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Power electronic control in electrical systems 401
at the low-voltage terminals of the transformer, i.e. 354/7 50:6 A at the
high-voltage terminals. (Note that this gives a total reactive power of
p
3 400 0:506 35:0 MVAr, which corresponds to the fraction
(s sin s)/p of the rated MVAr, since (s sin s)/p 0:350:)
(b) The phase delay angle is a 180 s/2 122:5 , so the peak phase
current (occurring at 180 )is
p p 3
^ 2 V LLrms 2 400 10
i ( cos a cos p)
X L 7 97:96
( cos 122:5 1) 382 A
27. (i) Explain with your own words and diagrams how the frequency and the
voltage are controlled in an isolated power plant with a local load. Assume
that the generator is a conventional wound-field synchronous machine driven
by a diesel engine.
(ii) Prove by means of a series of diagrams, or otherwise, that an unbalanced
linear ungrounded three-phase load can be transformed into a balanced, real
three-phase load without changing the power exchange between source and
load, by connecting an ideal reactive compensating network in parallel with it.
Assuming a delta-connected unbalanced load Y ab G ab jB ab , Y bc G bc
jB bc , Y ca G ca jB ca , derive expressions for the susceptances of the com-
pensating network.
(iii) An unbalanced delta-connected load draws the following power and reactive
power from a three-phase supply whose line±line voltage is 560 V:
200 kW between lines a, b
170 kW at 0.85 power-factor lagging between lines b, c
170 kW at 0.85 power-factor leading between lines c, a.
Determine the susceptances of a purely reactive delta-connected compensat-
ing network that will balance this load and correct its power factor to unity.
Also determine the resulting line currents.
(i) Frequency is controlled by the speed governor on the prime mover. Voltage is
controlled by the excitation in the generator.
(ii) See Figure 9.23. General result is
p
B gab B ab (G ca G bc )/ 3
p
B gbc B bc (G ab G ca )/ 3 (9:23)
p
B gca B ca (G bc G ab )/ 3
leaving G G ab G bc G ca in each phase of a wye-connected resulting net-
work.
2
2
(iii) In each phase P jQ VI V Y so Y (P jQ)/V so
3 2
in phase ab, Y ab (200 j0) 10 /560 0:638 j0 S G ab jB ab
3 2
in phase bc, Y bc (170 j105:537) 10 /560 0:542 j0:336S G bc jB bc