Induction Generator - Example Application

Example Application

We must use 10 hp, 1760 r/min, 440 V, 3 phase induction motor as an asynchronous generator. Full-load current of the motor is 10 A and full-load power factor is 0.8.

Required capacitance per phase if capacitors are connected in delta:

Apparent power S = √3 E I = 1.73 * 440 * 10 = 7612 VA
Active power P = S cos θ = 7612 * 0.8 = 6090 W
Reactive power Q = = 4567 VAR

For machine to run as an asynchronous generator, capacitor bank must supply minimum 4567 / 3 phases = 1523 VAR per phase. Voltage per capacitor is 440 V because capacitors are connected in delta.

Capacitive current Ic = Q/E = 1523/440 = 3.46 A
Capacitive reactance per phase Xc = E/I = 127 Ω

Minimum capacitance per phase:

C = 1 / (2*π*f*Xc) = 1 / (2 * 3.141 * 60 * 127) = 21 microfarads.

If load also absorbs reactive power, capacitor bank must be increased in size to compensate.

Prime mover speed should be used to generate frequency of 60 Hz:

Typically, slip should be similar to full-load value when machine is running as motor, but negative (generator operation):

Slip = 1800 - 1760 = 40 rpm
Required prime mover speed N = 1800 + Slip = 1840 rpm.

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