# Power Factor correction

1. May 25, 2013

### topcat123

The problem statement

Using the nomogram of FIGURE 10, calculate the size of
capacitor bank to give a final power factor of 0.95 for a single phase,
230 V, 50 Hz lighting installation which has a load current of 17.5 A and
operates at 0.8 pf.

The nonogram gives a value k = 0.44

Relevant equations

Qc = Pk

Xc = 1/2pifC

P = VI

The attempt at a solution

I have come up with two ways but i think the first is wrong as the current though the capasitor (If in parrallel) will not be the same as the load current.

Attempt 1

P = 230 x 17.5 = 4025W

Qc = 4.03 x 0.44 = 1.77 KVAr

P = I^2 x Xc

Xc = P/I^2 = 1.77 x 1000 x 17.5^2 = 5.783Ω

so C would come out at 550μF (as I said I think this is wrong)

Attempt 2

Power still = 4025W and Qc = 1.77 KVAr

but P = V^2/Xc so Xc = V^2/P

Xc = 230^2 x 1.77 x 1000 = 18.38Ω

so C would work out at 173μF

Just a bit of guidance would be appreciated.

2. May 25, 2013

### Staff: Mentor

Hi topcat123, Welcome to Physics Forums.

While I'm not that familiar with the usual methodologies used in Power Systems (like the nomogram you mention), I can suggest a more generic approach.

Since you are given the current magnitude and the power factor you can write the current phasor (complex or polar form). Knowing the voltage too, you can write the impedance or admittance of the load (Y = 1/Z = I/V).

Recalling that admittances add in parallel, you can find the admittance (of a capacitor) to add in order to bring the pf from 0.8 to 0.95; It's just a matter of reducing the imaginary component of the total admittance to the desired value, and capacitors have negative imaginary admittance...

3. May 26, 2013

### rude man

Ahem ....

4. May 26, 2013

### Staff: Mentor

Hah! Well caught. Dunno why that brain cell fired

5. May 27, 2013

### topcat123

So draw a phase diagram for 0.8 pf.
Work out the restive current.
Overlay the new 0.95 pf and work out the change in the reactive component.
from that i con work out the capacitor size.

Doing it this way i get 123μF

before i got 173μF

6. May 27, 2013

### Staff: Mentor

I think it would be better to work with admittances, since they add in parallel (presumably the capacitor bank will be placed in parallel with the load). I'm seeing a smaller value of capacitance for the result.

7. May 27, 2013

### topcat123

Ok

Z=V/I = 230/17.5 = 13.14Ω

cos θ = 0.8 so R = 0.8 x 13.14 = 10.512Ω

Xl = √Z^2 - R^2 = 62.16Ω

New pf = 0.95 = cos θ
θ = 18.19°
R tan 18.19° = Xl-Xc =3.455Ω

There for Xc = 58.71Ω

C=1/2πfXc = 54.22μF

8. May 27, 2013

### Staff: Mentor

Check your calculation of Xl. 62.16Ω looks too large.

9. May 27, 2013

### topcat123

yes Xl should be 7.88Ω forgot to √it.

So Xl-Xc = 3.455Ω
(1/3.455)-(1/7.88) = 1/Xc

Xc = 6.152Ω

C=1/2πfXc = 1/(2π x 50 x 6.152) = 0.000517F

Is this better?

10. May 27, 2013

### Staff: Mentor

One problem, while it is true that admittance is the inverse of impedance, Y = 1/Z, it is not true that the imaginary part of the admittance is equal to the inverse of the imaginary part of the impedance. That is, Im(Y) ≠ 1/Im(Z).

Start by finding Y = 1/Z. Then work with its components; you want to reduce its polar angle in the same fashion that you're trying to do for the impedance.

11. May 27, 2013

### topcat123

Z = V/I = 230/17.5 = 13.14Ω

Y = 1/Z = 0.076

G = Y cosθ = 0.076 x 0.8 = 0.061

Bl =√z^2 - G^2 = 0.076^2 - 0.061^2 = 0.045

new pf 0.95

cosθ = 0.95 θ = 18.19°

Bl-Bc = G tan 18.19° = 0.020

Bc = 0.025

Xc = 1/Bc = 40Ω

which makes C = 79.5μF

Hope i have it this time...

12. May 27, 2013

### Staff: Mentor

That looks MUCH better! I think if you hang on to a few more digits of accuracy for intermediate values your final result will be closer to 81.6 μF.

Cheers!

13. May 27, 2013

### topcat123

Thanks for the help gneill.

14. May 27, 2013

### Staff: Mentor

No worries, I'm glad to help!