Copper Bus Ampacity - DC

[mityeltu]

In one of many CDBI self assessment findings it was determined that some of the calcs for our vital battery system were missing DC ratings for the battery board buses. I have searched all over the place for what I can call reliable data for copper bus DC ratings and have found nothing.

Everything I find relates AC ratings.

The manufacturer of this equipment (ITE) is no longer available for comment, so I can't go back to the vendor. Does anyone know of a reliable source of DC ratings for copper bus?

 

[ron]

This one is for Al http://www.substationsupply.com/images/PDF's/Aluminum Bus Bar Alloy 6101 Ampacity Chart...pdf

I would think CU is similar in its relationship between AC /DC and you can see the AC rating is a little less than the DC rating. So you can use the AC ratings you found as conservative http://www.stormcopper.com/design/Ampacity-Quick-Chart.htm

 

[mityeltu]

Unfortunately, part of the problem is thewierd dimensions of my bus (1" x 1.25"). I have been looking for a method of calculating the current density, but what I have found makes me want to throw my hands up.

What I am leaning towards at this moment is taking data from http://www.copper.org/applications/busbar/ampacity/bus_table1.html and interpolating the value I need by the surface area. Using this method, I find that the ampacity is 740A. Does this method make sense?

 

[broadgage]

I would say that the DC ampacity is the same as the AC ampacity.

The ampacity of a cable, busbar, or other conductor is limited to avoid it becoming unduly heated and being damaged, or damaging connections, supports , terminals etc.

An AC current rating may be assumed to be an RMS rating, unless stated otherwise. A definition of RMS current is "that AC current that produces the same heating effect as a DC current"
Therefore for all practical purposes DC amps = AC amps, as regards busbar ratings.

AC ratings of large cables with ferrous armouring or covering are often less than DC ratings, due to inductive heating, this is of no consequence for busbars.

AC interupting ratings for fuses or breakers are usually higher than DC ratings, but again this does not affect busbar ratings.

 

[Besoeker]

Unfortunately, part of the problem is thewierd dimensions of my bus (1" x 1.25"). I have been looking for a method of calculating the current density, but what I have found makes me want to throw my hands up.

Yes, odd dimensions.
Busbars usually have quite unequal dimensions for height and thickness. Very non-square section as a rule. This gives a larger periphery for a given area. For example a square bar 1" by 1" you would get and area of 1 sq inch with a periphery of 4 inches. The same area in 2" by 0.5" gives a periphery of 5 inches. This equates to better cooling and thus the bar can be run at a higher current for the same temperature rise. Or need less coper for the same current which of course, makes them cheaper.

What I am leaning towards at this moment is taking data from http://www.copper.org/applications/busbar/ampacity/bus_table1.html and interpolating the value I need by the surface area. Using this method, I find that the ampacity is 740A. Does this method make sense?

I wouldn't extrapolate from that and I don't know if you can. but then I do my own calculation for rating copper bar.
I think that your 740A might be a bit conservative but I'm from UK and rules are a bit different here.
That said, a common figure bandied about is 1000A/sq inch.

 

[steve66]

Unfortunately, part of the problem is thewierd dimensions of my bus (1" x 1.25"). I have been looking for a method of calculating the current density, but what I have found makes me want to throw my hands up.

What I am leaning towards at this moment is taking data from http://www.copper.org/applications/busbar/ampacity/bus_table1.html and interpolating the value I need by the surface area. Using this method, I find that the ampacity is 740A. Does this method make sense?

What are you using for temp rise?

Current tends to flow on the surface (even at DC), so I wouldn't try interpolating for the added thickness of your buss. But I would say you can claim the same ampacity as a 0.5" X 1" bus bar they show in your link.

 

[Speedskater]

Current tends to flow on the surface (even at DC), so I wouldn't try interpolating for the added thickness of your buss. But I would say you can claim the same ampacity as a 0.5" X 1" bus bar they show in your link.

Why would DC current tend to flow near the surface? "Skin Effect" is caused by the self-inductance of the conductor, so it's greater with higher frequencies and larger cross-section conductors. Does uneven heating have anything to do with it?

 

[mityeltu]

I'm using 65?C rise to maximize my ampacity rating. I have absolutely no vendor data on the installed bus. The vendor manual is scant to say the least. The interpolation is the midpoint between the .25 x 1 and the .25 x 1.5. This gives me the 740A. Interpolation using Excel and AC ampacity vs surface area gets me within 1% error of this value.

I don't believe I can say that AC and DC curent ratings would be equal because of the nature of the beasts. AC has the nice zero crossing. If I look at average current which would equate to the DC, they will not be equal.

Integrate the sine wave over a period of 0 to 2PI and then devide by 2PI to get average current. This is, as I understand it, the equivalent DC for the given AC current. That tells me that a bus rated at 1000A AC will be rated at 637A DC. This is bad for what I'm trying to do, but I don't see any way around this. From paragraph 1 this means, again as I understand it, that my bus is rated at 471A. This is bad.

Looking at heating (I2R), using the RMS of the AC current and the equivalent DC (as calculated above), the AC will produce a greater heating. If copper bus is sized based partially on the heating effects of the current, then I can take a small credit for this, but probably not enough to cover my bum.

When my plant goes black and everything is loaded onto the batteries, my bus has to be able to handle 658A (for the first minute). After this, load shedding takes care of most of the problems. But for that minute, things are going to be toasty. Given the size of the bus, I don't see any way to justify saying it is rated appropriately.

 

[broadgage]

I believe that the AC and the DC ratings will be the same.
The heating effect of an AC current is in proportion to the RMS current, and not in proportion to the average current.
100 amps AC (RMS) will have exactly the same heating effect as 100 amps DC.

Many types of fuse are listed for both AC and DC circuits, they operate at the same current if used on AC or DC. If the DC current had a greater heating effect, then the fuse would blow at a lower current on DC, which is not the case.

Incandescent lamps work just as well on AC as on DC, they get equally hot, and therefore emit equal light on AC or DC.
If DC had a greater heating effect, then the lamp would be much brighter, and have a much shorter life on DC, this is not the case.

My 12 volt soldering iron works equally well from a 12 volt vehicle battery or from a 12 volt transformer, it does not get hotter on DC.

 

[Speedskater]

"Integrate the sine wave over a period of 0 to 2PI and then devide by 2PI to get average current. This is, as I understand it, the equivalent DC for the given AC current. That tells me that a bus rated at 1000A AC will be rated at 637A DC."

But, but you started with the RMS value of the AC current not the peak value. The reason that RMS value for AC current and Voltage is almost always used is that it's the equal/equivalent of the DC value. In any conductor the DC resistance is always equal to or less than the AC resistance/impedance.

 

[mityeltu]

@speedskater
You're right. I used 1000A as RMS. That makes my resulting calculation incorrect. Thank you. That actually makes my DC rating 900A after converting to peak curret by multiplying by sqrt(2). Thank you. This gives the bus I'm working with, after interpolation of the 740A, a DC rating of 664A. That's better.


@broadgage
I'm not sure that's correct. If the I2R is calculated using RMS, it will be higher than the DC equivalent. The DC equivalent will be lower than the RMS. This should make it better as the amount of heat generated from I2R losses is less with DC.

I understand your example of the fuses, but I don't know how to recitfy that. Am I making a calculation error? There are big differences between AC and DC. All I know to do is to make them equivalent to one another. I can't see that AC RMS is the same as DC. Is there something I have forgotten here?

 

[Speedskater]

Once again, in a resistive circuit the AC RMS current is the equivalent of the DC current. That is 1000 Amps AC RMS is the equivalent of 1000 Amps DC. The RMS part of the AC value is the square root of two of the peak value.

 

[broadgage]

AC (RMS) IS the same as DC.
Indeed an earlier definition of RMS was "equivalent in heating value to DC"

To prove this by calculation is tedious, but entirely possible.
The resistance of the conductor may be regarded as fixed.
Calculate the actual current at say 100 points on the sine wave, then calculate the heat produced at each of these points in time (current squared times resistance), then add up the 100 figures and divide the result by 100 to obtain the average.
The result will be almost the same as the square of the steady DC current (any small difference is due to calculating at 100 points, rather than an infinite number of points)

RMS means Root Mean Square, or put simply, take take the square of the current at each point on the sine wave, then calculate the mean or average of all these squares, then extract the square root of this average.

To avoid tedious calculations, in the real world AC voltages and currents are almost allways RMS values, and NOT peak or average values, unless specificly stated otherwise.
Any reference to the safe current capacity of any conductor, or indeed the safe working voltage, may be assumed to be the RMS value.

 

[iceworm]

... Is there something I have forgotten here?

Yes.

... I don't believe I can say that AC and DC curent ratings would be equal because of the nature of the beasts. AC has the nice zero crossing. If I look at average current which would equate to the DC, they will not be equal.

... Integrate the sine wave over a period of 0 to 2PI and then devide by 2PI to get average current. This is, as I understand it, the equivalent DC for the given AC current. ...

For an AC sine wave with no offset, "Average Current" has no meaning - has no value - literaly. The integeral evaluates to ZERO.

That's why the RMS value is used. As already said, this integeral evaluates the equavelent heating value as an equal value DC current. And, yes, the RMS defining integral requires you use the AC peak values.

Also note the RMS voltage times the RMS current gives the Average Power. This concept of "RMS power" is strictly an advertising gimmick.

Back to your design analysis:
I can't see your application from my side of your keyboard. However here are some questions/oberservations - most have already been mentioned.

1. For DC, the current density is uniform across the cross sectional area. The heating effect should be slightly less than an equivalent AC RMS.

2. Why are you limiting the temperature to 65C? The conductor temp is normally limited by the termination equipment temperature spec. It's unlikely the terminating equipment is less than 75C and possibly is 90C.

With the limited information we have, 700A DC on 1.25sq-in sounds fine to me.

ice

 

[BJ Conner]

Anderson Electrical Data Book

Anderson Electrical Data Book

The best source of the data you need. It's PFD make a copy for your library.


http://www.hubbellpowersystems.com/literature/connectors/AEC-41.pdf

 

[jim dungar]

The best source of the data you need. It's PFD make a copy for your library.


http://www.hubbellpowersystems.com/literature/connectors/AEC-41.pdf

Great reference. It has hardly changed since the 1955 version I have (i.e. some sections have been dropped).

 

[Speedskater]

To make things more complicated, most budget meters measure the average value of AC Volts and Amps. The meter then does some math and displays the effective value (RMS) of the AC Volts and Amps readings. This works fine for sine waves, but the less like a sine wave, the more error in the read-out.