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Cable Short Circuit Withstand Capacity 2

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timm333

Electrical
Jun 4, 2009
121
Can someone please explain what does the graphs at the following link mean? These graphs indicate that the 8 AWG cable can withstand only “6 kA” short circuit current at one cycle. But in actual practice we use 8 AWG cable all the time for 50 HP motors where the available short circuit current is far more than 6 kA.

 
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8 AWG can withstand just over 9 kA for one cycle based on that graph. [reading]


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Hi timm333

The graph that shows 6kA for one cycle is an aluminium conductor with a thermoset insulation rated for 105 deg C, what is your conductor material, insulation material and continous temperature rating as these will make a difference.

desertfox
 
Yes, but even for copper conductor on page 2, the 8 AWG cable can withstand only 9kA, which is far less than the actual available short circuit currents.

For example, if a motor is connected to 2000 kVA transformer at 480 V secondary, the available short circuit current will be about 45 kA. The minimum size cable that will withstand 45 kA on page 2 is #1. So what if we have to connect a 1 HP motor to this transformer. Shall we have to use a minimum of #1 cable for the 1 HP motor? Hard to believe.
 
Available short circuit current is also limited by cable attached, particularly on the lv side. There was a recent thread on a similar subject.

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Is there some easy way to calculate how much short circuit current is limited by 8 AWG cable?
 
Is there some easy way to calculate how much short circuit current is limited by 8 AWG cable?

Yes. Convert 50 kA to source impedance in ohms. You would know the voltage. (Look up basic formulas, if needed.) Add the impedance of the source to cable impedance in ohms and divide the total impedance by phase voltage (L-N). OR use a software.

At 480V, 50 feet of #8 cable brings the 50 kA below 9 kA and 40 feet brings it below 7 kA. A quick check also shows that an ordinary 40A MCCB will open fast enough at 5.7 kA and below, to protect #8 conductor. However even a 100A, class J fuse will protect the #8 conductor for 0.9 seconds.

There a few other things to consider. 50 HP motor at what voltage? . Per NEC (in conjunction with 110.14), #8 is generally only good for 40A. Even many local AHJ requires using 60 deg C column for sizes #1 and below. I personally do it as a routine.

Yes, you do need to consider short circuit withstand rating of the cables while sizing it. Especially, when you are tapping a #8 wire right at a 50 kA bus, as you describe or use current limiting fuse/breakers. I would think this is very rare. In most applications, NEC conductor sizing is found to be adequate for thermal withstand rating but there can be exceptions and you need to check it when in doubt.

Hope this helps.


Rafiq Bulsara
 
Rafiq pretty much hit the nail on the head.

If the conductor is protected by a fuse or molded-case circuit breaker sized in accordance with the NEC, it's probably OK. It's always good to verify by plotting the time-current curve for the cable damage on the same graph as your overcurrent protective device trip characteristics. The are a number of computer program that can do that for you.

Alan
“The engineer's first problem in any design situation is to discover what the problem really is.” Unk.
 
First of all the I"k in secondary of 2500 kVA 480 V transformer 6% [uk%] at the end of 100 ft 8 awg copper cable it is only 4.3 KA.
Second: one cycle fault clearing it is not possible. You need at least 70 milliseconds.
So, for 4.3 KA and 0.07 sec 8AWG will be fair enough. If the cable length will be less than 100 ft it will be a problem, indeed.
 
Short circuit current of 4.3 kA is at the end of the 100 feet 8 AWG cable. It will still be 50 kA at the start of the cable.

I agree that 8 AWG cable will be ok. I just want to understand the reason why the cable is rated for 4.3 kA and not for 50 kA.
 
If the short-circuit occurs downstream the cable-in the motor, for the above case- the cable has to remain in condition to continue the supply. If the short-circuit occurs in the cable itself you have to replace the cable. The upstream equipment shall be protected then, only.
 
In low voltage, short circuit is seldom the factor for selection of cable sizes. It is always the ampacity & voltage drop which rules the roost. Usually the impedance of the cable is good enough to drop the short circuit values to very low values within a few mts from the MCC / LV bus. Also, the LV circuit breakers / Fuses are good enough for fast interruption. In fact the current limiting type of MCCBs and fuses interrupt within half a cycle. The ICEA log-log characteristics should be used for MV cables. Cable sizing in LV may only be valid for short circuit for Power CB fed cables where intentional delay may be set to coordinate with downstream devices.
 
I agree, but unfortunately I am still confused what will happen if the fault occurs on the cable within the first few meters of the MCC/LV Bus; in theory the cable will be damaged before the opening of breaker.
 
From what cherry200 said, inverse time characteristics of your cable protection should be good enough to get your cable well protected. If you have set tripping to cover the cable length, any fault a few meters from the supply terminal will surely initiate earlier trip signal as the fault current will be greater than your setting. FWIW, if you follow your code, you are required to protect your cable cable from mechanical damage; hence faults along the cable length would be prevented.
 
If you have a fault in the first few meters out from th MCC/LV bus the cable is damaged. Period. Doesn't matter how fast the clearing time is; the only way to have the fault is due to cable damage.
 
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