Conductor Sizing

[electricaleng30]

I've been looking for some formulas to help me calculate conductor sizing, derated values, ampacity and so on. For example, say I have an 800A distribution panel, and I have an 800A MCCB that could be used at 100% of it's rated value (800A plug), what would be the size of the conductor that I would require for this? Whether it be single conductors, the type of insulation, multiple conductors, Nu-Al, Copper and so on. I was reading another thread that was saying how bad electrical contractors were for disputing mathematical calculation and following the code and we have lived that situation many times before. One formula that I have used in the past is:

cross section = (rho x (Power x lenght)) / Uv x admissible voltage drop.

There are all kinds of calculators on the net, but seems like I have yet to find a good information that would make it clear and simple as to what conductors values should be used when running a new feed. For that application above, we went with parallel runs of 750MCM Nu-Al 3 conductors (free air application, in a cable tray suspended from the ceiling). If anyone has documentation, literatures, even good books on sizing conductors, please send them my way. Thanks a lot.

 

[dpc]

This is determined by your local electrical code and/or the UL requirements for the circuit breaker. It's not something you can calculate or even apply logic to. It is what it is.

The breaker requires a certain minimum size conductor, but larger conductors may be required due to voltage drop, thermal derating or other factors.

It will be different if you are in the US, or in Canada, or elsewhere.

 

[wareagle]

You need to look at the terminals on the back of the breaker
and get the temperature rating of the terminals.. It should be 75C or 90C. It will likely be 75C. IF so you will need a conductor to handle the calculated load using the 75C rating of the conductor. Depending on the load calculation
the conductor may be 2-500 kcm copper/phase = 760 amps
or 2-600 kcm/phase = 840 amps. IF the terminal rating is 90C
then you would use the 90C rating of the conductor and
2-400 kcm/phase = 760 amps or 2=500 kcm/phase = 860 amps. You could also use smaller parallel conductors that match the load.

 

[waross]

You may want to take a code course in your area. I hoe that I have misunderstood some of your statements.
An 800 amp frame breaker with an 800 amp plug is only acceptable for 100% load if it is specifically approved for 100% loading. The ability to procure and install an 800 amp plug in an 800 amp breaker does not by itself make the breaker suitable for 100% loading.

For a good explanation of the code, if you are in Canada, buy the
CE Code HANDBOOK
An Explanation of Rules of the CE Code, Part I
It will give you some insight into the factors that were considered when the Code ampacity tables were developed.
It will give you a greater respect for the Codes and in the long run make your job easier.
Don't redesign the wheel. The ampacity tables were developed by competent engineers. The tables are proven. Any values at odds with the tables are suspect until proven.
You may be assuming more liability than you want to when you short change the code minimum requirements.

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

 

[raptor6458]

I agree with dpc and waross, you need to familiarize your self with your local electrical code. also familiarize with clients specs (i.e single core/multi core, XLPE/SWA/PVC,etc), derating values are publish by cable manufacturers and are based depending on their area (IEC, NEC, etc) and I believe are given free.