Handbook of Electrical Engineering

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

Similarly the phase ohmic reactanceXmrpis:-

Xolp=

(

Phase voltage

Phase reactive power

) 2

=

4000 × 4000

3 × 94. 68 × 103

= 56 .32 ohms per phase

Convert this impedance to the motor per-unit base.

The 1.0 pu motor kVA per phase=Smrp= 199. 36

The 1.0 pu motor impedance per phase=Zmrp

Zmrp=

(

Phase voltage

Phase VA

) 2

=

4000 × 4000

3 × 199. 36 × 103

= 26 .75 ohms per phase

Hence the per-unit motor running resistance isRolppu:-

Rolppu=

Rmrp

Zmrp

=

30. 4

26. 75

= 1 .136 pu per phase

And the per-unit motor running reactance isXolppu:-

Xolppu=

Xmrp

Zmrp

=

56. 32

26. 75

= 2 .105 pu per phase

WhereRolppuandXolppuare 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.

Rmr+jXmr=(Rmrppu+jXmrppu)

(base kVA) (motor rated voltage)^2

(motor kVA) (system base voltage)^2

=

( 1. 136 +j 2. 105 )( 3125 )( 4000 )^2

( 3 × 199. 36 × 103 )( 4181. 8 )^2

= 5. 4324 +j 10 .065 pu

h) Motor running conditions (suffix ‘s’)

Rated current to each phase=

Rated input VA

√

3 ×Rated motor voltage

Pmrp=

598. 08

√

3 × 4000

× 103 = 86 .33 amps

Starting current= 5 ×Rated current= 431 .63 amps