A group of small appliances on a 60 Hz system requires 20 kVA at 0.85 pf lagging when operated at 125 V (rms).

A group of small appliances on a 60 Hz system
requires 20 kVA at 0.85 pf lagging when operated at
125 V (rms). The impedance of the feeder supplying
the appliances is 0.01 + j0.08 ohms. The voltage at the
load end of the feeder is 125 V (rms).

a) What is the rms magnitude of the voltage at the
source end of the feeder?

b) What is the average power loss in the feeder?

c) What size capacitor (in microfarads) across the
load end of the feeder is needed to improve the
load power factor to unity?

d) After the capacitor is installed, what is the rms
magnitude of the voltage at the source end of
the feeder if the load voltage is maintained at
125 V (rms)?

e) What is the average power loss in the feeder
for (d)?

Any help before Friday May 6th at 9 am will be greatly appreciated, Thank You.

The impedance of the line is Sqrt(0.01^2 + 0.08^2) =0.0806 ohms
The current for the 20 KVA is I =20000 / 125 = 160 amps
The votage drop in the line is E = I * R = 160 * 0.0806 = 12.896 volts
This means the source voltage = 125 + 12.896 = 157.896 volts
The power loss in the feeder initially is W = I^2 * R = 160^2 * 0.0806 =2.06336 watts

I used an application I wrote for the following data:
The impedance of the load is 0.78125 ohms
The resistance of the load is 0.40625 ohms
The total circuit impedance is 0.78125 + 0.0806 = 0.86185 ohms
The total resistance of the circuit is 0.40625 + 0.01 = 0.41625 ohms
The total circuit reactance is 0.6673174 + 0.08 = 0.7473174 ohms
The inductive reactance of the load is 0.6673174 ohms
The inductance in henries = 0.0018 for the total circuit

The capacitive reactance for unity power factor must equal the inductive reactance at the same frequency. From the application this answer is 3975.097 microfarads

The KW of the 20KVAload at 0.85 power factor = KVA * power factor = 20 * .85 = 17 KW
The current for 17 KW = 17000 / 125 = 136 amps
The line loss after power factor correctio is W = 136^2 * .01 = 0.018496 KW
The voltage drop in the line after power factor correction will be E=I*R=136 *.01 = 1.36 volts


I did this on the fly. Take it or leave it. This is for single phase power.

If you are interested in my application (for free) I'll send you a installable application on your regular email.


TexMav