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),

V

√

3 × 1170. 0 ( 0. 0296 × 0. 35 + 0. 0108 × 0. 9368 )× 100

400

= 506. 625 ( 0. 01036 + 0. 01012 )

= 10 .374%

Therefore,

Vr

400

√

3

( 1. 0 − 0. 10374 )

= 206 .98 volts/phase

From (9.2),

(OA+AB+BC)^2 =( 206. 98 + 12. 121 + 11. 837 )^2

= 230. 9392 = 53333. 05

And

(DF−BE)^2 =( 4. 423 − 32. 443 )^2

= 28. 022 = 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.