• posted

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However our resistive/inductive load bank has been withdrawn from service.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 instead of KW?

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

Regards Boxie

• posted

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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

• posted

kVA=sqrt(kW^2+kVAR^2) as Don Kelly notes, the Pythagorean Theorem

The angle between the voltage and current is given as theta=arccos(pf) and is negative for lagging pf (the current lags the voltage).

kW=kVA*cos(theta) which is just kW*pf kVAR=-kVA*sin(theta) note the minus sign.

Inductive reactive power is positive when it is in the same direction as (positive) real power.

For a .8pf, the ratios of a 3,4,5 right triangle work just fine.

pf usually doesn't carry a sign so you have to figure out whether the rating is inductive or capacitive by some other means. 99.9999% of the time a synchronous machine is rated for inductive load... lagging pf.... overexcited operation, i.e., excitation above what would be required for unity pf operation at the same stator current. I've never seen a generator nameplate with a leading pf rating.

In order to _fully_ test the machine at its rating (60kVA @ .8pf (lag I assume)) you need to load it with 48kW and 36kVAR inductive at a terminal voltage of 200V and frequency of 400Hz. Actually, to fully test you would need to prove acceptable operation at some +/- deviation in voltage and/or frequency. The numbers should be in the standards to which the machine is built. If this is a production test, hopefully the purchaser has provided some guidance as to what constitutes an acceptable test.

Loading at 60kW, 200V will give you the full stator current and test (more or less) the stator but it will not load rotor fully and, therefore, not load the excitation system or cooling system fully. I'm assuming the common arrangement of field=rotor and armature=stator.

If the load has high harmonic content, all bets are off.

What is a GPU?

• posted

As the others have noted, 48kW at 1.0 pf is not the same current on the generator as 48kW at 0.8 pf.

But it *is* the same amount of load for the prime mover of this generator.

So, this begs the question, "What exactly are you trying to test?"

If you want to see if the generator will overheat when subjected to rated current flow, then this new load bank isn't going to cut it. You *could* load the unit to 60 kW @ 1.0 pf and that would subject the windings to rated current. But it might be more than the prime mover can handle (see below).

If you want to see if the generator output voltage will remain 'in-spec' when fully loaded, you really need a load bank that can absorb some VARS and load the unit to the 'nameplate' ratings (presumably, 60 kVA @ 0.8 pf). Voltage regulation is affected a lot by the power factor. A unit that can maintain voltage at 1.0 pf may not be able to do so when in service at 0.8 pf.

If you're trying to see if the prime mover for the thing can supply the power, then you're fine loading to 48 kW.

We have done 'load tests' on diesel-generators and the goal was to verify the engine capability. Then we had to keep the *real* load on the engine for an hour to ensure it warmed up to proper temperature and didn't get fouled with soot.

But your testing may be different. "What exactly are you trying to test?"

daestrom

• posted

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.

To answer your question, we actually fully load test to see if the 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 a major overload takes place!

Regards Boxie

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