Unclear formula for radiative energy transfer

Click For Summary
The discussion centers around the formula for radiative energy transfer, specifically the equation for heat flux density between two bodies with different temperatures and emissivity factors. The participants clarify that the equation applies when two surfaces face each other with a narrow gap, allowing heat transfer from one to the other. The derivation involves considering both the emitted heat flux and the reflected heat flux from each surface, incorporating their emissivity and reflectivity. Understanding these factors is crucial to accurately applying the formula and deriving it correctly. The conversation concludes with an acknowledgment of the importance of reflectivity in the derivation process.
kontejnjer
Messages
71
Reaction score
21
Right, so I was going over the formula sheet for my upcoming exam in thermodynamics, and I've stumbled upon this formula:

q=\frac{\sigma(T^{4}_{2}-T^{4}_{1})}{\frac{1}{\epsilon_1}+\frac{1}{\epsilon_2}-1}

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 T_1 and T_2 with emissivity factors \epsilon_1 and \epsilon_2, 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.
 
Physics news on Phys.org
That's the formula for the heat flux Q between two surfaces at different temperatures T1 and T2 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 Q1'=σε1T14, but also reflects a heat flux Q1''=(1-ε1)Q2, where Q2 is the heat coming from surface 2 and (1-ε1) is surface 1's albedo. So we get

Q1 = Q1' + Q1'' = σε1T14 + (1-ε1)Q2

and a similar equation for Q2. Solve those equations for Q1 and Q2 and finally subtract them to find Q = Q1 - Q2 to find the net heat flux, and if all goes well, you should get to your mystery equation.
 
Last edited:
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!
 
I do not have a good working knowledge of physics yet. I tried to piece this together but after researching this, I couldn’t figure out the correct laws of physics to combine to develop a formula to answer this question. Ex. 1 - A moving object impacts a static object at a constant velocity. Ex. 2 - A moving object impacts a static object at the same velocity but is accelerating at the moment of impact. Assuming the mass of the objects is the same and the velocity at the moment of impact...

Similar threads

  • · Replies 7 ·
Replies
7
Views
1K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 10 ·
Replies
10
Views
4K
  • · Replies 0 ·
Replies
0
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 7 ·
Replies
7
Views
697
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 10 ·
Replies
10
Views
3K