3 transformer bank loading

[jboyd428]

reference to thread238-143892


Also, when a 3-transformer bank is operating 3-wire wye primary 3-wire delta secondary a single phase secondary load divides 2/3 into the transformer that directly connects to the single phase load and the other 1/3 of the current through the other 2. Why is this?

 

[waross]

I always thought that the current divided 50%-50%.
The two out of phase transformers form an open delta. The open side of the open delta becomes a virtual transformer with the same basic characteristics as the in phase transformer.
respectfully

 

[JensenDrive]

I can show an analysis in the sequence component approach, but not in the ABC world.
The single phase load on the delta secondary is a phase to phase load. Note this is similar to a phase to phase fault, and recall from phase to phase fault analysis,
I1=Ifault/sqrt3
I2=-I1
I0=0
Assume Iload, single phase = 1.732<0. The equations above will give:
I1=1<0
I2=1<180
I0=0
Assume the primary leads by 30deg. Recall also that if positive sequence leads, then negative sequence lags. Hence, current on the Wye (primary) side is:
I1=1<30
I2=1<150
I0=0
Convert to the ABC world, and this is:
Ia=1<90
Ib=2<-90
Ic=1<90
Want a real easy ABC<>012 calculator/conversion program? Go to:

http://users.ece.utexas.edu/~grady/

and look for "ABC012".

 

[jboyd428]

Thanks for the post.