# 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?