Handbook of Electrical Engineering

276 HANDBOOK OF ELECTRICAL ENGINEERING

11.5.2 Calculation of fault current – rms symmetrical values

From sub-section 11.5.1 the emfE(E′′,E′orE) and appropriate reactanceX(Xd′′,X′dorXsd)are
known. Hence the symmetrical fault currentIfmay be easily calculated:

If=

E

X

per unit

For example:

A 6600 V, 4.13 MVA generator hasX′′d= 15 .5%,X′d= 23 .5% andXsd=205%

At full load with a power factor of 0.8 lagging the corresponding emfs are therefore:

E′′= 1 .1pu,E′= 1 .156 pu and E = 2 .77 pu

The rms fault currents are therefore:

If′′=

1. 1

0. 155

= 7 .097 pu (2564 amps)

If′ =

1. 156

0235

= 4 .919 pu (1776 amps)

If=

2. 77

2. 05

= 1 .351 pu (488 amps)

A typical oil industry power system can be approximated as shown in Figure 11.6. The major-
ity of oil industry systems are of the radial distribution type, with feeders radiating away from a

Figure 11.6 One-line diagram of an equivalent power system that has its own dedicated generators.