
Power Distribution 199
SOLUTION
Letter code G motor draws 5.6 to 6.3 kVA/hp on starting as per Table5.3. Therefore,
the worst-case starting current is 6.3 × 100 × 1000 ÷ (√3 × 460) = 791 A at θ =
cos
–1
0.35 = 69.5° lagging. The phase voltage at the source is 480 ÷ √3 = 277.1 V.
Using exact Equation (3.13), the voltage at motor terminals is
277.1∠ 0° – 791∠ –69.5° × (0.01 + j 0.02) = 259.4∠ 0.2° V/ph
Voltage drop magnitude = (277.1 – 259.4) = 17.7 V/ph
% V
drop
= (17.7 ÷ 277.1) × 100 = 6.39 %.
By approximate Equation (3.15), V
drop
= 791 (0.01 × 0.35 + 0.02 × 0.937) =
17.6 V/ph, a close match with the exact value of 17.7 V.
By approximate Equation (8. ...