Ahh, not totally, if you could clarify a bit after you get some rest that would be great, thanks very much for your help.400Kva*0.6+40Kw=280Kw
280Kw/0.9=311Kva
cos-1(280/311)=25.7 degrees
311sin(25.7)=135Kvar
There's probably an easier way of doing this but it's to early in the morning. Do you understand what I have done to work it out?
The purpose behind the question is to make sure the apprentice knows and understands the theory and math involved in AC theory. Particularly the difference between va,var,watts and the application of power factor. It also eludes to another method in improving power factor.Gawg there are some stupid questions out there.
And it is totally beyond me why someone would know the answer. Mind you I have had several glasses of wine tonight.
Also I'd like to know where that plant is.
The apparent power of the plant is 400Kva at a power factor of 0.6. So 400*0.6=240Kw. There is also another resistive component that consumes 40Kw of power. 240Kw+40Kw=280Kw in total. Remember Resistances will always simply add together, where as reactance(inductive and capacitive) will add up and/or cancel out. With your power triangle the true power will always remain constant (assuming you don't add any more resistive loads) as the power factor changes.Ahh, not totally, if you could clarify a bit after you get some rest that would be great, thanks very much for your help.