Page 407 - Power Electronic Control in Electrical Systems

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386 Examples, problems and exercises

half is similar, with V m at one end and E r at the other. The voltages E s and E r are
both equal to E and the reactive compensator maintains V m ˆ E.
For the sending-end half of the line we can write
E 2 d
P ˆ sin
X L /2 2
2
so that 1540 ˆ 500 /24 sin (d/2), giving d/2 ˆ 8:50177 . This is the transmission angle

over each half of the line, so the total transmission angle is d ˆ 17:0 . By symmetry,

the current phasor is at right-angles to the voltage-drop phasor joining V m and E s ,
so that the power-factor angle between I and V m is f ˆ d/4 ˆ 4:25089 . The same

power-factor angle is between E s and I, but it is a lagging angle at the sending end
and leading at the mid-point. This indicates that the line inductance is absorbing
2
reactive power at both points. The reactive power absorbed by X L /2 is I X L /2/phase.
p p

Since P ˆ 3V LL I L cos f, I L ˆ 1540/( 3 500 cos 4:25089 ) ˆ 1:78314 kA,
2
so that the total reactive power absorbed is 3I X L /2 ˆ 228:93143 MVAr in each half
of the line, or 457.86286 MVAr over the whole line. Of this, 25% is absorbed at the two
ends and 50% (228.93143 MVAr) at the mid-point. However, the shunt capacitance
6
2
2
at the mid-point is generating 2 E B c /4 ˆ 2 500 (4000 10 /4) ˆ 500 MVAr.
Therefore,thenetreactivepoweratthemid-pointis500 228:93143 ˆ 271:06857 MVAr,
and this must be absorbed by the compensator. The compensator must therefore have
2
aninductivesusceptanceB g ˆ 271:06857/500 ˆ 0:00108 S,orareactanceX g ˆ 922:2
.
At the sending end, the generator absorbs 271:06857/2 ˆ 135:5 MVAr, and the same
at the receiving end.
Problem 13 ± alternative solution (ii)
We can treat the line as a `long line'and use the distributed-parameter equations
p 6
based on Z 0 , y,etc. Thus, Z 0 ˆ (X L X C )where X L ˆ 48
and X C ˆ 1/4000 10 ˆ
2 p p
250
,so Z 0 ˆ 109:545
. Also the electrical length is y ˆ (X L /X C ) ˆ (48/250) ˆ
0:43818 radians ˆ 25:10575 . Taking just one half of the line, with equal voltages

at both ends, we can use the equation
E 2 d
P ˆ y sin 2
Z 0 sin
2
2
i.e. 1540 ˆ 500 /(109:545 sin 25:10575 /2) sin (d/2), giving d ˆ 16:87 . (Note the slight

difference from the 17:0 obtained with the `lumped-parameter'method in solution

(i) above.)
We can now find the reactive power required at each end of each half of the line:
2
2
E ( cos d cos y) 500 ( cos 8:43346 cos 12:55288 )

Q s ˆ ˆ ˆ 137:46 MVAr
Z 0 sin y 109:545 sin 12:55288
with the same at the receiving end and twice this value at the mid-point, i.e.
2
274.93 MVAr. The compensator must have a susceptance equal to 274:93/500 ˆ
0:00110 S/phase, i.e. a reactance of 909:35
/phase. Again note the slight difference
from the 922:2
/phase calculated with the lumped-parameter model.
2
Note that X L is proportional to the line length, whereas X C is inversely proportional to the line length, so
p p
that Z 0 ˆ (X L X C ) ˆ (x L x C ), where X is total reactance and x is reactance per unit length.