Page 294 - Handbook of Electrical Engineering

280 HANDBOOK OF ELECTRICAL ENGINEERING

It is customary to assume that all the generators are operating and that they are heavily loaded.

In which case the emf E behind the sub-transient reactance X is about 5 to 10% above the rated
g
d
terminal voltage, hence assume E is 1.1 pu. This emf drives the fault current around the circuit. In

g
IEC60909 the elevation in driving emf, or voltage, is given in Table I as ‘factor c’ and discussed in
Clause 6 therein.
The contribution of fault current I from the generators is,

g
E
g pu
g
I =
X
d X tg
+ S base
S gen S tg
1.1
=
X 0.08
d
+ S base
S gen K tg S gen
1.1S gen
(11.1)
g
I = 0.08

X +
d S base
K tg
The contribution from the high voltage motors is found as follows.
It may be assumed that the average ratio of starting current to rated current (I s /I n )of the
motor is,
I s
= 6.0 pu for high voltage motors
I n
Consequently the sub-transient impedance Z of the motors is,
hm
1.0
Z hm = = 0.167 pu (at S hm )
6.0
For typical high voltage motors the starting power factor is between 0.15 and 0.2 lagging,
hence assume 0.2. The sub-transient impedance becomes,

Z hm = 0.033 + j0.164 pu
The equivalent impedance Z tm of the motor unit transformers is 0.06 pu at a total capacity
of S tm .
Z tm = 0.0 + j0.06 pu
The emf E behind the motor sub-transient impedance is the air-gap emf and will in practice
hm
be slightly less than 1.0 pu, hence it is reasonable and conservative to assume it to be 1.0 pu.
The contribution of fault current I from the main switchboard motors is
hm

E hm
I pu
hm =
Z hm Z tm
+ S base
S hm K tm S hm
1.0S hm
(11.2)
=
j0.06
0.033 + j0.164 + S base
K tm