Transformer Bank Rating

Professional Engineer & PE Exam Forum

Help Support Professional Engineer & PE Exam Forum:

This site may earn a commission from merchant affiliate links, including eBay, Amazon, and others.

cdcengineer

Sharp as a piece of beach glass
Joined
Dec 28, 2009
Messages
1,809
Reaction score
25
Location
Colorado High Country
Alright folks, I've been causing myself some brain damage on this problem..

Consider a single phase transformers rated 500 kVA, 33 - 2.4 kV, with internal impedance Z=+j1.0 ohms (low side).

If (3) of the above transformers are connected in a 3-phase transformer bank, wye on the high side and delta on the low side, what is the transformer bank rating?

I came up with the Bank rating of 57.2Y - 2.4Delta kV, 1.5MVA

My problems is trying to come up with the X rating.... I'm getting 0.33333 on the low side (Delta), but this conflicts with the solution of 0.086.

Is this solution wrong or in pu unit, or am I hurting my brain for no reason?

Thanks

 
Here's what I get:

Voltage:

High side is wye connected so VLine=sqrt(3)*VPhase = sqrt(3)*33kV = 57.2 kV

Low side is delta connected so VLine=VPhase=2.4 kV

MVA:

A balanced three-phase bank will be rated three times the rating of the individual transformers.

3 * 500 kVA = 1.5 MVA

Impedance:

Their answer appears to be in per unit. On a p.u. basis, the transformer impedance is equivalent between a single transformer and the bank.

Zpu = ZActual/ZBase

ZBase = kVBase2/MVABase = 2.42/0.5 = 11.52 ohms

Zpu = 1.0/11.52 = 0.0868

 
Back
Top