Basic Hydraulic Principles - Dynatech

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CHAPTER

1 Basic Hydraulic Principles 1.1 General Flow Characteristics In hydraulics, as with any technical topic, a full understanding cannot come without first becoming familiar with basic terminology and governing principles. The basic concepts discussed in the following pages lay the foundation for the more complex analyses presented in later chapters.

Flow Conveyance Water travels downhill from points of higher energy to points of lower energy (unless forced to do otherwise) until it reaches a point of equilibrium, such as an ocean. This tendency is facilitated by the presence of natural conveyance channels such as brooks, streams, and rivers. The water’s journey may also be aided by man-made structures such as drainage swales, pipes, culverts, and canals. Hydraulic concepts can be applied equally to both man-made structures and natural features.

Area, Wetted Perimeter, and Hydraulic Radius The term area refers to the cross-sectional area of flow within a channel. When a channel has a consistent cross-sectional shape, slope, and roughness, it is called a prismatic channel. If the flow in a conveyance section is open to the atmosphere, such as in a culvert flowing partially full or in a river, it is said to be open-channel flow or free-surface flow. If a channel is flowing completely full, as with a water distribution pipe, it is said to be operating under full-flow conditions. Pressure flow is a special type of full flow in which forces on the fluid cause it to push against the top of the channel as well as the bottom and sides. These forces may result from, for example, the weight of a column of water in a backed-up sewer manhole or elevated storage tank. A section’s wetted perimeter is defined as the portion of the channel in contact with the flowing fluid. This definition is illustrated in Figure 1-1.

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Figure 1-1: Flow Area and Wetted Perimeter

The hydraulic radius of a section is not a directly measurable characteristic, but it is used frequently during calculations. It is defined as the area divided by the wetted perimeter, and therefore has units of length. The hydraulic radius can often be related directly to the geometric properties of the channel. For example, the hydraulic radius of a full circular pipe (such as a pressure pipe) can be directly computed as: R=

A Pw

or Rcircular =

where

R A Pw D

= = = =

π ⋅ D2 4 D = π ⋅D 4

hydraulic radius (m, ft) cross-sectional area (m2, ft2) wetted perimeter (m, ft) pipe diameter (m, ft)

Velocity As shown in Figure 1-2, the velocity of a section is not constant throughout the crosssectional area. Instead, it varies with location. The velocity is zero where the fluid is in contact with the conduit wall.

Figure 1-2: Velocity Distribution

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Basic Hydraulic Principles

Chapter 1

The variation of flow velocity within a cross-section complicates the hydraulic analysis, so the engineer usually simplifies the situation by looking at the average (mean) velocity of the section for analysis purposes. This average velocity is defined as the total flow rate divided by the cross-sectional area, and is in units of length per time.

V = Q/ A where

V Q A

= = =

average velocity (m/s, ft/s) flow rate (m3/s, ft3/s) area (m2, ft2)

Steady Flow Speaking in terms of flow, the word steady indicates that a constant flow rate is assumed throughout an analysis. In other words, the flow velocity does not change with respect to time at a given location. For most hydraulic calculations, this assumption is reasonable. A minimal increase in model accuracy does not warrant the time and effort that would be required to perform an analysis with changing (unsteady) flows over time. When analyzing tributary and river networks, storm sewers, and other collection systems in which it is desirable to vary the flow rate at different locations throughout the system, the network can often be broken into segments that can be analyzed separately under steady flow conditions.

Laminar Flow, Turbulent Flow, and Reynolds Number Laminar flow is characterized by smooth, predictable streamlines (the paths of single fluid particles). An example of this type of flow is maple syrup being poured. In turbulent flow, the streamlines are erratic and unpredictable. Turbulent flow is characterized by the formation of eddies within the flow, resulting in continuous mixing throughout the section (see Figure 1-3).

Figure 1-3: Instantaneous Velocity Distributions for Laminar and Turbulent Flow

Eddies result in varying velocity directions as well as magnitudes (varying directions not depicted in Figure 1-3 for simplicity). At times, the eddies contribute to the velocity of a

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Computer Applications in Hydraulic Engineering given particle in the direction of flow, and at other times detract from it. The result is that velocity distributions captured at different times will be quite different from one another, and will be far more chaotic than the velocity distribution of a laminar flow section. By strict interpretation, the changing velocities in turbulent flow would cause it to be classified as unsteady flow. Over time, however, the average velocity at any given point within the section is essentially constant, so the flow is assumed to be steady. The velocity at any given point within the turbulent section will be closer to the mean velocity of the entire section than with laminar flow conditions. Turbulent flow velocities are closer to the mean velocity because of the continuous mixing of flow, particularly the mixing of low-velocity flow near the channel walls with the higher-velocity flow toward the center. To classify flow as either turbulent or laminar, an index called the Reynolds number is used. It is computed as follows: Re =

where

4VR

Re = V = R = ν =

ν

Reynolds number (unitless) average velocity (m/s, ft/s) hydraulic radius (m, ft) kinematic viscosity (m2/s, ft2/s)

If the Reynolds number is below 2,000, the flow is generally laminar. For flow in closed conduits, if the Reynolds number is above 4,000, the flow is generally turbulent. Between 2,000 and 4,000, the flow may be either laminar or turbulent, depending on how insulated the flow is from outside disturbances. In open channels, laminar flow occurs when the Reynolds number is less than 500 and turbulent flow occurs when it is above 2,000. Between 500 and 2,000, the flow is transitional. Example 1-1: Flow Characteristics A rectangular concrete channel is 3 m wide and 2 m high. The water in the channel is 1.5 m deep and is flowing at a rate of 30 m3/s. Determine the flow area, wetted perimeter, and hydraulic radius. Is the flow laminar or turbulent? Solution From the section’s shape (rectangular), we can easily calculate the area as the rectangle’s width multiplied by its depth. Note that the depth used should be the actual depth of flow, not the total height of the cross-section. The wetted perimeter can also be found easily through simple geometry. A = 3.0 m × 1.5 m = 4.5 m2 Pw = 3.0 m + 2 × 1.5 m = 6.0 m

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Basic Hydraulic Principles

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R = A / Pw = 4.5 m2 / 6.0 m = 0.75 m In order to determine whether the flow is likely to be laminar or turbulent, we must determine the Reynolds number. To do this, first find the velocity of the section and a value for the kinematic viscosity. V = Q / A = 30 m3/s / 4.5 m2 = 6.67 m/s From fluids reference tables, we find that the kinematic viscosity for water at 20°C is 1.00 × 10-6 m2/s. Substituting these values into the formula to compute the Reynolds number results in Re = (4 × 6.67 m/s × 0.75 m) / (1.00×10-6) = 2×107 This value is well above the Reynolds number minimum of 4,000 for turbulent flow.

1.2 Energy The Energy Principle The first law of thermodynamics states that for any given system, the change in energy (∆E) is equal to the difference between the heat transferred to the system (Q) and the work done by the system on its surroundings (W) during a given time interval. The energy referred to in this principle represents the total energy of the system, which is the sum of the potential energy, kinetic energy, and internal (molecular) forms of energy such as electrical and chemical energy. Although internal energy may be significant for thermodynamic analyses, it is commonly neglected in hydraulic analyses because of its relatively small magnitude. In hydraulic applications, energy values are often converted into units of energy per unit weight, resulting in units of length. Using these length equivalents gives engineers a better “feel” for the resulting behavior of the system. When using these length equivalents, the engineer is expressing the energy of the system in terms of “head.” The energy at any point within a hydraulic system is often expressed in three parts, as shown in Figure 1-4: ƒ Pressure head p γ ƒ Elevation head z ƒ Velocity head V 2 2 g where

p

γ

z V

= = = =

pressure (N/m2, lbs/ft2) specific weight (N/m3, lbs/ft3) elevation (m, ft) velocity (m/s, ft/s)

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Figure 1-4: The Energy Principle

Note that a point on the water surface of an open channel will have a pressure head of zero, but will have a positive elevation head higher than that of a point selected at the bottom of the channel for the same station.

The Energy Equation In addition to pressure head, elevation head, and velocity head, energy may be added to a system by a pump (for example), and removed from the system by friction or other disturbances. These changes in energy are referred to as head gains and head losses, respectively. Because energy is conserved, the energy across any two points in the system must balance. This concept is demonstrated by the energy equation: p1

γ

where

p

γ

z V g HG HL

+ z1 +

= = = = = = =

p V2 V12 + H G = 2 + z2 + 2 + H L γ 2g 2g

pressure (N/m2, lb/ft2) specific weight of the fluid (N/m3, lb/ft3) elevation above a datum (m, ft) fluid velocity (m/s, ft/s) gravitational acceleration (m/s2, ft/s2) head gain, such as from a pump (m, ft) combined head loss (m, ft)

Hydraulic Grade The hydraulic grade is the sum of the pressure head (p/γ) and elevation head (z). For open channel flow (in which the pressure head is zero), the hydraulic grade elevation is the same as the water surface elevation. For a pressure pipe, the hydraulic grade represents the height to which a water column would rise in a piezometer (a tube open to the atmosphere rising from the pipe). When the hydraulic grade is plotted as a profile along the length of the conveyance section, it is referred to as the hydraulic grade line, or HGL.

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Basic Hydraulic Principles

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Energy Grade The energy grade is the sum of the hydraulic grade and the velocity head (V2/2g). This grade is the height to which a column of water would rise in a Pitot tube (an apparatus similar to a piezometer, but also accounting for fluid velocity). When plotted in profile, this parameter is often referred to as the energy grade line, or EGL. For a lake or reservoir in which the velocity is essentially zero, the EGL is equal to the HGL.

Energy Losses and Gains Energy (or head) losses (HL) in a system are due to a combination of several factors. The primary cause of energy loss is usually the internal friction between fluid particles traveling at different velocities. Secondary causes of energy loss are localized areas of increased turbulence and disruption of the streamlines, such as disruptions from valves and other fittings in a pressure pipe, or disruptions from a changing section shape in a river. The rate at which energy is lost along a given length of channel is called the friction slope, and is usually presented as a unitless value or in units of length per length (ft/ft, m/m, etc.). Energy is generally added to a system with a device such as a pump. Pumps are discussed in more detail in Chapter 6. Example 1-2: Energy Principles A 1,200-mm diameter transmission pipe carries 126 l/s from an elevated storage tank with a water surface elevation of 540 m. Two kilometers from the tank, at an elevation of 434 m, a pressure meter reads 586 kPa. If there are no pumps between the tank and the meter location, what is the rate of head loss in the pipe? (Note: 1 kPa = 1,000 N/m2.) Solution Begin by simplifying the energy equation. Assume that the velocity within the tank is negligible, and that the pressure head at the tank can be discounted because it is open to the atmosphere. Rewriting the energy equation and entering the known values, we can solve for head loss. The velocity can be calculated using the flow rate and pipe diameter. Q = 126 l/s × (1 l/s / 103 m3/sec) = 0.126 m3/s A = π × (0.6 m)2 = 1.13 m2 V = Q/A = 0.126 m3/s / 1.13 m2 = 0.11 m/s V2/2g = (0.11 m/s)2 / (2 × 9.81 m/s2) = 0.0006 m (negligible) Neglecting the velocity simplifies the energy equation even further, and we can now solve for head loss as HL = 540 m – 434 m – (586,000 N/m2) / 9810 N/m3 = 46.27 m

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The rate of head loss (or friction slope) can now be computed as Friction slope = 46.27 m / (2 × 1000 m) = 0.023 m/m, or 23 m/km

1.3 Orifices and Weirs The energy equation serves as the foundation for calculating the flow through and over hydraulic structures based on the size of the opening associated with the structure and the difference in energy on either side of it. The flow exiting the structure can be calculated by solving the energy equation for velocity, V2, and multiplying the resulting formula by the flow area and a coefficient to account for different hydraulic and physical variables. These variables include: head loss, the shape and nature of the opening, the contraction of the flow after it leaves the structure, and countless indefinable variables that are difficult to measure but produce quantifiable effects. Two common devices for which equations are derived in this manner are weirs and orifices. They are important not only because of their widespread usage in the industry, but also because the equations that describe them serve as the foundation for mathematical descriptions of more complicated hydraulic devices such as drainage inlets and culverts.

Orifices Orifices are regularly shaped, submerged openings through which flow is propelled by the difference in energy between the upstream and downstream sides of the opening. The stream of flow expelled from the orifice is called the jet. When the jet exits the orifice, adverse velocity components cause it to contract to a point after which the flow area remains relatively constant and the flow lines become parallel (see Figure 1-5). This point is called the vena contracta.

Figure 1-5: Cross-Sectional View of Typical Orifice Flow

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Basic Hydraulic Principles

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Orifices and the orifice equations have the following applications: ƒ Regulating the flow out of detention ponds ƒ Regulating the flow through channels in the form of radial and sluice gates ƒ Approximating the interception capacity of submerged drainage inlets in sag (see Chapter 3) ƒ Approximating the flow allowed through a submerged culvert operating under inlet control (see Chapter 4) ƒ Measuring flow Example 1-3: The Orifice Equation For the structure in Figure 1-6, derive the orifice equation for an orifice of area A.

Figure 1-6: Orifice Example

Solution First, start with the energy equation from Section 1.2:

p1

γ

+ z1 +

V12 p V2 + H G = 2 + z2 + 2 + H L γ 2g 2g

List known variables and assumptions:

ƒ The datum is at the centerline/centroid of the orifice ƒ ƒ ƒ ƒ

p1/γ = H Point 2 occurs at the vena contracta Elevation heads, z1 and z2, are equivalent The velocity in the tank at point 1 is negligible

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Computer Applications in Hydraulic Engineering ƒ The jet is open to the air, so the pressure at point 2 is 0 ƒ There is no head gain Taking these known variables and assumptions into account and solving for V2, the energy equation becomes: V2 = 2 g ( H − H L )

To find the flow exiting the structure at point 2, multiply both sides of the equation by the orifice area, A. AV2 = Q = A 2 g (H − H L )

where

Q = discharge (m3/s, ft3/s)

The point of discharge is the vena contracta, where the flow area is usually contracted from the original orifice area. Also, computations can be simplified by eliminating the head loss term, HL. Both of these variables are accounted for by applying an orifice coefficient, C, to the right side of the equation. The final form of the orifice equation becomes: Q = CA 2 gH

where

C = orifice coefficient

When dealing with storm sewer design, the orifice coefficient is generally about 0.6. For more in-depth information on orifice coefficients for different situations, see Brater and King’s Handbook of Hydraulics (1996).

Weirs Weirs are notches or gaps over which fluid flows. The lowest point of structure surface or edge over which water flows is called the crest, whereas the stream of water that exits over the weir is called the nappe. Depending on the weir design, flow may contract as it exits over the top of the weir, and, as with orifices, the point of maximum contraction is called the vena contracta.

Figure 1-7: Front View of Common Weir

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Basic Hydraulic Principles

Chapter 1

This contraction can be counteracted or suppressed by designing the weir such that its shape conforms to the shape of the channel. This type of weir is called a suppressed weir. With a contracted weir, the crest and nappe vary from the channel to such a degree that a significant contraction of flow area does occur. In addition to suppressed and contracted weir types, weirs are also distinguished as either sharp-crested or broad-crested. A sharp-crested weir has a sharp upstream edge formed so that the nappe flows clear of the crest. Broad-crested weirs have crests that extend horizontally in the direction of flow far enough to support the nappe and fully develop hydrostatic pressures for at least a short distance. Weirs can also be distinguished by their shapes. The most common shapes are shown in Figure 1-8. The effects of weir shape and other factors previously mentioned are accounted for with modifications to the weir equation (derived in Example 1-4), such as adjustments the weir coefficient. Table 1-1 contains information on coefficients for VNotch weirs.

Figure 1-8: Standard Weirs Table 1-1: V-Notch Weir Coefficients of Discharge — English Units Weir Angle, θ (degrees) 22.5 30 45 60 0.5 0.611 0.605 0.569 0.590 1.0 0.593 0.590 0.583 0.580 1.5 0.586 0.583 0.578 0.575 2.0 0.583 0.580 0.576 0.573 2.5 0.580 0.578 0.574 0.572 3.0 0.579 0.577 0.574 0.571 Derived from: Van Havern, Bruce P. Water Resources Measurements, American Water Works Association, 1986 Head (feet)

90 0.584 0.576 0.572 0.571 0.570 0.570

120 0.581 0.575 0.572 0.571 0.570 0.570

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Computer Applications in Hydraulic Engineering Weirs have the following applications: ƒ Serving as emergency spillways for regulating high-return event flows overtopping dams and detention ponds ƒ Regulating the flow in channels ƒ Measuring flow ƒ Approximating the flow over roadways acting as broad-crested weirs when flow exceeds a culvert’s capacity (see Chapter 4) ƒ Approximating the interception capacity of unsubmerged drainage inlets in swales (see Chapter 3) ƒ Approximating the flow allowed through an unsubmerged culvert operating under inlet control (see Chapter 4) Example 1-4: The Weir Equation Derive the weir equation for the rectangular weir with a crest of length L and head H, which discharges from a free outfall as shown in Figure 1-9.

Figure 1-9: Weir Example

Solution Begin with the energy equation: p1

γ

12

+ z1 +

p V2 V12 + H G = 2 + z2 + 2 + H L γ 2g 2g

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List the known variables and assumptions: ƒ The datum is at the crest of the weir ƒ P1/γ = H ƒ Point 2 occurs at the vena contracta ƒ The elevation heads, z1 and z2, are equivalent ƒ Assume the velocity in the tank is negligible ƒ The exiting stream pressure at point 2 is 0 ƒ There is no head gain After applying the known variables and assumptions and solving for V2, the energy equation becomes:

V2 = 2 g ( H − H L ) To find the flow, multiply both sides by the flow area, which in this case is the length of the weir, L, multiplied by the height of the head, H. AV2 = Q = LH 2 g ( H − H L )

where

L = length of weir (m, ft)

To account for head loss, contraction, and other variables, a weir coefficient, C, is applied. Unlike the orifice coefficient, the weir coefficient generally takes into account the constant 2g. Finally, the weir equation becomes:

Q = CLH where

3

2

C = weir coefficient (m1/2/s, ft1/2/s)

Unlike the orifice coefficient, the weir coefficient is not unitless. Care has to be taken that the correct coefficient is applied when using a specific unit system. Typical coefficient values for different weir shapes, as well as variations in the equation, can be seen in the previously shown Figure 1-8.

1.4 Friction Losses There are many equations that approximate the friction losses associated with the flow of a liquid through a given section. Commonly used methods include: ƒ ƒ ƒ ƒ

Manning’s equation Chézy’s (Kutter’s) equation Hazen-Williams equation Darcy-Weisbach (Colebrook-White) equation

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Computer Applications in Hydraulic Engineering These equations can be described by a generalized friction equation: V = kCR x S y

where

V C R S x, y k

= = = = = =

mean velocity flow resistance factor hydraulic radius friction slope exponents factor to account for empirical constants, unit conversion, etc.

The lining material of the flow channel usually determines the flow resistance or roughness factor, C. However, the ultimate value of C may be a function of the channel shape, depth, and fluid velocity.

Manning’s Equation Manning’s equation is the most commonly used open channel flow equation. The roughness component, C, is typically assumed to be constant over the full range of flows and is represented by a Manning’s roughness value, n. These n-values have been experimentally determined for various materials and should not be used with fluids other than water. Manning’s equation is:

where

V =

k 23 12 R S n

V k n R S

= = = = =

mean velocity (m/s, ft/s) 1.49 for U.S. customary units, 1.00 for SI units Manning’s roughness value hydraulic radius (m, ft) friction slope (m/m, ft/ft)

Chézy’s (Kutter’s) Equation The Chézy equation, in conjunction with Kutter’s equation, is widely used in sanitary sewer design and analysis. The roughness component, C, is a function of the hydraulic radius, friction slope, and lining material of the channel. The Chézy equation is: V = C RS

where

V C R S

= = = =

mean velocity (m/s, ft/s) roughness coefficient (see following calculation) hydraulic radius (m, ft) friction slope (m/m, ft/ft)

The roughness coefficient, C, is related to Kutter’s n through Kutter’s equation. Note that the n-values used in Kutter’s equation are actually the same as Manning’s n coefficients.

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U.S. Standard Units 0.00281 1.811 41.65 + + S n C= 0.00281    41.65 + n S   1+ R

where

C n R S

= = = =

S.I. Units C=

0.00155 1 + S n 0.00155    23 + n S  

23 +

1+

R

roughness coefficient Manning’s roughness value hydraulic radius (m, ft) friction slope (m/m, ft/ft)

Hazen-Williams Equation The Hazen-Williams equation is most frequently used in the design and analysis of pressure pipe systems. The equation was developed experimentally, and therefore should not be used for fluids other than water (and only within temperatures normally experienced in potable water systems). The Hazen-Williams equation is:

V = kCR 0.63 S 0.54 where

V k C R S

= = = = =

mean velocity (m/s, ft/s) 1.32 for U.S. customary units, or 0.85 for SI units Hazen-Williams roughness coefficient (unitless) hydraulic radius (m, ft) friction slope (m/m, ft/ft)

Darcy-Weisbach (Colebrook-White) Equation The Darcy-Weisbach equation is a theoretically based equation commonly used in the analysis of pressure pipe systems. It applies equally well to any flow rate and any incompressible fluid, and is general enough to be applied to open channel flow systems. In fact, the ASCE Task Force on Friction Factors in Open Channels (1963) supported the use of the Darcy-Weisbach equation for free-surface flows. This recommendation has not yet been widely accepted because the solution to the equation is difficult and not easily computed using non-computerized methods. With the increasing availability of computer solutions, the Darcy-Weisbach equation will likely gain greater acceptance because it successfully models the variability of effective channel roughness with channel material, geometry, and velocity.

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Computer Applications in Hydraulic Engineering The roughness component in the Darcy-Weisbach equation is a function of both the channel material and the Reynolds number, which varies with velocity and hydraulic radius. 8g RS f

V=

where

V g f R S

= = = = =

flow velocity (m/s, ft/s) gravitational acceleration (m/s2, ft/s2) Darcy-Weisbach friction factor (unitless) hydraulic radius (m, ft) friction slope (m/m, ft/ft)

The Darcy-Weisbach friction factor, f, can be found using the Colebrook-White equation for fully developed turbulent flow, as follows: Free Surface

Full Flow (Closed Conduit)

 k 2.51 = −2log +  12 R Re f f 

1

where

k = R = Re =

   

 k 2.51 = −2log +  14.8R Re f f 

1

   

roughness height (m, ft) hydraulic radius (m, ft) Reynolds number (unitless)

This iterative search for the correct value of f can become quite time-consuming for hand computations and computerized solutions of many pipes. Another method, developed by Swamee and Jain, solves directly for f in full-flowing circular pipes. This equation is: f =

where

f k D Re

1.325  5.74   k + 0.9  log e  3 . 7 D Re   

= = = =

2

friction factor (unitless) roughness height (m, ft) pipe diameter (m, ft) Reynolds number (unitless)

Typical Roughness Factors Typical pipe roughness values for each of these methods are shown in Table 1-2. These values will vary depending on the manufacturer, workmanship, age, and other factors. For this reason, the following table should be used only as a guideline.

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Basic Hydraulic Principles

Chapter 1

Table 1-2: Typical Roughness Coefficients Material Asbestos cement Brass Brick Cast-iron, new Concrete: Steel forms Wooden forms Centrifugally spun Copper Corrugated metal Galvanized iron Glass Lead Plastic Steel: Coal-tar enamel New unlined Riveted Wood stave

Darcy-Weisbach Roughness Height k (mm) k (ft) 0.0015 0.000005 0.0015 0.000005 0.6 0.002 0.26 0.00085

Manning’s Coefficient n 0.011 0.011 0.015 0.012

HazenWilliams C 140 135 100 130

0.011 0.015 0.013 0.011 0.022 0.016 0.011 0.011 0.009

140 120 135 135 − 120 140 135 150

0.18 0.6 0.36 0.0015 45 0.15 0.0015 0.0015 0.0015

0.006 0.002 0.0012 0.000005 0.15 0.0005 0.000005 0.000005 0.000005

0.010 0.011 0.019 0.012

148 145 110 120

0.0048 0.045 0.9 0.18

0.000016 0.00015 0.003 0.0006

1.5 Pressure Flow For pipes flowing full, many of the friction loss calculations are greatly simplified because the flow area, wetted perimeter, and hydraulic radius are all functions of pipe radius (or diameter). Table 1-3 presents the three pipe friction loss equations that are commonly used to design pressure pipe systems. There is much more information presented about pressure piping systems in Chapter 6, including further discussion on pumping systems, minor losses, and network analysis. Table 1-3: Three Pipe Friction Loss Equations Equation Darcy-Weisbach

Q (m3/s); D (m)

Sf =

0.083 fQ 2 D5

Q (cfs); D (ft)

Sf =

1.852

Hazen-Williams

Manning

Sf =

10.7  Q    D 4.87  C 

Sf =

10.3(nQ ) 2 D 5.33

0.025 fQ 2 D5

Q (gpm); D (in.)

Sf =

1.852

Sf =

4.73  Q    D 4.87  C 

Sf =

4.66( nQ) 2 D 5.33

0.031 fQ 2 D5 1.852

Sf =

10.5  Q    D 4.87  C 

Sf =

13.2( nQ) 2 D 5.33

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Computer Applications in Hydraulic Engineering

Example 1-5: Pressure Pipe Friction Losses Use the FlowMaster program to compare the head loss computed by the Hazen-Williams equation to the head loss computed by the Darcy-Weisbach equation for a pressure pipe having the following characteristics: 12-in diameter cast iron pipe (new) one mile in length with a flow rate of 1,200 gallons per minute (with water at 65°F). Solution Although there are no elevations or pressures given, these values are not needed to determine the head loss in the pipe. Setting up FlowMaster to solve for the “Elevation at 1” allows us to use zero elevation and zero pressure assumptions and fill in the rest of the pipe characteristics. For the Hazen-Williams equation, a C coefficient of 130 is assumed. This value results in 18.8 ft of head loss (which agrees with the computed 18.8-ft elevation at point 1). Using Darcy-Weisbach, a roughness height of 0.00085 ft is assumed. The solution indicates a head loss of 18.9 ft, which is only a 0.1-ft difference from the value predicted by HazenWilliams. Discussion If the same system is analyzed with 2,000 to 3,000 gallons per minute of flow, however, the difference in head loss between the two equations becomes almost 10 feet. Why such a big difference? For starters, the two methodologies are completely unrelated, and the estimated roughness coefficients were taken from a list of approximate values. If the Hazen-Williams equation is used with a roughness value of 125, the results are much closer. This difference should emphasize the fact that models are only as good as the data that is input into them, and the engineer needs to fully understand all of the assumptions that are being made before accepting the results.

1.6 Open-Channel Flow Open-channel flow analysis is more complex than pressure flow analysis because the flow area, wetted perimeter, and hydraulic radius are not necessarily constant as they are in a uniform pipe section under full-flow conditions. Because of this considerable difference, additional characteristics become important when dealing with open-channel flow.

Uniform Flow Uniform flow refers to the hydraulic condition in which the discharge and cross-sectional area (and therefore velocity) are constant throughout the length of the channel. For a pipe flowing full, the only required assumptions are that the pipe be straight and have no contractions or expansions. For an open channel, additional assumptions include:

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© 2002 Haestad Methods, Inc., 37 Brookside Rd., Waterbury, CT 06708. All rights reserved.

Basic Hydraulic Principles

Chapter 1

ƒ The depth of flow must be constant (that is, the hydraulic grade line must be parallel to the channel bed). This depth of flow is called normal depth. ƒ Because the velocity is constant, the velocity head does not change through the length of the section; therefore, the energy grade line is parallel to both the hydraulic grade line and the channel bed. In channels that are prismatic, the flow conditions will typically approach uniform flow if the channel is sufficiently long. When this occurs, the net force on the fluid approaches zero because the gravitational force is equal to the opposing friction forces from the channel bottom and walls. Example 1-6: Uniform Flow A concrete trapezoidal channel has a bottom width of 4 m and 45-degree side slopes. If the channel is on a 1-percent slope and is flowing at a depth of 1 m throughout its length, how much flow is being carried (use Manning’s equation)? How much flow would the same channel carry if it were a rectangular channel 4 m wide? Solution Because the channel is flowing at the same depth throughout, we can assume that normal depth has been achieved (that is, the friction slope is equal to the channel slope). We will assume a Manning’s n of 0.013 for concrete. From the trapezoidal geometry, we can easily calculate the area and wetted perimeter, and then the hydraulic radius, as follows: A = (4 m × 1 m) + 2 × (0.5 × 1 m × 1 m) = 5.00 m2 Pw = 4 m + 2 × (1 m × 20.5) = 6.83 m R = A / Pw = 5.00 m2 / 6.83 m = 0.73 m Manning’s equation for velocity can then be solved. The discharge can be computed as V = (1.00/0.013) × 0.732/3 × 0.011/2 = 6.25 m/s Q = V × A = 6.25 m/s × 5.00 m2 = 31.2 m3/s To answer the second part of the question, we simply repeat the steps for a rectangular section shape. A = (4 m × 1m) = 4 m2 Pw = 4 m + 2 × (1m) = 6 m R = 4 m2 / 6 m = 0.67 m V = (1.00/0.013) × 0.672/3 × 0.011/2 = 5.87 m/s Q = 5.87 m/s × 4 m2 = 23.5 m3/s

© 2002 Haestad Methods, Inc., 37 Brookside Rd., Waterbury, CT 06708. All rights reserved.

19

Computer Applications in Hydraulic Engineering As we would expect, this discharge is less than the discharge of the trapezoidal section.

Specific Energy and Critical Flow Of course, channels do not always flow at normal depth. If they did, it would make the engineer’s task quite simple. A more in-depth look at non-uniform flow is presented in Chapter 7, but this chapter will continue by focusing on another important concept — specific energy. For any flow section, the specific energy is defined as the sum of the depth of flow and the velocity head. E = y+

where

E y V g

= = = =

V2 2g

specific energy (m, ft) depth of flow (m, ft) mean velocity (m/s, ft/s) gravitational acceleration (m/s2, ft/s2)

If we assume the special case of an infinitely short section of open channel (with essentially no friction losses and no change in elevation), we see that the general energy equation can be reduced to an equality of specific energies. In other words, E1 = y1 +

V2 V12 = y2 + 2 = E2 2g 2g

Recall that the velocity of the section is directly related to the area of flow, and that the area of flow is a function of channel depth. This means that, for a given discharge, the specific energy at each point is solely a function of channel depth and more than one depth may exist with the same specific energy. If the channel depth is plotted against specific energy for a given flow rate, the result is similar to the graph shown in Figure 1-10.

Figure 1-10: Specific Energy

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© 2002 Haestad Methods, Inc., 37 Brookside Rd., Waterbury, CT 06708. All rights reserved.

Basic Hydraulic Principles

Chapter 1

As this figure shows, a depth exists for which the specific energy is at a minimum. This depth is called the critical depth. If the velocity is higher than critical velocity (that is, the depth is less than critical depth), the flow is considered supercritical. If the velocity is lower than critical velocity (the depth is greater than critical depth), the flow is subcritical. The velocity at critical depth is equal to the wave celerity—the speed at which waves will ripple outward from a pebble tossed into the water. A unitless value called the Froude number, F, represents the ratio of actual fluid velocity to wave celerity. The Froude number is computed as follows: F=

where

F D A T V g

V gD

= = = = = =

Froude number (unitless) hydraulic depth of the channel, defined as A/T flow area (m2,ft2) top width of flow (m, ft) fluid velocity (m/s, ft/s) gravitational acceleration (m/s2, ft/s2)

By definition, when the flow is at critical depth (that is, the velocity is equal to the wave celerity), the Froude number must be equal to 1. The equation can therefore be rewritten and re-factored to form the following equality: A3 Q 2 = T g

where

A T Q g

= = = =

flow area (m2, ft2) top width of flow (m, ft) channel flow rate (m3/s, ft3/s) gravitational acceleration (m/s2, ft/s2)

This equation can now be used to determine the depth for which this equality holds true, which is critical depth. For simple geometric shapes, the solution is relatively easy to determine. However, quite a few iterations may be required to find the solution for an irregularly shaped channel such as a natural streambed. In fact, several valid critical depths may exist for irregular channels. Example 1-7: Critical Depth What is the critical depth for a grassy triangular channel with 2H:1V side slopes and a 0.5% slope when the flow is 3.00 m3/s? If the channel is actually flowing at a depth of 1.2 m, is the flow critical, subcritical, or supercritical?

© 2002 Haestad Methods, Inc., 37 Brookside Rd., Waterbury, CT 06708. All rights reserved.

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Computer Applications in Hydraulic Engineering

Solution For such a simple geometry, we can quickly create a relationship between the flow area, top width, and depth of flow: T = 4y, A = 0.5 × T × y = 0.5 × 4y × y = 2y2 m2 Inserting these values into the previous equation for critical depth, we can algebraically solve for the channel depth: (2y2)3 / 4y = Q2 / g 8y6 / 4y = Q2 / g 2y5 = Q2 / g y5 = Q2 / 2g y = (Q2 / 2g)1/5 = [ (3.00 m3/s)2 / (2 × 9.8 m/s2) ]0.2 = (0.46 m5)0.2 = 0.86 m The critical depth for this section is 0.86 m. The actual flow depth of 1.2 m is greater than critical depth, so the flow is subcritical.

1.7 Computer Applications It is very important for students (and practicing engineers) to fully understand the methodologies behind hydraulic computations. Once these concepts are understood, the solution process can become repetitive and tedious—the type of procedure that is wellsuited to computer analysis. There are several advantages to using computerized solutions for common hydraulic problems: ƒ The amount of time to perform an analysis can be greatly reduced. ƒ Computer solutions can be more detailed than hand calculations. Performing a solution manually often requires many simplifying assumptions. ƒ The solution process may be less error-prone. Unit conversion and the rewriting of equations to solve for any variable are just two examples of mistakes frequently introduced with hand calculations. A well-tested computer program helps to avoid these algebraic and numeric errors. ƒ The solution is easily documented and reproducible. ƒ Because of the speed and accuracy of a computer model, more comparisons and design trials can be performed. The result is the exploration of more design options, which eventually leads to better, more efficient designs. In order to prevent an “overload” of data, this chapter deals primarily with steady-state computations. After all, an introduction to hydraulic calculations is tricky enough without throwing in the added complexity of a constantly changing system.

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© 2002 Haestad Methods, Inc., 37 Brookside Rd., Waterbury, CT 06708. All rights reserved.

Basic Hydraulic Principles

Chapter 1

The assumption that a system is under steady-state conditions is oftent perfectly acceptable. Minor changes that occur over time or irregularities in a channel cross-section are frequently negligible, and a more detailed analysis may not be the most efficient or effective use of time and resources. There are circumstances when an engineer may be called upon to provide a more detailed analysis, including unsteady flow computations. For a storm sewer, the flows may rise and fall over time as a storm builds and subsides. For water distribution piping, a pressure wave may travel through the system when a valve is closed abruptly (the same “waterhammer” effect can probably be heard in your house if you close a faucet quickly). As an engineer, it is important to understand the purpose of an analysis; otherwise, appropriate methods and tools to meet that purpose cannot be selected.

1.8 FlowMaster FlowMaster is an easy-to-use program that helps civil engineers with the hydraulic design and analysis of pipes, gutters, inlets, ditches, open channels, weirs, and orifices. FlowMaster computes flows and pressures in conduits and channels using common head loss equations such as Darcy-Weisbach, Manning’s, Kutter’s, and Hazen-Williams. The program’s flexibility allows the user to choose an unknown variable and automatically compute the solution after entering known parameters. FlowMaster also calculates rating tables and plots curves and cross-sections. You can view the output on the screen, copy it to the Windows clipboard, save it to a file, or print it on any standard printer. FlowMaster data can also be viewed and edited using tabular reports called FlexTables. FlowMaster enables you to create an unlimited number of worksheets to analyze uniform pressure-pipe or open-channel sections, including irregular sections (such as natural streams or odd-shaped man-made sections). FlowMaster does not work with networked systems such as a storm sewer network or a pressure pipe network. For these types of analyses, StormCAD, WaterCAD, or SewerCAD should be used instead. The theory and background used by FlowMaster have been reviewed in this chapter and can be accessed via the FlowMaster on-line help system. General information about installing and running Haestad Methods software can be found in Appendix A. FlowMaster replaces solutions such as nomographs, spreadsheets, and BASIC programs. Because FlowMaster gives you immediate results, you can quickly generate output for a large number of situations. For example, you can use FlowMaster to: ƒ Analyze various hydraulic designs ƒ Evaluate different kinds of flow elements ƒ Generate professional-looking reports for clients and review agencies

© 2002 Haestad Methods, Inc., 37 Brookside Rd., Waterbury, CT 06708. All rights reserved.

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Computer Applications in Hydraulic Engineering

1.9 Tutorial Example The following solution gives step-by-step instructions on how to solve an example problem using the FlowMaster computer program (included on the CD-ROM that accompanies this textbook) developed by Haestad Methods. Problem Statement Using Manning’s equation, design a triangular concrete channel with equal side slopes, a longitudinal slope of 5%, a peak flow capacity of 0.6 m3/s, and a maximum depth of 0.3. Also, design a concrete trapezoidal channel with equal side slopes and a base width of 0.2 that meets the same criteria. Create a cross-section of each channel and a curve of discharge versus depth for each channel. Assume the water is at 20°C. Solution ƒ Upon opening FlowMaster, click Create New Project in the Welcome to FlowMaster dialog. Enter a filename and click Save. ƒ Select Triangular Channel from the Create a New Worksheet dialog and click OK. ƒ In the Triangular Channel dialog, select Manning’s Formula from the Friction Method pull-down menu. Enter a label for the worksheet and click OK. ƒ Select Global Options from the Options menu and change the unit system to System International, if it has not already been done, by selecting it from the pull-down menu in the Unit System field. Click OK to exit the dialog. If you changed the unit system, you will be prompted to confirm the unit change. Click Yes. ƒ The worksheet dialog should appear. Because discharge, channel slope, and depth are given, the variable you need to solve for is the side slopes of the channel. Select Equal Side Slopes from the Solve for: menu at the top of the dialog. ƒ Enter the channel slope (you can change the units to percent by double left-clicking the units), depth, and discharge into the appropriate fields and select the Manning’s n for concrete. Click Solve. The equal side slopes should be 1.54 H:V. ƒ To design the trapezoidal section, first click Close on the triangular section worksheet to save it. Then click Create… at the bottom of the Worksheet List. ƒ Select the Trapezoidal Channel and OK. Repeat the same steps as before to design the triangular channel. The equal side slopes should be 0.80 H:V. Click Close to exit the worksheet. Creating Channel Cross-Sections ƒ Open the triangular section worksheet by highlighting it and selecting the Open button ƒ Click the Report button on the bottom of the triangular channel worksheet. Select Cross Section…

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© 2002 Haestad Methods, Inc., 37 Brookside Rd., Waterbury, CT 06708. All rights reserved.

Basic Hydraulic Principles

Chapter 1

ƒ Type a report title and click OK. Figure 1-11 provides a graphical representation of both the triangular and trapezoidal channel designs. Click Print if you want to print a copy of the cross-section. Click Close to exit the report. Finally, click Close to exit the triangular section worksheet. The drawing for the trapezoidal section cross-section is created in the same way from its worksheet.

Figure 1-11: Triangular and Trapezoidal Channel Designs

Creating Discharge versus Depth Curves ƒ Open the trapezoidal section worksheet by highlighting it and selecting the Open button. ƒ Because discharge needs to be on the y-axis (the ordinate) of the graph, you need to change Equal Side Slopes in the Solve For: field to Discharge. ƒ Click Report… at the bottom of the worksheet. Select Rating Curve from the dropdown list. ƒ In the Graph Setup dialog, select Discharge from the pull-down menu in the field labeled Plot. Select Depth from the pull-down menu in the field labeled vs. ƒ To scale the plot properly, make the minimum depth 0 m and the maximum depth 0.3 m. Choose an increment based on how smooth you want to make the curve. An increment of 0.1 will give the plot 3 points, an increment of 0.01 will give the plot 30 points, etc. Click OK. Click Print Preview at the top of the window, and then click Print to print out a report featuring the rating curve you have just created. Click Close to exit the Print Preview window. Click Close again to exit the Plot Window. ƒ Perform the same procedure for the triangular section. Your curves should match those in Figures 1-12 and 1-13.

© 2002 Haestad Methods, Inc., 37 Brookside Rd., Waterbury, CT 06708. All rights reserved.

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Computer Applications in Hydraulic Engineering

Figure 1-12: Comparison of Discharge versus Depth for the Trapezoidal Channel Design

Figure 1-13: Comparison of Discharge versus Depth for the Triangular Channel Design

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© 2002 Haestad Methods, Inc., 37 Brookside Rd., Waterbury, CT 06708. All rights reserved.

Basic Hydraulic Principles

Chapter 1

1.10 Problems Solve the following problems using the FlowMaster computer program. Unless stated otherwise, assume water is at 20°C. 1.

2.

The cross-section of a rough, rectangular, concrete (k = 0.2 × 10-2 ft) channel measures 6 × 6 ft. The channel slope is 0.02 ft/ft. Using the Darcy-Weisbach friction method, determine the maximum allowable flow rate through the channel to maintain one foot of freeboard (freeboard is the vertical distance from the water surface to the overtopping level of the channel). For these conditions, find the following characteristics (note that FlowMaster may not directly report all of these): a) Flow area b) Wetted perimeter c) Hydraulic radius d) Velocity e) Froude number A 450-mm circular concrete (n = 0.013) pipe constructed on a 0.6-percent slope carries 0.1 m3/s. a)

Using Manning’s equation and normal depth assumptions, what are the depth and velocity of flow?

b) What would the velocity and depth be if the pipe were constructed of corrugated metal (n = 0.024) instead of concrete? 3.

A trapezoidal channel carries 2.55 m3/s at a depth of 0.52 m. The channel has a bottom width of 5 m, a slope of 1.00 percent, and 2H:1V side slopes. a)

What is the appropriate Manning’s roughness coefficient?

b) How deep would the water be if the channel carried 5 m3/s? 4.

Use Manning’s equation to analyze an existing brick-in-mortar (n = 0.015) triangular channel with 3H:1V side slopes and a 0.05 longitudinal slope. The channel is intended to carry 7 cfs during a storm event. a)

If the maximum depth in the channel is 6 in, is the existing design acceptable?

b) What would happen if the channel were replaced by a concrete (n = 0.013) channel with the same geometry? 5.

A pipe manufacturer reports that it can achieve Manning’s roughness values of 0.011 for its concrete pipes, which is lower than the 0.013 reported by its competitors. Using Kutter’s equation, determine the difference in flow for a 310-mm circular pipe with a slope of 2.5% flowing at one-half of the full depth.

© 2002 Haestad Methods, Inc., 37 Brookside Rd., Waterbury, CT 06708. All rights reserved.

27

Computer Applications in Hydraulic Engineering 6.

A grass drainage swale is trapezoidal, with a bottom width of 6 ft and 2H:1V side slopes. Using the friction method you feel is appropriate, answer the following questions: a)

What is the discharge in the swale if the depth of flow is 1 ft and the channel slope is 0.005 ft/ft?

b) What would the discharge be with a slope of 0.010 ft/ft? 7.

A paved highway drainage channel has the geometry shown in the following figure. The maximum allowable flow depth is 0.75 ft (to prevent the flow from encroaching on traffic), and the Manning’s n-value is 0.018 for the type of pavement used.

Figure for Problem 7

a)

What is the capacity of the channel given a 2% longitudinal slope?

b) Create a rating curve to demonstrate how the capacity varies as the channel slope varies from 0.5% to 5%. Choose an increment that will generate a reasonably smooth curve. 8.

Using the Hazen-Williams equation, determine the minimum diameter of a new cast iron pipe (C = 130) for the following conditions: the upstream end is 51.8 m higher than the downstream end, which is 2.25 km away. The upstream pressure is 500 kPa, and the desired downstream pressure and flow rate are 420 kPa and 11,000 l/min, respectively. What is the minimum diameter needed? Assume pipes are available in 50-mm increments.

9.

2,000 gallons of water per minute flow through a level, 320-yard-long, 8-in-diameter cast iron pipe (C =130, k = 2.5908e-4 m) to a large industrial site. If the pressure at the upstream end of the pipe is 64 psi, what will the pressure be at the industry? Is there a significant difference between the solutions produced by the Hazen-Williams method and the Darcy-Weisbach method?

10. Develop a performance curve for the pipe in Problem 9 that shows the available flow to the industry with residual pressures ranging from 20 psi to 80 psi (assume the source can maintain 64 psi regardless of flow rate). Create similar curves for 10-in and 12-in diameter pipes and compare the differences in flow.

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© 2002 Haestad Methods, Inc., 37 Brookside Rd., Waterbury, CT 06708. All rights reserved.

Basic Hydraulic Principles

Chapter 1

11. Using the Darcy-Weisbach equation, find the minimum-sized of circular corrugated metal storm drain (k = 1.0 × 10-3 ft) that will carry 1.5 cfs with a maximum depth of 6 inches. The drain carries water down a hill 3 ft high to a pond with a free outfall 75 ft away. What pipe size should be used? Assume pipes are available in 3-in increments. What would the maximum capacity of this pipe be? What would the capacity of the pipe be when it is flowing full? 12. A channel with the cross-section shown in the following figure has a Manning’s coefficient of 0.040 from station 0 to station 3 and 0.054 from station 3 to station 8. The flow through the channel is 13 m3/s, and the water surface is 1.7 m high. Find the following: a)

Weighted Manning’s coefficient

b) Slope of the channel c)

Top width

d) Wetted perimeter e)

Flow regime (supercritical or subcritical)

Figure for Problems 12 & 13

13. A stream with the cross-section shown in the previous figure has a flow rate of 5 m3/s. The stream has a longitudinal slope of 0.002 m/m and a natural stony bottom (n = 0.050, stations 0 to 8). a)

Using Manning’s equation, what is the water surface elevation of the stream?

b) What is the maximum capacity of the channel? c)

How would the capacity of the channel be affected if you were to pave the center of the channel (n = 0.013) between stations 3 and 5?

14. A rectangular concrete channel with a width of 1 m and a height of 0.5 m is on a slope of 0.008 m/m. Design a concrete circular channel for which the depth is half of the diameter and the flow area is the same as that of the rectangular channel. Which channel is more efficient and by how much?

© 2002 Haestad Methods, Inc., 37 Brookside Rd., Waterbury, CT 06708. All rights reserved.

29

Computer Applications in Hydraulic Engineering 15. A weir is placed in a rectangular channel to measure the flow. The discharge from the rectangular channel enters a trapezoidal channel with a stony bottom. The trapezoidal channel is 0.50 m wide at the base with 2:1 (H:V) equal side slopes. The weir is a sharp-crested, v-notch weir with a crest 0.43 m above the channel bottom, a weir coefficient of 0.58, and a notch angle of 1.57 radians. The height of the water above the weir is 0.70 m, and the depth of water in the trapezoidal channel is measured to be 0.40 m. What is the flow rate? What is the slope of the trapezoidal channel (using Manning’s formula)? If the discharge is increased until the elevation of the water surface in the trapezoidal channel reaches 0.61 m, what will the headwater elevation be at the weir? 16. The outlet structure on a pond is used to regulate the flow out of the pond for different storm events. An outlet structure must be designed to discharge 2.20 m3/min when the water surface elevation in the pond reaches 1.52 m, and 6.29 m3/min when the water surface elevation reaches 2.60 m. The outlet structure will be a circular orifice and a sharp-crested rectangular weir combination, with the centroid of the orifice at an elevation of 0.90 m and the weir crest at an elevation of 2.50 m. Both will discharge to free outfall conditions. Assume an orifice coefficient of 0.6. Find the orifice diameter needed to supply the correct discharge when the water surface reaches the first specified elevation. What will the discharge from the orifice be when the water surface reaches the second specified elevation? Find the width of the weir needed to supply the extra discharge necessary to meet the requirement. Use Manning’s formula where necessary. 17. An approximately trapezoidal, clean, natural stream carries the discharge from a pond down a 0.001 slope. The maximum depth in the channel is 0.5 m. The channel has equal side slopes of 3.0 (H:V) and a bottom width of 1.0 m. The pond discharges water through a circular orifice into the channel. The centroid of the orifice is located 1.0 m above the bottom of the channel. Assume an orifice coefficient of 0.6. Design the orifice to discharge the maximum flow rate possible without exceeding the maximum allowed depth in the channel when the water surface in the pond reaches 4.6 m above the channel bottom. Use Manning’s formula when necessary.

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© 2002 Haestad Methods, Inc., 37 Brookside Rd., Waterbury, CT 06708. All rights reserved.

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Basic Hydraulic Principles - Dynatech

CHAPTER 1 Basic Hydraulic Principles 1.1 General Flow Characteristics In hydraulics, as with any technical topic, a full understanding cannot come wi...

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