Page 228 - Handbook of Electrical Engineering

212 HANDBOOK OF ELECTRICAL ENGINEERING

The motor.

The starting current is:-

I = 6.5 × 180.0 = 1170.0amps

The power factor is:-
cos Ø = 0.35, therefore sin Ø = 0.9368

Solution:
From (9.1), assume the sending voltage is constant at 400 volts.

AB = 1170.0 × 0.0296 × 0.3500 = 12.121 volts/phase
BE = 1170.0 × 0.0296 × 0.9368 = 32.443 volts/phase
EF = 1170.0 × 0.0108 × 0.9368 = 11.837 volts/phase
DF = 1170.0 × 0.0108 × 0.3500 = 4.423 volts/phase

From (9.3),
√
3 × 1170.0(0.0296 × 0.35 + 0.0108 × 0.9368) × 100
V
400
= 506.625(0.01036 + 0.01012)
= 10.374%

Therefore,
400
V r √ (1.0 − 0.10374)
3
= 206.98 volts/phase

From (9.2),
2
(OA + AB + BC) = (206.98 + 12.121 + 11.837) 2
2
= 230.939 = 53333.05
And
2
(DF − BE) = (4.423 − 32.443) 2
2
= 28.02 = 785.12

Hence the inequality in (9.2) is valid and the solution is accurate.

Since the volt-drop is less than 20% the motor will accelerate to full speed without difficulty.