Page 312 - Handbook of Electrical Engineering

298 HANDBOOK OF ELECTRICAL ENGINEERING

In order to reduce the total current to 30,000 amps, each generator needs to contribute 6000 amps.
Therefore the following condition must be satisfied,

E I 1 1.1 × 1049.8
I fg = = = 6000 amps
V(X d + X r ) 1.0 × (0.15 + X r )

Transposing gives,
1.1 × 1049.8
X r = − 0.15 = 0.0425 pu
1.0 × 6000
b) Case B: Star-connected reactors
From Figure 11.12 it can be seen that the left-hand side and right-hand pairs of generators con-
tribute the same amount of fault current, because the system is symmetrical. The combined

impedance of a pair of generators is X /2. The combined impedance of two pairs of generator
d
and their shared reactors is,

X d X r
Z pairs = +
4 2
The fault current contributed by the four outer generators is,

E I 1
I fg4 =
VZ pairs
The contribution from the centre generator is 7698.2 as found in a).
The total fault current is again 30,000 amps and is given by,

E I 1 1
I f 5 = + 7698.2 = 30000.0
V Z pairs + X r
 
1

= 1.1 × 1049.8   + 7698.2 amps
1.0  0.15 X r
+ + X r
4 2
Transposing gives,
X r = 0.00952 pu

c) Case C: Delta-connected reactors
This case is similar to the star case of b) and the equivalent delta reactance values are simply
three times those of the star reactance values.

Hence, X r = 0.02856 pu.
It is interesting to note that in this case there will be no current flowing in the reactor
that couples the two outer busbars. However, this reactor cannot be omitted because it serves its
purpose when faults occur at the outer switchboards.
d) Comparison of cases
As a rough estimate it may be assumed that the cost of a reactor is directly proportional to its
current rating and its value of reactance.