Page 293 - Handbook of Electrical Engineering

FAULT CALCULATIONS AND STABILITY STUDIES 279

transformers is simply the arithmetic sum of their individual ratings S ti .

n

S te = S ti
i=1

The equivalent impedance Z te of the transformers may be found from,

S te
Z te =
n
S ti

Z ti
i=1

11.5.2.6 Worked example
Three transformers feed a load from a main switchboard. Their ratings and impedances are,

Transformer No. 1 S ti = 10 MVA

Z ti = 0.008 + j0.09 pu
Transformer No. 2 S t2 = 15 MVA
Z t2 = 0.009 + j0.1 pu
Transformer No. 3 S t3 = 25 MVA
Z t3 = 0.01 + j0.12 pu

The total capacity S te = 10.0 + 15.0 + 25.0
= 50.0MVA
n
S ti
10.0 15.0 25.0
= + +
Z ti 0.008 + j0.09 0.009 + j0 + 1 0.01 + j0.12
i=1
= 40.432 − j465.93
50.0
Z te = = 0.0092 + j0.1065 pu
40.432 − j465.93

11.6 CALCULATE THE SUB-TRANSIENT SYMMETRICAL RMS FAULT
CURRENT CONTRIBUTIONS

The method adopted below is based upon the principles set out in IEC60363 and IEC60909, both
of which describe how to calculate sub-transient and transient fault currents, and are well suited to
oil industry power systems. The method will use the per-unit system of parameters and variables.
Choose the base MVA to be S base .