Power factor correction.

1. Oct 22, 2008

billmccai

1. The problem statement, all variables and given/known data

This is the circuit:

http://img129.imageshack.us/img129/3566/pfcog8.jpg [Broken]

Assume voltage is in rms and f is 50Hz.

I have to do three things.

First, calculate and compare true power vs apparent power in the load.

Second, find the value of a capacitor placed in parallel that would bring the LOAD to unity power factor.

Finally, calculate the real power in the circuit (both transmission and load impedances) with and without this capacitor, and compare them.

2. Relevant equations

3. The attempt at a solution

I didn't have a problem with the first part.

The load impedance is (8 + j3) ohms.

Power worked out to be 6212.33 + j2367.12. So true power was 6312.33 W and apparent was 6741.57

I'm having trouble with the second part. I've been told that I need to use:

Zcorrected load = Zc || (Zr + ZL)

which becomes Zcorrected load = [Zc * (Zr + ZL)] / [Zc + (Zr + ZL)]

But I'm not sure how to manipulate this to give me the value of Zc.

And I can't even start the third part until I've worked this out.

Last edited by a moderator: May 3, 2017
2. Oct 22, 2008

jhicks

find Zr + Zl first. A corrected load is one such that (Zr + Zl) || Zc has no imaginary component. Stick a capacitor on to cancel out the imaginary part added by Zr and Zl and you're good to go. Since it's two inductive loads you need to correct, you can assume it's a capacitor with impedance 1/(j*2*pi*f*C).

3. Oct 22, 2008

billmccai

I'm meant to bring the pf to unity at the load only, so i'm assuming i just forget the line impedance for now. In this case, i thought Zr + Zl just refered to (8 + j3), which is my load impedance. I don't know how to work out what value for Zc will cancel out the imaginary part in the formula Zcorrected load = [Zc * (Zr + ZL)] / [Zc + (Zr + ZL)].

Last edited: Oct 22, 2008
4. Oct 22, 2008

jhicks

You have to have the impedance have 0 phase (i.e. the voltage and current through the load have the same phase). Find the load impedance as it is now. What is the imaginary part? Set this equal to some impedance created by a capacitor with capacitance C and solve for it.

5. Oct 22, 2008

billmccai

Still not following.

My load impedance is 8 + 3j, right? I understand that I need to get rid of the imaginary part.. i just don't get how I mathematically determine the correct C value. I know it's going to be -(something)j ohms. But how to i manipulate Zcorrected load = [Zc * (Zr + ZL)] / [Zc + (Zr + ZL)] to spit out this value.

6. Oct 22, 2008

jhicks

Oh I finally see what you meant by "Zr" and "Zl". There's no such capacitor that I can tell that will satisfy that equation which implies placing it in parallel with the entire load. You have to place it either in series with the load or in parallel to the inductor. Sorry for the misunderstanding.

7. Oct 24, 2008

jhmz

this might help:
http://img516.imageshack.us/my.php?image=pfsd0.jpg
im doing the same question and am in same situation as op

My load impedence is 3 + j3
i started out using:
$$\frac{1}{Z_{L}}=\frac{1}{Z_{C}}+\frac{1}{Z_{PL}}$$
where $$Z_{PL}$$= Impedance of previous RL load which was 3+j3
and $$Z_{L}$$ is the impedance of the new load with capacitor
the frequency is 50Hz

i have got it down to:
$$Z_{L}=\frac{3+j3}{(j100\pi)(3) + (j100\pi)(j3)}$$
im not sure if this is right at all tho and where i go from here if it is correct, i cant find any similar examples in the textbook or online as other power correction questions use the apparent,reactive and power factor formulas to work out the power factor correction but the pf never has to equal 0 in these examples, which is what unity power factor is? The reactive power equation cannot be used for calculating the unity power factor as sin(0) = 0

so this equivalent impedance way seems to be the only method of calculating the size of the capacitor for power factor correction