Radiative heat transfer from cold to hot object?

[electricpete]

I’m not a big heat transfer guy. Just wondering about something.

Let’s say I have two rectangular sheets in space: call them A and B.

Question 1: is it possible to have
EmissivityA = 0.8
EmissivityB = 0.2
AbsorbtivityA = AbsorbtivityB = 0.5

==================

Now let’s also assume we have absolute temperature of B greater than A
Tb > Ta

and we also have
EmissivityA*Ta^4 > EmissivityB*Tb^4
i.e.
RadiatedHeatFluxA > RadiatedHeatFluxB

Assuming question 1 had answer yes, then absorbtion would be the same. It seems like the net heat transfer among these two objects is from cold to hot (and to environment)

Question 2: Is it possible to have thermal energy radiatively transferred from cold to hot object as described above?


=====================================
(2B)+(2B)' ?

 

[MintJulep]

Heat goes from hot to cold.

Are you using the right type of T? Need to use absolute units of temperature for all radiation calcs.

 

[electricpete]

T is defined as absolute as stated in original post.

Let’s assign some values to temperature for sake of discussion.
Ta = 293K
Tb = 313K

Combining with assumed values for emssivity
Ea*Ta^4 = 0.8 *293^4=5.90E+09
Eb*Tb^4 = 0.2*313^4 = 1.92E+09

Flux emitted from A towards B is higher than flux emitted from B towards A.
With similar absorption coefficients, on first glance it suggests heat transfered from A to B.

However, reflection not accounted for. Maybe that is the problem.


=====================================
(2B)+(2B)' ?

 

[MintJulep]

To calculate the radiant heat exchange between two bodies you need the areas and view factor for the surfaces exchanging heat.

You are using equations for radiation to an infinite black body.

 

[electricpete]

Let me address your concerns with typical for simplifying a plate problem:
Let's s say two large sheets meaning their length and width is much bigger than the distance between them. The outer sides (not discussed above) are enclosed by reflective insulator, so we are only concerned with the inner sides where the two plates face each other. The areas of the two plates are assumed equal, so we do not need to know the actual area to know which way heat flows.

It is not treated as blackbody, there is an emissivity less than 1 included. I think the key will be including reflection and making sure all the assumed coefficients (reflection, absorption and emissivity) are self consistent.


=====================================
(2B)+(2B)' ?

 

[ione]

I think Kirchhoff' law of gray body works here: I mean poor "absorbers" are also poor "emitters"

 

[Compositepro]

For a given object emissivity equals absorptivity. That is a basic principle. So the answer to question 1 is: no, it is not possible.

E and A can vary by wavelength.

 

[IRstuff]

That relationship ONLY applies at single wavelengths. The only way to get different emissivity and absorptivity is if the temperature difference between the two objects is huge.

For example, Earth and Sun. The Sun is ostensibly a 5778K blackbody, whose peak wavelength, per Wien's Displacement Law is about 500 nm, i.e., roughly in the green range. Thus, for Earth, its absorptivity is equal to 1 minus its albedo. However, the temperature of the Earth is roughly 300K, which puts the peak emission wavelength at about 9659nm, in the LWIR regime. This is reflected in the energy balance equation for the Earth:

http://www.shodor.org/master/environmental/general/energy/application.html

So, in answer to Q1, it's possible, but not for the temperatures stipulated. If the temperature difference was more like a factor of 2 or greater, the emissivities and absorptivities could be possible, or could be constructed. Even at that, the emissivities and absorptivities would need to be expressed as functions of wavelength, and you'd need to do the energy balance wavelength by wavelength.

TTFN

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[zekeman]

"Question 1: is it possible to have
EmissivityA = 0.8
EmissivityB = 0.2
AbsorbtivityA = AbsorbtivityB = 0.5"

No, assuming gray body radiation

emissivity= absorptivity
Ea=Aa
Eb=Ab
subscript a refers to body A
subscript b refers to body B

at a particular temperature for each body and since the bodies are at different temperatures,
Ea is not equal to Eb since they are temperature and material dependent moreover

The proof is well documented in the literature .

Now your inequality

EmissivityA*Ta^4 > EmissivityB*Tb^4

is not a statement of the difference in heat loss from either body since it does not include the multiple reflections between the bodies which gets complicated.It turns out (if you need the proof , I could furnish it) that, for this case, the effective emissivities of each body is Ea*Eb which thankfully validates the second law of Thermodynamics.