Page 582 - Handbook of Electrical Engineering

CALCULATION OF VOLT-DROP IN A CIRCUIT CONTAINING AN INDUCTION MOTOR 575

This compares well with I go found in the rigorous case.

E o = V go + I oo Z g = 1.0 + j0 + (0.448 − j0.2386)(0.02 + j0.25) pu
= 1.0686 + j0.1072 pu, which has a magnitude of 1.0740 pu.

Which is within 0.01% of the rigorous case.
s) Running conditions

The motor starter is closed and the generator emf is 1.0740 per-unit.
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

The total load resistance on the SWBD is R ln where:-

R o × R mn
R ln = = 1.5821 pu
R o + R mn
The total load reactance on the SWBD is X ln where:-

X o × X mn
X ln = = 2.9586 pu
X o + X mn
The series equivalent resistance is R ogn :-
2.9586 × 2.9586 × 1.5821
R ogn = = 1.2303 pu
2.9586 + 1.5821 2
2
The series equivalent reactance is X ogn :-
2.9586 × 1.5821 × 1.5821
X ogn = = 0.6579 pu
2.9586 + 1.5821 2
2
The total impedance seen by the generator emf E o is Z gn :-

Z gn = R g + R ogn + j(X g + X ogn )

= 0.02 + 1.2303 + j(0.25 + 0.6579)
= 1.2503 + j0.9079 pu
The current in the generator I gn is:-

(1.0686 − j0.1072)(1.2503 − j0.9079)
I gn = E o =
2.3875
= 0.6004 − j0.3502 pu