Page 568 - Handbook of Electrical Engineering

CALCULATION OF VOLT-DROP IN A CIRCUIT CONTAINING AN INDUCTION MOTOR 561

iv) Motor control centre data.
Rated voltage = 4.16 kV
Standing load = 500 kW at 0.85 PF lagging (cos φ ol ).

v) Motor feeder cable
Conductor size = 70 mm 2
◦
Conductor temperature = 75 C
Specific resistance R km = 0.343 ohms per km
Specific reactance X km = 0.129 ohms per km
Rated voltage = 5kV
Rated frequency = 60 Hz
Route length = 1500 m

vi) Motor
Rated voltage = 4kV
Rated efficiency = 95%
Rated power factor = 0.88 lagging
Starting current = 5 times the full-load current
Starting power factor = 0.25 lagging

Convert the data to the system base values.

a) For convenience choose the system base kVA and voltages to be:-

System base kVA = Generator kVA = 3125
System base voltage at the switchboard = 13, 800 volts
System base voltage at the MCC = 13, 800 × transformer ratio
4000
= 13, 800 × = 4181.8 volts,
13, 200
b) The system base value of the generator impedance R g + jX g is the same as that for the generator
kVA base.

R g + jX g = R a + jX d = 0.02 + j0.25 pu

c) Convert the transformer impedance to the system base values.

(base kVA) (trans pri voltage) 2
R c + jX c = (R pu + jX pu ) 2
(trans kVA) (base pri voltage)
(3125)(13, 200) 2
= (0.7 + j6.0) = 0.00635 + j0.05446 pu
(3150)(13, 800) 2

d) Switchboard (SWBD) parallel circuit components.