Where Should Radiation Be Considered in Heat Transfer Through a Wall?

[aforl]

Studying for a test, and this problem stumped me because of the confusing nature of the question... Not sure where radiation shld come into play. Any help appreciated! THANKS!

1. Homework Statement

The walls of an apartment are made of two layers of mat with a heater layer in between.
Thickness, thermal conductivity, emissivity of each surface and convective heat transfer coefficient in still air are given above.
The outdoor temperature is at -10^oC, and the indoor temperature is controlled by the heater layer.
Assume the apartment has 4 side walls, each 10m wide and 3m high, and heat loss through ceiling and ground are ignored.
Also assume inner surface of the wall (facing indoor) are perfectly insulated, but do not assume convection and radiation at outer surface is negligible.

(a) Sketch the (steady-state) temperature profile across the wall, from the outer surface to the inner surface.
(b) The heater is generating heat at 250W/m². What is the steady state temperature of the Mat B layer? To estimate the radiation heat transfer coefficient at the outer surface of the wall, assume temperature at surface of Mat A is close to -10^oC.

2. Equations
Conduction across plane (without heat generation):
q=kAΔT/L
T(x) = T_surface1 - ΔTx/L
R_t = L/kA
Conduction across plane (with heat generation):
d²T/dx² + q/k = 0
T(x) = T(x) = T_surface1 + qL²/2k (1 - x²/L²)
Thermal Resistance
Rconv = 1/hA
Rrad = 1/εσA(Ts + Tsur)(Ts²+Tsur²)

The Attempt at a Solution

(a) Assume temperature within heater layer is constant (can I do so?),

I drew them linear because without heat generation, the profile is linear. The slope in mat A is also steeper due to the k/L ratio compared to mat B. (Is this alright?)

(b) So to solve this, I drew a thermal circuit. However I'm not sure where to place the radiation resistance. Please help me verify if I drew this correctly.
The question says assume surface of A is -10C, wouldn't that means no radiation and convection?

Please help! Any help appreciated thanks so much !

 

[Chestermiller]

Here are my thoughts on this:
1. The temperature within B is constant, since the inner wall is insulated.
2. The temperature within the heater layer is not constant. It is highest at the side toward B and decreases monotonically toward A, but not linearly.
3. There reason they say that the outside surface is close to -10 is so that you can linearize the equation for the radiative heat transfer. So there actually is radiation and convection at the outer surface.
4. Your diagram of the resistances is basically correct.

Chet

 

[CWatters]

I might be wrong but...

There appears to be no mechanism for heat to be lost from the room - no ventilation and the loss from floor and ceiling is to be ignored? If that's correct the room side of the heater mat cannot be hotter than the room. If it was heat would flow into the room and raise the temperature. Likewise if the room side of the heater is colder than the room heat will flow out of the room and the temperature will fall.

So I would argue that the room side of the heater mat must be at room temperature. So the temperature in B will be constant. I think you can forget about Rb. All the heat generated by the matt must flow outwards once the temperatures are stable.

So I'm thinking it must be something like this (although Chester points out it wouldn't fall in a linear manner in the heating mat).

 

[Chestermiller]

I might be wrong but...

There appears to be no mechanism for heat to be lost from the room - no ventilation and the loss from floor and ceiling is to be ignored? If that's correct the room side of the heater mat cannot be hotter than the room. If it was heat would flow into the room and raise the temperature. Likewise if the room side of the heater is colder than the room heat will flow out of the room and the temperature will fall.

So I would argue that the room side of the heater mat must be at room temperature. So the temperature in B will be constant. I think you can forget about Rb. All the heat generated by the matt must flow outwards once the temperatures are stable.

So I'm thinking it must be something like this (although Chester points out it wouldn't fall in a linear manner in the heating mat).

View attachment 75112

Yes. That's what I was getting at when I said that "The temperature within B is constant."

Chet

 

[rezeew]

Dear student,

Thank you for reaching out for help with heat transfer. I understand that this problem has been confusing for you, but I will do my best to provide some guidance.

Firstly, it is important to understand the different modes of heat transfer involved in this problem. Conduction is the main mode of heat transfer within the walls of the apartment, as heat is transferred through the materials from the outer surface to the inner surface. Convection and radiation, on the other hand, occur at the outer surface of the wall.

In order to solve this problem, you will need to use the equations provided and apply them to each layer of the wall separately. This means considering conduction in the two mat layers and convection and radiation at the outer surface.

For part (a), you have correctly drawn the temperature profiles for each layer, assuming a constant temperature within the heater layer. However, it is important to note that the temperature at the outer surface of the wall will not be -10oC, as there is heat generation within the heater layer. This will result in a higher temperature at the outer surface, which will then decrease as heat is transferred through the layers.

For part (b), it is not necessary to draw a thermal circuit. Instead, you can use the equations provided to calculate the temperature of the Mat B layer. You will need to consider both conduction and radiation at the outer surface, as well as the heat generation within the heater layer.

Overall, it is important to carefully consider the different modes of heat transfer involved in this problem and apply the appropriate equations to each layer of the wall. I hope this helps. Good luck with your test!