Page 578 - Handbook of Electrical Engineering

CALCULATION OF VOLT-DROP IN A CIRCUIT CONTAINING AN INDUCTION MOTOR 571

Hence the terminal voltage of the generator V gs is:-

E o Z ogs (1.0687 + j0.1068)(0.3631 + j0.6375)
V gs = =
Z gn s 0.3838 + j0.8892
= 0.8151 − j0.0087 pu, which has a magnitude of 0.8152 pu.

Similarly the voltage of the MCC V ls is:-

(0.8151 − j0.0087)(0.3050 + j0.7874)
V gs Z ols
V ls = =
Z cn S 0.3114 + j0.8418
= 0.7660 − j0.0199 pu, which has a magnitude of 0.7669 pu.
Similarly the motor voltage V ms is:-

(0.7660 − j0.0199)(0.2390 + j0.9257)
V ls Z msl
V ms = =
Z msl + Z cm 0.2390 + j0.9257 + 0.00844 + j0.003175
= 0.7626 − j0.0140 pu, which has a magnitude of 0.7627 pu.

n) Calculate the percentage volt-drops
The customary method of defining volt-drop is in percentage terms as follows:-

No-load voltage − Loaded voltage
Volt-drop in percent = × 100%
No-load voltage
Where the no-load voltage is the service voltage that exists before the change in load is
applied and the loaded voltage is the service voltage during the application of the change in
load. For example, when a motor is being started there are two aspects to consider. Firstly, the
situation at the motor terminals since this determines the ability of the motor to create enough
torque during the starting period and, secondly, at the MCC since this influences the performance
of existing loads and their contactor coils. Other parts in the power system could be examined in
a similar manner, e.g. at the generator terminals and its switchboard. The motor example above
may be used to illustrate these comments:-
• Motor terminal volt-drop in percent.
No-load voltage = pre-disturbance voltage at the MCC.
Loaded voltage = voltage at the motor terminals at starting or running of the motor.

V lo − V ms
Volt-drop at starting% = × 100%
V lo
0.9936 − 0.7627
= × 100% = 23.24%
0.9936
V slo − V mn
Volt-drop at running% = × 100%
V lo
0.9936 − 0.9552
= × 100% = 3.86%
0.9936