Heat transfer at electrical lug connections.

[dbecker]

Hello.


I am faced with a real problem. I need to perform an analysis on electrical lug connections used in very large circuit breakers.

The lug connections use about 1" thick solid copper bus and terminate to 4 large DLO 313kcmil cables. Each side of bus has two lugs, a total of 4 per end of bus.

Now comes the problem. I need to assume some heat generation. If I am given the current and voltage that this unit (per bus connection) will be seeing, what and how do I assume contact resistance and Q heat generation?

Also, where should I apply this Q? I am assuming I shall apply this Q right at the interface of the two surfaces (Bus/Lug) in the form of heat flux at those surface areas.

If my assumptions are correct, please confirm and if not, please advise.

Thank you and I love this forum.


- Dan

 

[electricpete]

correct me if I'm wrong but isn't contact resistance unnaffected by contact area given a constant surface pressure?

Contact resistance in ohms is inversely proportional to area. Double the area... halve the resistance in ohms.

As I stated above, it appears the value plotted in the link is supposed to be (R*A) in units of ohm-mm^2. That would be a number that depends on contact pressure materials, etc, but not area. To back out the resistance of a given geometry, divide that number by area.

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[hacksaw]

bdecker,

The contact resistance you are using might be okay for freshly made connection not for typical industrial application. If you were running 28 W junction heat that results in something on the order of 0.5C/cm temp gradient along the buss.

 

[dbecker]

Yea hacksaw the contact wattage output is like 20 watts according to my calculations. This is very low, I think the calculation is wrong perhaps.

I guess my data that points to contact resistance being 12 micro-ohms is way to small. I have to talk to the test engineer about this. That is a very small resistance for a seemingly high temperature lug connection.

I know for a fact that even freshly made lug connections get hot by at least a 20-30C rise. And that is without 6 months of oxidation.

I think I need to get more accurate information.

So far everyones input has helped my greatly.

Also, electricpete; check out desertfox's website. It said

"It has been shown above that the contact resistance is dependent more on the total applied pressure than on the area of contact. If the total applied pressure remains constant and the contact area is varied, as is the case in a switch blade moving between spring loaded contacts, the total contact resistance remains practically constant."


Or am I misunderstanding?

Thanks for your time all.

 

[electricpete]

also, electricpete; check out desertfox's website. It said

"It has been shown above that the contact resistance is dependent more on the total applied pressure than on the area of contact. If the total applied pressure remains constant and the contact area is varied, as is the case in a switch blade moving between spring loaded contacts, the total contact resistance remains practically constant."


Or am I misunderstanding?

imo you are misunderstanding because the copper.org terminology is very poorly presented if not downright wrong.

Case A: Area=1mm^2, Applied Force = 50N, Pressure = 50N/mm^2
Case B: Area=2mm^2, Applied Force = 100N, Pressure = 50N/mm^2

Case B has got to have half the resistance (in ohms) of case A. That is the reality.

I think when your link talks about "resistance", it is not talking about resistance in ohms. Look right below the paragraph you quoted they give the equation Ri = C / p^n where Ri[sic] is "resistance of contact"[sic] and n is between 0.4 and 1. But below that is a graph with resistance as vertical axis which is intended to represent the equation... you can tell by the shape (Y~1/X^n). If the vertical axis was resistance in ohms, then we would see ohms on the vertical axis, but we don't. There is some units of area in there mm^2 in there written as microhm/mm^2 [sic].. As pointed out previously, the mm^2 belongs in the numerator not the denominator.

The quantity "Ri"[sic] they are trying to describe would be the inverse of conductance per area. i.e. Ri[sic] = 1/[G/A] where G=1/R. It can also be written as Ri[sic]=R*A and the units are micro-ohm*mm^2.

Sorry for getting carried away with the sic but it's hard to describe it without using their terminolgoy.

In a knife switch it is not the pressure that is constant, it is the total force that is constant as the contacts slide further closed, not the pressure. With constant total force and increasing area the the pressure goes down. Change in pressure cancels out change in area to give constant resistance.

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[desertfox]

Hi dbecker & electricpete

Indeed figure 16 the graph of "Resistance against Pressure" could be explained better,I looked in an old copy of "Copper for Busbars" which dates back to 1965 and I found that same graph, but underneath it the following was written:-

:-"Nevertheless it is convient,when illustrating in graphical form the variation of contact resistance with pressure to reduce both these quantities to a "unit area" basis, bearing in mind that both axes really represent average quantities taken over the whole "apparent" area of contact.

Further on it goes onto say that to obtain the total joint resistance then:-

divide the ordinates by the contact area .

So these two small statements can make you look at the graph in a completely new light.

desertfox

 

[desertfox]

hi dbecker/electricpete

Here is another reference about the resistance of bolted connections, this might make it clearer.
One thing I should mention with respect to joint area is that we calculate the average pressure on the joint not the actual pressure which by the nature of the metallic surfaces is in reality unknown.

http://books.google.co.uk/books?id=EkStW7v8VPkC&pg=PA220&lpg=PA220&dq=resistance+

of+electrical+busbar+joints&source=bl&ots=FpdQ_
jmVti&sig=jJCOJT0PL1SuXCiGQa-eJ0Z7x4c&hl=en&ei=eYGXS8GrNaj40wTd8rjqCw&sa=X&oi=
book_result&ct=result&resnum=3&ved=0CAoQ6AEwAjgy#v=
onepage&q=resistance%20of%20electrical%20busbar%20joints&f=false

It would be helpful if the OP could provide more specific details about the failure, as I have requested several times now, without more information the thread can't get much further.
The way I look at this discussion about resistance is:-
take a one metre length of conductor whatever its its area, now double the area of that conductor keeping the length constant and you will halve the resistance, conversely double the length and keep the area constant and the resistance doubles.
In the bolted joint however my interpretation is that if the actual pressure on the joint is not changed significantly by a slight increase or decrease in area of the conductors, then the resistance will not change, part of the reason I say that is because the force distribution is mainly centred around the fasteners ie washers, nuts and bolt heads, once outside these area's the force tends to drop off significantly.
I could by wrong but these are my thoughts.

desertfox

 

[IRstuff]

A backdoor approach is to use the known data, i.e., the known temperature rise and the known current to estimate the apparent resistance.

Since all of the power is eventually dissipated into the air, we can write:

power = area*htc*known_temperature_rise

let:
htc = 2.5 W/m^2-K (could be as as high as 10 W/m^2-K)
area = 4sides*1inch/side*2.5inchlength+4cables*700mm/cable(diam)*pi*2.5inchlength (assume that total heated length is divided by 2 to get the average length)

Then, power(=680W)/(1300A)^2 = resistance

Obviously, lots of swags here, but, for 60°C rise, the apparent resistance is on the order 0.4 milliohm, which is considerably higher than the "12 micro-ohm" measurement.

Anyway, food for thought...

TTFN

FAQ731-376

 

[electricpete]

It has been shown above that the contact resistance is dependent more on the total applied pressure than on the area of contact. If the total applied pressure remains constant and the contact area is varied, as is the case in a switch blade moving between spring loaded contacts, the total contact resistance remains practically constant.

Another nitpick on the copper.org terminology above... throughout the entire page they use “pressure” to indicate force per area. Here they are quite obviously using the term “pressure” to indicate force. One clue are the words “total applied” in front of pressure. Another clue is physics (the sentence is only correct if we interpret pressure to mean force).

The copper.org page assumes uniform contact pressure across whatever area we’re talking about and I made the same assumption in my discussion. I agree with desertfox the actual contact stress distribution in bolted connection may not be uniform.

I’m a little unclear about the big picture of this thread. It has already been identified sulfide corrrosion is the reason for the gradual resistance increase? (rather than relaxation of bolt tension for example). I would think as mentioned grease would exclude the contaminants. I would also think there has got to be some kind of plating that could be resistant. Why is it that you need a thermal model? The role of temperature in this sulfide corrosion would be affecting the reaction rate? Is there a threshhold temperature of interest that you are looking at?


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[dbecker]

electricpete, yes we have a threshold temperature that must not be exceeded. That is the point of all of my inquiries, how to better model the contact resistances so I can account for all the heat dissipated and in turn calculated component temperatures.

I am really after finding lug metal temperatures, that is the ultimate goal. I am using a 3D CFD model to compute local HTCs on the lug surface. All I need now are heat generation rate in the form of heat fluxes on my contact areas for my ANSYS model.

Give me a day to read over the last few responses, it's getting pretty hairy!

Thanks again,

 

[hacksaw]

electricpete,

the sulfide issue was specific instance unrelated to original post, but was mentioned to provide context for my own first hand experience relating to the op's query. interesting problem to say the least.

 

[electricpete]

Aha. I was definitely cornfused on that ponit. Thanks for straightening me out hacksaw.

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[JKEngineer]

dbecker -

I just saw this thread, so I'm coming to the conversation a little late.

By way of background, I do both thermal imaging and heat transfer analysis using FEA and CFD methods. I don't use ANSYS, so am not familiar with the details of its setup.

Here are some comments: (all include "in my opinion")
1. If you are looking at a failure or over temperature event driven by issues at the connection then the resistive heating in the remainder of the conductors due to their resistivity or shape changes will be minor by comparison. Yes, systems under load heat up, but failing systems whether due to loose connections, dirty connections, or corroded connections heat up a lot more. You should be able to ignore everything that contributes to normal load heating and look only at the unusual, which would be the increased resistance at the joint / part interface.

2. Using the heat generation of 20.28 watts on the interface is what I would do. I have done a sample analysis using that method, which I have used as a basis for papers at SPIE Thermosense and at EPRI's IRUG conference. I think it gives an excellent estimate of what will be going on. A couple of images from the paper are included in the brochure that is at the link I provide below.

3. You did not specify whether the component is air-filled, oil-filled, or under vacuum. If air or vacuum, radiation will have some effect, but most of the parts will be low emissivity and will not partake in significant radiative exchange. However, you need to assess whether nearby parts with higher emissivity are getting sufficiently hot by conduction or convection to be significant participants in radiation to the inside of the enclosure. If the unit is oil filled, then radiation will only be a factor on the outside of the enclosure to the surroundings.

4. Your most recent posting indicates that you are using the CFD model to assess heat transfer coefficients. That makes it sound like you are then doing manual calculations or non-CFD FEA calculations of the heat transfer. I would not use that route. Having CFD capability allows you to model the interior flow which will affect the heat removal from the interior parts and will also affect the temperature patterns being developed on the enclosure. These latter patterns may impact how the overall component dumps its heat to the surroundings. [I realize as I write this that using HTC for the parts means you are not dealing with an evacuated part.] I would suggest using the CFD for the full solution, including natural convection, especially on the interior. You might be able to use a film coefficient (plus radiation if appropriate) between the outside of the enclosure and the surroundings. If the system is oil filled, you will need density vs temperature properties for the oil. (BTW, if you have that data, I would appreciate your sharing it. I ran into an analysis that needed it, but could not find much.) If you wanted to, you could shorten the analysis time by doing pure heat transfer calcs in FEA without flow. I would suggest, in that case, that you increase the thermal conductivity of the fluid (air?) in the component by a factor of 2-3 to allow for the convective component and treat the problem as a conduction only exercise.

5. Since you question the resistance measurement, one approach would be to do a parametric study of T(lug) as the dependent variable vs heat generation rate at the contacts. This would be an implied function of the resistance of the contact. The test data of 12 microohms is a resistance, not a resistivity. (I say that because of all the discussion above about units and areas, etc. (not all of which I followed)) I would start with the 1300 amps you cite above, giving I2R= 20.28 watts as you state. You can then develop a curve of lug T vs. assumed resistance where the resistance is used to calculate the heat generation at the interface. 20watts may not seem like a big number, but it can do a lot of heating in a small volume.

I hope this helps.
Jack


Jack M. Kleinfeld, P.E. Kleinfeld Technical Services, Inc.
Infrared Thermography, Finite Element Analysis, Process Engineering

http://www.KleinfeldTechnical.com

 

[JKEngineer]

This is a link to a brochure on my website (which does not have an on-web link) that has two images from the FEA analysis I did of heating in an electrical component. This is strictly as a demonstration of the output obtained using the heat generation on the mating faces. BTW, this was done without CFD, so the pattern of heat on the component surface is not accurate. I also used the method of increasing the k of the air in the component to represent convective heat transfer.

http://www.kleinfeldtechnical.com/ktsinfo/FEAforThermographersBrochure.pdf

Jack


Jack M. Kleinfeld, P.E. Kleinfeld Technical Services, Inc.
Infrared Thermography, Finite Element Analysis, Process Engineering

http://www.KleinfeldTechnical.com

 

[unclesyd]

I ran across these while looking for a paper used by our Electrolytic Cell group concerning bus bars connections. I'm still looking for the paper.

http://www.informaworld.com/smpp/content~content=a713847350&db=all

http://www.genisim.com/download/tms2006.pdf

http://www.hsa.ei.tum.de/Publikationen/2002/02_st_1.pdf

 

[dbecker]

Thank you JKEngineer!

A couple things, the enclosure is air filled at ambient air pressure. There are vents and fans to help cool the lugs. I have the boundary conditions for the fans and vents and will be using these as my CFD BC's. The CFD is going to compute for my surface HTCs and from there I can apply them in my FEA (ANSYS) model and compute surface temps.

The only reasy I need heat generation or heat flux at the contact interface is for the ANSYS model to converge, I need the heat gen as boundary conditions for the ANSYS model.

Whoever is familiar with FEA, this is going to be a steady state solution with HTCs applied on all surfaces of the lug and no radiation assumed (T^4-T^4) is very low.

I will use 20.28W as I calculated before thanks for confirming that with me JKEngineer.

I will get back to you on this when I finish studying those links.

Thanks for all the help everyone.

 

[unclesyd]

The washer.

http://www.energy.siemens.com/us/po...topics-application-notes/techtopics16rev0.pdf

 

[dbecker]

Hello everyone. I am back. I am at the site looking at the circuit breaker.

desertfox, I spoke to the field engineer and he said the failure was partly due to a 130C temp rise at the lug. Not entirely due to that (long story). I mentioned earlier that the failure was ultimately caused by something out of the scope of this project (something that had to do with the breaker itself).

BUT! I must still calculate the lug temps because the field engineer is not going to take another chance and wants metal temp predictions.

So, I gathered more information.

The breaker is an AC breaker, 3 phase, 50-60Hz.

Current = 1300 amps
Voltage = 1380 V
8 cables per pole, 4 going in and 4 going out per pole
A total of 24 cables
A total of 6 bus bars with 4 lug connections each

Each busbar connection is made of copper, thickness were stated earlier as were overlap distances.

I am getting information that because this is AC, that the current density is going to be much higher due to skin effect, and that assuming DC is a less conservative (more dangerous) approach because it will underestimate the rise in current density associated with AC.

Please get back to me, I am grateful for all your input.

 

[desertfox]

Hi dbecker

I have recently calculated torque loadings in busbar joints and calculated stresses for temperature rises a bit less than you quote but the calcs highlight failure possibilities.
Now if you had a 130 degs temp rise it sounds to me that there was a fault on the system and that would certainly produce high stresses at the joint.
To save me trawling back through the posts can you give the section sizes of copper thats bolted together and not just the thickness (ie 50 x 10 and a 32 overlap), also how many bolts and what size and torque loading, are you using belville washers or just plain?
If you provide this info I can do some calcs for you.

desertfox