Calculating temperature and heat flow in a wall

[bohire]

Homework Statement

Consider the following multi-layer wall consisting of an interior insulation layer and an external light-weight concrete layer. Calculate the heat flow thorugh the structure and the temperature at the interface between the two materials.

Layer 1: Thermal insulation; Thickness (d): 100mm; λ: 0.033W/mK
Layer 2: Light-weight concrete; Thickness (d): 100mm; λ: 0.14W/mK

Homework Equations

Heat flow: Q=K*(T₁-T₂) [W] (But no temperature is given in the question)
Conductance: K=λ*A/d

The Attempt at a Solution

I don't know how to get there but the answer is Q=40.1W and T_m=-4.3C°.
I don't know which equations to use because there is no temperature nor wall area given in the question.

 

[Svein]

Your formula for conductance is wrong, it should be λ/d (dimensional analysis). So the thermal resistance is d/λ, giving R_θ1=0.1/0.033°K/W≈3°K/W and R_θ2=0.1/0.14°K/W=0.714°K/W. Thus the wall has a thermal resistance of 3.714°K/W⇒thermal conductivity of ≈0.27W/°K.

 

[bohire]

Your formula for conductance is wrong, it should be λ/d (dimensional analysis). So the thermal resistance is d/λ, giving R_θ1=0.1/0.033°K/W≈3°K/W and R_θ2=0.1/0.14°K/W=0.714°K/W. Thus the wall has a thermal resistance of 3.714°K/W⇒thermal conductivity of ≈0.27W/°K.

Thank you Svein but do you know where I go from there? I still need to find the heat flow and temperature between the materials.

 

[Chestermiller]

Actually, your equation for the conductance was correct, and Svein was not correct. He was mistaking Q for the heat flux rather than the total heat flow.

The key to this problem is recognizing that the same rate of heat flow Q passes through both wall layers. So you can do each wall separately, and then combine the results to get the heat flow Q and then the temperature T_m between the materials. Let layer 1 be to the left, and let layer 2 be to the right. The two temperatures across wall 1 are T_L and T_m, and the two temperature across wall 2 are T_m and T_R. Write your heat conduction equation (algebraically) for each of these layers, using the same Q for each. Now, for each layer, solve for the temperature difference across the layer. Then add the temperature differences together to eliminate T_m. This will give you an equation that you can use to solve for Q. Let's see what you come up with.

Chet

 

[Svein]

Actually, your equation for the conductance was correct, and Svein was not correct.

OK. The last time I did heat flow was in 1964.

 

[bohire]

Thank you for your answer Chet, this is how far I get
K₁=λ₁*A/d₁=0.033A/0.1=0.33A
K₂=λ₂*A/d₂=0.14A/0.1=1.4A

Q=0.33A(T_L-T_M)=1.4A(T_M-T_R)
Not sure if I'm supposed to do the rest or how I'm supposed to eliminate T_M
1.4/0.33≈4.24
T_L-T_M=4.24(T_M-T_R)
... T_M=(T_L+4.24T_R)/5.4
Then I'm pretty much stuck..

 

[Chestermiller]

Thank you for your answer Chet, this is how far I get
K₁=λ₁*A/d₁=0.033A/0.1=0.33A
K₂=λ₂*A/d₂=0.14A/0.1=1.4A

Q=0.33A(T_L-T_M)=1.4A(T_M-T_R)
Not sure if I'm supposed to do the rest or how I'm supposed to eliminate T_M
1.4/0.33≈4.24
T_L-T_M=4.24(T_M-T_R)
... T_M=(T_L+4.24T_R)/5.4
Then I'm pretty much stuck..

Didn't they tell you what TL and TR are equal to? How do they expect you to do the problem if they don't tell you the inside and outside wall temperatures?

Chet

 

[bohire]

Didn't they tell you what TL and TR are equal to? How do they expect you to do the problem if they don't tell you the inside and outside wall temperatures?

No and that is exactly why I'm struggling to solve it

I also tried this
from Q=0.33A(T_L-T_M)=1.4A(T_M-T_R)
I get T_M=T_L-Q/0.33A And T_M=Q/(1.4A)-T_R
⇒T_L-Q/0.33A=Q/(1.4A)-T_R ... ⇒ Q=(A/3.74)(T_L-T_R)
But the problem with the missing temperatures still remains.. Do you think that the question might be fault or impossible to solve because no temperature is given?

 

[Chestermiller]

No and that is exactly why I'm struggling to solve it

I also tried this
from Q=0.33A(T_L-T_M)=1.4A(T_M-T_R)
I get T_M=T_L-Q/0.33A And T_M=Q/(1.4A)-T_R
⇒T_L-Q/0.33A=Q/(1.4A)-T_R ... ⇒ Q=(A/3.74)(T_L-T_R)
But the problem with the missing temperatures still remains.. Do you think that the question might be fault or impossible to solve because no temperature is given?

Undoubtedly.

Chet

 

[bohire]

Undoubtedly.

Chet

Well that's pretty irritating.. I also see now that all the other exercises in the same chapter do give us interior and exterior temperatures. Anyway thank you for your time and have a good day.

 

[CWatters]

The temperatures aren't the only thing missing.

How can the answer be 40.1W if they don't give you the area of the walls? I believe the answer should be in W/m^2 (Watts per square meter).

 

[bohire]

The temperatures aren't the only think missing.

How can the answer be 40.1W if they don't give you the area of the walls? I believe the answer should be in W/m^2 (Watts per square meter).

I have no idea but I doubble checked and 40.1W is the answer given. All that is given in the question is in my original post. I guess Carl-Eric Hagentoft (author of the book) is to blame for coming up with a nonsense exercise.

 

[CWatters]

Not unusual for there to be errors in books. Is there an earlier problem that might have provided the temperatures and wall area?

 

[bohire]

Not unusual for there to be errors in books. Is there an earlier problem that might have provided the temperatures and wall area?

Yes, all the other ones in that chapter did.