Short Circuit Amperage

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wplanore

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I have a 75 kva transformer, 480-208y120, 3 phase ,5.5%Z. I know that there is 19,963 sca at the primary. I calculated that there is 3,501 sca at the secondary. Am I correct. I used the point -to-point method. Am I correct? Just want to check my calculation.
Thanks
 
If you want to post your calculation, we can comment more intelligently. I'm coming up with 3496 amps, so you're probably right, or maybe we are both wrong.

Jim T
 
wplanore said:
I have a 75 kva transformer, 480-208y120, 3 phase ,5.5%Z. I know that there is 19,963 sca at the primary. I calculated that there is 3,501 sca at the secondary. Am I correct. I used the point -to-point method. Am I correct? Just want to check my calculation.
Thanks
Click to expand...
3498. Close to your answer.
 
Short Circuit

Short Circuit

The formula that I used is in the Cooper Bussmann Bulletin EPR-1

f= I(pri) x V(pri) x 1.732 (%Z) /100000 x KVA

M= 1/1 + f

I(sec) = V(pri)/V(sec) X M x I(pri)

Thanks for your comments

Wayne
 
Weird...the spreadsheet has an entry box for available utility fault current. I didn't check the formulas to verify if it was calculating the impedance of the primary side.

But, even when doing it by hand, I get the following.

high-side impedance = 0.024 ohms

Transformer Impedance = 0.055 ohms

(480/208)*(0.024 + 0.055)*19,963 = 3639 amps at the secondary.

Still a considerable difference...I welcome your thoughts on how I'm doing this incorrectly.
 
bcorbin said:
Weird...the spreadsheet has an entry box for available utility fault current. I didn't check the formulas to verify if it was calculating the impedance of the primary side.

But, even when doing it by hand, I get the following.

high-side impedance = 0.024 ohms

Transformer Impedance = 0.055 ohms

(480/208)*(0.024 + 0.055)*19,963 = 3639 amps at the secondary.

Still a considerable difference...I welcome your thoughts on how I'm doing this incorrectly.
Click to expand...

To change sides of the transformer, i believe the multiplier for z is the voltage ratio squared. Also the transformer impedance is 5.5% not 5.5 ohms.

I calculate the high side impedance to be .0138 ohms at 480 volts or .0026 ohms at 120/208. I find the transformer z to be .0317ohms at 208 volts and the total system impedance to be .0026+.0317 or .03433 ohms.

120/.03433=3496 amps fault current.

Jim T
 
For transformer of this rating, it is best to treat the high voltage side as an infinite bus. As systems change, so does the fault current and in the future it could easily become an infinite bus.

Therefore, the 3ph fault calc becomes quite simple:

75KVA/0.055 = 1364KVA

1364KVA/(208xsqrt3) = 3785 A (low voltage side)

This would be the worst case. At any rate the minimum, rating for AIC is (I believe) 10,000A, and therefore you are quite within the rating.

In some cases 1ph to ground can be slightly higher then 3ph. But this is not common, and even if it was it would be only slightly higher.
 
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