Question

(a) Sketch a graph of starting current, and torque against the speed of rotation for a single cage motor. (6)

(b) A 230 V motor, which normally develops 10kW at 1000 rev/min with an efficiency of 85%, is to be used as a generator. The armature resistance is 0.15 Ohm and the shunt field resistance is 220 Ohm. If it is driven at 1080 rev/min and the field current is adjusted to 1.1A by means of the shunt regulator what output in kW could be expected as a generator, if the armature copper loss was kept down to that when running as a motor. (10)

Answer 1

Part (a)

Graph of starting current and torque against speed for a single-cage induction motor:

Starting current: at standstill (speed = 0) the starting current is high (5 to 8 times full-load current). As the motor accelerates the current falls, and at synchronous speed it would be zero (in practice small no-load current). The current curve falls from a high value at zero speed to a low value near synchronous speed.
Torque: at standstill the starting torque is moderate (about 1.5 to 2 times full-load torque). As speed increases the torque rises to a maximum (pull-out torque) at a speed corresponding to the slip for maximum torque, then falls to zero at synchronous speed. The torque-speed curve rises from the starting value, peaks, then drops to zero at synchronous speed.
The two curves are plotted against speed from 0 to synchronous speed.

Part (b)

230 V motor, 10 kW at 1000 rev/min, efficiency 85%, used as a generator:

As a motor: input power = 10/0.85 = 11.765 kW. Line current = 11765/230 = 51.15 A.
Shunt field current (motor) = 230/220 = 1.045 A. Armature current (motor) = 51.15 - 1.045 = 50.1 A.
Armature copper loss (motor) = Ia^2 Ra = 50.1^2 x 0.15 = 2510 x 0.15 = 376.5 W.
Back e.m.f. (motor) E = V - Ia Ra = 230 - 50.1 x 0.15 = 230 - 7.5 = 222.5 V.
As a generator driven at 1080 rev/min with field current 1.1 A:
E.m.f. is proportional to speed and flux. Flux is proportional to field current (assumed linear). E_g = E_m x (1080/1000) x (1.1/1.045) = 222.5 x 1.08 x 1.0526 = 252.9 V.
Armature copper loss kept the same as when running as a motor (376.5 W): Ia^2 x 0.15 = 376.5, so Ia = sqrt(376.5/0.15) = sqrt(2510) = 50.1 A.
Terminal voltage of generator V = E_g - Ia Ra = 252.9 - 50.1 x 0.15 = 252.9 - 7.5 = 245.4 V.
Load current = Ia - field current = 50.1 - 1.1 = 49.0 A.
Output power = V x I_load = 245.4 x 49.0 = 12.02 kW.

So the expected generator output is about 12 kW.

✓ Verified Model Answer (Text Solution)

Graph of starting current and torque against speed for a single-cage induction motor:

230 V motor, 10 kW at 1000 rev/min, efficiency 85%, used as a generator: