Short Circuit Calcs not adding up

Status
Not open for further replies.
J

JasonLSelf

Member
I want to make sure I am doing this short circuit calc correctly so if somebody could review my logic I would appreciate it.

According to the Bussman Plan Reviewers guide for AIC calcs the "Second transformer in a system" calculation goes as followed:

f=Primary side fault current * Primary Voltage * 1.732 (%Z) /100,000 * kVA
M=1/1+f
Secondary side fault current = (Primary Voltage/Secondary Voltage)*M*Primary side fault current.

If I do this math and have 20,345 available fault current at the primary side of the transformer (250kVA) here is what my math returns:
F=20345*480*1.732*7/100,000*250=266395.8
M=1/1+F=3.753
Available fault current = 480/208*3.753*20345=.1762

I find it hard to believe that .1762 is all the available fault current at the secondary side of this transformer, especially when my handy cheat sheet for transformer fault current tells me that a 250kva 208V secondary xfrmr has 36,737 available fault current at the secondary side.

I think I am missing the mark somewhere, any help figuring out where would be greatly appreciated.
 
JasonLSelf said:
I want to make sure I am doing this short circuit calc correctly so if somebody could review my logic I would appreciate it.

According to the Bussman Plan Reviewers guide for AIC calcs the "Second transformer in a system" calculation goes as followed:

f=Primary side fault current * Primary Voltage * 1.732 (%Z) /100,000 * kVA
M=1/1+f
Secondary side fault current = (Primary Voltage/Secondary Voltage)*M*Primary side fault current.

If I do this math and have 20,345 available fault current at the primary side of the transformer (250kVA) here is what my math returns:
F=20345*480*1.732*7/100,000*250=266395.8
M=1/1+F=3.753
Available fault current = 480/208*3.753*20345=.1762

I find it hard to believe that .1762 is all the available fault current at the secondary side of this transformer, especially when my handy cheat sheet for transformer fault current tells me that a 250kva 208V secondary xfrmr has 36,737 available fault current at the secondary side.

I think I am missing the mark somewhere, any help figuring out where would be greatly appreciated.
Click to expand...

EDIT: While I was figuring and typing, ENGY posted the same answer as I got.

I don't have the guide, but here are some things you could check.

You may be missing some parentheses in your formulas, and I can't get the numbers that you get in your arithmetic.

Your: F=20345*480*1.732*7/100,000*250=266395.8 should perhaps be

F=20345*480*1.732*7/(100,000*250) = 4.736

The 100,000 term is the factor that scales the KVA term in the denominator to the same units as Volts and Amps in the numerator, and divides the % term to make it a decimal fraction.

Then; it appears that M=1/1+F=3.753 might be:

M=1/(1+f) = 1/(1+4.736) = 0.1743

Your: Available fault current = 480/208*3.753*20345=.1762

might be: (480/208)*0.1743*20345 = 8183 Amps

That is still not the 36,737 that your handy cheat sheet says, but it is in the right direction.
 
I'm not sure what you have. Are you saying it is a 100KVA transformer with 7% Z? And 480V, 3 phase? Is that right?


I only get 1718 amps. It is almost as easy to set these calculations up in an Excel spreadsheet as it is to do one calculation. Then each time you have to redo them, it is easy.

Steve
 
Now I see: 250KVA 480V to 208V transformer. I get 8193 amps, basically the same as Dave and Bob.

Steve
 
Using the OP formula:
JasonLSelf said:
f=Primary side fault current * Primary Voltage * 1.732 (%Z) /100,000 * kVA
M=1/1+f
Secondary side fault current = (Primary Voltage/Secondary Voltage)*M*Primary side fault current.
Click to expand...

F=(20345*480*1.732*7)/(100,000*250)=4.7359
M=1/(1+4.7359)=0.1743
Available fault current = 480/208*0.1743*20345=8183.4

Using a speadsheet with no intermediate rounding: 8185.25
 
To get the short circuit current on the load side of a transformer, you have to know what the impedance of the transformer is. If it is for a new installation then you might not have the nameplate data, but IEEE gives you standard values the manufacturers design too.

With that said, a 250KVA transformer is only possible of passing approximately 4.35MVA of short circuit MVA (@5.75% impedance), which translates to approximately 12,100 A @ 208V. That is assumming that the high voltage side is considered an infinite bus. Depending on the actual short circuit contribution on the high voltage side, means the low voltage fault current could be something less, or equal to the 12,100 A @ 208V.

However, for design, you must use the conservative number becasue you cannot control the amount of high voltage side contribution that may occur in the future. This all, of course, is assuming that the contribution from the loads on the 208 V side is negligible.

IMO, following the bussmann calc is a mistake, unless you can validate the calculation/formula. From the approx. 8200 A, I'd have some reservations using it.
 
JasonLSelf said:
I find it hard to believe that .1762 is all the available fault current at the secondary side of this transformer, especially when my handy cheat sheet for transformer fault current tells me that a 250kva 208V secondary xfrmr has 36,737 available fault current at the secondary side.

I think I am missing the mark somewhere, any help figuring out where would be greatly appreciated.
Click to expand...

Your cheat sheet could be based on maximum available fault current at the 250 kva transformer secondary based on a specified %Z and assuming infinite primary fault current available.
 
wirenut1980 said:
Your cheat sheet could be based on maximum available fault current at the 250 kva transformer secondary based on a specified %Z and assuming infinite primary fault current available.
Click to expand...

Right. In this case, the cheat sheet must be assuming ~ 1.8% impedance rating.
 
kingpb said:
To get the short circuit current on the load side of a transformer, you have to know what the impedance of the transformer is. If it is for a new installation then you might not have the nameplate data, but IEEE gives you standard values the manufacturers design too.

With that said, a 250KVA transformer is only possible of passing approximately 4.35MVA of short circuit MVA (@5.75% impedance), which translates to approximately 12,100 A @ 208V. That is assumming that the high voltage side is considered an infinite bus. Depending on the actual short circuit contribution on the high voltage side, means the low voltage fault current could be something less, or equal to the 12,100 A @ 208V.

However, for design, you must use the conservative number becasue you cannot control the amount of high voltage side contribution that may occur in the future. This all, of course, is assuming that the contribution from the loads on the 208 V side is negligible.

IMO, following the bussmann calc is a mistake, unless you can validate the calculation/formula. From the approx. 8200 A, I'd have some reservations using it.
Click to expand...

I don't think the problem lay in the bussman formula itself. I think the problem with the original calculation was a misapplication of the formula by not grouping some of the calculations within appropriate parentheses.

If you take your numbers above and substitute the OP's given 7% impedance rating, the short circuit current would become a bit over 9,900 amps with unlimited current on the primary side.

As I mentioned previously, I originally used a slightly different method to come up with essentially the same answer as the bussman formula. Here's the other calculation:

The available short circuit current of a secondary transformer is it's full load amps / (it's impedance rating + (it's kVA / the available primary short circuit kVA))

In the OP's example

Full load amp = kVA / (voltage * 1.732) = (250 * 1000) / (208 * 1.732) = 693.95 amps
impedance rating = 7% = 0.07
primary short circuit kVA = (short circuit amps * primary V * 1.732) / 1000 = (20,345 * 480 * 1.732) / 1000 = 16,914

so the available short circuit current becomes:

693.95 / (0.07 + (250 / 16,914)) = 8185

It's just a different slice of the same loaf of bread.


I do agree that a more conservative number should be considered, since the primary short circuit current may change.
 
Last edited:
Status
Not open for further replies.
Top