# Power Factor Correction from 0.6 to 0.9

Thread Starter

#### Fnordgasm5

Joined Mar 2, 2013
10
Here's my problem;

I am given values for the AC voltage, AC frequency, the load in kW, and a lagging power factor of 0.6 and I am told to calculate the value of capacitor I would need to add to change the power factor to 0.9. I'm following the steps below but when I use the values I come up with to check my work it comes out wrong and I don't know what I'm doing wrong.

Here's what I'm doing;

1) I am assuming since the value of the load is given in kW and not VA that the value is real power (P). Using this and the power factor I am calculating apparent power S, reactive power Q, and the power factor angle.

2) I am assuming as the real power P is a component of the resistive inductance of the load that it will not change when a capacitor is added.

3) I am using the required power factor value 0.9 to calculate the power factor angle and using that along with the real power P value to calculate new values for apparent power (S2) and reactive power (Q2).

4) I am subtracting the new reactive power Q2 from Q1 to obtain the reactive power (Q3) that the capacitor will need to provide.

5) Using Q3 I am then calculating the impedance and then the capacitance of the capacitor placed in parallel.

6) As a sanity check I am calculating the impedance of the load and using that value to calculate total P, S, and R of the circuit and I get completely different results.

My suspicion is that I'm messing up my sanity check somewhere but I'm really not sure.

Any help would be most appreciated.

#### drc_567

Joined Dec 29, 2008
1,107
... not quite following your procedure completely. However, the one thing that jumps out is the placement of the compensating capacitor in parallel with the inductance component. Try using the capacitance in series. The reactive VARS are additive, in an algebraic sense.
... See if that improves the results. These problems start to make sense once you realize that the calculations involving both watts and vars require trigonometric considerations, while VARS only and watts only are just additive.
Power Factors

Thread Starter

#### Fnordgasm5

Joined Mar 2, 2013
10
The problem does specify that the capacitor is placed in parallel with the load. As far as I know the value of the reactive VAR required shouldn't change if the capacitor is in series or parallel but the inductance, and therefore the capacitance, will.

To be fair, I have been kinda sparse with the details so to be clear I understand that I am calculating the apparent power as a vector in polar form (S at an angle of θ) and then converting it to rectangular form ( S= P +jQ ) for ease of addition and subtraction.

#### drc_567

Joined Dec 29, 2008
1,107
... The link shown here has an answer given for the parallel capacitor,
( paragraph e ), but the method used to get that number is not shown.
A similar problem

Thread Starter

#### Fnordgasm5

Joined Mar 2, 2013
10
I have identified where I was going wrong. I wasn't calculating the total impedance of parallel elements correctly during my sanity checks correctly.

Thanks for your help.

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