.,-"{................,,,,......,,....'_

......... _o...

I ,

_._.ft.,,_._'"'" ____ .__ ._.. _ _ _ _ _ _ •__

40

~

__ .........._ _ _ _

._~,.~.

_ _ _ _ .___ "._._...

Introduction to Engineering Calculations

........_

~

Chap. 1

TABLE 1.9 ULTIMATE ANALYSIS OF PETROLEUM CRUDE Weight ~~ Type Pennsylvania Humbolt, Kan. Beaumont, Tex. Mexico Baku, U.S.S.R.

Spgr

At 'C

C

H

0.862 0.921 0.91 0.97 0.897

IS

8S.5 85.6 85.7 83.0 86.S

14.2 12.4 11.0 11.0 12.0

15

N

o

S 0.37 0.70

2.61 1.7 1.5

SOURCE: Data from W. L. Nelson, Petroleum Refinery Engineering, 4th ed., McGraw-Hili, New York, 1958.

Since petroleum and petroleum products represent complex mixtures of hydrocarbons and other organic compounds of various types together with various impurities, if the individual components cannot be identified, then the mixture is treated as a uniform compound. Usually, the components in natural gas can be identified, and thus their individual physical properties can be looked up in the reference books mentioned above or in the Appendix in this text. As will be discussed in Chap. 3 under gases, many times the properties of a pure gas when mixed with another gas are the sum of the properties of the pure components. On the other hand, liquid petroleum crude oil and petroleum fractions are such complicated mixtures that their physical properties are hard to estimate from the pure components (even if known) unless the mixture is very simple. As a result of the need for methods of predicting the behavior of petroleum stocks, empirical correlations have been developed in recent years for many of the physical properties we want to use. These correlations are based upon the °API, the Universal Oil Products characterization factor K, the boiling point. and the apparent molecular weight of the petroleum fraction. These parameters in turn are related to five or six relatively simple tests of the properties of oils. Some of the details of these tests, the empirical parameters, and the properties that can be predicted from these parameters will be found in the API Technical Data Book and in Appendix K.

1.7 Technique of solving problems If you can form good habits of problem solving early in your career, you will save considerable time and avoid many frustrations in all aspects of your work. in and out of school. In solving material and energy balance problems, you should (a) Read the available information through thoroughly and understand what is required for an answer. Sometimes, as in life. the major obstucle is to find out what the problem really is.

Chap. J CRUDE

N

o

s 0.37 0.70

2.61 1.7 1.5

'!Eering, 4th ed., McGraw-Hili,

\rc.

1.7

Technique of Solving Problems 41

(b) Determine what additional data are needed, if any, and obtain this information. (c) Draw a simplified picture of what is taking place and write down the available data. You may use boxes to indicate processes or equipment, and lines for the 1I0w streams. (d) Pick a basis on which to start the problem, as discussed in Sec. 1.3-6. (e) If a chemical equation is involved, write it down and make sure it is balanced. By this time you should have firmly in mind what the problem is and a reJsonably clear idea of what you are going to do about it; however, if you have not seen exactly how to proceed from what is available to what is wanted, then ~ou should

1

:::Ilt complex mixtures of ~Jus

types together with =1 be identified, then the :::e components in natural --:Jroperties can be looked ~cppendix in this text. As :~ PH' . 's of a pure gas . :Jert .he pun! com:.and pc
m your career, you will :.11 aspects of your work, nalance problems, you :Jughly and understand : life, the major obstacle

(f) Decide what formulas or principles are governing in this specific case and what types of calculations and intermediate answers you will need to get the final answer. If alternative procedures are available, try to decide which are the most expedient. If an unknown cannot be found directly, give it a letter symbol, and proceed as if you knew it. (g) Make the necessary calculations in good form, being careful to check the arithmetic and units as you proceed . (h) Determine whether the answer seems reasonable in view of your experience with these types of calculations.

i

Problems which are long and involved should be divided into parts and atlached systematically piece by piece. If you can assimilate this procedure and make it a paI"t of yourself-so that ! ou do not have to think about it step by step-you will find that you will be able to materially improve your speed, performance, and accuracy in problem ,olving. The major difference between problems solved in the classroom and in the plant lies in the quality of the data available for the solution. Plant data may be of poor quality, inconclusive, inadequate, or actually conflicting, depending on the accuracy of sampling, the type of analytical procedures employed, the skill of the technicians in the operation of analytical apparatus, and many other factors. The ability of an engineer to use the stoichiometric principles for the calculation of problems of material balance is only partly exercised in the solution of problems, even of great complexity, if solved from adequate and appropriate data. The remainder of the test lies in his ability to recognize poor data, to request and obtain usable data, and, if necessary, to make accurate estimates in lieu of incorrect or insufficient data." "8. E. Lauer, The Material Balallce, Work Book Edition, 1954, p. 89.

ad SA.

.4 MEWS

QW

, 'fiM,..i ,

42 Introduction to Engineering Calculations

Chap. J

1.8 The chemical equation and stoichiometry

hept;

As you already know, the chemical equation provides a variety of qualitative and quantitative information essential for the calculation of the combining weights of materials involved in a chemical process. Take, for example, the combustion of heptane as shown below. What can we learn from this equation?

+

+ 8 H,O

I

molee of he

(1.25)

6.023 :-

First, make sure that the equation is balanced. Then we can see that I mole (not Ibm or kg) of heptane will react with II moles of oxygen to give 7 moles of carbon dioxide plus 8 moles of water. These may be lb moles, g moles, kg moles, or any other type of mole, as shown in Fig. 1.13. One mole of CO, is formed from each -\- mole of C,H 16 • Also, I mole of H,O is formed with each t mole of CO,. Thus the equation tells us in terms of moles (not mass) the ratios among . reactants and products. Stoichiometry (stoi-ki-om-e-tri)'8 deals with the combining weights of elements and compounds. The ratios obtained from the numerical coefficients in the chemical equation are the stoichiometric ratios that permit you to calculate the moles of one substance as related to the moles of another substance in the chemical equation. If the basis selected is to be mass (Ibm' kg) rather than moles, you should use the following method in solving problems involving chemical equations: (a) Calculate the number of moles of the substance equivalent to the basis using the molecular weight; (b) change this quantity into the moles of the desired product or reactant by mUltiplying by the proper stoichiometric ratio, as determined by the chemical equation; and (c) then change the moles of product or reactant to a weight basis. These steps are indicated in Fig. 1.14 for the above reaction. You can combine these steps in a single dimensional equation, as shown in the examples below, for ease of slide rule calculations; however, the order of operations should be kept the same. An assumption implicit in the above is that the reaclion takes place exactly as written in the equation and proceeds to 100 percent completion. When reactants, products, or degree of completion of the actual reaction differ from the assumptions of the equation, additional data must be made available to indicate the actual status or situation.

ofC,

C 7 H' 6

II 0, ----->- 7 CO,

EXAMPLE 1.22 Stoichiometry In the combustion of heptane, CO, is produced. Assuming that it is desired to produce 500 Ib of dry ice per hour and that 50 percent of the CO, can be converted into dry ice, how many pounds of heptane must be burned per hour? 18From the Greek s/oicheion, basic constituent, and me/rein, to measure.

molc~

I g mol ofC, I kg mC

ofC, lIb mo

ofC, 1 ten rn

of C, ;00011 ofC, 100 I

--=_. ._. . _*""...__..

Chap. J

I_.tm_lIIt!f_:_....._.."..

'iii'.. ',.:/oijl,..."""sl/i£-u'"l-ilir_'.I1001"....

'$...._ '.. ...'..'.. ' _ . _.. _ _' _ _ _

..... '.0--... ' ... , . . . . ._ _. . . .

+

+

7CO, Qualilali.·e informalion

heptane

reacts

oxygen

with

-riety of qualitative and ,ne combining weights .=ple, the combustion is equation?

(1.25) ve can see that I mole ·vgen to give 7 moles of ales, g moles, kg moles. mole of CO, is formed 'led with each .~ mole of mass) the ratios among combining weights of numerical coefficients :nat . ,it you to cal'es 0 ler substance lSS (It. m , .:g) rather than ':ng problems involving of the substance equiv'e this quantity into the - by the proper stoichioand (c) then change the ~ steps are indicated in , steps in a single dimenase of slide rule calcula:ete same. ~Iion takes place exactly "ent completion. When :ual reaction differ from 51 be made available to ~

to give

reacts I molecule "f heptane with

6.023 x 10" molecules ofC,H16 I g mole of C,H'6 I kg mole of C,H16 lib mole ofC,H16 t ton mole

ofC,H 16 1(100)g of C,H16

+

+

II moleCUles of oxygen 11(6.023 x 10") molecules of 0, II g moles of 0,

to give

---+

---+

"'

II kg moles of 0,

---+

+

Illb moles of 0,

---+

+

II ton moles of 0,

-->-

+

11(32) g of 0,

water

and ,'8 molecules of water

7 molecules of carbon dioxide 7(6.023 x lO'l) molecules of CO, 7 g moles of CO, 7 kg moles of CO, 7 Ib moles orca, 7 ton moles of CO, 7(44) g of CO,

8(6.023 x 10") molecules ofH,O

+ +

100 g

30S g

352 g

S g moles ofH,O 8 kg moles ofH,O

+

81b moles ofH,o

+

8 ton moles ofH,O 8(1S) g ofH,O

+

~

~

"---v-

144 g

-

452 g 452 kg 452 ton 4521b

452 g 452 kg 452 ton 4521b Fig. 1.13. The chemical equation.

Basis: 10.01b C,H 16 Component

C,H16

A,

co, H,O

C,Ht6

that it is desired to :he CO, can be converted j per hour?

and

Quantilalil·e information

lib mole

~ming

carbon dioxide

Mol. wi 100.1 32.0 44.0 IS.O

71b mole

+

110,

10.01b C,H 16 100.1 Ib C,H'6 = 0.100 Ib mole C,H'6-->Ib mole C,H t6

7CO,

+

0.700 Ib mole CO, I Ib mole CO, 44.0 Ib CO,

SH,O =

30 S Ib CO . ,

10.0 Ib C,H16 yields 30.Slb CO, j

W ..

t("

·urc.

1

Fig. 1.14. Stoichiometry.

j

!

43

,I 1

i

44

Chap,

Jlltroduction to Engineering Calculations

j

Solutioll:

Basis: 500 Ib dry ice (or I hr) Mol. wt heptane

= 100 Chemical equation as in Fig, 1.13

500 Ib dry ice 100 Ib C,H 16 = 3251b C H lib mole C,H 16 ' ,. Since the basis of 500 Ib dry ice is identical to I hr, 3251b of C,H,. must be burneJ per hour, Note 'that pounds are converted first to moles, then the chemical equation is applied, and finally moles are converted to pounds again for the final answer,

EXAMPLE 1.23 Stoichiometry Corrosion of pipes in boilers by oxgen can be alleviated through the use of sodium sulfite. Sodium sulfite removes oxygen from boiler feedwater by the following reaction: 2 Na,SO,

+ O 2 ~ 2 Na,S04

How many pounds of sodium sulfite are theoretically required to remove the oxygen from 8,330,000Ib of water (10' gal) containing 10.0 parts per million (ppm) of di,· solved oxygen and at the same time maintain a 35 percent excess of sodium sulfite" See Fig. EI.23.

H20: 8,330,000Ib

H20: 8,330,000Ib

10 ppm 02

no oxygen

Fig. EI.2J. SOIUlioll: Additional data: mol. wt of Na2S0, is 126,

Chemical equation: 2 Na,SO,

+ 0, ~ 2 Na,SO.

Basis: 8,330,000 Ib H,O with 10 ppm 0, or 83,31b 0, 8,330,000 Ib H 2 0

10 Ib 0, = 833 Ib 0, ,(1,000,000 - 10 Ib 0,) lb H,O ' • elfecth'cly sume as 1,000,000

8,330,000 Ib 1-1,0

I Ib mole 0, 21b mole's Na,SO, 32 Ib 0, I Ib mole 0, '-::':-7iS*,,+I.:;:,3:,::5 I

= 8851b Na,SO,

.. Chap. I

......

.

.

rie"t

$

t

tt'

h

•.

The Chemical Eqllation and Stoichiometry 45

Sf'" 1.8

I.\A:'IIPLE 1.24 Stoichiometry A limestone analyzes CaCO, MgCO, Insoluble

.:lle C,H 16 -:lOle CO 2

92.89 5.41 1.70

(a) How many pounds of calcium oxide can be made from 5 tons of this lime·

stone? (b) How many pounds of CO 2 can be recovered per pound of limestone? (e) How many pounds of limestone are needed to make I ton of lime?

., H t'. must be burned :the chemical equation Ihe final answer.

Sollitioll: Read the problem carefully to fix in mind exactly what is required. Lime will mclude all the impurities present in the limestone which remain after the CO, has t-cen driven off. Next draw a picture of what is going on in this process. See Fig. r 1.24.

::Jugh the use of sodi urn ::ne following reaction: remove the oxygen JTIi11ion (ppm) of dis.::ess r "um sulfite? "IO

Limestone

lL-----~~~uble ..

.330,000Ib sn

}Lime

Fig. Et.24 .

To complete the preliminary analysis you need the following chemical equations:

+ CO, MgCO, - - * MgO + CO 2 CaCO,

-->-

CaO

Additional data are Mol. wt

CaCO, 100

MgCO, 84.3

CaO 56.0

MgO 40.3

CO 2 44

Basis: 100 Ib limestone This basis was selected because Ib =

CIb 0,

camp. CaCO, MgCO, Insol. Total

83.31b 0,

·ie's Na,SO, mole O 2

Ib = % 92.89 5.41 1.70 100.00

%.

Ib mole 0.9289 0.0641

-~

0.9930

CO 2 (1b) Ib 52.0 40.9 2.59 2.82 1.70 56.3 43.7

lime CaO MgO Insol. Total

l'\otc that the total pounds of products equal the 100 Ib of entering limestone. Now to calculate the quantities originally asked for: (a)

-

52.0 Ib CaO 5 ton CaO produced = TIOO=Tilb:-:Cst:-:o"'n'Ce+--'-:--::-+'---'-'--" = 5200 Ib CaO

. . . ." " " " "." ..,. . . . . . . . . .- _ . _ - '.................., - .'.'W' " ...."

·~-~-"

...- - ......

.

.....--......!I9W_.'' i,"',___..,...._,.....,,"......""":h"""'.~

~'-.*''''.J......'"''_'''-~i

Chap.}

46 Introduction to Engineering Calculations (b)

43.7 lb CO 2 0437 lb CO 2 recovere d = 100 Ib stone =. ,or, lIb mole CO 2 0.9930 Ib mole CaCO, + MgCO, 100 Ib stone lIb mole CaCO, + MgCO,

I

44lb CO 2 I Ib mole CO 2

(c)

. . lImestone required

100 Ib stone 2000 ItI ton

= 56.31b lime

-

0.437 lb

= 3560 Ib stone

In industrial reactions you will rarely find exact stoichiometric amounts of materials used. To make a desired reaction take place or to use up a costly reactant, excess reactants are nearly always used. This excess material comes out together with, or perhaps separately from, the product-and sometimes can be used again. Even if stoichiometric quantities of reactants are used, but if the reaction is not complete or there are side reactions, the products will be accom· panied by unused reactants as well as side-products or reactants. In these circumstances some new definitions must be understood: (a) Limiting reactant is the reactant which is present in the smallest stoichiometric amount. (b) Excess reactant is a reactant in excess of the limiting reactant. The percent excess of a reactant is based on the amount of any excess reactant above the amount required to react with the limiting reactant according to the chemical equation, or

%excess = o

. moles in excess .. 100 moles reqUired to react wah IImamg reactant

where the moles in excess frequently can be calculated as the total available moles of a reactant less the moles required to react with the limiting reactant. A common term, excess air, is used in combustion reactions; it means the amount of air available to react that is in excess of the air theoretically required to completely burn the combustible material. The required amount of a reactant is established by the limiting reactant and is for all other reactants the corresponding stoichiometric amount. Even if only part of the limiting reactant actually reacts, the required and excess quantities are based on the entire amount of the Iimit. ing reactant. Three other terms that are used in connection with chemical reactions have less clear-cut definitions: conversion, selectivity, and yield. No universally agreed upon definitions exist for these terms-in fact, quite the contrary. Rather than cite all the possible usages of these terms, many of which conflict, we shall define them as follows: (c) Conl'ersion is the fraction of the feed or some material in the feed that is converted into products. What the basis in the feed is and intQ what

I \ ,

'de x

Chap. 1

MgCO, .... Ib CO, mole CO,

=

0.437 lb

~:me

::niometric amounts of :cr to use up a costly 35 material comes out -and sometimes can :-.:.1S are used, but if the ~ducts will be accom=ctarits. In these cir:J.lhe smallest stoichio=iting reactant. The ::noup' C any excess :-.:. th. 19 reactant __--'--;--: 100 :.:J.g reactant

::!led as the total avail;:d to react with the ~ used in combustion Teact that is in excess :lurn the combustible :Jiished by the limiting :mding stoichiometric :;;.r actually reacts, the =-e amount of the Iimit-

=ical reactions have 1-/0 universally agreed ~ontrary. Rather than :mfiict, we shall define

.,erial in the feed that leed is and into what

.riOf'. "f

.... = -~AK t

,.eu

t

The Chemical Equation and Stmchiometry 47

\((. 1.8

products the basis is being converted must be clearly specified or endless confusion results. Conversion is somewhat related to the degree of completiol! of a reaction, which is usually the percentage or fraction of the limiting reactant converted into products . (d) Selectil'ity expresses the amount of a desired product as a fraction or percent of the theoretically possible amount from the feed material converted. Often the quantity dellned here as selectivity is called efficiency, conversion efficiency, specificity, yield, ultimate yield, or recycle yield. (e) Yield, for a single reactant and product, is the weight or moles of final product divided by the weight or moles of initial reactant (P Ib of product A per R Ib of reactant B). If more than one product and more than one reactant are involved, the reactant upon which the yield is to be based must be clearly stated. The employment of these concepts can best be illustrated by examples.

FXA:\IPLE 1.25 Limiting reactant and incomplete reaction Antimony is obtained by heating pulverized stibnite with scrap iron and drawing ,'If the molten antimony from the bottom of the reaction vessel:

Sb,S, -+- 3Fe --+ 2Sb -+- 3FeS 0.600 kg of stibnite and 0.250 kg of iron turnings are heated together to give 0.200 kg ,,[ Sb metal. Calculate: (a) (b) (c) (d) (e) (f)

The limiting reactant. The percentage of excess reactant. The degree of completion (fraction). The percent conversion. The selectivity (percent). The yield.

SO/lIIion: The molecular weights needed to solve the problem and the g moles forming the rasis are

Sb,S, Fe Sb FeS

g 600 250 200

mol. wt 339.7 55.8 121.8 87.9

g mole 1.77 4.48 1.64

1

1

1

(a) To find the limiting reactant; we examine the chemical reaction equation and note that if 1.77 g mole of Sb,S, reacts, it requires 3(1.77) =, 5.31 g mole of Fe, \\hcrcas if 4.48 g mole of Fe react, it requires (4.48/3) = 1.49 g mole of Sb 2 S, to be

I j

I ,j

.. $

*.;;:;4

;

---- ----------._,------.-.. 48 Introduction to Engineering Calculations

Chap.

J

available. Thus Fe is present in the smallest stoichiometric amount and is the limitin1 reactant; Sb,S, is the excess reactant. (b) The percentage of excess reactant is: - 1.49)(100) =18. 8 % excess Sb "S %o excess = (1. 771.49 0

(c) Although Fe is the limiting reactant, not all the limiting reactant reacts, We can compute from the 1.64 g moles of Sb how much Fe actually does react: 1.64 g mole Sb 1 3 g mole Fe 2 g mole Sb

2 46 I F . g mo e e

=

If by the fractional degree of completion is meant the fraction conversion of Fe to FeS, then fractional degree of completion

= ~::~ = 0,55

(d) The percent conversion can be arbitrarily based on the Sb,S, if our interest is mainly in the stibnite: 1.64 g mole Sb 1 g mole Sb,S, = 0 82 I Sb S ' 2 g mole Sb . g mo e "

% conversion of Sb,S,

to Sb

= ~:~;(100) =

46.3 %

(e) The selectivity refers to the conversion of Sb,S, (we assume) based on the theoretical amount that can be converted, 1.49 g mole: selectivity =

~:!~(IOO)

=

55 %

([) The yield will be stated as kilograms of Sb formed per kilogram of Sb,S, that was fed to the reaction: yield

= 0.200 kg Sb = 1.- kg Sb 0.600 kg Sb,S,

3 kg Sb,S,

EXAMPLE 1.26 Limiting reactant and incomplete reactions Aluminum sulfate can be made by reacting crushed bauxite ore with sulfuric acid, according to the following equation: AI,O,

+ 3H,SO. - > Al,(S04)3

-+- 3H,O

The bauxite ore contains 55.4 weight percent alumirnrum oxide, the remainder bein, impurities. The sulfuric acid solution contains 77.7 percent H,S04, the rest being water. To produce crude aluminum sulfate containing 1U"98 Ib of pure aluminum sulfate. 1080 Ib of bauxite ore and 2510 Ib of sulfuric acid $IOlution are used. (a) Identify the excess reactant. (b) What percentage of the eXCeSs reactant was llIrScd? (c) What was the degree of completion of the rC'.!tction?

*_...__......ili'«... • ________________ .... ,, _ __

.._ , ._ _ . _ _ _.. • ............'IIir"...• _ _ ~_~'.... u

The Chemical Equatioll alld Stoichiometry 49

Chap. I :nount and is the limiting

.';".'ution:

The pound moles of substances forming the basis of the problem can be computed Ji ~

f.1110\\S:

I7981b AI,(SO,), I Ib mole AI,(SO,), 342 • I Ib AI,(SO _ 4 )3 1080 Ib bauxite 0.5541b AI,O, I Ib bauxite

-:ning reactant reacts. We ::lUa lIy does react:

=

5261b I . mo e

I Ib mole AI,O, 101.91b AI,O,

= 5 871b mole .

::ie Fe ::rion conversion of Fe to .-f

= 0.55

(a) Assume that AI 2 0, is the limiting reactant. Then 5.87 x 3 = 17.611b mole II:SO, would be required, which is present. Hence H,S04 is the excess reactant. (b) The AI,(SO,), actually formed indicates that: ~.~6Ib

Ine Sb 2 S3 if our interest is

mole AI,(S04)'

31b mole H 2 SO,

_

====-=..::===hITlbc-::m~0:;IC:-e:c:A,:li-'27i(S:;,0~4)C', - 15.781b mole H,S04 was used

15.78(100) 19.88

= 794% .



.!

0

1

(c) The fractional degree of completion was =

5.26 _ 090 5.87 - .

46.3%

'we?

1

\ based on the rX:\:\IPLE 1.27 TIle meaning of yield

:Jer kilogram of Sb 2 S, that

J

Your assistant has rushed into your office with a glowing tale of a yield of 108 I't'rcent in the manufacture of phthalic anhydride from xylene by the reaction C,H,(CH,h

+ 302 -->- C,H,(CO),O + 3H,O pbthylic anbydride

Sh,)uld you compliment him or deflate his accomplishment?

'S

Solutioll; If the claim is that 1081b of phthalic anhydride (Ph) are produced per 100 Ib ,{ \ylcne (X), the selectivity is not especially good.

c:xite ore with sulfuric acid.

JOO Ib X -H 2 0

,xide, the remainder being :nt H 2 S0 4, the rest being of pure aluminum sulfate, :m are used.

1081b Ph

II II

1

Basis: 100 Ib xylene Ib mole X = 0944 Ib I X 1061b X . mo e

1 1

Ib mole X == 0730 Ib I PI 148 Ib Ph . mo e 1

1

(I00)0.730Ib mole X_ 77 3% 0.9441b mole Ph . 0

J

If the claim is that 108 percent selectivity is achieved, the claim must be erroneous.

1 You should remember that the chemical equation does not indicate the true mechanism of the reaction or how fast or to what extent the reaction will

1 j

i!

I

, !

"

we

30.,

_)&.

,we..;;;:;;:

< •

SO Introduction to Engineering Calculations

Chap.

J

take place. For example, a lump of coal in air will sit unaffected at room tem· perature, but at higher temperatures it will readily burn. All the chemical eqt:.;. tion indicates is the stoichiometric amounts required for the reaction an,; obtained from the reaction if it proceeds in the manner in which it is written Also remember to make sure that the chemical equation is balanced before usin; it.

1.9 Digital computers in solving problems High-speed digital computers have made a considerable impact on problclr solving both in industry and in education. There is no question that a gooe fraction of today"s engineering graduates (who may well be working as engineer, in the year 2000) will see the usc of computers in their technical work expan,~ dramatically. Consequently, you should know how to use computers in your computations and if possible apply your knowledge to solving some of the more complicated problems in this book. Many routes of approach are used in eng:· neering education to familiarize students with the role of the computer in solvin, engineering problems. On the one hand, library computer programs ("canned" programs) can be employed to solve specific types of problems, absolving you 0' most of the programming chores. On the other hand, you can prepare the com· plete computer code to solve a problem from scratch. Problems marked with all asterisk in this book indicate problems that are especially appropriate for com· puter solutions by canned programs, and at the end of each chapter are a f"11 problems for which the complete computer program is to be written. Wide differences in computers and the proliferation of programming languages hah' made the inclusion of detailed computer codes in this text less than particularJI meaningful except for one or two solved examples. You should not attempt to justify the use of a computer as a time-savin~ device per se when working with a single problem. It is much better to look on the computer and the program as a tool which will enable you, after some expcr:· ence, to solve complex problems expediently and which may save time anl: trouble for a simple problem. but may not. Modern. high-speed computers can eliminate repetitive, timc-consuming routine calculations that would take year, to complete with a slide rule or a desk calculator. Also, computers can essentialJI eliminate numerical errors once the programs are verified. Typical problenl' encountered in this text that are most susceptible to computer solution arc (:1' the solution ofsimllltaneous equations, (b) the solution of one or more nonline:lf equations, (c) statistical data fitting. and (d) various types of iterative calculJ' tions. Typical industrial problems routinely solved are: (a) Heat and mass transfer coefficient correlations. (b) Correlation of chemical structure with physical properties. (c) Analysis of gaseolls hydrocarbon mixtures.

Chap, I ~~cted

at room ternrhe chemical equa_ ~ the reaction and which it is writien, _:anced before using

:mpact on problem ;:estion that a good '0rking as engineers :-_nical work expand computers in your ::cg some of the more ,::n are used in engi::omputer in solving -:-ograms ("canned" ::5. al'ng you of =:1 p, the comems m", "ed with an :-Jropriate for COnt:: chapter are a few be written. Wide :::ng languages have :ss than particularly ~~r

as a time-saving

:n better to look on

.:.. after some experi::-lay save time and ~eed computers can Co, would take years :llers can essentially Typical problems :,er solution are (a) =or more nonlinear ~'f iterative caicula--

:-perties.

---""'"-.~

-

...

Digital Computers in Solving Problems 51 (J)

Particle size analysis.

(e) Granular fertilizer formulation. (f) Evaluation of infrared analysis. If) Optimum operating conditions. (h) Comprehensive economic evaluations. (i) Chemical-biological coordination and correlation. (j) Vapor-liquid equilibrium calculations. (k) Mass spectrometer matrix inversion. (I) Equipment design. (m) Preparation of tables for platinum resistance thermometer.

One of the most significant changes in the use of computers has been in the of the word solution. It is widely recognized that a mathematical closed(,'rrn solution, while unquestionably pleasing aesthetically and relatively uni'.crsal in application, may not necessarily be interpreted easily. Curves and ,:raphs may constitute a more desirable format for a solution, but a computer ,1nJ a program essentially consist of a procedure whereby a numerical answer ,an be obtained to a specific problem. Repetitive solutions must be obtained to rrepare tables, graphs, etc .• but these, of course, can be prepared automatically '.\ Ilh an appropriate computer code. The use of a guaranteed canned computer program saves the user consider,lhle time and effort relative to programming his own code. Every major comruter center has, or should have, an extensive subroutine library, often stored on an on-line library tape, for effecting extensive numerical calculations and/or ;;mulation. Of course, such programs may still be fallible. When an unexpected answer appears, the user of the program must know enough about the problem 10 be able to account for the discrepancy. Often the trouble lies in the interpretation of what the program is intended to do, or it is caused by improper entry of parameters or data. If you can avoid these difficulties, you can concentrate on stating the problem correctly and estimating the expected answer; the mechanICS of obtaining the answer would be left to the computer. Considerably more complex and realistic problems can be solved, furnishing a deeper understanding "f the principles involved, if the latter are not obscured by concern with the rroblem-solving technique, If you do use the computer, you will be forced to be far more precise than ~ llU have been accustomed to being in the past. Because of the nature of the coml'(lter languages, i.e., the necessity for very precise grammar and punctuation, It is unusual to solve completely correctly an engineering problem on the first ;lpproach. Experience has shown that, even in the most carefully prepared programs, there will be a certain percentage of errors, which may include pro~ral11l11ing logic, actual coding, data preparation, or the card punching. For this rcason, processing of a problem requires prior program verification in which a (omputer run of a sample problem is executed for which the results are known, call be estimated, or can be calculated by manual methods. To facilitate such a ~,caning

1

I I

f

,1

52 Introduction to Engineering Calculations

Chap. 1

check-out as well as to guard against undetected and possible future data errors. a good program should include self-checking features strategically located in the program. If physically impossible conditions are created, the calculation should be stopped and the accumulated results printed out. You no doubt will require many tries before achieving success. The turn-around time, i.e., the time elapsed between submission of a program to the computer and its return for checking and possible resubmission in the case of error, must be fairly short if problems are to be solved in a reasonable amount of time. The use of remote time-sharing terminals can assist in the speedy resolution of programming errors. You will also be forced to think more logically and in greater detail than you have been accustomed to doing, if you do your own programming, about the problem and how to solve it. The computer is a rigid task master that requires precision in the statement of the problem and the flow of information needed to effect a solution. No ambiguity is permitted.

( h.ll'. I

Ran,.

t.,

\~

i tr-:'\l.

..

\\hit'.~r: \\·.d~h;1~'

~

\\ t\h.Irl~' 'h·(,r.i"~

.,, .

I. n,ln!t'\:\ Yn'~.

,

I ){'l'll)\ ,!l 1'1"'0,

.l. P.I~C. (

Puhl.

'\ l:

mcnt I';

WHAT YOU SHOULD HAVE LEARNED FROM THIS CHAPTER I. You should have memorized the common conversion units. 2. You should be able to convert units from the American engineering system to the cgs and SI systems and vice versa with ease, and understand the significance of g " 3. You should understand moles, molecular weights, density, specific gravity, mole and weight fraction, temperature, and pressure, and be able to work problems involving these concepts. 4. You should have well in mind the' proper approach to problem solving and be able to effectively put into practice the principles discussed in the chapter. 5. You should know how to apply the principles of stoichiometry to problems involving chemical reactions.

rr 11..-('

... 1'.111 I. \I ( 19~11 ~, (j'n.I\,

"J, :'0"

6. t\. 1!lit.. r '; t \01 i \11.11 I

1.1. i :,

I..

General I. Benson, S. W., Chemical Calculations, 2nd cd., John Wiley, New York, 1963.

3. Henley, E. J., and H. Bieber, Chemical Engineering Calculations, McGraw-Hili. l-:ew York. 1959.

.c.

Hougen, O. A., K. M. Watson, and R. A. Ragatz, Chemical Process Principles. Part I. 2nd cd., John Wiley, New York, 1959.

!I. liltlcjohn. C. E., and F. G. Meenaghan. An Ifllrodllction to Chemical Engineering, Nostrand Reinhold, New York. 1959.

".,1

<

PRO£l!LM

SUPPLEMENTARY REFERENCES

2. Considine, D. M., and S. D. Ross. cds .• Handbook of Applied Instrumefllation, McGraw-Hili, New York, 1964.

, J

I

~

I'

111

Chap. I

,:e future data errors. f:ically located in the ,e calculation should ,0 doubt will require l.e., the time elapsed , return for checking ely short if problems remote time-sharing -:mg errors. Y Oll will , than YOll have been ,Qut the problem and requires precision in m needed to effect a

t,

H

c-ah' ,

tr f#

MWr (J't

Problems

It '

53

6. Ranz, W. E., Describing Chemical Engineering Systems, McGraw-Hill, New York, 1970. , Whitwell, J, c., and R. K. Toner, Consamtion of Mass and Energy, Ginn/Blaisdell, Waltham, Mass" 1969. K. Williams, E. T., and R. C. Johnson,

Stoichiometry for Chemical Engineers,

\lcGraw-Hill, New York, 1958.

Units and Dimensions I. Danloux-Dumesnils, M., The Metric System, Oxford University Press, Inc., New

York, 1969. 2. Donovan, F., Prepare NolV for a Metric Future, Weybright and Talley, New York,

1970.

:5 CHAPTER

':sity, specific gravity,

m

Chap. I

~,

..:nits. :n e~ 'ng system :mdeL "J the signi-

:M2eeiWC

Page, C. H., and P. Vigoureux, The International System of Units (SI), NBS Spec. Publ. 330, Jan. 1971. (For sale by the Superintendent of Documents, U.S. Government Printing Office, Washington, D,C. 20402, SD Catalog No. 13: 10: 330, price 50 cents.)

4. Paul, M. A., "The International System of Units," J. Chem. Doc" V. II, p. 3 (1971) . ~.

O'Day, E. F., Physical Quantities and Units, Prentice-Hall, Englewood Cliffs, N.J., 1967.

6. Klinkenberg, A., "The American Engineering System of Units and Its Dimensional Constant go" Indus. Engr. Chem., 61, no. 4, p. 53 (1969).

:md be able to work nroblem solving and :::ussed in the chapter. :llDmetry to problems

.,'. New York, 1963,

:1Jplied Instrumentation, ..;iations, McGraw-Hili, ::cal Process Principles. Chemical Engineering,

PROBLEMS" 1.1, Electrically driven automobiles have been proposed as one solution to the problem of excessive automobile emissions in large cities. The Ghia Rowan is a sleek !talo-American prototype, an electrically driven system which converts the kinetic energy produced by the car's natural momentum into chemically stored battery energy via the car's braking system. The vehicle is roughly 3 m long, weighs 596 kg, and has a range of about 320 km at a speed of 41-43.5 mi/hr. Powered by two 9-hp electric motors placed under the rear seats and connected to the control box, the Rowan has no ditTerential, transmission system, or clutch assembly, Rewrite the specifications given so that they are all consistent in (a) the Sl system and (b) in the American engineering system. 1.2. Hollywood (AP)-Blonde Jan Eastlund from Europe, a loser in last year's Miss Universe pageant, came up a winner this week with a studio contract and a starring role in her first movie. Jan speaks English with a slight accent, and is barned by Our units of measure. Asked for her measurements, she answered, ·';o.ly bust is 92, waist 58, and my hips 92," A studio press agent gave an imitation of a '9.-\n asterisk designates problems appropriate for computer solution, Also refer to the "'",puter problems after Problem 1.104,

"!

-

.... -.~~---- ..... -.-.--.....

. p.:

Chap. I

. below 760 mm Hg. ::Jg: .

.m

,

:Hg.

::Jute the number of

2

Material Balances

:nade. c:n new page .

..a fluid as a function :=ts. Equation (1.10) " dense than water),

Conservation laws occupy a special place in science and engineering. Common statements of these laws take the form of "mass (energy) is neither created nor destroyed," "the mass (energy) of the universe is constant," "the mass (energy) of any isolated system is constant," or equivalent statements. To refute a conservation law, it would be sufficient to find just one example of a violation. However, Lavoisier and many of the scientists who followed in his path studied chemical changes quantitatively and found invariably that the sum of the weights of the substances entering into a reaction equaled the sum of the weights of the products of the reaction. Thus man's collective experience has been summed up and generalized as the law of the conservation of matter. The only "proof" we have for this law is of a negative type-never in our previous experience has any process been observed that upon thorough examination (including consideration of the precision of the data) has been found to contradict the principle. Of course, we must exclude processes involving nuclear transformations, or else extend our law to include the conservation of both energy and matter. (Insofar as we shall be concerned here, the word process will be taken to mean a series of physical operations on or physical or chemical changes in some specified material.) In this chapter we shall discuss the principle of the conservation of matter and how it can be applied to engineering calculations, making use of the background information discussed in Chap. I. Figure 2.0 shows the relations between the topics discussed in this chapter and the general objective of making material and energy balances. In approaching the solution of material balance problems, we shall first consider how to analyze them in order to clarify the method and 71

;;""p;;_,."....._ ......__

"''''..........'''; _ _u",·""*_ _... _ _.,._ . _ ........._ _.....-.,.,."'"·F."._ _ _... u""!"'_........,. .."......... ' "'.M_ _....._ ......

.~,

.H

M #4R

72 Material Balances

Chap. 2

Sec. 2.1

2.1

Material balane,

To take into account the law of the conservation < balance is nothing more inventory of mass for a ' of the material balance 3i reaction:

Problem Statement

{

Material and Energy Balances

Fig. 2.0. Heirarchy of topics to be studied in this chapter (section numbers are in the upper left-hand corner of the boxes).

the procedure of solution. The aim will be to help you acquire a generalized approach to problem solving so that you may avoid looking upon each new problem, unit operation, or process as entirely new and unrelated to anything you have seen before. As you scrutinize the examples used to illustrate the principles involved in each section, explore the method of analysis, but avoid memorizing each example by rote, because, after all, they are only samples of the myriad problems which exist or could be devised on the subject of material balances. Most of the principles we shall consider are of about the same degree of complexity as the Jaw of compensation devised by some unknown, self-made philosopher who said, "Things are generally made even somewhere or some place. Rain always is followed by a dry spell, and dry weather follows rain. I have found it an invariable rule that when a man has one short leg, the other is always longer!" In working these problems you will find it necessary to employ some engineering judgment. You think of mathematics as an exact science. For instance, suppose that it takes I man I 0 days to build a brick wall; then I 0 men can finish it in 1 day. Therefore 240 men can finish the wali in I hr, 14.400 can do the job in I min, and with 864,000 men the wall will be up before a single brick is in place! Your password to success is the famous IBM motto: THINK.

accU~u~atiOn} wlthm the system

_ -

{inre th

rClU

systc bound,

In Eq. (2.1) the generat;('1 loss by chemical reaction should understand in read a time interval of any (.k~ differential time. Equation (2.1) red ue,'l (or usage) of material \\tt n acC!

and reduces further to L: within the system, If there is no flow in

3t)cI I

(2, I) reduces to the ba,;( "

within an enclosed isoLlttd Inherent in the form'.:! 3 system for which the b.l": or whole of a process a, ,.,: 2.1 shows a system in Wh,,:1.

,I

,

"

u"

F

:::t,

I·j~.

!,I

*_...-....,

.._ ...._ _,...ooI·.*liiiri_......"',....., .. ' ...._rfilif......·...'.·_ _...rtlii"...._ ..t ..t"',.,Ail!:, ..'t.. • .. r't... ·M ..s_e,.·_ _·.. ,." ....... f 11"_ _......•.. · .,.,.............

...iJj

Chap. 2

Material Balance 73

Sec. 2.1

2.1 Material balance To take into account the flow of material in and out of a system, the generalized law of the conservation of mas~ is expressed as a material balance. A material I>alance is nothing more than an accounting for mass flows and changes in inventory of mass for a system. Equation (2.1) describes in words the principle of the material balance applicable to processes both with and without chemical reaction:

{ aCc~7t~li~tiOn} the system

t~~~~~h}

{t~~~~~~}, {ge:~~~:~on}

_ { system boundaries

system ' boundaries

the

-

system

{con~~t~f:iOn} the system

(2.1) In Eq. (2.1) the generation and consumption terms in this text refer to gain or loss by chemical reaction. The accumulation may be positive or negative. You should understand in reading the words in Eq. (2.1) that the equation refers to a time interval of any desired length, including a year, hour, or second or a differential time. Equation (2.1) reduces to Eq. (2.2) for cases in which there is no generation (or usage) of material within the system, accumulation

2~ed

c:"mloy some -::ience. For -'lell 10 men 14,400 can ~'re a single :ro:: THI:-IK.

(2.2)

and reduces further to Eq. (2.3) when there is, in addition, no accumulation within the system,

=

each new 10' anything -.:Jinstna1:e the :.-S. but avoid =-"ILiy samples :'C!! subject of - :)ut the' same :'!! unknown, =ewhere or . ollows rain. =:g, the other

= input - output

input

= output

(2.3)

If there is no flow in and out of the system and no generation (or usage), Eq. (2.1) reduces to the basic concept of the conservation of one species of matter within an enclosed isolated system. Inherent in the formulation of each of the above balances is the concept of a system for which the balance is made. By system we mean any arbitrary portion or whole of a process as set out specifically by the engineer for analysis. Figure 2.1 shows a system in which flow and reaction take place; note particularly that

, ~ ~.--.------"

/

I Combustion r uel_~' chamber

\

Nozzle

\

I

. ,_Combust,on gases

~

\

'"

,

- -- -

Oxygen

.~

Sys tern boundary

Fig. 2.1. A now system with combustion.

1

j

74

Chap. 2

Material Balances

the system boundary is formally circumscribed about the process itself to call attention to the importance of carefully delineating the system in each problem you work. In nearly all of the problems in this chapter the mass accumulation term will be zero; that is. primarily steady-state problems will be considered. (See Chap. 6 for cases in which the mass accumulation is not zero.) For a steadystate process we can say, "What goes in must come out." Illustrations of this principle can be found below. Material balances can be made for a wide variety of materials, at many scales of size for the system and in various degrees of complication. To obtain a true perspective as to the scope of material balances, examine Figs. 2.2 and 2.3. o

Atmosphere Nz

..~:::'"~w , , ('.

o.

"'_

,~

3,800,000.000

"-,1,, . -~("~:j

[

•...,. ,-

~_J .,..,1",

..

<;I'v. 'I,.',

.("~"<

.

l .... ""'< .. l'

,t ••

~

Fig. 2.2. Distribution and annual rates of transfer of nitrogen in the biosphere (in millions of metric tons).

Figure 2.2 delineates the distribution of nitrogen in the biosphere as well as the annual transfer rates, both in millions of metric tons. The two quantities known with high confidence are the amount of nitrogen in the atmosphere and the rate of industrial fixation. The apparent precision of the other figures reflects chiefly an effort to preserve indicated or probable ratios among different inventories. Because of the extensive use of industrially fixed nitrogen, the amount of nitro-

'"--

.----~.

gen uvuil:!l· the utlllO,P: analysis ()f prodUcts. C deterioratic) We she quantities i, statement. f COUnts, pop I n the r design. in th cc)ntroi. and nd from so~ \'t soybeans In the desi~r \11 sorts (;f quite a few l

-

----.

..............

-,rocess itself to (;all 2m in each problem mass accumulation will be considered. =zero.) For a steadyillustrations of this

Material Balance

Sec. 2.1

Chap. 2

75

~:::=--'i~:-=::::::::=;~::::::=-----..J!212.. not h"rVflsted;in-

rR.;;;:;';;-:'~-l+-~~-1----'--' =:::-~'~'~O::;:;::=1 ,t!':.;ldlloQn

.:

manure etc:.

kiillrves.ted bV

seclS. buds. dls_

firH.etC.. , -_ _":===r-===::':~;Q..~falm . . . ast. ease. ro(.

l09glng debris

_150

husks, straw, VInes. culls.

primings. roots, stalks. elc.

=aterials, at many ::Jiication. To obtain :ne Figs. 2.2 and 2.3.

firewood

L----r,;;--..;;:~____..:;::_.~=1==r-=::'..:2E.--. losses to spoilage &. verm.n

balk.. scrap. sawdust, cl'ups.

IIgnm waste. elc

-/-----\-_.=.!._ .. storage losses & food,p1oceSSlnlll

-,ospheric

waste

··:on

Irash. scrap. sawdust

consumption

-luther. glue. tob.lceo • Of~IlIC

• sol\lenls

chem.cals • brewmg and distilling • soap ostilileh Ifof non-food use}

~~:==~~~=~~~~~~=~~~~~jErags.

trash gar..... - 50 reSpiration. bage. sewage ......- - - - - - - - - - - - - - - - - - - - - - - - - + - t a n n e r y wastes. chemical wastes . • I~

Fig. 2.3. Production and disposal of the products of photosynthesis. (Taken from A. V. Kneese, R. U. Ayres, and R. C. d'Arge. Economics and Ihe Enl'ironmcnt, Resources for the Future, Inc., 1755 Massachusetts Ave., Washington, D.C., 1970, p. 32.)

~ TIl

the biosphere

::uhere as well as the quantities known '::Jsphere and the rate ::!!urc' --fleets chiefly i\'entories. .. nt of nitro,,'0

gen available to land plants may significantly exceed the nitrogen returned to the atmosphere by denitrifying bacteria in the soil. Figure 2.3 resulted from an analysis of the organic wastes arising from the processing of food and forest products. Changes in any of the streams can pinpoint sources of improvement or deterioration in environmental quality. We should also note in passing that balances can be made on many other quantities in addition to mass. Balances on dollars are common (your bank statement. for example) as are balances on the number of entities, as in traffic counts, population balances, and social services. In the process industries, material balances assist in the planning for process design, in the economic evaluation of proposed and existing processes, in process control, and in process optimization. For example. in the extraction of soybean oil from soybeans, you could calculate the amount of sol\"<":01 required per ton of soybeans or the time needed to fill up the filter press, and lise this information in the design of equipment or in the evaluation of the economics of the process. All sorts of raw materials can be used to produce the same end product and quite a few different types of processing can achieve the same end result so that

0,** ..

Chap. 2

76 Material Balances

case studies (simulations) of the processes can assist materially in the financial decisions that must be made. Material balances also are used in the hourly and daily operating decisions of plant managers. If there are one or more points in a process where it is impossible or uneconomical to collect data. then if sufficient other data are available, by making a material balance on the process it is possible to get the information you need about the quantities and compositions at the inaccessible location. In most plants a mass of data is accumulated on the quantities and compositions of raw materials, intermediates, wastes, products, and by-products that is used by the production and accounting departments and that can be integrated into a revealing picture of company operations. A schematic outline of the material balance control in the manufacture of phenol is shown in Fig. 2.4.

I \

~

Process mass and energy bolance BOl'S: 2:,000 Ib, [email protected]{'()F

CII,m.col,

"omt

"

1

',040

60

Ho z C0 3

1,100

H,O

13,960

" "

N020

BT~

'f

(,H,

2,000

6O

HZSO.

2.000

6O

II

"

SerVtcn

HtOl X Ie'

I

i Weier

I

!O"~c'-f"td

I Recycle (lb.)

R'CJ'C't lIb) CoPldtn- Water

~

EVODorolor

Q,

21,000 9

"

27,000

""'~

00

(D'Ptcl-f,red ~fOI)

00

FUSion

pots

""",~

Centrifugal 1,000

"

H,SO ..

6,.

" 60

00

~'d'f'~~

'"

18,000 40,000

._-

J ;~ STeam

-Tolaa --

1,000

0"

200

21,000

"0

100

7,000

,28

'00

27,000

1.62

010

0"

000 000

Hame

lb

I! 2i~i;JI'--~iS2,OO~",f,2,:leOI80il

st.ll

i

.. _-

Vacuum

stili

Ie

! ~I

: (' :

,

Sulfo-

"

H,O

'f

[>,>0",15

(,H,

44,7

8h.Xl0'

',000 8

noter

.0

Chem,cal,

lote

heat)

7,000

6O

I~.

Melte~

... 22,1

Sleom

0" Stry,cu

,

,

I

I,

tll-(

,,

n I I

....

t

0<0

'00

000

2,433 ~~2S0~ ",0

11~~ia-1,009

Na2S0,

0,B~8

",0

H,O

4'I,e

i00,000

le,Ooo

'06

"0 2,400

~"ooJ'000

10-'<'

,

C,tiSOH

I-Z'JJO~

Fig. 2.4. Material (and energy) balances in the manufacture of phenol pre. sented in the form of a ledger sheet. Taken from Cilelll. Eng., p. 177 (April 1961). by permission.

-

"'0

__

...,...,.

I

~_a

_ _,. -'._. ..to,.;

..

.._ .......'IO$1i 5

"1II1......·.-..·_.""'".....·~....'..h....' ...• ..M.. +...Ff:!...-.-"l:,A-5~~r".."4'' ' """""v",,oliloif"..,,,OIi.u'-."'_',,"tiilileU,,,,,,,·"''..HiO'_,Ii.,•••"II •~ •_ _-

_1W ..........

.' .. . .

~.

77

Material Balallce

Chap. 2

'-Jally in the financial EXAMPLE 2.1 - operating decisions process where it is _':Ient other data arc :5 possible to get the ~s at the inaccessible :1 the quantities and ~~ts, and by-products -:11S and that can be A schematic outline ':101 is shown in Fig.

Water balance for a river basin

Water balances on river basins for a season or for a year can be used to check predicted ground water infiltration, evaporation, or precipitation in the basin. Prepare J water balance, in symbols, for a large river basin, including the physical processes Indicated in Fig. E2.1 (all symbols are for I yr).

Atmosphere Water, Vapor, Ice = A, River Flow =

R'--r---U.l~--"""::::==:=;~S"""'

__..,,.-__L_

Ground Water = G,

SF

"

Fig. E2.1.

Nome

Solutioll: Equation (2.2) applies in as much as there is no reaction in the system. Each term in Eq. (2.2) can be represented by symbols defined in Fig. E2.1. Let the subscript I, designate the end of the year and II designate the beginning of the year. If the system is chosen to be the atmosphere plus the river plus the ground, then the accumulation is (SA', - SA'') + (SR', - SR'') + (SO" - So,')

'04 --

'90

20.

The inputs are AI + RI the material balance is

..

0.2!!

,

1.£.2

12.

(a)

0<0

,

..

'~'_I2

.. ,.. ,.0

+ GI

(SA', - SA'')

= 0.72

River Flow = Rz Ground Water = Gz

.., 3tu. X10'

Atmosphere Water, Vapor,lce = Az

(AI - A 2 )

and the outputs are A2

+

(SR', - SR'')

+ (R I

-

R2 )

+

+ R2 + G2.

Consequently,

(SO" - So,,)

+ (G I

- G 2)

If the system is chosen to be just the river (including the reservoir), then the material balance would be

,

2,433

LO'

tI,OOO

",0

1,009

NC2S03

4.858

",'

2,400

"H~OH

N~2S0~

"0

21,700

·f phenol pre· -:_ Eng., p. 177

TOlol5

(b)

(SR', - SR',) = (R I -I- P) - (R 2 accumulation

I I

input

+output E + W)

and analogous balances could be made for the water in the atmosphere and in the ground.

One important point to always keep in mind is that the material balance is

a balance on mass, not on volume or moles. Thus, in a process in which a chemical reaction takes place so that compounds are generated and/or used up, you will have to employ the principles of stoichiometry discussed in Chap. I in making a material balance. Let us analyze first some very simple examples in the use of the material balance.

----'

...... ,

.....,.,...

..

...._'. .IfA!....";""._III1!11,4'1AI"""x . .... ,!!'!.:...

--~....... ''''''''.-.-....... - ... ~ .... f ;;,... ...... , •.,....,'"'_'"........."';p"'P"£""._iI!l!4~_,._,A .. ~"'''"'_"xllO:_z¥"a_

..

.........,"", ,--~

78

Chap. 2

Material Balances

(I"

EXAMPLE 2.2

\

Material balance

Hydrogenation of coal to give hydrocarbon gases is one method of obtaining gaseous fuels with sufficient energy content for the future. Figure E2.2 shows how a frcc· fall fluid bed reactor can be set up to give a product gas of high methane content. Inerts

Product Gas H2 +CO + CH. +CO 2

Raw Coal In

400°C

t

J

t

t

Combustion Goses N2+ CO 2 Inerts

1.\,\\11"

!

t

I

t.tl II .,1

II

t

Steom

t

Air

Fig. E2.2.

Suppose, first, that the gasification unit is operated without steam at room temperature (25°C) to check the air flow rates and that cyclones separate the solids from the gases effectively at the top of the unit so that no accumulation (build up) of coal. or air, occurs in the unit. If 1200 kg of coal per hour (assume that the coal is 80 percent C, 10 percent H, and 10 percent inert material) is dropped through the top of the reactor, (a) How many kilograms of coal leave the reactor per hour? (b) If 15,000 kg of air per hour is blown into the reactor, how many kilograms of air per hour leave the reactor? (c) If the reactor operates at the temperatures shown in Fig. E2.2 and with the addition of 2000 kg of steam (H 2 0 vapor) per hour, how many kilograms per hour of combustion and product gases leave the reactor per hour.

t hi I,

.. "

I!

I II

l' r~·

.

Solution:

Basis: 1 hr (a) Since 1200 kg/hr of coal enter the reactor, and none remains inside, 1200 kg/hr must leave the reactor. (b) Similarly, 15,000 kgjhr of air must leave the reactor.

"~-...-...-~

..

"

Chap. 2

'.".IW*.",_

, too "t.

ow aer'd e

Material Balance 79

Sec. 2.1

(c) All the material except the inert portion of the coal leaves as a gas. Consequently, we can add up the total mass of material entering the unit, subtract the inert material, and obtain the mass of combustion gases by difference: C'btaining gase"'s how a free-

1200 kg coal

10 kg inert 100 kg coal

ane~tent.

entering material Coal Air Steam Total

EXAMPLE 2.3

=

120 k inert g

kg 1,200 15,000 2,000 18,200 - 120 = 18,080 kgjhr of gases

Material balance

(a) If 300 Ib of air and 24.0 Ib of carbon are placed in a reactor (see Fig. E2.3) at 600°F and after complete combustion no material remains in the reactor, how many pounds of carbon will come out? How many pounds of oxygen? How many pounds total?

24.01b C - - - , , - - - - 300 Ib Air

::!Un

'-----7

at: room- tem-

·."e:solids from-the ,d. of coal, or ::oatiS' 8{J pe:rcen t ::h the:' top' of the

Fig. E2.3.

um

(b) How many moles of carbon and oxygen enter? How many leave the reactor? (c) How many total moles enter the reactor and how many leave the reactor?

Sollilion: cc·tmmy. kilograms ~Z'and with the ,. many kilograms :or- pe:r hour.

(a) The total material leaving the reactor would be 324 lb. Since air has 21.0 percent 0, and 79.0 percent N" Basis: 300 Ib air 300 Ib air I Ib mole air 29.0 Ib air

21.0 Ib mole 0, 100 Ib mole air

241b C , I Ib mole C 12.0 Ib C ""l1ains inside., 1200

=

=

2.181b mole 0,

2 00 Ib I C . mo e

The oxygen required to completely burn 24.0 Ib orc is 2.00 Ib mole; consequently 2.18 - 2.00

=

0.18

.",, , ,-,

__ ' ..J_ _!"--~"

80

Material Balances

Chap. 2

Src.1.:

is the number of moles of unused oxygen, and this is equivalent to (0.18)(32) = 5.76 Ib of 0, as such leaving the reactor. In addition, 881b of CO, leave and 28.2 Ib N, = 230 Ib N 2.181b mole 0, 79.0 Ib mole N, 21.0 Ib mole 0, lib mole N, ' The moles of N i are 230lb N'lllb mole N, 28.21b N,

820lb I N . mo e ,

=

Summing up all input and exit streams, we have

10to

ill

Ib

lb mole

alit

Ib

lb mole

0,

70 230 24 324

2.18 8.20 2.0a 12.38

0, N, CO,

5.76 230 88 324

0.18 8.20 2.00 10.38

N, C Total

hah

There will be no carbon leaving as carbon because all the carbon burns to CO,; the carbon leaves in a combined form with some of the oxygen. (b) The moles of C and 0, entering the reactor are shown in part (a); no moles of C, as such, leave. (c) 12.38 total moles enter the reactor and 10.38 total moles leave. Note that in the example above, although the total pounds put into a process and the total pounds recovered from a process have been shown to be equal. there is no such equality on the part of the total moles in and out, if a chemical reaction takes place. What is true is that the number of atoms of an element (such as C. 0, or even oxygen expressed as 0,) put into a process must equal the atoms of the same element leaving the process. In Example 2.3 the total moles in and out are shown to be unequal:

moles ill 0, 2.18 8.20 N, 2.00 C 12.38

10101

°

0,:

0,

2.18 Total

N,: C:

N, C

componelll moles

= 0,

2.18 8.20 2.00

0, 0, in CO, Total N, C in CO,

= 0,

:2 2

P{0

\1urh 01 I g II P

·,t.~U If)

:hc tC\t

I..

b.II.1 nl'e I"

.1 ffne',I" \(lluthHl ,,<

We

.11

tll,lt "ill \ ' ",(" .. oDd, 1:,

although the atoms of in (expressed as moles of 0,) equal the atoms of 0 (similarly expressed as 0,) leaving:

component moles ill

a\1l'1

We 11, " lern. '1 !; tne forl1l. mathern.lt:

~rrll(;lt"' ..

moles oul 0.18 0, 8.20 N, 2.00 CO, 10.38

10101

(The

;'f'lhlcnl\ : lllnlpill.. .1!, \ ! \ .. t.ll

0111

0.18 2.00 2.18 8.20 2.00

Keeping in mind the above remarks for processes involving chemical reactions, we can summarize the circumstances under which the input equals the

'f: .: '

.'" he I T~~' h\

I"

'\!i

i

C

,

ttl j':

~'

Chap. 2 (0.18)(32) ~

=

5.76 Ib

Program of Analysis of Material Balance Problems 81

Sec. 2.2

output for steady-state processes (no accumulation) as follows:

and

.,Olb N,

:ole

{

Nmp'"'"'

1

18

20 )0 :18

. burns to CO,; the :Jar! (a); no moles of leave. :C.ds p. .0 a pto· shown to be equal, . out, if a chemical :.'ms of an element :ess must equal the ':.3 the total moles

d the atoms of 0 fes out

0.18 2.00 2.18 8.20 2.00 Ig chemical reacinpl" 'quaIs the

type of balance Total mass Total moles Mass of a pure compound Moles of a pure compound Mass of an atomic species Moles of an atomic species

total balances

balances

Equality required for input and outpul of steady-state process withollt with chemical chemical reaction reaction Yes Yes No* Yes No* Yes No* Yes Yes Yes Yes Yes

(The asterisk indicates that equality may occur by chance.) We now turn to consideration of problems in which the accumulation term is zero. The basic task is to turn the problem, expressed in words, into a quantitative form, expressed in mathematical symbols and numbers, and then solve the mathematical statements.

2.2 Program of analysis of material balance problems Much of the remaining portion of this chapter demonstrates the techniques of setting up and solving problems involving material balances, Later portions of the text consider combined material and energy balances. Since these material balance problems all involve the same principle. although the details of the applications of the principle may be slightly different, we shall fIrst consider a· generalized method of analyzing such problems which can be applied to the solution of any type of material balance problem. We are going to discuss a method of analysis of material balance problems that will enable you to understand, first, how similar these problems are, and second, how to solve them in the most expeditious manner. For some types of problems the method of approach is relatively simple and for others it is more complicated, but the important point is to regard problems in distillation, crystallization, evaporation. combustion, mixing, gas absorption, or drying not as being diff~rent from each other but as being related from the viewpoint of how to proceed to solve them. An orderly method of analyzing problems and presenting their solutions represents training in logical thinking that is of considerably greater value than mere knowledge of how to solve a particular type of problem. Understanding how to approach these problems from a logical viewpoint will help you to develop those fundamentals of thinking that will assist you in your work as an engineer long after you have read this material.

82

Chap. 2

Material Balances

If you want to make a material balance for a system, in general you have to have on hand information dealing with /lro fundamental concepts. One of these is the mass (weight) in all streams of the material entering and leaving the system and present in the system. The other information required is the composition of all the streams entering and leaving the system and the composition of the material in the system. Of course, if a chemical reaction takes place inside the system, the equation for the reaction and/or the extent of the reaction arc important pieces of information. Let us look at some problems in general by means of the "black box" technique. All we do is draw a black box or line around the process and consider what is going into the process and what comes out of the process. This procedure defines the system, i.e., the process or body of matter set out to be analyzed. As previously mentioned, we shall assume that the process is taking place in the steady state, i.e., that there is no accumulation or depletion of material. Even if the process is of the batch type in which there is no flow in and out, let us pretend that the initial material is pushed into the black box and that the final material is removed from the black box. By this hypothesis we can imagine a batch process converted into a fictitious flow process, and talk about the "streams" entering and leaving even though in the real process nothing of the sort happens (except over the entire period of time under consideration). After you do this three or four times, you will see that it is a very convenient technique, although at the beginning it may seem a bit illogical. Let us start with a box with just three streams entering or leaving (all three cannot enter or leave, of course, in a steady-state process). In the examples in Figs. 2.5 and 2.6, F stands for the feed stream, P stands for product, and W stands for the third stream or "change," but any other symbols would do as well. The procedure of solution depends to some extent on which of th~ compositions and weights you know. All the problems in Figs. 2.5 and 2.6 have sufficient information shown to effect their solutions, except Fig. 2.5(d). With respect to Fig. 2.S(a) we could use a total mass balance to compute the value of W:

F=P+ IV 100

=

60

+

IV

or

W=40

~J% f~--"

"\_

·~o

-'0

tte,').-

.

CC~'$!~;c"

:c'%

t~:··

·~O

I{.:.~:

!O

'lev-

Compo$j~10!"l ~'..Q

40

% EIC'.· ~.. H,O

10

',lee·,

CompOSlllr", ~~();o

EIC·,

40

HI\:

10

MoO'

(2.4)

To determine the composition of W, because no reaction takes place. we can write a mass balance for each of the components: EtOH, H 2 0, MeOH [w = mass ("weight") fraction]. Ollt

EtOH

+ WE.OH.Hl40) (0.05)(60) + wH,o,,(40)

(2.4b)

.= (0.15)(60) + w",oH, .. (40)

(2.41:)

(0.50)(100) ~c, (0.80)(60)

H 20

(0.40)(100) .~

MeOH

(0.10)(100)

(2.4a)

h.\ Jdinilh)t. pUll'!! by di!": h,Hl' (0

be 'r

arc not {'l'u;,f

$'1'"

Chap. 2

did ,,-j ""Y"i1f'tt$Xrlrr "ttcf'tltMWtiCtre Cl'iliflHn

!

Program of Analysis of Alalerial Balance Problems

Sec. 2.2

:n general you ha vc and leaving the .-equired is the com.::nd the composition :m takes place inside . of the reaction are

Mass= ? Composition 50% EtOH 40 HzO 10 MeCH

F=IOOtb

Q ~~~" " EtCH }

W



P=601b

t

ftM'hr

ftf -at

83

Parameters Known Unknown Moss Compo Moss Compo

~ition

-'11 concepts. One of ~ing

ntUtMW it n " t 11



Composition 80% EtOH 5 H2O 15 MeOH

2

6

1

2-



Composition 91 % EtOH 9 H2O

1

9t I 2

0

System Boundory (0)

"1e "black box" tech-

:-ocess and consider 2ess. This procedure Jut to be analyzed. .3 taking place in the of material. Even if .j out, let us pretend .-'1t the final material .::n imagine a batch . :lout the "streams" of the sort happen s .. Af' 11 do this -ech, altho·ugh .Jr leaving (all three in the examples in :Jr product, and IV 1S would do as well. A th" compositions .ave sufficient infor'alance to compute

Mass=?

Composition 50% EtOH 40 HzO 10 MeOH

F=IOOlb

W

t

·0

Composition {MeOH 22% H20 78 PMass=? System Boundary

1

j

(b)

Mass=? Composition 50% EtOH 40 HzO 10 MeOH

W

PMoss =?

F= 100 Ib

1

Composition { 0.5 % EtOH 92.5 H2O 2.5 MeOH

System Boundary

f

CompOSition 80% EtOH 5 H2O 15 MeOH

1 I

,

9

2

0

6

2

2-

I

(e)

Composition Moss = ? Composition 50% EtOH 40 H2O to MeOH

W

EtOH} H20 =? MeOH PMoss= ?

F= 100Ib

System Boundary

Composition 80% EtOH 5 H2O 15 MeOH

1

(2.4) (d) No Solution

tkes place, we can

• The third percent con be obtained by difference

1,0, MeOH [ev

t A stream with only two components has 0% for the third component

=,

Fig. 2.S. Typical material balance problems (no chemical reaction involved).

Because (2.5) 0)

(2.4a)

'I

(2.4b)

0)

(2.4c)

,.-.

-'~'.~,-~

....... ""'r

by definition, only two of the mass fr3ctions are unknown; the third can be computed by difTercnce. Consequently. only two of the component material balances have to be soll·cd simultaneously. and in this particular instance the equations are not coupled so that they can be solved independently.



;a

lSe4.&

41·

$ ~,;*,

'$ ,£

"a.mx,

:;4!4J!1 ,..-

:

Parameters

Un~nown KnCMn Moss Comp Moss Com~-

CH 4 clOD %

D

F=161b

Pc?

c~ =? N =?

I-----~

t

st

i 1

3'

z

System /'" Mole % Boundory --------- . {Oz 21% (64Ib) A" =274Ib Nz79%(210Ib)

H20=

?

pcr(Cnl \} only one

(0)

Mole% Oz 21% Air { Nz 79%

number (l the flllmh F0J" i1 written a' i.e., hydr"l mass (IIO! J balance. '1 ~ and oXYf'cn problem ,:, Also keel' "

P =tlb mele

16.3% COz 04 CO 43 Oz 79.0 Nz

H=?

1

9t

2

2'

1

all

2

0

(b)

5.5% CoO

yr----

Ca(OH)z + NaZC0 3 _...:.F_c-,-?_ _-,-_,

H20 10.5% Na zC0 3 System ~~~ Boundary

/~

CaC0 3 + 2 NaOH NaOH 2500 gal/hr 10% NoOH solution 0.5% NO ZC0 3

are conserv Figlll~

1..---- 50% WasJe Sludge CoC0 3

t% NaOI:l

(e)

• The thJfd percenJ can be ablarned by difference.

tComponents not listed are 9resEIlt

In

0%..

I

However. ,n must be ;hl the nat lire ( The w; As a gencIJ (a) [) ri (b) Pl.,! (e) See e;,d

Fig. 2.6. Typical material balance problems (with chemical reaction).

You should recognize that not all of the four mass balances, Eq. (2.4)(2.4c), are independent equations. Note how the sum of the three component balances (2.4a), (2.4b), and (2.4c) equals the total mass balance. Consequently. the number of degrees of freedom or the number of independent equations will be equal to the number of components. I n many problems you can substitute the total material balance for anyone of the component material balances if you plan to solve two Of more equations simultaneously. Since many of the problems you will encounter involving the use of material balances require you to solve for more than one unknown, you should remember that for each unknown you need to have at least one independent material balance or other independent piece of information. Otherwise, the problem is indeterminate. For example, a problem such as illustrated in Fig. 2.S(c) in which the compositions of ali the streams are known and the weights of tIVO streams are unknown requires two independent material balances for its solution. What to do when more. or fewer. independent material balance equations are available than unknowns is discllssed in Sec. 2.4. The columns on the right-hand side of Fig. 2.5 tabulate the number of 84

l

to be () 1'1

Fc? C=? -'--'-----1 System . . . - - _ / Boundary

2

!..nnwn ~t: The ;I,k' i'e obl;w \t ream Ii

(d, See qr.::

(el Se'c' tilHl

(f) !\t.d

·\fter this hI uf materi;Ji b As t'\i'!t

~Ol

(al A

Ii>

(b) A

1

U

all the' t probkr' one stream .,.

. ·f' .....t

s.'C.

Parameters

"nown ':S 5

M )iIjf'X

'.

Program of Analysis of Material Balallce Problems 85

known and unknown parameters. for both the total mass and the compositions. The asterisk designates that one of the three unknown compositions can always be obtained by using the equivalent of Eq. (2.5), Hence in a three-component stream of unknown composition. only two of the three mass fractions (or percents) really need to be evaluated. The dagger indicates that in a stream with only one or two components. the composition of the other components is known to be 0 percent. Note how for each of the first three figures in Fig. 2.5 the number of independent material balances (three in each case) does not exceed the number of unknown quantities. For problems involving chemical reaction. a total mass balance can be written as well as a mass balance for each atomic species or multiple thereof. i.e., hydrogen expressed as H o' For example. in Fig. 2.6(a) you can write a total mass (not moles) balance and a carbon. a hydrogen. a nitrogen. and an oxygen balance. The carbon balance might be in terms ofC, and the hydrogen. nitrogen. and oxygen balances in terms of H 2 , N" and 0,. respectively. Depending on the problem statement, you would not necessarily have to use each of the balances. Also keep in mind that in using balances on atomic species, both mass and moles are conserved. Figure 2.6(c) illustrates a case in which the chemical equation is given. However, in solving problems related to Fig. 2.6(a) and (b), chemical equation(s) must be assumed to apply. or else some information given or assumed about the nature of the reaction that takes place. The strategy to be used in solving material balance problems is very simple. As a general rule, before making any calculations you should

3·1

8+

all



ertTt 1m at

Unknown

Camp Moss Co",;

9+

J.2

e

2

2

2'

0

:action).

lllces, Eq. (2.4).nree component ~. Consequently, II equations will u can substitute erial balances if ~

use of material lOuld remember ~ndent material the problem is . 2.5(c) in which of two streams solution. What ns are available the number of

I I

I I

I

!

I

(a) Draw a picture of the process. (b) Place all the available data in the picture. (c) See what compositions are known or can be immediately calculated for each stream. (d) See what masses (weights) are known or can easily be found for each stream. [One mass (weight) can be assumed as a basis.] (e) Select a suitable basis for the calculations. Every addition or subtraction mllst be made with the material on the same basis. (f) Make sure the system is well defined. After this has been accomplished. you are ready to make the necessary number of material balances. As explained previously. you can write (a) A tolal material balance. (b) A component material balance for each component present. Not all the balances, however, will be independent. Problems in which the mass (weigh t) of one stream and the composition of one stream are unknown can be solved without difficulty by direct addition or

)$,W$;:::::

.iiiA t'h,_tt~"'~~

;a,

¥-

i 1

l j I /,

86 Material Balances

Chap. 2

direct subtraction. Problems in which all the compositions are known and two or more of the weights are unknown require some slightly more detailed calculations. If a tie component exists which makes it possible to establish the relationship between the unknown weights and the known weights, the problem solution may be simplified. (The tic component will be discussed in detail in Sec. 2.5.) When there is no direct or indirect tie component available, algebra must be used to relate the unknown masses to the known masses.

Sec. 2.3

(d) Exec, react; of air comp Even if only I'll CO and CO" bustion produ, excess 0, (a n

2.3 Problems with direct solutions Problems in which one mass (weight) and one composition are unknown can be solved by direct addition or subtraction, as shown in the examples below. There is no need to use formal algebraic techniques. You may find it necessary to make some brief preliminary calculations in order to decide whether or not all the information about the compositions and weights that you would like to have is available. Of course, in a stream containing just one component, the composition is known, because that component is 100 percent of the stream. In dealing with problems involving combustion, you should become acquainted with a few special terms: (a) Flue or Stack gas-all the gases resulting from a combustion process including the water vapor, sometimes known as wet basis. (b) Orsal analysis or dry basis-all the gases resulting from the combustion process not including the water vapor. (Orsat analysis refers to a type of gas analysis apparatus in which the volumes are measured over water; hence each component is saturated with water vapor. The net result of the analysis is to eliminate water as a component being measured.) Pictorially, we can express this classification for a given gas as in Fig. 2.7. ro convert from one analysis to another, you have to ratio the percentages for he components as shown in Example 2.9. (c) Theoretical air (or theoretical oxygen)-the amount ot air (or oxygen) required to be brought into the process for complete combustion. Sometimes this quantity is called the required air (or oxygen).

Flue gas, slack gas Or

weI basis

COl} CO Ol Nz

Note that the J air may also be

or

since O 2 enterill[: The precision <,f may not be the given in a proi": number of un ~ III

EXAMPLE 2,4

A man

15 percenl, i[" it chimney to rOJ, (' Suppose [ha [ \ (' I [he air suprJ) I' ."

Solution: Let us c.lk.;, assuming [Iu( ('"

Dry

flue gas

Orsot onol ysis

on SOz free basis

dry basis

ClHrt:"

He explains Ih,"

or

SOz:_ _ _ _J HlO Fig. 2.7. Comparison of gas analyses on different bases.

;.• 14
'\I, . . . .

__. . .' '.___...

......_ _tt_....._Illi'liOn_'_ _ _j'e .....

_lj<.....,.-.*.'''.,..._ _liI1t."_",,'1_ _ .IIoCjijI·_ _milO·n=_'. _ _ __ -

t . . .·• • • ~III.

_..-A ..

Problems wilh Direct SOIUliolls 87

Sec. 2.3

Chap. 2

(d) Excess air (or excess oxygen)-in line with the definition of excess reactant given in Chap. I, excess air (or oxygen) would be the amount of air (or oxygen) in excess of that required for complete combustion as computed in (c).

3Te known and two more detailed calJle to establish the ~ights, the problem "cussed in detail in -" available, algebra :lasses.

Even if only partial combustion takes place, as, for example, C burning to both CO and CO" the excess air (or oxygen) is computed as if the process of combustion produced only CO,. The percent excess air is identical to the percent excess 0, (a more convenient method of calculation):

% excess air = 100

excess air = 100 excess 0,10.21 reqUIred air reqUIred 0,/0.21

o

Note that the ratio 1/0.21 of air to 0, cancels out in Eq. (2.6). Percent excess air may also be computed as

~re

unknown can be :nples below. There : necessary to make - rher or not all the ;)uld like to have is --nent, the composi:1e stream. u should become :Jill] _"":'1

% excess air =

= 100

:)1 air (or oxygen) ombustion. Some:::en).

(2.7)

excess 0, 0, entering - excess 0,

since

0, en teri ng process = 0, required for complete combustion

baJ .....

!!as as in Fig. 2.7. _!1e percentages for

100 0, entering process -- 0, required 0, required

or

, process

Jm the combustion "is refers to a type _cere measured over :er vapor. The net Jonent being mea-

(2.6)

+ excess 0,

(2.8)

The precision of these different relations for calculating the percent excess air may not be the same. If the percent excess air and the chemical equation are given in a problem, you know how much air enters with the fuel, and hence the number of unknowns is reduced by one.

I I

EXAMPLE 2.4

Excess air

A man comes to the door selling a service designed to checle "chimney rot." He explains that if the CO 2 content of the gases leaving the chimney rises above 15 percent, it is dangerous to your health, is against the city code, and causes your chimney to rot. On checking the flue gas from the furnace he finds it is 30 percent CO 2' Suppose that you are burning natural gas which is about 100 percent CH, and that the air supply is adjusted to provide 130 percent excess air. Do YOLI need his service?

SOllllioll : Let us calculate the actual percentage of CO, in the gases from the furnace assuming that complete combustion takes place. See Fig. E2.4. The 130 percent

f~ci%--'~I_--'r-_.J~~~6 tAiroz 21% CH 4

+ 2 Oz

Nz 79% - CO 2 + 2 H2 0

Fig. E2.4.

1

..... 88

Material Balallces

Chap.

Ha'pt--

~

excess air means 130 percent of the air required for complete combustion of CH •. The chemical reaction is CH.

+

20,

~

CO,

+

2H,O

Basis: 1 mole CH. By picking a basis, the quantity of the CH. stream is fixed. The quantity of air is fixed by th~ specification of the excess air and is computed as follows: On the bas" of 1 mole of CH" 2 moles of 0, are required for complete combustion, or 20,

1.00 air 0.21 0,

9 54 I' . d =. mo es aIr reqUIre

composed of 2 moles 0, and 7.54 moles N ,. The excess air is 9.54(1.30) moles air, composed of 2.60 moles 0, and 9.80 moles N,. Summing up our calculations so far we have entering the furnace: air 9.54 12.4 21.94

Required Excess Total

0, 2.00 2.60 4.60

oc

12.4

N, 7.54 9.80 17.34

t:XA\H. A WI lh"1 6<1 ;1

Since 2 moles of the 4.60 moles of entering 0, combine with the C and H of the CH. to form CO, and H 2 0, respectively, the composition of the gases from the furnace should contain the following: CH. CO, H,O 0,

N,

moles 0 1.00 2.00 2.60 f"54 9.80 22.94

% 0 4.4 8.7 11.3 75.6 100.0

The salesman's line seems to be rot all the way through.

EXAMPLE 2.5

Excess air

Fuels for motor vehicles other than gasoline are being eyed because they generate lower levels of pollutants than does gasoline. Compressed ethane has been suggested as a source of economic power for vehicles. Suppose that in a test 20 Ib of C,H. arc burned with 400 Ib of air to 441b of CO, and 121b of CO. What was the percent excess air?

Solutio,,: C,H.

+ 30, --)- 2CO, + 2H,O Basis: 20 Ib C,H,

Since the percentage of excess air is based on the complete combustion of C ,f!. to CO, and H,O, the fact that combustion is not complete was no influence on the

Ir- . i..!

r ul ;' .".

or

".1:.,

t

'Jt1\J\ ", (

I: I I

({hap. 2 :::nntiustion of CH,.

Problems with Direct Solutiolls 89

Sec. 2.3 definition of "excess air." The required Oz is 20 Ib CzH'l1 Ib mole CzH, 281b C,H,

3 Ib moles 0, I Ib mole C,H,

=

2.141b moles 0,

The entering 0, is Thec.quantity of air )i1aw&, 0n the basis

-"lus.tinn,. or

400 Ib air lib mole air 21 Ib moles 0, 291b air 100 Ib moles air

9.li>l(.JL3m

2.90 Ib moles 0,

The percent excess air is 100

lS

=

=

12.4

excess 0, _ looentering 0, - required 0, x required O 2 required O 2

. _ 2.90 Ib moles O 2 - 2.141b moles O 2 100 %o excess air 2.141b moles 0,

=

35.5%

EXAMPLE 2.6 Drying A wet paper pulp is found to contain 71 percent water. After drying it is found that 60 percent of the original water has been removed. Calculate the following: :he:
(a) The composition of the dried pulp. (b) The mass of water removed per kilogram of wet pulp. Solutioll: Basis: 1 kg wet pulp Initially we know only the composition and mass (the chosen basis) of the wet pulp. However, we can easily calculate the composition (100 percent H 2 0) and mass of water removed; examine Fig. E2.6. H 2 0 removed

=

0.60(0.71)

0.426 kg

=

r

Thus in essence we now know two masses and two compositions. (J) Bya pulp balance the final amount of pulp in the dried pulp is 0.29 kg and the associated water is in - loss = remainder 0.71 - 0.426 they generate .has- been suggested ;120 Ib of CzH. arc as.1he..percent excess

=

0.284 kg H 2 0

Wet pulp pulp: 0.29 H20: 0.7t

/

---

I-__~(r-ll \

--- ....

Dryer

Dried pulp pulp

\

I

?

1

II-...;I_ _-i I I

',--1--_ . .' I

I :nnbustion of CzH, no:·i,,"···'nce on the

[

I

water balallce

System Boundary

:!C3l.lSe

~

H20 only

1

I

I ,

60% of original H20 Fig. E2.6.

-

"""'"

j .•

;.4,1$#1

<,J;

~"

-

_

.. ..

_-_..--..-..•.. 90

,~.--~....

Chap. 2

Material Balallces

(a)

or, alternatively, since 40 percent of the H 2 0 was left in pulp, 0.40(0.71)

=

Sec. 2.3

0.284 kg H 2 0

(2) The composition of the dried pulp is

Pulp (dry) H 20 Total

kg 0.29 0.284 0.574

%

(n) . .

50.5 49.5 100.0

(e):

EXAMPLE 2.7

Crystallization

A tank holds 10,000 lb of a saturated solution of NaHC0 2 at 60"C. You want to crystallize 500 lb of NaHCO, from this solution. To what temperature must the solution be cooled?

Since the

Solutiol/: A diagram of the process is shown in Fig. E2.7. the temp

System Bou~:r~,\ //

---

Initiol

-

.....

Final

"

Am

/

\

\

NaHCOJ

{NaHCOJ

I

: Saturated \ SOlution

,,

f-----i

H2 0

.... _-----

f-r------~

--

/

H2 0

}

t-ig. I.:.S Saturated Solution

I

complhlt S/1!U

The di,till.I~~ .

/

f, om the Fig. E2.7.

Additional data are needed on the solubility of NaHCO, as a function of tem· perature. From any handbook you can find solubility temp. eC) g NaHCO,/IOO g H 2 0 60 16.4 50 14.45 40 12.7 30 Il.l 20 9.6 10 8.15 Basis: 10,000 Ib saturated NaHCO, solution at 60'C We know both the weight of the initial solution and the weight of the crystals: we know the composition of both the initial solution and the crystals (100 ~~:-':aHCO ,). Thus the weight and composition of the final solution can be calculated by subtraction.

r"

Chap. 2

'S"

gt:

'W

$1

*' f1

-tt'r

r

t .-

"$

tv

,'5 ' 7

td'

t

tlk

Problems with Direct Soilltions 91

Sec. 2.3 (a) Initial composition:

l6.4lb NaHCO J 16.4 Ib NaHCO J .,- 100 lb H 2 0 = 0.141 or 100 - 14.1 = 85.9%HO

14.1 ~NaHCOJ

(b) Material balance (component balance on the NaHCO J): initial NaJ-lCO J - crystals NaHCO J 500 0.141(10,000) -

=

final NaHCO J 910lb

(c) Final composition and weight: H 2 0 = (0.859)10,000 = 8590 Ib NaHCO J = 910lb

at 60 0 e. You want .!!fIIperature must the

Total

9500lb

Since the final solution is still saturated and has 910 Ib NaHCO J 8590 Ib H 2 0

10.6 g NaHCO J 100 g H 2 0

the temperature to which the solution must be cooled is (using a linear interpolation) 30'C - [11.1 11.1 -

EXAMPLE 2.8

J

J}saturated Solution

1O.6l [10°C] 9.6~

=

27'C

Distillation

A moonshiner is having a bit of difficulty with his still. The operation is shown in Fig. E2.8. He finds he is losing too much alcohol in the bottoms (waste). Calculate the composition of the bottoms for him and the weight of alcohol lost in the bottoms.

Solution: The information provided gives us the composition and weight of feed and distillate. The composition of the bottoms can be obtained by subtracting the distillate from the feed .

.::s a function of tem-

~d~

Boundary_~--

"'f

I I I I

1000 Ib feed: 10% EIOH 90% H20

I I

_---------- -, Vapor

Cooling L-.~:fJ'"- Water

Reflux J--'----'---:/'--Distillote (Producil: / GO% EtOH Distillation / 40'", H20 Column / WI ' ItlO feed / /

/

Heat

/

/

Bottoms (waste)

~ight

of the crystals; :als (lOO%NaHCO J ). COllater' 'ubtraction.

Fig. E2.S.

4Ci ,_ ,

M#P

n

A

A<4 '

Chap. 2

92 !lfa/erial Balallces

\.

Basis: 1000 Ib feed alit

ill conlponenl

material balance

{EtOH H 2O Total

wI

0/ /0

10 90 100

dis/illale bOlloms (lb) (lb) (by sublraction)

Ib

100 900 1000

60 40 100

40' 860 900

>\1

• Alcohol lost. 1000 lb feed II

EXAMPLE 2.9

~~ f~s~~~~te =

100 Ib distillate

40lb EtOH 100 900 Ib bottoms

=

4.4% EtOH

860 Ib H,O 100 900 lb bottoms

=

95.6%H 0 2

Combustion

The same ethane as used in Example 2.5 is initially mixed with oxygen to obtain a gas containing 80 percent C ,H, and 20 percent 0, that is then burned with 200 percent excess air. Eighty percent of the ethane goes to CO" 10 percent goes to CO, and 10 percent remains unburned. Calculate tbe composition of the exhaust gas on a wet basis. Solulioll: We know tbe composition of the air and fuel gas; if a weight of fuel gas is chosen as the basis, the weight of air can easily be calculated. However, it is wasted effort to convert to a weight basis for this type of problem. Since the total moles entering and leaving the boiler are nO! equal, if we look at anyone component and employ the stoichiometric principles previously discussed in Chap. I together with Eq. (2.\). we can easily obtain the composition of the stack gas. The net generation term in Eq. (2.1) can be evaluated from the stoichiometric equations listed below. The problem can be worked in the simplest fashion by choosing a basis of laO moles of entering gas. See Fig. E2.9.

[,haust Gos

Fuel

Gas CZH G 80 Ib mole Oz 20 Ib mole 200% e,cess

3\JJI.1

SO Ih

111,"

By an

0\11

C02

CO C2HG O2 N2

AI(

To{:

the

o.

H20 From a

Fig. E2.9.

A.4#

\\.1

, Cbap.2

I

Problems with Direcl Sollllions 93

Sec. 2.3

I

.7/toms (/b) .. subtraction) 40'

i

860 900

Basis: 100 lb moles of fuel CzH. C,H.

-+ i02 ->- 2C0 2 + 3H 20 + ~02 ->- 2CO + 3H 20

The total 0 1 entering is 3.00 times the required O 2 (l00 percent required plus 200 percent excess). Let us calculate the required oxygen: 0, (for complete combustion): 80 lb mole C,H 6 3.51b mole O 2 = 280 lb mole 0 1 lb mole C,H 6 ' Required O 2 : 280 - 20

=

260 lb mole 0,

(Note: The oxygen used to completely burn the fuel is reduced by the oxygen already present in the fuel to obtain the oxygen required in the entering air.) Next we calculate the input of O 2 and N, to the system: 0, entering with air: 3(260 lb mole 0,) = 780 lb mole 0,

N. entering with air:

a

- xyge. ,otain :: willi 200 percent "s to CO. and ro

,.ust gas on a wet

. fuet gas. 5 chosen ;s wastm effort to '71oles entering and ct and employ the "T with Eq. (2.1), -:-ation term in Eq. cow. The problem of mtering gas.

=

780 mole 02179 Ib mole N, = 2930 Ib mole N, 21lb mole 0, Now we apply our stoichiometric relations to find the components generated within the system: 80 Ib mole C 2 H.j21b mole CO, I 0.8 I Ib mole C 2H, 80 lb mole C,H 6

3 lb mole H 2 0

I Ib mole C 2H6

=

1281b mole CO . 2

0.8 = 1921b

80lbmolec,H61 2lbmoieCO /0.1 lib mole C,H,

=

I H 0 ,

rna e

16lb mole CO

80lb mole c,H6131b mole H 2 0 10.1 I Ib mole C,H, = 241b mole H 2 0 To determine the 0, remaining in the exhaust gas, we have to find how much of the available (800 lb mole) 0, combines with the C and H. 80lbmoleC 2 H'j 3.5lbmo1cO, 10.8'_"'lb b t CO dHO 1 Ib mole C,H, I -- 4_", rna le O ,to urn a 2 an ,

]

CO 2 CO

C2H. O2 N2

H20

80lb moles C,H, 2.5lb moles 0'1'0.1 I Ib mole C,H,

=

20 Ib mole O 2 to burn to CO and H 2 0 244lb mole 0, total "used up" by reaction

By an oxygen (0 2 ) balance we get O 2 out = 780 lb mole ""- 20 Ib mole - 244lb mole = 556lb mole O 2 From a water balance we find H,O out = 1921b mole

+ 241b mole = 216lb mole H 20

--

,

l

!

r-·

"

~ EXAMPLE 2.9 Alternative manner of presentation Intial Component

=%

lbmole

C.H. C,H. C,H.

1}

0,

20

80

Ib mole 0, Required

Actual Ib mole 0, Used

;(80) = 280

i(64) = 224 ~(8) = 20

Ib mole formed of ----CO,

H,O

CO

128

192 24

16

C,H. left

Reactions C,H. C.H.

8 -20 260

100

244

128

216

16

+ iO, --> 2CO, + 3H,O + ~O. --> 2CO + 3H,O C,H.

-->

C,H.

8

3(260) = 780 780H = 2930 air fuel used 780 + 20 - 244 = 556

0, cnlering wilh air: N, eniering wilh air: 0, in exil gas:

Summary of Material Balances

Total mole in '1'= total mole out = totallb out: Totallb in balances on atomic species

bala/lus 011 compounds

:J

<


~ ~ ~ ~

3. -. :::s

..... 0

~

C in H, in 0, in N, in

Ib Ib Ib fIb Ib Ib

mole mole mole mole mole mole

C 2 f-{. in

l l

~.g ~

~.,.

g

=. =.

Ib mole Ib mole Ib mole Ib mole

= = = =

:=. t;

~

e: >-

=.' ..........

::so;::S(';)O::S~PJ""'-

=- =:

1~ i

~

_

.....

a- ~ -, ::s 0.

~

-

~

_

;:;3::j~o.O~:::s(1)_~~ 0

~

.~ ~

7. 7'

t")

-,

~ ~~

50

c;

C out: 2(80) = 128 + 16 + 2(8) H, out: 3(80) = 216 + 3(8) 0, out: 20 + 780 = 128 + i(216) + j;(16) + 556 N. out: 2930 = 2930

= Ib mole C 2 H. out CO 2 in = Ib mole CO: out CO in = Ib mole CO out H 2 0 in =~ Ib mole H,O out N, in = Ib mole N. out 0, in = Ib mole 0, out

g

ri ~ ~ 0 ~ 0 ~ _. ~ (1)

(1)

r:.

mole mole mole mole

_<_nQ:l(1)~riQ~ri~V) 0 (1) ::r c: ::s 0" :::s PJ 0. -. p,) (1) !::: -,::r' -- (1) ::J t:.J ::. 3 VI 3 @ ::J PJ 0 ~ C) VI ~ ~

2.. _

<

Ib Ib Ib Ib

= .0

::r

I\)

:,:,.

Vl

(")

0

::l

~ .... C1> ~

'1>'

~

"-' ?":

g-.

-1- Ib mole C,H. consumed: 80=8+72+0 0= 128 - 128 -Ib mole CO 2 generated: 0= 16 - 16 - Ib mole CO generated: 0= 216 - 216 -Ib mole H,O generated: 2930 = 2930 + 0 + Ib mole N, consumed: 800 = 556 + 244 + Ib mole 0, consumed:

...

0

0" C ::J

-

'"

'-) C(

0

3.- 30

g_

'"::J

0>

(")

~..,,,,

s: ., '""

~

:;.

<: ;-

1, ~ ~' ~' ~ ~ -Vl

(each compound has to be multiplied by its molecular weight)

-,

,

r::

r~

T

C

::l 0>

g

Cl.

aO> ~,

'"<' r,

~

::::0>

'"

tno>a-

0

~

O''"'l .., :r 3 ...

:z::(")(")ZO(")~

o

N

0 ON N

N

...

:z:: !:2 ~ ~

3 '< or 3 ~ ~

'".., ...~ '" 0 -, Cl. ~.

en::l

:r

r,

" "

~

_::l

::T

:>

...

!>faterial Balances Using Algebraic Techniques

Sec. 2.4

95

The balances on the other compounds-C 2 H., CO 2 , CO, N 2-are too simple to be formally listed here. Summarizing these calculations, we have

% ill

lb mole compOllent

C 2 H. O2

fuel

80 20

N2

air

780 2930

CO 2 CO H2 O

100

3710

exhaust gas

exhaust gas

8 556 2930 128 16 216 3854

0.21 14.42 76.01 3.32 0.42 5.62 100.00

On a dry basis we would have (the water is omitted) componelll

Ib mole

%

C 2 H. O2

8 556 2930 128 16 3638

0.23 15.29 80.52 3.52 0.44 100.00

N2

CO 2 CO

I I 1++

Once you have become familiar with these types of problems, you will find the tabular form of solution illustrated in Example 2.9, on the facing page, a very convenient form to use.

2.4 Material balances using algebraic techniques II

II

II

II

II

c c ,'c:.c: . .-.1"- 0 --.:: -,:> 0 ~ N -=-

-

M

~UUl:ZO

-'UOUOU

~OOOOO

.:: :: E E E E

~.D..c.D..o..o

------

As illustrated in Figs. 2.5 and 2.6, problems can be posed or formulated in different ways depending on the type of information available on the process streams and their respectivc compositions. The problems treated in the previous section were quite easy 1O solve. once the problem had been converted from words into numbers. because the missing information pertained to a single stream. Only simpk addition or subtraction was required to find the unknown quantities. Other types of material balance problems can be sol\cd by writing the balance formally. and assigning letters or symbols to represent the unknown quantities. Each unknown stream. or component, is assigned a letter to replace the unknown value in the total mass balance or the component mass balance, as the case may be. Keep in mind that for each unknown so introduced you will have to write one independent material balance if the sct of equations you form is to have a unique solution. If more than one piece of equipment or more than one junction point is involved in the problem to be solved, you can write material balances for each

.... --~----

Chap. 2

96 Material Balances

piece of equipment and a balance around the whole process. However, since the overall balance is nothing more than the sum of the balances about each piece of equipment, not all the balances will be independent. Appendix L discusses how you can determine whether sets of linear equations are independent or not. Under some circumstances, particularly if you split a big problem into smaller parts to make the calculations easier, you may want Stream 8 to make a material balance about a mixil/g point. As illustrated in Fig. 2.8, a mixing point Stream C Stream A is nothing more than a junction of three or mOTe streams and can be de,·gnated as a Fig. 2.8. A mixing point. system in exactly the same fa,i;ion as any other piece of equipment. Some brief comments are now appropriate as to how to solve sets of coupled simultaneous equations. If only two or three linear material balances are written. the unknown variables can be solved for by substitution. If the material balances consist of large sets of linear equations, you wiil find suggestions for solving them in Appendix L. If the material balance is a nonlinear equation, it can be plotted by hand or by using a computer routine, and the root(s), i.e., the crossings of the horizontal axis, located. See Fig. 2.9. If two material balances have to be solved, the equations can be plotted and their intersection(s) located. With many simultaneous nonlinear equations to be solved, the use of computer routines is essential; for linear equations, computer routines are quick and prevent human error. Making a balance for each component for each defined sys-

...

\(,

,

..

h{'

,~ "

~. ~ , '

' ':,'

.

IX \ \1/'/ I I),;,,'· , If"

"

(Ifl'{'t I,'

Llrll \

.~!id

the

t......·fil

I

t

n'..!·!\

fIx)

Sa/1I11,'"

In

If,,,

"I," ,. and,. \\

mi\\IIl~.

(0) Single Nonlinear Equation

ffll\\C\l'f. \,

",I

cQU.1tHln\

II Solution

x, (b) Two Nonlinear Equations

Fig. 2.9. Solulion of nonlinear malerial balances by graphical lechniques.

._.;i;:o:' Hf

Chap. 2

li

rtf

r

'*

tnftlt

Material Balances Using Algebraic Techniques 97

Set:. 2.4

'wever, since the _Dout each piece _dix L discusses ?pendent or not. :-es, particularly :0 smaller parts '. you may want .3.bout a mixing • a mixing point .:on of three or _esignated as a ~ashion as any

tern, a set of independent equations can be obtained whether linear or nonlinear. Total mass balances may be substituted for one of the component mass balances. Likewise, an overall balance around the entire system may be substituted for one of the subsystem balances. By following these rules, you should encounter no difficulty in generating sets of independent material balances for any process. Illustrations of the use of algebraic techniques to solve material balance problems can be found below in this section and in the next section (on tie components) .

EXAMPLE 2.10

Mixing

Dilute sulfuric acid has to be added to dry charged batteries at service stations in order to activate a battery. You are asked to prepare a batch of new acid as follows. A tank of old weak battery acid (H 2 S0 4 ) solution contains 12.43 percent H 2 S0 4 (the remainder is pure water). If 200 kg of 7.77 percent H 2 S0 4 are added to the tank, and the final solution is IS.63 percent H 2 S0 4 , how many kilograms of battery acid have been made? See Fig. E2.1O .

" sets of coupled "::es are written, ..3.terial balances :ms for solving .llion, it can be . i.e., the cross.llances have to located. With " of "Juter (lUi, pre- . . :-h delJued sys-

Fig. E2.10. Solutioll: In this problem one mass and three compositions are known; two masses are missing. To assist us, we shall arbitrarily label the masses of the two unknown solutions F and P. With two unknowns it is necessary to set up two independent equations. However, we can write three material balances, any pair of which are independent equations.

Basis: 200 kg of 77.7 percent H 2 S0 4 solution type of balance Total compollellf H 2 S0 4 H 20

out

in

F+ 200 F(0.1243) F(0.S757)

P

+ 200(0.777) = P(O.lS63) + 200(0.223) = P(0.SI37)

Use of the total mass balance and one of the others is the easiest way to find P:

+ 200(0.777) 2110kgacid F = 1910 kg acid

(P - 200)(0.1243)

P

~ -¢.¢

44

(.., ",I(%l ,


*g

=

."4if, ( 44;;'.#),'1..1 ."'.-

=

P(0.IS63)

(a) (b) (c)

98

Chap. 2

Material Balances

EXAMPLE 2.Il

Scc.2.4

Distillation

A typical distillation column is shown in Fig. E2.11 together with the known information for each stream. Calculate the pounds of distillate per pound of feed and per pound of waste.

Woter

1200lb

System boundary~f

I

,,I

f--'--!-"- Distillote 85% EtOH 15% H20

Feed

35% EtOH 65% H20

f-i-";';"~

Waste

Solllli" Comer other comp'

5% EtOH 95% H20

Fig. E2.1I. Solution: Inspection of Fig. E2.11 shows that all compositions are known, but three weights are unknown. Because only two independent balances can be written, one weight must be chosen as the basis for the problem to effect an algebraic solution.

Basis: 1.00 Ib feed in

type of balance total EtOH H 20 Solving for D with W

=

1.00 1.00(0.35) 1.00(0.65)

out

= = =

D D(0.85) D(O.l5)

Now \\ ( are five sire.: weight of ,'"

+W + W(0.05) +

W(0.95)

(1.00 - D),

1.00(0.35) D

= D(0.85) + (1.00 = 0.375 Ib/lb feed

Balance'S on , Let

- D)(0.05)

Since W = I - 0.375 = 0.625 Ib, D 0.375 0.601b W = 0.625 = -J-b-

total acc(onc

EXAMPLE 2.12

water

Multiple equipment

air

In your first problem as a new company engineer you have been asked to set up an acetone recovery system to remove acetone from gas that was formerly vented but can no longer be handled that way, and to compute the cost of recovering acetone. Your boss has provided )OU with the design of the proposed system. as shown in Fig. E2.12. From the information given. list the flow rates (in pounds per hour) of all the streams so that the size of the equipment can be determined, and calculate the composition of the feed to the still.

We n,·' (.'quaIH'Il\ ,

I

cump()l1 lT : is not crl\' ,. - i Olle .,.

nally

......

_----

St.ll<.'

Chap. 2

!.faterial Balances Using Algebraic Tecimiqlles

Sec. 2.4

99

Air only

the known of feed

~und

A

Water 1200lb/hr

Product Acetone 99 wt % Water I wt %

Still

Absorber

I----Woste Acetone 5% Woter 95%

Gos F ~A:-'i;;'r-:+-o-c-et:-o-n-e..J 0.5 mole % ocetone)

Steom Fig. E2.12.

Sollltion: Convert mole percent of the air-acetone feed to weight percent because all the other compositions are given in weight percent:

mol. wt acetone = 58,

mol. wt air = 29

Basis: 100 Ib moles gas feed

=eeweig.

composition Acetone Air

A

Ib mole 1.5 98.5 100.0

mol. wt 58 29

Ib 87 2860 2947

--

WI

%

2.95 97.05 100.00

Now we know the compositions (on a weight basis) of all the streams. There are five streams entering and leaving the entire system (excluding the steam), and the weight of only one is known, leaving four weights as unknowns. Basis: I hr Balances on complete system (overall balances): Let F = feed to absorber, Ib ) A ~~ air leaving absorber, Ib the four unknowns P = product, Ib W= waste, Ib F

acetone

F(0.0295)

water -cCl

to set up

"l"i}'vented "~ru:etone .

'''n in Fig.

, of:allthe .".composi-

+ 1200 = P + W + A

total

air

= P(0.99)

-; W(O.OS)

1200 = P(O.OJ)

+ W(0.95)

F(0.9705)

=

A

We now have four balances and four unknowns; however, only three of the equations are independent equations, as you can easily discover by adding up the component balances to get the overall balance. What is wrong? It seems as if there is not enough information available to solve this problem . One additional bit of information w'Juld be enough to solve the problem as originally stated. This information might be, for example,

-""

100

Chap. 2

Material Balances

.... Sec. 2.4 Aroun,

(a) The weight of entering gas per hour product per hour, or waste per hour. (b) The composition of the feed into the still. With an extra piece of pertinent information available, a complete solution is possible; without it, only a partial solution can be effected. Observe that making additional balances about the individual pieces of equipment will not resolve the problem, since as many new unknowns are added as independent equations.

or We could g,

EXAMPLE 2.13

Countercurrent stagewise mass transfer

In many commercial processes such as distillation, extraction, absorption of gases in liquids, and the like, the entering and leaving streams represent two different phases that flow in opposite directions to each other, as shown in Fig. E2.I3(a). (The figure could just as well be laid on its side.)

Entering stream containing A Ib/hr of solvent (high concentration of solute)

r'-.-.J...,\ \

I

8

X,

Also a,

2 as folloll S

or to genera 11th stage,

E,it stream containing 8 Ib/hr of solvent (high concentration of solute)

or for the (II

"'--r~"~80undOry for balance around stage t

If lie Ie steady-state. we can WIllI,,: Entering stream containing 8 Ib/hr of solvent (low concentration of solute)

Exit stream containing A Ib/hr of solvent (low concentration of solule)

[qual;,'"

rdatio(1"') lI f' : ,bjl"renee l'~:~;

A,8 ' Ib of stream less solute in the stream Ib of solvent

slope B. A ., ,.

Fig. E2.13(a). This type of operation is known as countercurrent operation. If equilibrium is attained between each stream at each stage in the apparatus, calculations can be carried out to relate the flow rates and concentration of products to the size and other design features of the apparatus. We shall illustrate how a material balance can be made for such type of eqllipment. The letter X stands for the weight concentration of solute in pounds of solute per pound of stream, solute-free. The streams are assumed immiscible as in a liquid-liquid extraction process. Around stage I the solute material balance is in

out

Alb hr

Bib I'\~f...!£ hr Ib B

or A(X~

- X1)

=

B(Xf - Xf)

(a)

Material Balances Using Algebraic Techniques

Sec. 2.4

Chap. 2

101

Around stage 2 the solute material balance is

ner hour.

out Alb X11b..LBlb hr IbA hr

in

Alb hr

possible; additional problem,

:5

X~lb

IbB

or

=

A(X1 - X1>

B(Xf - Xf)

We could generalize that for any stage number n A(X~_I -

::Jrption of -::J different E2.13(a).

X:) = B(X: - X:+ I )

(b)

Also a solute material balance could be written a bout the sum of stage 1 and stage 2 as follows: in out (c) A(X~) + B(Xf) = A(X1) + B(Xf) or to generalize for an overall solute material balance between the top end and the nth stage, A(X~ - X:) = B(X1 - X:+.>

.. utei

or for the (n - l)th stage, A(X~ -

X~_I)

= B(X1 - X:)

Changing signs, (d)

If we rearrange Eq. (d) assuming that A, B, X~, and X1 are constants (i.e., steady-state operation) and XA and XB are the variables as we go from stage to stage, we can write " -X .-1

B XB• A

+ (XA0 - AB XB)I

(e)

Equations (a}--(e) represent an unusual type of equation, one that gives the relationship between discrete points rather than continuous variables; it is called a difference equation. In Eq. (e) the locus of these points will fall upon a line with the slope BfA and an intercept [X J - (B/A)XfJ, as shown in Fig. E2.13(b).

,ibrium is ne carried -:Jer design . made for of solute .edimmis-

X~_I

I

x:

I

Intercept '

(x~ - ~

X8 _ Xfl

Fig. E2.I3(b).

tal

I I

''l'''_ _ _ _

._~

•. _, ___ "_,_

Chap. 2

102 Material Balances

,\(..

:

2.5 Problems involving tie components (elements) A tie component (element) is material which goes from one stream into another without changing in any respect or having like material added to it or lost from it. If a tie component exists in a problem, in effect you nn write a material balance that involves only two streams. Furthermore, the material blance has a particularly simple form, that of a ratio. For example, in Fig. 2.S(b) all the MeOH in stream F goes into stream IV. Consequently, streams F and IV can be related by a MeOH material balance: COM,DH,W _ F or COM,DH,F IV 0.10(100) = 0.22IV

or

IV =

~:~~(100)

To select a tie element, ask yourself the question, What component passes from one stream to another unchanged with constant weight (mass)? The answer is the tie component, Frequently several components pass through a process with continuity so that there are many choices of tie elements. Sometimes the amounts of these components can be added together to give an overall tie component that will result in a smaller percentage error in your calculations than if an individual tie element had been used. It may happen that a minor constituent passes through with continuity, but if the percentage error for the analysis of this component is large, you should not use it as a tie component. Sometimes you cannot find a tie component by direct examination of the problem, but you still may be able to develop a hypothetical tie component or a man-made tie component which will be equally effective as a tie component for a single material. A tie component is useful even if you do not know all the compositions and weights in any given problem because the tie component enables you to put two streams on the same basis. Thus a partial sollllion can be effected even if the entire problem is not resolved, In solutions of problems involving tie components it is not always advantageous to select as the basis "what you have." Frequently it turns out that the composition of one stream is given as a percentage composition, and then it becomes convenient to select as the basis 100 Ib (or other mass) of the material because then pounds equal percentage and the same numbers can be used to represent both. After you have carried out the computations on the basis of 100 Ib (or another basis), then the answer can be transformed to the basis of the amount of material actually given by using the proper conversion factor. For example, if you had 32.51b of material and you knew its composition on the basis of percentage, you then could take as your basis 100 Ib of material, and at the end of the calculations convert your answer to the basis of 32.5 lb. Tie component problems can. of course, be worked by algebraic means, but in combustion problems and in some of the more complicated types of industrial chemical calculations so many unknowns and equations are invol\'ed that the

EX,\\l1

I" feed t" \i~rlllh.,;

(r;lIe. i" ~uch. If

I" !

dtum d: protein

dry caL percent S,,/

f" f \" that thl':

c

Tn " ) mater;,,1 ;"

"-

.._=_-..'..·.___

~_

([hap. ]

*l1li ......c.....'"""'"", ....._" ......>1. ............................ ~.,.~ _ _ ,__ .,~ ..... ~_,"-

h .... · _ _.....

>., , _ _ _ _ ._.~.ri-_-

__ . _ .._. __n' .. ~~~~.~;;:>.:L

Problems Inl"olring Tie Components (Elements)

Sec. 2.5

103

use of a tie component greatly simplifies the calculations. Use of a tie component also clarifies the internal workings of a process. Consequently, the examples below illustrate the use of both tie-component and algebraic solutions. Become proficient in both techniques, as you will have to draw upon both of them in your career as an engineer.

-=. mother :'r wstfrom

_ a material ::1aI!ce has a

:Ib) all the W=be

EXAMPLE 2.14

.j

Drying

Fish caught by man can be turned into fish meal, and the fish meal can be used as feed to produce meat for man or used directly as food. The direct use of fish meal significantly increases the efficiency of the food chain. However, fish-protein concentrate, primarily for aesthetic reasons, is used as a supplementary protein food. As such, it competes with soy and other oilseed proteins. In the processing of the fish, after the oil is extracted, the fish cake is dried in rotary drum dryers, finely ground, and packed. The resulting product contains 65 percent protein. In a given batch of fish cake that contains 80 percent water (the remainder is dry cake), 100.0 lb of water is removed, and it is found that the fish cake is then 40 percent water. Calculate the weight of the fish cake originally put into the dryer.

:::rent Jla5ses The :;ms.wcr -:: a precess =times the ovr e ::.aICl. 5

Solution: From Fig. E2.14 we see that all the compositions and one weight are known and that there is a tie component of BOC.

::3.t a rnmo r

-:-or for the

WIOOlb /_----.------- ___ zsystem Boundary

~:)mponent.

- f tbeprob-:ment or a ::oonent for ::ow all the

--

, f

\

I

-

............. "

I

Wet Fish Cake A

:::JOnent en::!on can be

\

"

\I

I

' i

I

1\ 1/

I



\.........\11

O.SO H20 " 020 BOC*--, \"

ays advan_:.:t that the =:Jd then it -~ material :Je used to :e basis of ::asis of the . actor. For ::1 the basis :md at the

I

"

/ "

:

"

Rotary Drier

B urner

iCifJ

,'0:::!)

B Dry . - - - - Fish Cake

I

:

i L

O~H~ Tie Component __ -0.60 BOC * _________________________________ _

* Bane Dry Cake

Fig. E2,I4 .

To solve the problem by use of algebra requires the usc of two independent material balance equations since there are two unknowns-A and B. Basis: 100 lb water evaporated

~eans,

but __ industrial _j H

in Total balance BOC balance

:c_

(I>

+

O~

out

..lo<"'~"-"''''""t±""'_ _

Chap. 2

104 Material Balances

Sec. 2.5 Anothc

Notice that because the bone dry cake balance involves only two streams, you can set up a direct ratio of A to B, which is the essence of the tie component: B

=

0.20A =...!..A 3 0.60

Introduction of this ratio into the total balance gives A = jA

+ 100

jA = 100 A

=

150 Ib initial cake

To solve the same problem by use of the tie component, take as a basis 100 Ib of initial material (the 100 Ib of H 2 0 will not serve as a useful basis). Basis: 100 Ib initial fish cake

Again we I" rated:

initial - loss = final componellt H 20 BDC

% = lb 80 20 100

lb· ~--

100 tie --+

lb ? 20 ?

I 11 3.3.i

% 40 60 100

You (;1 one basi, t, is ncccssar)

·AII H 2 0.

To make either a total or a water material balance for this problem, all streams must be placed on the same basis (100 lb initial cake) by use of the tie component. Tie final cake to initial cakr by a tie componellt: 20 Ib""B9C.

100 lb final cake

...

"'100=;-;lb~in~i;-'tia""l.-'c':-:a"'k:-Ce+--;6"'O'I"-b-:;1IDC.==-- = 33.3 lb final cake/100 lb mIllal cake

Notice that the BDC is the same (the tie component) in both the initial and final stream and that the units of BDC can be cancelled. We have now used one independent material balance equation, the BDC balance. Our second balance will be total balance: 100 lb initial -- 33.31b final

=

hy

SOllll' ":

complHlel1

66.7 lb of H 2 0 evap./100 lb initial cake

An alternative way to obtain this same value is to use a water balance 20 lb -nBC. 140 Ib H,O in dry cake 100 Ib initial cake 160 Ib 15DC in dry cake

= 13.3 lb H 2 0 left/IOO Ib initial cake 66.71b H,O evap./lOO Ib initial

Ontbc'''': \\ itb thl' dllfcr,'111 of l.l~) j" , You

80 Ib initial H,O - 13.3 Ib final H 2 0

=

No\\ : compon",' !

\\l"·

(\1 ...... 11'0 ;

to

cak~

Next a shift is made to the basis of lOa Ib H,O evaporated as required by the problem statement: 100 Ib initial cake 1 100 Ib H,O evap. _ 150 lb' .. I k 66.71b H,O cvap. lmtla ca e

I " \ \11'1 ' 'I •. n~("nl.

,\

Chap. 2

."ms, you can

Problems [nl'olving Tie Components (Elements)

Sec. 2.5

105

Another suitable starting basis would be to use I Ib of bone dry cake as the basis . Basis: lIb BDC component

H2 O BDC

final H 2 0

initial HO a basis 100 Ib

initial fraction 0.80 0.20 -1.00

loss Ib 100

final fraction 0.40 0.60 1.00

--

=

H 2 0 lost

water balance

0.401b H 2 0 0.601b RDC =

0.80 Ib H 2 0 0.201b BDC

0.67

4.0

= 3.33 Ib H 2 0jllb BDC

Again we have to shift bases to get back to the required basis of 100 Ib of H 2 0 evaporated: lIb -sIX:. 100 Ib H 2 0 evap. 1.0 Ib initial cake _ 150 lb' 't' I k 3.331b H 20 0.2Ib"fiBC 1m 13 ca e

em, all streams .Ie component.

',1Iial cake nilial and final used one indence will be Iota I

You can see that the use of the tie component makes it quite easy to shift from one basis to another. If you now want to compute the pounds of final cake, all that is necessary is to write 150 Ib initial cake -IOOlb H 2 0 evaporated 50 Ib final cake Now that we have examined some of the correct methods of handling tie components let us look at another method frequently attempted (not for long!) by some students before they become completely familiar with the use of the tie component. They try to say that

0.80 - 0.40 = 0040 lb H 2 0 evaporated 100lb 0040 lb = 250 lb cake

,ial cake -'lance

cake

On the surface this looks like an easy way to solve the problem-what is wrong with the method? More careful inspection of the calculation shows that two different bases were used in subtracting (0.80 - 0040). The 0.80 is on the basis of 1.00 Ib of initial cake, while the 0040 is on the basis of 1.00 Ib of final cake. You would not say 0.80/ A -- 0040/ B = OAO/A, would you? From our early discussion about units you can see that this subtraction is not a legitimate operation.

required by the EXAMPLE 2.15

Dilution

Trace components can be used to measure flow rates in pipelines. process equipment, and rivers that might otherwise be quite difficult to measlIre. Suppose that the

Chap. 2

106 Malerial Balances

water analysis in a flowing creek shows 180 ppm Na2S0 •. If 10 Ib of Na2S0. are added to the stream uniformly over a I-hr period and the analysis downstream where mixing is complete indicates 3300 ppm Na,SO., how many gallons of water are flowing per hour? See Fig. £2.15.

to Ib

.

.

-it'

.-

Sr<. 2.5 is that the <1 .•

has to t'C ,letc' balance.

NozSO.

Sohing the

t,,'

Fig. E2.15.

Tie-componenl Solulion: Basis: 100 Ib initial solution (water

+ Na2S0.

EX:\:\!PLE 1

initially present)

After choosing a basis of 100 lb of initial solution, we know three compositions and one weight. A tie component, the water, is present. inilial Ib = % 99.982

composition H,O

filial

% tie component

0.018 100.000

Na,SO.

The s"llI' /100 g 11,0 I water at II~I ( much B•• (:"'U

99.670 0.330 100.000

99.9821b 1'1,0 0.33 Ib Na 2S0. 100 Ib initial solution 99.671b H 20 = 0.331 Ib Na2S0./I00 Ib initial solution The tie component of water comprises our first material balance. Now that the exit Na,SO. concentration is on the same basis as the entering Na,SO. concentration, by an Na,SO. balance (our second balance) we find that the Na2S0. added on the same basis is 0.331 - 0.0180 = 0.313 Ib/lOO lb initial solution

The m;,,· or 34 g 100 ~ I

Hthe 100 C· and the pl<,I<

,-

/ /

I

Basis: I hr

I

I gal "":"T..~,,--,,70:o-;,.,t-=-----;,...:.c;::c..-+8":'.""'35 Ib = 383 gal/hr

1 1

I

I

Algebraic SOIUlioll: Let x = totallb flowing/hr initially y total balance water balance

=

+

\

totallb flowing!hr after addition of Na,SO.

x 10 0.99982x x

\

=Y

= 0.99670y

+ 10 =

}

mass balances

0.99982x 0.99670

The water balance involves only two streams and provides a direct ratio between

x and y. One of the ditliculties in the solution of this problem using a water balance

.. -~..".--

,,

/

COo

Chap. 2 ",SO. are added :em where mixing - are flowing per

Problell/s Inm/ring Tic COII/ponents (Elell/cnts)

Sec. 2.5

107

is that the quantity rO.99982 I I 000313 LO.9967U - "= . has to be determined quite accurately. It would be more convenient to use an Na2S0. balance, 0.OOO18x + 10 = 0.00330y Solving the total balance and the Na2S0. balance simultaneously,

x = 3210 Ib/hr 3210 Ib 1 gal hr 8.35 Ib

EXAMPLE 2.16

?sent) .o-ee compositions

=

385 gal/hr

Crystallization

The solubility of barium nitrate at 100 c C is 34 g/100 g H 20 and at O°C is 5.0 g /100 g H,O. If you start with 100 g of Ba(NO,h and make a saturated solution in water at 100'C, how much water is required? If this solution is cooled to O°C, how much Ba(NO,), is precipitated out of solution?

Tic-componcnt Soilltion: Basis: 100 g Ba(NO,), The maximum solubility of Ba(NO,), in H,O at 100'C is a saturated solution, or 34 g/100 g H 20. Thus the amount of water required at 100°C is 100 g H,O 100 g Ba(NO,), = 295 H 34 g Ba(NO,), g , :ion

°

If the lOO'C solution is cooled to OT, the Ba(NO,), solution will still be saturated,

that the exit :::mcentration, by . .:led on the same J\\"

and the problem now can be pictured as in Fig. E2.16. The tie element is the water.

///

_-----_

........--System --...." Boundary

Original

t' ' I

100 g Ba(NO l J1

I

295 H 0g

\

2

I I

.\ Composition Known

Final

Weight Known

,, "-

' ....

I

Composition Known

I

5 g Ba(NO l J1ilOO g H20

\

I

___r '~ __ ...J_ Componenll

---

295 g H20

Weight Unknown

Crysto Is Composition

Known l

Weight Unknown

ratio between

. wi'

lance

Fig. E2.16.

108

Chap. 2

Material Balances 295 g H 2 0

1

7i('-("0";;'1 '

5 g Ba(NO,), • 14.7 g Ba(NO,lz in final solution 100 g H,O

original

final

lOOg Ba(NO,h -

In th" I", . two

crystals

14.7 g Ba(NO,h = 85.3 g Ba(NO,lz precipitated

Cn(~TIl~!: ,I"

(if we Chnt''>l' \. quest inn') ah'i,'

and air

Algebraic Solution: Let y

tic COl11pl,,1!1cn:

g final solution

=

H 20 balance

295

= (~)

Total balance

395

=

another

+y

y

= 295G~) =

x

=

395 - y

=

til' ""I':"

availabkd.lLII,· the oxidatill!l i. present. Theil \"

g H 20 Y 100 -7- 5 g final solution

x

\trl.'.II" ,

the data til d('._ from the k,t ,'.

x = g crystals

EXAMPLE 2.17

Sec. 2.5

310

85 g Ba(NO')2 crystals We can "" II;,· leaving thl' rr, outlet rnatnl.ll .

Combustion

The main advantage of catalytic incineration of odorous gases or other obnoxious over direct combustion is the lower cost. Catalytic incinerators operate at lower temperatures-500'-900C compared with 1100'-1500'C for thermal incinerators-and use substantially less fuel. Because of the lower operating temperatures. materials of construction do not need to be as heat resistant, reducing installation and construction costs. In a test run, a liquid having the composition 88 percent C and 12 percent H2 is vaporized and burned to a flue gas (fg) of the following composition: substanc~s

CO 2

In step ta) I,:!.· In the <11\ :

13.4

0, N,

In the

3.6

t .... '!

83.0 100.0%

To compute the volume of the combustion device, determine how many pound moles of dry fg are produced per 100 lb of liquid feed. What was the percentage of excess air used? See Fig. E2.17. System Boundory

. Test liqUId

~---,@..=--\~ . . . ---~ I

. C:C1C!y IIC ox!dJt,on unit' L

\''-i----W./' . .

H2O

(a,) L'sing Ii' (

(a,) Using Ih I'

Ig

(h)

(dry)

~f1t'\\

puted

Tie

Air

The~;'

,.,/

Fig. E2.17.

-.< ...,.,.,

(!

l

Chap. 2

.j

Problems Inrolring Tie Compollellls (Elements)

Sec. 2.5

109

Tie-Componelll Solulion: In this problem four streams are present all of whose compositions are known, twO entering and two leaving. We have the weights of three streams as unknown, (if we choose one weight as the basis). However, since we do not have to answer any questions about the water stream, two tie elements to relate the test fluid, dry flue gas, and air streams to each other would be sufficient to solve the problem. In examining the data to determine whether a tie element exists, we see that the carbon goes directly from the test fluid to the dry flue gas. and nowhere else, and so carbon will serve as one tic component. The N, in the air all shows up in the dry flue gas so N, can be used as another tie component. As explained before, there is no reason to convert all the available data to a weight basis, even though the moles of material entering and leaving the oxidation unit are not the same. Let us consider the moles of each atomic specie present. Then we have in lOa Ib of test fluid:

c: H = H,:

¥

=

7.331b mole 6.00 Ib mole

We can say the pound moles of C entering the process equal the pound moles of C leaving the process since the molecular weight is a constant factor in both inlet and outlet material and can be canceled out. Thus ::lbnoxious !'s or -mal ~-:Dcral

.lalion and

Ib C in Ib mole C in

=

Ib C out

==

Ib mole C out

In step (a) below both of these methods of attack are employed. In the dry flue gas Basis: 100 Ib mole dry fg

C'ercent H, 13.4lb mole C I

121b C . lIb mole C

=

161 Ib C

In the test fluid Basis: 100 Ib test fluid .'IInd moles .c of excess

88 Ib C II Ib mole C 121bC (a,) Using Ib C

=

=

733 Ib I C . moe

Ib C as the tie component,

lOa Ib mole dry fg 88 Ib C 54.61b mole dry fg --1"6"I'I;-b'C'.-'----"--f-I"'O"'O"I'b-:-t-=-es-,.t'n'u.."id = 100 Ib test fluid (a,) Using Ib mole C

=

Ib mole C as the tie component,

100 Ib mole dry fg I 7.33 Ib mole C 54.6 Ib mole dry fg 13.4lb mole C I 100 Ib test fluid '= 100 Ib test tluid (b) The N, serves as the tie component to tie the air to the dry flue gas, and since we know the excess 0, in the dry flue gas, the percentage of excess air can be computed. Basis: 100 Ib mole dry fg 83.0 Ib mole

N'II.OO Ib mole air

...

0.791b mole N, (105.0)(0.21) = 0, entering

--"'~Y"""W",,"'AJi"''''''''''_'''''-->'~-

=

105.0 Ib mole air 100 Ib mole dry fg

22.1 Ib mole 0,

--____,. . _,..,....~.. .-

.4!

..

~c;:: i1

110

Chap. 2

Material Balallces

Or, saving one step, 83.0 Ib mole N, 10.21 Ib mole 0, in air __ 22.1 Ib mole 0, entering 0.79 Ib mole N, in air IOU Ib mole dry Ig . %• excess air

100 =

excess 0, 0, entering - excess 0,

=

(100)3.6 22.1 - 3.6

194% =

.



To check this answer, let us work part (b) of the problem using as a basis Basis: 100 Ib test fluid

Sec. 2.5

for each milll wilh Ihe '';IT Igas and infl.: In this cxanw tical oxygl:11 .

A1xchralr An algcb Let

The required oxygen for complete combustion is

comp. C

H,

lb 88 12

lb mole 0, required 7.33 3 10.33

lb mole 7.33 6

Th.e excess oxygen is

= 1.97 Ib mole

54.6 Ib mole dry fg 100 Ib oil

One of 1/1 Since three lJ\ independent n that can be \I total

. - excess 0, 100 _ 1.97(100) -~ 190% %o excess air - required 0, -~ 10.33 . • COll1pOIlCIlJ

The answers computed on the two different bases agree reasonably well here, but in many combustion problems slight errors in the data will cause large differences in the calculated percentage of excess air. Assuming no mathematical mistakes have been made (it is wise to check), the better solution is the one involving the use of the most precise data. Frequently this turns out to be the method used in the check herethe one with a basis of 100 Ib of test fluid. If the dry flue-gas analysis had shown some CO, as in the following hypothetical analysis, CO, 11.9% CO 1.6 0, 4.1 N, 82.4 then, on the basis of 100 moles of dry flue gas, we would calculate the excess air as follows: 0, entering with air:

I

82.4lb moles N, 0.21 Ib mole O 2 = '1 91b I U.791b mole N, -. mo e

c H,

N, 0,

Note thai only two strc.lI· use the C and

°

2

Excess 0,: 4.1 - \6 = 3.3 Ib mole % excess air

= 10021.93~ 3.3

=

17.7%

Note that to get the true excess oxygen, the apparent excess oxygen in dry flue gas, 4.1 Ib moles, has to be reduced by the amount of the thcoretical oxygen not combining with the CO. According to the rcaclions C

+ 0,-+ CO"

C

+ fO,

--)0

CO

The rcmaindel . that the dry 11,. EXA:\fPI.E 1.1 \ A contipu

each other. Oil

Chap. 2

Sec. 2.5

Problems Involving Tie Components (Elements)

111

for each mole of CO in the dry flue gas, 1 mole of 0, which should have combined with the carbon to form CO, did not do so. This l: mole carried over into the flue gas and inflated the value of the true excess oxygen expected to be in the flue gas. In this example, 1.6 Ib moles of CO are in the flue gas, so that (1.6/2) Ib mole of theoretical oxygen arc found in the flue gas in addition to the true excess oxygen.

4%

Algebraic SOllilioll: An algebraic solution of the originally stated problem would proceed as follows: Let x = Ib of test fluid

:; basis

y = Ib moles of dry fg z

=

Ib moles of air

w = Ib moles of H 2 0 associated with the flue gas One of these unknowns can be replaced by the basis, say x = 100 Ib of test fluid. Since three unknowns will remain, it will be necessary to use in the solution three independent material balances from among the four component and one total balance that can be written. The balances are lOla/ 100 + 29. = 30.25y + 18w where 30.25 = average mol. wI. of fg componenl . :y w .ze d

~re~

=

y Ib mole dry fg

=

IV

c

100(0.88)

':Ie use of the ::heck here-

H,

100(0.12)

N2

0.79z = 0.83y

- hypothetical

O2

0.212 =

es

=-lIstaK... _ .. dye

IV

.L

,

excess air as

0.134 Ib mole C I I Ib mole dry fg

12 Ib C I Ib mole C

Ib mole H 2 0 II Ib mole H, 2.016 Ib H, I Ib mole HT,~O,.j-';:l:;'1b;:';:":n'-,0:";I'-e'-'H+-2 Ib mole H 2 0

0.51b mole O 2 lIb mole H 2 0 Y Ib mole dry fg I 0.036 III mole 0, I I Ib mole dry fg

+ 0.134 mole 0,

Note that the balances for the tie components C and N2 (as well as H,) involve only two streams, whereas the O 2 (and total) balance involves three streams. We shall use the C and N2 balances. Basis: 100 Ib of test fluid 88 y = (0.134)(12) 0.83

z

54.61b mole dry fg

= 57.5 Ib mole air

= 0.79 Y

0, entering

=

=

57.5(0.21)

=

12.1 Ib mole

The remainder of the calculations follow those described above. Can you demonstrate that the dry flue-gas analysis is slightly inaccurate? :ory flue gas, .[ combining

EXAMPLE 2.18

Distillation

A continuous still is to be used to separate acetic acid, water, and hen7cne from each other. On a trial run the calculatcd data werc as shown in Fig. E2.18. Data record-

'"

. ;; ::;:;:w.w;

.j

~r

_:01

112

System Boundary~/----,..--'-:::""~_ _ fO.9%

I

I

Aqueous { 80% acetic acid (HAcJ Solution

20% water (H 20)

Feed

21.7

i i

\

: {

\

\

\

H20 Bz

'-...

Mrtt

Take a new basis:

I

67.4lb Bz 75.91b HAc

I I

Algebraic SO/lit ion : To employ algebra material balance for ead

I

\

'M'

Waste

i

STILL

I :

Benzine (Bz) (data not available)

67.4

HAC}

¥fa

Sec. 2.5

Chap. 2

Material Balances

P

I

I

I

Product 350lb HAc/hr

Fig. E2.18.

ing the benzene composition of the feed were not taken because of an instrument defect. Calculate the benzene flow in the feed per hour. A quick inspection of the flow diagram shows that enough information about weights and compositions would be available to work this problem if only the benzene composition in the feed were known, bUt it is not. However, we have one added source of information not readily apparent on the surface, and that is that there are /IVa tie components from the feed to the waste: water and benzene. A choice now must be made in procedure: the problem can be calculated using the tie components or using algebraic techniques.

The equations are

Tie-Campanelli Soilition: Take a convenient basis of 100 Ib of waste (100 Ib of feed would involve more work since the benzene composition is unknown, but 100 Ib of aqueous solution80 percent HAc, 20 percent H,O-would be a sound basis). Let x = Ib of benzene in the feed/IOO Ib feed.

Any three will be indepm three equations involves r

Total bal,

HAc balal Bz balanc,

EXAMPLE 2.19

A chemical companl' amount of moisture and. job assignment is as purd you a contract for 10,.." 3.2 percent and a maxin,,, ery). You accept this c,":: coal as reported by yL'ur t The billed price for this ,', '. outside your plant. The ,," correct. Calculate the c,'"

Basis: 100 Ib waste composition

HAc H 2O Bz Total

feed Ib

waste

(100 - ,.-)0.80 (100 - x)O.20 x

100

prodllct

% = Ib

%

10.9 21.7 67.4 100

100

For an initial step let us calculate the quantity of feed per 100 Ib of waste. We have the water in the feed (21.7 Ib) and the benzene in the feed (67.41b); these two materials appear only in the waste and not in the product and can act as tie components. All that is needed is the HAc in the feed. We can use water as the tie component to find this quantity. 21.71b H,O I 0.80 Ib HAc 100 Ib waste 10.20 lb H 2 0

0=

=

Solution: What you rca II) r· l ' allowable content of 01,11 . of the combustible m"tel.·

86.81b HAc/IOO Ib waste

The product now is HAc in feed - HAc in waste 86.8 10.9

Econor

HAc in product 75.9

HAc balance

I!

I

COn1~;;

1\.t Oi'" "

Chap. 2

Problems Involving Tie Components (Elements)

Sec. 2.5

113

Take a new basis: Basis: 350lb HAc product", I hr 67.4 Ib Bz in feed 350 Ib HAc product = 311 Ib Bz/hr hr 75.9 Ib HAc in product Algebraic Solution: To cmploy algebra in the solution of this problem, we need an independent material balance for each unknown. The unknowns are W= waste,lb

c:t D

F= feed,lb

HAc/hr

x = Ib benzene/lOO Ib feed Basis: 1 hr '" 350 Ib HAc product of an instrument ::normation about _ only the benzene . one added source _ there are two tie '2'ice now must be :::Donp'

~r

The equations are

F= W+ 350

Total balance H 2 0 balance

F[0.20(l1~ -

HAc balance

F[0.80(;~ -

using

F[I~J

Bz balance ..l1d involve more _!Jeous solution= Jb of benzene in

x)] = W(0.217) x)] = W(0.109) + 350 = W(0.674)

Any three will be independent equations, but you can see the simultaneous solution of three equations involves more work than the use of tie components. EXAMPLE 2.19 Economics

waste. We have =se two materials ::omponents. All -:nponent to find

A chemical company buys coal at a contract price based on a specified maximum amount of moisture and ash. Since you have married the boss's daughter, your first job assignment is as purchasing agent. The salesman for the Higrade Coal Co. otTers you a contract for 10 carloads/month of coal with a maximum moisture content of 3.2 percent and a maximum ash content of 5.3 percent at $4.85/ton (weighed at delivery). You accept this contract price. In the first delivery the moisture content of the coal as reported by your laboratory is 4.5 percent and the ash content is 5.6 percent. The billed price for this coal is $4.85/ton-as weighed by the railroad in the switchyards outside your plant. The accounting department wants to know if this billing price is correct. Calculate the correct price. Solution: What you really pay for is 91.5 percent combustible material with a maximum allowable content of ash and water. Based on this assumption, you can find the cost of the combustible material in the coal actually delivered.

Basis: I ton coal as delivered composjliOfl

_.lance

Combustible Moisture plus ash

delil'ered 0.899 0.101 1.000

contract coal 0.915 0.085 1.000

*"=$

... ;10.1'"

Chap. 2

114 Material Balances

S,·c.2.6

The contract calls for $4.85 I ton contract 0.8~9 tondc~mb. ton contract 0.915 ton comb. ton e.

=

$4.76/ton del.

The billed price is wrong. If the ash content had been below 5.3 percent but the water content above 3.2 percent, presumably an adjustment should have been made in the billing for excess moisture even though the ash was low.

2.6 Recycle, bypass, and purge calculations Processes involving feedback or recycle of part of the product are frequently encountered in the chemical and petroleum industry as well as in nature. Figure 2.10 shows by a block diagram the character of the recycle stream. Figures 2.3 and 2.4 indicate some typical recycle streams in the world at large. As another example, in planning long space missions, all the food and water will have to be provided from stores on board the spacecraft. Table 2.1 lists the daily intake and output per man of solid and water. Since the average internal water consumption amounts to some 5.50 lb/man/day, this leaves an additional requirement of 4.69 lb that must be made up from stores or reclaimed from waste products containing water. Figure 2.11 shows the sources of water and how by the use of several recycle streams the 5.50 lbjday is proposed to be collected. Many industrial processes employ recycle streams. For example, in some drying operations, the humidity in the air is controlled by recirculating part of the wet air which leaves the dryer. I n chemical reactions, the unreacted material may be separated from the product and recycled, as in the synthesis of ammonia. Another example of the use of recycling operations is in fractionating columns where part of the distillate is refluxed through the column to maintain the quantity of liquid within the column.

.',

Ouff"!



,

.

"

hI!,"'

Process

,, ,

I'

//r'_ _ _ _ _R_e..;CY_C_le_R_ _ _ _---.""

"'" 3-----

Gross -,' Product

I

/'4····

,

'.'

p

\',

Separator ,

'-- -------1----------

Fig. 2.10. A process wilh recycie (Ihe numbers designate possible material balances-sec text).

Ttl

\,' \

--.-

... ...,.... ,

----,..,Chap. 2



Recycle, Bypass, alld Purge Calculatiolls

Sec. 2.6

115

O.221b

.'n del.

'''m!en! above 3.2 :oilling for excess

~

are frequently nature. Figure .:.:am. Figures 2.3 _'ge. As another :- will have to be =ne ~""" intake .::m, : con::mio.. ..:quire-:led from waste :.'.:er and how by . be collected. ==ple, in some ::::ulating part of ..:.:-eacted material _SIS of ammonia. :'TIating columns o maintain the :l.

coduc!

::: n"

, , O.591b

,, ,

\

~O

in Food O.151b

~

for Leak Make up O.231b

Fig_ 2.11. Water and oxygen recycle in a space vehicle. TABLE

2.1 MAN'S DAILY BALANCE

Total balance (lb)

Water balance (Ib)

Output

Urine (95~;H,O) Feces (75.8 %H,O) Transpired Carbon dioxide (1.63 Ib 0,) Other losses

3.24 0.29 2.20 2.24 0.14 8.11

3.08 0.22 2.20

Iflput

Food Oxygen Metabolic water Water

1.5 1.92

0.15

4.69 8.11

5.s0

0.66 4.69 5.50

Recycle streams initially may be confusing to you, but with a little practice in solving problems involving recycle you should thereafter experience little difliculty. The first point you should grasp concerning recycle calculations is that the process shown in Fig. 2.10 is in the steady state-no buildup or deplc-

----.-----------~ ... -.,------.'-~~~"

~.,

116 Malerial Balances

Chap. 2

tion of material takes place inside the process or in the recycle stream. The values of F, P, and Rare cOl1slal1l, The feed to the process itself is made up of two streams, the fresh feed and the recycle material. The gross product leaving the process is separated into two streams, the net product and the material to be recycled. In some cases the recycle stream may have the same composition as the gross product stream. while in other instances the composition may be entirely different depending on how the separation takes place and what happens in the process. Recycle problems have certain features in common, and the techniques you should use in solving these problems are the familiar ones previously encountered in this chapter in the use of material balances. You can make a material balance on a total basis or for each component. Depending on the information available concerning the amount and composition of each stream, you can determine the amount and composition of the unknowns, If tie components are available, they simplify the calculations. If they are not available, then algebraic methods should be used. A balance can be written in several ways, four of which are shown by dashed lines in Fig. 2.10: (a) About the entire process including the recycle stream, as indicated by the dashed lines marked 1 in Fig. 2.10. (b) About the junction point at which the fresh feed is combined with the recycle stream (marked 2 in Fig. 2.10). (c) About the process only (marked 3 in Fig. 2.10). (d) About the junction point at which the gross product is separated into recycle and net product (marked 4 in Fig. 2.10), Only three of the four balances are independent. However, balance 1 will not include the recycle stream, so that the balance will not be directly useful in calculating a value for the recycle R. Balances 2 and 4 to include R. It would also be possible to write a material balance for the combination of 2 and 3 or 3 and 4 and include the recycle stream. If a chemical reaction occurs in the process. the stoichiometric equation, the limiting reactant, and the degree of completion all should be considered before beginning your calculations. As discussed in Chap. I, the term COIlI"ersioll as applied to Fig. 2.10 may be either the fraction or percentage of the fresh feed which reacts or the fraction of the feed plus recycle. If 100 Ib of substance A in the fresh feed is converted on an overall basis into 40 1b of desired product, 301b of waste, 20 Ib of a secondary product, and 10 Ib of A pass through the process unchanged, the total conversion of A is 90 percent based on the fresh feed. Howc\,cr, the yield of primary and secondary products is only 60 Ib of products per 100lb of A. If, in addition, the recycle stream contains 100lb of A, the total con\'ersion of A on a once-through basis is only 45 percent. You can see that the basis on which the conversion and yield arc calculated must always be clearly specified, When the fresh feed consists of more than one

---

~.....

..... ---~.-.-~ ..,...

-,~,,""""!""-""''''''-''

Sec. 2.6

material. usually th compoun( You a simplifi( if such a " dure of sc perhaps It unknown and for \\ R, X, or

\1

way and, up, you 11" econOlll ic Some illu,

EXAMPI.I

A diS!

toluene mi contains Y toluene. TI

Ib/hr. A 1" withdrJ\\ n of the coltt ratio of th,

SO!IIIi. Sec 1, tie

COn1l'l'l

Chap. 2 2aI'h. The values ::.:ide up of two· =:1ct leaving the material to be =omposition as ~:on may be en· = what happens

the te\:'hniques '·1Jreviously en::eu can make a =iling on the :Jf each ~tream, TIS. If tie cum''= not available,

Recycle, Bypass, and Purge Calculations 117

Sec. 2.6

material, the yield and conversion must be stated for a single component, usually the limiting reactant, the most expensive reactant, or some similar compound. You should carefuly analyze a problem to ascertain whether there is a simplified method of solution (as illustrated in the following examples), but if such a simplification escapes you, or if there is none, then the standard procedure of setting up algebraic material balances will always be effective, although perhaps long and involved. One psychological stumbling block is the stream of unknown weight or composition which is found to be essential for the solution and for which you have no information. By labeling this stream with a letterR, X, or whatever-you can proceed to make material balances in the ordinary way and solve for the unknown stream. Of course, for each unknown you set up, you must write an independent equation, and so from the viewpoint of the economic use of your time, it is advisable to minimize the number of unknowns. Some illustrations follow.

,own by dashed EXAMPLE 2.20 Recycle without chemical reaction

::J.oined with the

=separated into -=ce I will not ::-ectly useful in ....ie R. It would :: of 2 and 3 or =mc equation, ~ De considered .= term conver. "~e of the fresh : 0 of substance =sired prod uct. .::ss through the ...::0 on the fresh conly 60lb of =;mtains 100lb .:5 percent. You =Jcul ., must one

A distillation column separates 1O,000Ib/hr of a 50 percent benzene-50 percent toluene mixture. The product recovered from the condenser at the top of the column contains 95 percent benzene, and the bottoms from the column contain 96 percent toluene. The vapor stream entering the condenser from the top of the column is 8000 lb/hr. A portion of the product is returned to the column as reflux, and the rest is withdrawn for use elsewhere. Assume the compositions of the streams at the top of the column (V), the product withdrawn (D), and the reflux (R) are identical. Find the ratio of the amount refluxed to the product withdrawn. Solutiol1:

See Fig. E2.20. All the compositions are known and two weights are unknown. No tie components are evident in this problem; thus an algebraic solution is mandatory.

\ R liquid _-,F_'"1-'-i tO,OOO Ib/hr 0.50 Bz 0.50 Tol

0

, __ '\) 0.95 Bz Liquid ~ 005 Tol / Boundary line for I balonce oround " condenser I

,"-----Boundary line for /' overall balance / /

'--",.--w 004 Bz L' 'd 0.96 Tol IQUI

Fig. n.20.

-----.-

----- ..

-.-.~

..

..' .". ..._.-----,..,.,

..

..

+

M

118

Material Balances

Chap. 2

By making an overall balance (ignoring the reflux stream for the moment), we can find D. With D known, a balance around the condenser will give us R. Basis; I hr

Sec. 2.6 Sollirioll: In this p:" l>ut this can \'C stream (\)11;.,' composition I,

Overall material balances,' Total material:

Ii'

F=D+ W 10,000

=

D

+W

(a)

Component (benzene):

+ WXw D(0.95) + W(0.04)

FXF = DXD 10,000(0.50) =

(b)

With ali •. for an over.I:1 . draw a pidllfc H 20 c"apc'f.,:·,

Solving (a) and (b) together, 5000

=

(0.95)(10,000 - W)

+ O.04W

W = 49501b/hr D = 50501b/hr Balance around the condenser,' Total material:

V=R+D 8000 = R

+ 5050

R = 29501b/hr R D

EXAMPLE 2.21

=

2950 5050

= 0.584

Recycle without chemical reaction

Data are presented in Fig. E2.21 (a) for an evaporator. What is the recycle stream in pounds per hour?

To

lh:h"r'

(Omp!.HlcIlL

l

(a) A I',

H20

(1)) A I.

----

,- ,,/

-- "

The '\

I I

10,000 Ib/hr 20% KN0 3 Solution

300°F

I

I

1/

R

I

Recycle 100°F

\50%

Saturoted Solution (

,KN03 \

'---4..j \

/ ,'.......

"

Boundary line "_ for overall balance

' ...

c Fig. E2.2I(a).

0.6 Ib KNO J ) Ib H20

1.ltle< " ,

rllUnh ,~~(.I' 1"'1.11 !

.:C

Chap. 2

.::rrnent), we can R.

(a)

(b)

Recycle, Bypass, and Purge CalclIlations

Sec. 2.6

119

Solution: In this problem the compositions are all known except for the recycle stream, but this can be easily calculated. On the basis of I Ib of water, the saturated recycle stream contains (1.0 Ib H,O, 0.61b KNO,) =" 1.61b total. The recycle stream composition is

0.61b KNO, I lib H,O 0 3751b KN'O 'ib I . '31 so utlon I Ib H 20 11.6 Ib solution~' . With all compositions known, a search for a tie component shows that one exists for an overall balance-the KNO,. Temporarily ignoring the interior streams, we can draw a picture as in Fig. E2.21(b). Now we can calculate the amount of crystals (and H 20 evaporated, if wanted). Basis: 1 hr := 10,000 Ib feed 10,000 lb F 0.20 Ib KNO, I Ib crystals _ 1 I I I lIb F 0.961b KNO, - _080 Ib, hr crysta s

feed

..;,0. '" '" Crystals

"

",,"

,0.20 KN0 3

"

0.80 H20

0.96 KN0 3 0.04 H20

Fig. E2.21(b).

:-ecyc1e stream

To determine the recycle stream R we need, in the absence of an additional tie component, either (a) A balance around the evaporator-or (b) A balance around the crystallizer. The latter is easier since only three rather than four (the recycle added to the feed is the fourth stream) streams are involved. Total balance on crystallizer: M=C+R M = 2080

::.6 Ib KN0 3 ) Ib H20

]'Ix",

0.5M

:e-r .S

+ H20)

+R

Component (KNO , ) balance on crystallizer:

= CXc + RXR = 0.96C + R(0.375)

Solving these two equations, 0.5(2080 -+- R) R

= 0.375R + 2000 =

76801b/hr

4" •

...

-

.

Chap. 2

120 Material Balances

Sec. 2.6 (a)

EXAMPLE 2.22

mole 01 IOOlb,

Recycle with a chemical reaction

In a proposed process for the preparation of methyl iodide, 2000 Ib/day of hydroiodic acid are added to an excess of methanol: HI

+ CH,OH -~ CH,I + H 2 0

If the product contains 81.6 percent CH,I along with the unreacted methanol, and the waste contains 82.6 percent hydroiodic acid and 17.4 percent H ,0, calculate, assuming that the reaction is 40 percent complete in the process vessel:

Solution: In this problem all the compositions are known (in weight percent) as well as one weight-the HI input stream. The unknown quantities are the weights of the CH,OH (M), the product (P) ,the waste (W), and the recycle stream. An overall balance about the whole process is all that is needed to calculate the M stream, but it will not include the recycle stream-see the dashed line in Fig. E2.22.

---

CH 30it",

,,

T, On the 0.96

(11

junctic about but th

,,

HI

0.968 --Ill

-j

(a) The weight of methanol added per day. (b) The amount of HI recycled.

----

R,'

reaCh)

\

rr-rl---

CH31 81.6% Product

CH 30H 18.4

2000lb/doy

\\·C'-C:ll

'1

HI

-------

HI 82.6% w

~

17.4

t

os e

fact II scq\lC.I

We c. Fig. E2.22.

We first need to convert the product and waste streams to a mole composition in order to make use of the chemical equation. A basis of 100 Ib of waste is convenient, although 100lb of product would also be a satisfactory basis. Basis: 100 Ib waste

anJ

MW

Ib = % 82.6 17.4 100.0

compo HI H,O

Ib mole 0.646 0.968

128 18

Ib

= % 81.6 18.4

MW 142

32

Ib mole 0.575 0.575

equivalent lb mole of CH,OH 0.575 0.575 1.150

~.

Ie;" r·

In

Basis: 100 Ib product

camp. CH,I CH,OH

Our ,.

I .

- -_ _ _

Clmp.2

10 D:r{day C!Yfhydro-

~._<..

*__.....&I10"=_ _

.tid."_ _ _

=t)1 a$ w~n as one =m€theCH,OH =lli limJance about it will Dot include

=

t_n.'' 'i......

t-·~'. .

11fW''... ?

·I/II!••IIiooio.......................·.. • ___, _ _ • .. ·i&io'"'I;.~

H .. ". ;. . . ._ _

Recycle, Bypass, and Purge Calculations

Sec. 2.6

121

(a) From the chemical equation we observe that associated with each pound mole of H 2 0 in the waste is lib mole of CH,L Consequently the entering HI is (per 100 Ib waste). Reacted 0.9681b mole H 2 0 100 Ib waste

=d methanol, and ~t HzO, calculate, vessel:

'_' '·-'IIoIUI...·_'..·....•..d'i>,]·..

W... O'..'''''T..

ili·' . .

I Ib mole CH,I lib mole HI 1281b HI 1241b HI I Ib mole H 2 0 I Ib mole CH,I Ib mole HI = 100 Ib waste

Total HI = reacted + unreacted = 124 On the same basis the CH,OH entering is,

+ 82.6

= 206.61b HI/IOO Ib waste

321b CH,OH 0.9681b mole H 2 0 lib mole CH,I 1.150 Ib mole CH,OH 100 Ib waste I Ib mole H 2 0 0.575 Ib mole CH,I lib mole CH,OH 61.91b CH,OH = 100 Ib waste Basis: 1 day 61".9""I"bc-C",H",,',,,;0-:-H _2ooo,-",lb_H_Iho;I0""0,I,-b",w,a"st,et1 day 206.61b HI 100 Ib waste

=

600 Ib CH OH/da ,y

(b) To obtain the value of the recycle stream R, a balance must be made about a junction point or about just the reactor itself so as to include the stream R. A balance about junction I indicates that (all subsequent bases are 1 day) 200

+ R = reactor feed

but this equation is not of much help since we do not know the rate of feed to the reactor. Similarly, for a balance on junction 2, 18,4

gross product = R

+P + W

we can compute the value of the product and waste but not that of the gross product. The only additional information available to us that has not yet been used is the fact that the reaction is 40 percent complete on one pass through the reactor. Consequently, 60 percent of the HI in the gross feed passes through the reactor unchanged. We can make an HI balance, then, about the reactor: (2000

+ R)(0.60)

=ie composition in ~is convenient,

400

= R

=

+ 800

0.40R

R = 1000 Ib/day Our material balance around the reactor for the HI is known aS2 once-through balance and says that 60 percent of the HI that enters the reactor (as fIresh feed plus recycle) leaves the reactor unchanged in the gross product (recycle plus product).

Two additional commonly encountered types of process streams are shown in Figs. 2.12 and 2.13: :mivalem Ib [! o/CH,OH 0.575 0.575

l.\sn

(a) A bypass stream-one which skips one or more 'SIlages of the process and goes directly to another stage. (b) A purge stream-a stream bled off to remove an a=mulation of inerts or unwanted material that might otherwise buiild! up in the recycle

..- ..._*-....,..-•.".,.".,. .,. ". .,. .---".~.'""'"....., ""}',~,

.,

il?t14i}4',.S;;:_lW$Iii·f"~

122

Chap. 2

Material Balances

Sec. 2.6 Solutio, Byexan bypasses the plant. All th process? W< or P):

Byposs B

Feed

Process

f - - - - Product

Fig. 2.12. Bypass stream.

(a) Ove

Total rr Recycle R )---Purge

Compo Feed

Process

)----Product

Using ( Fig. 2.13. A recycle stream with purge. stream. The purge rate is adjusted so that the amount of purged material remains below a specified level or so t hat the rate of } {accumulation

=

0

=

The overall tower; for t (b) Ba/, pentane to\\ Total IT

{rate of entering material} _ {rate of purge} and/or production and/or loss

Calculations for bypass and purge streams introduce no new principles or techniques beyond those presented so far. An example will make this clear.

EXAMPLE 2.23

Bypass calculations

Compo

In the feed-stock preparation section of a plant manufacturing natural gasoline, isopentane is removed from butane-free gasoline. Assume for purposes of simplification that the process and components are as shown in Fig. E2.23_ What fraction of the butane-free gasoline is passed through the isopentane tower?

Conseq

t

Isopenlone side sl,eo m (5) i-C sH'2

Debutonizer

(X)

IF)

to o Ibs IBulone

-

Anothc mixing point that streams (c) Ba/,

Isopenlane lower

Total n Compo

n-C~H'2

(2) . J - T'Ill ,nolurol

free

905-

alline plonl (P) 9lD% n-C sH'2 '«11% i-C sH,2

feed (I) n-C sH'2 80% i-C s H'2 20%

Equllic

Fig. E2.23.

__

.

.....••.•.. "

e1t

([hap. 2

-

Producl

Recycle, Bypass, and Purge Calculations

Sec. 2.6

123

Solulion: By examining the flow diagram you can see that part of the butane-free gasoline bypasses the isopentane tower and proceeds to the next stage in the natural gasoline plant. All the compositions are known. What kind of balances can we write for this process? We can write the following (each stream is designated by the letter F, S, or P): Basis: 100 Ib feed (a) Overall balances (all compositions are known; a tie component exists): Total material balance: in = out too = S + P Component balance (n-C,): in 100(0.80)

-fffadud

(1)

out

=

+ P(0.90)

S(O)

(2)

Using (2),

=

P

Jfpu"

~;material

-'~' at;

purge} .:nd/OI: loss

:rr:inciples. or tech:iris- clear.

=

100(0.80) 0.90

89lb

S = 100 - 89 = 11 Ib

The overall balances will not tell us the fraction of the feed going to the isopentane tower; for this we need another balance . (b) Balance arol/nd isopentane tower: Let x = lb of butane-free gas going to isopentane tower. Total material: out

in =

+ II-C,H 12 stream

x = II

(another unknown, Y)

(3)

Component (n-C,): natural gasoline, --)oses of simp Ii ficahat fraction of the

.... J!

x(0.80)

= y

(a tie component)

(4)

Consequently, combining (3) and (4), x

= II

x

=

+ 0.8x

55 lb,

or the desired fraction is 0.55

Another approach to this problem is to make a balance at points (I) or (2) called mixing points. Although there are no pieces of equipment at those points, you can see that streams enter and leave the junction. (c) Balance oral/lid mixing point (2): ~~

material into junction

material out

Total material: (lOO - x)

+y

= 89

+

=

(5)

Component (iso-C,): :nurol 90solon!

IPl

n-C~U

(100 - x)(O.20)

0

89(0.10)

Equation (6) avoids the use of y. Solving,

j--'

20 - 0.2x

=

8.9

x = 55 Ib, as before

(6)

124

Material Balallces

Chap. 2

After a little practice you can size up a problem and visualize the simplest types of balances to make.

Up to now we have discussed material balances of a rather simple order of complexity. If you try to visualize all the operations which might be involved in even a moderate-sized plant, as illustrated in Fig. 2.4. the stepwise or simultaneous solution of material balances for each phase of the entire plant is truly a staggering task, but a task that can be eased considerably by the use of computer codes. Keep in mind that a plant can be described by a number of individual, interlocking material balances which. however tedious they are to set up and solve, can be set down according to the principles and techniques discussed in this chapter. In a practical case there is always the problem of collecting suitable information and evaluating its accuracy, but this matter calls for detailed familiarity with any specific process and is not a suitable topic for discussion here. We can merely remark that some of the problems you will encounter have such conflicting data or so little useful data that the ability to perceive what kind of data are needed is the most important attribute you can bring to bear in their solution. By now, from working with simple problems, you should have some insight into the requirements for the solution of more complicated problems. In Chap. 5 you will encounter some of these more complex problems.

WHAT YOU SHOULD HAVE LEARNED FROM THIS CHAPTER I, You should be able to analyze a material balance problem in order to

(a) (b) (c) (d) (e)

Find out what the problem is. Draw a picture of the process. Put down all the compositions of the entering and leaving streams. Decide which masses (weights) are known and unknown. Find a tie component and/or set up algebraic mass balances involving the unknowns. (f) Solve for the required values. 2. You should understand the once-through balance for recycle problems involving chemical reactions.

SUPPLEMENTARY REFERENCES 1. Benson, S. W., Chemical Calculations, 2nd ed., Wiley, New York, 1963. 2. Henley, E. J., and H. Bieber, Chemical Engineering. Calculations, McGraw-Hili. New York, 1959. 3. Hougen, O. A., K. M. Watson, and R. A. Ragatz, Chemical Process Principks, Part I, 2nd cd., Wiley, New York, 1956.

--

"

.

: I

Chap. 2

, the simplest types of

::ther simple order of • :1 might be involved ::e stepwise or sim ule entire plant is truly \. by the use of com. a number of indivi:.IS they are to set up Iechniques discussed ~m of collecting suit:ler calls for detailed topic for discussion : will encounter have .:ty to perceive what JU can bri ng to bea r ::;TIS, 'lOuld have :'e CL ated prob:: complex problems.

Problems

Chap. 2

125

4.

Littlejohn, C. E .. and G. F. Meenaghan, An Introduction to Chemical Engineering, Van Nostrand Reinhold. New York. 1959.

5.

Nash, L., Stoichiometry, Addison-Wesley, Reading, Mass., 1966.

6.

Ranz, W. E .• Describing Chemical Engineering Systems, McGraw-Hill, New York, 1970 .

7. Schmidt, A. X., and H. L. List, Material and Energy Balances, Prentice-Hall, Englewood Cliffs, N.J., 1962. 8. Whitwell, J. c.. and R. K. Toner, Conscrl'ation of Mass and Energy, Ginn/Blaisdell, Waltham, Mass., 1969 . 9.

Williams, E. T., and R. C. Johnson. Stoichiometry for Chemical Engineers, McGraw-Hill, New York, 1958.

PROBLEMS' 2.1. The steady-state hydrology of the Great Lakes based on average flow values for the period 1900-1960 is shown in Fig. P2.!. Compute the return into lake St. Clair and the evaporation from Lake Ontario based on the data shown. Would the balance for I year be a steady-state balance? For I month? P·090

legerd'

'S CHAPTER

(R I BaSin Rl,J"Off

IPI

PreclpltatO"lon

lake

IE I £Vl!lpol'ahon frQ"Tl lake

.em in order to

Values -

'caving streams. ·;own. _5 balances involving

r...l.-1--1l:<.i.._.r-....J.1

JOOO

cis

])80

L._-.-.Jw;o.:;t---L,.....-,J SI

La~e RIver

Tecycle problems in-

Fig. Pl.I.

'.' York, 1963. lations, McGraw-Hill. .=al pr

"5

Principles,

2.2. By use of an overall steady-state material balance determinine whether or not the petrochemical process indicated in Fig. P2.2 has been properly formulated. The block diagrams represent the steam cracking of naphtha into various products, and all flows are on an annual basis. i.e., per year . 'An asterisk designates problems appropriate for computer solution. Also refer to the computer problem after Problem 2.68.

;;

.__

.........--___- ..

~_

-~-

.... '__.",.,..___

...._%"'¢%"""'.,,..,,__,

."W~,,.....

.""_~~

Gases, Vapors, Liquids, and Solids

3

,i !

L

In planning and decision making for our modern technology, the engineer and scientist must know with reasonable accuracy the properties of the fluids and solids with which he deals. If you are engaged in the design of equipment. say the volume required for a process vessel, you need to know the specific volume or density of thc gas or liquid that will be in thc vcssel as a function of temperature and pressure. If you are interested in predicting the possibility or extent of rainfall, you have to know something about the relation between the vapor pressure of water and thc temperature. Whatever your current or future job, you need to have an awareness of the character and sources of information concerning the physical properties of fluids. Clearly it is not possible to have reliable detailed experimental data on all the useful pure compounds and mixtures that cxist in the world. Consequently, in the absence of experimental information, we estimate (predict) properties based on modifications of well-established principles, such as the ideal gas law, or based on empirical correlations. Thus thc foundation of the estimation methods ranges from quite theoretical to completely empirical and thcir reliability ranges from excellent to terrible. In this chapter we shall first discuss ideal and real gas rclationships including some of the gas laws for pure components and mixtures of ideal gases. You will learn about methods of expressing the p- V- T properties of real gases by means of equations of state and. alternatively, by compressibility factors. Next we shall introduce the concepts of vaporization, condensation, and vapor pressure and illustrate how material balances are made for saturated and partially 146

"'~,"';'::<':'"

:

t:

i, ,., . j I. a !:d. ~. ttl ~ ~::s9':~O~

$:=:

f'

0

t~

II

n

__

:;.:

~

d .....

0. ;:. .- (=)' (') ~

o ;;: "' ..... _ ;:I

o..::3~ -,~c"

en

::s<~~-;.

;'og-cg;: (D~ ::;. =. ~ "r:1Vl~ (3 .B s

~: 0

-,~

-.'-

~c..~"'1~::S

0.. ~

.- -

~

-.

"O-gQOQe;~ ~ .., ?'l ~ ~ ..... .... ('p • ::s t:i. 'Q0 ('JZ '" ........ J !2. _

::!.

0..'

a.3~~~o -. ~ -.., ::s ::s

=

~ _.

~cr~~g9--

c'(']Q 0 ~

:::-.::"

("J

"<

~_

'::lcncn ..

---_. ---- -_._ .........__.._-_.

--1

d ~ ~

.., ..,"' @

=-"':1.8

::s -,

~

;,1

tP

- . ;:I

~

(1)'1;'"

~.Ideal

3 ?~

-

~

0

-

=..; (l)

('p

..,

0

"

'I

::.

=- ~"

"r:1

(b

-....

(")

'"

V'J'"

.....

l1.;lJ

t

............ .

I

\>

I

Phose Retations Belween Gases, Liquids and Sotlds

Definitions

I-

fl!;fJ. Gas DenSity.

Compressibility I.... - ~

Lo..

~ Humidity

Equations of State

Factors

r-

~ Phase

1-

~ Phase

..l..dJ

Vapor Pressure

sp. gr.

Saturation

~ Partial

I-

Rule

Diagrams

1-

Saturation; Relative Humidity

~ Phase

~".';"""""'~'""'"

-

r

r ~

~

Chapter I

~

Phenomena

i

Material Balances Involving

Material and Energy Balances

Background Information

....

f

2&J

--- _..

...

1-

If j

I Chapter 2 Material Balance Techniques

f-

I

{

t

Real ,- l1..lJ Mixtures

-.I

CA)

-

1--

~.

e- ...~

0

8 ::n_. "<:l(')oC::S ~ =; o· :;. 0 0 :;;;? 0.: ~ ;:I <> ;:l

__ _-.._-_

Gos Law

Mixtures

-.

S

I Real Gases I

'-

.. l:; ~

;:l . " I» '" 0"0-'_0 l' ~~=:r..c_O

0'0 '" 0' g

Itdeol Goses I

~ Perfect

-.

U tJ :J

:::. ;. :- =' ~ '" c o C1I tn 0 ..

Pressure .. Temperature .. Volume Relations for Goses

I-

I".,j

t:: t:: ~{JQ~O~"d~

~o~~o..~

!::.;:I

-'" -

o

'4,

'.

tl

II

F

Fig, 3.0, Hierarchy of topics to be studied in this chapter (section numbers are in the upper left .. hand corner of the boxes).

r

~

, t

J "' ...... ,.. ... "....

_c~,L.;_"_,_

::...._ •...-~

,,;,,"~._-;.,.'

_,.l-; .

,A.

"

148

Gases, Vapors, Liquids, alld Solids

Chap. 3

Sec. 3

saturated gases. Finally, we shall examine the qualitative characteristics of gas-liquid-solid phases with the aid of diagrams. Figure 3.0 shows the interrelationships among the topics discussed in this chapter and how they relate to the making of material and energy balances.

The f imm" pre,." occlir cula\<' tiom:' total,

3.1 Ideal gas laws In 1787, Jacques Charles, a French chemist and physicist, published his conclusions about the relationship between the volume of gases and temperature. He demonstrated that the volume of a dry gas varies directly with temperature if the pressure remains constant. Charles, Boyle, Gay-Lussac, Dalton, and Amagat, the investigators who originally developed correlating relations among gas temperature, pressure, and volume, worked at temperatures and pressures such that the average distance between the molecules was great enough to neglect the effect of the intermolecular forces and the volume of the molecules themselves. Under these conditions a gas carne to be termed an ideal gas. More properly, an ideal gas is an imaginary gas which obeys exactly certain simple laws such as the laws of Boyle, Charles, Dalton, and Amagat. No real gas obeys these laws exactly over all ranges of temperature and pressure, although the "Iighte~" gases (hydrogen, oxygen, air, etc.) under ordinary circumstances obey the ideal gas laws with but negligible deviations. The "heavier" gases such as sulfur dioxide and hydrocarbons, particularly at high pressures and low temperatures, deviate considerably from the ideal gas laws. Vapors, under conditions near the boiling point, deviate markedly from the ideal gas laws. However, at low pressures and high temperatures, the behavior of a vapor approaches that of an ideal gas. Thus for many engineering purposes, the ideal gas laws, if properly applied, will give answers that are correct within a few percent or less. But for liquids and solids with the molecules compacted relatively close together, we do not have such general laws. In order that the volumetric properties of various gases may be compared, several arbitrarily specified standard states (usually known as standard conditions, or S.C.) of temperature and pressure have been selected by custom. The most common standard conditions of temperature and pressure are Universal Scientific

32°F and 760 mm Hg (and their equivalents)

Natural Gas Industry

60°F and 14.7 psi a

The first set of standard conditions, the Universal Scientific, is the one you have previously encountered in Chap. I and is by far the most common in use. Under these conditions the following volumetric data are true for any ideal gas: I g mole = 22.4 I at S.c. I Ib mole

= 359 ft' at S.c.

I kg mole = 22.4 m' at S.c.

EXA\ (,

tions.

~

aPri;r(

feel. " tcmpn ~[aJld"

3 vcnrh ~ht.)\1, (> Jirl'ell scicn:! the j(/('

In :q'r' a tin.,-; sinni:,

Her,' " r.tOr,·" nl;l~ ht

" .. ,;.1 ~url'

SlI,,,'.d t t'f J)', I

Chop.)

ve characteristics of 3.0 shows the inter.:md how they relate

:Jublished his conclu=d temperature. He :n temperature if the calton, and Amagal, .:!iations among gas .3 and pressures such _c enough to neglect .:.ne molecules them-"!aJ gas. More propcertain simple laws No tp~! gas obeys ""sur ~ugh the cue. .aces obey '.·rer" gases such as StITes and low tem:::cs. under conditions , Jaws. However, at cor approaches that J i gas laws, if propC'w percent or less. ! '\'ely dose together,

. may be compared,
':-e are !:II'

equivalents)

Sec. 3.1

Ideal Gas Laws

The fact that a substance cannot exist as a gas at 32°F and 29.92 in. Hg is immaterial. Thus, as we shall see later, water vapor at 32°F cannot exist at a pressure greater than its saturation pressure of 0.18 in. Hg without condensation occurring. However, the imaginary volume at standard conditions can be calculated and is just as useful a quantity in the calculation of volume-mole relationships as thoubh it could exist. In the following, the symbol V will stand for total volume, and the symbol f- for volume per mole, or per unit mass.

EXAMPLE 3.1

Use of standard conditions

Calculate the volume, in cubic feet, occupied by 88lb of CO 2 at standard conditions . Solution:

Basis: 88 lb CO 2 88lb CO 2 I Ib mole CO 2 359 ft' CO 2 441b CO 2 lib mole CO 2 = 718

ftl

CO 2 at S.c.

Notice how in this problem the information that 359 ftl at S.c. = I Ib mole is applied to transform a known number of moles into an equivalent number of cubic feet. Incidentally, whenever you use cubic feet, you must establish the conditions of temperature and pressure at which the cubic feet are measured, since the term Uftl," standing alone, is really not any particular quantity of material.

3.1.1 Perfect Gas law. Boyle found that the volume of a gas is inversely proportional to the absolute pressure at constant temperature. Charles showed that. at constant pressure, the volume of a given mass of gas varies directly with the absolute temperature. From the work of Boyle and Charles scientists developed the relationship now called the perfect gas law (or sometimes the ideal gas law): pV = nRT (3.1) In applying this equation to a process going from an initial set of conditions to a final set of conditions, you can set up ratios of similar terms which are dimensionless as follows:

__ (~)(TI) (PI)(VI) V P2

..:me, is the one you ,,;1 common in use. 0: for any ideal gas:

149

2

-

112

T2

(3.2)

Here the subscripts I and 2 refer to the initial and final conditions. This arrangement of the perfect gas law has the convenient feature that the pressures may be expressed in any system of units you choose, such as in. Hg, mm Hg, N/m', atm, etc., so long as the same units are used for both conditions of pressure (do not forget that the pressure must be absolute pressure in both cases). Similarly, the grouping together of the absolute temperature and the volume terms gives ratios that are dimensionless.

---~.~---~,.,-.,------.- .....

--

,

• 1 1

1

__ .'_._.. _'_---_ ..•.

·"""",.,

,_...

.

150 Gases, Vapors, Liquids, and Solids

_--_ Chap. 3

... Sec. 3.1

Let us see how we can use the perfect gas law both in the form of Eq. (3.2) and Eq. (3.1).

EXAMPLE 3.2 Perfect gas law

EXAM}

An oxygen cylinder used as a standby source of oxygen contains 1.000 ft3 of 0, at 70'F and 200 psig. What will be the volume of this 0, in a dry-gas holder at 90°F and 4.00 in. H,O above atmospheric? The barometer reads 29.92 in. Hg. See Fig. E3.2.

Pro the \Val<.a vapor, higher" 23°C, A! Soh Fir, it into til by use" you can,

V,? 90°F 4 in. H20 gouge

'--

70°F 215 psio Fig. El.2.

So/ulion.· You must first convert the temperatures and pressures into absolute units:

+ 70 = 460 + 90 = 460

atmospheric pressure . 'fal

1m I

550 0 R

1.00

=

200 psig

+ 14.7 PSia! 29.92 in. Hg 14.7 psia

29.92 in. Hg

+

= 29.92 + 0.29 =

,"l

--.~'.

= 29.92 in. Hg = std atm = 14.7 psia

pressure -

final pressure

530'R

= 437 in. Hg

Notice 1\(

4 in. H,O 29.92 in. Hg 12 m. H 2 0 33.91 ft H,O ft H,O

EXA~1f'1

30.21 in. Hg

the iJ""I! if the no

The simplest way to proceed, now that the data are in good order, is to apply the laws of Charles and Boyle, and, in effect, to apply the perfect gas law. From Charles' law, since the temperature increases, the volume increases; hence the ratio of the temperatures must be greater than 1. The pressure decreases; therefore from Boyle's law, the volume will increase; hence the ratio of pressures will be greater than 1. Basis: 1 ft 3 of oxygen at 70'F and 200 psig 437 in. Hg 30.21 in. Hg = 15.0 ft at 90°F and 4 in. H,O gauge

y~qJ

":'7' .: Wett. t nt? t, i'W

Chap. 3

· form of Eq. (3.2)

Ideal Gas Laws 151

Sec. 3.1 Formally, the same calculation can be made using Eq. (3.2), since n,

EXAMPLE 3.3 · ens 1.000 ft' of O 2 ·!!as holder at 90'F .2 in. Hg. See Fig.

= n2

Perfect gas law

Probably the most important constituent of the atmosphere that fluctuates is the water. Rainfall, evaporation, fog, and even lightning are associated with water as a vapor or a liquid in air. To obtain some feeling for how little water vapor there is at higher altitudes, calculate the mass of 1.00 m' of water vapor at 15.5 mm Hg and 23°C. Assume that water vapor is an ideal gas under these conditions. Sollllion: First visualize the information available to you, and then decide how to convert it into the desired mass. If you can convert the original amount of water vapor to S.c. by use of the ideal gas law and then make use of the fact that 22.4 m' = I kg mole, you can easily get the desired mass of water vapor. See Fig. E3.3.

1.00 m3 15.5 mm

? m3 760 mm

23°C

O°C

::Jsolute units:

Fig. E3.3.

Basis: 1.00 m' H 2 0 vapor at 15.5 mm Hg and 23°C 1.00 m' =

437 in. Hg

Notice how the entire calculation can be carried out in a single dimensional equation.

Hg -~20

.'!:T,

is to apply the

EXAMPLE 3.4

Perfect gas law

You have 10 Ib of CO, in a 20·ft' fire extinguisher tank at 30'C. Assuming that the ideal gas laws hold, what will the pressure gauge on the tank read in a test to see if the extinguisher is full? See Fig. E3.4 .

_ vv.

·'" increases; hence ·=eases; therefore :~s will be grealer 20 1f3

10 Ib CO 2

10 Ib CO 2

147 psio

O°C

30°C

Fig. E3.4.

_ _ _""-

it

Chap. 3

152 Gases, Vapors, Liquids, and Solids Solutioll: Employing Eq. (3.2), we can write (the subscript I stands for standard 2 for the conditions in the tank)

(I

~onditions,

PI ~

14.7 psia

I Ib mole CO 2

303'K

~c....!:=-r=-':'::"":"=-"-I-'-':';44iT.:lb:':C"'O~2'1.--Ti:"::''':.-::-hCA7':7t7;-27;';3'',ii K

= P2 =

. 66 pSla

VI

Hence the gauge on the tank will read (assuming that it reads gauge pressure and that the barometer reads 14.7 psia) 66 - 14.7 = 51.3 psig.

We have not used the gas constant R in the solution of any of the example problems so far, but you can use Eq. (3.1) to solve for one unknown as long as all the other variables in the equation are known. However, such a calculation requires that the units of R be expressed in units corresponding to those used for the quantities p- V-To There are so many possible units you can use for each variable that a very large table of R values will be required. But, if you want to use R, you can always determine R from the p- V-T data at standard conditions which you have already memorized and used; R merely represents a pVjT relation at some fixed condition of temperature and pressure.

EXAMPLE 3.5 Calculation of R Find the value for the universal gas constant R for the following combinations of units: (a) For I Ib mole of ideal gas when the pressure is expressed in psia, the volume is in ft'/lb mole, and the temperature is in OR. (b) For 1 g mole of ideal gas when the pressure is in atm, the volume in em', and the temperature in "K. (c) For I kg mole of ideal gas when the pressure is in N/m 2 , the volume is in m'lkg mole, and the temperature is in OK.

Solution: (a) At standard conditions p

= 14.7 psia

V = 359 ft'/lb mole T

= 492'R

Then R

=

pV = T

14.7 psia 359 ftJ 492"R I Ib mole

10 73 (psia)(ft') . (R)(lb mole)

it

l'

r. I, I", ,

Chap. 3

Ideal Gas Laws

Sec. 3.1

153

(b) Similarly, at standard conditions,

ior standard .;.-onditions,

= I atm

p

V=

22,400 em' jg mole

T= 273°K ~'K . "K = pz = 66 psla

R -

pV _

I atm 22,400 em' _ 8206 (em 3 )(atm) T - 273"K I g mole . CK)(g mole)

(e) In the SI system of units standard conditions are

= 1.013

p

V= ;!auge pressure and that

_'f any of the example unknown as long as :r . such a calculation

c:

-ing to those used for au C' •• "se for each -- B: )U want to star" _ conditions -, represents a pV/T

pV _ T -

22.4 m 3 jkg mole

= 273°K

T R =

x 10' Njm'

1.013 x 10' N/m' 22.4 m 3 _ 8 30 273°K I kg mole - .

X

10'

(N)(m) ("K)(kg mole)

We want to emphasize that R does not have a universal value even though it is called the unirersal gas constant. The value of R depends on the units of p, V, and T. Similarly, you should realize that R is not a dimensionless quantity; i.e., there is no value of R = 21.9, but there is a value of

R

=

21.9 (in. Hg)(ft3) (Ib mole)CR)

EXAMPLE 3.6 Perfect gas law Calculate the volume occupied by 88 Ib of CO, at a pressure of 32.2 ft of water and at 15°C. ;lowing combinations

.u in psia,

Sollltion: See Fig. E3.6 .

the volume

tt 3 at 32.2 It H20 and tSOC

the volume in em', Fig. E3.6.

n', the volume is in Solution No. I (using Boyle's and Charles' laws): At S.C.: p = 33.91 ft H,O ~

V

rt'

= 359 Ib mole

T= 273°K Basis: 88lb or CO, ::c88::...:..:lb,-C::c0=,--,--'t--......--..:-rv;:--+..::3::...59::...:..:ft_'+=~33:.:.~91 = 798 ft' CO, 44mole Ib CO, 322 I Ib CO, I Ib mole . at 32.2 ft H,O and 15°C

"'6j--

*

,....4

rr

e

'."

154

Chap. 3

Gases, Vapors, Liquids, and Solids

Solurion No.2 (using gas constant R): First, the value of R must be obtained in the same units as the variable p, T. For 1 lb mole,

R

t't

Sec. 3.1

V, and

=pV T

and at S.c., p = 33.91 ft H 2 0

V=

359 fP/lb mole

T = 273'K

R

= 33.91 359 _ 44 6 (ft H,O)(ft')

273 -

. (Ib moleJCK)

SOllilior; The gas ideal gas la\\ the chemical in the cell.

Now, using Eq. (3. I), inserting the given values, and performing the necessary calculations, we get Basis: 88 lb CO,

V = nRT = _87i8;-;I-i--brC,.;,O~,,--+-,,-(44-,-,-.6_f-,--t;:.cHeo,O-,--,-,)(_ft--,-')+-_2_8-,,8_'K--,-p 441b CO, 32.2 ft H,O (Ib moleWK) lb mole CO, = 798 ft' CO 2 at 32.2 ft H 2 0 and 15°C

If you will inspect both solutions closely, you will observe that in both cases the same numbers appear and that both are identical except that in the second solution using R two steps are required to obtain the solution.

If the m we know the grams of NC The sum x gN(

EXAMPLE 3.7 Perfect gas law One important source of emissions from gasoline-powered automobile engines that causes smog is the nitrogen oxides NO and NO,. They are formed whether combustion is complete or not as follows: At the high temperatures which occur in an internal combustion engine during the burning process. oxygen and nitrogen combine to form nitric oxide (NO). The higher the peak temperatures and the more oxygen available, the more NO is formed. There is insufficient time for the NO to decompose back to O 2 and N, because the burned gases cool too rapidly during the expansion and exhaust cycles in the engine. Although both NO and nitrogen dioxide (NO,) are significant air pollutants (together termed NO.,), the NO, is formed in the atmosphere as NO is oxidized. Suppose that you collect a sample of a NO-NO, mixture (after having removed the other combustion gas products by various separations procedures) in a 100-cm' standard cell at 30'C, Certainly some of the NO will have been oxidized to NO, 2NO

+ 0, --+ 2NO,

during the collection. storage, and processing of the combustion gases so that measurement of NO alone will be misleading. If the standard cell contains 0.291 g of NO, plus NO and the pressure measured in the cell is 1265 mm Hg, what percent of the NO + NO, is in the form of NO' See Fig. E3.7.

--- -

.....

..... ...'''''.,~.~.~

.

The weight pc

and the mole

3.1.2 G defined as th foot, grams p volume varic' tioned, you, otherwise sr.' calculated ll\ the containc,;

Chap. 3

.s the variable p,

Sec. 3.1

Ideal Gas Laws

155

V, and V=100 cm 3 p =1265 mm Hg

Fig. EJ.7. Solution: The gas in the cell is composed partly of NO and partly of NO,. We can use the ideal gas law to calculate the total gram moles present in the cell and then, by using the chemical equation and the principles of stoichiometry, compute the composition in the cell. Basis: 100 em' gas at 1265 mm Hg and 30 e D

cuing the necessary cal-

R

= 760 mm Hg 273°K

n

=

_p_V

RT

that III Doth cases the the second solution

1 g mole

1000 em' 1I

= 624

X

.

10 4 (mm Hg)(cm') CK)(g mole)

1

.~

,

"

= __...:1:.:2:,:6.=.5i.m:::m c::-:H",g":-v=,.,-t..::1.=.0.=.0.=.c:.:m:...' = 0.00670 g mole 624 x 104 (mmHg)(cmJ) 30J"K .

(,K)(g mole)

If the mixture is composed of NO (MW = 30) and NO, (MW = 46). because we know the total mass in the cell we can compute the fraction of, say, NO. Let x = grams of NO; then (0.291 - x) = g NO,. Basis: 0.291 g total gas

.11

The sum of the moles is x g NO

automobile engines !Drmed whether comes which occur in an c.nd nitrogen combine "nd the more oxygen ne NO to decompose :.ng the expansion and 'xide (NO,) are signi· ill the atmosphere as

11 g mole NO 30 g NO

0.0333x

j

"fter having removed '"dures)' in a 100·em' oxidized to NO,

...1ses so that measure"ins 0.291 g of NO, what percent of the

, (0.291 - x) g N0'II g mole NO, ., 46 g NO,

+ (0.291

- x)(0.0217)

= 000670 .

= 0.00670

x = 0.033 g

The weight percent NO is

~:~~~ (100) ~ II % and the mole percent NO is

=

(0.0333)(.0. 033 ) (100) 17 % (0.00670) 0

3.1.2 Gas Density and Specific Gravity. The density of a gas is defined as the mass per unit volume and can be expressed in pounds per cubic foot, grams per liter, or other units. Inasmuch as the mass contained in a unit volume varies with the temperature and pressure. as we have previously mentioned, you should always be careful to specify these two conditions. Unless otherwise specified, the volumes are presumed to be at S.c. Density can be calculated by selecting a unit volume as the basis and calculating the mass of the contained gas.

I

1

156 Gases, Vapors, Liquids, and Solids

Chap. 3

,\,- .

EXAMPLE 3.8 Gas density What is the density of N 2 at 80'F and 745 mm Hg expressed in (a) American engineering units? (b) cgs u;lits? (c) Sf units? Solution: (a)

1ft' density = 0.0696 Ib/ft' of N 2 at 80°F and 745 mm Hg (b)

(c)

1m' density = 1.115 kg/m' of N2 at 80°F (300'K) and 745 mm Hg (9.94

X

10' N/m 2)

The specific gral'ity of a gas is usually defined as the ratio of the density of the gas at a desired temperature and pressure to that of air (or any specified reference gas) at a certain temperature and pressure. The use of specific gravity occasionally may be confusing because of the manner in which the values of specific gravity are reported in the literature. You must be very careful in using literature values of specific gravity. Be particularly careJul to ascertain that the conditions of temperature and pressure are known both for the gas in question and for the reference gas. Among the examples below. several represent inadequate methods of expressing specific gravity. (a) What is the specific gral'ity of methane? Actually this question may have the same answer as the question, How many grapes are in a bunch '! Unfortunately, occasionally one may see this question and the best possible answer is sp gr =

density of methane at S.C. density of air at S.C.

(b) What is the specific gravity of methane (H, .e, Loo)? Again a poor question. The notation of (H, 1.00) means that H, at S.c. is used as

.=

I '\,

Chap. 3

Idea/ Gas Laws

Sec. 3.1

157

the reference gas, but the question does not even give a hint regarding the conditions of temperature and pressure of the methane. Therefore the best interpretation is

cd in

_ spgr - de~n~s~it~Y70~f~mieTth~a~n~e~at~S~.C~. density of H, at S.c.

(c) What is the specific gravity of ethane (air = 1.00)? Same as question (b) except that in the petroleum industry the following is used: density of ethane at 60°F and 760 mm Hg sp gr

= density of air at S.c. (60"F, 760 mm Hg)

(d) What is the specific gravity of butane at 80°F and 740 mm Hg? No reference gas nor state of reference gas is mentioned. However, when no reference gas is mentioned, it is taken for granted that air is the reference gas. In the case at hand the best thing to do is to assume that the reference gas and the desired gas are under the same conditions of temperature and pressure:

J.0696Jb cmHg

,.115 g

sp gr

:nrn F

=

density of butane at 80°F and 740 mm Hg density of air at 80°F and 740 mm Hg

(e) What is the specific gravity of CO, at 60°F and 740 mm Hg (air = I.OO)? sp gr - density of CO,• at 60°F and 740 mm Hg density of air at S.c.

1.llSkg -ig (9.94 X ]04 N/m')

(f) What is the specific gravity of CO, at 60°F and 740 mm Hg (ref. air at S.C.)? sp gr

:-atio of the density 3r (or any specified .:e of specific gravity ·.....hich the values of ':ery careful in using : a ascertain that the the gas in question -:ral represent inade, this question may '~DeS are in a bunch? :::stion and the best

= same as question (e)

EXAMPLE 3.9 Specific gravity of a gas What is the specific gravity of N, at 80°F and 745 mm Hg compared to (a) Air at S.c. (32'F and 760 mm Hg)? (b) Air at 80'F and 745 mm Hg? SO/lilian:

First you must obtain the density of the N, and the air at their respective con, ditions of temperature and pressure. and then calculate the specific gravity by taking a ratio of their densities, Example 3.8 covers the calculation of the density of a gas, and therefore, to save space, no units will appear in the following calculations: (a)

=

0.06961b N,/ft' at 80°F, 745 mm Hg

Basis: I ft 3 of air at 32°F and 760 mm Hg .o(»)? A
:s used as

i .1

J

Basis: J ft3 N' at 80°F and 745 mm Hg

-=-t-i-Ti~~-h-""'t-'2::..::8

,", a'

1

I

1:~~ 1~:~ 13591

29

-,·0,0808 lb air/fl 3 at 32°F, 760 mm Hg

,

t

I

158 Gases, Vapors, Liquids, and Solids

Chap. 3

Sec

1<,

Therefore s

r pg

=

~lJ

0.0696 = 0 863 1b N,/ft' N, at 80°F, 745 mm Hg 0.0808' Ib air/ft J air at S.c.

rl', '.:

(, Hal·:'

tan~,

Note: specific gravity is not a dimensionless number.

(Ofln,.

(b) Basis: I ft' air at 80°F and 745 mm Hg

,0

I 149217451 540 760 359 129 -_ 0.0721 Ib/ft at 80 F and 745 mm Hg

(sp gr)N. = =

0.0696 0.721

=

A (\"" 400

0 965 1b N,/ft' N, at 80°F, 745 mm Hg . lb air/ft' air at 80'F, 745 mm Hg

0.965 Ib N,/lb air

I

spgr

=d

d. ir

=

I

492 540 492 540

745 28 760 359 745 29 = 760 359

;~ =

I' ".

COth-; ~'''';

Note: You can work part (b) by dividing the unit equations instead of dividing the resulting densities: N,

I.'

you! " 150·11' I

part: .. : excr I ,!:l put to",' ) I·:

the I" ." di\ il.!~> \.t

0.965 I b N,/Ib air

nent

:,j

The latter calculation shows that the specific gravity is equal to the ratio o/the molecular weights of the gases when the densities of both the desired gas and the reference gas are at the same temperature and pressure. This, of course, is true only for ideal gases and should be no surprise to you, since Avogadro's law in effect states that at the same temperature and pressure I mole of any ideal gas is contained in identical volumes.

3.1.3 Ideal Gaseous Mixtures. In the majority of cases, as an engineer you will deal with mixtures of gases instead of individual gases. There are three ideal gas laws which can be applied successfully to gaseous mixtures:

p,

Equation (3.3) is Dalton's law of the summation of the partial pressures.

E(jIl.I·



p"nc::

I,

n10k, \ ~ mi\l,

the

(a) Dalton's lall"s. Dalton postulated that the total pressure of a gas is equal to the sum of the pressures exerted by the individual molecules of each component gas. He went one step further to state that each individual gas of a gaseous mixture can hypothetically be considered to exert a partial pressure. A partial pressure is the pressure that would be obtained if this sallie mass of individual gas were alone in the same total I'olullle at the same ten:perature. The sum of these partial pressures for each component in the gaseous mixture would be equal to the total pressure, or

+ p, + p, + ... + P. =

cal" " rdr:lI,. , yoy \" ,I·

ynur:\1

(a) Dalton's law of partial pressures. (b) Amagat"s law of partial volumes. (c) Dalton's law of the summation of partial pressures.

P,

wl1,,-11 ,?

(3.3)

r" "

\\hoCI\

1.1 II

'

nll\!

If I' ( . .. ,f " the' ,

To illustrate the significance of Eq. (3.3) and the meaning of partial pressure, suppose that you carried out the following experiment with ideal gases. Two tanks of 150 ft' volume, one containing gas A at 300 mOl Hg and the other gas B at 400 mOl of Hg (both gases being at the same temperature of 70'F), are connected together. All the gas in B is forced into tank A isothermally. Now you have a 150-ft' tank of A + B at 7000101 of Hg. For this mixture (in the 150-ft' tank at 70'F and a total pressure of 700 0101 Hg) you could say that gas A exerts a partial pressure of 300 mOl and gas B exerts a partial pressure of 400 mOl. Of course you cannot put a pressure gauge on the tank and check this conclusion because the pressure gauge will read only the total pressure. These partial pressures are hypothetical pressures that the individual gases would exert and are equivalent to the pressures they actually would have if they were put into the same volume at the same temperature all by themselves. You can surmise that, at constant volume and at constant temperature, the pressure is a function only of the number of molecules of gas present. If you divide the perfect gas law for component I, p,v I = JI,RT" by that for component 2, p,V, = JI,RT" at constant temperature and I'o/wne, you can obtain

'mmHg

..,5 mm Hg

.·mmHg .'mm Hg ,5

Ideal Gas Laws 159

Sec. 3.1

Chap. 3

instead of dividing

Nz/lb air

!!J=!!.J.

_, to the ratio of the -'0 gas and the refer-

ese,

p,

only for ates that ntain~ .... identical j.

which shows that the ratio of the partial pressures is exactly the same numerically as the ratio of the moles of components I and 2_ Similarly, dividing the ideal gas law for component I by the gas law for all the molecules, p,v, = n,RT" you will get Dalton's law of partial pressures:

,

m.

!!J = !!J = mole fraction Pt

~S,

as an engineer _. There are three -1IXtures:

=

(3.5)

YI

(3.5a)

p, =y,P,

where i stands for any component. (b) Amagat's lal\'. Amagat's law of additive volumes is al1alogous to Dalton's law of additive pressures. Amagat stated that the total volume of a gaseous mixture is equal to the sum of the volumes of the individual gas components if they were to be measured at the same tern perature and at the total pressure of all the molecules. The individual volumes of these individual components at the same temperature and pressure arc called the parria/ l'O[Wlles (or sometimes pure compol1ellf volumes) of the individual components:

(3.3)

VI

'ressures .

-..."...~"',':

n,

Equation (3.5) shows that the ratio of the partial pressure of an individual component to the total pressure is exactly the same numerically as the ratio of the moles of the individual component to the total moles. With this principle under your belt, if the mole fraction of an individual gaseous component in a gaseous mixture is known and the total pressure is known, then you are able to calculate the partial pressure of this component gas by generalizing Eq. (3.5):

Qf a gas is equal of each compo"as of a gaseous S5ure. A partial 55 of individual reo The sum of ;ture would be

...-.-..',....'-""-.. .

(3.4)

n,

..."'_t...,.,..,_ _.....- _ • .,.,.......

-..'.~·""'_.t

+ V, + V, + ... + V. =

V,

............-.-..,. ...._ _......,...._ _...

"",~

_~

(3.6)

_ _ _ _....,__.,..•.,.,_ _ _ __

1

I ,,

,

I

1

I, \

I 'j

db

160 Gases, Vapors, Liquids, and Solids

. ;,.,:

Chap. 3

Reasoning in the same fash;on as in our explanation of partial pressures,

at the same temperature and pressure, the partial volume is a function only of number of molecules of the individual component gas present in the gaseous mixture, or (3.7)

"

,

and

!::.t = !!.! = Y = mole fraction V, n, I

(3.8)

which shows that the ratio of the partial volumes is exactly the same, numerically, as the ratio of the moles of components I and 2, or the ratio of the moles of component I to the total moles. Equation (3.8) states the principle, presented without proof in Chap. I, that volume fraction

= mole fraction = y,

(3.9)

for an ideal gas.

EXAMPLE 3.10 Partial pressures and \'olumes A gas-tight room has a volume of 10,000 ftl. This room contains air (considered to be 21 percent O 2 and 79 percent N,) at 70'F and a total pressure of 760 mm Hg. (a) What is the partial volume of O 2 in the room?

(b) (c) (d) (e)

What is the partial volume of N2 in the room? What is the partial pressure of O 2 in the room? What is the partial pressure of N, in the room? If all of the 0, were removed from the room by some method, what would be the subsequent total pressure in the room?

F'\.'( I

the oll,'m

Solution; See Fig. E3.10(a)

,

,1

l

1(1

L

? tt 3 02 tO,OOO tt 3 760mm 70°F Air

760 mm ond 70°F ? tt 3 N2 760 mm ond 70°F

<,

"

or I. a~J 2, 10,

Fig. E3.IO(a).

l

j ~)~

Basis: 10,000 ft ' of air at 70'F and 760 mm Hg Partial volumes can be calculated by multiplying the total volume by the respective component mole fractions [Eq. (3.8)]: (a) Vo• = (0.21)(1O,000)·~ 2.100 ft' 0, at 70'F, 760 mm (b) VN , = (0.79)(10,000) = 7,900 ft' N2 at 70'F, 760 mm total volume ~~ 10,000 ft' air at 70"F, 760 mm

EX""II'! I

It is at ':' ,\ ponen!.

Chap. 3 =rtial pressures,

:.:nction only of m the gaseous

Ideal Gas Laws 161

Sec. 3.1

Note how the temperature and pressure have to be specified for the partial volumes to make them meaningful. Partial pressures can be calculated by multiplying the total pressure by the respective component mole fractions [Eq. (3.5a»); the basis is still the same: (c) Po, (d) p",

(3.7)

= (0.21)(760) = 160 mm Hg when V = 10,000 ftl at 70°F = (0.79)(760) = 600 mm Hg when V = 10,000 ftl at 70°F total pressure = 760 mm Hg when V = 10,000 ft' at 70°F

(e) If a tight room held dry air at 760 mm Hg and all the oxygen were removed from the air by a chemical reaction, the pressure reading would fall to 600 mm Hg. This is the partial pressure of the nitrogen and inert gases in the air. Alternatively, if it were possible to remove the nitrogen and inert gases and leave only the oxygen, the pressure reading would fall to 160 mm Hg. In either case, you would have left in the room 10,000 ft 3 of gas at 70'F. See Fig. E3.IO(b).

(3.8)

esame, numeri.:io of the moles :-:=,iple, presented (3.9)

10,000 tt 3 p'?

10,000 tt 3 760 "'m 70'F

~s

a. ;idered -" of ,,~'" = Hg.

Air

70'F

Oz 10,000 tt 3 P'

?

70'F

Nz

Fig. E3.IO(b). :nod,. what would

For use in our subsequent calculations you should clearly understand now that the original room contained: 1. 7,900 ftl dry N, at 760 mm Hg and 70'F 2. 2,100 ft3 dry 0, at 760 mm Hg and 70°F 3. 10,000 ft' dry air at 760 mm Hg and 70'F (add I and 2) or 1. 10,000 ft3 dry 0, at 160 mm Hg and 70°F 2. 10,000 ft 3 dry N, at 600 mm Hg and 70"F 3. 10,000 ftl dry air at 760 mm Hg and 70'F (add 1 and 2) ~-:

by the respective

EXAMPLE 3.11

Ideal gas mixtures

A flue gas analyzes CO:, 14.0 percent; 0" 6.0 percent; and N" 80.0 percent. c

It is at 400 F and 765.0 mm Hg pressure. Calculate the partial pressure of each com-

ponent.

162

Gases, Vapors, Liquids, and Solids

Chap. 3

.\

.

Solution:

Basis: 1.00 kg mole fiue gas compo CO. O. N.

kg moles 0.140 0.060 0.800 1.000

p (mm Hg) 107.1 45.9 612.0 765.0

On the basis of 1.00 kg mole offiue gas, the mole fraction y of each component, when multiplied. by the total pressure, gives the partial pressure of that component. Thus, VCt' of flue gas represents four different things: V ft' V ft' Vft' Vft'

fiue gas carbon dioxide oxygen nitrogen

at at at at

765.0 mm Hg and 107.1 mm Hg and 45.9 mm Hg and 612.0 mm Hg and

400°F 400"F 400"F 400"F

3.2 Real gas relationships We have said that at room temperature and pressure many gases can be assumed to act as ideal gases. However, for some gases under normal conditions. and for most gases under conditions of high pressure, values of the gas properties that you might obtain using the ideal gas law would be at wide variance with the experimental evidence. You might wonder exactly how the behavior of real gases does compare with that calculated from the ideal gas laws. I n Fig. 3. I you can see how the pP product o~ several gases deviates from that predicted by the ideal gas laws as the pressure increases substantially. Thus it is clear that we need some way of computing the p- V-T properties of a gas that is not ideal. Essentially there are four methods of handling real gas calculations: (a) (b) (c) (d)

Equations of state. Compressibility charts. Estimated properties.' Actual experimental data.

('.

t"

,.

. t·

'

"

Even if experimental data is available, the other three techniques still may be quite useful for certain types of calculations, Of course, under conditions such that part of the gas liquefies, the gas laws apply only to the gas phase portion of the system-yoll cannot extend these real gas laws into the liquid region any more than you can apply the ideal gas laws to a liquid, "

'Computer programs are available to estimate physical properties of compounds and mixtures based on structural group contributions and other basic parameters; refer to E. L. Meadows. Chem, Eng, Progr .• vol. 61. p. 93 (1965).

Chap. 3

Real Gas Relationships

Sec. 3.2 1.50.

t.4D 1.30.

t.tD 1.00.

=mponent, when :..-omponent. Thus,

0..90.

"

"~~~"~ o.SD

~

g

----

l,..---" V ----::::::: ::::- ~ r--

.......

~

.Jl Y 0..70. ~ -"" 0.,60.

<::".

/

0.50. /.

0..40.

-=al conditions, :1e gas properties .:.>e variance with cnehavior of real .:lWS. In Fig. 3.1 :::"Ult predicted by -" it is clear that :::""1at is not ideal. . .::ulations:

, :Ies still may be =nditions such '::Jnase portion of .:.:uid region any

,.; compounds and refer to E. L.

:~;

V

r

c-

/

V

/

V

,/

/

0..30. :laSe.

/

/

"0

I

CD2~ D'C

./

L~

/' ./

CH 4, D'C

H2 •

.,P"

,/'

"

,/

v/ .,,-/ CH4.~DD'C

V

/

1.20.

H ~DD'C ....... 2·

V

/

163

0.20.

'-

0..10

o

0.

/'

/

tDD

20.0.

40.0. 30.0. Pressure (alml

60.0.

70.0.

Fig. 3.1. Deviation of real gas from ideal gas laws at high pressures.

3.2.1 Equations of State. Equations of state relate the p- V-T prop· erties of a pure substance (or mixtures) by semitheoretical or empirical rela· tions. By properly we shall mean any measurable characteristic of a substance, such as pressure, volume, or temperature, or a characteristic that can be cal· culated or deduced, such as internal energy. to be discussed in Chap 4. Experi· mental data for a gas, carbon dioxide, arc illustrated in Fig. 3.2, which is a plot of pressure, versus molal volume with absolute temperature (converted to 0c) as the third parameter. For a constant temperature the perfect gas law, pC- = RT(for I mole), is, on this plot. a rectangular hyperbola because = constant. The experimental data at high temperatures and low pressures is not shown but closely approximate the perfect gas law. However, as we approach the point C, called the critical point (discussed in Sec. 3.2.2), we see that the traces of the experimental data are quite different from the lines whi.:h would represent the relation pf' = constant. The problem, then. is to predict these experimental points by means of some equation which can, with reasonable accuracy, represent the experimental data throughout the gas phase. Over 100 equations of state have been proposed

pro

Chap. 3

164 Gases, Vapors, Liquids, and Solids

••

...... experimental data - - Von der Wools' equation - - - 1wo phose envelope

·

tOO

150

200

V. cml/g mole

250

350

Fig. 3.2. p·V·Tproperties of CO,.

to accomplish this task, of which a few are shown in Table 3.1. We would not, of course, expect the function to follow the experimental data within the two-phase region outlined by the area inside the envelope A, B, C, D, and E.

(a) I'an der Wool's equation. van der Waals' equation, first proposed in 1893, does not yield satisfactory predictions at low gas temperatures (see Fig. 3.2), but it does have historical interest and is one of the simplest equations of state. Thus it illustrates some of the typical theoretical development that has taken place as well as some of the computational problems involved in using equations of ~tate. Keep in mind that some of the equations listed in Table 3.1 are completely empirical while others, such as van der Waals' equation. are semiempirical; that is, although they were developed from theory, the constants in the equation or portions of the equation are determined by empirical methods. van der Waals tried to include in the ideal gas law the effect of the attractive forces among the molecules by adding to the pressure term, in the ideal gas law, the term

,

(3.10) where 1/

=

V

= volume

number of moles

a = a constant, different for each gas

--~

Chap. 3

Real Gas Relationships

Sec. 3.2 TABLE

3.t

EQUATIONS OF STATE

(for I mole)

Holborll:

van der Waals: (p

165

+ ;2)(V -

pV =

b) = RT

RT[t

+ B'p + C'p2 + ... J

Beattie·Bridgeman: Lorentz:

p

RT ' = -;c-(V + b) V2

pV =

a - --,,V2

p

RT e-O'.'v RT = _'--

7 = -RTBob

(V - b)

(V - b)

3.1. We would data within the B, C, D, and E.

p

.... ield satisfactory have historical :ustrates some of .:i as some of the ~. Keep in mind . empirical while :hat is, although ::n or portions of . of the attractive :.De ideal gas law,

_a_ TI/2V(V

+ b)

](V-b)=RT

=

b

= 0.0867 RT,

(1

Kammerlingh-Onnes:

w

RT +

4V + V2': + ~ +~ V' v'

RTBo - Ao -

i~

= bRT- a + T2 ~exp(-+-) V2

q =

p,

RT[1

V=

P=

R 2 T 2 ., 0.4278--'p,

a

pV =

Rf!~c

Benedict-Webb-Rubin:

TV2

Redlich-Kwong:

[p+

-

T2

= -.!!L _ _a_

350

+ aAo

lJ = RBobc

Berthelot:

p

4V + + + VJ ? V2

Re RTBo - Ao - T2

P=

Dieterici:

RT +

qexp(- J2)

= aa.

+ ~V + V2 ~ + ... J

He also tried to take into account the effect of the volume occupied by the molecules themselves by subtracting a term from the volume in the ideal gas law. These corrections led to the following equation:

1l2a) (V ( p + VI

lib) = IIRT

(3.11 )

where a and b are constants determined by fitting van der Waals' equation to experimental p-V-T data, particularly the values at the critical point. Values of the van der Waals constants for a few gases are given in Table 3.2. Let tiS look at some of the computational problems that may arise when using equations of state to compute p-V-T properties by using van der Waals' equation as an illustration. van der Waals' equation can easily be explicitly solved for p as follows:

(3.10)

p

=

IIRT 11 2a (V -- lib) - -V'

(3.12)

However, if you want to solve for V (or II), you can see that the equation becomes cubic in V (or II), or VJ _ (nb

+ nRT)V2 + n p

2

a V _ nJab = 0

p

p

(3.\3)

166

Gases, Vapors, Liquids, and Solids

Chap. 3

TABLE 3.2 VAN DER WAALS' CONSTANTS FOR GASES (From the data of the International Critical Tables)

Air Ammonia Carbon dioxide Ethylene Hydrogen Methane Nitrogen Oxygen Water vapor

a*

bt

aIm ( em' )' gmole

( em' ) gmole

x x x x x x x x x

36.6 37.3 42.8 57.2 26.6 42.8 38.6 31.9 30.6

1.33 4.19 3.60 4.48 0.245 2.25 1.347 1.36 5.48

10" 10" 10" 10" 10" 10" 10" 10" 10"

\

. ( Ib ft')2 . 'To convert to pSla mole ,multIply table value by 3.776 x 10-'.

tTo convert to lb ~ole' multiply table value by 1.60

X

10- 2 •

Consequently to solve for V (or n) you would have to (a) Resort to a trial-and-error method, (b) Plot the equation assuming various values of V (or n), and see at what value of V (or n) the curve you plot intersects the abscissa, or (c) Use Newton's method or some other method for solving nonlinear equations (refer to Appendix L). A computer program can be of material help in technique (c). You can obtain a close approximation to V (or /1) in many cases from the ideal gas law, useful at least for the first trial, and then you can calculate a morc exact value of V (or n) by using van der Waals' equation. EXAMPLE 3.12 van der Waals' equation

,~

A S.O-ft' cylinder containing 50.01b of propane (C,H.) staMs in the hot sun. A pressure gauge shows that the pressure is 665 psig. What is the temperature of the propane in the cylinder? Use van der Waals' equation. See Fig. E3.l2.

Solulion: Basis: 50 lb of propane

The van der Waals' constants obtained from any suitable handbook are ' ( Ib ft' a -_ 3.27 x 10.pSla mole b

=

)2

ft' 1.35 lb mole

n2a) ( P+J7l (V-nb)=nRT

,.

..

'.

Chap. 3

Real Gas Relationships 167

Sec. 3.2 665 psig

bt em' ) ·.gmole

5 ft3 50lb

C,H a

36.6

T·?

37.3 42.8

57.2 26.6 42.8 38.6 31.9

Fig. EJ.12.

All the additional information you need is as follows:

30.6

p

= 665 psig + 14.7 = 679.7 psia

.. 1073 (psia)(ft') . R In proper umts IS = . (Ib mole)('R) n

[679.7 .:md see at what

50lb

= 44lbjlb mole = 1.l371b moles propane

+ (1.137)2?5~?

x J04)J[5 - (1.l37)(1.35)] T

= 1.l37(1O.73)T

= 673°R = 213°F

_5sa, or :ving nonlinear

(b) Other equations of state. :{ou can obtain gas law, useful ::act value of V

In

the hot sun. of the

:~perature

are

Among the numerous other equations of state which have been proposed in addition to van der Waals' equation are those shown in Table 3.1. The form of these equations is of interest inasmuch as they are attempts to fit the experimental data with as few constants in the equation as possible. One form of the equation known as the \'irial form is illustrated by the equations of Kammerlingh-Onnes and Holborn. These are essentially power series in 1/1" or in p, and the quantities B, C, D, etc., are known as I'irial coefficients. These equations reduce to pI' c:, RT at low pressures. The vi rial form of an equation of state is the best we have today; however, it is not possible to solve theoretically by statistical thermodynamics for the constants beyond the first two. Therefore, fundamentally, these equations are semiempirical, the constants being determined by fitting the equation to experimental data. The Beattie-Bridgeman equation, which has five constants (exclusive of R), and the Benedict-WebbRubin equation, which has eight constants, are among the best we have at the present time. The two-constant Redlich-Kwong equation also appears to be quite good according to a study of Thodos and Shah.' Naturally, the use of these equations involves time-consuming calculations, particularly when carried 'G. Thodos and K. K. Shah, Ind. Eng. ellem., v. 57, p. 30 (1965).

168 Gases, Vapors, Liquids, and Solids

Chap. 3

out by hand rather than on the computer, but the results will frequently be more accurate than those obtained by other methods to be described shortly. In spite of the complications involved in their use, equations of state are important for several reasons. They permit a concise summary of a large mass of experimental data and also permit accurate interpolation between experimental data points. They provide a continuous function to facilitate thermodynamic calculations involving differentiation and integration. Finally, they provide a point of departure for the treatment of thermodynamic properties of mixtures. However, for instructional purposes, and for many engineering calculations, the techniques of predicting p- V- T values discussed in the next section are considerably more convenient to use and are usually just as accurate as equations of state.

3.2.2 Compressibility Factors. In the attempt to devise some truly universal gas law for high pressures, the idea of corresponding states was developed. Early experimenters found that at the critical point all substances are in approximately the same state of molecular dispersion. Consequently, it was felt that their thermodynamic and physical properties should be similar. The law of corresponding states expresses the idea that in the critical state all substances should behave alike. Now that we have mentioned critical state several times, let us consider exactly what the term means. You can find many definitions, but the one most suitable for general use with pure component systems as well as with mixtures of gases is the following: The critical state for the gas-liquid transition is the set of physical conditions at which the density and other properties of the liquid and vapor become identical. Referring to Fig. 3.3, note how as the pressure increases the specific volumes of the liquid and gas approach each other and finally become the same at the point C. This point, for a pure component (only), is the highest temperature at which liquid and vapor can exist in equilibrium. From a slightly different view a critical point is a limiting point that marks the disappearance of a state. Consider Fig. 3.2. When gaseous carbon dioxide is compressed at 25"C its pressure rises until a certain density is reached (D). On further compression the pressure remains constant and the gas condenses. liquid and vapor being in coexistence. Only after all vapor is condensed to liquid (B) does the pressure rise again. If the temperature is raised a little. the vapor reaches a higher density before it starts condensing; on the other hand. the condensation is completed at lower liquid density. Thus the coexisting phases have a smaller difference in density at the higher temperature. Above 30''C this density range diminishes very rapidly with temperature and finally. just above 30'C, the two densities at which condensation is initiated and completed coincide. Vapor and liquid are no longer distinguishable; the separating meniscus disappears. Below the

;

,.

.

fWn ' ..

Chap. 3

:requently be more ~:i shortly. ~tions of state are ~-;y of a large mass ::l between experi:::::ilitate thermody.:m. Finally, they ..:unic properties of :v engineering calc:d in the next sec"y just as accurate

nevise some truly :ending states was :ernt all substances .:m. Consequently, should be similar. ~ cr ,tate all

Real Gas Relationships 169

Sec. 3.2

.0

:::::

,;

u

.; E

:>

~

Absolute Pressure, psio

";5, let us consider . but the one most .~ as with mixtures

'Jf physical condi::nd vapor become

Absolute Pressure, psio

Fig. 3.3. The critical point for a pure substance (water).

:::e specific volumes 2e the same at the ~st temperature at :1rly different view .::rance of a state. ::TJpressed at 2YC ::1her compression end vapor being in :5 the pressure rise a higher density ._:lOn is completed :maller difference range diminishes :he two densities yapor and liquid c::ne" low the

critical temperature, the system can exhibit two coexisting phases; above, it goes from the dilute to the dense state without a phase transition. The critical point marks the highest temperature on the coexistence curve. Experimental values of the critical temperature (T,) and the critical pressure (p,) for various compounds will be found in Appendix D. If you cannot find a desired critical value in this text or in a handbook, you can always consult ReId and Sherwood' or Hakata and Hirata,' which describe and evaluate methods of estimating critical constants for various compounds. The gas-liquid transition described above is only one of several possible transitions exhibiting a critical point. Critical phenomena are observed in liquids and solids as well. For example, at low temperatures water is only partially miscible in phenol and vice versa. Thus, if equal amounts of water and phenol 'R. C. Reid and T. K. Sherwood, The Properties 01 Gases and Liquids, 2nd ed., McGrawHill, New York, 1965. 'T, Hakata and M. Hirata, J. Chen!. Engr. 01 Japan, 3, p. 5 (1970).

---.--~--~ --~.--,-~,

170 Gases, Vapors, Liquids, and Solids

-.. ---,~.-,-

..............

--.--------~,-

Chap. 3

...

<~,-""- 'dt... ' ';.t..

Sec

1."'

are shaken together, they will separate in two phases, one consisting of water with some phenol and the other of phenol with some water dissolved in it. If the temperature is increased, the concentration of phenol in water increases and also that of water in phenol. Finally (if the initial composition is chosen correctly) a critical temperature of the binary liquid mixture is reached at which the two phases become equal in composition, so that the interface disappears and the phases can no longer be distinguished from one another. Above the critical temperature, the system cannot coexist in two phases. A plot of the coexisting compositions against temperature would give a coexistence curve closely analogous to the coexistence curves for the gas-liquid transition. Another set of terms with which you should immediately become familiar are the reduced conditions. These are corrected, or normalized, conditions of temperature, pressure, and volume and are expressed mathematically as

T T=, T<

(3.14)

:<

(3.15)

p,

=

V

V=, V<

(3.16)

The idea of using the reduced variables to correlate the p- V-T properties of gases, as suggested by van der Waals, is that all substances behave alike in their reduced, i.e., their corrected, states. In particular, any substance would have the same reduced volume at the same reduced temperature and pressure. If a relationship does exist involving the reduced variables that can be used to predict p" T" and V" what would the equation be? Certainly the simplest form of such an equation would be to imitate the ideal gas law, or p,V, = yT,

(3.17)

where y is some constant. Now, how does this concept work out in practice? If we plot p, vs. T, with a third parameter of constant V,. (3.18) we should get straight lines for constant V,. That this is actually what happens can be seen from Fig. 3.4 for various hydrocarbons. Furthermore, this type of correlation works in the region where p V = nRT is in great error. Also, at the critical point (p, = 1.0, T, = 1.0), V, should be 1.0 as it is in the diagram. However, the ability of Eq. (3.18) to predict gas properties breaks down for most substances as they approach the perfect gas region (the area of low pressures somewhat below p, = 1.0). If this concept held in the perfect gas region

.'

Chap. 3

:1sisting of water ~lssolved in it. If water increases Jsition is chosen :-eached at which :rface disappears ·:cher. Above the • plot of the co:nce curve closely

Real Gas Relationships

Sec. 3.2 6. 0

J

~I

0

IU

1/

4.0

i

[

jf~ ~ 1.0

~ '?" y

II )Ir1 p--,"

>-' pr

V

V

I ./

/

,/

17

V

~ 3V

V

l)"

'l..~

~

./

I~

i

o!lr'

1

vi ~ V1 1

.)

V

Y

07

~

V· ~ l-' 1rt II '7

I

,$-

:1

~

rl

I

)

I

./ II U I pi

2.0

I

I

V

n" rl rI'

J

I

J

II

.I

'j II

.~

,

I

If

I

17

n- V-T properties

"

~

1

If

II

.J

,0

V

I

il

P, 3.0

I

II

)

've

1.1

j

/

(3.16)

(3.17)

I

II

~!

II

I

behave alike in substance would ::re and pressure. .::.t can be used to :ne simplest form

)

&)

(3.14)

(3.15)

1/

<:;>/

.;n.

oecome familiar cd, conditions of ~,atically as

V

~

171

V

i--'"

,o~

,g., ~

i

l.-- V

L.w V

>-'" P: P" --~ ri!'"

~

I..--" i--'"

~.g9... i--'"f"

Q Methane • Ethylene o Ethane • Propane

plot p, vs. T,

I--

o Penlone

• Heptane

(3.18)

o

I

1.0

1.2

1.4

"

1.6

1.8

T, lV what happens :rmore, this type .::., error. Also. at ;5 in the diagram. . breaks down for ::rrea' ')w pres,,;:rf regIOn o



Fig. 3.4. A test of the relation p,/T, = r/ V,.

for all gases, then for 1 mole

pV =

RT

should hold as well as Eq. (3.17) p, V, = "IT,

(3.la)

*

.....-...,-

ey

Chap. 3

172 Gases, Vapors, Liquids, and Solids For both of these relations to be true would mean that

(3.19) or p, V, RT,

=

constant

(3.20)

for all substances. If you look at Table 3.3 you can see from the experimental data that Eq. (3.20) is not quite true, although the range of values for most substances is not too great (from 0.24 to 0.28). He (0.305) and HeN (0.197) represent the extremes of the range. TABLE 3.3 EXPERIMENTAL VALUES OF p,f/,/RT, FOR VARIOUS GASES

NHJ Ar CO, He

0.243 0.291 0.279 0.305

HCN H,O N, Toluene

0.197 0.233 0.292 0.270

CH. C,H. C,H. C.H ••

0.290 0.285 0.277 0.264

C,H. CH,OH C,H,OH CCI.

0.270 0.222 0.248 0.272

A more convenient and accurate way that has been developed to tie together the concepts of the law of corresponding states and the ideal gas law is to revert to a correction of the ideal gas law, i.e., a generalized equation of state expressed in the following manner: pV= znRT

(3.21)

where the dimensionless quantity z is called the compressibility factor and is a function of the pressure and temperature:

z=

l{I(p, T)

(3.22)

One way to look at z is to consider it to be a factor which makes Eq. (3.21) an equality. If the compressibility factor is plotted for agiven temperature against the pressure for different gases, we obtain something like Fig. 3.5(a). However. if the compressibility is plotted against the reduced pressure as a function of the reduced temperature,

z = l{I(p" T,)

(3.23)

then for most gases the compressibility values at the same reduced temperature and reduced pressure fall at about the same point, as illustrated in Fig. 3.5(b). This permits the use of what is called a generalized compressibility factor, and Figs. 3.6(a) and (b) through 3.10 are the generalized compressibility charts or z-factor charts prepared by Nelson and Obert.' These charts are based on 'L.C. Nelson and E. F. Obert, Chern. Eng., vol. 61, no. 7, pp.203 .. 208 (1954). O. A. Hougen and K. M. Watson, Chemical Process Principles. J. Wiley. New York. 1943. also presenlS z-factor charts. as do many articles in the literature for specialized gases, such as natural gases .

.....

"

-"~'

Chap. 3

Real Gas Relationships 173

Sec. 3.2 1.4,---::--------, Temperature' lOa· C

(3.19)

(3.20) lhe ex perimen tal values for most :ld HeN (0.197)

H. 1,OH

H,OH .. '4

A

Z"i;

1.~==------:;~--f----1

Z

0.270 0.222 0.248 0.272

~d!t

gether , law IS to revert ; state expressed (3.21)

Pressure, atmospheres

Reduced pressure,P. (b)

(a)

Fig. 3.5(a). Compressibility factor as a function of temperature and pressure.

Fig.3.5(b). Compressibility as a function of reduced temperature and reduced pressure.

factor and is a

(3.22) "kes Eq. (3.21) !'craturc against ..s(a). However, o a function of (3.23) :od temperature in Fig. 3.5(b) . .. sibility factor, ssibility charts are based on ·4). O. A. Hougen pl .... resents as r,ascs.

"".3,

Fig. 3.6(a). General compressibility chart.

Chap. 3

174 Gases, Vapors, Liquids, and Solids

'"

f:

r'" ( ................,.,

.. r'

.

~

"',

,-

I •• 0."

I----j--+-+-+--+----j--i--+--*--j 0.01

... - _.-

0.02

"Fig.3.6(b). General compressibility chart, very low temperatures.

30 gases. Figures 3.6(b) and 3.7 represent z for 26 gases (excluding H 2 , He, NH 3 , H 2 0) with a maximum deviation of I percent, and Hz and H,G within a deviation of 1.5 percent. Figure 3.8 is for 26 gases and is accurate to 2.5 percent, while Fig. 3.9 is for nine gases and errors can be as high as 5 percent. For Hz and He only, Newton's corrections to the actual critical constants are used to give pseudocritical constants (3.24)

p; =

Pc

+ 8 atm

(3.25)

which enable you to use Figs. 3.6 through 3.10 for these two gases as well with minimum error. Figure 3. lOis a unique chart which. by having several parameters plotted simultaneously on it, helps you avoid trial-and-error solutions or graphical solutions of real gas problems. One of these helpful factors is the ideal reduced volume defined as

V,. (or V; in the Nelson and Obert charts) =

Vc. is the ideal critical volume,

~

(3.26)

l'e,

"

I'f f'"·'.>: lJ! . I I l' t'f t' \. \ i ! l

1.1 f

,:l.ll j.

~.

,i"

\~

\

I .' (

/

"

"

I .,,'. " i..1,·

or

(i _ RT, (", - Pc

r-

v

! .

•• tL_~_ .. _~.

:'. ! '

(3.27)

Both V" and c. are easy to calculate since T, and p, are presumed known. The development of the generalized compressibility chart is of considerable practical as well as pedagogical value because it enables engineering calculations to be made with considerable ease and also permits the development of thermodynamic functions for gases for which no experimental data are available. All you need to know to use these charts are the critical temperature and the critical

• ...-.-....,j~,.

Chap. 3

.--.. ~-~,.....-:

Real Gas Relationships 175

Sec. 3.2 ~ Comprt'uibllity foetor Z=pv/RT

I

I

l

I

I

I

I

I

I

I

,

,

f-~

"- .....

... ~

..... .-:atlll

L::,.

;;:xcIuding H 2 , He, . and H 2 0 within :-ate to 2.5 percent, S percent. For H2 :stants are used to (3.24) (3.25) gases as well with \'ing several para:Jd-error solutions ::Jful factors is the (3.26)

" (3.27) :lrned known. The =-_siderable practical ::alcl"··'ons to bc 'C'!en ermodyava. .c. All you 're and the critical

\

, b.

0.'

0,

Red\.ictd pressure, Pr

u

.,

•..

'...:,. "

~ 0·'

Fig. 3.7. General compressibility chan, low pressures.

pressure for a pure substance (or the pseudo values for a mixture, as we shall see later). The value:: = I reprcsents idcality, and the value z = 0.27 is the compressibility factor at the critical point. Quite a fcw authors have suggested that the generalized compressibility relation, Eq. (3.23), could provide better accuracy if another parameter for the gas were included in the argumel,lt on the right-hand side (in addition to p, and T,). Clearly, if a third parameter is used, adding a third dimension to p, and T" you have to cmploy a set of tables or charts rather than a single table or chart. Lydcrsen et aI.' developed tables of z that included as the third parameter the critical compressibility factor. z, c= pJ'JRT,. Pitzer' used a different third parameter termed the aCelltric factor, defined as being equal to -Inp" - I, where p" is the value of the reduced vapor pressure at T, = 0.70. Viswanath and Su' compared the z factors from the Nelson and Obert charts, the Lydersen-Greenkorn-Hl'ugen charts, and the two-parameter (T, and p,) Viswanath and Su chart with the experimental z's for T,'s of 1.00--15.00 °A. L. Lydorsen, R. A. Greenkorn, and O. A. Hougen, Uni.-ersityof Wisconsin Engineer"",1.' L'-lu'rimclIlal Stu/ivIJ Report no. 4. f\'1adison, \Vis., 1955. -K. S. Pitzer. J. Allier. CiI<'III. Soc .. V. 77, p. 3427 (1955). 'D. S. Yiswanath and G. J. Su, A.I.Ch.E. J., v. II, p. 202 (1965).

~

.

:: 1-+Ht--it---1

:,~\ \\,1\\\' , \.~ \\'

\

\

.;

~

" ~ ~

""~

~

• l-+-ffiHt--i o

:: " I

r-- ~ ~ ~

ori

1-+-it1itt-i

\

,

,

I\'l~ \

\

\ \

\

III \

E

'\r\ \i~

'. \ \

\

"

~\ l \~~

":l

" E

\1\, \

-

1,\1\,

,\ , \ ~

\ '~\

\

~

\~~ \

\:..,

"5 .;:

>,

~ to.

:E

.,

~

. '" 0

~ I-+-+-HH-i

~

~ ~

i'.

Co.

E 0

"

" 'J

C

:l 1--t---t---1ItIH

'J

0

00

..; ,;,

iZ :: 1-+-+--ll!1t1

176

~II·~· II

UJ. t I

Fig. 3.8. General compressibility chart, medium pressures.

t

$

I

r

~

\ ComprUSlblr.,y foetor, ZI'E)Y/RT

•• I I I I I I I I

I

t,

I

~.

f,'

}

.~

i"tl I r.::r

~m~~tt::tXf~'QOiif ".~

'ir,-OfoO

-

_11111111 m1111 gtt1ill1lllllllll~ I

0.0 I

-

...

:::l

10.,

I

I

J

I

!

20

l5

)()

Rf'duced preUUf., P r

Fig. 3.9. Generalized compressibility chart, high pressures.

l'

40

I

\

~.\

\

\

~--~4-~-~,~-4---»~\~--~--~4--+4\--t---~-H-4-4rl~~r--+·-----1~ \ \

,

\

I

\

\

~+-~~~---+~\~--~--~~\~--~-+--t-~\H---~--++-+-4-~~Mr-+------i~ , I ,

I

\ \ " \ ~~~~-4~+-1rt-4r44-\~\~4--+~4-++~1+~~~---1~ I' \ \ \ I \ \ '. (

I \ \ \1

i\.

.... N

o

o

~

178

Sec. 3.2

Rea/ Gas Relationships

179

and p;s of 0-40 for 19 gases with the following results:

ViswanathSu

NelsonObert

LydersonGreenkornHougen

415

355

339

0.6\

0.83

1.26

Number of points Average deviation (%) in z

.,;

;;; " u

~

It is seen that all these charts yield quite reasonable values for engineering purposes. Because the use of a third parameter does not necessarily yield gas calculations of greater accuracy and because the presentation of z values so as to include the third parameter is considerably more cumbersome, we shall not show the three-parameter tables here.

;;;

·0

"0.

'"

-=

.~

~

.;

. .'", &.

~

u

~

.q :0 .;;;

'"

~

0.

E

0 u

"0

"

EXAMPLE 3.13 Use of the compressibility factor

In spreading liquid ammonia fertilizer, the charges for the amount of NH, are based on the time involved plus the pounds of NH, injected into the soil. After the liquid has been spread, there is still some ammonia left in the source tank (volume = 120 ft'), but in the form of a gas. Suppose that your weight tally, which is obtained by difference, shows a net weight of 125 lb of NH, left in the tank as a gas at 292 psig. Because the tank is sitting in the sun, the temperature in the tank is 125'F. Your boss complains that his calculations show that the specifjc volume of the gas is 1.20 ft'llb and hence that there are only 100 Ib of NH, in the tank. Could he be correct? See Fig. E3.13.

.~

;;;

292 psig "

~

" " 0"

125' F V =120f1 3

...

0 ..;

.

Fig. D.13.

Ii;

So/ulioll:

Basis: lIb ofNH, Apparently your boss used the ideal gas law in calculating his figure of 1.20 ft'/lb of NH, gas: R - 1073 (psia)(ft') . (lb mole}l"R) T = 125'F

+ 460 =

+ 14.7 = 307 psia

p

=

/I

= 17 Ibllb mole

292

585'R

lib

V = It~T

= (I/17)(I3~~3)(585)

=

1.20 ft'llb

180

Chap. 3

Gases, Vapors, Liquids, and Solids

However, he should have used the compressibility factor because NH, d""" not behave as an ideal gas under the observcd conditions of tempcrature and pressure. Let us again compute the mass of gas in the tank this time using pV

=

'. .\.

znRT

""

What is known and unknown in the equation? p = 307 psia

V= 120ft'

z=? n

=h Ibmole

T= 585°R The additional information needed (taken from Appendix D) is ·T,

= 405.4°K ~

729'R

p, = 111.3 atm ~ 1640 psia

Then, since z is a function of T, and p" T, p,

T

= T., =

585°R 729'R

=L _

307 psia = 0 18i 1640 psia .

p, -

=

0.803

\\ <' "

From the Nelson and Obert chart, Fig. 3.7, you can read z ~ 0.845. Now calculated as

V can

be

fhl\\f\("!

h('\;tuv-

t'lhn P"

or

\ h.ltt\.

v

= 1.20 fP ideal 0.845 ft' actual = 1.01 ft'/Ib NH

Jb

I ft' Ideal

'

Note in the calculation above that we have added some hypothetical units to z to make the conversion from ideal cubic feet to actual cubic feet clear: IlbNH, 1 120ft' 1.01 ft'

=

'i"4,,~ It,-,'

I'l,lk

: ... ,.

JI91bNH J

Certainly 1191b is a more realistic figure than 100 Ib, and it is easily possible to be in error by 6 Ib if the residual weight of NH, in the tank is detcrmined by difference. As a matter of interest you might look up the specific volume of NI-I, at the conditions in the tank in a handbook-you would find that P ~o 0.973 ft'jlb, and hence the com· pressibility factor calculation yielded a volume with an error of only about 4 percent. , t ,'n\

EXAMPLE 3.14

Use of the compressibility factor

A single-stage-to-orbit vehicle with space-shuttle capabilities may be feasible if the mechanical strength of frozen propellants can be utilized-particularly frozen oxygen, if it could be vaporized at the required rate. During thc thrust pcriod, the solid oxygen cylinder would be melted from its bottom surface and the resulting liquid propellant conveyed to a conventional pump-fed rocket engine where the oxygen would be vaporized and combined with vaporized fuel.

~ I •

i

!,(" •

I~

Real Gas Relationships

Sec. 3.2

..:.ctot:because. NH 3 d,.ce, not temperature, and pressure, ",.us.ng:

181

Suppose that 3.500 kg of O 2 is vaporized into a tank of 0.0284-m3 volume at . 25'C. What will the pressure in the tank be; will it exceed 100 atm? Solution: See Fig. E3.14.

Healing Coils



3,500 kg O2

is'

Fig. E3.14.

Basis: 3.500 kg O 2 We do know from Appendix D that

Tc

. :::::: (

Pc

olo.w j( aan, be

= -1I8.8°C = 49.7 atm

or

-118.8

+ 273.1 = 154YK

However, this problem cannot be worked exactly the same way as the previous problem because we do not know the pressure of the O 2 in the tank. Thus we need to use the other parameter, V" (i.e., V; on the chart), that is available on the Nelson and Obert charts: , _. 0.0284 m31 32 kg _ , V (molallOll/llle) - 3.500 kg J kg mole - 0.259 m {kg mole Note that the molalroll/Ille ml/st be used in calculating V" since mole. Next,

oe!ic:al;unit&tcr z to'make

,.:.:r:

V = "

RT, = 0.08206(m')(atm) 154YK Pc (kg mole)('K) 49.7 atm

=0

~55

.-

V"

\

;

1,

is a volume per

• j

m' kg mole

Then . easily pussible to be in :nined by difference. As d, at the conditions in ~. and hence tlle com· , only about 4 percent.

V "

V = 0.259 = -,,--- = V"

0.255

Now we know two parameters, V" and, with T T,

=

T

T,

=

248'K 154.3'K

1.02

=

=

-25°C -+ 273

=

248°K,

1.61

From the Nelson and Obert chart, Figs. 3.8 or 3.10, p,

·:es may be feasible if ~particularly frozen .ne thrust period. the ~d the resulting liquid -ICre the oxygen would

UM*'.

=

1.43

Then p =

PrP~

= 1.43(49.7) = 71.0atm The pressure of 100 atm is not exceeded.

,f

·"' ....,.'..»."""4.....__..j"'......._,..'-'_4___

.,....+."'..._,~,_____

U ...""~ . . .,..... ' ' ' ' ' . _....._ ._ _ ~...,''"'_ _ _ _ _ _ _ _ ' ...'''_ ........... ~._....._..,."'.,''''.,..,.,~....._ . _. . . .

,,.,.... d_t'_ _.................... ' .. ' ___

~..L,,~

betMi'

n ·e'_;:.

Chap. 3

182 Gases, Vapors, Liquids, and Solids

EXAMPLE 3.15

Src. J.l

3.2.3 G .• t. poumh an,l,

Use of compressibility factor

Repeat Example 3.12, this time using the compressibility factor, i.e" the equation ofstatepV = zIlRT. "A 5-ft' cylinder containing 50.0 Ib of propane (C,H,) stands in the hot sun. A pressure gauge shows that the pressure is 665 psig. What is the temperature of the propane in the cylinder?"'

ture~ and 1'-," Thu, the q , aC'cur~H:y

h 'f

';I".

and a' Ihat rnr Idt",tI ' lO!!cthcr to 1", (I prove to he .' .,'I dcvelop met~· !

Solutioll: See Fig. E3.15.

,

propcrtIC"'! ",

dl,cu"cd he: ".

(a) Equal/F'

Takl', I.. n thc P;lrll.ri ;•.. , add the In,!" " the !<'t;r1 pIC

Fig. EJ.1S.

for con:pc)ff(';: . f

V

= zIlRT, we know that = 5 ft'

Il

= 441bjlb mole

p

= 665 psig + 14.7 ~ 680 psi a

In the equation p V

50lb

= 1.1351b moles C,H,

However, z as well as T is unknown. Since z is a function of T, and p" we should first calculate the latter two quantities (T, = 370'K, or 666'R; p, = 42.1 atm, or 617 psia, from Appendix D) but arc stopped by the fact that T is unknown. Another parameter available for use, to avoid a trial-and-error or graphical solution, is the factor zT, (or V; could be used, as in the last example):

zT= pV

or

IIR

zT., = 680 psia 5 ft '

for

('fC .•

ill

(psia)(ft') 666'R .. (I b moleJC R)

'J h" ted;:' ho\\c\rf. t~

= 0.42 From the Nelson and Obert chart, Fig. 3.10, at

zT, = 0.42

you can find T,

=

=~~~ =

InftIl.

pV zT, = IIRT,

1.1 J5 Ib mole 10 n

p,

COnlI'OIJO- i t?

} 1,10

1.02 and

T = T, T, = (370)(1,02) = 378'K

= (666)(1,02) = 680'R

1"

Chap. 3

_=Ior, i.e., the equation c::Jane (C)H,) stands in _. What is the tempera-

~/( .

Real Gas Relationships 183

.f.2

3.2.3 Gaseous Mixtures. So far we have discussed only pure com",'.:nt!s and their p- V- T relations. Most practical problems involve gaseous mix:,rc> and there is very little experimental data available for gaseous mixtures. I Ilu' the question is. How can we predict p- V- T properties with reasonable ,'"uracy for gaseous mixtures? We treated ideal gas mixtures in Sec. 3.1.3. .1!HI '3W. as we shall see later (in Chap. 4 for the thermodynamic properties). l!l.It fM ideal mixtures the properties of the individual components can be added 1"I:(ther to give the desired property of the mixture. But this technique does not i'f()\C to be adequate for real gases. The most desirable technique would be to dnclop methods of calculating p- V- T properties for mixtures based solely on the \'Topcrties of the pure components. Possible ways of doing this for real gases are ,bcusscd below.

(a) Equations of state. Take, for example, van der Waals' equation, Eq. (3.12). You can compute the partial pressure of each component by van der Waals' equation and then .1,1t! the individual partial pressures together according to Dalton's law to give I he tOlal pressure of the system,

/"r component A, (3.28)

for componenl B, (3.29) . T, and p" we should ~R; Pc = 42.1 atm, or

is unknown. Another

"1 tOfal,

PT =

RT[~-1V - nAb A

-

nB

V - Il s b s

1 [Il A, a -1- nsa , v' A n

+ ... 1

+ ',

.J

.1

1

(3.30)

l!", tl'Chnique provides a direct solution for PT if everything else is known; l"'''Clcr. the sOlution for Vor IlA from Eq. (3.30) is quite cumbersome. (h)

I ]

cle.,

.::i solution. is the factor

I

AI'erage COllstants.

A simpler and usually equally effective way to solve gas mixture problems '\ I"~ l"e average constants in the equation of state. The problem resolves itself , :" Ih" question, How should the constants be averaged to give the least error ", Ihe "holc~ For van der Waals' equation, you should proceed as follows. I ,'1 ' /0 (Usc linear mole fraction weight):

(3.31 )

~;qt

184 Gases, Vapors, Liquids, and Solids

*"

Suo J.:

Chap. 3

thing ;It",,,, n prc\ I"ll'" In 1\" \ l'n the .1"". critlLI) \.,' Thus the I' (\

a (Use linear square root average weight): (3.32) Jfyou are puzzled as to the reasoning behind the use of the square root weighting for a, remember that the term in van der Waals' equation involving a is n 2a

V2 in which the number of moles is squared. The method of average constants can be successfully applied to most of the other equations of state listed in Table 3.1. The average constants can also be calculated from the pseudoreduced constants described below. For example, for van der Waals' equation,

27 R2 T'; a---- 64

p,

\\ hcre ,,: al~o

heen :

t, ,qi

pseudo'I'! ,; mole

and

~

a\(.'I.!.·~·

critical p,':' I

live p\clld.·:! t (c) Mean compressibility factor. Another approach toward the treatment of gaseous mixtures is to say that =m can be called the mean compressibility factor. With such a relationship the only problem is how to evaluate the mean compressibility factor satisfactorily. One obvious technique that might occur to you is to make Zm a mole average as follows:

pV = zmnRT, where

If ~l)\1 alc \\ell kn,>'"

such as ,'" ,\ 'I

(3.33) Since z is a function of both the reduced temperature and the reduced pressure, it is necessary to decide what pressure will be used to evaluate p,. (I) Assume Dalton's law of partial {JI'essllres. For each component z is evaluated at T, and the reduced partial pressure for each gaseous component. The reduced partial pressure is defined as (3.34)

(2) Assume Amagat's law of pure componelll rolumes. For each component z is evaluated at T, and the reduced total pressure on the system. (d) Pseudocritical properties. Many weighting rules have been proposed to combine the critical properties of the components of a real gas mixture in order to get an effective set of critical properties (pselldocritical) that enahle psclidorcduced properties of the mixture to be computed. 9 The pseudo reduced properties in turn are used in exactly the same way as the reduced properties for a pure compound. and in effect the mixture can be treated as a pure compound. In instances where you know no-

... , 1'1", r '

.You should still add Newton's corrections for H, and He.

-'~""_"""'''_'''''''''~- __''''''''''''M' .~_~",--,.

.

.

...,.......

, . ,

.

nfib'

Chap. 3

(3.32) ::uare root weighting ~volving a is

·.erage constants can -~ listed in Table 3.1. 20reduced constants

"',

Real Gas Relationships

Sec. 3.2

thing about the gas mixture this technique is preferable to any of those discussed previously. In Kay's method pseudocritical values for mixtures of !;\ases are calculated on the assumption that each component in the mixture contributes to the pseudocritical value in the same proportion as the number of moles of that component. Thus the pseudocritical values are computed as follows: p~

= PcAYA

+ PcnYB + .. ,

(3.35)

T;

=

T,.YA

+ T'.YB + ...

(3.36)

p;

where = pseudocritical pressure, T; = pseudocritical temperature. (It has also been found convenient in some problems to calculate similarly a weighted pseudo-ideal-critical volume V; .. ) You can see that these are linearly weighted mole average pseudocritical properties. (In Sec. 3.8 we shall compare the true critical point of gaseous mixture with the pseudocritical point). Then the respective pseudoreduced values are

P', -.E.. - p~ :::nures is -ibili -"ear -=r to yuu

to say that -lor. With ressibility is to make

185

(3.37) (3.38)

If you are faced with a complicated mixture of gases whose composition is not well known, you still can estimate the pseudocritical constants from charts' 0 such as shown in Figs. 3.11 (a) and (b) if you know the gas specific gravity. Figure

.,

(3.33)

.~

ErJ~H-H

:::e reduced pressure, :ete Pr' =h component z is 3seous component.

~ (3.34)

~

(01

500

i

~

Dr each component

~

~m.

0-

/

450

E ,!:

./"

,e critical properties

350 0.5

~ctive

set of critical :'[ies of the mixture c!sed in exactly the _ and in effect the ~lere you know no-

~

V

V

0.6 07 0.8 09 1.0 Specific Gravily of Natural Gas (Air' 1.001

(bl

Fig. 3.11. Estimation of critical properties of natural gases. 10Naturai Gasoline Supply Men's Association, Engineering Data Book, Tulsa, Okla., 1957, p. 103.

1

I i

I1 j

1

1

I

to

~...,;,

Chap. 3

186 Gases, Vapors, Liquids, alld Solids

3.11 is good only for natural gases composed mainly of methane which contain less than 5 percent impurities (CO,. N,. H,S. etc.). Kay's method is known as a two-parameter rule since only p, and T, for each component are involved in the calculation of =. If a third parameter such as Z" the Pitzer acentric factor. or is included in the determination of the mean compressibility factor, then we would have a three-parameter rule. Reid and Leland II have shown that all the weighting rules can be obtained from a common base, and all reduce to Kay's rule as limiting cases with the proper assumptions. All the pseudocritical methods do not provide equal accuracy in predicting

r"

TABLE

3.4 PREDICTIO" OF p-V-TVALUES OF GAS DENSITY BY PSEUDOCRITICAL METHODS

..

Su . .?! 1'-1 '-T P''':''-

ct al. rn',''''' meters h\ ! h n(lt the 11'" 'I of the m.': ,( In \llY'I;,1 for trcatli1! that 110":0:' method. "11 t accur;tcy il[\ out in T.d 'C are quite.! :c point for t ilC

% Root-mean-square % Root-mean-square Method

deviation of density, 35 systems

deviation of density, CO 2 and H2S free systems

10.74

6.82

EXA~lrll

:

(I) Three-parameter Kay's rule:

T~

=L

at 90 atm 1':",

TclYi

2: Pc,)'i z~ = L ZCt)'j

P; =

T~ = L TclYi T~ _ ~ T~fYi I-~ pc PCI

6.06

I

=

L

5.85

Vcl)'i

Pc

T~

-

.r;;; V~ =

1L L 8

i

)'iYj

j

L)'t~

.;p;,

L

(h)

\ ,1 : J

~

(e)

\';1"

~

(d) \h. (e) \1.-.. (f) J"" ..°tJ

(3) Virial approach (Joffe's method /11):

T' = --{

(.J

(a) Pc:f(

(2) Empirical:

V~

methods

So/un, 'f~,

[(7 )1:3 +-E! (7 )1/3J' -E.!

Pc,

PCI

4.98

4.24

4.32

3.26

VCIJ'i

I : \\'

I

10 ~.Jnu·, l

Inp,J.'·

IIR. C. Reid and T. W. Leland. A.l.Ch.E. J., vol. II, p. 228 (1965).

/Jo('/";'" (':

~

Chap. 3 ,~thane

which contain

-nly p, and T, for each parameter such as ·;:JJnation of the mean ::Jeter rule. Reid and ..:lIned from a common ;: proper assumptions. ~~uracy in predicting :j

TNSITY

% Root-mean-square .co deviation of density, CO 2 and HzS rrl'C

systems

6.82

5.85

Sec. 3.2

Rml Gas Relationships 187

p- V-T properties, however, but most suffice for engineering work. Stewart'Z

ct al. reviewed 21 different methods of determining the pseudoreduced parameters by three-parameter rules (see Table 3.4). Although Kay's method was not the most accurate, it was easy to use and not considerably poorer than some of the more cOlllplex techniques of averaging critical properties. In summary, to evaluate all the various methods which have been presented for treating p- V-T relationships of gaseous mixtures, we would have to state that no one method will consistently give the best results. Kay's pseudocritical method, on the average, will prove reliable, although other methods of greater accuracy (and of greater complexity) are available in the literature, as brought out in Table 3.4. Some of the longer equations of state, with average constants, are quite accurate. All these methods begin to break down near the true critical point for the mixture.

EXAMPLE 3.16 p-V-TRelations for gas mixtures A gaseous mixture has the following composition (in mole percent): Methane Ethylene Nitrogen

CH. 20 CzH. 30 Nz 50 at 90 atm pressure and 100°C. Compare the molal volume as computed by the methods of (a) (b) (c) (d) (e)

Perfect gas law. van der Waals' equation plus Dalton's law. van der Waals' equation using averaged constants. Mean compressibility factor and Dalton's law. Mean compressibility factor and Amagat's law. (f) Pseudoreduced technique (Kay's method). Basis: 1 g mole of gas mixture Additional data needed are:

r

em' a(atm) ( -l g lila e

CH.

C 2H. N2 3.26

I

l

l

i

t

I ~

Solution: 4.24

!

2.25 4.48 1.35

X X X

R

10 6 10 6 10 6

=

( em' ) b g //lole

T,eK)

p,(atm)

42.8 57.2 38.6

191 283 126

45.8 50.9 33.5

8206 (cm')(atm) .

(g mole)l' K)

'zW. E. Stewart, S. F. Burkhart. and David Voo. Paper given at the A.I.Ch.E. meeting in Kansas City. Mo .• May 18, 1959. Also refer to A. Satter and J. M. Campbell. Soc. Petrol. Engrs. J .• p. 333 (Dec. 1963); and H. E. Barner and W. C. Quinlan. I&EC Process Design and D,,·elopmeflt. vol. 8. p. 407 (1969).

]88

Chap. 3

Gases, Vapors, Liquids, and Solids

(a) Perfect gas law: V =

n~T

=

1(82.~~(373)

340 em' at 90 atm and 373°K

=

(b) Combine van der Waals' equation and Dalton's law according to Eq. (3.30):

Substitute the numerical values (in the proper units):

~·~.56 + V ~·~7.1 + V ~·~9.3J 10 + 40.5 X 10 + 33.8 X 10

90 = 82.06(373) [V

- h[9

X

4

4

4

(I)

]

The solution to this equation can be obtained by trial and error or graphical means. For the first trial assume ideal conditions. By plotting the right-hand side of Eq. (1) against V until the total is 90, we find the equation converges at about V = 332 em3 at 90 atm and 373°K. (e) To use average constants in van der Waals' equation, write it in the following fashion: V3 -

where

=

0

a and b are the average constants. (a)'12 = 0.2a1:1l. + 0.301:;&. + 0.50/./,' ii = 2.30 X 10' atm ( cm

l

gmoe l

-

b

n Vl _

a;nl

(b + ~)nV' + (!)n'V -

[45.0

)'

= 0.2bcH • + 0.3b c ,Ii. + O.5b N , = 1 g mole (the basis)

+ (82'~b(373)J V' + [2.30 ~

=

'. cm l 45.0 g mole

10'J V _ [(2.30 x ;6')(45.0)J = 0

I"

, Ii'"

(

(2)

.

( t~

Let us work out a method of solving Eq. (2) for V differrnt from the method used in part (b): I(V)

=

Vl -

- V3

385V'

+ 2.56

+ 385V'

x 104 V - 1.15

X

= 2.56 x 10' V - 1.15

10' X

,

=0

10'

"

, I

(3)

Splitting this equation into two parts, I,(V) = 2.56 x 104 V - 1.15 [,(V)

X

(4)

10'

= - V' + 385V' = V2(385 -

V)

(5)

and plotting these parts (see Fig, E3.16) to see where they cross iis one method of obtaining the real root of the equation. You know V is in the vicin~tly of 340 em'.

A

L

.\

X aPiA Pl¥!$

t_~_~~'--"-

"

"j ~ ..

Real Gas Relationships 189

Sec. 3.2

jfdl~t(J: Eq;

(3:30):

j

'~PNJ

I

j

I

I

(a)

: 6ITOlf or.

graphical ,e I1i8Hnhand' side: of =attatlout: V = 132 rcciiriintliccfWlowing Fig. E3.16.

f,(V) 7.55 x 10' 7.04 x 10' (only 2 points required for straight line)

V(cm')

340 320 310

)(45.0)J = 0

(1)

om rh~·method. used

,=0 106

(1)

1

i

,

CH. C,H. N,

45.8 50.9 33.5

191 283 126

Compo

z 0.99 0.93 1.00

(z)(y)

CH. C,H. N,

l

0.2 0.3 0.5

18 27 45

0.394 0.535 1.340

1i

T,

1

1.95 1.32 2.95

0.198 0.279 0.500 0.977

Zmean

Then V =

:e method of obtain-

f 340~""'.

1

From the graph V ~ 316 em' at 90 atm and 373'K. (d) The table below shows how Dalton's law can be used to estimate a mean compressibility factor. T = 373 T,(,K) Compo p,(atm) y (90)y = p p, = ;,

(4) (5)

I

f,(V) 5.20 X 10' 6.65 X 10' 7.20 x 10'

=

RT ::.:::: (0.977)(82.06)(373) p 90 332 em' at 90 aIm and 373'K

Zmeln

-%-...

----...."....................._ ....- -......,...,.........~·~~-'. .-·~-~,.. ___~•..,."';K"""........,.. J:w....."'...;ft_,a ....""".. , .....

.4...."'A-~4t"" .... ,:;,.. , ",PP....,..".,._ ..

tr

:etltfi $'2

t

190

Chap. 3

Gases, Vapors, Liquids, and Solids

,,

(e) Combining Amagat's law and the z factor, p,(alm)

T,("K)

Y

90 p, = Pc

45.8 50.9 33.5

191 283 126

0.2 0.3 0.5

1.97 1.78 2.68

Compo

z

(z)(y)

CH. CzH.

0.97 0.75 1.01

0.194 0.225 0.505 0.924

Compo

CH. CzH.

Nz

Nz

Zmean

= 373

T. '

T,

1.95 1.32 2.95

I ".

f 1

J J

Vl

:r

f

V = 0.924{82.06){373) = 313 em' at 90 atm and 373°K 90 (f) According to Kay's method, we first calculate the pseudocritical values for the mixture by Eqs. (3.35) and (3.36).

CH. CzH. (45.8)(0.2) -+- (50.9)(0.3)

p; =

Nz (33.5)(0.5)

P"YA + P"Ys + P'cyc = + = 41.2 atm = T, .. YA + T,.ys + T,,)'c = (191)(0.2) -+- (283)(0.3) -i- (126)(0.5)

T;

=

186°K

I

,, : \,

;

'. ".1

(,I "

fl.',

Then we calculate the pseudoreduced values for the mixture by Eqs. (3.37) and (3.38): p 90 p; = p~ = 41.2

= 2.18,

, T 373 T, ~~ T; = 186 = 2.00

With the aid of these two parameters we can find from Fig. 3.8 that z

\



,~

IC"

.'

'" . hi

I!:\.'

=

0.965. Thus

del.. "i,l i

,

d( '..,

V = zRT = (0.965)(82.06)(373) = 328 em' at 90 atm and 373"K

EXAMPLE 3.17

Use of pscudoreduced ideal molal v'olurne

In instances where the temperature or pressure of a gas mixture is unknown, it is convenient, in order to avoid a trial-and-crror solution using the generalized compressibility charts, to compute a pseudocritical ideal volume and a pseudoreduced ideal volume as illustrated belo\\'. Suppose we have given that the molal volume of the gas mixture in the previous problem was 326 em' at 90.0 atm. What was the temperature? Basis: 326 em' gas at 90.0 atm and T"K CH. CzH.

N,

••· ... 4

( '. I

'.

" <•

•., f ,;. :

So/uliOIl: Compo

(':

\

90

p

T,eK) 190.7 283.1 126,2

p,(atm) 45.8 50.9 33.5

y 0.20 0.30 0.50

yT, 38.14 84.93 63.10 186.2

yp, 9,16 15.27 16.75 41.2

Chap. 3

90 p,

T

,

=

373 T,

Vapor Pressure 191

Sec. 3.3

Note we have used mole fractions as weighting factors to find an average and p; = 41.2 as in the previous example. Next we compute V;,:

1.95 1.32 2.95

', V "

=

V' ,

=

p;

RT; -- (82.06)(186.2) p; (41.2)

f

V;,

=

= ;; =

326 371

!~:~

=

=

T;

=

186.2

371 0 J/ 1 . cm gmoe

0.879

= 2.19

T; = 1.98. Then T = T; T; = (186.2)(1.98) =

From Fig. 3.8 or Fig. 3.10,

~critical

values for the

Nz (33.5)(0.5) ... (i

~qs.

(3.37) and (3.38):

2.00

:hat z = 0.965. Thus :md 373"K

~jxture

is unknown, ,mg the generalized md a pseudoreduced molal volume of the oat was the tempera-

yp, 9.16 15.27 16.75 tl.2

3.3

369°K

Vapor pressure

The terms vapor and gas are used very loosely. A gas which exists below its critical temperature is usually called a vapor because it can condense. If you continually compress a pure gas at constant temperature, provided that the temperature is below the critical temperature, some pressure is eventually reached at which the gas starts to condense into a liquid. Further compression does not increase the pressure but merely increases the fraction of gas that condenses. A reversal of the procec!ure just described will cause the liquid to be transformed into the gaseous state again. From now on, the word vapor will be reserved to describe a gas below its critical point in a process in which the phase change is of primary interest, while the word gas or noncondensable gas will be used to describe a gas above the critical point or a gas in a process in which it cannot condense. Vaporization and condensation at constant temperature and pressure are equilibrium processes, and the equilibrium pressure is caEied the vapor pressure. At a given temperature there is only one pressure at which the liquid and vapor phases of a pure substance may exist in equilibrium. Enther phase alone may exist, of course, over a wide range of conditions. By equilibrium we mean a slate in which there is no tendency toward spontaneous change. Another way to say the same thing is to say equilibrium is a state in whicrn all the rates of attaining and departing from the state are balanced. You can visualize vapor pressure, vaporization, amd condensation morc easily with the aid of Fig. 3.12. Figure 3.12 is an expande
1 j

I

1

I I

1 i

,!

1i , ; \

~

1 I

l -.,...,....--~

.....

vS

192

Chap. 3

Gases, Vapors, Liquids, and Solids

Su. 3.3

will be : at 170'F vapor at a constal Iy to a r temperal noticcah

A

760~~~---------------+-

500

i.e., evar

:F E E

..

I

<:(

Vapor (superheated)

180

J

5~~~L---------t~5~O------t~9~O~2~1~2-----------­ Sotid and vapor in equilibrium

T, of

Fig. 3.12. Vapor pressure curve for water.

and the term boiling point is taken to mean the "normal boiling point." The normal boiling point for water occurs when the vapor pressure of the water equals the pressure of the atmosphere on top of the water. A piston with a force of 14.7 psi a could just as well take the place of the atmosphere, as shown in Fig. 3.13. For example, you know that at 2 12°F water will boil (vaporize) and the pressure

liquid itt it had al at consli higher II dense at stances I Sun ment-, with th, or coml would ]1 the air.' vapor p peraturi goes U!" vapon/J linder- , water II

A' illustrai vaporil II on tf Th in cquil vapor I M-N-O subl ill)! thcrI1w Th rcgiof1~

Fig. :ll thrnlt

to the'

It mell, Fig. 3.13. Transformation of liquid water into water vapor at Constant pressure .

.~

,.,-,t"""....,'' ' j·...,.,"',."''".'' ',.~lpl::1:>,,!~:,t

...,................... , ;_~_ .....·~4W"',"'.._ _"...

d r.h

i,

satuLlt

.'.-'.",.,..'~

1

1m

Chap. 3

:g point." The northe water equals ',ilh a force of 14.7 nown in Fig. 3.13. .e) and the pressure ~)f

- at constant

Sec. 3.3

Vapor Pressure

nt ttt

193

will be 760 mm Hg, or I atm (point B). Suppose that you heat water starting at 170°F (point A) in an open pan-what happens? We assume that the water vapor above the pan is at all times in equilibrium with the liquid water. This is a constant-pressure process since the air around the water in {he pan acts similarly to a piston in a cylinder to keep the pressure at atmospheric pressure. As the temperature rises and the confining pressure stays constant. nothing particularly noticeable occurs until 212T is reached, at which time the water begins to boil, i.e., evaporate. It pushes back the atmosphere and will completely change from liquid into vapor. If you heated the water in an enclosed cylinder, and if after it had all evaporated at point B you continued heating the water vapor formed at constant pressure, you could apply the gas laws in the region B·C (and at higher temperatures). A reversal of this process would cause the vapor to condense at B to form a liquid. The temperature at point B would in these circumstances represent the del\' poil//. Suppose that you went to the top of Pikes Peak and repeated the experiment-what would happen then? Everything would be the Sltme (points D-E-F) with the exception of the temperature at which the water would begin to boil. or condense. Since the pressure of the atmosphere at the top of Pikes Peak would presumably be lower than 760 mm Hg, the water would start to displace the air, or boil, at a lower temperature. However, water still would exert a vapor pressure of 760 mm Hg at 212'F. You can see that (a) at any given temperature water exerts its vapor pressure (at equilibrium): (b) as the temperature goes up, the vapor pressure goes up; and (c) it makes no difference whether water vaporizes into air, into a cylinder closed by a piston, or into an evacuated cylinder-at any temperature it still exerts the same vapor pressure as long as the water is in equilibrium with its vapor. A process of vaporization or condensation at constant temperature is illustrated by the lines G-H-I or I-H-G, rcpectively, in Fig. 3.12. Water would vaporize or condense at constant temperature as the pressure reached point H on the vapor-pressure curve (also look at Fig. 3.14.) The p-Tconditions at which ice (in its common form) and water vapor are in equilibrium are also seen in Fig. 3.12. When the solid passes directly into the vapor phase without first becoming a liquid (line J-K as opposed to line LM-N-O), it is said to sublime. Iodine crystals do this at room temperature; water sublimes only below 3YF, as when the frost disappears in the winter when the thermometer reads 20'F. The vapor-pressure line indicates the separation of both solid and liquid regions from the vapor region and extends well past the conditions shown in Fig. 3.12 all the way to the critical temperature and pressure (not shown). Above the critical temperature, water can exist only as a gas. A term commonly applied to the vapor-liquid portion of the vapor-pressure curve is the word saturated. It means the same thing as vapor and liquid in equilibrium ,vith cach other. If a gas is just rcady to start to condense its first drop of liquid. the gas is called a saturated gas; if a liquid isjust about to vaporize, it is called a saturated liquid.

:

I

I

I J i

,1

194

Gases, Vapors, Liquids, and Solids G P' 900 mm

Chap. 3

H P' 500 mm

I

Fig. 3.14. Transformation of liquid water into water vapor at constant

temperature. These two conditions are also known as the dew point and bubble point, respectively. If you have a mixture of liquid and vapor at equilibrium (called a wet gas), both the liquid and vapor are said to be saturated at the specified conditions. The vapor-pressure line in Fig. 3.12 thus represents the state of a pure component designated by a number of special terms depending on what aspect of the state is of primary importance: (a) (b) (c) (d)

The The The The

saturated liquid. saturated vapor. bubble point. dew point.

The region to the right of the vapor-pressure curve in Fig. 3.12 is called the superheated region and the one to the left of the vapor-pressure curve is called the subcooled region. The temperatures in the superheated region, if measured as the difference (0 - N) between the actual temperature of the superheated vapor and the saturation temperature for the same pressure, are called degrees oj superheat. For example, steam at 500 c F and 100 psi a (the saturation temperature for 100 psia is 327.8°F) has (500 - 327.8) = 172.2°F of superheat. Another new term you will find used frequently is the word quality. A "wet" vapor consists of saturated vapor and saturated liquid in equilibrium. The weight fraction of vapor is known as the quality.

EXAMPLE 3.18

Properties of wet "apors

The properties of a mixture of vapor and liquid in equilibrium (for a single component) can be computed from the individual properties of the saturated vapor

~,

.

Chap. 3

S~c.

Vapor Pressure

3.3

195

and saturated liquid. The steam tables (Appendix C) are a good source of data to illustrate such computations. At 400"F and 247.3 psia the specific volume of a wet steam mixture is 1.05 ft 3 /lb. What is the quality of the steam? Solution: From the steam tables the specific volumes of the saturated vapor and liquid are

VI

=

V. =

0.0186 ft 3 /1b,

1.8633 ft 3 /1b

Basis: lib weI steam mixture Let

X

= weight fraction vapor. 0.0186 ft3 (J - x) Ib liquid 1 Ib liquid

+

0.0186 - 0.0186x 1.845x

1.8633 f(3 x Ib vapor 1 Ib vapor

+ 1.8633x =

= 1.05 ft3

1.05

= 1.03

x = 0.56

,'r vapor at constant

Other properties of wet mixtures can be treated in the same manner. and bubble point, respec:Qui)" 1 (called a wet It th, 'Jed conditions. ,late ot a pure component ... hat aspect of the state is

3.3.1 Change of Vapor Pressure with Temperature. A large number of experiments on many substances have shown that a plot of the vapor pressure (p*) of a compound against temperature does not yield a straight line but a curve, as you saw in Fig. 3.12. Many types of correlations have been proposed to transform this curve to a linear form (y = mx b); a plot of log (p*) vs. (lIT), for moderate temperature intervals, is reasonably linear:

+

log (p*)

-----y

, Fig. 3.12 is called the pressure curve is called d region, if measured as f the superheated vapor called degrees of superIration temperature for !perheat. Another new " "wet," vapor consists The weight fraction of

uilibrium (for a single of t~ '1turatcd vapor

=

m( ~ ) + b

(3.39)

~

mx+ b

Equation (3.39) is derived from the Clausius-Clapeyron equation (see Chap. 4). Empirical correlations of vapor pressure are frequently given in the following form (refer to Appendix G for values of the constants): log (p*) = - t :

c +B

(3.40)

where A, B, C = constants different for each substance t = temperature in °C Over wide temperature intervals the experimental data are not exactly linear as indicated by Eg. (3.39), but have a slight tendency to curve. This curvature can be straightened out by using a special plot known as a Cox chart. '3 The log of the vapor pressure of a compound is plotted against a special nonlinear temperature scale constructed from the vapor-pressure data for water (called a reference substance). "E. R. Cox, Ind. Eng. ellem" v. IS, p. 592 (1923).

it

Chap. 3

196 Gases, Vapors, Liquids, and Solids 247

400

I

350

6)

300 _ 250

. 200 . 150

tty

D

u on

3.7~

<>

'u

0.9~

Q.

on

~

100 ,/ LL • 75 i-;,-;- '\\<:>'~:1. O.I~~., ~. ~.,/ ~ .....:> 50 V'l~..~"() e ~",~ 25

.

,/

Q.

E

t!!

/

-25

f'f

/'

/

'"'"'II V II

V

:I

11

"'/

/

~ 4.J

/

<:)

~o

r' I

/j

I to

'" ~v! ,,'" 6

V

j .'

!J


V

1.0

V./ rv

~I

• Crticoll ~int

-50 0.1

~

l/

0

100

1000

10,000

Vapor pressure, psio (log scale)

Fig. 3.1S. Cox chart.

As illustrated in Fig. 3.15, the temperature scale is established by recording the temperature at a given vapor pressure of water for a number of vapor pressures. The numbers on the line for water indicate the vapor pressures of water for selected values (50, 100, 150, etc.) of temperature. The vapor pressures of other substances plotted on this same graph will yield straight lines over extensive temperature ranges and thus facilitate the extrapolation and interpolation of vapor-pressure data. It has been found that lines so constructed for closely related compounds, such as hydrocarbons, all meet at a common point. Since straight lines can be obtained on a Cox chart only two sets of vapor-pressure data are needed to provide complete information about the vapor pressure of a substance over a considerable temperature range. We shall discuss in Chap. 4 under the topic of the Clausius-Clapeyron equation other information that can be obtained from vapor-pressure plots. 3.3.2 Change of Vapor Pressure with Pressure. The equation for the change of vapor pressure with total pressure at constant temperature in a system is

= 1~1 (acp*)) apT T I

where

........

r· = PT

=

molal volume of saturated liquid or gas total pressure on the system

Under normal conditions the effect is negligible.

(3.41)

Chap. 3

Vapor Pressure

Sec. 3.3

EXAMPLE 3.19

197

Extrapolation of vapor-pressure data

The control of existing solvents is described in the Federal Register, v. 36, no. 158, dated August 14, 1971, under Title 42, Chapter4, Appendix B, Section 4.0, Control

of Organic Compound Emissions. Section 4.6 indicates that reductions of at least 85 percent can be accomplished by (a) incineration or (b) carbon adsorption. Chlorinated solvents and many other solvents used in industrial finishing and processing, dry-cleaning plants, metal degreasing, printing operations, and so forth, can be recycled and reused by the introduction of carbon adsorption equipment. To predict the size of the adsorber, you first need to know the vapor pressure of the compound being adsorbed at the process conditions. The vapor pressure of chlorobenzene is 400 mm Hg at JlO.O°C and 5 atm at 205'C. Estimate the vapor pressure at 245°C and at the critical point (359°C).

Solution: The vapor pressures will be estimated by use of a Cox chart. See Fig. E3.19. The temperature scale is constructed by using the following data from the steam tables: p. H 2 0 (psia) 0.95 3.72 11.5 29.8 67.0 247 680 1543 3094

10,000

~d

b) . __ vrding

of vapor pres,sures of water J[ pressures of nes over exten::lterpolation of :ed for closely 'n point. Since vapor-pressure lOr pressure of scuss in Chap. :ration that can

700 Q;

600

Ji :g.,

500

",,"

Q. (J')

-;;:: 300

• i::> .,

e 200

E ~

150 100 0.1

/1,

@

~9":

V?

""l iT

I ex:>

en

N

/!

1

1

:

1

I

I

~

: I

I

I

I

1

I

i

I I "-

I I

I

<.D

to

100

"

N

-0

1I

i : I

.!-... .l-en

- ,..,

ex:> . . , 0

<.D

1000

10,000

Vapor Pressure, psio (Log Scale)

(3.41 )

I j

I

~

"-

f

1 /I

71/ i /,

J

1 ,

""",)"

j

1

/

.. .. "c=>'~/ all:

11

1

/\

400

Q.

:e equation for aperature in a

t ('F) 100 150 200 250 300 400 500 600 700

,

Fig. EJ.19. The vapor pressures from I to 3100 psia are marked on the horizontal logarithmic scale. Next, draw a line representing the vapor pressure of water at any suitable angle on the graph so as to stretch the desired temperature range from the bottom to the top of the vertical axis. For each vapor pressure, the temperature is marked and a hori-

1 1 , ,I

I 1

198

Chap. 3

Gases, Vapors, Liquids, and Solids

zontalline drawn to the ordinate. This establishes the temperature scale (which looks almost logarithmic in nature). Now convert the two vapor pressures of chlorobenzene into psia, 400 mm 1 14 .7 mm psia = 774 . llOoe = 230'F 760 . pSIa

, 4

\

.

'.

... \ .

5 atm 14.7 psia _ 73 5 . 205°e = 4010F 1 atm . pSIa and plot these two points on the graph paper. Next draw a straight line between them, and extrapolate to 471°F (245°C) and 678'F (359'C). At these two temperatures read off the estimated vapor pressures: 150 psia 147 psia

Estimated Experimental

)f '"

.. .l

700 psia 666 psia

Experimental values given for comparison. An alternative way to extrapolate vapor pressure data would be to use the Othmer 14 plot as explained in Sec. 4.4.

3.3.3 Estimating Vapor Pressures. Miller" recommends the following equation to estimate vapor pressures based on the normal boiling-point temperature (T.) and the critical temperature and pressure: logp, = where

~, [I

-

G = 0.210

a

=

r; + k(3 + T,XI

+ 0.2000

T,.lbp, 1- T,.

~.,

Other methods of estimating vapor pressures are given by McGowan.'"

3.3.4 Liquid Properties. Considerable experimental data are available for liquid densities of pure compounds as a function of temperature and pressure. Gold" reviews 13 methods of estimating liquid densities and provides recommendations for different classes of compounds if you cannot locate experimental data. As to liquid mixtures, it is even more difficult to predict the p- V·T F. Olhmer and E. S. Yu, Indus. Engr. Chem. v. 60, p. 20 (1968). "d. G. Miller, J. Phys. Chem., v. 69, p. 3209 (1965). '61. C. McGowan, Reo. Trm'. Chim., v. 84, p. 99 (1965). I7P. l. Gold, Chem. Engr., p. 170 (Nov. 18, 1968).

14 0.

ii'

- TYl

k = (0/2.306 G) -- (I + ~,,) .' (3 + T,.XI - T,.)

T,. =

i·~-

" '"

Chap. 3

ture scale (which looks .:1to psia, !)OF

.,!!ht line between them, tWO temperatures read : 59°C)

-sia .)sia would be to use the

'mm!' orrr.

'he following-point

YJ

cGowan.16 d data are avail. temperature and ,ities and provides mot locate experipredict the p-V-T

.).

Saturatioll 199

51'(.3.4

properties of liquid mixtures than of real gas mixtures. Probably more experimental data (especially at low temperatures) are available than for gases, but less is known about the estimation of the p- V-T properties of liquid mixtures. Corresponding state methods arc usually used, but with less assurance than for gas mixtures.

!i 1

3.4 Saturation

~

How can you predict the properties of a mixture of a pure vapor (which can condense) and a noncondensable gas? A mixture containing a vapor behaves somewhat differently than does a pure component by itself. A typical example with which you are quite familiar is that of water vapor in air. Water vapor is a gas, and like all gases its molecules are free to migrate in any direction. They will do so as long as they are not stopped by the walls of a container. Furthermore, the molecules will distribute themselves evenly throughout the entire volume of the container. When any pure gas (or a gaseous mixture) comes in contact with a liquid, the gas will acquire vapor from the liquid. If contact is maintained for a considerable length of time, equilibrium is attained, at which time the partial pressure of the vapor (vaporized liquid) will equal the I'apor pressure of the liquid at the temperature of the system. Regardless of the duration of contact between the liquid and gas, after equilibrium is reached no more net liquid will vaporize into the gas phase. The gas is then said to be saturated with the particular vapor at the given temperature. We can also say that the gas mixture is at its dew point. As an illustration, assume that you put dry air at 14rF into a container in which liquid water is present. The initial total pressure on the air and water is to be 760 mm Hg. If you keep the total pressure on the container constant at 760 mm Hg (Figs. 3.16 and 3. I 7), eventually the water will vaporize, and water vapor will enter and mix with the air until the partial pressure of the water in

I I I 1

i iI

i

l

:

Conslanl Temperalure and Pressure (variable volume)

Conslanl Temperalure and Volume (variable pressure)

Ptotol .. 760~---::Po"'ir;':';"-~ 580 ~

EQui~,b~

i!' Q.

(salurotionl Time---

180

OL--------------Time ___

Fig. 3.16. Change of rartial and lolal pressures on vaporizalion of waler into air at constant temperature.

,.,............·""·..,.....

. ;,..,..,"'.P-"."..,.

-~-~-~

-~~~

..

... MO.,

200 Gases, Vapors, Liquids, and Solids

Chap. 3 160mm

760mm

."'0,

,\rr

14

760mm

If :l m,t\:dl, '

",?:,. "",',."

i

\.\ .lter .\ ';._'

::': Air

,'. + c',"

I

rLll'h,.·,,; \

::v~~~r':D . : (fully .~;:,

saturated r

"'ig.3.17. Evaporation of waler at constant pressure and temperature.

the air reaches approximately 180 mm Hg (the vapor pressure of water at 147°F). Regardless of the duration of contact between the water and the air (after this partial pressure of water vapor is reached), no more water vapor can enter the air. The air is saturated with respect to water vapor and cannot contain additional water. Of course the volume of the system will change if the total pressure is maintained at 760 mm Hg, or else some air will have to leave the system if the volume and pressure are to stay constant. If the volume of the system is fixed, the total pressure will rise to 760 + J 80 = 940 mm Hg. Assuming that the ideal gas laws apply to both air and water vapor, as they do with excellent precision, you can calculate the partial pressure of the air as follows at saturation:

P... + Pu,o = PIO"" P." + 180 mm = 760 rnm P.!, = 760 - 180 = 580 mm Hg

EXA'II'I I;,:

From Eq. (3.4), we know that Pair

=

PH,O

lh.tt

nair

\,1

H~

at

at constant temperature

af~ I~f"

7~S 11111.\

~,~

ll, '

;, way t IL\1

nUtO

or, from Eq. (3.7),

i -'}':-

S,11I4(;'"

Sec}

at constant temperature

I,'

arc

Then

We can generalize these expressions for any two components as follows:

!!.J = pz

PI = !!.l P, - p, nz

=

VI V, - V,

=

vV I z

(3.43)

and

V

PI =Vpz

or

(3.44)

Z

where the subscripts indicate component I, component 2, and total.

.... ,.

,""

lht: n'I''': . ~.ltur.Jt<:\t ','

1:

..

.

,)

Chap. 3

Sec. 3.4

Saturalion

201

If the temperature rises in an air-water mixture and the vapor phase is initially saturated with water vapor, the change in the partial pressure of the water and the air and in the total pressure, as a function of temperature, can be graphically shown as in Fig. 3.18. At Coostant Volume

At Constant Pressure

Air

="temperature. Water

'..:re of water at 14TF). =d the air (after this .:: vapor can enter the ::annot contain add i:e jf the total pressure ,:J leave the system if :nne of the system is

________ ~iquld Volume x 10 3 Temperature

Temperature

.Hg. . wa :Jress~

Fig. 3.18. The change of total pressure (volume) and partial pressures (volumes) of air and water in a saturated mixture with increasing temperat ure.

or, as they Jf the air as

EXAMPLE 3.20 Saturation What is the minimum number of cubic feet of dry air at 20°C and 738 mm Hg that are necessary to evaporate 13.1 lb of alcohol if the total pressure remains constant at 738 mm Hg? Assume that the air is blown over the alcohol to evaporate it in such a way that the exit pressure of the air-alcohol mixture is at 738 mm Hg. So/ulioll: See Fig. E3.20. Assume that the process is isothermal. The additional data needed

are P!j'Ohol at 20°C (68°F) = 44.5 mm Hg

(3.42) lS

Air 738mm

as follows:

(3.43)

f

l AlcoholJ 13.llb

738mm Saturated air-alcohol

mixture.

Fig. £3.20.

(3.44) Id total.

;

The most alcohol the air can pick up is a saturated mixture; any condition less than saturated would require more air. Basis: 13.1 Ib alcohol

I

I t ~

I

1

I

I :1 ~ !

1

I "1 j

202

Gases, Vapors, Liquids, and Solids

Chap. J

The ratio of moles of alcohol to moles of air in the final gaseous mixture is the same as the rallo of the partial pressures of these two substances. Since we know the moles of alcohol, we can find the number of moles of air: P,lcobol

=

\.

n.lcobol

Pair

nAir

From Dalton's law,

r"', "'. Pair

=

P.I,obol =

P,ot.1 -

P.tlcobol

44.5 mm

P.;, = (738 - 44.5) mm

13.1 Ib alcohol I Ib mole alcohol (738 - 44.5) Ib moles air 461b alcohol 44.51b moles alcohol .,..-;35_9,-f_t'rl""""",,.+.7,..6,,,0_m,,-m..:: _ 1800 f 3 . I Ib mole 738 mm t aIr at 20 0 e and 738 mm Hg

EXAMPLE 3.21

.-

Saturation 0

If 1250 em' of wet H. are saturated with water at 30 e and 742 mm Hg, what is the volume of dry gas at standard conditions? The vapor pressure water at 30'e is 32mmHg.

Solution: (a) Long Solution:

• f

Your first impulse in selecting a basis is probably to ask yourself the question, What do I have or what do I want? The answer might be that you have 1250 em' of saturated H. at 30 0 e and 742 mm Hg. See Fig. EJ.210

I

..

I"

Soturoted Mixture of H2

+ H20 l'ig. EJ.21.

Basis: 1250 em' H, saturated with water at 300 e and 742 mm Hg 1250cm' wet gas (742 - 32)cm' dry H, 742 em' wet gas

= 1051 em' dry H2 at S.c.

'"..,.....,..,.., ..,;;'."'*_""""'.....__"",,2"'*.............. " i'!'",Ii·~:r~.."..·'~~·-~-.,..·-'~----------'"·-.,.....·--·- . .,..--· . . . . . --"·~·-.,..."-~'"~--"',"~--'-"'.~ ..---.~.,...~....,...

f

Chap. 3 seous mixture is the .. Since we know the

Saturation

Sec. 3.4

203

(b) Shorler Sollllion: You can save a little time by recalling that

742 = 32

+ 710

In the given gas mixture you have any of the following: (a) 1250 em l wet H, at 30'C and 742 mm Hg. (b) 1250 em' dry H, at 30'C and 710 mm Hg. (c) 1250 em' water vapor at 30'C and 32 mm Hg. "les air ~ohol

Therefore, since the problem asks for the volume of dry H 2, you could take as a basis the dry H 2 , and then apply the ideal gas laws to the dry H 2 • Basis: 1250 em' dry H, at 30'C and 710 mm Hg

'lHg

:.

You can see that by choosing the second basis you can eliminate one step in the solution.

EXA:\IPLE 3.22 Saturation

A telescopic gas holder contains 10,000 ft' of saturated gas at 80'F and a pressure of 6.0 in. H ,0 above atmospheric. The barometer reads 28.46 in. Hg. Calculate the wcig.ht of water vapor in the gas. Sollllion: The total pressure of the saturated gas must be calculated first in in. Hg. See Fig. EJ.22.

Wet Gas

Fig. E3.22.

6.0 in. H ,0

29.92 in. Hg

-'--'---'--"-+.~--t"3"3"".9"I-'f't'H';,O

044'

= .

tn.

H

g

barometer = 28.46 in. Hg P'O'"

mmHg

= 28.90 in. Hg

P. = vapor pressure of H ,0 at 80"F

P, --y H 2 at S.c.

i

l i

42 m'- •J~, what is ::e v 30 c C is

lhe question, What .00 cm' of saturated

~

28.90

= P.

+

p",

= 27.87 + 1.03

.= 1.03 in. Hg

204

Gases, Vapors, Liquids, and Solids

Chap. 3

There exists

'.

at 28.90 in. Hg and SO°F 10,000 ftl wet gas 10,000 ft ' dry gas at 27.S7 in. Hg and 80°F 1.03 in. Hg and 80°F 10,000 ft' water vapor at

oR

= 80 + 460 = 540 R 0

.' 10,000 ft' H 20 vapor

18lb H 20 lIb mole H 20 = 15.71b H 20

"

EXAMPLE 3.23 Smokestack emission and pollution

(

"

A local pollution-solutions group has reported the Simtron Co. boiler plant as being an air polluter and has provided as proof photographs of heavy smokestack emissions on 20 different days in January and February. As the chief engineer for the Simtron Co., you know that your plant is not a source of pollution because you burn natural gas (essentially methane) and your boiler plant is operating correctly. Your boss believes the pollution-solutions group has made an error in identifying the stackit must belong to the company next door that burns coal. Is he correct? Is the pollution-solutions group correct? See Fig. E3.23.

"

, " f'''

I,

"

Fig. D.2J.

Solution: Methane (CH.) contains 2lb moles of H 2 per pound mole of C, while coal (see Example 4.24) contains 71lb of C per 5.61b of H2 in 100 lb of coal. The coal analysis is equivalent to

71lb C

111~2~~~ C = 5.91b moles C

5.61b HJJl:l mole H, = 2 8lb mole H, --r2.02 Ib H , ' •

or a ratio of 2.8/5.9 = 0.47 Ib mole H ,lib mole C. Suppose that each fuel burns with 40 percent excess air and that combustion is complete. We can compute the mole fraction of water vapor in each stack gas. Basis: I lb mole C

t: ,

t,

s~c. 3.4

Saturation

205

Nall/ral gas

eH. fuel compo

e Hz Air

composition of combustion gases, lb moles eoz excess Oz Nz HzO 1.0 2.0 10.5 0.80 1.0 0.80 10.5 2.0

lb moles 1.0 2.0

-F ~

JiI'-z-6"

.01e:J!i,>-O = 1S;1nr-}JI~D

Excess Oz: 2(0.40) = 0.80 N z : (2)(1.40)(79/21)

+ 20 z -~ CO, + 2H zO

=

10.5

The total pound moles of gas produced are 14.3 and the mole fraction H,O is 2.0 ~ 0 14 14.3 = . Coal

e + Oz -->- eoz

.:;.0,. &iterr plant' as I1~i1'

Si1lokestack .lei etlgineer; for the -:: I:JeoaP"''''''Vo-u burn :1!f cr '. Your

: aifYi. .stack-':11lctt1.1!>'the'pollu-'

fuel compo

Ib moles

e Hz

0.47

Air

composition of gas produced, lb moles eoz excess Oz N, H,O I 0.47 0.49 6.5 6.5 0.49 0.47

e Hz 1(0.40) - 0.40 (0.47)(1/2)(0.40) - 0.094 N z : 1.40(79/21)(1 + 0.47(1/2» = 6.5

Excess O 2 :

The total pound moles of gas produced are 8.46 and the mole fraction HzO is 0.47 "" 0 056 8.46 - . If the barometric pressure is 14.7 psia, then the stack gas becomes saturated at

Pressure Equivalent temperature: or c,. while: coal of coal. The coo I

~

2.81b mole H,

C·" fuel burns with ;ute-the mole frac-

natural gas 14.7(0.14) = 2.06 psia

coal 14.7(0.056) = 0.083 psia

127°F

ny mixing the stack gas with air, the mole fraction water vapor is reduced, and hence Ihe condensation lemperature is reduced. However, for equivalent dilution, the coalburning plant will always have a lower condensation temperature. Thus. on cold winter
206 Gases, Vapors, Liquids, and Solids

Chap . .'

burned properly. The sulfur contents as delivered to the consumers are as foUo\<, natural gas, 4 x 10- 4 percent (as added mcrcaptans); number 6 fuel oil, up to :' percent; and coal, from 0.5 to 5 percent. What additional steps would you take to resolve the questions that were original;:, posed?

3.5 Partial saturation and humidity In the foregoing section we dealt with mixtures of gas and vapor in which the g:" was saturated with the vapor. More often, the contact time required betwecc the gas and liquid for equilibrium (or saturation) to be attained is too long. an,: the gas is not completely saturated with the vapor. Then the vapor is not ic equilibrium with a liquid phase, and the partial pressure of the vapor is less thac the vapor pressure of the liquid at the given temperature. This condition is calle<: partial saturation. What we have is simply a mixture of two or more gases whic'. obey the real gas laws. What distinguishes this case from the previous exampk' for gas mixtures is that under suitable conditions it is possible to condense par: of one of the gaseous components. In Fig. 3.19 you can see how the partial Saturated Olr

----L-...... Portiol -!saturation

Canden>otian -1

Change in temperature

-,I

~

Conden~otion;; begins I ~'

Toto) pressure

Mi,;ture

~~Portiol

£ -Fully saturated gas

I

I I

I

~

pressure of alf Aif pV=nRT

Portial pressure of ..oter vapor _ _=,-=-":":",.--Woler pV = nRT

Temperature Fig. 3.19. Transformation of a partially saturated water vapor-air mixture into a saturated mixture as the temperature is lowered (volume = constant).

pressure of the water vapor in a gaseous mixture at constant volume obeys th~ ideal gas laws as the temperature drops until saturation is reached. at which time the water vapor starts to condense. Until these conditions are achieved. you can confidently apply the gas laws to the mixture. Several ways exist to express the concentration of a vapor in a gas mixture. You sometimes encounter weight or mole fraction (or percent), but more frequently one of the following: (a) Relative saturation (relative humidity): Sec. 3.5, I. (b) Molal saturation (molal humidity); Sec. 3.5.2.

·:..b~:·S..

n

Chap. 3 =onsumers are as follows: :coer 6 fuel oil, up to 2.6 .cstions that were originally

t

*

Partial Saturation and Humidity 207

Sec. 3.5

(c) "Absolute" saturation ("absolute" humidity) or percent saturation (percent humidity); Secs. 3.5.3 and 5.3. (d) Humidity; Sec. 3.5.2 and 5.3 . When the vapor is water vapor and the gas is air, the special term humidity ;Irrlies. For other gases or vapors, the term saturation is used. 3.5.1 Relative Saturation.

vapor in which the gas rime required between _uained is too long, and :~n the vapor is not in :Jf the vapor is less than This condition is called '\'0 or more gases which the previous examples . ·ssible to condense part '!n see how the partial ..l

eRS

=

pupo,

Relative saturation is defined as

=

relative saturation

(3.45)

PSltd

where P.opo, = partial pressure of the vapor in the gas mixture Put' = partial pressure of the vapor in the gas mixture if the gas were saturated at the given temperature of the mixture, i.e., the vapor pressure of the vapor Then, for brevity, if the subscript I denotes vapor,

VI/V, _..!'!.L - ~

eRS - I!J.. - PI/P, -

- pf -- pt/Pr -

Vsatd/ V , -

nutd

-

Ib~atd

(3.46)

You can see that relative saturation, in effect, represents the fractional approach to total saturation, as shown in Fig. 3.20. If you listen to the radio or TV and

Volume of H20 vapor if the air were saturated

~ure

Volume of oir Actual volume of H20 vapor

Fig. 3.20. A gas partially saturated with water vapor.

\"apor-air mixture

wered (volume =

hear the announcer say that the temperature is 70°F and the relatil'e humidity is 60 percent, he means that

P~IO(lOO) .ant volume obeys the is reached, at which lditions are achieved, "por in a gas mixture. ::-cent), but more fre-

= %eRJC

(3.47)

PH1D

"'lIh both the PH,O and the P~,o being measured at 70°F. Zero percent relative ';lturation means no vapor in the gas; 100 percent relative saturation means Ih;1l the partial pressure of the vapor is the same as the vapor pressure of the ,ubstance that is the vapor.

J-:XA:\IPLE 3.24

Relath'e humidity

The weather man on the radio this morning reported that the temperature this "((ernoon would reach 94°F, the reiati\'e humidity would be 43 percent, the barometer

1

j i

I,

208

Gases, Vapors, Liquids, and Solids

Chap. 3

29.67 in. Hg, partly cloudy to clear, with the wind from SSE at 8 mi/hr. How many pounds of water vapor would be in I mi' of afternoon air? What would be the dew point of this air? Soflll ion: The vapor pressure of water at 94°F is 1.61 in. Hg. We can calculate the partial pressure of the water vapor in the air from the given percent relative humidity; from this point forward, the problem is the same as the examples in the previous section.

= (1.61 in. Hg)(0.43) = 0.69 in. Hg = p, - Pw = 29.67 - 0.69 = 28.98 in. Hg

Pw Pal.

Basis: I mi' water vapor at 94°F and 0.69 in. Hg ~(52180mflt')'

t!

~'i 1

181b H,O lIb mole

- . T

= 1.51

x 10' Ib H,O

Now the dew point is the temperature at which the water vapor in the air will first condense on cooling at constant total pressure and composition. As the gas is cooled you can see from Eq. (3.47) that the percent relative humidity increases since the partial pressure of the water vapor is constant while the vapor pressure of water decreases with temperature. When the percent relative humidity reaches 100 percent, 100E! pi

= 100 /0 0

or

/

the water vapor will start to condense. This means that at the dew point the vapor pressure of water will be 0.69 in. Hg. From the steam tables you can see that this corresponds to a temperature of about 69°F.

3.5.2 Molal Saturation. Another way to express vapor concentration

,. I

~',

in a gas is to lise the ratio of the moles of vapor to the moles of vapor-free gas:

= molal saturation

lI .. po •

(3.48)

nVlpor-free illS

If subscripts I and 2 represent the vapor and the dl"Y gas, respectively, then for a binary system, (3.49) p, P2 =p,

+

II,

+ 1/, =

!!.1 = E..t = ~ = ~ II,

p,

V,

1/, -

II,

=

(3.50)

II,

p, p, - p,

V, V, - V,

(3.51)

By multiplying by the appropriate molecular IVeigrn1s, you can find the weight of vapor per weight of dry gas:

(nVIPor)(mol. wt\'lJ'or) _ wt"\
(3.52)

The special term humidity (X) refers to the pounds
,/

~~<,_,~T"ir t-re.!'tf>~'A,:........_"';'i.·'-:1ftt> ,~::_.:, ..•

Chap. 3

':SE at 8 mi/hr. How many ? What would be the dew

':e can calculate the partial :::1t relative humidity; from !es in the previous section.

><

._,,;,,--ti-~:;.:.>;;:: •. , ___ , .,;",_, ,;',ii~~~~;;"·"__A';i'·'Hi.'tt,; '.-z'o' let'

,

Partial SalUration and Humidity

Sa. 3.5

209

3.5.3 "Absolute" Saturation; Percentage Saturation. "Absolutc" saturation is defined as the ratio of the moles of vapor per mole of mporjree gas to the moles of vapor which would be present per mole of mporfree gas if the mixture were completely saturated at the existing temperaturc and lotal pressure: (

CiS = "absolute saturation" =

m.Hg ·9 in. Hg

moles vapor ) moles ~apor-free gas ""u,1 mo es vapor ) ( moles vapor-free gas ntu .. "d

(3.53)

Using the subscripts I for vapor and 2 for vapor-free gas,

181b H,O lIb mole = 1.51 x 10' Ib H,O ·.\"ater vapor in the air will :lsition. As the gas is cooled .\" increases since the partial cure of water decreases with ,00 percent,

percent absolute saturation =

(~Ltu'l (100) (~Ltu'l (100)

(~~t",,"d

Since PI saturated = pf and p, = PI

+ p"

=

(;~t""t'd

PI percent absolute saturation = 100 p, - PI = P~

pf

.)

PI

(p,p, -- PIpf) 100

(3.54)

(3.55)

p,- pf

., the dew point the vapor YOU can see that this corre-

~ess

vapor concentration moles of vapor-free gas:

Now you will recall that PI/pf = relative saturation. Therefore, percent absolute saturation = (relative saturatiOn)(P, - pf) 100 P. -PI

(3.56)

Percent absolute saturation is always less than relative saturation except at saturated conditions (or at zero percent saturation) when percent absolute saturation = percent relative saturation.

(3.48) EXAMPLE 3.25

:is, respectively, then for (3.49) (3.50) (3.51 )

Partial saturation

Helium contains 12 percent (by volume) of ethyl acetate. Calculate the percent relative saturation and the percent absolute saturation of the mixture at a temperature of 30'C and a pressure of 740 mOl of Hg. Solutioll: The additional data needed are P:'.Ac at 30°C

I'OU

-t

can find the weight

= 119 mm Hg

Using Dalton's laws,

(3.52) =

.vater vapor per pound Idity rh~rts in Sec. 5.3.

(from any suitable handbook)

PHe =

(740)(0.12) Pr -

= 88.9 mm Hg

PEtAc

= 740 - 88.9 = 651.1 mm Hg

I1 1

i!

! !

,.,...._""'ttM...."..·..........__...............__...........' ....

_._·"'..,~,:.:~iA'i4!~

210

Gases, Vapors, Liquids, and Solids

At 30°C the (a) Percellt re/atil'e saturatioll

Chap. 3

Src ..1.(,

=

100~

= 100

PElAc

88 9 . = 119

74.6~~

=

(b) Percellt absolute saturatioll

88.9 740 - 88.9 100 119 740-119

PEtAc

100 p, - PEtAc P:tAc P, - P:'tAc

=

88.9 651.1 100 119 621

= 71.1 %

Tl1i'i

equalilli} I

(a) (h)

EXAMPLE 3.26

Partial saturation

A mixture of ethyl acetate vapor and air has a relative saturation of 50 percent at 30'C and a total pressure of 740 mm Hg. Calculate the analysis of the vapor and the molal saturation. Solutioll: The vapor pressure of ethyl acetate at 30'C from the previous problem is 1/9 mmHg.

% relative saturation = 50 = P~tA' 100

(el The ,\c ....-t'tllrll('

atxlul

~

3.6 Malen; and

1",71

PEtAc

From the above relation, the PEtA, is PEtAc

(a)

= 0.50(119) = 59.5 mm Hg

"EtAc Il,

=

PEtA,

p,

= 59.5 = 0.0805 740

Hence the vapor analyzes EtAc, 8.05 percent; air, 91.95 percent.

The soluth"1 (·1 S.~lli()n. and \ ,I;~ in Chap. ~. II t"p of hg. 1 ~ to remove tll· I

(b) Molal saturation is IlEtAc

=

ndr

PEtAc

=

_LP",E;
PI -

Pair

PEtAc

59.5 740 - 59.5

= 0.0876 mole EtAc

mole air

EXAMPLE 3.27 Partial saturation The percent absolute humidity of air at 86"F and a total pressure of 750 mm of Hg is 20 percent. Calculate the percent relativc humidity and the partial pressure of the water vapor in the air. What is the dew point of the air? Y\):,I ..

Solutioll: Data from the steam tablcs are p~,o

To get the relative humidity,

M

at 86c F

PH.O!P~,o.

nl r,.I(tI~ ~ \ =

31.8 mm Hg

we need to find the partial pressure of the

thr dr!

1Il!~ ,':

\,)u~~ht.

t\ '

In,lude Ihe"

"\.

([liap;. J

Material Balances inl'olvillg Condensation and Vaporization

Src.3.6

211

water vapor in the air. This may be obtained from 750 ~ PH,O 100 31.8 750 ~ 31.8

Pt - pA20

This equation can be solved for PH,O 100

0.00885 = 6.65 (a)

.1V~~

percent rhe vapor and

m

'~vio\l"

0.00885PH,O = PH,O

PH,O

(b) ~lm. ~ 50'

~

PH,O 750 ~ PH,O

%
= 6.6 mm Hg

l0036i~8

= 20.7%

(c) The dew point is the temperature at which the water vapor in the air would commence to condense. This would be at the vapor pressure of 6.6 mm, or about 41°F.

- '''\ulem is rI9

3.6 Material balances involving condensation and vaporization The solution of material balance problems involving partial saturation, condensation, and vaporization will now be illustrated. Remember the drying problems in Chap. 2? They included water and some bone-dry material, as shown in the top of Fig. 3.21. To complete the diagram, we can now add the air that is used to remove the water from the material being dried.

751b H2 0

Woter

25 Ib Bone-dry leather

Bone-dry leother

J

HZO Vopor

~r~sure

of 750 mm of ::le partial pressure of

Bone-dry air

Bone-dry oir

Fig. 3.21. Complete drying schematic.

j

1

1 !

'ani

sure of rhe

You can handle material balance problems involving water vapor in air in exactly the same fashion as you handled the material balance problems for the drying of leather (or paper, etc,). depending on the information provided and ~ought. (You can find additional humidity and saturation problems which include the use of energy balances and humidity charts in Chap. 5.)

11

212

Gases, Vapors, Liquids, alld Solids

Chap. 3

We should again stress in connection with the examples to follow that if you know the dew point of a gas mixture, you automatically know the partial pressure of the water vapor in the gas mixture. When a partially saturated gas is cooled at constant pressure. as in the cooling of air containing some water vapor at atmospheric pressure, the volume of the mixture may change slightly, but the partial pressures of the air and water vapor remain constant until the dew point is reached. At this point water begins to condense; the air remains saturated as the temperature is lowered. All that happens is that more water goes from the vapor into the liquid phase. At the time the cooling is stopped, the air is still saturated, and at its dew point.

EXAMPLE 3.28 Material balance with condensation If the atmosphere in the afternoon during a humid period is at 90"F and 80 percent

cRX (barometer reads 738 mm Hg) while at night it is at 68'F (barometer reads 745 mm Hg), what percent of the water in the afternoon air is deposited as dew? Solutioll: The data are shown in Fig. E3.28.

,- r·

Doy

Night

H20 Vopor

H20 Vapor

Dry Air

Dry Air

H20 Liquid

J

Fig. EJ.2S. If any dew is deposited at night, the air must be fully saturated at 68°F. You can easily check the dew point of the day air and find it is 83°F. Thus the air first becomes saturated at 83°F. The partial pressures of air and wat~r vapor during the day and night are p. day = 36(0.80) = 28.8 mm p. night = 17.5(1.00) p, =P.i,

day:

P'i,

night:

P.i,

= 17.5 mm

+ P.

= 738 - 28.8 = 709.2 mm - 17.5 = 727.5 mm

= 745

This gives us in effect the compositions of all the streams. As a basis you could select I ft' of wet air, I Ib mole wet (or dry) air, or many other suitable bases. The simplest basis to take is Basis: 738 Ib moles moist day air because then the moles

= the partial pressures.

~

.."-

Ii. :' ';', ~ I:.

It(. .. ',



1

Chap. 3 :';es to follow that if ly know the partial ~:l.ially saturated gas c:.raining some water -=tay change slightly, ~:1 constant until the :-,se; the air remains is that more water c cooling is stopped,

Material Balances lnvot.'ing Condensation and Vaporization

Src.3.6

initial mixture (day) lb moles = camp. partial press. Air 709.2 H 20 . 28.8 Total 738.0

213

final mixture (night) lb moles = partial press. 727.5 17.5 745.0

tI,

cle.ment

On the basis of 738 Ib moles of moist day air we have the following water in the night air: 17.51b mole H ,0 in night air 1709.21b mole air 727.5 Ib mole aIr I

= 17.0 Ib mole H

0 2

28.8 - 17.0 = 11.81b moles H 2 0 deposited as dew :?t 90°F and 80 percent l barometer reads 745 .:lsited as dew?

100 ~~:~

=

41 % of water in day air deposited as dew

This problem could also be solved on the basis of Basis: lIb mole of bone-dry air (BDA) initial

final

=

change

28.81b moles H,O 709.2Ib moles BDA

17.5 Ib moles H,O 727.5 Ib moles BDA

=

change

0.0406

0.0241

_ 00165 Ib mole H 2 0 -. Ib mole DBA

~:~~~ 100 = 41 % of water vapor deposited as dew Note especially that the operation initial

final

28.8 17.5 738- - 745 is meaningless since two different bases are involved-the wet initial air and the wet final air. =mated at 68°F. You -"F. Thus the air first :::lt~r vapor during the

EXA:\IPLE 3.29

j

Dehydration

By absorption in silica gel you are able to remove all (O.72lb) of the H 2 0 from moist air at 60'F and 29.2 in. Hg. The same air measures 1000 ft 3 at 70'F and 32.0 in. Hg when dry. What was the relative humidity of the moist air? Solution: See Fig. E3.29.

60'F

H20 Vapor

f.-.-l

0.72 Ib H2O

and

tor dry) air, or many

j -1

29.2 in.

Hg

Dry Air

r--

Fig. E3.29.

1.000 ft3 Dry Air

at 70'F and 32.0 in. Hg

1 I

Ij I

214

Gases, Vapors, Liquids, and Solids

Chap. 3

The wet air now appears to have the following composition: lb moles

BDA H 20 0.72/18 Wet air

2.77

=

0.040 2.81

The partial pressure of the water in the moist air was the total pressure times the mole fraction water vapor:

I

29.2 in. Hg 0.04 2.8\

= 0415' H In. g .

Saturated air at 60°F has a vapor pressure of water of 0.52 in. Hg. Consequently, the relative humidity was

P:,o

=

PH,O

0.415 100 0.52

=

80% 0

EXAMPLE 3.30 Humidification One thousand cubic feet of moist air at 760 mm Hg and nOF and with a dew point of 53'F enter a process. The air leaves the process at 740 mm Hg with a dew point of I37"F. How many pounds of water vapor are added to each pound of wet air entering the process? Solution: See Fig. E3.30.

t

H20

/System Boundary

Air

Air

H20 Vapor

H20 Vapor

Dew Point 53 'F

Dew Point t3JOF

Initial

Final

Fig. E3.30.

Additional data are dew point t (OF)

p:!,o(mmHg) 53°F 10.3137°F 138.2-These give the partial pressures of the water vapor in the inirial and final gas mixtures.

11 ".

' . ,l"';'

t~

..

t

.-'

Chap. 3 ..32.0 in. Hg 10 moles dry air

Material Balances Involring Condensation and Vaporization

Sec. 3.6

215

We know the initial and final compositions of the air-water vapor mixture, since the partial pressure of the water vapor in the moist air is the pressure of water at the ,lew point for each case. Basis: 760 moles initial wet air initial moles 10.3 749.9 760

component Vapor BOA' Total 'Bone-dry ai r.

nressure times the mole

final pressure (mm) I3S.2 749.9 601.8 740

moles tie <--element

In the final gas we can say that there are 13S.2 moles vapor/60l.S moles BOA I3S.2 moles vapor/740 moles moist air . Hg. Consequently, the

601.S moles BOA/740 moles moist air Since we have a tie element of 750 moles of bone-dry air and want to know how many moles of vapor are present in the final air on our basis of760 moles initial wet air, 749.7 moles BOA 138.2 moles vapor in final _ 17' I . fi I 6OI.S moles BOA in final - mo es vapor In na

. nOF and with a dew 40 mm Hg with a dew J each pound of wet air

172 moles final vapor - 10.3 moles initial vapor

= 162 moles of vapor added

But these calculations were all based on 760 moles of initial wet gas. Therefore our answer should be 162 moles vapor added 0.213 moles vapor added I mole wet gas in 760 moles wet gas in = Now the molecular weight of water is 18; the average molecular weight of the original wet air has to be calculated:

compo BOA

ory

moles 750 10 760

H 20 -

Vapor

Final

mol. wt 29 IS

21,9301b 760lb mole

'm! 137°F

0.2131b mole H 2 0 I Ib mole wet gas

181b I Ib mole H 2 0

Ib

21,750 180 21,930

= 28 S .

lib mole wet gas 28.81b

= 0 133 Ib H 2 0 .

Ib wet gas

Comments: (a) If the original problem had asked for the moles of final wet gas, we could have used the tie element in the following manner: 750 moles BOA 740 moles wet gas out 602 moles BOA out = 921 moles wet gas out

.-_.---

_._ __ ..

,.,.,

:;,*,

*

..

i4

Ii •

'tr ,-

216

Gases, Vapors, Liquids, and Solids

Chap. 3

(b) If a basis of I mole of bone-dry air had been used, the following calculation would apply: Basis: 1 mole BDA in:

moles vapor = 10J mole BDA 749.7

= 00138

out:

moles vapor = 138.2 mole BDA 601.8

= 0229

.

.

Since there are more moles of vapor in the final air (per mole of bone-dry air), we know that vapor must have been added: moles vapor added = 0.229 - 0.0138

= 0.215 moles

The remainder of the calculations would be the same.

We have gone over a number of examples of condensation and vaporization, and you have seen how a given amount of air at atmospheric pressure can hold only a certain maximum amount of water vapor. This amount depends on the temperature of the air, and any decrease in the temperature will lower the water-bearing capacity of the air. An increase in pressure also will accomplish the same effect. If a pound of saturated air at 75°F is isothermally compressed (with a reduction in volume, of course), liquid water will be deposited out of the air just like water being squeezed out of a wet sponge (Fig. 3.22). This process has been previously described in the p-T diagram for water (Fig. 3.12), by line G-H-I.

psio . .

PH,o : POir

J

1~:;3 j[·t:;?<

Plolol'14.7

pSia

psio

PHz0' 0.11 PH,o ' 0.43 4.ir '14.6 Poir ' 58.4 :Soturated" Plolol ' 58.8 ~_..!'!I Plolol ' 14.7 .Salurated: "','; Air:' . liquid H2 0

N~0~\;W:

Saturated Air 75° F, I atm

Saturated Air 75° F, 4 atm

{:: .....

: pOrtialiy • :Saturated: ,;..'.'.'-Air':::·::-.

: . :<':>::. .,,:,:::.-":::;,::,':.-. "

Partia"y Saturated Air, 75° F,r---'------,

lotm Fig. 3.22. Effect of an increase of pressure on saturated air and a relurn to the initial pressure. For example, if a pound of saturated air at 75°F and I atm is compressed isothermally to 4 atm (58.8 psia), almost three-quarters of the original content of water vapor now will be in the form of liquid, and the air has a dew point of 75°F at 4 atm. Remove the liquid water, expand the air isothermally back to I atm, and you will find the dew point has been lowe~cd to about 36"F. Mathe-

-.

Chap. 3

iollowing calculation

S~r.

Phase Phenomena

3.7

217

matically (I = state at I atm, 4 = state at 4 atm) with z = 1.00 for both components: For saturated air at 75 F and 4 atm, D

For the same

~ir

saturated at 75°F and I atm, nH.D) (P~'D) _ 0.43 ( nair 1 = Pair I -- 14.3

-:me-dry air), we know 'olOles

Since the air is the tie clement in the process,

(n.) n I

:.llion and vaporiza:oheric pressure can ,lS amount depends :oerature will lower :.liso will accomplish ::rm' mpressed :Je 0 cd out of 'ig. 3.22). This pro- water (Fig. 3.12),

0.43 = 58.4 0.43 14.3

H,O

PH,D PH,D P.l,

=

+ P."

= 14.7

=

=

nH.D

11."

P."

~~{i\:W; -rtlolly

, :uroted

0.245

0.43 58.4

0.00737

From these two relations you can find that

= 0.108 psia = 14.6

P.o." =

"o~:i~M'i

=

14.3 58.4

24.5 percent of the original water will remain as vapor. After the air-water vapor mixture is returned to a total pressure of I atm, the following two familiar equations now apply:

PH,O

~oturoted:

=

14.7 psia

The pressure of the water vapor represents a dew point of about 36 D F, and a relative humidity of 100 P~'D = 0.108 100 = 25 % PH,D 0.43

3.7 Phase phenomena

; 75° F,r----'-----, ~

atm

We shall now consider briefly some of the qualitative characteristics of vapors, liquids. and solids.

cd a return to

:.ltm is compressed !e original content :las a dew point of thermally back to Jout 36°F. Mathe-

3.7.1 The Phase Rule. You will find the phase rule a useful guide in establishing how many propcrties, such as pressure and temperature, have tll he specified to definitely fix all the remaining properties and number of phases that can coexist for any physical systcm. The rule can be applied only to systems "1 "qllilibrilllll. It says

F=C-6'+2

ail

(3.57)

~

218

....~.

Gases, Vapors, Liquids, alld Solids

Chap. 3

where F = number of degrees of freedom, i.e., the number of independent properties which have to be specified to detemline all the intensive properties of each phase of the system of interest. C = number of components in the system. For circumstances involving chemical reactions, C is not identical to the number of chemical compounds in the system but is equal to the number of chemical compounds less the number of independent-reaction and other equilibrium relationships among these compounds. (}' = number of phases that can exist in the system. A phase is a homogeneous quantity of material such as a gas, a pure liquid, a solution, or a homogeneous solid. Variables of the kind with which the phase rule is concerned are called phase-rule I'ariables, and they are intensil'e properties of the system. By this we mean properties which do not depend on the quantity of material present. If you think about the properties we have employed so far in this text, you have the feeling that pressure and temperature are independent of the amount of material present. So is concentration, but what about volume? The total volume of a system is called an extensive property because it docs depend on how much material you have; the specific volume, on the other hand, the cubic feet per pound, for example, is an intensive property because it is independent of the amount of material present. I n the next chapter we shall take up additional properties such as internal energy and enthalpy; you should remember that the specific (per unit mass) values of these quantities are intensive properties; the total quantities are extensive properties. An example will clarify the use of these terms in the phase rule. You will remember for a pure gas that we had to specify three of the four variables in the ideal gas equation p V = nRT in order to be able to determine the remaining one unknown. For a single phase (J' ~ I, and for a pure gas C = I, so that

F=C--(J'+2= I - I +2=2

.• i

" ,; '~,

;

.

, f

,.

variables to be specified

How can we reconcile this apparent paradox with our previous statement? Since the phase rule is concerned with intensive propcrties only, the following are phase-rule variables in the ideal gas law:

I, .

I . .< •

pJ""> (specific molar volume) } T

3 intensive properties

Thus, the ideal gas law should be written as fo II 0 II'S :

pf" =,

RT

and once two of the intensive variables are fixed, the third is also automatically fixed. An il/l"Qrial/t system is one in which no variation of conditions is possible without one phase disappearing. An example with which you may be familiar

' - - 0. . . .,."........"' ....., ...,.,*..... , .........,~,~""'."""''''.'''."'_"'W,""",,"""C

c·· -.

?O:,;::

~'

;.

Chap. 3 ~- independent pro," all the intensive

:::lS!ances involving =:nber of chemical :::mber of chemical :::lction ana other .::s.

_ nhase is a homo'- iiquid, a solution, ::1cerned are called ","stem. By this we :::,lterial present. If :!1is text, you have of the amount of : The total volume :c~nd on how much :he c" hic feet per ::::de t of the :.2.ke ,Jditional Temember that the :ve properties; the :2.se rule. You will ::Jr variables in the ,he remaining one I, so that ''!

specified

::vious statement 0 ':lly, the following

liso automaticallY jitions is possible may be familiar

:J

Phase Phenomena 219

Srr.3]

.s the ice-water-water vapor system which exists at only one temperature ~O.OIT):

F=C-
F=C-
r

p-r

T-r

I

1

1

!

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i 'I

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,I

.. 220 Gases, Vapors, Liquids, and Solids

".' ....

Chap. 3

I

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1

1 I

1 1

m

ICE \',

\

\

I

'

\

\

: ""

I

, I

I

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: ,, I I

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........ 1

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CritiCol point I

~.,.:.c~

GAS

: I

VOLUME-

Fig. 3.23. p-V-Tsurface and projections for H 2 0.

tant features which are shown. The two-phase vapor-liquid region is represented by the heavy envelope in the p-f' and the T-)' diagrams. On the p- T diagram this two-phase region appears only as a single line because you are looking at the three-dimensional diagram from the side. The poT diagram shows what we have

Chap. 3

S~C.

Phase Phenomena 221

3.7

W

0:

=>

~

0: W

0-

::.

....w

SOLID-VAPOR VOLUME-

,, ,

I

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SOLID-LIQUID

t

/GAS

lsometricS

SOLID LlOUID

\

/'

/

~ Criticol ......- POint

LlOUID-VAPOR VAPOR

J. region is represented Ihep-Tdiagram this il ar \;ing at the she. at we have

Fig. 3.24. p-V-Tsurface and projections for CO 2 .

so far called the "apor-pressure curve, while the associated diagrams show that the "curve" is really a surface on which two phases coexist. each having the properties found at the appropriate bounding line for the surface. A point on the surface shows merely the overall composition of one phase in equilibrium with another, as, for example, at point A we have liquid and at point B we have vapor

222

Gases, Vapors, Liquids, alld Solids

Chap. 3

\"

;

and in between them no single phase exists. If we added the liquid at A to the vapor at B, we would have a two-phase mixture with the gross properties as shown by point M. You will notice that in the diagram the lines of constant temperature and pressure become horizontal in the two-phase region, because if you apply tl,e phase rule

p-r'

F=C-
3.7.3 Phase Phenomena of Mixtures. Presentation of phase phenomena for mixtures which are completely miscible in the liquid state involves some rather complex reduction of three-, four-, and higher-dimensional diagrams into two dimensions. We shall restrict ourselves to two-component systems because, although the ideas discussed here are applicable to any number of components, the graphical presentation of more complex systems in an elementary text such as this is probably more confusing than helpful. We shall

-.o? • \.

.it.I

Chap. 3

Phase Phenomena

Sec. 3.7

iiquid at A to the properties as ~,e lines of constant .:lse region, because

Experimental Orthabaric Data

=,OS5

T

two-phase region, ~;;: and temperature ~:ng in this connec.. : point; the critical .actually in the p-f' . ~ and a liquid phase -,j liquid. The tem. :::f!, water, air, and ""'lUre as that at the ',apor are In equlby t O.OOloC -:-re, Ig val ues -=,oint 101' the water and liquid water. ::"'-:1 by plotting con:}::cf!d to higher and "e becomes shorter =oint of zero slope). jection.lf carefully Y.: determined this .;. 3.25, which says :~mes) of the pure ::JD of temperature. ':his is not an exact ;:TY or critical speciextrapolated until c;-e quite difficult to

:1On of phase phe.uuid state involves :c-dimensional dia"o-component sys:::>Ie to any number systrms in an eleh We shall

223

, 1

1 Fig. 3.25. Law of rectilinear diameters used to find critical temperature and density .

lise hydrocarbons for most of our examples of two-component systems since a vast amount of experimental work has been reported for these compounds. It would be convenient if the critical temperature of a mixture were the mole weight average of the critical temperatures of its pure components, and the critical pressure of a mixture were simply a mole weight average of the critical pressures of the pure components (according to Kay's rule), but these maxims simply are not true, as shown in Fig. 3.26. The pseudocritical temperature falls on the dashed line between the critical temperatures of CO 2 and S02' while 100.----------------------------------.

90

--7--

--

For 60% S0240% CO2 mixture

!c' _ - -

Locus of pseudocritical points for mIXtures of COl and S02 computed by Kay's ru Ie

E c

I ~

1

! \

J

I

1

l I

I

I I

1 j

I

j curves

Fig. 3.26. Critical and pseudocritical points for a mi"turc.

224 Gases, Vapors, Liquids, and Solids

Chap. 3

' ..

the actual critical point for the mixture lies somewhere else. The dashed line as in Fig. 3.31 illustrates (for another system) the three-dimensional aspects of the locus of the actual critical points. Figure 3.27 is a poT diagram for a 25 percent methane-75 percent butane mixture. You can compare this figure with the one for a pure substance, water. as in Fig. 3.12; the p-T curve for pure butane or pure methane would look just Constonl specific volume lines

1800 Liquid

\/

~

Pmc ,,' Cricondemtherm

'':''';~=~-..J/

/

/

I

i

'" H

,,/

/'"

_---=O§"C.. .1;ritic~

Point

p, (psio)

1 Vapor

I

220 Two-Phose Region

Fig. 3.21. poT diagram for binary system: 25 percent CH. - 75 percent C.H 1c •

like the diagram for water (excluding the solid phases). For a mixture of a fixed composition, we have only three variables to consider, p, P', and T. Lines of constant specific volume are shown outside the envelope of the two-phase region. Inside the envelope are shown Jines of constant fractions of liquid, starting at the high pressures, where 100 percent liquid exists (zero percent vapor), and ranging down to zero percent liquid and 100 percent vapor at the low pres· sures. The 100 percent saturated liquid line is the bubble-point line and the 100 percent saturated vapor line is the dew-point line, The critical point is shown as point C. You will note that this is not the point of maximum temperature at which vapor and liquid can exist in equilibrium; the latter is the maximum cricondentherm, point D. Neither is it the point of maximum pressure at which vapor and liquid can exist in equilibrium because that point is the maximum cricondenbar (point E), The maximum temperature at which liquid and vapor can exist in equilibrium for this mixture is about 30° higher than the critical temperature, and the maximum pressures is 60 or 70 psi a greater than the critical pressure. You can clearly see, therefore. that the best definition of the critical point was the one which stated that the

--_._

.... ---

--

I

,

.

.,

*<

6t .....

Ckap.3

jashed tine as in :J aspects of the :': perrent butane "ubstance, water. . would look just -e lines

S~C.

Phase Phenomena 225

3.7

density and other properties of the liquid and vapor become identical at the aitical point. This definition holds for single components as well as mixtures. With the aid of Fig. 3.27 we can indicate some of the phenomena that can take place when you change the pressure on a two-component mixture at con~tant temperature or change the temperature of a mixture at constant pressure. If you proceed from A in the liquid region to F and then to B in the vapor region oy means of an isotherm'al process of some nature, you can go from 100 percent liquid to 100 percent vapor and notice a phase change. A similar type of analysis could have been applied to the p-Twater diagram (Fig. 3.(2) and the same conelusion reached, although the fraction of vapor and liquid cannot be found from Fig. 3.12. It is only when you cross the two-phase region that a change of phase will be noticed. If you go from A to G to H to B, the liquid will change from 0 to 100 percent vapor, but there will be no noticeable phase change. Figure 3.28 illustrates this transition for a mixture by a change of shading, and a similar diagram could be drawn for a pure component. A very interesting phenomenon called retrograde condensation (or vaporization) can occur in two-phase systems, a phenomenon which initially appears to be quite contrary to your normal expectations. If you start at point I in Fig. 3.29 (an enlarged portion of Fig. 3.27 in the vicinity of the critical region) you

7 8

Liquid

j

£

220 75 percent

::1ixture of a fixed . and T. Lines of =.r the two-phase :-.5 of liquid, start'0 percent vapor), ~r at the low pres:: line and the 100

this is not the exist in equilibNeither is it the erst in equilibrium . The maximum -:J for this mixture ::ximum pressures :rly see, therefore, :h stated that the

p

T Fig. 3.28. Change in phase characteristics for a mixture of two compo· nents .

T

f

Fig. 3.29. The two types of retrograde condensation.

I

.21

':1

----.---~

are in the vapor region. As you proceed to raise the pressure at constant temperalure. you will arrive at point 2, where you are on the saturated-vapor line, or the dew point line. Continuing to increase the pressure will, as you might expect, incrrase the amount of liquid formed. However, soon you will reach a point where J mJximum amount of liquid has formed (3), and increasing the pressure still further to points 4 and 5 will reduce the amount of liquid present, and eventually }ou will wind up back in the vapor region at 6. This phenomenon of retrograde condensation can happen, of course, only if the critical point, C, is located

I I

I

, ;a;;;;".

e .'

226

Gases, Vapors, Liquids, and Solids

Chap. J

above and to the left of D as shown on the diagram so that there is a '"bulge" in the \'icinity of D for the mixture. The interesting thing about this phenomenon is that you would expect. as the pressure increased at constant temperature. to continually increase the amount of liquid formed. whereas the opposite is actually true after point 3 is reached. Another type of retrograde phenomenon exists in a few systems when you change the temperature at constant pressure. This would be illustrated by the line 7,8,9, 10, II. and 12. where first. as the temperature increases to point 8 on the bubble-point curve, some liquid starts to vaporize. The liquid continues to vaporize until a maximum of vapor is obtained at 9, and then the quantity of vapor starts to decrease until you reach a point where you have all liquid again at II. Although the points of maximum percent liquid (or vapor) enveloping the regions of retrograde phenomena are shown in the diagram by solid lines, there are actually no phase boundaries at these points. The solid lines merely represent the locus of the region in which the maximum amount of liquid or vapor can exist (the point of infinite or zero slope on the lines indicating the fraction liquid). exaggerated on the diagram for the purpose of illustration. Formally, retrograde phenomena of these two types are known as (a) isothermal retrograde vaporization (proceeding from I to 6) or condensation (proceeding from 6 to I), or retrograde vaporization or condensation, respectively, of the "first type." and (b) isobaric retrograde vaporization (proceeding from 12 to 7) or condensation (proceeding from 7 to 12), or retrograde vaporization (condensation) of the "second type." Figure 3.30 shows the same phenomena on a p-f" diagram. Figure 3.30, like Fig. 3.29, is for a mixture of fixed composition. The constant-temperature lines illustrated are Te. the critical temperature, and Tmw the cricondentherm. The middle line, 6 to I, shows a process of retrograde condensation of the first kind. The shadowed areas represent the regions in which

liquid

6

'mo,O CricondenlhermTwo-Phose Region

_ Region of retrograde "'" phenomena

Fig. 3.30. Retrograde condensation on a p- V diagram.

Chap. 3

Sa. 3.7

Phase Phenomena 227

: there is a "bulge" this phenomenon :.:nt temperature. to _:is the opposite is -=1

" systems when you ? illustrated by the ::1creases to point 8 :Ie liquid continues :len the quantity of .'lve all liquid again :Jor) enveloping the ::>y solid lines, there :.:es merely represent lquid or vapor can ":ne fraction liquid), ::Jrmally, retrograde :rrograde vaporiza:.: frc . to I), or - n .. pe," and "7) or \..vJ1densation :ldensation) of the "..f> diagram. -nosition. The con:!erature, and Tmu , ':ess of retrograde :Ie regions in which

Temperature

I j

Mole per cent Ethane

Fig. 3.31. p-T and T-x diagrams for ethane-heptane mixtures of different compositions.

retrograde phenomena take place. A T-f" diagram is not illustrated since it is not particularly enlightening. The diagrams shown so far represent two-component systems with a fixed IHerall composition. You might wonder what a diagram would look like if we were to try to show systems of several compositions on one page. This has been done in Fig. 3.31. Here we have a composite p-T diagram, which is somewhat awkward to visualize, but represents the bubble-point and dew-point curves for \,Irious mixtures of ethane and heptane. These curves in essence are intersections I,f planes in the composition coordinate sliced out of a three-dimensional system and are stacked one in front of the other, although in two dimensions it appears that they overlie one another. The vapor-pressure curves for the two pure components are at the extreme sides of the diagrams as single lines (as you might expect). Each of the loops represents the two-phase area for a system of any ~pecific composition. An infinite number of these slices are possible, of course. The dotted line indicates the envelope of the critical points for each possible composition. Although this line appears to be two-dimensional in Fig. 3.31, It actually is a three-dimensional line of which only the projection is shown in the figure. Another way to handle and illustrate the phase phenomena for the twol'I'mponent systems we have been discussing is to use pressure-composition l1i;lgrams at constant temperature or, alternatively, to use temperature-compositi,'n diagrams at constant pressure. A temperature-composition diagram with pressure as the third parameter is illustrated in Fig. 3.31 for the ethane-heptane \) ~tell1. Discussion of the phase diagrams for multicomponent systems (ternary .tnd higher) is beyond our scope here.

I

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J

]

l ,

j

~

1

j I

i

228

Gases, Vapors, Liquids, and Solids

Chap. 3

WHAT YOU SHOULD HAVE LEARNED FROM THIS CHAPTER I. You should be familiar with the ideal gas laws, know under what circumstances they are applicable, and know how to apply them in problems. 2. You should be able to distinguish under which circumstances a gas acts as an ideal gas and under which circumstances it acts as a real gas, and be able to use, as appropriate, (a) the ideal gas equation, p V = nRT; (b) the compressibility factor together with pV = znRT; or (c) an equation of state. 3. You should know how to treat ideal and real gas mixtures, and in particular be able to apply Kay's rule. 4. You should be able to use the generalized compressibility chart with ease, and be able to calculate the parameters required for its use. S. You should know what the following terms mean: vapor gas vapor pressure critical point critical temperature critical pressure pseudocritical temperature pseudocritical pressure condensation vaporization sublimation equilibrium normal boiling point saturated liquid

saturated vapor bubble point dew point superheat subcooled quality saturated air relative humidity absolute humidity isometric isobaric isothermal retrograde condensation

6. You should be able to estimate vapor pressures as a function of temperature. 7. You should know how to solve material balance problems involving vaporization, condensation, saturation, drying, and humidification. 8. You should be able to apply the phase rule to simple systems. 9. You should be familiar with the qualitative phase behavior of pure components and mixtures and also be able to draw typical diagrams illustrating this behavior.

SUPPLEMENTARY REFERENCES 1.

American Petroleum Institute. Division of Refining, Technical Data BookPetrolcum Refining, API, New York, 1970.

2. Edmister, Wayne C. Applied Hydrocarbon Thermodynamics, Gulf Publications Co., Houston, 1961, Chaps. 1-4.

., , I

I :

, I

.

'.

,j

Chap. 3

Problems 229

Chap. 3 3.

Henley, E. J., and H. Bieber, Chemical Engineering Calculations, McGraw-Hili, New York, 1959.

4.

Hougen, O. A., K. M. Watson, and R. A. Ragatz, Chemical Process Principles, Part I, 2nd ed., Wiley, New York, 1956.

5.

Natural Gas Processors Suppliers Association, Engineering Data Book, 10th cd., Tulsa, Okla., 1970.

6.

Reid, R. c., and T. K. Sherwood, The Properties of Gases and Liquids, 2nd cd., McGraw-Hili, New York, 1966.

,

and in particular

7.

Wexler, A., Humidity and Moisture, 4 vols., Van Nostrand Reinhold, New York, 1965.

1 ,J

.lIy chart with ease,

8.

Williams, E. T., and R. C. Johnson, Stoichiometry for Chemical Engineers, McGraw-Hili, New York, 1958.

I

S CHAPTER under what circum-:l in problems. :rances a gas acts as :::-eal gas, and be able '?.T; (b) the com pres'on of state. C::S,

...lse.

References on Phase Phenomena I.

Ricci, J. E., The Phase Rule and Heterogeneous Equilibrium, Van Nostrand Rein· hold, New York, 1951.

2. Stanley, H. E., Introduction to Phase Transitions and Critical Phenomena, Oxford University Press, London, 1971.

PROBLEMS"

:lsation :on of temperature. :, involving vapori-

n. ems. '.>r

of pure compoillustrating

~grams

.ical Data Boo"Gulf Publications

3.1. A given amount of N2 has a volume of 200 em' at 15.7°C. Find its volume at O°C, the pressure remaining constant. 3.2. A gas measured 300 em' at 27.5'C, and because of a change in temperature, the pressure remaining constant, the volume decreased to 125 em'. What was the new temperature in DC?

3.3. Under standard conditions, a gas measures 10.0 liters in volume. What is its volume at 92°F and 29.4 in. Hg? 3.4. Since 1917, Goodyear has built over 300 airships, most of them military. One of the newest is the America, which has a bag 192 ft long, is 50 ft in diameter, and holds about 202,000 ft' of helium. Her twin 210·hp engines produce a cruising speed of 30-35 mi/hr and a top speed of 50 mi/hr. The 23·ft gondola will hold the pilot and six passengers. Blimps originally were made of rubberimpregnated cotton, but the bag of the America is made of a two·ply fabric of Dacron polyester fiber coated with neoprene. The bag's outer surface is covered with an aluminized coat of Hypalon synthetic rubber. Assuming that the bag size cited is at I atm and 25 c C, estimate the temperature increase or decrease in the bag at a height of 1000 m (where the pressure is 740 mm Hg) if the bag volume does not change. If the temperature remains 25'C, explain how you might estimate the volume change in the bag. I. An asterisk designates problems appropriate for computer solution. Also refer to the computer problems after Problem 3.129.

J

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'~';"':"

';'essllre.'ofl5D P'" " find, the 'volunoc . ' ~ Y. How,do)o'J ,.

Energy Balances

,aw'usiog'tfle'e'r., ' ,M ilata:takcn .... n .

4

',:' "/I/crelnen/ill

,':'vd[.ofbatrOOI1. ,,~' ~ ~ 4.2 4.s 4.9 <; c 5.5 6,~

6.5

7.f' 9(

7.0 9.0 10.0 11.9 13.5

10.[ 11.S 13.~ j.:' ~

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-:.-.:yd(l~~~-knsr~i~ ,

,".ltlalrt; Wha \:\\ :' :::tlfe T.ilr lll'ihe:b ,5l;e

Table 3,1 l

alpJ

i j,)I'.mud:' T'<')wer will be used in the world 10 years from now?-IOO years from

HL'rIX.lsts' yield a considerable range of disagreement, but one common : :alne:,;' ~'U Ihe forecasts is that the trend in energy usage is uP. and up sharpiy, " Jilu,t:"cl.,~d in Fig. 4.1, for the United States. The combination of growmg ',"eTc" cO'i.~\lmption per capita with population growth calls for substantial ".:f~;,ses, ;1;:] lotal energy consumption that dwarfs our current productive "::'iilll(,~' ilf !,ublic expectations make these forecasts come true. the economic (,',:',allO,,' .eli the sources of power and the design of new types of power proouc. ,,~~ ;;.n;,J,lts will require that engineers continue to playa vital role in our ;;;>\\.','

.~

·.......... 1';:;(\11'

·r~.

.'I'11c;~:;·.r1

facet of power generation is its relation to our environment. The 9f power by combustion or by nuclear fission by its very nature . :<,10,-;,1\ ~'.""e form of pollution. Fossil- and nuclear-fueled power plants na\'e ., "'''';'C''~ "" land. water, and air resources during the production. processm;, .,,:,;,.,." "~l!J disposal of the products of the generation of energy. Table ... 1 ' C ' T n ' d Jhesc effects. With electrical power usage doubling every 9 years. It ".' ,~".)., 'x Ihat the effect on the environment of the generation of energy , ,·C'·:"-':,' ."~ p)",idcrable concern. 'J t'· '.i:·,.. ),,:~~ publicly acceptable. effective. and yet economical conversion of , : c" .,:,,',,{,; Into energy and to properly utilize the energy so generated. ~ ou ""~,Jnd Ihe basic principles underlying the generation, uses. and trans."" ,.'~ ~'N~rf!Y in its different forms. The ans,!!crs to questions such as. Is ,~n~r:J\e.. ,,1

"f,two:nonidc:t1 ,',,'" ... itfboth:graOl . oJ pa'nd:7:l,,' , A.,B o, eD, a, h.,

, Is'thi:moldr,I,: '~:be.erl :tabui.Jlc.: v. 4(;., no.: L':. r ::~edtlc '."oJ WI''''

percent.

""'""i'k, '<'~ An Ellergy Mode! for the United State", Ihe U.S. Department of the :,' ·.''''''~nt Prinring Office, Washington, D,C., 1968.

253

I

I 1 I i

l

I I 1

, t ) I

''I!

1 ~

~

I

Definitions

Potential Energy

I Kinetic I W rk Energy 0

Heat

I Internal I Enthol py Energy



l

~

~ ,,;

;a~

F

.~

'~:1

Reversible Processes

.- ~ ;

F

.---

I

r

~

,1

FE Mechanical Energy Balance

'i,1

1- F

Eotholpy Changes

I

+

~ Heat of Solution and Mixing

1

Heat of Reaction

,.,

I

I Balance I Techniques Materia!

L4,7-~ Flame Temperature

Material and Energy Balances

-

;

1'. ~,

l

r

Phose Transitions

I

I

"1

1

r

I

. t~~

Chapter 3

Gases, Liquids, Solids

-

t

Chnpi~rJ

Background ' Information

~

t Experimental Dolo

I

<=~,----"-.

t

tf.

Heat Copccity

• ~ I

Chapter?

r [4,4J

I

1=

Energy Balance

l4.2 1

(.:

t

,

Fig. 4.0. Heirarchy of topics to be studied in this chapter (section numbers ~Ht'

j,

.'

,

4.,

in the upp("r kft-hand c.:orncr of the

,; "

i

h(l--:C,",),

"

.

~

r

-;

:' S ,n

.'

...i~

"".ff

Concepts and Units

Sa. 4.1

255

~~-----.-------r------'-------r------.-------'~

1

1 1

I 1

1 \

Yeor

j

Fig. 4.1. Forecast of future energy consumption.

1

thermal pollution inherently necessary, What is the most economic source of f uei, What can be done with waste heat, and other related questions can only arise from an understanding of the treatment of energy transfer by natural processes or machines. I n this chapter we shall discuss energy balances together with the accessory background information needed to understand and apply them correctly. Before laking up the energy balance itself, we shall discuss the types of energy that are ,,[ major interest to engineers and scientists and some of the methods that are u,cd to measure and evaluate these forms of energy. Our main attention will be devoted to heal. work. enthalpy, and internal energy. Next. the energy balance "ill be described and applied to practical and hypothetical problems. Finally, we 'haJJ discuss the generation of energy through the mechanisms of reaction and hil\\" energy generation fits into industrial process calculations. Figure 4.0 shows loe rdationships among the topics discussed in this chapter and also in pre\ious chapters.

·

1 1

i 1 1

4.1 Concepts and units When you think about energy. what sort of image comes to mind? Perhaps it is an electric generator whirring at the bottom of a dam, or perhaps a bullet speed-

A .... ;, 44#'

,.

PF.d:::;;a ,

4$

1M .,3

"

1 I I

~

"" TABLE

4.1

IMPACT OF POWER PLANTS ON THE ENVIRONMENT

Effect on Land Source of Energy Coal

Effect on Water

Production

Processing

Utilization

Production

Processing

Utilization

Production

Processing

Disturbed land,

Solid waste disposal

Ash, slag disposal

Acid mine drainage, silting of

Leaching of waste piles

Increased water temps.

Mine fires

Dust

mme wastes

Oil

Oil spills, brines

Utilization Sulfur oxides, nitrogen

oxides, particulate matter

streams

Pipelines, spilis

i

Effect on Air

Increased water temps.

Carbon monoxide,

nitrogen oxides.

hydrocarbons

Natural gas

Pipelines

Uranium

Disturbed land

Increased water temps. Disposal of radioactive

material

,

1

Disposal of Increased radiowater active

temps.,

material

accidents

Waste /Iaring

I (;

1:

,

~

t ~r

Accidents

f;':

,

Concepts and Units 257

Su.4.1

....o

ing through the air. If it is the latter. you instinctively feel that the bullet in motion is different from one at rest. You know a ball of iron in the sun is in a different state than one in the shade. If a gas confined in a cylinder at high pressure is allowed to expand. it can do work against a rdarding force. and at the same time the temperature of the gas may fall. However. the same change in the properties of the gas can be brought about in other ways which seem to have \ery little to do with gas expansion; for example. the temperature of the gas might be reduced by placing the cylinder on a block of ice. We are able to interpret and explain all these and many other phenomena in terms of the physical quantity energy. In Sec. 4.5 we shall relate different types of energy by means of an energy balance. In this section we shall discuss the units used to express energy and shall define several common types of energy. Unfortunately, many of the terms described below alre used loosely in our ordinary conversation and writing. whereas others are so fundamental to our thinking that they defy exact definition. You may feel thal you understand them from long acquaintance-be sure that you really do. Table 1.1 lists the units used to express energy in va1rious systems; you can tind the conversion factors for these and other units in Appendix A. The calorie is roughly defined as the amount of energy required to raise the temperature of I g of water I' C at a pressure of I atm. Similarly. the Bri:uish thermal unit (Btu) is the amount of energy required to raise I lb of water l' F. However. since the heat capacity of water varies with temperature. it is necessary to be a little more precise in specifying what type of calorie and British thermal unit (Btu) is under consideration (there arc four or five common kinds). The type of calorie that forms the basis of the thermochemical properties of substances in this text is the thermochemical calorie (equal to 4.184 J). A second type of calorie is the llllemaliol/al Steam Table (/.T.) calorie (equal to 4.1867 J). The British thermal unit is defined in terms of the calorie: 1.8 cal I Btu u;:=-gYou can see that these calories differ by less than one part per thousand, so that for calculations made to slide-rule accuracy (but not for p'~ecise measurements) there is hardly any reason for you to consider the diffcJ:'ences between them. Certain tcrms that have been described in earlier chaplers occur repeatedly in this chapter; these terms are summarized below with some elaboration in view of their importance.

E .2 c:

E

::>

(a) System. Any arbitrarily specified mass of material or segment of apparatus to which we would like to devote our attention. A system must be defined by surrounding it with a system boundary. A system enclosed by a boundary which prohibits the transfer of mass across the boundary is termed a closed system, or I/ollflow system. in distinction to an opell system. or flo\\' system, in which the exchange of mass is permitted. All the mass or apparatus external to

--.....- ...,...,...,- - -.....- -....- ....-*. . . . . . . . ."''".,..... - .....,..,..----~ .......-,.~,.,..,.,."....,.~-.~....,--~..----,

1

1 I \

1

i 1 i

l )

I

I

I

I I

--~,

258 Energy Balances

Chap. 4

I I

the defined system is termed the surroundings. Reexamine some of the example problems in Chap. 2 for illustrations of the location of system boundaries. YOll should always draw similar boundaries in the solution of your problems, since this will fix clearly the system and surroundings.

(b) Proper'.\". A characteristic of material which can be measured, such as pressure, volume. or temperature~or calculated, if not directly measured, such as certain types of energy. The properties of a system are dependent on its condition at any given time and not on what has happened to the system in the past. An extensive property (variable, parameter) is one whose value is the sum of the values of each of the subsystems comprising the whole system. For example, a gaseous system can be divided into two subsystems which have volumes or masses different from the original system. Consequently, mass or volume is an extensive property. An intensive property (variable, parameter) is one whose value is not additive and does not vary with the quantity of material in the subsystem. For example, temperature, pressure, density (mass per volume), etc., do not change if the system is sliced in half or if the halves are put together. Two properties are independent of each other if at least one variation of state for the system can be found in which one property varies while the other remains fixed. The set of independent intensive properties necessary and sufficient to fix the state of the system can be ascertained from the phase rule of Sec. 3.7.1. (c) State. Material with a given set of properties at a given time. The state ofa system does not depend on the shape or configuration of the system but only on its intensive properties.

.,

,,', ", ('!. t I -.

't

-'

'j

1"

; ;!

'<..

.

, < ·,t ! r \

.,

,..

'H. B. Callen, Thermodynamics, Wiley, New York, 1960, p. 7.

...

\ ~ !

Next we shall take up some new concepts, although they may not be entirely new, because you perhaps have encountered some of them before. (a) Heat. In a discussion of heat we enter an area in which our everyday use of the term may cause confusion, since we are going to use heat in a very restricted sense when we apply the laws governing energy changes. Heat (Q) is commonly defined as that part of the total energy flow across a system boundary that is caused by a temperature difference between the system and the surroundings. Heat may be exchanged by conduction, convection, or radiation. A more effective general qualitative definition is given by Callen': A macroscopic observation [of a system] is a kind of"hazy" observation which discerns gross features but not fine detail. ... Of the enormous number of atomic coordinates [which can exist). a very few with unique symmetric properties survive the statistical averaging associated with a transition to the macroscopic description [and are macroscopically observable]. Certain of these surviving coordinates arc mechanical in

\ .. \' (

!,

¥,

,

Chap. 4 .:!ine some of the example system boundaries. You , of your problems. since can be measured. such as . directly measured. sllch ::n are dependent on its ~ned to the system in the ·'·hose value is the sum of _,ie system. For example. which have volumes or 1'. mass or volume is an .ClOse value is not additive ~ubsystem. For example • . do not change if the

nature [such as volume). Others are electrical in nature [such as dipole moments] .... It is equally possible to transfer energy to the hidden atomic modes of motion [of the atom] as well as to those modes which happen to be macroscopically observable. An energy transfer to the hidden atomic modes is called heat. To evaluate heat transferquantitatil·e!y. unless given" priori. you must apply the energy balance that is dicussed in Sec. 4.5. and evaluate all the terms except Q. Heat transfer can be estimated for engineering purposes by many empirical relations, which can be found in books treating heat transfer or transport processes. (b) Work. Work (W) is commonly defined as energy transferred between the system and its surroundings by means of a vector force acting through a vector displacement on the system boundaries

where F is the direction of dl. However this definition is not exact inasmuch as (I) The displacement may not be easy to define . (2) The product of F dl does not always result in an equal amount of work. (3) Work can be exchanged without a force acting on the system boundaries (such as through magnetic or electric effects).

one variation of state while th~ other remains ·ary fficient to fix Jase of Sec. 3.7.1.

~ey

may not be entirely , before. 'nich 0 u r everyday use .0 use heat in a very changes. Heat (Q) is )ss a system boundary ~m and the surround· or radiation. A more ,y" observation

the enormous ewwith unique 5sociated with acroscopically mechanical in

JFdl

W=

• 51

a given time. The state , of the system but only

Concepts and Units 259

Src.4.1

Since heat and work are by definition mutually exclusive exchanges of energy between the system and the surroundings. we shall qualitatively classify work as energy that can be transferred to or from a mechanical state. or mode, of the system, while heat is the transfer of cnergy to atomic or molecular states, or modes, which are not macroscopically observable. To measure or evaluate \lork quantitative/y by a mechanical device is difficult. so that, unless one of the energy balances in Sec. 4.5 or 4.6 can be applied. in many instances the value of the work done must be given a priori. (c) Kinetic Energy. Kinetic energy (K) is the energy a system possesses because of its velocity relative to the surroundings. Kinetic energy may be calculated from the relation

K=

tmv'

(4.1)

or

K = tv'

(4.1 a)

where the superscript caret (~) refers to the energy per unit mass (or sometimes per mole) and not the (otal kinetic energy as in Eq. (4.1) (d) Potential Energy. Potential energy (P) is energy the system possesses l>ecause of the body force exerted on its mass by a gravit<:tional field with

(ElM

.M

4

n,a

3 'f .

260

Energy Balances

t

Chap. 4

respect to a reference surface. Potential energy can be calculated from

P

= mgh

(4.2)

lL_

Su.4.1

(el In/ana! '. !noiccu Ia r. ;ll,'" l'l.'::

m1Cr()\C(1pl~

or

P=gh where the symbol per mole).

(~)

(4.2a)

again means potential energy per unit mass (or sometimes

with \\h,,:, 1f1lerna! cncr~\

C'I.t

t;!.h.~f()~COpll'.l i:

\.

To cak"!.l: can he: mC11,UIl,i that It is;In n.l, :

EXAMPLE 4.1

Energy changes

l'<:'dcscrihc:J

'1l

,I",

A lOO-lb ball initially on the top of a 15-ft ladder is dropped and hits the ground. See Fig. £4.1. With reference to the ground, determine the following:

9

hy takin!! the 10"1,

By dcfinition ",

reference

~ymhol

surface

C,. I ,,' .

that the

\C(,",: "

~(qucntly.

d,.I· .' (4.3) as h)\\" . ,

Fig. E4.1.

(a) What is the initial kinetic and potential energy of the ball in (ft)(lb,)?

initial K

=

1mv' = 0 ft

... IP lmtla

=

mg

h

=

100 Ibm gsec' 15 ft ---"'-t-'("ft")("I'b'-m')-+-g'(seci)(lb,)

+1500(ft)(lb,)

(b) What is the final kinetic and potential energy of the ball?

l\ote that Y"'" tntcln,,1 rnrI ,', cncr~:y. IIl\i(.i (1,)11\

\\hH:h

final P = 0 K, - K,

=

im(v~

- t'j)

H'r\' often [':

=0

P, = mg(h2 - hI) = 0 - (+ 1500) = -1500 (ft)(lb,) ViI

A decrease in potential energy has occurred. (d) If all the initial potential energy of the ball were somehow converted into heat, how many Btu would this amount to?

1500 (ft)(lb r )!

1 Btu I 778 (ft)(Ib,)

=

I 93 Bt . u

The system has not been specified in this example. Where would you place the system boundary?

~

.....

1\ ~:l.'"

"111l\ \."I,d";

(c) What is the change in kinetic and potential energy fot the process?

-

':

rnth.oI\·,

(f) In:!.; (.,

final K = 0

P2

'.,

hCft'

I'

1\

I,· . '

t

Chap. 4 . calculated from (4.2) (4.2a) . llIlit mass (or sometimes

c'pped and hits the ground. ~

S~C.

Concepts and Units 261

4.1

(e) Internal Energy. Internal energy (U) is a macroscopic measure of the molecular, atomic, and subatomic energies, all of which follow definite microscopic conservation rules for dynamic systems. Because no instruments c~ist with which to measure internal energy directly on a macroscopic scale, internal energy is calculated from certain other variables that can be measured macroscopically; such as pressure, volume, temperature. and composition. To calculate the internal energy per unit mass (0) from the variables that can be measured, we make use of a special property of internal energy, namely, that it is an exact differential (because it is a poil1l or stale property, a matter to he described shortly) and. for a pure component, can be expressed in terms of the temperature and specific volume alone. If we say that

following:

0= OCT, V) by taking the total derivative, we find that

dO = (ao) dT + (aq) dV aT r av T

(4.3)

By definition (aO/aT),' is the heat capacity at constant volume. given the special symbol C•. For most practical purposes in this text the term (at af')T is so small that the second term on the right-hand side of Eg. (4.3) can be neglected. Consequently, changes in the internal energy can be computed by integrating Eq. (4.3) as follows: C!.IIe bal! in (ft)(Ibf )?

U,-U,= A

A

JT' C.dT

(4.4)

l

I1 1 1 1

T,

Note that you can only calculate differences in internal energy, or calculate the internal energy relative to a reference state, but not absolute values of internal energy. Instead of using Eg. (4.4), internal energy changes are usually calculated from enthalpy values, the next topic of discussion.

ball ?

(f) Enthalpy. In applying the energy balance YOll will encounter a variable which is given the symbol H and the name enthalpy (pronounced en'-thal-py). This variable is defined as the combination of two variables which will appear very often in the energy balance

'r the process?

(4.5)

lOW

converted into heat,

-e would you place the

where p is the pressure and V the volume. [The term enthalpy has replaced the now obsolete terms "heat content" or "total heat," to eliminate any connection whatsoever with heat as defined above in (a).] To calculate the enthalpy per unit mass. we use the property that the enthalpy is an exact differential. For a pure substance, the enthalpy can be expressed in terms of the temperature and pressure (a more convenient varia ble for enthalpy than volume). If we let H

=

H(T,p)

iii

,...'"

---~~'''''--~'-~'''''',",i

"4,.4¥4 .; 4

'_$$

__

1

, !i \i

\ i

-----~.----..:--------=~-.--

262 Energy Balances

Chap. 4

...

..'~

_ _" . •

....._ _ _ . • •_ _ _ _b.- .........

~

S~C.

4.1

by taking the total derivative of H, we can form an expression corresponding to Eq. (4.3):

dH = (aH) aT

dT -':p

(aH) ap

dp

(4.6)

T

T

(aHlaT)

By definition p is the heat capacity at constant pressure, given the speciai symbol Cpo For most practical purposes (aH,ap)T is so small at modest pressures that the second term on the righ t-h and side of Eq. (4.6) can be neglected. Changes in enthalpy can then be calculated by integration of Eq. (4.6) as follows:

(4.7) However, in high-pressure processes the second term on the right-hand side of Eq. (4.6) cannot necessarily be neg'lected, but must be evaluated from experimental data. One property of ideal gases that should be noted is that their en thai pies and internal energies are functions of temperature only and are not influenced by changes in pressure or volume. As with internal energy, enthalpy has no absolute value; only changes in enthalpy can be calculated. Often you will use a reference set of conditions (perhaps implicitly) in computing enthalpy changes. For example. the reference conditions used in the steam tables are liquid water at 32'F and its vapor pressure. This does not mean that the enthalpy is actually zero under these conditions but merely that the enthalpy has arbitrarily been assigned a value of zero at these conditions. In computing enthalpy changes. the reference conditions cancel out as can be seen from the following: initial state of system enthalpy = H, - Ii", Net enthalpy change

final state of system enthalpy = Ii, - Ii",

= (ii, - Ii",) - (Ii, - ii",) = ii, - Ii,

(g) Point, or State. Functions. The variables enthalpy and internal energy are called poinl junctions, or state rariables. which means that in their differential form they are exact dilTerentials. 3 Figure 4.2 illustrates geometrically the concept 3To test if a differential dS is exact, where dS = R dx

+Z

dy

form the partial derivatives (aRjay), and (az/ax)y, and verify whether

of st'lte variable. In pr,', ," shown by the wi!!gly IJ!~" I. ~ny other route, ano ,', ,: " shown by the wiggl) Iinl' ,: final states of the S)\le'''' then at constant tl'mrcr .. that takes place fir't ,11, , I,'ng. as the end POlilt I' . ." I hat of an airplanc r . York but is detoured !. (,)qS him the same" I destination. The !!a'''!J:'" in analogous fashion }-.,' . w~ deal, may vary dq'~':' rq:ardless of path. Ill' '>cw York, the 1'"",10. h,lck. Thus !:J.li 0 II" , . :l!ld back to state I af'."·

All the inten,j,,· 1'" , point function, ,11:d . Ihat we can say. I'M n

Me

If so, dS is exact; if not, dS is not exact. The statement that dS is an exact differential is completely equivalent to saying that S is independent of path. As an example, applying the test to Eq. (4.6) yields iJ

dP or

as expected.

r.

(aM ? a (a!J) aT)p = aT iJp.T

·\I"a)·, keep in "lind!' \..dnd,ltcd hy 1,)!li;tf \tatC'.

t~I~II)::!'

rCf.ndk"

Before f"fllltli.I:' tlel.tllthc cakula!I"" , .

~i.~

'.. __.___________ __ .~

+.~.

___ ,. ___________ • ____ ..

_____ _

Concepts and Units 263

S~C. 4.1

}'tr~onding

_~

to

(4.6) ;

i

~v¢n

the special ;;loot pressures =tC'd.Changes ,Hows:

i l

(4.7) ,t·hand side of Item experi· J j§ Ihat their 'Y and are not J

;lly

change~

in

of conditions , the referen,e 1$ 1

)fCS'

:::~,

.ions .~ of lero at ::e conditions tyslem

~

fi",

fl,

. ternal energy :-Ir differential 'ytheconcept

Flg.4.2. Point function.

of state variable. In proceeding from state I to state, 2, the actual process is shown by the wiggly line. However, you may calculate flH by route A or B. or any other route, and still obtain the same net enthalpy change as for the route ,hown by the wiggly line. The change of enthalpy depends only on the initial and final states of the system. A process that proceeds first at constant pressure and then at constant temperature from J to 2 will yield exactly the same flH as one that takes place first at constant temperature and then at constant pressure as long as the end point is the same. The concept of the point function is the same ;IS that of an airplane passenger who plans to go straight to Chicago from New York but is detoured because of bad weather by way of Cincinnati. His trip ,{)sts him the same whatever way he flies. and he eventually arrives at his dcstination. The gasoline consumption of the plane may vary considerably. and in.an:llogous fashion heat (Q) or work (W). the two "path~ functions with which we deal, may vary depending on the specific path chosen, while flH is the same regardless of path_ If the plane were turned back by bad weather and landed at 'cw York, the passenger might be irate, but at least he could get his money \'>;lc!;. Thus 1:J.H = 0 If a cyclical process is involved which goes from state Ito 2 .1nd ba.c!; to SUite J again. or

fdH=O All the jmensive properties we shall work with, such as P. T. 0, p, if. etc., point functions and depend only on the state of the suhstance of inleresl,so Ih,11 We can say, for example, that

liTe

fdT=O :renlial is COO1mlying the 1•• (

TdO=

0

A!\\'ays keep in mind that the values for a difference in a 'Point function can be "
\ ,I

Il 1 \ I

I I !

1\ ,

..

...

._.----_~--,-.~"O'''..__._~·._.

__ -

-.-...,.~

___ .".._ _ _ _.,..,.",...._ _ _...._ ...,... , -------

lou;,.

264 Energy Balances

Chap. 4

such calculations. The discussion will be initiated with consideration of the heat capacity Cpo

4.2 Heat capacity The two heat capacities have been defined as

Sre.4.2

,-.lpacity consiq, . perature differer.,( The heat el:.J represented figUL" orthe absolute te: is 7ero. As the tc , point is reached. ! arc sho\\ n abo ('r: ,

(a) (b) To give these two quantities some physical meaning, you can think of them as representing the amount of energy required to increase the temperature of a substance by 1 degree, energy which might be provided by heat transfer in certain specialized processes. To determine from experiment values of C p (or C,), the enthalpy (or internal energy) change must first be calculated from the general energy balance, Eq. (4.24), and then the heat capacity evaluated from Eq. (4.7) [or (4.4)J for small enthalpy changes. We can provide a rough illustration of the calculation ofC p using data from the steam tables. We shall take the enthalpies for water vapor at I psia and at 300 0 and 350°F. Clearly the temperature difference is far too large to precisely approximate a derivative, but it should suffice under the selected conditions to give a Cp valid to two significant figures:

Cp

=

AH =

(l217.3 - 1194.4) = 22.9 Btu I lb

(aH) aT

~ = AT

(A H)

p

22.9

p

= 50

Fi~.

4.3. If· I

Btu

= 0.46(1b)(A O F)

In line with the definition of the calorie or Btu presented in the previous section, we can see that the heat capacity can be expressed in various systems of units and still have the same numerical value; for example, heat capacity may be expressed in the units of cal kcal Btu (g mole)(°C) = (kg mole)(,C) = (Ib mole)CF) Alternatively, heat capacity may be in terms of cal Btu (g)CC) = (lb)CF)

or

J

Notc that

\ ~ _I:

I Btu 4.1841 (Ib)CF) = (g)('K)

r.

. . ,', .. : ~

and that the heat capacity of water in the SI system is 4184 J/(kg)CK). These relations are worth memorizing. Note that in each system of units the heat

I

r •'-, : .,

'

__ 1tt_......-...__

Clwp.4

::tion of the heat

.-

....-.-.----

.•..

_--------

Sec. 4.2

Heal Capaciry

265

capacity consists of energy divided by the product of the mass times the temperature difference. The heat capacity over a wide temperature range for a pure substance is represented figuratively in Fig.4.3, where the heat capacity is plotted as a function of the absolute temperature. We see that at zero degrees absolute the heat c3pacity is zero. As the temperature rises. the heat capacity also increases :lntil a certain point is reached at which a phase transition takes place. The phase transitions are shown also on a related poT diagram in Fig. 4.4 for water. The phase transiPhose Transitions

:.hin1c of them as =perature of a -ransfer in certain :" C, (or C,), the ;rom the general ~ed fr"'~ Eq. (4.7) .dta from at 1 psia and at 2rge to precisely :red conditions to

Cp liquid

0

T

0

Gas

j

(OK)

Fig. 4.3. Heat capacity as a function of temperature for a pure substance.

_ USI •. _

c

B--

Ice

~ C>

-8 I I

I I

Vopor

I I

I :i in the previous -arious systems of ,1 capacity may be Cp -

I I I I Gos

V I

I

~

T{OK).

T(OK)

Cp

I I I I I

j

Liquid I

~

Gas

~

T(OK)

I J

Fog. 4.4. Heat apacity and pha,.. transitions.

J/r'°K). These o . the heat

tion may take place between two solid states, or between a solid and a liquid state, or between a solid and a gaseous state, or between a liquid and a gaseolls ,tatc. Figure 43 shows first a transition between solid state I and solid state II, then the transition between solid state II and the liquid state, and finally the transition between the liquid and the gaseolls state. Note that the heat capacity IS a continuolls function only in the region between the phase transitions; con-

1 i

I

i

,

I

I 1

i

I

"ft;

266

Energy Balances

Chap. 4

sequently it is not possible to have a heat capacity equation for a substance which will go from absolute zero up to any desired temperature. What the engineer does is to determine experimentally the heat capacity between the temperatures at which the phase transitions occur. fit the data with an equation. and then stop and determine a new heat capacity equation for the next range of temperatures between the succeeding phase transitions. Experimental evidence indicates that the heat capacity of a substance is not constant with temperature. although at times we may assume that it is constant in order to get approximate results. For the ideal monoatomic gas. of course. the heat capacity at constant pressure is constant even though the temperature varies (see Table 4.2). For typical real gases, see Fig. 4.5: the heat capacities shown are for pure components. For ideal mixtures, the heat capacities of the individual components may be computed separately and each component handled as if it were alone (see Sec. 4.8 for additional details). TABLE

4.2

HEAT CAPACITIES OF IDEAL GASES

05..(.4.2

Most of arc empirica:' i~nction of I nample, or "hcre Ihe t<'ll hcit. degree,

or a form 'I; Kelvin or de Cenligrade ,1 Ihc Icmper .• 11 cr 11l()(k'Lt' Iypes rep,,·,e I ill' (;l'f, pi / "Ii/icd I hl",~

0\

Approximate Heat Capacily (C p) High Temperature Room Temperature (Translational. (Translational Rotational, and Rotational and Vibrational Degrees of Degrees of Freedom) Freedom Only)

Type Molecule Monoatomic PoIyatomic, linear Polyatomic, nonlinear

n R

=

(l'llst;1I1h .'1

IIInc delenll ini"ormaIiP" , prc"u[c, I' I (,rlhe [ckler

!R

jR

(3n - !)R

~R

(31l - 2)R

4R

number of atoms per molecule in Sec. 3.1.

.'il',·(ll, (

= gas constant defined

he.ll cap.I,'I'

I he

(,'(lJ!1n"p

,,,,,,~n"d "

Temperalure, OF _~~~

t

16

~ 15

... !

~

__

14

~~13 ~ OJ 12

./

/'

8G" - ~ 10

,'" 7 ;3 6

-

--1--::::--..-

~ ~ 9 1.1: -I E 8 0_

,...- f/

I

CO 2

I

I

---r-- ' /

i

, ,1-+

I

!

I

i

60

I I

I

!

-.

H2 ~N2' CO

I

I

50

1

I

0: I

i

I

" ,=

\.lfMClty

I': '"

0

~ 0 E ~

40

_'!:.' ~

0

~

! I i

1

30

1\\'11', I ic' '

200 400 600 800 1000 1200 1400 1600 1800 2000 2200 2400 2600 2800 3000 3200 3400 3600

Temperarure, °C til)

Fig. 4.5. Heat capacity curves for the combllstion gases.

\,

.~!:hpu~~h 11;(

I

I

Air

J I

, H2 O

!

,

1

I i !

o

'~\lillcrJ(.d

=MOO~~3600~004~4~~~~~ro

,

(I,'

dpprl)\I'!;,

\

---_._---_. Chop. 4 ~tion

Heat Capacity 267

Sec. 4.2

for a substance What the :apacity between the .::ata with an equation. .:) for the next range of

Most of the equations for the heat capacities of solids. liquids. and gases arc empirical. We usually express the heat capacity at constant pressure C p as a function of temperature in a power series, with constants a, b, c, etc.; for example,

,. of a substance is not .::me that it is constant ~lOmjc gas, of course. .:lUgh the temperature _5; the heat capacities neat capacities of the =d each component

or

~mperature.

~lails).

_t Capacity (Cp)

Room Temperature rr ... -:tnslational otational ~rees of J'reedom Only) ~R ~R

4R

__~~__~-r60

~

---+--+--H-+ 50

"0 E

Cp

= a + bT+ cP

Cp =

a+ bT+

cT-'!

C p = a -:- bT - cT-> or a form such that we are dividing by T. then it is necessary to use degrees Kelvin or degrees Rankine in the heat capacity equations, because if degrees Centigrade or Fahrenheit were to be used. we might be dividing at some point in the temperature range by zero. Since these heat capacity equations are valid only over moderate temperature ranges. it is possible to have equations of different types represent the experimental heat capacity data with almost equal accuracy. The task of fitting heat capacity equations to heat capacity data is greatly simplified these days by the use of digital computers. which can determine the constants of best fit by means of a standard prepared program and at the same time determine how precise the predicted heat capacities are. Heat capacity information can be found in Appendix E. The change of C, with pressure at high pressures is beyond the scope of our work here. Details can be found in several of the references listed at the end of the chapter. Specific heat is a term similar to specific grarily in that it is a ratio of the heat capacity of one substance to the heat capacity of a reference substance.' The common reference substance for solids and liquids is water. which is assigned the heat capacity of 1.0 at about I 7'C. Since the heat capacity of water is approximately unity in the cgs and American engineering systems, the numerical values of specific heats and heat capacities are about the same :dthough their units are not. For example, the units of the ratio of the heat (:lpacity of substance A to that of water are CPA

_

CPR,O -

"'"""

Btu!(ib)(OF) Btuf(JbH,o)('F)

~

40

-+-f-+-H30

l

EXAMPLE 4.2 Heat capacity equation The heat capacity equation for CO: gas is C p = 6.393

!O 3000 3200 3400 3600

.gases.

= a + bT

where the temperature may be expressed in degrees Centigrade, degrees Fahrenheit, degrees Rankine, or degrees Kelvin. If C p is expressed in the form of

w

=1::::~~tj

Cp

+ IO.IOOT

x 10- 3

-

3.405P

X

10- 6


I 1 1

1 I

I 1

ret

268 Energy Balances

-k.,

Chap. 4

with C p expressed in cal/(g mole)CK) and T in oK. Convert this equation into a form so that the heat capacity will be expressed over the entire temperature range in (a) Cal/(g moleWC) with Tin °C and 11 Tin 11 'c. (b) Btu/(Ibmole)(OF) with Tin of and I1Tin 11°F. (c) Cal/(g moleWK) with T in of and I1T in I1'K. (d) J/(kg mo!eWK) with T in oK and I1T in I1°K.

~

d#'stt

S~C.

4.2

and convert it ic: C Btu '(Ib mole)(tl.

+ 8.240

>: 10'

- 3.405 ;" 10'

Solution: Changing a heat capacity equation from one set of units to another is merely a problem in the conversion of units. To avoid confusion in the conversion. you must remember to distinguish between the temperature symbols that represent temperature and the temperature symbols that represent temperature difference even though the same symbol often is used for each concept. In the conversions below we shall distinguish between the temperature and the temperature difference for clarity. (a) The heat capacity equation with T in °C and 11 Tin 11°C is C

cal _ 6 393 cal 11 11oK p (g mole)(11 0c) - . (g mole)(11 OK) 111°C

-

=

= 8.896

+ 10.100

x

I

+ 8.240 x

J



+ 273)'K

+ 273)2°K 2

+ 2.757

- 3.405 x 1O- 6 Ttc - 1.860 x 1O- 3 T. c - 0.254 3 1O- To c - 3.405 x 1O- 6 Ttc IO-Qoc

(d)

'(kg mole)('" I

3405 x 10-6 cal 111 OK (To c . (g mole)(11 °K)eK)2 111°C

6.393

the equation dc\ e

C

01 10 3 cal 1 111 °K (To c . 000 x - (g mole)(11 °K)(OK) 1110C

+1

(e) Since

EXA:\IPLE .1..1

(b)

Show that C.

Bt u I 1 A OK C = 6.393 .,-----,c:.:a~_o=+;.::...c,=.=:=-~,=-::.,.+...,::..,Ll=.== P (Ib mole)(11 OF) (g mole)(11 OK) 1.8 11°F

Solution: The hc.lI (,'1'

ror any gas, - 3.405 - 3.405 x 10- 6 [273 + (T' F - 32)/1.8]2 = 6.393 + 2.575 + 5.61 x lO- l T' F - 0.222 - 0.964 x 1O- 3 T' F - 1.05 x 1O- 6 Ttp = 8.746 + 4.646 x lO- l T. p - 1.05 x 10-'Ttp

For the idea/,r.· •.

Note: In Sec. 1.4 we have developed and made use of the expression T' K

=

273

Ii

+ TOFI~ 32

If the equation for part (a) were available. the solution to part (b) would be simplified by employing it. For example, we use cal Cp(g molc)(I1 0 C) = 8.896 + 8.240 x lO- l To c - 3.405 x 1O- 6 T. c

\{l

t h.1t

Chap. 4 .=uation into a form :.:nure range in

Heat Capacity 269

Sec. 4.2 .lnd convert it into

Btu

C, (I b mol e)( ~ C F)

_ 3.405 . 10-6 X

- another is merely :::::e conversion. you -="'1at represent tern· ::::re difference even conversions below ::rure difference for

8.896 cal

-(g....::.m:.::o.:.le.:.)(,:,~c-:,-=C,-)+--,'-c'--''-'''''::':'::-,+~:':'::,--l--~:'''=''

=

cal 1454 g molesl I Btu 1 I ~cC I[(TF - 3~\I,8F'C2 (g mole)(~cC)('C')i lIb mole 1252 call1.S ~FI = 8.746 + 4.646 x 1O- 3 T'F - 1.05 >: 1O- 6TCF

(c) Since Btu (lb th~

I

cal (g mole)(~'K)

mole)(~ 'F)

equation developed in part (b) gives the heat capacity in the desired units. (d)

lS

j,

I kg . i (T. c

+ 10.1000

+ 273)"K

_ 3405 •

= 2.90

X

cal

_

(g mole)(~'K)(-K)

1 4 ,184

1 cal

J 1 1000 g ! T' K I kg i

4.184 J 1000 g , P I kg I 2 10' + 42.27T'K - 1.425 x 1O- TCK 10-6

cal

(g

X

X

10-3

mole)(-K)2(~'K) I I cal

I I

K

: 10' - 'c

: 1O- 3 T. c - 0.254 EXA!\!PLE 4.3

Show that C p = Cv

. 11 OK .811OF -'3 _.

Heat capacity of the ideal gas

+ R for the ideal monoatomic gas .

Solution: The heat capacity at constant volume is defined as

+ T' F1.8 - 32JoK

(aO) aT

cv =

(a)

f

For any gas, C p

32)/1.8]2

(aB) aT = (ao) aT

For the ideal gas, since

=

[ao +aTa(pVl] + p(ai') aT =

p

p

pV =

(c)

v

RTwe can calculate

R

(d)

p

(b) would be sim-

'0

(b)

0 is a function of temperature only, =

~nd from

p

p

(~~). (~~)r = C :nression

[ao aT + pav]

= p

I

that

C.

1 j

= c, + R

1

I, ....

.I:

270 Energy Balances

Chap. 4

4.2.1 Estimation of Heat Capacities. We now mention a few ways by which to estimate heat capacities of solids, liquids and gases. For the most accurate results, you should employ actual experminental heat capacity data or equations derived from such data in your calculations However. if experimental data are not available, there are a number of equations that you may use which give estimates of values for the heat capacities. (a) Solids. Only very rough approximations of solid heat capacities can be made. Kopp's rule (1864) should only be used as a last resort when experimental data cannot be located or new experiments carried out. Kopp's rule states that at room temperature the sum of the heat capacities of the individual elements is approximately equal to the heat capacity of a solid compound. For elements below potassium, numbers have been assigned from experimental data for the heat capacity for each element as shown in Table 4.3. For liquids Kopp's rule can be applied with a modified series of val:.Jes for the various elements, as shown in Table 4.3 TABLE

4.3

VALUES FOR MODIFIED

Kopp's

.hr!i:

:\

RULE

Atomic Heat Capacity at 20'C in cal/(g atom)('C) Element C H B

Si

°or S F

P All others

EXAMPLE 4.4

Solids

Liquids

1.8 2.3 2.7 3.8 4.0 5.0 5.4

2.8 4.3 4.7 5.8 6.0 7.0 7.4 8.0

6.2

Use of Kopp's rule Basis: 1 g mole Na,SO. ·IOH,O 6.2 5.4 4.0 2.3

= = =

=

12.4 cal/(g atom)(,C) 5.4 56.0 46.0 119.8 cal/(g mole)('C)

The experimental value is about 141 cal!(g mole)(,C), so you can see that Kopp's rule does not yield very accurate estimates .

--------,......

.

~~~----,.--~--

(1 ;.1

(

Determine the heat capacity at room temperature of Na,SO. ·10H,O. Na 2 x Six o 14 x H 20 x

.hr:,' \"

:

. '(.

f

:'

---".,. Chap. 4 :=lention a few ways by gases. For the most _ neat capacity data or ',\'ever, if experimental ~2.t you may use which

lleat capacities can be =n when experimental '"p's rule states that at :ndividual elements i, :::>ound. For elements ::7imental data for the :..:Juids Kopp's rule can ;::iements, as shown in

T)

::s

Heat Capacity 271

Src.4.2 (b) Liquids

(I) Aqueous solutions. For the special but very important case of aqueous 'olutions, a rough rule in the absence of experimental data is to me the heat '.lrJcity of the water only. For example, a 21.6 percent solution of NaCI is assumed to have a heat capacity of 0.784 cal/(g)('C); the experimental value at 25T is 0.806 calj(g)('C).

(2) Hydrocarbons. An equation for the heat capacity of liquid hydrocarbons ;md petroleum products recommended by Fallon and Watson' is C,

+ 0.128

= [(0.355 X

where 0 API

=

1O- t)[0.05K 3

sp gr

x 10- 2 'API)

+

+ (0.503 + 0.117

x 10- 2 CAPI)

0.41)

(~614;j600F)

- 131.5 and is a measure of specific gravity.

t = OF K = The Universal Oil Products characterization factor which has been related to six easily applied laboratory tests. This factor is not a fundamental characteristic but is easy to determine experimentally. Values of Krange from 10.1 to 13.0 and are discussed in Appendix K. C, = Btuj{lbWF)

(3) Organic liquids. A simple and reasonably accurate relation between C, and molecular weight is

C,

=

kMa

where M is the molecular weight and k and a are constants. Pachaiyappan ct aI.' give a number of values of the set (k, a). For example,

compounds Alcohols Acids Ketones Esters Hydrocarbons, aliphatic

,sO.·JOH 2 0.

k 0.85 0.91 0.587 0.60 0.873

a -0.1 -0.152 -0.0135 -0.0573 -0.113

Gold' reviews a number of estimation methods for liquids and indicates their precision, (e) Gases and Vapors (I) Petroleum rapors. The heat capacity of petroleum vapors can be estimated from'

C , :m can see that Kopp's

1

=

(4.0 -

sliT + 670)

6450

'J. F. Fallon and K. M. Watson, NaIl. Pelrol. News, Tech. Sec, (June 7, 1944). 'V. Pachaiyappan. S. H. Ibrahim. and N, R. Kuloor. Chern. Eng., p. 241 (OCI 9, 1967). 'r. I.. Gold. Chen!. Eng., p. 130 (April 7, 1969), 'B.hlke and Kay, Ind. Eng. Chern., v. 21, p. 942 (1929).

II ! 1

II \

272

Energy Balances

Chap. 4

where C p is in Btu/(lb)CF), Tis in of, and s is the sp gr at 60°F/60°F. with air as the reference gas.

\.-,



-I.

(~~:

l;'

j",

(2) Kothari and Doraiswamy9 recommend plotting

Cp

.\

= a + b log,o T,

Given two values of Cp at known temperatures, values of C p can be predicted at other temperatUles with good precision. The most accurate methods of estimating heat capacities of vapors are those of Dobratz 'O based on spectroscopic data and generalized correiations based on reduced properties I I. Additional methods of estimating solid and liquid heat capacities may be found in Reid and Sherwood, I I who compare various techniques we do not have the space to discuss and make recommendations as to their use.

/

,

I( "

.

4.3 Calculation of enthalpy changes 1 hen

without change of phase Now that we have examined sources for heat capacity values, we can turn to the details of the calculation of enthalpy changes. If we omit for the moment consideration of phase changes and examine solely the problem of how to calculate enthalpy changes from heat capacity data, we can see that in general if we use Eq. (4.7), tiff is the area under the curve in Fig. 4.6:

H. dH = tiH = fT. C

J

HI

A

A

Tt

p

dT

II

1\ ,,, "~I

"h'le (,.

~~~

','1

.......-.-. (. II

I.,'

,'

I~

~~

1101

", \

~tlO

•I

lOil ':(~1

.,

~(k)

, !

.~

, ,

(..( It I

rOd M~l II( ~)

Fig. 4.6. Calculation of enthalpy change.

I (k).j) 111"110,)

9M. S. Kothari and L. K. Doraiswamy. Hydrocarbon Processing and Perroleum Refiner. v. 43, no. 3, p. 133 (1964). lOCo J. Dobratz, Ind. Eng. Chern .• v. 33, p. 759 (1941). I 'R. C. Reid and T. K. Sherwood, The Properties o/Gases and Liquids. 2nd ed .• McGraw· Hill, New York, 1966.

J \~ "'.)

I.:"' ~_ I

,

~ '1i$i@JI.>t%.AiJ!

BA;a; "

a

,,%

J':;W:;A

.(t;t

_J

'"

'';1'" I', ~ I

"'"

•• I

I'

-Chap. 4

,O°F, with air as

----------.---------Calculation of Enthalpy Changes withollt Change of Phase

Sri". 4.3

(The technique of graphical integration is illustrated in Example 6.6.) If our heat capacity is expressed in the form C p = a + bT + cP, then

I:J.H =

f'

(a

+ bT +

T,

:] be predicted at :nethods of est i:1n spectroscopic

" :m3cities may be ., we do not have

cP)dT = a(T2 - T I)

1

b ' - TD + 2(n

I

(4.8)

+.£.m 3

JI

TI)

I

lfa different functional form of the heat capacity is available, the integration can ,till be handled readily. It is also possible to define a mean heat capacity, which, if known, provides .l quick and convenient way of calculating enthalpy changes. The mean heat capacity, C." is defined as the enthalpy change divided by the temperature difference for this change, or

C

_ H2 - HI T2 - TI

J

(4.9)

Po -

Then it is possible to compute an enthalpy change, if C., is available, as

I:J.H = C.,I:J.T = C• .(T2 - T I)

/e car ''1 to the It con:ne now ,_ ~alculaie l!eneral if we use

TABLE 4.4a MEAN HEAT CAPACITIES OF COMBUSTION GASES [cal!(g mole)CC)] (To convert to Ji(kg mole)CK), mUltiply by 4184.) Reference Temperature: OC; Pressure: I atm

0 IS

2nd ed., McGraw·

(4.10)

where C., is between TI and T 2'

'C

i Petrolellm Refiner.

J.

273

25 100 200 300 400 500 600 700 800 900 1000 1100 1200 1300 1400 1500

N2

O2

Air

H2

CO

CO2

H 2O

6.959 6.960 6.960 6.965 6.985 7.023 7.075 7.138 7.207 7.277 7.350 7.420 7.482 7.551 7.610 7.665 7.718 7.769

6.989 6.998 7.002 7.057 7.154 7.275 7.380 7.489 7.591 7.684 7.768 7.845 7.916 7.980 8.039 8.094 8.146 8.192

6.946 6.949 6.949 6.965 7.001 7.054 7.118 7.190 7.266 7.340 7.414 7.485 7.549 7.616 7.674 7.729 7.781 7.830

6.838 6.858 6.864 6.926 6.955 6.967 6.983 6.998 7.015 7.036 7.062 7.093 7.128 7.165 7.205 7.227 7.260 7.296

6.960 6.961 6.962 6.973 7.050 7.057 7.120 7.196 7.273 7.351 7.428 7.501 7.570 7.635 7.688 7.752 7.805 7.855

8.595 8.706 8.716 9.122 9.590 10.003 10.360 10.680 10.965 11.221 11.451 11.68 11.85 12.02 12.17 12.32 12.44 12.56

8.001 8.009 8.012 8.061 8.150 8.256 8.377 8.507 8.644 8.785 8.928 9.070 9.210 9.348 9,482 9.613 9.740 9.86

I J 1

I

.,

I, j I

.CleRCE: Page 30 of Reference 12 in Table 4.6.

,

I i

~

274

Energy Balallces

then

Chap. 4

If the heat capacity can Cpo is

be expressed as a power series.

f' CpdT = f' T~TJ

C p ., =

+ bT + cT'. The ,:

(a

+

bT

+ cP)dT

:...!.-TL'- - - - - - - _

. t,dT _ a(T2 -

Cp = a

Sre. 4.3

-

T 2 -T, T,)

-+- (bI2leT; -

(4.11)

Tn -+- (cI3)eTl - Tn

T2 - T,

By choosing O°F or DoC as the reference temperature for Cpo' you can simplify the expression for Cpo' Tahles 4.4a and 4.4b lists mean heat capacities for combustion gases based on the same reference point

(O'C) for two different tempera-

ture scales.

As 'I>

/:;.Ii"" .. TABLE 4.4b MEAN HEAT CAPACITIES OF COMBUSTION GASES [Btu/(Ib molelCF)] Reference Temperature: 32'F; Pressure: I atm

and

OF

N,

O2

Air

H2

CO

CO 2

H 2O

32 60 77 100 200 300 400 500 600 700 800 900 1000 1100 1200 1300 1400 1500 1600 1700 1800 1900 2000

'6.959 6.960 6.960 6.961 6.964 6.970 6.984 7.002 7.026 7.055 7.087 7.122 7.158 7.197 7.236 7.277 7.317 7.356 7.395 7.43:1 7.471 7.507 7.542

6.989 6.996 7.002 7.010 7.052 7.102 7.159 7.220 7.283 7.347 7.409 7.470 7.529 7.584 7.637 7.688 7.736 7.781 7.824 7.865 7.904 7.941 7.976

6.946 6.948 6.949 6.952 6.963 6.978 7.001 7.028 7.060 7.096 7.134 7.174 7.214 7.256 7.298 7.341 7.382 7.422 7.461 7.500 7.537 7.573 7.608

6.838 6.855 6.864 6.876 6.921 6.936 6.942 6.961 6.964 6.978 6.981 6.984 6.989 6.993 7.004 7.013 7.032 7.054 7.061 7.073 7.081 7.093 7.114

6.960 6.961' 6.962 6.964 6.972 6.987 7.007 7.033 7.065 7.101 7.140 7.182 7.224 7.268 7.312 7.355 7.398 7.439 7.480 7.519 7.558 7.595 7.631

8.595 8.682 8.715 8.793 9.091 9.362 9.612 9.844 10.060 10.262 10.450 10.626 10.792 10.948 11.094 11.232 11.362 11.484 11.(.0 11.71 11.81 11.91 12.01

8.001 8.008 8.012 8.019 8.055 8.101 8.154 8.210 8.274 8.341 8.411 8.484 8.558 8.634 8.712 8.790 8.870 8.950 9.029 9.107 9.185 9.263 9.339

SOURCE: Page 31 of Reference 12 in Table 4.6.

EXAMPLE 4.5

Calculation of !Hi using mean heat capacity

Calculate the enthalpy change for I Ib mole of nitrogen (N,) which is heated at constant pressure (I atm) from 60 F to 2000 F.

nol

C,

'J hi' fl

cr .it\'~ ...

: '"

fllO"'! bl.~ (~.

mlln!>.-!;'.II

Wh.,t " !I

, !

(

"

,"

G ,-W".

Cp = a

Chap. 4

I

+ bT + cT',

t

(4.1l )

~p.,

you can simplify

:21 capacities for comTWO

different tempera-

:>N GASES

j

!

I f

'M

l'tn

Calculation of Enthalpy Changes withollt Change of Phase 275

Sec. 4.3

Solution: The enthalpy data can be taken from Table 4.4b; reference conditions, 32'F:

C,m = 7.542 Btu!(Ib molc)(,F) C pm = 6.960 Btu!(Ib mole)(,F)

Basis: lIb mole of N, AB,000-60 = AB,ooo - A B 60 = 7.542 Btu 1(2000 - 32'F) _ 6.960 Btu 1(60 - 32'F) (Ib mole)('F) (lb mole)(-F) = 14,740 Btu/lb mole

j

\

As shown above, if the reference conditions are not zero (OOK, O'F etc.), the A8,ooo-60 is C,.....(T2000 - T",) - C p ...(T60 - T",)

~

and not

1

Cp.ttnT2000 -

C p _ ea T60 .

. :m

EXAMPLE 4.6 C02

H 20 8.001 8.008

.S

8.012

8.793 9.091 9.362 9.612 9.844 10.060 10.262 10.450 10.626 10.792 10.948 11.094 11.232 11.362 11.484 11.60 11.71 11.81 11.91 12.01

8.019 8.055 8.101 8.154 8.210 8.274 8.341 8.411 8.484 8.558 8.634 8.712 8.790 8.870 8.950 9.029 9.107 9.185 9.263 9.339

1

Calculation of AH using heat capacity equations

CO 2 CO O2 N2

;

1

9.2 L5 7.3 82.0 100.0

j

What is the enthalpy difference for this gas between the bottom and the top of the stack if the temperature at the bottom of the stack is 550°F and the temperature at the top is 200'F? Ignore the water vapor in the gas. Because these are ideal gases, you can neglect any energy effects resulting from the mixing of the gaseous components.

1

Sollltion: Heat capacity equations from Table E.I in Appendix E (T in OF; C, mole)('F» are

j

+ 0.7624 x 1O- 3 T - 0.7009 = 7.104 + 0.7851 x 1O- 3 T - 0.5528 = 8.448 + 5.757 x 1O- 3 T - 21.59 x = 6.865 + 0.8024 x 1O- 3 T - 0.7367

N2 :

C p = 6.895

x IO-'P

02:

Cp

x IO-'P

CO 2 :

Cp

CO:

Cp

1O- 7 P

+ 3.049

=

Btu/(Ib

j

x 1O- lo P

x 1O-'T2

11,' multiplying these equations by the respective mole fraction of each component, and then adding them together, you can save time in the integration.

+ 0.7624

N,:

0.82(6.895

02:

0.073(7.104

+

;-: 10-] T - 0.7009

X

10- 7 P)

0.7851 x 1O- T - 0.5528 x IO-'P) 3

"

I

Basis: 1.00 Ib mole gas

) which is heated at

)

The conversion of solid wastes to innocuous gases can be accomplished in incinerators in an environmentally acceptable fashion. However, the hot exhaust gases must be cooled or diluted with air. An economic feasibility study indicates that solid municipal waste can be burned to a gas of the following composition (on a dry basis):

I

t

i l

" 216 Energy Balances

Chap. 4

CO,:

0.092(8.448

+ 5.757 x IO- J T - 21.59 x 1O- 7 Tz + 3.059 x

CO:

0.015(6.865

+ 0.8024 x

CPO ..

=

7.049

AH = J'OO C

+ 1.2243

A

5.50

p

IO- J T - 0.7367

x IO- J T - 2.6164 x 1O- 1 P

dT = f'oO (7.049 + 1.2243 ,SO

.

+

'O P) l

+ 0.2815

X

1O- lo P

= -2465 - 160.6

+

10

f l' r

[(200)4 _ (550)4]

(l \ ';.

f'\.11 h \ ~\ ,_

Hlnl

j \l U

We shall now briefly discuss the use of enthalpy tables and charts to calculate enthalpy changes. The simplest and quickest method of computing enthalpy 4.5a

l.l ~l 1

fat h.I:;'\

ENTHALPIES OF COMBUSTION GASES

(Btu/lb mole) Pressure: I atm

oR 492 500 520 537 600 700 800 900 1000

1100 1200 1300 1400 1500 1600 1700 1800 1900 2000 2100 2200 2300 2400 2500

N, 0.0 55.67 194.9 313.2 751.9 1450 2150 2852 3565 4285 5005 5741 6495 7231 8004 8774 9539 10335 11127 11927 12730 13540 14350 15170

SOURCE:

0, 0.0 55.93 195.9 315.1 758.8 1471 2194 2931 3680 443 5219 6007 6804 6712 8427 9251 10081 10918 11760 12610 13460 14320 15180 16040

t

In I{r!,' "t' I "l1d,")

= -2612 Btu/lb mole gas

TABLE

:~

\.("c:\j';"r

7

10-

oS hI;

(.tr.i .. I', \

_ 2.6164 x 10- [(200), _ (550)'] 3 X

\ 41 r,l~ l' \

d.ILI I,:

1O- IO P)dT

1.2243 x IO- [(200)' - (55W] 2

4 13.8 - 0.633

\ "j

! '. ~,) ('\ .;,

x lO- l T - 2.6164 x 1O-1 P J

+ 0.2815

!un)~( '.

~.... ", L.

+ 0.2815 x

= 7.049[(200) - (550)]

IO-

x 1O- 7 P)

Air

H,

0.0 55.57 194.6 312.7 751.2 1450 2153 2861 3579 4306 5035 5780 6540 7289 8068 8847 9623 10425 11224 12030 12840 13660 14480 15300

Page 30 of Reference 12 in

Table

0.0 57.74 191.9 308.9 744.4 1433 2122 2825 3511 4210 4917 5630 6369 7069 7789 8499 9219 9942 10689 11615 12160 12890 13650 14400 4.6.

CO

0.0 55.68 194.9 313.3

752.4 1451 2154 2863 3580 4304 5038 5783 6536 7299 8012 8853 9643 10440 J.J243 12050 12870 13690 1452IJ 15350

CO,

0.0 68.95 243.1 392.2 963 1914 2915 3961 5046 6167 7320 8502 9710 10942 12200 13470 14760 16070 17390

187}0 :!OO70

21430 22800 24180

,

0.0 64.02 224.2 360.5 867.5 1679 2501 3336 4184 5047 5925 6819 7730 8657 9602 10562 11540 12530 13550 14570 15610 16660 17730 18810

.

·-.., :.

H,O

~

·,

,, I,' ·

,

·, ;

,

.

tl

...

j'

rart t .

Chap. 4

+3.059 x to-lOP) -2)

.28t5 x to-lOP ,64 x to-'P 515

x to-lOP) dT

')'] 'J)')

:50)4]

c" and charts to calcu, .: computing enthalp:

.~--------------------------------------------------.--------

Calculation of Enthalpy Changes without Change of Phase 277

Src.4.3

changes is to use tabulated enthalpy data available in the literature or in reference t>ooks. Thus, rather than using integrated heat capacity equations or mean heat .:apacity data, you should first look to see if you can find enthalpy data listed in tahles as a function of temperature. Tables 4ja and 4.5b list typical enthalpy dJta for the combustion gases. Some sources of enthalpy data ar~ listed in Table .t6. The most common source of enthalpy data for water is the steam tables; ~cc Appendix Cl2 The reference by Kobe and associates (Table 4.6, 12) lists heat cJpacity equation, enthalpy values, and many other thermodynamic functions for over 100 compounds of commercial importance. Some of this information c;Jn be found in the Appendix. Computer tapes can be purchased providing information on the physical properties of large numbers of compounds as described in Reference 13 in Table 4.6, or from the Institution of Chemical Engineers, London. Remembering that enthalpy values are all relative to some reference state, you can make enthalpy difference calculations merely by subtracting the final enthalpy from the initial enthalpy for any two sets of conditions. Of course, if

TABLE

4.5b

~:::s

'K

CO, 0.0

68.95 243.1 392.2 963 1914 2915 3961 5046 6167 7320 8502 9710 10942 12200 13470 14760 16070 17390 18730 20070 21430

273 291 298

H 2O 0.0 64.02 22-1.2 360.5 867.5 1679 2501 3336 4184 5047 5925 6819 7730 8657 9602 10562 11540 12530 13550 14570 15610 16660 17730 18810

300 400 500

£'00

;00 800 "<)()

I(.)()

1100 I~OO

11110 ~

.'00 1<(Xl

17<0 >\'~ :~f;O

=("I)() :~~(l

'-'10 i(\\O

n10

N,

0,

Cmmusnos (cal./g mole)

ENTHALPIES OF

Air

H2

0.0

0.0

0.0

0.0

0.0

125.3 174.0 187.9 883.2 1588 2301 3024 3766 4532 5299 6088 6888 7700 8518 9356 11458 13600 15770 17940 20140 22350 26800 31280

126.0 175.0 189.0 896.9 1628 2383 3161 3959 4773 5601 6439 7288 8145 9009 9880 12083 14320 16600 18910 21250 23620 28450 33380

125.1 173.7 187.6 883.6 1592 2312 3044 3795 4569 5346 6142 6950 7768 8593 9434 11550 13700 15880 18080 20300 22530 27040 31590

123.4 171.6 182.5 873.6 1575 2275 2978 3684 4394 5112 5838 6575 7320 8076 8842 10821 12830 14900 17020 19190

125.5 174.1 188.0 884.2 1590 2310 3047 3800 4571 5357 6157 6968 7790 8621 9459 11582 13740 15910 18110

21380

25820 30480

203:!O

22530 27000 31500

I j

CO,

0.0 156.7 217.9 235.7 1172 2200 3300 4459 5667 6916 8200 9513 10852 12211 13590 14980 185\0 Z:!l00

25750 29-140 33150 36890 44430 52050

H 2O 0.0

144.2 200.3 216.4 1024 1853 2707 3589 4499 5440 6411 7412 8440 9494 10573 11675 14520 17480 20520 13630 26790 30000 36520 43120

~ I '~ee R. W. Haywood, Thermod},namic Tahles in SI C\ferric) Units, Cambridge University """'. london, 1968, for steam tables in metric units .

..

1

,j

GASES

CO

I

;!,

1

Il

!j

'j

Il

.

278

Energy Balallces

t

Chap. 4

the enthalpy data are not available, you will have to rely on calculations involving heat capacities or mean heat capacities. If a pure material exists in solid, liquid, and gaseous form, a chart or diagram can be prepared showing its properties if sufficient reliable experimental information is available. Examples of such charts are shown in Fig. 4.7 and in Appendix J; sources of such charts are listed in Table 4.7. Many forms of charts TABLE

4.6

..

SOURCES OF ENTHALPY DATA

1. American Petroleum Institute, Division of Refining, Technical Data Book-Petroleum Refining, 2nd cd., API, \Vashington D.C.. 1970. 2. American Petroleum Institute. Research Project 44, Selected Values of Physical alld Thermodynamic Properties of Hydrocarbons and Related Compounds, API, Washington, D.C., 1953. 3. Chermin, H. A. G., Hydrocarbon Processing & Petroleum Refining, Parts 26-32, 1961. (Continuation of Kobe compendium for gases; see item 12 below.) 4. Egloff, G., Physical COllstants of Hydrocarbons. 5 vols., Van Nostrand Reinhold, New York. 1939-1953. 5. Hamblin, F. D., Abridged Thermodynamic and Thermochemical Tables (5.1. Units), Per· gamon Press, Elmsford, N.Y. (paperback edition available). 1972. 6. Hilsenrath, 1., and others. "Tables of Thermal Properties of Gases," SGtional Bureau of Stalldards (U.S.). Circ. 564, 1955. (Additional single sheets issued as ;>;BS-NACA sponsored tables.) 7. International Atomic Energy Agency, Thermodynamics of Nue/ear Afaterials, Vienna. Austria, 1962. 8. Journal of PhYSical and Chemical Reference Data, American Chemical Society, 1972 and following. 9. Keenan, 1. H., and F. G. Keyes, Thermodynamic Properties of Steam, Wiley, New York, 1936. 10. Keenan, 1. H., and 1. Kay, Gas Tahles. Wiley, New York, 1948. II. Kelley, K. K., various U.S. Bureall of .1 fines Bulletins on the Thermodynamlc Properties of Substances, particularly minerals; 1935-1962. 12. Kobe, K. A .. et al. "Thermochemistry of Petrochemicals." Reprint no. 44 from the Petroleum Refiner. A collection of 0 sefles of 24 articles covering 105 ditTerent substances that appeared in the Petrolellm Rejlncr. 13. Meadows, E. L.. "Estimating Physical Properties," Chem. Eng. Progr., v. 61. p. 93 (1965). (A.I.Ch.E. Machine Computation Committee report on the preparation of thermodynamic tables on tapes.) 14. Perry, J. H., Chemical Engineers' Handbook. 4th cd .. McGraw-Hili, New York, 1964. 15. Ross, L. W .. Chem. Eng .. p. 205 (Oct. 10. 19(6). 16. Rossini. F. K .• and others, "Tables of Selected Values of Chemical Thermodynamic Properties," .~·alional Bureau of Standard, (C.S.). Circ. 500. 1952. (Revisions are being issued periodically under the Technical Note 270 series by other authors), 17. Sage, B. H .• and W. N. Lacey. Some Properties of the Lillhter Hydroearoons. Hydrogen Sulfide, and Carbon Dioxide. American Petroleum Institute, ]\;ew York. 1955. 18. Stull. D. R .. and H. Proplict. JAXAF Thanloc/rcmicai Tah/e5, 2nd cd., Gl)\I.:rnmcnt Printing Office No. 0303-0872, CI3, 48: 37 (AD-732 043). 1971. (Data for over 1000 compounds.) 19. Zeise, H., Thermodmamik. Band III [ Toballen. S. Hirzel Verlag. Leirzig. 1954. (Tabb of heat capacities. entha!pies. entropies, free energies, and equilibrium constants.)

~,

..

.... '

~------~.--

........------""~--~'----.

__--,--,---- ---------......

..

.-

Chap. 4

-=.alculations involva chart or diaexperimental in Fig. 4.7 and in .=y forms of charts -::TIl,

~able

=

!

-Jl1ra Book-Petroleum

I

1

.""s of Physical and Ther· 31, Washington, D.C., .:-:1!.

1 u c

ParIs 26-32, 1961-

" :5 i"

.0

:strand Reinhold, l"cw

~

.;::

..:zi>/es (S.I. Units), Per· ~s.'"

:S5U

';~ma/

Bureau 'B$-J-;ACA

,~ m ~

,,;:, Q.

.::ar kIQt~rjals,

Vienna,

:=aI Society, 1972 and

'0 .c

C

w

E

" ;;;,

.::!

""

1• 1j

;., 0.

.:::

C

"" " ~

" ~

0..

,..: oj

=odynamic Properties

l

';;i

~ ~

:=n, Wiley, New York.

1

oil

~

:c:1nt no. 44 from the :~':5 different substances

1

l

1·, j

I I

! ,1 •

" ..., v. Ill, p. 93 (1965). =ron oftherrnodynamic

I

.. '. New York, 1964.

I

::oical Thermodynamic _. (Revisions are being authors). ·-drocarbons. Hydrogen York, 1955. :'2.-.- Government Print-2 for over 1000 C('m-

i.eipzig. 1954. (Tables :;urn· (;onstanrs.)

279

Wi

Sa. 4 ..1 4.7 THERMODYNAMIC CHARTS SHOWI"G E",HALPY DATA FOR PVRE CO"POlJ"DS [For mixtures, see V. F. Lesavage et al. Ind. Eng. Chem., v. 59, no. II, p. 35 (1967).] TABLE

Compound Acetone Acetylene Air Benzene J, 3-Butadiene i-Butane n-Butane n-Butanol

I

I

-- ! I

Butanol, tert. n-Butene Chlorine Ethane Ethanol Ethyl ether Ethylene Ethylene oxide n-Heptane n-Hexane Hydrogen sulfide Isopropyl ether Mercury Methane Methanol Methyl ethyl ketone Monomethyl hydrazine Neon Nitrogen n-Pentane Propane n-Propanol Propylene Refrigerant 245 Sulfur dioxide Combustion gases

Hydrocarbons

Reference 2 V. C. Williams, Am. Inst. Chem. Engrs. Trans., v. 39, p. 93 (1943); A.I. Ch. E. J., v. I, p. 302 (1955).

1

a (/;"r: he pint ted ~

C. H. Meyers, J. Res. Natl. Bur. Std., A., v. 39, p. 507 (1947).

c\amplc,I':'

1,3 1,3,4 L. W. Shemilt, in Proceedings 0/ the Conference on Thermodrnamic Transport Properties 0/ Fluids, London, 1957, Institute of Mechanical Engineers London (1958). F. Maslan, A.I.Ch.E. J., v. 7, p. 172 (1961). 1 R. E. Hulme and A. B. Tilman, Chem. Eng. (Jan. 1949). 1,3,4 R. C. Reid and J. M. Smith, Chern. Eng. Progr., v. 47, p. 415 (1951). 2 1,3 J. E. Mock and J. M. Smith. Ind. Eng. Chern., v. 42, p. 2125 (1950). E. B. Stuart, et aI., Chern. Eng. Progr .• v. 46, p. 311 (1950).

tcmpcr
anyt\\0I11 II

1

he seen

SinCC

cnthalp\' ". Jrd",n. YOU ... :.il

to specify I '.'f

fix thc

.
I

Ihal Spc,d"'fJ

pcrlic\

r,

"I:)

ones. u\u.IL"

1

prc"urcl: :'" rntC'rn~JI

en," , til

h~

,\ltrh1'!

':t

J. R. West, Chern. Eng. Progr., v. 44, p. 287 (1948). 2 General Electric Co. Report GET-1879A, 1949. 1,3,4 J. M. Smith, Chern. Eng. Progr., v. 44, p. 52 (1948).

\\ III find II

2

w;tnl.

amon~

the

~tll"t"nrr. 1'1 l..

,

[nth.I':'.c

F. Bizjak, and D. F. Stai, A.I.A.A. J., v. 2, p. 954 (1964).

alilcd met J", 1 r,Jlltrthuth':

Cryogenic Data Center, National Burean of Stanaards. Boulder. Colo. G. S. Lin, Chern. Eng. Progr., v. 59, no. II, p. 69 (1963). 1,3,4 1,3.4 L. W. Shemilt, (see ref. for butanol). 1,3 R. L. Shank, J. Chern. En/? Data. v. 12. p. 474 (1967). J. R. West. and G. P. Gillsti.i. Phl"S. CollaidChem .. v. 54. p. 601 (1950). H. C. Hottel, G. C. Williams. and C. N. Satterfield. Thermod!,nonll( Charts for Combustion Processes (I) Text. (II) Charts, Wiley, New York. 1949. J

Res',ll,

",I(

.\·"/U//l'" r,'r tl:r '_.l-"j .tt

L. N. Cajar, et aI., Thermod"namic Properties and Reduced Correlations/or Gases. Gulf Publishing Co .. Houston. Texas. 1967. (Series of articles "hich appeared in tho magazine }{ydrocarbon Processill{? from J 962 to 1965.) 2. P. T. Eubank and J. \1. Smith, i. Chem. Eng. Data. v. 7. p. 75 (1962). 3. W. C. Edmister, Applied H"drocarbon Thcrmodl'llamics, Gulf Publishing Co .. Houston. Texas, 1961. 4. K. E. Starling et aI., }f,'drocarbon Processinf(, 1971 and following; to be published as a book. Charts available from Gulf Publishing Co., Houston, Texas, separately.

:1 k.'

I.

1 '.\ 1\ ",.

• 4;1 , .~ ( : "

,-ora.

'(.to I

I

"

"')

!

: 1 "J4.,' •

280

..

,._- _---...

f

-------------------------~-

Calculation of Enthalpy Changes without Change of Phase 281

Ser.43 !'UllE COMPOU"OS

are a\'ailable. Some types that may occur to you would have as coordinate axes

. p. 35 (1967).]

pvs.H p vs. J:' pvs. T HVS.SI3

39, p. 93 (1943); A.I.

307 (1947).

. -p

on Thermodynamic

-,,-ritute of Mechanical

1949). .47, p. 415 (1951).

:.2. p -

.~

(1950).

I (

- (1964). =rds, Boulder, Colo. (1963).

Since a chart has only two dimensions. the other parameters of interest have to he plotted as lines of constant value across the face of the chart. Recall, for example, that in the pJ". diagram for CO" Figs. 3.2 and 3.24, lines of C0nstant temperature were shown as parameters. Similarly, on a chart with pressure and enthalpy as the axes, lines of constant volume and/or temperature might be drawn. You should have noted for pure gases in the illustrated problems that we had to specify two physical properties such as temperature and pressure to definitely fix the state of the gas. Since we know from our previous discussion in Chap. 3 that specifying two properties for a pure gas will ensure that all the other prorcrties will have definite values, any two properties can be chosen at will. The "nes usually selected are the ones easiest to measure (such as temperature and pressure); then the others (speciiic volume, density, specific enthalpy. specific rnternal energy, etc.) are fixed. Since a particular state for a gas can be defined by any two independent properties, a two-dimensional thermodynamic chart can he seen to be a handy way to present combinations of physical properties. Although diagrams and charts are helpful in that they portray the rebtions among the phases and give approximate \alues for the physical properties of a ~uhstance, to get more precise data (unless the chart is quite large), usuaily you will find it best to see if tables are available which list the physical properties you want. Enthalpies and other thermodynamic properties can be estimated by generalilcd methods based on the theory of corresponding states or additi\c bond cnntributions.'4.1 S.I 6 J:X:\:\IPLE 4.7

Calculation of enthalpy change using tabulated enthalpy \'alues

Recalculate Example 4.5 using data from enthalpy tables.

SolI/lion: 967). m., v. 54, p. 601 (19'll) !~rfield, iJrermod.\'fW n :.,

ems, Wiley, New YOll

For the same change as in Example 4.5. the data from Table 4.5a are at 2000'F ¢ 2460'R: tili = 14.880 Btll.!lb mOle) 'f _ 3"F at 60°F <> 520'R: tiff = 194.9 Btu/lb mole r~. temp. - Basis: lib mole N,

tiff = 14,880 - 194.9 = 14.865 Btu/lb mole -relations/or Gases. Gu:: ",':hich appeared in thO

- (] 962). 1Jolishing Co., ~;n

. Ie

HOll::.tl':~,

~ publishl~li .l~ ~parately .

! IS is ('lllrOI'.l'-this diagram is called a .\fvllier diagram. HIiL)dcrson. Greenkorn. and Hougen. op. cif. ~. r. J. Janz. Estimation of Thermodynamic Properties of O',;onic Compounds. Academic I'".". SOW Yor~. 1967. !~F. K. Landis. 1v1. T. Cannon, and L. N. Canjar. Hydrocarbon Processing. v. 44, p. ]54

,.f,,, ).

tX-4 £

44@

. . ~_~_1",, _ _ _ _ _~"""''''''''~·..--~--,,..,.. .,.,'--, .•-~.-

i

I I

i I

7'7"'"

282

Ellergy Balallces

EXAMPLE 4.8

Chap. 4

we

.\Cc. 44

fnr' .q"'l

Use of the steam tables

tLc mIL.

What is the enthalpy change in British thermal units when I gal of water is heated from 60' to 1150'F and 240 psig'

.n

d.ll,)

\apor r:(

Solutioll: Basis: lIb H 2 0 at 60'F From steam tables (ref. temp. = 32'F),

il il

You (~!ll ! ,)rJin.](\ c In: :i; follo\ll" "

= 28.07 Btu/lb = 1604.5 Btu/lb

at 1150'F and 240 psig (254.7 psia)

h.li = (1604.5 - 28.07) = 1576.4 Btu/lb h.H = 1576(8.345) = 13,150 Btu/gal

1<11,'111 /:, ,;I

Note: The enthalpy values which have been used for liquid water as taken from the steam tables are for the saturated liquid under its own vapor pressure. Since the enthalpy of liquid water changes negligibly with pressure. no loss oi accuracy is encountered for engineering purposes if the initial pressure on the water is not stated.

4.4 Enthalpy changes for transitions

\()Phl . . i'~.:\l 4,41

":OIllJXH",J

\\here .:\/;!

In making enthalpy calculations, we noted in Fig, 4.3 that the heat capacity data are discontinuous at the points of a phase transition. The name usually given to the enthalpy chilnges for these phase transitions is lalent heal changes, latent meaning "hidden" in the sense that the substance (for example. water) can absorb a large amount of heat without any noticeable increase in temperature. Unfortunately, the word heal is still associated with these enthalpy changes for historical reasons, although they have nothing directly to do with heat as defined in Sec. 4.1. Ice at oce can absorb energy amounting to 80 cal!g with out undergoing a temperature rise or a pressure change. The enthalpy change from a solid to a liquid is called the heat of jiJsion, and the enthalpy change for the phase transition between solid and liquid water is thus 80 cal/g. The heal of rapori::arion is the enthalpy change for the phase transition between liquid and vapor, and we also have the !zeat of sublimarion, which is the enthalpy change for the transition directly from solid to vapor. Dry ice at room temperature and pressure sublimes. The !1ii for the phase change from gas to liquid is cailed the heal ofcondensation. You can find experimental values of latent heats in the references in Table 4.6, and a brief tabulation is shown in Appendix D. The symbols used for latent heat changes vary, but you usually find one or more of the following employed: h.ii, L, }., A. To calculate the enthalpy change for the vaporization of water using steam table data, the liquid or gaseous water should be heated or cooled to the dcw point and vaporized at the dcw point. Keep in mind that the enthalpy changes

ftC

I, 4.4 ~.J,yll

~

tk'.,j

,.dUl·~i.,~d

\\h~fC

.\Ii

,

.\ I'

,

,

,

eti-

Chap. 4

pi! of water is heated

_S4.7 psia)

:l water as taken from '-:lr pressure. Since the . :l Joss of accuracy is

:ne water is not stated.

1E1tI11ll/py Changes for :TWltsiliOIlS ::zs3

Sec. 4.4

for vaporization given in .the steam ·tables.arcfor water under it~·v.aporpressureat the indicated .temperatUT.c. N.aliollai Bureau ojStalldarcll Circular500pTovidesthc data at 25"C that we .can .use to .compare lheheatof vaporization of water.at its vapor presSUTeand at Illl:m: ~C

ImmHg

25 25

23.75 700.

kcaljgmole 10.514 10.520

4.4.1 Heat of F:usion. Tbeheat tlf :fusion for many elements and compounds can be expressed :as

. '9-1]

f

af water using stearn or cooled to the de" !be enthalpy changes

18,936

You can $Cf: that the .effect of preS~UTe 'is ~ui1e small and may be neglected for ordinary engineer.ing calculations. an 1he :absence of experimental values for the latent heats of transition, the following :approximate methods 'will provide a rough .estimme of the molar kilent .heats. Rejd :and :Sherwood, The Properties.£)f Gases and Liquids, give more sophisticated methods. 17

{' 2-3 AJ!! = constant = . 5-7

:=e heat capacity data :::arne usually given to heat changes, latent =mple, water) can c.::ase in temperature. .enthalpy changes for :l with heat as defined :J!g without undergo· . change from a solid ::nange for the phase The heat oj rapori~a­ !en liquid and vapor. ~:.alpy change for the =perature and pres:mid is called the heal ::eats in the references D. The symbols used :lOre of the following

Btu/lbmole 18,925

for ,elements for inorganic compounds for organic compounds

where

=

8.75

+ 4.5171og lo T •

l

(4.13)

/iii, .• = molar heat of vaporization at T. in cal/g mole T.

=

normal boiling point in 'K

(b) Clausius-Clapeyron Equation. The Clapeyron equation itself is an exact thermodynamic relationship between the slope of the vapor-pressure curve and Ihe molar heat .of vaporization and the other variables listed below:

/iii,. dp* _ dT T(I\ - f',) \I

(4.14)

here p. = the vapor pressure T = absolute temperature AH. = molar heat of vaporization at T f", = molar volume .of gas or liquid as indicated by the subscript g or I "Also see S. H. Fishtine. Hydrocarbon Proassing. v. 45. no. 4, p. 173 (1965). "W. Kistyakowsky. Z. Physik u. Chem .• v. 107, p. 65 (1923).

1 i

4.4.2 Heat of VapoTization. (a) Kistyakov.:sky's equation. Kistyakow,ky" developed Eq. (4.13) for nonpolar liquids, and it provides quite accurnte \alues Dflhe mDlar heat of vaporization for these liquids:



1

1

where W, = molar heat of fusion in ca1/g moIe Tf = melting point in OK

Ii;!,.

i

I

1 1I

1

I

1,

284

Energy Balances

Chap. 4

Su.4.4

Any consistent set of units may be used. If we assume that (I) V, is negligible in comparison with (2) The ideal gas law is applicable for the vapor:

r.,

V.

A redu(cd ! reduced prc".J SC\'eral of the, .

= RTlp*

(c) Rcr..·r,·,

~: =~:T~T

estimate the n .. ' at its boiling PI: known liquid '.,

Rearranging the form,

dlnp* d(llT)

=

2303 dlog ,op* . dOlT)

= _~Hv

(4.15)

R

You can plot the log,o p* vs. liT and obtain the slope -(~HJ2.303R). (3) ~Hv is constant over the temperature range of interest, integration of Eq. (4.15) yields an indefinite integral

~H,., B log ,. p * -- -2.3RT'

(4.16)

or a definite one log,. pf =

pt

~H,. (~ _ ~)

2.3R T2

(I) Du;'n,:'

against the t~n':> For exampk. ,: : stance (wa tcr) .",' temperatures (1: 1 rnatelya str;II," : To indicate h lion of comp""!:

(4.17)

T,

Either of these equations can be used graphically or analytically to obtain ~Hv for a short temperature interval.

If lhe vapor pr," rnenl of Eq.

I~ ";

EXAMPLE 4.9 Heat of vaporization from the Clausius-Clapeyron equation Estimate the heat of vaporization of isobutyric acid at 200"C. Soluliol1: The vapor-pressure data for isobutyric acid (from Perry) are pressure (111111)

le1l1p. ('C)

pressure (aI1l1)

100 200 400 760

98.0 115.8 134.5 154.5

I 2 5 10

lemp. COC) 154.5 179.8 217.0 i50.0

Basis: I g mole isobutyric acid Since ~fjv remains essentially constant for short temperature intervals, the heat of vaporization can be estimated from Eq. (4.17) and the vapor-pressure data at 179.8'C and 217.0"C. 179.8°C <> 452.8 c K; 2 _

~fj,

217°C

¢

490 0 K

Given the slopc "i used to ca1cul.ll'· : (2) O,;'",,'r;, Duhring plot c\,: each other at 0;:,:' against log" ( i", very wide tem!,," J

r III

"5 - (2.30J)(1.987)L490 - 452.SJ ~jfv = 10,700 cal/g mole = 19,500 Btu/lb mole

log,.

or in terms of ti:,

19

0. F.

Ot!w~". ~.

ftthniquC' and a \. 10

D. F. Othmcr

S, )'u, /"d.

:;~,

IJ:!;. Chcm., v.

6\1, ' Il

----.....-

"" ..... ."...,....,.",

i

LI1(', ( ;. "

Chap. 4

S~C.

Enthalpy Changes for Transitions

4.4

285

A reduced form of the Clapeyron equation, i.e., an equation in terms of the reduced pressure and temperature which gives good results is described in ~e\'eral of the references at the end of this chapter. (c) Reference substance plots. Several methods ha\'e been developed to estimate the molal heat of vaporization of a liquid at any temperature (not just at its boiling point) by comparing the AH, for the unknown liquid with that of a known liquid such as water.

(I) Duhring plot. The temperature of the wanted compound A is plotted against the temperature of the known (reference) liquid at equal vapor pressure. For example, if the temperatures of A (isobutyric acid) and the reference sub~tance (water) are determined at 760. 400, and 200 mm pressuJre, then a plot of the temperatures of A vs. the temperatures of the reference substaoce will be approximately a straight line over a wide temperature range. To indicate how to apply the Duhring plot to estimate the heat ofvaporization of compound A, we apply the Clapeyron equation to each substance:

(4.15)

JJ./2.303R). .:.::rest, integration of

(4.16)

-AH'A (4.17) ..leally·

d(lnp!) = d(lnp;',)

"btain AH,

R d(\/T A )

-AH,,,,

(4.18)

; I

1

1 i.j

~ j

1

R d(lIT",)

If the vapor pressures of A and the reference substance are the same, rearrangement of Eq. (4.18) gives ron equation

d(_I) T dT"

~ tiH"A.

dT",

rer

=

T;ef dTA

n

= AH,,,,

or in terms of the slope of the Duhring plot, ·emp. eC) 154.5 179.8 217.0 i50.0

dT", _ A R,...

n" n

(4.19) dT" - Ak·", Given the slope of the Duhring plot, AH."" and T", and T A > Eq. (4.19) can be used to calculate AH, A'

(2) Olllll1er plot. The Othmer plot 19 is based on the samt' concepts as the Duhring plot t'xcept that the logariThms of vapor pressures art' plotted against each other at equal tempera/lire. As illustrated in Fig. 4.8. a plot of the log 1 0 (p!) against 10gIO (p:;,) chosen at the same temperature yields a straight line over a \cry wide temperature range. By choosing values of vapor pressure at equal

re intervals, the heal :ssure data atl79XC

"d. F. Othrner, Ind. Eng. Chem., \'. 32, p. 841 (1940). For a complete re"jew of the t('(hnique and a comp3rati\'c statistical analysis among \";!rious prcdictin~' methods, refer t,· D. F. Othmer and H. N. Huang, Ind. Eng. Chern .. \'. 57, p. 40 (1965); D. F. Othmer and E. S Ylt. Ind. Eng. Chern .. v. 60, no. I, p. 22 (1968); and D. F. Othrner and H. T. Chern. Ind. Ir,f. Chern., v. 60, no. 4, p. 39 (1968).

mole

1

I I1

'~li!

- _ . _ _ _ _- .. . . . . , . _......,...~_ _ _ _ _ 4 _ _~_ _· _ ."



,.-

286

Energy Balances

I? _______

..... -..,~

Chap. 4

I

L____

1

3 --

I 1

1

5-.-----

~ '1 -----

~-

I I

I

f.ref (OC)'2

fA (OCJ'2

*...o ~ "-

1_ _ _ _ -

g

...J

-----I

3

16

! 14

15 L0910

Fig. 4.8. Othmer plot

p'~

a~d

vapor-pressure curves.

temperature (TA = T",) in Eq. (4.18). we obtain

~lln p!) = = m = slope of Othmer plot as in Fig. 4.8. (4.20) ( n Pi'.,) 6,H, ,,' The Othmer plot works well because the errors inherent in the assumptions made in deriving Eq. (4.20) cancel out to a considerable extent. At very high pressures the Othmer plot is not too efiective. Incidentally, this type of plot has been applied to a wide variety of thermodynamic and transport relations, such as equilibrium constants, diffusion coefficients. solubility relations, ionization and dissociation constants, etc., with considerable success. Othmer recommended another type of relationship that can be used to estimate the heats of vaporization:

f:..lf'A

1n P'A * -- AH, t;.fj"" T,,,, I * -T n p,,,, U 'rer c

,\1

C i1

~()()

'I hell

(4.21)

A

p:

where is the reduced vapor pressure. Equation (4.21) is effective and gives a straight line through the point r:, = I and T", = l. The Gordon method plots the log of the vapor pressure of A versus that of the reference substance at equal reduced temperature. EXAMPLE 4.10

Fr0l1l1

·Ih(' Ml\\~, \ .dw:'. II I rCl/11l

.II

,j

(r '.: .;

.?O() (.

Use of Olhmer plot

Repeat the calculation for the heat of vaporization of isobutyric acid at 2()()OC using, an Othmer plot. "hcrt- :\ I·

Solulion:

Data are as follows: temp.

ee)

154.5 179.8 217.0 250.0

"............

.\ !

p~o (a/m)

1 2

5 10

log p* 0 0.3010 0.698

1.00

- - - - -....- ,....;,............ , ...- . . "...... , ",p,,...- - - - - - ..., ...."' .... "

p~,o (a/m)

log p*

(I\,~,r

5.28 9.87 21.6 39.1

0.723 0.994 1.334 1.592

.Ij'p!I(.n;,'·

1Y!'Y~,.~~.~7',-',.., ,._--.

--

: ' I;

\'

:11· 1

,

,-'--€:hap.4

Sec.4A

Enthalpy Clumges for Transitions 1X7

110

O.B

(0.80-020)

j /'

.4.8.

(.UO)

;!Cas .-\.1: .

'ODS

I

, 1

~

nigh :1"i! crlf plot has ~ri=,.guch as ~.:mii!lIrian and

Fiig.lE4.10. !Fl'~m

j

Othmer iJllat, !Fig. !':4'] D

~Iope

1A.~";
020 = 1.15

1

J.41 -0.90

"

::n b"e used to

I

At 20CnC '(391"F),:Emm the steam tables,

(&33:8 Btu!lb)

j

Then

Mi",o = ye and gives a - method plots ,lance at equal

(13<,000)(1.15) = 17,:50 Btu!lb mole

The Jmswer:in this case is lower than that in Example 4.9. Without the experimental value. it is difficult to 'say what the proper answer is. However. allhe normai boiling pornt {154 'C).liii';mis known to be 17.700 Btu, lb mole. ,and il.H.,;~ should De lower at 200'r so that 17,:50 Btuilbmole'seems to be the more satisfactory value.

1

i I

I '(d) '£nrp;rical'relatioll oJWatsan.':: o Watson fOllnd J:mpirically that

Alf., = .t:..H",

(I - T")O.,, I

00'

T,.

whcre;o.ff", = heat of vaporization of,a pure liquid at T, 1:::.H" = :heat ,of vaporization of the same li,quid lit T, c p.

-:"23

Gold" 'Summarizes numerous estimation methods and delineates their ranges of application.

"94 ~34

59

-'OK. 'M. ·Watson.ind. Eng. Chcm .. v. 23, 11- 360 (1931); v. 35,p. 398 (1943). "P. 'I.·Gold. Cllem. Eng., 'p. 109. (Feb. 24, 1969).

f

l , j

j j

..

288 Energy Balances

~

Chap. 4

~ystem

4.5 The general energy balance Scientists did not begin to write energy balances for physical systems prior to the latter half of the nineteenth century. Before 1850 they were not sure what energy was or even if it was important. But in the 1850s the concepts of energy and the energy balance became clearly formulated. We do not have the space here to outline the historical development of the energy balance and of special cases of it, but it truly makes a most interesting story and can be found elsewhere.",23.24.".2. Today we consider the energy balance to be so fundamental a physical principle that we invent new kinds of energy to make sure that the equation indeed does balance. Equation (4.22) as written below is a generalization of the results of numerous experiments on relatively simple special cases. We universally believe the equation is valid because we cannot find exceptions to it in practice, taking into account the precision of the measurements. It is necessary to keep in mind two important points as you read what follows. First, we shall examine only systems that are homogeneous. not charged, and without surface effects, in order to make the energy balance as simple as possible. Second, the energy balance will be developed and applied from the macroscopic viewpoint (overall about the system) rather than from a microscopic viewpoint, i.e., an elemental volume within the system. The concept of the macroscopic energy balance is similar to the concept of the macroscopic material balance, namely, {

Accumulation Of} energy within the system

=

{tranSfer of energy} into system through system boundary energy genera,} wlthm system

+ { tlOn

~

~

{tranSfer of energy out} of system through system boundary

{energy con,} sumptlOn within system

:lA. W. Porter, Thamodrnamics. 4!h ed., Methuen, London, 1951. 23D. Roller. Tire Earl.\' Dere/opmefll altlre Concepts of Temperature and Heat, Harvard University Press, Cambridge, Mass., 1950. l4T. M. Brown, Am. J. Pln'sics, v. 33, no. 10, p. I (1965). "J. Zernike, Cizem. Wl'Ckh/ad., v. 61, pp. 270-::74, 277-279 (1965). z.L. K. Nash, J. Chem. EdIlC., v. 42, p. 64 (1965) (a resource paper).

'~-....."""''''.- - . . . "...., - - _...., .....,.,..-

i#""jO,.~..,~\c:.,..,.,.,_.~7"'ry;;:"~'_:;';'~~~~:; ",-

............ , _ _ _ _ _ _.",4.4............. _ ...._ ,......."',U..

I

can be ! text. elle system h trons, cr conceptl decay 1.:.1 Fr~r:

(4.22).\\ .

in Eq. (4 tinal tlnh~ The sure! without : notation i at the end

F~ (~ ht I

,, ,,

(4.22)

While the formulation of the energy balance in words as outlined in Eq. (4.22) is easily understood and rigorous, you will discover in later courses that to express each term of (4.22) in mathematical notation may require certain simplifications to be introduced. a disclission of which is beyond our scope here, but which have a quite minor influence on our final balance. We shall split the energy (El associated with mass. mass either in the system or transported across the system boundaries, into three categories: internal energy (U), kinetic energy (K), and potential energy (P). In addition to the energy transported across the

,

S ..... 4.5

_ _ _, ....

Chop. 4

_~ i !y~s

prior to 'cere not sure what ::oncepts of energy ::ot have the space and of special ~n be found elseDe SQ. €un41mental 10 make iwre that 'nuen below is a _. rel;at.ively simple ~..alil~ we cannot _,on «l·r the measure-

.:.=

what folnot charged . ...ance: as simple as _ aT fmm the ;ro~ ;rosco pic

5)'$lcmboundariC5 associated with ma5S flow into anrl out oflhe system. energy can be transferred by hcal (Q) and work (1-1"). For most {)flhe purposes of this lext, -energy generation {)r loss will be zero. However. energy transferred 10 the >;·stem by.an external electric or magnetic field. or by the slowing dOlm ·of neutrons, energy transfer not comeniently included in Q or IV. may be in.clurl.ed wnceplually in the energy balan.ce asa generation term. Similarly. rarlioactive decay can be viewerl as a generation term . Figure 4.9 shows the various types of energy 10 be acr:ounted for in Eq. (-4.22), ...;hile Table4.81ists lh~spccific indivirluallerms v.-h1cb.areto be employed in Eq. (4.22). As to th~ notation. th~ subscr.ipls1, and 1= refer 10 the initial.and 'final time :periorls over which the :accumulation is 10 be evaluated. Wilh 12 > 1,. The:superscript caTel (A)=ans that thesyrnbol stands for energy per unit mass.; wrtbo.ul lhe caTet, the symbol means cenergy of the lolal mass present. .other no:ta:til:l.nis:c",ident from Figs. 4.9 and 4.10and can be found in the notati.on list at ihe
VO'lill Irea;d

... ~ODs,.

SYSTEM

i€tITlWt~

'M as~

f1WI'!Jy

'Energy

€mmt2?M=II-~ .2 , .

tlE=I'2'-I" '/::'m = m0, -ml,

'trnernn:£ner.gy

;tntemo/ 'Ener9Y

1:.1J = .v,Z -U "

!/:;u = lJ'Z-1J'1 :!TlfY'

>I< inelicfner.gy

lIlut}

l'xK= 'K,z -I
-:nrough :lry

'KineliJ:Energy

(4.22)

!/:;-K= TV

--Kt ,

Z

'RotenICo/Energy

'Potentio/Energy

!/:;'P=P'2 -0,

!/:;P= P'2 -0 ,

.as outlined in Eq. ~ later courses that cay require certai n nd our scope here. We shall split the -rransported across ; U), kinetic energy sported across the

'e ellld Hear. Harvard

i.

Jliig.·";9 .. Enccgy ba/ancc over.a :timcperiod 'for flow and nonf\uw systems.

. .,.

-'~

____ _ _"""""""_·r_ _ ...",,._,_______ ~

~_

...... '

290 Energy Balallces

Chap. 4

Sec. 4.5

With the aid of the symbols from Table 4.8, the general energy balance. Eq. (4.22), can be written as follows. (An alternative formulation of the balance in terms of differentials is given in Chap. 6, where emphasis is placed on the instantaneous rate of change of energy of a system rather than just the initial and finai states of the system; the formulation in this chapter can be considered to be the result of integrating the balance in Chap. 6.):

m,,(O + K + h. - m,.(O + K + h. = (0, + K, + i>,)m, - (0. + K. + i»m, + Q - W + p,v,m, - pJ'.m,

.-((cumul,I;1 Tl!,C I'!

I

Infer:1 I\.lne: ~ 1\)((.,(

(4.23)

\fat I

J."nrr;:J' d i , ' np('(1Jl" Ifltet Kl:1t

I't':('j .\fan

Sr( IIral I"f' Xn kO'~ .i,~ MCllI,I\'I;';~

i

\Vor~ !,\

"I'~ k

fr \' '~;

whcrc 1\ . <

,

Reference plone for potentiol energy

/'

t

Fig. 4.10. General process showing the system boundary and energy transport acrOss the boundary.

Ii (I

~

I,

Heol-terrestrial Radiotion

<

/I'

il "',)1,· Ih.lI

--~--Kinetic energy ..... - ........... of wind ATMOSPHERE

'q : (.~

i j

\1.(

I

)f

t".t!.II'~C

'.:1

:'1! 1.11 "., '

~

I

t.,\rnr \ \'-, \ ,

EARTH

Fig. 4.11. Energy balance on the earth and its atmospher",. The thickness of the arrows arc roughly proportional to the magnilude of Ihe energy Iransfers.

,

I'

\ 1

",

'- ,,1 \ "

'

\'

. ';:

I'

~'

"

.

q

- ---- ........

. •rr

...... ..

(;hap.4

~"'-

..

-.--.--..•...----....-

TABLE

4.8

-

. ..

LIST OF SYMBOLS USED 'N THE GE"ERAl E"ERGY BAlA:-
Accumulation term Type 0/ energ},

Allime I,

Internal Kinetic Potential

(4.23)

-.

The General Energy Balance 291

Sec. 4.5

~al energy balance. ._:ion of the balance .5 is placed on the :-. just the initial and ::>econsidered to be

•..

AI time t1-

V"} E'l

v•• }

m.,

mit

K't P"

X'I p.

Mass Energy accompanying mass transport Type 0/ energy

Transport in

Internal Kinetic Potential Mass

Et!

Transport out

V,

V2

K,

K2

P,

P2

m,

m2

Nel heal inpullo system Net 'K'ork done by system on surroundings

Mechanical work or work by moving parts: Work to introduce material into system. less work recovered on removing material from system:

Q

w

or in simpler form, which is easier to memorize,

IlE = E., - E., = -1l[(H + K + P)mj

+ Q-

W

(4.24)

where Il = difference operator signifying Ollt minus in, exit minus entrance, or final minus initial in time :nergy trans·

I l

!

--

-"'i-iERE

!< thickness ltude of the

I

!

E=U+K+P H= u+pf Q = heat absorbed by the system from the surroundings (Q is positive for heat entering the system) W = mechanical work done by the system on the surroundings (W is positive for work going from the system to the surroundings) Note that the expression IlO .!.- Il p f' = IlH has been employed in consolidating Eq. (4.23) into (4.24). so that the significance of the enthalpy term in the energy t>alance manifests itself. A more rigorous derivation of Eq. (4.24) from a microscopic balance may be found in a paper by BirdY Keep in mind that a system may do work, or have work done on it, without SOllle obvious mechanical device slIch as a turbine. pump. shaft. etc., being present. Often the nature of the work is implied rather than explicitly stated. For example, a cylinder filled with gas enclosed by a movable piston implies that the surrounding atmosphere can do work on the piston or the reverse; a balch fuel l{'11 docs no mechanical work. unless it produces bubbles, but does deliver a ,Urrcnl at a potential difference; and so forth. "R. B. Bird, Reprint nO. 293. Uni,·. Wis. Eng. Expt. SIO. Repl., 1957.

I

1

l

I

292

Energy Balances

Chap. 4

Suo -15

Now for a word of warning: Be certain you use consistent units for all terms; the use of foot-pound, for example, and Btu in different places in Eq. (4.24) is a common error for the beginner. The terms p/>, and p,(> 2 in Eq. (4.23) and Table 4.7 represent the so-called "pV work," or "pressure energy" or "flow work," or "flow energy" i.e., the work done by the surroundings to put a unit mass of matter into the system at I in Fig. 4. IO and the work done by the system on the surroundings as a unit mass leaves the system at 2. If work is defined as

w=

s:

Fdl

then, since the pressures at the entrance and exit to the system are constant for differential displacements of mass, the work done by the surroundings on the system adds energy to the system at I:

w, =

s:

F, dl

=

S:' p, dV = p,(V, -

0)

= p,V,

where V is the volume per unit mass. Similarly, the work done by the fluid on the surroundings as the fluid leaves the system is W 2 = pi' 2' a term that has to be subtracted from the right hand side of Eq. (4.23). In most problems you do not have to use all of the terms of the general energy balance equation because certain terms may be zero or may be so small that they can be neglected in comparison with the other terms. Several special cases can be deduced from the general energy balance of considerable industrial importance by introducing certain simplifying assumptions: (a) No mass transfer (closed or batch system) (m, IlE

II

= Q-

=

m2

=

0)

(4.25)

W

Equation (4.25) is known as theftrst law of thermodynamics for a closed system. (b) No accumulation (IlE = 0), no mass transfer (Ill, = m 2 = 0):

I

(4.26)

Q= W (c) No accumulation (IlE = 0), but with mass flow:

;

Q- W (d) No accumulation, Q

=

=

M(H

0, W

=

+ K + P)mj 0,

/:iN ,~

K=

°

0,

P=

(4.27) 0, (4.28)

Equation (4.28) is the so-called "enthalpy balance." (e) No accumulation (!l.E~' 0), no mass transfer (1/1, = m 2 = 0): Q~dV

(4.29)

Take, for example, the flow system shown in Fig. 4.12. Overall, between sections 1 and 5, we would find that !l.P ~, 0. In fact, the only portion of the

~.,#.#.~,.



ij)

,.

*P,M

"':;;:;;"'''''4

system "Ih' ~, olher sec'I"'f) may be llc.~i Belween se,! After rcadlll(' linl/ing 10 ,:, your abtlll! I, Sorne '!"~ worth rcme", (a) iJo[/" (0) Adu;

we

11)

adLlh (I) T (2) (I ;I[

(3) II I rI

(e) 1",1'.11 (d) /s"":C

Occasi,,!:;. hrat. Scnsible 1 rdaenec ICIl"" eXcluding all:lermed Ialcnl ), Onc furil,c' prescnled ILl' " 111 surface Clll': ,: I hc\c more <' h, l"prcsslon h! '{

presently I., "',' 'plil inlo 1I1ll',I'

Chap. 4

S~C.

The General Energy Balance 293

4.5

4

-:sunl units for all terms;

5

:>\aCe3. in Eq. (4.24) is a , represent the so-called tnergy" i.e., the work ::ltO the system at I in ~;.::ndings as a unit mass

"W

I

I

1

IW I 2

I I I 5

Fig. 4.12. Flow system. t

'svstem are constant for ~ surroundings on the

p,P,

=

_:,k done by the fluid on ~. 2' a term that has to

l'

.;:1~ \t'.-~ of the general

ay be so small :: Ie. . Several special ; considerable industrial

::!T'

:;)ns: (4.25)

modynamics for a closed

(4.26) x;

(4.27)

0, (4.2~)

;Janee." '11, "" 1112

= 0): (4.2
4.12. Overall. between ;nc only portion of th,'

"'s........ .

.. i£j@¢ ....., ...._ _ ..

'''--.'-~'

system where AP would be of concern would be between section 4 and some other section. Between 3 and 4. AP may be consequential. but between 2 and 4 it may be negligible in comparison with the work introduced by the pump. Between section 3 and any further downstream point both Q and IV are zero. After reading the problem statements in the examples below but before continuing to read to solution. you should try to apply Eq. (4.24) yourself to test your ability to simplify the energy balance for particular cases. Some special process names associated with energy balance problems are worth remembering: (a) Isothermal (dT = O)---constant-temperature process. (b) Adiabatic (Q = OJ-no heat interchange. i.e., an insulated system. If we inquire as to the circumstances under which a process can be called adiabatic, one of the following is most likely: (I) The system is insulated. (2) Q is very small in relation to the other terms in the energy equation and may be neglected. (3) The process takes place so fast that there is no time for heat to be transferred. (c) Isobaric (dp = O)-
,

1

j j

,-- ...-_.......... 294

Energy Balances

Chap. 4

We turn now to some examples of the application of the general energy balance.

EXAMPLE 4.11

!ire. 4.5

l-C" heat is

Te.

~.{.,mc ~orl

h ..

EX:\~fPLE

Application of the energy balance

Changes in the heat input that spacecraft encounter constitute a threat both to the occupants and to the payload of instruments. If a satellite passes into the earth's shadow the heat nux that it receives may be as little as 10 percent of that in direct sunlight. Consider the case in which a satellite leaves the earth's shadow and heats up. For simplicity we shall select just the air in the spacecraft as the system. It holds 4.00 kg of air at 20 c C (the air has an internal energy of 8.00 ;. 10' Jikg with reference to fixed datum conditions). Energy from the sun's radiation enters the air as heat until the internal energy is 10.04 ~ 10' J;kg. (a) How much heat has been transferred !O the air~ (b) If by some machinery in the spacecraft 0.110 " 10' J of work were done 011 the air over the same interval, how would your answer to part (a) change?

4

Air is \,.' :10,5 Btu 11') c"(!t vc:Jocit~ n' for the

1

COtl1r"(

Svlution This i, ,', i ..:'~4} can be u

Solution: (a) The air is chosen as the system, and the process is clearly a closed or batch system. Everything outside the air is the surroundings. See Fig. E4.1 1. Using Eq. (4.24)

pressor (ll r terms

4.IfC

of •:~'1

~---------

Fig. E4.1I.

Nrxt. we sh .• ::;-. with the absence of K and P in the ac:umulation term inside the tank, and with no mass flow from the system to the surroundings or the reverse, we have (for W = 0)

AU = u" - U,' = Q - W Basis: I kg air

0" - 0" = (10.04 x 10') - (8.0 x 10')

= 2.04

x 10' Jjkg

so that Q = 2.04 x 10' Jjkg. Note that the sign of Q is positive, indicating that heat has been added to the system. Basis: 4.00 kg of air

Q

! '-el/"-:

lhr

r~,!

(~('rk Ilm('~.

2.04: 10' J 1 4 .00 kg = 8.16 x 10' J g

(b) If work is done on the air (by compression or otherwise), then the work term is not zero but W = -0.110 x 10' J

1\,-\ "I'll 4 11

Note that the sign on the work is negative since work is done 011 the gas. In this second

\ r (' •• ",I

case,

Q = AU -l- W

= (8.16

x 10') - (0.11 x 10') = 8.05 x 104 J

:. ;.l.;

1\1

/I

C,(:~.

,.1..

Cfrap . .,

. : tI:r.e general energy

S~C.

7ne Germ-a] Energy Balar:re 29:5

4.5

Less heat is required !han before to Teach the same value fOT the internal energy because ".,me work has been done on the gas. E.XA\IPLE 4.12 Application of the ~ balance

.:.:mte a threat both to ~:l5se5

into the earth's ..: of that in direct sun· =dow and heats up -cstenL. Itbolds 4.00 kg ·,,·ith reference to fixed :c :as beat antil the inter· =n5ferred to the air: =k done on the =ange?

Air is being compressed from 1 atm and 460 0 R (where it has an enthalpy of :10.5 Btu(lb) to 10 atm and 500'R (where it has an enthalpy oi 219.0 Btu Ib). The exit .elocity oflhe air from the compressor is 200 ftlsec. What is the horsepowerre
w=

-, ,

v2 ~ 200 iI/sec

=ly a d.
_4.11. Usiltg Eq. (4.24) Fig. E4.l2-

pressor (AE = 0 and m, = n11), Q terms are of importance. Then

= 0 (no heat exchange), and

no potential energy

Basis: I Ibm air

fir

= -A(n + ~')

AH = 219 - 210.5 = 8.5 BtU/Ibm l"ext, we shall assume that the entering velocity of the air is zero, so that tank,. and with no we have (for W = 0)

::!Ie

A~

T

1 Btu I 32.2 (ft)(lb~) 778 (ft)(lb,) 2 (sec')(lb,J = O.80Blu/lbm

sec'

fir = -(8.5

10' 1/kg

=. indicating that heat

(200 ft)1

=

+ 0.8) =

-9.3 BtU/Ibm

(VOle: The minus sign indicates work is done on the air.) To convert to power (work/time),

Basis: 200 lb airjhr J 'l.

_ 9.3 Btu 200 Ib 1 hp hP Jb hr 2545 Btu hr

then the work term rXA:'I1PLE 4.13

=

gas. In this second

5 x 10']

= 0.73 hp

Application of the energy balance

A steel ball weighing 100 Ib is dropped 100 it to the ground. What are the values IV ~nd the change in U. K, and P for this process? What is the change in the \,·1.11 energy of the universe (system plus surroundings)?

1,1 (l ~nd

,---

.---"".".,--.....,."~"...,........,......

4

...._ _ _...._'......."'.~"¢)"""J

_,~.~

_.........,.,..,.,.AAf'.· ,*..~. :._~ ____"'--_____.

ill·s..........

296 Energy Balances

Chap. 4

Solution: See Fig. E4.I3.

Fig. E4.13.

Let the system be the ball and the surroundings be everything else. This is a batch (nonflow) process. The general energy balance reduces to

[CO" + Kr, + Pr,) - COr, + Kr, + Pr,)]m = Q -

W

After the ball hits the ground and is at rest. we know that its velocity is zero, so that

11K = O. We can let the ground be the reference plane for potential energy. Conse-

i i -( t, . ,:; •

quently,

m(P" - P,.) = =

m(g/g,}(h 2

-

h,)

(100 Ibm)(g/g,)( -100 ft)

= -

10,000 (ft)(lbf )

If Q and Ware zero, then Vr, - Vr, may be calculated. If Q and Ware not zero. then V" - Vr, is dependent on the values of Q and W, which are not specificall, stated in the problem. If Vrr - Vr, were known, then Q - "1/ could be determined. Without more information or without making a series of assumptions concerning Q. W, and V" - V r,. it is not possible to complete the requested calculations. The change in the total energy of the universe is zero.

T) r ~ .'



.'

'" '

i .

1 ~'.~

EXAMPLE 4.14 Application of the energy balance Water is being pumped from the bottom of a well 150 ft deep at the rate of 200 gal/hr into a vented storage tank 30 ft above the ground. To prevent freezing in the winter a small heater puts 30.000 Btuihr into the water during its transfer from the well to the storage tank. Heat is lost from the whole system at the constant rate of 25.0l1() Btu/hr. What is the rise or fall in the temperature of the water as it enters the storage tank,assuming the well water is at 35'F? A 2-hp pump is being used to pump the water. About 55 percent of the rated horsepower goes into the work of pumping and the rest is dissipated as heat to the atmosphere.

Solution: Let the system consist of the well inlet. the piping, and the outlet at the storage tank. It is a flow process since material is continually entering and leaving the system. See Fig. E4.14.

'~

,4&

';1

~

i- ,

1

Chap. 4

The General Energy Balance 297

Sec. 4.5

Fig. E4.14.

..!ling else. This is a batch

The general energy equation reduces to Q - W = A[(H -'-

,- w velocity is zero, so tha t Jotential energy. Conse·

'and Ware not zero. are not specificaii\ could be determined. ",-nptions concerning Q. :;alculations.

,~h

(a)

The term ll.E is zero because no accumulation of energy takes place in the pipe.

Basis: J hr operation The total amount of water pumped is

200 gal hr

'00 (ft){lb,)

i + PJm]

II

hr 1 8.33 Ib J gal

=

1 ,

16661b/hr

!

The water initially has a velocity of zero at the well inlet. The problem does not specify the size of pipe through which the water is being pumped so it is impossible (., calculate the increase in kinetic energy experienced by the water as it leaves the ;,:pc at the storage tank. Unless the pipe is exceptionally small. however, the kinetic energy term may be considered negligible in comparison with the other energy terms m the equation. Thus The potential energy change is

"P- =mg ... "h

mL>

a..aep at the rate of 200 Drevcnt freezing in the s transfer from the \I~ll ;onstant rate of 25,(X\) :J.S it enters the stora;:~ sed to pump the water. : of pumping and ttle

=

1666 Ibm 32.2 ft I 180 ft I hr

sec' "

"'" f ) lb )Ih 32.2 (ft)(Ib m ) = """,000 ( t ( , r

(sec')(lb,) 300.000 (ft)(lb.)'hr 778 (1t)(lbrJ, Btu

=

The heat lost by the system is 25.000 BtU,hr while the heater puts 30,000 Btulhr

= 5000 Btu/hr

1 he rate of work being done On the water by the pump is

: outlet at the storase Jd leaving the system,

w=

2 hp 10.55 i 33.000 (ft.)(lb)

Bttu

I (min Jl h'--'p';-)--'--r--,--'-'-t"'7"'"7/;'"(""'ff"'tt)"'\I""b'),

= -2800 Btu/hr

~·04AJ@!4;,·.~-

.-~"'C:""'

1

I I

I

386 Bt Ih u r

C'.:" the system. The net heat exchange is Q = 30,000 - 25,000

I

__""'___"""___""'_""'''''''j'''_____'''''__' ' '_',' ',fX__,.,.,...__.'..._'."I¢"",.....- -.......... : ~,#_~"...

I

".%, •. 'Ii

298 Energy Balances

Chap. 4

Src.4.5

AH can be calculated from Eq. (al:

= AH =

5000 - (- 2800)

AH

+ 0 + 386

7414 Btu/hr

The energy

We know from Eq. (4.7) that, for an incompressible fluid such as water, However, since c O~" jf - pV, Because the temperalt~re range considered is small, the heat capacity of liquid water may be assumed to be constant and equal to 1.0 Btu/(Ibl{'F) for the problem. Thus

Aii =

CpAT

=

(I.O)(AT)

Since no inform doubt be ncgliglr

Basis: I Ib H 2 0 Now

AH = 741~rBtu 16~~ Ib

(a) The heat (b) The entl: The AIf,

= 6.36 Btu/lb

= (I.O)(AT)

AT=:::. 6.4°F temperature rise

(c) Th,. enthl

EXAMPLE 4.15 Energy balance Note th,lt enthalr, ( prodU(h r

Steam that is used to heat a batch reaction vessel enters the steam chest, which is segregated from the reactants. at 250'C saturated and is completely condensed. The reaction absorbs 1000 Btu/lb of material in the reactor. Heat loss from the steam chest to the surroundings is 5000 Btu/hr. The reactants are placed in the vessel at 70'F and at the end of the reaction the material is at 212 F. If the charge consists of 325 Ib of material and both products and reactants have an average heat capacity of C p = 0.78 Btu/(Ib)(CF), how many pounds of steam are needed per pound of charge? The charge remains in the reaction vessel for I hr. 0

lillm .... (d) In addill."

Solution: Analysis of the process reveals that no potential energy. kinetic energy, or work terms are significant in the energy balance. The steam flows in and out, while the reacting material stays within the system. Thus there is an accumulation of energy for the material being processed but no accumulation of mass or energy associated with the steam. See Fig. E4.15(a).

Ica\-'in~ tilt

can be

ment

O(loss)' -5000 Btu/hr System Boundor y, , "..-:---

I

\ I

I I

I

I .':c-'-t-'-.:-:..:..o...j-Y J

".....

Soturoted Steom 250'C

Soturoted Condensate

Fig. E4.IS(a).

("

LI

'n

{p

t:

....-....-.---

...

Chap. 4

Sec. 4.5

The Gellenli Et,ergy Balance 299 Basis: I hr of operation System: Reaction vessel, including steam chest

The energy balance reduces to

(Ur,

: such as water,

-

Vr

-6Hstea r:. i- Q

)muerial =

However, since enthalpy is more convenient to work with

(a)

than~nternal

energy, we let

{; = ii - pV, ::apacity of liquid water . ior the problem. Thus

[(ii., - if,,) - (p,//" - p" P',,lllll = Q - A.1l".. m Since no information is available concerning the pV differet1llce, and it would no doubt be negligible if we could calculate it, it will be neglecte(t,. and then (if" - ii,JIIl = -AH""m

+Q

(b)

(a) The heat loss is given as Q = - 5000 Btu/hL (b) The enthalpy change for the steam can be determined fmm the steam tables. The AH"p of saturated steam at 2S0'C is 732.2 Btu/lb, so that

Ib

AH""m = - 732.2 Btu/lb (c) Th" enthalpy of the initial reactants, with reference to zero enthalpy at 70o P, is /).H, ,

IIl(Aii)

=

I

70

JI!

70

=0

Cp dT

Note that the selection of zero enthalpy at 70'F makes the calculation of the enthalpy of the reactants at the start quite easy_ The enthalpy of the final products relative to zero enthalpy at 70'F is

", the steam chest. which " cor -Iy condensed. ~!!ai

=

om the steam

.=d 1, ... ,~ vessel at 70 F ::narge consists of 325 I b '::ce heat capacity of Cc ,:1' per pound of charge '?

AH" =

212

JI!

J

70

CpdT= mCp(212 -70)

_ 325 Ib 0.78 Btll '(212 - 70rF hr (/blC F)

= 36000 B 'h ,

tUI

1 1

r

/).Hmot"'" = 36,000 Btu/hr

(d) In addition to the changes in enthalpy occurring in the material entering and leaving the system, the reaction absorbs 1000 BtU/lb. This quantity of energy can be conviently thought of as an energy loss term, or alternately as an adjustment to the enthalpy of the reactants.

. kinetic energy, or work ';s in and out, while the -accumulation of energ" nr energy associated witll

AH = - I~ Btu 1 3:;;S/b = -325,000 Btu!hr Introducing these numbers into Eq. (b). we find that 36 oooBt~ = (737.2 Btu )(S~team) 'hr Ib steam hr Btu - 5000 _ 325 ooo~~':! hr 'hr from which the pounds of steam per hour. S. can be calculated as

1 (c)

S = 366.000 Btu II Ib steam = 4961b steam hr I 737.2 Btu hr "I

366.000 Btu LLlb steam I / hr hr 1737.2 Btu 1325 Ib charge

Z01',-

= I 53 .!£..stea~ .

1

Ib charge

--.,.. ",_;,..,.."""'_......., _......._'!"'II#_ _ ;" ......., _....,.....,.- ............................. ,L",.... '", ......_ ,.. · "•., - . ............, . - - - -.. , .... ,.,....----

,--~

300 Energy Balances

-

....

Chap. 4

If the system is chosen to be everything but the steam coils and I ines. then We would have a situation as shown in Fig. E4.15(b). Under these circumstances we could talk about the heat transferred into the reactor from the steam chest. From a balance on the steam chest (no accumulation), Q = !J.H" .. m

51? System I

I

°system

I

~

+

I

°syslem II = ~ ,1 j~,'.

Steam Chest System 11

Saturated 250°C

;~

1

1,,1 ,

Saturated Condensate Fig. E4.1S(b)

Both have negative values, but remember that QSJuem I

= -

QSY:Jlem II

so that the value of Qm"," 1 is plus. The energy balance on system J reduces to

f·o( :-, .

(H" - H,,)nr

=

(d)

Q

where Q = "mm ,b", -". (-5000 Btu). The steam used can be calculated from Eq. (d); the numbers will be identical to the calculation made for the original system.

EXAMPLE 4.16

Energy balance

, •r

-

Ten pounds of water at 35"F, 4.00 Ib ice at 32'F, and 6.00 Ib steam at 250'1' and 20 psia are mixed together. What is the final temperature of the mixture? Ho\\ much steam condenses? Solution; We shall assume that the overall batch process takes place adiabatically. See Fi,. E4.16. The system is 20lb of H,O in various phases. The energy balance reduces to (superscripts are s ~o steam. I\' = water, i = ice)

20C,. - (600:, -+ 100~

+ 40:.> = 0

(a)

--

I

1 1 ,("

1'.

Chap. 4

Sec. 4.5

The General Energy Balance 30t

.= coils and lines, then we =se circumstances we couid = chest. From a balance

Fig. E4.16.

If U is replaced with

H-

pV for convenience in the calculations, we get

20H" - (6H;, + lOH~ + 4H;,) = 20(pVJ" - 6(pV):, - lO(pV);' - 4{pV);,

(b)

The last two terms on the right-hand side of Eq. (b) cannot be more than 1 Btu at the very most and can safely be neglected. Because of the phase changes that take place as well as the nonlinearity of the heat capacities as a function of temperature. it is not possible to replace the enthalpies in Eq. (b) with functions of temperature and get a linear algebraic equation that is easy to solve. The simplest way to solve the problem is to assume the final conditions of temperature and pressure and check the assumption by Eq. (cl (c) If the assumption proves wrong, a new assumption can be made, leading to a solution to the problem by a series of iterative calculations: Basis:

(d)

can be calculated from .:;,de for the original system.

41b ice at 32'F 10 lb H,O at 35'F { 61b steam at 250"F and 20 pSJa

Since the enthalpy change corresponding to the heat of condensation of the steam is quite large, we might expect that the final temperature of the mixture would be near or at the saturation temperature of the 20 psia steam. Let us assume that Tf;n,' ..~ 228'F (the saturated temperature of 20 psia of steam). We shall also assume that all the steam condenses as a first guess. Using the steam tables (ref.: zero enthalpy at 32"F, saturated liquid), we can calculate the following quantities to test our assumption: Blu

= 7008.0

:1d 6.00 Ib steam at 250 F ,lure of the mixture? Ho\\

10 Af'iw = 10 Ib13.02 Btu " lb

= 30.2

The heat of fusion of ice is 143.6 Btu1lb:

:.ace adiabatically. Sce fi!' ", energy balance rcdu,''''

• =0

(al

4 AH' "

= _ 4 Ib1143.6 Btu

-574.4

lb =

-453.2

= 6010.6

...

302 Energy Balances

Chap. 4

----. Sa. 4.6

The left-hand side of Eg. (c) is

=

20 flH h



So(ul iOfl:

20 Ib 1 196. 2 Btu Ib

= 3924.0 I Btu 778 (ft)(lb,) ~~ -1.2

=

total left-hand side of Eg. (cl

3922.8

Evidently our assumption that all the stearn condenses is wrong. since the \'alue of the left-hand side of Eq. (el is too small. With this initial calculation we can foresee that the final temperature of the mixture will be 228 F and that a saturated steam-liqt:id water mixture will finally exist in the vessel. The next question is, How much steam condenses? Let S = the amount of steam at 228'F:

The formation balance. As a unit n' with only modest br ing air (Q = 0), nor ~~ 0). Both kinetic . reduces to Now

C

and

S[H" - pV]"po, ..;- (20' - S)[H" - pVlBQUid = 6010.6

(flH,.)".o, = 1156.\ Btu.!lb

=

196.2 Btullb

(pV)".o,

=

(19.7)(12)'(20.08)/(778) = 73.2 Btu/lb

(PV)IiQUid

~ 0.0

(AH,,)IiQUid

S(l\56.1 - 73.2) -i- (20 - S)(l96.2)

=

6010.6

S = 2.351b In effect, 6.00 - 2.35 = 3.65 Ib of steam can be said to condense. If there had been only I lb of steam instead of 6 Ib to start with, then on the first calculation the right-hand side of Eg. (c) would have been 1168.0

+ 30.2

- 574.4 - (1/6)(453.2)

=

Assume that an ideal,: pressure change with ... error the change in L' roughly 3° t04'FjlOJ(! the mountain, the e1"." to be for a cloud to f.·r bottom of the peak Oc. humidity can be for :\ , , ton (altitude 6388 fl), "I is 40 percent. A more complieJI'," is the formation of con!

548.3 Btu

Hence lower temperatures would have been chosen until the left-hand side of Eq. (c) was small enough. Only liquid water would be present.

4.6 Reversible p(( mechanical C,'

EXAMPLE 4.17 Application of the energy balance The peaks of Mount Everest. Mount Fujiyama. Mount Washington, and other mountains are usually covered with clouds even when the surrounding area h3s nice weather. It is clear that water condenses because the temperature is lower, but why doesn't water condense elsewhere at the same altitude? Is the cloud cap stuck b: some force? See Fig. E4.l7.

The energy balance (,i cnergy without inq uin r experience with mach,') transformation are no . cannot be transferred I: the reverse direction. ) mechanical work, T01 ~ccond law of thcrrnl,d. As applied 10 11,,; '<'cond law of thern1
(a) The so-called Fig. FA.17.

~~"""'-r_._ _ _ _ _ _ ,..,........_~_--._......_ _ _.....-

_______ • "'_''''''''_''''''_''''''''_.~'_~~''''''lI'''"- .....",...,--~.".

potential

I

encr~':

Rel'ersible Processes and the Mechanical Energy Balance 303

Sec. 4.6

Chap. 4

SO/lltion: The formation of cloud caps can be predicted with the aid of the general energy balance. As a unit mass of warm moist air, our system, moves up the side of a mountain with only modest breezes, it cannot exchange energy by way of heat with the surrounding air (Q = 0). nor does it exchange a significant amount of water vapor (m, = m, ~ 0). Both kinetic and potential energy effects can be neglected, so that Eq. (4.24) reduces to

= 3924.0 1 Btu __ _ , 78 (ft)(lb,) -- _1._-_

= 3922.8 since the value of the ::>n we can foresee that the =ated steam-liquid water -,.. much steam condenses:

:C.:Jg,

~ow

AO = and

6010.6

.:::rt with, then on the first

ieft-hand side of Eq. (el

~

r,

~ pdV

Assume that an ideal gas expands so that pV = RT. Then, from information as to the pressure change with altitude plus the fact that C,. = 0.25 cal!(g moleWK). by trial and error the change in temperature can be determined as the air rises. The cooling is roughly 3° to 4' F/l 000 ft of elevation. During the descent of the air on the other side of the mountain, the cloud will evaporate. You can also calculate how high a peak has to be for a cloud to form for a given relative humidity and temperature of air at the bottom of the peak or, alternatively, given the peak height. what the highest relative humidity can be for a cloud not to form. As an example of the latter. for :-'1t. WashingIOn (altitude 6388 ft), the maximum percent relative humidity of 70'F air at sea level is 40 percent. A more complicated but related problem that we do not have the space to go into is the formation of contrails by supersonic aircraft.

_ Btu/lb

3.3 Btu

~

C,dT~C,(T,-T,)

T,

- Jr.

W=

.=

T.

J

f 4.6 Reversible processes and the mechanical energy balance

Washington, and other =ounding area has m,'" Cllture is lower, but \\In :::ioud cap stuck by sorn,·

l I

The energy balance of the previous section is concerned with various classes of energy without inquiring into how "useful" each form of energy is to man. Our experience with machines and thermal processes indicates that all types of energy !fansformation are not equally possible. For example. it is well known that heat cmnot be transferred from a low temperature to a high temperature. but only in the reverse direction, nor can internal energy be completely converted into mechanical work. To account for these limitations on energy utiliz.J.tion, the ~crond law of thermodynamics was eventuaJly developed as a general principle. As applied to fluid dynamic problems. one of the consequences of the '«'ond law of thermodynamics is that two categories of energy of different 'quality" can be envisioned: (a) The so-called mechanical forms of energy, such as kinetic energy, potential energy, and work, which are completely convertible by an ideal

. tz g »



., j

j

~

i

\

I

i

~

j 1

304

Energy Balances

Chap. 4

(frictionless, reversible) eng me from one form to another within the class. (b) Other forms of energy, such as internal energy and heat, which are not so freely convertible. Of course, in any real process with friction. viscous effects. mIXIng of corn· ponents, and other dissipative phenomena taking place which prevent the corn· plete conversion of one form of mechanical energy to another. allowances \\ ill have to be made in making a balance on mechanical energy for these "losses" Hl quality. To describe the loss or transfer of energy of the freely convertible kind 10 energy of lesser quality, a concept termed irreversiblily has been coined. :\ process which operates without the degradation of the convertible types of energy is termed a reversible process; one which does not is an irreversihlc process. A reversible process proceeds under cpnditions of differentially balanceJ rorces. Only an infinitesimal change is required to reverse the process, a concerl which leads to the name reversible. Take, for example, the piston shown in Fig. 4.13. During the expansion process the piston moves the distance x and the volume of gas confined in Ihe YI

.A

..

\cf. 4.6

: r ,,:110n 1c\\

Jlnhalan(c

"1.

i'U)\,:C\S

U:n('

(1\((,'<",

.'.I;d

~()

"c '-

(\lh.ln~:cr.

,

'""c

I'Ll,: t'\llher )',1:\ ',1

rcl'rr<,.rnl'

.';;r :n,I\Ij;i,

i !:rn. Ill,ln',

I'."

\ 1, .ll'l

rlrl11(,I):'1 ,.!

(qU.lth'll

_x_

"hrH~

I.

\:\'0 I'~

v

j'f .:\

Yz

I ,;; J .1 ~ :

Fig. 4.13. Gas expansion.

piston increases from VI to V 2 • Two forces act on the reversible (frictionb, I piston; one is the force exerted by the gas, equal to the pressure times the area.'1 the piston, and the other is the force on the piston shaft. If the force exerted bl the gas equals the force F, nothing happens. If F is greater than the force of tl,·, gas, the gas will be compressed, while if Fis less than the force of the gas. the ~.I' will expand. In the latter case, the work done by the expanding gas and the piston will b.' W = F dx. The work of the gas would be 11'= j P dV if the process \\ Ci' reversible, i.e., if the force F divided by the piston area were differentially less .Ii all times than the pressure of the gas, and if the piston were frictionless; but In.l real process some of the work done by the gas is di.ssipated by viscous dYe, I , and the piston will not be frictionless so that the w(~'rk as measured by i r .. , will be less than f p dV. For these two integrals to be ~ual, none of the a\'~iI.II'" energy of the system can be degregated to heat or irnl.ernal energy. To adll"'; such a situation we shall have to ensure that the Il1JI0Vemenl of the pisl"n I'

f

1

;--';0 TC.

110

h.~

~.,:

!",

f

~

I ( \ fIt, ,~

\\,

• I

...

1\""'11

.,

i i'

--_..

,...........:-..... " (fJiap. 4-

) anotHer within the ..!

hcat;.wHichamnot

~S;. mixing: 01: com· .:~. prevenr the. com· :ITer.;. allowances wi II i.cllrthese:"losses" in

cnnl.\ro1li We: kin d to lJeen. aoi ned.. A :mrver.tibJe.: types· of .:r ill' aw irreversible .,1ero:ntiaJIY balanced :1
'::-~.a:conc.cpl

"TiltH' ti/\u: wpnnsion ~ mlUltum:t il11 the

Rnersible Processes and the .\fechanical En"l!rY Balance

Sec. 4.6

frictionless and that the motion of the piston proceeds under on1!y a differential imbalance of forces 50 no shock or turbulence is presenl. Na1urally such a process would take a long time to complete. No real process which involves friction. shock, finite temperature differ· ences, unrestricted expansion, viscous dissipation. or mixing cam be reversible. Jnd so we see that the dropping of a weight from a tower. the operation of a heat exchanger, and the comtusion of gases in a furnace are all irrevcrsible processes. Since practically all real processes are irreversible you may wonder why we bother paying attention to the theoretical reversible process. The reason is that It represents the best that can be done, the ideal case, and gives us a measure of our maximum accomplishment. In a real process we cannot do as well and so :lrc less effective. A goal is provided by which to measure our cffectiveness. And then. many processes are not too irreversible. so that there is not a big discrepancy between the practical and the ldeal process . A balance on mechanical energy can be written on a microscopic basis for an elemental vDJume by laking the scalar product of the local velocity and the cquatiDn of motion."· After inlegration over the entire .olume of the system the .\lcad.l~Sla1e mechanical enef'Ry balance (for a system with mass interchange with the surrounding.;) becomes. on a per unit mass basis,

A(K + Pg + j"" V tip ;,

.::rsibfe (ffictionless I area of "' e force exerted tw ... n the force of the ; ohlle gas. the gas

.;~e t~mes·the

.d the'pisWrr wifl be the proces~ were ..!!tkt-~nti:a[[y less at .:;tiolll.:s!; uu tin 3 :>)' viscous effects. .::asured by J F dl ~e orthe availahlc ·nergy. To achic\c : of the piston is

305

-:--

IV + Ev

= {)

(4.30)

where E, represents the Joss of mechanical energy. i ..e., the irreversible conver· sion by dhejlowing fluid of mechanical energy to internal energy. a term which must in each individual process be evaluated by experiment (or. as occurs in practice, by use of already existing experimental results for a similar process). Equation (4.30) is sometimes calleD the Bernoulli equation. especially for the reversible process for 1-vhieh E,. = O. The mechanical energy balance is best applied in fluid-Jlow calculations when tbe kinetieand potential energy terms and lhe wOTk are of importance. and tbe friction losses can be evaluated from h:mdbooks v,:ilh the aiD nftriclion factors or orifice coefficients.

EXA~IPLE

4.18

Application ·of the mechanical
Calculate the work per minute requi:red to pump 1 lb ..of water per minute from ! 00 ps.iaand SOP 10 ] 000 psia :and 100' F.

Snlutiol1: lfthe process.is .assumed 10 be reversible so 1hat E,. =0. if the pump is 100 percent rflicient. and if no kinetic or po.tcntial energy terms are inmlved in the process. Eq. \4.30) r.educes to

-

W= ~'R.

1l. Bir.d. up. cit.

SP'p, Vdp "

306 Energy Balances

Chap. 4

From the steam tables. the specific volume of liquid water is 0.01608 ft',lb m at BOT and 0.01613 ft' /Ib m at 100' F. For all practical purposes the water is incompressible. and the specific volume can be taken to be 0.0161 ftJilb m • Then, Basis: I min W = _I Ibm nun

flOOO 0.0161 100

d 'P

1 Ibm 0.0161 ft' (1000 - 100) Ib r I (12 in.), [

min Ibm Btu = _2.6S mm

in.'

1ft

I Btu 1778 (rt)(lb t ) (e)

Tb:·. ( ., 1\, ~

About the same value can be calculated using Eq. (4.24) if Q = K = P = O. because the enthalpy change" for a reversible process for I Ib of water going from 100 psia and 100'F to 1000 psia is 2.70 Btu. However. usually the enthalpy data for liquids other than water are missing, or not of sufficient accuracy to be of much value. which forces the engineer to turn to the mechanical energy balance. One might now well inquire for the purpose of purchasing a pump-motor. say. as to what the work would be for a real process, instead of the fictitiOUS reversible process assumed above. First, you would need to know the efficiency of the combined pump and motor so that the actual input from the surroundings (the electric connection) to the system would be known. Second, the friction losses in the pire. valves. and fittings must be estimated so that the term t . could be reintroduced into Eq. (4.301. Suppose, for the purposes of illustration. that t,. was estimated to be. from an appropriate handbook. 320 (ft)(Ib r J:lb m and the motor-pump efficiency was 60 percent (based on 100 percent efficiency for a reversible pump-motor). Then,

=~F:L!...b"i-;~CT.:--rt-,,=1mm b'3m = .:.;1

0.41 Btu/min

pIC

(d) In

";

f·l\

'll r I ~ «(.~

)

SIn, , 11 (" I; V.ll:

I: : ~

"

d"(

"f t"! t

~

I': '. ~, " If) I

ct '.

\\ ;:!\

111l' '·t;

'Ik

"

'("\

(:,

b)

"I'

Remember that the minus sign indicates work is done on the system. The pump motor must have the capacity 3.09 Btu I min

EXAMPLE 4.19

Calculation of work for a batch process

One pound mole of N, is in a horizontal cylinder at 100 psia and 70'F. A I-in.' piston of5-lb weight seals the cylinder and is fixed by a pin. The pin is released and the Nl volume is doubled. at which time the piston is stopped again. What is the work done by the gas in this process?

\ ~

.

Solution: (a) Draw a picture. See Fig. E4.19. (b) ls the process a flow or nonflow process? Since no material !caves or enters the cylinder, let us analyze the process as a nonflow system.

I ,"II

"From J. H. Keenan and F. G. Keyes, Thermodynamic Properties 0/ Steam, Table J. Wiley, New York, 1936.

',,"

--...........-_ _.....- - - - -_ _.~ ••"".,.........~_..,.~._,.~......._ _ _ _ "''"'_ .C"":-,,....,.-~_~~~~~.'.-,~" . .•~-

\'

Chap. 4

Reversible Processes and the "fechanical Energy Balance 307

Sec. 4.6

0.01608 ft'/lb m at 80 F incompressible, and

/lniliOI Syslem

:::7 is

r

1~~-;;9 f~3!

lip '100 PSIO'I it===T!===== i I

11~1~~~j

I Btu

Fig. E4.19.

778 (ft)(lb,)

if Q = K = P = O. :ll water going from 100 me enthalpy data for ::..::y to be of much value . .:.iance. =::tg a pump-motor. S'". :ne fictitious reversibic ::::Jency of the combincJ -:- ~s (the electric con nee_ In tpo -;oe, valves. ar,J :-:ldl 0 Eq. (4.301 . :: to om an appro''';:-Jency was 60 percent Then, ,.i)

Btu/min

estern. The pump motor

(c) This process is definitely irreversible because of friction between the piston (and the gas) and the walls of the cylinder, turbulence in the the gas. the large pressure drops, etc. (d) In selecting a system to work with. we can choose either the gas itself or the gas plus the piston. Let us initially select the N~ gas as the system. with the surroundings being everything else including the piston and the c\ iinder. (c) Since the gas expansion is a nonftow process, Eq. (4.30) does not apply. Even if you had an expression for the unsteady-state mechanical energy balance (which we have not discussed s'nce it involves the concept of entropy). it would not be possible to calculate the work done by the gas because the process is irreversible and the E, term wodd not be known. As might be expected, a direct application of IV = r F dl fails for this problem because the pressure of the gas does not necessarily equal the force per unit area applied by the piston to the gas, i.e., the work cannot be obtained from IV = p dV. (f) Let us change our system to include the gas, the piston, and the cylinder. With this choice, the system expansion is still irreversible, but the pressure in the surroundings can be assumed to be atmospheric pressure and is constant. The work done in pushing back the atmosphere probably is done almost reversibly from the viewpoint of the surroundings and can be closely estimated by calculating the work done on the surroundings:

i

W

=

-2hp p

Assuming that z ::lsia and 70°F. A I-in.: ~e pin is released and the ~in. What is the work

=

I"F dl IV. p dV = pA V =

I,

VI

14.7Ib, 144 in. 2 2120lb r in. I ft 2 = ft2

= 0.995 and is constant, the initial volume of gas is

ft' 530'R 114.71 0 .995 V 1 = 1 359 Ib mole 492~j{ t 100

=

569 ft'/lb I . mo e

and AVof the gas is 2VI - VI = VI = 56.9 ft'llb mole so that the volume change of the surroundings is -56.9 ft'. Then the work done on the surroundings is lIb mole

-::::l!erial leaves or enter> ,\'Stem. .-;"r:rlies of Steam, Table ~.

The minus sign indicates that work is done on the surroundings. From this we know that the work done by the alternative system is W., • t • m

=

-W.u"ouodin ..

-~",,,....---------,.....--~

= -(-120,000) = 120,000 (ft)(lb,)

.___

308

Energy Balances

Chap. 4

(g) This still does not answer the question of what work was done by ... , gas. because our system now includes both the piston and the cylinder as well as the gas. You may wonder what happens to the energy transferred from the gas to the piston itself. Some of it goes into work against the atmosphere. which we have just calculated; where does the rest of it go? This energy can go into raising the temperature of the piston or the cylinder or the gas. With time, part of it can be transferred to the surroundings to raise the temperature of the surroundings. From our macroscopic viewpoint we cannot say specific· ally what happens to this "Iost" energy. (h) You should also note that if the surroundings were a vacuum instead of air, then no work would be done by the system, consisting of the piston plus the cylinder plus the gas, although the gas itself probably still would do some work.

4.7 Heat of reaction So far we have not discussed how to treat energy changes caused by chemical reactions, a topic that is frequently called thermochemistry. The energy change that appears directly as a result of a reaction is termed the heat oj reactioll, AH.... a term which is a legacy of the days of the caloric theory. ("Heat of reaction" in terms of modern concepts is best thought of as the energy released or absorbed because of the reaction that may be transferred as heat in certain types of experiments which are delineated below but may also end up as internal energy or as another form of energy.) The energy released or absorbed during a reaction comes from the rearrangement of the bonds holding together the atoms of the reacting molecules. For an exothermic reaction, the energy required to hold the products of the reaction together is less than that required to hold the reactants together and the surplus energy is released. To take account of possible energy changes caused by a reaction, in the energy balance we incorporate in the enthalpy of each individual constituent an additional quantity termed the standard heat (really enthalpy) of formation, AH;, a quantity that is discussed in detail below. Thus. for the case of a single species A without any pressure efrect on the enthalpy and omitting phase changes, (4.31) For several species in which there is no energy chanr,e on mixing we would havc (4.32) where i designates each species. ", is the number of moles of species i. and s is the total number of species. If a mixture enters and leaves a system without a reaction taking place. we would find that the same species entered and left so that the enthalpy change in

..

",,+.A4

...>......

~_

"S.'_

......

Sec. 4.7 either the flow or the ferent with the modif because the terms tha would cancel. For ex: output enthalpy woul, AHoutpu:

=

11

and the input enth~lp:

AH,oput =

IJ

We observe that f../( terms that we havc de However, if are:: differ, and the terms example. suppose tha t 4 leave. Then

To determine the the experimenter USt);: energy, potential ener,' process (a bomb cal,,, process in an expcrim'." of Eq. (4.33) are Zcr<' . term) version of Eg. (~ and

In other words, th,' lllent appears as heal '.' is observed is termed " , II' '" 0, and the ten'I'''': 'I reams, can also b~ 1> .'1

"de of Eq. (4.33) is ;,':.' calorimeter of co 11> 1:1 ". The caleulati,)ll' : ,"I hough the dTt'L' I ,'I '''lder most cond,lh'!1' 'i ;~~~!kc the n('c('~sar: ,-'. There are certi\lll' "

*

AIioo."

Chap. 4 ,'K

was done by ... ,

gdl.

';::1 the cylinder as well

J-;

transferred from tic .against the atmosphere. :' it go~ This energy COl" ::vlinder or the gas. \\n:l c. to raise the temperature :-n we cannot say specific. 71')'

Hmt of R.em::t.ion 309

5ec.4.7

either lru: flow DT the :accumulation terms in Eq. (424) would not beauy different with tru: modification described above than ...·.hat WI: have m.ed hefon:, "~cau!'C tru: :terms that account rOT the beat ,of reaction in tbe energy ,balance \loula :ca=1. FOTnampll:, fOT lru: .case of Iwospt:cicsin a flow system, the ourputl:nthalpy would be ...... 0

r

..

0

AHnut.ut='ll/1Hfl 7'll:PJif2

cut , + J;:r'T-zer (n,C p, 711:iCp2)dT

and lht: irrprrt =thalpy would be , a vacuum instead of air. ::1g of the piston plus tne .itill would do some work.

Mhrput

=

'n,M!, -+ 'll::!lii/2 -+ .s::,'0n,Cp, + 1I:1.Cp2),dT

.ubSCT¥1: that .t:UiQu,,"ut - .t:Ui,rrput would l1:. Then !'vI:

1.t.'TlTIS that

~es

caused by chemical The energy change .' j thf' heat of reactioll. .:ori ry. ("Heat (If ;' as J1ergy released 2!Ted as heat in certain . also end up as internal '~:1 or absorbed during J :::-mg together the atoms . 1ne energy req ui red to _;.at required to hold the ·~ry.

J;.ff=t -

A.H'n= IQn,tlii';,

t

-+: 7T.'Td

(4.31 ) :1

mixing we would h:nc dT

(4.32)

:es of species i, and s is "action taking place. \I e : the enthalpy change in

{n,Aii~,

+ .tY1i1~J

1(..1133)

:T;Ycf

il-o ktermin.e tru: -.valuesof the 'S.tandar.dheats (~nthalpies)af formation, ce-Kperinlenteru5ually .scl.crts :either :asimple ilow prD.cess .witbout .kinetic mergy, 'Po.tential -energy, arwor·k ~ffcc(s (a :flow .caloriln.etCTl, or a -&imple ,batch 70=55 {a :b.ornb !:alarimeter) ..in ·whicbcto .conduct.the Tcaction. For .the flow rrOl!C!lS'in:mJ celCperiment in which :tru:scnsibleheat :terms an the Tight ,hand side .11 :Eq.(4.33) :ar1: zcroandno work is .done, lru: steadystatc (no .acc.umulation ternl) w1:rsian
.= by a reaction, in the ::aividual constituent an =:nhalpy) of formati(lll. . for the case of a singk -,V and omitting phase

-+ m~~... ) -

n", lenjC~'J + :11jC.".,).dT - J"',n (n,Cp, + :1~Cp2)dT

@ = £\[.tlni] !A'IJ~xn

= ( I ·products

= llH"Q '~34)

·n ~ '/\[]"t, -rnc;tant5

]n.otftcnwards,the.energy change caused by thereac(it~1l in s.uch.an experiment :appears as heat transferred to or from ·the system, and the 'value of Q that " uoserved ,jstermed thc heat of reaction. A flowcalorimet.:r in which Q = 0, II' =0, :and :thetempcrature change is 'measured between the input .and output ,( reams, .can ;also.bc .used to .compute the b!:at of reaction, because .the left hand 'IJe of Eq. (4.33) is zero. Refer to ,Sec.4.7-3 for.ananafysis of the :batch bomb ,:dorimcter ofconstant volume. The calculations that we shall'make here will all he for 10w 'pressures,and, .1!though the cflect of pressure UP01! heats of reaction is Telatively negligible under lllost.conditions, if exceedingly high pressures arc encountered, you should ~:"kc the necessary corrections as explained in most texts 01! thermochemistry. There.are certain conventions and symoolswhich you should always keep in

rt

310 Energy Balances

Chap. 4

mind concerning thermochemical calculations if you are to avoid difficulty later on. These conventions can be summarized as follows: (a) The reactants are shown on the left-hand side of the equation, and the products are shown on the right-for example, CHlg)

+ H,O(I) ~ CO(g) -:- 3H,(g)

(b) The conditions of phase, temperature, and pressure must be specified unless the last two arc the standard conditions, as presumed in the example above, when only the phase is required. This is particularly important for compounds such as H ,0, which can exist as more than one phase under common conditions. If the reaction takes place at other than standard conditions, you might write CH.(g, 1.5 atm)

'iii

.'Icc. 4.7

for all COlli!" then the the experience a ~tandard he (If react ion I A form. forms I mol, "f reaction 1 mental) stan standard hea

EXAMPLE 4

+ H,O(l) ~ CO(g, 3 atm) + H,(g, 3 atm)

(c) Unless otherwise specified, the heats of reaction, the enthalpy changes. and all the constituents are at the standard state of 25'C (7TF) and I atm total pressure. The heat of reaction under these conditions is called the standard heat of reaction, and is distinguished by the superscript 0 symbol. (d) Unless the amounts of material reacting are stated. it is assumed that the quantities reacting are the stoichiometric amounts shown in the chemical equation. Data to compute the standard heat of reaction are reported and tabulated in two different but essentially equivalent forms:

So/uthln.

In the

Ir .•

7"ai>lIlatcd da/.J

!>"th If ,(~) ,,; .'\ppenuI\ I " tills (' . .)mrl\lll~~i ;1 hove),

The \ ,I

tlllllhc

IC.I(tll ~i

or by somc (",i'

(a) Standard heats of formation (b) Standard heats of combustion We shall first describe the standard heat of formation (llfI;) and afterward the standard heat of combustion (lll(), (The superscript C denotes standard stoIC. I The units of both quantities are usually tabulated as energy per mole. such as kilocalories per gram mole. The "per mole" refers to the specified referencc' substance in the related stoichiometric equation. 4.7.1 Standard Heat of Formation. This is a special type of heat e1 j' reaction. one for the formation of a compound from the ekments. The initi,iI reactants and final prod ucts mllst be stable and at 2YC and I atm. The reactie1 n does not necessarily represent a rcal reaction that would proceed at constant temperature but can be a tictitious process for the formation of a compound fronl the elements, By defining the heat of formation as zero in the standard state felf each element. it is possible to design a system to express the heats of formatio"

F,\,.,\II'U: J.; SUpf'tl'-C :;.

'n . ·!~f.d d,Il.!

l

,

'1 hl\ '.1 ('I',rl~~) 1!lX-I.l'",) ~llcd ~

N

(I I

/1

,\ i \. r n",~'r ~'\ ' , ( ,,1 ( , I

Chap. -4 :: avoid difficulty later

.:.!le equation, and the

- H2 (g)

=e must

be specitled

. as presumed in the _. This is particular"

= exist as more than

Heat of Reaction 311

Sfr.4.7

for all compounds at 25°C and I atm. If you use the conventions discussed above, then the thermochemical calculations will all be consistent. and you should not experience any confusion as to signs. Consequently, with the use of well-defined standard heats of formation. it is not necessary to record the experimental heats of reaction for every reaction that can take place. A formation reaction is understood by convention to be a reaction which forms I mole of compound from the elements which make it up. Standard heats of reaction for any reaction can be calculated from the tabulated (or experimental) standard heats of formation values by using Eq. (4.34), because the standard heat of formation is a state (point) function.

=n takes place at other EXAMPLE 4.20 Heats of formation

,+ H,(g, 3 atm) -::!le enthalpy changes. _ of 25'C (77' F) anJ " th('<~ conditions is =ai! . the super· ~. it is assumed that :-:1ounts shown in the

::::lorted and tabulakd

What is the standard heat of formation of HCI(g)? Solulion: In the reaction at 25°C and I atm, iH,(g)

l1frf (~) g mole

Tabulated dolo:

0

- 22063

'

hath H,(g) and CI,(g) would be assigned 11if'; values of 0, and the value shown in Appendix F for HCl(g) of -22.063 cal/g mole is the standard heat of formation for thIS compound (as well as the standard heat of reaction for the reaction as written above). The value tabulated in Appendix F might actually be determined by carrying out the reaction shown for HCl(g) and measuring the energy liberated In a calorimeter, or by some other more convenient method.

EXAMPLE 4.21

. ~) and afterward the =:::lotes standard statl!.! -::y per mole. such as ::.e specitled reference

0

+ iCI,(g) ---+ HCI(g)

Indirect determination of heats of formation

Suppose that you want to find the standard heat of formation of CO from experimental data. Can you prepare pure CO from C and 0, and measure the energy liberJtcd? This would be far too difficult. It would be easier experimentally 10 find the energy liberated for the two reactions shown below and add them as follows: Basis: I g mole CO

11 if~x.(experimenlal) =ial type of heat "I' ~lements. The initi,JI .:. I atm. The reaction ::>roceed at constant . of a compound from ::1e standard state f,'r :,e he3 1 < of formati,'n

A:

B' B:

C(f3)

+ 02(g) _ ........ CO,(g)

CO(g) --'C(f3)

~02(g) .~...

+ -jO,(g) -

l1fi;"

A-B

-94.052 kcal g mole -67.636 kcal g mole

CO,(g)

....... CO(g)

~ (-94.052) - (-67.636) ~

Iliff

=

-26.416 kea! g mole

lhe energy change for the overall reaction scheme is the desired heat of formation per n'ole of CO (g). See Fig. £4.21.

Su.4.7 You, algebrai.: 1 ~ubtractccl

rules of pr. calculatwr;

To sin at 25'-(, aTh as H:O(~) into accour

t,ll"" ~ -94052 Fig. £4.21.

EXAMPLE 4.22 Calculation of the heat of reaction Calculate!:J.fi~xD for the following reaction of 5 moles of NH,:

4NH,(g)

+ 50,(g) ----+ 4NO(g) + 6H zO(g)

Solution: Basis: 4 g mole NH,

tabulated data

NH,(g)

O,(g)

NO(g)

HO,{g)

-11.04

o

+21.60

-57.80

!:J.fi,} per mole at } 25°C and I atm (kcal/g mole)

We shall use Eq. (4.34) to calculate I1H7xn for 5 g moles of NHJ:

!:J.H:xo = (4(21.60) + 6( -57.80)] - (5(0) -216.24 kcal/4 g mole

=

I1H

O

UD

or

+ 4( -1104)] l1H:xn = -54.0 kcal/g mole NH,

= (-54.0) kcalj5 g mole NH J

g mole

NH,l

= -270.0 kcal

Practically. I cai/g mole \I There;1' tion. A g()od 500, and i h ' heals of f.ll" heat of flm:' dlcmic;d Illc"

and Dora1,\1.· RemembL'r t:: for exothcm I

EXAMPLE 4.23 Heat of formation with a phase change If the standard heat of formation for H,O(I) is --68.317 cal/g mole and the heat of evaporation is -'-10,519 cal/g mole at 25°C and I atm, what is the standard heal of formation of H 2 0(g)? Solution: Basis: I g mole H 2 0

4.7.2 St ..He the

~tandard

,t.,.,

"Jlh the

,[.I1.j

We shall proceed as in Example 4.21 to add the known !Chemical equations to yield the desired chemical equation and carry out the same opcrations on the heats of reaction. For reaction A, Lifi:xn = .2: bit} pToducL~ - L if; re"a-Ctant,: A:

B: A

H,(g) + iO,(g) ----+ H,O(l) H,O(l) ---->- H 2 0(g)

+ B:

!:J.fi:XOA

H,(g)

(b) 111(' :::


I1fi:xn = -68.Jn7 kcal'g mole = .,. 10.5R9 kcal;g mole

for c\:

I1fi,',p

+ iO,(g) -~ H,O(g)

+ I1fi~,p= I1fi:xoH

,O(i)

= I1fi;H,Oii) = -57.79/$ kcal'g mole

St'I.'\li!;

chemical cil,::

)OK." \ ... { I'I('~L

"B("II":.II 'e

312

~"~."

Heat of Reaction

Sec. 4.7

313

You can see that any number of chemical equations can be treated by algebraic methods, and the corresponding heats of reaction can be added or subtracted in the same fashion as are the equations. By carefully following these rules of procedure, you will avoid most of the common errors in thermochemical calculations. To simplify matters, the value cited for tJ.H"p of water in Example 4.23 was at 25"C and I atm. To calculate this value, if the final state for water is specified as H 2 0(g) at 25'C and I atm, the following enthalpy changes should be taken into account if you start with H 2 0( I) at 25'C and I atm:

H 20(I) 25°C, I atm

l"H,

tJ.H"p at 25°C and I atm

H 2 0(I) 25°C, vapor pressure at 25°C I I

H2

flHI '" lJ.H.... p at the vapor pressure

or

water

• 0(g) 25°C, vapor pressure at 25°C

l"H'

H2 0(g) 25°C, I atm H0 2 (g) )

Kcal/g mole NH3 .Okcal

Practically, the valve of tJ.H"p at 25°C and the vapor pressure of water of 10,495 cal!g mole will be adequate for engineering calculations. There are many sources of tabulated values for the standard heats offormation. A good source of extensive data is the National Bureau of Standards BlIfletin 500, and its supplements, by F. K. Rossini. A condensed sct of values for the heats of formation may be found in Appendix F. If you cannot find a standard heat of formation for a particular compound in reference books or in the chemical literature, tJ.fi~. may be estimated by the methods described in Verma and Doraiswamy 3o or by some of the authors listed as references in their article . Remember that the values for the standard heats of formation are negative for exothermic reactions.

mole and the heat Ihe standard heat of

4.7.2 Standard Heat of Combustion. Standard heats of combustion are the second method of expressing thermochemical data useful for thermochemical calculations. The standard heats of combustion do not have the same standard states as the standard heats of formation. The conventions used with the standard heats of combustion are

'",::mical equations «l ..:nons on the heats of

(a) The compound is oxidized with oxygen or some other substance to the products CO,(g). 11,0(1), HCI(aq)," etc . (b) The reference conditions arc still 25'C and I atm. (c) Zero values of tJ.FI: are assigned to certain of the oxidation products as, for example, CO,(g), H,O(l). and HCl(aq).

'g

;':.ants·

kcal/g mole " kcal/g mole

]OK. K. Verma and L. K. Dorai,wamy. Ind. Eng. Chern. Fundamentals, v. 4, p. 389

. kcal/g mole

\19(5) .

"IlCl(aq) represents an infinitely dilute solution of Hel (sec Sec. 4.8).

~~"'.A ...

314

Energy Balances

Chap,

.f

(d) If other oxidizing substances are present. such as S or N,. or if Ci. ;" present, it is necessary to make sure that states of the products ;,,: carefully specified and are identical to (or can be transformed intolli,: final conditions which determine the standard state as shown , .. Appendix F. The standard heat of combustion can never be positive but is always negal/I <: A positive value would mean that the substance would not burn or oxid,,": . In the search for better high-energy fuels. the heat of combustion of ca,: element in terms of energy per unit mass is of interest. You can see from I,' 4.14 tbat the best fuels will use H. Be or B as building blocks. With everythirli' -;;;

"""

H

" 0"

5

"•.... ....

q

~

3'" '0 .0

"a; 3 =>

Be

~

2
2

~

~

=>

.0

E u

"

'0

Atomic number

Fig. 4.14.

else held constant, the greater the heat of combustion per unit mass, the lon;cr the range of an airplane or missile. For a fuel such as coal or oil. the standard heat of combustion is knolln .!' the healing ralue of the fuel. To determine the heating value, a weighed samrie" burned in oxygen in a calorimeter bomb. and the energy given oif is detected i·· measuring the temperature increase of the bomb and surrounding apparal l ., Because the water produced in the calorimeter is in the vicinity of r<)," temperature (although the g3S within the bomb is saturated with water vapor ... this temperature), essentially all the water formed in the combustion proel'" . condensed into liquid water. This is not like a real combustion process In . furnace where the water remains as a vapor and passes up the stack. Clll>: quently we have two heating values for fuels containing hydrogen: (a) the gr<" or higher, heating value, in which all the water formed is condensed into I:·

"'of

$'fmts'rt

t%

rtf'

([hap. 4

:h as' ~. or No.' or if Cl, is statl::S- of the products are :1 b~ r~ansformed into) the :1dard gt'ate as' shown in jve but is always negative. :lUld not bum or oxidize, ,:::at of combustio n 0 f each .;51. You can see from Fig. :1g bilJcks. With everything

iReal,of-Reactioll 315

S,-C, 4.7

!:quid state. :and (b) the net. or lower. heating value, in which .all the water :','rmed Temains in the va'por'state, The value you determine in the calorimeter is ::~C gross heating value,: this is .the one reported .along with the fuel analysis and ,hould bepn:sumed unlesstbe.data state speclticallYlhat the'net heating value is being Teported. You can .estimate the heating value of.a coal within .about 3 percent from Ihc Dulong formula": The hlgher heating v.alue (HHV) in British thermal units per pound = n4,544C

+

62,028

(R - .~ ) + 4050 S

",here iC = "
!H -

~=

w.eighr,fr.action n.ethydrogen

= itliXtal IW.ei:ght fr.acti.on ,hyrlr.ogen-

t w.eight fr.actionoxygen

The vJilu.es cdr iC, 11, 5, .an!! D.can be taken from the fuel -OT flue-gas analysis. If the ,heatinE'¥alue of.a fuel is known .an!! the C.an!! S .are known, the approximale VAlue
EX.\M'J"[JE -4.2'1

'Hcltting ,\lIlue ;of-coal

Coal :gasification .consists of the .chemical transformation of solid coal into gas. Thehea:tit\g¥alllcs,of coal differ, but the highcrthe heating vallJC. the higher the value of thegns 'Produced (essentially methane, caTbon monoxide, hydrogen, etc.), The foUOO\',rngcOO.'llhas.a ,r.cp.ortedheating\:alue of 12,810 Btu/lb45 received, Assuming that this is 'the gr.ossheating "\lalue, .calculate lbe net .heating value. fIJDITIP·

%

rc

7~,iO

lH2

$:6 J:6 1..7 <6.1 13.0

'N,

= per unit mass, the longer

lNetS

of amrbustion is known as :: value; a weighed sample is ~rgy given oif is detected by ::::d mrrounding apparatus. 1>, in the vicini!y of room ,:'U1~ated with water vapor at -: the combustion process is -. combustion process in a -,::sses up the stack, Conse· eng hydrogen: (a) the gross. -med is' condensed into the

(()-2

Ash

nOQ,o

Solutioll: The
lBaSis':

~OOlb .coal

ns received

The water formed un combustion is S!6:Ib'H'lnb'moleH211IbmoleH20 181bH,O =501bHO 2.02lbH, I I Ib mole H, lib mole H,O 2 "H. H . .Lowry, .ed., Clremi.my oJCoa/ Utilizatioll, Wiley, New York, 1945, Chap, 4.

........... ____

316 Energy Balances

Chap. 4

The energy required to evaporate the water is

The net heating value is

= 12,300 Btu(lb

The value 1020 Btu/lb is not the latent heat of vaporization of water at nOF (1050 Btu/lb) but includes the effect of a change from a hedting value at constant volume to one at constant pressure (-30 Btu/lb) as described in a later section of this chapter. A general relation bctween the gross heating value and the net heating valuc is net Btu/lb coal

= gross Btu/lb coal -

..... tf'

S,'c.4.7 How Can data? The PH' of combustion equations (B " The details arc

SOlbHzO 1020 Btu _ 510 Btu 100 Ib coal Ib H 2 0 - Ib coal 12,810 - 510

'~

91(% total H by weight)

C:I 0/

A:

B: C:

-A

H:I

+ 2B 20

Standard heats of reaction can be calculated from standard heats of combustion by an equation analogous to Eq. (4.34): AH~XD =

-[L; t,.H~ products

-

= -[L: n prod tJ.H c prod -0

L; H~ reactants] '" ~o L..J nrelct AH c react

(4.35)

Note: The minus sign in front of the summation expression occurs because the choice of reference states is zero for the right-hand side of the standard equations.

EXAMPLE 4.25

Calculation of heat of reaction from heat of combustion data

What is the heat of reaction of the following reaction: C,H,OH(I) + CH,COOH(I) ethyl alcohol acetic acid

-->-

C,H,OOCCH,(I) ethyl acetate

+ H,O(1)

.~

!!.I,

EXA:'IIPl.E 4.:'

The fOncHl 1'Clow at 25 C !i

I. C,II,I"I 2. C,I/,II" 3, H:ISI 4. H,OIII 5. C (cll.1::,. 6. C (p.l;':' Calculate the I,,'

Solution:

Basis: I g mole C,H,OH Tabulated data

C,H,OH(I)

CH,COOH(I)

C,H,OOCCH,(1)

1'1,00)

(a) The

qi'

(h) T))<'

':.i'

(() The .'

II,: ,.,' i

Af)~ per mole at} 25°C and I atm (kcal/g mole)

AU;"

-326.70

-208.34

-538.75

o

So/uriofT

ell Rtf":,, "due.,

We use Eq. (4.35):

AB;.. = -[-538.75 - (-326.70 - 208.34)] = +3.72 kcal/g mole EXAMPLE 4.26

Calculation of heat of formation from heats of combustion

Calculate the standard heat of formation of C,H,(g) (acetylene) from the standard heat of combustion data. Solution: The standard heat of formation for C,H,(g) would be expressed in the followin, manner (basis: I g mole C,H,):

2C(fi)

+ H,(g) ---.- C,H 2 (g)

AB'} = ?

( II C.II {."",

,I (,

(1)

.: 'II

('>I

I ~)

",

,1,

Chap. 4

:0 Btu -:' coal

.b :=ation of water at ,7°F (1050 -'f! value at constant volume to

_ .Later seclion of this chapter. ~ and the net heating value is rotal H by weight)

:-arn standard heats of com=oat•

.J

(4.35)

.-:: expression occurs because -..:land side of the standard

:eat of combustion data

::.: - ;3IJ(I)

+ H 2 0(1)

:~te

Heal of Reaclion 317

Src.4.7

How can we obtain this equation using known standard heat of combustion data? The procedure is to take the chemical equation which gi,·es the standard heat ()f combustion of C,H,(g) (Eq. A below) and add or subtract other known combustion equations (B and C) so that the desired equation is finally obtained algebraically. The details are shown below: I!H~ (kcaljg mole) A: C,H,(g) + 2i-0,(g) -----+ 2CO,(g) + H,O(I) -310.615 B: C(fJ) + O,(g) ~ CO,(g) -94.052 C: H 2 (g) + i-O,(g) ~ H,O(!) -68.312 -A +2B + C: 2C(fi) + H,{g) ~ C,H,(g)

I!H~.. = AH, = -(-310.615) + 2(-94.052) + {-68.317} = -:-54.194 gmo kcal le EXAl\fPLE 4.27

Combination of heats of reaction at 25°C

The following heats of reaction are known from experiments for the reactions below at 25°C in the standard thermochemical state: Afj~An rxn kcaljg mole -29.6 I. C,H.(g) + H,(g) ~ C,H.(g) 2. C,H.(g) + 50,(g) ~ 3CO,(g) + 4H,O(l) -530.6 -68.3 3. H,(g) + fO,(g) -----+ H,O(!) 4. H,OC!} -----+ H,O(g) + 10.5 (I!H~.p) 5. C (diamond) + O,(g) -----+ CO,(g) -94.50 6. C (graphite) + O,(g) -----+ CO,(g) -94.05 Calculate the following:

c-i,OOCCH,(1) H,O(I} -538.75

0

-';-3.72 kcal/g mole

(a) The standard heat of formation of propylene (C,H. gas). (b) The standard heat of combustion of propylene (CJH. gas). (c) The net heating value of propylene in Btu/ft' measured at 60°F and 30 in. Hg saturated with water vapor. SolUlion: (a) Reference state: 25'C, I atm; C(p) (graphite) and H,(g) arc assigned zero '·Jlues. Desired: 3C(p) + 3H,(g) -~ C,H.(g)

llH, = ? o:<:ats of rombustion

.=tylene) from the standard

:= expressed

in the followin,

+ H 2 (g) C ,H,(g) + 50,(g) (3) -4[H,(g) + lO,(g) (6) -3[C(PJ -+ O,(g) (I) C,H,(g)

~

(2)

~

(7)

--+

C,H,(g) 3CO,(g) -;.- 4H,0(J) H,O(I)]

~-~ CO,!g))

= (I) + (2) + (3) -+ (6): C,H.(g) --+ 3C(P>

I!H} C,H,!,)

; =?

_ ....

, ,e

=

-+ 3H,(g) +4.8 kcaljg mole

kcaljg mole I!HI = --29.6 llH, = -530.6 -4AH, = +273.2 -311H. = +282.2

llH:.n

=

-4.8

, i

I 1 j I I

l

I, I

!

318 Energy Balances

Chap. 4

(b) Reference state: 25'C, I aIm; CO 2 (g) and H,O{l) have zero values. Desired:

C]H,(g)

(I) C,H,(g) + rUg) (2) C]H,(g) -'- 50 ,(g) (3) -[H,(g) -'- ~O,(g)

~

+

~O,(g) --+

--~

C]H,(g) 3CO,(g)

-->

H ,0(1)]

3C0 2 (g)

+ 3H,O(l) kcal/g mole IiH, = -29.6 IiH, = -530.6 -IiH] = -'-68.3

+ 4H,0(I)

+ (3): + ~Oz(g) - + 3CO,(g) + 3H,O(l)

II . Ii

.~

(8) = (1) -:- (2)

C,H,(g)

IiH~"

= -491.9

IiH~ C,II,(iJ = -491.9 keal/g mole

;l

C]H,(g) -i- t02(g) --+ 3C0 2 (g)

(8) C,H,(g) + ~O,(g) --+ 3CO,(g) (4) 3H,O(l) --+ 3H,Olg) (9) = (8)

+

C,H,(g)

.'

/lC.l! ,·1 ,

!.1~C (." t

(e) Net heating value. Desired:

\\':: " \om~I;\~

\lU!

+ 3H,0(g) kcal/g mole IiH, = -491.9 3&H. = +31.5

+ 3H 2 0(1)

,hI t;.

! L.1I 11:( ~ 1::-

.!

Lr .1:

t!:('1,,('

L\~

(4):

+

~O,(g) --+

3 CO 2 (g) -'-- 3H,O(g)

&Hnn

IiH;

= -460,400 cal/g mole

IiH,

= (-460,400)(1.8) = -829,000 Btu/lb mole

= -460.4

rn .a t,,··

At 60'F, 30 in. Hg, saturated with water vapor, vapor pressure of H 2 0 at 60'F = 13.3 mm Hg pressure on gas = (760) ( 30.00) 29.92 - 13.3

= 763 - 13.3 = 749.7 molal volume

=

heating value =

~

3591760 I (460 -'- 60) 750'1 (460 .,.- 32)

750 mm Hg ]

= 384 ft lIb mole

82;8~0 = 2160 Btu/ft] measured at 6O'F and 30 in. Hg satd.

k,t n,'1 :' 1!Ir

ri:l;' oJ'

rt\'f

l! Be

t,;>!~)\. f'" ' ,

f,).

II ".

= 82;7~ = 2185 Btulft]

.

'

One of the common errors made in these thermochemical calculations is h' forget that the standard state for heat of combustion calculaticns is liquid \Valli, and that if gaseous water is present. a phase change (the heat ot' vaporization .'f heat of condensation) must be included in the calculations. If the 11nal prod uet ," Example 4.25 had been H,O(g) rather than H 2 0(I). we would have incorporal e,:

:

1 ~\n~ I! ' -:

= 379 ft'llb mole

net heating value

",

Ilor ~'. 11\<," hr,l~

II,: (

At 60°F, 30 in. Hg, dry, molal volume

4.73 Volum"

"

j

.«hap. 4

; have: zerO" values.

No:al/lpno/e = --4ffi. 9 'Y1BJ.. = +J15

/lB.

_ih'

--'---'~-., .-.~---

... ,

..

,

------.-.--

Su.4.7

Heat

0/ Reaction 319

the phase transition for water as follows:

- 3.HiO(l) kenl/emo le &HI = -29.6 I:S.H1. = -5:3U.6 -·M, = +68.3

---_._--_.

+ CH,COOH(l) ->- C H,OOCCH,(l) --'- H,o(l)

A:

C 2 H,OH(I)

B:

H,O(I) --+ H 2 0(g) !J.H,," = + 10.52 C,H 2 0H(l) -'- CH,COOH(l) ---+ C 2 H,OOCCH,(l) -:- H 2 0(g) !J.H;.. = (+ 3. 72) + (l0.52) = -+- 14.24 kcal/ g mole

2

!J.H~XQ

A

+ B:

= +3.72

With adequate data available you will find it simpler to use only the heats of combustion or only the heats of formation in the same algebraic calculation for a heat of reaction. Mixing these two sources of enthalpy change data, unless you take care, will only lead to error and confusion. Of course, when called upon, you should be able to calculate the heat of formation from the heat of combus· tion, or the reverse, being careful to remember and take into account the fact that the standard states for these two types of calculations are different. Since the heat of combustion values are so large, calculations made by subtracting these large values from each other can be a source of considerable error. To avoid other major errors. carefully check all signs and make sure all equations are not only balanced but are written according to the proper convention.

4.7.3 Heats of Reaction at Constant Pressure or at Constant Volume. Let us first consider the heat of reaction of a substance obtained in a bomb calorimeter, such as in a bomb in which the volume is constant but not the pressure. For such a process (the system is the material in the bomb), the general energy balance, Eq. (4.24), reduces to (with no work, mass fiow, nor kinetic or potential energy effects)

!J.U

60 C F and 3D in. Hg saId.

Qp - Q. ft'

~-mil:a1

calculations is ll' .:::ulations is liquid wal<:r. .~ heat of vaporization Of ::5. If the final product in would have incorporat~d

=

U" - U"

= Q.

(4.36)

Here Q, designates the heat transferred from the bomb. Next, let us assume that the heat of reaction is determined in a steady-state flow calorimeter with IV = O. Then if the process takes place at constant pressure the general energy balance reduces to: !J.H = Qp Here Qp designates the heat transferred from the flow calorimeter. If we subtract Qp from Q,. and introduce H = U + pT?, we find that

=

!J.H - (U" - U,,)

=

(U 2

-

UI) - (U" - U,,)

(4.37)

+ (m 2P/'2 -

mIPl i\) Suppose, furthermore, that the internal energy change per unit mass for the batch. constant volume process is made identical to the internal energy difference ~twccn the outlet and inlet in the flow process by a suitable adjustment of the :cmpcrature of the water bath surrounding the calorimeter. Then

(V" - U,,) = (U 2

-

UI)

1

i j

1

1

I j

I I i

1

! !

i

Q, - Q. = (m,P2V, - mIPJ'r I )

(4.38)

320

Energy Balances

Chap. 4

To evaluate the terms on the right-hand side of Eq. (4.38), we can assume for solids and liquids that the LlmpJ- change is negligible and can be ignored. Therefore, the only change which must be taken into account is for g2.ses present as products and/or reactants. If, for simplicity, the gases are assumed to be ideal, then at constant temperature

Sec. 4.7 Since reported value of - 26. 7

mpV =nRT Llm pi""

=

4.7.4 Inc

Lln RT

and

Qp - Qu = Lln (RT)

(4.39)

Equation (4.39) gives the difference between the heat of reaction for the constant pressure experiment and the constant volume experiment.

EXAMPLE 4.28 constant volume

Difference between heat of reaction at constant pressure and at

should calcula: aetualJy forme, the portion of I absorb some en of the reaction. include it in the tants or produc: or not, you mll'

4.7.5).

Find the difference between the heat of reaction at constant pressure and at constant volume for the following reaction at 25"C (assuming that it could take place): C(s)

EXAMPLE 4.29

+ ±02(g) ~ CO(g)

An iron pye;: dirt, rock, etc.) is I ing to the reactio'

Solution:

Basis: I mole of C(s) Examine Fig. E4.28. The system is the bomb; Q bomb.

= + when heat is

B: Bomb

absorbed by the in order to produll uct (cinder) which per kilogram of l'

ystem

Solillion: The material I, standard heat of r,

0'-

COlorimeter

!

Fig. E4.28.

< c:

From Eq. (4.39), Qp - Q, = Lln(RT).

f~

Lln=I-±=+±

q

since C is a solid. Qp - Q,

=

iRT

If the measured heat evolved from the bomb was 26,711 cal, Qp

=

Qu

+ 296 =

-26,711

+ 296 =

-26,416 cal

The size of the this correction is relatively insil!nificant compared to either the quantit' Qp or Q,.. In any case, the heat of reaction calculated from the bomb experiment" LlH un.

=

Q •. = -26,711 II

I

= 0.5(1.987)(298) = 296 cal

1

26,711 cal gmole

Required 0,: Enlering 0,: Entering N,: The solid ",aSf c pute the amount (':

',,*,,*",

Heat oj Reaction 321

Sec. 4.7

Chap. 4

Since reported heats of reaction are normally for constant pressure "rocesses, the value of -26,711 would not be reported. but instead you would report that

c. (4.38), we can assume ::lIe and can be ignored. .count is for gases prcscnt

AH

(4.39)

Qp

n

=

-26.416 I

=

26.416 caL g mole

4.7.4 Incomplete Reactions. If an incomplete reaction occurs. you should calculate the standard heat of reaction only for the products which are actually fonned from the reactants which actually reacl. In OlDer words, only the ponion of the reactants which actually undergo some change and liberate or absorb some energy are to be considered in calculating the overall standard heat ofthe reaction. ]f some material passes through unchanged, then you shoule! not include it in the standard heat reaction calculations (however. when the reactantsor products are at conditions Olher than 2YC and I atm. whether they react or not, you must include them in the enthalpy calculations as explained in Sec. 4.7.5).

\

- reaction for the constant ".ent.

:n c:onstant pressure and

= In.

~ases are assumed to b~

or

3t

c constant pressure and at

::lg tl>" " could take place}:

EXAMPLE 4.29 Inromplete reactions An iron pyrites ore containing 85.0 percent FeS1 and 15.0 ~rcent gangue (inen din, l'ock, etc.) is roasted with an amount of air equal to 200 percent excess air according 10 the ifeaction

=

heat is absorbed by the

4FeS z

+ nO l

->-

+ 850,

2Fc:O,

in or.der1Opmduce SO,. All the gangue plus the Fe10, end up in the solid waste product (cinder) which analyzes 4.0 percent FeS 1. Determine the standard beat of reaction per il..i]ogram ",f orc.

Solution: The material balance for the problem must be worked out prior to determitllng the standard heat of reaction. Basis: J.OO kg ore Ore Compo

kg

!FeS l

Gangue = 296 cal ~al,

-26,416 cal

=pared to either the quantl!\ -rom the bomb experiment " ~11

r-'

\

=

~~

mol.

:85.0 15.0 lOO.O

iR.equired 0,:

'0.708 ·kg mole FeS1

Entering 0,:

(I

Entering N,:

5.&2

"1

120

I II

kg moles 0:

kgmoJes 0.708

H kg moles reS:

=

1 94 k moles . g

°

2

+ 2)(1.94) = 5.81 kg moles 0 1

Gn

I J

= 21.9 kg moles N1

ij

The solid wa'tc comprises Fe10,. gangue. and unburned FeS:. \\'e need 10 con

pute the umountof .unburned FeS1 on the busis of 100 kg of ore. Let

1

.f

= kg Fe~

i

1 I

322 Energy Balances

Chap. 4

that does not burn. Then the cinder comprises

Fc,O,

material> enthalple (4.24) tOi process h

kg x 15

component FcS, Gangue

("'8""5_-----'-x:...)_k",g_F-.e~S-;.,...:,b,;.,u_rn_e-=d+-2.,.:m-"-ol,e,,,s-";F",,e-,-;'O;:.. =. .'.f--.'1...:,6::..0.:.:.k7g-=F.;e~,.:;0~, = ( 2 )(85 _ x) 120 kg FeS, 4 moles FeS2 kg mole Fe20, 3

kg mole FeS2

=

x

+

x 15 -;- (2/3)(85 - x)

x = 2.90 kg FeS2 (2/3)(85 - x)

(a) (b) \ (c) \ (d) I a

Thus 0.040

Sl'C. 4.7

= 54.7 kg Fe,O,

Com

The composition of the solid waste is

Solid waste compo FeS, FC20, Gangue

kg 2.90 54.7 15

mol. wt 120 160

kg moles 0.0242 0.342

Finally, the FeS, that is oxidized is (85 - 2.90) kg FeS,

II

klLmole FeS, = 0685 k i F S 120 kg F eS 2 • g rna e e 2

Corresponding to 0.685 kg mole of FeS, oxidized: 0.685 kg mole FeS, 0.685 kg mole FeS,

8 kg moles SO, 4 kg moles FcS,

III

I 37 k

=.

I SO d d g mo es 2 pro uce

kg moles 0, _ 1 88 k I 0 d g mo es 2 use 4 kg moles FeS, --.

Tabulated heats of formation in kilocalories per gram mole are Comp.:

,

FeS2(C)

02(g)

Fe20,(C)

t.H,: -42.520 0 -196.500 t.}{~,. = (1.37)( -70.960) + (0.342)( -196.500) = - J 35.292

S02(g) -70.960 - (0.685)( -42.520)

In ('111 1'1 l)\ i slate at \\~; and I all1. stream en',,; changes ;IT (' state h) I lj l!su;llIy II j' "senslhk h' usc to dcl··:

<;il

kcal

t.ii~xn = 1.353 kcal/kg are

Note that only the moles reacting and produced are used in computing the heat of reaction. The unoxidi7cd FcS, is not included. nor is the N,. The 0, used contributes ~ value of zero to the !:J./li" because the !:J.Hi of the 0, is zero.

4.7.5 The Energy Balance when the Products and Reactants Are Not at 25C. You no doubt realize that the standard state of 2ST for the heats of formation is only hy accident the temperature at which the products enter ;mel the reactants kave a process. In most instances the temperatures of the

()).

"t,,·,!

(h) I

\(

In ~

(c)

. \ 'i:

r,·[

\n.\ r'!U\: \ i, If 111 t h.' ,

I

I

H~1Z1 of Reartion

See. 4.7

m

materials entering and leaving will be higher or lower than 25'C. However, since enthalpies (and hence heats of reaction) are point functions. you can use Eq. (4.24) logether with Eq_ (4_341, or Eq. (4.35), 10 answer 'lues.tions about the process being analyzed. Typical questions might be

(a) What is the heat of reaction at a temperatllre other lhan 25°C, but still ;at :\ ;ann?

(b) What is the temperatuTe of an incoming or exit stream? (c) What is the temperature of the reaction? {d) U-ow much material mllst be introdllced 10 provj.de a specified :amo.unt ~frheal tr.ansfeT ?

(lo11Side-r :the ;process ilLustr.ated in Fig. 4.15, illA

fOT

\\\:hiCh lhe l1"e.actian is

+ Hir"B -->-
"If

lIIotes' OO(f24;!: O::l4;!:

'Reactor 'RelIct io n :at T .~

IIilg·lIJ13.

rr ?JDIiucts .Iff)

lfuoce~s ··with :r.cactinn.

;In -emplQ)ling Eqs.(4.14)aniJ (4.34), yo.usho.uld .alwaysfirst :choose a :reference

" mule-'are. SOi£g}

-1ff%O

(0:685.,( -42S21l)

m OOffipuling:the-heat of -_The,OplSed. contributes zero.

state :lit·whichthe .hent 'of:reaclion is known. This us.ually turnsouttohe 2YC a-ndl ·litm. -;rhcnthe :next-stcpisJo £alculate :theenthalpy .changes [or each -strea-m:etitering:and .leaving relative to .this re(erellce :Slale. Finally,.the ,enthalpy changes.ar-e'summedup, including the heat of:react ian .calculated in the standard state :by :Eq. (4.:34), ,and.thcother terms.in Eq. ,(4.14) introduced if-pertinent. USllally Jicl'rovesccom:cnicl1t in .calculating the enthalpy .change 10 .compute the "sensible heat" .changes ,and the heat oCrcaction separately. Methods you _can usc ;to cdeterminc-.the'~sensible heat" ~Hvalues Jor tilte .indi.\:idualstrcams:are '.(a) CObtain :the.enthalpy -.values fr.om a set ofpubHshed -tables, .e.g., .the 'Steam :tablesor:tables sllch as in Appendix D, :(b) :Use -tJ.H= :'C• .,(T, -- T"r)- £ ...(1',- T"r'»..:as :previously .discussed :inSec. -4:2, (c) AnalyticalJy,;graphically, or numerically, .find!

:t;.'H:= JT':CpdT .

;:sand:Reacta-nts Are

cd- statc- of rs·-C for the at which the produc[s - the-temperatures of the

1 ~ ! J

1

1

T,

for.each .component individually, .using .the Ifespecti'ic heat .capacity 'equations. Any phase-changestaking:place among the reactants oJrproducts not accounted

\,

1

I'M in the chemical-equation must also betaken into acamunt in thc,calcu!ations.

\ i

324 Energy Balances

Chap. 4

Sec. 4.7

Let us first demonstrate how to calculate the heat of reaction at a temperature other than 25°C. By this we mean that the reactants enter and the products leave at the same temperature-a temperature different from the standard state of 25°C. Figure 4.16 illustrates the information flow for the calculations assuming a steady-state process (f1£ = 0), no kinetic or potential energy changes, and W = O. The general energy balance, Eq. (4.24), reduces to

Then we d,

and

Q = f1H or (4.40)

Simplified, QA + bB at T

t:.H", at T

cC

+ dO

Rea ctants

T

f>H R '

J(QC

PA

Furthermor

at T . Pradu cts T

+ bCp B)dT

f>Hp '

Tref

Reference Temperature

QA + bB at Tref

T

cC

+ dCp

p C

)dT 0

where we Ie If the i:

+ dO

at

f>Hrxn at Tref

J7;,f(cC

Tr,f

QA-

-cC

bB-

-dO

(f1JJ

where Cis 1 Finall) T

Fig. 4.16. Heat of reaction at a temperature other than standard conditions.

or

in terms of the notation in Fig. 4.16. By definition. Q, as calculated by Eq. (4.40). is equal to the "heat of reaction at the temperature T" so that f1H"'r

as the case 11 perature T,.. Equati,;

=

f1H"'r .. , -;. f1Hp - f1HR (4.41) To indicate a simplified way to calculate f1l!p - f1H R, suppose that the heat

capacity equations are expressed as C p = cx

+ PT + yP

(4.42)

Then, to obtain f1Hp -' f1lf R • we add up the enthalpy changes for the products and subtract those for the reactants. Rather than integrate separately, let us consolidate like terms as follows. Each heat capacity equation is mu)tiplied by the proper number of moles: aCpA = a[cx A -+- P.,T -+- YAP] (4.43a)

+ PoT -,.

yeP]

(4.43b)

-+- PeT·J. YeT'] dCpD = d[cx D + PDT -+- l'DT']

(4.43c)

bCp " = b[cxs cCpe

= C[CXc

where Now, if MI,

1 hen you eel other temr'" .. hangcs is [.>

(4.43d)

"'-,..

........... ~-..,---~.

Chap. 4

_: of reaction at a tempera:-,1S enter and the products :-.1 from the standard state ::lr the calculations assum_~ntial energy changes. and _'=5 to

(4.40)

Sec. 4.7

Hear of Reaction 325

Then we define a new term !J.C, which is equal to original expression

equh-alent

new term

cCpe

+ dCpD -

(aCN --+- bC'B) = !J.C,

and

+ drJ. T(cPe + dPD) (cae

(aa .. -+- bas)] = !J.a

D) -

(4.44)

(ap" --+- bPs)] = T!J.P

P(rye --+- dYD) - (ay" --+- bI'B)] = P!J.I' Simplified, !J.C, can be expressed as !J.C,

=

+ !J.PT + !J.'lP

!J.a

(4.45)

Furthermore,

J(cCpc + dCpo)dT Tret

-dO =:0.

IT

(!J.a --+- !J.PT --+- !J.I'P) dT

Tn

= !J.a(T -

TR )

+ !J.,f (P -

+ ~Y (T'

TJ.)

(4.46) -

TI)

where we let TR = T"r for simplicity. If the integration is carried out without definite limits,

(!J.Hp - !J.HR )

-cC

!J.C, dT =

Tft

T

llHp'

fT

(!J.Hp - !J.HR ) =

- Products

=

f

(!J.C,) dT

=

M.T --+- !J.,f P --+-

~T'

--+- C

(4.47)

where C is the integration constant. Finally !J.H". at the new temperature Tis T

!J.H"'T = !J.HUDTn

-+ !J.fX(T -

TR )

+ !J.,f (P -

TJ.)

+ ~Y (P -

Tj) (4.48a)

!J.I'TJ - C 3

(4.48b)

or

standard conditions.

calculated by Eq. (4.40), so that _"H R (4.41) ~HR' suppose that the heat

AHrxnr --

35

Ll.

+ uO: A T--"---' 2 !J.P T 1

--"--I

as the case may be. Using Eq. (4.48a) and knowing !J.H", at the reference temperature T R • you can easily calculate !J.H"," at any other temperature. Equation (4.48b) can be consolidated into

AHrUT --.uAH0 T' LlCX AT'-;- T !J.P T 1

(4.42) --; changes for the products .:llegrate separately, let us equation is multiplied by

AHrxnTu

Ll

Ll.

T

,

!J.YTJ T

(4.49)

where

!J.Ho

=

(!J.H mTR --+- C)

Now, if !J.H,.. is known at any temperature T. you can calculate !J.Ho as follows: (4.43a)

(4.50)

(4.43b) (4.43cl (4.43d)

Then you can use this value of !J.Ho to assist in the calculation of !J.H", at any other temperature. Probably the easiest way to compute the necessary enthalpy changes is to use enthalpy data obtained directly from published tables. Do not

--.,..

326 Energy Balances

Chap. 4

forget to take into account phase changes. if they take place, in the enthalpy calculations.

Sec. 4.7

Then, with

EXA;\IPLE 4.30 Calculation of heat of reaction at a temperature different from standard conditions An inventor thi.1ks he has developed a new catalyst which can make the gas phase reaction CO 2 -:- 4H z -->- 2H zO -i- CH. proceed with 100 percent conversion. Estimate the heat which must be provided or removed if the gases enter and leave at 500'C. Solution: In effect we need to calculate heat of reaction at 500'C from Eq. (4.41) or Q in Eq. (4.40). For illustrative purposes we first use the technique based on Eqs. (4.49) and (4.50).

Basis: 1 g mole CO,(g) CO 2 (g) + 4H2(g) ~.~ 2H 2 0(g) -:- CH.(g)

_ b.Ho ( kcal ) kg mole

f

Mfuo ...•x

=

=

94,052

0

57,798

17,889

or 43,925 h

EXA;\IPLL standard CUI' Repc;it Table 4.5b, Solulif'1 The he"

[-17,889 - (2)(57,798)] - [(4)(0) - 94,052] -39,433 caljg mole CO,

First we shall calculate b.Cp:

Cpeo , = 6.393 -:- 10.100 x IO-'T - 3.405 x IO-'P

= 6.424 + 1.039 x 10-' T - 0.078 x

Tin OK

IO-'P

T in OK

+ 3.464 x IO-'T - 0.483 x IO-'P C pCH • = 3.204 + 18.41 x IO-'T - 4.48 x IO-'P b.rx = [(1)(3.204) + (2)(6.970)] - [(1)(6.393) +- (4)(6.424)1

Tin OK

CpH,

CpH,o = 6.970

From Eq,

(4

Tin OK

= -14.945

b.p = [(1)(18.41) -:- (2)(3.464)](10-') - [(1)(10.100) = 11.087 X 10-'

b.)' = [(1)(-4.48)

+ (2)(-0.483)](10-')

+ (4)(1.039)](10-')

- [(1)(-3.405)

+ (4)(-0.078)](10-')

= -1.729 x 10-' llC, = -14.945 + 11.082 x IO-'T - 1.729 x IO-'T'

)\;otc th"t : L rrfc:rcrH.:c 1-.:, c\.lInplc.

I',~"

.,\ Iho\<:

lj',~'~

Next we find b.H 0, using as a reference temperature 298°K,

llHo

=

b.p, ll)' b.H"n". - llrxT - 2 T- - 3-T' 1\.\"1'11

= -39,443 - (-14,945)(298) - 11.082 x 10-' (298), 2

(- \, r', \ .' .

(-1.729/< 10-')(298)'

her .tt

= -39,443 + 4450 - 493 + 15 = -35,431 cal

''''-"

- - - -............. - -•• ?',........._

*_........._-"',......._____....."'. i"'!'·N_ _.., ._--,,.;

..

e", . .,." , _ . . . .-

. . . . . . ." ..,.,. "",,,,,;;:a;(AWJ&_ . .. ",,,,. :,.• ,......~--

,~Il_

r

Ji',~,d: !' (·t~

--1,

Heat 0/ Reaction

Sec. 4.7

Chap. 4

Then, with !i.Ho known, the !i.H ... at 77rK can be determined.

~ley take place, in the enthalpy

"H rJtluu'-", -- L.l "H 0

£l.

.ll!

327

=

a temperature different from

-L

"NT'T 2 !i.P T'.. -;--, T !i.'1 T 3

,L,l"",

-35,431 - 14.945(773)

+ 11.082 i'

3

10- (773)2

1. 729 x 10- 6 (773)3

3

vst which can make the gas phase

= CH.

-43,925 cal

or 43,925 kcal/kg mole CO 2 must be removed.

heat which must be provided EXAMPLE 4.31 Calculation of heat of reaction at a temperature different from standard conditions

at 5{)()'C from Eq. (4.41) or Q "e technique based on Eqs. (4.49)

- lH 20(g)

Repeat the calculation of the previous example using enthalpy values from Table 4.Sb and Appendix D.

Solution: The heat of reaction at 25'C and 1 aIm from the previous example is

+ CH.(g)

57,798

17,889 j

- IV

/~

~

94,052]

Tin oK

10- 6 T2

JO-6P 1O- 6T2

I

!i.H (kcal/kg mole); reference is O"C

Tin oK Tin OK

kcal

A

AH"n",'K = -39,433 kg mole

I

temperature

CO 2

25°C 5{)()OC

218 5340

H 20 4254

210 5730

From Eq. (4.41),

Tin oK

'O-6T2

1(6.424)1

= -39,433

+ [(1)(5730

+ (4)( -0.078)](10- 6 )

~ .729 X

10- 6 P

+ 2(4254 218) + 4(3499 -

- 210)

- [(1)(5340 -

'/ + (4)(1.039)](10- 3 ) 3.405)

200

172 3499

200)] 172»)

kcal = 44,235 kg mole CO 2 ~ote that the enthalpics of the products and of the reactants arc both based on the refcrence temperature of 25"C. The answer is not quite the same as in the previous nample, because the heat capacity data used in Example 4.30 were not quite the same .IS those used in calculating the !i.H values in the tables.

::'98°K.

3

.I02 X 10- (298)' 2

EXAMPLE 4.32

Carbon monoxide at 50'1' is completely burned at 2 atm pressure with 50 percent air which is at 1000'1'. The products of combustion leave the combustion chamher at 8{)()~F. Calculate the heat evolvcd from the combustion chamber expressed as HTltish thermal units per pound of CO entering.

nl"CSS

-- 35,431 cal

Application of the energy balance to a reaction

i

t

\

_"", .....r"'-"-',",'. . . .___...·.........tt"',t_

328

I"

"t5

Energy Balances

Chap. 4

Solution:

i

"tw:w

Sec. 4.7 (b)

Basis: I Ib mole of CO CO(g)

+ iO,(g) ~ CO,(g)

Material balance:

Amount of air entering: I Ib mole air I Ib mole CO 0.5 Ib mole 0, 1.5 Ib mole 0, used Ib mole CO 1.0 Ib 0 , mole needed 0.21 Ib mole 0, = 3.57 Ib mole air

(c)

Amount of 0, leaving, unreacted: (0.21)(3.57) - 0.5 = 0.25 Ib mole 0, Amount of N, leaving: (0.79)(3.57) = 2.821b mole N,

(d)

Enthalpy data have been taken from Table 4.5a and Appeooix D. The heat of reaction at 25'C (77'F) and 1 atm from Example 4.21 is -61.636 cal g mole of CO or -121,745 Btu/lb mole of CO. We can assume that the slightly higher pressure of 2 atm has no effect on the heat of reaction or the enthalpy values. See Fig. E4.32. Entering Moterials

1000 -------------Leaving Malerials

800 -----

------'----------.--'-t:.HNz

t:.Hoz

1,0 F

77° F 1-..--~--<......--O---+---4-Reference TIrmperolure

50 _l~Hco

Fig. E4.32.

AH (Btu/lb mole); reference is 32°F temperatllre

CO

air

0,

N,

CO,

50'F 77°F 800'F IOOO'F

125.2 313.3

312.7

315.1 5690

312.2 5443

39~,2

= IlHn.nn F + I:lHprooucts AHUD77 F = -121,745 Btuflb mole Q

(a)

80126

6984 IlHre,cunts

,

~-"""-

____

~-""--'-~

~'.'I'!' 'i' I

_ _'''''_ _ _ _ _ _ '''_ _ _'''''_ _

__

''<'--..

"\.,,,~~"-- _ _--

4.7.6 Ten what is called t:' inside the proce i.e., there is n.' is taking place, present, such' a' In calculation, reaction temper may dictate b, adiabatic flame' been calculated' 1920T. The ae! The adiah process. \Vc ca" The adiabatic r. must be spcciti," i ical combusti(Hl ' which can he I,' oxidants, and ,.prohlems arc "t" ( materials. micr,") of dectricity lI':' To calcuLI:,' energy libcr~tcd I .n hy the cnte"I', raise the tempe!,

._-_._---_._------- ... ---_.-

---.-~-

Chap. 4

Su.4.7 fb)

Heat

I1H.,odu
=

or Reaction

329

F

CO 2 N, (I )(8026 - 392.2) -i-- (2.82)(5443 - 312.:n 0,

+ (0.25)(5690' -

315.1)

= 7633.8 -1-- 14,480 -1-- 1343

= 23,457 Btu/lb mole

lIb mole air cd 0.21 Ib mole O 2

Air

(e)

(d)

CO I1H77F -+- I1H,o'F - I1H77F = (3.57)(6984 - 312.7) -1-- (1)(125.2 - 313.3) = 23,800 - 188.1 = 23,612 Btu/lb mole

I1H"""nu = I1H, 000 F

Q

= =

:uI Appendix D. The heat of

: is -67,636 caljg mole of ct the slightly higher pressure .llilpy values. See Fig. E4.32.

~~ference

Temperalure

CO 2 "J

392.2

:,

8026

:::tants

_.*--,.,........_._.,.

-

-121,745 + 23,457 - 23,612 -121,900 Btu/lb mole

-121,900Btu Illb mole CO Ib mole CO I 28 Ib CO

=

-4360B

tu l

ilbCO

4.7. 6 Temperature of a Reaction. We are now equipped to determine what is called the adiabatic reactiol1 temperature. This is the temperature obtained Inside the process when (a) the reaction is carried out under adiabatic conditions, i.e., there is no heat interchange between the container in which the reaction is taking place and the surroundings, and (b) when there are no other effects present, such as electrical effects, work, ionization, free radical formation, etc. In calculations of flame temperatures for combustion reactions. the adiabatic reaction temperature assumes complete combustion. Equilibrium considerations may dictate less than complete combustion for an actual case. For example, the adiabatic flame temperature for the combustion of CH. with theoretical air has t>,,'rn calculated to be 2010'C; allowing for incomplete combustion, it would be InOT. The actual temperature when measured is I 88S'C. The adiabatic reaction temperature tells us the temperature ceiling of a process. \Ve can do no better, but of course the actual temperature may be less. The adiabatic reaction temperature helps us select the types of materials that :nllst be specified for the container in which the reaction is taking place. Chem,,~I combustion with air produces gases at a maximum temperature of 2500'K "(lIch can be increased to 3000 K with the use of oxygen and more exotic p\idallts, and even this value can be exceeded although handling and safety rrohlcms are severe. Applications of such hot gases lie in the preparation of new :··.I!rnals, micromachining, welding using laser beams. and the direct generation ,·t ekctricity using ionized gases as the driving fluid. To calculate the adiabatic reaction temperature, you assume that all the cnl'l!'Y liberated from the reaction at some base temperature plus that brought " by the entering stream (relati\e to the same base temperature) is available to 'J·,C the temperature of the products. We assume that the products leave at the

,

1,, 1

Chap. 4

Sre . .,

temperature of the reaction. and thus if you know the temperature of the products, you automatically know the temperature of the reaction. In effect, for this adiabatic process we can apply Eq. (4.40). Since no heat escapes and the energy liberated is allowed only to increase the enthalpy of the products of the reaction, we have

To fin,

330

Energy Balances

j\.

to

.1

70.: A

(4.51) "energy pool"

Any phase changes that take place and are not accounted for by the chemical equation must be incorporated into the "energy pool." Because of the character of the information available. the determination of the adiabatic reaction temperature or flame temperature may involve a trial-and-error solution; hence th~ iterative solution of adiabatic flame temperature problems is often carried out on the computer. EXAMPLE 4.33

Adiabatic flame temperature

Calculate the theoretical flame temperature for CO burned at constant pressure with too percent excess air. when the reactants enter a~ 200'F. Solution:

CO(g)

~1.I"C

;

+ !O,(g) ---->- CO,(g)

Basis: t g mole CO(g); ref. temp. 25'C (77'F) (200°F

93.3°C)

¢

Material balance:

4.11

entering reartallfs compo moles

CO(g) 02(req.) O,(xs) O,(total)

exit products comp_ moles

CO 2(Y O,(g)

1.00 0.50 0.50 1.00 3.76 4.76

N, Air

N,(g)

1.00 0.50 3.76 Inr: \;1

dcnt c' pI,'. ,\

!1H"n at 25'C

=

-67.636 cal

Note: For the purposes of illustration. in this example the mean heat capacities have a reference temperature of 25'C rather than OCC_ Then, for Eq. (4.51). bli""un": reactants

I'

compo

1110les

bT

Cpo

CO(g) Air

1.00 4.76

68.3 68.3

6.981 6.993 ~

blip,"du, ..

bli

476 2270 bliIfl. = 2746 cal

= 2746 + 67,636 = 70,382 cal

"

S~c.

Chap. 4

Heats 0/ Solution and Mixing 331

4.8

To find the temperature which yields a IlH,.od"," of 70,382 cal, the simplest procedure is to assume various values of the exit temperature of the products until the ~ IlHp -- 70,382. Assume TFT (theoretical flame temperature) = 1800'C:

the temperature of the pro: the reaction. In effect, for this

'W

.~!d

is allowed only to increase ::2ve

IlT compo CO 2 O2 N2

(4.51 )

::accounted for by the chemical ~·ol." Becau~ of the character ::f the adiabatic reaction tern· '-and-error solution; hence the :::-oblems is often carried out on

:':0 burned at constant pressure : a~ 200'F.

Assume TIT

! i

Droducts moles ~I

1.00 0.50 3.76

II I ~j

. cal =ple the mean heat capacities

i

I

= 1500'C:

I

= 1500 - 25 = 1475°C moles 1.00 0.50 3.76

compo CO 2 O2 N2

l

Il

Cpo IlH 12.94 23,000 8.349 7,400 7.924 52,900 ~ IlH p = 83,300 cal

moles 1.00 0.50 3.76

IlT

I

= 1800 - 25 = 1775'C

Cpo 12.70 8.305 7.879 L IlHp

=

IlH 18,740 6,120 43,600 68,460 cal

I ~

Make a linear interpolation: TFT = 1500 -'- 70.382 - 68.460('00) , 83,300 - 68,460 J TFT = 1539"C

¢

= 1500 -'- 39 ,

2798'F

4.8 Heats of solution and mixing So far in our energy calculations, we have been considering each substance to be a completely pure and separate material. The physical properties of an ideal solution or mixture may be computed from the sum of the properties in question for the individual components. For gases, mole fractions can be used as weighting values, or. alternatively, each component can be considered to be independent of the others. In most instances so far in this book we have used the latter procedure. Using the former technique, as we did in a few instances, we could write down, for the heat capacity of an ideal mixture, (4.52) or, for the enthalpy,

IlHm ,xtu" = xAIlIlA

.. =

IlH 476 2270 2746 cal

].382 cal

+.

xBI'J.Hn

+.

xcl'J.Hc

+

(4.53)

These equations are applicable to ideal mixtures only. When two or more pure substances are mixed to form a gas or liquid solulIOn. we frequently find energy is absorbed or evolved upon mixing. Such a \"lution would be called a "real" solution. The total heat of mixing (Illfm ;""l ':;" to be determined experimentally, but can be retrieved from tabulated nperimental (smoothed) results, once such data are available. This type of

1

l

It

Chap. 4

332 Energy Balances

energy change has been given the formal name heat of solution when one substance dissoves in another; and there is also the negative of the heat of solution. the heat of dissolution, for a substance which separates from a solution. Tabulated data for heats of solution appear in Table 4.9 in terms of energy per mole for consecutively added quantities of solvent to solute; the gram mole refers to the gram mole of solute. Heats of solution are somewhat similar to heats of reaction in that an energy change takes place because of differences in the forces of attraction of the solvent and solute molecules. Of course, these energy changes are much smaller than those we find accompa:Jying the breaking and combining of chemical bonds. Heats of solution are conveniently accommodated in exactly the same way as are the heats of reaction in the energy balance.

Sec. 4.8

The exprc 5 moles ot of HCI. T !11oles of II The ., the last C() increment' tion (the, change dc, products h just the ne Fig. 4.17, a

TABLE 4.9 HEAT OF SOLUTION OF HCI (at 25°C and I atm)

Composition HCJ(g) HCI·!H,O HCI·2H 2 O HCI·3H 2 O HCHH 2 0 HCI·5H 2 O HCI·8H 2 O HCI·IOH1O HCI·15H,O HCI·25H,O HCI·50H,O HC1·IOOH,O HCI·200H 2 O HCI·500H,O HCI·loooH,O HCI·5000H,O HC1·50,OOOH,O HC1·=IJ,O

Total Moles H20 Added to 1 mole HCI

Incremental Step (cal/ g mole)

Integral Heat of Solution: The Cumulative - l1ir (cal/g mole)

6268 5400 1920 1040 680 1000 300 359 305 242 136 85 76 39 59 35 16

6,268 11,668 13,588 14,628 15,308 16,308 16,608 16,967 17,272 17,514 17,650 17,735 17,8J1 17,850 17,909 17,944 17,960

-l1B' for Each

0 1 2 3 4 5 8 10 15 25 50 100 200 500 1,000 5,000 50,000

${)URCE: National Bllreall 0/ Standards Circlliar 500 (Reference 15 in Table 4.6).

The solution process can be represented by an equation such as the following: HCJ(g)

+ 5H,Q - > HCJ:5H,Q

HCl(g)

+

or

becomes m. s talldard ill I ( HCI. In the solution dat; you can easi i ca used by ;1<

EXA\IPLE, CaicuiJI( Solution' We trC;!1

A: H:

Al):o'n

0=

5H 2 0

._->

HCI(5H,O)

-15,308 cal/g mole HCI(g)

A ., 11.

It is itnl'l

"--......

......

,,,

Chap. -I

,:utio" when one sub-

;' the heat of solution. a solution. Tabulated : energy per mole for '-am mole refers to the :r to heats of reaction ::lces in the forces of these energy chan.ges -;::aking and combinin; ::Jmmodated in exactl) :.alance.

-_ -.-._-..

H('>Q:/s ,of Solution and Mixing .333

Sec. 4.8

The expression HCl(5H:O) means that 1 mole of HCI has been dissolved in 5 moles of water. and the enthalpy change for the process is -15.308 cal'g mole "f He!. Table 4.9 shows the heat of solution for various cumulative numbers of "wles of water added to I mole of He!. The standard inlegral heat of SOlulioll is the cumulative Afi~oln as shown in the last column for the indicated number of molecules of water. As successive mcrements of water are added to the mole of He!. the cumulativc heat of solution (the integral heat of solution) increases. but the incrcmental enthalpy change decreases as shown in Table 4.9. Note that both the reactants and products have to be at standard conditions. The heat of dissolution would be Just the negative of these values. The integral heat of the solution is plotted in Fig. 4.17, and you can see that an asymptotic value is approached as the solution

-hHsOolution

Integral Heat of Solution: The Cumulative . if' (caljg mole) 6,268 11,668 13,588 14,628 15,308 16,308 16,608 16,967 17,272 17,514 17,650 17,735 17,811 17,850 17,909 17,944 17,960 .0

15 in Table 4.6).

::la.tion such as the fol·

17,960 col/g mole

HCl'n H20

n-

Fig. 4.17. Integral heat of solution of Hel in water.

hecomes more and more dilute. At infinite dilution this value is called the siallriarri integral heat of so/ution at infinite di/utioll and is - 17.960 calg mole of HCI. In the Appendix are other tables presenting standard integral heat of ~olution data. Since the energy changes for heats of solution are point functions, you can easily look up any two concentrations of HCI and find the energy change caused by adding or subtracting water.

LXAMPLE 4.301

Heat of solution

Calculate the standard heat of formation of Hel in 5 g moles of water. Soilitioll:

We treat the solution process in an identical fashion to a chemical reaction: kcal/g mole

A: Air; B:

A g)

= -22.063

t1ii:01n =

+ B: Mi;

=

-15.308 -37.371

~H,(g) -'- ~CI,(g) HCI(g) -,- 5H,0

=

Hel(g)

=

HCI(51!,O)

~~~~----~--------------

!H,(g)

+ ~CI,(g)

5H,O

'=

HCI(5H,0)

It is important to remember that the heat of formation of H ,0 itself docs not

e"t'

334

Energy Balances

Chap. 4

enter into the calculation. The heat of formation of Hel in an infinitely dilute solution is

l!..f)'j

=

-22.063 - 17.960

=

Sec. 4.

dCJ(

-40.023 kcal,g mole

Another type of heat of solution which is occasionally encountered is the partial molal heat of solution. Information about this thermodynamic property can be found in most standard thermodynamic texts or in books on thermochemistry, but we do not have the space to discuss it here. One point of special importance concerns the formation of water in a chemical reaction. When water participates in a chemical reaction in solution as a reactant or product of the reaction. you must include the heat of formation of the water as well as the heat of solution in the energy balance. Thus, if HCI reacts with sodium hydroxide to form water and the reaction is carried out in only 2 moles of water to start with, it is apparent that you will have 3 moles of water at the end of the process. Not only do you have to take into account the heat of reaction when the water is formed, but there is also a heat of solution contribution. If gaseous HCI reacts with crystalline sodium hydroxide and the product is I mole of gaseous water vapor, then you could employ the energy balance without worrying about the heat of solution eifec!. If an enthalpy-concentration diagram is available for the system in which you are interested, then the enthalpy changes can be obtained directly from the diagram by convenient graphical methods. See Chap. 5 for illustrations of this technique.

J( :;(1

3n 40

50 100

::CXI ~'('

I

St;" iollow,:

EXAMPLE 4.35 Application of heat of solution data

The I duty reqil

An ammonium hydroxide solution is to be prepared at 77'F by dissolving gaseous NH, in water. Prepare charts showing

lion.

(h) P t\d,-~ S.1111 T'

(a) The amount of cooling needed in Btu to prepare a solution containing I Ib mole of NH, at any concentration desired. (b) The amount of cooling needed in Btu to prepare 100 gal of a solution of any concentration up to 35 percent NH,. (c) If a 10.5 percent NH, solution is made up without cooling. at what temperature will the solution end up?

Solutioll: Heat of solution data have been taken from NBS Circular 500. (a)

Basis: 171b NH,

=

Reference temperature

lib mole NH,

=

77'F "" 2ye

To convert from kilocalories per gram mole to British thermal units per pound mole, multiply by 1800.

.............

--_.P. . . . . . ",.. . . __.,.,•., ; ,. _. .,._. .

- ...........,.,....................."',..,'.........,."..., -~."'Q«...- -....¥.....

*...."'**...."".,,...""_f,,.~I~~J~·

_~.3i

.,

Chap. 4

t I

~>

_ _ _ _ _L

. .:Jly encountered is the :5modynamic property ~ in books on thermo·

Ii

;-e. of water in a chern· in solution as a c~e heat of formation of :>alance. Thus, if Hel ..::.::tion is carried out in .-u will have 3 moles of = take into account the 2150 a heat of solution .,urn ')xide and the :.lid Jy the energy .~::tion

description IH,O IH 2 0 2H 2 O 3H 2 O 4H 2 O 5H,O IOH,O 20H,O 30H,O 40H,O 50H 2 O l00H,O 200H 2 O ooH 2 O

~_~

__

state g aq aq aq aq aq aq aq aq aq aq aq aq aq

-dfJ~olrl Btu/lb mole 0 12,700 13,700 14,100 14,300 14,450 14,700 14,800 14,800 14,800 14,850 14,850 14,900 14,900

-MI'} Btl//lb mole 19,900 32,600 33,600 34,000 34,200 34,350 34,600 34,700 34,700 34,700 34,750 34,750 34,800 34,800

kcal/g mole 11.04 18.1 18.7 18.87 18.99 19.07 19.23 19.27 19.28 19.28 19.29 19.30 19.32 19.32

WI %NH 3 =

I

%NH 3 for I H 2 0:

weight ~·~NHJ

100 48.5 32.0 23.9 19.1 15.9 8.63 4.51 3.05 2.30 1.85 0.94 0.47 0.0

=

Ib NH,(100) Ib H 2 0 -, Ib NH3

181(;~1~)17

= 48.5%

The heat of solution values shown in Fig. E4.35(a) are equivalent to the cooling duty required. (b) Part (b) of the problem requires that a new basis be selected, 100 gal ofsolution. Additional data concerning densities of NH.OH arc shown in Table E4.35. Sample calculations are as follows:

solution containing I lb d .' l!lil of a solution of an) NOll':

. Ib/IOO I _ (sp gr soln)(62.4lblft 3)(1.003)(I00) enslty, . ga 0.48 gal,f!')

1.003 is the specific gravity of water at 77'F.

density @ 320~-;;NH3 = (0.889)(62i~~·003)(I00) cooling req'd} Btu/IOO gal

= 741 Ib/IOO gal

= (Ib moles NH3)(_fl.HO 100 gal

Btu ) .01'lb mole NH3

cooling req'd } for 100 gal =.(13.94)(13,700) = 191,000 Btu/IOO gal soln 32.0 ~~ N H 3 soln 1its per poun,l

,~~'.::~;}~-~~;¢'df~Ah·

Weight percents have been computed as follows:

t

:;le

w

MJ~ol ••= [-32,600 - (-19,900)] = -12,700 Btu/lb mole NH3

I

500.

__

For example,

::-r the system in which 2ined directly from the ..:or illustrations of this

-r

,_,~,

Standard heats of solution have been calculated from the cumulative data as follows:

:l.

:.:mg, at what temperature

~_,.

Heats of Solution and .\fixing 335

-fl.li';

~:on

'F by dissolving gaseous

_ _ , _ _ • __ ._c~"_,,._,_,_

Sec. 4.8

an. infinitely dilute solu· gmole

&-.~

lh~se data are portrayed in Fig. E4,35(b). Basis: 100 gal of solution at concentrations and densities shown

Chap. 4

240,(

200,( e

12,000

0

on

::I:

0

.,

0

'" Z ~

160,(

."

~ .... :>

10,000

.- It 0,0

.0

:::::>

CD

.,.

<-

CD

eo :;::

..

.'0

BOOO

~

."

:>

e

0

0

.... 0

.0

80,0

0

u

6000

.,

40,0

::I: I

4000

2000

00

(c) Lb mole HzO/ib mole NH3

tJ.E

(0)

Fig. E4.35(a). TABLE

E4.35

-~fJ~"ln sPKr·

%NH,

@4'C

32.0 23.9 19.1 15.9 8.63 4.51 3.05 2.30 1.85 0.94 0.47 0.0

0.S89 0.914 0.929 0.940 0.965 0.981 0.986 0.990 0.992 0.995 0.998 1.000

density, Ib/lOO gal

741 761 774 784 805 819 826

cooling r('q'd

IhNH,

Ih moles NH )

BIll

per 100 gal.

100 gal

loogal

Ihmoie:-;H,

BIll

148 124.7 69.4 37.0

13.94 10.70 8.70 7.32 4.08 2.18

13.700 14,100 14,300 14,450 14,700 14,800

191.000 151,000 124,000 106,000 60,000 32,200

19.0

1.12

14,SOO

16,600

237 182

830

7.8

0.46

14,850

6,800

834

0

0

14.900

0

336

..''''--..-~~.~~.~----~

l. You Shl)U:,j equation. '1 JnC'an~ ~lIh~

'SOURCE: N. A. L1nge, Handhook of Chmlislry, SIh ed., Handbook Publishers Inc, Sandusky, Ohio, 1958.

.....

WHAT YOU

''''''

>a

i.e .. heal"]'

knoll' II h" be omitte,

...-----Nt

Chap. 4

What You Should Hare Learned frol1l This Chapter 337

24~000r-------------------------------------,

... o

o o

...

72,000 Btu

c:

g

<.>

35 Concentrotion, wt % NH3 (hI Fig. E4.35(b).

200

Basis: 100 gal soln at 10.5 % NH,

(c)

.1£ ~.1H = mC,.1T = 72,000 Btu/lOO gal sp gr soln

m

"iH,

per 100 fa!. Btu Bill Ib mole NH,

191,000 151,000 124,000 106,ON 60,000 32,200

14,800

16,600

14,850

6,800

14,900

0

"d., Handbook Publishers I fl •.

= 4.261 J/(g)CC)

= 1.02 Btu/(lb)('F)

coolilIg ret/',I

13,700 14,100 14,300 14,450 14,700 14,800

= 0.955

= (0.955l(62i~(~ .(03)(100) = 800 Ib

C p 10.5% NH, so In

-A.fI~o'n

[cf. Fig. E4.35(bl]

AT.1H = mCp

L>

Tn • d

I, I

72,000

= (800)(1.02) =

= 77 + 88

88'F

= 165'F

WHAT YOU SHOULD HAVE LEARNED FROM THIS CHAPTER

I

I

I. You should know what the energy balance is in words and as a mathematical equation. You should be able to explain what each term in the equation means and be able to apply the equation to practical or simulated problems, i.e .. be able to state what each term is in I ight of the conditions in the problem, know what assumptions can and cannot be made. and know what terms can be omitted.

1

1

i

• .1. ,; 4)1.,;

~r"'U'

• ..... ~"'4

J!$¥

'iF

#/.4;44

m.:;;

it

$

P

w

(,.

p,

.-----_._' Chap. 4 :m should be able to set jow and non flow pro~:1 be considered to be

(/wp.4

Problems 339

S.

Yesavagc, V. F., ct aI., "Enthalpies of Fluids at Elevated Pressures and Low Temperatures," Illd. Eng. Chern., v. 59, no. II, p. 35 (1967).

9.

Zcrmansky, M. W., and H. C. Van Ness, Basic Engineering Thermodynamics, McGraw-Hili, New York, 1966.

energy and heat and JOcate enthalpy or heat :;:xh data are not avail-

'ld how it differs from a "C:1emistry, and especially '" heats of reaction with -.,-ious temperatures, and __ 2.ts -"mbustion from

"f

-,:Je heaLs of solution or :ype. -: temperature for simple or not and whether or - -proportions,

,;,s, Gulf Publishing 0',.

.'lieulalions, McGraw-IIill. ,:emieal Process Prineirl"I. ·~rmical Process Pril1Cil'it s.

~Dnding States Principk."

. :lined,

":ni:

Dover, New YI.'rJ.... il1ccrillg Thern:,"

PROBLEMS'2 4.1. Explain specifically what the system is for each of the following processes; indicate the portions of the energy transfer which are heat and work by the symbols Q and W, respectively: (a) A liquid inside a metal can, well insulated on the outside of the can, is shaken very rapidly in a vibrating shaker. (b) Hydrogen is exploded in a calometric bomb and the water layer outside the bomb rises in temperature by I ec. (c) A motor boat is driven by an outboard-motor propeller. (d) Water flows through a pipe at 10 ftfmin, and the temperature of the water and the air surrounding the pipe are the same. 4.2. Draw a simple ske~ch of each of the following processes, and, in each, label the system boundary, the system, the surroundings, and the streams of material and energy which cross the system boundary: (a) Water enters a boiler, is vaporized, and leaves as steam. The energy for vaporization is obtained by combustion of a fuel gas with air outside the boiler surface. (b) The steam enters a rotary steam turbine and turns a shaft connected to an electric generator. The steam is exhausted at a low pressure from the turbine. (c) A battery is charged by connecting it to a source of current. (d) A tree obtains water and minerals through its roots and gives off carbon dioxide from its leaves. The tree. of course, gives off oxygen at night. 4.3, Draw a simple sketch of the following processes; indicate the system boundary; and classify the system as open or closed: (a) Automobile engine. (e) A river. (f) The earth and its atmosphere. (b) Water wheel. (c) Pressure cooker. (g) An air compressor. (h) A coffee pot. (d) Man himself. 4.4. Are the following variables intensive or extensive variables? Explain for each. (a) Pressure. (b) Volume. (c) Specific volume. (d) Refractive index . (e) Surface tension. "An asterisk designates problems appropriate for computer solution. Also refer to the ''''mpliler problems after Problem 4.111.

Combined Material and Energy Balances

5

Now that you have accumulated some experience in making energy balance, and in the principles of thermochemistry, it is time to apply this knowledge 1<1 situations and problems involving both material and energy balances. You h;\\: already encountered some simple examples of combined material and encr~' balances, as, for example, in the calculation of the adiabatic reaction temper:!' ture, where a material balance provides the groundwork for the writing of c,;' energy balance. In fact, in all energy balance problems, however trifling lhl" may seem, you must know the amount of material entering and leaving the' process if you are to apply successfully the appropriate energy balance eqt;.i' tion(s). In this chapter we shall consider both those problems which require you Ic' make a preliminary material balance prior to making an energy balance :!11,1 also problems which require you to solve simultaneous material and enCL' balances. In line with this discussion we shall take up two special types ofch:!f>. enthalpy-concentration charts and humidity charts, which are prepared hy c,'''--bining material and energy balances and which are useful in a wide Variel) ,: practical problems. Figure 5.0 shows the relationships among the topics w ;< discussed in this chapter and the relationships with topics in previous chapler.

362

1

j,



H' _

.... )., ,,Q,i,,,,'...,,"""_ _""

Sil11ldlalleolls Use of Material alld Ellergy Balallces 363

Sec. 5.1

I I Grooh.coi A.ds and Techn.ques I

~Enlholpy ConcentroflOn Charts

. -cal and .::"ances

5

Chapter 2 5t

Maferial Solonce Techniques

i Chapter 1 Background Information

J

V

~

Humidity

Charts

Chopler 3

I

Simultaneous Solution of Motenol and Energy Bolonces

1

~Comple,

Problems

Gases , liquids I Solids

~

Chapter 4 Energy Balances

I I

.m 1 ,!, energy balances . to apply this knowledge to

~

:l energy balances. You have .: Dined material and energy adiabatic reaction temperaJwork for the writing of an ~'lems, however trifling they 1al entering and leaving the ':Jriate energy balance equa-

,'DIems which require you to ,',lIlg an energy balance and 'neous material and energy .two special types of chart'. which are prepared by comuseful in a wide variety of :lpS among the topics to be wpics in previous chaptcrs.

5.1

For any specified system or piece of equipment you can write (a) A total material balance. (b) A material balance for each chemical component (or each atomic species, if preferred).

,

f

,Pi

* )/,'·...,16. . . ._"',....

Simultaneous use of material and energy balances for the steady state

1

• , ,9ft,

Flg.5.0. Hierarchy of topics to be studied in this chapter (section numbers are in the upper left-hand corner of the boxes) .

You can use the energy balance to add one additional independent overall equation to your arsenal, but you cannot make energy balances for each individllal chemical component in a mixture or solution. As we mentioned previously, yOU need one independent equation for eaC'h unknown in a given problem. The energy balance often provides the extra piece of information to help you resolve an apparently insuperable calculation composed solely of material balances. We can illustrate any given system or piece of equipment in the steady Hate by drawing a diagram such as Fig. 5.1. Figure 5.1 could represent a boiler, a distillation C'olumn, a drier, a human being, or a city, and takes into account processes involving a chemical reaction .

364 Combined Material and Energy Balances

Chap. 5

Be.

Workout W

If'!~

be Alb - - . ;

Energy Llberoted from Reochon, Solution, elc

~-~

Dlb

Bib

Heot in

Clb

0

Boo over

Fig. 5.1. Sketch of generalized flow process with chemical reaction.

= H, =

Let x,

weight fraction of any component enthalpy per unit mass of any component with respect to some reference temperature

If the subscript numbers 1,2, 3, etc., represent the components in each stream, you can write the following material balances in algebraic form:

balance Total: Component I: Component 2: etc.

in A AxA , AxA •

DI

out

+ B + Bx = + BXB, = B,

C CXc, CXc.

+ D + Dx D, + DXD,

p"

As previously discussed, the number of independent equations is one less than the total number of equations. In addition, an overall energy balance can be written as (ignoring kinetic and potential energy changes)

Prr. '] (

I Prr.

For a more complex situation, examine the interrelated pieces of equipment in Fig. 5.2. Again the streams are labeled, and the components are 1,2,3, etc. How should you attack a problem of this nature? From an overall viewpoint, the problem reduces to that shown in Fig. 5.1. In addition, a series of material balances and an energy balance can be made around each piece of equipment I, II, and Ill. Naturally, all these equations are not independent, because, if you add up the total material balances around boxes I, II, and III, you simply have the grand overall material balance around the whole process. Similarly, if you add up the' component balances for each box. you have a grand component balance for the entire process, and the sum of the energy balances around boxes I, II. and III gives the overall energy balance. Following Fig. 5.2 (on p. 365) is a list of the material and energy balances that you can write for it.

~"""-

-~'-<'

~~~~~~~ .,.~'""..,~---,......,"~-, ~~

...,-..,.,-....., " ,....

~~:.oY_«~_~~

______ , ...n-"".......""",,,-

,"f"-

'1 (

I Note t he ~, these b,,',

1:'(.\\11'11 A d:-,!

"hh)rnh(."::! \ nlumn I'" ~ I percellt I,

Chap. 5 Boundary line for individual equipment

balonceX~: - ~

Clb

I

_-

------:-= ----.., V

I

I'

"

\

\

\

r+-t-'-

I

F-t-+<"'i

I \ \ .... -

,'

Dlb

" ,......

~

Boundary line for overall balance - "mical reaction.

,~-

-_/ /

- - - _________ /

i

! -:1t with respect to some -~1Jonents

in each stream, :-aic form:

: uations is one less than

Fig. 5.2. Sketch for interrelated processes and streams.

f balance Entire process: Total Component Energy Process I: Total Component Energy

Jen as (ignoring kinetic

Process II: Total Component Energy

pieces of equipment in

Process III: Total Component Energy

-:l

H--4--w I I

\

--

1,2,3, etc. How should viewpoint, the problem _~i material balances and c.;)uipment J, II, and III . ...:use, if you add up the ., simply have the grand .-.:arly, if you add up th~ --::oonent balance for the _ud boxes J, II, and III . p_ 365) is a list of the

in

Qn

+ Qm +

FXF,

F IlHF

out

+

F=D+W Fx F , = DXD, + WXw, F IlHF = D IlHD + W Mfw i,

F+R+Y=V+L Rx R, + YXy, = VXv, + Lx L ,

+ R IlHR +

Y IlHy = V IlHv

+

I

I ~

l

L IlHL

~

V=R+D VX v, = RXR, + Dx D,

QII

+ V IlHv =

R IlHR

1

+ D IlHD

1

L=Y+ W Lx L , = YXy, -1- WXw,

Qm

-+ L IlHL =

Y IlHy

I

+ W IlHw

Note the symmetry among these equations. The methods of writing and applying these balances are now illustrated by an example.

I l

EXAMPLE 5.1

Simultaneous material and energy balances

A distillation column separates 10.000 Ibihr of a 40 percent benzene. 60 percent chlorobenzene liquid solution which is at 70T. The liquid product from the top of the column is 99.5 percent benzene, while the bottoms (stream from the reboiler) contains I percent benzene. The condenser uses water which enters at 60-'F and leaves at 140' F,

365

AQ-A

...

t" ... .s;.;:, ... ,-

1

366 Combilled .~farerial alld Energy Balal/ces

Chap. 5

while the reboiler uses saturated steam at 280"F. The reflux ratio (the ratio of the liquid overhead returned to the column to the liquid overhead product removed) is 6 to I. Assume that both the condenser and reboiler operate at I atm pressure, that the temperature calculated for the condenser is 178"F and for the reboiler 268'F, and that , . the calculated fraction benzene in the vapor from the reboiler is 3,9 weight percent (5.5 mole percent). Calculate the following: (a) (b) (c) (d)

Stc.5 ((

10

be

J

The pounds of overhead product (distillate) and bottoms per hour. The pounds of reflux per hour. The pounds of liquid entering the reboiler and the reboiler vapor per hour. The pounds of steam and cooling water used per hour.

Solution: (a) Figure E5.l will help visualize the process and assist in pointing out what additional data have to be determined. First we have to get some pertinent enthalpy or heat capacity data. Then we can make our material and energy balances. (The exit streams have the same composition as the solutions in the condenser or reboiler.) (b) Convert the reboiler analysis into mole fractions.

(ll I

Olcrall

Basis: lOOlbB

camp. Bz

Ib

CI

99

lb moles 0.0128 0.88 0.8928

mol. wt 78.1 112.6

I

mole/r. 0.014 0.986 1.000

Olcral:

W

~

Condenser

~

I

leI i

60°F

-0,'f40°F

t

P Product 0.995 Bz 0.005 CI

R Ff0,000 Ib/hr 70·F 0.40 Bz 0.60 CI

H2O

Still

(f)



11>!.11 :

~ AlW Cf 112,6

Bz 78.1

I

I Reboiler

L

00000

280°F Saturated Steam S

Fig. ES.l.

f

Os '.+

8 Boltoms om Bz 0.99 CI

\ \ {' h,l\ :,~ , l ,.~', 1, II ;, ' : ~ ,

-,t' , teNr. 1'"46 " t'

. . . .H

e

"-'

([hap. 5 -'tjo~ther.atjo·of the

(c) The heat capacity data for liquid benzene and chlorobenzene will be assumed to be as follows' (no enthalpy tables are available):

liquid

Jdlli:t r.emovcd) is 6 to I.

; atm. pressure. that the rehoiJer 268'F, and that .;:er i& r,9, weight percent ~

:WlllS'

70 90 120 150 180 210 240 270

per hour.

"ur. ire pointing out what

0.31 0.32 0.335 0.345 0.36 0.375 0.39 0.40

J1HnporhatiOtH

0.405 0.415 0.43 0.45 0.47 0.485 0.50 0.52

140 135 130 126

170 166 160 154

I , I

F=P+B

+B

10,000 =P Overall benzene balance: Fxp =

PXp

+ BXB

10,000(0.40) = P(0.995)

!

10,000(0.40)

I

P(0.995) -I- (10,000 - P)(O.OI)

"'-0 B = 6040 Ib/hr "'---0

(e) Material balances around the condenser:

~

= 6

V, - p Product

=

+ B(O.OI)

P = 3960 Ib/hr

I

or

R

=

6P

= R -I- P =

=

6(3960)

23,760

+

= 23,760 Jb/hr --®

3960 = 27,720 Ib/hr

(f) Matcrial balances around the reboiler:

a99S 8z a005 CI

Bz

(d) Overall material balances: Overall total material balance:

!

---()c=-

Btu/lb

CI

Basis: 10,000 Jb feed/hr

=apm:it;y ditta. Then we .IDS·Ilav-e: tire: same com-

molirfr··

C p , Btu/(lbWF) CI Bz

Temp.eF)

,eboiler \lapOe per hour.

-lSI:

Simultaneous Use of Material alld Energy Balances 367

Sec. 5.1

I

Total:

L

= B + V,

BXB

+ V,x ...

Benzene: LXL

=

L=6040+V,

- 8 Bottoms

om

Bz

0.99 CI

f

LXL

=

6040(0.01)

+

V.(0.039)

\\'e hav'c three unknowns and only two illdependent equations. We can write additional equations around the still, but these will not resolve the problem since we would still Ix- left with one unknown more than the number of indep('lldcllt equations. This is the ,(age at which energy balances can be used etfectively. 'Data estimated from tabulations in the Appendix of Process Heat Transfer by D. Q. Kern, McGraw-Hili, New York, 1950 .

.-.--....,...,.~-~.

368

Combilwd Malerial and Energy Balances

Chap. 5

(g) Overall energy balance: Let the reference temperature be 70'F; this will eliminate the feed from the enthalpy calculations. Assume that the solutions are ideal so that the thermOdynamic properties (enthalpies and heat capacities) arc additive. No work or potential or kinetic energy is involved in this problem. Thus

We know all the terms except Q""rn and Q'ood .... ' and still need more equations. (h) El)ergy balance on the condenser: This time we shall let the reference temperature equal 178'F; this choice simplifies the calculation. because then neither the R nor P streams have to be included. Assume that the product leaves at the saturation temperature in the condenser of 178'F. With the condenser as the system and the water as the surroundings we have

and, since

Qsnu:m

= -

Il.H",ater =

QcoDdenJcr

QsurroundiDiu dHcondenser =

Vi -AH.. po",,,,on) = 27,720[170(0.995) Solving, Qc = -4.71

+

140(0.005)]

=

5.89

X

lb (BtU) _ - 3960hr 46.9 1b

=

fl78 Cpp dl 70

10 4 Ib HzO/hr

-<--@

+

(BtU) 6040 lb hr 68.3 1b

. . = J268 CpBdl 70

Btu/lb

Btu/lb

Ails

Bz

CI

AI'g

Bz

Cl

Avg

36.2

46.9

88.1

68.0

68.3

Q....m

=

5.31 x 10· Btulltr

f

6

r

=

5.31 x 10 Btu/hr 923 Btujlb

<760Ib/h

= ..

r

~d

-<

\::3'

(j) Energy balance around the reboiler:

Q" .. m

---

..

..

a difference

In pract i.

tcdious, a

From the steam tables, AH .. p at 280'F is 923 Btu/lb. Assume that the steam leaH'S at its saturation temperature and is not subcooled. Then d 'h

If we had q

Btu + 4.71 x 10• hr

An assumption that the P stream was pure benzene and the B stream was pure chi orobenzene would be quite satisfactory since the value of Qc is one order of magnitude larger than the "sensible heat" terms

'~--....

5.1

WCPH,o(lz - 'I)

47.0

Ib steam use

Encrgy b"

W(I)(140 - 60) = Qw.", = - Q"ndoo",

In the preceding equation the f C p dl was determined by first graphically integrating p dl to get Ai! for each component and then weighting the AH values by their respective weight fractions:

=

CJr.l,

lOr, ~ut as opro"', The\ or can be ~ the o\"cr J I

-t1Hwater

JC

..

rchl)l:~

hcat

Material h

Qwatcr

(i) Amount of steam used: We shall use the overall energy balance from step (g).

Allp

Ihe

10· Btu/hr (heat evolved), and the cooling water used is

X

W

Q" .. m

We I-

.'J.li r . If.-

surroundings (water)

system (condenser) !!HcondcDser =

Src.5.1

+ L(AlIL ) + V.CMI,..> + B(M1 8 )

........"..."~,~-,,,,,,"---------,~-,-----,,,,--,-,~-,~------,,--~-'

computer I large-scale,,'opc her,: An im how to rlh ,,,Iutioll. 1,\ scnse to pr ,Ind the Ill!

rule gi,'Cs

t

:ntct1sirc \.f

lIOn that ,Ir of degrees'

th.......

Chap. 5 C2!ure be 70'F; this \\':11 ::.at the solutions are ide.l! - atpacities) are additi'e. :-:>blem. Thus - F

7'

J7.

CPT

dt

~

liHF =0 need more equations. the reference tempera. ::ause then neither the R :: leaves at the saturation . 'Ine system and the water ::...1

~t

-,..-..._------

Simultaneous Use of Material and Ellergy Balallces 369

Sa. 5.1

We know that Q""m = 5.31 X 10" Btu/hr. We do not know L or the value of :;Ii ,. However, even if liB L is not known, the temperature of stream L entering ,,:c rcboiler would not be more than 20'F lower than 268'F at the very most, and the heel! capacity would be about that of chlorobcnzene. We shall assume a difference of :0 F, but you can note in the calculations below that the effect of including liB L, J.' opposed to ignoring it, is minor. The value of V. is still unknown, but all the other values (liB"" B, liB.) are known (l[ can be calculated. Thus. if we combine the energy balance around the reboiler with :'1C overall material balance around the reboiler, we can find both L and V•. Reference temperature: 268'F Energy balance: S.31 x

J06~: + (Llb)(0.39(1~~~F»)(-20'F) =

....ter)

V.[(0.99XI26) + (0.OIXI54)]

+ B(O)

Material balance:

water

L=6040+V. 5.31 x 10" - (6040

-I,) ....;;::. water

= -

::ooli

'Zv balance from step (g). 4.71 x 10" ~~u

dt

Btu/lb Avg

o

68.3

'Stream was pure chlor," :me order of magnitude

:::Je that the steam

V.

k'l\~'

X

L = V.

=

39,300

+ 7.8V•

(

1

+B

= 126.3 V.

10" = 126.3 Vb

= 5.26 ;4 10" = 39,300 lblhr

+ 6040 =

45,340 Iblhr

@

,i



'1

1

If we had ignored the enthalpy of the L stream, then V. would havl!' been V

::r graphically integrating ..:.he liB values by thm

V.)(7.8)

5.31 x 10" - 0.047

Qcondenscr

!r used is

+



=

5.31 X 10" 126.3

=

I j

42 100 Iblh ' r

" difference of about 7.1 percent.

In practice, the solution of steady-state material and energy balances is quite tedious, a situation which has led engineers to the use of digital and hybrid mmputer routines. Certain special problems are encountered in the solution of !~rgc·scale sets of equations, problems of sufficient complexity to be beyond our ,("pc here. An important aspect of combined material and energy balance problems is how to ensure that the process equations are determinate, i.e., have at least one ~olution. We can use the phase rule (discussed in Sec. 3.7-1) plus some common ~nse to provide a guide as to the number of variables that must be specified "nd the number of variables that can be unknown in any problem. The phase ("Ie gives the relationship between the number of degrees of freedom and the "'l'/Tsi1'1' ,"ariables such as temperature, pressure, specific volume, and composi· llvn that are associated with each stream. According to Eq. (3.57), the number <'f degrees of freedom for a single phase at equilibrium is

F=C-l+2=C+!

j

1

j

1

J i, .,, j

'I

j; I

r"_lW_...._ _ .. __........'"·...'....... ~·

·~

~_"

,.~

370

"_#_..,. . .

__"'._........._ ......_,_ _.....

,,.~

<... it'h.

to -et+>i~"

Combined Material and Energy Balances

tt r

Chap. 5

However, to completely specify a stream, we need to know how much material is flowing, so that in addition to the C + I intensive variables, one extensive variable must be given, for a total of Nt = C+ 2

where Nt is the total number of variables to be specified for a single stream, Each independent material and energy balance that is written reduces the degrees of freedom in a problem by one, Consequently, in any problem you can add up the total number of variables and subtract the number of equality constraints and specified input conditions to get the difference, which is the number of variables that have to be specified if the problem is to be determinate. If a chemical reaction takes place, the number of independent material balances that can be made is equal to the number of atomic species, not C, the number of components. We shall now illustrate the application of the above concept to several simple cases.

EXJ,MPLE 5.2

Determining the degrees of freedom in a process

We shall consider five typical processes, as depicted' by the respective figures below, and for each ask the question, How many variables have to be specified, i,e .. what are the degrees of freedom (d,L) to make the problem of solving the combined material and energy balances determinate, All the processes will be steady-state ones. (a) Simple slream jllllClioll, sillgle phase [Fig. E5.2(a)]. We assume that Q = 0 and Iof' = 0, and that the pressure and temperature of P, and P, are the same as that

~P,

z-·

~

Fig. ES.2(a).

of Z. the input stream. The statement of the total number of variables, constraints, and degrees of freedom can be summarized as follows. Total number of variables (3 streams): Number of independent equality constraints: Material balances: Energy balance: Tp , = T z ; Tp , = Tz :

Total no. d.L

~!.'It_';:;"j

,4

4.

'"

,

_dii~



..

=

3(C

+ 2)

3(C

+ 2) C I

2 2

C - 2C - 7

:--;ote that" rOl1cnls

an~

the tall, Wecar and the cor c, 2 _. I III

~!aterial

ba

Energy bala Definitions:

If Z, xz, T z . tion only of that fi P, and that if PI is, balance can ponent mat, If the ill same as Z a being C -~ I number of il' I. The coml Conseq uent I: the same as t (b) Mix,

Pz

Pp,=Pz;Pp,=P z : Input conditions in Z specified:

Sec. 5.1

=C- I

+2

.

'M'r

. - - -.....

([hap. 5

..-now how much material variables,. one extensilc

~ect for.

a single stream. .:.: is; wrinen reduces thc . ill any problem you can :mmber. of equality can· :=e;. whjch. is the number .~ tQ1 fur determinate. If a =.lIlllter.iaL balances that =.:;. IIlJ!t CC;. the num ber of

oy the' respective figurcs , haW' .~ be specified. i.e .. -:-: of I the combined wil. .cady·state OIlCS. \\
\laterial balances: Z

= PI +P 2

ZXz = P1XPt

ZYz

+ P2XPI

= PIYP, + P 2 YP,

[ncrgy balance: Definitions:

+ YP, = xp, + YP, = Xz + Yz =

1

1 1

If Z, xZ, T z , and pz are specified in the input, then Hz is known because H is a function only of T,p, and x. We have as given T p , = T z = T p, and pp, = pz = pp" so that lip, and Hp, depend only on xP, and xp,. respectively. You can see by inspection lhat if PI is specified, representing the 1 degree of freedom available, the total material ~Jlance can be solved for P and then the (nonlinear) energy balance and one-component material balance can" be solved simultaneously for Xp, and Xp" If the initial assumption had been that the compositions of PI and P, were the ',lme as Z and we reduced the total d.f. by an additional factor of 2(C - I). there t-cmg C - 1 compositions to be specified. we would be making an error because the number of independent equations would not be C -r I anymore but would simply be I. The component material balances and energy balance would all be the same. Consequently the number of degrees of freedom would be 3(C

+ 2)

- 1 - 2 - 2 - (C

+ 2)

- 2(C - 1) = 1

the same as before. (b) Mixer [Fig. E5.2(b»). For this process we assume that W

of variables. constraints. J(C + 2)

C 1 2 2

11-11

/

~,

I

C+2 .- 1

--...

'(lIe that although C -+- I material balances can be written corresponding to C com· ;""ncnts and one total balance. only C of the balances are independent and included ,n the tally above. We can illustrate the relation between the count for the degrees of freedom above and the constraints by considering a two-component system for which d.f. = C - I 2 -1 = 1:

annc.ept. to several

!"Oem&'

........

Simultaneous Use of Material and Energy Balances 371

Sa. 5.1

xp, .:..DO\IIf

~-----

Fig. ES.2(b) •

= O.

•!

372

Chap. 5

Combined Material and Energy Balances

Sec. 5.1

Total number of variables (3 streams + Q): Number of independent equality constraints: Material balances: Energy balance: Total no. d.f. = 3(C

+- 2) +

+ 2)

3(C

-+- I

Ihc system. I wc mean th.! tlon is kno\\ component.

C I

I - C - I = 2C

+6

./ ,

(c) Heal exchanger [Fig. E5.2(c)]. For this process we assume that IV

Z,

----IN-.rw-.rw~--

Z2

=

O.

1\

P, P2

o

I n general yo,

Fig. ES.2(c).

Total number of variables (4 streams + Q): Number of independent equality constraints: Material balances (2 separate lines): Energy balance: Total no. d.f.

=

4(C

+ 2) +

4(C

+ 2) +

I

2C I

I - 2C - I = 2C

+8

I

(d) Pump [Fig. E5.2(d)]. Here Q = O.

z

IT-P

You call ferent simpk freedom. In 1 double coun, interconnectil lion.

Fig. ES.2(d)

Total number of variables (2 streams + Wi: Number of independent equality constraints: Material balances: Energy balance: Total no. dJ. = 2(C

+ 2) +

Other choice,

I - C - I

2(C

+ 2) + I C I

EXAMPLE 5.

= C +4

we could carr \ sidcred as a 'l"i

(e) Two-phase well-mixed tank (stage) at equilihrium [Fig. E5.2(e)] (L. liquid phase; V, vapor phase). Here W = O. Although two phases exist at equilibrium insido

Consider I

lor, an equilici Ihe total d.f. redundant COli

The net d.f. LI you can sec f'~ Fig. ES.2(e).

We can k, /. is actually"

Chap. 5 Simultaneous Use of Afaterial and Energy Balallces 373

Sec. 5.1

3(C

2):

+ 2) +

I we mean that both phases are at the same temperature and pressure and that an equatic>n is known that relates the composition in one phase to that in the other for each component.

C I 1= 2C

I

Ihe system, the streams enlering and leaving are single· phase streams. By equilibrium

:as:

+6

cess we assume that W

,.

=

Total number of variables (4 streams -~ Q): Number of independent equality constraints: Material balances: Energy balance: Composition relations at equilibrium: Temperatures in the two phases equal: Prcssures in the two phases equal:

o.

Total no. d.r.

=

4(C

+

4(C

+

2)

+

I

C I C I I

2) -+- I - 2C - 3

=

2C

+6

In general you might specify the following variables to make the problcm determinate:

'J: -:ts:

Input stream L,: Input stream 1'0: Pressure: Q:

4(C+2)+1 2C I

- 1

Total: Other choices are of course possible.

= 2C + 8

p

!

I 2(C

+ 2) + 1

C+2 C+2 1 1 2C+6

I

You can compute the degrees of freedom for combinations of like or different simple processes by proper combination of their individual degrees of freedom. In adding the degrees of freedom for units, you must eliminate any double counting either for variables or constraints and take proper account of interconnecting streams whose characteristics are often fixed only by implication.

i I

C 1

-1=C-I-4

-ium [Fig. E5.2(e)] (L, liqUlJ .!Ses exist at equilibrium ins Ice

! I

EXAMPLE 5.3

Degrees of freedom in combined units

Consider the mixer.separator shown in Fig. E5.3. Superficially it would seem that we could carry out an analysis as follows to combine the units. For the mixer con~idcrcd as a separate unit, from the previous example, d.f. = 2C + 6. For the separalor, an equilibrium unit, we find from the previous example that d.f. = C + 4. Hence the total d.r. = 2C -1- 6 -+- C + 4 = 3C -I- 10. However, we have to eliminate certain redundant counts of variables and constraints for the combination: Eliminate variable Z: Eliminate Qs: Total: Eliminate one energy balance: Net change:

C+2 I

C+3 I

C+2

The net d.r. for the entire lInit would be 2C + 8, but this count is incorrect because )ou can see from the previous example that the correct number of d.r. = 2C .;- 6. We can locate the source of this discrepancy if we note for the mixer that stream L is actually a two-phase stream; hence N, = C -' 1, not C + 2. The proper analysis

l

1!

,

, 1 J

I

374

Combined lIfa/erial alld Energy Balances

Chap.

j

Z--+---I

Z

Fig. ES.3. Degrees of freedom in combined units. is as follows: For the mixer:

Total no. variables: Constraints (as before): Net: For the separator: Total no. variables: Constraints (as before): Net: Eliminate redundant counting:

singletwophase phase streams stream 2(C -+- 2) , I(C + I)

QM

+ I = 3C + 6 C+I 2C + 5

2(C

+ 2) -+-

I(C

+ I) + I = 3C + 6 2C + 3 C+3

Drop Qs: Eliminate Z: Together:

C+ I

C+2

Consequently the total degrees of freedom are (2C

-+- 5) + (C + 3) - (C + 2) = 2C -+- 6

as in Example 5.2. We do not have the space to illustrate additional combinations of simr:' units to form more complex units, but Kwauk' prepared several excellent rahk, summarizing the variables and degrees of freedom for distillation column'. absorbers, heat exchangers, and the like.

5.2 Enthalpy-concentration charts An enthalpy-concentration chart is a convenient graphical method of rcpre' senting enthalpy data for a binary mixture. If available,' such charts arc u;clc: 'M. Kwauk, A.l.Ch.£. J., v. 2, p. 240 (1956). 'For a literature survey as of 1957 see Robert Lemlich, Chad Gouschlich, and RonJiJ Hoke, Che",. Eng. Data Series, v. 2, p. 32 (1957). Additional references: for CCI., sec M. \1

·f- +*'ii.

·4

),_.

W ¥-¥

"Ag, .4,TW .•'''.II!'M

~~'TlII"_ _""'."'A""'_''''.i+_~*'''''''M''''''

__' ' ' #$''__.........:.......

ett

'*'

~.-"

'r

,"

..,

~.'-'"""~

Enthalpy-Concentration Charts 375

Sec_ 5.2

Chap. 5

.-,

in making combined material and energy balances calculations in distillation, crystallization, and all sorts of mixing and separation problems. You will find a few examples of enthalpy-concentration charts in Appendix I. At some time in your career you may find you have to make numerous repetitive material and energy balance calculations on a given binary system and would like to use an enthalpy-concentration chart for the system, but you cannot find one in the literature or in your files. How do you go about constructing such a chart?

~jned

units.

, ·,e am - I)

QM + 1 = 3C

+6

C+ 1 2C

+5

- I) -II = 3C 2C

+3

+6

C+3

I I! ,

I

j

1 -I

I C+2

2C-I6

":Jal combinations of simple c:::!red several excellent tabks ::1 for distillation columns.

" graphical method of repre_::lIe,' such charts are useful

. Chad Gottschlich, and Ronald " references: for CCI., sec M. M.

5.2-1 Construction of an Enthalpy-Concentration Chart. As usual, the first thing to do is choose a basis-some given amount of the mixture, usually I Ib or I Ib mole. Then choose a reference temperature (Ho = 0 at To) for the enthalpy calculations. Assuming I Ib is the basis, you then write an energy balance for solutions of various compositions at various temperatures: (5.1) where AHm,Xlu
AHA' AHB

= the enthalpy of lIb of the mixture

the enthalpies of the pure components per lb relative to the reference temperature (the reference temperature does not have to be the same for A as for B) AHru •• ", = the heat of mixing (solution) per lb at the temperature of the given calculation =

In the common case where the AH m'.'" is known only at one temperature (usually 7r F), a modified procedure as discussed below would have to be followed to calculate the enthalpy of the mixture. The choice of the reference temperatures for A and B locates the zero enthalpy datum on each side of the diagram (see Fig. 5.3). From enthalpy tables such as the steam tables, or by finding t~110F

(HA - HA,) =

f"

CPA

dt

you can find AHA at 7rF and similarly obtain fl.HB at 77'F. Now you have located points A and B. If one of the components is water, it is advisable to choose 32°F as to because then you can use the steam tables as a source of data. If both components have the same reference temperature, the temperature isotherm will intersect on both sides of the diagram at 0, but there is no particular advantage in this.

Krishnaiah et al.. J. Chern. En? Da!a. v. 10. p. 117 (1965); for EtOH-EtAc. see Robert Lemlich. Chad Gottschlich. and Ronald Hoke. Bri!. Chern. Eng .• v. 10. p. 703 (1965); for methanol,oluene. see C. A. Plank and D. E. Bu-rke. Hydrocarbon Processing, v. 45. no. 8. p. 167 (1966); for acetone-isopropanol. see S. N. Balasubramanian. Bri!. Chern. En?. v. II. p. 1540 (1966); fJr alcohol-aromatic systems. see C. C. Reddy and P. S. ~Iurti. Bri!. Chem. En?. v. 12. p. 1231 (1967); for acetonitrile-water-ethanol. sec Reddy and Mufti. ihid., v. 13. p. 1443 (1968); and for alcohol-aliphatics. see Reddy and Murti, ibid.• v. 16, p. 1036 (1971).

lJ 1I.,IO*"

w .., ·

...__ ...... " ..........._ _.-'.. '-...<"',..... _<,. <

Ghap.5 -c a plot of enthalpy vs< tern.

<-----_._--"' Sec. 5.1

Enthalpy·Concentration Charts

379

Step 3: Plot the enthalpy values for the saturated pure vapor (dew point), points C and D. A supplementary graph again is required to assist in interpolating enthalpy values .

. 7 psia =d pure C 7 (358°F). Fix the ~lY plotting the temperatur~. «:nalpy of the pure C. at 146· F , lO mole. These are marked in

p =100 psia

:>f liquid between A and B ",erimental measurements are :c-
/F 0/

=d. ]n this way the bubble·

,~

~

E

/

.D

,:ATI!'\

'. 100

.....::>

PSIA

/

i

+-

CD

/

. Btu/ •.. ",ole

,/

~

n ':>',/ '/..':--' ,/

IlfiL ~~ AHc.xc, _:1c,XCr

0 :254 490 710 .100 .515

qo

.:.420 "~OO

100 ,360 .500 ('00 :40 '00

"70

,/

,/

Ilfic ,xc, + liHCtxCT 19,250 18,250 17,280 16,300 14,600 12,650 11,000 8,300 6,240 4,700 3,480 2,440 1,540 740 366 0

19,250 18,504 17,770 17,010 15,700 14,165 12,870 10,720 9,040 7,800 6,840 5.940 5,140 4,480 4,166 3,970

,/,/

...... ...:: .......... .

.....

..........................

J 00

QI Q2 Q3 0.4 0.5 Q6 Q7 Q8 Q9 1.0 Y or x, mole fraction n - butane (C4 ) Fig. ES.4.

Step 4: Fill in the saturated·vapor curve between C and D by the following calculations:

Sec. 5.2 378

Combined !>faterial and Energy Balances

Ghap.5

Intermediate values of enthalpy were interpolated from a plot of enthalpy vs. temperature prepared from the data given on p. 377.

Step 3: P points C and enthalpy valul

Basis: 1 lb mole mixture Reference conditions: 32°F and 14.7 psia Step 1: Plot the boiling points of pure C, (l46'F) and pure C 7 (358°F). Fix the enthalpy scale from 0 to 30,000 Btu/lb mole mixture. By plotting the temperalurcenthalpy data on separate graphs you can find th2t the enthalpy of the pure C, at 146'F is 3970 Btu/lb mole and that of C 7 at 358'F is 19,250 Btu/lb mole. These are marked in Fig. E5.4 as points A and B, respectively. Step 2: Now choose various other compositions of liquid between A and B (the corresponding temperatures determined from experimental measurements arc shown in the table), and multiply the enthalpy of the pure component per mole by the mole fraction present:

30,OC

25,OC

...

Cl>

No heat of mixing is present since ideality was assumed. In this way the bubblepoint line between A and B can be established.

~

20,OC

><

E .0

.......

H-x

CALCULATIOl'iS FOR

C,-C 7

LIQUID (SATURATED) AT

:::>

100 PSIA

of-

(l)

...o

f>HL ~~ f>Hc.xc.

Temp., of

Mole fro

xc.

xc.

f>H c•

Mfc•

358 349 340.5 331.5 315 296 278.5 248 223.5 204 188 173.8 160.5 151.0 147.8 146.0

0 0.02 0.04 0.06 0.10 0.15 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 0.95 1.0

1.0 0.98 0.96 0.94 0.90 0.85 0.80 0.70 0.60 0.50 0040 0.30 0.20 0.10 0.05 0

13,200 12,700 12,220 II ,750 11,000 10,100 9,350 8,050 7,000 6,200 5,600 5,000 4,500 4,150 4,000 3,970

19,250 18,650 18,000 17,380 16,200 14,900 13,750 II ,850 10,400 9,400 8,700 8,120 7,700 7,400 7,310 7,280

f>Hc.xc. f>Hc.xc, 0 254 490 710 1,100 1,515 1,870 2,420 2,800 3,100 3,360 3,500 3,600 3,740 3,800 3,970

19,250 18,250 17,280 16,300 14,600 12,650 11,000 8,300 6,240 4,700 3,480 2,440 1,540 740 366 0

1 5,00

"::1::.:;'

Enthalpy, Btu/lb mole

+ aHCTxcr 19,250 18,504 17,770 17,010 15,700 14,165 12,870 10,720 9,040 7,800 6,840 5,940 5,140 4,480 4,166 3,970

10,00

500'

Step 4: F,:I calculations:

tv

380

Combined Material and Energy Balances

d,..'"

Chap. 5

Sec. 5.2 Enthalpy, Btu/lb IIl
IlHy = Mole fro

Temp., of

Ye.

358 349 340.5 331.5 315 296 278.5 248 223.5 204 188 173.8 160,5 151.0 147.8 146,0

0 0.084 0.160 0.241 0.378 0.518 0.630 0.776 0.857 0.907 0.942 0,968 0.986 0.996 0,999 1.000

fl.fic,Yc,

IlHe.)'e. IlHc,yc.

YeT

IlHc.

IlHe.

1.0 0,914 0.84 0759 0,622 0.482 0.370 0,224 0.143 0,093 0,058 0,032 0.014 0.004 0.001 0

17,500 17,150 16,850 16,500 16,050 15,800 14,980 14,100 13,420 12,950 12,500 12,100 11,800 11,500 11,450 11,400

29,250 28,700 28,100 27,550 26,700 25,800 24,900 23,600 22,550 21,700 21,100 20,600 26,100 19,720 19,600 19,540

29,250 26,200 23,600 :!O,860 16,600 12,420 9,210 5,290 3,220 2,020 1,225 659 282 79 20 0

0 1,440 2,700 3,970 6,060 8,050 9,425 10,950 11,500 11 ,750 11,780 11,700 11,620 11,450 11 ,430 11,400

+ Iliic.)"c, 29,250 27,640 26,300 24,830 22,660 20,470 18,635 16,240 14,720 13,770 13,005 12,359 11 ,902 11,529 11,450 11,400

Slep 5: Draw tie lines (the dashed lines) between the bubble· point and dewpoint lines (lines E-F, G-H, etc.). These lines of constant temperature show the equilibrium concentrations in the vapor and liquid phases. The experimental data were selected so that identical temperatures could be used in the above calculations for each phase; if such data were not available, you would have to prepare additional garhs or tables to use in interpolating the isobaric data (data at 100 psia) with respect to the mole fraction Y or x. If you wanted to draw the lines at even temperature intervals, such as 140,160, 180°F, etc., you would also have to interpolate. Step 6: Draw isothermal lines at 140, 160°F, etc., in the liquid and vapor regions. Presumably these will be straight lines since there is no heat of mixing. Since: ~+~=I

IlHL

~)

= IlHm • xtu " = xc.IlHc. + xe,IlHc. = xc,IlHc. + (I

(b)

Because IlHe. and Ilife• are constant at any given temperature, IlHe, - Ilil c, is constant, and Eq. (b) is the equation ,)f a straight line on a plot of lliJ m ."",, vs. Xc.' The 140'F isotherm is fixed by the points land K in the liquid region, and the 360°F isotherm is fixed by points Land M in the vapor region:

l(C.) K(C,)

3800 7100

360°F, vapor Ilfly (Blu/mole) (C.) (C,)

EXAMPLE 5.5 A One hundred r to give a 10 percen at 32°F are required

Use the steam tabl . I as your source n at 32°F for liquid . pure caustic ha\'in~

Solulioll: The data rcqu! FrO)

NaOH cone. 73 10

We can make am.!

- xc.)IlHc.

= (IlHc. - IlHe.)xc. + IlHe.

140°F, liquid Ilil L (BIU/mole)

The 240°F isotherm up to the saturatedin the vapor region i

The tie element in total H 20 added I

100 1\1

loTt;

Next we make an (

17,550 29,300 To differentiate k

Chap . .1 Enthalpy-Collcentralion Charts 381

Sec, 5.2 .:i1b mOle

..

l1Hy = 11 He. Yc. l1He,ye. + l1He,Yc, 29,250 26,200 23,600 20,860 16,600 12,420 9,210 5,290 3,220 2,020 1,225 659 282 79 20 0

29,250 27,640 26,300 24,830 22,660 20,470 18,635 16,240 14,720 13,770 13,005 12,359 11,902 11,529 11,450 11,400

Dub' . nt and dew.:ratu i the equilib::xperimental data were _'ve calculations for each '!'epare additional gaphs !)sia) with respect to the :1 temperature intervals, . :ate. .;J uid and vapor regions. .. j mixing, Since: (a)

The 240'F isotherm (and others higher than 140'F) in the liquid region is real only up to the saturated-liquid line-the dotted portion is fictilious. Portions of isotherms in the vapor region lower than 360'F would also be fictitious, as, for example, at 240"F.

EXAMPLE 5.5

Application of the enthalpy-concentration chart

One hundred pounds of a 73 percent NaOH solution at 350"F are to be diluted 10 give a 10 percent solution at 80°F. How many pounds of water at 80"F and ice

at 32"F are required if there is no external source of cooling available? See Fig. E5.5.

Feed 73% NoOH

T ~4 cb

Product 10% NoOH

Fig. E5.5.

Use the steam tables and the NaOH-H,O enthalpy-concentration chart in Appendix I as your source of data. (The reference conditions for the latter chart are H = 0 at 32"F for liquid water and H = 0 for an infinitely dilute solution of NaOH, with pure caustic having an enthalpy at 68"F of 455 Btu/lb above this datum.) Sollllion: The data required to make a material and an energy balance are as follows: From the steam tables M'j of liquid H,O = 48 Btu/lb Mf of ice = -143 Btu/lb (or minus the heat of fusion)

From Appelldix 1 NaOH cone.

73 10

temp., of MI, BIII/ib 350 468 80 42

We can make a material balance first. (b)

_cure, l1ife• -l1ife, is .Jt of Il.Hmir..lurc VS. Xc~. iiquid region, and the 360°F, I'apor ..J.Hy (BIll/mole) 17,550 29,300

Basis: 100 Ib 73 percent at 350"F The tie element in the material balance is the NaOH, and using it we can find the total H 2 0 added (ice plus H 2 0): 100 Ib product 173 Ib NaOH 10 Ib NaOH 100 Ib feed

= 730 Ib product/iOO Ib feed

less 100 Ib feed gives 630 Ib H 2 0 addedflOO lb feed Next we make an energy (enthalpy) balance, or To differentiate between the ice and the liquid water added, let us designate the ice

, 1

1 !., I,

~"..'-''''f

_ _ _......_ _ _ _".. , ......

~~,

382

..;.;..-

rlW Ok

·"'·rt___.......·......*i__, _

..'",,,,,,..·""'.....

~' .":«~,.h:~~,;~",>",W""iil!!_-1e

Combined Materia/ and Energy Ba/ances

Chap, 5

Sec. 5"

as x Ib and then the H 2 0 becomes (630 - x) Ib: in

out

H 20

73 % NaOH solution

------------.

ice

46,800

1-

-------.

100 Ib 1468 Btu -'- (630 - x) Ib 148 Btu -'- x Ib Ib ' Ib'

+ 30,240 -

191x

10% ;'\aOH solution

---------.

143 Btu _ 730 Ib 42 Btu Ib Ib

= 30,660

x = 243 Ib ice at 32'F liquid H 2 0 at 80°F

= 387 Ib

EXAMPLE 5.6 Application of the enthalpy-concentration chart Six hundred pounds of 10 percent NaOH at 200'F are added to 400 Ib of 50 percent NaOH at the boiling point. Calculate the following: (a) The final temperature of the solution. (b) The final concentration of the solution. (c) The pounds of water evaporated during the mixing process. SO/iltion,' Basis: 1000 Ib final sol ution Use the same NaOH-H 2 0 enthalpy-concentration chart as in the previous problem to obtain the enthalpy data. We can write the following material balance: camp. NaOH H 2O total

10 % solution -,- 50 % solution 60 200 540 200 600 400

= final SO/lilian 260 740

1000

wI

%

26 74 100

Next, the energy (enthalpy) balance (in Btu) is 10% solution 50% solution 600(152) T 400(290) 91,200 -I 116,000

fino/ solution i1H 207,200

207,200 Btu o~ 207 Btu1lb 1000lb ' On the enthalpy-concentration chart for NaOH-H 2 0, for a 26 percent NaOH solution with an enthalpy of 207 Btu:!b. you would find that only a two-phase mixture of ("I saturated H 2 0 vapor and (b) NaOH-H 2 0 solution at the boiling point could exist. To get the fraction H 2 0 vapor, we have to make an additional energy (enthalp), balance. By interpolation. draw the tie line through the point x -~ 0.26. H ,~ 207 (make it parallel to the no and 250 F tie lines). The final temperature appears front Fig. E5,6 to be 232'F; the enthalpy of the liquid at the bubble point is about 175 Btu/lb. The enthalpy of the saturated water vapor (no NaOH is in the vapor phase) from the steam tables at 232' F is 1158 Btuilb. Let x = Ib H,O evaporated .

$Q4J4>i.

'"

'*'"

4

..W).... ;::;

_ '" 4$.

PAZ

problems

can be sol this techn method; :

charts at "

Note that the enthalpy of the 50 percent solution at its boiling point is taken from the bubble-point curve at x = 0.50. The enthalpy per pound is

.. _

5.2One ofth

. ...

-:;t

Any with the; B, and C:

___ "" _______

'_'''~

__ '_

._.~._....i

Sec. 5.2

Elllhalpy-COllcentralioll Charls

383

Chap. 5 Interpoloted Tie line

t,

out 10% l\aOH solution ~ .3 Btu 730lb 42 Btu -J-= Ib

--------------

360

'" o

bH (Btu lib)

'"

°

'Tl

-...;;;:;:.:::i::<~7=--Finol

temperoture of the solution ~ 232°F

x

~

0.26

x--Fig. ES.6 •

.,n chart ::ce added to 400 Ib of 50 pcr·

Basis: 1000 Ib final solution x(IIS8)

983x

process.

::lg

260

740 -000

- x) 175 = 1000(207)

= 32,000

x = 32.61b H 2 0 evaporated

c[ as in the previous problem :material balance: . sallliioll

+ (1000

wi

%

26 74 100

.alUlioll

H

.200

5.2-2 Graphical Solutions on an Enthalpy-Concentration Chart. One of the major advantages of an enthalpy-concentration chart is that the same problems we have just used as examples, and a wide variety of other problems, can be solved graphically directly on the chart. We shall just indicate the scope of this technique, which in its complete form is usually called the Ponchon-Savaritt method; you can refer to some of the references for enthalpy-concentration charts at the end of this chapter for additional details. Any steady-state process with three streams and a net heat interchange Q with the surroundings can be represented by a simple diagram as in Fig. 5.4. A, B, and C are in pounds or moles. An overall material balance gives us

-"oiling point is taken from ..:nd is

A+B=C

,



1

1

1 j I

(5.3) j

1 1

26 percent NaOH solutl<'" _ two.phase mixture of 1.1 i nailing point could c'"'' . .!!tional energy (cnthal;" , :Joint x ~" 0.26, Ii :,'" ,emperature appears (,<-,:; ~~bble point is about 1-' .)H is in the vapor ph.!" i '--1 1 0

,.,

..

f"atcd.

~----~",,-"

\ 1----8

A--ooJ

c Fig, S.4. A typical process with mass and heat interchange.

...... -g .' ..' .

.,.,.,~_

",,'..__r _ t.._

_~_r:"!

."-'·"'}*_..........

.........._ _ _ _

M .. • _ _. .t_~'""""

______. . . . . _ ...•_..'__.....

384 Combined Material and Energy Balances

Chap. 5

and a component material balance gives AXA

Ob',

+ BXB =

(5.4)

CXc

An energy balance for a flow process (neglecting the work and the kinetic and potential energy effects) gives us' A

Q=f.t.H

;!I

or Q

+ AHA + BHa =

(5.5)

CHc

First we must choose some basis for carrying out the calculations. and then Q is dependent on this basis. Thus, if A is chosen as the basis, Q" will be Q per unit amount of A, or

Q..

=Q A

(5.6)

.tnd h.1I11

AI~c

Similarly, if B is the basis, (5.7) Thus

(5.8) Let us choose A as the basis for working the problem, and then Eq. (5.5) becomes AQ..

+ AHA + BHa =

CHc

A(H..

+ QA) + BHa =

CHc

or

On

fl .. ' x(.~.\·,

~U)

(5.9)

(5.10)

Similar equations can be written if B or C is chosen as the basis. (a) Adiabatic processes. As a special case, consider the adiabatic process (Q = 0). Then (with A as the basis) the energy balance becomes AH..

+ BHB =

CHc

=

(A

+ B)Hc =

AHc

+ BHc

or or

(5.11 ) Combining material balances, Ax"

+ BXa =

CXc = (A

+ B)xc =

AXe

+ Bxc

r.leh Ihe P'

or

n:ll",

or A Xc - X B B = x .. - Xc

(5.12)

:lnd

!:

!llll',!

'In what follows J{ is always the enthalpy relative to some reference value, and suppress the caret (A) and the symbol A to make the notation more compact.

~..,.

~<

- .... ~-~-...........---..,..."""''''.,. . . . . .·-"'........~.,.·'"'~,..,..,.......__"'r..' _4,,"".._ _#4"""',...........1 G,.,-."""*_~,..:;.:'I1"W...i_;t_...._#..""" ~,,,,,,,,

r 1,,",,11 .1 r LII I

Chap. 5

Enthalpy-Colleentration Charts 385

Sec. 5.2

Obviously then, (S.4)

.e work and the kinetic and

(5.13) Similarly, solving the energy and material balances for

A and

C instead of

A and B, we can get (S.S) ~t

the calculations, and then the basis, Q" will be Q per

A _ Xc -- X B __ He - HR C - X" - XB - H" - HB

and in terms of Band C we could get a like relation. Rearranging the two righthand members of Eq. (5.13), we can find that (5.13a)

-

Xa

Xc -

Xn

Xc

(S.6)

(5.14)

Also, from Eq. (S.14), (5.14a)

(S.7) (5.8)

and then Eq. (5.5) becomes

On an enthalpy-concentration diagram (an H-x plot) the distances He-HR' H[Ha, and H,.-He are projections on the H axis of a curve, while the distances xe-xa, X,,-Xa , X,,-X e are the corresponding projections on the X axis (see Fig. 5.5) The form of Eqs. (5.13a) and (S.l4a) is such that they tell us that they are

(5.9)

(5.10)

as the basis . . lder the adiabatic process :Jee becomes H

(S.lI) Fig. 5.5. Projections of the line BeA on the l/ and x axes.

(5.12 1 CJrne reference value, and "(

. marc compact.

each equations ofa straight line on an H-x plot and that these lines pass through the points A, B, and C which are represented on an H-x diagram by the coordinates A(li", x,,), B(lia , x a), and C(He, x,,), respectively. Since Eqs. (5. IJa) and (5.14a) both pass through point C, a common point, and have the same slope (lie - Ha)f(x" .- x R), mathematical logic tells us they must be equations of the same line, i.e., the lines are actually a single line passing through B,C, and A. as shown in Fig. 5.5. The symmetry of this type of arrangement leads to usc of the lera arm rule .

386 COII/bined Material and Energy Balances

Chap. 5 Sec. 5.2

+

If A B = C, then point C lies on a line between A and B, and nearer to the quantity present in the largest amount. If more B is added, C will lie nearer B. Whcn two streams arc subtracted, the resulting stream will lie outside the region of a line between the two subtracted streams, and nearer the one present in the larger amount. Thus if C - B = A, and C is larger than A, then A will lie on a straight line through C and B nearer to C than B. Look at Fig. 5.5 and imagine that B is removed from C. A represents what is left after B is taken from C. The il/rerse lever arm principle tells us how far to go along the line through A and B to find C if A·;· B = C, or how far to go along the line through C and B if C - B = A. In the case of A B = C, we saw that the weight (or mole) ratio of A to B was given by Eq. (5.13). From the symmetry of Fig. 5.5, you can see that if the ratio of A to B is equal to the ratio of the projections of line segments BC and AC on the If or x axes, then the amount of A -:- B should be proportional to the sum of the line segments (xc - x B ) (x" - xc), or (Hc - H B ) (H" - Hcl. Furthermore, the amount of A should be proportional to the measured distance BC and the amount of B proportional to the distance AC. Also the ratio of A to B should be

a basis agam,

Sl

Briefly, I Clie on and C[(/

+

+

+

A

7i =

distance Be distance AC

(5.15)

This relation is the inverse lerer arm rule. The distances BC, AC, and AB can be measured with a ruler. As an example of the application of Eq. (5.15), if A were 100 lb and the measured distances were

BC= 27.5mm AC = 49.0mm AB

=

76.5mm

then

.i = B

100 B

= BC

= 27.5 mm

AC

49.0mm

= 0.561

, Pro, Material

and

B= The sum of A

+B=

100 = ;78lb 0.561

Energy PI

C is 100 -I- 178 = 278 lb. Alternatively,

.i = Be;, = 27.5 mm = a 360 C

AB

76.5 mm

A

C

100

= 0.360 = 0.360

i

I

.

Thuj

=- 278 Ib

(b) Nonadiabatic processes. Let us now consider the circumstances under which Q is not equal to zero in Eq. (5.5). This time, for variety, let us take C as

and, re;11

A similJI

, .it- .P

iii""

.) .,

#$

,;;

. t .10

.4\11 e,

'lNiii

P'.!it

b, ,;;

... :c;;

,~.....,..p"",,!

' .. F.J

>... ..j:'

b

Chap. 5 Enthalpy-Concentration Charts 387

Sec. 5.2

:lnd B, and nearer to -"ed, C will lie nearer will lie outside the 'arer the one present =~r than A, then A B. Look at Fig. 5.5 jeft after B is taken

a basis so that Q = CQc. Then we have two pairs of equations to work with again, AH"

+ BHs = C(Hc Ax" + BXB = CXc

Qc)

(5.16) (5.4)

Briefly, Eqs. (5.16) and (5.4) tell us that on an H-x diagram points A, B, and C lie on a straight line if the coordinates of the points are A(HA , x,,), B(HB , x B ), and C[(Hc - Qcl, xcl, respectively,' as illustrated in Fig. 5.6 Since the coordi-

:,ng the line through line through C and :Je weight (or mole) :)f Fig. 5.5, you can , projections of line _'f A B should be p) (x" - xc), or should be propor-:Jroportional to the

+ +

(5.15) TJj can be

'.'ere 100 Ib and the

,

i I

x Fig. 5.6. Inverse lever arm rule applied to a nonadiabatic process.

I

'Proof of these relations is analogous to the previous development. Material balances: Ax" + BXB = CXc = (A + B)xc = Axc + Bxc

A(x" - xc)

=

B(xe -

(a) (b)

XB)

Energy balance:

AH"

+ BHB = =

C(He - Qc) = (A + B)(He - Qc) A(He - Qc) -1- B(He - Qcl

A[H" - (He - QclJ

=

B(lle - Qcl - HB

(e) (d)

Thus

A _ Xe -

Ii -~umstances under '. let us take C as

x" -

XR _,

(He - Qc) - !fa

Xe - HA - (He -

Qc)

(c)

and, rearranging. HA - (fie XA

Xc

Qcl

=

(lie -

Qcl - fiR

Xc -

A similar equation can be obtaincd for the ratio A/e.

Xs

(f)

.......

"1 t

'#

M

.. ·--.:.~".iI_"·"'M"'·

_ ..............·..a...........;,;"......·_*"'...__..•·..tl'..• _ _ _ _ _........_ _......·..."'b.......

~

Sec. 5.2 388

Combined Material alld Energy Balances

Chap. 5

H He (H8 +08 )

H8 L....-'-'--!---'''-'-:L-_--:L

r

r

Figs. 5.7 and 5.8. Inverse lever arm principle for different bases.

nates of C are really fictitious coordinates and the enthalpy projection of C is (He - Qd, we shall designate the point through which the line passes [(He - Qe ), xcl by the more appropriate notation CQ' If A or B had been chosen as the basis, we would have set up diagrams such as Figs. 5.7 and 5.8 A different series of figures could be drawn for other typical problems, as, for example, C being split up into two streams with heat being absorbed or evolved. Additional details related to more complex problems will be found in the references on enthalpy·concentration charts at the end of the chapter.

To obtain t as in Fig. I

EXAMPLE 5.7 Graphical use of enthalpy-concentration charts Do Example 5.5 again by graphical means.

Solution: Plot the known points A and C on the H·x diagram for NaOH. The known points are as follows (with the given coordinates or data): solution Initial Final Added

x 0.73 0.10 0.00 0.00

(A) (C)

(D) (E)

H, Btu/lb 468

T, OF

350 80 80 32 (ice)

We me,

What we shall do is remove solution A from the final solution C to obtain solution B, i.e., subtract C - A to get B as in Fig. E5.7a. This is the reverse of adding A to B to get C. Draw a line through A and C until it intersects the x = 0 axis (for pure water). The measured (with a ruler) ratio of the lines BC to AC is the ratio of73 percent NaOH solution to pure water. Basis: 100 lb 73 percent NaOH solution at 350°F (A) 100 lb (A) Ib H 2 0 (E)

0~i~9 =

Ib H 2 0

=

B!; AC

=

=

2.85 units ~o 0 159 .

17.95 units

630 lb ice plus liquid water (B)

EXAMPLE Do bed

Sollltiv!1 See Fi~.

_ _ _..... _ ....... 'f...""' _ _ ..... .

'~~

EllIhalpy,COllcclltratioll Charts 389

Sec. 5.2 Chap. 5 .-----IAlft,....A)

I

A(468,0.73)

468

I

I

/[HC, xcll I

,

.,tH,+Oal, Ie]

I I

'"00

-;:;

I

0

'"

0

"

I I I

'"'"0

I

I I

."

=>

cD

~

x '::-:Ill

I I I I I

bases.

J

0.73 ~Ipy



~

projection of C

j

To ice ot -143 Btu/lb

.:ch the line passes ,'r Bhad been chosen

'j

Fig. ES.7(a).

5.7 and 5.8 ',vpicaI pL'oblems. as,

• bcing absorbed or will be found in of f 1pter.,

::IDS

To obtain the proportions of ice and water, we want to subtract D from B to get E as in Fig. E5.7(b).

4B 0' H20 at BO°F

r

H

8 Mixture af ice and water [O+EJ

I1

NaOH. 'Fhe known

,

-143 E' Ice at 32°F

"'F

i

Fig. ES.7(b). We measure the distances DE and BE. Then,

=

,xis- ~for pure water).

=

=

3.65 units 5.95 units

=

Ib H 2 0 CD)

= =

630(0.613)

=

3861b

630 - 386

=

244 Ib

Ib icc (E)

2f7JpercentNaOH (AJ

BE DE

Ib H 2 0 (D) Ib total (B)

:: tn, obtain solution of adding A, to B

Graphical use of enthalpy·concentration charts

Do Example 5.6 again by graphical means. So/utioll: See Fig. E5.8.

..,-"',---"".,,,,,.

-~.

l 1,

j

1

1 EXAMPLE 5.8

3)

0 613 .

'n~

390

t.i"

t&

at

..

t ....

e!!



r.

iVt

-'

gee 1 t

Combined Material and Energy Ba/ances

.-';($

t'

t

'n'irtti't

~,~

Chap. 5

Sec. 5.2

H

mol

He

Fig.

~

The gove is added and i

x Fig. ES.S. Graphical use of enthalpy charts. or

Basis: 1000 lb of total solution Again plot the known data on an FI-x chart. Point C, the sum of solutions A and B, is found graphically by noting by the inverse lever arm rule that distance BE distance BA

lb A lbA +B

=

400 1000

=

Connect poin (see Fig. E5.9i

0.4

Now measure BA on Fig. E5.8 and plot BC as O.4(BA). This fixes point C. Sir,ce C is in the two-phase region, draw a tie line through C(parallel to the nearby tie lines) and measure the distances CD and ED. They will give the H 2 0 evaporated as follows: CD

ED =

lb H 2 0 evaporated 1000 lb mixture

lb H 2 0 evaporated

~

331b

The final concentration of the solution and the temperature of the solution can be read off the H-x chart as before as

x

= 26%

FI

=

(FIA - (

To reach Ao ~ plus value h.l·

EXAMPLE 5.9 Graphical use of enthalpy-concentration charts with processes involving heat transfer An acetic acid-water mixture is being concentrated as shown in Fig. E5.9(a). How much heat is added or removed?

SO/lilian: First we have to prepare an enthalpy concentration chart cry plotting the data for the acetic acid· water system from Appendix I. The coordinates of all the points. A. B, and C. are known. These points can be plotted on the FI-x chart, and then some basis chosen. A, B, or C.

.•

l.

*' ;

*' ,45 14_ """

'HI . . ::;;:::,&14

¥.

If point A Q h and QA WGU I.If we COli

or

, ~,~';:¢'.'~

I

Chap. 5

Sec. 5.2

Enthalpy-Concentration Charts

.I

391

Basis: 1 Ib mole A

,r - - - - - A mole % H20' 80% (soturoled voporl

c



-0'7

mole % H20 ' 50% He ' 360 Slullb mole

l' - - - - 8

mole % H20' 0% H8 ' 7200 Sfu/lb mole

Fig. ES.9(a). Graphical use of enthalpy-concentration charts with processes involving heat transfer. The governing material and energy balances in this case are (assuming that Q is added and is plus)

=

CXc _1.5.

Q + CHc

=

Ax"

+ DXB

AH" + DHB

=

AQ" + CHc

or . the sum of solutions A =nn rule that . .4

A(H.. - Q .. )

+

DHB

=

CHc

Connect points Band C with a line. and extend the line toward the right-hand axis (see Fig. E5.9(b». What we are looking for is the point A" located at the coordinates

tc. , to the nearby ,'e the H z0 evaporated as

1.0

x

:: of the solution can be

Fig. ES.9(b)

IS

"'ith processes imolving

shown in Fig. E5.9(a).

:!ort

II = (II.. - Q,,) and x = 0.80 on the extension of the line BC through Band C. To reach AQ from A. we must subtract Q,,; by measuring on Fig. E5.9(c), we find a plus value has to be subtracted from H A , or

Q.. = +24,000 Btu/lb mole A

If point AQ had fallen above A. then heat would have been removed from the system and Q" would have been negative. If we convert our answer to the basis of I Ib mole of C_ we find that

.i. = BC

by plotting the data

. 'n. These points can be

cc.

(heat added)

C

AB

~= 0.5 - 0 = 0625 0.8 - 0 .

or 1 C = 0.625 = 1.60 Ib mole

.~'-

,.

Sec. 5.3

392 Combined Material and Energy Balances

Chap. 5

You \ and reIatil

(a) Tl ex C

w

1

in th

ca

wi

Fig. ES.9(c).

(b) TI in Basis: I Ib mole C 24,000 Btu

Qc = Ib mole A

1.00 Ib mole A 1.60 Ib mole C = 15,000 Btu/lb mole C

As an alternative solution, we could have selected C as the basis to start with and used the equations CXc = AXA + BXB Q

+ CHc = AHA + BHB =

C(Hc

+ Qd

In this case, join A and B and find point Co at x = 0.50 on the line AB. Measure CoC; it is 14,900 Btu, which is close enough to the value of 15,000 Btu/lb mole C calculated above. A similar calculation might have been made with B as the basis, by joining points A and C and extending to the x = 0 axis in order to get Bo. QB would be measured from Bo to B on the x = 0 axis.

In Chap. 3 we discussed humidity, condensation, and vaporization. In this section we are going to apply simultaneous material and energy balances to humidification, air conditioning, water cooling towers, and the like. Before procecding further, you should review briefly the sections in Chap. 3 dealing with vapor pressure. Recall that the humidity X is the pounds of water vapor per pound of bOllcdry air (some texts use moks of water vapor per mole of dry air as the humidity) or, as indicated by Eq. (3.52), (5.17)

."

<$

(itN,

'iP5$€U$QJ ,

.".'. . ., .__,".'__"' ' ' '., . .

W~~_

h,1 (d)

n Ile (01

tiD Su btl th,

5.3 Humidity charts and their use

"-

wi (c) TI

,,q.........,........,.___......'''...................,•..,.;_"""'__

PS!

ed nll.

btl As the to cool urll equals the' We say th.! temperatu:;

...

-

---------~

Humidity Charts and Their Use 393

Sec. 5.3 Chap. 5

You will also find it indispensable to learn the following special definitions and relations. (a) The humid heat is the heat capacity of an air-water vapor mixture expressed on the basis a/lib 0/ balle-dry air. Thus the humid heat Cs is (S.18)

!

where the heat capacities are all per pound and not per mole. Assuming that the heat capacities of air and water vapor are constant under the narrow range of conditions found for air conditioning and humidification calculations, we can write

I

0.9 1.0

Cs

=

0.240

+ O.4S(JC)

(S.19)

where Cs is in Btu/CF)(Ib dry air). (b) The humid miume is the volume of I Ib of dry air plus the water vapor in the air,

v= ::>mole C =e '

I

3S9 ft' lib mole .

start with

I Ib mole air T' F -1- 460 291b air 32 -1- 460

359 ftl

-r lIb mole = (O.730T· F -1-

_::he line AB. Measure .CIOO Btu/lb mole C cal::oasis, by joining points --' B would be measured

yaporization. In this energy balances to -=-:Jd the like. Before =:hap. 3 dealing with ~

per pound of bone:lir as the humidity)

(5.17)

"

lIb mole H~2-=-O+.+-,--~+JC...:.,;.lb'-----,H:..£2-=-O 181b fLO Ib air

336)(2~ + ~~)

(S.20)

where V is in ftl/lb dry air. (c) The dr}~bulb temperature (TDn ) is the ordinary temperature you always have been w,ing for a gas in cF. (d) The wet-bulb temperature (TWB) you may guess, even though you may never have heard of this term before, has something to do with water (or other liquid, if we are concerned not with humidity but with saturation) evaporating from around an ordinary mercury thermometer bulb. Suppose that you put a wick, or porous cotton cloth, on the mercury bulb of a thermometer and wet the wick. Next you either (I) whirl the thermometer in the air as in Fig. 5.9 (this apparatus is called a sling psychrometer when the wet-bulb and dry-bulb thermometers are mounted together). or (2) sct up a fan to blow rapidly on the bulb at 1000 ft'l min or more. What happens to the temperature recorded by the wetbulb thermometer? As the water from the wick e\'aporates, the wick cools down and continues to cool until the rate of energy transferred to the wick by the air blowing on it equals the rate of loss of energy caused by the water c\'aporating from the wick. We say that the temperature at equilibrium with the wet wick is the wet-bulb temperature. (Of course. if watcr continues to evaporate, it e\'entually will all



',"

"",<>

Hi"

II

J, J

1

Ii

~,

__ ~..... _ .. ' ..........·•_ _ r ,,'""" _ _....."'&...._'''_~,'''._

394

........._

"-""·"'a;x_..•...._·,........ "'... --..··_0_ _....·_·....._ _ _111'&_._4_

Chap. 5

Combined Materiol and Energy Balances

Sec. 53

disappear, and the wick temperature will rise.) The final temperature for the process described above will lie on the 100 percent relative humidity curve (saturated-air curve), while the so-called wet-bulb line showing how the wet-bulb temperature changes on approaching equilibrium is approximately a straight line and has a negative slope, as illustrated in Fig. 5.lD.

Air (Energy in)

(c) (d)

With an: chart an OfT scales sh, mixture. itself. Til

.

~

.

~ H20

Evaporating (Energy Loss)

Fig. 5.9. The wet-bulb temperature obtained with a sling psychrometer.

Adiabatic sol uration line and Wet bulb temperature line

Temperalure - -

Fig. 5.10. General layout of the humidity chart showing the location of the wet-bulb and dry-bulb temperatures. the dew point and dew-point temperature, and the adiabatic saturation line and wet-bulb line.

Now that we have an idea of what the various features portrayed on the humidity chart (psychrometric chart) are, let us look at the chart itself. Fig. 5.11

(a) and 5.11 (b) (inside back cover). It is nothing more than a graphical means of presenting the relationships for and between the material and energy balances in water vapor-air mixtures. Its skeleton consists of a humidity (JC) - temperature (TD . ) set of coordinates together with the additional parameters (lines) of (a) Constant relative humidity (10-90 percent). (b) Constant moist volume (humid volume).

We humidit) of H 2 0) : only at n, significall! enthalpy' air and II, ing US\! of following

Consolida:

You v. equilibriur' water. Ra:1 mental idl".: of water all ture and 1:1 changes. T'

6 For a d references at :

'See G.

j

Chap. 5

Humidity Charts alld Their Use

SL'e. 5.3

temperature for the :lve humidity curve :lg how the wet-bulb )ximatcly a straight

"rmometer

-Wet Wick

(c) Adiabatic cooling lines which are the sanlc (for water vapor only·) as the wet-bulb or psychrometric lines. (d) The 100 percent relative humidity (identical to the 100 percent absolute humidity) curve or saturated-air curve. With any two values known, you can pinpoint the air-moisture condition on the chart and determine all the other required values. Off to the left of the 100 percent relative humidity line you will observe scales showing the enthalpy per pound of dry air of a saturated air-water vapor mixture. Enthalpy corrections for air less than saturated are shown on the chart itself. The enthalpy of the wet air, in BtuJlb dry air, is (5.21)

~~oroting __ J55)

. ,,"chrometer.

395

We should mention at this point that the reference conditions for the humidity chart are liquid water at 32°F and I atm (not the vapor pressure of H 2 0) for water. and OCF and I atm for air. The chart is suitable for use only at normal atmospheric conditions and must be modified' if the pressure is significantly different than I atm. lf you wanted to, you could calculate the enthalpy values shown on the chart directly from tables listing the enthalpies of air and water vapor by the methods described in Chap. 4, or you could, by making use of Eq. (5.21), compute the enthalpies with reasonable accuracy from the following equation for I Ib of air:

----

t!..fi = 0.240(T' F

-

0)

C p(A T) for air

+

-

JC[I075

+ 0.45(T'

heat of vaporization of water at 32

32)]

-----------F -

(5.22)

Cp(AT) for

water vapor

1

1

Consolidating terms:

:!" location of the -m and dew-point :nd wet-bulb line.

:lres portrayed on the chart itself, Fig. 5.11 :: a graphical means elf :md energy balances 1n :dity (JC) - tempera. parameters (lines) e,f .2

t!..fi

=

0.240T' F

+ JC(l061 + 0.45T.

F)

(5.23)

You will recall that the idea of the wet-bulb temperature is based on the equilibrium between the rates of energy transfer to the bulb and evaporation of water. Rates of processes are a topic that we have not discussed. The fundamental idea is that a large amount of air is brought into contact with a lillie bit of water and that presumably the evaporation of the water leaves the temperalure and humidity of the air unchanged. Only the temperature of the water changes. The eq uation of the wet-bulb line is

(5.24)
'See G. E. McElroy, U.S. Bur. Mines Rept. /m'est., No. 4165, Dec. 1947.

1,

,I I

Sec. 5.3 Chap. 5

396 Combined Material and Energy Balances

makeup where he = heat transfer coefficient for convection to the bulb k~ = mass transfer coefficient AN". = latent heat of vaporization 3C = humidity of moist air T = temperature of moist air in OF

0.:

The equation for the wet-bulb lines is based on a number of assumptions, a discussion of which is beyond the scope of this book. However, we can form the ratio

(3CWB - 3C) (TWB - T)

-

3C) T) = -

Cs

Recirculoled Water T' conslonl 'Tsoturoted

Tsoturoted

Makeup water Fig. 5.12. Adiabatic humidification with recycle of water.

.~

....,

._-

p...,.....................,. . . . -~

'.

-

~...........,. ." ' _.....""~._..,"'_,. _ "i'II-' _ _.. -,.,.'".... __ ,.;III!. _.'!'I'_.

,=..."...". ................... -, ,. ~_.

"""0"",,

'""'. _ _ "".,,~

,.,"

which is t; Notie;

~

.....

because C tion. for II For other I Only t cause 3C s j, Thus, for 't the humid it linear, will described c The ad entering air (I or 2 peru the deviatil) can emplo) can be used bulb proCC\ usual matcr from the hl.1 content of :

moisture ad!

content rises) while a little bit of the recirculated water is evaporated. At equilibrium, in the steady state, the temperature of the air is the same as the temperature of the water, and the exit air is saturated at this temperature. By making an overall energy balance around the process (Q = 0), we can obtain the equation for the adiabatic cooling of the air. The equation, when plotted on the humidity chart, yields what is known as an adiabatic cooling line. Employing a version of Eq. (5.22) with the equilibrium temperature of the water. Ts. taken as a reference temperature rather than 0 or 32°F. we get, ignoring the small amount of

'~''''''_''''''''''''''J!",a",. . ,.......".. = ....._ ....

r

(5.26)

AN".

For other substances, the value of hJk~ can be as much as twice that of water. Another type of process of some importance occurs when an adiabatic cooling or humidification takes place between air and water that is recycled as in Fig. 5.12. In this process the air is both cooled and humidified (its water

Cooling Tower

This can

(5.25)

For water only, it so happens that hJk~ - Cs , i.e., the numerical value is about 0.25, which gives the wet-bulb lines the slope of

(3C WB (TWB

II

~

~~

ance for lice and 5.13. You ea in the rrfen' the therml'" 8Microli il l University \1:;

.

M

>

-- ----..

C StM"

"'~-----

Sec. 5.3

Hllmidity Charts and Their Use

397

Chap. 5

makeup water or assuming that it enters at Ts. ~

bulb

enthalpy of air entering

enthalpy of water ",por in air entering

enthalpy of air leaving

-,umber of assumptions, However, we can form

=

enthalpy of water mpor in air l<~a\"ing

0.240(Ts - T~) -1- Xs[,MI". T

= Ali". .i,

H,O ..

0.240

T.JJC s - X. i ,)

+ 0.45X. i ,

(Xs - X) = _

(Ts - T. i ,)

(5.26) as twice that of water. '=urs when an adiabatic J wat~· that is recycled 'nd ' 'fied (its water "1

j

I

I I

!

evaporated. At equilib:rie same as the tempera· c:nperature. By making an 'can obtain the equall"n " plotted on the humi,lit\ ~. Employing a version ,)1' ':'leT, Ts. taken as a rcfcl--;oing the small amounl "i'

-'-

Ts

(5.28)

which is the equation for adiabatic cooling. Notice that this equation can be written as

the numerical value is

IS

(5.27)

-7- 0.45(Ts - Ts)J

This can be reduced to

(5.25)

.' of water.

H,O .. Ts

t

Cs

(5.29)

Aliv •• .. Ts

because Cs = 0.240 -1- 0.45JC and TWB = Ts. Thus the wet-bulb process equation, for water only, is essentially the same as the adiabatic cooling equation. For other materials these two equations have different slopes. Only two of the quantities in Eq. (5,28) are variables. if Ts is known, because Xs is the humidity of saturated air at Ts and Ali". H,O.t Ts is fixed by Ts. Thus, for any value of Ts. you can make a plot of Eqs. (5.25) and/or (5.29) on the humidity chart in the form of X vs. T'i,' These curves, which are essentially linear, will intersect the 100 percent relative humidity curve at Xs and Ts, as described earlier. The adiabatic cooling lines are lines of almost constant enthalpy for the entering air-water mixture, and you can use them as such without much error (lor 2 percent). However, if you want to correct a saturated enthalpy value for the deviation which exists for a less-than-saturated air-water vapor mixture, you can employ the enthalpy deviation lines which appear on the chart and which can be used as illustrated in the examples below. Any process which is not a wetbulb process or an adiabatic process with recirculated water can be treated by the usual material and energy balances. taking the basic data for the calculation from the humidity charts. If there is any increase or decrease in the moisture content of the air in a psychrometric process. the small enthalpy effect of the moisture added to the air or lost by the air may be included in the energy balance for the process to make it more exact as illustrated in Examples 5.11 and 5.13. You can find further details regarding the construction of humidity charts in the references at the end of this chapter. Tables are also available listing all the thermodynamic properties (p, JC, All, and 5) in great detail.' Although

j

j

1

j

j

1

1 j

1

i

j 1

r-,

'Microfilms of Psychromelric Tables, 1953. by Byron Engelbach are available from University Microfilms, Ann Arbor. Mich.

..

,,~

,.~,.,---"",-

........,J..-

....... ' - - _.........• _ _ _....._

398

'"_w..__·•..·....

_._.....,~."'.»tS ......_ _.. ' .,.'W~,

..

l!nrt

wt)'tM¥

"t'

m

t

Combilled Alarerial alld Ellergy Balallces

b* M

Chap. 5

we shall be discussing humidity charts exclusively, charts can be prepared for mixtures of any two substances in the vapor phase, such as CCI. and air or acetone and nitrogen, by use of Eqs. (5.17)-(5.28) if all the values of the physical constants for water and air are replaced by those of the desired gas and vapor. The equations themselves can be used for humidity problems if charts are too inaccurate or are not available.

et

Sec. 5. (c

(d (e

EXAMPLE 5.10 Properties of moist air from the humidity chart List all the properties you can find on the humidity chart for moist air at a drybulb temperature of 90'F and a wet-bulb temperature of 70'F. Solulion: A diagram will help explain the various properties obtained from the humidity chart. You can find the location of point A for 90"F DB (dry bulb) and 70°F WB (wet bulb) by following a vertical line at T D • = 90°F until it crosses the wet-bulb line for 70'F. This wet-bulb line can be located by searching along the 100 percent humidity line until the saturation temperature of 70'F is reached, or, alternatively, by proceeding up a vertical line at 70'F until it intersects the 100 percent humidity line. From the wet-bulb temperature of 70'F, follow the adiabatic cooling line (which is the same as the wet-bulb temperature line on the humidity chart) to the right until it intersects the 90'F DB line. Now that point A has been fixed, you can read the other properties of the moist air from the chart. See Fig. E5.10.

EXAM M much I point (' So

A, lineof,

/

(J1JC, 1'( line at I TLI

Fig. ES.IO. (a) Dew po';,I. When the air at A is cooled at constant pressure (and in effect at constallt humidity), as described in Chap. 3, it eventually rcaches a temperature at which the moisture begins to condense. This is represented by a horizontalline, a constant-humidity line, on the humidity chart, and the dew point is located at B, or about 60' F. (b) Rclatire humidity. By interpolating between the 40 percent (J1JC and 30 percent (J1JC lines with a ruler, you can find that point A is at about 37 percent (J1Jc.

~•. ~ WI ..

utt

•. ;:;

.,., " _ 4

r -,",""

."

I;"

A ,SA I

, 4tu::;e

P#4 ,

• .... A

R

..

.

Also ;,' added .,

--Chap. 5 c::.n be prepared for ::1. and air or ace.les of the physical 'red gas and vapor. ~s if charts are too

':>r

Humidity Charts and Their Use 399

Sec. 5.3

(c) Humidity (X). You can read the humidity from the right-hand ordinate as 0.01 J2lb H,O/Ib air. (d) Humid rolume. By interpolation again between the 14.0- and the 14.5-ft' lines. you can find the humid volume to be 14.097 ft'Jlb dry air. (e) Enlhalpy. The enthalpy value for saturated air with a wet-bulb temperature of 70°F is tiH = 34.1 Btu/lb dry air (a more accurate value can be obtained from psychrometric tables if needed). The enthalpy deviation for less- thansaturated air is about -0.2 Btu/lb dry air; consequently the actual enthalpy of air at 37 percent
moist air at a dry-

EXAMPLE 5.11 from the humidity bulb) and 70"F WB crosses the wet -bul b :Clong the 100 percent -",d, or, alternatively. : 00 percent humidity -:;:: c('~"~o; line (which 'oart : right until ","ou , ad the other e:.j

Heating at constant humidity

Moist air at 50°F and 50 percent
Solution: As shown in Fig. E5.11. the process goes from point A to point B on a horizontal line of constant humidity. The initial conditions are fixed at Tn. = 50°F and 50 percent

T

Fig. ES.ll. OlJC. Point B is fixed by the intersection of the horizontal line from A and the vertical Ilnc at 100'F. The dew roint is unchanged in this process and is located at Cat 32.5°F. The enthalpy values are as follows (all in Btu/lb dry air):

-:>ressure (and in etfc(t ...:lily reaches a temr~rJ' - represented by J hl'll" '="''"l3.rt. and the dew rl'ln~

-""nt
point

till",.

oH

tiH" •• ol

A B

16.15 28.57

-0.02 -0.26

16.13 28.31

Also at A the volume of thc moist air is 12.92 ft'llb dry air. Consequently the heat .JJcJ is (Q= tiii) 28.31 16.13 = 12.18 Btu/lb dry air. 12.18Btu Ilbdryair _ 09.PB If"" I .. Ib dry air 12.92 IV - . - tu t 100tia mOist air

.......

-.~-

...

,

..

_~--"""''''--._i

It_ _ _..............·a'_!1l· ...&....."_,;;,iIo....._'-___ •••'. .' .' ......._

400

0"'..............'-= ..· ....

.............._ _. .'.or ... ·"'._ _,~ti_ _ _. .·.... _ _-. . . . ." . .'.......·""...

Chap, 5

Combined Material and Energy Balances

Sec. 5.3 Humidity v;,

EXAMPLE 5.12

Cooling and humidification

One way of adding moisture to air is by passing it through water sprays or air washers. See Fig. £5.12(a). Normally. the water used is recirculated rather than Difference:



111~ \ ~ II/I \\

!3!. /

I

f\

EXAMPLE 5 A proce" sprays sounds below the de"

Make-up H20 Fig. ES.12(a).

wasted. Then. in the steady state. the water is at the adiabatic saturation temperature which is the same as the wet-bulb temperature. The air passing through the washer is cooled. and if the contact time between the air and the water is long enough. the air will be at the wet-bulb temperature also. However. we shall assume that the washer is small enough so that the air does not reach the wet-bulb temperature; instead the following conditions prevail: T DB • OF 100 Entering air 70 80 Exit air Find the moisture added per pound of dry air. Sollllion: The whole process is assumed to be adiabatic. and. as shown in Fig. £5.12(b), takes place between points A and B along the adiabatic cooling line. The wet-bulb temperature remains constant at 70°F.

trick. If the enl( heat has to be ,. exit air is at 56 Sollllion.· From Fig,

Look at Fig.

~

E~

80'F WO°F T

Fig. ES.12(b).

The cooling dul:

Chap. 5

Humidity Charts and Their Use

Sec. 5.3

401

Humidity values are

X Ib H 20 :hrough water sprays or :.:; recirculated rather than

• Ib air 0.0135 0.0088

B A

Difference: 00047 Ib H 2 0 . bl dry air

EXAMPLE 5.13

added.

Cooling and dehumidification

A process which takes moisture out of the air by passing the air through water sprays sounds peculiar but is perfectly practical as long as the water temperature is below the dew point of the air. Equipment such as shown in Fig. E5.13(a) would do the

=

Cold Wafer

mmxation temperature =sTrw ". " ()ugh the washer .:::..--r enough, the air ~ ..at the washer is ~e; instead the

Cooler

1 Bleed H20 (excess)

shown in Fig. E5.12(b). c-..:;oling line. The wet-bulb

j

Fig. ES.13(a).

_3

trick. If the entering air has a dew point of 70°F and is at 40 percent
1

Solution: From Fig. 5.II(b) the initial and final values of the enthalpies and humidities are

\

2

AB(Ib dry Btu) air

B

A III

H0) Ib dry air

X (grainS

41.3 - 0.2

62

= 41.1

23.0 - 0 = 23.0

Look at Fig. ES.13(b) to locate A and B. The grains of H 2 0 removed are III - 62

= 49 grainsflb dry air

The cooling duty is approximately 41.1 - 23.0

= 18.1

Btuflb dry air

I

'+"'11';

402

Chap. 5

Combilled Malerial alld EJ:ergy Ba/allces

Sec. 5.3

T

Fig. ES.I3(b).

In the upper left of the humidity chart is a little insert which gives the value of the small correction factor for the water condensed from the air which leaves the system. Assuming that the water leaves at the dew point of 54 of, read for 49 grains a correction of -0.15 Btu/lb dry air. You could calculate the same value by taking the enthalpy of liquid water from the steam tables and saying,

=iTi;--1-,7;~-,,:,:,~+4:.:9~gc-:roi'a.:.:i e:-,ct:=e.=d .. I Ib n:,:s",r..:,e dry air Je;:'

= 0 . 154 Btu II b d ry air.

,

I c

I

Combined material and energy balances for a cooling tower

You have been requested to redesign a water-cooling tower which has a blower with a capacity of 8.30 x 10' ft'/hr of moist air (at 80°F and a wet-bulb temperature of 65°F). The exit air leaves at 95°F and 90°F wet bulb. How much water can be cooled in pounds per hour if the water to be cooled is not recycled, enters the tower at 120'F, and leaves the tower at 90'F? So/ulioll: Enthalpy, humidity. and humid volume data taken from the humidity chart are as follows (see Fig. E5.14) A

B

0.0098

0.0297

X (grains H2O)

69

208

Aii(Ib dry Btu ) air

30.05 - 0.12 = 29.93

55.93 - O. \0 = 55.83

13.82

14.65

Ib dry air

Ib dry air

A(

ft3 ) V Ib dry air

"'"' ~""""~__~R"".""¢"""'~'___""'"~.........,.....L_"'.."'."'' '.....,,.,.,,........_,....' "'._........_.,"'....."...~..

_",.'.,)A"""....,........_ ...._"".,'. . .'~,'

;:p_ _ , ...

The c 1 atm)

!

air in air out H 2 0 out 41.1 - 23.0 = 0.15 = 17.9 Btu/lb dry air

XCb H 2 0 )

I !I

The energy (enthalpy) balance will then give us the cooling load:

EXAMPLE 5.14

The coolie process.

""lit;rA";&I!!fi!

while that I for liquid, vapor pre, its vapor I water. Fo: into aCCOl: The lc (a)

,\1 U Til

(b) Ell

si' 1M'tt it

ae

"~,,

Chap. 5

Humidity Charts and Their Use 403

Sec. 5.3

I !

I

A

Air

B

80'F '65'F

T08 ' TW8

T

Fig. ES.14.

=n gives the value of the ':nich leaves the system. ::-or 49 grains a correction ny taking the enthalpy

The cooling water exit temperature can be obtained from an energy balance around the process. Basis: 8.30 x 10' ftl/hr moist air

::llib dry air

The enthalpy of the entering water stream is (reference temperature is 32°F and I atm) !Hi = CPH.ot::.T = 1(120 - 32) = 88 Btu/lb H 2 0

c :a cooling tower

C.-er which has a blower .:: wet-bulb temperature :c·w much water can be ~':cIed, enters the tower

8.30 x 10' ft' lb dry air = 600 . 10' lb d . /h . A ryalr r 13.82 ft'

while that of the exit stream is 58 Btu/lb H 2 0. [The value from the steam tables at J20°F for liquid water of 87.92 Btu/lb H 2 0 is slightly different since it represents water at its \'apor pressure (1.69 psia) based on reference conditions of 32'F and liquid water at its vapor pressure.] Any other datum could be used instead of 32°F for the liquid water. For example, if you chose 90°F, one water stream would not have to be taken into account because its enthalpy would be zero. The loss of water to the air is 0.0297 - 0.0098 = 0.01991b H 2 0/lb dry air (a) Matl'rial balance for water stream:

Let

W

Then (If' - 0.0199) =';n

the humidity chart

= lb H 2 0 entering the tower in the water stream per Ib =

dry air Ib H 2 0 leaving tower in the water stream per Ib dry air

(b) Ellergy balallce (ellthalpy balance) aroulld the entire process:

B

air and water in air entering 29.93 Btll 16.00 x 10' lb dry air Jb dry air

J.0297 208 88 Btll , JbH 2 0

..1_

0.10 )4.65

= 55.83

air and water in air leaving 55.83 Btll 6.00 x 10' lb dry air = lb dry air

404

Chap. 5

Combined Alalerial and Energy Balances

water stream leaving + 58 Btu (W - 0.0199) Ib H 2 0 6.00 x 10' Ib dry air Ib H 2 0 lb dry air 29.93

+ 88W =

(W -

= 0.825 lb H 2 0/1b dry air 0.0199) = 0.8051b H 2 0/1b dry air

55.83

+ 58(W -

0.0199)

W

The total water leaving the tower is O.8051b H 2 0 6.00 x I~/b dry air Ib dry air

=

4.83 x 10' Ib/hr

5.4 Complex problems In this book we have treated only small segments of material and energy balance problems. Put a large number of these segments together and you will have a real industrial process. We do not have the space to describe the details of any specific process, but for such information you can consult the references at the end of this chapter. It is always wise to read about a process and gain as much information as you can about the stoichiometry and energy relations involved before undertaking to make any calculations. Do not be unnerved by the complexity of a large-scale detailed plant. With the techniques you have accumulated while studying this text you will find that you will be able to break down the overall scheme into smaller sections involving a manageable number of streams that can be handled in the same way you have handled the material and energy balances in this text. Perhaps you can find a tie element (or make one up) that will permit you to work from one unit of the process to the next. Example 5.15 briefly outlines, without much descriptive explanation, how you should apply your knowledge of material and energy balances to a not-too-complex alcohol plant. Problems in which simultaneous material and energy balances appear can often be solved using special-purpose digital computer programs intended for the design and operation of chemical plants. Such comput~ programs include techniques of solving the material and energy balances for an almost endless combination of pumps, compressors, heat exchangers, distillation col LImns. reactors, and so forth. To be able to solve the balances for even one of these pieces of equipment, you need to have a bank of data for the stream properties. Most programs furnish the necessary physical data and make provision for you to provide any missing information. Assuming that subprograms have been prepared and are available that simulate the behavior of individual pieces of process equipment and produce numerical outputs for varying inputs and process parameters, as the user your main task is to designate the interconnections between the units and the choice of the subprograms to represent each

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{,--;

TABLES,I

'~1

.;~

.' ~

Name APACHE CHEOPS

"

~:;

I

I1

General Electric Co" Bethesda, Md. Shell Development Co., Emeryville, Calif.

CHESS

University of Houston

CHIPS FLOWTRAN

Service Bureau Corp, LB,M,

GEMCS

McMaster University Hamilton, Ont., Canada 1.e.I., Ltd" Runeorn, England

NETWORK 67

~

Where Developed

Monsanto Co .. St. Louis, Mo.

PACER

Purdue University and Dartmouth College

REMUS

University of Pennsylvania

SWAPSCO

Stone and Webster, Boston. Mass. Bonner and Moore Assoc., Houston, Texas

UOS

Reference

I t

R, R, Hughes et aI., 6th World Petrol, Con/. Proc" Frankfurt, Sec, VII, paper) 7, p, 93 (1963) R, L. Motard, H. M. Lee, and R, W, Barkley, CHESS-Chemical Engineering Simulation System, Tech Publishing Co., Houston, )968 R. L. Rorschack and R. E. Harris, "Process Simulation Made by Computer," Oil and Gas J" p. 62 (Aug. 17, 1970) A, I. Johnson, GEMCS Manual, Dept. Chem, Eng., McMaster Univ" Hamilton, Ontario, 1969 S, M, Andrew, "Computer Flowsheeting Using Network 67; an Example," Trans, blSt. Chem, Eng" v. 46, p, TI23 (1967) e. M. Crowe et aI., Chemical Plant Simulation, Prentice-Hall, Englewood Cliffs, N,J" 1971 P. T. Shannon et aI., "Computer Simulation of a Sulfuric Acid Plant," Chem, Eng, Progress, v, 62, No.6, p, 49 (1966) P. E. Ham, Users' !Hanua! for Remus (Rouline lor Executive }.fulti-Unit Simulation), School of Chern. Eng" Univ, of Pennsylvania, Philadelphia, 1969 T, Utsumi, "Compu(er Package Aids Systems Engineering," Oil and Gas j" p, )()() (June 8, 1970)

A

o

!JI

j ...,..," ....,,_.. ~_~~"""'-"'-'.""k"ib'~~.<_-",_~.~ ...<4,~:l'~.;~~~;,;,:~;.+,_,~"'_.,~_;,,~, .."- ....""'~_"-.

,"._cil,"", .. ~'1"";·_~~ __,__ ,';';;;:.'-~'::M~-;~~K~-'~:",j·~,,,, __ ,.....· .. ,<

L.,I

f

..

406

Combined !'.fa/erial and Energy Balances

Chap. 5

unit. In some instances you can also select the subroutine to be lIsed to solve the large sets of nonlinear equations that must be solved simultaneously. Table 5.1 lists some of the programs available that are user-oriented. Evans et al.· give a very lucid state of the art survey of heat and material balanCe programs, and additional references can be found at the end of this chapter.

EXAMPLE 5.15

---........ Sec. 5.4

first fraction further fract column. Bot: of the colum and physical venience:

A more complex problem

The Blue Ribbon Sour Mash Company plans to make commercial alcohol by a process shown in Fig. E5.15(a). Grain mash is fed through a heat exchanger where it is heated to 170'F. The alcohol is removed as 60 percent (weight) alcohol from the 80% Water tO% Alcohol tO% Organic material

60% Botto 95% , Botto

Overhead

Overhead t76'F Vapor

S/I

Fecd

t72 'F Vapor

Make a c 10,000 Ib/hr

To Vanillin Plant

Reflux

176'F

1--'==-1 liquid

Reflux

(a) Deter

160'F

1-"':":=::'::"-1 Liquid

(I) (2) (3) (4) (5) (6)

eO'F Heat Exchanger

m

Bottoms 210'F

Liquid 170'F Saturated Steam 176'F

Condensate

H20

eO'F

Fig. ES.lS(a). 9L. B. Evans et aI., Chern. Eng. Progress, v. 64, no. 4, p. 39 1968).

IOO'F

H20

0 RI

Bt 01 Rc

Be (b) Calcul (e) Deterr (d) Calcu! II in t-' 130'F. SO/II/ion: Let us call

16*

rt

$

Complex Problems 407

Sec. 5.4

Chap. 5

first fractionating column; the bottoms contain no alcohol. The 60 percent alcohol is further fractionated to 95 percent alcohol and essentially pure water in the second column. Both stills operate at a 3-to-1 reflux ratio and heat is supplied to the bottom of the columns by steam. Condenser water is obtainable at soeF. The operating data and physical properties of the streams have been accumulated and are listed for convenience:

::. be used to solve the ..cltaneously. Table 5.1 .:1. Evans et aLQ give .:::.iance programs. and :-'ler.

operating information and data =-:nmercial alcohol by a :::eat exchanger where it c::ght) alcohol from the

stream Feed 60% alcohol Bottoms (I) 95% alcohol Bottoms (II)

"verhead : OF Vapor

state Liquid Liquid or vapor Liquid Liquid or vapor Liquid

b.p. eF) 170 176 212 172 212

heat of Cpo Btu/(Ib)COF) vaporization liquid "apor (Btu/lb) 0.96 950 0.85 0.56 675 1.00 970 0.50 0.72 0.48 650 1.00 0.50 970

Make a complete material balance on the process and Reflu.

;

IGO°F



-=='--'; liquid

iI

. -eam

.,zO

tOO°F

H20

30°F



(a) Determine the weight of the following streams per hour: (I) Overhead product, column (I). (2) Reflux, column (I). (3) Bottoms, column (I) . (4) Overhead product, column (II). (5) Reflux, column (II). (6) Bottoms, column (II). (b) Calculate the temperature of the bottoms leaving heat exchanger Ill. (c) Determine the total heat input to the system in Btu/hr. (d) Calculate the water requirements for each condenser and heat exchanger II in gal/hr if the maximum exit temperature of water from this equipment is 130°F. Solution: Let US call the alcohol A for short and the organic matter M.

I

:8).

;

I 1 I

r

)iMAM.4

..." \( ..,ea

>

."lI;;;;

--------

...

_......._.. ..

. ".•-<.',,,.,")'0,'"_.;..._ ......

$. . ., ._

...... ,-_--,----

i

.... ~

STEP

I:

MATERIAL BALANCES

Basis: 10,000 Ib/hr mash (a) Material balance arol/nd column I and condenser:

Feed

A { HzO M

In (0.10)(10,000) = 1,000 Ib (0.80)(10,000) = 8,000 (0.10)(10,000) = 1,000

Ollt

l,oooIb

{~ZO

1000) _ 1000 ( 0.60

{~ZO

(8000 - 667)

Overhead vapor

{~20

4(1000) 4(667)

Bottoms

{~zO

Product Bottoms

=

667 7,333 1.000 10,ooolb

10,000 Ib

(b) Material balance around column 1 only: Feed

Ezo

Reflux

{~zO

(3)(1000) (3)(667)

1,000 Ib 8,000 1,000 3,000 2,000 15,000 Ib

(c) Material balance around column II. and condenser: I,OOOlb Feed 667

{~zO

4,000 Ib 2,667 7,333 1,000 15,000 Ib

{~20 (be: -

(d) Overhead 1000) product. Bottoms (H 2O (667 - 50)

1,6671b

~

,

I

1,000 Ib =

50 617 l,6671b

i \

f J

i

(d) Material balance on column II CUlling reflux and overhead vapor:

In

'\

1

r ...... 1

J;\

Ollt

1.00011:>

,t

J:ccJ

r

I.t 1(11/ (I,

H2 O

I.

{~%O (1000 O

II

Overhead

667

product Bottoms {H 2

-1000) 0.9S (667 - SO)

1,6671b

,-,-

---"" _......". .......

..---" .......~.

,.-",' ...-,

-

...

.

..

...........-----"".. ,"'''"..-.. ' ...' ..-~.. _,.".

=

~O

617 1,6671b

~

{ (d) Material balance on column II cUlling reflux and overhead vapor:

In

Feed (e) Reflux

STEP

2:

{~20 {~20

A Vapor { H 2 0 Bottoms (H 2 0

I,ooolb 667 3,000 150 4,8171b

3(1000) (3)(50)

Out (4)(1000) (4)(50) (667 - ~O)

4,000 Ib 200 617 4,8171b

!

L

ENERGY BALANCES

I

(a) Energy balance on heat exchanger III [Fig. E5.15(b)):

fO,OOO Ib

Reference Temperature , 80°F

fO,OOO Ib

Feed 800F

f700F

Bottoms

,

Fig. ES.15(b).

In Feed (10,000)(0.96)(80 - 80) Bottoms (8333)(1.0)(210 - 80)

.~

I

= =

0 Btu 1,084,000 Btu 1,084,000 886,000

Out Feed (IO,ooo)(0.96XI70 - 80) Bottoms (833)(1.0)(T - 80) = 8333T - 666,000 + 864,000 = 8333T

T = 106.5'F ,

= (8333T -

864,000 Btu 666,000) Btu

®

.a..

~

,

f',

,".>'........ "'._~_...:.......,_

.~.4. ______ . . . ~_" ... _"'._;..,.

,.~_.-..------.........,:,_~~,' .. , .. ~_ _ _ .~_~..... ~""'_ ..... ,.....:i;., _ _ _ ..;,,_~,;...,;.~ _~.~~~~-.lOo .. .............:.~~_.,..;.,_._

... _-'-'."......,

I

/

f

1l 1 ~

~

'" .I>.

t

(b) Energy balance around heal exchanger III inlel, and overhead vapor and reflux line from I:

1

Reference temperature

1

=

176'F 0111

In

Feed Reflux Steam

(10,000)(0.96)(80 - 176) (5000)(0.85)(176 - 176)

o

Bottoms to vanillin plant (8333)(1.0)(106.5 - 176) = -579,000 Vapor (6667)(0.85)(176 - 176) + (6667)(675) = 4,500,000

tJ.Hs , (tJ.Hs, - 921,000)

3,921,000

= =

-921,000

"l

tJ.Hs, - 921,000 = 3,921,000 tJ.Hs , = 4,842,000 Btu/hr (c) Energy balance around heat exchanger I outlet and overhead vapor alld reflux from II:

Reference temperature

Reflux Steam

=

172'F OUI

In Feed to II (1667)(0.85)(176 - 172) tlH varorization--.: steam@176"P

5,650

--------

+(1667)(675) (3150)(0.72)(160 - 172) (tJ.Hs.

+

(4200)(0.72)(172 - 172) + (4200)(650)

Vapor

1,125,000 (617)(1.0)(210 - 172) Bottoms -27,200 tJ.Hs, 1,103,450) tJ.Hs. + 1,103,450 = 2,748,400 tJ.Hs. = 1,645,000 Btu/hr

= 2,725,000 23,400 2,748,400

(d) Tolal energy input into syslem:

CD

tJ.Hs , + tJ.Hs. + steam heater = 4,842,000 + 1,645,000 + 1,125,000 = 7,612,000 Btu/hr •

(

(e) Energy balance on condenser

t

r:

w=

! !J

Ib H20/hr;

Reference temperature = 176'F

In Vapor Co,'lin~

(6667)(0.85)(176 -- 176) H ,0

(IV)! 1.0)(HO

17(,)

+ (6667)(675)

Ollt

=

4,500,000 I)f','"

Condensate

,..

(6667)(0.85)(176 -

17~)

.

n

I

I,

liq',

I \,

' II",' ,')11

"Ii

LI .. IUO

Ill)

(d) Total energy input into system: Il lis , + Mf s ,

2,748,400

+ 1,103,450)

(AH s•

I

11,1 i)(I.IJ)VIO

All,.

Steam

1

Illdl"''',

,:,;,.'(1()

I .. ')

AH•• + 1,103,450' .8,400 AHs. = 1,645,000 Btu/hr

+ steam heater =

4,S42,OOO

+

1,645,000

+

1,125,000

= 7,612,000 Btu/hr •

®

\

I

(

\

ie} l.nr'A·.. IMlanc(' on condrflJrr I:

Reference temperature

W = lb H20/hr;

Vapor Cooling H 20

In (6667)(0,85)(176 - 176) (W)(1.0)(80 - 176)

=

176c F alit

+ (6667)(675) =

50W

Condensate Cooling H 20

4,500,000 -96W (-96W + 4,500,000)

= 4,500,000

or

Woo

4,5~OOO

=

(6667)(0,85)(176 - 176) (W)(1.0)(l30 - 176)

o

=

-46W -46W

!

90,000 Ib H 20/hr

L.

-fd.\ .~

90,000 Ib H20/ hr = 10800 I H O/h ( 8,3451b/gal ,ga 2 r

I

(f) Energy balal/ce on condenser II:

W

Vapor Cooling H 20

In (4200)(0,72)(172 - 172) (W)(1.0)(SO - 172)

= Ib H20/hr;

Reference temperature

=

172"F 0111

+ (4200)(650) =

sow =

Condensate (4200)(0,72)(160 - 172) = Cooling H20 (W)(1.0)(172 - 130)

2,725,000 -92W (-92W + 2,725,000)

2,761,000

w~-

or

-36,000 -42W (-42W - 36,000)

2,761,000 = 55,200 Ib H20/hr J:rt..

55,200 Ib H 20 = 6620 gal H 20/hr 8,3451b/gal

+-------0

(g) Energy balance on heat exchanger If:

4 i

Condensate Cooling H20

J

I

l ,

Reference temperature = 80°F

W = Ib H20/hr;

0.;.

In (1050)(0,72)(160 - 80) (W)(1.0)(80 - 80)

0111

= 60,500

Product Cooling H 20

o

(1050)(0,72)(100 - 80) (W)(1.0)(130 - 80)

60,500

-""

50W

W

= 9071b H 2 0/hr

15,120 SOW SOW

+ 15,120 = or

60,500

109 gal H 2 0/hr '"

@

+ 15,120

..

...... _ '.. , t_ _.........._ _ _ _ _ _ ......._·'.....' _ ....''"'','''...... _ .........-_·"""'T,.::'-'~· r,f.::Jiooh ......J" '5 ~-il

'

_

----

-'W's w+b

·t . . . . .-

Ft· -

.«..._ _ _ _ _. . . .""'" " ' . ." " " 111YZ.... ·'l"i'("*'.. -._

412 Combined Material and Energy Balances

Chap. _,

Chap. 5

SUPPl

WHAT YOU SHOULD HAVE LEARNED FROM THIS CHAPTER 1. By the end of this chapter you should have perfected your techniques in using material and energy balances separately or in combination to such an extent that you should be able to analyze any type of process and write and solve the appropriate balances. 2. You should know how to use humidity charts and enthalpy-concentration charts as aids in the solution of problems.

I.

(

2.

( I

3. L

4. p,

1. A 19

2.

G 19

3. GI 4. Lc ed

S. MI

*1. Ba, Me *2.

Bn

·3.

Ell

**4.

Lei Scr

*S.

Me'

··6.

Otl'

*7.

Rol ed.,

·S.

Wh,

I.

Badl

Me<

.,"_..

Chap. 5

:HAPTER ::hniques in using :0 such an extent ::-ne and solve the

Supplementary References 413

Chap. 5

SUPPLEMENTARY REFERENCES Combustion of Solid, Liquid, and Gaseous Fuels 1.

Gaydon, Flames, 3rd ed .• Chapman & Hall, London, 1970.

2.

Griswold, John, Fue/s, Combustion and Furnaces, McGraw-Hill, New York, 1946.

3.

Lewis, W. K., A. H. Radasch, and H. C. Lewis, Industrial Stoichiometry, 2nd ed., McGraw-Hill, New York, 1954.

4.

Popovich, M., and C. Hering, Fuels and Lubricants, Wiley, New York, 1959.

::>y-concentration

Gas Producers and Synthetic Gas 1.

Arne, Francis, "Manufactured Gas," Chem. Eng., pp. 121-123 (Mar. 24, 1958).

2. Griswold, John, Fuels, Combustion and Furnaces, McGraw-Hill, New York, 1946. 3. Gumz, Wilhelm, Gas Producers and Blast Furnaces, Wiley, New York, 1958. 4.

Lewis, W. K., A. H. Radasch, and H. C. Lewis, Industrial Stoichiometry, 2nd ed., McGraw-Hill, New York, 1954.

S.

Mills, G. A., Enl'ironmental Sci. & Technology, v. 12, no. 5, p. 1178 (1971).

Enthalpy-Concelllration Charts .1.

Badger, W. L., and J. T. Banchero, Introduction to Chemical Engineering, McGraw-Hill, New York, 1955.

·2.

Brown, G. G., et aI., Unit Operations, Wiley, New York, 1950.

·3.

Ellis, S. R. M., Chem. Eng. Sci., v. 3, p. 287 (1954).

··4. ·5.

Lemlich, Robert, Chad Gottschlich, and Ronald Hoke, Chem. Eng. Data Series, v. 2, p. 32 (1957). McCabe, W. L., Am. Inst. Chem. Eng. Trans., v. 31, p. 129 (1935).

··6. Othmer, D. F., et aI., Ind. Eng. Chem., v. 51, p. 89 (1959). '7.

Robinson, C. S .• and E. R. Gilliland, Elements of Fractional Distillation, 3rd ed., McGraw-Hill, New York, 1939.

'8.

White, R. R., Petroleum Refiner, v. 24, no. 8, p. 101; v. 24, no. 9, p. 127 (1945).

Humidity Charts alld Calculations 1. Badger, W. L., and J. T. Banchero, Introduction to Chemical Engineering, McGraW-Hill, New York, 1955. 'Concerned with H-x chart calculations. ··Conccrned with l/-x chart construction.

~'.I1'

''''''"''''#__'''·''f»........'f"".",,'___.....

___

··d·. .. .. ," ....... _ ....._ " ' " "...." ' " " _ " - '...;0 ........ ' ....,.._

414

2.

....................._

....._

....._ _ _ _ _ _ _ ~.

_ - - - - " '...i£-;

Combined Material and Energy Balances

Chap. 5

McCabe, W. L., and J. C. Smith, Unit Operations of Chemical Engineering, McGraw·Hill, New York, 1956.

Chap. 5

PROBLE!,

3. Treybal, R. E., Alass Transfer Operations, McGraw-Hill, New York, 1955.

4.

Woolrich, W. R., and W. R. Woolrich, Jr., Ai,. Conditioning, Ronald, New York, 1957.

5.1. Lime Fig. :

Industrial Process Calculations 1. Hougen, O. A., K. M. Watson, and R. A. Ragatz, Chemical Process Principles, Part I, 2nd ed., Wiley, New York, 1954. 2.

Lewis, W. K., A. H. Radasch, and H. C. Lewis, Industrial Stoichiometry, 2nd ed., McGraw·Hill, New York, 1954.

3. Nelson, W. L., Petroleum Refinery Engineering, 4th ed., McGraw-Hill, New York, 1958. 4.

Shreve, R. N., The Chemical Process Industries, 2nd ed., McGraw· Hill, New York, 1956.

S.

Williams, F. A., Combustion Theory, Addison.Wesley, Reading, Mass., 1965.

Solution oj Material and Energy Balances Via Digital Computers 1. Andrew, S. M., British Chem. Eng., v. 14, p. 1057 (1969). 2.

Beckett, K. A., "Computer-aided Process Design," Chem. Eng., no. 228, p. 163 (1969).

3.

Hutchison, H. P., and G. F. Forder, Chemical Plant Network Specification Using a Visual Display, Symposium Series No. 23, The Institution of Chemical Engineers, London, 1967, p. 169.

4.

Kenny, L. N., and J. W. Prados, A Generalized Digital Computer Program for Performing Process Alaterial and Energy Balances, University of Tennessee, Knoxville, 1966.

S.

Lee, W., and D. F. Rudd, A.l.Ch.E. J., v. 12, p. 1184 (1966).

6.

Sargent, R. W. H., Chem. Eng. Progress, v. 63, no. 9, p. 71 (1967).

with 1 that c heat,

5.2. A vel (carb, to CC The 1< or Ca ratio,

(a) Tl (b) Tl

Your

5.3. A fcc, ccn!r., the CI

Our

ofm

Unsteady-State Material and Energy Balances

6

from

( ( (

In previous chapters all the material and all the energy balances you have encountered, except for the batch energy balances, were steady-state balances. i.e., balances for processes in which the accumulation term was zero. Now it is time for us to focus our attention briefly on unsteady-state processes. These are processes in which quantities or operating conditions change with time. Sometimes you will hear the word transient state applied to such processes. The unsteady state is somewhat more complicated than the steady state and in general problems involving unsteady-state processes are more difficult to handle mathematically than those involving steady-state processes. However, a wide variety of important industrial problems falls into this category, such as the startup of equipment, batch heating or reactions, the change from one sct of operating conditions to another, and the perturbations that develop as process conditions fluctuate. In this chapter we consider only one category of unsteadystate processes, but it is.the one that is the most widely used. The basic expression in words, Eqs. (2.1) or (4.22), for either the materi~tl or the energy balance should by now be well known to you and is repeated as Eq. (6.1) below. However, you should also realize that this equation can be applied at various levels, or strata, of description. In other words, the engineer can portray the operation of a real process by writing balances on a number of physical scales. A typic!l illustration of this concept might be in meterolof!Y. where the following dilferent degrees of detail can be used on a descending scak

444

' " . . ,....... ""....."" . .~, ••-.--.. .............._

-.""*_0__....__-.-.~......_..,..........,...."".__ ...............,......._.__......,""'"",..,..,"'.........,"'......,..,,..,..._

.... i

in de, of b:l: scop" 1 sequc' inder lion ~j throe. he sui arc

CI..:

into 11 ~lJ)pllt·

\.'r en!..' eli :-.,\!

,.\ b]I.I!h

lion, \\

,f-'-'''''''~'''''­

-"_...------.-----4

Ullsteady-State Material alld Ellergy Balances 445

Chap. 6

of magnitude in the real world:

~r;al

Global weather patte~n Local weather pattern Individual clouds Convective flow in clouds Molecular transport The molecules themselves

6

Similarly, in chemical engineering we write material and energy balances from the viewpoint of various scales of information:

, f

!I

I I

::laIc. /ou have ensteady-slale balances. .~rm was zero. Now it ·slale processes. These . ::ms change wilh time. ed to such processes. :ne steady state and in ::lore difficult to handle ·,sses. However, a wide category, such as the .:.:lange from one set of :::1at develop as process : category of unsteady~d.

for either the material vou and is repeated as .. :: this equation can b.: ,er words, the engineer _lances on a number of .:ght be in meterology. :l on a descending scale

(a) (b) (c) (d) (e)

Molecular and atomic balances, Microscopic balances, Multiple gradient balances. Maximum gradient balances, and Macroscopic balances (overall balances)

in decreasing order of degree of detail about a process.' In this chapter, the type of balance to be described and applied is the simplest one, namely, the macroscopic balance [(e) in the tabulation above]. The macroscopic balance ignores all the detail within a system and consequently results in a balance about the entire system. Only time remains as an independent variable in the balance. The dependent variables, such as concentration and temperature, are not functions of position but represent overall averages throughout the entire volume of the system. In effect, the system is assumed to be sufficiently well mixed so that the output concentrations and temperatures are equivalent to the concentrations and temperatures inside the system . To assist in the translation of Eq. (6.1) ACCUmulatiOnj jTransport into j jTransport out j or depletIOn '. system through of system _. system - through system within the j system boundary boundary

G~neratiOn} + { wlthm system

{consumptiOn} --. wllhtn system

(6.1 )

into mathematical symbols, you should refer to Fig. &.1. Equation (6.1) can be applied to the mass of a single component or to lb' total amount of material or energy in the system. Let us write each of the terms in E4. (6.1) in mathematical symbols for a very small time interval AI. Let the accumulation be positive IAdditional information together with applications concerning these variolls types of

balances can be found in D. M. Himmelblau and K. B. Bischolf, Process AI/alysis alld Simulalion, Wiley, New York, 1968.

..P!'",_ _ -'"'?f......_-.. ........"""*..'..'_ _ ~~..~......

446

. &1, ....

_''''j'¥_r_........... ' ·.'..

·tt7.. !t_......_ ........

~_·. .

':If........M ___ ' .' ..........' ................. ~'"

Unsteady-State Material and Energy Balances

Chap. 6

Chap. 6

Finally, tr a chemical reae

Transport

Transport

In

Out 2

where;"

=

net rea(

Introductic (6.3) and (6.4) c

Component mal

p"Vk"

Fig. 6.1. A general unsteady-state process with transport in and out and internal generation and consumption.

in the direction in which time is positive, i.e., as time increases from t to t + 11t, Then, using the component mass balance as an example, the accumulation will be the mass of A in the system at time t + 11t minus the mass of A in the system at time t, Accumulation =

p"Vla", - p"VI,

.CClli

Total material t

Energy balance:

where 1

= mass of component A per unit volume V = volume of the system

p"

.ccumuhtio

Note that the net dimensions of the accumulation term are the mass of A. We shall split the mass transport across the system boundary into two parts. transport through defined surfaces S, and S,. whose areas are known, and transport across the system boundary through other (undefined) surfaces. The net transport of A into and out of the system through defined surfaces can be written as Net flow across boundary via S, and S, = P"vS 11tls , - p"vS 11tls, where v = fluid velocity in a duct of cross section S S = defined cross-sectional area perpendicular to material flow

Again note that the net dimensions of the transport term are the mass of A. Other types of transport across the system boundary can be represented by Net residual flow across boul1dary = IV" 111

jj = rate of iii = rate of Q = rate of IT" = rate of IV = rate of through p = total n, The other notal: Chaps. 2 and 4: now all expres,c. If each side

!!..'

where IV" = rate of mass flow of component A through the system boundaries other than the defined surfaces S, and S, 'The symbol II means that the quantities preceding the vertical line are evaluated at time + At, or at surface S I, or at surface S 1. as the case may be.

It or time I

''"''- -_,..._ ......_ ......__..,, ___,.....__.... ~-"'.

where

_M"'~"'"*"_.~--

__..,.,..._..,...-.. . . . _.4

4.

44It

_

*

.......c::o_........=-. ,' •.

~,.~.,.

Similar relaliom the limit of each

·Chap. 6

,

..... .4-~.-"""~"" .

....,

Unsleady-Slale "falerial and Energy Balances 447

Chap. 6

Finally, the net generation-consumption term will be assumed to be due to a chemical reaction r,,:

_Cvstem Boundary

Net generation-consumption = ;" AI where;" = net rate of generation-consumption of component A by chemical reaction

Transport Out

Introduction of all these terms into Eq. (6.1) gives Eq. (6.2). Equations (6.3) and (6.4) can be developed from exactly the same type of analysis.

2

Component material balances: P"Vk+A' -

·.an in and out and

p"Vk =

accumulation

::::reases from I to t + !J.t. ,e, the accumulation will ~ mass of A in the system

PAVS AIls, - PAVS AIls. transport tbrou.h defined bouad.ries

+ wA AI + ;,,!J.I trauport

thrOUlb otber boaadarles

,eDen·

tion or consumplion

(6.2)

Total material balance:

pVI.. A, - pVI, = pvS!J.t Is, + pvS AIls. + w At accumulation

transport through defined boundarn:'

transport through

(6.3)

other

klindaric:s

Energy balance:

Ek+A,-EI,=I(ii+ ~2 +gh)mMls, -I(ii+ '; +gh)nl!J.lls. volume

accumulation

transport tbrousb defined boundaries

(6.4) .:: are the mass of A. . Doundary into two parts. '::.as are known, and trans· :=iined) surfaces. The net :-_'led surfaces can be wnt·

.: At Is,

1Vork.

tbrouib

other

boundaries

where

iJ = rate of energy transfer accompanying Ii' nl = rate of mass transfer through defined surfaces Q = rate of heat transfer

TV = rate of work done by the system

- p"vS !J.t Is.

:- to material flow term are the mass of A. ::n be represented by =

beat

transport

w,,!J.t

Ii'

=

p

=

rate of total mass flow through the system boundaries other than through the defined surfaces S, and S2 total mass, per unit volume

The other notation for the material and energy balances is identical to that of Chaps. 2 and 4; note that the work, heat, generation, and mass transport are now all expressed as rate terms (mass or energy per unit time). If each side of Eq. (6.2) is divided by !J.t, we obtain (6.5)

,ne system boundaries

::-:Clcal -:ma}

. evaluated at

lin"

Similar relations can be obtained from Eqs. (6.3) and (6.4). Next, if we take the limit of each side of Eq. (6.5) as !J.I ~ 0, we get d(';hV) = -!J.(PAVS)

+ IVA +;A

(6.6)

,

J



nt

'"

r

I'N'

e )

448

t....,..,x"

r

. .tt' '?

e- 'xc 7

'$

.... r,

*'''tef'

Chap. 6

UI/steady-Slale Malerial alld Ellergy Balallces

Chap. 6

Similar treatment of the total mass balance and the energy balance yields the following two equations:

d(pV) dt

~7) =

-ll[ ( Ii

= -ll(pvS) + IV

simultancOl

(6.7)

+ v; + gh) m] + Q - W + fJ

(6.8)

The relation between the energy balance given by Eq. (6.8), which has the units of energy per unit time, and the energy balance given by Eq. (4.24), which has the units of energy, should be fairly clear. Equation (4.24) represents the integration of Eq. (6.8) with respect to time, expressed formally as follows:

E" - E r, =

in the same

the objecti\

Jr.r, (-1l[(1i + K + P)mj + Q• + fJ -

W}dt

The quantities designated in Eq. (4.24) without the tilde (-) are the respective integrated values from Eq. (6.9). To solve one or a combination of the very general equations (6.6), (6.7), or (6.8) analytically is usually quite difficult, and in the following examples we shall have to restrict our analyses to simple cases. Ifwe make enough (reasonable) assumptions and work with simple pr!Jblems. we can consolidate or eliminate enough terms of the equations to be able to integrate them and develop some analytical allswers. Numerical solutions are also possible. In the solution of unsteady-state problems you liave to apply the usual procedures of problem solving initially discussed in Chap. I. Two major tasks exist:

Then, assun function of (or more) g' If dl al can use Eq. for some recr equation is ' use of signs, simple unste mathematic" texts dealing EXAMPLE (,

A tank h, Water TUns if' same rate. If t' tank uniform . that the densit SOIUlion: We shall s Slep I: D

(a) To set up the unsteady-state equation. (b) To solve it once the equation is established. After you draw a diagram of the system and set down all the information available, you should try to recognize the important variables and represent them by letters. Next, decide which variable is the independent one, and label it. The independent variable is the one you select to have one or a series of values, and then the other variables are all dependent ones-they are determined by the first variable(s) you selected. Which is chosen as an independent and ,which as a dependent variable is usually fixed by the problem but may be arbitrary. Although there are no general rules applicable to all cases, the quantities which appear prominently in the statement of the problem are usually the best choices. You will need as many equations as you have dependent variables. As mentioned before. in the macroscopic balance the independent variable is time. In mathematics. when the quantity x varies with time, we consider dx to be the change in x that occurs during time dl if the process continues throui!h the interval beginning at time I. Let us say that I is the independent variable and x is the dependent one. In solving problems you can use one or a combinatil'n ofEqs. (6.6), (6.7), and (6.S) directly, or alternatively you can proceed. as sholln in some of the examples, to set up the differential equations from scratch exactly

• .b-:tO'l~.

,$It

U$l _......

,""'.w:

';:;

..

;;,~

41

, ...



uP I#.

Step 2: n the independ( ,.. be the dependc dent variable. [ Slep 3: \\': tion: Slep 4: It ! balance on the J to be isotherm.,

---

lit ....'! ••

b WPM"

;'

'M

Chap. 6 ,;c

energy balance yields the (6.7) (6.8)

-:1.

(6.8), which has the units

..::n by Eq. (4.24), which has 4.24) represents the integra::Jally as follows:

") + iJ -

.,,-"'.'-""

_.--l

t

JV}dt

'lilde ( - ) are the respective ::leral equations (6.6). (6.7). Ihe following examples we :-~ make enough (reasonable) consolidate or eliminate :::'le th,.- and develop some 3sib .. D lie. J apply the usual Chap. 1. Two major tasks

=

Unsleady-Slale Malerial and Energy Balances 449

Chap. 6

in the same fashion as Eqs. (6.2}-(6.4) were formulated. For either approach the objective is to translate the problem statement in words into one or more simultaneous differential equations having the form

~; =

(6.10)

[(x, t)

Then, assuming this differential equation can be solved, x can be found as a function of t. Of course, we have to know some initial condition(s) or at one (or more) given time(s) know the value(s) of x . If dt and dx are always considered positive when increasing, then you can use Eq. (6.1) without having any difficulties with signs. However, if you for some reason transfer an output to the left-hand side of the equation, or the equation is written in some other form, then you should take great care in the use of signs (see Example 6.2 below). We are now going to examine some very simple unsteady-state problems which arc susceptible to reasonably elementary mathematical analysis. You can (and will) find more complicated examples in texts dealing with all phases of mass transfer, heat transfer, and fluid dynamics. EXAMPLE 6.1

;'j"

Unsteady-state material balance without generation

A tank holds 100 gal of a water-salt solution in which 4.0 Ib of salt are dissolved. Water runs into the tank at the rate of 5 gal/min and salt solution overflows at the same rate. If the mixing in the tank is adequate to keep the concentration of salt in the tank uniform at all times, how much salt is in the tank at the end of 50 min? Assume that the density of the salt solution is essentially the same as that of water.

Solulion: We shall set up the differential equations which describe the process from scratch. Slep I: Draw a picture, and put down the known data. See Fig. E6.1. -:::'11 all the information avail=abIes and represent them ::::dent one, and label it. The ::]e or a series of values. anJ -:.are determined by the fir,! ::nependent and which as a - Dut may be arbitrary. ;\1.::ases, the quantities \\ hie:) .:rfe usually the best choice'. ..::::ent variables. c=e the independent variable . with time, we consider ,iI _.c process continues !lnoli,'" 2; independent variahk :lJld , use one or a combina: i ,':' you can proceed. as sh,)\\ Ii c:.:ltions from scratch CXih: t1 :,

5 gal/min

5 gal/min pure H10 (Olb soil/gal)

fOO gal 4 Ib 5011

Fig. E6.1.

Slep 2: Choose the independent and dependent variables. Time, of course, is the independent variable. and either the salt quantity or concentration in the tank can be the dependent variable. Suppose that we make the mass (quantity) of salt the dependent variable. Let x = lb of salt in the tank at time I. Slep 3: Write the known value of x at a given value of t. This is the initial condition: at I = 0, x = 4.0 Ib Slep 4: It is easiest to make a total material balance and a component material balance on the salt. (No energy balance is needed because the system can be assumed to be isothermal.)

;

j J

! I

'1

*

!

z··

450

.\ '"M

ti'

'Mt

ok q

He

H

bt

....,

~"""

'_tf. .__.....'"_.....__

t"'.......t'''._M......

___

Unsteady-State Material and Energy Balances

Chap. 6

Chap. 6

The initial c(

Total balance:

=

accumulation [mlol Ibl/+ M

-

[m lol Ib]t

5 gal tit min = :-m"""'in"+.,....;;-i-'---;;-!-!'-'-;!"T'-"-l'::':':...:.o=

5 gal min

The sok

titmin = 0

EXAMPLE

This equation tells us that the flow of water into the tank equals the flow of solution out of the tank if PH,O = P.ol. as assumed. Otherwise there is an accumulation term. Salt balance:

== in

accumulation

5 gal mm

tit min

I

A square time requirec' Solution. Step I: I

[x IbJ. + At - [x Ib]t = 0 - ~."'"""",":-rl----

Dividing by At and taking the limit as tit approaches zero, lim [xlt+M - [xli

At

A/-O

=

-0.05x

or dx dt = -0.05x

(a)

Notice how we have kept track of the units in the normal fashion in setting up these equations. Because of our assumption of uniform concentration of salt in the tank. the concentration of salt leaving the tank is the same as that in the tank, or x Ib/l 00 gal of solution. Slep 5: Solve the unsteady-state material balance on the salt. By separating the independent and dependent variables, we get dx = -0.05 dt x This equation is easily integrated between the definite limits of

1=0,

x =4.0

1=50,

x = the unknown value Xlb x

I

dx

In 4~ = -2.5,

4J

=

Slep 4: r: time til, the ;tank is in the

= -0.05 ISO dt

x

4.0

0

In

4J =

where

2.5

x = 1~~2

12.2,

=

0.328 Ib salt

An equivalent differential equation to Eq. (a) can be obtained directly from the component mass balance in the form of Eq. (6.6) if we let PA = concentration of salt in the tank at any time; in terms of Ib/gal: d(PA V) __

dt

-

Slep 2: S independent \ dent variable. height of the Step 3: \.

An overall

(:..5.£;ga::.:.I-l-"-'Ao.:I;:.b _ 0) nlln

The depl. Ib, is

01.'

gal

If the tank holds 100 gal of solution at all times, V is a constant and equal to 100. so that (b)

Although tih The accumu!.,

----.. ---------.Chap. 6

Chap. 6

Unsteady-State Material and Energy Balances 451

The initial conditions are

=

P" = 0.04 The solution of Eq. (b) would be carried out exactly as the solution of Eq. (a). at

cmin

ll. t min = 0

t

0,

EXAMPLE 6.2 Unsteady-state material balance without generation A square tank 4 ft on a side and 10ft high is filled to the brim with water. Find the time required for it to empty through a hole in the bottom I in. 2 in area .

the flow of solution .:;.n accumulation term.

:.:IS

Solution: Step 1; Draw a diagram of the process, and put down the data. See Fig. E6.2(a). min

!

l to

1

salt. By separating the

~g ~i:~ J

f

Fig. E6.2(a).

1

II

Hole

Step 2: Select the independent and dependent variables. Again, time will be the independent variable. We could select the quantity of water in the tank as the dependent variable, but since the cross section of the tank is constant, let us choose h, the height of the water in the tank, as the dependent variable. Step 3: Write the known value of h at a given value of t; at

of

l

o

(a)

::lion in setting up these ;on in the tank, =ne 1 ,. x lb/lOO gal

j

1

t

= 0,

1

I

h = 10ft

Step 4: Develop the unsteady-state balance(s) for the process. In an elemental time I1t, the height of the water in the tank drops 111r. The mass of water leaving the tank is in the form of a cylinder 1 in. 2 in area, and we can calculate the quantity as

JI 1, \

lin.21Ift2Iv*ftIPlbll1lsec= v*l1t lb 144 in.' sec ft' p 144

j

where

P = the density of water v* = the average velocity of the water leaving the tank

salt obtained directly from let PA = concentration

The depletion of water inside the tank in terms of the variable h, expressed in lb, is

I l

16 ft 21 p l~ 1h ft 1 _ 16 ft 21 p l,b h ft ft 1~6t fl

= 16P I1h

:!Stant and equal to 100. (1))

= in

1

1

An overall material balance indicates that accumulation

f ~J I

-

I

out

16p I1h - 0 _ Pl)· I1t " 144

(a)

Although I1Ir is a negative value, our equation takes account of this automatically. The accumulation is positive if the right·hand side is positive, and the accumulation

452

Unsteady-State },[aterial and Energy Balances

Chap. 6

is negative if the right-hand side is negative. Here the accumulation is really a depletion. You can see that the term p, the dcnsity of water, cancels out, and we could just as well have made our material balance on a volume of water. Equation (a) becomes Ah v' At

=

-

Chap. 6

where c is 3! dynamics) h;; 32.2 ft/sec 2), into Eq, (b) 1

(16)(144)

Taking the limit as IJ.h and At approach zero, we get

or

dh v' dl - - (16)(144)

(b)

Unfortunately, this is an equation with one independent variable, I, and two dependent variables, hand v'. We must find another equation to eliminate either h or v' if we want to obtain a solution. Since we want our final equation to be expressed in terms of h, the next step is to find some function that relates v' to h and I, and then we can substitute for ". in the unsteady-state equation. We shall employ the steady-state mechanical energy balance for an incompressible fluid, discussed in Chap. 4, to relate v' and h. See Fig. E6.2(b). Recall from Eq. (4.30) with W = 0 and E, = 0 that

Equation (g)

I

and

to yield 8,

(c)

EXAMPLE 6 A small st moles of a mi.' tional mixture liquid in the s is related to x.

Fig. E6.2(b).

We assume that the pressures are the same at sections CD and (g) of the system consisting of the water in the tank. Equation (c) reduces to v2

v2

-

~

+g(h z - h,)

=

0

(d)

how long will i state (Uequilib! E6.3.

where the exit velocity through the l-in. 2 hole at boundary @ ", = the velocity of the water in the tank at boundary CD If t', ~ 0, a reasonable assumption for the water in the large tank at any time, at least compared to "2, and the reference plane is located at the l-in.2 hole, V2

=

t'i = t'2

-2g(O -- h,)

=

= ";2g"

2gh

(e)

Because the exit-stream flow is not frictionless and because of turbulence and orifice effects in the exit hole, we must ~orrect the vaillc of" given by Eq. (e) for fric· tionless flow by an empirical adjustment factor as follows: V2

"

~-b*,,"IN_:ew:te~ ........,

",. . . '"

i,

..

=

c";2gh = v'

(f)

Fe!

Co' Prc

Bl

_ _ H . ,"_~ ..

"6.i!.'SiW-hh

f

'

..

-~,-~

.

Chap. 6

:nul at ion is really a depletion. CiS out, and we could just a\ cater.

Unsteady-State Afaterial and Energy Balances 453

Chap. 6

where c is an orifice correction which we could find (from a text discussing fluid dynamics) has a value of 0.62 for this case. In the American engineering system (g = 32.2 ft/sec 2 ), v' = 0.62,J2(32.2)h = 4.97,J7i ft/sec. Let us substitute this relation into Eq. (b) in place of v'. Then we obtain dh

4.97(h)1/2

dt

= -

(16)(144)

or dh -464}j17! = dt

(b) :~ndent

variable, f, and two ::luation to eliminate either" . .llal equation to be expressed . elates v' to hand t, and then

(g)

Equation (g) can be integrated between att=O

h=IOft

and h

= Oft

:..oalance for an incompressible -2(b). Recall from Eq. (4.30)

at

= B,

f

-464

J

o

the unknown time

dh hll2

=

10

to yield B, (c)

B = 464

EXAMPLE 6.3

J

IO

dh

o}jl7!

f'

dt

0

1=

= 464 [ 2Jh

10

2940 sec

Material balance in batch distillation

A small stilI is separating propane and butane at 275°F and initially contains 10 lb moles of a mixture whose composition is x = 0.30 (x = mole fraction butane). Additional mixture (x, = 0.30) is fed at the rate of 5 lb moles/hr. If the total volume of the liquid in the stiIl is constant, and the concentration of the vapor from the stiIl (x D ) is related to Xs as follows:

-0

Xs

•.,j

® of the system consisting (d)

XD

=

1+ Xs

how long will it take for the value of Xs to change from 0.30 to 0.40? What is the steadystate ("equilibrium") value of Xs in the stilI (i.e., when Xs becomes constant)? See Fig. E6.3.

at boundary ® ~::lUndary


.. cge tank at any time, at least ,-in.' hole, (e)

oecause of turbulence and t' given by Eq. (e) for fri,(f)

·...,.",.44*

Feed IF

'O.~30~:b~~~

Composition: Propone 0.70 Butane 0.30 Fig. E6.3.

" " _ . at

.im«!'~-'"t.","

.....-

__

.........._c.., _ ..

~*_.""" ~,_

454



. . . _.,.f_. . . .."...""'.."____. . ."'. . . . . .-...._..............___.....___ .0/", _

"""~"

--

~"

-

~

.-

-- _

_

___

"O"-W'

:

##1

"....:'r.

Chap. 6

Unsteady-State Material and Energy Balances

Solution: Since butane and propane form ideal solutions, we do not have to worry about volume changes on mixing or separation. Only the material balance is needed to answer the questions posed. If t is the independent variable and Xs the dependent variable, we can say that Butane balallce (C.): The input to the still is

Chap. T

aCCum

The

Vi

EXA;\:

5 moles feed 0.30 mole C'I!lt hr hr mole feed

A amoun solutio of com compo conti ill Assum

The output from the still is equal to the amount condensed,

5 moles condo Xo moles C. !It hr hr mole condo The accumulation is 10 moles in still Xs moles C, J __ 10 moles in still Xs mol~s CH. mole in stIIlj<+A' mole III still

~- •• ~
=

lO!lx s

t

Our unsteady-state material balance is then

=

accumulation in out 10 !lxs = 1.5!lt - 5xo !It

or, dividing by !It and taking the limit as !It approaches zero,

d:t =

O.lS - O.Sxo

As in the previous example, it is necessary to reduce the equation to two variables by substituting for Xo: Xs XD = 1 + Xs Then

~s = O.lS

- 1

Sit'

~\s(O.S)

dxs

=

time

t (1

St(,

dt

O.IS _ O.Sxs

1 +xs The rat,

The integration limits are at t t

f

o. 40 0.30

f

o. 40 0.30

= =

Xs = 0.30 Xs = 0.40

0,

e,

dxs O.IS - [O.Sxs/(1

( I + Xs ) dXs 0.15 - 0.35xs

=

e= [-

Xs

where f..

9

=

+ xs)] I

f°dt = e

nleasurc 0.40

0.35 - (0.35)2 In (0.15 - 0.35xs )]0.30

e = 5.85 hr

If you did not know how to integrate the equation analytically, or if you only had experimental data for Xo as a function of Xs instead of the given equation, you could always integrate the equation graphically as shown in Example 6.6 or integrate numerically on a digital computer.

'-., .",~

p

+« 8M. - . II

¥:»o;:j

a. U;# ;ac;:"

would j'. l'\c, system):

&,*

..su

or

- _--------,-<"""'-.

..

Chap. 6

~ave

to worry about

:ce 1S needed to answer ~.

dependent variable.

..

Unsteady-State Material and Energy Balances 455

Chap. 6

The steady-state value of Xs is established at infinite time or, alternatively, when the accumulation is zero. At that time, 5x s 0.15 = I0.+ Xs or Xs = 0428 . The value of Xs could never be greater than 0.428 for the given conditions.

EXAMPLE 6.4 Unsteady-state chemical reaction A compound dissolves in water at a rate proportional to the product of the amount undissolved and the difference between the concentration in a saturated solution and the concentration in the actual solution at any time. A saturated solution of compound contains 40 g/iOO g H,O. In a test run starting with 20 Ib of undissolved compound in 100 Ib of pure water it is found that 51b are dissolved in 3 hr. If the test continues, how many pounds of compound will remain undissolved after 7 hr? Assume that the system is isothermal.

_., CH. still

=

_::1

10Axs

Solution: Step 1: See Fig. E6.4.

ilill

-:n to two variables by

Solution contoining undissolved compound

Fig. E6.4.

time

Step 2: Let the dependent variable x = the pounds of undissolved compound at f (the independent variable). Step 3: At any time t the concentration of dissolved compound is (20 - x) Ib compound 100 Ib H,O

The rate of dissolution of compound according to the problem statement is ~

ra t e hr

=

k( Ib) (40 Ib compound _ (20 - x) Ib compound)

x

JOO Ib H 2 0

100 Ib H,O

where k is the proportionality constant. We could have utilized a concentration measured on a volume basis rather than one measured on a weight basis. but this would just change the value and units of the constant k. Next we make an unsteady-state balance for the compound in the solution (the system): accumulation

_c::ally, or if you only '" given equation. you =ple 6.6 or integrate

. . 4Ji Si, ,11.

'*

,,," -,.,...,

dx dt

= in =

out

0 _ 0

+

generation

+ kx( 40

JOO

or dx _ kx ('0 dt - 100 -

+ x)

_ 20 JOO

X)

v.

456

Chap. 6

UI/steady-State Material al/d EI/ergy Balal/ces

Chap. 6

Step 4: Solution of the unsteady·state equation: dx x(20 + x)

k

= 100 dt

This equation can be split into two parts and integrated,

dX_~=~dt x x+20 5 or integrated directly for the wnditions 0,

Xo = 20

=3,

x, = 15

10 =

1,

12 =

7,

Since k is an unknown, one extra pair of conditions is needed in order to evaluate k. To find k, we integrate between the limits stated in the test run:

JXI> 20

dx

I"

I 3

20

x

dx

k

+ 20 =T

X]" -= [I n x + 20'0

odt

k -(3 - 0) 5

or

k = -0.257 (Note that the negative value of k corresponds to the aCtual physical situation in which undissolved material is consumed.) To find the unknown amount of undissolved compound at the end of 7 hr, we have to integrate again:

I

X, dx _ 20

[

x

(b) Na,SO,

IX> ~ = -0.257 I' dt 20 X

In_x_Jx, x + 20 '0

+ 20

5

[

0

Xlb II

= -0.0257(7 _ 0) 5

x, = 10.71b

EXAMPLE 6.5

Step 3 Step 4 (a) Total b,

or

Overall unsteady-state process

A 15 percent Na2S04 solution is fed at the rate of 12lb/min into a mixer which initially holds 100 Ib of a 50-50 mixture of Na,SO. and water. The exit solution leaves at the rate of 10 lb/min. Assuming uniform mixing, what is the concentration of Na,S04 in the mixer at the end of 10 min? Ignore any volume changes on mixing. Solutiol/: ]n contrast to Example 6.1, hoth the water and the salt concentrations, and the total material in the vessel, change with time in this problem. Step I: See Fig. E6.5. Step 2: · Na, SO· h k . . (lbIbNa,so.) I.e t x = f factIOn 4 III t e tan at tIme t III t~tal y = totallb of material in the tank at time

t

Step 5: 1 equations Will in Eq. (b) and

To get y, we

)1

Hte'

mit'·' --

<'

fl·

Chap. 6

Chap. 6

Unsteady-State Material and Energy Balances 457

10lb/min

121b/min 1.8 Ib NO~S04

10.2 Ib H20 f '0, 50 Ib NO~S04 50 Ib H20

Fig. E6.S.

:..ed in order to eval ua te k. :.:Ie test run:

Slep 3: At time I = 0, x = 0.50, y = 100. Slep 4: Material balances between time I and t (a) TOlal balance:

dt

=

accumulation

+ At:

in

out

_ [ ] _ A = 12lb IAI min _ 10 lb IAt min [] Y'+M y, Y min min or ical situation in .nt of undissolved

dy =2 dt

(a)

(b) Na,S04 balance: accumula tion

[X lb Na,SO.

::x.:.lb:rr=-Nia::,.::sFo",4+,-Y.:.I::.b.:.to::.t:=.:allJ _ [ Ib total ,+A,

dt o

=

y lb total]

Ib total

in

, out

= 1.81b Na,S041At min _ 10 lb total x lb Na,S04 A I min ---n~1~ln~-1--'ICb~t~07ta~I~~~~~

min

or

d(xy) = I 8 - 10 dl . x ..:: lb/min into a mixer which water. The exit solution :.. what is the concentration - volume changes on mixing. :.j

Step 5: There are a number of ways to handle the solution of these two differential equations with two unknowns. Perhaps the easiest is to differentiate the xy product in Eq. (b) and substitute Eq. (a) into the appropriate term containing dy/dt: d(xy) _

dy

d t - x dt

::;ajt concentrations, and the

.em.

f'

~a,S04)

+

dx

y dt

dy

=

f'

y = 100

4,

:c:t

l

0

j

2dt

y - 100 = 2t

total

, j

i

To get y, we have to integrate Eq. (a): 100

. :J

(b)

+ 21

"~,;'

___

...

"""~-:..t_.",,,

IjIote ............ • . . . . ._

...

458

_"1_.........

_."-t.....______

;,,,t..._ · _..

* .....· _.•.. . . . . . . . ._ _.....

!. •._ _. . . ._ _ _ II._~.;.,.

Chap. 6

Unsteady-State Material and Energy Balances

Then

d~:) =

x(2)

+ (100 + 2t)~; =

~;(l00 + 2t) = 0.>0

dx 1.8 - 12x = .

1 I 1.8 - 12x - 12 n 1.8 _ 6.0

proper unit to 90°F? \\ Additio

1.8 - 12x

flO

x

f

1.8 - lOx

Chap. 6

0

dt 100 + 2t

I I 100 + 20 n 100

Solutio, Thepr time; the d,

= '2

x = 0267 Ib Na,SO~ . Ib total solutIOn An alternative method of setting up the dependent variable would have been to let x = Ib Na,SO. in the tank at the time t. Then the Na,SO. balance would be

dx Ib Na,SO. dt min

1.81b Na,SO. min

=

10 Ib total x Ib Na,SO. min y Ib total

dx lOx dt = 1.8 - 100 + 2t

or dx

10

Tr+ 100 + 2t X

= 1.8

The solution to this linear differential equation gives the pounds of Na,SO. after 10 min; the concentration then can be calculated as x/yo To integrate the linear differential equation, we introduce the integrating factor e ftd ','10+0.2'):

f

xeI (d'/10+0.2') =

+ 0.21)' = 1.8 0, and C = 35 + 10'; x(10

at x

= SO, t

=

(e f (d'/10+0.2t1)(1.8) dt

x

=

1.5(10

+C

f (10 + 0.2t)' dt + C 35 x 10'

+ 0.2t) + (10 + 0.21)'

at 1= 10 min;

x

35 x 10'

= 1.5(12) + (12)'

= 32.1 A good chn makes the ir

The concentration is

=-- = y

the same as because the steady state. Our ne: At. Let Ts

32.1 = 0 267 Ib Na,SO. 100 + 20 . Ib total solution enthalpy of I input sIre,"

EXAMPLE 6.6

Unsteady-state energy balance with graphical integration

Five thousand pounds of oil initially at 60'F are being heated in a stirred (perfectly mixed) tank by saturated steam which is condensing in the steam coils at 40 psia. If the rate of heat transfer is given by Newton's heating law, i.e., •

dQ

Q = dr

= "(T",,", -

TOil)

where Q is the heat transferred in Btu and" is the heat transfer coefficient in the

transfer heat

enthalpy of I output stre,' enthalpy change inside tank

·4-·~< ~.------'-'."« <.~

Chap. 6

Unsteady-State Material and Energy Balances 459

Chap. 6

proper units, how long does it take for the discharge from the tank to rise from 60 to 90°F? What is the maximum temperature that can be achlcved in the tank? Addilional dala:

lOx

entering oil flow rate discharge oil flow rate h

= 1018 lb/hr at a temperature of 60°F =

10181b/hr at a temperature of T

= 291 Btu/(hr)(OF) = 0.5 Btu/(lb)(OF)

Cpo,' Solution: The process is shown in Fig. E6.6(a). The independent variable will be I, the time; the dependent variable will be the temperature of the oil in the tank, which is

,'ariable would have been ,alSO. balance would be

xlb Na:SO •

.v Ib total

=~~'~~I~=:J

1018 Ib/hr Tin' GO°F-I.:

:Jund~

:::rat(

. 'alSO. after 10 ::ar differential

tOt8 tb/hr

------ Tout' T

Temperature of saturated steam 1$' 26rF

+c

Fig. E6.6(a).

- C

the same as the temperature of the oil discharged. The material balance is not needed because the process, insofar as the quantity of oil is concerned, is assumed to be in the steady state. Our next step is to set up the unsteady-state energy balance for the interval fl.t. Let Ts = the steam temperature and T = the oil temperature: accumulation

=

input - output

A good choice for a reference temperature for the enthalpics is 60°F because this makes the input enthalpy zero. en enthalpy of the input stream

=m Integration .:ned in a stirred (perfectl;'

steam coils at 40 psia. Ir ....... ,

:·.:-ansfer coefficient in

th~

transfer heat enthalpy of the output stream enthalpy change inside tank

hr

I

I

lCT _ T) = 291 Btu (267 - T)°F A: hOr ) I S (hr)«F)

I I

10I81b 0.5 Btu (T - 60tF I AI hr hr (Ib)p·")

[ 1

I

r-

input (Btu)

output (Btu)

5000 Ib 0.5 Btu (T - 60)OFl (lb)("F)

- J.. ".

accumulation (Btu)

Chap. 6

Chap, '

The rate of change of energy inside the tank is nothing more than the rate of change of internal energy, which, in turn, is essentially the same as the rate of change of enthalpy, i.e., dE/dl = dU/dl = dH/dl, because d(pV)/dl ~ O. We use Eq. (6.8) with jj = O. We assume that all the energy introduced into the motor enters the tank as (V:

The at

460 Unsleady-Slale Afalerial alld Energy Balances

The ar, Theta

dT dl

WHA,

= 44.1 -

f

0.32T

dT ".i7""',--::.::.".,,,,,,, 60 44 . 1 0.32T 90

n

= f" d

I

0

=

ing un, necess,'

(J

An analytical solution to this equation gives (J = 1.52 hr. For illustrative purposes, so you can see how equations too complex to integrate directly can be handled, let us graphically integrate the left-hand side of the equation. We set up a table and choose values of T. We want to calculate 1/(44.1 - 0.32T) and plot this quantity vs. T. The area under the curve from T = 60' to T = 90'F should be 1.52 hr. T

0.32T

44.1 - 0.32T

60 70 80 90

19.2 22.4 25.6 28.8

24.9 21.7 18.5 15.3

I

44.1 - 0.32T

= ifJ

0.0402 0,0462 0.0540 0.0655

SUPPL Hin Ne' 2. Jen; Aca 1.

3.

Mal ChCl

4. Shei 2nd

These data are plotted in Fig. E6.6(b).

I

PROBt.

0.07

1

~

I ! I

r+ 0,06

,

,

J, '

.,. 0.05 I

,...0.04

f-1"

I

.r,

/

......

,J

, I I

, !

I

i

' .1

~

I

-+--i

!

+-:::::tl

I

I

I

,:

I

70

80 T

Fig. E6.6(b).

me' is t

I ,

6.2. T!;

I

i

I

I

atr

to,

rrtIJ

This area conrains 311 small squares I

6.1. A

fer

i

I, 1 • 'J '

I

, ,

0.03 60

f-br

Each small square' (1)(0.002) '0.002 hr

cor T(\l

If!' th,;

i

90

of 6.3. A 1 6.t)

------

'" ([hap. 6

=, the cate of change

Problems 461

Chap. 6 The area below

the: tate of change of -:-:=y introduced into the

tP

= 0.030 is (90 - 60)(0.030 - 0)

The area above

tP = 0.030 but under the curve is (311)(0.002)

The total area

:u

= 0.90 hr

= 0.622 hr

= 0.90 + 0.622 = 1.52 hr.

.~,

:910

WHAT YOU SHOULD HAVE LEARNED FROM THIS CHAPTER This chapter was intended to acquaint you with the basic concepts underlying unsteady-state material and energy balances and to enable you to set up the necessary differential equations for very simple physical systems .

.=iillistrati'i'e" purposes, be' handled, let _~ ul"a: liable and choose :.::is quantjty V~. T. The

. ::y

=

32 hr.

Q.1.l. IVJ..

SUPPLEMENTARY REFERENCES 1. Himmelblau, D. M., and K. B. Bischoff, Process Analysis and Simulation, Wiley, New York, 1968. 2. Jenson, V. G., and G. V. Jeffreys, Mathematical Methods in Chemical Engineering, Academic Press, New York, 1963.

Marshall, W. R., and R. L. Pigford, The Application of Differential Equations to Chemical Engineering Problems, University of Delaware, Newark, 1942.

(1):0462 EIC0540

3.

(I)~0655

4. Sherwood, T. K., and C. E. Reid, Applied Mathematics in Chemical Engineering, 2nd ed., McGraw·Hill, New York, 1957.

PROBLEMS r-I:;~ I

./

l-

i I

90

'*J

£%

6.1. A defective tank of 1500 ft 1 volume containing 100 percent propane gas (at I

atm) is to be flushed with air (at 1 atm) until the propane concentration is reduced to less than I percent. At that concentration of propane the defect can be repaired by welding. If the flow rate of air into the tank is 30 ft 'lmin, for how many minutes must the tank be flushed out? Assume that the flushing operation is conducted so that the gas in the tank is well mixed. 6.2. The catalyst in a fluidized bed reactor of 200·ft 3 volume is to be regenerated by contact with a hydrogen stream. Before the hydrogen can be introduced into the reactor, the 0, content of the air in the reactor must be reduced to 0.1 percent. Ifpure N, can be fed into the reactor at the rate of20 ft 3 /min. for how long should the reactor by purged with N, ? Assume that the catalyst solids occupy 6 percent of the reactor volume and ~hat the gases are well mixed. 6.3. A plant at Canso, Nova Scotia, makes tish·protein concentrate (FPC). It takes 6.61b of whole fish to make I Ib of FPC. and therein is the problem-to make money, the plant must operate most of the year. One of the operating problems

... "". -)"~' ,.'"e,:

,""'"1fV""....)t"',...,..,..,t....,......,.,.,.....""7"..,..","'..... "' -

fiO

"AjipeltdixA.

\,p-lir

------1------

- -

'1.'73

X

"10- 1

3.715

v

10- 1

1.114

X

10- 6

3.653

X

10-$

3.766:< "10- 7

5.0505 x "10-;

lfiru

9.478 ,- 10-'

-.

- 9:196 L285

!

-Joule

--~.~

, 10-

0.-2390

3

10-l

0.1241

1.J56

1-he values. f,," of'their isol ..

"4.184

__

I

A,ctfni,im ~Iumlnitim

:;.t.:tne'riCium ;A'ntlmon,

·~·I'-g6n

.

..);rsco"nk

'A'-stali"ne TBfirtum Js(irk.eli-um IBetylli-um

,

, ~~-~~-~l~

),rm

naim 1.J'j16

X

:3j42

X

10- 3

'\0"

«(},9869

!7'6d,0

[ :i5~~~lt(tC'_

:1

'9_812

x' 10- 6

IsiSmuth fsoron I8tomine 2

I<;admi-um

!

1:333 '<10'

3,'387 " '10'

,

:

X

\(]alif6!'niwn "Carbon IC~rilim

'-Chlorine 'Chromium ~obalt

-1.000 :< 10' 'LOI'3

cta:csium 'CakH..im

CCopper

10'

i

tcurium [Dysprosium IEinstein~um

iEfbj.um !EUropium 'Fefmium IFkwrinc !Francitim

)


(Germanium 'Gold 'Hafnium

!Helium

J

;Holmium ;Hydrugen 'hlditim Iodine Iridium Iron tKrypt6n '.Lanthanum ·Lawrenc;unl

·t.ead l.lithium ·l.utetium Mar.nesium Man~
Mende! .. , iu:-: -SOURer.. (' ;

.--..--

Appendix. C

Steam Tables Absol.. :

Temp. Fahr.

t,

SOURCE: Combustion Engineering. Inc.

Absolute pressure atmospheric pressure ~ vacuum. Barometer and vacuum columns may be corrected to mercury at 32'F by subtracting 0.00009 X (I - 32) ,< column height, where 1 is the column temperature in ·F. One inch of mercury at 32 F ~ 0.4912 Ibjin.' Example: Barometer reads 30.17 in. at 70F. Vacuum column reads 28.26 in. at 80 F. Pounds pressure = (30.17 - 0.00009 38 x 30.17) - 28.26 - 0.00009 x 48 x 28.26) ~ 1.93 in. of mercury at 32' F. Saturation temperature (from table) = 100 F. ;-:::c

TABLE C.I SATURATED STEAM: TEMPERATURE TABLE

Absolute prl'Ssurt'

Temp. Fahr.

Enthalpy

SpecIfic volume

Sat. Lbfin.! p

In. Hg 32'F

liquid

,',

Sal. Vapor

Evap.

,',

rjg

Sat. liquid h,

Evap. hfo

Sat. Vapor h,

140 142 144 146 148

0.t806 0.1957 0.2120 0.2292

0.01601 0.01602 0.01602 0.01602

3305.7 3060.4 2836.6 2632.2

330.5.7 3060.4 2836.6 2632.2

0 2.01 4.03 6.04

1075.1 1074.9 1072.9 1071.7

1075.1 1076.0 1076.9 1077.7

ISO 152 154 156 158

40 42 44 46 48

0.1217 0.1315 0.1420 0.1532 0.1652

0.2478 0.2677 0.2891 0.3119 0.3364

0.0160:? 0.0 I 60:? 0.01602 0.01601 0.01602

2445.1 2271.8 .:!112.2 1965.5 1829.9

2445.1 2271.8 2112.2 1965.5 1829.9

8.05 10.06 12.06 14.07 16.07

1070.5 1069.3 1068.2 1067.1 1065.9

1078.6 1079.4 1080.3 1081.2 1082.0

160 162 164 166 168

SO 52 54 56 58

0.1780 0.1918 0.2063 0.2219 0.2384

0.3624 0.3905 0.4200 0.4518 0.4854

0.01602 0.01603 0.01603 0.01603 0.01603

1704.9 1588.4 1482.4 1383.5 1292.7

1704.9 1588.4 1482.4 1383.5 1292.7

18.07 20.07 22.07 24.07 26.07

1064.8 1063.6 1062.5 1061.4 1060.2

1082.9 1083.7 )084.6 1085.5 1086.3

170 172 174 176 178

60 62 M 66 68

0.2561 0.2749 0.2949 0.J162 0.3388

0.5214 0.5597 0.6004 0.6438 0.6898

0.01603 0.01604 0.01604 0.01604 0.01605

1208.1 1129.7 1057.) 989.6 927.0

1208.1 1129.7 1057.1 989.6 927.0

28.07 30.06 32.06 34.06 36.05

1059.1 1057.9 1056.8 1055.7 1054.5

1087.2 1088.0 1088.9 1089.8 1090.6

180 182 184 186 188

70 72 74 76 78

0.3628 0.3883 0.4153 0.4440 0.4744

0.7387 0.7906 0.8456 0.9040 0.9659

0.01605 0.01606 0.01606 0.01607 0.01607

868.9 814.9 764.7 71S.0 674.4

868.9 814.9 764.7 718.0 674.4

38.05 40.04 42.04 44.03 46.03

1053.4 1052.3 1051.2 1050.1 1048.9

1091.5 1092.3 1093.2 1094.1 1094.9

190 192 194 196 198

80 82 84 86 88

0.5067 0.5409 0.5772 0.6153 0.6555

1.032 1.101 1.175 1.253 1.335

0.01607 0.01608 0.01608 0.01609 0.01609

633.7 595.8 560.4 527.6 497.0

633.7 595.8 560.4 527.6 497.0

48.02 50.02 52.01 54.01 56.00

1047.8 1046.6 1045.5 1044.4 1043.2

1095.8 1096.6 1097.5 1098.4 1099.2

90 94 96 98

0.6980 0.7429 0.7902 0.8403 0.8930

1.421 1.513 1.609 1.711 1.818

0.01610 0.01611 0.0161 I 0.01612 0.01613

200 202 204 206 208 210

46S.4 441.7 416.7 393.2 371.3

468.4 441.7 416.7 393.2 371.3

58.00 59.99 61.98 63.98 65.98

1042.1 1040.9 1039.S 1038.7 1037.5

1100.1 1100.9 I 10l.R 1102.7 1103.5

212 215 220 225

100 102 104 106 J08

0.9487 1.0072 1.0689 1.1338 1.2020

1.932 2.051 2.176 2.308

0.01613 0.01614 0.01614 0.01615 0.0)616

350.8 JJ 1.5 313.5 296.5 280.7

67.97 69.96 71.96 73.95 75.94

1036.4 103S.2 10]4.1 103.1.0 1032.0

1104.4 1105.::

2.4·47

350.8 33 I.S 313.5 296.5 280.7

1 lOt.. I 1107.0

J 107.9

h = enthalpy. Btu/lb.

472

v)l4l!$Q. ;q

UO U2 U4 U6 138

0.0886 '0.0961 O.I04t 0.1126

r = specific volume, ftl/lb.

';Nt4M

120 122 124 126 128

32 J4 36 38

92

--- .., 44~

110 112 114 116 118

. .4

}4kN+i,ep; 9

.....



; ;us.

19MW;>

.i

,Pf

q

$

, ;nap?

230 235 240 245 250

(0

..

1

7 ~

R S

9 q

Iil

10 I' II 1.-; I

~

I', I.'

"

14

I' I ~. I~I ~

'

.

2~

-'.

-.1---"---..--.-...._ _ _ _...."ll.. I...

·'~40._

..

'<-'-

..J

--.---..--~--""---~,-~"

TABLE Col (CONT.)

earn Tables

Specific yolume

Absolutt pressure

Temp. Fahr.

Evap. h"

Sat. Vapor h,

265.1 251.6 238.5 226.2 214.5

17.94 19.93 81.93 83.92 85.92

1030.9 1029.7 1028.6 1027.5 1026.4

1108.8 1109.6 1110.5 J 111.4 1112.3

203.45 193.16 183.44 174.26 165.70

203.47 193.18 183.46 174.28 165.12

87.91 89.91 91.90 93.90 95.90

1025.3 1024.1 1023.0 1021.8 1020.1

1113.2 1114.0 1114.9 1115.7 1116.6

In. Hg

p

32'F

"

1.214 1.350 1.429 1.512 1.600

2.594 2.749 2.909 3.078 3.258

0.01611 0.01617 0.01618 0.01619 0.01620

265.7 251.6 238.5 226.2 214.5

1.692

3.445 3.640 3.846

Liquid

Enthalpy

Sat. Liquid h,

Sat. Lb/in.!

Evap.

r"

Sat. Vapor

"

• al 32' F by subtracling - :rature in 'F.

110 112 114 116 118

_,.26 in. at SOT Pound, ~S x 28.26) = 1.93 in. of

120 122 124 126 128

1.889 1.995 2.105

4.286

0.01620 0.01621 0.01622 0.01623 0.01624

130 132 134 136 138

2.221 2.343 2.470 2.603 2.142

4.522 4.110 5.029 5.300 5.583

0.01625 0.01626 0.01626 0.01621 0.01628

151.55 149.83 142.59 135.73 129.26

151.51 149.85 142.61 135.15 129.28

91.89 99.89 101.89 103.88 105.88

1019.5 1018.3 1017.2 1016.0 1014.9

1117.4 1118.2 1119.1 1119.9 1120.8

140 142 144 146 148

2.887 3.039 3.198 3.363 3.536

0.01629 0.01630 0.01631 0.01632 0.01633

123.16 111.31 111.88 106.12 101.82

123.18 111.39 111.90 106.74 101.84

101.88 109.88 111.88 113.88 115.81

1013.7 1012.5 101l.3 1010.2 1009.0

1121.6

J077.7

ISO 152 154 156 158

3.116 3.904 4.100 4.305 4.518

5.818 6.181 6.511 6.841 1.199 1_566 1.948 8.348 8.165 9.199

0.01634 0.01635 0.01636 0.01631 0.01638

91.18 92.19 88.62 84.66 SO.90

97.20 92.81 88.64 84.68 80.92

111.81 119.87 121.81 123.81 125.81

1001.8 1006.1 1005.5 1004.4 1003.2

1125.1 1126_6 1121.4 1128.3 1129.1

1018.6 1019.4 1080.3 )081.2 1082.0

160 162 164 166 168

4.139 4.910 5.210 5.460 5.120

9.649 10.12 10.61 11.12 11.65

0.01639 0.01640 0.01642 0.01643 0.01644

17.31 74.00 70.79 61.16 64.81

11.39 14.02 10.81 61.18 64.89

121.81 129.88 131.88 133.88 135.88

1002.0 1000.8 999.1 998.5 991.3

1129.9 1130.1 1131.6 1132.4 1133.2

110 112 174 176 118

5.990 6.212 6.565 6.869 1.184

12.20 12.77 13.31 13.99 14.63

0.01645 0.01646 0.01641 0.01648 0.01650

62.12 59.50 51.01 56.64 52.39

137.89 139.89 141.89 143.90 145.90

996.1 995.0 993.8 992.6 991.4

1134.0 1134.9 1135.7 Ill6.5 1131.3

180 182 184 186 188

1.510 1.849 8.201 8.566 8.944

15.29 15.98 16.70 11.44 lUI

0.01651 0.01652 0.01653 0.01654 0.01656

50.26 48.22 46.28

62.14 39.52 51.03 54.66 52.41 50.28 4S.24 46.30

42.61

44.45 42.69

141.91 149.92 151.92 153.93 155.94

990.2 989.0 981.8 986.6 985.3

Ill8.1 1138.9 1139.7 1140.5 114l.3

190 192 194 196 198

9.336 9.144 10.168 10.605 11.051

19.01 19.84 20.10 22.51

0.01657 0.01658 0.01659 0.01661 0.01662

40.99 39.38 37.84 36.38 34.98

41.01 39.40 31.86 36.40 35.00

151.95 159.95 161.96 163.91 165.98

984.1 982.8 981.5 980.3 919.0

1142.1 1142.8 1143.5 1144.3 1145.0

200 202 204 206 208 210

11.525

12.010 12.512 13.Dll 13.568 14.123

23.46 24.45 25.47 26.53 27.62 28.15

0.01663 0.01665 0.01666 0.01661 0.01669 0.01610

33.65 32.37 31.15 29.99 28.88 21.81

33.61 32.39 31.11 30.01 28.90 21.83

161.99 110.Ql 112.02 174.03 176.04

911.8 976.6 915.3 974.1 912.8 911.5

1145.8 1146.6 1147.3 1148.1 1148.8 1149.6

212 215 220 225

14.696 15.591 11.188 18.915

29.92

0.01612 0.01614 0.01611 0.01681

26.81 25.35 23.14 21.15

26.83 25.37 23.16 21.17

186.10 IS8.14 193.18

910.3 968.3 965.2 961.9

1150.4 1151.4 1153.3 1155.1

230 IJ5 240 245 250

20.18 22.80 24.97 21.31 29.82

0.01684 0.01688 0.01692 0.01696 0.01700

19.311 17.761 J6.307 15.010 13.824

19.388 11.118 16.3:4 15.021 13.841

198.22 203.28 208.34 213.41 218.48

958.7 955.3 952.1 948.1 945.3

1156.9 1158.6 1160.4 1162.1 1163.8

Enthalpy ~at. ~iQuid

h,

0 2.01 4.03

6.0-'

E\'ap. h" 1075.1 1014.9 .... '2.9 .. 1

J075.1

)076.0 )076.9

8.0, JO.06

j~9.3

::.06 :4.07 i6.07

1068.2 1061.1 1065.9

18.07 :0.Q1

1064.8 1063.6

::.01 ':4.07

1062.5 1061.4 1060.2

1082.9 1083.1 1084.6 1085.5 1086.3

.'0.05

1059.1 1051.9 1056.8 1055.1 1054.5

1087 .~ 1088.0 1088.9 1089.8 1090.6

,-8.05 "·.04 . :.04 -'.03 -0.03

1053.4 1052.3 1051.2 1050.1 1048.9

109L5 1092.3 1093.2 1094.1 1094.9

"::'6.07

:8.01 .-,0.06 ::.06 ·4.06

··'.02 _ ':.02 .":.01 .... 01

: :.>.00 _ ~'_OO ".99 ! .98 _'.98 ,.98 -.97 -.96 06 05 "4

,.5

Sat. Vapor h,

1041.8 1046.6 1045.5 1044.4 1043.2

1095.8 1096.6 1097.S 1098.4 1099.2

1042.1 1040.9 1039.8 1038.1 1031.5

1100.1 1100.9 1101.8

1036.4 1035.2 1034.1 1033.0

J 104.4 1105.:

lon.a

IIO~.7

1103.5

1106.1 1107.0 1107.9

1.188

4.062

21.59

v

..

-.-

Steam Tables 473

Appendix. C

ABLE

....

co;

specific volume, fi'/lb .

44.43

h

s:<

enthalpy, Btu/lb.

118.06 180.Q1

1122.4

1123.2 1124.1 1124.9

1

~

f;

" r,~

'.J

.~ I

,1

.!

J

-.1 >~

474 Appendix C TABLE

Ab~olule

Specific

preSSure

Temp. Lb/in.%

Fahr.

In. Hg 32"F

p

C.I (CONT.)

Sat. Liquid

Sat. liquid

Sat. Vapor

,',

Evap.

IZ.735 I I. 754 10.861 10.053 9.31J

11.771 10.878 10.070

,',

1','1

Sal. Varor

Evap.

h,

h"

h,

942.0 938.6 935.3 931.8 928.2

116:".t}

•. no

223.56 228.65 2J3.74 238.84 243.94

12.1S2

255 260 265 270 275

32.53 35.43 38.54 41.85 45.40

0.01704 0.01708 0.01713 0.01717 0.01721

280 285 290 295 300

49.20 53.25 57.55 62.1J 67.01

0.01726 0.01731 0.01735 0.01740 0.01745

8.634 8.015 7.448 6.931 6.454

8.651 8.032 7.465 6.948 6.471

249.06 254.18 259.31 264.45 269.60

924.6 921.0 917.4 913.7 910.1

1173.7 1175.'1

305 310 315 320 325

72.18 77.68 83.50 89.65 91>.16

0.01750 0.01755 0.01760 0.01765 0.01771

6.014 5.610 5.239 4.897 4.583

6.032 5.628 5.257 4.915 4.601

274.76 279.92 285.10 290.29 295.49

906.3 902.6 898.8 895.0 891. I

1181.1 1182.5 1183.9 1185.3 1186.6

330 335 340 345 350

103.03 110.31 117.99 126.10 134.62

0.01776 0.01782 0.01788 0.01793 0.01799

4.292 4.021 3.771 3.539 3.324

4.310 4.039 3.789 3.557 3.342

300.69 )05.91 311.14 316.38 321.64

887.1 883.2 879.2 875.1 871.0

1187.8 1189.1 1190.3 119U 1192.6

355 360 365 370 375 380 385

143.58 153.01 162.93 173.33 184.23 195.70 207.71

0.01805 0.01811 0.01817 0.01823 0.01830 0.01836 0.01843

3.126 2.940 2.768 2.607 2.458 2.318 2.189

3.14' 2.958 2.786 2.625 2.476 2.336 2.201

326.91 332.19 337.48 341.79 348.1 I 353.45 358.80

866.8 862.5 858.2 853.8 849.4 844.9 840.4

1193.7 1194.7 1195.7 1196.6

390 395 400 405 410 415 420

220.29 233.47 247.25 261.67 276.72 292.44 308.82

0.01850 0.01857 0.01864 0.01871 0.01878 0.01886 0.01894

2.064 1.9512 1.8446 1.7445 1.6508 1.5630 1.4806

2.083 1.9698 1.8631 1.7632 1.6696 1.5819 1.4995

364.17 369.56 314.97 380.40 385.83 391.J0 396.78

835.7 831.0 826.2 821.4 816.6 811.7 806.7

1199.9 1:200.6 1201.2 1201.8 120:2.4 1103.0 1203.5

425 430 435 440 445 450

3l5.91 343.71 362.27 381.59 401.70 422.61

0.01901 0.01910 0.01918 0.01926 0.01934 0.01943

1.4031 1.3)03 1.26 J 7 1.1973 1.1 367 1.0796

1.4221 1.3494 1.2809 1.2166 1.1560 1.0990

402.28 407.80 413.35 418.91 424.49 430.1 (

801.6 796.5 791.2 785.9 780.4 774.9

1203.9 1204.3 1204.6 1104.ti 1204.9 1205.0

455 460 465 470 475

444.35 466.97 490.43 314.70 539.90

0.0195 0.0196 0.0197 0.0198 0.0199

1.0156 0.9745 0.9262 0.8808 0.8379

1.0451 0.9941 0.9459 0.9006 0.8578

435.74 44 1.42 447.10 452.84 458.59

769.3 763.6 757.8 751. 9 745.9

1205.0 1205.0 1204.Q J.:!04.7 1204.5

480 485 490 495

SOO

566.12 593.28 621.44 650.59 680.80

0.0200 0.0201 0.0202 0.0203 0.0204

0.7972 0.1585 0.7219 0.6872 0.6544

0.8172 0.7786 0.7421 0.7075 0.6748

464.37 41'O.lti 476.01 481. 90 4S7.80

739.8 733.6 727.3 720.8 714.2

J ~03.j 1202.7 1102.0

50S 510 SIS 520 525

712.19 744.55 777.96 812.68 848.37

0.0206 0.0207 0.Q208 0.0209 0.0210

0.6230 0.5932 0.5651 0.5382 0.5128

0.6436 0.6139 0.5859 0.5591 0.5))8

493.8 499.8 505.8 511.9 518.0

707.5 700.6 693.6 686.5 679.2

530 535 540

885.20 923.45 962.80

0.0212 0.0213 0.0214

0.4885 0.4654 0.4433

0.5097 0.4867 0.4647

.$1-4.2 5.10.4 536.6

671.9 664.4 656.7

,,= specific Volume. (t'/lb.

..

Absoll-

Enthalpy

~'olume

h

,~

Ilb7.J IlfllI.O 1170,Il 117~.1

1176,7

1178.2 1179.7

1197.5

1198.4 1199.2

12()4.~

I~03.S

J:!OI.J 1200.4

1199.4 J 198,4 1197.~

1196.1 1194.1\ 1193.3

enthalpy. Btu:lb .

'.+E ;;;... *' ...,.--.. ~--------....,...,...-~--------...",....,. .$

'-.

tt;

;;p:

""'1$(,

;;;

!4Qz

Q 4

.J.

443&

eaec.4

¢.

.4

q::::

,'T'"""""

Temp. Fahr.

545 550 555 560 565 570

IIii 1 Il

Ii II

":3

575 580 585 590 595

I:

IJ' :~.

I .; ~

600 605 610 615 620 625 630 635 640 645

I'

16' 1!l~ I"

"

J7~ 18~

,

19: 19 _~ 2u~,

21.-

650 655 660 665 670

2'+-: 253

675 680 685 690 695

2" I ~ :270J 2fl'l :2S" : 29',,;

700 705

3(1'.)· 31e::

705.34'

J':C:

22\' 2" 2~1'

·Critical temp!:!

_0 ....".. ' "'_~'''''''''''~~

r-----

.....

-----,--~-

Steam Tables 475 TABLE C.I (CONT.)

:~uid

".56 _•. 65

Evap.

5dt. Varor

h"

h,

~.74

>.84 c 94

942.0 938.6 935.3 931.8 928.2

Iltl5.~

1167.3 1169.0 1170./}

1172.1

924.6 921.0 917.4 91l.7 910.1

1173.7 J 175.1

1181.1

'5.49

906.3 902.6 898.8 895.0 891.1

1182.5 1183.9 1185.3 1186.6

h).69 :-5.91 i.14 :6.38 _1.64

887.1 883.2 879.2 87l.1 871.0

1187.8 1189.1 1190.3 1191.5 1192.6

.'.6.91 :2.19 7.48 '2.79 .'J 11 'J.4S

866.8 862.l 8l8.2 851.8 849.4 nQ.9 4

1193.7 1194.7 1195.7 1196.6 1197.5 1198.4 1199.2

09.06

,,18 UI 4.45 ".60 , 4.76 . -9.92 .5.10 ~J.29

.\~.80

-·U7 :·Q.S6 ';4.97 ;0.40 ;;5.83 1.30 ~5.78

.1.0

826.2 82\.4 816.6 811.7 806.7

1176.7

117X.2 1179.7

1199.9 1200.6 1201.2 1201.8 1202.4 1 ~OJ.O 1203.5

801.6 796.5 791.2 785.9 780.4 774.9

1203.9 1204.3 1204.0 120. Ui

1205,0 1205.0 120.;.")

·~.59

769.1 763.6 757.8 751.9 745.9

4.37 '0.18 ,6.01 ·1.90 ":7.80

739.8 733.6 727.3 720.8 714.2

J 204.2 120.\.:-: 1::::0.\.-' 1202.7 1202.0 1201.3 1200.4

~8.0

707.5 700.6 693.6 686.5 ' 619.2

: ••2 .J.4 0.6

671.9 664.4 656.7

1196.1 t 19.t.!! I 193. J

:1.28 C1.80 : ].35 ,8.91 ~4.49

30.1 ( :5.74 ." 1.42 ..'7.10 ",84

'3.8 -'9 .. 8

J5.8 ; 1.9

Specific

Absolute pressure

Enthalpy _L

1204.~

1105.0

1:.'04.7

1,04.5

1199.4

1198.4 1197.2

Temp. Fahr.

In. Hg 32'F

Lbtin.t p

Sal. Liquid

,',

~'o{u,"e

Enthalpy Sat.

Evap. rl!1

Vapor

",

545 550 555 560 565 570

1003.6 1045.6 1088,8 11lJ.4 1179.1 1226.7

0.0216

0.4221

0.0218

0.4021

0.0119 0.0221 0.0222 0.0224

0.3830 0.3648 0.3472 0.3304

575 580 585 590 595

1275.7 1326.1 1378.1 1431.S 1486.5

0.0226 0.0228 0.0230 0.0232

0.1143 0.2989

0.0234

0.2840 0.2699 0.2563

0,3070 0.2931 0.2797

600 605 610 615 620

1543.2 1601.S 1661.6 1721.4 1787.0

0.0236 0.0239 0.0241 0.0244 0.0247

0.2432 0.2306 0.2185 0.2068 0.19l5

625 630 635 640 645

1852.4 1919.8 1989.0 2060.3 2133.l

0.0250 0.0253 0.0256 0.0260 0.0264

650 655 660 665 670

2208.8 2286.4 2366.2 2448.0 2532.4

675 680 685 690 695

Sat. Liquid hI

0.4438 0.4239 0.4049 0.3869 0.3694 0.3528

542.9 S49.3 S5S.7

0.3369

582.1

0.3217

588.9 595.7

Sat. Evap. h"

Vapor

648.9 640.9 632.6 614.1 615.4 606.S

1191.8 1190,2 1188.3 1186.3 1184.2 1181.9

_.7

597.4 588.1 578.6 568.8 558.7

1179.5 1177.0 1174,3 1171.4 1168.4

0.2668 0.2545 0.2426 0.2112 0.2202

616.8 62:4.1 631.5 638.9 "";.l

S48.4 537.7 526.6 515.3 50l.7

1165.2 1161.8 IIS8.1 1154.2 IllO.2

0.1845 0.1740 0.1638 0.1539 0.1441

0.2095 0.1993 0.1894 0.1799 0.1705

654.3

670.4 678.7 687.3

491.l 478.8 465.5 452.0 437.6

1145.8 1141.0 Illl.9 1130.7 1124.9

0.0268 0.0273 0.0278 0.0283 0.0290

0.1348 0.1256 0.1167 0.1079 0.0991

0.1616 0.1529 0.1445 0.1362 0.1281

696.0 705.2 714.4 724.5 734.6

422.7 407.0 390.l 372.1 35l.J

1118.7 1112.2 1104.9 1096.6 1087.9

2619.2 2708.4 2800.4 289l.0 2992.7

0.0297 Om05 0.0316 0.0328 0.0345

0.0904 0.0810 0.0716 0.0617 0.0511

0.1201 0.1115 0.1032 0.0945 0.0856

745.5 757.2 710.1 784.2 801.3

332.8 310.0 284.5 2l4.9 219.1

1078.3 1067.2 1054.6 1039.1 1020.4

700 705

1094.1 3199.1

0.0169 0.0440

0.0389 0.0157

0.0758 0.0597

g23.9 870.2

171.7 77.6

995.6 947.8

705.34·

3206.2

0.0541

0

0.0541

910.3

0

910.1

·Critical temperature.

,

t,. =

specific volume, fll/lb.

562.2 568.8 575.4

602.6

662.2

h '-"' enthalpy, Btuilb.

r-

h,

:~

.,. ,.

.:. ~

I

.,, j

J,

NO .

.3Z.376.

L.OGARITHMIC

©~~

NORMAL.

QRAPH PIA-PER



99

(;0/78

Y\O.

.-/c)J"

l'C)

2.2.. -, he Utml ~IIII'

(VICidd III ,

r

t-l-t-+-t+H+r-I-H-H-I+t-P-I+t+I+I+IHl~ II

99

-

'1111!j - -f-I, -u; ~ -f--

Ii ,11 -

-

~

'ry)

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'7

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i

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30

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Hodel No.: _ _ _ _zz- C,#E

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Croll-Reynolds Job No.':

~o/78

OPERATING CONDITIONS A.

Gas Inlet Rate Capacity:

B.

Gas Inlet Temperature:

C.

Gas Inlet Pres sure:

D.

Gas

Composition:

t::S%

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Liquid Inlet Rate: ____________________ ________________________gpm 1. ________________________________________________gpm 2. ~c~'~"~GL_

At Liquid Inlet: __________________~/~~~___________________=~~psi~

F.~Pressure

1.

x spec if ic g r a v it y : ________________________________ 2.

----------------------------------------psig x spec if i c g r a vi t y : ________________________________

G.

Liquid Composition: 1.

2.

II.

H.

Maximum Liquid Temperature:

1.

[ :d £

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-------~--~~-~-------

Ac ro s s S ys tern : ______..3~__...,AL-',e.t.~~~____~I~I~l;;;1R'1~(~€l~S~)Fw:;=E!;~·

PERFORMANCE DATA A.

% Removal efficiency of

B.

% Removal efficiency of

C.

%

D.

% Removal efficiency of

Removal efficiency of

Based upon above operating conditions.

O.7:?A ?

OPERATING CONDITIONS AND PERFORMANCE DATA Croll-Reynolds Job No::

Hodel No.: _ _~Zc2,----,,~,,-,~:..:c~=-_ __ 1.

60178

OPERATING CONDITIONS A.

Gas Inlet Rate Capacity: ______"'-/o"'-_---=3""s:""'---_ _ _ _ acfm

B.

Gas Inlet Temperature:

C.

Gas Inlet Pres sure:

D.

Gas

Composition:

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Liquid Inlet Rate: ________________________gpm 1. _______________ ___________________________________________________ gpm 2. ~c~'~"U~~

F.APressure At Liquid Inlet: ________________~~~~~_________________=~~ps~ 1.

2.

x specific gravity: --------------------________________________________________________psig x specific gravity:

G.

--------------------

Liquid Composition: 1.

2.

II.

,..c-

H.

Maximum Liquid Temperature:

1.

b itt E t Ac ro s s S y stem: ____---.:-3~_"fAL-"c;c~~i/'_·_ _ '11::;i'1~1::Sl[:i:I~('i\11O~!il;=J)J"=:::w::;:s,;;i:i. ..

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PERFORMANCE DATA A. B.

Removal efficiency of % Removal efficiency of

C.

%

D.

%

Removal efficiency of % Removal efficiency of

Based upon above operating conditions.

O.Z~A

.

CROLL - REYNOLDS COMPANY, INC. CHEMICAL

a.

751 ENVIRONMENTAL CONTROL EOUIPMENT

MECHANICAL ENGINEERS CENTRAL P.

O.

BOX

AVENUE 668

VACUUM EOUIPMENT

WESTFIELD, N . .J. 07091

E.JECTOR & VACUUM PUMP SYSTEMS THERMO COMPRESSORS

WET SCRUBBING EOUIPMENT:

VACUUM REFRIGERATION SYSTEMS

DUST, GAS, FGD, NOX TEL. 201-232~4200 TWX NO. 710-997-9642

SOLVEI-n REMOVAL AND SPECIAL SYSTEMS COOLING TOWERS AND f"ACKINQ MATERIAL

SURFACE 0. 8AROMETRIC CONDENSERS

December 23, 1983

'~'

,.

Western Research Institute University of Wyoming Research Corp. P. O. Box 3395 Laramie, WY 82071

(

Attention: Mr. T. Owen Subject: Your Order No. 40555 Our Order No. 60178 Gentlemen: Enclosed are two (2) certified prints and one (1) certified reproducible of our drawing FSB-20640 Rev. 1 showing the No. 22-6HE Fume Scrubber we are furnishing under the subject order. Please note that this drawing shows the location of mist eliminator pad, per your request. Also, per your request, we have enclosed four (4) copies of Bete Fog Nozzle, Inc. drawing 2NF-4699 (2 pages). Manufacture of this equipment is proceeding according to the enclosed. If there are any questions, please contact our Mr. Roger Kirchheim as soon as possible. Very truly yours, CROLL-REYNOLDS COMPANY, INC.

'-1

\...../, .

//f:/yj /)(.

.z!

(Mrs.) M. Partelow Enc.

DESIGNERS AND MANUFACTURERS SINCE 1917

<.

/~a4l&-c-c/c-

WESTERN

R~SEARCH

INSTITUTE

T NTEROFF ICE HF'HlRANDlTH

June 20. 19H/. TO:

Andy

Loni"

FRC1H:

Tom Owen,

SUII.JEcr:

Drawillg for Venturi Scrubber Throat Valve Control

Environment~l

Engineering Division

Plense find attached a copy of the proposed design for the venturi scrubber throat valve' cuntrol ring. PIL'
Thomas E. Owen C'c:

Place

i),rui'oI\J of 0;\:5

FIolA)

-

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\ 1 1 spaces in shaded area For Use By Purchasing Or Receiving Only

Iqs

UNIVERSITY OF wxqMlNG RESEARCH CORPORATION ~~ 6ilfiEQUtSITtON

I'

VENDOR BUSINESS CODE CODE

~O ~;Z¢
Suggested Vendor, Address, Phone No., Co n ta c t

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Authorjzer __________~------------------------------------------------------------~ Signature Dat8 Receiver Signatu re Date

All spaces in shaded area For Use By Purchasing Or Receiving Only

<:

Date of Request

Date Required

Deliver To : INamel

Unit

Item No.

Qty .

Un it

Pr ice

4

, Requisitioner

Signature

Approver

Signature

, Date " >

-'-

" " "

Date

Account

CROLL - REYNOLDS COMPANY, CHEMICAL

a

751 ENVIRONMENTAL CONTROL EQUIPMENT WET SCRUBBING

INC.

MECHANICAL ENGINEERS CENTRAL P.

O.

BOX

AVENUE 666

VACUUM EQUIPMENT

WESTFIELD, N. oJ. 07091

EJECTOR & VACUUM PUMP SYSTEMS

THERMOCOMPRESSORS

~OUIPMENT:

VACUUM REFRIGERATION SYSTEMS

DUST,GAS,FGO,NQX

SURFACE & BAROMETRIC CONDENSERS

TEL.201-232-4200 TWX NO. 710-997-9642

SOLVENT REMOVAL AND SPECIAL SYSTEMS COOLING TOWERS AND PACKING MATERIAL

December 7, 1983

university of Wyoming Research Corp. P. O. Box 3395 Laramie, WY 82071 Attention: Mr. T. Owen Subject: Your Order No. 40555 Our Order No. 60178 Gentlemen: Enclosed are four (4) certified prints of our drawing FSB-20640 showing the No. 22-6HE Fume Scrubber we propose to furnish under the subject order. Print approval has been received. Manufacture of this equipment is proceeding according to the enclosed. If there are any questions, please contact our Mr. Roger Kirchheim as soon as possible. Also enclosed are four (4) copies each of the following: Operating Conditions and Performance Data Performance Curve Proposal Data Sheet Inst. 179

Operating Instructions

Information concerning the spray nozzles will be forwarded as soon as possible after receipt from supplier. Very truly yours, CROLL-REYNOLDS COMPANY, INC.

~~, '7;(.. ~~~~AJ (Mrs.) M. Partelow

Enc.

DESIGNERS AND MANUFACTURERS SINCE 1917

Unl...

THii PRINT CONTAINS PAOPRJETARY INFORMA·T noN WHICH LlUST NOT BE USED FOR COM PET. TrvE PURPOSES OR IN ANY WAY DETRIWENT"l , TO IETI FOG NOZZLE, INC. PRINT MUST 8E "E· ',' TURNED ON REQUEST.

0\11._

NOled, U. . Stand.rd

DimentiOnal Tol.rano... 8oal. dlm.nllon. or traction'

Two.-Pl*o. Decimal.

Th_ilc.O.olmai.

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BETE FOG NOZZLE, INC.

-..".~~:.~6\df£C\ ~Q..

GREENFIELD, MASS.

'l/e,

~"'f ot:;o

90

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DATE I

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DESCRIPTION

DRAWING

DRAWN: !il;. ,I CHECKED: IDATE: IZ/i2l~ DATE:

No.2.Nf -~c:::;~

REV.

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·I I00

U~ p~!NT COHr..'JNS rnOrR,ETAR¥ IHFOAW ... ~N WH~N WUo T Nor BE USEe FOR COMPEn.. " PURPOSES OR IN ANY WAY DETRIMENTAL. ) BfT! ~ .NOZZLE. ,NC. PRINT MUST BE RE· JRi,,"onn"yu,.,.

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BETE FOG NOZZLE, INC.

\ZA~' Dr'\tA

GREENFIELD, MASS.

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Unl ••• Otta.NI•• Noted. U.. 8landard Dlmen,lonal Tolerano, •• .±. I i •• " , - p . . " . O . " ! , , , . ! . · .±. .010:

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DESCRIPTION

I

OPERATING INSTRUCTIONS HIGH ENERGY VENTURI SCRUBBER

Operating conditions for Order No.

60178

Required gas pressure drop ~~~.~P~=~3~P~S~I~_____________ "

w.

G.

Required liquid pressure ________~1~5L_________________~/\~ psi, Required liquid flow

gpm

0.6

GAS IN

LIQUID IN

------+t+f--+---l---+-+I-I~--

REMOVABLE SPRAY NOZ ZLE

VENTURI THROAT

TO SEPl\RATOR

Page 1 #179

OPERATING PRINCIPLES

In operation, the venturi scrubber uses the energy from a high velocity gas stream to entrain and atomize a liquid stream. The high atomization of the liquid and the extreme turbulence in the venturi throat gives this scrubber its exceptionally high ability to remove small micron or submicron dust particles. The effectiveness of this unit is dependent upon the gas velocity in the throat. This can be easily measured through the pressure drop which the unit requires. If the pressure drop is not as shown on the above information, the unit will not function properly. This particular venturi scrubber is a fixed throat unit. The design gas velocity should be maintained at all times. The liquid recycle rate and pressure are also important. Once these values are obtained, the unit will perform satisfactorily. Little or no maintenance is required on the venturi scrubber and separator. There are no internal moving parts which could wear or come out of alignment. The only causes of difficulty would be a build up of solid material inside of the unit. If the gas pressure drop continues to build up or the gas flow decreases, please check the throat area for the possibility of a solids build up. This can be mechanically removed if necessary. The spray nozzles are removable through the flanges provided. They should be inspected occasionally to assure that they are not eroded or that corrosion has not ruined the orifice. Aaight amount of wear will not significantly affect performance of this type of scrubber. If the unit does not appear to be functioning properly; however, these nozzles should be checked and particular care taken of the possibility of plugging due to the presence of solids. The nozzles are designed to be of nonplugging design; however, excessive solid loads could still cause difficulties. The separator for this unit is covered under a separate set of operating instructions which should be referred to.

Page 2 #179

~

PROPOSAL

Customer Reference ____ 4

CROLL-REYNOLDS HIGH ENERGY SCRUBBER SYSTEM

°

Gas Inlet/Outlet _ _ _ _--'2'--'_'_ _ _ _ _ _ _::.2_"_ _ __ 1/2 " _ _ _ _ _ _ _ __ Scrubbing liqUid Inlet _ _ _-='--.:::'--_ Separator Tank: Not Included Tank Diameter: 6" liqUid Storage YES 0 Tank Connections Drain Manway: None

TO: ____Wc.:...:::e-=s:..,:t::-ern Research Institute ____U=.::.:n"'i:...:ve,.:ersity of Wy.

Res.

C.R. Inquiry_ _ _ _ __

5=-:::5-=5'----_ _ _ _ _ _ _ _ __

Corp.

NO

Included ~ Tank Height

0

4 ' 6"

~

Capacity , Fill _ _ _ , Overflow

1"

Gallons

Separator Type: _....!tco1e'---"s"'h'--.!M~i,;;s'-'t~E~1-"i""m'!.,1~·n<.!3a""t~o~r,--_ _ __ Other

MATERIALS OF CONSTRUCTION ITEM

MATERIAL

~16L Stainless J')~eel_ _ _ _ __ Venturi:_ Separator: _ _ _ "_ ... _ _"_.._____. __ "__ . _______

MODEL NO.: 22-6HE QUANTITY: ___ -'O"-'-n""e~_ _ _ _ __ PRICE PER SYSTEM: ---,$"",2,-,-,-"'.6.."-4o~_ __ F.O.B.: TERMS:

Shops -

Teterboro,

NJ

Net 30 lC

APPROXIMATE WEIGHT Shipping: _ _ __

Operating:

CONSTRUCTION DESIGN

ESTIMATED DELIVERY: _ _--'1::-0"--_ Weeks after drawing approval Weeks for drawings

o

Spray Nozzles: _3.l6.. ..S±a in 1 ess.. . S.teel __,,___.__ ,,_

ITEM IN STOCK DELIVERY: _ _ _ _ Weeks after receipt of order

Deliveries are quoted based on standard shop schedules at the time of quote. Special delivery requirements may be accommodated on an individual basis.

COMMENTS: Since Croll-Reynolds Co. has no control over the mixture or concentration of corrosive elements which may come in contact with the equipment supplied, no guarantee is expressed or implied concerning materials of construction for corrosion resistance. Customer should make every effort to assure suitability prioF to purchase.

In general accordance with: PS 1569 0 ASME Code!l(l ASME Code Stamped 0 or vessel designed for pressure/vacuum 15 PSIG 1_ _ _ __ Wall Thicknesses 1/8" Min. Head Thicknesses Bottom/Top,_=1.'.,/...:8,--'_'--=-:M:=i:::n::.,:.; 1/8" Min. Hold down lugs lifting lugs

Included ell: Included 0

Not included 0 Not included ell:

OPERATING CONDITIONS Gas Inlet Rate Capacity: 1 0- 3 5 §cfm o Gas Inlet Temperature 110-140 P 170 0 p Design Pressure: 3-15 PSIG (Atm. Press.-580 MM Hg.Abs Gas Composition Particulate

G:ts Containing Hydrocarbon

& Mist

liquid Inlet Rate: _ _-'O"'.o..:6_____ gpm Pressure at liqUid Inlet_--=1::c5=----___ ps~ Differential

x Specific Gravity: liqUid Composition: _ _ _ _w:::a"'t::.e::::r=--___________ Maximum liqUid Temperature: _ _---"l,..O'-'O'--Op-"-_ _______ Pressure Drop Across System:

Proposal by:

Date:

M. I. Holzman 10/3/83

3 PSIG

l"oI,(es) ','1.0.

PERFORMANCE DATA _ _ _900:5 " --__ % Removal efficiency of 0.75 Micron _ _ _ _ _ _ % Removal efficiency of _ _ _P_ar_t_i_c_ul_a_t_e_ _ _ _ _ _ _ % Removal efficiency of % Removal efficiency of _ _ _ _ _ _ _ __ _ _ _ _ _ _ % Removal efficiency of __________

Based upon above operatinq conditions

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PRINT C ERTIFIED CORRECT: Y OUR ORDER NO. ITEM NO. OUR ORDER NO. , 017~ ENGINEERING DEPT. DATE: /~- 7-~3

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CROLL-REYNOLDS COMPANY, INC.

WESTFIELD, N.J.

1\\0. E . 8 0 CT 140LES N ALL I="LAf\.IGf'D COI\.IN'S flR.E f'Q:JA .... L-Y SPA{...t-:O STI<.ADOLE. NATURAL.

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REMARK~

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TI-lRU INCREM£NT5 01='

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VENTURI 1'0 8£ 5lJPPO TED By C.U:::,TOME:'R.! I "

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- - c::;=======;:::t:::;=:======:J ::::.::::_..i...__*,--_',-P R INT CERT IFIED CORRECT.. Y OUR ORDER N O. ¥o5'rf IT EM NO. OUR O R D E R NO.~O/7~ E NGINEERING DEPT. DATE: /~ - 7- ~3

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CROLL. EYNOLDS COMPANY , INC. WESTFIELD, N.J.

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November II, 1983 Roger Kirchheim Croll-Reynolds Company, Inc. p. O. Sox 668 Westfield, NJ 07091 Dear Mr. Kirchheim: As per our telephone conversation on November 11, 19S3. please find enclosed an approved copy of your drawing FSB-20640 showing the No. 22-6HE Fume Scrubber. This approval is based on your acknowledgment that the design parameters we discussed will not change the design of the Fume Scrubber. The deSign specifications and a revised purchase order will be transmitted under separate cover from our Purchasing Office. If you have further questions, please contact me at 307-721-2341. Sincerely,

Thomas E. Owen cc:

~lace

1-1. r·1cTernan

Purchasi ng \~~f'

bee:

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\,!YOi·:II:G RES[f,RCH CORPOPiIT]ON [>urchasing Office' 1',0, GO ), 339 5, University Station LilriJlJ1ie, Wyoming 82071 (307) 721-2272 PURCHASE ORDER

U;I!VLf.~S[TY (F

PURCHASE ORDER NUM!3ER _ _~5_5~ __ . . The purcha se order number mils t appenr on rlocwnents and paclages.

vrrmop.: Croll-Reynolds Attn: Mike Holzman P.O. Box 668 Westfield, NJ 07091 (201) 232-4200 Qi-de r

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of

Page

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1

SHIP TO: WESTERN RESEARCH INSTITUTE Univer sity of Wyoming Research Corp , 9th & Lewi s Lal-amie, WY 82070

DaTe

- - i F,O,B, IhiPYl"il _ ~.9L05L8.L ,___ ;J,eterboro, NJ ___-'--_*._S_e_e ]ternF Articles/Sel'v}~i"_s_ . _ __

--1.-, ~;-Si-gn -a-n~

fabricate a small variable throat venturi scrubber and cyclone separator for the North Site I Retort Facili ty 10-ton Retort Bui ldin g addition, Mode l 226HE, The Venturi is to be des i gned to opera te over the range of the following parameters: I Gas Flow 10-35 scfm 20 -23 scfm (average) Ga 5 Tel'lpera ture 110-140 oF Gas Pressure 3 psig (average), 15 psig (maximum) Barometric Pressure 580 nln Hg (ilverage) Upstream Pipe Interface 2 in, schedule 40 Downstream Pipe Interface 2 in. schedule 40 Liquid/Gas Ratio Range 10:1 to 35:1 Liquid In jectio n Method Jets Scrubbing Liquid Caustic or amine solut ons I'li th some inorganic sa ts present and moderate tb high lev els of ammonia i Gas BTU Content 130-175 BTU/SCF Dia meter of separator 6", 316 stainless steel Couplings between the scrubber and the separator of a standard pipe size and thread such that the venturi can

I LL TO:

\1, ', L'. l'n Rescarch Inst i tute Uni'/L'rJity of Wyo l"ing Rese'll'ch C--,rporJtiul1 hUn: rinance Officc 1' , 0, () X 3395 , Il niv cr sily SLG' 10il

Lilt'a:, ["

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y r'I I'()IJIl'

ea.

$2640,00

, $2640,00 I

I

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___>{(/Ul (I 'd¥.idt~ Pam C. Lebeda, Contracting Officer

PageZo f 2 40555

be placed 1n different physical relationship to the separator. *Delivery Schedule: Drawings - 3 weeks ARO Scrubber &Separator - 9 weeks after receipt of approved drawings.

CONFIRMING ORDER PLACED WITH MIKE HOLZMAN BY PAM LEBEDA, 10/5/83.

/....-."

L __ .

r .

November 11, 1983

Roger Ki rchheim Croll-Reynolds Company, Inc. 751 Central .I\venue P. O. Box 668 Westfield, NJ 07091 Dear Nr. Kirchheim: This letter is an addendum to our PUrchase Order No. 40555. Please reference you Order No. 60178 for a venturi scrubber and cyclone separator. As per our telephone conversation, we are transmitting the revised design specifications below. Design Specifications Gas Composition (mole percent) Nitrogen Carbon Dioxide Hydrogen Water Vapor Hydrocarbons

65% 25% 3% 2% 5%

Gas Flow Rate 10-35 scfm

Temperature Operating Design Pressure (1oerating Design

3 psig -3 psig to 15 psig

Scrubbing Liquor Composition (Any of the following) L~ater

Caustic Solution - 5% Ammonia Solution - 5% Amine Solution - 15% Scrubbing Liquor Temperature 40

0

r to 140°F

letter to Kirchheim Page 2 November 11, 1983 Barometric Pressure 580

IlIll

HG

liquid to Gas Ratio 10 to 35 gallons per 1000 scf Approximate BTU Content of the Gas 30 BTU per scf Other Specifications Vendor to specify particulate removal efficiency. Vendor to supply WRI with operating data verses particulate removal efficiency, if available. Vendor to guarantee specified particulate removal efficiency for specified operatng conditions. Vendor to specify liquid inlet nozzles and provide specification details to WRI including nozzle make and orientation. Vendor to supply required pressure verses atomization data for nozzles. Vendor to specify variable throat specifications and supply recommended operating conditions to WRI. Vt'!ndor to specify the required liquid level in the separator. Vendor to supply correlations of throat valve position verses pressure drop to i~RI, if possible. The materials of construction for all components will be stainless steel. The type of stainless steel will be specified by the vendor. A two inch mist eliminator in the separator is to be considered standard equipment. Couplings between the scrubber and the separator to be flanged to allow rotation in 90° increments.

Letter to Kirchheim Page 3 November II, 1983 These revised design specifications mainly involve the acquisition of information and are not substantive changes in the design parameters. Please contact me if you have questions regarding these specifications or if these changes will cause a change in the purchase price so that we can submit a revised purchase order. Sincerely,

Thomas E. Owen cc:

Purchasing J. Weber ~ McTernan VB. P1 ace A. long

bee:

Hrtter

>~S File

TE'J:ja

VENTURI SCRUBBER AND CYCLONE DESIGN CRITERIA Gas Flow

10-35 scfm 20-23 scfm (average)

Gas Temperature

110-140°F

Gas Pressure

3 psig(average), 15 psig (maximum)

Barometric Pressure

580' mm Hg (average)

Upstream Pipe Interface

4 in. schedule 40

\/"",

Downstream Pipe Interface

-'

,)

,i'

4 in. schedule 40

Liquid/Gas Ratio Range

10:1

Liquid Injection Method

Jets

Scrubbing Liquid

Caustic or amine solutions with some inorganic salts present and moderate to high levels of ammonia.

Gas BTU Content

to 35:1

"

--tj-6-175 BTtJ/SCF /

f(P

SPECIAL FEATURES

Couplings between the scrubber and the separator of a standard pipe size and thread such that the ventrui can be placed in different physical relationship to the separator.

L

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SIZE

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LETTER

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DIMENSIONS

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Apr .4PPNI7Y1HAT£ CROll-REYNOLDS COMPANY, INC. WESTFIELD. N.J. PR.OPOSED c;: C-...o.

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stomer Refere nce _ _ _ _4!.....=.()~S_=_S_S_________

PROPOS!\L

CROLL-REYNOLDS HIGH ENERGY SCRUBBER SYSTEM

C.R. ' Inqu iry' _ _ _ _ __

Gas Inlet/Ou tlet: _ _~J",,--_"-:--zr___ I_=..2=--'I _____

.kif

Scrubbing Liquid Inlet: _ _ _-'ilL-_ _ _ _--.,-_ _ _ _ __

0

Separator Tank: Not Included 1/

Tank Diameter: " liquid Storage: YES 0 Manway:

'0 :

none

Separator Type:

g

NO

Tank Connections: Drain

Included

Ii'

LJ 1_ /' f f Tank Height: _-'L'--..'!(o'--_ _ Capacity _ __ _ Gallons

/N, Fill _ _ _ , Overflow _ __

Ill} esLII

P/t0f e

l 'h

Other:

MATERIALS OF CONSTRUCTION ITEM

MATERIAL

Venturi : _ _ _..>.3""/-"C'--'L'--'S""'-' ,~ < _ _ _ _ _ _ _ _ _ _ __ _ Separator: _ _ _

JlODEL

C

£

;)J.. L'-!=._ _ __ N O .:~---==---..-.:~'1

_ _---C(j~I]t1... ('_--:--:-_ ' RICE PER SYSTEM: 2. r; '10

} UANTITY~

_ __ I

_ __

: .0.8.: _ _.....-I..< .s~h~,'f~c.s~-_~Ti. ~(~f,.::Jr:(:2.p~c:c..:,rc~J,!/II =J_ rERMS: Net 30 0'"

:STIMATED DELIVERY: /0 W eeks after drawing

:=. .s~ t(fI1 ~(:...-

3/ ( 5 .S

Spray Nozzles:

APPROXIMATE WEIGHT Shipping: _ _ __ _ _ _ _ _ Operating : _ _ _ _ _ __

CONSTRUCTION DESIGN In general accordance with :

approval

_ _ _ _ W eeks for drawings

PS 15·69 0

ASME Code

Wall Thicknesses :

] ITEM IN STO CK D ELIVERY: _ _ _ _ Weeks after receipt of order )eliveries are quoted bas ed on standard shop schedules at the ime of quote. Special delivery requirements may be accommo· jated on an individual basis.

ri

ASME Code Stamped 0 l 0 f5 fJ

/_-=--=-__

or vessel designed for pressure/vacuum {"

Head Thicknesses Bottom/ Top

fI1 i1 ,

1: /.1 1'11 11

~

Hold down lugs

Included

Lifting lugs

Included 0

I

I It

-g

/'11 1 '1

No t Included 0 Not Included

e:(

OPERATING CONDITIONS 10 - sS, :kIm Gas Inlet Temperature , I/O - /'/0 ~;1 Pressure, 3 - I'::'~"s , j Iq/I,,? = S 8'O 'i1~ Ga s Inlet Rate Capaci ty,

COMMENTS:

3i nce Croll-Reynolds Co. has no control over the mixture or ;oncen tration of corrosive elements which may come in contact Nith the equipment supplied, no guarantee is expressed or mplied concerning materials of construction for corrosion 'esistance, :::ustomer should make every effort to assure suitability prio. to )urchase.

Ga s CompoS ition .

J)~/!/ ( . /c, 1t

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(,04/1

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,

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M/..st

Ltc;Ul d Inlet Rate: 0 • (, gpm Pres sure at Liquid Inlet ,_---'/....:S_ __

pSlg

x Speclftc Gravity' liquid Composition _~w""'-"Gu.~-"e'"-r_ _ __ _ _ __ _ __ MaXimum liquid Temperature:

_ .L)-=o'-'o"'--°LE________

Pressure Drop Across System ' _.-:3:;f~.:;,'.{/.;:j\_-~..l~a~liIiliiA.'IIN.~)oIilAIfJ,:€G ,

P roposal by : Date:

?J.;J ~~--.

1:/3/ t-3 I

PERFORMANCE DATA % Removal efficiency of --=:0:<.::.. . -'.7_.s---,-/V7.:..c.:.:/c"::.!.r-=Cf:;."'~~ _ _ _ ___ % Removal efficie~cy of pqr//(vj",f(!

'10,-

,

_ _ _ _ __ % Removal efficiency of _ _ _ _ _ __ _ _ __ _ __ % Removal efficiency of _ _ __ _ __ _ _ _ _ _ _ _ % Removal efficiency of _ _ _ _ _ _ __ Based upon above operating conditions