Question

You're the operator of a 15,000 V rms, 60 Hz electrical substation. When you get to work one day, you see that the station is delivering 6.10 MW of power with a power factor of 0.940.

What is the rms current leaving the station?

How much series capacitance should you add to bring the power factor up to 1.0?

How much power will the station then be delivering?

Answer 1

V_rms=15000 V,

P = 6.1MW = 6.1x10⁶ W

PF = cos? = 0.94 = R/Z, ? = cos^-1(0.94)= 19.948⁰

Now, P = V_rmsI_rms cos?

so 6.1x10⁶ = 15000xI_rmsx0.94

I_rms = 6.1x10⁶ /15000x0.94 = 432.624 A

Now Z = V_rms/I_rms = 15000/ 432.624 = 34.67 ?

and R = Z cos? = 32.59 ?

We know that

tan? = (X_L-X_C)/R where X_L = ?L, X_C = 1/?C

therefore, (X_L-X_C)/R = tan?

(X_L-X_C) = Rtan? = 32.59 tan(19.948⁰)=11.83 ?

To get power factor 1.0, ?’ = 0,

and tan?’ = 0

Suppose we add a capacitor C_new in series to the existing one.

Then 1/C_eq= 1/C + 1/C_new

and X’_C=1/?C_eq=1/?C + 1/?C_new = X_C + 1/?C_new

therefore if tan?’ = 0 that means X_L-X’_C = 0

so, X_L-X_C – 1/?C_new = 0

so, 1/?C_new = X_L-X_C= 11.83 ?

so, C_new = 1/(11.83?) = 1/(11.83x2?f) = 224 ?F

Since ?’ = 0, Z’ = R and cos ?’ = 1,

and P’ = (V_rms)²/Z’ = 6.9 MW

Ans: rms current = 432.6 A, required capacitance = 224 ?F, New power = 6.9 MW