How do you figure KWH savings by turning off lighting loads at night when you know amperage?
jim d asked:
If I have a facility that runs on 3phase 120/208 and I am charged .07cents per KWH and I shut off appx 250amps at night for lighting loads for 8hours how much money do I save a year and whats the formula, and does it matter being 3phase if all 3phases aren’t shut off equally proportioned?
If I have a facility that runs on 3phase 120/208 and I am charged .07cents per KWH and I shut off appx 250amps at night for lighting loads for 8hours how much money do I save a year and whats the formula, and does it matter being 3phase if all 3phases aren’t shut off equally proportioned?
if 250 amps is the sum of the currents in all 3 phases, then wattage is 120 x 250 watts. Divide by 1000 to get kW, and multiply by 8 hours to get kW-hrs.
Then multiply by cost per kW-hr (7¢ is more likely than 0.07¢)
. [who] billrussell42