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## Main Question or Discussion Point

Hello,

I have a light AC fan, that I wish to correct its power factor.

According to my measurements, it has around 0.7 LAG power factor.

Here are my digital measurements:

Vsupply = 204VAC

I(load) = 0.237Amps

phi = 43 , pf = 0.7 LAG

My target power factor (pf) is to be 0.85 , and to neutralize (reasonably) the lagging power factor, I inject it with leading VAR supplied by the capacitor bank.

The supply here, have a maximum value of 230VAC , so I got capacitor with Vrated = 450VAC > Sqrt(2)*230

For the capacitance this is the formulas/algorithm I've used to get C:

Q1 = Vrms * Irms * pf1 * tan(43)

Q2 = Vrms * Irms * pf1 * tan(32) //since cos^-1(32) = ~0.85

Qcap = Q1 - Q2 = Vrms ^ 2 / Xc

Xc = Vrms ^2 / Qcap = 1 / 2 * pi * f * C //hence C is calculated

Results:

My questions are the followings:

Thanks

I have a light AC fan, that I wish to correct its power factor.

According to my measurements, it has around 0.7 LAG power factor.

Here are my digital measurements:

Vsupply = 204VAC

I(load) = 0.237Amps

phi = 43 , pf = 0.7 LAG

My target power factor (pf) is to be 0.85 , and to neutralize (reasonably) the lagging power factor, I inject it with leading VAR supplied by the capacitor bank.

The supply here, have a maximum value of 230VAC , so I got capacitor with Vrated = 450VAC > Sqrt(2)*230

For the capacitance this is the formulas/algorithm I've used to get C:

Q1 = Vrms * Irms * pf1 * tan(43)

Q2 = Vrms * Irms * pf1 * tan(32) //since cos^-1(32) = ~0.85

Qcap = Q1 - Q2 = Vrms ^ 2 / Xc

Xc = Vrms ^2 / Qcap = 1 / 2 * pi * f * C //hence C is calculated

Results:

__*Note*:__I connected two 100K 2W in series across capacitor terminals, as bleeding resistor for electrical safety**1)**I branched a 6uF 450VAC in parallel with the load, I noticed that the load drew more current of 0.4Amps where originally it draw 0.2Amps //*This is referred to over-compensation***2)**I decreased the 6uF to 4uF , still were an over-compensation, but lighter than before, and the motor drew 0.3Amps**3)**I kept on decreasing the capacitance until 1.6uF there I noticed that the motor got back drawing its original amount of current**4)**Still decreasing, until 0.5uF what I noticed is that the current is as was at original ~0.237Amps and no improvement in reducing this current, not reducing the phase angle and pf improvementMy questions are the followings:

**A)**Is power factor correction applicable for such a small 60W load?**B)**What is the value of capacitance C to improve its power factor?**C)**After revising formulas, why I am not having an answer that intersect with real hardware working?Thanks