Temperature Rise in Copper Busbars

Are the sections you need to estimate the rise.

I am a mechanical engineer, and when I get involved with occasional electrical issues in the US I will use the CDA site for info. See if this link helps you:

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

Hi Ried,

I'm a mechanical engineer too, working with Switchgear. I searched the web (and this very forum) for some clarity on how the Electrodynamic forces apply in a Copper busbar arrangement, with a current of 125 Amps. Is there a way to figure this out on Ansys (like a mech). There are 3 parallel busbars, carrying 125 Amp, AC current, they have a thickness of 4 mm and width of 12.5 mm. How is the energy (power loss) calculated? We equated the actual I2R value to the formula - 7.66T^1.25/L^.25. But the temperature rise we get from this formula is only about 3 K, whereas practically the thermocouple should show a change of about 70K. What could we be doing wrong in this scenario. I am not giving up until we find a way to crack this.

My application was DC bus bar ampacity. Sorry, but I don't have good suggestions for temp rise modeling.

I agree with agrima.joshua. Since the busbars are enclosed the calculation has to be done in other way.Here is an extract from ABB Switchgear Manual :

4.4.4 Temperature rise in enclosed busbars Busbars in medium and low-voltage substation design are often installed in small compartments or in conduits. For this reason they are subject to more severe thermal conditions than busbar configurations in open installations.Therefore it is not possible to select the busbar cross sections directly from the load tables in Section 13.1.2. Because of the number of parameters influencing the temperature of enclosed busbars (such as position of the busbars in the conduit,conduit dimensions, ventilation conditions, ambient temperatures), the permissible current load must be calculated for the specific configuration.

The heat network method has proven useful for this calculation; Fig. 4-32 b.

Heat flows are generated by electric power losses.

Symbols used: Indices used:

a Heat transfer coefficient D Forced cooling

A Effective area K Convector

P Heat output S Radiation

R Equivalent thermal resistance O Environment

Dt Temperature difference 1 Busbar

D Throughput of circulating 2 Inside air

cooling medium (D = V/t) 3 Enclosure

C Radiant exchange number

T Absolute temperature

cp Specific heat

r Density

That I think WILL work for me. Thank you so much 7anoter4. Unfortunately, right now I am at eh design office so I cannot really download the ABB switchgear manual from any filesharing site. Would it be possible for you to mail me the PDF that you took that extract out of, on the following email address

CR4 Admin - email address removed

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I'm afraid I cannot do it. But you could download it from:

http://www.google.co.il/url?sa=t&rct=j&q=abb%20switchgear%20manual%20chapter%204&source=web&cd=2&cad=rja&ved=0CCQQFjAB&url=http%3A%2F%2Fdevxstudiom.org%2Fabb%2Fabb%2Bswitchgear%2Bmanual%2Bchapter%2B4.html&ei=d1KCUICYB6O80QXb6oHYBA&usg=AFQjCNG27zDmsq791Vw9rqUevMYmKceX4g

My office server does not allow access to any filesharing sites. So I cannot download it right now. But thanks anyway.

The combined radiation and convection heat transfer coefficient is about 1.1 mW/(sqcm.DegC tempdiff between surface and surroundings).

From the above: Del T = 1000 x Wattage loss x /(1.1 x area of bus bar in sqcm) will give an approximate value for the rise in temp of the bus bar. Any conduction of heat by the connections at either end will significantly lower this since the wattage loss gets reduced.

Sir,

Here is a "back-door" approach to the question: Your posts apparently give two differrent dimensions for the bus bar's cross-section, either 12.5x3mm or 12.5x4mm. If it is the former, the area is approx. 37.5 mm²; if the latter, it is approx. 50.0 mm². Going to the NEC, table 8 at the back, the former is midway between #2 and #1 AWG while the latter is about 2/3 the way up from #1 to #1/0. From the ampacity tables for insulated wires in conduit, article 310-16, interpolation would yield a 45 degC rise at about 122A (for 12.5x3) or 142A (for 12.5x4). This table is for insulated cables in raceway with an ambient of 30 degC. True, the orientation and configuration of the busbars will cause different amounts of heating, but this should at least give you a way to doulbe-check the answers your other calculations give.

A similar question was asked a few months ago. At that time I gave the web site address for a very good set of engineering formulae on bus bar design, including temperature rise, required bracing, and magnetic forces under fault conditions. Don't have that address at the present time--sorry.

--JMM

Thank you Jmueller,

Actually our design was initially 12.5X3 mm cross section. However, we decided to increase the cross section area to 4 mm instead. And I did see the thread on busbar heating from a few months back. Are you talking about the copper.uk link? I did find the empirical formulae but they do not help much as they don't account for the fact that the assembly is a closed assembly (its a Distribution Board).

Are you talking about this table? http://www.houwire.com/products/technical/article310_16.html

PS- I don't have to be addressed as a sir, if at all you were addressing me. I am a mechanical engineer. Female.

If there are 3 bus bars [one per phase] of 3*12.5 -as jmueller said-the cross section area will be 37.5 mm^2 CU ,then the

resistance at 65oC will be 1/58*(234.5+65)/(234.5+20)/37.5=5.41/10^4 ohm/m or 0.75.41/10^4 for 700 mm busbar.

Skin effect and proximity effect negligible Tair=35 oC

The electrical losses will be Peff=3*125^2*5.41/10^4*.7=17.75 w [load factor 100%]

The temperature rise of the inside surrounding air conveying these losses outside will be:

DT=Peff/ak/Am ak=convection constant approx. 4.5 w/m^2/oK Am=heat-dissipating outer surface of enclosure=0.72 m^2

DTair=17.75/4.5/0.72=5.47 oC

Using ABB Switchgear Manual Table 13-12 a combined heat dissipation factor [convection+radiation] average will be 10 w/m^2/oC

Using this factor for indoor location- but still not enclosed we'll get:

DTbus=17.75/3/10/((12.5*2+3*2)/1000*0.7)= 27.27 oC [one busbar of 700*12.5*3 mm CU]

DTBus to EXT=DTbus+DTair=27.27+5.47= 32.74 oC

Following ABB ch.4 procedure: ak= 4.5 Ak12=0.7*(0.1+2*0.0125)=0.0875 m^2 AK23=0.693 m^2

RK12=2.54 RK23= 0.206 RK30=0.206

as13=C13*(Tbus^4-Tencl^4)/( Tbus-Tencl) ; C13=epso*epsr; epso= 5.6704/10^8 Stefan-Bolzman Constant

epsr=1/(1/0.63+1/0.29) [ copper bar oxid.-bare steel ]

Tbus=65 Tencl=40 o C as13=1.556 As=0.7*(0.1+2*0.0125)=0.0875

RS12=1/as13/As13=1/1.556/.0875=7.346

The same way as30= 1.3485 and As30=0.693 RS30=1.07

The problem is RD20=1/Cp/ro/D if Cp=1004 ;ro=1.293 D=vol-air/sec=air-speed*section of flow; spd=5 m/sec Area=0.05*0.7=0.035 m^2

D= 0.175 and RD20=0.0044.

Now we have to solve the power flow network.

Let's say P1,P2 and P3 are virtual circular powers- in such way Pi*Ri=DTi.

If we shall calculate the voltage drop around each circle we shall get:

Req=RK30*RS30/(RK30+RS30)

P2*(RK12+RK23+RS13)-P3*RK23=P1*RS13

P3*(RD20+Req+RK23)-P2*RK23=P1*Req

But P1=Ptot=17.75 w

Then P2=13.23 w ;P3= 15.11 w

Now DT=P1*RS13-P2*RS13+P1*Req-P3*Req= 32.67 oC

Neglecting RD20[RD20=10^4 for instance] DT=38.55

If the actual temperature rise it is 70 oC it could be possible an other small heat loss from vicinity penetrates this compartment.

I have a similar question. please answer me if the ambient temperature is 40c and a current of 240 A is passed through a copper bus bar that its cross section is 5x25, what is the temperature rise ?

If there are same dimensions of the compartment and no other heat losses would enter the room-as lighting bulb or relay coil- then 20 to 25 dgr.C could be the [bare]busbar temperature rise.