How do I calculate heat transfer through a wall with multiple layers?

[kris24tf]

Hello, I have a test coming up and this is a question that will be on it...

The wall of a house is composed of solid concrete with outside brick veneer and is faced on the inside with fiberboard. The fiberboard width is 2 cm, the concrete width is 15 cm, and the brick width is 7 cm. If the outside T is -10 degrees, and the inside T is 20 degrees, how much E is conducted through the wall with dimensions 3.5mx5m in 1 hour?

I want to try something along the lines of k1A(T-T1)/d1=k2A(T2-T1)/d2 or something along those lines, but I know I have to use a T3, but I'm not sure how to find the third T. If you or anyone could help me set that up, that would be great.

I know to use the equation delta Q/ delta t= kAT/d but I also know I have to find the temperature between the brick and fiberboard. I just don't knwo how to find it. I've tried everything, so any clear guidance would save my day...

 

[SpeeDFX]

I remember the the equation for a problem like this was something like..

power = [(area)(change in temp)]/[sumation of (length/conductivity)] , where the lengths are the thickness of each individual layer, and then the "conductivity" is the conductivity of that particular layer.

sorry, I'm kinda too tired to derive how they came about right now. I might do it later

 

[kris24tf]

Thanks, anytime before tomorrow at 11 would be great, but I appreciate that advice...

PS: I believe that answer is supposed to be 4.1 or something... I can't remember what,

 

[mezarashi]

This problem actually does not require you to find the temperature inbetween the conductive layers. We can make use of resistive networks. Also, we assume no convection effects at the surfaces.

$$\frac{dQ}{dt} = \frac{T_1 - T_2}{R}$$

where R represents the total resistance of the resistive network. We see here that because all the layers are in series (i.e. the heat must pass through all of them to get through to the other side), we can write:

$$R_t_o_t_a_l = R(concrete) + R(brick) + R(fiber)$$

If you recall, $$R = \frac{L}{kA}$$Now to also address your problem with finding the intermediate temperatures, you can do this through assuming equililbrium (dT/dt = 0) such that the heat through each suface is equal (or else there would be a temperature change and violate our assumed equilibrium).

You can write the equations of heat going through each and equivalate them, giving you enough equations to solve for the intermediate temperatures. For example:

$$\dot Q_1 = \frac{k_1A}{L}(T_1-T_2)$$
$$\dot Q_2 = \frac{k_2A}{L}(T_2-T_3)$$
$$\dot Q_3 = \frac{k_3A}{L}(T_3-T_4)$$

and so on, with

Q1 = Q2 = Q3 = ...

 

[kris24tf]

That last equation helped a lot! Thanks.