Surface to Air transfer 2

[coconutalley]

I have busbars that carry current and I want to do
a rough order of magnitude calculation of the temperature of the busbar without measuring it [or using any thermal software]. Does anyone know how to do this?
The following variables are known:
1. Ambient temperature, say Ta
2. Surface area of material that interfaces with air, A
3. Material thermal conductivity,k
4. Amount of heat or power (W) input.
This is in still air. Neglect radiation.

I know that I could use the first order formula for heat
Q = (k*A/L)*(Ta - T), T=temperature of busbar = ?
Thanks

 

[IRstuff]

Similar form, except that you should be using h_c*area, where h_c is the effective convective/radiative heat transfer coefficient and area is the effective area.

Your expression for thermal conductivity, while correct, uses the wrong area, since it would actually be the cross-sectional area of the bar.

TTFN

Eng-Tips Policies FAQ731-376

 

[coconutalley]

Thanks.
So, in other words, the expression should be:
Q = h_c*area*(Ta - T)?
Is there a typical h_c coefficient or is there a reference somewhere? What is "area" supposed to be?

 

[IRstuff]

They are commonly listed in heat transfer textbooks such as:

http://web.mit.edu/lienhard/

http://www/ahtt.html

Review the chapters pertaining to convective cooling. You should also consider whether radiative cooling has an effect.

TTFN

Eng-Tips Policies FAQ731-376

 

[coconutalley]

Thanks,
I did a google and found some answers too.

http://www.engineersedge.com/heat_transfer/convection.htm

The convection coefficient used is 18 BTU/hr*ft^2*F.
The Area is the surface area, in this case the busbar.
In this case, the temperature rise works out to about 5 C
from ambient. I will do actual test to see how far away
this figure is. I'm guessing it'd be off by 10% to 20%, but at least it gives me a ball park figure.

 

[IRstuff]

Don't know where they got that number, but you should have read some more. Convection coefficient is HIGHLY dependent on air velocity.

Still air coefficients are more like 1.25 BTU/hr-ft^2-ºF

TTFN

Eng-Tips Policies FAQ731-376

 

[25362]

A simplified equation (Holman's Heat Transfer) for laminar free air convection from a horizontal cylinder is

h[sub]c[/sub] = 1.32([Δ]T/d)[sup]0.25[/sup]​

Where [Δ]T = (T-Ta) is in [sup]o[/sup]C
d, diameter, is in m.
h, W/(m[sup]2[/sup].[sup]o[/sup]C)

If you know Q = h[sub]c[/sub][×]Area[×][Δ]T, a few iterations would give you the value of T.

Are you sure there is no radiation effect ?

There are nomographs to estimate h[sub]r[/sub] to be added to h[sub]c[/sub] in the above equation, again depending on T and Ta.

 

[sailoday28]

Assume the bar to be rectangular (Brick shape) and very long. Thermal conductivity, constant and a constant Nusslet/Biot number hl/k as constant.

T- Tamb at center = q'''*L^2/k)[ 1/Biot +0.5 ]

T- Tamb at outer surface = (q'''*L^2/k)[ 1/Biot]
where q''' is heat generated per unit volume
obtained from I^2?R


d^2Theta/dx^2 + q'''/k= 0 (1)
where q''' is heat generated per unit volume (2)
theta= T - Tambient (3)

Center of bar is x=0 x=L is outer surface

d theta/dx =0 at x=0 (4)

integrate (1) dtheta/dx = -(q'''/k)x (5)

integrate (5) theta = -(q'''/k)(x^2)/2 + C (6)

at x=L -k(dtheta/dx)|L =h(theta)|L (7)
where |L means at L

substiture 5 and 6 at L into 7
q'''L = h [ C - (q'''/k)(L^2)/2 ] (8)

Solve for C
(Lq'''/(kh) + (q'''/k)(L^2)/2 = C (9)

substitute C from 9 into 6

theta = -(q'''/k)(x^2)/2 + (Lq'''/(kh) + (q'''/k)(L^2)/2
(10)

Simplifying
theta = (q'''*L^2/k)[ 1/Biot +0.5 -(x/L)^2 /2 ]

Therefore T- Tamb at center = q'''*L^2/k)[ 1/Biot +0.5

T- Tamb at outer surface = (q'''*L^2/k)[ 1/Biot]


Regards






d theta/

 

[JKEngineer]

I agree that the value you cite of 18 BTU/h ft2 F is way too high, a more likely value, as cited by IRStuff, is of the order of 1.25 BTU/hft2, or lower. ASHRAE often cites values of this order at low velocity for air, but they usually include a component for radiation. ASHRAE adds that component in order to keep the relationships linear. For busbar, with high reflectivity, I agree with your decision to neglect radiation.

You could probably treat the interior of the busbar as isothermal. Alternatively, you could also defend treating this as a 1D problem -- transfer through the copper from its interior to the surface with a k, x component, transfer at the surface to the bulk air T with a convection coefficient. The solid phase transfer would need a term to represent the heat generation due to the resistivity of the copper and the electrical load.

If the busbar is in a limiting enclosure, or is ganged with several additional bars as it would be in a switch enclosure, or if there are exterior drivers for air flow, then analyzing a single busbar as an isolated, natural convection case may be inaccurate.

My approach, since I have access to CFD software, would be to model the system. (If of interest feel free to contact me.)

Jack

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

http://www.KleinfeldTechnical.com

 

[sailoday28]

JKEngineer (Chemical)"You could probably treat the interior of the busbar as isothermal. Alternatively, you could also defend treating this as a 1D problem -- transfer through the copper from its interior to the surface with a k, x component, transfer at the surface to the bulk air T with a convection coefficient."

How would your approach be different than my post?

Regards

 

[JKEngineer]

sailoday:
Not sure whether it would be or not. I will admit to not having followed yours carefully enough to tell.
Jack

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

http://www.KleinfeldTechnical.com

 

[kenvlach]

Busbar calculations have been done countless times over 100+ years.
See

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

& Copper for Busbars

http://www.cda.org.uk/megab2/elecapps/pub22/index.htm

Especially,
3. Current-carrying Capacity of Busbars
Methods of Heat Loss
Heat Generated by a Conductor.

Some differences in the types of electricity and codes, e.g.,
BS 159, ANSI C37.20, BS 5486 (equivalent to IEC 439...

 

[coconutalley]

Thanks to all for informative replies.
I thought to neglect radiation only because the busbar
will have low resistance and would generate low heat.

Here's some specifics if somebody wants to take a crack
at calculations:
Dimensions: L = 20" x 1" by 0.25"
Heat Dissipation: 13 Watts
It will be in still air [though there is a power supply
that blows air nearby.
Ambient temperature: 40C

In a month or two, I will hopefully be able to verify.

 

[JKEngineer]

The very interesting links (hence the star) provided by kenvlach point out that radiation can actually be a significant factor depending on the actual emissivity, or surface condition, of the bus bar(s).

If they are in "like new" condition the emissivity will be low and radiation will be less important. However if the surfaces are dirty or oxidized, or even treated to raise the emissivity, then radiation becomes significant. My earlier agreement on neglecting radiation was based on "like new" shiny bus bars.

Jack

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

http://www.KleinfeldTechnical.com

 

[IRstuff]

I get a delta_T of 63ºC with 8W/m^2-K, which should be somewhat generous.

This results in about 1 watt of radiated emission, even with 0.07 emissivity. With something like 0.40 emissivity, the radiated power is 6 W, so radiated should be included if oxidation of the busbar is a factor.

When radiated is included 0.07 emissivity results in a delta of 58ºC, but with a 0.4 emissivity, the delta drops to 44ºC

TTFN

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