Radiation heat transfer between parallel plates

In summary: A## - area of the plates, ##\varepsilon_{1}## and ##\varepsilon_{2}## - their emissivities, ##c## - specific heat, ##m## - mass of one plate.This temperature difference is way too low. I performed a numerical simulation and the result is that the plate with initial temperature ##T_{1}=293.15 \ K## heats up to ##309.5977 \ K## while the plate with initial temperature ##T_{2}=373.15 \ K## cools down to ##310.2834 \ K##. The problem is also that with this analytical
  • #1
FEAnalyst
342
144
TL;DR Summary
Is it possible to calculate temperature change knowing the radiative heat flux exchanged between two paralel plates?
Hi,
the approximate (not accounting for plate size and separation distance) formula for heat flux exchanged via radiation between two parallel plates is:
$$q=\frac{\sigma (T_{1}^{4}-T_{2}^{4})}{\frac{1}{\varepsilon_{1}}+\frac{1}{\varepsilon_{2}}-1}$$ where: ##\sigma## - Stefan-Boltzmann constant, ##T_{1}## and ##T_{2}## - initial temperatures of the plates, ##\varepsilon_{1}## and ##\varepsilon_{2}## - their emissivities.

But is there a way to calculate the final temperatures of the plates (in steady state, when the amount of heat known from above equation will be exchanged)?
 
Engineering news on Phys.org
  • #2
FEAnalyst said:
But is there a way to calculate the final temperatures of the plates ...
If the plates are isolated in a perfect thermally insulated environment, then their temperatures will equilibrate. The mass and thermal capacity of the plate material will probably be important in determining the amount of energy to be shared, and so the final temperature.

If there are thermal losses from the plates, then it becomes a more complex problem of energy flow that, for fixed parameters, will reach a steady state. The thermal capacity of the plate material will probably not be so important.
 
  • Like
Likes Dullard
  • #3
I went back to this topic and thought that I could use these two simple formulas to calculate the temperature difference using the heat flux obtained with the equation from the first post: $$Q=q \cdot A$$ $$\Delta T=\frac{Q}{c \cdot m}$$ where: ##Q## - heat transfer rate, ##A## - area of a single plate, ##c## - specific heat, ##m## - mass of a single plate.

Here are the inputs I used to test this approach: $$\sigma=5.67037442 \cdot 10^{-8} \ \frac{W}{m^{2}K^{4}}$$ $$T_{1}=293.15 \ K, T_{2}=373.15 \ K, \varepsilon_{1}=0.02, \varepsilon_{2}=0.02$$ $$A=0.01 \ m^{2}, c=1300 \ \frac{J}{kg \cdot K}, m=0.054 \ kg$$
With these values, I got the following results: $$q=\frac{5.67037442 \cdot 10^{-8} \cdot (293.15^{4}-373.15^{4})}{\frac{1}{0.02}+\frac{1}{0.02}-1}=-6.8748 \ \frac{W}{m^{2}}$$ $$Q=-6.8748 \cdot 0.01=-0.0687 \ W$$ $$\Delta T=\frac{-0.0687}{1300 \cdot 0.054}=-0.001 \ K$$
This temperature difference is way too low. I performed a numerical simulation and the result is that the plate with initial temperature ##T_{1}=293.15 \ K## heats up to ##309.5977 \ K## while the plate with initial temperature ##T_{2}=373.15 \ K## cools down to ##310.2834 \ K##. The problem is also that with this analytical approach I can't really predict the steady-state temperature of the plates (they reach almost the same temperature that I would like to find analytically). Is there another way to calculate this?
 
  • #4
Asymptotically they will reach the exact same temperature (somewhwere intermediate depending upon their relative heat capacities)
Please do the algebra before you put in numbers. Everyone will be happier.
 
  • #5
hutchphd said:
Asymptotically they will reach the exact same temperature (somewhwere intermediate depending upon their relative heat capacities)
And it will likely be around ##310 \ K##, according to the simulation. But is there a way to predict it analytically?

hutchphd said:
Please do the algebra before you put in numbers. Everyone will be happier.
The three formulas that I used are provided in my posts, before the actual calculations. The first one is in the original post in this thread so it may be less visible. I should also add that the calculations are correct in terms of numbers, I double-checked them using math software. However, these formulas are not sufficient here, apparently.
 
  • #6
That makes no sense. Algebra is not broken. Require the energy flux to be equal and solve for T.
 
  • #7
hutchphd said:
Require the energy flux to be equal and solve for T.
Could you elaborate on this? I will be very grateful. I'm still learning the heat transfer stuff since it's not my area of everyday work. I did something similar - calculated the heat flux using the formula from this website: https://web.mit.edu/16.unified/www/FALL/thermodynamics/notes/node136.html
then converted it to heat transfer rate and solved for the temperature difference. What exactly should I equalize? I have a single equation for both plates and it includes their initial temperatures. That's the only difference between them since they have the same area, mass and emissivity.
 
  • #8
hutchphd said:
Require the energy flux to be equal and solve for T.
For a closed system, the temperature of final equilibrium is not rate dependent.
Without phase changes, the average KE of the molecules is proportional to temperature.
The mass of the plates; m1 and m2 .
The temperature of plates; T1 and T2 .
Total KE is therefore; m1*T1 + m2*T2 ;
Final temperature; ( m1*T1 + m2*T2 ) / ( m1 + m2 ) ;
If m1 = m2 ; Final temperature; T = (T1 + T2) / 2 = arithmetic mean.
 
  • #9
As a check on understanding the OP can repeat the calculation allowing for different heat capacities.
It always amazes me how well these models work.
 
  • #10
Baluncore said:
Final temperature; T = (T1 + T2) / 2 = arithmetic mean.
Is it really that simple and not necessary to do the actual heat transfer calculations? It would be nice but in this case I get ##\frac{293.15+373.15}{2}=333.15 \ K## while the simulation yields ##310 \ K##.
 
  • #11
Use algebra.
 
  • #12
hutchphd said:
Use algebra.
I can use the three formulas that I have and algebra to get a single formula: $$\Delta T=\frac{A \cdot \varepsilon_{1} \cdot \varepsilon_{2} \cdot \sigma \cdot \left( T_{2}^{4}-T_{1}^{4} \right)}{c \cdot m \cdot \left( \varepsilon_{1} \cdot \left( \varepsilon_{2} - 1 \right) - \varepsilon_{2} \right)}$$ but it will give me the same result: ##-0.001 \ K##.
 
  • #13
FEAnalyst said:
s it really that simple and not necessary to do the actual heat transfer calculations? It would be nice but in this case I get 293.15+373.152=333.15 K while the simulation yields 310 K.
What is the "actual" calculation?? You are not understanding.
In equilibrium for your system the net heat flow is zero. The equations all tell the same story $$(T_2^4-T_1^4)=0$$.
I believe the crux of your issue may be that each plate is always both absorbing and emitting. This is always the case in the real world. If your calculation allows radiation to leak out of the two-plate system, then one must include the rest of the universe. Eventually these two plates wll be at 3K Like the Cosmic Background
I have no idea what the "actual" calculation uses for the system and I think perhaps you do not either. Such is the nature of blind calculation.
 
  • #14
Heat transfer is still a bit of a mystery to me. But I'm trying to learn how this stuff works. I agree that the case discussed here is quite abstract and it doesn't take into account any heat exchange with the environment. Just the heat flow between these two plates, like in the example on the MIT website linked above. I know that the final temperatures of both plates will be equal when the system reaches a steady-state. The problem is that with the equations shown before I can't get the right value of the final temperature (##T_{1}## and ##T_{2}## are just the initial temperatures). I can only calculate the net heat flux using the first equation.

So getting to the point - can this final temperature of ##310 \ K## (according to the simulation) be calculated analytically for this system? Or is it not possible and the assumptions would have to be changed?
 
  • #15
FEAnalyst said:
s it really that simple and not necessary to do the actual heat transfer calculations? It would be nice but in this case I get 293.15+373.152=333.15 K while the simulation yields 310 K.
I am not following. What is the "simulation" to which you refer?
 
  • #16
hutchphd said:
I am not following. What is the "simulation" to which you refer?
I performed a finite element analysis study. The two plates were modeled in 3D and placed in front of each other. I defined a radiation condition between them, assigned materials and initial temperatures. I can share some screenshots if needed. I’m not sure if the result of this analysis is correct but that’s what I want to check using the analytical calculation.
 
  • #17
The results of the simple analytical arguments are correct. The final temperature will depend simply upon the initial size, temp and heat capacity of the otherwise isolated subsystems. If you solved a different problem then you will get a different answer. If you want to specify exactly how you did the calculation, we can look for the problem.
I don't want to examine the entrails (no screenshots of code or output for me thanks).
 
  • #18
FEAnalyst said:
I performed a finite element analysis study. The two plates were modeled in 3D and placed in front of each other.
I would want to know the boundary conditions used. What was the temperature of the boundary? Does energy radiate from the back of the plates, across the boundary of the simulation, or is the simulation isolated, so conservation of energy holds.
Screenshots will NOT help.
 
  • Like
Likes hutchphd
  • #19
Hi @FEAnalyst, let's try a Gedanken.

  1. Assume a cubical metal container without a top.
  2. Inside the container you create a fabric wall dividing the volume in half.
  3. You pour different amounts of water in each half.
    • The different water levels represent the different thermal energies (temperatures) of your two identical plates
    • The fabric wall represents the emmisivity of your two plates

As you can see, the water level in the two sections will eventually equal each other.

The amount of time to equilibrate depends on whether the fabric is thin, open weave nylon (high emmisivity), or tent canvas (low emmisivity).

If one partition of the container is larger than the other, that would represent a larger thermal mass ( different plate sizes, or different specific heats).

The metal container is the equivalent of your plates being perfectly isolated from the surrounding environment. If the plates are not isolated, that would be represented by using a porous container rather than a metal one, perhaps a cotton knit.

Hope this helps a little!

Cheers,
Tom
 
  • Like
Likes hutchphd
  • #20
Baluncore said:
I would want to know the boundary conditions used. What was the temperature of the boundary? Does energy radiate from the back of the plates, across the boundary of the simulation, or is the simulation isolated, so conservation of energy holds.
Screenshots will NOT help.
The only condition that I explicitly defined was the interaction between the two surfaces of the plates that face each other - they were allowed to exchange heat by radiation. More precisely, they were assumed to belong to a cavity (a closed one because for an open cavity I would also need ambient temperature and this is not included in my analytical solution). All the other surfaces had no boundary condition defined and thus the natural boundary condition (##q=0##) was imposed by the solver. So the surfaces were considered adiabatic. Apart from that, I also defined the initial conditions - the initial temperature for each plate (its whole volume).

I should add that the software accounts for view factors when solving surface-to-surface radiation problems. More complex analytical formulas also take these parameters into account. I omitted them because the plates are parallel, rectangular and aligned and I assume that they are infinite (even though they are not, of course) so the view factor is ##1##. The simulation also calculates the view factor value for whole surfaces and in this case, it was close to ##1##.

Tom.G said:
lets try a Gedanken.
That's indeed a very interesting analogy, thank you for sharing this. It may indeed help with my further considerations.
 
  • Like
Likes hutchphd
  • #21
Well this does sound odd and I don't see where the issue is. Can you do some simple "black box" testing?
If you start the plates at the same T do they end up there?
If you radically change the geometry of the plates do the results change (they should not change for an isolated system).
etc.
In fact the results here should depend not at all on the detail of radiative transfer: any energy exchange will lead to the same final T if the two body system is thermally isolated and energy conserved
 
  • #22
FEAnalyst said:
More precisely, they were assumed to belong to a cavity (a closed one because for an open cavity I would also need ambient temperature and this is not included in my analytical solution).
To reach equilibrium at 310 K = 37 °C, that must be a body cavity.
 
  • Haha
Likes hutchphd
  • #23
hutchphd said:
Well this does sound odd and I don't see where the issue is. Can you do some simple "black box" testing?
If you start the plates at the same T do they end up there?
If you radically change the geometry of the plates do the results change (they should not change for an isolated system).
etc.
In fact the results here should depend not at all on the detail of radiative transfer: any energy exchange will lead to the same final T if the two body system is thermally isolated and energy conserved
I wanted to carry out some additional tests but now (likely after updating to the newest version) the software doesn't let me run even the original scenario because, as I've mentioned, the cavity is defined as closed while in fact it's open and the software detects it based on view factors. I could just change its definition to open but this requires the specification of ambient temperature. The advantage is that it could make this scenario more realistic than such a fully isolated system. And this inability to solve the original problem in a new version of the software may suggest that the previous result was incorrect (non-physical). But the disadvantage is that I don't know how to solve this analytically with the new assumption. Any ideas? Is it even possible when the ambient temperature has to be considered?
 
  • #24
Totally confused. This software comes from where?
 
  • #25
hutchphd said:
Totally confused. This software comes from where?
It's Abaqus, a general-purpose finite element analysis solver. I can make different assumptions in the analysis (add some boundary conditions, maybe use gap radiation in contact with separation instead of cavity radiation) but the problem is that I don't have a reliable analytical solution for these different assumptions. The books only provide a formula for heat flux in a basic case with an isolated system consisting of two parallel plates exchanging heat by radiation.
 
  • #26
The thermodynamics do not depend upon the radiative transfer and vice-versa. You insist upon conflating them, using software that is inadequate to do so. Why is this a useful enterprise?

.
 
  • #27
This software can easily handle such a problem but the thing is that it's hard to ensure compatibility between numerical analysis and analytical solution - the assumptions are often different for various reasons (usually the analytical solution is simplified and doesn't take into account all the effects considered by the simulation but it's not always the case - sometimes it's just a matter of tweaking available parameters in the software).

I've found an interesting article discussing a similar benchmark problem (and other scenarios) performed in a different FEA software (Comsol) and compared with the analytical solution: https://www.researchgate.net/publication/309291144_Surface_to_Surface_Radiation_Benchmarks

The paragraph "3.2 Radiation between two plates" discusses an equivalent case. Maybe I will have to compare just the fluxes, without verifying final temperatures of the plates. Or consider a different problem - enclosed cavity (like in other cases discussed in the article) - this may solve my issue with result mismatch.
 
  • #28
I don't know what your point is. If you simplify the problem the analytic result is trivial and exact. If you make it more complicated you need FEA.
But if your software does not reproduce the analytic result in the limit, there is something wrong with the software or your understanding of it. The rest is detail.

(I have no idea why the reference is germane)
 
  • #29
hutchphd said:
(I have no idea why the reference is germane)
The article is from a German conference but it's written in English.

I will try with other benchmark problems, as I've mentioned above.
 
  • #30
FEAnalyst said:
I will try with other benchmark problems, as I've mentioned above.
Can you make a box to contain the FEM experiment that is made from perfect, double-sided, broad spectrum mirrors?
 
  • Love
Likes Tom.G
  • #31
FEAnalyst said:
is there a way to calculate the final temperatures of the plates (in steady state, when the amount of heat known from above equation will be exchanged)?
##T_{final}## given ##T_1## and ##T_2## ? Stefan-Boltzmann equation : juggling, thereof.

For the two values given : 293.15K , 373.15K , the results of that (radiative) and the heat-conduction equation mentioned differ by ~2%.

Do the algebra first. Waaaay less messy.
 

1. What is radiation heat transfer between parallel plates?

Radiation heat transfer between parallel plates is the transfer of thermal energy through electromagnetic waves from one plate to another, without the need for a medium or physical contact.

2. What factors affect radiation heat transfer between parallel plates?

The factors that affect radiation heat transfer between parallel plates include the temperature difference between the plates, the surface properties of the plates, the distance between the plates, and the presence of any intervening materials.

3. How is radiation heat transfer between parallel plates calculated?

The calculation of radiation heat transfer between parallel plates involves using the Stefan-Boltzmann law, which states that the rate of heat transfer is proportional to the fourth power of the temperature difference between the plates and the emissivity of the plates.

4. What is the difference between radiation heat transfer and conduction or convection?

Radiation heat transfer is different from conduction and convection in that it does not require a medium for heat transfer. Conduction involves the transfer of heat through a solid material, while convection involves the transfer of heat through a fluid medium.

5. How can radiation heat transfer between parallel plates be minimized?

To minimize radiation heat transfer between parallel plates, the plates can be coated with a material that has a low emissivity, or a reflective material can be placed between the plates to reduce the amount of radiation that can pass through. Increasing the distance between the plates can also decrease the rate of heat transfer through radiation.

Similar threads

  • Mechanical Engineering
Replies
9
Views
1K
  • Mechanical Engineering
Replies
34
Views
4K
  • Mechanical Engineering
Replies
1
Views
903
  • Mechanical Engineering
Replies
10
Views
2K
  • Mechanical Engineering
Replies
3
Views
588
  • Mechanical Engineering
Replies
1
Views
2K
  • Mechanical Engineering
Replies
7
Views
1K
  • Mechanical Engineering
Replies
1
Views
1K
Replies
9
Views
1K
  • Mechanical Engineering
Replies
6
Views
6K
Back
Top