Stephan's Law: Net Power Loss/Gain Through Radiation

[Godwin Kessy]

He says that Power net=EKA(T⁴-T_o⁴)
Where Power net= net power loss or gain through radiation
E- Emmisivity of a body
K- Stefans constant
A-Surface Area of the radiating body
T-Temperature of a body
T₀- Temperature of the surrounding

And Power net=Power gained - Power loss

Now a question!
How can we say that heat gained by the body is also given by the emissivity of the body and its surface area while its the surrounding that's radiating to the body!
What I mean is, we where supposed to use the emissivity of the surrounding etc...

 

[vertices]

He says that Power net=EKA(T⁴-T_o⁴)
Where Power net= net power loss or gain through radiation
E- Emmisivity of a body
K- Stefans constant
A-Surface Area of the radiating body
T-Temperature of a body
T₀- Temperature of the surrounding

And Power net=Power gained - Power loss

Now a question!
How can we say that heat gained by the body is also given by the emissivity of the body and its surface area while its the surrounding that's radiating to the body!
What I mean is, we where supposed to use the emissivity of the surrounding etc...

If the object is in thermal equilibrium, the absorptivity must equal the emissivity (otherwise the object would heat up/cool down indefinitely). So if the emissivity is E, the absorptivity also equals E. (This assumes the object in question can be approximated as a blackbody, which conserves energy in the way I just described - in general, the emissivity is less than the absorptivity).

 

[Yeti08]

I think the question that you're asking is why you aren't using the properties of the surroundings in your equation for net radiative heat transfer, so that is what I will try to address.

To begin with, the Stefan-Boltzmann law defines the blackbody emissive power by

$$[1]\ \ E=\sigma T^{4}$$,​

Grey body emissive power can be defined as

$$[2]\ \ E=\epsilon \sigma T^{4}$$,​

and heat transfer between blackbodies i and j is

$$[3]\ \ q= \sigma A_{i} F_{ij} \left(T_{i}^{4}-T_{j}^{4}\right)$$​

However, you can't just add an $$\epsilon$$ to the last equation and say it is for grey body heat transfer, as you seem to be alluding to. When absorptivity is less than one, reflections become an issue which make things a bit more complicated. For two enclosed surfaces (or surfaces forming an enclosure) the heat transfer is defined by

$$[4]\ \ q=\frac{\sigma \left(T_{i}^{4}-T_{j}^{4}\right)}{\frac{1-\epsilon_{i}}{\epsilon_{i} A_{i}}+\frac{1}{A_{i}F_{ij}}+\frac{1-\epsilon_{j}}{\epsilon_{j} A_{j}}}$$​

You can see that when the emissivities are 1 this becomes the above equation for blackbodies.

--------------------

q - heat transfer
E - emissive power
A - surface area
$$\sigma$$ - Stefan-Boltzmann constant
$$\epsilon$$ - emissivity
F - view factor
subscripts i and j - for objects i and j



EDIT:
The equation

$$[5]\ \ q= \epsilon_{i} \sigma A_{i} F_{ij} \left(T_{i}^{4}-T_{j}^{4}\right)$$​

will work for the case where an object (i) is in a very large enclosure (j). When $$A_{i}/A_{j}\approx 0$$ equation [4] reduces to equation [5]. This is because the enclosure is acting as a cavity in which almost all its radiation is going to itself so multiple reflections will eventually be re-absorbed making it act like a true blackbody. If you ever get to try out an IR camera the cavity effect is an interesting phenomenon that is easy to find.