Page 574 - Handbook of Electrical Engineering

CALCULATION OF VOLT-DROP IN A CIRCUIT CONTAINING AN INDUCTION MOTOR 567

Motor feeder cable R cm = 0.00844 pu
X cm = 0.003175 pu
Motor running R mr = 5.43024 pu
X mr = 10.065 pu
Motor starting R ms = 3.8244 pu
X ms = 0.9875 pu
j) Rigorous solution
The sequence of calculations is as follows:-
• Initial conditions, using suffix ‘o’.
• Running conditions, using suffix ‘n’.
• Starting conditions, using suffix ‘s’.
• Compare the calculated voltages and find the volt-drops.
• Design comments.
k) Initial conditions
The motor starter is open and the generator terminal voltage is 1.0 per-unit.
Hence,
V go = 1.0 + j0.0pu.

Find the initial values of I c and V l ,i.e. I co and V lo , noting that I m = 0.0
At the MCC the parallel load is R ol in parallel with X ol .
Convert the parallel load into a series load of R ol1 + jX ol1 .
The formulae for this conversion are:-
2
R ol X ol
R oll = pu per phase
2 2
R ol + X ol
2
X ol R ol
X oll = pu per phase
2 2
R ol + X ol
Where
R ol = 6.2498 and X ol = j10.085
Hence,
Z oll = R oll + jX oll = 4.5156 + j2.7985 pu
The impedance seen at the SWBD is
Z ol1 + Z c = 0.00635 + 4.5156 + j(0.05446 + 2.7985)

= 4.5220 + j2.8530 pu
1.0 + j0.0
V go
Z oll = =
Z oll + Z c 4.5220 + j2.8530
= 0.1582 − j0.0998 pu
V go Z oll
V lo = = 0.9936 − j0.0080 pu,
Z oll + Z c