OT: thermal radiation

jolio ST

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Oct 2004
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I have a little problem in the mathematics department. I am designing a control panel and I have a couple of relays and monitoring devices, lamp indications etc, devices that dissipates heat. The client wants to know the temperature of heat-radiating devices when this panel is placed in a room-temperature area. I have heard of the calculation based on Stefan-Boltzmann's law, and used an online calculator to do the job. but I am curious about the actual deriviation of the formula. I was wondering is there anyone here who has done this calculation before? Can you explain how the calculation is being done?šŸ¤žšŸ»

regards
Sherine T.
 
You will probably want to obtain a Heat Transfer textbook that explains radiation heat transfer. It's not very simple to explain/ derive as it uses absolute temperatures raised to the fourth power, surface integrals to define intensity, and then there are the surface properties of emittance, reflectance, absorbtion, and transmission to be considered.

Most times you have to just use simplifications for a specific application as the mathematics gets very cumbersome in a hurry.
 
Oh..That's alright. I wasn't able to substitute in values and calculate it out on paper at first, a bit of a mis-calculation on my part. The formula for heat radiation with relation to surrounding temperature is as follow:
P = eĻƒA(T^4-Tc^4)
Where P is the energy emitted by the device; e is the emissivity rating ranging from 0 to 1; Ļƒ is Boltzmann's constant of (5.6703 x 10^-8); T is the temperature of the emitting device, and Tc is the temperature of the surrounding area where the device is located.
The missing value is T for this case, so say if I have a AC-to-DC power converter rated at 300W, mounted on a board with surface area of 4 sq meters. it's not an ideal blackbody therefore I estimated the emissivity to 0.8; and surrounding temperature to be approximatly 40 deg.C, which has to be converted to deg.kelvin.
So I would have 300W = 0.8*(5.6703*10^-8)*4[T^4-(313.15K)^4]; which would give:
300W = 1.814496*10^-7(T^4-(313.15K)^4)
(T^4-(313.15K)^4) = 300W/1.814496*10^-7
(T^4-(313.15K)^4) = 1653351674.51
T^4 = 1653351674.51 + (313.15)^4
T^4 = 11269688443.8
T = sq4(11269688443.8)
T = 325.820K
T = 52.67 deg.C

T^4-Tc^4 is due to heat lost to surrounding temperature.

Regards
Sherine T.
 
I would think that there is a lot more going on heat transfer-wise than just radiation.

Heat is being conducted to the panel through the components being mounted on the panel while it is also convected to the air within the panel. The air in the panel convects to the box, then there is the conduction through the box walls to the outside surface where there is conduction to whatever the box is mounted to, convection to the ambient air and, of course, radiation to everything else.

I always use a program to figure this stuff out.

It does remind me of a story from my first job where they used to make heating systems for trains. The engineers had done a complete heat transfer analysis for the heating of a train care and actually had one fitted with temperature sensors to check their calculations. Things seemed pretty good except that on some cold clear nights they noticed that the temperatures inside the train car (with no heat on) would actually be slightly below ambient temperature. Everything was calibrated properly and all the equipment checked out ok.

When they added in the heat transfer due to radiating heat into space they verified the measured temperature... go figure.
 

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