Calculating transformer fault currents HI/LO side 1

[gobblerhuntr]

I am trying to learn some more about fault and fault currents. If a utlilty has a known fault current on the high side of a power transformer (161kv) of x amount of amps three phase. With the nameplate information given (kva,voltage,impedance) what is the forumla to determine the fault current on the low side bushings (13kv)?? I have looked for examples and all I find are ones that pertain to the utility as being bolted. All the new arc flash rules have got me wanting to learn a little more about these procedures. I understand it may not be as easy as a b c, but I would appreciate any direction someone could give for me to find the proper formulas and procedures for the calculations. I have an example in mind of a 20 MVA, 8.2%, 161:13kV transformer.

 

[dpc]

Think of the utility as one impedance (between your transformer and an infinite source) and the transformer as another impedance. The fault current on the low side of the transformer is going to be a function of the two impedances in series, so you add them together.

If you take the utility fault current and divide it by the line-to-ground voltage, that gives you the utility impedance.

The trick is the fact that the transformer also transforms impedances in addition to voltages and currents.

Power engineers use the per-unit system for this stuff. You should be able to Google up some good tutorials on per-unit.

 

[jimgineer]

"If you take the utility fault current and divide it by the line-to-ground voltage, that gives you the utility impedance."

DPC, Am I missing something?

V=IZ not I/V = Z

I really need to brush up on this stuff but is that correct?

 

[jimgineer]

OK, got it ---

First you need to calculate the available FLA on the secondary. This is going to be, in your case based on the info given:

20MVA/.0138MV = 1449A

Then take 1449A/.082 %Z

And you get:

17674A



( I assumed the 13kV you had listed was 13.8kV on the secondary side)

That should do it

 

[kh2]

To solve this problem you have to know the utility (source) 3-phase fault current or MVA. Assume that the source 3-phase fault is 2500 MVA. Then use the transformer MVA as the MVA base.
1) Source (utility) p.u. impedance = MVA base/SC MVA source = 20/2500 = 0.008.
2) Transformer p.u. impedance = 8.2/100 = 0.082
3) Total Zp.u. = 0.008 + 0.082 = 0.090
4) Transformer Secondary fault current = I(secondary)/ total Zp.u.
5) Transformer secondary current = 20 MVA x 1000/(1.73 x13 kV) = 889.28 Amps.
6) Transformer fault current at 13 kV side = 889.28/0.090 = 9,880 Amps.

Hope this will help.