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Power electronic control in electrical systems 383
Electrical length y phase angle between E s and E r at the surge-impedance
load.
jd P r jQ r sin y
(ii) E s E s e V r cos y jZ 0
V r
. . . use this to derive
2
E (cos d cos y)
Q s Q r s
Z 0 sin y
2 2
(iii) (a) Surge impedance load P 0 V /Z 0 500 /50:4 4960 MW P max
0
P 0 / sin y 4960/ sin 8:7 32 793 MW.
(b) P/P 0 sin d/sin y 0:47, so sin d 0:47 sin 8:7 ,i.e. d 4:07674 .
2
Q s Q r 500 ( cos 4:07674 cos 8:7 )/(50:4 sin 8:7 )
294:3 MVAr ± absorbing at both ends.
11. (i) What are the functions of reactive compensation applied to electrical trans-
mission systems?
(ii) What are the differences between passive and active compensators? Give
examples of both.
(iii) By means of a sketch showing V r /E s vs. P/P 0 , illustrate how the receiving-
end voltage of a transmission cable can be maintained within a narrow range
near 1.0 p.u. by means of switched shunt compensating devices. E s is the
sending-end voltage, P is the power transmission, and P 0 is the natural load.
(i) (a) to produce a flat voltage profile at all levels of power transmission;
(b) to improve stability by increasing the maximum transmissible power
P max ; and
(c) to provide the most economical means for meeting the reactive power
requirements.
(ii) Passive compensation fixed or switched reactors and capacitors
Active compensation continuously variable devices: e.g. thyristor-control-
led reactors, synchronous condensers, AVRs used with turbine-generators;
`FACTS'devices.
(iii)
L C
V r R
S
E s
C
Uncompensated
L
P
Uncompensated
L Uncompensated C P o
Fig. 9.6