Y-Delta Fault Current Calculation, two 50's and one 75

[steelerfan28655]

One of our customers has requested fault current data for the secondary side of their transformer bank. They have a Y-Delta, 240 volt, three phase, four wire service, with two 50 kVA and one 75 kVA, 7200 to 120/240 transformers. The center tap of the 75 kVA provides the load side neutral. My question is, how to calculate the fault current at the secondary spades. I have read different practices including the worst case scenario of treating all three transformers as if they were 75 kVA.
The single phase to ground primary fault current is 4326 amps. Three phase to ground is 5710 amps.
High side: R1=0.3853, X1= 1.2038, R0=0.8923, X0=2.3132
Z= 1.65% for the 50 kVA's and 1.6% for the 75 kVA.

I have done many of these for single phase services and for grounded wye three phase services. However the split secondary of the different size transformers has me wondering how to incorporate the differences in impedances into a common secondary bus.

Any help would be appreciated. My boss has already denied my request for a subscription to the IEEE library.

Happiness is a way of travel, not a destination.

 

[magoo2]

What you're asking is a little tricky to calculate.

Dave Smith of Power Technologies Inc. wrote a program to do this some years ago. It’s a DOS-based program which I managed to hold onto and it still works.

Lighting transformer is connected between b-n on primary and B-C on secondary.

Maximum fault currents are given for the following fault conditions:

Fault Amps
A-N 11,951
B-N 13,256
C-N 11,850
A-B 14,988
B-C 17,727
C-A 15,219
A-B-C 19,849
A-B-C-N 19,887

This may feed you, but it doesn't teach you how to fish! I'll try to see if I have any write-up on the methodology involved.

Note that you don't need to know the available phase-to-ground fault current since the neutral of the transformer bank is floating.

 

[JensenDrive]

I have never had the pleasure of facing a problem such as yours, but my first reaction is to use the brute force method: Convert impedances to real ohms, draw up a circuit network. For the xfmrs, I might divide the reactance and resistance between the primary and secondary in proportion to the turns ratio. You would need to know how to develop and solve simultaneous equations to show what happens when you short the secondary, phase to phase and phase to ground.

 

[nmsuEE]

I also have a similar problem. I would like to calculate the fault currents for different combinations of lighting and power transformers. Where can I find the method for these calculations?

 

[steelerfan28655]

All of the consulting engineers that we use suggested that we treat the bank as three transformer with ratings equal to the transformer with the highest rating. This will, of course, be too high, but should allow some leeway as far as safety factor.

Happiness is a way of travel, not a destination.

 

[nmsuEE]

I believe thats what we have done in the past, but I'm interested in comparing the actual results with those obtained by assuming three transformers of equal size.

 

[kellyo3]

I assume all three are in parallel? Why can't you select a base MVA and use that to determine each transformer %Z in pu based on the base MVA. Then you take the parallel comination of the three impedances and calculate the fault at the secondary relative to 240 volts. You would also need a source impedance on the primary side of the bus, which would be in series with the 3-transformer parallel impedance combination. Use the total fault impedance to determine your fault kVA and get the fault current from there. If the 50kVA transformers are not center tapped on the delta side then these do not contribute to a line to ground fault on the secondary. Am I missing something else?