Calculating short circuit of parallel generators

[Noooorm]

I have (6) 1.25MVA generators feeding a paralleling switchgear. To determine the switchgear short circuit rating, do I calculate the short circuit of each individual generator and multiply by (6) generators? Or do I consider the system as (1) 7.5MVA generator with an equivalent parallel Xd" (1/Xd" total = 1/Xd"1 + 1/Xd"2 + ...)?

 

[jim dungar]

I have (6) 1.25MVA generators feeding a paralleling switchgear. To determine the switchgear short circuit rating, do I calculate the short circuit of each individual generator and multiply by (6) generators?...

Good idea.

Actually I put six of them into my software (PTW by SKM) and let it do itss thing.

 

[Noooorm]

Could you? I don't have the software. They're 400V, 50hZ. I guess go with the worst case, 0.09 for X"d. I'll see which answer it's closest to.

 

[rcwilson]

I have (6) 1.25MVA generators feeding a paralleling switchgear. .... calculate the short circuit of each individual generator and multiply by (6) generators? Or do I consider the system as (1) 7.5MVA generator with an equivalent parallel Xd" (1/Xd" total = 1/Xd"1 + 1/Xd"2 + ...)?

Calculate for one generator and multiply by 6.

I short circuit = FLA/Xd" = [1.25 x 1000/(.480 x 1.732) ] /0.2 = 1504 /0.2 A = 7517 Amps per generator.
7517 Amps x 6 = 45,106 Amps

Or using the MVA method:
Assume typical Xd" = 0.2, Short Circuit MVAsc = 1.25 MVA/ 0.2 = 6.25 MVA,
I short circuit = MVAsc x 1000 / (kV x1.732) = 7517 amps (assuming 480 V).
Total short circuit = 7517 amps x 6 = 45,106 Amps.

Same answer if we multiply MVAsc of one generator by 6 and calculate amps.
6.25 x 6 = 37.5 MVA total short circuit power,
I = 37.5x 1000/(0.480 x 1.732) = 45,107 Amps.

Since the impedance Xd" is per unit or per cent and not in actual ohms we don't need to calculate the parallel impedance. Example:

Six each 1.25 MVA units = 7.5 MVA all with Xd"=0.2. MVAsc = 7.5/0.2 = 37.5 MVA, same answer as above.

 

[charlie b]

I think you will get the same answer either way. If the generators were not identical, however, a method closer to your second one would be needed. But my preference is to use SKM, and leave the messy math to someone else. :happyyes:

 

[rcwilson]

400V will change you short circuit current to 480/400 x 45,107 = 54, 128 if your generators have the typical 0.2 Xd". I think most generators in this size have Xd" = 0.12 - 0.25 per unit ( 12-25%).

Don't forget to add in any contribuiton form motor loads.

 

[Noooorm]

Thank you! You are gentlemen and scholars.

 

[steve66]

I think you will get the same answer either way. If the generators were not identical, however, a method closer to your second one would be needed. But my preference is to use SKM, and leave the messy math to someone else. :happyyes:

That was my first though too, but it doesn't seem to work. Consider (2) 1KW generators, z = 0.2. That's 5000 amps SC per genrators, or 10000 amps total.

Using the second method, you would have a 2 KW generator, and z =0.2 in parallel with 0.2 would be 0.1. 2000/ 0.1 = 20,000 amps. Unless my math is wrong, that answer is too high.

For the second generator, we are both doubling the capacity, and halving the impedence, giving a 4 fold increase, when the real fault current is only going up by a factor of 2.

 

[rcwilson]

The generator impedance is expressed in percent or per unit. It is the measured or calculated impedance based on the generator nameplate data. That number is not the actual real world ohms that we could measure with an ohmmeter (if there was such a thing as an AC ohmmeter).

If 0.2 was real world ohms, then adding them like parallel resistances is correct.

But since it is a percentage or per unit number we don't change it when we are paralleling identical units.

Said another way, the per unit impedance of two paralleled 1 MVA generators with impedance = 0.2 pu is still 0.2 per unit. The math works because on individual units, it's 0.2 per unit of 1 MVA and on two parallel units it's 0.2 of 2 MVA.

If you want to do it in real world ohms just convert to ohms:

MVA = kV^2 / Z. (Power = voltage squared/ ohms impedance). or Z= kV^2 /MVA.

Zgen = (0.400kV x 0.400kV)/ 1.25 MVA = 0.1280 ohms. This is the 100% or 1.0 per unit impedance. If Xd"= 0.2 pu, then Zgen = 0.2 x 0.1280= 0.0256 ohms

Short circuit current for one 1.25 MVA generator at the OP's 400V is limited only by the generator impedance of 0.0256 ohms. Isc = 400V / (0.0256 x 1.732) = 9,021 amps. Six generators would be 6 x 9021= 54,126 Amps.

With six generators in parallel the impedance for a short circuit would be the parallel impedance of six generators = 1/ (1/0.0256 + 1/0.0256 + 1/0.0256 + 1/0.0256 + 1/0.0256 + 1/0.0256) = 0.0256/6

Z= 0.00427. Isc = V/Z where V= phase- neutral volts = 400 /(1.732 x .00427) = 54,126 amps.


I probably confused some of us, but I was trying to explain the per unit, per cent concept.