Power Factor Correction - Annual Energy Benefits

Ibnul Jaif Farabi

Joined Nov 27, 2016
11
Hi, I'm trying to solve this problem ---->>>

A 3-phase feeder supplies a spot load connected to its end. The reactive current of the load is 50 A and its power factor is lagging. A PFC bank is installed next to the load to provide partial compensation. The bank current is 28.0 A. Compute the annual $benefit from the power loss reduction. Data given: Feeder resistance: 2.16 Ohm Loss factor: 0.40 Energy cost rate: 0.05$/kWh

I did this in this way ---->>> According to mine calculation, the answer is $649.992, but it's wrong as the answer should be$ 2288.76.

Are there any suggestions?

KL7AJ

Joined Nov 4, 2008
2,229
Did you figure for all three legs?

Ibnul Jaif Farabi

Joined Nov 27, 2016
11
Did you figure for all three legs?
I just used these formulas, as the reactive current = 50 A, PFC current = 28 A

As resistance = 2.16 ohm

So change in power loss = (50)^2 * 2.16 - (28)^2 * 2.16 = 3.71 kW

Annual energy loss = 0.40 * 3.71 * 8760 = 12999.84 kW

Energy savings in dollars = 12999.84 * 0.05 = 649.992

I've made an error in somewhere. Can you please point it out?