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.
Read more about this topic: Induction Generator
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