Short Circuit Calc - General Question 1

[SoFloJoe]

Hi,

I am doing a simple short circuit calc for a 4 plex. The fault current available is 15625 symmetrical amperes at the 120/240 volts.

I selected my bussman C value based on the 750 kcmil wire at 21,387

My calc looks as follows:
F = 2 x 75ft x 15625 / (21,387 x 240) = 0.45661
M = 1 / (1 + 0.45661) = 0.6652
SCA = 16,625 X 0.68652 = 10,727 A

So my question is about the AIC rating. Should I use a minimum rating of 10K AIC? I do not think there is a 11k AIC rating? Otherwise I would go with the 22k AIC.

Also when I change the distance lets say to 50 ft. It works inversely which goes against my logic. The distance is actually less than 75ft. With this inverse relationship shouldn't the distance be a minimum and not maximum? Right now I say that the conductor length is maximum 75ft.

Thank you for your help,

 

[waross]

The C value is the reciprocal of the impedance per foot.
Divide the C value by the total length of cable to get the reciprocal of the cable impedance.
Line to neutral short circuit current is taken as 1.5 times line to line short circuit current.
Adding the cable impedance to the transformer impedance assumes that the cable and the transformer have the same X/R ratio.
A reduced neutral will increase the cable impedance.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[waross]

This link may help.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[SoFloJoe]

Thank you for your help. Yes I am aware of the bussman pdf and am using that.

I am just curious as to how it relates to the AIC rating. Since the relationship of the length to the available short circuit calc is inverse it goes against logic. Meaning I would think the AIC rating would need to be increased with increased length of wire but it actually goes down. So I should not say max length of 75ft but rather minimum length of 75ft to meet the criteria.

 

[xnuke]

The SCA rating drops as the length of wire increases because there is more impedance between the source of the fault current and the end of the longer wire. If you want to be conservative when calculating SCA, choose a shorter length of wire so the SCA goes up.

You cannot use a device with an AIC rating less than the SCA, so don't even think of using a 10 kAIC with a SCA of 10,727 A. You must go to the next higher AIC available.

xnuke
"Live and act within the limit of your knowledge and keep expanding it to the limit of your life." Ayn Rand, Atlas Shrugged.
Please see FAQ731-376 for tips on how to make the best use of Eng-Tips.

 

[cuky2000]

Follow XNuke advises using a protective device with the next higher AIC available.
Another alternative is to increase the length of the feeder to be used as SC damping element. However, this could be a costly or impractical proposition.

Any reason to use the "C" Values for Cond for 5kV “C=21387”?.

See below for additional details.

 

[SoFloJoe]

Thanks for the responses.

5kv is what the primary service voltage is at the utility. So it made sense to match that, or am I mistaken here?