Page 575 - Handbook of Electrical Engineering

568 HANDBOOK OF ELECTRICAL ENGINEERING

which has a magnitude of 0.9936 pu.

Find the initial emf, E o , of the generator.
At the SWBD the parallel load is R og in parallel with X og .
Convert the parallel load into a series load of R ogl + jX ogl .

R og = 3.4722 pu and X og = j7.1696 pu,

hence Z ogl is:-
Z ogl = R ogl + jX ogl = 2.8125 + j1.3622 pu.

The initial load current I ogo is:-

(1.0 + j0.0)(2.8125 − j1.3621)
V go
I ogo = =
2
Z ogl 2.8125 + 1.3621 2
= 0.288 − j0.1395 pu.
The total initial generator current I go is:-

I go = I ogo + I co = 0.1582 − j0.0998 + 0.288 − j0.1395
= 0.4462 − j0.2393 pu.

Hence,

E o = V go + I go Z g = 1.0 + j0 + (0.4461 − j0.2393)(0.02 + j0.25) pu
= 1.0687 + j0.1068 pu, which has a magnitude of 1.0741 pu.

l) Running conditions
The motor starter is closed and the generator emf is 1.0741 per-unit.
Assume the rated impedance for the motor since this will give a worst-case running
impedance for it. (The 500 kW motor will be over-sized in any case with respect to the driven
machine by about 10% and so the actual impedance will be about 10% higher than the rated
impedance.)
The parallel impedance of the running motor is Z mn :-

R mn = 5.4324 and X mn = j10.065 pu

The series impedance of the running motor is Z mnl :-

Z mnl = R mnl + jX mnl = 4.2069 + j2.2706 pu

Now add the feeder cable impedance in series to obtain the total series impedance between
the MCC and the motor. Call this total impedance Z mnlc .