Page 572 - Handbook of Electrical Engineering

CALCULATION OF VOLT-DROP IN A CIRCUIT CONTAINING AN INDUCTION MOTOR 565

Similarly the phase ohmic reactance X mrp is:-

2
Phase voltage

X olp =
Phase reactive power
4000 × 4000
= = 56.32 ohms per phase
3 × 94.68 × 10 3
Convert this impedance to the motor per-unit base.

The 1.0 pu motor kVA per phase = S mrp = 199.36
The 1.0 pu motor impedance per phase = Z mrp
2

Phase voltage
Z mrp =
Phase VA
4000 × 4000
= = 26.75 ohms per phase
3 × 199.36 × 10 3
Hence the per-unit motor running resistance is R olppu :-

R mrp 30.4
R olppu = = = 1.136 pu per phase
Z mrp 26.75
And the per-unit motor running reactance is X olppu :-

X mrp 56.32
X olppu = = = 2.105 pu per phase
Z mrp 26.75
Where R olppu and X olppu are parallel components representing the motor during the full-
load running condition. Convert this impedance to the system base at the motor system voltage
of 4181.8 volts.

(base kVA) (motor rated voltage) 2
R mr + jX mr = (R mrppu + jX mrppu ) 2
(motor kVA) (system base voltage)
(1.136 + j2.105)(3125)(4000) 2
=
3
(3 × 199.36 × 10 )(4181.8) 2
= 5.4324 + j10.065 pu
h) Motor running conditions (suffix ‘s’)

Rated input VA
Rated current to each phase = √
3 × Rated motor voltage
598.08 3
P mrp = √ × 10 = 86.33 amps
3 × 4000
Starting current = 5 × Rated current = 431.63 amps