Electric Power Generation, Transmission, and Distribution

In this model the line-to-line voltages inEq. (21.181)will change during each iteration until conver-
gence is achieved.

3. Constant current loads
   In this model the magnitudes of the currents are computed according to Eq. (21.179) and then held
   constant while the angle of the voltage (d) changes during each iteration. This keeps the power factor of
   the load constant.

ILab¼jjffILab dabuab

ILbc¼jjffILbc dbcubc

ILca¼jjffILca dcauca (21:182)

4. Combination loads
   Combination loads can be modeled by assigning a percentage of the total load to each of the above three
   load models. The total delta current for each load is the sum of the three components.
   The line currents entering the delta connected load for all models are determined by

ILa

ILb

ILc

2

4

3

(^5) ¼
10  1
 110
0  11
2
4
3
5
ILab
ILbc
ILca
2
4
3
(^5) (21:183)
In both the wye and delta connected loads, single-phase and two-phase loads are modeled by setting the
complex powers of the missing phases to zero. In other words, all loads are modeled as three-phase loads
and by setting the complex power of the missing phases to zero, the only load currents computed using
the above equations will be for the nonzero loads.
21.1.7 Shunt Capacitor Models
Shunt capacitor banks are commonly used in a distribution system to help in voltage regulation and to
provide reactive power support. The capacitor banks are modeled as constant susceptances connected in
either wye or delta. Similar to the load model, all capacitor banks are modeled as three-phase banks with
the kVAr of missing phases set to zero for single-phase and two-phase banks.
21.1.7.1 Wye Connected Capacitor Bank
A wye connected capacitor bank is shown in Fig. 21.27. The individual phase capacitor units are
specified in kVAr and kV. The constant susceptance for each unit can be computed in either Siemans
or per unit. When per unit is desired, the specified kVAr of the capacitor must be divided by the base
single-phase kVAr and the kV must be divided by the
base line-to-neutral kV.
The susceptance of a capacitor unit is computed by
Bactual¼
kVAr
kV^21000
Siemans (21:184)
Bpu¼
kVArpu
V^2 pu
per unit (21:185)
where kVArpu¼
kVAractual
kVAsinglephasebase
(21:186)
Vpu¼
kVactual
kVlinetoneutralbase
(21:187)

• 
  


• 
  


• 
  

  −
  −
  −
  Van
  Vcn
  Vbn
  jBa
  jBb
  jBc
  ICa
  ICb
  ICc
  FIGURE 21.27 Wye connected capacitor bank.