Page 303 - Handbook of Electrical Engineering

FAULT CALCULATIONS AND STABILITY STUDIES 289

Hence

P load = P hm1 + P hm2 + P lm + P hs + P ls
= 90.0 MW

Q hm1 = 19.84 MVA r ,Q hm2 = 14.17 MVA r
Q lm = 3.10 MVA r ,Q hs = 13.65 MVA r

Q ls = 0.5 MVA r
Q load = (Q hm1 + Q hm2 + Q lm + Q hs + Q ls ) 1.015

= 52.03
√ 2 2
S load = (P load + Q load ) = 103.96 MVA

The operating power factor PF load is,

P load 90.0
PF load = = = 0.8657 lagging
S load 103.96
The generator MVA is S gen which equals 37.5.
Choose the base MVA to be S base = 100.
Assume all five generators are operating when the three-phase zero impedance fault occurs.

Calculate the rms symmetrical fault currents for the generators and each type of load.
a) The generators and unit transformers

5 × 1.1 × 34.0

I = = 0.0 − j9.639 pu
g
0.08
j 0.13 + 100.0
1.25
• The high voltage motors and unit transformers.
These consist of two groups S hm1 and S hm2 , let their total be S hm .

1.0 × (40.23 + 28.74)
I
hm =
j0.06
0.033 + j0.164 + 100.0
1.10
0.6897
=
0.033 + j0.164 + j0.0545
= 14.1184(0.033 − j0.2185)
= 0.4659 − j3.0849 pu

• The high voltage variable speed drive motors.
These can be ignored as sources of sub-transient current.