What is the required KVA rating of a capacitor bank used to correct a .67 lagging PF to .95 PF in a 2400KVA load?

Question

I keep getting 1252.95, but I am unsure of the steps I am taking to solve the problem. It would be helpful to see the steps someone else uses to either confirm my answer, or see where I am going wrong. I am also confused how to determine the KVA rating. Do I just round, so 1300KVA rated capacitor based on my current answer?

Answer 1

Here is the step by step procedure, as we know:

S*cos(θ) = P, and S*sin(θ) = Q:

P1 = S1*cos(θ1)

P2 = S2*cos(θ2)

Assuming the same active power required, and we only add pure capacitor without affecting the active power, meaning P1 = P2, therefore:

S1*cos(θ1) = S2*cos(θ2) => S2 = S1*cos(θ1)/cos(θ2):

As noted in the question: cos(θ1) = 0.67, and cos(θ2) = 0.95

S2 = S1*cos(θ1)/cos(θ2) = 24000*0.67/0.95 = 16926 kVA

cos(θ1) = 0.67 => sin(θ1) = 0.74

Q1 = S1*sin(θ1) = 24000*.74 = 17816 KVAr

cos(θ2) = 0.95 => sin(θ2) = 0.31

Q2 = S2*sin(θ2) = 16926*.31= 5247 KVAr

The added capacitor should be sized to change the reactive power from Q1 to Q2, which means

Q1-Q2 = 12569 KVAr

If you really need to size the capacitance value [which is not needed here]:C = delta_Q/(2*pi*f*V^2)