Question

A 3-phase 50Hz 2-pole 440V (L-L rms) induction motor drive has a rated speed of 2955 rpm and rated torque of 35 Nm. This motor drives a load with a torque characteristic that increases linearly with speed such that it equals the load torque at the rated speed. It draws a current of 2.1 A(rms) per phase at the rated voltage at a power factor of 0.81 (lagging). The stator resistance Rs = 2.0 ohms. At the rated operating speed, calculate the amplitude of the phase voltage (Va,braking) needed produce a regenerative braking torque that is 0.8 times the rated torque in magnitude.

Answer 1

The torque increases linearly with speed. T=KN The load torque is equal to rated torque at rated speed. 35=K (2955) K = 0.0118 The torque-speed relation is, T =0.0118N Determine the speed at 0.8 times the rated torque. To = 0.0118N = 0.8x35 N = 2372.9 rpm Determine the power at this torque. P=T:( 200 )=(0.5)(35)( 20 * 2072.9 P=6957.7 W Write the formula for the power. P=31°R. (4+5) N -N S= N v _120f 120x50 = 3000 rpm N = P 2 =3000 rpm 3000 - 2955 SE 3000 =0.015

The stator current is, ( 440 2.1= 2+ R 0.015 R2 = 1.784 22 Substitute the values in the power equation. P=31]R, (4-) 6957.7=31] (1.784)(-0.0055 12 = 4.45 A The expression for power is also equal to, P = 3V I cos P. 6957.7 Pa to akting 31, cosø 3x 4.45x0.81 Va, trading = 643.5 V (rms)