Power factor calculation
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There doesn't seem to be enough information here to provide an answer without making a lot of assumptions that may be way off. What is the source of this problem and do they give a solution?
cableguy
Has never sniffed a stink bug
Ha, I just worked this problem a couple hours ago. I had no idea either.
Apparently the "no load" current is understood as being all VAR - so it's j28 amps, and since that's the "no load", that's the 'constant' part of the running motor.
This means at full load, you have a phasor magnitude of 92 amps, with a j28 reactive component, along with an 87.x "real" portion.
You then cut the "real" portion in half - 43.x amps - and still have a magnitude j28 reactive. The cosine of the inverse tangent of 28/43.x will be your half load power factor.
I didn't catch the concept of it either until I looked at the solution. It's a weird problem. As long as you can "assume" that the no load current is a constant lossy j28, I guess it works out....
Apparently the "no load" current is understood as being all VAR - so it's j28 amps, and since that's the "no load", that's the 'constant' part of the running motor.
This means at full load, you have a phasor magnitude of 92 amps, with a j28 reactive component, along with an 87.x "real" portion.
You then cut the "real" portion in half - 43.x amps - and still have a magnitude j28 reactive. The cosine of the inverse tangent of 28/43.x will be your half load power factor.
I didn't catch the concept of it either until I looked at the solution. It's a weird problem. As long as you can "assume" that the no load current is a constant lossy j28, I guess it works out....
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LMAO
Well-known member
So basically, you are saying that no load current is all reactive and, as you increase the load, you add to real current (which is 0 at no load)? And reactive current stays constant throughout?
Where did you guys find this problem? just curious...
cableguy
Has never sniffed a stink bug
Yes, the no load current is all reactive in this case - at least that's what the solution states. Me, I never made that "cognitive leap" when I looked at the question.
I'm taking the Testmasters Power PE course in Houston, and it was in the problems section of the manual.
LMAO
Well-known member
To solve this problem you have to make some assumptions; it is true that most of current at no load is reactive but you have to assume that at no load, slip is zero which makes rotor copper loss zero but you still have stator copper loss which is not zero at no load even if slip is zero.
I doubt a problem like this make its way to PE, I hop not.
I doubt a problem like this make its way to PE, I hop not.
I hope it does not make it to the PE but best be prepared than sorry u didn't.
I am still struggling to understand the algerbraic soloution.
Can you clarify what u meant by {"This means at full load, you have a phasor magnitude of 92 amps, with a j28 reactive component, along with an 87.x "real" portion.
You then cut the "real" portion in half - 43.x amps - and still have a magnitude j28 reactive}
Where did u get 87? and 43?
real portion in half would be 92/2 = 46 right?
I am still struggling to understand the algerbraic soloution.
Can you clarify what u meant by {"This means at full load, you have a phasor magnitude of 92 amps, with a j28 reactive component, along with an 87.x "real" portion.
You then cut the "real" portion in half - 43.x amps - and still have a magnitude j28 reactive}
Where did u get 87? and 43?
real portion in half would be 92/2 = 46 right?
At full load you're measuring the same reactive component as at no load i.e. 28A + whatever real component gets you a vector sum of 92A. The real component is Sqrt (922 - 28 2) = 87.6 amps. This is full load of 87.6 + j28 = magnitude of 92.
Now you make the second assumption that efficiency is linear from full load to 50% load and therefore the real component of current is halved 87.6/2 = 43.8 and the reactive component still stays at 28 amps. Then just figure angle of current.
IMHO, the assumption that at no load the motor has zero real loss (which means it is dissipating no heat when you touch it) is a stretch... and the flat efficiency is a bit of a stretch although it mostly falls apart for loads less than 50%. I don't think it is a great problem.
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Fixed word choice... constant efficiency would enable real current to be proportional to load, i.e. 1/2 half full load real current at 1/2 actual shaft load.
cableguy
Has never sniffed a stink bug
DK got it correct. The solution in my manual has it drawn out with a power triangle. One triangle with the "full load" current with 87.x, 92, and j28, and another triangle with the "half load" of 43.x, j28, etc.
