The purpose of this write up is to help the reader estimate the sExample before you hire the PE to do the official calculation.

All electrical systems are susceptible to short circuits . The short circuit currents can produce considerable thermal and mechanical stresses in electrical distribution equipment. Therefore, it’s important to protect personnel and equipment by calculating short-circuit currents during system upgrade and design. Because these calculations are life-safety related, they’re mandated by 110.9 of the NEC.

In the example show in Fig 1 our goal is to find the irrupt rating for the two breakers (1600A and 250A) and also find the withstand rating of the two panels (main and sub panel).

Fig (1) |

First will calculate the full load current for the 1MVA transformer

I F.L = P / (1.73 * V L-L ) ; where P is the transformer power rating in VA, V L-L is the line to line RMS voltage at the secondary side of the transformer

I F.L= 1,000,000/ 1.73*480= 1,202 A ; the I F.L is the full load current of the transformer

Now to find the short circuit rating of point 1 (i.e if we bolt all the three wires of the secondary side of the transformer how much current we will see at that point)

I S.C. for 3 Ph= I F.L /Z% ; where Isc is the short circuit current , Z% is the transformer impedance which is usually can be obtained from the local utility company

I S.C. 3 ph for point 1 = 1202 /0.05 = 24,506 A or 24.5 KA (this is the available short circuit current at point 1)

The short circuit current at point 2 will be lower than point 1 due to the line impedance ( line impedance for the cables will impose the current)

The calculation below will show the short circuit current at point 2

From the drawing ; cable length =L= 50 ft.

F = (1.73 * L* I S.C. 3 ph) / (C *n*V L-L) ; F for three phase fault ——–[1]

F = (2 * L* I S.C. L-L) / (C *n*V L-L) ; F for line to line fault ——–[2]

F = (2 * L* I S.C. L-N) / (C *n*V L-N) ; F for Line to Neutral fault ——–[3]

In our case we will use for this example equation #1

C= the conductor factor , value can be obtained from table (1) by locating the conductors size in the proper configuration

n= the number of parallel runs

L= length in ft

Table (1) These values are equal to one over the impedance per foot based upon resistance and reactance found in IEEE std. 241 |

F= (1.73 *50 *24,506)/(28,303*5*480)= 0.031206

M= 1/(1+F) =0.969738 ; where M is the multiplier

I S.C. 3 ph for point 2= 24506* M= 24506*0.969738= 23,764 A or 23.7 KA (this is the available short circuit current at point 2)

Its worth to mention is that the further the main panel from the transformer the lower is the short circuit current but you will have bigger voltage drop due to the larger line impedance.

The calculation below will show the short circuit current at point 3 (similar to point 2 calculation)

F = (1.73 * L* I S.C. 3 ph at point 2 ) / (C *n*V L-L)

F= (1.73 *100*23,764)/(16673*1*480) = 0.51370

M= 1/(1+F) = 1/1.51370 = 0.6606

I S.C 3 ph for point 3 = 23,764 *0.6606 = 15,699 A or 15.7 KA

I hope you find this example and the short circuit calculation useful. please let me know if you have any questions

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