Capacitor Calculation for Power Factor Improvement ( induction motor )

On Nov 23, 2009 by Sunil Saharan At 11:18 AM | | 0 comments »
Capacitor calculation or capacitance calculation for a capacitor bank to improve power factor is very important topic for an electrical engineer. Capacitor bank help to reduce reactive power. Here is one numerical problem on capacitor or capacitance calculation for an induction motor.
Numerical Problem : -
A single phase 220 volts, 50 Hz, motor takes a supply current of 50 Amperes at a power factor of 0.5 lagging. The motor power factor has been improved to 0.9 lagging by connecting a condensor in parallel. Calculate the capacitance of the capacitor required ?
Solution : -
Please note that current power factor is 0.5 it indicates that motor is induction motor.
Data extracted from above problem : -
Motor Type : - 1 phase, Induction motor,
Voltage Applied : - 220 volts,
Current from Supply : - 50 Amp,
Supply Frequency : - 50 Hz,
Current Power Factor =cos ¤=0.5 ( lagging ),
¤ = cos-10.5=60¤.
Active Component of current, say Ia = I * cos ¤ = 50 * 0.5 = 25 Amp,
One thing you must remember regarding power factor improvemet Improving power factor does not change active power consumed in the appliance.
Since the voltage also same so only current will be changed after power factor improvement. So only total current will be changed but active component will be same i.e. 25 Amp.
Tan ¤ = tan 60 = 1.732,
Reactive component of current, say Ir1 = I tan ¤ = 25 * 1.732 = 43.3 Amp,
Let the capacitor of C capacitance is connected in parallel with motor then,
New Power Factor = 0.9,
cos ¤1 = 0.9,
tan ¤1= 0.4843
New reactive component = I tan¤1 = 25 * 0.4843 = 12.1 Amp,
Thus, current through capacitor = 43.3 - 12.1 = 31.2 Amp,
As we know,
Ic = V/Xc = 2*pi*f*C*V,
C = 451 * 10-6 F

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