short circuit calculation

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elicio

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Mr. bennie & charlie I will try again ask yhe question about short circuit.
I have 7.5 hp motor connected to MCC at 380 meters. The voltage system is 480V (Line to Line), maximun short circuit at the MCC is 39.55 MVA and i need to calculate the appropriate cable by tree methods, current capacity, voltage drop and short circuit level. I have already calculated this cable and is AWG #4. But i have a doubt, where do i have to calculate the short circuit ? at the motor?s connection or near the MCC.

The motor?s feeder is a continuous (1x3c #4 AWG) by tray and does not hace any connection betwen the ends.

The NEC have some article where define that is necesary the cable calculation by short circuit method?
 
Re: short circuit calculation

Go to your first post. You need not continue to add posts to the forum If you have additional question, go to the origional post and ask them.
 
Re: short circuit calculation

Originally posted by elicio:?I need to calculate the appropriate cable by three methods, current capacity, voltage drop and short circuit level.
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Why? Who says so? Who (or what book, or what law) do you think is asking you to calculate using three different methods?

The NEC does require that you install a cable that has enough ampacity (or, using your phrase, ?current capacity?) to handle the load. The NEC suggests, but does not require, that the cable be sized large enough to create a voltage drop no more than 3% along a feeder and 5% all the way to the load. The NEC has no requirement for addressing short circuit levels as a part of the process of selecting a cable size.

The NEC have some article where define that is necesary the cable calculation by short circuit method?
Click to expand...
No it does not, as I have told you before.

By the way, I agree with Bob. So I'm closing the other threads, and refering back to this one.
 
Re: short circuit calculation

elicio, I will take a shot at this. I often use voltage drop to determine cable size rather than the minimum code requirement. The point is as Charlie has stated is it?s not required by the NEC and in fact could get you in trouble for short distances.

I will assume you are trying to keep the voltage drop to less than 5% The formula I use is: CM=(2KIL)/VD

Where;
CM= Circular Mill area of copper cable
K= constant of copper = 11.1 to 12.9. I will use 11.1
I = Load current = 11
L = 1250 feet in place of 380 meters
VD = Voltage drop = 5% of 480 = 24

Answer is 12718 circular mills or next size up is 16,510 or an 8 AWG.

When you use this method you must cross check the answer with 310.16. Since this is a motor using 11 amps the required minimum is a 14 AWG. So an 8 AWG far exceeds any code requirement. And if you want to use a 4 AWG go right ahead your voltage drop will be reduced to 7.3 volts.

Is this the answer you are looking for? :confused:

[ September 25, 2003, 11:05 PM: Message edited by: dereckbc ]
 
Re: short circuit calculation

Just a long shot: This could be a misunderstanding. Someone came up with the idea to relable the "National Electrical Code" "International Electrical Code", and I think it's sometimes referred to as IEC today?

That's arrogant and confusing, sine there is already a well known organisation called IEC which has an electric code. This code does contain requirements to do short circuit calculations, but the NEC is not based on the IEC standards.
 
Re: short circuit calculation

Yes, gentlemans, thanks you for your cooperation, you are very kind,
Regards,
Elicio
 
Re: short circuit calculation

Using the data of the previous post's and the Bussmann point to point method, I get:

39550 Available Short Circuit Current (ASCC)
480V
motor FLA 11 Amps
1250 feet to motor
4 AWG Copper
55 Amps of motor contribution

Nonmagnetic raceway, the ASCC at the motor would be 830.3 Amps

Magnetic raceway, the ASCC at the motor would be 826.2 Amps.
 
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