Also is there a rule of thumb on how much the inrush will be based off of the constant amps the unit pulls
FLA (full load amps) * 1.25 = Inrush, sort of. Actually, 125% should cover a little more than the actual inrush. As the motor gets older and a little stickier, it may require more inrush, hopefully never exceeding the 125 %, until the day it actuall locks up and quits working. Also, you could do 125 % on the motor, to allow for inrush current, and total it with the total of the other circuits, doodads, etc. Take that sum total and make it 80% of whatever feeds it. The reciprocal of that 80% is 125%. So,
(((motor fla)*1.25) + (total other loads))*1.25 = equals ample current supply, with of course, fuses where you need them.
If this same setup was supplied by a smaller trans. say 500VA, what would be the out come?
Ideal Transformer Law
Power in = Power out. Example 75 kVA xfmr 480v P / 208 V S. Primary current is 90 A. Secondary current is 208 A. (figures are rounded) You can use that example to see what minimum size xfmr you can get away with. Also, by looking at this, you can see why the load side wire in this type of xfmr is bigger than the line side wire. And vice versa, a voltage step-up xfmr would have smaller wire on the load side.
[This message has been edited by rws (edited 12-07-2006).]
