Page 570 - Handbook of Electrical Engineering

CALCULATION OF VOLT-DROP IN A CIRCUIT CONTAINING AN INDUCTION MOTOR 563

Convert to system base values.

MCC load power
MCC load kVA = S ol =
MCC load power factor
P ol 500 × 1000 3
= = = 588.235 × 10 VA
cos φ ol 0.85
588.235 × 1000 3
MCC load kVA/phase = S olp = = 196.078 × 10 VA
3
500 × 1000 3
MCC load power/phase = P olp = = 166.667 × 10 VA
3
√ 2 2
MCC load reactive power/phase = Q olp = (S olp − S olp )
3
= 103.29 × 10 VAr
2
Phase voltage

Ohmic resistance per phase = R olp = =
Phase active power
2

V ol 4160 × 4160 6
= = × 10
P olp 3 × 166.67 × 10 3
= 34.61 ohms per phase
2
Phase voltage
Ohmic reactance per phase = X olp =
Phase reactive power
2

V ol 41, 600 × 4160 6
= = 3 × 10
Q olp 3 × 103.29 × 10
= 55.848 ohms per phase
Convert to the system base impedance values.

(load base kVA) (system base kVA)
System impedance in per-unit = Ohmic impedance = 2
(system base voltage) (load base kVA)

Hence
(34.61 × 588.235 × 1000)(3125 × 1000)
R ol = = 6.2498 pu
2
(4160) (588.235 × 1000)
and
(55.848 × 588.235 × 1000)(3125 × 1000)
X ol = = 10.085 pu
2
(4160) (588.235 × 1000)
These are the parallel elements of the load in per-unit at the system base.
f) Motor feeder cable. Convert to system base.

2
A70 mm three-core 5 kV cable has an ohmic impedance of 0.343 + j0.129 ohms per kilo-
metre per phase and a current rating of 250 A. Hence the total ohmic impedance is 0.5145 +
j0.1935 ohms per phase.