Discussion in 'Electrical Engineering' started by Boxie-Boy, Aug 2, 2006.

1. ### Boxie-BoyGuest

This is really a simple question for those up on ac theory. However I,m
have difficulty getting my head around this problem.

We have a 60KVA 200V 400Hz aircraft GPU. We also used to have a
resistive/inductive load bank. We use the 3,4,5 triangle theory for
applying the correct load @ 0.8PF ie 36KVAR (3 reactive load) + 48KW (4
resistive) = (5)60KVA. Interms of amps per phase it equates to
(60000/115)/3 = 174Amps. We never had a problem with this method.

However our resistive/inductive load bank has been withdrawn from
service. We now have a purely resistive load bank with the load selector
dials marked in KW.

OK what our people are doing now is selecting 48KW on the load bank of
which they believe is full load. I know the full load current for a
60KVA (48KW) machine is 174A, but selecting 48KW is about 140A per
phase. This in my opinion is NOT full load.

Here is the strange thing, if i input 60KW on the load bank the line
currents are 174A. bank on!

So am i correct in saying that at unity power factor KW=KVA? Should the
manufacturers of the load bank have marked up the input dial in KVA

Can anyone put the math in simple terms or correct my theory?

Regards
Boxie

2. ### Don KellyGuest

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You are correct. However the KW marking is quite correct for a purely
resistive load bank- just remember that
60KVA at unity power factor (resistance load) =60KW
The alternator (I assume that this is what you mean by GPU) may be rated at
60KVA 0.8 pf lag. The rated output current is 174A but at this power factor,
the field current is at its rated value. At unity pf for a given voltage,
the field current is lower- not a problem except that you are not testing
the ability of the field to provide sufficient excitation at 0.8pf but this
doesn't appear to be what you are interested in.

Testing at 48KW unity pf is definitely NOT testing at rated KVA

As to the math- Instead of a 3,4,5 triangle with an angle between the
Power(4) and reactive(3) the angle is something else. At 0.9 pf , 60KVA the
triangle at P=54 (0.9*60)KW , reactive =root(60^2- 54^2) =26KVAR. (by
pythagorus theorem)
At unity PF 60KVA, P=60*1 =60 and the reactive becomes root (60^2-60^2) =0
so KW=KVA

3. ### Boxie-BoyGuest

Thanks for the info, it has really helped.

For info a GPU is a Ground Power Unit that is designed to supply
aircraft power for maintenance/starting.

Engine can provide the necessary power and that ir runs at correct
speed. ( Frequency 400Hz +/- 10%)

We also we check the voltage regulation is maintained within 200Vac +/-
10% and that the Ammeters are reading correctly. And that overlaod
protection is functioning, however this last one has to be done
carefully! If at all. My argument is if this is not tested and the
protection circuit is inop or out of spec, then it could be expensive if  