# Study Help

1471 8
Another question that I am having trouble with...

An industrial plant has an inductive load of 400KVA operating at a power factor of 0.6 lag and a resistive load of 40KW. A synchronous motor is used to correct the PF to 0.9 lag. Assuming the sychronous motor produces no watts what would the corrected plant KVARS be.

The correct answer is listed as 135.5KVAR but I'm not sure where to go with this, any help would be appreciated, thanks.
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#### Aussielec

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400Kva*0.6+40Kw=280Kw
280Kw/0.9=311Kva
cos-1(280/311)=25.7 degrees
311sin(25.7)=135Kvar

There's probably an easier way of doing this but it's to early in the morning. Do you understand what I have done to work it out?

#### ponyboy

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Jesus. What kind of apprenticeship is this. Just multiply your distance by 2 and bend it 30°. Christ

#### daveEM

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Gawg there are some stupid questions out there.

And it is totally beyond me why someone would know the answer. Mind you I have had several glasses of wine tonight.

Also I'd like to know where that plant is.

#### macaber

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400Kva*0.6+40Kw=280Kw
280Kw/0.9=311Kva
cos-1(280/311)=25.7 degrees
311sin(25.7)=135Kvar

There's probably an easier way of doing this but it's to early in the morning. Do you understand what I have done to work it out?
Ahh, not totally, if you could clarify a bit after you get some rest that would be great, thanks very much for your help.

#### Aussielec

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Gawg there are some stupid questions out there.

And it is totally beyond me why someone would know the answer. Mind you I have had several glasses of wine tonight.

Also I'd like to know where that plant is.
The purpose behind the question is to make sure the apprentice knows and understands the theory and math involved in AC theory. Particularly the difference between va,var,watts and the application of power factor. It also eludes to another method in improving power factor.

#### bkmichael65

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(400,000 x .6) + 40,000 = 280kW/.9=311kVA
sq root of (311k sq - 280k sq)=135kVAR

#### Aussielec

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Ahh, not totally, if you could clarify a bit after you get some rest that would be great, thanks very much for your help.
The apparent power of the plant is 400Kva at a power factor of 0.6. So 400*0.6=240Kw. There is also another resistive component that consumes 40Kw of power. 240Kw+40Kw=280Kw in total. Remember Resistances will always simply add together, where as reactance(inductive and capacitive) will add up and/or cancel out. With your power triangle the true power will always remain constant (assuming you don't add any more resistive loads) as the power factor changes.

So now the power factor has been changed to 0.9, but remember our resistive load is still 280Kw(we have added additional reatance to improve power factor in the way of the synchronous motor). 280Kw/0.9= 311Kva which is the new apparent power consumed by our plant.

I've simply then used the inverse cosine feature to find the angle between apparent power( the hypotenuse of the power triangle) and true power(adjacent). Cos-1(280/311)=25.7

Then I've just transposed the equation sin(theta)=opposite/hypotenuse(because we're trying to find the opposite side of our power triangle which is vars). So sin(25.7)*311Kva=135Kvar.

As I said there's probably a much easier way of doing this, that I'm sure someone else will come up with.

Hope this helps.

#### BlackHowling

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Instead of using the cosine and sine functions you could always use Kva(squared)-watts(squared=kvars(squared) then sqrroot the vars

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