Page 60 - Power Electronic Control in Electrical Systems

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50 Power systems engineering ± fundamental concepts

from which it can be shown that
Q s ˆ E s I( sin d cos f ‡ cos d sin f)
E s sin d E s cos d E r
ˆ E s sin d ‡ E s cos d
X s X s (2:22)
E s E r cos d
ˆ E s
X s
Note the symmetry between this expression and the one for Q r in equation (2.20).

Example 1
Suppose E s ˆ E r ˆ 1:0p:u: and P ˆ 1:0p:u: 11 The transmission system has
X s ˆ 0:1p:u: and R s is negligible. We have sin d ˆ PX s /E s E r ˆ 1 0:1/1 1 ˆ 0:1
so d ˆ 5:739 and cos d ˆ 0:995. Then Q s ˆ 1:0(1:0 1:0 0:995)/0:1 ˆ‡0:050 p:u:

and Q r ˆ 1:0(1:0 0:995 1:0)/0:1 ˆ 0:050 p:u. Thus the receiving end is gener-
ating reactive power and so is the sending end. The power factor is lagging at the
sending end and leading at the receiving end. The phasor diagram is shown in Figure
2.5b (not to scale).

Example 2
Suppose E s ˆ 1:0p:u: while the receiving-end voltage is reduced to E r ˆ 0:95 p:u:,
with P ˆ 1:0p:u: and X s ˆ 0:1p:u: Now sin d ˆ 1:0 0:1/1:0 0:95 ˆ 0:105 and

d ˆ 6:042 , slightly larger than in example 1 because E r is reduced by 5%. Also Q s ˆ
1:0(1:0 0:95 cos 6:042 )/0:1 ˆ 0:553 p:u: and Q r ˆ 0:95 (1:0cos 6:042 1:0)/0:1

ˆ 0:422 p:u: Since Q s is positive, the sending-end generator is generating VArs. Q r
is also positive, meaning that the receiving-end load is absorbing VArs. The phasor
diagram is similar to that shown in Figure 2.17 (but not to the same scale).
Notice that a 5% reduction in voltage at one end of the line causes a massive
change in the reactive power flow. Conversely, a change in the power factor at either
end tends to cause a change in the voltage. A 5% voltage swing is, of course, a very
large one. Changes in the power tend to produce much smaller changes in voltage;
instead, the load angle d changes almost in proportion to the power as long as d is
fairly small (then sin d d).
The transmission system has an inductive impedance and therefore we would
expect it to absorb VArs. If we regard jX s as another impedance in series with the
load impedance, we can treat it the same way. The current is obtained from
I sin f ˆ (E s cos d E r )/X s ˆ 0:444 p:u:, and I cos f ˆ E s sin d/X s ˆ 1:053 p:u:
Therefore I ˆ 0:444 ‡ j1:053 p:u: ˆ 1:143e j22:891 p:u: (with E r as reference phasor).
2 2
The voltage drop across X s is jX s I and the reactive power is I X s ˆ 1:143
0:1 ˆ 0:131 p:u: Note that this equals the difference between Q s and Q r .
2
We could have made a similar calculation in example 1, where I X s ˆ 0:1. Again
this is Q s Q r . In Example 1 also jQ s jˆjQ r j, which means that each end of the line
is supplying half the reactive VArs absorbed in X s .

11
The per-unit system is explained in Section 2.13. If you aren't familiar with it, try to read these examples
as practice in the use of normalized (per-unit) values. In effect, they make it possible to forget about the
units of volts, amps, etc.