How to calculate Ampacity of a Rectangular Copper Bar 2

[nonsinusoidal]

Does anyone know how to calculate the ampacity of a rectangular copper bar. Please see the attached drawing for details. Please note that the cupper bar is going to be utilized as a terminal point for conductors up to 600 volts. The terminations will have two hole lugs as as shown on the drawing. Any suggestions to accomplish this task are highly appreciated.

 

[CM3PhaseDesign]

It's going to be around 4000A, however it does depend upon your maximum temperature, and the ambient temperature.

Also is it AC or DC ?

 

[mcgyvr]

http://www.copper.org/applications/busbar/homepage.html

see the tables at the bottom..

 

[nonsinusoidal]

Hello CM3PhaseDesign,

Can you expand on your response? It will be an AC power source. Maybe you can explain how 4000 amps came about.. Thank you

 

[Zogzog]

mcgyvr, thanks for the link, bookmarked.

 

[waross]

The old rule of thumb was 1000 Amps per square inch.
4000 Amps seems a little high for 8" x 1/4" (2 square inches).
The tables in mcgyver's post are a little higher (about 25%) than 1000 Amps per square inch.

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

 

[edison123]

1000 amps/sq. inch still rules in India.

Muthu

http://www.edison.co.in

 

[asco1]

Per UL891 Table 23 & 24:
less than 3000A: 1000A per sq. in.
3000A - 6000A: 800A per sq. in

This is what we typicaly use in our switchboards. Some city codes require 800A per sq. in. everywhere.

Your .25 x 8.0 bar will need to be 20" (or double up so that you have 20" or more of .25 cu)

 

[waross]

Your .25 x 8.0 bar will need to be 20" (or double up so that you have 20" or more of .25 cu)

The original poster has not stated a current. A response suggested that the bar would carry 4000 Amps. Most of us would not use that figure.
However this application is not as simple as many of us would assume. There are five parallel cables feeding in and five parallel cables feeding out.
Each cable pair has 16/5 = 3.2 inches of bar. This may be considered as five individual cable splices each with a connecting strap 3.2 inches wide by 1/4 inches thick.
That is .8 square inches and 800 amps per cable pair.
Five cables times 800 Amps per cable equals the 4000 Amps suggested by CM3PhaseDesign.
The direction of current flow is such that we should be using the 16 inch dimension rather than the 8 inch dimension to calculate the ampacity.

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

 

[asco1]

Waross, you are correct in you assessment....The 8" dimension is not to be taken into consideration.

At 800A per sq in, you will need 20" of .25 cu to carry the 4000A. (the 8" dimension can be anything)

 

[rcw retired EE]

With 5 cables in and 5 out, could the bars be configured with the connections back-to-back using the same bolts to hold the incoming and outgoing cable lugs to either side of the bar? The current flow would be through the 1/4" dimension of the bar and not along the bar. Some equalizing current would flow along the bus between the pairs of cables, but most would be through the bar.

That arrangement ends up looking like a low voltage transformer bushing with 10/phase cables, five on each side. 1/4" may be too small for structural reasons.

 

[waross]

asco1;
The question was not how much copper but how to calculate the ampacity.
By your standard the ampacity would be 3200 Amps.
In India and other places the ampacity would be 4000 Amps.
There are other methods of calculating bus ampacity that would consider the relatively high circumference to area ratio of a 1/4" x 16" bus bar and possibly rate it higher than 4000 Amps.

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

 

[cuky2000]

a) How critical is to determine the ampacity of a rectangular copper bar?
b) Will be sufficient to know that the copper plate is rated with larger ampacity than the cables for this application?

Assuming the cable lugs are rated with the right size, if the copper plate area is larger than the sums of the cable lugs areas, I can’t see how the copper plate could be rated with less ampacity than the cables.

 

[waross]

Hi cuky2000;
We haven't seen you in this form for awhile. Good to have you back.
Bill

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

 

[cuky2000]

Hi Bill and all,
Thanks for your warm welcome back. I am engaged in a job that doesn’t let me a lot of breaks.
I promise myself and my wife to slow down, but as many of our colleagues, we keep going and going. Any advice how to slow down? Seen this is a common down syndrome to be an engineer.
Thanks Hi Bill and all,

Thanks for your warm welcome back. The reason (excuse) of been less active in this forum is that I am engaged in multiple jobs that doesn’t allow for breaks.

I promise myself and my wife to slow down, but as many of our colleagues, we keep going and going. Any advice how to slow down?

Seen this is a common down syndrome to be an engineer.

Thanks

 

[RNRPE]

Yes, I do. In fact my company OrionMagnetics, LLC is currently developing a software to thermal rate (current carrying capacity)equipment such as rigid circular bus, circuit breakers (medium and high voltage), bare overhead conductor, CT's, etc.

 

[nonsinusoidal]

I am astonish by the diverted responses. I am back to where I started.......

 

[edison123]

Then may be you didn't read or understand the posts. How much are you paying for these helpful tips again ?

Muthu

http://www.edison.co.in

 

[waross]

How much ampacity do you need?
Are you bound by any codes?
1000 Amps per square inch is a historical standard.
The link given by mcgyvr Gives tabulated values based on temperature rise and heat radiation from different shape ratios. The link also gives the calculation method which is what you originally asked.
asco1 gives you UL891 Table 23 & 24 at 1000Amp and 800 Amp and mentions that some codes use 800 Amps for all installations.
All of the answers are valid. You have to know where you are for starters. If you want diversity look at the ampacity tables in the NEC.


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

 

[RNRPE]

One of the best resources for calculating ratings for bare overhead conductor, rigid circuilar bus, and rectangular bus are IEEE Standards 738-2006 and 605-1998. These standards provide heat balance equations that are easily used for rectangular bus.

As was mentioned previously, there is slight difference between AC and DC rating, depending on the size of the bus, since Skin Effect may have to be taken into consideration. If the bus is to be used indoors, then the Solar Heat Gain is 0 and there will be no benefit from forced convection cooling, i.e. wind.