Isolation Transformer Calculation Query

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aggybooya

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Sunderland, UK
Hi all

Wondering if anyone can help with a question I've been asked by a customer. Not even sure if it's a valid question but here goes:-

Their site have some 415/415V Delta-Star Isolation Transformers. They want to know how to calculate the primary current (in this case at the Delta connections) from the secondary current if the load is unbalanced. For example - L1 = 100A, L2 = 50A, L3 = 100A. Assume a load PF of unity.

Is there a way to actually do this or is this one of those moments where I'm going to end up red faced? :ashamed:

Any help would be much appreciated.

Thanks

A.T.
 

charlie b

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Lockport, IL
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Retired Electrical Engineer
There is a way. But the math is "hard." That is another way of saying I no longer remember how to do it. Sorry. :(
 

charlie b

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Oh, and I forgot to add, "welcome to the forum." To that I will add a request that you waive in the direction of Morpeth. My daughter and her Geordie husband live there. They are Sunderland fans.
 

aggybooya

Member
Location
Sunderland, UK
Ah, well that confirms that it is a valid question at least. It's also a long time since I've had to get involved in such complex calculations.

Perhaps there is someone else on here who can help....
 

GoldDigger

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Placerville, CA, USA
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Retired PV System Designer
I agree that it can be done.
Each wye line current will transform to current in both the directly corresponding delta winding and the series combination of the other two primary windings. The current will divide in a 2:1 ratio because of the winding impedances.
Then add the three sets of currents as vectors.
Easy to say and harder to do the math.

Tapatalk!
 

david luchini

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Connecticut
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Engineer
Ah, well that confirms that it is a valid question at least. It's also a long time since I've had to get involved in such complex calculations.

Perhaps there is someone else on here who can help....

It's not too complicated. You know the load current on each phase equals the current in each winding on the star side. You know the current in each winding on the delta side is the same current reduced by the winding ratio. The line currents on the delta side would be the difference of the currents in the delta side windings. Assuming a DY1 connection, Ia=At-Bt, Ib=Bt-Ct and Ic=Ct-At, where At=delta side A winding current, Bt=delta side B winding current and Ct=delta side C winding current.

So for load currents: Ila=100<0, Ilb=50<-120 and Ilc=100<120...

you know that delta winding currents will be At=57.83<0, Bt=28.92<-120 and Ct=57.83<120.

So, Ia=At-Bt = 57.83<0 - 28.92<-120 = 76.51<19.11
......Ib=Bt-Ct = 28.92<-120 - 57.83<120 = 76.51<-79.11
......Ic=Ct-At = 57.83<120 - 57.83<0 =100.17<150.00
 
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