Page 400 - Power Electronic Control in Electrical Systems

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Power electronic control in electrical systems 379

With E ˆ V
2 2

X s Q X s P
2
V ˆ V ‡ ‡
V V
2
2
4
V ˆ (V ‡ X s Q) ‡ (X s P) 2
2V 2
2
2
; Q ‡ Q ‡ P ˆ 0
X s
p 
2
whence Q ˆ S S P 2
2
Now S ˆ 4:16 /0:9 ˆ 19:288 MVA (all three phases) and P ˆ 1:40 MVA, so the net
reactive power required to make E ˆ V is
p 
2
Q ˆ Q load ‡ Q g ˆ 19:2884 19:2884 1:40 2
ˆ 19:2884 19:2375 ˆ 38:5259 or 0:0509 MVA
The correct solution is 0:0509 MVA so that with Q load ˆ 0:70 MVAr we get
Q g ˆ 0:7509 MVAr. Then the capacitor current is
0:7509 10 6
I g ˆ p ˆ 104:2A
3 4160
p
The capacitor reactance must be (4160/ 3)/104:2 ˆ 23:047
/phase (assuming wye
6
connection), and therefore at 50 Hz the required capacitance is 10 /(2p 50
104:2) ˆ 138 mF.
4. 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 compensating
network that will balance this load and correct its power factor to unity. Also
determine the resulting line currents.
General result is
p
B gab ˆ B ab ‡ (G ca G bc )/ 3
p
B gbc ˆ B bc ‡ (G ab G ca )/ 3
p
B gca ˆ B ca ‡ (G bc G ab )/ 3
leaving G ˆ G ab ‡ G bc ‡ G ca in each phase of a wye-connected resulting network.
2
2

In each phase P ‡ jQ ˆ VI ˆ V Y so Y ˆ (P jQ)/V so
3
2
in phase ab, Y ab ˆ (200 j0) 10 /560 ˆ 0:638 ‡ j0 S
2
in phase bc, Y bc ˆ (170 j105:357)/560 ˆ 0:542 j0:336 S
2
in phase ca, Y ca ˆ (170 ‡ j105:357)/560 ˆ 0:542 ‡ j0:336 S