J
I don't think 3600 kvar is the correct answer. I may be reading the question wrong.
Take the origional information
4500 kva at 0.60 PF. This equates to
2700 kw and 3600 kvar. If you improve the PF to 0.90 then the
KVA = KW/pf = 2700 kw/0.90 = 3000 kva.
KVAR = sqrt(KVA² - KW²)= sqrt(3000²-2700²)
KVAR = 1306 kvar. So the new kvar must be
reduced to 1306 kvar. That is a reduction of 2294 kvar.
The sync motor must be adding 3600 kvar -
1306 kvar = 2294 kvar.
Robbie
If you do not load the sync motor then it will not add the 746 kw. However if you
do load the motor then you would have to add that load to the new reduced load of 3000 kva and as you say thing would change.
jfwayer
What was the answer for the sync motor PF?
[This message has been edited by Bob (edited 10-19-2005).]
