short circuit calculation

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billsnuff

billsnuff

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15 KVA xfmr = 480 vac / 120 vac 125A

15 * 1000 = 15000/120 = 125

100/2%=50

125 * 50 = 6250A

L-N f = 2 * 30 * 125 = 7500/4760 * 120 = 571200 7500/571200 = 0.0131

M = 1/1+0.0131=0.9870

6250*.987=6169A I s.c. sym RMS

4*125=600A I sym mtr contrib

I total s.c. sym RMS = 6169 + 500= 6669A

is this correct since there is no E l.l. (120 vac not 240 vac)
 
You have not detailed your data. I understand the 15 kva is a 480/120 volt transformer. Is the %Z = 2%? If so the 15000/120 volts = 125 amps.
125 amps/%Z = 125/.02 = 6250 amps. I can follow the motor contribution figures. Motor contribution is the LRA of the motors. Many cases it is disregarded.
 
bob, thanks for the reply, i am adding data to a plant one-line using Cooper Bussmann.

1 ph I l.l. = (kva x 1000) / E l.l.

Multiplier = 100 / %Z

I s.c. = I l.l. x Multiplier

1ph L-N fault f= (2 x L x I l-n)/(C x E l-n)

C = conductor from table

Multiplier M = 1 / (1 + f)

I s.c. sym RMS = I s.c. x M

I sym mtr = 4 x I l.l.
 
billsnuff said:
just had a brain cramp cause this was 120vac secondary and not 240 vac.

so, Impedance line, line (I l.l.) = 120vac in this case, i guess......:confused:
Click to expand...
I have seen this method of making the calculation but not familiar enough to comment.
If you had a 480/240 ground center tap so that you had secondary of 120/240 volts, a line to neutral fault would be higher that a line to line fault.
I believe a multiplier of 1.5 x L-L fault is used to calculate the line to neutral fault. However if you have a secondary of just the 120 volts then the 6250 is correct.
 
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