FN(~Pl) = minfN(PN)+ FN_{~lpl) which is the required solution.
179
Unit Commitment
E 5.1 Obtain the economic schedule for the two units, the production costs of which are given as follows, to supply a load of 3MW, in steps of 1MW CI
=
0.8 p~ + 25
PI
C 2 = 1.2 p~ + 22 P2 Use dynamic programming method.
Solution: With usual notation 0.8 x 32 + 25 x 3
7.2 + 75
F I(3)
=
fP)
F2(3)
=
min [fiO) + fP), f 2(1) + FP), Fi2) + FI(I), F2(3) + FI(O)]
=
min [0 + 82.2, 1.2 + 22 + 3.2 + 50,4.8 + 44 + 0.8 + 25, 10.8 + 66 + 0]
=
=
=
82.2
= min [82.2, 76.4, 75.6, 76.8] = 74.6 The most economical combination is unit 1 supplying I MW and unit 2 supplying 2 MW. E 5.2 If only a load of 2MW is to be supplied, how are the units to be committed in E5. I.
Solution: F I(2)
= f l (2) = 0.8
x 22 + 25 x 2 = 53.2
Min [ fiO) + F I(2), fiO + FI(I), fi2) + FI(O)] = min [ 53.2, 1.2 + 22 + 0.8 + 25, 1.2 x 4 + 22 x 2 + 0] = min [ 53.2, 49, 48.8] F2(2) = fi2) + FI(O). Unit 2 supplies the total load for economy From the above two examples unit commitment schedule may be now prepared based on the two results. F2(2)
=
Load range
IMW 2 MW 3MW
Unit in service
2 2 2 and 3
For more accuracy the step size can be reduced to 0.5 MW, 0.2Mw or even 0.1 Mw. The smaller the step, size, the larger the calculations needed.
Operation and Control in Power Systems
180
Questions 5.1
Explain the unit commitment problem.
5.2
What are the various constraints in unit commitment problem? List them.
5.3
Explain the priority  list method for unit commitment.
5.4
Discuss the dynamic programming method to solve unit commitment problem in a power system.
5.5
What is the cost function for unit commitment? Explain
Problems P S.l Write down the recurrence functional relation for scheduling generation among 5
units to meet a load demand of IOMW in steps of 2MW. PS.2 Use dynamic programming method to determine the most economical units to be committed to supply a load of 6MW. There are three units with the following data C 1 = O.8 P1 2 + 22P1 C2
= O.85Pi + 21P2
C3 = O. 8P32 + 20P3
The maximum and minimum capacities of each unit are 5MW and I MW respectively. PS.3 Use dynamic programming method to determine the most economical units to be committed to supply a load of 9MW. There are four units with the following data.
The maximum and minimum limits for each unit are 6 MW and I MW respectively.
Unit Commitment
181
PS.4 Given the input output cost curves Fl (PI) = 21 + 6P1 + 0.00 I2P12 Rs/kcal
F2 (P2 )=2I+7P2 +O.OOIPi Rs/kcal FJ (PJ ) = 17 + 8PJ + 0.00 II pJ2 Rs/kcal The maximum and minimum values are 30:s P1:S 300 MW 40:s P2:S 400 MW 25:s P3:S 250 MW
Obtain unit commitment schedule for a load demand of 600MW.
6
LOAD FREQUENCY CONTROL
In a power system the load demand is continuously changing. In accordance with it the power input has also to vary. If the input  output balance is not maintained a change in frequency will occur. The control offrequency is achieved primarily through speed governor mechanism aided by supplementary means for precise control. At the outset, the speed governor mechanisms and its operation will be presented.
Governor: The power system is basically dependent upon the synchronous generator and its satisfactory performance. The important control loops in the system are :
(i) frequency control, and (ii) automatic voltage co"trol.
In this chapter the frequency control will be discussed. Frequency control is achieved through generator control mechanism. The governing systems for thermal and hydro generating plants are different in nature since, the inertia of water that flows into the turbine presents additional constrains which are not present with steam flow in a thermal plant. However, the basic principle is still the same; i.e. the speed of the shaft is sensed and compared with a reference, and the feed back signal is utilized to increase or decrease the power generated by controlling the inlet valve to turbine of steam or water as the case may be.
183
Load Frequency Control
6.1
Speed Governing Mechanism
The speed governing mechanism includes the following parts. Speed Governor: It is an error sensing d'evice in load frequency control. It includes all the elements that are directly responsive to speed and influence other elements of the system to initiate action. Governor Controlled Valves: They control the input to the turbine and are actuated by the speed control mechanism. Speed Control Mechanism: It includes all equipment such as levers and linkages, servomotors, amplifying devices and relays that are placed between the speed governor and the governor controlled valves. Speed Changer: It enables the speed governor system to adjust the speed of the turbo generator unit while in operation.
6.2
Speed Governor
A simple schematic representation of the governor is shown in Fig. 6.1.
t
XB
B
txc c
.J J
Fig. 6.1
Speed Governor
The pilot valve v operates to increase or decrease the opening of the steam inlet valve V. Let XB and Xc be the changes in the position of the pilot valve v and control valve V responding to a change in governor position. XA due to load. When the pilot valve is closed XB= 0 and Xc == 0, (Le.,) the control valve is not completely closed, as the unit has to supply its noload losses. Let roo be the noload angular speed of the 9'rbine. As load is applied, the speed falls and through the linkages the governor operates to move the piston P downwards along with points A and B. The pilot valve v admits oil under n and lifts it up so that the input is increased and speed raised. If the link Be is removed then the pilot valve comes to rest only when the speed returns to its original value. An "isochronous" characteristic will be obtained with such an arrangement where speed is restored to its preloaddisturbance value. This is shown in Fig. 6.2 and Fig. 6.3.
Operation and Control in/Power Systems
184
c
8
Fig. 6.2 Isochronous Governor
tIOO%t==============(A) Isochronous
Speed
I
(8)
Drooping
o
50%
100%
Power Fig. 6.3 Governor Characteristics
With the link Be, the steady state is reached at a speed slightly lower than the no load speed giving a drooping characteristic for the governor system. A finite. value of the steadystate speed regulation is obtained with this arrangement. For a given speed changer position, the per unit steady state speed regulation is defined by Steady state speed regulation
N N =
~
r
Where
No = Speed at no  load N r = Rated speed N = Speed at rated load
The isochronous and drooping characteristics are shown in Fig. 6.3.
185
Load Frequency Control
6.3
Steady State Speed Regulation
The opening of the control valve is directly'related to the power input to the unit so that
P
Xc
P,
C
=
..... (6.1 )
Where C is the opening of the control valve corresponding to rated output Pr and Xc is the opening while the turbine delivers an output ofP. The displacement at A, xA is proportional to the change in speed so that
xA = K (No  N) =K~N
..... (6.2)
Defining the lever ratios
and
AB _I BC  ,
..... (6.3)
BC _I AC  2
..... (6.4)
AB _I AC  3
..... (6.5)
~_ AB_I Xc BC
,
so that
XA
= I,
XC
..... (6.6)
P .C P,
= II 
~N=~=!i.~.C= CII.~ k
k P,
k
P,
..... (6.7)
The steady state speed regulation ..... (6.8)
thus, the steady state speed regulation is directly proportional to the output power. R has the dimension of HzlMW.
6.4
Adjustment of Governor Characteristics
For a given governor R can be changed by varying the lever ratio II and the no load speed No The droop of the governor characteristics can be changed from curve B to B I as in Fig. 6.3 by changing the lever ratio II' Generally in order to coordinate a generating unit with the rest of the system, the lever ratio II is adjusted. This is done occasionally and when the machine is cold.
Operation and Control in Power Systems
186
When the machine is in operation the no load speed No can be adjusted by varying an external force acting on the governor thrust sleeve at S so that the characteristic is shifted parallel to itself as shown by curves C and D. This is achieved by the speed changer. See Fig. 6.4.
I Speed
        
C 8 81 D
Power
Fig. 6.4 Action of speed changer
6.5
Transfer Function of Speed Control Mechanism From Eqn. (6.2).
xA = K (N o N) = K I (roo  ro)
=
k ! flro
..... (6.9)
If the point C is fixed, then ..... (6.10) ..... (6.11) 8
A
Fig. 6.5
If C were to move with A fixed then from Fig. 6.6. ~_ AS_I Xc  AC  3 Xs =
13
, Xc Z
c
Load Frequency Control
187
A
c
B
Fig. 6.6
As the point C will be shifted upwards by Xc due to the load increment, the actual movement of B would be .,
xB=12xA13xc
..... (6.12)
The opening of the pilot valve, xBdetermines the rate of oil flow into the cylinder. The movement of the control valve, Xc is proportional to the total oil admitted under the piston P. Therefore, Xc = k
fXB ;
where k is a constant
..... (6.13)
Taking Laplace transform on both sides. k
Xc (S) = XB(S) s Taking Laplace transform of Eqn. (6.12),
..... (6.14)
XB(S) = 12 X A(S) 1)Xc(S)
..... (6.15)
substituting for XB(s) in (6.14), SXc(s) = 1 X (s)I X (S) k
2
A
)
C
..... (6.16)
..... (6.17)
..... (6.18)
..... (6.19)
Operation and Control in Power Systems
188 I =Ts k 13
is the time constant of the speed control mechanism and is of the order of several milliseconds. The transfer function of the speed control mechanism G sc
6.6
(S) = XC
..... (6.21 )
Transfer Function of a Power System
Under steady state operating conditions the synchronous angular frequency O)s = 21tfs
where fs is the synchronous frequency (i.e.) the rated frequency.
ilo =
Change in angular position of the generator rotor corresponding to an increased load demand il PD on the generator.
0= Os + ilo radians
. .... (6.22)
d d d 0) = (D) = (os +ilo) = 0)5 +ilo dt dt dt
..... (6.23)
0) = O)s + ilO) rad / sec
..... (6.24)
Let
f
..... (6.25)
Where
I d M= iloHz 2n dt
=
fs + ilf
..... (6.26)
The kinetic energy stored in the machine W=
~ 10)2 2
=
~ 1(21tf)2 MJ 2
..... (6.27)
The kinetic energy at synchronous speed is
Then
I , Ws. = 1(2nfs)MJ 2 .
..... (6.28)
W=WS[f:)' MJ
..... (6.29)
..... (6.30)
Load Frequency Control
== Ws (
189
1+2M) M fs for smal I
fs
..... (6.31 )
The rate of change of kinetic energy is the increase in power
~(W) = 2Ws ~(M) dt
fs
..... (6.32)
dt
W Defining per unit inertia constant H = _s Pr d 2H d .  (W) =  (M)p.u. takIng Pr = 1.0 pu dt fs dt
..... (6.33)
Further, all types of composite loads experience a change in power consumption with frequency. Defining the load damping factor.
D  oPo
 i\f p.u. MW/Hz
..... (6.34)
Where PD is the load demand in p.u.; the change in load demand in this case is then D. M
p.u.M.W
For a small step change in load demand by written as ~PG ~Po =
~
PD the power balance equation can be
2H d DM +  (M) fs dt
..... (6.35)
Taking Laplace transform of the above equation
[~PG (S)  ~Po (S)]= D~F(S) + 2H .S.~F(S) fs
..... (6.36)
..... (6.37)
..... (6.38)
Operation and Control in Power Systems
190
..... (6.39)
where
K
P
1 = Hz/p.u.M.W.
. .... (6.40)
D
..... (6.41 )
and
The transfer function relating the frequency change to the change in inputoutput power may be then written as
Kp ..... (6.42)
Gp(S)= I+STp Fig. (6.7) shows the block schematic of Eqn. (6.39)
/    ....... dF(S)
Fig. 6.7
The complete block schematic of speed governing system is shown in Fig. (6.8).
6.7
Transfer Function of the Speed Governor
Let m be the mass of the flying masses c f be the friction control and C s be the spring control of the governor system. dP D
ro Fig. 6.8 The Block Diagram of Speed Governing System
The equation of motion of the governor is
191
Load Frequency Control
..... (6.43) Taking the Laplace transform on both sides of Eqn. (6.43). (mS 2 + crS + cs) xA(S) == ks~w(S) Rearranging ks ~(j)(S) = mS + crS + Cs
..... (6.44)
== Gg(S)
..... (6.44)
x A (S)
2
Where Gg(S) is the transfer function of the speed governor. v=
~~
== natural frequency of oscillation of flying masses with no damping
s== 2~ksm
== damping constant
XA(S)== G(S) .
6.8
~
..... (6.45)
(j) (S)
Governing of Hydro Units
The governing system of hydro turbines has to meet complex requirements because of the destabilising effect of the inertia of water. Hydrogovernors are provided with temporary droop compensation to prevent over travel of the gate motion. The setting of temporary droop and decay time constant Tr of the compensator determine the dynamics of response following changes in load. The speed drop for a hydro unit is defined by ..... (6.46)
p.u. speed drop ==
, where
N I == extrapolated speed corresponding to zero gate opening N2 == extrapolated speed corresponding to 100% gate opening
and
NR
=
rated speed.
192
Operation and Control in Power Systems
The temporary droop provided by a dashpot mechanism consisting of a spring, needle valve and piston is shown in Fig. (6.9). Governor permanent droop r~~__.c
Needle valve
/
A'I
LO1
I
Temporary droop mechanism
~p"otvaJve
o
D
Fig. 6.9 Hydro turbine speed governing mechanism
The dash pot (temporary droop mechanism) is represented by the transfer function .
..... (6.47) where s is the complex frequency
0'
+ j{().
The value of the gain 0 ranges from 0.2 to 1.0. The rest or decay time constant Tr lies between 2.5 and 25s. The block schematic of a typical hydro governing system is shown in Fig. 6.10. The value of k lies between 0.03 to 0.06.
Pilot valve & servomotor
PRef
+
Rate limits
K
Gate servo
Position limits
Penstock turbine
Feedback for permanent droop
+ Dash pot for temporary droop
Fig. 6.10 Block schematic of Hydroturbine speed governing system
Load Frequency Control
6.9
193
Penstock Turbine Model
Consider the penstockturbine system shown in Fig. 6.11, having an effective length of I:, m discharging water to the turbine at a velocity V mls operating at a head of H m. Following a load change, consider the following p.u. changes in the variables:
L H
________ J Fig.6.11 Penstock Turbine mode
l\H
=
change in head
l\N = change in speed l\X
=
change in turbine gate opening
l\Q
=
change in flow
l\T
=
change in turbine torque
It can be proved that, p.u.change in head(l\H) = p.u.change in flow(l\Q) where
2 1 exp[2JeS] 1 + exp[ 2TeS]
..... (6.48)
Te = elastic limit of the penstock Z = normalized penstock impedance
And
s
=
(J
+ jro, the complex frequency.
Using first order Pade's approximation x
1
2 e x =  x 1+2
Eqn. (6.48) becomes. l\H l\Q
= _Te Zs =  T w S
where T w is defined as water starting time.
..... (6.49)
194
Operation and Control in Power Systems The changes in flow (discharge) and torque can be represented by AQ = allAH + a l2 AN + a l3 AX
..... (6.50)
AT = ~IAH + ~2AN + ~3AX
..... (6.51 )
where all' a 12, a 13 , ~I' ~2 and ~3 are constants. The turbine torque change due to speed changes can be neglected in comparison to other changes since speed change is relatively small. Therefore, Eqn. (6.50) becomes AQ
allAH + a 13 AX
=
AT = a21 AH + a23 AX Using Eqn. (6.49) and (6.50).
.. ... (6.52) For an ideal, loss less turbine with valve opening Xo all = 0.5 Xo a 21 = 1.5Xo a 13 = 1.0 ~3 =
AT so that
AX
1.0 =
1 (1.5X o + O.5Xo)TwS I+O.5XoTwS
=
1 Xo TwS 1+0.5XoTwS
At full load in p.u. Xo= 1.0 Therefore,
AT = AX
ITw s == ITw s ==GPT(S) I+O. 5Tws l+(T;S)
..... (6.53)
Eqn. (6.53), is the classical turbine penstock transfer function. The water starting time,
Tw' varies from 0.5 to 4 s and can be calculated from the relation.
T == LV w
Hg
..... (6.54)
where g is the acceleration due to gravity (6.8 m/s2).
Load Frequency Control
195 ,_ _ _ _ _ _, Change in gate KG (I +STr ) position
Change in speed
(I + STI )(\ + ST2 )
Fig. 6.12 Approximate transfer function for hydroturbine speed governor
The block diagram in Fig. 6.10 excepting penstock turbine model can be represented by the linear approximate model shown in figure 6.12 where KG=11k and TI and T2 are given by Ts 1
2
T ,T
k
="2± (T2
)l
2
.; TA
and
1 T =G Ks
Typical values for Tr and 0 can be obtained using the relations Tr
5Tw 0= 2. 5Tw and 2H where H is the inertia constant of the turbinegenerator on the machine MVA base. =
change in gate position
Reference Ka tl+STrJ
6F(S)
(I +STI )(1 + ST 2 ) Turbine & Penstock Fig. 6.13 Block diagram of a hydro turbine speed governing system
A complete block diagram for representing the speed  governing system, turbine and penstock for dynamic studies is shown in Fig. (6.13).
196
Operation and Control in Power Systems
6.10 Modal for a Steam Vessel Steam inlet piping or steam chest can be considered as a steam vessel with steam input and steam output Fig. 6.14. For changes in steam valve position the dynamics can be represented by
    . . . Steam outlet Steam Vessel
Steam
Input Fig. 6.14 Steam Vessel
Where W is weight of steam in kg and Om and OOlIt are the rate of flow of steam in kg/sec at inlet and outlet of the steam vessel respectively. Assuming that the flow rate is proportional to the pressure. p
Oout 0 P 0
..... (6.55)
o
Where Po and
0 0 are the steady state pressure and rate of flow of steam in the vessel. ..... (6.56)
Assuming constant temperaLure in the vessel
..... (6.57)
d(V) dp
= dp ~
dt
Where V is the volume of the vessel in m3 and v is the specific volume of steam (m 3/kg) in the vessel.
Load Frequency Control
197
=
V.~(~)dP dp
v dt
= V ~(~). dQout .~ dp v dt Qo = V
~(~)~.~Qout dp v Q dt o
Let
~.V~)
T=
Q. _ Q In
..... (6.59)
dp v
Qo
out
=
..... (6.58)
T dQ out
dt
..... (6.60)
Taking Laplace transforms of Eqn. (6.60). Qin (S)  Qout (S) = T S Qout (S) Qout (S) I =Qin (S) I +ST
..... (6.61 )
The transfer function of the steam vessel is G
Sv
I 1+ ST
(S)
..... (6.62)
6.11 Steam Turbine Model In normal steady state the turbine power Pr keeps balance with the electromechanical airgap power PG' resulting in zero acceleration and a constant speed or frequency. Changes in Pr or PG i.e., ~PT or ~P G or both will upset the above balance. If the difference ~PT  ~PG is positive the turbine generator unit will accelerate, if negative it will decelerate. The Turbine  Generator power increment ~PT depends entirely upon the valve power increment ~P v and the response characteristics of reheat and nonreheat type of turbines are different. However, it is possible to express the turbine  generator dynamics in terms of turbine transfer function. The turbine  generator transfer function is given by G
(S) TG
= ~PTG (S) ~Pv (S)
For a simple nonreheat type turbine the model is given by a single time constant
Operation and Control in Power Systems
198
K TO G TO (S) = 1 + ST TO
..... (6.63)
Where KTO is the gain constant and TTG is the time constant of turbine  generator unit.,
6.12 Reheat Type Steam Turbine Model Reheat turbines have more than one time constant. Fig. 6.15 shows a tandem compound single reheat turbine system configuration. Its block diagram is shown in Fig. 6.16. The time constants TCh' TRh and TCo represent the delays due to the steam chest and inlet piping reheater and cross over piping respectively Fig. 6.16.
Valve Position
Control Valve & Steam
Stage Stage
Chest Legentd
condenser
H.P. = High pressure stage turbine I.P. = Intermediate pressure stage turbine L.P. = Low pressure stage turbine Fig. 6.15 Tandem Compound Single Reheat System
+
+
l+sTCo
Fig. 6.16 Block Diagram Representations
KHP• KIP and K LP are the fractions of the portions of the total turbine power developed in various stages. The steam chest and inlet piping to the first turbine cylinder, the reheater and the cross over piping down introduce time delays between valve movement and the change in steam flow. The time constants are associated with the entrained steam in the components mentioned.
LOlld Frequency Control
199
If a two stage steam turbine with a reheat unit is considered then the dynamic response will be influenced by (i) the entrained steam between the steam inlet valve and first stage of turbine and (ii) the storage action in the reheater which causes the outlet of the L.P. stage to lag behind that of the H.P. stage.
Inlet valve
(a)
(b)
Fig. 6.17 Two Stage Reheat Unit
The turbine transfer function is characterized by two time constants. For ease of analysis it can be assumed that the turbine is modeled by a single equivalent time constant. The value of T T lies between 0.2 to 2.5 sec in most of the cases. This is shown in Fig. 6. I 7(b).
6.13 Single Control Area In the previous sections models for turbine  generator, power system and speed governing systems are obtained. In practice, rarely a single generator feeds a large area. Several generators connected in parallel, located also, at different places will supply the power needs of a geographical area. Quite normally, all these generators may have the same resl'onse characteristics for changes in load demand. In such a case, it is possible to define a control area, grouping all the generators in the area together and treating them as a single equivalent generator. For small load changes all these generators swing in unison. Putting together, the various models derived so far a single control area or simply an area can be conceived as shown in Fig. 6.IS. From Eqn. (6.20),
From Eqn. (6.42), Kp
OpeS)
=
and from eqn. (9.39) L1F(S)
I +STp
= [L1 Po(S) 
LI PD(S)]
[1 +K;TP ]
Note that ~Po(S) = ~PTO(S) as generator power change which is also the turbine power change and also the turbine generator unit power change.
200
Operation and Control in Power Systems
Also, from eqn. (6.63) GTG
= KTG /1
+ STTG 1 R Ll PD(S)
Xd S) LlPref
Ks
K TG
1+ STs
I+STTG
+ +

Kp I +STp
Ll PTG Ll PG
Ll/lS)
Fig. 6.18 Block diagram of a single area system
6.14 The basics of Load Frequency Control The following basic requirements are to be fulfilled for successful operation of the system:
1. 2. 3. 4.
The generation must be adequate to meet all the load demand The system frequency must be maintained within narrow and rigid limits. The system voltage profile must be maintained within reasonable limits and In case of interconnected operation, the tie line power flows must be maintained at the specified values.
When real power balance between generation and demand is achieved the frequency specification is automatically satisfied. Similarly, with a balance between reactive power generation and demand, voltage profile is also maintained within the prescribed limits. Under steady state conditions, the total real power generation in the system equals the total MW demand plus real power losses. Any difference is immediately indicated by a change in speed or frequency. Generators are fitted with speed governors which wi II have varying characteristics: different sensitivities, dead bands response times and droops. They adjust the input to match the demand within their limits. Any change in local demand within permissible limits is absorbed by generators in the system in a random fashion. An independent aim of the automatic generation control is to reschedule the generation changes to preselected machines in the system after the governors have accommodated the load change in a random manner. Thus, additional or supplementary regulation devices are needed along with governors for proper regulation. The control of generation in this manner is termed loadfrequency control. For interconnected operation, the last of the four requirements mentioned earlier is fulfilled by deriving an error signal from the deviations in the specified tieline power flows to the neighboring utilities and adding this signal to the control signal of the loadfrequency control system. This last requirement will be discllssed in detail in the next chapter.
201
Load Frequency Control
Should the generation be not adequate to balance the load demand, it is imperative that one of the following alternatives be considered for keeping the system in operating condition: I. Starting fast peaking units. 2. Load shedding for unimportant loads, and 3. Generation rescheduling. It is apparent from the above that since the voltage specifications are not stringent. load frequency control is by far the most important in power system control. The block schematic for such a control is shown. in Fig. 6.19.
2
Actual output
Sensor
Fig. 6.19 Block diagram for load frequency control
In order to understand the mechanism of frequency control, consider a small stepincrease in load. The initial distribution of the load increment is determined by the system impedance; and the instantaneous relative generator rotor positions. The energy required to supply the load increment is drawn from the kinetic energy of the rotating machines. As a result, the system frequency drops. The distribution of load during this period among the various machines is determined by the intertias of the rotors of the generators partaking in the process. This problem is studied in stability analysis of the system. After the speed or frequency fall due to reduction in stored energy in the rotors has taken place, the drop is sensed by the governors and they divide the load increment between the machines as determined by the droops of the respective governor characteristics. Subsequently, secondary control restores the system frequency to its normal value by readjusting the governor characteristics.
6.15 Flat Frequency Control In this method, a master machine is charged with the task of mainta,ining the system frequency. All the remaining machines carry constant loads. The capacity of the machine which controls
202
Operation and Control in Power Systems
the frequency should be between 5 and 10 % of the entire generating capacity of the system for effective frequency control. In small, independent systems, old and inefficient machines may be assigned to frequency regulation leaving new and efficient machines to supply the load demand. Modern power systems are so large that it is impossible to design a single central control system that would handle the overall control job. It is extremely useful to take into account the weak links in the system and then apply control through decomposition. The demarcation of load frequency control and Mvar voltage control characteristics is one such decomposition. Geographical and functional decomposition are successfully applied to power systems and this leads to the concept of area control. A modern power system can be divided into several areas for load frequency control. Each control area fulfils the following: I. The area is a geographically contiguous portion of a large inter connected area, which adjusts its own generation to accommodate load changes within its precincts. 2. Under normal conditions of operation, it exchanges bulk power with neighbouring areas. 3. Under abnormal conditions of operation, it may deviate from pre determined schedules and provide assistance to any neighbouring control area in the system. 4. It is expected, in addition, to partake with the other areas in the system in a suitable manner in the system frequency regulation. The rotors of all generators in a control area swing together for load changes. Thus, a coherent group of generators within a geographical region may constitute a control area which is connected to other similar areas by weak tie lines as shown in Fig. 6.20. Inter tie with area I
Inter tie with area 3 Inter tie with area n
Inter tie with area 2
Fig. 6.20 Single control area inter ties
6.16 Real Power Balance for Load Changes It has been already m.entioned that when an imbalance occurs between real power generation and load demand, frequency deviation takes place. Consider a single area operating in steadystate and let it undergo a step change ~PD in load demand. Assuming that the generation
Load Frequency Control
203
changes by an amount ~p G following the load change due to governor action, the power imbalance is ~p G  ~p D' There are three possible modes in which this power can be absorbed. For positive power imbalance: 1. The area kinetic energy increases (i.e. the kinetic energy of all the generator rotors in
the area increases), 2. The load demand increases (due to load characteristics) and
3. The power outflow from the area to other interconnected areas increases, if such interconnections exist.
6.17 Transfer Function of a Single Area System Under steady state operating conditions. Transfer functions for a single control area can be derived exactly in the same way as for the system in sec. (6.6). IDS = 27tfs where IDs and fs are synchronous angular speed and rated frequency respectively. If ~() is the change in angular position of the equivalent generator representing the area corresponding to an increased load ~PD in the area. 0 = Os + ~o rad Then, d
d
.
d~o
C=(o)=(L +CL)=C s + dt dt s . dt
IDs + ~ID f =fs+M
..... (6.64)
=
Let
I d M=  27t dt The area kinetic energy
~oHz
Where
W = \J2 I ID2
=
..... (6.65)
Y2 I (2 7t t)2 MJ
Where I is the moment of intertia of the area. Also, the kinetic energy at synchronous speed IDs is Ws = \J2 I (2 7t t)2 MJ
Thus,
W
r:::.
~ Ws (~
J
MJ
for small M ( M) fs fs
Ws I + 2 
..... (6.66)
204
Operation and Control in Power Systems The rate of change of kinetic energy is the increase in area power which is
~(W)= 2Ws ~(M) dt
fs
dt
Defining per unit inertia constant M = W/Pr d 2M d  (W) =  (M)p.u. dt fs dt All types of composite loads experience a change in power consumption with frequency. Defining the load damping factor D.
_ oPD
0
8f" p.u. MW/HZ
..... (6.67)
Where PD is the load demand in p.u; the increase in load demand in this case is then D.6f p.u. MW, For a small step change in load demand, 6f p.u. MW. For a small step change in load demand, 6P D the power balance equation takes the form.
.. ... (6.68) Taking the Laplace transform of the above equation [6PG (S)  6PD (S)]= D6F(S) + 2H S6F(S) fs or
..... (6.69) Where Kp
= 110
HzJp.u. MW 2H Tp == f 0 s
205
Load Frequency Control
The transfer function relating the frequency change to the change in input/output powers may be designed by Gp(S) so that
Kp I +STp
Gp(S) =
..... (6.70)
The block schematic for a single area system is shown in Fig. 6.21 (a) An entire control area may be represented as in Fig. 6.21(b). ~PD(S)
.PGe,jJ 1~fI~F·~S) (a)
1"+ EqUivalent turbme ~P o(s) generator (b)
Governing system
~F(s)
PO\\ier system
Fig. 6.21(a, b) Block diagram of a single area system
6.18 Analysis of Single Area System Consider'the single area system shown in Fig. 6.21 (b). For a step load change in the system, the following equations can be written
[L\P o(S)  L\PG(S)] Gp(S)
=
L\ F(S)
I _ L\P (S) RG ST (S)L\F(S) G where
GST = (
Solving for L\F(S)
L\F(S) = 
KTG
x
1+ STrG
Ks
)
1+ STs
G p (S)
1+ (
~ )G p(S)G
L\P For a step load change, L\Po(S) = __ 0
S
Substituting the value of L\PoCS)
L\P (S) o ST (T)
206
Operation and Control in Power Systems
Applying the final value theorem, the steady state error Mss would be ~fss
Lim ~
]
= S~O LSM'(S) = 
D
M'o K K
+ TG
S
R The product KrG Ks can be made unity by properly selecting the units for the inputoutput quantities to the combined block G ST(S) so that ~fss
~p
= _ _0_
1 R
D+The quantity (D + llR) is defined as area frequency response characteristic or area frequency regulation characteristic (AFRC) and denoted by ~. It may be noticed that as R becomes smaller ~ fss approaches zero.
Reference Power Setting In Fig. (6.18) reference power is indicated at the input summer to the system. This determines the starting point of the governor characteristic. Under static conditions setting KsKrG = 1. We obtain
Changing the reference power setting will also change the turbine generator output in a proportional manner. For instance if the machine connected to an infinite bus in which case M = 0 we have the direct relationship. M'ref= M'TG For a fixed setting of the speed changer the steady state increase in power output from the turbine generator M'TG is directly proportional to the frequency drop. As in this case M'ref = 0
1
M' = ~f TG R R has the units hertz per MW or hertz pr unit MW as the case may be.
207
Load Frequency Control
106
Set to 50% load
60
40
20
80
100
120
Percent of rated output
Fig. 6.22 Speed Changer Setting
If 0
=
0, then AFRC
P=
=
I/R = the area speed regulation coefficient.
The time constants Ts and T T are much smaIler compared to T p and as such, an approximate solution to the response of the system could be obtained by neglecting them. Then ~F(S)
Gp(S)
=
1+(
~)Gp(S)
Kp I +STp
~PD
S
~PD
1+(~)I+K;Tp
S
Where A = Kp + R
B =Tp R Partial fractions for the expression in the bracket can be obtained as follows:
C
.
0
=+S A + BS S A + BS S
And at
A B S=_· 0=
=
O·, C
I
At
= A
B'
A
Operation and Control in Power Systems
208 So that
M(t) == VI [~F(S)]
~PDP [2. S
S+
I
KpP
JI
Tp
..... (6.71 ) The response is shown in Fig. 6.23. L\f( Hz)
""T"+...
F(s)
L\PD (Hz)
J3
Fig. 6.23 Uncontrolled single area systemresponse to step load change
6.19 Dynamic Response of Load Frequency Control Loop It has been shown that the load frequency control system posses inherently steady state error for a step input. Applying the usual procedure, the dynamic response of the control loop can be evaluated so that the initial response also can be seen for any overshoot. For this purpose considering the relatively larger time constant of the power system the governor action can be neglected, treating it as instantaneous action. Further the turbinegenerator dynamics also may be neglected at the first instant to derive a simple expression for the time response. It has been proved that
~F(S)==
p
I G l+G s G TG G p
R
~PD(S)
209
L(}ad Frequency Control For a step load change of magnitude k. . k ~PD(S)
=S
Neglecting the governor action and turbine dynamics
~F(S) = _
Gp
k
I+~G R p
S
Kp ) ( = I+STp (
I
)S k
I Kp 1+'R 1+ STp
Applying partial fractions
(~ ~"ll
Kpk 
S[S + Tp + RTp
6.20 Control Strategy The uncontrolled system is subject to a steady state error for step load changes. To reduce this error, consider first the introduction of a negative feedback signal from the frequency deviation, i.e. Let
~Pc =
K)
~
.... (6.72)
F(S)
where K) is the gain for the proportional control. The response would then be
..... (6.73)
In the steady state the frequency error would be
=
M ss
Kp
I(
K+
I)
R)
~P
=_ 0
~PD
I (I
)
 +  + K
Kp
R
I
..... (6.74)
210
Operation and Control in Power Systems
If K] is made very large then only ~fss reduces to zero. This, in other words, means that R should be made equal to zero, which is not desirable. Proportional control is not suitable for reducing the steady state error to zero. It is a well known fact in control theory that integral control will improve the steady state performance. Let
..... (6.75)
So that
..... (6.76)
The negative sign signifies that the control signal ~ Pc has to be increased for reduction in frequency. The equations for the system in Fig. 6.24 are: L\F(s)
L\F(s)
Fig. 6.24 Proportional and Intelgral Control
..... (6.77) Also, if Ks KTG is adjusted to be unity
I )( 1 J=~PG ( &+lfR)~F(S)( S l+STs I+STTG Substituting eqn. (6.78) in eqn. (6.77) and solving for ~F(S) ==
for a step load change
~F(S)
~PD(S)Kp(l + STT)(1 +STs)S (l + STs)(l+ STT)(1 + STp)S + Kp(K2 + Sf R)
~PD(S) =  ~PDfS,
~fss ==
..... (6.78)
Lim S~O
[S.~F(S)
]
and the steady state error obtained as
Load Frequency Control =
211
SKp ~PD (\ + STT )(S + STs )S S (\ + STs)(1 + STT )(1 + STp)S + Kp(K2 + SI R)
=
0
..... (6.79)
By making use of integral control strategy, the steady state error can be eliminated. For the reasons explained before, TT and T s cam be neglected for an approximate analysis for the response. For a step load disturbance.
S(~pp )
F(S) =
. ~PD
S 2 + S(I +Kp)  I +K2Kp R Tp Tp
..... (6.80)
S
The response ~f(t) depends on the nature of the denominator expression, i.e. the characteristic equation. S2 + 2 80J n S + OJ n2 Where
and
OJ = n
il =
JKTpK 2
1+~ R 2Tp
P
is the natural frequency
J Tp
K2Kp
=damping ratio
The gain for critical damping can be obtained by setting 8 equal to unity.
or
. .... (6.81)
For values of K2 greater than that given by eqn. (6.81), the response is oscillatory due to under damping. For values ofK2 less then the critical value response is monotonically decreasing without oscillations or as in Fig. 6.25. The selection of the gain for the controller should be such that the following specifications are satisfied.
Operation and Control in Power Systems
212
Without integral control
Fig. 6.25 Response to integral control with different gain settings
1. The control loop must be stable. 2. The frequency error should return to zero following a step load change. The deviation in the transient state must also be minimized. 3. The integral of the frequency error should not exceed a certain value (say 150Hz or 3 seconds)
6.21 PID Controllers From the above analysis, it is clear that proportional integral and derivative control strategy can be applied for load frequency control. While proportional control is inherent in the feedback through the governor mechanism itself, derivative control when introduced improves transient performance and ensures better margin of stability for the system.
where K3 is the gain of derivative controller. The response to such a control is
By proper choice of K J' be satisfied.
K~
and K3 all the specifications for the system performance can
213
Load Frequency Control
E 6.1 A \00 MW generator has a regulation parameter R of 5%. By how much will the turbine power increase if the frequency drops by 0.1 Hz with the reference unchanged.
Solution: Actual change in frequency = 5% of 50Hz = 0.05
x
50 = 2.5Hz
R = 2.5Hzl100Mw = 0.025 HzlMw If
M = 0.1 Hz, the increase in turbine power ~P=
I
I
 M= x(O.I) =4 MW 2 0.025
The turbine power increase = 4MW. E 6.2 A 100 MW generator with R = 0.02 HzlMW has its frequency fallen by 0.1 Hz. If the turbine power remains unchanged by how much the reference power setting be changed.
Solution: The signal to increase the generation is blocked. Thus at the input summing point the reference power setting must be changed. Such that
~ Pref (i.e .. )

~ Pref 
~R
M= 0
I
R M=
1 0.025 xO.l = 4 Mw
E 6.3 Two generators with ratings 100 MW and 300 MW operate at 50Hz frequency. The
s.ystem load increases by 100 MW when both the generators are operating at about half of their capacity. The frequency then falls to 49.5Hz. If the generators are to share the increased load in proportion to their ratings what should be the individual regulations? What should be regulations if expressed in , per unit Hertz/per unit megawatt?
Solution:
1 AP= M R
~P = 2 J
I M R2
M=0.5 Hz Power is shared in proportional to their ratings
214
Operation and Control in Power Systems
Hence
100 = 25 MW 400
~P
= 100 I
x 
~P
= 100 2
x 
R
=  0.5 = 0.00667
300 = 75 MW 400
75
I
If regulation is expressed in p.u. then with f = 50Hz
R2
= 
0.00667 50
300
x
I
=
0.04 pu HzlPu MW
Both have the same value, even though based on their individual ratings, they have different regulation. E 6.4 Determine the primary load frequency control loop parameters for a control area having the following data: Total rated area capacity Pr = 1000 Mw Normal operating load
=
500 MW
Inertia constant H = 4.0 sec Regulation R = 2.5 Hzlpu MW
Solution: Load damping
OP 500MW D = Of = 50Hz = 10 MW/Hz
[Here, the load damping is assumed linear and percentage change is assumed to be the same] In per unit
10 1000
=
D
2H Tp = fsD K
p
1 D
= 
= 0.01 pu MW/Hz 2x4.0
= 50x 0.01
=
16 sec
1 = O.oI~  = 100 Hzlpu MW
215
Load Frequency Control
E 6.5 Determine the area frequency response characteristic and the static frequency error for a system with the following data, when I % load change occurs? D
=
0.0 I pu Mw/Hz
R = 2.5Hzlpu MW
Tp = 16 sec
Kp = 100 Hzlpu MW
Solution: Area frequency response characteristic
~
1
=D+ R
1 = 0.01 + 2.5 = 0.41 MW/Hz
M= M =~
100 x 0.41
=0.02439Hz
E 6.6 In the example E6.5, the governor is blocked so that it does not change the generation. In that case what would be the steady state frequency error?
Solution: When the governor is not acting, the feedback loop is not existing. In such a case R is infinite. I
~=D+ R =D=O.OI puMW/Hz
Hence,
M
~f =  
~
= 0.01 = I Hz 0.01
Frequency falls by 1 Hz (i.e.) f = 50  I = 49 Hz. It may be noted that with the generator acting the frequency from E6.5 is 50  0.02439
= 49.9756 Hz
The importance of feedback through governor mechanism can be understood from the above. E 6.7 A 100 MVA synchronous generator operates initially at 3000 rpm, 50Hz. A 25MW load is suddenly applied to the machine and the steam valve to the turbine opens only after 0.5 sec due to the time lag in the generator action. Calculate the frequency to which the generated voltage drops before the steam flow commences to increase to meet the new load. The value of the stored energy for the machine is 5kWsec per KVA of generator energy. Also calculate the value of H constant for the generator.
216
Operation and Control in Power Systems
Solution: Stored energy
= 5KW per/kVA
(i.e.,)
=
Load increase
=25 MW
500 MW sec pr 100 MVA
Energy required to supply this load for 0.5 sec
=
25 MW sec
Frequency at 500 MW sec stored energy = 50Hz Frequency fall
M
=
M
i1f 50
==
f
25xO.5 MW 500
sec
i1f = 50 x 25 x 0.5 = 1.25Hz 500 Frequency falls to 50  1.25 H constant
=
=
48.75 Hz
stored kinetic energy at rated frequency = 5MW  sec x I OOMV A = 5 sec machine rating MV A I OOMV A
E 6.S Given the following parameters, obtain the frequency error. Plot it when a step load disturbance of (i) I % and (ii) 2% occur in the system. Tp = 22 sec
R = 2.5 Kp= 100
Solution: The expression for i1F(s)
Gp = I~i1PD(S) l+GsGToG p R
Neglecting the turbine dynamics and governor action (G s GTO :::: 1.0)
k
i1F(S) = I +STp
k]
I +~_~ R 1+ STp
s
= 0.01 and k2 = 0.02
simulating the transfer functions in .MATLAB the response is obtained and shown in Fig. E(6.8).
Load Frequency Control
217
Fig. E6.8 Step load frequency error characteristic without supplementary control 1. k = 0.01 2. k 0.02 (governor action and turbine dynamics are neglected
=
E 6.9 Show the effect of governor action and turbine dynamics, if they are not to be neglected in E6.8 given Ts = 100 msec and TTG = 0.5 sec.
Solution: For this the exact frequency error is used
LlF(S) =
I Gp
1+
:::::
LlPD (S)
RGSGTGG p
LlPD(S) = O.Olpu;
Ks KTG
.
_ Ks G s '"""1 +STs
G TG 
K TG I+STTG
1
The response is shown in Fig. E(6.9). It can be seen that greater the step load change, larger the error and the governor action and turbine dynamics does not cause any change in the response in the steady state, except for a transient deviation at the beginning of the distribution.
218
Operation and Control in Power Systems
Fig. E6.9 Step load frequency error characteristic without supplementary control 1. k 2. k 0.02 Ts 100 msec and TTG 0.5 sec
=
=
=
=0.01
E 6.10 An isolated power station has the following parameters Turbine time constant
T t = O.5sec
Governor time constant
Ts
=
0.2sec
Governor inertia constant H
=
4 sec
Governor speed regulation R
=
R per unit
The load varies by 0.8 percent for a I percent change in frequency, i.e. D = 0.8 . The governor ~peed regulation is set to R = 0.05 pu. The turbine rated output is 250 MW at nominal frequency \>f 50Hz. A sudden load change of 50MW(dP L = 0.2 pu) occurs. (a) Construct the SIMULINK block diagram and obtain the frequency deviation response. (b) Set integral gain to 7 and obtain the frequency deviation response and compare both the responses.

          
Load Frequency Control
219 t.
Solution: To create a SIMULINK block diagram presentation select new (model) file from FILE menu. This provides an untitled blank window for designing and simulating a dynamic system. Copy different blocks from the simqIink libraries or other previously opened windows into the new window by depressing the mouse button and dragging. I.
Open the continuous library and drag the transfer function block to the window. Double click on transfer function to open the dialog box. Enter the numerator and denominator values (the coefficients in the descending powers of s, if any power of s is missing, enter zero) of the transfer function .
2.
Open the math library and drag the sum block in to the window. Open the sum dialog box and enter +  under list of signs.
3.
From the math library drag the Gain block into the model file right click on the gain block and click on the Flip option to rotate the gain block by 180degrees.
4.
Open the source library and drag the step input block to window. Double click on it to open its dialog box and set up the step time(step duration), initial and final values(which will be same) to represent the step input.
5.
Open the sink library and drag scope to window to observe the response. By using the left mouse button, connect all the blocks.
Before starting simulation, set the simulation parameters. Pull down the simulation dialog box and select parameters. Set the start time, stop time and for a more accurate integration, set the maximum step size. In this example the parameters for all the blocks for the system in the fig. are initialized. Open mfiIe and enter the parameter values. The following mfiIe has to be run prior to the simulation.( of model file) Open new mfiIe and enter the parameter values as shown below. Tg
=
0.2;
T t = 0.5; H= 5; D
=
0.6;
R = 0.05;
%for integral control Kj = 7; Save the mfiIe under parameters and run the file. SIMULINK block diagram and results for LFC By using above procedure construct the simulink block diagram for the load frequency control of isolated power system as shown in Fig. E 6.1O(a).
220
Operation and Control in Power Systems
Constant
Governer
Turbine
Inertial.load
Step l/R
'<KI4' Fig. E 6.1 O(a) Block diagram Model of Load Frequency Control (Isolated Power System)
Pull down the file menu and use save as to save the model under AFC. Start the simulation. Double click on the scope, click on the auto scale, the result is displayed as shown in Fig. E 6.1 O(b). Add integral controIIer block for above system and save the file under LFC 1. Start the simulation. Double click on the scope, click on the auto scale, the result is displayed as shown in figure below.
Fig . E6.10(b) Response of LFC 1. Uncontrolled response 2. Integral control response
Load Frequency Control E 6.11
221
A single control area system with the following data experiences a sudden load change of3%. Kp= 100 Tp= 25 R = 2 Hzlp.u. MW
with integral control using a gain of 10 obtain the frequency plot with time and show that the frequency deviation is reduced to zero.
Solution: Using MATLAB, the solution is obtained and shown in Fig. (E 6.11). Kj=IO Kp = IOO Tp = 25 R =2 num = Kp/Tp d l = (I1Tp) + (K/ (R
* Tp))
d2 = Kj * K/Tp
Transfer Fan
Fig. (E 6.11)
den = [I d l d 2]
222
Operation and Control i!!.!Qwer Systems
6.22 The optimal Control Problem A dynamic system is described mathematically by
x = f(x,u, t)
..... (6.82)
Where x denotes the state vector and u the control vector Let it be desired to minimize the performance index or cost functional I
..... (6.83)
J = fL(x,u,t)dt
°
Pontriagyn's method for optimal control involves in creating an augmented cost function for minimization. . L ' (x,x,u, t) = L(x,u, t) + At[f(x,u, t)  x
..... (6.84)
Applying EulerLagrange equations, we obtain
.
au ax . au x=aA A.=
0=
Where
..... (6.85) ..... (6.86)
au au
u = L (x,
..... (6.87) u, t) + At [f(x,
U,
t)]
..... (6.88)
and A. denotes Lagrange multipliers. Simultaneous solution to eqn. (6.85, 6.86, 6.87) subject to boundary conditions gives optimal control Uo. But, the equations are non linear and time varying and hence even for simple systems the solution will be difficult to obtain.
6.23 The Linear Regulator Problem Consider the linear dynamic system
~(t) = [A](t)X(t) + [B][t]!:! (t)
..... (6.89)
Let the cost functional be
I I 12 J = _XT (t 2 )HX(t 2 ) + fX(t)Q(t)X(t) + u T(t)P(t)u(t)]dt 2 21J
..... (6.90)
Where Hand Q are real symmetric positive semi definite matries and P is a real symmetric positive definite matrix. Again t2 is fixed, but X(t 2) is free. For simplicity, omitting the arguments .
..... (6.91 )
223
Load Frequency Control For optimality
..... (6.92) ..... (6.93)
..... (6.94)
From the above
VO =PI(t)B T(t)90(t)
..... (6.95)
substituting this VO in equation (6.92) XO (t) = A(t)X(t)  B(t)pI (t)B T(t)9° (t)} XO (t) = Q(t)X(t)  AT (t)9° (t)
..... (6.96)
..... (6.97)
or
..... (6.98)
where + is called state transition matrix. ..... (6.99)
From the boundary conditions we obtain H XO(t 2)
=
)..0 (t 2) = HXO (t 2)
and hence,
+21 XO(t)+ +12 AO(t)
..... (6.100)
It can be derived from the above that A° (t) = [+22  H+ 12 ]1 [H+ 12  +22]x° (t) R(t) XO (t) VO(t) = _P 1B T(t)Ao(t) =
From eqn. (6.95)
= _pl(t) =
..... (6.101 )
BT(t) R(t) XO(t)
_[Kt (t) XO(t)]
The optimal control is a time varying linear combination of system states.
..... (6.102)
Operation and Control in Power Systems
224
6.24 Matrix Riccati Equation Consider eqn (6.10 1)
°
A (t) == R(t)Xo (t) Differentiating
i Get) == R(t)Xo(t) + R(t)Xo (t)
..... (6.103)
Substituting XO(t) and AO(t) from eqn. (6.92) and (6.93) == Q(t) X(t)  AT(t) AGet) == R(t) X(t) + R(t) A(t) X(t)  R(t) B(t) pl(t) BT(t) AO(t)
..... (6.104)
Eliminating A(t) and rearranging the terms R(t) B(t) pl(t) BT(t) R(t) XO(t)  R(t) A(t) X(t)
..... (6.105)
 R (t) XO(t)  Q(t) XO(t)  AT(t) R(t) X (t) == 0
..... (6.106)
since the above eqn. must be satisfied for at XO(t) R(t) + R(t)A(t) + AT (t)R(t)  R(t)B(t)pl(t)B T(t)R(t) + Q(t) • The above equation is called matrix Riccati equation. R is an n
x
..... (6.107)
n symmetric matrix.
n(n + 1) 2 first order differential equations only need be solved to get the elements ofR matrix, so that the elements of feed back gain K(t) can be computed further from equation (6.102) For controllable dynamic systems H = 0, A, B, P and Q are time invariant matrices and as ex) R(t) ~ RO, a constant matrix. The matrix Riccatti equation becomes
S~
ROA + ATRo+ Q  ROBpl BT RO = 0 In the steady state R (t) ~
..... (6.108)
o.
The gain elements from K matrix can be determined as constants after the elements of RO matrix are determined. This involves solution to simultaneous nonlinear algebraic equations. There are several methods available for the solution of which Klineman's method is simple and elegant.
6.25 Application of Modern Control Theory The load frequency control (LFC) problem can be viewed as a dynamic optimization problem. Integral square error criterion can be applied and controllers can be designed using optimal control theory. Controller design using this theory was first proposed by Fosha and Elgerd. The results obtained by using this theory contradict the existing practice in the selection of the bias parameters and the integral controller gains. The system equations are put in the state variable form denoted compactly by X==AX+Bu+Fd
..... (6.109)
225
Load Frequency Control
where A.B and F are the system, input distribution and disturbance distribution matrices respectively and X, U and d are the state, control and disturbance vectors respectively. By a suitable transformation, eqn. (6.24) is transformed into the form.
X' =A'X' +B'U'
..... (6.110)
The LFC problem is then treated as a linear regulator problem and the solution is obtained by solving algebraic Riccatic equation.
6.26 Optimal Load Frequency Control Single Area System The states to be minimized are the frequency deviation Mand the time integral of the frequency deviation f~f.dt. The contrl signal is to be weighted and included in the cost function so that too large a control is not required. The cost function selected is
f 2"1 ('T~
00
J=
Qx +
UT
)
Pu dt
..... (6.111)
o
Where Q and P are the weighting matrices for the state and control vectors and X and u. Consider the block diagram for the single area system shown in Fig. 6.26. rII/R~.....,
~...+t.F(s)
Fig. 6.26 Block diagram for single area system.
As indicated in section (6.18) Ks'
KrG is set equal to one.
The following states are defined XI =
filf(t)dt
~ = ~
Thus,
=
integral of frequency error
f(t) = frequency deviation
x3
= ilPTdt) = change in turbo generator output
x4
= ~X v(t) =
xT =
[J~f
valve displacement at speed governing system.
ilf LlPTG
Let u be the control vector and d
~Xv ]
= LlPd
be the disturbance vector.
Here, the control is a single input scalar u, and the disturbance d is the change in the load demand ilPd
226
Operation and Control in Power Systems
The perfonnance index is
qll XT Qx = {x j x_} { 2 q2l Since only ~f and
fMdt
..... (6.112)
are to be minimized.
Identifying the states and the state variable system model, we obtain
Xl (t) Since
Xj(t)
= X2 (
t)
= fM{t)
..... (6.113)
dtand ~(t)
= ~f(t)
From the block diagram
M(s)
= [
M>TG (s) M>p {s)J
(I
..... (6.114)
+KiT ) p
Cross multiplying and rearranging S ~ F(s)
= __ 1
Tp
~F(s) + Kp ~PTG (s) Tp
Kp M>D (s) Tp
..... (6.115)
In the time domain, using the state variables
. ()
1 Tp
Kp Tp.
Kp Tp
..... (6.116)
~PTG(S)=( 1+ST1 l~Xv (S)
..... (6.117)
x2 t =x2+x3~' d
Also, from the block diagram
TG
As before, STTG ~PTG(S) = ~xv(s)  ~PrG(s) and in time domain
..... (6.118)
X2 (t) =_Ix4(t) __I_ X 3 (t) TTG Jro Finally, from the block diagram
..... (6.119)
~X v (s) = i.e.,
(_1) [U(S)~M(S)] I+ST5
R
1 STs ~(s) = ~ ~ (s) + U(s)   M(s) R
..... (6.120)
..... (6.121)
In the time domain ..... (6.122)
Load Frequency Control
227
Putting eqn. (6.113, 6.116, 6.119 and 6.122) in matrix from, the state variable model is obtained as 0 XI
x2 (t) x3 (t) x 4 (t)
Kp
I
0
=
0
0
(t)
0
0
__1_
_1_
TTG
TTG
0
Ts
__1_
0
0
1p
Tp
RTs
I
0
0 [ x2 Xl
0
x4
I
+ 0 + x3 Ts
kp T p d
0
..... (6.123)
0
redefining the state and control variables.
=
XI
Xl
"2 = x 2 ,
=
x3
X3
+ LlPD
x4 = X4 + LlPD U + u' + LlPD
..... (6.124)
Eqn. (6.112) can be reduced to
.'
Xl X2 X3
0 0
= 0
1 1 Tp 0
Tr
.'
x4
0 Kp 1Tp I
0
RTs
0
0 0
Tr 1
[~}
0 0 0 I
u'
..... (6.125)
Ts
Ts
The optimal control to minimize the performance index given in equation (6.111) can be determined using the solution technique described for the linear regulator problem. It is required to solve the algebric matrix Ricoati equation.
Q + ATR + RA RBpl BT R
=0
..... (6.126)
for the elements of the R matrix which is positive defmite and symmetric. Kleinman's method may be used to solve the matrix Riccati equation. An initial feedback gain vector KI is selected such that the matrix ~ = [A  B K I] has eigenvalues with negative real parts. Then the matrix equations.
RT + ~ Al + Q + ~T P KI = 0
Ai
are to be solved for the elements of RI" The new gains are computed using K(l) = pl BT R I
I
The procedure is repeated till convergence is obtained for the elements of R I .
228
Operation and Control in Power Systems
The optimal control UO
= pl BT RX(t) = 
KT X(t)
..... (6.127)
can be calculated. E 6.12 For the single area system with the following data, determine the optimal control. T p = 0.04s
R = 2Hzlp.u. MW
TT = O.Ss
Kp = 100 Hzlp.u. MW
Ts=O.IS
Pn
Assume
=
0.01 p.u.
Q[~ ~1
Solution: Substituting the parameters in the algebraic matrix Riccati equation and solving the equation using, Kleinman's method with an initial vector of K j = [1 1 1 1] The values of Ki converge to K
=
[1.0000
1.1368
1.7092
0.2976]
The optimal control is shown in Fig. E 6.12. 8 7 6
5
r ';'
:::
4 3
2
~ 0
::s
0 1 1.0
o
2.0 Time (s)
3.0 .
Fig. E 6.12 Optimal control
4.0
Load Frequency Control
229
6.27 Optimal Control for Tandem Compound Single Reheat Turbine Generator System The model for tandem compound single reheat turbine is discussed in Chapter 3 : The system is shown in Fig. 6.27.
Fig. 6.27 Tandem compound single reheat turbine system
The state  variable model of this system is given by
x = AX+Bu+ Fd with the initial conditions X(O) = 0 where
XT
=
[Xl
X2
= (f~f M 0
~PRH
~PG
0
1 1 Tp
0
0
X4
X3
0 Kp Tp 1 Tco
A= 0
0
0 0
0 0
Xs
X6]
~PCH
~XV ]
0
0
0
0
0
0
TJ 1
T2
TCH
TRH
TRH
FHP
0
1
0
The constant T 1 and T2 are given in chapter 3. 1 F FIP FIP  = LP +Tl Tco Tco TRH
0
0
1 TCH
0
.... (6.128)
0
TCH 1 Ts
230
Operation and Control in Power Systems
BT = [0
0 0 0 0
FT = [0 
~:
;s
1 1
0 0 0 0
u=APc and d=AP D The above equations are transformed into the form Xl =AX I +Bu l using the transformation 0 0 X=X'+
d d d d
and u~1 + d A quadratic cost function =
r
±(X'T.QX·
+u'TRu'~t
is selected and the algebraic matrix Riccati equation is solved as before to obtain the optimal gains using the equations. QRB pl BTR + RA +ATR= 0 u opt = _pl BT RX
and
=_LTX
..... (6.129)
The results are plotted in Fig. E.6.
E 6.13
Compute the optimal load frequency control for a single thermal power system with tandem compound singlereheat steam turbine with the following data R = 20s Ts = O.ls TCH = 0.25s FJP = 0.4
TRH = 7.5s Tco = 0.45s FHP = 0.3 FLP = 0.3
Kp = 2 p.u. Hz / p.u. MW Tp = 20s D = 0.01 p.u.
Load Frequency Control
231
Plot the variations of the controlled states with time
Solution: Q and P matrices are selected as follows 1 0 0 0 0 0 0 1 0 0 0 0 Q=
0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 ; P=1 0 0 0
The solution is obtained using a computer program for the algorithm explained in 6.26. The feedback gains, starting from initial values of unity for each, converge to the following solution.
LT = [1.0 0.6617 0.0269 0.3037 0.0047 0.0177] Plots of fM and ,M, the two optimally controlled states are shown in Fig. E. 6.13 (a) and (b)
o
10
5
15
Time (5) I x 103
.... ..::: 2 x I0 3
1
3xI03
Fig. E6,13 Optimal response (a) Optimal response of time error to step load change 4xI04
....~t
,.
5
10
Or+~_r+_~
+
15
4xI04
8x 104 12x 104
Fig. E6.13(b) optimal response of frequency error to step load change
232
Operation and Control in Power Systems
6.28 Optimal Control of Hydro Speed Governing System Consider the system shown in Pig. 6.20. The state variable mode for the system is X =AX +Bu+ Cu + P.d with the initial conditions XeO) = 0 where XT = [XI X 2 X3 X 4 Xs =
0 Kps Tps
1 1
0
Tps
A= 0
0
0
0
0 0
0
0
0
0
0
0
0
1
1
O.5Tw 0 0
0. 5Tw 0 0 K3

0

KI
K2
BT =[0 0 0 0 0 T C =[0 0 0 0 0 pT = [0
_ Kps Tps
K:z, K3 K4 Ks are given by  Tps TR K1TpTGTRTpS _ Kps K2 Tp TG Tps cr K 3 =TpTGTR K4 = TR(cr+o)+TG . TpTGTR 1
1
Tp
TR
=+
TpT~TG 1 Tp~Gl
0 0 0
u = ~ Pc = control input and d = ~ PD = stepload disturbance.
Ks
X6 ]
U~f ~f ~PG ~PGV ~Ppv ~ppv] 0
The constants Kl'
..... (6.130)
Kps Tp TR Tps
0. 5Tw 1
0
0
K4
1
Ks
Load Frequency Control
233
The state and control variable are transfonned using the relation, Xl
Xl
X2
X X
X3
U
=
0
2 3
X4
X
X5
X5
X6
X6
= u' + d
0
4
d +
..... (6.131)
d 0 u' TpTG
(j
The transfonned equations are
where
j( = A'X' + B' .u' A'=A
and
B'T
=
[0 0 0 0
TP~J
The cost functional to be minimized is selected as
J=
~ r(X'TQX' +u'Tpu')dt
Selecting Q and R matrices are : 1 0 0 0 0 0 0 Q=
1 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
andP=l the algebraic matrix Ricccati equation Q = R B' pl B'T R + R A' +A'T R = 0 is solved for the elements of R. The optimal control is obtained using u = _pl B'T R X' opt =_LTX
..... (6.120)
234
Operation and Control in Power Systems The states with optimal control are determined from ..... (6.133)
E 6.14
For the hydro system shown in Fig. 6.14.1; using the following data, compute the optimal control Distributor volve and Pilot volve and servomotor
Vet min
Turbine and penstock
Power system
Permanent speed droop
Speed governing system
Fig. ES.14.1 Hydro speed governing system
+
Time (s)
o
~
~ 5
..,l: ..,
4
8
12
16
20
0.004
0.008
(i) Optimal response (ii) Uncontrolled response
E
E=
0.012
1
0.016
(a)
Fig. ES.14.2 Uncontrolled and optimal response of hydro speed governing system (a) uncontrolled and optimal response of time. error
235
Load Frequency Control TG = 0.2 s Tp = 0.04s Tw = 0.6 s Tps = 20s TR = 3s (J = 0.05 8 = 0.2
KPS = 2 p.u. Hz! p.u. MW Ll PD = 0.01 p.u.
The 100 % load conditions on the plant. ., Time (s) 0.0004
+
0 0.0004
t
0.0008
E
0.0012
:::'
0.0016
+
0.0020
(i) Optimal response (ii) Uncontrolled response
0.0024
Fig. E6.14.2 (b) uncontrolled and optimal response of frequency error
Solution: Following the procedure outlined in Sec. 6.20, the optimal feedback gain components are computed as
LT =[1.0 3.4061 0.078 0.5393 0.1656 0.0063] The uncontrolled and optimal response are shown in figure E6.14.2 (a) and (b).
6.29 A Review of Optimal Control The optimal control determined by eqn. (6.116) is quite often impractical due to the following reasons: 1. The optimal control is a function of all the states of the system. In practice, all the states may not be available. The inaccessible states or missing states are required to be estimated. 2. It may not be economical to transfer all the information over long distances.
236
Operation and Control in Power Systems
3. The control which is a function of the states in turn is dependent on the load demand. Accurate prediction of the load demand may be essential for realizing the optimal controller. 4. The optimal control is also dependent on the weighting matrices in equation (6.111) and is thus not unique.
6.30 Load Frequency Control with Restrictions on the Rate of Power Generation The optimal load frequency control discussed in section 6.25 does not include the effect of the limits on the rate of change of power generation. If these limits were not included in the control, there will be a tendency for the system to chase large momentary disturbances. This causes undue wear and tear of the controller. Even a telemetering outage may cause problems to the system in operation. Several methods are proposed to include the effect of generation rate constraints in the computation of the control. If the generation rate constraints denoted by P are introduced into the state vector, the system order will be changed. Instead of augmenting them, it is suggested that while solving the state equations, it may be verified at each step whether or not the generation rate constraints are violated. Another simple way of considering the generation rate constraints is dual mode control strategy. Separating the state vector X into those states which are to be rate limited as XI and the rest by xr, the state space representation of the system becomes
..... (6.134)
the generation rate constraints are given by ..... (6.135) max
That is
IAn Xd+AIr Xr +BI u~ XI
The choice of U(t) may be made using max
Blu(t) = KXI
AnXI AJrX r
..... (6.136)
where K is a ( m X m) diagonal matrix with sgn. (Xj ), j = 1, 2, ..... m and m is the number of rate limited state variables out of n state variables. The dual mode control may be represented by eqn. (6.134) and (6.136)
237
Load Frequency Control max
UO(t) = K,X; ifx'J < Xlj
1
= Bi' [ Kx,max LX
j = 1,2, ............ ,n
;ifxiJ >
max
X'j
;j
= 1,2,...........n
..... (6.137)
..... (6.l38)
The generation rate constraints result in larger deviations in area control errors. As the rate at which generation can change in the area is constrained by the limits power import via tie lines becomes imperative. Under generation rate constrained conditions, the selection of governor speed regulation cofficient R requires careful consideration. In practice, a low value, of the order of2 to 4%, is chosen for R. With ,a proper, supplementary control, the steady state error can be reduced to zero, whatever may be the value ofR. However, it is desirable that a proper value of R be selected so as to give the best dynamic response. Improper selection of R may lead to instability whatever may be the integral controller gain settings. In systems with hydrothermal combination, the generation rate in the hydro area generally remains below the safe, permissible generation rate, and as such the rate constraints for generation at all the hydro plants can be omitted. The presence of governor deadband introduces oscillations in the dynamic response. It has been reported that the governor dead band does not influence the selection of integral controller gain settings in the presence of generation rate constraints.
6.31 Load Frequency Control using Output Feedback The optimal control derived in Sec. 6.22 requires the availability of all the states. However, if some of them are not available, reconstruction of those states requires either the Lueneberger observer or Kalman filter. This may not be feasible for various reasons. Controllers based on output feedback are proposed to overcome this problem. It has been pointed out later in chapter 7 that area control error for each area is used to implement the tie line bias control strategy. Correspo_~ding to this practice, if the area control error is taken as the output, then the load frequency control problem may be considered as an output zeroing problem in the presence of persistent disturbances. An additional output equation Y=HX is required to be considered in conjunction with the system Eqn. (6.134). The design of output dependent controller is not an easy task as it leads to an equivalent parametric optimization problem. Nevertheless, making use of matrix minimum principle, the controller can be designed. To simplify the computation, minimum norm methods are proposed.
238
Operation and Control in Power Systems
6.32 Load frequency control ana J!.conomlc dispatch Whenever load changes the initial frequency correction is achieved by the speed governor and this control can be treated as primary load frequency control. The adjustment offrequency by the primary control loop may take a few seconds. After the speed governor response is over, the steady state frequency error is reduced to zero so that system frequency is maintained constant by the integral controller action. This will be after the primary control action is over and can take a time period up to one minute. This i~the secondary control. The adjustment of frequency error to zero by changing the gener ion schedules that are determined earlier by economic criterion again requires readju tment of generation. This tertiary control can be implemented by using economic dispat h computer which works on the cost characteristics of various generating units in the area. The speed changer settings are once again operated in accordance with economic dispatch computer programme.
Load Frequency Control
239 QUESTIONS
6.1 Explain the necessity of maintaining a constant frequency in power system operation. 6.2 With a neat diagram, explain briefly different parts of a turbine speed governing system. 6.3 Derive the model of a speed governing system and represent it by a block diagram. 6.4 With a block diagram explain the load frequency control for a single area system. 6.5 Derive the model of a speed governing system and represent it by a block diagram. 6.6 With first order approximation explain the dynamic response of an isolated area for load frequency control. 6.7 Discuss the importance of combined load frequency control and economic dispatch control with a neat block diagram 6.8 Discuss in detail the importance of load frequency problem. 6.9 Distinguish between load frequency control and economic dispatch control. 6.10 A synchronous generator supplies power to a synchronous motor via a transmission network. Find equivalent inertia constant of a machine connected to infinite bus. 6.11 Explain how modem control theory can be applied to load frequency control 6.12 Describe how optimal control can be determined in case of LFC problem. 6.13 What are the limitations of optimal control theory ?
240
Operation and Control in Power Systems PROBLEMS
P 6.1 Two generators rated 200MW and 400MW are operating in parallel. The droop characteristics of their governors are 4% and 5% respectively from no load to full load. Assuming that the generators are operating at 50Hz at no load, how would a load of 600Mw be shared between them? What will be the system frequency at this load? Assume free governor operation. Repeat the problem if both the governor have a droop of 4%. P 6.2 A 100MVA asynchronous generator operates on full load at a frequency of 50Hz. The load is suddenly reduced to 50MW. Due to time lag in the governor system, the steam valve beings to close after 0.4 secs. Determine the change in frequency that occurs in this time. Given H = 5 KW s/KYA of generator capacity. P 6.3 Two generators rated 200Mw and 400MW are operating in parallel. The droop characteristics of their governors are 4% and 5% respectively from no load to full load. The speed changes are so set that the generators operate at 50Hz sharing the load of600Mw in the ratio of their ratings. If the load reduces to 400Mw, how will it be shared among the generators and what will the system frequency? Assume free governor operation. The speed changers of the governors are reset so that the load of 400MW is shared among the generators at 50Hz in the ratio of their ratings. What are the no load frequencies of the generators?
P 6.4 In the single area system shown below determine (a) The steady state frequency error with t.P c = 0 (b) Critical gain K of the integral control of t.P c =  fKM
P 6.5 A 500Mw generator is operating at ~ load of20Mw. A load change of I % causes the frequency to change by 1%. If the system frequency is 50Hz determine the value of load damping factor in per unit.
•
7
CONTROLOF INTERCONNECTED SYSTEMS
Power Systems came into existence in 1880s and from that time onwards the systems have grown enormously in both size and complexity. For better performance and reliability of operation and control, there were many significant developments in generation, transmission and distribution. The concept of energy control centers emerged in 1970's. Computer aided analysis and computer based control have been proposed in this context.
7.1
Interconnected Operation
Power systems are interconnected for economy and continuity of power supply. For the interconnected operation incremental efficiencies, fuel costs. water availability, generation limits, tie line capacities, spinning reserve allocation and area commitmen'ts are important considerations in preparing load dispatch schedules. In this chapter the power control of interconnected system is presented.
7.2
Flat Frequency Control oflnterconnected Stations
Consider two generating stations connected by a tie line as in Fig. 7.I(a). For a load increment on station B, the kinetic energy of the generators reduces to absorb the same. Generation increases in both the stations A and B, and frequency will be less than normal at the end of the governor response period Fig. 7.1(b). The load increment will be supplied partly by
242
Operation and Control in Power Systems
A and partly by B. The tie line power flow will change thereby. If a frequency controller is placed at B, then it will shift the governor characteristic at B parallel to itself as shown in Fig. 7.1(c) and the frequency will be restored to its normal value fs' reducing the change in generation in A to zero.
feN) f(N)
•
A
1
~
1
,iPA
1
J ,iP
I.
~_..J._
B
P
B
1
1
14
I...
+'~
o
~t
(b)
(a)
P t+t~
{}
(c)
Two interconnected stations
(b)
Uncontrolled system with load increment on Station B
(c)
Frequency controller located at Station B
Fig. 7.1 Two station system
If the load increment comes on station A, then as before, initially the generation in both A and B changes to absorb the additional load, while finally the additional load is absorbed by B only. Station A absorbs none of its load changes in the steady state. It is possible that, in interconnected operation, a given station can be made to absorb the load changes occurring elsewhere in the system so long as the controlling station has capability to absorb the change. The same analysis can be extended to a two area system.
Assumption in Analysis: The following assumptions are made in the analysis of the two area system: 1. The overall governing characteristic of the operating units in any area can be represented by a linear curve of frequency versus generation. 2. The governors in both the areas start acting simultaneously to changes in their respective areas. 3. Supplementary control devices act after the initial governor response is over.
Control of Interconnected Systems
243
The following time instants are defined to explain the control sequence: to is the instant when both the areas are operating at the scheduled frequency and tie line interchange and load change takes place. t l is the instant when governor action is initiated at both A and B. t2 is the instant when governor action ceases. t3 is the instant when regulator action begins. t4 is the instant when regulator action ceases. Consider a load increment in area B. From Fig. 7.2(a), it is clear that at the end of the governor response, the tieline schedule is upset and frequency is less than normal. If now a frequency controller is provided in area B, which shifts the governor characteristic upwards, parallel to itself, so as to provide the required control action, generation in B meets its own load change. Tie line schedule is maintained. Change in generation in A is also reduced to zero. Consider now the controller action to load change in area A with the controller located in area B as before. The response is shown in Fig. 7.2(c). f
f
Load change inB
Load change inB f
0
PA
PB
t4  t3  t2  tI   to
To A
0
PA
t3 t2 ~
~
To B
To A
(a) Unregulated case
~
I I I I I I I I I I I 1II'"'
~To
(b) Controller at B
Fig. 7.2 Flat frequency control (a) Load increment in area B  no controller (b) Load increment in area B and frequency controller in area B
B
PB
244
Operation and Control in Power Systems
~~f Load change inB
I
o
PB
I I I I I I
,,j
To A ~
0
~
To B
The line flows
(c) Load increment in area A and frequency controller in area 8
While the initial governor response is the same as for the previous case, the action of the controller in B will force the generation in area B to absorb the load increment in area A. When the controller begins to act at t 3, the governor characteristic is shifted parallel to itself in B ti II the entire load increment in A is absorbed by B and the frequency is restored to normal. Thus, in this case while the frequency is regulate
7.3
Flat TieLine and Flat Frequency Control
Since a flat frequency controller cannot provide the required control, let us consider another controller, a tie line power flow controller located in the area. As an example, let a frequency controller be provided in A and a flat tieline controller be located in area B. The following four possible cases are considered according to the sequence of operation of the controllers in areas A and B. 1. Load change in A: Sequence of operation A, B, A, .. . 2. Load change in B: Sequence of operation B, A, B, .. . 3. Load change in A: Sequence of operation B, A, B, .. . 4. Load change in B: Sequence of operation A, B, A, .. .
245
Control of Interconnected Systems
Case 1 : Let a load increment take place in area A. It is assumed that following the initial governor response, the frequency controller in A acts first at t3 and its control action ceases at t4 . By the time the tieline controller at B becomes effective at t5, it finds the system in normal condition. both frequency and tie line deviations are reduced to zero, and further regulating action is unnecessary. This action is shown in Fig. 7.3(a). f
®
Flat frequency controller
PAl
I
Tie line frequency controller
0
tslliilI
t4 
t3 ,
t2
T
1
1
1
T
t1 
PB
1
______ L __
 +
l+,
1
1
11



to~~~~~~
To A
~
0
I.~
To 8
Fig. 7.3 Flat time line and frequency control frequency controller in area A and tieline controller in area B. (a) Load increment in area A  sequence of operation A  B  A  etc.
Case 2: Let the load increment occur in area B. The tie  line controller in B acts first following the initial governor response at t3 and increases the generation in B ti II the load increment is completely absorbed. At the instant t5, the frequency controller in A finds that the conditions are normal and no regulating action is needed. The load increment in B is absorbed by the generation in B itself. The case is illustrated in Fig. 7 .3(b). Case 3 : Load change occurs in A and the controller in B acts first. The initial governor response which begins at the instant t1 ceases at t2. The tie line controller in B begins and completes its action before the frequency regulator at A takes control. During this period of control action, i.e. t3t 4 by the tie line regulator the frequency decreases further than it was at the end of the governor action period while the tie line power schedule is maintained.
246
Operation and Cont:ol in Power Systems f
I I
I
o
Flat frequency controller
Tie line frequency controller
~:==~I::J1======
t3  
t2 
t)+
I I I I I
=====llS'  + 
i
.1

~
_L
I
~t
to~4~~++
To A
~
0 i~~ To B Tieline flows
Fig. 7.3 (b) Load increment in area B  sequence of operation BABetc.
At t5 the frequency controller in A takes over and restores the frequency to its normal value. It can be seen from Fig. 7.3(c) that the tie line schedule is again upset but by a larger margin than before. Considering subsequent cycles of control action by the controllers in B and A, it can be seen by similar reasoning that stable operating conditions cannot be attained for the sequence of regulator action conceived in this case. f
®
Fig. 7.3 (c) Load increment in area A  sequence of operation BABetc.
247
Control of Interconnected Systems
Case 4 : Finally consider the load increment to have taken place in area B. Let us assume that the frequency controller in A starts and completes its action the tieline  controller in B starts functioning. The response to governor and controller action is showf.l in Fig. 7.3(d). f
To A .   0 i.~ To B
Fig. 7.3 (d) Load increment in area 8  sequence of operation A8A etc
The flat frequency controller at A incre?,ses the generation in A to correct the frequency by shifting the characteristic parallel to itself upwards; The generation in A and hence the tieline power from A to B increases. At ts' when the tieline controller takes over the control, it will find excessive power flow from A to B and to keep it at the scheduled value, it increases the generation in B. While doing so, the generation in A gets reduced and the tie line power flow to B is also reduced. Finally, when the frequency controller begins to act in A at t8 neither the tieline power schedule nor the frequency is normal. The system starts oscillating and no regulation is feasible. It can be seen from the above analysis that if one of the regulators is sluggish in any area where changes in power demand take place, than other regulators, flat tieline flat frequency regulator strategy cannot assist in getting the desired control.
7.4
TieLine Bias Control
The regulators described in Sec. 7.3 can attain the desired control in so far as the local load changes are considered. However, they fail to respond to changes in remote areas. One way to overcome this problem is to make the flat tieline controller responsive to frequency changes also. That is, the flat tieline controller is to be biased so that we obtain a tieline bias controller.
248
Operation and Control in Power Systems
With this type of controller, the regulator does not act to hold a constant tieline interchange regardless of system frequency, but instead allows the tie line schedule to deviate from the normal in adjustable proportion to frequency deviations from the rated value. In principle, the tie line bias controller in an area endeavours to take action for changes in load in its own precincts, while taking no cognizance of the same for other interconnected areas. I The characteristics of tieline bias controller and flat frequency controller are shown in Fig. 7.4.
i
Lower generation
i
Raise generation
0
p~
(a) Tieline bias controller
Lower generation Raise generation
0
p~
(b) Flat frequency controller
Fig. 7.4 Tie  line bias and flat frequency controller characteristics
The point of intersection of both the characteristics is the only point of operation where the system, frequency and tie line schedule can be maintained at the normal values. As in the previous control strategy, all the four possible cases of operation are discussed. Let area A be equipped with a flat frequency controller and area B be provided with a tieline bias controller.
Case 1 : Consider a small load increment in Area A. Let the flat frequency controller in area A act first. The response to governor and controller action is same as that shown in Fig. 7.5(a). Following the initial governor response, the frequency controller in A begins and completes its action before the tieline bias controller at B comes into operation. Both the frequency and tieline power schedules are maintained at scheduled levels by the controller in A. No further corrective action is needed by the tieline bias controller in B.
Case 2 : Let the load increment take place in B with the tieline bias controller in B acting first. The response to governors and controllers would again be the same as given in Fig. 7.5(b). Following the initial governor response, the tieline bias controller in B acts during t3  t4 an9, by shifting the governor characteristic in B, both the tie line power and frequency deviations are corrected. No further regulating action is required from the controller in A.
Case 3: Let the load change occur in area A. The controller in area B acts first, followed by the frequency controller in area A. If the governor characteristic is prevented from being shifted by the action of the tieline regulator in B during the interval t3  t4 the
Control of Interconnected Systems
249
frequency controller in A can restore the frequency to its normal value during the period t5  t6 so that the controller at A acts for changes in load in area A only and absorbs all of them. This can happen if the slope of the ticline bias controller in B is same as the slope of the governing characteristic at A at the point of operation under consideration. Under such a condition, the response to controller action is shown in Fig. 7.S(a). f
®
I
o
I
l~~
11
t2,tl 
to
To A ~ 0 i~~ To B
Fig. 7.5 Tieline bias control in area A and tieline bias controller in area B (a) Load increment
In
area A  controller
In
A acts first
Case 4 : Finally, let the load increment occur in B while the controller in A initiates action first following the governor action. It r.an be seen from Fig. 7.S(b) that during the period tl  t2, the governors in A act following the load increment in area B. The frequency falls and the tie line power flows from A to B. During the period t3  t4 , the flat frequency controller in A acts, generation in A increases and more power flows to B via the tieline. The frequency is normal. During t5  t6, the tielil}e bias controller in B acts to set the tieline deviation to zero. The governor characteristic is raised in B. Frequency increases but tieline flow from A to B is reduced. Neither the frequency nor the tieline schedule are normal at this stage. However, during the period t7  t8, the flat frequency controller in A again takes over the control and the governor is set at a lower value in A. Generation in A is reduced until the frequency is normal and tieline power flow deviation is zero.
From the above analysis it can be inferred that each controller operates effectively to control load changes in its own area. For load changes in other interconnected areas, the tieline bias controller can be set to refrain from taking improper action while the frequency regulator still persists in its tendency to sacrifice the local area requirement to overall requirement.
250
OP.eration and Control in Power Systems Since the frequency controller in A responds unnecessarily to changes in load in B, it may be replaced by another tieline bias controller. This results in complete tieline bias control.
o __ 
To A
®
f
~
t~~
To B
7.5 (b) Load increment in area 8  controller in A acts first
7.5
Complete TieLine Bias Control
With tieline bias controllers located in both the areas the control action is complete in all the cases. For the case w1lere the tieline bias controller in A initiates action for local load changes, the response is shown in Fig. 7.6(a). The response to controller action is similar for load changes in B when the controller in B initiates regulating action. f
I I
t4_
t3__
I I
o
+_1. __ _ 1
t
t 2 __ tJ to~~~~~~ 0 ~ To 8
Tieline flows (a)
Fig. 7.6 Complete tieline bias control  tie line controllers in both the areas A and 8 (a) controller action in A for local load changes
Control of Interconnected Systems
251
For the case when a controller initiates action for nonlocal load changes the controllers will not act if the slope of the controller characteristic is matched to that of the governing characteristic. The response characteristics are illustrated for a load change in area A while the controller in B is the first to initiate action Fig. 7.6(b). During this period the controller in B is prevented from regulating the system. The . incremental changes are corrected by the controller in A only, during t5  t6 . The control action is quite satisfactory. From the above analysis, it is seen that the governor action in an area is not opposed by supplementary control action in the same area. f
®
to~~~~~
A"
0
~B
Tieline flows
Fig. 7.6(b) Load change in A  controller in B acts first
The governor action is not changed till the controller in the area, where a load change has occurred, becomes effective. This avoids unnecessary change in generation, frequency or tieline power. The controller in the area, where load change occurs, acts in such a manner that the area absorbs its own load change. Only a single shift is necessary to the governor characteristic to restore both frequency and tieline power to normal. A smooth, cooperative regulation is thus achieved with a tieline biastieI ine bias control scheme for the two area system. For successful operation, it is imperative that the control characteristic of each tieline bias controller must have the same slope as the governor characteristic of its own area. Fig. 7.7 shows the three types of controller characteristics for comparison.
It can be seen from Fig. 7.7 that flat tieline control characteristic and flat frequency control characteristic are only limiting cases of tieline bias control characteristic with zero bias and infinite bias respectively. A tie line bias controller with larger bias (less slope) compared to its area governor charlatanistic will be more sensitive to load changes in the outside system than the governor and acts like a flat frequency controller.
252
Operation and Control in Power Systems
I
f
f
f Raise
Lower
gen
gen
Raisegen
'..:.... p
A+
I
Lower gen
Lower gen Governor characteristic
'        .. P ~p
~B
(a) Flat tieline controller
gen
(b) Flat frequency controller
A~
~B
(c) Tieline bias controller
Fig. 7.7 A comparison of controller characteristics
If the tieline bias controller has larger slope (smaller bias) than the governor characteristic, it acts like a flat tieline controller. It is therefore necessary to match the speed governor characteristic and the tieline bias control characteristic for satisfactory control.
In
I'
I"
Fig. 7.8 Mismatching of tieline bias control characteristics
Further, as the tieline bias controller acts to correct the deviations in the tieline power flow schedules, the tieline interchange must be metered accurately and then a correcting strategy be applied. If the tieline bias controller characteristic is not matched to the concerned area governor characteristic, the effect could be seen from the following analysis. Let the governor characteristics of areas A and B be represented by curves I and 2 as shown in Fig. 7.8. The tieline bias controller characteristic is assumed to coincide with the governor characteristic in area A (i.e. curve I) but the characteristic of the controller in area B is assumed to be curve 3.10 is the actual tieline power flow. Due to mismatching of the curves in B, the interchange increases to " and the frequency falls from the synchronous frequency fs to f '. In order to control the frequency, the characteristic' is shifted upwards to I' and the curve 3 to 3'. The interchange increases to I" while the frequency is regulated. After this discussion mathematical analysis of the two area system will b~ now presented.
253
Control of Interconnected Systems
In chapter 6, single control area and its response to changes in load, both uncontrolled and controlled is discussed. Let us consider now two control areas connected by a tieline. Each control area will have the same frequency and the same frequency deviations for the entire area. The tieline is considered weak in the sense that both the areas can have frequency deviations different from each other in uncontrolled mode (i.e) M, and f1f2 respectively.
7.6
Two Area System  TieLine Power Model Tie line (X)
Area I
Area 2
I'D 12 Fig. 7.9
Consider two inter connected areas as shown in figure operating at the same frequency
fl while a power
P,~ flow from area I to area 2 let
IV,I and IV 21
be the voltage magnitudes at
8? and 8~ voltage phase angles at the two ends of the tieline while P,~ flows from area I to area 2 then, pO
_IVlllv2ISin(~o (0) X \u, 2
..... (7.1 )
12 
where X is the reactance of the line. If the angles change by f1o,and f10 2 due to load changes in areas I and 2 respectively. Then, the tieline power changes by f1P12
=
Iv,ollv~1 X
f1P 12
0
0
COS(OI  oJ (f1P,  f1P 2 )
..... (7.2)
f1P 1 = 1
MW/radian ..... (7.3) f10 1 M2 M as synchronizing coefficient of the tieline or "stiffness coefficient" of the line, denoted Defining
'=
by~
..... (7.4)
consider
f1(j) =
(i.e.)
21tM
~f18 dt
=
~f1o dt
I d
M=f18Hz 21t dt
254
Operation and Control in Power Systems In otherwords
fM dt rad = 21t fMI dt rad
L\8 = 21t Hence
L\8 1
and
L\8 2 = 21t fM2 dt rad
..... (7.5)
From equation (7.4) LlPI2 = 21tT°{JM1dt 
fM 2dt} MW
..... (7.6)
Taking Laplace transform on both sides
o
21tT L\P12 (S) =   [L\FI (S)  L\F2 (S)] S
..... (7.7)
Block schematic of the above equation is shown in Fig. (7. I 0) If the two areas are rated at Prl and Pr2
P P
_rl
then if
r2
=a 12
~.:I~~
L\P 1(s)
_...J.t__.~
L\P 2(S)
_:,I_ _ _ _
. . t.,li..rI .
L\P 12(S)
~
2n:TO
Fig. 7.10 Block diagram for tieline power
7.7
Block Diagram for Two Area System
The block diagram for a two area system can now be developed. Each area can be represented by a block diagram as in the case of a single area system, but with suffixes I and 2. the block diagram for the tieline power deviation can be used to inter connect both the areas as shown in Fig. (7:1 I).
255
Control of Interconnected Systems
I
6T1(s) RI
6PRefl
6P Ref2
I
6T2 (s) R2
+
Fig. 7.11 Block diagram for 2area systems
7.8
Analysis of Two Area System
Steady State Response: Consider the speed changer positions as fixed so that Ll P ref, I = Ll P ref, 2 = 0 let the loads in both the areas change by LlP DI = kl and LlP D1 = k2, step changes in the two area respectively. Let MV be the final or steady state frequency deviation due to load
changes. Simialary LlP102 let be the change in the tieline power flow. From the Fig. 7.11
..... (7.8)
..... (7.9) where LlP~G,1 and LlP~G,2 are the steady state changes in turbinegenerator outputs. Also, from the same Fig. 7.11, at the summing point of load and tieline powers, we get
°
0)
I ( ~M kl LlP12 Kpi = M
°
..... (7.10)
256
Operation and Control in Power Systems
(i.e.)
since
(t similarly for area 2 (
J
MO kl = D1M o
~2 M
O 
k
I
+,1PI~
.... b.ll)
I . I
2) =D 2M + ,1PI~ O
..... ~7.12)
solving the above equations (7.11) and (7.12) we obtain
,1fo = ( kl + k2
and
Ll
where
~I
~I + ~2
JHz
..... (7.13)
ApO __ ApO _ ~lk2 ~2kl 12  Ll 21 
..... (7.14)
~I +~2
1 =D 1 + 
..... (7.15)
RI
1 = D2 + R2 which are defined as the area frequency response characteristics. Let both the areas be identical and
~2
Then,
DI
=
RI
= R2 = R
..... (7.16)
D2 = D
~I = ~2 = ~
The deviations become
and
,1fo=k l +k 2 Hz 2~ ,
...... (7.17)
,1pO __ ,1po _ k2  kl MW 12 21 2 p.u.
..... (7.18)
If the area ratings are different and
~
=
a 12 then,
Pr2
and
..
Control 0/ Interconnected Systems
1.57
If a load disturbance occurs in only one of the areas, it is clear that with kl k2
=
=
0 or
0 the frequency derivation ~fl is only half of the steady state error that would have
occurred had there been no inter connection. Thus, with several systems inter connected, the steady state frequency error would be reduced. Also from the tieline power derivation it can be observed that half of the load change in either area will be supplied by the other area, which demonstrates the importance of emergency assistance in inter connected or pool operation.
7.9
Dynamic Response
From the block diagram of Fig. (7.11) the following equation can be written
..... (7.19)
~P21 (S) = ~P21
(S)
There are four equations with four variables, ~fl' ~f2' ~P12 and ~P21 to be determined for given
~PDl
and
~PD2.
The dynamic response can be obtained; even though it is a little bit
involved. For simplicity assume that the two areas are equal. Neglect the governor and turbine dynamics, which means that the dynamics of the system under study is much slower than the fast acting turbinegovernor system in a relative sense. Also assume that the load does not change with frequency (D,
=
D2 = D = 0).
We obtain under these assumptions the following relations
258
Operation and Control in Power Systems since D
=
0
equations (7.19) and (7.20) simplify to ~P12 (S) ]  f
1 [  R ~FI (S) 
~PDl (S)
1 [  R ~F2 (S) 
~PD2 (S)  ~P21 (S) ]  f

O
2SH
=~FI (S)
..... (7.21)
= ~F2 (S)
..... (7.22)
O
2SH
subtracting (7.22) from (7.21)
~ ~FI (S) + ~R ~F2 (S)  ~PDl (S) + ~PD2 (S)  ~P12 (S) + ~P21 (S)] ~ = ~F (S)  ~F (S) 2SH 2
[ R
I
1 ] f ~P12 (S) [ R ~F2 (S) + ~FI (S) + ~PD2 (S)  ~PDI (S)  2~PI2 (S) 2SH = 2nT O
°
since Therefore,
~P
12
(S)[
Sfo + 2fo _S_] 2nRT0 2SH 2SH +I 2nTO
=[~PD2 (S)  ~PDI (S)]f O 2SH
..... (7.23)
Control of Interconnected Systems
259
..... (7.~4)
The denominator is of the form
(S2 +2KS+ro 2)=(S+K)2 +(ro 2 _K2) where setting
..... (7.25) Note that both K and ro 2 are positive. From the roots of the characteristic equation we notice that the system is stable and damped. The frequency of the damped oscillations is given by roo. Since Hand fO are constant, the frequency of oscillations depends upon the regulation parameter R. Low R gives high K and high damping and vice versa. IfR ~ a; K ~ 0 is the condition for nogovernor action and there will be undamped oscillations. We thus conclude from the preceding analysis that the two area system, just as in the case of a single area system in the uncontrolled mode, has a steady state error but to a lesser extent and the tie line power deviation and frequency deviation exhibit oscillations that are damped out later.
260
Operation and Control in Power Systems
E 7.1 Two control areas have the following characteristics
Area I;
Rl 0
=
0.011 p.U
1=
0.85 p.u.
Base MVA = 1000 Area 2;
R2 = 0.018 p.u O2 = 0.95 p.u.
Base MVA = 1000 A load change of200 MW occurs in area I. Determine the new steady state frequency.
1lution:
kl
= 0.2pu ; k2 = 0
I I PI =0 1 +=0.85+=90.9091+0.85=91.7591 Rl 0.011
1 1 = 0.95 +   = 55.55 + 0.95 = 56.405 R2 0.Ql8
P2 = O 2 + 
~fO
=
kl
=
0.2
(131 +P2) 91.7591+56.405
=~=0.0013683 148.16
f= 50  (0.0013683 x 50) = 49.93Hz
E 7.2 In example E. 7.1, determine the tieline power flow deviation.
Solution:
~Ptleline =
= (~fl)
k 1P2 A
A
PI +P2
P2 = 0.0013683 x 56.405
= 0.0771789 p.u MW = 77.1789 MW
p.u. Hz
261
Control of Interconnected Systems
E 7.3 In E 7.1 if the load disturbance occurs simultaneously also in area 2 by 100 MW determine the frequency and tieline power changes.
Solution: K2 = 0.1 pu
MO = kl +k2 =
0.2+0.1
=0.0020248puHz
~I + ~ . 91.7591 + 56.405
~f>= 0.10124 Hz
f= 50  0.10124 = 49.89876 == 49.9 Hz
~p.
tIe  hne
= ~lk2  kl~2 rt
rt
PI +P2
91.7591
X
0.1 0.2 X 56.405 148.16
=
9.17591 11.2810 148.16
2.10509 _ 0.014208 u Mw 148.16 P =  14.2082 MW E7.4 Two interconnected areas A and B have capacities 2000 MW and 750 MW respectively. The speed regulation coefficients are 0.1 p.u. for both the areas on their own area ratings. The damping torque coefficients are 1.0 p.u. also on their own base. Find the steady state change in system frequency when a load increment of 50 MW occurs in area A. Find also the tie line power deviation. System frequency is 50 Hz.
Solution:
 .!A  2000  2.67
aI2 
Pr2
750
I D = 1.0 p.u. MW/puHz = 50 p.u. MW/Hz R = O.lpu = 0.\ x 50 = 5Hz/p.u. MW ~I = ~2 = ~ = D + ~
1
R=
1 50 +
\
5'
= 0.02 + 0.02 = 0.22
o k2 + a l2 k l f = ='"'~2 + a12~1
50Mw 1 kl = 2000puMW = 40 puMw;k2 =0 1
2.67xMO= 40 = 0.06675 = 0.08267Hz 0.22 + 2.67 x 0.22 0.8074
262
Operation and Control in Power Systems
== 0.00681 x 2000 == 13.62MW
E7.5 In problem E 7.4, if the tie line connection is lost while it is carrying 50MW load, what will be frequency in system A.
Solution: The system A will be now an isolated area. The frequency drop
~f
== 
°
~PDA ~
==  40 == 0.1 13636Hz
0.22
E7.6 For a two identical area system the following data is given. Determine the frequency of oscillations when a step load disturbance occurs. Speed regulation coefficient == R == 4 Hz Ip.u. MW Damping coefficient D == 0.03 pu MW1Hz System frequency
=
50Hz
The tie  line has a capacity of 0.1 p.u. The power angle is 30° just before the occurrence of the load disturbance.
Solution: fo 50 K ==   ==. == 0.625 4RH 4x4x5 TO == 0.1 cos 30° == 0.0866 0)2
== 2nTofo == 2n x 0.0866 x 50
H 0)
= 2.333
0)0
=5.443
5 K 2 = 0.3906
== ~5.443  0.3906 ==
~5.052375 == 2.2477489rad I sec
However, if the damping coefficient is not to be neglected K == ~ [D + ~] == 0.625 + ~ x 0.03 == 0.625 + 0.075 == 0.7; k 2 == 0.49 4H R 4x5 In this case ro = ~5.443  0.49 = ~0.4953 == 2.2255rad I sec
Control 0/ Interconnected Systems
263
E7.7 A power system consists of two areas interconnected by a tie line which has a capacity of 500MW and is operating at a power angle of 35°. If each area has a capacity of 5000MW and the speed regulation coefficients for both the areas are also the same and are equal to R = 2HzJpuMW determine the frequency of oscillation of the power for step change in load. The intertia constants are also the same for both the areas and are equal to H = 5sec.
Solution: TO
500 = cos35° = 0.819 x 0.1 = 0.0819 5000
0)0 =
[21t X50XO.0819]_( 50 )2 5 4x2x5
=./ 5.148 
1.5625 = ./3.5855 =1.8935rad / sec E7.8 In the above problem, if a step load change of85MW occurs in one of the areas determine the tie line power deviation.
Solution: The two areas are equal, the load change will be shared equally by both the areas.
A power of 85/2 change occurs.
=
42.5 MW will flow from the other area into the area where a load
E7.9 Two power stations A and B operate in parallel. They are inter connected by a short transmission line. The station capacities are 100MW and 200MW respectively. The generators at A and B have speed regulation 3% and 2% respectively. Calculate the output of each station and the load on the interconnector if, (a)
the load on each station is 125MW,
(b)
the loads on respective bus bars are 60MW and 190MW and
(c)
the load is 150MW at the station A bus bar only.
Solution: P'+P 2 = 125+125 = 250MW
Case: (a)
Regulation
NoN
fof
"=
PI
100
1f 0.03
P 200
1 f 0.02
=
 2=  
264
Operation and Control in Power Systems 0.0003P I = (I  f) = 0.0001P 2 that is,
3P I = P2 P I +P2 =2S0
Solving
PI = 62.S MW and P2 = 187.SMW A
62.5MW
~1t~
B
0+1~
__ 62'_5
125MW
187.5MW
125MW
Case (b) : the loads on each bus bar is determined by the speed regulation characteristics.
3 PI
=
P2
PI + P2 = 60 + 190 = 2S0MW Solving
PI = 62.S MW and P2 = 187.S MW
A 62.5MW
B
~f+~_2'50+i~ 60MW
187.5MW
190MW
Case (e) : There is load of ISOMW at A only
3 PI = P2 PI + P2 = ISO Solving
PI = 37.S P2
=
112.SMW
A
37.5MW
B
r;:::'\ ~ I 12.5 M W I r;:::'\ '01I~I+1'0 150MW
112.5MW
265
Control of Interconnected Systems
E7.10 The turbines in a power station A have a ulliform speed regulation of 2.5% from no load to full load. The rated capacity of the! generators connected to the bus bars total 60MW and the frequency is 50Hz. Station B has total generating capacity of 35MW at the bus bars and has a speed regulation of 3.0% connected through an induction motor generation set rated 10MW. Which has1a fupload slip of 3.0%. There are loads of 32MW and 24.5MW connected to station A ~nd B respectively. Find the load on the inter connecter cable at this operating condition.
Solution: A
60MW
IOMW
+p
B
6?+~~M_G,set1~+l6?
35MW
24.5MW
32MW
Let P MW flow from A to B Total load on A
=
(32+P)MW
Total load on B
=
(24.5 P)
Percentage drop
Percentage drop
In
speed at A
=
(32 + P) 2.5 =~ 60
.
d spee at B
=
3 (24.5  P) .5 ''35
.
In
3P Percentage drop in speed of induction motor generator set = 100 Percentage drop in speed at B  percentage drop in speed at A = percentage drop in speed of induction motor generator set 3.5(24.5  P) 35 (24.5  P) 10
2.5(32 + P) 3P =10 60
3P 32 + P =10 24
''  
solving
106P = 268 268 P = 106
=
2.5283MW
266
Operation and Control in Power Systems P flows from A to B Total load as
A = 32 + 2.5283 = 34.5283MW
Total load as
8 = 24.5  2.5283
=
21.9717MW
7.10 TieLine Bias Control Implementation From the preceding discussion it is clear that in interconnected operation each area in normal steady state must control the changes in such a fashion that it absorbs its own load change.
The tie line bias control strategy presented can be summarized: all inter connected areas must contribute their share to frequency control in addition to taking care of their own net interchange. The control error for each area can be now defined as a linear combination offrequency and tie line power errors. Area control error for area I; ACE 1 =
~P12
and area control error for area 2; ACE 2 =
+ 8 1~f1
~P21
. .... (7.26)
+ 8 2 ~f2
Hence in the block diagram of Fig. 7.11 now the
..... (7.27)
~p ref commands
can be defined . ..... (7.28)
and
..... (7.29)
where 8 1 and 8 2 are the frequency bias parameters for areas I and 2 while K·1 and K; are the integral controller gain constants. Note that under steady state conditions when eqn. (7.26) ~P12 + 8 1~f1 = 0 and
~P21
~p 12
and
~fl, ~f2
become zero, from
+ 8 2 M2 = 0 convey that the controller action in the final
stage is independent of 8 1 and 8 2 . In fact, even one of the 8's, either 8 1 or 8 2 can be zero. It is suggested that only one 8 can be selected as equal to the area frequency response characteristic
P= 0
I
+  to give R
satisfactory performance. The complete tieline bias control of a two area system is shown in Fig. 7.12. p
If area ratings are different (I) block is replaced by ( a 12 )
=
(fr2
Control
0/ Interconnected Systems
267
.~B,~.~
I
~F,(s)
R,
Fig. 7.12 Tieline bias control of a two area system
The selection of integral controller gains K'I and K; must be such that too large a value should not lead to chasing of minor deviations of noconsequence.
7.11 The Effect of Bias Factor on System Regulation Let a and b be the ratios of generation of areas A and B in terms of the total generation, i.e. a + b = I p.u. Let Psc be the total system generating capacity per cycle and correspondingly, let PAC and PBC be the generating capacity per cycle of A and B. BA is the tieline bias setting at area A expressed in generation capacity / cycle and Bs is the imposed regulating characteristic for the whole system with a tieline bias controller in A only, with no controller in B, the ratio of bias setting at A
~o
the generation characteristic at A, r, is given by
Psc = PAC a + PBC (I  a) = PAC . a + PBC . b
..... (7.30)
Bs = BA a + PBC (I  a) = BA . a + PBC . b
..... (7.31)
268
Operation and Control in Power Systems
The change in natural frequency governor characteristic is
~
fn following the load change as dictated by the
=~PD P sc fb is the system frequency deviation due to the controller, then M
n
If ~
M =~PD S B s from eqn. (7.30) and (7.31) M = _ _ _~_P=D_ _ n PACa + PBC (1 a)
..... (7.32) ..... (7.33)
and
The improvement change in frequency deviation, as a percentage of the initial disturbance is ..... (7.34)
(PAC  BA)a (BA  Pnc)a + PBC
Since, r = BA
,
/ PAC
and assuming
100
..... (7.35)
= P BC '
a(r 1) xl 00% 1 + a(r I)
..... (7.36)
~f = (b I)(r I) x 100% r  b(rI)
..... (7.37)
L\f
or
=
PAC
x
The percentage change in frequency is independent of ~ PD' and depends only on a, b and r. When r = I; ~ f = 0, therefore ~fn=~fs
The initial ungoverned frequency deviation is equal to the subsequent governed frequency deviation.
Control of Interconnected Systems
269
7.12 Scope for Supplementary Control During operation, the system, frequency, tieline flows and other derived quantities will be continuously swinging with periods generally of a few seconds. Considering the time lags in measurements, the prime mover governor characteristics, etc. supplementary controllers cannot respond to rapid swings offrequency and tieline power. Neither can they effectively regulate these changes nor is it expected of them. It is desired that supplementary controllers should regulate the frequency and tieline deviations about a base that corresponds to the prescheduled values. In a reactive sense, the supplementary control is a steadystate control. A high speed dynamic analysis of the system "and area characteristics are required only in connection with improving system stability. .
7.13 State Variable Model for a Three Area System In chapter 6 state variable mode for a single area system and application of optimal control has been discussed. Modern control theory can be applied very elegantly for multi area system. The block diagram representation of a three area system is shown in Fig. 7.13. The folIowing equations are written directly from the block diagram.
Fig. 7.13 Three area system
270
Operation and Control in Power Systems
..... (7.38)
Control of Interconnected Systems
271
Taking the inverse Laplace transform of equation (7.38)
~fl =_I_(~fl +Kpl~PGI Kpl~PDI Kpl~Pne.l) Tpi
~f3
= }( M3 + Kp3~PG3 
Kp3~PD3  Kp3~Ptle,3)
P3
~PGl =l(~PGl +~XVl) TTl
..... (7.39)
~Ptlel =21tTI02(MIM2)+21tTI03(MIMJ
~Ptie.2 = 21tT~1 (M 2  Ml ) + 21tT~3 (~f2  ~fJ ~Pl1e,3 = 21tT~1 (M3  M 1 )+ 21tT302 (M)  M 2 )
The above equations are put in the form
X= AX+Bu+Fd
...... (7.40)
where XT
=
[M1 M2 M3 ~PGI ~PG2 ~PG3 ~PVl ~PV2 ~PV3 ~Ptie, I ~Ptle, 2 ~Pl1e, 3]
T
u
T
d
= [~P CI ~PC2 ~PC3] = (~PDl
~PD2 ~PD3)
The matrices A, Band F are given by
1
0
0
0
0
Kpl
0
TPl 0
1
0
0
0
0
0
0
0
KPl
0
T
0
0
0
0
0 0
0
0
0
0
0
1
0
0 0
1
0
0
0
0
0
Kp2
0
0
TPl Kp3
0
0
Kp3 Tp3
0
0
0
0
0
1
0
0
0
0
1
0 0
0
"=' ~
0
0
0
1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
..... (7.41)
1
0
1
R2 TS2 0
R T
U
0
0
0
0
0
1
~~2 1
R3 TS3
Q s C
0
0
0
0
1
0
TS3
I
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
.,1= .... ...
= = 1= = =n = .... .,=
Tn TSl
P B
A
0
TT2
RITSl 0
0
Tn
Tn 
0
Tp3 1
TT2
A=
0
TP2
Tn 0
_ Kpl TPl
Tp2 0
0
TPl Tp2
0
0
j
=...= ~
= ~ ., ~
00
'< r.I
.... a
~
r.I
273
Control of Interconnected Systems Where,
A= 21t(TI02 + TI03) B= 21t(Tfl + Tf3)
c= 21t(T~1 + T3 P = 21tTI2°  =  21tT ° Q
02 )
I3

°
R = 21tT2I
S = 21tT23° 
0

°
T = 21tT31
and
U = 21tT32
B=
0
0
0
0
0
0
0
0
0
0
0 0
0
0
0
0
0
0
0
0
1
TSI 0
0
TS2 0
0
0
0
TS3 0
0
0
0
0
0
0
..... (7.42)
Operation and Control in Power Systems
274 Kpi Tpi
F
0
0
0
_ Kp2 Tp2
0
0
0
0
0
0
0
0
0
Kp3 Tp3 0 0 0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
~
..... (7.43)
The equations can be written down for any number of interconnected areas in a similar manner.
7.14 State Variable Model for a Two Area System In the case of two area system ~Ptle,2 = a 12 Plle,1
P a I2 =_ll Pr2 The system matrices time become 1 0 TpI and
0
0
0
Kpi Tpi
0
0
0
Tp2 0
A=
0
0 0
Tst
0
TSI 0
0
TS2

0
0

0
21tTl~
Kp2 Tp2 Kpi Tpi I
_ Kp2 Tp2 Kp2 a 12 TP2 0 0
TS2
0
RITSI
0
0
0
TSI R2 TS2  21tTl~
0 0
0 0
0
0
0
TS2 0
0
275
Control of Interconnected Systems
oro 0 0
BT
0
0
TSI
000 0
0
TS2
r FT =
0
KPI :PI
:]
0 0 0
o 0]
_ Kp2 0 0 0 o 0 Tp2 The static variables, control variations are disturbance variables reduce to XT
= [M1 M2 LlPGI LlPG2 LlX vl LlX V2 LlPtie,d
u = [LlPCI ] LlPC2 d = [LlPDI ] respectively. LlPD2
7.15 State Variable Model for a Single Area System The state variable model for the single area system is a single input, single output model with no tieline power. The matrices of the system equations are Tp
0
Ts RTs
0
0
B=
Tp I
0
A=
~
0 1
Ts _ Kp Tp 0 F= 0 XT = [M LlPG LlX v] u = LlPc d = LlP D
Ts 1 Ts
276
Operation and Control in Power Systems

E 7.11 Solve the state space equations for a three area system with the following parameters and plot the frequency, generation and tieline power deviations for a step load disturbance of 0.01 p.ll. in area 1. RI = R2 = R3 = 2.SH:zJp.u. MW TpI = Tp2 = Tp3 = 20s TTl = TT2 = TT3 = 0.3s TSI = TS2 = TS3 KpI
=
=
0.08s
Kp2 = Kp3 = 120H:zJp.ll. MW
= T32 = T31 = 0.07 p.ll. MW/rad
TI2 = TI3 = T23 = T21 PDI = O.Olp.ll. MW
PD2 = PD3 = 0.0 Solution: The state variable mode for the system using eqn. (7.27) is given by
~1\
=
~f2
= 20 [ M 2 +
~f3
= 20 [ M3 + 120(~PG3 )120(0.0) 120(~Ptie,3)]
·
1 20 [ MI + 120(~PGI )120(0.01) 120(~Ptle,I)] 1
120(~PG2)  120(0.0)  120(~Ptle,2 )]
1
1[
~XVI = ~XVI
·
0.08
I[
I]
+O.OMI 2.5
I]
~XV2 = ~XV2 +0.0~f2
0.08
· . 0.081[
2.5
I]
~XV3 = ~XV3 +0.0~f3
.
2.5
Control of Interconnected Systems
277 ~t( s)
2
0
4
3
5
6
7
8
9
10
II
12
13
14
7
8
9
10
11
12
13
14
 0.004
N  0.008 ~
'
l
0.012  0.016  0.020 (a)
I
I, .
;\j}v •
(",
~ \~ /,,
.~p G3
.,...............,. ~PG2 ,,~PG,
0
2
3
4
5
6
~t( s)
(b)
Fig . E 7.4.1 Response of a three area system to step load change in one area: (a) variation of frequency error
(b) variation of generation changes
~Ptie'
= 27t(O.07)(M,  M 2 ) 
27t(O.07)(M,  M2 )
~Ptie2
= 27t(O.07)(M2  M, ) 
27t(O.07)(M2  M 3 )
~Prie3 = 27t(O.07)(M3 
M,)  27t(O.07)(M,  M 2 )
The equations are solved on digital computer using the Runge  Kutta method. The results are plotted: The riot is silm,vn in Fig. E 7.4.I(c).
278
Operation and Control in Power Systems
0.007 0.006 0.005 0.004 0.003
r .!:!
c:
l
0.002 0.001 0 2
D.OOI
3
4
5
6
7
8
9
10
II
12
13
14
~t(s)
D.002 D.003 D.OO4 D.OOS D.006 D.007 D.008 D.009
(\
\ I \I ,';1 \ I
\ 1 ,)
"
I I\ iI '\ // \
II\! \! I",
'\j
'\
/
r'\'.. . //"......,~ ~ ~.~

6Plie 1
D.OIO
E 7.4.1 (c) Variation of tie line power
E7.12 Given the following parameters pertaining to a two area system in appropriate units Ttl
0.28
Tt2
0.3
Tpi
18
Tp2
20
RI
2.4
R2
2.5
TSI
0.1
TS2
0.1
Kpi
120
KP2
100
TI2
0.06
Obtain the frequency deviations in both the areas and tie power deviation when a load change of llP DI == 0.0 I p.u. MW occurs.
Control of Interconnected Systems
279
Solution: The system is simulated in MATLAB the frequency deviation is shown in Fig. E 7.S (a) and the tieline power deviation in Fig. E 7.S (b).
Fig . E 7.5(a)
Fig. E 7.5(b)
280
Operation and Control in Power Systems
E7.13 For the system given in E 7.5 obtain the controlled response of the system with tieline bias control.
Solution: For tie line bias control the bias coefficients are computed form I I I 8 1 =°1 +  =  + RI KPI RI
I R2
and
1 KP2
1 R2
8 2 = 0 2 +  =  + 
An integral control gain of KI = 0.5 is used for both the areas. The tieline power deviation is shown in Fig. E 7.6(a) and the frequency response in Fig. E 7 .6(b).
\ A):OO5'
\
\
\
Fig . E 7.6 (a)
Fig. E 7 . ~ (b)
Control of Interconnected Systems
281
E7.14 In the system shown in example 7.5, if load disturbance of L1PDI
= L1P D2 =
occur
simultaneously obtain the freciuency response and tie line power deviation.
Solution: The response of tieline deviations and frequency deviation are shown in Fig. E 7.7(a) and E 7.7(b).
;:
;,
"::0.005 0,01
_·it
:..().Qj~
0.025 0.03
0.035 '''2 '34''5'67"8''9"0 0
Fig . E 7.7 (a) 8 x 10'"
~
6 4
I \
MI
/''\,
0
I \ /' /
2 : \
I
2
4
6 ~
10
0
~.
\) v
2
3
4
5
Fig . E 7.7 (b)
6
7
8
.9
fO
282
Operation and Control in Power Systems
E7.1S For the system in E (7.5) obtain tieline bias controlled frequency and tieline power deviation with integral controller gains Kfl
=
K'2
=
0.5.
Solution: The responses are shown in Fig. E 7.8(a) and E 7.8(b).
Fig. E 7.8 (a)
Fig. E 7.8 (b)
Control of lnterc~nnected Systems
283
The MATLAB programme and the simulation setup are given below. Ttl
0.28;
Tt2
0.3;
Tpl
18;
Tp2
20;
RI
2.4;
R2
2.5;
Tsl
0.08;
Ts2
0.1;
Kpl
120;
Kp2
100;
Tl2
0.06;
Gain K
Transfer Fon
Transfer Fon I
simout I To Workspace I Scope 3
Gain I ~~K~~~~~
Fig. Uncontrolled Case
284
Operation and Control in Power Systems
Gain ~<K~~,
Bl Gain 3
simout I To Workspace 1
Gain I ~CK~~~~~,
Fig. With tieline bias control
7.16 Model Reduction and Decentralised Control Decentralized control has been developed to avoid the necessity of too many feedback loops in the controller structure. Optimal controller design requires that every sensor output affect every actuator input. In decentralized control, restrictions are placed on information transmission between sensors and actuators.
Fig. 7.14 Decentralized control
285
Control of Interconnected Systems
l
2
Considering Fig. (7.14) XI is used to determine the vector u while X alone is used to determine vector u2 • This is complete decentralized control.
Model Reduction by Aggregation: Consider
Let
X(t) = AX(t)+ Bu(t)
..... (7.44)
2:(t) = F2(t)+ Gu(t)
..... (7.45)
It is required that eqn. (7.45) represent the same dynamic system rep'resented by eqn. (7.44), given that the number of variables in 2 is much less that those in X. For dynamic exactness 2(t) = CX(t)
..... (7.46)
It is necessary then that the equations 'FC and
=
CA
..... (7.47) ..... (7 .48)
G=CB
are satisfied. The aggregate state vector 2(t) must be a linear combination of certain modes of X(t). The eigenvalues of F are the eigenvalues of A corresponding to those modes of X(t) which are retained in 2(t). In other words, a model reduction is achieved by retaining dominant modes. Many a time, it is advantageous to define 2(t) corresponding to physical variables. Then 2(t)
=
HX(t)
..... (7.49)
This may be termed as modal approximation. Eqn. (7.44) may be rewritten as
XI(t)] [ X2 (t) 
[All lOA 21
..... (7.50)
where 10 is a small, positive parameter. For 10 = 0, the system decomposes into two completely independent sub systems. The two sub stems thus obtained may be treated each separately for simulation, analysis and control design. The dimensionality of the problem is then reduced. In multi area operation, the decision makers in each area may use a detailed model of his area and a dynamic equivalent of the remainder of the system. Thus, the same system appears in different forms to different controllers. By singular perturbation technique, it is possible to devise control strategy for fast and slow dynamic models of the system. Multi parameter perturbation method may be used for periodic coordination of all the subsystem controllers using Pareto control strategy.
286
Operation and Control in Power Systems
Decentralized suboptimal controller Consider a system defined by an ndimensional state vector X. It is desired to replace it by a reduced order model with mdimensional measurement vector Z. Then each Z may be considered to be a linear combination of the states X so that
Z = MX
..... (7.51)
If UO is an appropriate control for the original system, it is structured from
UO=KX
..... (7.52)
In view of the relation (7.51) UO must also be a linear combination of Z. Therefore
Uo= FZ The elements ofF can be obtained using various concepts in control theory. For example, if minimum norm concept is utilized then Fj
=
K M; (M) M; tl M) j
=
1,2, ...... , m
..... (7.53)
In eqn. (7.53) K is the full state variable feedback gain matrix. For a two area power system each area may be controls u l and u2 which depend upon the vectors ZI and Z2 given by
ZJ = (LlfILlPoILlXEILlP"e,1 fACE)]
..... (7.54)
zi = [Llf2LlPG2LlXE2LlP"e,dACE2]
..... (7.5~)
Remote measurements are thus avoided. In fact, area controllers are influenced only a little by remote measurements.
Control
0/ Interconnected Systems
287
Questions 7.1.
Explain how the tieline power deviation can be incorporated in twoarea system block diagram?
7.2.
What is a tieline? Explain
7.3.
Derive the relation between steady state frequency error and tieline derivation for step load disturbances in both areas
7.4.
What is area frequency response characteristic. Explain it in the context of two area systems.
7.5.
What are the advantages of inter connected operation of power systems? Explain.
7.6.
Explain how state variable representation can be obtained for a two area system. Write down the equations.
7.7.
Sketch and explain the block schematic of a twoarea system.
7.8.
What are the features of the dynamic response of a two area system for step load disturbances?
7.9.
Explain tieline bias control applied to a twoarea system.
7.10.
What are the considerations in selecting the frequency bias parameter.
288
Operation and Control in Power Systems Problems P 7.1
Two interconnected control areas have the foIlowing characteristics Area I Rl
O.Olpu
=
0 1 = 0.08 p.u. Base MVA = 500 Area 2 R3
=
0.02pu
O2 = 1.0 p.u.
Base MVA = 1000 A load change of 100MW occurs in area l. Determine the new steady state frequency and tieline power derivation. P 7.2
In P 7.1 if a load change of 200MW occurs in area two determine the frequency derivation and tie line power derivation.
P 7.3
In P 7.1 if a load change of 150MW in area I and a load change of 250MW in area 2 occur simultaneously determine the frequency and tie line power derivation.
P 7.4
Given Rl = 2.4Hzlpu MW R2
2.6Hzlpu MW
TTl
0.28s
TT2
0.31s
TSI
0.075 S
TS2
0.08s
Kpl
100Hz / pu MW
Kpl
120Hz / pu MW
Tl2
0.07 p.u. MW / rad
0.07 p.u. MW / rad T21 Solve the state equations for a two area system and plot the frequency, generation and tie line power deviations for a step load disturbance. P 7.5
Repeat problem 7.4 using MATLAB compare the result.
P 7.6
Repeat problem 7.5 with PDI = O.Olpu MW in area I and PD2 = 0.01 pu MW in area 2.
P 7.7
Repeat problem 7.3 with PD2 = 0.03 p.u. MW. What are your conclusions from the study of the results of problem P7.6 and P7.7.
8
VOLTAGE AND REACTIVE POWER CONTROL
Industrial and domestic loads, both, require real and reactive power. Hence, generators have to produce both real and reactive power. Reactive power is required to excite various types of electrical equipment as well as transmission network. The reactive power requirement of consumers arises mostly from the lagging vars n.eeded to supply magnetizing current to transformers and induction motors.
8.1
Impedance and Reactive Power
In the transmission network, requirement is the difference between that absorbed in the series inductance (J2X) and that produced in the shunt capacitance (y2B). There is a level of loading at which the leading vars of charging current balance the lagging vars of the inductive lines; called the natural or surge impedance loading of the system. This natural load for a transmission line is given approxirtlately as
(~) p.u.,
In the case of cables the shunt capacitance is higher
and series inductance is lower. As a consequence of which the natural loading is higher. However, this limit is generally above the thermal limit of the cable. The natural impedance loading for overhead lines is shown in Fig. 8.1 as a function of system voltage on logarithmic scale. The conductors are bundled two together for 220kY and four for 380kY.. TaSte 8.1 shows the reactive power requirements of overhead lines, cables and transformers. The reactive power compensation of the transmission system depends on the load and its power factor. When the
Operation and Control in Power Systems
290
ro'
104 Irrttrr,,
10 I
1''''''''
10
20 30
100 220
1000
(kV) .
Fig. 8.1 Natural impedance loadings as a function of system voltage
line is operated at noload, the full charging power occurs and would result in considerable increase in voltage unless some compensating device is used. With full compensation at noload, the line may be operated at any partial load between noload and fullload with the voltage not exceeding the permissible limits: Table 8.1 Var requirements of overhead lines, cables and power transformers Equipment
Mvar requirements noload
Inductive,  Capacitive Full load
Over head lines: 400kV,4 x 0.4 sq. inch 100 miles 275 kY. 2 x 0.4 sq. inch 50 miles 132 kV. 1 x 0.175 sq. inch 20 miles
105.0
+1251.0
22.0
+174.0
1.6
+10.0
540
487.0
170
155.0
14.5
13.6
400t275kV,750MVA
+0.5
275/132kV, 240MVA 132133kV,'9omVA
+OJ
+90.0 +48.0
+0.2
+20.5
Under taken Cables: /
400kV,2 x 3.0 sq. inch 10 miles 275kV, 1 x 3.0 sq. inch 10 miles 132kV,1 x 0.5 sq. inch 5 miles
Transformers:
Voltage and Reactive Power Control
291
The loss in a line is given by,
P =K L
2
p £ y2 COS2
q,
where P is the power transmitted at voltage Y over a line of length e at a power factor cosq,. Higher voltages are selected for transmission to keep the losses in an economically justifiable relationship to the power P. In view of the inverse square relationship, reduction of reactive power becomes an essential factor for obtaining efficient operation of the high voltage lines. The reactive power requirement of overhead lines are shown in Fig. 8.2(a) & (b).
i... E
0 0
0
Inductive 600
/
700 kv/
400 /
200 0
0.5
_:;::=;:;:::P'"
200

/
/
./ ./
/:::.... "'400kv ~1.0
1.5
2.0 p
Capacitive (a)
r Cl..
1.0 0.8 0.6 0.4 PSIt).2 0 0.2 0.4 0.6
2.0
(b)
Fig. 8.2 Reactive power requirement of overhead lines (4 conductor bundle) (a) Variation with line voltage (b) Variation with line length
In order to supply quality service to customers reliably and economically, voltage and (or) var control play (s) a leading role. Such a control has to be exercised at all the power system. i.e. right from the generating point to the consumer terminals.
Operation and Control in Power Systems
292
I
Rapid changes in voltage (flicker) can result due to some industrial loads such as a* furnaces, arc welders, and wood chippers. Fig. 8.3 shows the real and reactive power demand following the daily load cycle supplying a composite system load which may create relatively large variations in voltage if control is not exercised. Also, there may be cyclic and noncycli¢ loads that create voltage disturbance at both transmission and distribution levels.
5
r
r
4
5 4
3
3
2
2
10
10
30
20
Real power (MW)
~
Reactive power
20
30
(Mvar)~
Fig. 8.3 Arc furnace consumption
In addition, events such as planned line switching, un planned line trips, planned and unforeseen generator trips and equipment failure may produce voltage and var variations. Unless proper voltage support is given at strategic locations in the system, the aforesaid events may result in loss of stability and possible loss of service to a large number of consumers. For long distance transmission of power, the use of HVDC transmission has proved economical in certain cases. The var demand of DC terminals varies usually from 060% of the MW rating of the DC lines as power transfer is varied over its full range. When a fault takes place on the nearby AC system, the var demand of the DC link may reach a high value and unless compensated may produce large AC voltage variations. Continuous control can be achieved by means of synchronous compensators installed at line ends and/or in the intermediate substations. The use of shunt connected controllable var compensation to improve the power transfer capability and stability is an acknowledged fact. From Fig. 8.4 it can be seen that the theoretical maximum power transfer takes place at a power angle of 0'2 = 90.
P'2=  xl2
Fig. 8.4 Power transfer with
,,0 intermediate voltage support
Voltage and Reactive Power Control ..
293
With an intermediate, controllable, shunt var compensator, the angle could be increased, in principle to 1800 across the line (Fig. 8.5).
controllable shunt var compensator
Fig. 8.5 Power transfer with intermediate var compensator
8.2
System Voltage and Reactive Power
Consider the system shown in Fig. 8.6. The voltage at bus 2 is related to the voltage at bus I by the relation. Bus 2
Bus I
I
..
I
P + 1Q
Fig. 8.6 Power flow across a short line
Also, So that
V 2 = VIIZ VI' I· = P + jQ
..... (8.1 ) ..... (8.2)
I = P  jQ V;
..... (8.3)
=P
jQ VI
since V I is the reference phasor. From eqn. (8.1) and (8.2)
V2
=VI(p~~Qlz
Neglecting the line resistance
P
Q
V 2 =V I J' XLX V V L I I
..... (8.4)
Operation and Control in Power Systems
294
Eqn. (8.4) is illustrated by the phasor diagram in Fig. 8.7. It can be inferred from eqn. (8.3) or Fig. 8.7 the voltage level is influenced largely by the reactive power drop QX L VJ
,
PX L since the quadrature component   does not materially affect the voltage profile (both the
VJ
drops are only small fractions of the bus voltage magnitudes).
V2
Fig. 8.7 Phasor representation of eqn. (8.3)
8.3
Reactive Power Generation by Synchronous Machines
Synchronous generators are able to produce both lagging and leading vars. Overexcitation of a generator field produces vars while under excited field causes vars to be absorbed. At lagging power factors, the limit on var generation is imposed by either rotor heating (due to maximum excitation current limit) or by stator heating (thermal MVA loading limit of the stator) consideration. Turbine output limit
Rotor heating limit
0.4 Lead
0.2 0 0.2 0.4 0.6 0.8 p.u. Mvar Lag
1.0
Fig. 8.8 (a) Operating chart for nonsalient pole synchronous machine
Generators are invariably fitted with automatic voltage regulators which maintain the thermal voltage at its normal value by adjustment of excitation. The operating charts for salient and non salient pole synchronous machines are shown in Fig. 8.8.
295
Voltage and Reactive Power Control
Turbine output __ limit
Theoretical stability limit
!...
Stator heating limit
Rotor heating limit
... 0.8
0.6 0.4 Lead
0.2 0 0.2 0.4 0.6 p.u.MVAr Lag
Fig. 8.8 (b) Operation chart for salient pole machine
8.4
Effect of Excitation Control
Consider a synchronous machine with terminal voltage VI' The direct axis rotor angle with respect to a synchronously revolving axis is 8. The voltage due to excitation acting along the quadrature axis is Eq and Eq' is the transient voltage along this axis. If a load change occurs and the field current, If' is not changed then the various quantities mer..ioned change with P, the real power as shown in Fig. 8.9(a).
i
00
8
,
Eq
>"'cr
E~
'""
VI
'cr
""
r Fig.8.9(a)
In case the field current If is changed such that the transient flux linkages along the qaxis Eq' proportional to 'P f' the field flux linkages is maintained constant, the power transfer could be increased by 3060% greater than in case (a) and the quantities are plotted for this case in Fig. 8.9(b).
1
8
00
E~
>"
'CT
'"" ""
q
VI
'iT
p
Fig.8.9(b)
296
Operation and Control in Power Systems
If the field current If is changed along with P simultaneously so that Vt is maintained constant then it is possible to increase power d~livery by 5080% more than case (a). This is shown in Fig. 8.9(c).
i
E'
;;:
Vt
q
00
.i:T L.1.l
.,.
L.1.l
P Fig.8.9(c) Variation of Voltages
It can be concluded that excitation control has a great role to play in power system operation.
8.5
Voltage Regulation and Power Transfer
If the fall of terminal voltage is assumed to be linear, then the graphs of machine terminal voltage with load P can be represented by the relation.
E = Eo (\  KP) Where Eo is the nonload terminal voltage and K is a coefficient of regulation (Fig.8.1 0).
t o
P(MW)+
Fig. 8.10 Variation regulation
Since
EV. s: P =smu X
..... (8.5)
= Eo V sin8 Eo V KPsin8
"
X
X
where 0 is the angle between E and V. Substituting Pm for
T
E V
and solving for P
Voltage and Reactive Power Control P=
297
Pm sino 1+ PmKsino
which is a maximum at 0 = 90° having the value P max
=
Pm
1 K + Pm
under ideal conditions, where there is perfect control of excitation, K is zero. If the machine is represented by a certain voltage E behind a certain reactance Xe then Fig. 8.11 illustrates the effect of various types of regulators on its performance. It can be proved for an ideal, voltageactuated automatic regulator that the maximum value of the gain is
K vmax =
Xd  xd , Xd
..... (8.6)
where xd and x~ are direct axis synchronous and transient reactances of the machine respectively. There is also a minimum value for the gain to ensure stability at large load angles. For better performance, forced regulation using the derivation of current and / or voltage is recommended. The effect of different types of regulations is shown in Fig. 8.11. 2.0
1.8 1.6 1.4 1.2 1.0 0.8 0.6 0.4 0.2
Forced regulator xe
g
0
Fig. 8.11 Effect of regulators
8.6
Exciter and Voltage Regulator
The function of an exciter is to increase the excitation current for voltage drop and decrease the same for voltage rise. The voltage change is defined
C1 V~(Vl  Vref ) where VI is the terminal voltage and Vref is the reference voltage.
298
Operation and Control in Power Systems Exciter ceiling voltage:
It is defined as the maximum voltage that may be attained by an exciter \yith specified conditions of load.
Exciter response:
It is the rate of increase or decrease of the exciter voltage. When a change in this voltage is demanded. As an example consider the response curve shown in Fig. 8.12. Response
Exciter build up :
= IOOV = 250V / sec OAsec
The exciter build up depends upon the field resistance and the charging of its value by cutting or adding. The greatest possible control effort is the complete shorting of the field rheostat when maximum current value is reached in the field circuit. This can be done by closing the contactor shown in Fig. 8.13.
200
o
0.2
0.4
Seconds +
Fig. 8.12
tt: Stator
Alternatorfield Pi/ot Exciter
Main Exciter Fig. 8.13
299
Voltage and Reactive Power Control
When the exciter is operated at rated speed at no load, the record of voltage as function of time with a step change that drives the exciter to its ceiling voltage is called the exciter build up curve. Such a response curve is show in Fig. 8.14. 0
C
To ceiling voltage
EFD
i a
d
0''
e
. seconds Fig. 8.14 Exciter Buildup curve
Line ac represents the excitation system voltage response. Response ratio
Cd p.u. V /sec Oa(O.5)
..... (8.7)
Table 8.2 Typical ceiling voltages Response ratio
0.5 1.0 1.5 2.0 4.0
Conventional exciter
1.25 1.4 1.55 1.7

1.35 1.5 1.65 1.8
...
SCRexciter
12 1.2  1.25 1.31.4 1.45 1.55 2.02.1
In general the present day practice is to use 125V excitation up to IOMVA units and 250V systems up to 100MVA units. Units generating power beyond IOOMVA have excitation system voltages variedly. Some use 350V and 375V system while some go up to 500V excitation system.
8.7
Block Schematic of Excitation Control
A typical excitation control system is shown in Fig. 8.15. The terminal voltage of the alternator is sampled, rectified and compared with a reference voltage, the difference is amplified and fed back to the exciter field winding to change the excitation current. An excitation system using ampJidyne is shown in Fig. 8.16.
Operati'on and Control in Power Systems
300
Exciter generator
P.T
Voltage
Amplifier
Fig. 8.15 Excitation control Pilot exciter
Ac Main Exciter
Slip Rings
Synchronous Generator
Automatic 1'
.1 Voltage Magnetic Amplifier
Regulator
Fig. 8.16 DCType Excitation Systems
8.8
Static Excitation System
In the static excitation system, the generator field is fed from a thyristor network shown in Fig. 8.17. It is just sufficient to adjust the thyristor firing angle to vary the excitation level. A major advantage of such a system is that, when required the field voltage can be varied through a full range of positive to negative values very rapidly with the ultimate benefit of generator voltage regulation during transient disturbances. The thyristor network consists of either 3phase fully controlled or semi controlled bridge rectifers. Field suppression resistor dissipates energy in the field circuit while the field breaker ensures field isolation during generator faults.
,
Voltage and Reactive Power Control A compact AC static excitation scheme is shown in Fig. 8.18. Field breaker
CT
Field suppression resistor
control Reference
Thyristor network
Fig. 8.17 Static Excitation Scheme
Thyristor Bridge
Synchronous Generator
Fig. 8.18 AC StaticType Excitation Systems
8.9
Brushless Excitation Scheme Generator
Field
'1000' PT
Gate control
Fig. 8.19 Brushless excitation scheme
301
302
Operation and Control in Power Systems
In the brushless excitation system of Fig. 8.19, an alternator with rotating armature and stationary field is employed as the main exciter. Direct voltage for the generator excitation is obtained by rectification through a rotating, semiconductor diode network which is mounted on the generator shaft itself. Thus, the excited armature, the diode network and the generator field are rigidly connected in series. The advantage of this method of excitation is that the moving contacts such as slip rings and brushes are completely eliminated thus offering smooth and maintenancefree operation. A permanentmagnet generator serves as the power source for the excitor field. The output of the permanent magnet generator is rectified with thyristor network and is applied to the excitor field. The voltage regulator measures the output or terminal voltage, compares it with a set reference and utilizes the error signal, if any, to control the gate pulses of the thyristor network. A detailed brush less excitation scheme is shown in Fig. 8.20. Ac main exciter
Synchronous Generator
Automatic Voltage Regulator Fig. 8.20 AC AlternatorType Excitation System
8.10 Automatic Voltage Regulators for Alternators For an isolated generator feeding a load the automatic voltage regulator (AVR) functions to maintain the bus bar voltage constant. However, on dynamic, interconnected systems the AVR has the following objectives I. To keep the system voltage constant so that the connected equipment operates satisfactori Iy 2. to obtain a suitable distribution of reactive load between machines working in parallel and 3. to improve stability. The last of the above functions is not within the scope of this book. Under normal working conditions, it is not difficult to maintain generator terminal voltage at any specified level. However, with sudden disturbances like load fluctuations, the AVR is required to reduce the magnitude and duration of voltage variations as effectively as possible.
Voltage and Reactive Power Control
303
The magnitude of the voltage dip is primarily determined by the transient reactance of the machine while the duration of the dip depends on the time constant of the generator and rapidity of regulation. Static regulators are more useful in this context as they have lesser time delays. The AVR which senses the terminal voltage and adjusts the excitation to maintain a constant terminal voltage also maintains the reactive output at the required level since the latter depends on the effective voltage difference between generator terminals and its point of connection to the main system. Thus. while AVR is allowed to maintain the voltage on the low voltage side, a change in reactive output to suit a change in system conditions is obtained by tap changing a generator transformer. With several generators synchronized on a single low impedance bus section, one generator with the AVR may be permitted to maintain bus bar voltage while the excitation of the rest are used to distribute properly the vars among the generators.
8.11 Analysis of Generator Voltage Control
+o~~
Reference voltage
",...0 R
Amplitier
~_r_r_O
'.Lr'"IO
Y B
+ Alternator
Exciter
Rectitier t                ' '            1 and tilter t            '
Fig.8.21 Excitation System
Consider the excitation system shown in Fig. 8.21 the block diagram representation is given in Fig. 8.22. Load +
v
Vref
v
Amplltier
Exciter
Generator
Fig. 8.22 Excitation System Block Schematic
304

Operation and Control in Power Systems
From the block diagram
The amplifier is generally assumed to be instantaneous in action so that Ae A, the output from the amplifier is KAAe where KA, is the amplifier gain. The input to the exciter field will be absorbed by the resistance and inductance of the exciter field Re and Le respectively so that Ae A= ReAie + Le
~(Aie) dt
where Aie is the change in exciter field current. If I ampere change in field current produces K) volt change in the output, then Ae r = K)Aie The transfer function of the exciter can be obtained as follows: Ae r Ae A
K)Aie AieRe + Le
~(Aie) dt
In the frequency domain, taking Laplace transform
AEr{s) AEA (s)
K) Re + Le S
=
_~L Re
I+S~
~ I +STe
..... (8.8)
Re
Ke Hence Ge = is the transfer function of the exciter with the time constant I +STe
The value of the amplifier time constant may be of the order of 0.02 to 0.1 sec while Te may be 0.5 to 1.0 sec for conventional machines. At a later stage it will be shown that the exciter block is more conveniently represented I
by KE
+ STE
in accordance with IEEE recommendation.
The input voltage signal Ae r to the generator field, when applied to the circuit results in the following Kirchoff's voltage equation.
Voltage and Reactive Power Control
305
where Rf and Lff are the alternator field resistance and self inductance respectively and flif is the change in the field c~rrent. Taking Laplace transform. flEr(S) = (R f + LffS)fllf(S) If the output voltage changes by fllVI then L\I I' (s) =
.fi
ffiL fa
fllVI where Lfa is the mutual inductance between the field and stator
phase winding. Hence, the transfer function for the generator block will be
..... (8.9)
TdO =
where Tgf is the direct axis open circuit time constant also denoted more commonly by Rf seconds.
LIT /
The voltage regulator loop can be represented by the block diagram shown in Fig. 8.23.
Amplitier
Exciter
Generator
Load
Fig. 8.23 Voltage regulator block diagram
The three cascaded transfer function blocks GA' Ge and Ggf can be combined into a single block.
so that the feed back control loop can be further simplified as in Fig. 8.24.
Operation and Control in Power Systems
306
~IVI
~IVI
Fig. 8.24 Simplified Block Diagram
8.12 Steady State Performance Evaluation The control loop must regulate the output voltage VI so that the error is made equal to zero. It is also imperative that the response must be reasonably fast, yet not cause any instability problem. The steady state voltage error ,1e ss is given by Mss =
L\lvl~ef
= ,1I Vlo _ ref
 L\ Vss
G(O) ,1IVIO 1+ G(O) ref ~
where G(O) is the value of G(s) as s
0 (i.e.) the steady state value
= 1+ G(O)  G(O) ,1IVlo 1+ G(O) ref =
1 ,1IVlo 1+ G(O) ref
G(O)=~~
KgI' =K (I + 0) (1 + 0) (1 + 0)
1 ,1e ss =  I+K
..... (8.10)
Larger the overall gain of the forward block gain K smaller is the steady state error. But too large a gain K can calise instability.
8.13 Dynamic Response of Voltage Regulation Control Consider ,1V(t)=L 1[,1V
ref
(S) G(S) ] 1+ G(S)
The response depends upon the roots of the characteristic eqn. 1 + G(S) =
o.
Voltage and Reactive Power Control
307
As there are three time constants, we may write the three roots as SI' S2 and S3' A typical root locus plot is shown in Fig. 8.25. jro
Kc
(J

S3
S2
SI
0
Fig. 8.25 Root  Locus Plot
From the plot, it can be observed that at gain higher than Kc the control loop becomes llnstable.
8.14 Stability Compensation for Voltage Control Since at higher gains the voltage control loop tends to oscillate and even may become unstable. it requires phase lead compensation or derivative control. As there are three time constants contributing to phase lag, a phase lead compensator with transfer function. Gc (S) = I + STc will alter the open loop transfer function to G VR (S )
_ 
K(I + STc)
=(I + STA)(1 + STe)(1 + STgf )
..... (8.11 )
It will not affect the steady state accuracy as K is kept constant
8.15 Stabilizing Transformer Consider the excitation system with a stabilizing tralJSformer as shown in Fig. 8.26. KA = Error amplifier gain Ksp TST
~
stabilizing transformer gain and time constants.
A stabilizing transformer is introduced in the exciter circuit as shown in Fig.8.26(a, b). The input voltage Ve causes a current feed back loop.
Taking Laplace transform
i ST to flow through the resistance and inductance of the
308
Operation and Control in Power Systems Field
Error amplifier I t     j Power L
amplifier
,.........~r.
0
,....; rt.t
a d
'"'tt_t_
Main excitor
Fig. 8.26 (a) Circuit
V Ref(S)
+
Fig. 8.26 (b) Block diagram
Ve(S) = R f IsT(S) + L f S IsT(S) (R f + S Lf ) IsT(S)
=
If the mutual inductance is M for the stabilizing transformer then,
VST (t)
" Taking Laplace transform
=
M ~ i,t (t) dt .
VST = MS 1st (S)
Hence SM
VST(S)
SM
Ve(s)
R f + LfS
Rf 1+
lLs Rf
L
Letting
Rf f
M
= Tst and R = K st f
Vst(S)
SK
  =    =st 
Ve(S)
1 + TstS
..... (8.12)
Voltage and Reactive Power Control
309
It may be noted that since the secondary of the stabilizing transformer is connected to the input terminals of an amplifier, it draws zero current. Thus, the stabilizing transformer improves the response of the excitation system giving derivative control.
8.16 Voltage Regulators The heart of the excitation system lies in the voltage regulator. It is a device that serves the output voltage change (some times current or both) and provid~s corrective action to take place. The early voltage regulators were all of electromechanical type. Electronic voltage regulators came in 1930's. They provide smoother and faster voltage control than the electromechanical type. Another development resulted in the use of rotating amplifiers. Magnetic amplifiers were also used for voltage regulators. Later versions of voltage controllers used solid state active circuits. Whatever may be the hardware used, there was consistent effort to provide faster response, without dead band, backlash, and absolutely reliable. Excitation systems which include voltage regulators also are classified into continuously acting type and non continuously acting type. All the systems have a potential transformer and rectifier to sample the output a.c. voltage and provide d.c. signal to the control loop. Another component is a comparator that compares the d.c. signal with reference value and provides an error signal. An amplifier amplifies the error signal and brings it to the operational level. The exciter output is manipulated by the amplified error signal to suitably modify the output voltage of the generator. As a simple proportional feed back results in a steady state error, a rate feedback or derivative control is proposed by Kron in 1954 and this was discussed in section (8.15).
Computer representation of excitation system For synchronous machines, several models were developed and linear representations of the machine using state space methods needed suitable models for excitation system for computer simulation to perform dynamic studies. IEEE has suggested certain standard models for different types of excitation system.
I. Type 1 System: It is a continuously acting regulator and exciter system. It will be explained in detail later. 2. Type 1 S system: This is a special case of continuously acting system. In this system excitation is obtained through rectification of the terminal voltage while in type I system regulator voltage has maximum and minimum limits, in this case, the maximum regulator voltage is made proportional to Vt' the terminal voltage of the generator.
310
Operation and Control in Power Systems 3. Type 2 System: In this system rotating rectifiers are used and is brush less. Altematorexciter and diode rectifiers are rotating on the same shaft and hence has no slip/ rings. Being brush less, the excitation voltage (E FD) is not available for feed back. In this there are two damping loops giving good performance.
4. Type 3 System: This type of model is evolved for systems where both current and voltage signals are used for feed back.
5. Type 4 System: All the previously described systems are continuously acting type with high gain and fast action. Several of the earlier systems are non continuous acting type. [n such cases presence of dead zone results in open loop operation. They are also slow due to friction and inertia. Rheostat systems are examples for such an operation. [n type 4 systems now there are two speeds of operation depending upon the magnitude of the voltage error. Type 4 system hence is nonlinear and as such is not represented in state variable form. Westinghouse and General electric have developed such excitation systems.
8.17 IEEE Type 1 Excitation System Consider Fig. 8.27. Let V f be the voltage across the field winding and if the current through it Afand also the flux linkages in the field winding. . .... (8.13) consider the nonload magnetization curve shown in Fig. 8.27 for the exciter voltage build up. slope of airgap I
line =
Rag
VexO
Fig. 8.27 In Engineering Units
. coe ffi' AB · t he saturatm Se IS JClent B. 0 ) The excIter voltage v 0ex == f("Jf ' Je\
..... (8. [4)
Voltage and Reactive Power Control
311
o
. Vex S 0 If =  + eVex Rag
where
Rag
..... (8.15)
is the resistance given by air gap line.
The saturation may be represented by
..... (8.16) an exponential relation, A and B to be determined.
. .... (8.17)
from eqn. (8.2) ,
,
..... (8.18)
rf
[
.
10
d
~ (
0)
dv~x
= 1+RagSe(lf)JVex +olI.f Vex  Rag dVex dt
In per unit system (see Fig. 8.28) Vf
Let
( p.u. ) _ ~ [1
Rq
+
R S (. )10
ag e If JVex p.U.+
AB B
dAf{V ex ) dv ex (p.lI.) ~~
~
..... (8.19)
III
Operation and Co.tro. in Power Systems V xpu A
V'pu
slope of airgap line= I /
,
/
8
I+..:'i~
,
,
~epu (V xpu~
It
.,
Fig. 8.28 In Per Unit
+____....,..._.. Em
n
Rag
Pu ) SePLI ( \ex
Ero=v ex p.u.
Fig. 8.29
Then the transfer function for the exciter block can be represented by Fig. (8.29). IEEE type 1 excitation system representation is shown in Fig. 8.30.
Fig. 8.30 IEEE Type I Excitation System
313
Voltage and Reactive Power Control
8.18 Power System Stabilizer It can be proved that an voltage regulator in the forward path of the exciter generator systems will introduce a damping torque and under heavy loading conditions this damping torque becomes negative. This is a situation where dynamic instability may calise concern. Further, it has been also pointed out that in the forward path of excitation control loop, the three time constants introduce a large phase lag at low frequencies just above the natural frequency of the excitation systems. To overcome this effect and improve damping, compensating networks are introduced to produce torque in phase with the speed. Such a network is called "power system stabilizer" (PSS) Fig. 8.31 shows the block schematic of PSS at the appropriate point in the control loop.
~ ~(lJ()r~N
vs I~EFD
L..._ _....J
Regulator and Exciter Fig 8.31 Connecting Power System Stabilizer
Power system stabilizer is generally shown as a feedback element from the shaft speed and has a transfer function of the form
..... (8.20) A washout circuit for reset action is used to eliminate steady effect after a time lag To
(4 sec to 30 sec). This control circuit ensures that there is no permanent offset in the terminal voltage due to a prolonged error in frequency. The later may occur due to prolonged over load or islanding situation. A lead compensation pair is used at two stages with center frequencies
21n ~TIT2 and
2~ ~T3 T4
. This part of the stabilizer circuit will improve the phase lag.
A filter section is generally added as shown in Fig. 8.32 so that undesirable frequency components are suppressed and thus eliminate the possibility of undesirable interaction. Limits are placed so that the output signal of the PSS is prevented from driving the excitation into heavy saturation. The output signal of the PSS is fed as a supplementary input signal V pss for the regulation of the excitation system. Design of the PSS circuit is beyond the scope of this book.
314
Operation and Control in Power Systems Limiter
PhaseCompensation
Washout
Filter
~
1+ sTI
l+sTl
K,s
1+ ST2
I + sT.
I +sT"
VPSS }l~
V x = V'l'li Regulator and Exciter 1..
~________~
KSF
f~
TSF
Fig. 8.32 Power System Stabilizer
8.19 Reactive Power Generation by Turbo Generator
~"
The reactive power generation capability of a steam turbogenerator is dependent on the type of construction. Unit construction: In this type, where there is no steam interconnection, the generator can absorb and produce reactive power while supplying the rated real power to the load. A reduction of active power load on the generator increases both its reactive absorption and production capability. However, the active power scheduling is quite often based on an optimal criterion as discussed in Chapter 6, overlooking which may add to the costs. The level to which real power loading may be reduced is mostly determined by considerations of boiler performance. Minimum permissible loadings usually fall in the range of 50 to 70% of rating. Steam range construction: In the steam range construction having steam interconnection between generators, the minimum loading constraint imposed by the boiler is of less significance since steam from one boiler can be used to supply several generators. As an example, iffour sets are supplied by a single boiler, then with three sets drawing steam just for gland ceiling, maintaining vacuum and cooling while the fourth set at the partial load, a wide range of reactive generation or absorption capability can be obtained. However, this arrangement is not in use now.
8.20 Synchronous Compensators There are rotating machines which are connected to the system at appropriate places only for the purpose of controll ing vars. They are synchronous motors operating on load with adjustable excitation. They are normally fitted with voltage regulators which operate when the voltage deviates by a certain predetermined percentage of system voltage. These machines are also capable of giving dynamic reactive power compensation, i.e. they make available the required.reactive power within a short time. The prime consideration for installing synchronous compensator is its flexibility of operation under all load conditions.
Voltage and Reactive Power Control
315
This aspect justifies its use in certain systems even though the initial cost is high. Another important consideration for using the synchronous compensator is the occurrence of sudden voltage dips due to sudden short circuits or overloads. Under such conditions the synchronous compensator performs better than other apparatus. With additional excitation windings through which excitation can be forced momentarily to counteract the transient effects, the synchronous compensator can even improve stability. The synchronous compensator may be supplied from the tertiary winding of the transformer as shown in Fig. 8.31, so that additional cost on transformer is not incurred.
8.21 Reactors Inductive reactors absorb reactive power and may be used in circuits, series or shunt connected, While series connected reactors are used to limit fault currents, shunt reactors are used for var control. Reactors installed at line ends and intermediate substations can compensate upto 70% of charging power while the remaining 30% power at noload can be provided by the under excited operation of the generator.
~IY~ Bus
I
7
Ll.y ~
I
Bus
(b) Shunt reactors in parallel
(a) Shunt reactor with taps
Fig. 8.33 Shunt Reactors
With increase in load, generator excitation may be increased with reactors gradually cutout. Fig. 8.33 shows some typical shunt reactor arrangements.
S.22 Capacitors Capacitors produce vars and may be connected in series or shunt in the system. Series capacitors compensate the line reactance in long overhead lines and thus improve the stability limit. ;However, they give rise to additional problems like high voltage transients, subsynchronous resonance, etc. Shunt capacitors are used for reactive compensation. Simplicity and low cost are the chief considerations for using shunt capacitor. Further, for expanding systems additions can be made. Fig. 8.34 shows the connected of shunt capacitors through the tertiary of a transformer.
Operation and Control in Power Systems
316
HV~
Bu~
~p11 LV
~
I
I
~
Bus
I
I
Shunt capacitors parallel
In
Fig. 8.34 Shunt Capacitors
Table 8.3 Comparision of Shunt Capactor and Synchronous Condensers Criteria
Shunt capacitors
Synchronous condenser
Initial cost Losses Variation of reactive components of current Over load Installation Increase of rating
1.0
1.0 to 1.2
Less than 1.5% In steps only with possible surges of voltage and current Not possible Simple Possible by adding more units
1.5 %t04 % Continuous
..~
Effect ofhannonic voltage Possibility of resonance
Possible with reduced reacting Relatively difficult 17:. Not possible unless another unit is installed No effect.
8.24 TapChanging Transformers Tapchanging transformers with variable transformation ratio can calise substantial change in the flow of vars. The tapchanging transformers when used in radial lines maintain voltage at their secondary terminals or at load terminals within limits. When used in tie lines, the tapchanger can regulate vars substantially. In case of weaker tie lines, active power may also change to some extent. Lastly, when used in networks or loops, circulating vars can be controlled by tapchanging transformer. Consider a tie line with sending end and receiving end, both having tapchanging transformer connected to them. Let Vs and Vr be the voltages at either end at nominal conditions. Let ts Vs = actual voltage at sending end and tr Vr = actual voltage at receiving end Vs  Vr = VI _. V2 == (lR cos
311
Voltage turd Reactive Power Control PR+QX
Hence
Vr
Hence
ts Vs == tr Vr +
PR+QX V r
To ensure that the same overall voltage prevails allover the line minimum are used so that ts tr = I
ranac of taps
Substituting this condition and solving for ts
t; [1  ( PRVI+VsQ X
II ~ =
Vs
..... (8.21 )
8.25 TapStaggering Method At low loads. the reactive losses in transforme.rs are also low. The surplus ge~tjd by the system at such low loads can be absorbed by increased reactive power losses in transformers using tap staggering. Consider the pair of transformers shown in Fig. 8.35 connected between HV and LV buses.
IN,~
LV~ Bus
Fig 8.35 Tap Staggering
318
Operation and Control in Power Systems
If the taps on the transformers are staggered, a quadrature current circulates around the transformers and carries additional reactive (I2X) losses. With more number of transformers operating, the losses could be increased by this method.
8.25 Voltage Regulation and Short Circuit Capacity Consider the short line voltage regulation IR IX . I1Y=cos
Yr
Y,
where Yr is the receiving end voltage and
11 Y = (R + JX). Ir
= (R + JX) P  jQ Y = (PR
~QX)+ {PX ~QR)
when a three phase fault occurs at the load bus, the short circuit volt amperes into the bus Sse = Y . Isc = Y .
Y Zse
(R + jX)
Y"
R+jX= Sse
y2 R
Sse
cos
y2 and
X
Sse
sin
y2
y2.]1
I1Y = Pcos
Y"
]1
y2 . + J.[ PSIl1<1>sc Qcos
Voltllge and Reactive Power Control
319
It can be shown that the magnitude of voltage regulation given by I R coscp + I X sincp or
PR+QX V as the imaginary component gives only phase shift. Hence, the real part of
equation for d V above gives the change in voltage magnitude Hence, d V = [Pcoscpsc + Psin CPsc ]V
Sse
Sse
For small changes in P and Q
Once again, remembering that the real power does not affect the voltage magnitude much dV= IdQ' h. sm'l'~c V
S<;c
.
Under short circuit conditions the current is almost lagging by 90° to the voltage Hence,
sin CPse
Thus
= I.
dV
=dQ
V
Sse
..... (8.22)
Per unit change in voltage magnitude is equal to the rate of change in reactive power to the short circuit capacity of the bus. It may be noted in this context that for a system with X » R, the inphase voltage drop would be approximately equal to (I sin cpr)X, An inphase voltage compensator or booster would control the reactive power flow in the system. The quadrature voltage drop would be approximately equal to (I sin CPr)' The quadrature voltage compensator would control the active power flow in the system. The ratio of
X
R is very high for EHV and UHV lines.
320
Operation and Control in Power Systems
8.27 Loading Capability of a Line There are four kinds of limitations identified for the loading of a transmission line. (i) (ii) (iii)
(iv)
thermal limit dielectric limit stability limit natural loading limit.
Of all the above limits natural loading limit is the lowest limit and thermal limit is the highest. (i)
Therma/limit: This limit depends upon Atmospheric condition including ambient temperature and wind conditions Condition of the conductor and Ground clearance
Normal loading at the design and planning level is determined by loss evaluation. Any increase in line capacity with thermal consideration should also consider transformers connected at either end also. (ii) Dielectric limit: Line design on the basis of insulation requirement is always on the conservative side. Line voltage can always be increased by 10% for increased power transfer. This requires proper consideration of transients and arresters in operation.
Stability limit: There are several stability considerations for increased transmission capability. They are Transient stability Dynamic stability Steady state stability Frequency collapse . Voltage collapse • Sub synchronous resonance
8.28 Compensation in Power Systems Electrical power demand is growing at a great rate day by day and the generation is, in general, not able to cope up with the demand. Several ways of increasing the power demand. Several w~ys of increasing the power generation are investigated including many non conventional modes. Again, transmission of increased power over the existing lines is considered to meet the increasing demand as laying of new Iines and acquisition of right of way are too expentive in developed areas. This necessitated implementation of compensation in power systems. For many years series and shunt compensators are in use. Electrical energy cannot be stored in bulk quantities. There must always exist a balance between generation and demand.
321
Voltage and Reactive Power Control
8.29 Load Compensation Utilization of reactive power to improve voltage profile and power factor is termed as load compensation. Load compensation thus refers to improving power quality. Reactive power can be injected by installing various sources such as reactors and capacitors in shunt or in series with the system at appropriate places. This has been discussed in section (8). Another aspect to be considered is that as far as possible negative sequence currents are to be avoided so that power loss is reduced. This is achieved by operating the system in balanced condition.
Line compensation: Surge impedance loading (SIL) of a line is the power delivered by a line to a purely resistive load equal to its surge impedance. SIL =
Vc
J¥
MW
=:0
VC Rc
Power transmitted by a line is usually expressed in terms of this power. When surge impedance is the load impedance the voltage profile of the line is flat. This is not practicable. However, it is possible to modify the characteristics of the line by using such elements as capacitors, inductors or synchronous machines. These elements can be connected in series or in parallel. Shunt compensation is similar to load compensation. Shunt connected capacitors improve the power factor, reduce the reactive power and thus increase the real power transmission. However, this method has limitations and the size of the condenser banks increase to a prohibitive level. Further they are to be switched off under Iight load conditions as otherwise the voltage rise may not be acceptable. Series compensations using capacitors cancel a part of the line reactance and increase the maximum'power transfer. Such a compensation win reduce the power angle for a given power transfer and enables increased loading. Excessive series compensation using capacitors may result in resonance. Sub synchronous resonance at frequencies less than 50Hz occur with series compensation if care is not taken. This is discussed later in the chapter.
Line compensation results in (i) minimization of Ferranti effect (ii) elimination of the need for under excited operation of generators and (iii) enhanced power transfer capability From the receiving end power circle diagram, we have
'IVsllVRI IAIIV~I = cos(/3  8)    cos(/3  a) R IBI IBI
P with usual notation.
PR is a maximum when 0
=
/3
..... (8.23)
322
Operation and Control in Power Systems
..... (8.24) similarly
..... (8.25) For a tieline with series impedance
B=Z=R+jX= ~R2+X" D = 1.0 (1=0 ,1=0 and
f3
and
p.s
=
Tan i X/R
.. max
differentiating
PR.max
= "
VSVR vi R 1 + 0 R 2 + X " R  + X
~
with respective to X
4( VR)" _I Vs
when
vs =V R''X=
"J'3R
..... (8.26)
Voltage and Reactive Power Control
In overhead lines
X
R
323
ratio is much greater than this optimum value needed to maintain
the same voltage at both the ends. For 132kV lines
X
R
ratio" is from 2.S to 3.S; for 27S kv lines
/
it is about 8 and for 400kv lines it is as high as 16. Hence to maintain voltage profile
X
R
ratio
has to be decreased. This can be effectively done by series capacitor compensator so that line inductance is compensated. Nevertheless, series capacitors are to be protected against over currents and over voltages during fault conditions. E 8.1 Two substations A and B are interconnected by a line having an impedance of (0.03 + jO.12) p.u the substation voltages are 33 Lt) kv and 33 LOo kv respectively. In
phase and quadrature boosters are installed at A. Determine their outputvoltage ratings and MVA ratings in order to supply SMVA at 0.8p.f lagging at substation B.
Solution: Let
Base MVA
=
Base KV
33kv
VA
=
=
SOO
1.0 L2° p.u and VB
Let current through the inter connector
=
=
1.0 LOo p.u
I
Let load current at B = IB SMVA IB = SMVA (0.8  jO.6) = (0.8  jO.6) p.lI. Without boosting voltage drop in the line (0.03 + jO.12) (0.8  jO.6)
=
Z IB
=
(0.096 + jO.078)p.u.
=
voltage drop in the line with boosters = VA VB = 1.0 L20 1.0 LOo =
0.99939 + jO.0348994  1.0
=
(0.00061 + jO.0348997) p.lI.
Hence, in phase component of boosted voltage = + 0.096  (0.00061) =
0.09661 p.u.
Quadratic component of boosted voltage
324
Operation and Control in Power Systems = 0.078  0.3349 = 0.0431 p.u. Rating of the in phase booster = 0.09661 x 1.00
= 0.09661 p.u. = 0.09661
x
500 = 48.3 MVA
Rating of the quadrature booster = 0.0431 x 1.00
= 0.0431 p.u. = 0.0431
x
500 = 21.55 MVA
E 8.2 Two substations A and B operating at II kv 3phase are connected by two parallel lines. 1 and 2. Each line has a 11/132 kv transformer and a 132/11 kv substation. Each line has an equivalent impedance of ZI = 0.2 + .i0.4 and Z2 = 0.2 + jO.6 ohms per phase which includes both the transformers and the line, referred to II kv side. (a)
If the bus bar A is at II kv and is sending 30Mw at 0.8 p.f leading, find the individual currents into each transformer and the powers at the station A.
(b) If the transformer at A in line 2 is fitted with tappings on the II kv side. what percentage tapping would be required to make each line carry equal reactive powers? What would be the power sent by line 2 in this case?
Solution:
~
0.2 + j 0.4
~
Line2 0.2 + j 0.6
Line I
1II132KV
A
Fig. E 8.2
Let base voltage be II kv Let
Base MVA
=
30.0 0.8
= 37.5MV A
Total current flowing through both the lines
Base impedance
~~
132111KV
=
37.5 (0.8 + jO.6)  (0.8 + jO.6)p.u. 37.5
=
100 L36.87° (since II kv = 1p.u.)
Ilkv x Ilkv = 3.2270hms 37.5
~~
B
315
Voltage and Reactive Power Control 2 = 0.2 + j0.40 = (0.0627)0.124d u , 3.227 P 0.1386 L63°.43
=
22 = 0.2 + jO.60 = 0.0627 + jO.186 = 0.1961 L71.S6 op.u. 3.227 Hence, . Z· ,+ Z· ,= 0.5+ jl.O = 0.1 24+ JO.3101 p.u. 
3227
= 0.334 L68.20 p.u.
I, = 0.1961 x 1.0 L(71 0 .S6+36 0 .8768 0 .2)=0.S84L40 o.23p.u. 0.334
°
°)
°
12 = 0.1386 x 1.0 L (0 68 .43 + 36 .87  68 .2 = 0.41 SL32 .Ip.u. 0.334 3
37.5 x 10 = 1968A mpers r::; '\"3 x II
Base current
I, =11SSL40.23° (I968xO.S87L400.23)
12 =S17L32.1O° (1968x0.41SL32°.IO) The per unit MVA line I S, = P, + jQ, = 1.00 LOo x 0.587 L40.23° = (0.448  jO.379) p.u. The per unit MVA from line 2 S2
=
P2 + jQ 2 = 1.00 LO O x 0.415 L32.1 0 = (0.352  jO.220S) p.u.
P,
=
0.448
x
37.5 = 16.8 MW
P2 = 0.352
x
37.5
=
13.2 MW
Q,
x
37.5
=
14.21 MVAR (leading)
=
0.379
Q 2 = 0.2205
x
37.5
=
8.27 MVAR (leading)
(b) The tap changer at station A on the sending end transformer in line 2 will alter the reactive power loading. Total load
=
(P, + P2 + jQ, + jQ2) = (30  j22.S) MVA
Equal reactive powers means Q, "" Q 2 = 
2~.S
= 11.25MVAR
326
Operation and Control in Power Systems P 1 = 16.8 MW and
P2
The change in voltage
=
13.2 MW
~V
is given by the approximate expression,
~ V = ZI = PR + QX V
= tV _ V = PR + QX V where t is the tap changer setting ratio
(t  I) V =
(t  I)
=
PR+QX V
PR+QX
V2
t = 1 + PR+QX 2 V P = P2 = 0.352 p.u. (= 13.2 MW/37.5) Q 2 = 0.30 p.u. (= I I .25MVAR/37.5) X2
=
0.186 p.u.
R] = 0.062 p.u. t = 1+ 0.352 x 0.062 + ( 0.3)0.186 1.00 x 1.00 percentage tap change
= 1 0.034 = 0.966
= 96.6  100 = 3.4 %
Power sent through line 2 S2 = P2 + j Q 2 = (0.352  jO.30)pu S1 = S  S2 = (0.8  jO.6)  (0.352 j0.3) = (0.448  jO.3)p.u.
S1 = P 1 + j Q 1 = 16.8  j 11.25 MVA S2
= P2 + j Q 2 = 13.2 j 11.25 MVA
P 1 = 16.8 MW P2 = 13.2 MW Q 1 = I 1.25 MVAR Q;! = 11.25 MVAR
Voltage and Reactive Power Control
327
E 8.3 A load of (15 + j 10) MVA is supplied with power from a generating station from a line at II Okv 3phase 50Hz. The line is of 100km length. The line is represented by 1t model with the parameters R = 26.4 ohms X = 33.9 ohm B = 2!9 x 106
Voltage at t~e generated in 116kv. Determine the power supplied by the generating station.
B/2
Fig. E 8.3
SR = PR + j OR VR =
Reactive power through
=
IS + jlO 3 MVA/Phase
~ = 63.5KV
B
2"
at receiving end
OCR = 63.5 x 63.5 x
219 X 106
2
6
x 10 var
= 883062.75 = 441531.37VAR = 0.44151MVAR 2 . . Receiving end power
IS. + jl 0'_ .0.441S3 S = PR + OR = . 3 . J = (S
+ j 2.8918) MVA
Real power loss in the line = 12 R • 6 SR (S + j2.89I8)I0 I= = '......;;,.VR (63.5)2 x 10 3
JS2 +2.8918 2 xl0 6
 =
63.Sx10 3
90.96
328
Operation and Control in Power Systems 12R == (90.96)2 x 26.4YA ==0.2184313MW 3
12X == reactive power loss == 52 +2.89: xl0 x33.9==(90.96)2 x33.9YA 63.5 Reactive power in the sending end capacitance 8/2 Oes == y2 Xes Ys ==
116
.J3
== 66.9746 6
Oes == 66.9746 2 x 219x 10 x 10: == 0.491 I 72.87MYAR 2 10 Power at the sending end Ss == S + 12 R + 12 X + OCR + Qes == (5 + j2.89) + 0.218 + jO.279  jO.04415 == (5.218 + j 1.9574) MYA
power consumed from the station 3 x Ss == 3 [5.218
tl j 1.9574]
MYA == (15.54 + j5.8722)MYA
E 8.4 A short line having an impedance of (2 + j3) ohm interconnects two power stations A and 8 both operating at II kv; equal in magnitude and phase. To transfer 25MW at 0.8 power factor lagging from A to 8 determine the voltage boost required at plant A.
Solution: 3
I ==
25 x 10 == 1640.2476L36.870 x II x 0.8
.J3
== 1640.2976 (0.8  jO.6) == (1312  j984.1486)A
voltage drop in the short line == (1312.2  j984.1486) (2+j3) == 2624.4  j 1968.297 + j3936.6 + 2952446 == 5576.8458 + j 1968.3 This voltage drop must be compensated by the booster so that the voltages are maintain the same at both the ends Voltage boost == (5576.85 + j 1968.3) volts.
8.29 Static Compensators In Recent years reactive compensation of charging power is made feasible with the application of 3phase, thyristor, power controller circuits with automatic control functions
Voltage and Reactive Power Control
(a)
329
(b)
(c)
Fig. 8.36 Idealized var compensators
The term static var compensator is applied to a number of static var compensation devices for use in shunt reactive control. These devices consist of shunt connected, static reactive element (linear or nonlinear reactors and capacitors) configured into a var compensating system. Some possible configurations are shown in Fig. 8.36. Even though the capacitors and reactors in Fig. 8.36 are shown connected to the low voltage side of a down transformer, the capacitor banks may be distributed between high and low voltage buses. The capacitor bank often includes, in part, harmonic filters which prevent tharmonic currents from flowing in the transformer and the high voltage system. Filters for the 5th and 7th harmonics are generally provided. Fig. 8.37 shows one type of static var compensator. The thyristor controlled reactor (TCR) is operated on the low voltage bus. In another form of the compensator illustrated in Fig. 8.38 the reactor compensator is connected to the secondary of a transformt"T. ~
HYBus Step down transformer
Capacitor and filter bank
Fig. 8.37 Thyristor controlled reactor compensator system
330
Operation and Control in Power Systems
With this transformer, the reactive power can be adjusted to anywhere between 10% to the rated value. With a capacitor bank provided with steps, a full control range from capacitive to inductive power can be obtained. The reactor's transformer is directly connected to the line, so that no circuit breaker is needed. The primary winding is star connected with neutral grollnded, suitable to the thyristor network. The secondary reactor is normally nonexistent, as it is more economical to design the reactor transformer with 200% leakage impedance between primary and secondary windings. The delta connected tertiary winding will effectively compensate the triplen harmonics. The capacitor bank is normally subdivided and connected to the substation bus bar via one circuit breaker per subbank. The regulator generates firing pulses for the thyristor network in such a way that the reactive power required to meet the control objective at the primary side of the compensator is obtained. The reactor transformer has a practically linear characteristic from no load to full load condition. Thus, even under sllstained over voltages, hardly any harmonic content is generated due to saturation. The transformer core has non ferromagnetic ·gaps to gil'e the required linearity.
The following requirements are to be borne in mind while designing a compensator. 1. Reaction should be possible, fast or slow, whenever demanded. No switching of capacitor should take place at that time to avoid additional transients in the system. Commutation from capacitor to reactor and vice versa should be fast. 2. No switching of the capacitors at the high voltage bus bar, so that no higher frequency transients are produced at EHV level. 3. Elimination of higher harmonics on the secondary side and blocking them from entering the system. In a three phase system the thyristor controlled inductors are normally delta connected as shown in Fig. 8.39 to compensate unbalanced loads and the capacitors may be star or delta connected. Transmission line
HV Bus
r
rJMY~ Reactor transformer
,
Fig. 8.38 Reactor  transformer compensator
331
Voltage and Reactive Power Control
8.31 Steady State Performance of Static var compensators In the thyristor controlled reactor, the inductive reactance is controlled by the thyristors. For a limited range of operation the relationship between the inductive current iL and the applied voltage V is represented in Fig. 8.40. As the inductance is varied, the susceptance varies over a range within the limits BLmln and BLmax (corresponding to XLmax and XLmln ) while the voltage changes by v volts.
Fig. 8.39 Fixed capacitor, thyristor controlled inductor type var compensator
The use of a fixed capacitor in parallel with the controlled reactor results in the chaltacteristics shown in Fig. 8.41. The currents can be controlled to have any desired value over the range selected from leading to lagging.
v
o Fig. 8.40 Thyristor controlled reactor charactenstics
Corresponding vars are produced or absorbed. Fig. 8.41 shows the effect of changing the oonduction angle 0 from 1800 to a small value 0cThe current flowing in the inductance would be different in each half cycle, varying with the .conduction angle such that each successive half cycle is a smaller segment of a sine wave. The fundamental component of inductor current is then reduced to each case. Quick control can be exercised within one half cycle, just by giving a proper step input to the firing angle control
332
~
Operation and Control in Power Systems 62
Block diagram representation of a static var compensator is given in Fig. 8.42. The net change in the reactor current IR drawn by the static var compensator supplied through the system impedance Zs
=
i1 V 1 i1! I
v
v
/
~
/ /
(Lag) +
+(Lead)
+(Lead)
(Lag)
+
Applied voltage
Reactor currents Fig. 8.41 Effect of conduction angle on reactor currents
Following an instantaneous change in firing angle e, the admittance BL of the inductor and therefore the current, 'I will settle down to a new value in about one cycle or less. The impedance Zs is the entire system impedance viewed by the compensator. The thyristor controlled reactor can be represented by the transfer function
\ e ST1 GR(S)=1 +T2S
where TI and T] are of the order I ms and 4 ms. For practical purposes, the compensator can be assumed to have a pure gain of unity with the present thyristor technology. The AVR can be represented by
Transformer Fig. 8.42 Block diagram of static var compensator
333
Voltage and Reactive Power Control Gv(S) = _k_ (I +T4 S) I+T3S (I+T5S)
..... (8.27)
The time constant T) lies between 50 and ISO ms. The compensator block diagram for dynamic studies is shown in Fig. 8.43.
VE (2asin2a) j = jX : E=CY. C = 2n'
E
Fig. 8.43 Compensator block diagram for dynamic studies
Static var compensators when installed reduce the voltage swings at the rolling mill and power system buses in drive system applications. They compensate for the average reactive power requirements and improve power factor. Electric arc furnaces impose extremely difficult service requirements on electrical power systems since the changes in arc furnace load impedance are rapid. random and non symmetrical. The three phases of a static var compensator can be located independently so that it compensates for the unbalanced reactive load of the furnace and the thyristor controller will respond quickly in order to minimize the voltage fluctuations or voltage flicker seen by the system. Lme
Furnace Input ~oltageand
;=~=::!E~=~::; Light pulses
current)
Magnetic interface
Fig. 8.44 Arc furnace applications
334
Operation and Control in Power Systems
Thus, the furnace characteristics are made more acceptable to the power system by the static var compensator. Fig. 8.44 shows the application of the static var compensator to an arc furnace installation for reactive power compensation at the HV bus level.
8.32 Overvoltages on Sudden Loss of Load Over voltages caused by load rejection are basically caused by a voltage rise across the source reactance which may be aggravated by over speed causing a decrease in the line capacitive reactance and increase in inductive reactance. There are two types of over voltages following sudden loss of load. The first occurs with the first cycle or two after the load rejection takes place. The second type, dynamic in nature, may last a few seconds and is caused by higher machine internal voltages and over speed. This type is, to a large extent influenced by the existing conditions before the load rejection takes place, the type of voltage regulator and the compensating devices. The magnitude of the first type can be determined from: I. The constant flux linkage voltage in the equivalent circuit of the rotor direct and quadrature axes. 2. The voltage rise components across the transient reactances of the two axes due to linecharging current. The second type of over voltage can be computed from the relation.
.. ... (8.28)
where Es is the sending end voltage E'(t) the voltage behind transient reactance as a function of time
Qc' the reactive MVA generation of the line at rated voltage and frequency and, SC' the short circuit MVA for fault at sending end (based on transient reactance). If the rejected load forms a large portion of the total load there will be a rise in frequency, Qc and E!(t) and decrease of Sc with a cumulative effect to increase the overvoltage. To avoid self excitation: the line reactive generation should be within
_ Qc 
O.9Pn ?
n(Xd+X t )
where n is the over speed in p.u.,
Xt the transformer reactance in p.lI. Xd the direct axis reactance in p.u. and Pn the nominal MVA of the generating capacity at the sending end.
..... (8.29)
Voltage and Reactive Power Control
335
If the excitation system has negative field current capability and AVR is provided then
where Xq is the quadrature axis reactance in p.u. The over voltages resulting from self excitation are dangerous not only for the system but also to machines and transformers connected to the sending end. This conditions must be avoided by providing the excess ofline reactive vars with shunt compensator or rapid switching I off of lines. Further, over voltages caused by load rejection prove to be worse when switching is performed on the transformer secondary at the receiving end. Due to voltage increase, the transformer gets over excited and then saturated with subsequent injection of harmonic voltages into the system. Then, with some combination of system inductance and line capacitance, due to Ferro resonance, extremely high voltage will be produced for 0.5 to LOs, resulting in equipment damage. Shunt reactor compensation is quite effective in limitings the over voltages due to load rejection. Such compensation has given 50% reduction in over voltages for line to ground faults. When synchronous compensator is installed, it automatically absorbs the reactive power from the system which is a function of the compensator reactance and interval voltage. The time constant of the machine, however, limits the rate at which the compensator inductive current can be increased. The static var compensator on the other hand can change over to its maximum current in the inductive range very rapidly since the compensator output increases as the square of the voltage. Further, the overload capacity of the static var compensator in the inductive range can be increased at additional costs much lower than those required for synchronous compensators.
8.33 Voltage Dips Voltage dips take place either due to faults or switching in of other motors connected to the same bus. A voltage dip causes the synchronous condenser to act in accordance with its internal voltage behind the sub transient and transient reactance and immediately starts feeding the corresponding capacitive reactive power into the system. In fact, with modern excitation control equipment with an appropriate ceiling voltage, it is possible to maintain a high capacitive current for a few seconds to support the system voltage. Static var compensator, on the other hand, as it consists of passive elements only, while reacting very quickly fails to support the system voltage since the capacitive current decreases as the square of the voltage. At large voltage dips, it is observed that the synchronous compensator is superior in performance to static var compensator.
Operation and Control in Power Systems
336 Design
0/ Filter Circuit
The maximum harmonic content permitted in the delta connected system as per IEEE specifications are for the 5th harmonics  5% of fundamental and
for the
7th
harmonics  2.5% of fundamental
Since the compensator circuit is delta connected, third harmonic is not considered. I fundamental
=
260A
15th harmonic
=
260
(~) 100
=
13.0 A
17th harmonic
=
2.5 ) 260 ( 100
=
6.5 A
Filter for fifth harmonic At resonance
I ro ==
~LFCF
n
The impedance offered by filter circuit at power frequency is lro 2 L FC F
roC F 1 and this should be much greater than the impedance offered by CM which is CM, i.e. 0)
we require lro 2 L FC F > _1_
roC F

roC M
Let this be greater be 20 times the value
1 C F == (nroY LF
2
LF = 1316 xOIO 6 x31421x24 ==0.OO643H C F ==
,
1
~
(5 x 314) x 6.43 x 10'
== 0.00006303 == 63.03J.LF
Voltage and Reactive Power Control
337
Filter for seventh harmonic: The current for
7th
harmonic = 6.5A nf = 350Hz nro
I
= ;::==
~LFCF
Again assuming a factor of 20 LF =
Therefore,
CF =
20 6 7 = 0.000 16H = O. 16mH 1316xlO (314) x48 I 7'
(7x314t xO.16IxI0
,=0.001284F=1284f.lF
8.34 Subsynclrronous Resonance Series capacitor compensation is the most technically feasible and economically viable choice for increasing the transmission capability ofEHV power systems. However, series compensation brings in certain problems related to dynamic performance of the system. One such problem is sub synchronous resonance. It is an electrical system condition where the electrical system exchanges energy with a turbinegenerator at one or more frequencies below the synchronous frequency of the system. This interchange may be a Iittle damped or undamped. The exchange of energies takes place at the natural frequencies of the respective systems. For the electrical system, the natural frequencies are higher compared to the mechanical system. The range lies between 15Hz and 45Hz for steam turbo generator whereas it is 10Hz and below for a hydro generator. For the electrical system, the modes of oscillation depend upon the number of circuit configurations that can be made through switching. Depending on the system operating conditions and mechanical moment of inertia, the electromechanical system has a frequency of oscillation between I and 2 Hz. There are three types of sub synchronous oscillations that have been identified. The apparent negative resistance (rotor resistance viewed from armature terminals) characteristic of the generator at sub synchronous frequencies causes the generator to act as an induction generator and this effect is called the induction generator effect. If this negative resistance exceeds the network resistance for a specific combination of inductance and capacitance, self excitation may result anhe resonant frequency.
Operation and Control in Power Systems
338
Resonance may take place due to the torsional interaction between electrical and mechanical systems. This may occur when the electrical resonance frequency is near the complement of the torsional resonant frequency of the turbogenerator shaft system (the difference of synchronous frequency and the turbine mechanical natural frequency is the complement of the torsional resonant frequency). A small voltage induced in the armature by rotor oscillation under these conditions can result in large sub synchronous currents. This current will produce oscillatory component of rotor torque and if the mechanical damping is small, the electromechanical system may experience growing oscillations. Finally, when the transmission system contains a series capacitor, the transient electrical torque, under fault or switching operations, may contain large amplitude frequency components close to resonant frequency of the shaft system. This can result in high shaft torques which may be seriously damage the turbo generator units. Consider a series compensated transmission system, let L1, Lg and Lt be the inductances of the line, generator and transformers respectively and let Cs be the series capacitance then the resonant frequency.
otherwise,
=(j)
s
~c X
..... (8.30)
L
where fs is the rated frequency; Xc and XL are the capacitive and inductive reactances calculated at the rated frequency. Since the compensation is generally less than 100% from eqn. (8.30) it can be seen that the resonant frequency is always less than the rated frequency. Several methods are suggested to prevent sub synchronous resonance. One technique is the use of filtering and damping. A blocking filter may be inserted in series with the generator stepup transformer. Otherwise, an appropriately designed reactor may be connected in parallel with an existing series capacitor to form a blocking filter. Other types of filters are also suggested in literature. Relays are also used to detect excessive mechanical system. System switching method can also be implemented to avoid sub synchronous resonance. When a system configurations occurs which can the generator can be isolated from the series capacitances that create problems. This is accomplished by switching the machine off from the compensated system and switching on to an uncompensated line. Unit tripping also can be implemented if all conditions that cause transient torques are completely predetermined.
Voltage and Reactive Power Control
339
E 8.5 A load of (66 + j60) MVA at the receiving end is being transmitted via a single circuit 220KV line having a resistance of 21 ohm and reactance of 34 ohm. The sending end " voltage is maintained at 220kV. The operating conditions of power consumers require that at this load voltage drop across the line should not exceed 5%. In order to reduce voltage drop standard single phase 660V, 40 kVAR capacitors are to be switched in series in each phase of the line. Determine the required number of capacitors, and rated voltage neglect the losses.
Solution: Three phase load
= (66 + j60) MVA 66 + j60 3
Per phase load
=
(22 + j20) MVA
The impedance
Z = R + jX = (21 + j34) ohm
Phase voltage
v
= P
220 == 127.02kV
fj
Without series capacitor 6. V == PRR + OR X
= 22 x 21 + 20 x 34
VR Permitted drop in voltage == 
5
100
=
== 462 + 680
127.02
= 8.9907KV
127.02
5% of 127.02
x 127.02 = 6.351kV
Let Xc be the capacitive reactance to bring down the drop from 8.9907kV to 6.351 kV 6.351 = PRR+(XXC)QR VR
22x21+20(34 X d 127.02
Solving for Xc Xc = 16.65 ohm Given that 40kVAR capacitor at 0.66kV are to be switched. I' c
= 40kVAR = 60.606A 0.66kV
V R IR
~
(22 +j20) MVA
= ~222 +20~
I R
xl0
127.02x10 1
6
.J884 x 10 127.02
1
= 234.07 Am s
p
340
Operation and Control in Power Systems capacitors per phase in parallel =
n
234.07 = 3.862 60.606
The number of capacitors must be either 4 or more than 4. Xc
Vc
=
Ie
=
660V 60.606A
= 10.890hm
Each capacitor has 10.89 ohm reactance. Four in parallel makes Xc
parallel =
10.89 4 ohm
Since total Xe has to be 16.65 ohm The series units ns will be 10.89 x ns = 16.65 4 Hence
= 16.65 x 4 = 6.1157
n s
10.89
As ns must be a whole number n,
=
7
Installed capacity of the capacitor bank
Qc
=
Rated voltage of the capacitor bank
=
0.66 kV
x
7 == 4.62 kV
E 8.6 Find the capacity of a static VAR compensator to be installed at a bus with ± 5% voltage fluctuation. The short circuit capacity == 7000MVA.
Solution: i1Q == capacity of the compensator Sse == Short circuit capacity
i1 V == Voltage fluctuation i1 V = i1Q Sse i1Q == i1 V . Sse == ± 0.05
x
7000 == ± 350 MVAR
Capacity of the static VAR compensator = ± 350 MVAR
Voltage and Reactive Power Control
341
E 8.8 A 400kV line is fed through a 132!400kV transformer from a constant 132kV supply. At
the load end of the line another transformer of nominal ratio 400!132kV is used to reduce the voltage. The total impedance of the line and transformers at 400kV is (50+j 100) ohm. Both, transformers are equipped with tap changing so arranged that the product of the off nominal setting is unity. If the load on the systems is 250MW at 0.8 p.f lagging. Calculate the settings of the tap changer required to maintain the voltage of the load bus at 132kY.
132/400 250MW 0.81g Fig. E 8.8
Solution: Power per phase (P) = 250 = 83.33MW 3 250 0.6 Reactive power per phase = x  = 62.5MVAR 0.8 3 132 VR = VS = Jj = 76.21247 kV
=
76.21247
x
103 V
we have
132 V =V = kV R
S
Jj
when the secondary is maintained at 132kV the primary is kept at 400kY.
Operation and Control in Power Systems
342
Since total impedance is referred to 400kY side we obtain
=1 _ 4166.5 + 6250 1_~6.5 53333.333 53333.333
1  0.1953093
ts = JI.2427 136 = 1.1 147706 ts
=
I = 0.8970455 1.1147706
   = 1.2427136
0.8046906
Voltage and Reactive Power Control
343
Questions I. Discuss in detail about the generation and absorption of reactive power in power system components. 2. Explain reason for variations of voltages in power systems and explain anyone method to improve voltage profile. 3. Explain clearly what do you mean by compensation ofline and discuss briefly different methods of compensation. 4. What i~ a static compensator? Explain with diagrams working principle of various types of static compensators. 5. Explain with diagrams, the operation of a fixed capacitor and thyristor controlled reactor. 6. Discuss in detail about the generation and absorption of reactive power in power system components. 7. Discuss in detail about the generation and absorption of reactive power system components. 8. Describe the effect of connecting series capacitors in the transmission system. 9. (a) What does on mean by load compensation? (b) With neat diagrams discuss shunt and series compensation (c) What are the specifications of lead compensator? 10. Discuss the advantages and disadvantages of different types of compensating equipment for transmission systems. 11. Discuss in detail the voltage stability problem in power systems.
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Operation and Control in Power Systems
Problems P 8.1 Find the regulation and efficiency of an SOlan, 3phase, 50cis transmission line delivering 24,000 kVA at a power factor ofO.Slagging and 66kV to a balanced load. The conductors are of copper, each having resistance 0.12 ohm per km, 1.5 c, outside diameter, spaced equiIaterally 2.5m between centers. Neglect leakance and use the nominal7t method. P 8.2 A 3phase, overhead line has resistance and reactance of 6 and 20 ohm respectively per phase. The sending end voltage is 66kV while the receiving end voltage is maintained at 66kV by a synchronous phasemodifier. Determine the kVAr of the modifier when the load at the receiving end is 75MW at power factor 0.8 lagging; also the maximum load that can be transmitted. P 8.3 A long line from a hydroelectric station operating at 132kV feeds a 66kV system through a transformer. The load taken from the 66kV windings of the transformer is 50MVA at power factor O.S lagging. A tertiary winding on the transformer feeds a synchronous condenser at II kV. If the power factor at the receiving end of the line is to be unity, calculate the rating of each of the three windings. Neglect losses. P 8.4 Find the rating of synchronous compensator connected to the tertiary winding of a 132kV star connected, 33kV star connected, Ilk V delta connected three winding transformer to supply a load of 66MW at O.S power factor lagging at 33kV across the secondary. Equivalent primary and tertiary winding reactances are 32 ohms and 0.16 ohm respectively while the secondary winding reactance is negligible. Assume that the primary side voltage is essentially constant at 132kV and maximum of nominal setting between transformer primary and secondary is I: I. P 8.S A single circuit three phase 220kV line runs at no load. Voltage at the receiving end of the line is 21 OkV. Find the sending end voltage, if the line has resistance of 20.5 ohms, reactance of 81.3 ohms and the total susceptance as 5.45 x 104 mho. The transmission line is to be represented by 7tmodel. P 8.6 A )ower system is operating at 1000Mw, 132kV, 50Hz, with O.S p.f. lagging in parallel with another line at 750MW, I 32kV, 50Hz with 0.707 p.f. lagging. Both are interconnected at the station and when the compensating device is on, the overall power factor is improved to 0.9 lagging. Suggest suitable capacitors, shunt and series. Individual loads and combined load are to improve power factor to 0.9 in all c~ses.
9
INTRODUCTION TO ADVANCED TOPICS
Modern power systems with super power plants are highly complex and sophisticated with huge capital investment. Further, advanced technologies are used in the design and operation of these systems. For efficient and reliable operation of these systems, control action is envisaged with the following organs: I. 2. 3. 4.
Data acquisition and control Computers, Man  machine interface, and Software and human operators.
During the earlier days, control action was appl ied to generation dispatch and supervisory control. Modern systems demand a comprehensive and integrated approach to monitoring and controlling of power flows for economic and secure operation. This brought into the picture a third and equally important aspect of control, viz, security control. The human operator is finding it more and more difficult to take instantaneous decisions in cases of serious and complex situations. Generation control has slowly shifted from analog to digital methods during the sixties. Station and system supervisory control too has shifted to digital computer masters from hardwired masters. The man  machine interface consisted of strip chart records, loggers, indicating lights, anunciators, console push button panels, etc. Black and white or colour CRT display is
Operation and Control in Power Systems
·346
already in vogue. However, the entire control configuration has undergone a radical change with the introduction of system security requirements.
9.1
FACTS Controllers
In chapter 8 static var compensators are discussed. Subsequent developments lead to the emergence of Flexible AC Transmission System (FACTS) controller. Hingorani and Gyugi are pioneers in this area. FACTS controllers is a fast developing area with potential applications. The basic principle of the flexibility in the transmission of power can be understood easily from the following example Fig. 9.1.
Line I
1+ Load Line 2
Fig. 9.1 Transfer of Power
The power flow from the generator to the load through the parallel lines is dependent upon the impedances or reactances when resistances are neglected. If Xl = X2 ' the transfer of power is 50% on each of the two lines. If X2 = 2X I then the p.ower transfer will be in the ratio of 2 : I. Thus the higher impedance line carries less load and the lower impedance line may even get over loaded. If one of the lines, say line 1 is a HYDC line then power flow is electrically controlled and the above problem is eliminated. Further, with HYDC the stability problem also is controlled because of the speed of control. However, HYDC is expensive in a relative sense. A FACTS controller can over come the problem discussed earlier by controlling the impedance, or phase angle. A FACTS controller can control the power flow in any manner that is desired. It is possible to inject desired voltage in series with the line.
Types of FACTS Controllers Generally FACTS controllers are divided into four categories. They are briefly discussed.
9.1.1 Series Controllers They are either variable reactor or capacitor or a power electronic based variable source so as to serve the need. The drop of voltage across the variable impedance can be considered as series injected voltage (Fig. 9.2).
Introduction to Advanced Topics
347
H~
Line
Fig. 9.2 Series Controller
9.1.2 Shunt Controller The shunt controller, also, may be either a variables source or a variable impedance or a combination connected in shunt as shown and they all inject current into the line (Fig. 9.3). Line
f Fig. 9.3 Shunt Controller
9.1.3 Series  Series Controllers Two different types of configurations are feasible. A combination of separate series controllers operated in a coordinated manner in the network or other wise, it could be a unified controller wherein series controllers provide independent series reactive power control as well provide real power transfer through a power link. This is shown in Fig. 9.4.
I I
~
l I
A. C. Line 1
D. C. Power link for real power exchange
I I
~
l I
A. C. Line :2
Fig. 9.4 Series  Series Controllers
In FACTS terminology the term unified is used to indicate that the D.C. terminals of all controller converters are connected together fm real power transfer.
348
Operation and Control in Power Systems
9.1.4 Series  Shunt Controllers In the same way as series  series, this could be also a combination of separate series and shunt controller  controlled in a coordinated manner or a unified power controller with series and shunt elements. The basic scheme is shown in Fig. 9.5.
v. Line
D. C Power link for
f
Real power exchange
Coordinating unit
Fig. 9.5 Series  Shunt Controllers
9.1.5 Power Flow Control There are two types of converters  voltage sourced and current sourced, using gate turn off devices. The voltage sourced converter consists of a gate turn off device paralleled with a reverse diode. A D.C. capacitor is used as a voltage source. The current sourced converter has a gate turn off device in series with a diode. It has a D.C. reactor as a current source. Voltage sourced converters are preferred in FACTS controllers. Sequential switching of the device will give A.C. voltage from the D.C. source. They are called thus static synchronous compensator. It is operated as a shunt connected static var compensator whose output voltage can be controlled independent of A.C. system voltage. Such a device is also called STATCOM. The structure is shown in Fig. 9.6.
o Fig. 9.6 Static Synchronous Compensator
Introduction to Advanced Topics
349
9.1.6 Static Var Compensator(SVC) It is a device that supplies capacitive or inductive current so as to maintain, voltage at any bus. This has been already discussed in Chapter 8. The device is shunt connected. In a similar manner a thyristor controlled reactor (TCR), shunt connected has the effective reactance varied in a continuous manner by conduction control of the thyristor. TCR naturally belongs to the class of SVC. If the reactance is varied in a stepped manner instead of continuous mode, the device becomes thyristor switched reactor TSR. Several shunt connected reactors may be switched in or out. Likewise, thyristor switched capacitor (TSC) is operated such that shunt capacitor units are switched in  and out depending upon the need. It should be noted that in this case the thyristors are operated without firing angle control. They operate with full or zero conduction. A thyristor controlled resistor (TCBR) with shunt connection can be use to brake a generator during accelerating period when a fault occurs. They are called then thyristor controlled braking resistors. In the same way as STATCOM a static synchronous series compensator (SSSC) operates with a voltage sourced converter or current sourced converter giving output voltage in series with the line. Thyristor controlled series reactor (TCSR) thyristor controlled series capacitor (TCSC) thyristor controlled switched series reactor (TSSR) are similar devices in series mode of operation.
9.1. 7 Unified Power Flow Controller This is a combination of series and shunt connected static compensators (STATCOM and SSSC). They are coupled through a common D.C. link. The controller permits bidirectional flow of active power between both the series output terminals of the SSSC and the shunt output terminals of the STATCOM. Without any external electric energy source they provide simultaneously real and reactive series line compensation. UPFC is also capable of providing independently controllable shunt reactive compensation. This is schematically shown in Fig. 9.7. There are several more devices such as thyristor controlled phase shifting transformer (TCPST), inter phase power controller (lPC) Thyristor controlled voltage limiter, thyristor \ controlled voltage regulator etc.
9.1.8 Advantages due to FACTS devices The following are some of the advantages that can be obtained by using FACTS controller. I. Power in lines is controlled in any desired manner. 2. Line capacity is increased, practically upto thermal limits 3. By raising transient stabilitv limit, system security is enhanced
350
Operation and Control in Power Systems ~~~~~Line
~
~
I sssc
0 =}= D. C. Link
0
Statcom
Fig. 9.7 United Power Flow Controller
4. Up gradation of lines is easier.
5. Reduced reactive power flow, thereby permitting greater active power flow. 6. Reduced cost of energy received due to enhanced line capacity.
9.2
Voltage Stability
Voltage stability is an integral part of the power system response and is an important aspect of system stability and security. Voltage instability has been detected well before the onset of angle instability in many cases. If the problem is not corrected it can lead to voltage collapse and system wide disturbance. The loss of synchronism of generators is called angle instability. Voltage stability also called load stability is a subset of overall stability of a power system and is a dynamic problem that occurs due to monotonically changing voltages. In industrialized areas increase of load demand is met without a corresponding increase in transmission capacity leading to severe problem including voltage stability. Voltage stability is the ability to maintain the voltage so that when load is increased load power will increase and so both power and voltage are controllable. A power system at a given operating state and subjected to a given disturbance is voltage stable ifvoltage near loads approaches post disturbance equilibrium values. Following voltage stability, a system undergoes voltage collapse if the post disturbance equilibrium voltages near loads are below acceptance limits. Both voltage stability and collapse may occur in a time period of a fraction of a second to a few minutes.
Consider the plot of voltage V at a load bus as a function of load power P shown in Fig. 9.8 as ABCDE.
Introduction to Advanced Topics
351
v A _"
B
v I '1
,
  i  
, C )Nose pomt
1/1 1/ 1
v
Yo
1
)/iu
1
// ~~~~
E
PeR
P
Fig. 9.8 P. V. Diagram at the Load Bus
It can be observed that there are two voltages Y I and Y2 for a constant power load. The high voltage solution is stable while the other is unstable. The maximum loadability is determined by point C and the part CDE is uncontrollable. R+j:. VS~,VR~ I
...
Line
l.oad P +.IQ
Fig. 9.9 Transmission Line with load
Mathematically the voltage YR can be expressed by (see Fig. 9.9).
/
V =[ 2;Ys ±1~(2QXY(r _4X 2(p2 +Q2)] R
..... (9.1 )
from whi6h it is clear that Y R is a double valued function. For differe~ power factors the PY characteristic can have the shape shown in Fig. 9.10.
v /' Nose of the curve
/
08 (load)
L _ _ _ _ _ _ _ _~~_ _ _ _ _ _ _ _ _ _
E
P
Power
Fig. 9.10 P. V. Curves for Different Power Factors
352
Operation and Control in Power Systems
Static analysis is sufficient to assess voltage stability, but for accurate prediction dynamic analysis must be carried out. The probability of voltage instability increases as the system is operated close to its maximum loadability limit. Environmental and economic expansion of transmission network and for obvious reasons distant location of the generators from the load centers all result in over loading of the existing networks. The present trend is to optimally utilize the inher\!nt margins available with flexible A.C. transmission system controllers. Reactive power compensation close to the load centers as well as at the critical buses in the network is essential to over come voltage instability. The location size and speed of control of FACTS controllers determine the maximum benefit that can be obtained without getting into voltage instability.
9.3
Power Quality
Any amount of power can be drawn from and delivered to an infinite bus without affecting (i) the magnitude of bus voltage (ii) the frequency of bus voltage.
The short circuit ratio at a load bus (SCR) is defined as SCR = Short circuit current at rated voltage with load short circuited rated load current It can be seen that SCR is iaversely proportional to the Thevenin's impedance of the network as seen from the bus. If SCR is low the load bus is said to be connected to weak power grid and vice versa. Power quality (PO) problem is likely to occur at that bus. One aspect of PO can be selected in terms of phase voltages V R' V) and V B as
1
[
V V: = VB
sin rot
J2 V sin( rot 
231£ )
..... (9.2)
(2 )
sin rot + 31£
under all conditions, V and ro remaining constant. Similarly the harmonic content of the bus voltage is also an indication of PO. Electric power quality can be loosely defined as a measure of how well electric power can be utilized by customers. Power utilization is degraded when wave shapes are irregular, voltage regulation is poor, harmonics and flicker are present or even when there are momentary events that distort the usually sinusoidal voltage wave. All the above conditions result in degradation of power quality. The introduction and the wide spread use of high power semi conduction devices at all levels of power system operation, and utilization have made non sinusoidal load currents quite co~mon.
Introduction to Advanced Topics
353
9.3.1 Power Quality Index The harmonic components of voltages and current are used as power quality indices. Let i(t) be the load current that is periodic 11' 12,
.....
are the fundamental and harmonics of i(t).
For all general purposes, the standard is taken as total harmonic distortion (THO) defined by I
a
THO
=
L'1=2 II
..... (9.3)
In regard to frequency, high frequency phenomenon is termed noise and the low frequency phenomenon is named flicker. As has been already mentioned rectifier loads are the main source of harmonic load current. Oifferent rectifier connections such as single phase bridge and 3phase bridge rectifiers contribute to harmonic distortion and various factors such as THO, displacement factor, power factor are clearly defined. It may be noted that displacement factor is the cosine of the angle between fundamental voltage and fundamental current while, power factor is the ratio of real or active power to apparent power V RMS JRMS '
9.3.2 Voltage Sags A short duration magnitude reduction in voltage is called voltage sag. The time period may be of the order of few seconds. The cause for a sag is sudden, short duration increase in current. Such an event occurs during motor starting, transformer energizing, faults etc.' For example, the transient voltage when a big motor is started will have the shape of all the phase voltages shown in Fig. 9.11. The voltage falls by a large amount with gradual recovery over a few cycles. 200
., ~
215
'"Cl
210
.0
205
C>()
~ ~
~ ~
195 190 0
2
11
3 Time (Cycles)
 Fig.
9.~1
Voltage sag due to motor starting
354
Operation and Control in Power Systems
It can be seen that the voltage drops are practically the same in all the three phases. In case of magnetizing inrush current of transformers, the inrush current is not the same in all the phases. This is shown in Fig. 9.12 second harmonic and fourth harmonics contribute more to harmonic distortion. II 10.9 10.8
~ (l)
~
~
[/J
:::E
ex:
10.7 10.6 10.5 10.4 10.3 10.2 10.1 . 10 0
5
15
10
20
25
Time (Cycles)
Fig. 9.12 Voltage sag due to transformer energizing
In case of faults, the most common type of fault is single line to ground fault. The voltage falls more in two phases and recovers sharply after a few cycles. This is shown in Fig. 9.13 the most severe faults are due to short circuits and faults involving earth. 11.5 II
~
>
10.5
(l)
~
~
10
:::E
9.5
[/J
ex:
9 8.5
0
2
4
6
8
10
12
Time (Cycles)
Fig. 9.13 Voltage sag due to a fault
14
16
•
355
Introduction to Advanced Topics
Transformer c6nnections affect the impedance of fault. For example a phase to phase fault for a star connected load gives a similar sag as a singlephase fault for a delta connected load. Further a voltage sag is not exactly recorded the way it occurs at the load due to intervening components such as transformers which effect the voltage sags.
9.3.3 Rectifier Loads Most electronic equipment (such as personal computers) with power rating less that I kw. Use single phase diode bridge rectifiers. The output of the bridge rectifier is filtered using a capacitor filter or an LC  filter. The diode bridge rectifiers are often followed by switch mode power supplies. The switch mode power supply converts the unregulated D.C. link voltage to a well regulated output voltage. For higher power applications (greater than 1 KW) either 3phase bridge of six pulse type or twelve pulse type are used. The single phase case is shown in Fig.
A.C
SMPS
mains
Output
Fig. 9.14 Single phase rectifier load
9.14. The voltage sag depends upon the energy storage device used and the switch mode power supply (SMPS). If the sag in the A.C. main voltage is of considerable magnitude and duration, the D.C. link voltage may fall below the minimum value for which it is desired. This results in the shut down of the .electronic load connected to the output termals of SMPS.
9.3.4 Flicker Light flicker perception is said to be a physiological process in which the eye and the brain participate. Voltage fluctuations occur due to mainly electric arc. furnace loads. These occurs because of voltage and current harmonics due to inherent non linearity of the arc characteristic. Light flicker is caused mainly due to reactive power flow from the system at the instant of short circuit created by the arc. Any step taken in regard to control of reactive power will control light flicker. Furnaces' of ratings about I OOMVA and above cause the problem. Again when wind turbines are connected to system they are frequently switched on and off around the cut  in speed then voltage fluctuations occurs. Another major source for harmonics in addition to electronic loads is adjustable speed drives as they use diodes, SCRs, power transistors, switches etc to chop wave forms to
356
Operation and Control in Power Systems
control power. Total harmonic distortion of voltage less than 5%' is acceptable but THO above 10% requires, positive steps to reduce the harmonics.
9.3.5 Power Acceptability or Voltage Tolerance The voltage tolerance or more commonly power acceptability curves are loci dra»,n in the bus voltage duration tim~ plane. There curves indicate the tolerance ofa load to withstand momentary low and high voltage events. The International Electrochemical Commission (IEC) calls the same curves as equipment immunity curves. Computer business equipment manufactures association (CBEMA) has defined its power acceptability curve but later in 1996 the Information Technology Industry Council (ITIC) announced its power acceptability curve which superceded CBEMA curve. This is shown in Fig. 9.15. 250r~~~~

~ ~
200 ................................. ;................................................................. . : Overvoltaga Conditions f50 ............................. '1' ...... "' .......................... ,............................
Q)
~
~ ~
m
:0)
100 ·..........
·t·. · · · . . . . . ;'0j.(5'....................................,.................. ,'_. :!O
·T..........· · . · ·. . ·....·........ · ........ . ·'·· . · . ·.
50 ............................. =0
.!: ~
g> tI1
..c:: U
o
Acceptable
Power I'r'
~
~;BR~at~ed!L======::!:===~ ]Volta~ge::...._.]""":;~'
(Jj;
E:
50 ............................ ~; ....... ·....r·..""'.. ;,,;,,: .. ·.:..:.;·"~ .. ·;... ....; .. ·.;...·..""' .. ;..::·~"'......;.;.;.;..:..:..:..;.;:.;, ~i
Undervoltage Conditions
100LO.OO~01~O.OO1=O~.O1~~O.=1===1====1=O===1=OO==~1000 Time (Seconds) , Fig, 9.15 ITIC Power acceptability curve
9.3.6 Solutions to Power Quality problem There are several power electronic solutions to power quality problem. A dynamic voltage restorer, which is a voltage source inverter CVSt) injects a voltage in series with the distribution line. This is shown in Fig. 9.16. Three phase voltages generated by the VSI can be injected into the line through a special coupling transformer. The terminals of each of the three primary windings are brought out individually and are connected in series with the lphase lines. On the secondary side the thre'e windings are connected either in star or in delta and the three terminals of the, VSI are connected'to the three secondary terminals.
Introduction to Advanced Topics
357
VSource
Fig. 9.16 Dynamic Voltage Restorer
Point of Common Coupling (PCC) is the point in the inter connected power system where loads are connected to the network. It is the point at which the load interacts with loads and the network itself. For example the pec for typical residential applications is the distribution transformer secondary. Where the distribution transformer serves a single customer, the primary of the transformer is the PCe.
vpee =
V, (t) + VDVR (t)
..... (9.4)
The DVR will maintain near perfect power quality conditions at the PCe. The DVR is useful for sags, harmonics and flicker. There are many other devices such as unified power quality conditioner (UPQC) which is a combination of STATCOM and DVR for even better performance.
9.4
Data Base for Control
Computer control of modern power systems is proposed to improve economy, to maintain quality and for better security. Such a control is feasible, primarily if all meter readings and other information pertaining to the operating state of the system are processed in real time into a more useful form so that control decisions are made using it. For power systems in normal operating state, bus voltage magnitude and phase angles are the state variables as they completely determine the state of the system in steadystate (quasistatic state, infact). The vector of voltage magnitudes and phase angles is called the static state vector. State estimation is the mathematical process of obtaining the state vector from the variable information. A knowledge of the state vector enables the system operator to take important decisions. The state estimator, an algorithm based on load flow and statistical estimation theory, is of immense use in calculating security and contingency evaluation. At an advanced level, it can also be used for load frequency control and economic dispatch.
358 9.5
Operation and Control in Power Systems State Estimation
Since loads and generation change continuously, the power system never attains a steady state. For the purpose of mathematical analysis, however, a steady state operating point over a short interval of time may be considered where the changes in system variables are within reasonable limits.
Power System Variables The variables of interest for network monitoring are (i) voltage magnitudes and phase angles, real and reactive power flows at each network bus and (ii) real and reactive power flows in each line. Meters for measuring watts, vars and volts can be placed at generator buses, EHV lines, tie  lines and load buses and the readings can be telemetered in real time to the control centers. However, only some of the measurements are available immediately while the rest are either delayed as is the case ofinformation from a neighbouring, interconnected system or not available (like data from the low voltage distribution side). Further, the available data may not be perfect due to errors in the meters. Some of the meter readings may also be missing. Let n be the dimension of the state vector X and m be the dimension of the measurement vector Z. Since the slack bus is fixed, we have Nb voltage magnitudes and (NbI) voltage phase angles where Nb is the number of buses.
XT
=
[Vi' V 2'
.....
VNb ' 82
.....
8 Nb ]
..... (9.5)
Knowing X, all other quantities like active and reactive power flows, line flows, etc can be computed. In practice, all the values of X cannot be directly metered. The measurement vector Z and the state vector X are related by the nonlinear equation.
Z = h(X) + 11
.. ... (9.6)
where h is the measurement noise. By exercising option in the selection of variables for metering. Five types of measurement vectors can be identified.
Case I swing bus.
Measuring real an,d reactive powers PI and Q I at all nodes, except P, at the ZT =
[P 2, P3,
.....
PNb' Q, ..... QNb]
and it is a (2 Nb  I) dimensional vector. PI and Q I are given by equation (2.11) and (2.12) Case 2.' Real and reactive powers at all buses except Pi at slack bus and voltage _ magnitudes at all buses may be measured in which case PNb ' Q I ..... QNb Vi .... · V Nb ] is a (3Nb  I) dimensional vector. ZT =
Here
ZT
[P 2
.....
Case 3: Real and reactive power flow in the lines Pij and Qij at both ends ot'each line may be measured.
Introduction to Advanced Topics ZT  [P
IJ'
359
0] IJ
and it is a 4 N e dimensional vector where N e is the number of lines. Case 4: All the variables of case 3 and voltage magnitudes at all buses may be measured yielding. ZT
[PII' 011' VI' V~ ..... V Nh ]
=
an (N I + N h) dimensional vector. Case 5: All variables may be measured giving ZT
[PI
=
0
1
PI~ Ol~ VI ~\]
Which is (4N h  I + 4 N () dimensional.
Modelling of uncertain(v: Gaussian distribution is assumed for mathematical modelling of the uncertainty in measurements . ..... (9.7)
The uncertainty is due to . I. Instrumentation error,
2. Operational uncertainty. and 3. Incompleteness of the model Least Squares Estimation Using Eqn. (9.6) n
The optimal estimate
J
=
Z  h (X)
..... (9.8)
X is that value which minimizes the cost functional =
nT R I n = [Z  h(X)T] R I {Z  heX)]
..... (9.9)
Where R is the covariance matrix of the observation noise n. The minimization may be carried out using any of the standard techniques including steepest descent method, Fletcher  Powell method, etc.
Linear Estimation Theory In this case, the measured variables are assumed to be linear functions of the state variables. The measurement equation (9. I) is then written as Z=AX+n ..... (9.10) where A is the m
x
n measurement matrix
Z is the m  dimensional vector of noise measurements comprising real and reactive powers, and voltage
360
Operation and Control in Power Systems X is the ndimensional state vector and n is the noise vector It is desired to minimize J = (Z  AX? RI (Z  AX) so that the optimal estimates
X is obtained.
..... (9.11 )
In practice, three cases may arise.
Case I : m > n; the estimated equations are ..... (9.12) The estimate covariance matrix is Rx=(ATNIAt l
..... (9.13)
The noise covariance matrix
Nx = E (1111 T)
..... (9.14)
The number of basic arithmetic operations required for the computation of the optimum estimate area of the optimum estimate are of the order ofmn'. The difficulty in this case is that the maximum Rx tends to be singular making inversion difficult.
Case 2: m = n: The estimation equations are X = AI Z
..... (9.15) ..... (9.16)
In this case, the computational problems are fewer since A is a square non singular matrix and Rx is determined from R, = AI N(ATt l
Case 3: m < n; while an estimate of 11 can be determined in principle, he estimate is neither unique nor does it have much relevance to state estimation in power systems.
9.6
Power System Security
The operating modes of a power system have already been discussed in Chapter I. In the 'normal' mode, both the load and the operating constraints are satisfied and adequate spinning reserve is available. In the 'alert' state the load requirements are met with, but an adequate spinning reserve is not available. A system is in 'emergency' mode when the operating constraints are not completely satisfied. There may be two types of emergency modes. The system may be in 'steady state emergency' when the steady state operating constraints like bus voltage limits are violated. On the other hand, the system may reach 'dynamic emergency mode of operation when, for example, stability operating constraints are violated. In either case, the normal load requirements cannot be satisfied. In the 'restorative mode' control action is initiated, but the load constraints are not completely satisfied. Under this condition, the system may be in a completely or partially shutdown state.
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The above classification of the states of the power system will facilitate in distinguishing between the different types of controls to be applied to the system. It can now be stated that the objective of system security control is to keep the power system in normal mode of operation and prevent it from entering into either emergency or restorative mode. The system state is continuously assessed via data acquisition systems making use of telemetered data, redundant data and by predicting the missing or doubtful data whenever a metering channel becomes erroneous or its accuracy is beyond the normal specified range. This job is referred to as security monitoring. During security assessment, a series of fast computations are made to examine the effect of various credible contingencies and those that are not so. If some of the assumed contingencies result in unsatisfactory performance from the point of view of security, a corrective strategy is called for, to determine the best corrective action either by special calculations within the computer or from results obtained from offline studies. Continuous monitoring for security and sending command signal for corrective action whenever necessary is 'referred to as security control. Steady state security analysis is performed to determine whether there will be a new steady state operating point at which the system will settle after the postfault oscillations have been damped out. The post contingency steadystate solution will have to be checked for violation of operational constraints. If it violates the constraints, the contingency is declared to be insecure and necessary corrective action is to be initiated. Steady state security may be defined as the ability of the system to operate steadystatewise within the specified limits of safety and supply, following a contingency in the time period, after the fast acting automatic Control devices have restored the system balance, but before the slowacting controls like transformer tappings, human decisions, etc, have responded. For steady state security analysis the following contingencies may be considered: I. Loss of a generating unit 2. Sudden loss of a load, 3. Sudden change in flow in an intertie, 4. Outage of a transmission line, 5. Outage of a transformer, and 6. Outage of a shunt capacitor or reactor. The outage may be either network outage or power outage. Fast and decoupled load flows are employed for security analysis of systems.
9.7
Steady State Security Assessment
The steady state security assessor is a computer program that uses real time data for analyzing the current state of the system for steady state security and is thus essentially an online process. For this, an approximate load flow algorithm is needed. Consider the inputoutput model given by
362
Operation and Control in Power Systems F (X, u, d) = 0
..... (9.17)
where X is the vector of dependent variables u is the vector of control variables and d is the vector of disturbance variables For a given disturbance vector, it is required to determine the output X corresponding to a given input u as shown [n Fig. 9.17.
u
F (X. R d)
·1
• X
f
Fig. 9.17 System representation
Eqn. (9.14) and bus system. There are may be denoted by the bus impedance matrix,
Fig. 9.17 correspond to 2n real nonlinear power equations for any n2n bus power injections (nreal power and nreactive power) which vector S. The network parameters are given by the bus admittance or designated by P.
The steady state security assessment problem is concerned with the analysis of the system described by the vector triplet (X, S, P). Given a base triplet (Xo, so, pO) along with contingency details, pI, i = I, 2, ...... , nw;
SI, i = nw + I, ....... nw + pO where nw is the number of network outages and po is the number of power outages, it is then required to determine the postcontingency states.
XI' i
=
I, 2, ....... , (nw + po)
The steady state security assessor is a computer program that determines the postcontingency states. The program is also interlinked with several other programs at the computer control center.
9.8
Application to Outage Studies
Referring to the fast decoupled load flow algorithm given in chapter 2.
B'~b == ~p
IVI
The above equations may be represented as
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Introduction to Advanced Topics
..... (9.18) ..... (9.19)
so that
The outage of a line (neglecting charging capacitance) joining buses i and j can be reflected in K by modifying two elements in row i and two in row j. The new outage matril' can be presented as
K' = [K + BA AT]
..... (9.20)
where B is the line susceptance and A is a column vector with AI = + I and AI = I and zeroes in other locations.
It can be shown that K'I = K I  h L AT K I
..... (9.21 )
h=[B'+ATLr l
..... (9.22) ..... (9.23)
yl = K'IC
..... (9.24)
with L = K I A and The solution vector
gives the postoutage state of the system.
Complete system security involves the following: I. System prediction: This is done by computer programs that predict the system load conditions in advance for a predetermined time interval (say, an hour or day),
2. System contingency evaluation: This involves programs that detertninethe state of the system consequent to generation, line or transformer outages.
3. System corrective strategy This is aimed at taking all possible decisions to keep the system in steady state secure condition. It also involves the necessary steps to be taken to overcome both unexpected and anticipated problems, either in the present state of operation or in future:
4. Automatic control: This eliminates the necessity of the operator taking initiative for all corrective actions.
9.9
Pattern Recognition Methods
System security can be evaluated through pattern recognition techniques also. The approach is quite different from conventional methods. Pattern recognition methods essentially involve the formation of a training set with pattern vectors, feature extraction and defining a security function.
Pattern Vector Each operating condition ofa power system is determined by real and reactive power generation, line flows, loads voltage and voltage magnitudes and phase angles. All these variables from measurements constitute the compollents of a pattern vectur X = (XI' x~ ..... , xn ).
Operation and Control in Power Systems
364 Training Set
Every conceivable operating condition of the system gives rise to a corresponding pattern vector. The set of all possible pattern vectors constitutes the training set.
Pattern Classification Each of the patterns belonging to the training set are classified as either secure or insecure through offline studies. Secure patterns contain all operating conditions that are secure while insecure ones contain such conditions in system operation that are insecure.
Feature Extraction The actual number of variables used for representing an operating condition may be too many. But in practice, only a few of them may considerably influence the security of the pattern. It is desired to select a relatively small number of such variables called feature variables and the process of selection is called feature extraction. Feature extraction can be simplified by applying operational experience and engineering judgment in addition to any statistical criterion. The feature vector is defined by
F = (fp f2' .......
~n);
m< n
..... (9.25)
All variables that are redundant may be removed from the pattern. Variables with high correlation coefficients may also be eliminated.
Security Function A security function may be defined as a linear function of the feature variables, S (F)
=
Sjl + SI fl + S2 f2 + .. :.. + Sm fm
..... (9.26)
The security function S (F) > 0 if F is secure and S (F) < 0 if F is insecure. Different methods are available to determine the constants So' SI' ..... Sm' For example, least squares method or optimal search method can be used successfully to determine the coefficients So' Sm' More sophisticated and mathematically complex functions may be defined in place of Eqn. (9.26). After the security function is identified, its validity may be checked using patterns of known class.
Security Evaluation Once a valid security function is determined, then any operating condition represented by the corresponding feature vector may be used, and from Eqn. (9.23) the operating state may be classified either as secure or insecure. The success of the method depends on the proper selection of the feature variables and security function.
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365
Pattern recognition methods may not have much relevance for steady state security evaluation. However, it appears that for transient security evaluation, pattern recognition techniques may provide the answer in future for situations where the fast acting automatic control devices have restored the system balance and the slow acting controls like transformer tappings, human decisions, etc. have not yet responded.
9.10 Power System Control Centres Increase in unit sizes, growth of interconnections and the need to maintain the system in normal mode require sophisticated control, instrumentation and protection: The multiplicity of monitoring instruments in the control room and their distance apart make the observation of more than a few vital ones almost impossible, especially during the intense activity of plant startup. The operators are called upon to visualize the implications of a variety of changing plant parameters and take critical decisions. These requirements led to the development and application of more advanced solidstate modular electronic instruments, computer based direct digital control and data processing systems. A modern power system control centre has the following functions to perform: 1. Automatic generation control, 2. Economic load despatch 3. Automatic voltage (reactive power) control, 4. Online load flow, 5. Online short circuit. 6. State estimation, 7. Security monitoring, 8. Steady state security analysis, 9. Supervisory control, 10. Automatic trouble analysis, 11. Emergency control like load shedding, generator t ripping, and 12. Automatic circuit restoration, etc. The computer system involves dual configuration with external interfaces to monitor the data. The first one is a process computer linked by telechannels to various generating and substations for data acquisition. The second one is a larger one where major calculations are carried out and is linked to the process computer. For realtime computer control of power systems, the following basic components are needed: 1. System wide instrumentation, 2. High speed digital telemetry, 3. Central processing unit (CPU), 4. Memory and bulk storage,
Operation and Control in Power Systems
366 5. Interactive display, and
6. Software (operating and application). The real time computer is designed to perform data acquisition, storage and retrieval, data processing, interactive display and remote signaling and control. It consists of modems and interfaces, CPU, memory and bulk storageand inputoutput devices like display devices, card reader, printer, etc. Fig. 9.18 shows a functional block diagram of a realtime computer.
nrsPI A Y
1<11111111
rI
MODE.MS'~
DATA •
FROM
THE SYSTEM
r4~
BULKSTORAGE
Fig. 9.18 Functional Block Diagram of a real time computer
The data base consists of static data, dynamic data and software. The static data consists of the details of lines, transformers. generators, etc. The dynamic data includes line flows. voltage levels, breaker condition, generation and demand. The software includes operating software, application software and support software. The operating software is the system software for real time operation compiling routines, file management, etc. The application software consists of programs written for power system operation and control, which were listed earlier. The support software consists of diagnostics: debugging, maintenance and testing programs. Computer control can take any of the following forms: I. Offline computer control, 2. computerassisted control. and 3. online computer control. For planning and operating computation, which are carried out at infrequent intervals, offline computer calculations suffice. The results are to be updated at regular time intervals. ; In the computerassisted control scheme, necessary data is transmitted at regular intervals to the computer located in a central control station and its decisions are communicated to the human operator. For shorttime processes extending up to a few minutes, the system is under continuous control of direct acting devices. Under emergency conditions. the online computer which processes the data continuously sends commands to the direct acting devices according to a preplanned strategy to prevent danger to the system.
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Introduction to Advanced Topics
The type of computer control, the configuration and the peripherals used are ultimately decided by the cost that the utility industry is willing to invest.
9.11 Level Decomposition in Power Systems Power systems are characterized by strong hierarchical order. Accordingly, control strategy can be devised to fit into the hierarchical structure advantageously as indicated in table 9.1. Table 9.1 Hierarchical Structure in Power System System Monitoring and control Level First Level Generating stations and sub stations Local control center Second Level Sub transmission and transmission networks Area load dispatch center Third level Transmission system State load dispatch center Fourth level or top level
[nterconnected power systems
Regional control centers
Local Control Centre A number of control functions can be performed locally at power generating stations and substations using local equipment and automatic devices. Some of the typical control applications are :
1. Local monitoring and control, 2. Protection, 3. Auto reclosure, 4. Voltage regulation, 5. Capacitor switching, 6. Feeder synchronization; 7. Load shedding in the event of necessity, and 8. Network restoration.
Area Load Dispatch Centre A group of generating stations and substations, along with the associated network and loads may be considered as a unit for control under an area load dispatch centre. The area control centre receives information and processes it for appropriate control action.
State Load Dispatch Centre The minutetominute operation of the power system at the state level may be carried out by the stale load dispatch centre. It may have the following functions:
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368
I. System generation and load monitoring and control, 2. System wide, state monitoring and control, 3. Circuit breaker state monitoring and control, 4. Load shedding and load restoration, 5. Supervisory control for lines and equipment states, 6. System alarm monitoring and corrective action, and 7. Planning and monitoring of system operations.
Regional Load Dispatch Centre The regional load dispatch centre may be regarded as a coordinating and monitoring centre for state level dispatch centres with the following main objectives: I. Integrated operation of state level dispatch centres, 2. Operation and maintenance schedules for the generating plant, 3. Operation and maintenance schedules for maximum capacity utilization, 4. Monitor and control interstate power transactions, and 5. Monitor and control interregional power transactions.
9.12 Network Automation Network automation reduces the job of the human operator to some extent. It involves monitoring the process and remote control. Whatever may be the level of automation, all important emergency decisions are required to be taken by the human operator. The input data for the network automation may be obtained from important load centres through microprocessor or computerbased remote terminal units. The remote terminal units may perform the following functions: 1. Obtain the system data in analog form, 2. Convert the analog data into digital form, and 3. Transmit the digital data to area load dispatch centre. The data so transmitted will contain information like circuit breaker status, isolator position, etc. which will be indicated only when the condition changes as well as regularly sampled information about the network variables such as line flows (MW, Mvar), frequency (Hz), voltage levels (kV), currents (Amps), etc. The information processed will also include generator or transformer tripping, battery failure, etc. Also other functions that the remote • terminal units may perform include limit or threshold value monitoring, fluid level processing, data logging, etc. . The data may be used for taking appropriate control action with the system of computer control planned.
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369
@ Oisk~
Di5k~
~ Con!iOl~
Console
o Iflone
Drive tor dlspbys
.dollng laclhly Visual
di~plQy
unll
Fig. 9.19 Computer control in Power system
The data may be classified as real time, static and long term data. The real time data consists of measurands, alarms, etc and is stored in core memory. Static data consists of the background picture, text data files, etc. and is normally stored on a secondary, fast access memory. Long term data may be stored on secondary, slowaccess memory and consists of data for postmortem analysis, measurands, etc
The manmachine interface may consists of the following: I, Alphanumeric  technological key boards. 2. Visual display units. 3. Mosaic boards 4. Digital displays 5. Loggers, and 6. Strip chart recorders A typical computer control set up is shown in Fig. 9.19.
9.13 Load Prediction Accurate prediction of load demand is required for service reliability and efficient operating performance of a power system. If the system is predominantly thermal, this is all the more
Operation and Control in Power Systems
370
important in order that adequate generating capacity to supply the load demand, maintain system security and supply the necessary spinning reserve is economically scheduled. In practice, spinning reserve is provided as a backup for either the loss of the largest unit or for the loss of transmission capacity which renders unavailable the greatest amount of generating capacity. But, one should not lose sight of the possibil ity ofinaccuracy in system load prediction consequent to which deficit operating reserve capacity may result. For plant loading schedules in thermal systems, load prediction up to two hours in advance is necessary while for unit commitment schedules prediction up to 24 hours is sufficient. Also, at all stations and control centres, shorttime prediction is needed for storage and. display of advance information. Based on this information, predictive security assessment of the system is made. This also helps to contain the rates of change of generator outputs within their permissible limits. For the implementation of economic scheduling of generation using digital computers, detailed estimates of the future load demands are essential In order to allow sufficient time for the calculation and implementation of the generator schedules. Whatever method is envisaged for the calculation of such economic schedules consistent with the security and spare requirements of the system, the schedules should be calculated every 15 or 30 minutes and each economic schedule should be a predictive one, for at least about 30 minutes ahead of the event. It is then obvious that the predictions are to be revised frequently in the light of any fresh information so as to minimize the estimation errors. Peak load demand forecasts are useful in determ'ining the investment required for additional generating and transmission capacities required. Forecasts for planning require data extending over several previous years. Meaningful forecasts can be obtained with lead time of 3 to 5 years. Generally, load demand is assumed to have two salient features helpful in prediction. These are the longterm trend component dependent on economic growth, seasonal changes, etc. and the weather sensitive fluctuations in daily and hourly load demands. Thus, the overall load demand has a predictable trend component superimposed on which is an erratic variation attributable to weather fluctuations. Summer, winter and monsoon periods will have definite patterns of load consumption. Also, the load demand varies during the day continuously, following a pattern with definite peaks at certain times of the day.
The weather sensitive component depends on the following meteorological factors:
1. Temperature,
2. 3. 4. 5.
cloudiness, wind velocity, visibility, and precipitation.
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371
9.14 Load Prediction using Matereoiogicai Data Weather Weighting Method In this method, the load is separated into two main components. The first component is a base load which is of fixed value and the second a variable component which is a function of the weather conditions. Estimates can be made 24 hours ahead, using the weather, forecast. The temperature base for weighting the effect of the predicted temperature on the load is the normal, mean temperature of the month. The normal, mean temperature of the month has zero weight. Similarly the change in consumers demand due to cloudy weather may be assumed to vary in direct proportion to the degree of cloudiness. This in turn may be expressed by an illumination index with fair, clear sky corresponding to zero weight. The base load is determined from past records, proper weighting of the elements of the weather will be attained only after several trials. The method of prediction stabilizes after this trial period. It may be noted that the base loads for week days and weekend will generally be different for any hour. Using these base loads, a load estimate based on the best available weather forecast can be made using proper weighting of meteorological factors like temperature, cloudiness, wind, velocity; etc. In thermal stations, a boiler may take about 3 hours of time to reach full operating condition from the cold. Thus, peak load estimates In systems with predominantly thermal generation are more critical, since they are required in time to bring any necessary generating capacity online. The feasibility of this method lies in the consistency of the calculations of the base load. The base loads are to be revised every year as load demand changes continuously.
Multiple Regression Methods The meteorological information can be reduced to a number of spec!fic factors like: I. Temperature (T) 2. Cooling power of the wind(W), 3. Illumination index(L) 4. Rate of precipitation (P), etc In deriving the functional relationship between the variation of demand and the specific metrological factors to which it is sensitive, it may be assumed that the weather sensitive component of the demand may be expressed as the sum of functions of the respective meteorological factors.
Linear Regression Techniques The first obvious approximation which can be made is to assume a linear functional relationship. It may be assumed that the basic demand curve is closely approximately by a step function with a step length of one week.
372
Operation and Control in Power Systems The data may be fitted into a regression equation of the form Y
=
a + b l T + b2 W + b) L + b4 P + F(t) + d
where Y is the demand at a fixed time each day; T, W,L, P are the corresponding specific meteorological factors; a, b 1, b2, b3, b4 are constants d is the day of the week correction and F(t) is a polynomial function of the ith week. The regression coefficients b I, b2, b3, b4 and d are to be estimated. In the above equation, more factors may be added or some may be removed, if necessary. In the linear regression technique the data points are fitted into a linear equation of the type. Y = ao + a l X where Y is the dependent and X the independent variable. In the least squares curve fitting method, we minimize the square of the deviation of the dependent variable from its actual or observed value. Let Y I = ao + al XI If n is the number of observations or data points
Dividing throughout by n gives
LY
LX,
I
=ao+a,
n
since Y and
Y
n
X
L' and L' are the mean values ofY and X respectively n n
Y = a o + a, X and
X are the average values. y,
=
Y,  Y
and Then, subtracting Y from the both sides of the basic equation Y,  Y = a o X,  Y =a o +a,X, au a,X, ao  ao + a l XI = a, X. i.e. a coordinate transformation is effected. The individual deviations become i.e.
YI =
Id, I= Iy,  a ,x ,I and
D= Ld~ = ,;\
is the error term.
L(YI alx}
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373
To minimize D we get the condition dD
=O=2L>,(Y, a,x,) or
da,
.
LX, Y,  a,
I
x~
=
0
form which i.e., having found a" from the equation Y = a 0 + a, X we get ao as
Also
For the case where Y is a function of several sets of variables XI' X2, multiple regression coefficients ai' a~, ..... ak can be derived.
........ ,
Let where Y is the most probable value of Y. then if,
Y= Y  Y x
=
X X
we have
Y = a,X,
Let
D
+a 2 X 2 + ..... +a k Y. so that ao is eliminated.
yf
=
(Y
=
I[y(ao+a,X, +a 2 X 2 + ..... +a k x k ))2
Ifwe differentiate this with reference to ai' a2, ak
X k the
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Operation and Control in Power Systems
 L x IY+ a IL x ~ + a 2 L x I X2 +........... + a k L x I Xk = 0 ie
Rewriting as
 Lx 2 y+a l Lxl X2+a2Lx~ +........... +akLx l xk=O
ajLX~ +a 2 Lx jx 2 +a 3Lx 1X3 +................. + akLxlXk = LXIY
alLx l X2 +a2Lx~ +a 3Lx 2 X3 +................. +a kLx 2 Xk = aiL XI Xk +a 2L x 2 Xk +a 3
LX Y 2
L X3X +................. + a kL X~ = LXkY k
or in matrix equation form
IX XI IAI
=
I
XY
I
the matrix equation is to be solved for A. Using the technique indicated above, the coefficients in eqn. (9.1) can be determined. The terms of F(t) can be determined using the method of polynomials of least squares. It was found that a good fit to a year's data could be obtained by fitting polynomials up to the sixth degree.
Least Squares Polynomials If we write 2
~
3
Y=a O +a IX+a 2 X +a3 X for a cubic form, the individual deviations d, can be expressed as
?,
d 1= YI aOaIX I a2X~ a 3 X;
and D the sum of the squared deviation is ~ d I2 = (Y D = L.. I

a0

a IX I

I~I
Minimizing with respect to ao' ai' a2 and a3
a 2 X I2

a, X I3)2
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Introduction to Advanced Topics which can be rewritten as LY, ::: nao +aILX, +a~LX~ +a,LX; LX,Y, =aoLX, +aILX~ +a 2 LX;' +a,LX~
In the matrix form
LX,
LX; LX, LX~ LX; LX; LX,2 LX;' LX~ LX; LX; LX~ LX~ LX~ n
LX~
I
LY,
al a,
LX,Y,
a,
LX;Y, LX3y , ,
a,
from which the values ofao, ai' a 2 and a 3 can be calculated. Extending the equation to sixth order will yield a good fit for F(t) for an year's data.
Nonlinear Regression Methods If the relationship between demand and the specific meteorological factors on which it depends are nonlinear, the linear functions introduced in the regression analysis earlier will give only average effects over the range covered by the data. We cannot assume any specific mathematical model for nonlinear functional relationship in advance, since its nature is not known in advance. We can write S = fl(T) + fiW) + f3 (L) + f4 (P) + F(t) + d Where the constant a has been included in the basic demand curve F(t). As a first approximation, the form of the unspecified functional relationships fl(T), fiW), f 3(L), f4(P) can be taken as straight lines whose slopes have the values of the corresponding linear regression coefficients. First, all the weather sensitive components are completely removed. F(t) + d = x  fl (T)  f2 (W)  f3 (L)  f.j (P) Plotting a graph of the above relations, we can make an estimate of F(t) +d. Grouping these estimates according to the day of the week and noting the difference between their respective means and the grand mean, approximation to the day  of the week adjustments are to be made. Then F(t) is known, giving the basic demand curve. From this initial estimate and the data of the meteorological variables. T. W etc. the linear approximations are successively replaced by curvilinear relationships and the proces~ is carried on until no further improvement is obtained. For a daily load estimation, the basic demand for the following day can be read from the final graph of F(t) and to this can be added the dayofthe week correction. A further weather correction can be made by applying the weight determined from the meteorological response curves using the best available weather forecasts.
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Operation and Control in Power Systems
9.15 Spetral Expansion Method In this method, only past load data is required. It is well  known that the pattern of load demand tends to repeat every 24 hours. The load demand curves may be divided to given ensemble of time series. The problem then becomes one of predicting a nonstationary process given by an ensemble of sample functions. Consider the load curve for any working day, divided into the three 8 hour periods or four 6 hour periods. These parts may be made to have an overlap of, say, 2 hours for continuity in prediction. Then the load on the mth working day at the nth instant of time can be denoted by pmn' For example, we can have six part load curves for the time period of, say, 12 noon to 6 pm for the six working days of a week, this forms an ensemble of six functions. For Sundays a similar ensemble can be constructed. It is possible to represent the load by the relation. Pmn
=
Amn + fj(T m) 8 mn + f2(L m) Cmn
. .... (9.27) + f3 (W m) Dmn +...... . Where fj(T m)' f 2(L m) fiW m) are functions of temperature Tm' illumination Lm, and wind velocity Wm respectively. 8 mn , Cmn and Dmn are coefficients used to given weightage to the weather parameters depending on the time of the day. Amn is the base load. The sample functions Pmn are thus assumed to belong to a linear manifold generated by the weighting functions of their linear combinations. As a preliminary step, it is required that from the load all trend components are eliminated. Finally, the residual component can be expanded into time series using spectral theory of stochastic processes. Equation (9.27) may be rewritten as ..... (9.28) where Pwet) and Pit) represent the trend components of weekly and daily patterns of load demand. Pmn(t) is the residual load component. The component Pw(t) is given by ..... (9.29) where nd is the number of days in a week. Pct
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Introduction to Advanced Topics
where nw is the number of weeks for which data is available. The residual component Pmn(t) can be expressed as a time series. If N is the total number of days for which data is available, the auto correlation function for the residual component is given by I N R(t:r) = Ly,(t)y,(t) N ,=1
..... (9.3 I)
A characteristic function fit) is defined by the integral equation
l' R(t, t)f. (t)dt = b
k
f, (t)
..... {9.32)
where T is the total time period over which the prediction is valid. The residual load component is then expressed as K
Pmn (t)
= L a k bUk (t) + e(t) k=1
The coefficients ak can be determined for the best fit using any standard criterion. The coefficients bk are determined by eqn. (9.32) and e(t) is the error term. The number of terms in the time series, K, can be determined using the minimum, mean  square error criterion given by e 2 (t) = R(t.t)  b k fk (t)
where e{t) is the average error.
9.16 Prediction by Scaling a Standard Load In this method, a particular day is taken as reference and the load curve for the reference day is taken as standard. The load on the day o~ prediction is compared with the standard load curve. Each measured load of the day is divided by the corresponding value of the load on the standard load curve at that instant. Consider the load curve for the previous hours. Let this be divided into four quarter hourly periods. The load for the next one hour can be predicted as exponentially weighed average of the previous four ratios. The ratio at the (n + k)th instant is estimated from the relation r(n+k) =
Ix 4
1 x
3
Lrn_mx
n
..... (9.33)
111=0
where x is a weighting parameter determined empirically, the value of which varies between zero and unity. The predicted load can be obtained by mUltiplying the predicted ratio r(n+kJ by the corresponding value of the load on the standard load curve.
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Operation and Control in Power Systems
9.17 Short  Term Load Forecasting Using Exponential Smoothing From the available data, the seasonal and weekly variations are removed. The remaining load may be represented as a linear combination of known functions of time and noise component. pet)
= AT
F(t) + net)
..... (9.34)
The elements A are to be assumed constant during the period offorecasting even though they change slowly. Using known fitting functions F(t) and current coefficients A(T). forecasted loads can be computed by extrapolating equation pet + t)
= AT(T) F(t + t)
..... (9.35)
Where pet + t) is the forecast with the lead time t. The elements of A(T) can be computed using weighted least squires criterion. For this, the expression
J=
f P [peT  j)  A
 j)
is minimized. The constant
r
j
T (T). F(T
~
has value between zero and unity.
j=O
As an example, the time function may be a Fourier series given by m
pet) = A +
L aj sin ro, t + b, sin ro, t ,=1
If the time period is taken as one week then.
ro·= ,
27t N (7)(24)'
where N is a positive integer. The selection of the actual number of frequencies is dependent on the auto  correlation function and power spectrum for the observed hourly dat,. The coefficients "a,", "b," and Ao are to be revised each hour
9.18 Peak Power Demand Prediction The fluctuating values of peak power demand may be assumed to be controIled by probabilistic laws in time. The monthly peak demand series, for example, contains three additive components\ the trend Pp seasonal Ps and noise Pw Thus the peak demand is given by
PMO (T)
=
PT (t) + Ps (t) + PN (t)
The trend component shows a continuous change in the demand, normaIly an increase. The seasonal component consists of periodic variations, the pattern of which increases in amplitude yearbyyear. Using the stochastic theory of stationary and nonstationary processes, the peak demand can be predicted.
379
Introduction to Advanced Topics 9.19 State Estimation in Load Forecasting
Real time control of power systems requires accurate short term forecasting of load demand. State estimation in forecasting will be quite useful for such a purpose. The model for forecasting demands upon the lead time. Fer a very short lead time, less than 10 minutes or so, the system load and the observations may be represented by
\
~t) =
pet + Z(t)
pet) + net)
pet) + K(t)
=
..... (9.36)
Where pet) is the load at the instant t n is the system noise Z(t) is the observation at the instant t and k(t) is the observation noise. For short term load forecasting, with lead times of about ten minutes to an hour, the load may contain an increment ~P(t). The weather conditions may not play any role, since in such a short time they may not vary so much as to change the consumption pattern. The model could be represented in thts case by ~t) =
pet + ~P(t
Z(t)
+ =
~t)
pet) +
~P(t)
= pet) +
+ n 1(t)
n2 (t)
..... (9.37)
pet) + K(t)
The expected or average values of the noises is zero. E[n(t)] = E[n 1(t)] = E[nlt)] = ErK(t)] = 0 Further
E [net) K(t)]
=
0
..... (9.38)
Rewriting eqn. (9.37) for the forecasted model P
..
.
. .... (9.39)
pet + ~t) = pet) + K(t)[Z(t + ~t)  pet)
where 0 < K(t) < I The value of K that minimizes the variance of forecasting error is K(t +
~t) =
R(t) K(t) +Q(t)
..... (9.40)
Where R(t) and Q(t) are the variance of observation noise and system noise respectively. The forecasted load demand is
.. pet + ~t)
= pet) + K(t + ~t)[Z(t + ~t) 
.
pet)
. .... (9.41 )
for the short term model
.
.
pet + ~t) = [I  K(t + ~t)][P(t)] + K(t + ~t)Z(t + ~t)
..... (9.42)
the procedure is basically not different from the exponential smoothing method, but the correcting gain K can be sequentially evaluated if the noise variances are known in advance.
Operation and Control in Power Systems
380
9.20 Generating Capacity Reliability and Qutage Probabilities The availability of enough generating capacity to meet the load demand is the foremost priority at the planning level that ensures satisfactory operation of the power system. The ap~ication of probability methods enables quantitative prediction of the system reliability. The results obtained by probability techniques will only provide a criterion for evaluation of the situation rather than give a guarantied result.
Rules for combination of event probabilities All events such that occurrence of anyone event is not influenced by the occurrence of any other event, are called independent events. The probabi Iity of occurrence of any two independent events is given by the product of the probabilities of occurrence of each event separately. On the other hand, if two events cannot happen both at the same time, then they are said to be mutually exclusive. The probability of occurrence of mutually exclusive events is given by the sum of the probabilities of the occurrence of each event in the mutually exclusive event group. Binomial distribution may be used to calculate the probabilities for power generation reserve capa;;ily calculations. For any event, if s is the probability of success and f is the probability of failure, then for n trials n
(s + f)n
=
L nC, Sr fnr r~O
The binomial coefficients can be determined from Pascal's triangle
I
I
4
2
I
3
3
6
etc. 4
Application to Power Plant Capacity Calculations Consider a power station with two units each of 60MW. Let the forced outage rate or the probability of any set being out of service be 0.05. Then the availability rate or the probability of the set being in service is obtained as (10.05) = 0.95. For the plant under consideration three states of operation may be listed: 1.
Both the units are in services
2.
only one unit is in service, and
3.
both the units are out of service
381
Introduction to Advanced Topics
The probability for the existence of each of the above states can be computed using the binomial expansion with n = 2. The results are shown in Table 9.2
Table 9.2 Two  unit System Outage Probabilities Probability
Capacity Available MW
State
Capacity Outage MW
I
0
120
(0.95)(0.95) = 0.9029
2
60
60
2(0.95)(0.05) = 0 0950
3
120
0
(0.05){0 05) = 0 0025
"
1.0000
Total
It may be noted that the probabilities are computed from the relation (s + f)~
=
S2 + 2sf +f
In a similar manner. the probability of occurrence of various outages for a 3unit power plant can be computed. Consider a power station with three identical units. each of capacity 110MW. The forced outage rate for each unit is 0.03. the availability rate, therefore, for each Ilnits is 0.97. The probability of occurrence of various states in operation are computed and I. .... ted in Table 9.3
Table 9.3 Three  Unit System Outage Probabilities State
Capacity Outage MW
Capacity Available MW
I
0
120
I
0
330
2
110
220
Probability
(0.95)(0.95) = 0.9029 (0.97)3
=
0.912673
3(0.97)2 (0.03) = 0.084681 3(0.97)2 (0.03)2
0.002619
3
220
110
4
330
0
(0.03)3 = 0.000027
Total
1.000000
=
The probabilities are computed from the relation (S+f)3 = S3 + 3s2f
+ 3sf + f3
Finally, taking the example of a 4unit power plant, each with a rating of 65Mw and forced outage rate of 0.04, the probabilities of occurrence of the five possible states of operation are listed in Table 9.4. (S+t)4 = S4 +4s 3f +6s 2(2+4s(3+[4 outage probabilities for dissimilar units in a plant or system with different forced outage rates can be calculated extending the basic principle further.
Operation and Control in Power Systems
382
Consider a power system with two similar units, each of 60MW capacity and another generating unit and another generating unit of II OMw. The forced outage rate for the 60Mw unit is 0.03, while the same for the II OMW unit is 0.1.
Table 9.4 Four unit System Outage Probabilities State
Capacity Outage MW
Capacity Available MW
Probability
I
0
260
(0.%)4 = 0.84934000
2
65
195
4(0.96)3(0.04)=014155000
3
130
130
6(0.96)2(004)2 = 0.00884730
4
195
65
4(0.96) (0.04)3 = 0.00024576
5
260
0
(0.04)4 = 0.00000250
Total
1.00000000
To begin with, the probabilities are computed separately as before. These are shown in Table 9.S
Table 9.5 Outage probabilities for different type of units Two  60 MW sets Capacity out Mw
One 1I0MW
Probability
Capacity out MW
Probability
0
(0.97)" = 0.9409
0
090
60
2(0.97)(0.03) = 0.0582
110
0.10
120
(0.03j2 = 0.0009
Total
1.0000
Total
1.00
The results of both the calculations are combined now to find the outage probabilities for the combined system (Table 9.6)
Table 9.6 Outage Probabilities for System with Dissimilar Units Capacity out MW
Probability
0
(0.9409) (0.9 = 0.846810
60
(0.0582) (0 9) = 0.523800
110
(0.1) (0.9409) = 0.940900
120
(00009) (0.9) = 0.000081
170
(0.0582) (D. I ) = 0.005~~9
230
(0.0009) (0.1) = 0.000090
Total
1.000000
383
Introduction to Advanced Topics Loss of load probability
Loss ofload probability can be obtained by combining the system load duration curve with the capacity outage probability. It may be pointed that capacity outage may not necessarily result in loss of load. This will be demonstrated in the example that follows. It can be seen from Fig. 9.20 that any capacity outage less than the reserve capacity will not result in loss of load. Installed capacity
.+
1 ~
Reserve capacity
Load duration curve
~:
''I
"0 '
'"
0'
....ll L _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _
~
~
Time (for which load exceeded the value) .....
I
Fig. '9.20 Application of probability to loss of load , Installed capacity Reserve capacity
..... .....
..... .....
ime (for which load exceeded the value)
Fig. 9.21
dified model of capacity on maintenance Installed capacity Capacity on maintenance
1
Reserve capacity
~
6r~==~~
alo
Load duration curve
....l
o Time (for which load exceeded the value) _ Fig. 9.22 Alternative model for capacity on maintenance
384
Operation and Control in Power Systems
If Pk is the probability of the outage Lk and tk is the period of time units for which the outage causes loss of load. The total expected loss of load for the time period is n
E(t) =
L Pk tk
time units
k=l
All outages in excess of the reserve capacity will result in loss of load for different time periods and each outage Pk contributes a system load loss of Pk tk time units. For a daily load duration, curve, the time period is hours. For annual load curves, daily peaks may be used. The effect of capacity on maintenance can be considered in two ways. This is shown in Fig. 9.20 & 9.21. Consider, as an example for the loss of load probability calculation, a system with four generating units, each rated at 65 MW. The forced outage rate for each generator unit is 0.04. the load duration curve is a straight line with a maximum of 190MW and a minimum of 40MW. Since the installed capacity is 260MW, the reserve capacity is 260 J 90=70MW. The outage probability table is shown in Table 9.7
Table 9.7 Cumulative Probabilities for a 4unit System State Capacity MW
Capacity available MW
Individual probability
Cumulative probability
I
0
260
0.84934000
1.00000000
2
65
195
0.14155000
0.15064000
3
130
130
0.00884730
0.00909560
4
195
65
0.00024576
0.00024830
5
260
0
0.00000256
0.00000256
As may be noted from Table 9.7, the cumulative probability decreases as the capacity outage increases. Installed capacity
130
f
MW
Load duration
~
6 '0
195
MW
t2 = 83.3 h
'"
.9 40 ~ o
p...
o Time (for which load exceeded the value) ..100% Fig. 9.22 Loss of load with time
Introduction to Advanced Topics Since the reserve capacity is 70MW, state 2 does not Involve any loss of load. State 3 results in a loss of load for a period of 40% of time and state 4 has loss of load for 83.3% of time. This is shown in Fig. 9.22. The expected load calculations are shown in Table 9.8.
Table 9.8 Expected Load Loss calculations State Capacity MW
Capacity available MW
Individual probability
Time tk in per cent
Cumulative probability
I.
0
260
0.84934000
0

2.
65
195
0.14155000
0

3.
130
130
0.00884730
40
0.353890
65
0.00024576
83.3
0.020466
Total
0.374350
4.
195
If the load duration curve has a time period of about 261 days excluding 52 weekends (Sundays and Saturdays) from 365 days, then the expected load loss is (0.37435)(261) = 0.97705 days 100 the maintenance of the desired level of reliability on the operation of the system is dependent 0 the spinning capacity reserve. In the case of static reserve capacity, the available generation is the entire installation capacity; while for a spinning reserve capacity, it is generally assumed that there is sufficient installed capacity available and only the time delay in bringing in a replacement is considered. Thus, in spinning reserve capacity calculations, outage replacement rate is used in place of forced outage rate for probability evaluation. The outage replacement rate is defined as the product of unit failure rate and delay time for replacement unit. However, this is valid only if the outage replacement rate is much less than unity.
Operation and Control in Power Systems
386
Questions I.
What are FACTS controllers? Explain
2.
What are the various types of FACTS devices? Explain their working principle briefly
3.
Explain the working of a dynamic voltage restorer
4.
What is state estimation. What is the significance of it in power systems operation
5.
Explain the various methods of load prediction
6.
Why load prediction is necessary in power system operation? Explain
7.
Explain the term voltage stability. How can it be over come? Discuss
8.
What do you understand by the term "power quality"? Explain
9.
Discuss the various factors involved in power quality problem
10.
Explain any method for the computation of loss of load probability.
II.
Ho~
12.
Explain "Security" in relation to power system operation.
13.
What is contingency evaluation is connection with power system security.
14.
What are energy control centers? Explain.
reliability studies are important in power system operation? Explain.
Objective Questions I.
Consider a power system with three identical generators. The transmission losses are negligible. One generator (GI) has a speed governor which maintains its speed constant at the rated value, while the other generators (G2 and G3) have governors with a droop of 5%. If the load of the system is increased, then in steady state
[]
(a) generation of G2 & G3 is increased equally while generation of G I is unchanged. (b) generation ofGI alone is increased while generation ofG2 & G3 is unchanged (c) generation of G I, G2 and G3 is increased equally (d) generation ofGI, G2 and G3 is increased in the ratio 0.5:0.25:0.25 2.
Gausss  Seidel iterative method can be used for solving a set of (a) linear differential equations only (b) linear algebraic equations on Iy (c) both linear and nonlinear algebraic equations (d) both linear and nonlinear differential equations
3.
A power station consists oftwo synchronous generators A and B of ratings 250 MVA and 500 MVA with inertia 1.6 p.u. and 1 p.u., respectively on their own base MVA ratings. The equivalent p.u. inertia constant for the system on 100 MVA common base is
(a) 2.6
4.
(b)
0.615
(c)
1.625
(d)
9.0
The incremental cost characteristics of two generators delivering 200MW are as follows
_I = 2.0 + O.OIPI dPI For economic operation, the generations PI and P2 should be: (a) PI = P 2 = 100 Mw (c) PI
= 200MW, P2 = OMw
(b) PI = 80MW, P2 = 120Mw (d) PI = 120MW, P2 = 80Mw
388 5.
Objective Questions A power system has two synchronous generators. The Governer  turbine characteristics corresponding to the generators are P J = 50(50  f), P2 = 100(51  f) Where f denotes the system frequency in Hz, and P J and P2 are respectively, the power outputs (in MW) of turbines 1 and 2. Assuming the generators and transmission network to be lossless, the system frequency for a total load of 400MW is [] (a) 4705Hz
6.
(b) 48.0Hz
(c) 4805Hz
(d) 49.0Hz
The bus impedance matrix of a 4bus power system is given by
Z BUS
jO.3435 jO.2860 jO.2723 jo.2277] jO.2860 j0.3408 jO.2586 jO.2414  jO.2723 jO.2586 jO.2791 jO.2209 [ jO.2277 jO.2414 jO.2209 jO.2791
A branch having an impedance ofjO.2 ohm is connected between bus 1 and the reference. Then the values of Z22,new and Z 23,new of the bus impedance matrix of the modified network are respectively [ ]
7.
(a) jO.5408 ohm and j0.4586 ohm
(b) jO.1260 ohm and jO.0956 ohm
(c) jO.5408 ohm andjO.0956 ohm
(d) jO.1260 ohm andjO.1630 ohm
A power system consists of 300 buses out of which 20 buses are generator buses, 25 buses are the ones with reactive power support and 15 buses are the ones with fixed shunt capacitors. All the other buses are load buses. It is proposed to perform a load flow analysis for the system using Newton  Raphson method. The size of the Newton  Raphson Jacobian matrix is [ ] (a) 553
8.
x
553
(c) 555
x
555
(d) 554
x
554
Ifa generator of250 MVA rating has an inertia constant of6MJIMVA, its inertia constant [ ] on lOOMVA base is (a) 15MJIMVA
9.
(b) 540 x 540
(b) lOo5MJ/MVA
(c) 6MJ/MVA
In load  flow analysis, the load at a bus is represented as (a) a constant current drawn from the bus (b) a constant impedance connected at the bus (c) constant real and reactive powers drawn from the bus (d) a voltage  dependent impedance at the bus
(d) 2.4MJ/MVA [
389
Objective Questions
10. A transmission line has equal voltages at the two ends, maintained constant by two sources. A third source is to be provided to maintain constant voltage (equal to end voltages) at either the midpoint of the line or at 75% of the distance from the sending end. Then the maximum power transfer capabilities of the line in the original case and the other two cases respectively will be, in the following ration. [] (a) 1 : 1 : 1
(b) I : 2 : 1/0.75
(c)I:2:4
(d) I : 4: 16
11. The inertia constant ofa I OOMVA, 50Hz, 4pole generator is 10MJ/MVA. If the mechanical input to the machine is suddenly raised from 50MW to 75MW, the rotor acceleration will be equal to [ ] (a) 400kV
(b) 260kV
(c) 80kV
(d) 40kV
12. Economic scheduling of generation is performed for a period of
(a) Y:z hr
(b) 24 hr
(c) I sec
(d) I week
13. The input  output characteristic is plotted (a) Fuel input Vs power output
(b) Fuel input Vs time
(c) Fuel rate V s power output
(d) Fuel rate Vs energy output
14. Hear rate curve is plotted in (a) kcal/kwhr Vs KW
(b) Kcal/hr Vs KW
(c) kcallKW Vs Kw
(d) Kcal Vs KW
15. Incremental production cost and incremental fuel cost are (a) both the same
(b) IPC > IFC
(c) IPC < IFC
(d) They are not related
16. The incremental transmission loss for a 2plant system is given by
17. Consider the following constraints (i) PaPoP=O
(iii) p2 + Q2 :s; (S rated?
(ii) Qa  Q O Q = 0 (iv) pmm :s; P :s; pmax
The equality constraints are (a) (i) only
(b) (i) (ii) and (iii)
(c) (i), (iii) and (iv) (d) (i) and (ii)
390
Objective Questions
18. Penalty factor is given by
]
OPL) (b) dF ( 1 OPL (a) dF ( 1 +  ) dPj oPi dP j oPI
(c) (
OP)
1+_L
(d)
oPi
(1 oP
OPL
)
I
19. IfR = 0.04 HzJMw by how much will the turbine power decrease, iffrequency rises by O.IHZ [ ] (a) 0.024Mw
(b) 1I0.024MW
(c) 0.0024Mw
(d) 4.17MW
20. What is the load damping factor D for an area operated at 1000Mw, 50Hz (b) 50
(a) 100
(d) 20
(c) 25
21. If the power system has a transfer function
Kp 1 +STp
and given that the load damping
D = 25MW/Hz. What is the value of Kp (a) 0.04HzJMW
(b) OAHzJMW
(c) 0.25HzJMW
(d) 25HzJMW [
22. Area frequency response characteristic is
1
(a) D+R
I (b) R +D
D
(c) R
D (d)  R
23. A speed governing system has D = 20MWlHz. What is the frequency drop for 1% load increase if the area is rated for 2000MW [ ] (a) 1 Hz
(b) 1.5 Hz
(c) 0.5 Hz
(d) 0.2Hz
24. Steady state frequency error in a single control area can be eliminated by (a) proportional control
(b) derivative control
(c) integral control
(d) proportional plus derivative centro I
25. In a single uncontrolled area, area control error is defined by (a) fddt
d (b) M dt
(c) df
(d)  .!:df
R
26. When integral control is applied, if the control gain Kr is greater than critical gain, the response is [ ] // (a) over damped (c) un damped oscillatory
(b) damped oscillatory
. I( d)
unstable operation
391
Objective Questions 
27. Which of the following statements is correct? (a) primary load frequency control is performed by integral control (b) secondary load frequency control is performed by integral control Cc) speed governor action always keeps the speed constant with load changes
Cd) economic load scheduling is always required to be performed before load frequency control 28. Turbine generator dynamics and speed governor actions affect the performance of the
frequency response with load
[
]
(a) seriously (b) only negligible action at t = 0+ (c) may cause trouble if governor is fast in action (d) may cause trouble if the rating of the turbine is very high 29. The time constant of the power system Tp is defined by
2H
(a) Tp = of D
( b) T = 2fo p
HO
(c) Tp
f~
20
=of H
(d) 2HD
30. Which of the following correct?
]
(a) load frequency control has serious affect on voltage profile (b) if a generator is connected to an infinite bus the frequency change with load is negligible (c) turbine power increment for load change does not depend upon the response characteristic of the turbine (d) all the generating stations in any country can be treated as belonging to a single control area 31. Given til = 50Hz, 0 = 20Mw/Hz, area rated power = 2000Mw and H = 5 sec the area time constant Tp is [ ] (a) 5 sec
(b) 10 sec
(c) 15 sec
(d) 20 sec
32. Unit commitment is (a) Economic schedule among different units (b) Planning of generating units for future load increase Cc) Optimal combination of units for operation at anyone time Cd) To choose proper units from thermal, hydro and nuclear plants
392
Objective Questions
33. Which of the following is valid constraint for unit commItment
( c)
I. pI< pmax I
I J I
I
(d) Vmtn < pm3x 2 + Qmax 2
pmlO :os;
I
34. Which of the following is a valid dynamic programming equation (b)
Fn(~Pi)fn(Pn)+Fn_{~PI)
(d)
F,(t,P}
m+N(PN)+
F,_.(~P,)]
35. Dual variables are
[
]
(a) Lagrange multipliers (b) Kuhn  Tucker multipliers (c) both Lagrange and Kuhn  Tucker multipliers (d) the problem variables
36. The dependent variables in optimal load flow studies are
IVI and 0 = 0.02, B22 = 0.04; B12 
(a) P and Q
37. Given
Bll
(c) P and
(b)
B21 =
IVI
] (d) P,Q,
IVI
and 0
0.001 2
tv
load
What are the transmission loss when lOMw is supplied by plant 2 to the load [
(a) 2Mw
(b) 4Mw
(c) O.OOIMW
]
(d) zero
38. 0.04 regulation in speed for a 100Mw rated generator means (i) 2.0Hz drop in frequency (ii) 40Hz drop in frequency
(iii) R = 0.02 HZ/Mw (a) (i) is only correct
(b) (i) and (ii) are both correct
(c) (ii) and (iii) are both correct
(d) (i) and (iii) are both correct
[
]
Objective Questions
393
39. In a power system during load flow studies (i) there are four variables at each bus (ii) the voltage magnitude at one bus is specified and this bus is called voltage controlled bus (iii) at all the load buses P and Q are specified for the loads
(iv) power losses are 'assigned to swing bus. Which of the above is I are correct? (a) '(i) and (ii)
(b) (ii) and (iii)
(c) (i) (iii) and (iv)
(d) (i) (ii) (iii) and (iv)
40. In load flow studies
(i) convergence of iterative method depends upon the diagonal dominance in YBUS matrix (ii) sparsity of Z  bus matrix is well exploited
(iii) for large well conditioned system with nbuses the number of iteration required is approximately n2 • (iv) Z  matrix methods is very sensitive to choice of slack bus. Which of the above is I are true, [ ] (a) (i) and (ii)
(b) (ii) and (iv)
(c) (iii) and (iv)
(d) (i) and (iii)
41. In load flow solution by NR method (i) the rectangular coordinates method is slower compared to polar  coordinates method (ii) the Jacobian in case of rectangular coordinates is not symmetric
(iii) the Jacobian with usual formulation in polar coordinates is symmetric (iv) the NR method converges in 25 iterations which of the following is correct (a) (i) and (iv) are true
(b) (ii) and (iv) are true
(c) (i), (ii) and (iv) are true
(d) (i), (ii) and (iii) are true
42. Economic dispatch problem is related to (i) selection of generating units for loading (ii) planning for generating system expansion (iii) determine profit from energy sales
(iv) ascertain cost of energy production (a) (i) and (iii)
(b) (i) and (iv)
(c) (ii) and (iii) ,
(d) (iii) and (iv)
394
Objective Questions
43. In the emergency mode of operation (a) only system frequency is maintained (b) only maximal load demand is being met (c) both Ca) and (b) are maintained (d) only system frequency and voltage are maintained 44. Electrical stiffness of the transmission line is given by
L\P (a) L\cS
L\P M
(b) 
L\Q (c) L\cS
L\Q
(d)
M
45. Transmission capacity can be increased by
(i) reducing the effective reactance of the line (ii) increasing the voltage levels
(iii) by suing bundled conductors (iv) by increasing the load angle d to some extent (a) (i) and (iii) are the same (b) (i) (ii) and (iii) increase the power transfer (c) (i) and (iv) do not mean the same control (d) bundled conductors infact decrease the transmission capacity 46. The natural load of a transmission line can be expressed by (a)B.X\'
(b)B/X
(c)X/B
] (d)B+X
where B is the line susceptance and X is the line reactance of the line being considered 47. The transmission losses in a line are (a) directly proportional to voltage Y ,(b) inversely proportional to voltage Y (c) directly proportional to y2 (d) inversely proportional toy2 48. Rapid changes in system voltage cause (a) Flicker
(b) Yoltage dips
(c) corona
(d) loss of synchronism
395
Objective Questions 49. Shunt connected controllable var compensators will (i) improve power transfer capability (ii) improve stability (iii) improve power transfer but deteriorate stability
(iv) improve stability but reduced power transfer capacity (a) (iii) is correct
(b) (iv) is correct
(c) (i) and (ii) both are correct
(d) (i) is only correct all other
are wrong.
50. Voltage response of an exciter expressed as (a) Volts / sec (b) volts / ampere (c) field amperes per output amperes (d) change in field voltage to change in output voltage 51. The exciter voltage for a 100MVA set is about (a) IOOV
(b) 440V
(c) IKV
(d) 250V
52. As per IEEE standard, the exciter block is represented by
~
(b)
(a) 1+ STe
I +STe
]
53. The voltage regulator block diagram shows an (i) inherent phase lag
(ii) inherent phase lead
(iii) inherent steady state error
(iv) inherent instability at larger gains
(a) (i) and (ii) are correct
(b) (ii) and (iii) are correct
(c) (ii), (iii) and (iv) are correct
(d) (i) iii) and (iv) are correct
54. Under sterdy state condition, if K is the over all state error .1ess is (a).1e
I
=ss 1+ K
(c) .1e ss is independent ofk
forwar~
path gain. Then, the steady [ ]
I (b).1e ss =K
(d) .:1e ss
I
=1+ K
Objective Questions
396 55. A stabilizing transformer (i) introduces phase lead (ii) introduces phase lag
(iii) increases stability margin (iv) has its secondary drawing negligible current in a voltage regulator (a) (ii) is correct
(b) (i) and (ii) are both correct
(c) (ii) and (iv) are correct
(d) (ii) (iii) and (iv) are correct
56. A power system stabilizer improves (i) phase lag (ii) improves damping (iii) produces torque in phase with speed
(a) (i) and (ii) only are correct
(b) (ii) and (iii) only are correct
(c) (i) and (iii) only are correct
(d) (i) (ii) and (iii) are correct
[
57. Which of the following is false (a) synchronous motors operate with adjustable excitation
(b) synchronous compensators cannot given dynamic reactive power compensation (c) by supplying a synchronous compensator from a tertiary winding, need for separate transformer can be avoided. (d) A synchronous compensator can even improve stability 58. Per unit change in vol.tage magnitUde is equal to ~Q
(a) 
·Sse
S
(b) J£.
(d) l:!.Q + Sse
~Q
where ~Q is the change in reactivepower and Sse is the short circuit capacity. 59. Which of the following are true? (i) the quadrature voltage compensator would control active power flow in the system (ii) an inphase voltage booster would control the reactive power flow in the system (iii) the natural loading limit is the lowest limit for power transfer
(iv) thermal limit is the highest limit for power .transfer (a) (i) and (ii) are only correct
(b) (ii) only is correct
(c) (i) (ii) and (iii) only are correct
(d) all are correct
397
Objective Questions
60. Line compensation (i) increases Ferranti effect (ii) requires under excited operation of generators
(iii) reduces power transfer capabi Iity
(iv) is never used in power system operation (a) (i) and (ii) are false
(b) (ii) and (iii) are false
(c) (i) (iii) and (iv) are false
(d) all are false
61. When sending voltage is equal to receiving end voltage, for maximum power at receiving end . [ ] (a) X = J'jR
(b) R = J'jx
62. For a 400kv line the ratio (a) 1.5
X R
X (c) R = 
R (d) X = 
(c) 10
(d) 16
2
2
is about
(b) 6
63. In Var compensators using thyristors (a) filters are not necessary (b) the dominant harmonic is second (c) filters are needed for 5th and
7th
harmonic
(d) capacitors are switched on the h.v. side only 64. Static var compensators
J
(i) reduce voltage swings at the rolling mills (ii) improve power factor (iii) compensate the unbalanced reactive load of arc furnace loads
(iv) can control the fundamental component of current flowing in an inductor (a) (i) and (ii) are correct
(b) (ii) and (iv) are correct
(c) (i), (ii) (iii) and (iv) are correct
(d) all are correct
65. Series  capacitor compensation (i) produces undamped or slightly damped oscillations (ii) produces oscillations the frequency of which are of the order of 103 to 106 K.Hz. which causes serious problem (iii) may produce electromechanical oscillations of frequency I to 2 Hz
(iv) will not produce any self excitation. Which of the above is true? (a) (i) and (ii)
(b) (ii) and (iv)
(c) (iii) and (iv)
(d) (i) and (iii)
398
Objective Questions
66. The power flow from area 1 to area 2 is giv~n by with usual notation
67. Synchronizing coefficient of a line is given by AP (a) Ts = A8 (c) TsAP = [Ao l  Ao 2] 68. Which of the following is correct
(i) M = _I ~ A8 Hz 21t dt
=21t fAfdt rad
(ii) M
(iii) M = _I f Afdt 21t (a) (i) and (ii) (b) (i) only
(iv) M = 21t fMdt (c) (i) and (iv)
(d) (i) (iii) and (iv)
69. The inertia consists of two groups of machines which do not swing together are MI and M2• The equivalent inertia constant of the system is : .[ ] (b) M,  M2 ifMI > M2
(a) MI + M2
~
M,M 2
(c) M +M ,
(d) "M I M 2
2
70. If AP OJ = K\ and AP 02 = 0, in a two area system, the steady state frequency error is
71. What is the tieI ine power deviation if AP D:! =
Is and AP DI =0 for a two area system
72. For two equal areas, for a load disturbance of AP D in an area, the steady state frequency [ ] error is (a)
AP
_D
2~
~ (b) 2AP
D
2~
(c) AP
D
(d) A~PD 2
399
Objective Questions
73. If two equal areas with tie line power interchange of P12 experience a load change f1P D in one of the areas, the tie line power deviation is [ ] L\PD (a) 2
(b)
• n. L\PD 1'1
2
(c)
n. L\PD
1'_1
")
....
(d) L\P o
74. Which of the following is true (i) A synchrono\ls generator at sub synchronous frequencies acts as an induction generator (ii) At sub synchronous frequencies the rotor resistance viewed from the armature ter.minals is negative. (iii) At sub synchronous frequencies resonance may occur between electrical and mechanical system (iv) Series compensation causes sub synchronous resonance at very low frequencies (a) (i) is true
(b) (i) and (ii) are true
(c) (i), (ii) and (iii) are true
(d) (i), (ii), (iii) and (iv) are true
Answers to Objective Questions 51.
30.
(b) (b) (b) (a) (b)
31.
(d)
56.
32.
57.
34.
(e) (e) (d)
35.
26.
58.
(d) (e) (d) (a) (d) (d) (b) (a)
59.
(d).
(c)
60.
(d)
36.
(b)
61.
37.
(d)
62.
72.
(a) (d) (e) (d) {d) (a) (a) (a) (e) (a) (a) (a)
73.
(a)
74.
(d)
12.
(b) (b) (b) (d) (b) (b) (b) (a) (e) (a) (a) (a)
13.
(a)
38.
(~)
63.
14.
(a)
39.
64.
15.
(b)
40.
16.
(d) (d) (d) (d) Cd) (a) (a) (a) (e) (e)
41. 42.
(d) (d) (d) (d)
43.
(c)
68.
44.
(a) (d) (b) (d) (a) (e) Ca)
69.
1. 2. 3. 4. 5.
6. 7. 8.
9. 10. 11.
17. 18. 19. 20. 21. 22. 23. 24. 25.
27. 28. "29.
33.
45. 46. 47. 48. 49. 50.
52. 53. 54. 55.
65. 66. 67.
70. 71.
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2.
Rao, S.S., Optimisation: Theory and Application, Wiley Eastern Ltd., New Delhi. 1978.
3.
Kirk, Donald E., Optimal Control Theory: An Introduction, Prentice Hall Inc., Englewood Cliffs, New Jersey, 1970.
4.
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5.
Kirchmayer, L.K., Economic Control of Interconnected Systems, John Wiley and Sons, New York, 1959.
6.
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7.
Taylor, Openshaw (Ed.), Power System Plant, George Newnes Ltd .. London. 1955.
8.
Stagg Glenn and A.H. EIAbiad, Computer Methods in Power System Analysis McGrawHill B09k Co. Ltd., New York, 1968.
9.
Blindleistung, VDE Verlag, GmBH, Berlin 1963.
w.,
10.
Handschin E. (Ed.), Real Time Control of Electric Power Systems, Elsevier Publishing Co., Amsterdam 1972.
11.
P.M. Anderson, Power System Control and Stability, A.A. Fauad, The Iowa State University Press.
12.
Paul.e. Krause, Analysis ofEleotrical Machines, McGraw Hill Book co.
13.
V. Venikov, Transient Processes in Electrical Power System, Mir Publishers, Moscow
14.
Hadi Saadat, Power System Analysis. McGraw Hill International Edition .
15.
PSR Murty, Power System Operation and Control, Tata  McGraw HiU 1984.
.
402
References
Periodicals and Conference Proceedings : 1.
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2.
Funk, G. "Die Spannungsabhangigkeit Von Drehstrom lasten" Elektrizitats wirtschafi,68,1969.
3.
Murty, P.S.R., "Load modelling for power flow solution," Journal ofIE(India), 58, Pt. EL3, Dec., 1977.
4.
IEEE Committee Report, "Dynam ic Models for steam and hydro turbines in Power system studies", IEEE Trans. PAS May 1969.
5.
Dsewenski, A.l., "Uber den Einfluss der spannungsenderung auf die wirkund Blindleistung der Verbraucher", Energietechnik, 12 Nov., 1953.
6.
Anschutz, H., "Frequenzund Spannungsschwankungen in Drehstromnetz und ihre Auswirkungen," ETZ13, Dec. 1949.
7.
Woodwards J.L. "Hydraul ie turbine transfer function for use in governing studies", IEEE Proc, lIS No.3, Mar. 1968 (Correspondence).
8.
Ramey, R.
9.
Hale, H.W. and J.W. Ward, "Digital computer solution of power flow problems", AlEE Trans. PAS, 75, June 1956.
10.
Glimn A.P. and G.W. Stagg, "Automatic calculation ofload flows", AlEE Trans. (PAS), 76, June 1956. '
11.
Br.own H.E. et ai., "Power flow solution by impedance matrix iterative method", IEEE Trans (PAS), 82, Apr. 1963.
12.
H.E. Brown, Zmatrix algorithms in load flow programs, IEEE Trans (PAS), 87 March 1968.
13.
Hart, C.E. and W.F. Tinney, "Power flow solution by Newton's method", IEEE Trans (PAS), 86, 1967.
14.
Stott. B. "Decoupled Newtonload flow", Trans (PAS), 91, Sep/Oct. 1972.
15.
Peterson N .M. W.F. Tinney, and D. W: Bree, "Iterative linear ac power flow solution for fast approximate outage studies", IEEE Trans. (PAS), 91, Sep./Oet. 1972.
16.
Stott, Band O. Alsae, "Fast decoupled load flow", IEEE Trans. (PAS), 93, May/ June 1974.
References
403
17.
Tinney W.F. and J. W. Walker, "Direct solution ot spares network equations by optimally ordered triangular factorisation", Proc. IEEE, 55, Nov. 1967.
18.
Stott, B, "Load flow calculation methods", Proc. IEEE, 62, July 1974.
19.
Noakes, F. and A. Arismonandar, "A Bibliography on optimum operation of power systems 19191959", IEEE Trans (PAS), 82,1963.
20.
Baker, A.D. and S.O. Navarro "The. use of cubic cost curVes in the economic loading of electric power systems", IEEE Trans. (PAS), 84,1965.
21.
Smith, H.M. and S. Y. Tong, "Minimizing' power transmission losses by reactive volt amperes control", IEEE Trans (PAS), 82,1963.
22.
Bauman, R."Power flow solution with optimal reactive flow"., Archiv fur Elektrotechn ik, 48,1963.
23.
Dopazo, J.F. O.A. Klitin, G. W. Stagg andM. Watson, "An optimization technique for real and reactive power allocation", Proc [EE; 45, 1967.
24.
Carpentier, J. "Contribution a I'etude du dispatching economique'Bulletin de la Societe fransaise des Electriciens, 3 Aug. 1962.
25.
Carpentier, J. and J. Siroux, "L'optimisation de la production a L'Electricite de France", Bulletin de la Societe Frallcaise des Electriciens, 1963.
26.
Pescon, J.,D.S.Piercy,W.F. Tinney an~ O:J. Tveit, "Optimum control of reactive power flow," IEEE Trans. (PAS), 1968.
27.
Shen, C.M. and M.A. Laughton, "Determination of optimum power system, operating conditions under constraints", Proc. IEEE, 116, Feb. 1969.
28.
Dommel, H. W. and W.F.Tinney, "Optimal power flow solutions", IEEE Trans (PAS) 87, 1968.
29.
EI Abiad, A.H. and FJ. Jaimes, "A method fOT optimum scheduling of power and voltage magnitude, IEEE Trans (PAS) 88, Apr. 1969.
30.
Fletcher, R. and MJ.D. Powell, "A rapidly convergent descent method for minimization," Computer journal, June 1963.
31.
Bird, J.1.,' 'Load dispatching on the CEGB System," Paper presented at Symposium on Optimal power system operation, Univ. of Manchester, Sep. 1969.
32.
Lowery, P.G. "Generating unit commitment by dynamic programming," IEEE Trails (PAS) 85,1966.
33.
Udo, M., "Optimisation of startup and shutdown of thermal units", Proe. Third PSCC 1969 Rome paper 0.S.7.
404
References
34.
Kerr, R.H., J.L. Scheidt, A.J. Fontana and J.K. Wiley, "Unit Commitment" IEEE Trans (PAS), 85, 1966.
35.
Hara, K. Kimura M. and N. Honda, "A method for planning economic unit commitment and maintenance of thermal power systems", IEEE Trans. (PAS) 85,1966.
36.
Nagrath, I.J. and S.K. Agarwal,"Optimum scheduling of hydrothermal systems", Proe. lEE, 1) 9, Feb. 1972.
37.
Nagrath, I.J., G. Dayal and D.P. Kothari, "Optimum scheduling ofhydrothennal system taking transmission loss into account", Jour. IE (India), 52. Oct. 1971.
38.,
Drake, J.H., L.K. Kirchmayer, R.B. Mayall, and H. Wood, "Optimum operation ofa hydrothermal system," Trans. AlEE (PAS), 82, Aug. 1962.
39.
Bonaert, A.P. AEIAbiad and Koivo "Optimum scheduling of hydrothermal power systems", Proc. Seventh PICA Con/., Boston, 1971.
40.
Murty, P.S.R., "Effect of system loads on optimal load flow solutions," Jr. (India), 58,.Pt. EL3, Dec. 1977.
41.
Hochstetter Werner, "Properties of static compensators for power supply systems", Siemens Review, Aug. 1977.
42.
Achenbach, H., W. Hanke and W. Hochstetter, "Controllable static reactive power compensators in electrical supply systems", Second IEAC Symposium on Control in Power Electronics and ElectricaJ Drive, Dusseldorf, Oct~ 1977.
43.
Tyll, Heinz, '''Use of static compensators in transmission system", Symp. On Reactive Power Compensation, Rio de Janeiro, Brazil, May 1979.
44.
Fouad, A.A., and K. T. Khu, "Subsynchronous resonance zones in the IEEE bench mark power system, IEEE Trans. (PAS) 97, May/June 1978.
45.
Kilgore, L.A., L.e. Elioh and E.R. Taylor, "The prediction and control of self excited oscillations due to series capacitors in power systems", IEEE Trans. (PAS), 90, May/June 1971.
46.
Hergert, J.e. and e. W. Park, "Parameter identification and verification of low order boiler models", IEEE Trans. on PAS. PAS75, No.4, July/Aug. 1976.
47.
Cohn, Nathan "Power flow controlBasic concepts for interconnected systems", Mid West Power Conference, USA, 1950.
48.
Elgerd, 0.1. "Trends and philosophies of power systems control", Alterton Conference on Circuits and Systems, Oct. 1973.
[£
References
405
49.
Cohn, Nathan, "Fundamentals for control of generati<;m and power flow on interconnected systems", Invited Talk, North American Power Systems Interconnection Committee Meeting, Las Vegaes, Navada, USA. Sep. 1970.
50.
mgerd, 0.1. and C.E. Fosha, "Optimum megawattfrequency control ofmultiar~ electric energy system", IEEE Trans (PAS), 89, Apr. 1970.
51.
Fosha, C.E. and 0.1. Elgerd, "The megawattfrequency control problem: A new approach via optimal control theory", IEEE Trans (PAS) 89. Apr. 1970.
52.
Murty, P.S.R. and P. V. Ramana Rao, "Optimal load frequency control and system modelling". Jour IE (India), Pt. EL4, 57, Feb. 1977.
53.
Ramana Rao P. V. and P.S.R Murty, "Identification an\t verification of reduced order steam reheat turbine model", Jour. (IE), Pt. ME6, May 1981.
54.
Ramana Rao, P. V. and P.S.R Murty, "Modelling, Identification and optimal control of hydro governor system", Jour. IE (India), 60, Pt. EL2, Oct. 1979.
55.
Venkateswarlu K. and A.K. Mahalanabis, "Optimum load frequency control with . restrictions on the rate of power generation, Jour I.E. (India) Pt. EL4, 58, Feb. 1978.
56.
Venkateswarlu K. and A.K. Mahalanabis. "Load frequency control using output feedback", Jour. IE (India), Pt EL4 58, Feb. 1978.
57.
Nanda, J. M.L. Kothari, and P.S. Satsanghi, "Automatic. Generation control reheat thermal system considering generation rate constraint and governor dead band'., Journal IE (India), Pt EL6, 63. June 1983.
58.
Liacco Thomas E.Dy. et aI., "Automation of CEI system lor security", PICA Boston, USA, 1971.
59.
Bhaskaran K., "Technology trends in power plant operation", Second Symp. on Power plant dynamics and control BHEL (R&D), Hyderabad, Feb. 1979.
60.
Liacco, Thomas E. Dy "Real time computer control of power systems", Proc. A IEEE, 162, July 1974.
61.
Larson, R.E. et aI., "State estimation in power systems", IEEE Trans. (PAS) 89,1970.
62.
Kuppurajulu, A., "Digital Computers for on line control", NPSC, Electrical Engg. Dept., Osman ia University, Dec. 1981.
63.
Choudhari, S.H. "Turbine instrumentation and automation", NPSC, Elec. Engg., Osmania University, Dec. 1981.
406
References 64.
Gupta, P.G., "Stochastic approach to peak power demand forecasting in Electric Utility systems", IEEE Trans. on PAS PAS90 No 2, Mar./Apr. 1971.
65.
Christiannse, W.R., "Shortterm load forecasting using exponential smoothing", IEEE Trans on PAS PAS90 No.2, Mar'! Apr. 1971.
66.
Mathewan, P.D. and H. Nicholsan, "Techniques for load prediction in the electricity supply industry", Proc. lEE (London), tiS (power) Oct. 1968.
67.
Farmer, D.E. and MJ. Potton, "Development of online prediction techniques with results from trials in the south West Region of the CEG", Proc lEE Londo!"!, 115, Oct, 1968.
68.
Toyoda. J, MoShing Chev and Yukiyoshi Inoue, "An application of state to estimation to shortterm load forecasting partI "Forecasting modelling PartII, Implementation, IEEE Trans. On PAS, PAS  89, No 7, Sept/Oct 1970.
69.
Larson, Robert E, Willam, F, Tinney and John Peschan, "State estimation in power systems partI theory and feas.ibility", IEEE Trans. on PAS, PAS. March 1970.
70.
Larson, Robert E., W.F. Tinney, L.P. Hajdu and D.S. Piercy "State estimation in power systems PartII: Implementation and applications", IEEE Trans on PAS 80, March 1970.
Lecture Notes and Others: I.
Real time control of power systems for security, Summer school notes, sponsored by ISTE, College ofEngg. Guindy, Madras, May/June 1976.
2.
General Electric's static var control, General Electric Brochure, Switchgear Business Dept. USA.
3.
Advances in power systems, Winter school notes, sponsored by ISTE, Regional Engg. College, Warangal, 196777.
4.
P.Y. Ramana Rao, Parameter estimation and model reduction of power plant components using identification techniques for power system control, Ph.D. thesis, Kakatia Un iv., Warangal, May 1981.
5.
P.S.R. Murty., Optimal load flow studies in interconnected networks, (German), Doctoral thesis, Technical University, Berlin.
In.dex E
A Active Power Scheduling 103
Effect of Bias Factor ,267
Analysis of Single Area System 205
ElectricalStiffness 7
Arc furnace 293, 332,333
Emergency Mode 6
Area Load Dispatch Centre 367
Energy Control Cente~ 7
B
Energy Scheduling 121 Equal Incremental Cost Metho
B. coefficients 98,99
Evaluation oft for Computation 107
Basics 200
Excitation Control 295,296
Brushless excitation 300
Exciter 297,298
Brushless Excitation Scheme 301
Exciter BUjId Up 298
Bus Impedance Formation 29
Exciter Ceiling Voltage 298
c Capacrto~
Exciter Response 298 Exponential Smoothing 378,379
7,315,321
F
Commitment schedules 171, 173 Compensation 289,292,320,321
FACTS Controlte~ 7,346
Complete TieLine Bias Control 250,251,266
Fast Decoupled Methods 27
Constraints 94,95
Flat Frequency Control 201, 241, 244, 251
Constraints for Plant 173
Flat TieLine and Flat Frequency Control 244
Cost Function 171,173, 180
Flicker 352, 355
D Decentralised Control 284 Decoupled Methods 27 Dynamic Programming 175, 177, 179, 180 Dynamic Response 199,208,237,257,287,306
Fuel ordering 5
G Gauses  Seidel Iterative Method 13 Generating Capacity Reliability and Outage Probabi 380
408
Index
Generating Costs at Thermal Plants 93
LoadPrediction 7,370,371
Governor 182, 183
Loadability 351,352
Governor Controlled Valves 183
Loading Capability 320
Gradient Method for Optimal Load Flow 156
Loss of Load Probability 383,384
H
M
Heat Rate Characteristic 88
Matrix Riccati Equation 224
Hydro Speed Governing System 232,235'
Merit Order Method 97
Hydro Thermal Scheduling 5, 120, 125
Modal for a Steam Vessel 196
I IEEE Type 1 Excitation System 310
ModelReduction 284,285 Modem Control Theory 224,239 MUltiple Regression Methods 371
Impedane of Reactive Power 289
N
Incremental Production Cost Characteristics 89 Incremental Water Rate Characteristics 91
NetworkAutomation 368
Induction Regulators Static VarCompensators 7
Newton Raphson method 16, 71
Input Output Curves 87
NcnlinearProgramming 157
Interconnected Operation 241,242, 266
Nonlinear Regression Methods 375
K
Normal mode 6
o
Kleinman's method 227
L
Optimal Control 222,223 Optimal Load Frequency Control 225
Lagrange Function 156, 158, 159
Optimal Ordering 25,26
Lagrange Multipliers 97
Output Feedback 237
Least Squares Estimation 359
Overvoltages 334
Least Squares Polynomials 374
p
Level Decomposition in Power Systems 367 Linear Estimation Theory 359
Pattern Recognition Methods 363
Linear Regression Techniques 371
Penalty Factor 106
Load Compensation 321
Penstock Turbine Model 193, 195
Load flow analysis 8, 73
PID Controllers 212
Load Flow Solution Using Z Bus 29
Plant Ordering 5
Load Frequency Control 183,200,201,202,208,212
Plant Scheduling Methods 96 Power Acceptability or Voltage Tolerance 356
409
Index Power Quality 7,352,353
Speed Changer 183,184,186,206
Power Quality Index 353
Speed Control Mechanism 183, 186, 188
Power system 182, 188,201, 208, 230
Speed Governing Mechanism 183, 192
Power System Control Centres 365
Speed Governor 182, 183
Power System Security 360,386
Spetral Expansion Method 376
Power System Stabilizer 313
~tability Compensation 307
Power Transfer 292, 295, 296, 321
Stabilizing Transformer 307
Preventive mode 6
State Estimation 7,357, 358,360
Priority  List Method 174
State Estimation in Load Forecasting 379
Pumped Storage Plant 120
State Variable 269,274,275
PVbus 14,22
State Variable Model 269, 274, 275
R Rate of Power Generation 236 Reactive Power Control for Loss Minimization 155 Reactors 7,315,321 Real power balance 200, 202 Rectangular coordinates 16, 17, 20, 22 Rectifier loads 353,355 Regional Load Dispatch Centre 368 Reheat Type Steam Turbine Model 198
Static Compensators 328,343 Static Excitation System 300 Static Load Model 32, 33 Static Var Compensator (SVC) 349 Steady State Security Assessment 361, 362 Steady State Security Assessor 361,362 Steady State Speed Regulation 184, 185 Steam Turbine Model 197, 198 Subsynchronous Resonance 337 Supplementary Control 242,251,269 Synchronizing Coefficient 7
Restorative mode 6
s
Synchronous Compensators 292,314 System Voltage and Reactive Power 293
Security Evaluation 364,365
T
Security Function 363,364 Series  Series Controllers 347
Tap Changers 7
Series  Shunt Controllers 348
Tapchanging transformers 3 16
Series Controllers 346, 347
TapStaggering Method 317
Short Circuit Capacity 318, 319, 340
The Incremental Fuel Cost Characteristic 88
Short Term Hydro Thermal Scheduling 125
The Incremental Heat Rate Characteristic 88
Shunt Controller 347
The Linear Regulator Problem 222
Single Area System
The Optimal Control Problem 222
203,205,225,254,259,269,275
The Polar Coordinates Method 19
Single Control Area 199,203,221
Three Area System 269,276
Sparsity of Network 22
Thyristor Controlled Resistor (TCBR) 349
410
Index
u
Thyristor Controlled Series Capacitor (TCSC)
349 Thyristor Controlled Series Reactor (TCSR) 349 Thyristor Controlled Switched Series ReaCl\Jr (TSSR) 349
Unit Commitment 171,171,177, Unit Commitment by Dynamic Programming
177
v
Thyristor Switched Capacitor (TSC) 349 TieLine Bias Control 247,250 Transfer Function 186, 188, 190, 192 Treatment of Generator Buses 19 Triangular Decompostion 23 Two Area System 266, 288 Type 1 S system 309 Type 1 System 309 Type 2 System 310 Type 3 System 310 Type 4 System 310
Voltage Dips 315,335 Voltage instability 350,352 Voltage Regulation 296,300,'306, 318 Voltage Regulator 297,302,305 Voltage Regulators 294,302, 309 Voltage Sags 353,355 Voltage Stability 7,350
w Weather Weighting Method 371