# Heat Transfer fourier's law Problem

1. Mar 19, 2013

### guppygould

Q2​A large wall of thickness 5cm and thermal conductivity k = 0.65W/mK has a front surface with an emissivity of 0.8. At this surface there is radiation exchange with the surroundings and convection heat transfer to the air. The air and surroundings are at 22˚C, and the convection heat transfer coefficient is 20W/m2K. If the front surface temperature of the wall is 152˚C what is the rear surface temperature?

fourier's law, Stefan Boltzmann law, appropriate convection law. Those are my guesses but I'm not entirely sure

i don't really have much of one because we aren't given the area of the wall but any input is much appreciated!
-Leo

2. Mar 19, 2013

### voko

Does the numerical value of the area make any difference? Assume it is some A.

3. Mar 19, 2013

### guppygould

Whatever answer I get won't be numerical if I assume that, but thanks!

4. Mar 19, 2013

### voko

In fact you should be able to work out the number. Show what you are doing.

5. Mar 19, 2013

### Staff: Mentor

You don't need to know the area if you work with heat fluxes (W/m2). The radiate and convective fluxes from the front surface to the surroundings are in parallel with one another, and are in series with the conductive flux through the wall thickness.

Write down equations for the radiative and convective fluxes from the wall surface to the surroundings as functions of the front surface temperature. Write down an equation for the conductive heat flux through the wall thickness as a function of the temperatures at the front and rear surfaces of the wall. Write down an appropriate relationship between the three fluxes, given what I said above.

6. Mar 19, 2013

### guppygould

I don't have a computer currently so my ipad is all I ave and it's less than ideal for typing formulas but basically just subbing in numbers to Fourier's law and what I think are the other appropriate formulae but I come unstuck because I am trying to find out the final temperature and I don't have it to sub into Fourier's law. Hold the phone; I think I may have cracked it!

7. Mar 19, 2013

### guppygould

Thank you very much for the last response -I missed it while I was typing on my iPad!

8. Mar 19, 2013

### guppygould

Am I missing anything or are the only equations to use Fourier's law, Stefan-Boltzmann Law & convective heat flux?

9. Mar 19, 2013

### voko

You have a steady state situation. The power conducted through the plate must be equal to the power radiated and taken away by convection. What other equations you might need?

10. Mar 19, 2013

### guppygould

I don't know, that's why I asked.

11. Mar 19, 2013

### voko

12. Mar 19, 2013

### guppygould

I've broken physics or done the calcs wrong -my final temperature is cooler than the ambient temperature! (5.7) -more likely the latter. Any help would be much appreciated!
-Leo

13. Mar 19, 2013

### voko

I am very sorry, but I cannot read most of what is written there.

14. Mar 19, 2013

### Staff: Mentor

Well, like voko I can't see it all, but, at the very least, it should be T - 152, not 152 - T in your equation.

I can't see why you can't write out what you did, even if you have an I pad. What's the problem?

Chet
P.S. If you're asking people for help, it's only common courtesy to present what you did in a way so that they can easily read it. Otherwise, you are wasting their valuable time.

Last edited: Mar 19, 2013
15. Mar 19, 2013

### guppygould

Sorry for the calcs, I can't because they don't have many symbolic characters (or at least I can't find them) but thank you so much for trying to read my scrawl!

16. Mar 19, 2013

### voko

You don't have to use any symbols, you can just use words if you want.

For example: RadiatedPower = emmisivity*sigma*Area*T^4. ConvectedPower = convCoeff*Area*(T - Tamb). And so on.

17. Mar 19, 2013

### guppygould

Ah, yeah true

18. Mar 19, 2013

### Staff: Mentor

What's wrong with the symbols and the word processing capabilities provided by Physics Forums?

19. Mar 19, 2013

### guppygould

I can't see them on the iPad app

20. Mar 19, 2013

### Staff: Mentor

Then, please accept my apology for being so impatient with you.

Incidentally, I calculated the heat flux resulting from convective heat transfer at the front wall to the surroundings, and it came out to be 2600 W/m2. Even without the radiative flux contribution, this would result in a temperature for the rear wall of 200 C higher than the front wall. The radiative flux would only add to this. Incidentally, in calculating the radiative flux, shouldn't you subtract the flux radiating back from the surroundings at 22C?

Also, WELCOME TO PHYSICS FORUMS.

CHET