neutral load calculations

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dok

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NYC
Is the formula for the neutral load calculation of a three phase 4 wire wye system the same formula used to calculate the?neutral load for a three phase 4 wire delta system with a high leg
the square root of a squared + b squared + c squared minus (a+b)+(a+c)+(b+c)?
 

roger

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Re: neutral load calculations

The end of your formula should be (a x b)+(b x c)+(a x c)

Roger
 

don_resqcapt19

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Re: neutral load calculations

The neutral current in a 3 phase high leg delta system is the same as in a single phase system. Your formula is for the neutral current in a 3 phase wye system, but it is not correct. The
last part should be -[(a*b) + (b*c) + (a*c)].
Don
 

dok

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Location
NYC
Re: neutral load calculations

Thanks for the correction, however the way I wrote the formula and the way you told me is the right way are the same and will always result in exactly the same answer
 

dok

Member
Location
NYC
Re: neutral load calculations

So in a high leg 3 phase 4 wire delta system the neutral carries the highest imbalance between any 2 phases?
 

bob

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Location
Alabama
Re: neutral load calculations

Dok
Thanks for the correction, however the way I wrote the formula and the way you told me is the right way are the same and will always result in exactly the same answer
are you saying that [(a*b) + (b*c) + (a*c)]gives the same answer as [(a+b)+(a+c)+(b+c)]?
I don't think so. Assume a,b and c = 10. Check it out.
 

don_resqcapt19

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Re: neutral load calculations

So in a high leg 3 phase 4 wire delta system the neutral carries the highest imbalance between any 2 phases?
You can't use the high leg for line to neutral loads so just forget about it. Treat the other two hots and the grounded conductor as if it were a single phase 120/240 volt system.
Don
 

dok

Member
Location
NYC
Re: neutral load calculations

I stand corrected. I mistakenly had plus symbols (+) and not multiply symbols (x) in the second half of the formula.I thought you were referring to the order in which I wrote them. Thanks Bob and Don. Also for Don, I thought using phase A @120v. to phase C @ 120v. would be the same as using phase B @ 240v to neutral. I thought this type of service was originally used in factories where there were many single phase and three phase motor loads.
 

roger

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Re: neutral load calculations

In the system you describe there would never be 240v to neutral.

It's called a center tapped Delta, and only one winding is used with the grounded conductor, i.e.
points B&C, the conection point of the winding or windings (point A) not tapped is not part of the gounded portion of the delta configuration.

This point would read 208 to the grounded conductor.


Look at the delta diagram below

;)


Roger

[ November 24, 2005, 10:40 AM: Message edited by: roger ]
 

rattus

Senior Member
Re: neutral load calculations

Roger, I agree. It is a neutral relative to L1 & L2, but not relative to the delta.
 

charlie b

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Lockport, IL
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Re: neutral load calculations

Originally posted by dok:. . . the way I wrote the formula and the way you told me is the right way are the same and will always result in exactly the same answer
Other than the inadvertent ?plus? where it should have said ?times,? I would say they are the same. But that is because I read it as intending everything after the word ?minus? to have been included in the stuff you are subtracting.

I see this,
Originally posted by dok: minus (a+b)+(a+c)+(b+c)
and I do not attempt to apply any mathematical ?Distributive Property? or ?Associative Property.? You are not looking at a mathematical formula. You are looking at an English sentence that contains some math symbols. It would have been more clear if there were a set of brackets around the three terms. But I understood what was meant.
 
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