Page 247 - Offshore Electrical Engineering Manual

234 CHAPTER 5 Conductor Sizing, Load Flow and Fault Calculation

Summing the reactances,
1
″
X = = 0.0912 pu
d
1/0.195 + 1/0.1715
1
″
X = = 0.104 pu
q
1/0.265 + 1/0.1715
1
″
X = = 0.0327 pu
o
1/0.09 + 1/0.0515
1
R n = = 2.63 pu
1/4.06 + 1/4.06
Hence,
″
″
X + X = 0.0912 + 0.104 = 0.1952 pu
d
q
X + X = 0.912 + 0.104 + 0.0327 = 0.2279 pu
″
″
d q
(
| Z 1 | = X + X + X ″ ) 2 + (3R n ) 2
″
″
d q o
2
2
Z 1 = (0.2279) + (3 × 2.03) = 6.09 pu
Therefore the system fault currents (neglecting motor contribution) are
17.7
Three ‐ phase fault current = = 16.97 kA
3 × 6.6 × 0.0912
17.7
Phase ‐ to ‐ phase fault current = = 13.74 kA
6.6 × 0.1952
3 × 17.7
Earth fault current = = 0.762 kA
6.6 × 6.09
This method of calculation provides low values which, although not suitable for
providing fault currents for the selection of switchgear, are useful for relay setting, as
neglecting motor fault contributions provides a safety margin. The extra fault current
provided by motors should significantly reduce the operating times of all overcurrent
and earth fault protection devices, provided the current transformer saturation is not
a problem. A check should be made, however, that generator cable reactances do not
reduce prospective fault currents by more than a negligible amount. This reduction is
likely to be significant for the smaller generators operating at LVs.

SWITCHBOARD FAULT CURRENTS
The prospective fault currents for medium-voltage switchboards can be obtained
in a similar manner by summing generator and transformer reactances as follows.
Take the base again as 17.7 MVA, and take as an example the generator switchboard