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Unclear formula for radiative energy transfer 
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#1
Apr2714, 03:29 PM

P: 42

Right, so I was going over the formula sheet for my upcoming exam in thermodynamics, and I've stumbled upon this formula:
[itex]q=\frac{\sigma(T^{4}_{2}T^{4}_{1})}{\frac{1}{\epsilon_1}+\frac{1}{\epsilon_2}1}[/itex] with a description that (I think) translates as heat flux density. I'm currently puzzled as to where this equation is applied, as our assistant didn't say anything about it or how it was derived in the first place. I'm guessing that we have two bodies with temperatures [itex]T_1[/itex] and [itex]T_2[/itex] with emissivity factors [itex]\epsilon_1[/itex] and [itex]\epsilon_2[/itex], so q is supposed to be the energy (heat) transferred in unit time over a unit surface from one body to the other, but I still haven't the slightest idea as to how this equation is derived. Are the shape of the bodies and their mutual position relevant here? What are the conditions under which the equation is applicable? I'm somewhat puzzled so any help would be much appreciated. 


#2
Apr2714, 06:17 PM

Thanks
P: 1,948

That's the formula for the heat flux Q between two surfaces at different temperatures T_{1} and T_{2} facing each other with a narrow gap in between (narrow enough that all the heat coming from one reaches the other).
To understand that equation you must realize that the surface 1 not only produces a heat flux Q_{1}'=σε_{1}T_{1}^{4}, but also reflects a heat flux Q_{1}''=(1ε_{1})Q_{2}, where Q_{2} is the heat coming from surface 2 and (1ε_{1}) is surface 1's albedo. So we get Q_{1} = Q_{1}' + Q_{1}'' = σε_{1}T_{1}^{4} + (1ε_{1})Q_{2} and a similar equation for Q_{2}. Solve those equations for Q_{1} and Q_{2} and finally subtract them to find Q = Q_{1}  Q_{2} to find the net heat flux, and if all goes well, you should get to your mystery equation. 


#3
Apr2814, 07:46 AM

P: 42

Ahh, I see it now, I completely forgot about the reflectivity of both surfaces so my original derivation didn't match up, thanks a lot!



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