New answers tagged energy-efficiency
AS WE KNOW 1 TR = 3.516 KW FOR 5TR IT IS= 3.516X5 =17.58 KW CURRENT FOR 430 V SUPPLY = 17.58X1000/(1.73*430*0.9) ASSUMING pf 0.9 =17580/669.51 =26.2 AMP (APPROX)
That is fundamentally correct for the time the lights are on, yes. It will depend on the specific LEDs and ballasts chosen (the ballast may or may not be part of the fixture, and some cheesy outfits try to give the lumens per watt INTO the LED itself, ignoring the watts wasted in the ballast converting from AC to DC.) @Tester101's "wire heating" strawman ...
Top 50 recent answers are included