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Power electronic control in electrical systems 97
since E s V r . This means that the generator at the sending end behaves exactly like
the shunt reactor at the receiving end in that both absorb the same amount of
reactive power:
E 2 E 2 1 cos y
Q s Q r s s (3:35)
X Z 0 sin y
The charging current divides equally between the two halves of the line. The voltage
profile is symmetrical about the mid-point, and is shown in Figure 3.12 together with
the line-current profile. In the left half of the line the charging current is negative;
at the mid-point it is zero; and in the right half it is positive. The maximum voltage
occurs at the mid-point and is given by equation (3.2) with x a/2
y Z 0 y E s
V m V r cos sin (3:36)
2 X 2 cos (y=2)
Note that V m is in phase with E s and V r , as is the voltage at all points along the line.
For a 300 km line at 50 Hz, y 18 and with E s V 0 1:0p:u:, the mid-point
voltage is 1.0125 p.u. and the reactive power absorbed at each end is 0:158 P 0 . These
values should be compared with the receiving-end voltage of 1.05 p.u. and the
sending-end reactive-power absorption of Q s 0:329 P 0 in the absence of the reac-
tor. For continuous duty at no-load with a line±line voltage of 500 kV, the rating of
the shunt reactor would be 53 MVAr per phase, if Z 0 250
.
Equation (3.36) shows that at no-load the line behaves like a symmetrical line as
though it were two separate open-circuited lines connected back-to-back and joined
at the mid-point. The open-circuit voltage rise on each half is given by equation (3.36)
which is consistent with equation (3.21) when P 0.
3.5.1 Multiple shunt reactors along a long line
The analysis of Figure 3.12 can be generalized to deal with a line divided into n
sections by n 1 shunt reactors spaced at equal intervals, with a shunt reactor at each
end. The voltage and current profiles in Figure 3.12 could be reproduced in every
section if it were of length a and the terminal conditions were the same. The terminal
voltages at the ends of the section shown in Figure 3.12 are equal in magnitude and
phase. The currents are equal but opposite in phase. The correct conditions could,
therefore, be achieved by connecting shunt reactors of half the reactance given by
equation (3.32) at every junction between two sections, as shown in Figure 3.13. The
shunt reactors at the ends of the line are each of twice the reactance of the inter-
mediate ones. If a is the total length of the composite line, replacing a by a/n in
equation (3.32) gives the required reactance of each intermediate reactor
Z 0 sin (y=n)
X (3:37)
2 1 cos (y=n)
In Figure 3.13, the sending-end generator supplies no reactive current. In practice it
would supply, to a first approximation, only the losses. If the sending-end reactor was
removed, the generator would have to absorb the reactive current from the nearest
half of the leftmost section. Each intermediate reactor absorbs the line-charging
