Heat Transfer fourier's law Problem

[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

 

[voko]

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

 

[guppygould]

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

 

[voko]

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

 

[Chestermiller]

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.

 

[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!

 

[guppygould]

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

 

[guppygould]

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

 

[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?

 

[guppygould]

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

 

[voko]

Well, start with just these then.

 

[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

 

[voko]

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

 

[Chestermiller]

View attachment 56877I'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

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.

 

[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!

 

[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.

 

[guppygould]

Ah, yeah true

 

[Chestermiller]

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

 

[guppygould]

I can't see them on the iPad app

 

[Chestermiller]

I can't see them on the iPad app

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

 

[guppygould]

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

Don't be silly -you're the one helping the needy after all!
I'll check in the morning, it's late where I am but I'll be sure to let you know how I get on.
And thank you very much!
-Leo

 

[guppygould]

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. Incidentally, in calculating the radiative flux, shouldn't you subtract the flux radiating back from the surroundings at 22C

Sorry for asking so many questions but how exactly did you go about calculating the convective heat flux? And with regards to the radiative flux; maybe -I wasn't sure which temperature was which in the equation

-Leo. Yeah, I'm a mere mortal, not a physics God!

 

[Chestermiller]

Sorry for asking so many questions but how exactly did you go about calculating the convective heat flux? And with regards to the radiative flux; maybe -I wasn't sure which temperature was which in the equation

-Leo. Yeah, I'm a mere mortal, not a physics God!

$Convective Heat Flux = 20(\frac{W}{m^2K})(152-22)(K)=2600\frac{W}{m^2}$

I'll let you figure out how to get the radiative heat flux from a surface at 152C to surroundings at 22C.

Notice that, in these equations, the area of the surface is not included. In heat transfer, fluxes are usually expressed per unit area.

 

[guppygould]

That's great thankyou! I see why the areas aren't needed now!

-Leo

 

[guppygould]

Incidentally, in calculating the radiative flux, shouldn't you subtract the flux radiating back from the surroundings at 22C?QUOTE] I think I've calculated the radiative flux of the wall by (152)^4(0.8)(Stefan-Boltzmann constant)=24.2 which sounds wrong to me. Originally I converted the 152C to Kelvin which gives an answer that seems more likely but I'm not sure really. Yeah, I'm struggling with this

 

[Chestermiller]

Incidentally, in calculating the radiative flux, shouldn't you subtract the flux radiating back from the surroundings at 22C?QUOTE] I think I've calculated the radiative flux of the wall by (152)^4(0.8)(Stefan-Boltzmann constant)=24.2 which sounds wrong to me. Originally I converted the 152C to Kelvin which gives an answer that seems more likely but I'm not sure really. Yeah, I'm struggling with this

No. You were right in the first place. You should be using absolute temperature. It should be ((425)^4 - (295)^2)(0.8)σ.

 

[guppygould]

No. You were right in the first place. You should be using absolute temperature. It should be ((425)^4 - (295)^2)(0.8)σ.

That's lovely thank you very much, I think I've cracked it. Which is a big thing seeing as it was so much of a problem to me!

 

[guppygould]

Scratch what I said about cracking it; I get some ginormous minus number!

 

[Chestermiller]

Lets see your work. What did you get for the convective and radiative fluxes?
Did you use T - 152 for calculating the conductive flux instead of 152 = T? Let me guess. You didn't and you got something like -150 C, rather than something like 450 C.

Chet

 

[guppygould]

Radiative heat flux as 17470 & convective heat flux as 2600 and I've tried both ways with the temperatures because I had it in mind from one of your earlier posts. Thanks for all your help and comments CHET I can see I'm probably getting to be a pain now. Is the correct equation for the conductive heat flux Q=-K(deltaT) though?

 

[Chestermiller]

Radiative heat flux as 17470 & convective heat flux as 2600 and I've tried both ways with the temperatures because I had it in mind from one of your earlier posts. Thanks for all your help and comments CHET I can see I'm probably getting to be a pain now. Is the correct equation for the conductive heat flux Q=-K(deltaT) though?

q = -k (ΔT/Δx)

Also, recheck your radiative flux calculation. It seems too high by a factor of about 10.

 

[guppygould]

Ok I will do tomorrow! Thanks again
Leo

 

[guppygould]

And as if by magic I have an answer that looks reasonable, I can't thank you enough for your help CHET!

 

[timmayy]

Hi, I've been working through this same question and found an answer of 439 centigrade - which seems far too high for me, i assumed the back wall would be a lower temperature than the front wall, and so must be less than 152 (the front wall) and larger than 0 obviously

 

[timmayy]

my working is as follows:
Q_conduction - Q_convection - Q_radiation = 0

Q_conduction = (0.65 x (425 - T₂)) / 0.05 = 5525 - 13T₂ (the 425 being 152 + 273 to convert into kelvin)

Q_convection = 20 x (152 - 22) = 2600

Q_radiation = 0.8 x 5.67x10^-8 x (425⁴ - 295⁴) = 1136.36

so,
(5525 - 13T₂) - 2600 - 1136.36 = 0
∴
1788.64 = 13T₂
137.58 = T (kelvin)
-135.41 = T (celsius)
which is clearly wrong