Hi there,
I have a circuit consisting of four (single phase) induction motors. Each motor develops 15kW @ 80% efficiency. Operating at 0.75 lagging power factor. The total setup is supplied 240V @ 60 Hz.
I need to work out how to correct the power factor to 0.8 lagging (total supply) using a shunt capacitor. I'm having a bit of trouble, but I've given it a go and I'd just like to check my answer.
Power out = 15kW
Total Power out = 15 * 4 = 60 kW
Power in = 15/0.8 = 18.75kW
Total input power = 18.75 *4 = 75kW
Current drawn by each motor: 18.75kW/(200 * 0.8) = 117.2 A
Supply current = 117.2 * 4 = 468.8 A
Before Power Factor Correction:
Motor power factor = 0.75 lagging
Total Active Power = VI*0.75 = 240 * 468.8 * 0.75 = 84.4 kW
Total Reactive Power = VI sin 41.4 = 240 * 468.8 * sin 414.4 = 74.4 kvar
After Power Factor Improvement:
Connection a cap doesn't change the P = 84.4 kW
New power factor (lets call this phi') = 0.8 lagging
Total reactive power = Ptanphi' = 84.4 *tan 36.86 = 63.2 var
Cap reactive power Qc = 63.2 - 74400 = -74336.8 var
Thus, Xc = V^2 / |QC| = 240^2 / 74.336.8 = 0.775 ohms
C = (2*pi*f*Xc)^-1 = 1/(2 * pi * 60 * 0.775) = 0.0034 F = 3.4 mF
So, for each motor, adding 3.4/4 = 0.85 mF of capacitance across the motor will correct the power factor to 0.8 lagging.
Feedback is appreciated, thanks.
I have a circuit consisting of four (single phase) induction motors. Each motor develops 15kW @ 80% efficiency. Operating at 0.75 lagging power factor. The total setup is supplied 240V @ 60 Hz.
I need to work out how to correct the power factor to 0.8 lagging (total supply) using a shunt capacitor. I'm having a bit of trouble, but I've given it a go and I'd just like to check my answer.
Power out = 15kW
Total Power out = 15 * 4 = 60 kW
Power in = 15/0.8 = 18.75kW
Total input power = 18.75 *4 = 75kW
Current drawn by each motor: 18.75kW/(200 * 0.8) = 117.2 A
Supply current = 117.2 * 4 = 468.8 A
Before Power Factor Correction:
Motor power factor = 0.75 lagging
Total Active Power = VI*0.75 = 240 * 468.8 * 0.75 = 84.4 kW
Total Reactive Power = VI sin 41.4 = 240 * 468.8 * sin 414.4 = 74.4 kvar
After Power Factor Improvement:
Connection a cap doesn't change the P = 84.4 kW
New power factor (lets call this phi') = 0.8 lagging
Total reactive power = Ptanphi' = 84.4 *tan 36.86 = 63.2 var
Cap reactive power Qc = 63.2 - 74400 = -74336.8 var
Thus, Xc = V^2 / |QC| = 240^2 / 74.336.8 = 0.775 ohms
C = (2*pi*f*Xc)^-1 = 1/(2 * pi * 60 * 0.775) = 0.0034 F = 3.4 mF
So, for each motor, adding 3.4/4 = 0.85 mF of capacitance across the motor will correct the power factor to 0.8 lagging.
Feedback is appreciated, thanks.