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Economic Operation of Power System
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317 views20 pagesModule 4 PSA-2 18EE71
Economic Operation of Power System
Notes for Power system analysis 2 VTU subject 18EE71
Uploaded by
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© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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/ 20
CHAPTER
Economic Operation
of Power Systems
————
15.1 INTRODUCTION
The sizes of electric power systems are increasing rapidly to meet the energy requirements. A
number of power plants are connected in parallel to supply the system load by interconnection
of power stations. With the development of integrated power systems (that is, grid systems) it
becomes necessary to operate the plant units most economically. The economic scheduling of
generators aims to guarantee at all times the optimum combination of generators connected to
the system to supply the load demand, The economic dispatch problem involves two separate
steps namely the unit commitment and the on-line economic dispatch. The unit commitment is
the selection of units that will supply the anticipated load of the system over a required period
of time at minimum cost as well as provide a specified margin of the operating reserve, known
as the spinning reserve. The function of the on-line economic dispatch is to distribute the load
among the generating units actually paralleled with the system in such a manner as to minimize
the total cost of supplying the minute to minute requirements of the system.
The main factor controlling the most desirable load allocation between various generating
units is the total running cost. The operating cost of a thermal plant is mainly the cost of the
fuel. Fuel supplies for thermal plants can be coal. i The other
it |, natural gas, oil, or nuclear fuel. The
costs such as costs of labour, supplies, maintenance, etc., being difficult to be determined a4
approximate, are assumed to vary as a fixed percentage of the fuel cost. Therefore, these costs
ae included in the fuel cost. Thus, the operating cost of a thermal plant, which is mainly
fuel cost, is given as a function of i i ion ji 2 i n
it sicnt goucaton, n of generation. This function is defined as a nonlinear funclio!
For optimal operatior , a
For optimal operation of thermal system, our problem is to find the generation of dilly
units so that the total fuel cost is a minimum subj isfyi i sical
there are two types of constants, namely eget Satisfying certain constraints. Fe
ly, equality constraints and inequality constrain's-
—
384 Cheetos 18 Economie Operation of Power Systems 385
15.2. INCREMENTAL FUEL COST
‘The input-output curves of generating units of thermal plant are important. 10 describe the
efficiency of the plant. A typical input-output curve of a unit is shown in Fig. 15.1. It is an
experimental curve plotted with abscissa as output power P; in MW and ordinate as fuel (heat)
input in joules per hour (Ih) of the ith unit. The ordinates of the curves may be converted to
fuel cost in C) Rvh by multiplying the fuel input by the cost of the fuel in rupees per joule.
This curve is shown in Fig. 15.2 where Pimin is the minimum loading limit below which it is
not economical to operate the unit and Prax is the maximum active power output of ith unit.
‘The majority of generating units have a nonlinear generation cost function C;. We shall assume
that the variation of fuel cost of each generator (Ci) with the active power output (Pi) is given
by a quadratic polynomial, We can write
C= y+ Bi Pity PF -(15.2.1)
fuel cost of generator i
P= power output of generator i
| Os Bi, and y are constants, cy includes salary and wages, interest and depreciation and is
independent of generation. Bj represents basically the fuel cost and 7; is a measure of losses in
the system. Usually Bj; dominates,
Fuel input in Jih
0 Fein Fax
Pin Fnax
Power output P, in MW Power output , in MW
Fig. 15.1. Typical input-output curve of a Fig. 15.2. Fuel-cost curve.
generating unit.
The slope of the cost curve at a point M is given by
AG
tn = 3
where A Cis the increase in fuel cost corresponding to an increase of power output A Pi. The
incremental fuel cost for a generator for any given electrical power output is defined as the
limiting value of the ratio of the increase in cost of fuel in Rs/h to the corresponding increase
electrical power output in MW when the increase in power output tends to zero. 386 Electrical Power Systems
Incremental fuel cost for the ith generator is given by
ar -= (152.2)
GOr= MM ap: armor
forall exceptj=1
The incremental cost is equal to the slope of
approximated as a quadratic polynomial (with posit
=i d
GO ap aP;
(IO); = Bi + 27; Pi RS/MWh ++(15.23)
Equation (15.2.3) is of the form y= mx +c. ,
Thus the plot of (JC); versus Pris a straight line. In other words, the incremental cost curves
are linear (with positive coefficients). A plot of incremental cost versus power output is called
incremental cost curve. It is shown in Fig. 15:3.
the fuel cost curve. If the cost curve is
ve coefficients) as in Eq. (15.2.1), we have
(i+ BiPi+ HPP)
RsiMWh
Incremental fuel cost
2 Pima Pirax P(MW)
Fig. 15.3. Incremental cost curve for generator i, (a) Actual incremental cost curve; (b) Approximated
(linear) incremental cost curve.
For better accuracy when the cost is represented by a nth degree polynomial, incremental
fuel cost may be represented by short line segments from point to point. Alternatively, in the
inverse form output power P; may be expressed in terms of polynomials of (IC); as given by
P= Gi0 + a1 (IC) + 2 (IO) +... (5.24)
15.3 ECONOMIC DISPATCH NEGLECTING TRANSMISSION LOSSES
Consider n generators in the same plant or close enough electrically so that the line losses ma¥
be neglected. Let Ci, C2, .... Cn be the operating costs of individual units for the corresponding
power outputs Pi, Po, ..., Pr respectively. If C is the total operating cost of the entire syste™
and Pp is the total power received by the plant bus and transferred to the load, then
C=C 42+. G=L CG (15.3)
=1 . —aaeoX—
3 387
Economic Operation of Power system:
Chapter 15
Bu (15.3.2)
Pr=Pit+ Prt... + Pa=
Our problem can be stated as follows :
.
(15.3.1)
Minimize C=) Ci
ist
subject to the constraint
Pr- >, Pi=0
im
Equation (15.3.2) shows that
instant must be met by the total generation. It is to be noted
constraint. -
This is a constrained minimization problem. We shall use the classical Lagrangian multiplier
technique to solve it. For this purpose first we form the augmented or Lagrange cost function
C’ defined by
cActy (15.3.3)
where fis the equality constraint equation given by
w+(15.3.2)
if transmission losses are neglected, the total demand Pa at any
that Eq, (15.3.2) is the equality
PR=y, Pi 8 f(Piy Po, oy Pr) =0 (15.3.4)
ist
and 2 is the Lagrange multiplier. Combination of Eqs. (15.3.3) and (15.3.4) gives
CA cHalPe-D Pi (15.3.5)
int
Equation (15.3.5) can be solved for minimum by determining the partial derivative of the
function C” with respect to variable P; (plant generation) and equating it equal to zero.
5
ac 80 4 m0 is
opi Pi ap; PDP FO I=L Qn
or $C 22 440-1) =0
aC
B +-(15.3.7)
Since C;, is a function of P; only, the partial derivatives become full derivatives, that is,
a> dP, ++(15.3.8) 388 Electrical Power Systoms
‘Therefore, the condition for optimum operation is
aC _ dQ _ Gn +=(15.3.9)
dP, dPy Py
since 4 ig the inceemental cost generation (IC); for generator (*) Eq. (15.3.9) shows thy
oP, cane .
the criterion for most economical division of load between units iin Plant Corban unis
close enough electrically so that the line losses may be neglected) is tt its Must
operate at the same incremental fuel cost. This is known as the princiP’ of equa criterion
or the equal incremental cost-loading principle for economic operalion. z should be noted tha
nn equations of Eq, (15.3.9) and the constraint equation (15.3.4) are sufficient for determining
(n+ 1) unknowns P}, P2, ..., Pa and 2.
Example 15.1 The fuel costs of a two-unit plant are given by
C1=100+2P)+0.005 3; Cy=200+2P2+0.01 PH
where P; and P2 are in MW. The plant supplies a load of 450 MW. Find economic load
scheduling of the two units and the incremental fuel cost. Neglect losses.
Solution
The incremental fuel cost of the first generator is given by
oy = SE =2 420.005 Ps =2+0.01 Py Rs/MWh
1
‘The incremental fuel cost of the second generator
(Ca=22=2+002P2
aP,
For optimum load division the two incremental costs should be equal, that is,
Oi = (IC
240.01 P1=2+0.02 Pp
P\=2P)
Since the total load is shared by the two generators
P\+P2= 450
Combination of Eqs. (E15.1.1) and (E15.1.2) gives
2P2+P2=450, P2= 150 MW and P;=300 MW
The incremental fuel cost of generator 1 corre:
(E15.1)
«(E15.1.2)
sponding to load P; = is given b
(IC) =2+0.01 Py =2 +0.01x300=5 ea icone
The incremental fuel cost of generator 2 corres
sponding to load P= 1 js given by
2+0.02x150=5 RuMWn 2= 150 MW is give
Hence the it .
© ineremental fuel cost of the plant for imost economic operation is Rs 5 pet MWh
n is Rs
(UC). =2+0.02 P2 = 389
Chaptor 15 Economic Operation of Power Systems
i f two units
Example 15.2 The incremental fuel costs in rupees per MWh for a plant consisting ©
are given by
acy
dPy =0.16 Pi +30
aCe _
a2 = 0.20 P2 +25
‘Assume that both units are operating all the time throughout the year. ‘The maximum and
minimum loads on each unit are 200 MW and 50 MW respectively. If the total load varies
between 100 MW and 400 MW find the load division between two units as the system load
varies over the full range.
Solution
For minimum load P; = 50 MW on unit 1
Ct 0.16 x50-+30=38 RIMWh
dP;
For minimum load P2=50 MW on unit 2
C2 _ 0.20 x50+25=35 R/MWh
dP
Thus, when the load is 100 MW, it is shared equally by the two units but unit 1 operates
at a higher incremental fuel cost than unit 2. The incremental fuel cost of unit 1 is Rs 38 per
MWh and that of unit 2 is Rs 35 per MWh. With the increase of the plant output, the more
load should be added to unit 2 till JC of unit 2 also becomes Rs 38 per MWh. Until that point
is reached, the /C of the plant is determined by unit 2 alone. When the load on the plant is 100
MW, each unit operates at its minimum load of 50 MW with plant JC = 35 Rs/MWh.
When P78
0.20 Po + 25 = 38
or P2=65 MW
Since the load on unit 1 at JC=38 is 50 MW, the total load delivered at equal incremental
costs of Rs 38/MWh is (50+65)=115 MW. From this point onwards, increase the load on
each unit so that the units operate at the same incremental costs and these operating conditions
are found by assuming successively higher values of IC and calculating P1,.P2 and total load.
This process is continued till the load on unit 1 reaches its upper limit of 200 MW. Then the
load on unit 2 is increased keeping the load on unit 1 equal to 200 MW corresponding to its
upper limit of 200 MW. At this stage the two units will not operate at the same IC, and unit
1 will determine the plant JC. The load on unit 2 is increased till it also reaches its upper limit
of 200 MW so that the total load on both the units together becomes 400 MW. The results are
shown in Table 15.1. 390 Electrical Power Systems
Division for Example 15.2
‘Table 15.1. Load
Pa Plant output (Pi+ Pi)
MW
ay PI
35 50 50 100
38 50 65 115
40 62.5 5 1375
45 93.75 100 193.75
50 125 125 250
55 156.25 150 306.25
60 187.5 175 362.5
62 200 185 385
63 200 190 390
64 200 195 395
65 200 200 400
- =S—=SsréseOOSOe_ONN
Example 15.3 Determine the saving in fuel cost in Rs/h for the economic distribution of toa!
Iead cf 115 MW between the two units of plant described in Example 15.2 compared with eal
distribution of the same total load.
Solution
From Table 15.1 it is found that for equal incremental costs, unit 1 should supply 50 MW «od
anit 2 should supply 65 MW corresponding to a total of 115 MW. For equal sharing each uit
will supply 57.5 MW.
‘The increase in cost of unit 1 is given by
515
15 5s
AG =f acre) (0.16 P) +30) dP; = jos sor,
50
016
2
(57.5 ~ 50°) + 30 (57.5 - 50) = 64.5 + 225 = 289.5 Rs/h
Change in cost of unit 2 is given by
15 15 :
AG f dQ,= He
a= J, dCs fr (0.20 P, +25) ap,
0.20 A +25P2|
65
“i 2 _ gat
departure from economic loading is ae eee eel
AC=AC\+A CG) =2895 ~ 279.375 = 10.125 Rshh
ue” Chapter 15 Economic Operation of Power Systems 391
Thus, the saving in fuel cost due to economic loading is Rs 10.125 per hour. If itis assumed
that the system load remains constant throughout the year, the total yearly saving
= Rs 10.125 x 365 x24 = Rs 88695
This saving justifies the need of economic loading of units and the necessity of using devices
for controlling the loading of each unit automatically.
15.4 TRANSMISSION LOSS AS A FUNCTION OF PLANT GENERATION
Consider a simple power system consisting of two generating plants and one load as shown in
ig. 15.4.
Plant? = Plant 2
A ¢ 8
fiete
—1L-o
Load
Fig. 15.4. A.simple system connecting two generating plants to one load.
Let Rac, Rac, and Rep be the resistance of the lines AC, BC and CD respectively. For the
given system we can write the transmission losses as
Pi= 3h P Rac+31 bP Ract31h+hl Reo +(15.4.1)
If we assume that 1; and Jz are in phase,
Int+hl=lhl+thl ++(15.4.2)
Pp=31h P (Rac+ Rep) +31 b P (Rac+ Rep) +61 Wh I Rep »(15.4.3)
Let P} and P2 be the three-phase power output of plants 1 and 2 at power factors of
cos 1 and cos gz, and Vi and V2 be the bus voltages at the plants.
__ Pi _ Pe
Inl= BTV Teos qi’ \bl= BTV Te0s
Substituting these values of fi and Jp in Eq. (15.4.3) we get
Ppa PtRACtRO_ 4 op, Py Rep Rect Ro (45.4.4)
og
"Tv Peos? 1 TViTI Val cos 91 608 G2 *1V2 Pos? ga
Equation (15.4.4) can be written as
Pr= PTB +2P1 Pr Biot P3Bx
15.4.5) 392 Electrical Powor Systoms
B Ract+ Rep
IVP cos? @1
Where By =—Rac*+ Rev _ (15.46)
V2 cos? @2
_ Rep
2 TVi IT Val (cos 1) (C05 a) ;
The terms By1, Bi2, and Bz2 are called loss coefficients or B coafetns,
i ett in kilovolts and line resistan
If the voltages in Eq. (15.4.6) are line-to-line voltages in istances axe
in ohms, the units for the loss coefficients will be in reciprocal megawatts (MW), Then,
Eq. (15.4.5), with three-phase powers in megawatts, Pz, will be in megawatts also. If ally.
Quantities are in per unit, the coefficients will also be in per unit.
It is seen that the loss coefficients depend on source voltages and power factors. The source
Voltages and power factors depend on and vary with system operating conditions. However, 3
Coefficients are constants. It is sufficiently accurate to calculate B coefficients for some average
Operating conditions and use these values for economic loading for all the load variation,
However, for large load variations or for major system changes, several sets of loss coefficients
are used,
Example 15.4 For the system shown in Fig. 15.4, the voltage at bus C is Ve= 1.0/0* pu, Te
currents in lines AC and BC are 1.05/0° pu and 0.9/0° pu respectively. The line impedances
are as follows :
Zac= (0.05 +j0.20) pu, Zac=(0.04+j0.16) pu, Zcp=(0.03 + 0.12) pu.
Find the loss coefficients and the transmission loss,
Solution
Va=Ve+ Zac li =1/0° + (0.05 +) 0.2) (1.05 /0°)
= 1.0525 +j0.21 pu=Va cos gn+j Va sin gy
Va cos (pa = 1.0525
Va=Ve+ Zac la= 1/0° + (0.04 +} 0.16) (0.9 /0°)
= 1.036 +j 0.144 pu= Vg cos op +, Vp sin 9B
Va cos p= 1.036
Power output of plant 1
Pi=Re [Va Ii] = Re [(1.0525 +j 0.21) x 1,05 70°
Power output of plant 2
P2=Re [Va]
= 1.1051 pu
= Re [(1.036 + 0.144) x 0.9 /0°] = 0.9324 pu
Ract Rep _ 0.05 +0.03
Bu= = 0.05 + 0.03 _
(Wacos ga)? (1.0525)? 0.072218 pu
Br Reo o.
= ).03
= Wpcoro) Pree
(Va-c0s @a) (Va c08 pa) ~ (1.0525) x (1.036) = 0.027513 pu 393
Chapter 15 Economic Operation of Power Systems
Bin = Bact Rep. _ 004+ 0.03 _o.0652196 pu
(Vacosgs)> (1.036)
Transmission loss
PL = Pi Bu + P3 Bn + 2P PoBi
= (1.1051)? x 0.07218 + (0.9324)? x 0.0652196 + 2 x 1.1051 x 0.9324 x 0.027513
= 0.2015944 pu
Check
Transmission loss = sum of PR losses in lines AC, BC and CD
=i Ract BRac+ (h+hy Reo
= (1.05)? x 0.05 + 0.97 x 0.04 + (1.05 +.0.9)? x 0.03 = 0.2016 pu
15.5 GENERAL LOSS FORMULA
Ideally, the exact power flow equations should be used to account for the transmission loss in
the system. However, it is a common practice to express the transmission loss in terms of active
power generations only. This approach is commonly known as the loss formula or B-coefficient
method. The simplest form of loss equation for a system of k plants is given by
kok
P=), (Pm Brn Pr) (15.5.1)
n=l
ransmission loss
Pm = active power generation at mth plant
.ctive power generation at nth plant
Brn = Bum :
‘The coefficient Bmn are commonly known as loss coefficients or B-coefficients. The formula
given by Eq, (15.5.1) is called George's formula.
For a two-plant system,
2 2 2
P=, (Pm Bna Pr) = Dy (Pm Bm Pi + Pm Bro P2)
mal asl msl
= Py By Pi + Pi Biz Po + P2 Bai Pi + P2 Bro Po
= PHB +2P) P2 Biot PiBn (15.5.2)
For a three-plant system,
P= P} Bu + Ph Bua + P3 Bsa + 2P1 Po Bio-+ 2P2 P3 Bas + 2P3 Pi Bis (15.5.3)
‘The matrix form of the transmission-loss equation is given by
Pi=P" BP (15.5.4)
where P” is the transpose of P. 394 Electrical Power Systems
For a total of k sources,
, By Bia: Bik
5 Bu Br Bx
P= : and B= / “gg
Pt Be Bro Bue
15.6 OPTIMUM LOAD DISPATCH CONSIDERING TRANSMISSION LOSSES
7 i the loads are different, the
When the distances of generating plants from the loa i 5 COSt of ify
transmission losses will affect the economic distribution. Consider n generating plans
Ci, Ca, «sss Cn be the fuel costs of individual plants for the corresponding electrical power
of P1, Po, "» respectively. Let Pr be the total power received by the loads (that is, thehy
demand) and Pz the total transmission losses. For n plants, the total fuel cost is
n
C=C + Crt... + Cr= Ci lS
i=.
The total input to the network from all plants is
2
P=Pi+P2+...+Pa= > Pi ol l580)
Since the total demand and the transmission losses must be met by the total generatint
that instant,
PR+PL=P\+Pot...+Pp
a
yr «Als
We P ,
e shall sssime that the total transmission loss PL is a function of generation. Tha!
PLE @(P1, Pa, ..., Pr)
wll
The optimal load dis
patch problem ideri as fol
, Considering transmission losses can be stated
Minimize C= va
ABs
im
subject to the constraint
hn
Pr+PL-¥ Pixg °
ist
Equation (15.6, 6) i
9-9) Is the active
Power balance equari
quation, Chapter 15 Economic Operation of Power Systems 395
Making use of Lagrangian multiplier 2, the augmented cost function C* is defined as
CS CHAlPe+PL- > Pi ee?)
iat
For economical load dispatch,
Since the cost of a unit depends on the output of that unit only C; is a function of P; only,
and therefore, the partial derivatives become full derivatives and |
and
+-(15.6.9)
The partial derivative (@P:/dP)) is known as the incremental transmission loss (ITL)
associated with generator i. It gives the extra system loss incurred by an increment of active
power injection by generator i. The partial derivative (8C;/@Pi) is the incremental fuel cost
(IQ); of generator i.
Equation (15.6.9) represents a set of n equations with (n +1) unknowns. Here n generators
are unknown as 2 is also unknown. These equations are called coordination equations because
they coordinate the incremental transmission losses with the incremental cost of production.
Equation (15.6.9) can be written as
1 dc |
QPL) dP;
i
«-(15.6.10)
«-(15.6.11)
The factor Li, is called the penalty factor for plant i. It depends upon the location of the
plant, The larger the incremental transmission loss, the larger is the penalty factor. 396 Electrical Power Systems
Equation (15.6.10) can be written as
dCs aCn_
wap ips alee, “4
Equation (15.6.12) shows that for minimum See Fa aoe fu a Cost of en hn
sp is ints in the system. In oth ln
He, it ty factor is the same for all plan ; n ieee
maid by is Pena Jaen ned when each plant is operated such tha the penit
ed
incremental costs are equal. .
Equation (15.6.9) can be written in the alternative form as
(Oi=AU-CATL)\ f= 1,2). “S$.
Special case , , /
Equation (15.6.12) gives the criterion for optimum loading of plants when the transmissi
are considered. If the losses are neglected, P, =0 and
losses
dCi_
ap;-*
Cy _ aCe _
vr dP, dP,
This relationship is already obtained in Eq. (15.3.9).
Example 15.5 On the system consisting of two generating plants the incremental costs in pes
per megawatt hour with P; and Pz in megawatts are
aC _ , aa_
Pi =0.15 Pi} + 150 ; dP, = 925 P2+175
‘The system is operating on economic dispatch with P; = P;=200 MW and PL 292 Find
2
the penalty factor of plant 1,
Solution
On economic dispatch,
-AELSS!)
_ Pi=P2=200 MW
Substitution of the values of P,, Pp,
Li (0.15 x 200 + 150)
Ly in Eq. (B15.5.1) gives
= 1.25 (0.25 x 200 + 175)
180L; = 281.25, 2, = 281.25 _
1180 = 15625 Chapter 15. Economie Oporation of Powor Systems 397
Example 15.6 A power s:
ir is being dispatched
ystem has (wo generating plants and the power is being dispatches
economical
with Py = 150 MW and P)= 275 MW, The loss coefficients are +
Bus 0.10% 107 Mw!
Br =-0.01 x 107 Mw"
B= 0.13 107 MW,
To raise the total load on the system by | MW will cost an additional Rs 200 per hour. Find
(2) the penalty factor for plant 1, and (b) the additional cost per hour to increase the output of
plant 1 by | MW,
Solution
(a) For a system with two plants,
PL= PI By +2P) P2 By + P3 Bap
oP.
Dp, 2P1 Bu + 2P2 Birt 0
Py
= 2x 150x 0.1 x 107 +2 x 275 (- 0.01 x 10) = 0.3 - 0.055 = 0.245
Penalty factor for plant 1
ts
19, 1=0.245
oP;
(b) Incremental cost of the system is given by
2 = Rs 200 per MWh
l= = 1.3245
dC _
We have Ly Py =A
dCi
1.3245 dP) = 200;
51 Rs/MWh
Example 15.7 In a two-plant system, the entire load is located at plant 2, which is connected
‘o plant 1 by a transmission line, Plant 1 supplies 100 MW of power with a corresponding
transmission loss of 5 MW. Calculate the penalty factors for the two plants,
Solution
For a two-plant system,
PL= PIB + 2P1P2Bi2 + P3B22 .-(EI5.7.1)
Since all the load is at plant 2, varying P2 docs not affect the transmission loss Pt. Thus,
from Eq. (E15.7.1)
PL= PIB
5 = PRB\, = (100)°By = 10°81, 398 Electrical Power Systems
Bu=—y
10
and * ae =a
= 2P,By, =2x 100
Penalty factor for plant 1
1
beh
oP1
PL _
Now, 9, Py 0
penalty factor for plant 2
1
=e!
OP2
45.7 ITERATIVE METHOD OF SOLVING COORDINATION EQUATION
‘The coordination equation for nth plant is
Cn OP
cue (L511
| a, tap, ae
Let us assume that the plants have the quadratic cost functions of the form
Oy On P+ Bn Path 51)
Uche= = Qn Pn + Ba Rs/MWh (1573)
__ It is seen that for a quadratic cost function, the incremental cost curve ((IC)s vs Pe cane]
is a straight line given by Eq, (15.7.3). In this equation
{ia = slope of incremental production cost curve
Bn = intercept of incremental production cost curve on (IC)n axis.
From Eq. (15.5. BD
oP,
Bsa Pa 54
‘Thus the ee Eq. (15.7.1) can be written as
k
On Pa + Bn +22, Bran Pm = 2,
m=. stems 399
Chapter 15 Economic Operation of Power SY’
Collecting all coefficients of Pn we get
k
= 2D 2Bnan Pn Bat
m=1
men
Pr (On + 2A Brin)
Solving for P, we get
a ¢
a
1-38 -Y) 2BraP
mal
men
On
X + 2Ban
Pa= (15.7.5)
Since the incremental transmission loss depends on power outputs from all the plants, the
coordination equation (15.7.5) cannot be solved directly. It is well suited for solution by iterative
method, The following procedure is used :
1. Assume a suitable value of A= Ao. This value should be greater than the largest intercept
of the incremental cost of the various units. Calculate Pi, P2, ...5 Pn based on equal
incremental costs. .
2. Calculate the generation at all buses with the help of Eq, (15.7.5) keeping in mind that
the values of powers to be substituted on the right-hand side of Eq. (15.7.5) during
zeroth iteration correspond to the values calculated in step 1. For subsequent iterations
the values of powers to be substituted correspond to the powers in the previous iteration.
However, if any generator violates the limit of generation that generator is fixed at the
limit violated.
3, Check if the difference in power at all generator buses between two consecutive
iterations is less than a specified value, otherwise go back to step 2.
4, Calculated the losses using the relation
kok
PL=, YL PmBnn Pn
mal n=l
5. Calculate
k
AP=| >) Pn-Pr-P
n=l
6. If AP is less than a specified value €, stop calculations and calculate cost of generation
with the values of powers. If A P pare ty bi
Br dag La
a
But > P= Pr
isl
ot (15.1)
We calculate 2. from Eq. (15.7.7) and substitute in Eq, (15.7.6) to get individual Ps
This method does not require iterative sol transmissio!
ire i issi
He quire iterati lution for a system where ission Losses
Starting values of 4, that is, 4 and
u ; 1: : 0) : te
Procedure may'be obtained from Eq. (137) ee Poe
1
HP =F OB) for 21,2, ..4n
rave of Power Systems 401
Chapter 15 Economic Operation
Example 15.8 A system consists of two plants connected by a transmission line as shown in
Fig. 15.5. The load is at plant 2 Ifa load of 125 MW is transmitted from plant 1 to the load,
there is a loss of 12.5 MW. Determine the generation schedule and the load demand if the cost
of the received power is Rs 70 per MWh. Assume that the incremental costs of the two plants
are given by
ac dC
ap, = 0.25 Pi : ==
dP; 1 +40 Rs/MWh ; a,
Solve the problem using (a) coordination equations; (b) penalty factor method.
0.20 Po +50. Rs/MWh.
Plantt 4 Plant 2
Load
Fig. 15.5. Illustrating Example 15.8,
Solution
Since the load is at the bus of plant 2, therefore, the line loss will not be affected by variation
of Po, Thus,
By. = Bz = 0 and Bn =0
For a 2-plant system
Py = PI Bu + P3 Bn +2P1 Pr Br
When P)=125 MW, P.=12.5 MW,
we have 12.5 = (125) Bu +0+0
ot Bu= 2328x104 MW?
(125)
4 aPr
Therefore, PL=8x 10~ P} ; 5+ = 16x10 Pi
OP
We are also given 2=70 Rs/MWh.
(a) Use of céordinates equation
The coordination equation for plant 1 is given by
dCi, . aPL
ay, Pu _y
a a,
' aC a ang Pt
Substituting the values of Py, » Wand Pr we have
0.25 Py +40 +70 16 x 10% Pi =70
(0.25 +.0.112) P1 = 30, Pi = 82.8729 MW
The coordination equation for plant 2 is given by 402 Electrical Power Systems
da, BP
ap; **aps~*
or (0.20 P2+50+0= 70, P2= 100 MW
The line loss is given by
Py=8x 10" P}= 8x 10x 82.8729)" = 5.494 MW
‘Therefore, the total load, Pe= Pi + P2— PL = 82.8729 + 100 - 5.494 = 177.3789 yw
(b) Use of penalty factor
The penalty factor for plant 1 is given by
1
(1-16x 107 Pi)
For optimal dispatch
ac 1 =
ay, 1 ___x 0.25 P+ 40)=70
apy an16x 10" Py)
or 0.25 Py +40= 70-70 16 x 104 P1
30
Pi 25+ 0.12 = 828729 MW
Penalty factor for plant 2,
or
l=
For optimal dispatch,
dry,
Lap, hi 1% (0.20P2+50)=70
=20L
P2= 599 = 100 MW
It is seen that both the methods give identical results. This is due to the fact thatthe pe
factor has been derived from the coordination equation.
EXERCISES
1, The fuel costs of two generators are given by
« fon ISPL+O1PERSh; ——Cp=1.8425P2+0.1 PB RSs
oes ‘mand on the generators is 250 MW, find the economic loadin
SEirceeseeccmmree—--___s—s—“‘