capicitor placement
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Loss saving equation based technique was proposed in [7]. In [8]heuristics and greedy search technique based solution was pro-posed. Fuzzy reasoning based method was proposed in [9]. Simu-lated annealing was proposed in [15] and Genetic Algorithmbased solution has taken in [10,24] respectively. Interior pointbased solution was proposed in [11,14]. Extended Dynamic Pro-gramming Approach was proposed in [12], Plant Growth
Simulation Algorithm and using of loss sensitivity factor wasproposed in [13], heuristic search and node stability based method
on than the solu-predicted one ismethod t
acitor placproblem.
Problem formulation
For a distribution network, the loss associated with the reactivecomponents of branch currents can be written as
PLr Xni1
I2ri Ri 1 Corresponding author. Tel.: +91 9734248638.E-mail address: manas202006@yahoo.com (M. Mukherjee).
Electrical Power and Energy Systems 62 (2014) 9094
Contents lists available at ScienceDirect
n
.e lin distribution system. Modied discrete PSO based solution wasproposed in [3,20]. In [4,5], the capacitor placement was formu-lated as a mixed integer non-linear problem. [6,16,17] proposesParticle Swarm Optimization (PSO) based capacitor placement.
load variation thus may lead to an inferior solutition where probability of load variation over theconsidered. The present paper thus proposes auncertainty of the load variation in the caphttp://dx.doi.org/10.1016/j.ijepes.2014.04.0040142-0615/ 2014 Elsevier Ltd. All rights reserved.o takeementmination of the size, location and type of such capacitors for a dis-tribution system is a complex optimization problem and requiresinformation regarding the load variation of the system with time.
Different solution techniques had been presented by manyresearchers in the past for solving the problem of placing capacitor
coarse picture of the probable situation. The actual scenario maywell deviate the predicted one by a considerable margin. Thusinstead of load representation by a number of denite load levels,probabilistic variation of loads would be a better representation.The capacitor placement decision based upon the xed pattern ofIntroduction
Shunt capacitors are used in distrreactive power. If they are connectedload terminal voltage can be mainlimit and the line loss and total sythe load demand on distribution syeffective compensation, capacitorsswitchable in nature, where a minimkept connected to the system (xcapacitors are switched in or out as tsystems as a source ofroper location and size,within the acceptableost can be reduced. Asay vary with time forbe of xed as well asapacitor kvar is alwaysacitor) and additionaldemand varies. Deter-
was proposed in [18], and bacterial foraging solution was proposedin [21]. Hybrid honey bee colony algorithm based solution wasproposed in [23] Uncertainty was taken into account in [19].
In all of the solution techniques load demand was assumed tofollow a denite pattern-represented by a number of xed loadlevels. In reality however, the load demand is quite uncertainand depends upon many factors in such a way that it is impossibleto predict the actual load before the actual occurrence. Load fore-casts, based upon historic records of load variation can predict aInterval arithmeticSolving capacitor placement problem convariation
M. Mukherjee , S.K. GoswamiElectrical Engineering Department, Power System Section, Jadavpur University, Kolkata
a r t i c l e i n f o
Article history:Received 18 June 2013Received in revised form 28 March 2014Accepted 2 April 2014Available online 13 May 2014
Keywords:Capacitor placement problemLoad variationUncertainty
a b s t r a c t
The paper reports on the souncertainty in the variatioload variation as determinfuzzy interval arithmetic tbase levels. Both xed andtems are presented.
Electrical Power a
journal homepage: wwwdering uncertainty in load
32, India
ion of the capacitor placement problem in distribution system consideringf loads. Solution techniques available in the literature generally considerc. In the present paper uncertainty in load variation is considered usingique. Load variations are represented as lower and upper bounds around
itchable capacitors have been considered and results for standard test sys-
2014 Elsevier Ltd. All rights reserved.
d Energy Systems
sevier .com/locate / i jepes
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Nomenclature
PLr active power loss of the system associated with thereactive components of branch currents for original sys-tem
Iri reactive component of branch current of the ith branchfor original system
Irinew reactive component of branch current of the ith branch
for compensated systemRi resistance, respectively of the ith branchIril, Iriu lower and upper limit of Iri respectively
Ic reactive current drawn by the capacitorPLrcom active power loss of the system associated with the
reactive components of branch currents for compen-sated system
S loss savingVm magnitude of voltage of bus m before compensationk number of capacitor busesQc capacitor sizeVc voltage magnitude vector of capacitor bus
M. Mukherjee, S.K. Goswami / Electrical Power and Energy Systems 62 (2014) 9094 91where Iri and Ri are the reactive component of branch current andresistance, respectively of the ith branch.
But, actually Iri is not of xed value. Because, the load variationin any power system cannot be truly represented by a single loadcurve. Conventional way of representing load variation by a singleload curve basically represents the mean of load variation. A betterrepresentation would be to use a curve like Fig. 1, where instead ofrepresenting by a mean variation, the range of variation is shown.So in the load duration curve, each load level is represented by arange of load levels (like Fig. 2) rather than a single load. So it isbetter to represent Iri as
Iri Iril; Iriuwhere Iril and Iriu are lower and upper limit of Iri respectively.
Because of this variation in this pattern of the loads, the loss PLrshould be considered as an interval quantity instead of xed quan-tity. Therefore, in capacitor placement problem every quantityshould be considered as an interval quantity. For this purpose basicoperation of interval number is to be known which is described inthe next section.
Interval arithmetic
An interval number X = [xl, xu] is the set of real numbers x suchthat xl 6 x 6 xu; xl and xu are known as the lower limit and upperlimit of the interval number, respectively. A rational number k isrepresented as an interval number K = [k, k].
Let X = [xl, xu] and Y = [yl, yu] be the two interval numbers.Then addition, subtraction, multiplication and division of thesetwo interval numbers are dened as below [22]:
X Y xl yl;xl yu 2
X Y xl yu;xu yl 3X Y minxl yl;xl yu; xu yl;xu yu;maxxl yl;xl yu;xu yl;xu yu 4
Fig. 1. Load curve considering load variation.X Y X Y1 5where
Y1 1=yu;1=ylif 0 R yl;yu 6Also, the distance between these two interval numbers is
dened as [24]:
qX;Y maxjx1 y1j; jx2 y2j 7For power system application, calculations involving complex
numbers, rather than real numbers are needed. Hence, in the nextsub-section, basic operations involving complex interval numbersare presented.
Complex interval number
Any complex number Z = X + iY; where i is the complex opera-tor, is said to be a complex interval number if both its real part(X) and the imaginary part (Y) are interval numbers. Hence, X canbe represented as X = [x1, x2] and Y can be represented asY = [y1, y2], where, x1, y1 are the lower limits and x2, y2 are theupper limits, respectively. The conjugate of a complex intervalnumber is given by Z = X iY: Let Z1 = A1 + iB1 and Z2 = A2 + iB2be two complex interval numbers. Then the addition, subtraction,multiplication and division of these two complex interval numbersare dened as [22]
Z1 Z2 A1 A2 iB1 B2 8
Z1 Z2 A1 A2 iB1 B2 9
Z1 Z2 A1 A2 B1 B2 iA1 B2 A2 B1 10
Z1 Z2 C iD 11
where C = (A1 A2 + B1 B2) (A22 + B22) and D = (A2 B1 A1 B2) (A22 + B22).
Fig. 2. Load duration curve considering load variation.
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oweIt is to be noted that, Eqs. (8)(11) can be evaluated by applyingthe fundamental operations as dened in Eqs. (2)(5).
Solution technique of the capacitor placement problem
The solution of the capacitor placement problem has two com-ponents the location of the capacitor and the size of capacitor ateach location. It is obvious that, location of buses must be a xednumber. So to nd the location, the method described in [7] is usedwhere only the base load will be considered. Once the location isknown, interval load will be used to nd the size of the requiredcapacitor.
Determination of location
A radial distribution system with n branches is considered here.Let a capacitor C is placed at busm (except the source bus) and a bea set of branches connected between the source and capacitorbuses. The capacitor draws a reactive current Ic and as it is a radialnetwork it changes only the reactive component of current ofbranch set a. The current of other branches (Ra) is remainunchanged. Thus, the new reactive current Irinew of the ith branchis given by
Inewri Iri Di Ic 12where
Di 1 if branch i 2 a 0 otherwise
Here Iri, is the reactive current of the ith branch in the original sys-tem obtained from the load ow solution. The loss PLrcom, associatedwith the reactive component of branch currents in the compensatedsystem (when the capacitor is connected) can be written as
PcomLr Xni1
Iri Di Ic2 aRi 13
The loss saving S is the difference between Eqs. (1) and (13) andis given by
S PLr PcomLr Xni1
2Di Iri Ic Di Ic2 Ri 14
The capacitor current Ic, that provides the maximum loss savingcan be obtained from
@S@Ic
Xni1
2Di Iri Ic Di Ic2 Ri 0 15
Thus the capacitor current for the maximum loss saving is
Ic Xni1
Di Iri Ri !, Xn
i1Di Ri
!
Xi2a
Iri Ri ! X
i2aRi
!,16
The corresponding capacitor size is Qc Vm Ic 17Here Vm is the magnitude of voltage of bus m before compensation.The above steps are repeated for all the buses (except the root bus)to get the highest possible loss saving for a singly located capacitor.The bus for which highest loss saving is obtained is termed as can-didate bus. When the candidate bus is identied and compensatedusing Eq. (17), the above technique is again used to identify the next
92 M. Mukherjee, S.K. Goswami / Electrical Pand subsequent candidate buses. That will provide only the loca-tions where the capacitors are to be placed. Obviously capacitorsobtained from Eq. (17) are local optimal value. So they are not use-ful when more than one capacitor is to be placed. So the size of mul-tiple capacitors for optimal location is to be determinedsimultaneously and the procedure of nding optimal sizes isdescribed in the following sections.
Size of capacitors for interval load
Here also the method described in [7] is used with a differentmanner. First load ow solution is done taking interval load [1,2].This will provide lower and upper limit of all the branch currents.Let the followings are considered:
k = number of capacitor busesIc = k-dimensional vector consisting of capacitor currents.
Actually Ic = [Icl, Icu], where Icl and Icu are the lower an upperlimit of Ic respectively.
aj = set of branches from the source bus to the jth capacitor bus(j = 1, 2, . . ., k)D = a matrix of dimension n k
The elements of D are considered as
Dij 1; if branch i 2 a 0; otherwiseWhen the capacitors are placed in the system, the new reactive
component of the branch current is given by,
Inewri Ir D Ic 18As Ir and Ic are interval number, Irinew will be also interval
number.The loss PLrcom associated with the new reactive currents in the
compensated system is
PcomLr Xni1
IriXkj1
Dij Icj !2
Ri 19
The loss saving S obtained by placing the capacitors is the differ-ence between Eqs. (1) and (19) and is given by
S Xni1
2Iri Xkj1
Dij IcjXkj1
Dij Icj2 !" #
Ri 20
The optimal capacitor currents for the maximum loss saving canbe obtained by solving the following equations:
@S@Ic1
0@S
@Ic2 0
. . .
. . .
@S@Ick
0 21
After some mathematical manipulations, Eq. (21) can beexpressed by a set of linear algebraic equations as follows:
A Ic B 22where A is a k x k square matrix and B is a k-dimensional vector. Theelements of A and B are given by
X X" #
r and Energy Systems 62 (2014) 9094Ajj Ajjl;Ajju i2aj
Ri;i2aj
Ri 23
-
Ajm Ajml;Ajmu X
i2aj\amRi;
Xi2aj\am
Ri
" #24
where Ajjl and Ajju are the lower and upper limit of Ajj; and Ajml andAjmu are lower and upper limit of Ajm and both are same as branchparameters are considered to be xed.
Bj Bjl;Bju Xi2aj
Iril Ri;Xi2aj
Iriu Ri" #
25
where Bjl and Bju are lower and upper limit of Bj respectively. Also Iriland Iriu are lower and upper limit of branch reactive current
would have a lower and upper limit.
Solving interval load ow, lower and upper limits of B matrix isobtained as
result is summarized in Tables 3 and 4.
lower load level. So the switched capacitor placement problem issolved starting from the lowest load level and the capacitorinstalled at a lower load level will be considered as xed capacitorsfor all the higher load levels. The method is applied for the same 10bus system. The load duration data are given in Table 5. It isassumed that the substation voltage is 1.05 p.u. at peak load condi-tion and 1.0 p.u. during the remaining periods [4]. The result issummarized in Tables 6 and 7.
From the results interval of bus voltages, required capacitorkVAR, system cost of the compensated are obtained by which allthese thing are whether in an acceptable limit or not can bedetermined
Table 2Possible choice of capacitor sizes and cost/kVAR.
Sl. No. 1 2 3 4 5 6kVAR 150 300 450 600 750 900($/kVAR) .500 .350 .253 .220 .276 .183
Sl. No. 7 8 9 10 11 12
Table 3Required capacitor for 10-Bus system.
Bus no. Capacitor size(kVAR)
Lower limit Upper limit
M. Mukherjee, S.K. Goswami / Electrical Power and Energy Systems 62 (2014) 9094 93Numerical results
The proposed method is tested for 10 bus system. The load andbus data is given in Table 1 which is considered as base load. Costof energy is taken as 0.06 $/kW h and capacitor cost is obtainedfrom Table 2[9]. In this work the nearest value of the capacitor cor-responding to Appendix-2 is taken, so that the cost of the capaci-tors can be calculated. For interval load, 5% of the base load istaken.
Using Eqs. (14) and (17), corresponding loss saving is calculatedfor each and every bus except the source bus i.e. Bus No. 0. Then itis noticed that highest loss saving can be achieved if Bus No. 4 iscompensated with a capacitor of size 3998.5 kVAR and total losssaving of 81 MW is obtained. So this bus is compensated withthe capacitor of 3998.5 kVAR. After that repeating the same pro-cess it is observed that highest loss saving is achieved if Bus No.8 is compensated with capacitor of 852.09 kVAR and total loss sav-ing of 14 MW is obtained. After that no such signicant loss saving
Table 1Data for 10-Bus system.
From bus To bus Impedance Load connected at to bus
R (ohm) X (ohm) kW kVAR
0 1 0.1233 0.4127 1840 4601 2 0.0140 0.6051 980 3402 3 0.7463 1.2050 1790 4463 4 0.6984 0.6084 1598 18404 5 1.9831 1.7276 1610 6005 6 0.9053 0.7886 780 1106 7 2.0552 1.1640 1150 607 8 4.7953 2.7160 980 130respectively.Only the branch resistances and reactive currents in the original
system are required to nd the elements of A and B. The capacitorcurrents for the highest loss saving can be obtained as
Ic A1 B 26As A and B are both interval quantity, R.H.S. of this equation is
interval quantity. So the value of Ic will be in the form [Icl Icu],where Icl and Icu are lower and upper limit of capacitor currentrespectively.
Once the capacitor currents are known, the optimal capacitorsizes can be written as
Qc Vc Ic 27Here Vc is the voltage magnitude vector of capacitor buses, whosevalue is like [Vcl, Vcu], where Vcl and Vcu are lower and upper limitof Vc respectively. So Qc will be also an interval number which8 9 5.3434 3.0264 1640 200
Bus 0 is the substation node, the voltage of which is xed at 23 kV.This method can be also applied to solve the switched capacitorplacement problem. It is assumed that the variation of the load isconformal [4,5], the capacitor kVAR required in a particular loadlevel should be at least equal to that required at the immediateApplication to switched capacitor placement problemconsidering load uncertaintyBlowerlimit 0:01400:0361
p:u: and Bupperlimit
0:1130:0265
p:u:
So the lower and upper limit of required capacitor value nearerto Appendix-2 is obtained as 2700 kVAR and 3300 kVAR for Bus No.4 and 750 kVAR and 1050 kVAR for Bus No. 8 respectively. Thecan be achieved using this process. So it can be concluded thatoptimal location is 4 and 8.
Then A matrix is formed as
A 0:0060 0:00600:0060 0:0428
p:u:
kVAR 1050 1200 1350 1500 1650 1800($/kVAR) .228 .170 .207 .201 .193 .187
Sl. No. 13 14 15 16 17 18kVAR 1950 2100 2250 2400 2550 2700($/kVAR) .211 .176 .197 .170 .189 .187
Sl. No. 19 20 21 22 23 24kVAR 2850 3000 3150 3300 3450 3600($/kVAR) .183 .180 .195 .174 .188 .170
Sl. No. 25 26 27kVAR 3750 3900 4050($/kVAR) .183 .182 .1794 2700 33008 750 1050
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instead of sequential solution. This may be done by using robustoptimization technique which is being pursued by the authors atpresent.
Table 4Comparison between original system and compensated system (10 Bus).
Original system Compensated system
Active powerloss (kW)
Lowerlimit
Upperlimit
Active powerloss (kW)
Lowerlimit
Upperlimit
629 974 521 941
94 M. Mukherjee, S.K. Goswami / Electrical Power and Energy Systems 62 (2014) 9094Annual cost ($) Lowerlimit
Upperlimit
Annual cost ($) Lowerlimit
Upperlimit
105,672 163,632 88,240 158,900
Table 5Load duration data for the test system.
Load level 1 2 3
Per unit load 0.3 0.6 1.1Load duration (h) 1000 6760 1000
Table 6Required capacitor for 10-Bus system for different load level.
Load level Bus No. Capacitor sizes(kVAR)
Lower limit Upper limitConclusion
The present paper reports a new formulation of the capacitorplacement problem considering uncertainty in the variation of dis-tribution system load. Unlike the conventional approaches of con-sidering the load variation by simply using a number of load levels,the present paper represents the load variation by upper and lowerbounds of the loads at different load levels and introduces intervalarithmetic to incorporate the effect of such variation in the solu-tion of the capacitor placement problem. As the basic aim of thepaper is to introduce the interval arithmetic technique in thecapacitor placement problem, the solution approach had beenmade simple by separating the problem of placing and sizing ofcapacitors. The capacitor locations are rst selected based uponthe sensitivity of capacitor introduction at the location. Once loca-tions are determined, the size is then found out. The problem how-ever is not decoupled and should rather be solved simultaneously
1 4 750 9008 150 150
2 4 1650 18008 450 450
3 4 3000 34508 750 1050
Table 7Comparison of system cost between original system and compensated system (10Bus).
Original system Compensated system
Lower limit Upper limit Lower limit Upper limit
$95,809 $126,820 $88,900 $116,990References
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Solving capacitor placement problem considering uncertainty in load variationIntroductionProblem formulationInterval arithmeticComplex interval number
Solution technique of the capacitor placement problemDetermination of locationSize of capacitors for interval load
Numerical resultsApplication to switched capacitor placement problem considering load uncertaintyConclusionReferences
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