# Wye/Delta motor connections



## LMAO (Oct 17, 2010)

I bumped into another problem; an AC motor which is normally Delta connected draws 367A starting current; what would be the starting current if it is started in Wye configuration?

I thought the answer should be 367x3=1101, because Wye impedance is one third of delta impedance. But the answer in the book has it other way around; it shows to be 367/3=122.3.

Any thought?


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## cableguy (Oct 17, 2010)

What you're looking at there is a Wye-start / Delta Run motor. This reduces startup current.

Check out this bulletin:

http://www.usmotors.com/Service/Bulletins/Issue7-May03.pdf

The bottom of page 2 has your answer.


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## mull982 (Oct 17, 2010)

LMAO said:


> I bumped into another problem; an AC motor which is normally Delta connected draws 367A starting current; what would be the starting current if it is started in Wye configuration?I thought the answer should be 367x3=1101, because Wye impedance is one third of delta impedance. But the answer in the book has it other way around; it shows to be 367/3=122.3.
> 
> Any thought?


I agree with your answer. Zwye = 1/3 Zdelta so I would think that the impedance in a wye configuration would be 3 times less and therfore the current would be 3 times greater.

I know however that startup current is less in a wye configuration however I dont see this relationship through the delta/wye impedance configurations?


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## cableguy (Oct 17, 2010)

If you look at the diagrams in the PDF I linked, and think about it...

Line current in the Y is going to be the line-to-line voltage, divided by sqrt(3) (to get line-to-neutral), divided by Z of the winding.

Line current in the delta is going to be sqrt(3) times line-to-line voltage divided by the Z of the winding.

So effectively, the line current for this delta is going to be 3 times the Y connection.

Let's throw some made up numbers at it.

Line-to-Line = 480V

Z = j1

|Iy|=(480/sqrt(3))/j1 = 277 amps

|Idelta|=sqrt(3)*(480/j1)=831 amps

Note that the impedance isn't changing - they are moving the windings around with the motor control circuit, but the windings are the same no matter what.


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## LMAO (Oct 18, 2010)

cableguy said:


> If you look at the diagrams in the PDF I linked, and think about it...
> Line current in the Y is going to be the line-to-line voltage, divided by sqrt(3) (to get line-to-neutral), divided by Z of the winding.
> 
> Line current in the delta is going to be sqrt(3) times line-to-line voltage divided by the Z of the winding.
> ...


I got it, thanks.


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## mull982 (Oct 18, 2010)

LMAO said:


> cableguy said:
> 
> 
> > If you look at the diagrams in the PDF I linked, and think about it...
> ...


I agree with your calculations here but I still think that the equivelent wye impedence is 1/3 of that of the delta impedance. So I would think that the impedance is lowering but at the same time the voltage is lowering so maybe thats what causes the current to be lower?


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