Electric Power Generation, Transmission, and Distribution

Solution

The sequence impedance matrix is

zseq



¼

0 : 7735 þj 1 : 9373 0 0

00 : 3061 þj 0 : 6270 0

000 : 3061 þj 0 : 6270

2

4

3

5

Performing the reverse impedance transformation results in the approximate phase impedance matrix.

zapprox



¼½ŠAs zseq



½ŠAs^1 ¼

0 : 4619 þj 1 :0638 0: 1558 þj 0 :4368 0: 1558 þj 0 : 4368

0 : 1558 þj 0 :4368 0: 4619 þj 1 :0638 0: 1558 þj 0 : 4368

0 : 1558 þj 0 :4368 0: 1558 þj 0 :4368 0: 4619 þj 1 : 0638

2

4

3

5

Note in the approximate phase impedance matrix that the three diagonal terms are equal and all of the
mutual terms are equal.
Use the approximate impedance matrix to compute the load voltage and voltage unbalance as
specified inExample 21.1.
Note that the voltages are computed to be balanced. In the previous example it was shown that when
the line is modeled accurately, there is a voltage
unbalance of almost 1%.

21.1.4 Step-Voltage Regulators

A step-voltage regulator consists of an autotrans-

former and a load tap changing mechanism. The

voltage change is obtained by changing the taps of

the series winding of the autotransformer. The

position of the tap is determined by a control circuit

(line drop compensator). Standard step regulators

contain a reversing switch enabling a+10% regu-

lator range, usually in 32 steps. This amounts to a

5 =8% change per step or 0.75 V change per step on

a 120 V base.

A type B step-voltage regulator is shown in

Fig. 21.13. There is also a type A step-voltage regu-

lator where the load and source sides of the regula-

tor are reversed from that shown in Fig. 21.13.

+ +

+

+

+

+

Z+

Z+

Z+

Ia

(Z 0 −Z+)/3 (Ia+Ib+Ic)

Ib

Ic

Vcg

Vbg

Vag

Vcg

Vbg

Vag

FIGURE 21.12 Approximate line segment model.

NNNNNN

• –

+

+

S R

L

L

Control

PT

Control

CT

Shunt

Winding

Reversing

Switch

Series Winding

V source

V load

Preventive

Autotransformer

FIGURE 21.13 Type B step-voltage regulator.