Reconfiguration and restoration of distribution systems
7.4 Capacitor application in distribution systems
Capacitor location in distribution networks is a very important option to reduce electrical losses and therefore it has been widely used. Capacitors not only help to save losses but also play an important role in power factor correction and in the improvement of the voltage profile especially in long feeders. Even with the ben-efits of capacitor location in distribution systems, their location has to be analyzed carefully not only for the high costs involved, but also for the overvoltages which capacitors may produce in networks with harmonic circulation when resonance conditions may occur.
A capacitor is a device consisting essentially of two electrodes separated by a dielectric insulating material that is capable of supplying magnetizing kVAR to the system. The capacitive reactance has the nature of a negative inductive reactance.
This property is utilized in electrical circuits to compensate the effects of inductive reactance and the lagging reactive kVA of inductive loads.
Capacitors can be classified as series or parallel according to the type of con-nection they have. Series capacitors are connected in series with lines to compen-sate for inductive reactance. Shunt capacitors are connected in parallel with lines to compensate the reactive power or current required by an inductive load.
Example 7.1 Consider a simple 11 kV radial line transmitting power by an over-head system to a lagging power factor load. Figure 7.5 presents the system diagram and the equivalent circuit.
Analyze the sending and receiving-end conditions for each of the following three cases:
(a) Without capacitors: The line-to-line receiving-end voltage is assumed con-stant at 11 kV. All calculations are referred to 11 kV base voltage.
(b) With shunt capacitors: Three single phase capacitor units, each having a reactance of 45.7W, are connected between phase and neutral adjacent to the load. Calculate the current taken by each capacitor, IC, IR, and ES.
(c) With series capacitors: In this case a capacitor unit is connected in series with each phase, the reactance of each unit being 5W. Calculate IRand ES. For the case (a):
The line-to-line receiving-end voltage is assumed constant at 11 kV. All cal-culations are referred to 11 kV base voltage.
IR¼ ffiffiffi4:27 MVA
Referred to 11 kV
XTR= 1 Ω/ph Referred to 11 kV
Load 4.27 MVA
at 0.85pf
Figure 7.5 Diagrams for example 7.1
Taking ERas reference vector:
Figures 7.6 and 7.7 show the vector diagram for the current case. That is the base case for the other points.
For the case (b):
Three single phase capacitor units, each having a reactance of 45.7 W, are connected between phase and neutral adjacent to the load.
Then, current taken by each capacitor:
IC¼ 11ffiffiffi; 000 p3
45:7ffð90Þ A
¼ 139ffð90Þ A; leading the applied voltage by 90 The inductive component of the load current
¼ 264 sinð31:78Þffð90Þ
¼ 139ffð90Þ lagging the applied voltage by 90
ER
Figure 7.7 Power diagram for the base case
ER = 11 kV
Figure 7.6 Voltage and current vector diagram for the base case
Therefore, the current taken by the capacitor will neutralize the inductive component of the load current, and the actual current taken from the line will be only 10.5 A, in phase with the applied voltage:
IR ¼ 224:42ffð0Þ
In this case a capacitor unit is connected in series with each phase, the reac-tance of each unit being 5W.
IR¼ 264ffð31:78Þ
For the purpose of comparison the relevant sending- and receiving-end con-ditions for the three cases are tabulated in Table 7.1.
The following can be seen:
1. Both shunt and series capacitors reduce the voltage drop and also the MVAR demand at source of supply.
2. The MVAR rating or the shunt capacitor bank for the three phases
¼ 3IC2XC106MVAR
¼ 3ð224:42Þ245:7 106kVAR
¼ 6:9 MVAR
The MVAR rating of the series capacitor bank
¼ 3ð264Þ25 106MVAR
¼ 1:05 VAR
Thus, for the same improvement in voltage regulation, the series capacitor bank is much smaller than the shunt capacitor bank. At the same time, it should be noted that series capacitors reduce MVAR demand at source far less than shunt capacitors do.
3. Shunt capacitors reduce the line power losses by reducing the receiving-end current.
Table 7.2 shows a comparison between the shunt capacitor and the series capacitor.
7.4.1 Feeder model
A general model is considered here, which corresponds to a distributed load along the feeder with the possibility of having a concentrated load at the end whose value
Table7.1Comparisonofresultsforexample7.1 TypeofsystemLinevoltageVoltage dropPowerfactorSending-endpowerReceiving-endpowerActive power Sending end(kV)Receiving end(kV)Sending endReceiving endActive power (MW) Reactive power (MVAR) Apparent power (MVA)
Active power (MW)
Reactive power (MVAR)
Apparent power (MVA) CaseA,without capacitors13.25112.35kV (20.4%)0.84lag0.85lag5.13.3lag6.14.32.7lag5.10.8MW (18.6%) CaseB,withshunt capacitors11.65111.65kV (5.9%)0.986lag1.04.470.76lag4.534.3Nil4.30.17 (3.9%) CaseC,withseries capacitors11.48110.48kV (4.4%)0.864lag0.85lag4.532.65lag5.34.32.7lag5.10.23 (5.35%)
can also be set to zero, as proposed by Neagle and Samson in reference. If the total feeder current at the substation end is I1and the concentrated load at the other end takes a current I2, the expression for the current along the feeder depends on the distance from the substation according to the following equation:
iðxÞ ¼ I1 ðI1 I2Þx ð7:8Þ
If a factor p¼II21is introduced, the equation becomes:
iðxÞ ¼ I1½ðp1Þx þ 1 ð7:9Þ
The profile corresponding to that expression is shown in Figure 7.8. If the load is uniformly distributed along the feeder and there is no concentrated load at the end, I2, and therefore p, is equal to zero, and (7.9) becomes:
iðxÞ ¼ I1ð1 xÞ ð7:10Þ
Likewise, if the load is concentrated at the end of the feeder, p = 1, the expression for the current is:
iðxÞ ¼ I1 ¼ I2 ð7:11Þ
Table 7.2 Comparison between shunt capacitor and series capacitor
Shunt capacitor Series capacitor
Size 45.7W 5W
Voltage regulation Good Good
Loss reduction Very good Nothing
Power factor Very good Fair
Stability Good Very good
Cost Reasonable High
LagLead
I1
I2
x
Figure 7.8 Current profiles for feeders with uniformly distributed loads
7.4.2 Capacitor location and sizing
The reduction in losses by using network reconfiguration can be further enhanced by placing capacitors along the feeders. The following sections will develop theories for such applications which have been proposed in several works. The different magnitudes, i.e., power, energy, impedance, and current, will be dealt with in pu values.
The power loss dissipated in a circuit with a total current I, a resistance value R, and a power factor anglef can be expressed as:
P1¼ I2R¼ ðIcos fÞ2Rþ ðIsin fÞ2R ð7:12Þ If a capacitor with a current Icis installed, the reactive part of the current will be compensated and therefore the new total losses (P2) can be expressed as:
P2¼ I2R¼ ðI cos fÞ2Rþ ðI sin f ICÞ2R ð7:13Þ Therefore, the loss reduction can be found as:
DP ¼ P1 P2¼ 2IRICsinf IC2R ð7:14Þ
Equation (7.14) shows that only the reactive component of the load current is required for power loss reduction studies. Therefore in the rest of this section, that component will be referred to as I1R.
If a feeder has a uniformly distributed loading whose current is that given by (7.9), the total losses due to the reactive component without capa-citors will again be referred to as P1 and can be found with the following equation:
P1¼ 3ð1
0
I1Rð1 þ ðp 1ÞxÞ2
Rdx ð7:15Þ
where:
p¼ I2R=I1R
I1R Reactive current at the substation end I2R Reactive current at the feeder end
x Distance along the feeder ranging from 0 to 1 The result is given in the following equation:
P1¼ RI1R2ðp2þ p þ 1Þ ð7:16Þ
7.4.3 Reduction in power losses with one capacitor bank
Figure 7.9 shows a distribution feeder with one capacitor bank installed at a dis-tance a from the source.
If the capacitor bank takes a current ICas shown in Figure 7.9, the new total power losses P2, due to the reactive current including the capacitor bank, are calculated as:
P2¼ 3 The result of this expression is:
P2¼ RI1R2ðp2þ p þ 1Þ 3RI1R2 ðp 1ÞIC
The power loss reduction due to the capacitor installation is found by sub-tracting (7.18) from (7.16) as follows:
DP ¼ P1 P2¼ 3RI1R2 ðp 1ÞIC
The optimum sizing and location of one capacitor bank to reduce power losses is found by taking partial derivatives of (7.19) with respect to ICand the distance a as follows:
Figure 7.9 Distribution feeder with one capacitor bank
@DP
By equating (7.20) and (7.21) to zero and solving simultaneously for ICand a, the following values are obtained:
IC¼2
3I1R ð7:22Þ
a¼ 2
3ð1 pÞ ð7:23Þ
The maximum power loss reduction with one capacitor is obtained by repla-cing (7.22) and (7.23) into (7.19). The corresponding expression is:
DPMAX¼ R I1R2
8
9ð1 pÞ ð7:24Þ
Equations (7.23) and (7.24) are valid if p 1/3. For larger values of p, the magnitude of a becomes greater than 1 or even negative, which does not have physical meaning. Therefore for p greater than 1/3, the capacitor bank should be located at the end of the feeder which corresponds to a value of a equal to 1. If this value of a is introduced in (7.19), the value of the capacitor current is:
IC¼1
2 I1Rðp þ 1Þ for p > 1=3 ð7:25Þ
Likewise, if the value of ICgiven in (7.23) and a = 1 are introduced in (7.19) the reduction in losses when a capacitor bank is installed and p> 1/3 is given by the following equation:
DP ¼3
4 R I1R2 ðp þ 1Þ2 for p> 1=3 ð7:26Þ
7.4.4 Reduction in power losses with two capacitor banks
Figure 7.10 shows a distribution feeder with two capacitor banks, one installed at a distance a and the other at a distance b from the source.
Initial losses:
P1¼ 3ð1
0
½Ið1 xÞ2Rdx¼ I2R ð7:27Þ
Losses if capacitors are connected:
P2¼ 3ða
Therefore the loss reductions are expressed as:
DP ¼ 3Rðað2IIC IICa 3IC2Þ þ bð2IIC IICb IC2ÞÞ ð7:29Þ It can be demonstrated that the optimal values of a and b are:
a¼ 13IC
2I ð7:30Þ
b¼ 1IC
2I ð7:31Þ
7.4.5 Loss reductions with three capacitor banks
DP ¼ 3Rðað2IICIICa5IC2Þþbð2IICIICb3IC2Þþcð2IICIICcIC2ÞÞ ð7:32Þ The optimum localization of capacitors:
a¼ 15IC
2I ð7:33Þ
i(x)=I(1−x)−2IC
IC
IC
i(x)=I(1−x)−IC
1.0 pu
1 0
C1 C2
a
b
i(x)=I(1−x)
Figure 7.10 Distribution feeder with two capacitor banks
b¼ 13IC
2I ð7:34Þ
c¼ 1IC
2I ð7:35Þ
7.4.6 Consideration of several capacitor banks
The expressions presented in (7.19), (7.29), and (7.32) can be generalized for any feeder with n sections to find the power and energy loss reductions when capacitors are located in different places.
For more than one segment on the feeder, expressions (7.19), (7.29), and (7.32) are still valid, although new techniques that consider heuristic strategies have been proposed. Under these circumstances the expressions can be used by putting a = 1, which is then directly applicable to one segment at a time. Therefore when a capacitor is installed downstream at a segment n, the reduction in the power losses are given as a function of that capacitor current, and of currents at the beginning and at the end of the segment, respectively, as follows:
DPm¼ P1m P2m¼ 3 Rm
ðI1Rmþ I2RmÞ ICn ICn2
ð7:36Þ
For the segment where a capacitor is located, the expressions for power and energy have already been developed. General expressions are obtained by considering the effect of one capacitor at a time for every section toward the source as follows:
DP ¼ DPnþXn1
m¼1DPm ð7:37Þ
That equation means that for a capacitor installed in a segment n, the power loss reduction is: Likewise if a number of k banks are considered:
DP ¼ Xk
7.4.7 Capacitor sizing and location using software
It is impractical to perform the calculation of the optimal number and size of capacitor banks by hand. A number of software packages are available which normally request the following information:
● maximum size allowed (kVAR) for each installation,
● available places (poles),
● maximum number of banks,
● step in kVAR of the bank,
● load factor, and
● existing banks.
Figure 7.11 corresponds to a real distribution system with 12 feeders. The losses of the system are high and the power factor low. Therefore possibilities to improve the system are considered by installing capacitor banks in the most critical feeders.
One of the feeders is considered to determine the number and size of capacitor banks which is shown in Figure 7.12. Two conditions of load factors are analyzed:
0.6 and 0.9. Two locations are allowed of banks having steps of 50 kVAR. A maximum of 2 MVAR has been established.
After running the analysis with a software package, two banks are considered for each loading factor as shown in Table 7.3. The bold letters in the Loss column correspond to the losses of the system before the capacitors were used. An
Figure 7.11 Distribution system to illustrate application of capacitors
economic analysis not shown in this example is recommended to define the option to select from the two presented in the results.
7.5 Modeling of distribution feeders including VVC equipment