Power Factor Correction - Single Phase Induction Motors

Discussion in 'Homework Help' started by Zaraphrax, Oct 1, 2009.

1. Zaraphrax Thread Starter Active Member

Mar 21, 2009
47
3
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.

2. t_n_k AAC Fanatic!

Mar 6, 2009
5,448
790
I have C(total) = 455.63uF

Method used:

Original System - uncompensated

Total Power = 75 kW @ pf=0.75 lagging
[phase angle = -acos(0.75) = -41.41°]

Total Reactive Power = 75*tan(-41.4°) = -66.1438kVar

Compensated System

If the new pf is 0.8 then new phase angle = -acos(0.8) = -36.87°

The resultant Total Reactive Power = 75*tan(-36.87°) = -56.25kVar

Change in Reactive Power = +9.84kVar

Compensating (Capacitive) Reactive Power required = 9.894 kVar = ωCV^2

Therefor C =9.894 x 10^3/(2*pi*60*(240)^2)= 455.63uF
(or 113.91uF per motor)

Hopefully someone else will provide a third opinion which matches either of our answers.

I think you may have gone astray in the motor current calculation ....

If the total real power is 75 kW and the power factor is 0.75 then the apparent power is 100 kVA. Each motor then draws 25 kVA which equates to a motor current of 104.167 Amps at 240 V.

3. Zaraphrax Thread Starter Active Member

Mar 21, 2009
47
3
I think you might be right actually (not enough coffee for me today obviously). That'd seem to make more sense than what I had. Let's see if someone else can confirm. Thanks for response.