Parallel Capacitance - Power Factor Question

[Jason-Li]

Homework Statement

A 50 kW load operates from a 60 Hz 10 kV rms line with a power factor of 60% lagging. Determine the capacitance that must be placed in parallel with the load to achieve a 90% lagging power factor.

The Attempt at a Solution

For Z at 0.6pf
S=P/pf = 50000/0.6 = 83333.33VA
I = P/Vs = 83333.33/10000 = 8.333A
R=p/I^2 = 50000 / 8.333^2 = 720Ω
Z=R/pf = 720/0.6 = 1200Ω
Xc=√(1200^2-720^2) = 960Ω
CosΦ=0.6 Φ=53.13°
Z1=720+j960Ω or 1200∠53.13°Ω

so for Z at 0.9
CosΦ=0.9 Φ=25.842°
Z2=r/pf = 720/0.9 = 800Ω
Xc=√(800^2-720^2) = 348.711Ω
Z2 = 720+j348.711Ω or 800∠25.842°Ω

Hence Z required = Z1 - Z2 = 720+j960 - 720+j348.711Ω = 0+j611.289Ω
so Xc = 611.289Ω
then C = 1/(2πfXc) = 1 / (2π*60*611.289) = 4.339μF

My main problem with this answer is I have seen several other answers equalling 1.126 μF using other methods such as https://www.physicsforums.com/threads/capacitance-power-factor-question.591792/ however I don't understand why either method don't give the same answer? Any help would be much appreciated.

 

[magoo]

I think you started out OK with the 0.6 PF portion except that what you call Xc is actual inductive or Xl. Nonetheless, your Z1 reflects this result.

I like to keep a circuit diagram in mind, like the following

I also like to work with the power quantities rather than impedances.

So you start off with 50 kW and that doesn't change for the resistive portion.

Initial apparent power is 50/0.6 or 83.3 kVA and the reactive power is 50/0.6 * 0.8 or 66.7 kvar.When you get to the 0.9 pf situation, the resistive or 50 kW portion doesn't change so the new reactive power is 50/0.9 * 0.436 or 42.5 kvar inductive.

To achieve the reduction in kvar from 66.7 to 42.5 kvar, we add parallel capacitance. So the capacitance required must make up the difference of 22.2 kvar. With your 10 kV voltage source, what capacitance is required to give you 22.2 kvar? I'm not sure it is the same answer that you showed but I haven't checked that yet.

 

[Jason-Li]

Hi Magoo thanks for the fast response!

So Xc = V / Qc = 10000^2 / 222222 = 4500Ω

C = 1/ (2πfXc) = 1 / 2*π*60*4500 = 0.589μF

Still doesn't seem correct? Still not sure why my original answer doesn't work either if I'm honest

 

[anorlunda]

Look at @magoo's schematic.

That shows capacitance in parallel with the load resistance. You calculated the entire current as flowing through the resistor as if the load's capacitance was in series with the resistor.

S=P/pf = 50000/0.6 = 83333.33VA
I = P/Vs = 83333.33/10000 = 8.333A
R=p/I^2 = 50000 / 8.333^2 = 720Ω

 

[Jason-Li]

@anorlunda ah that makes sense as I wouldn't be able to calculate the I properly thanks!

So would my initial response be correct?

So Xc = V / Qc = 10000^2 / 22222 = 4500Ω C = 1/ (2πfXc) = 1 / 2*π*60*4500 = 0.589μF

Just referring to other sources and they have used the value of 42.5kVAr rather than 22.2kVAr for calculating Xc?

 

[gneill]

Just referring to other sources and they have used the value of 42.5kVAr rather than 22.2kVAr for calculating Xc?

I think that some values are getting mixed up. Can we agree that for the original circuit that the apparent power is correctly calculated to be $S = 83.33\;kVA$, and that the reactive power is $Q = 66.67\;kVAr$?

Then, taking the pf = 0.9 case we look to find the desired reactive power $Q_d$.

$S_d = \frac{P}{0.9} = 55.56\;kVA$ {The apparent power for the pf 0.9 case}

$Q_d = \sqrt{S_d^2 - P^2} = 24.23\;kVAr$

In the following diagram is the situation laid out in the form of power triangles:

The $\Delta Q$ is what the what the added capacitance will need to "consume".

$\Delta Q = Q - Q_d = 42.45\;kVAr$

So your "other sources" have it right.

 

[Jason-Li]

@gneill Thanks for that very thorough explanation, definitely one of those moments when it just clicks now.

So 'rerunning' the calc.

Xc = V / Qc = 10000^2 / 42450 = 2355.712Ω C = 1/ (2πfXc) = 1 / 2*π*60*2355.712 = 1.126μF

The help is much appreciated! Also thank you @magoo & @anorlunda .