# Heat transfer through layered wall

An electrical heater is placed between two layers of insulation. The three layers (the heater and the two layers of insulation) have the same thickness. The figure below shows different temperature profiles at steady-state. The air temperature at the front- and backside of the heater is $$\ce{T_{front}}$$ resp. $$\ce{T_{rear}}$$. Heat transfer in the air film adjacent to the heater is negligible.

a) Which profile corresponds to that of a turned on heater?

b) Which profile corresponds to that of a turned off heater?

c) Which profile is impossible?

d) Which of the three layers has the lowest thermal conductivity? Which layer has the highest?

I have no answer for a) or b).

c) My reasoning is that since a heater is a conductor, the k-value needs to be high, resulting in a fairly straight line in the temp. profile. Hence, (C) is impossible. Is this correct thinking?

However, I don't know what it means when a temperature profile has that kind of curvature for one of the layers, such as in (A) and (B), so if someone could please explain that, I would appreciate it.

d) Based on the fact that low k-values give a steeper line, I would say that layer 3 has the lowest thermal conductivity. And because the heater is a conductor it should have the highest thermal conductivity.

Also, how can you tell if the heater is turned on or off from looking at different temperature profiles? I am having a very difficult time understanding this and would appreciate if someone could make this a bit clearer.

I apologize if these types of questions are considered off-topic, but I really want to know how to determine these things and we have not been provided with any answers nor explanations.

I would start with guessing which one represents the "off" state. I would say (D), since the temperature slopes evenly from one air interface to the other, and is fairly constant in the heater layer. Based on this diagram you can also conclude as you did that layer 3 has the lower conductivity. And the slope in layer 2 indicates that it has a finite (not infinite) conductivity higher than that of 1 or 3.

Based on observations made in (D), you can conclude also that (B) represents the "on" state. As you explain, (C) is not sensible unless heating originates at the interface between layers 1 and 2. (A) also does not make sense, because it suggests that there is a heat sink between 1 and 2 (where is that heat going?).

• I'm sorry, I don't quite understand the difference between (A) and (B). – katara Feb 26 '19 at 11:43
• In (A) the temperature reaches a minimum between layers 1 and 2. This is not physically possible in a steady state. In situation (A) heat will flow to the lower T region between 1 and 2, increasing T. – Buck Thorn Feb 26 '19 at 12:02
• How there can be a layer with T higher than that at both sides? – Alchimista Feb 26 '19 at 15:10
• @Alchimista I don't understand your question, what do you mean? – Buck Thorn Feb 26 '19 at 15:37
• Forget sorry. I took the heater at one side :)) – Alchimista Feb 26 '19 at 16:03

D definitely corresponds to the profile with a turned-off heater, and the lowest conductivity layer is layer 3, as @Night Writer has correctly pointed out.

Since the heater is a heat source, the temperature profile within the heater layer is going to be parabolic. You can show this by solving the steady state heat conduction equation for the case of a heat source within the material. So, on this basis, profiles A and B are possible. However, in profile A, heat is flowing into the interface between layers 1 and 2 from both sides. This is impossible for steady state operation. So profile B is the profile for the turned-on heater, and profile A is impossible.

Profile C is also impossible since heat is flowing out of the interface between layers 1 and 2 on both sides, which is inconsistent with steady state operation.

So profiles A and C are both impossible.