Page 576 - Handbook of Electrical Engineering

CALCULATION OF VOLT-DROP IN A CIRCUIT CONTAINING AN INDUCTION MOTOR 569

Z mnlc = R mnlc + jX mnlc = R mnlc + R cm + j(X mnlc + X cm )

= 0.00844 + 4.2069 + j(0.003175 + 2.2706)
= 4.2153 + j2.2738 pu
The total load on the MCC consists of the static load Z ol1 (series components) in parallel with
the cable and motor Z mnlc (series components). The total impedance Z oln is therefore:-

Z oll × Z mnlc
Z oln = R oln + jX oln = = 2.1828 + j1.2590 pu
Z oll + Z mnlc
The impedance seen at the SWBD for the cable, motor and MCC load is Z cn :-

Z cn = Z oln + Z c = 2.1828 + j1.2590 + 0.00635 + j0.05446

= 2.1891 + j1.3135 pu
This impedance is in parallel with that of the local load Z og on the SWBD. The total equivalent
load on SWBD is Z ogn where:-

Z ogl × Z cn
Z ogn = R ogn + jX ogn = = 1.2341 + j0.6746 pu
Z ogl + Z cn
Hence the total impedance seen by the generator emf E o is Z gn :-

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

= 0.02 + 1.2341 + j(0.25 + 0.6746)
= 1.2541 + j0.9246 pu
The current in the generator I gn is:-

1.0687 − j0.1068
E o
I gn = = = 0.5928 − j0.3519 pu
Z gn 1.2541 + j0.9246
Hence the terminal voltage of the generator V gn is:-

E o Z ogn (1.0687 + j0.1068)(1.2341 + j0.6746)
V gn = =
Z gn 1.2541 + j0.9246
= 0.9689 − j0.0344 pu, which has a magnitude of 0.9695 pu.
Similarly the voltage of the MCC V ln is:-

(0.9689 − j0.0344)(2.1828 + j1.259)
V gn Z oln
V ln = =
Z cn 2.1891 + j1.3135
= 0.9556 − j0.0504 pu, which has a magnitude of 0.9570 pu.
m) Starting conditions
The motor starter is closed. Repeat the procedure as for l) the running conditions, but with the
starting impedance using the suffix ‘s’ for starting.