Handbook of Electrical Engineering

FAULT CALCULATIONS AND STABILITY STUDIES 279

transformers is simply the arithmetic sum of their individual ratingsSti.

Ste=

∑n

i= 1

Sti

The equivalent impedanceZteof the transformers may be found from,

Zte=

Ste

∑n

i= 1

Sti

Zti

11.5.2.6 Worked example

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

Transformer No. 1 Sti=10 MVA

Zti= 0. 008 +j0.09 pu

Transformer No. 2 St 2 =15 MVA

Zt 2 = 0. 009 +j0.1 pu

Transformer No. 3 St 3 =25 MVA

Zt 3 = 0. 01 +j0.12 pu

The total capacity Ste= 10. 0 + 15. 0 + 25. 0

= 50 .0MVA

∑n

i= 1

Sti

Zti

=

10. 0

0. 008 +j 0. 09

+

15. 0

0. 009 +j 0 + 1

+

25. 0

0. 01 +j 0. 12

= 40. 432 −j 465. 93

Zte=

50. 0

40. 432 −j 465. 93

= 0. 0092 +j0.1065 pu

11.6 Calculate the Sub-Transient symmetrical RMS Fault Current Contributions

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 beSbase.