Does the Order of insulating layers matter?

[Kaguro]

TL;DR

If a system is covered by two layers of different heat conductivities, would their order matter in how much net heat is lost?

Given a system with some heat capacity. It is at any temperature T, and it is covered by two layers A and B. Say layer A is poor conductor of heat and B is comparatively better.

If we let the system sit in a matrix of lower temperature for some time (long enough for heat to pass through A but not long enough that our system comes in equilibrium with the surrounding temperature..), does the final temperature of system depend on ordering of layers A and B?

 

[Chestermiller]

Are you saying that the thermal inertial of A and B is significant, or just the thermal inertia of the system? Is the boundary of the system flat (slab) or is the boundary curved?

 

[Kaguro]

I would like to know what happens with each of the cases you mentioned.

First let's consider that the heat capacities of A and B are not significant. What in that case?

And if their heat capacities are significant, then what?

Also, why would the shape of the boundary matter here? The heat released should be proportional to the exposed surface area and not on the shape.

 

[Chestermiller]

Let's say the system is a cylinder (i.e., curved surface), and the thicknesses of the insulating layers are equal, and on the same order as the radius of the pipe. Then, in the case where the thermal inertias of the insulating layers are negligible, the same amount of heat flows through the insulating layers per unit time, but the inner layer has more resistance, because the heat is distributed over a smaller average surface area. So, switching the order of the layers will affect the rate of heat transfer. OK so far?

 

[Vanadium 50]

If I am trying to keep in, at say 20 C, and outside it's 0 C, suppose one material X has a high heat transmission near 20 C and low everywhere else, and the other Y has a high heat transmission near 0 C and low everywhere else. If I have X near the interior and Y near the exterior it will cool faster than the reverse.

However, such materials are unusual and laboratory curiosities. They usually (possibly always) involve a phase change.

Google "heat diode" for more.

Apart from this, @Chestermiller is right - geometry is the driving factor.

 

[Richard R Richard]

Given geometric conditions of equality on a flat surface

$$A_1 = A_2 = A$$

Y

$$x_1 = x_2 = x$$

For the same temperature difference the inversion of the layers results in the same rate of transfer of energy or heat.

$$Q = \dfrac {A (T_i-T_e) k_1k_2}{x (k_1 + k_2)} = \dfrac {A (T_i-T_e) \mu}{x}$$

$$\mu = \dfrac {k_1k_2}{k_1 + k_2} = \dfrac {1}{\dfrac {1}{k_1} + \dfrac {1}{k_2}}$$

The difference lies in the fact that the temperature at the junction interface between materials is subjected to different temperatures, tm is higher if the material with the highest thermal conductivity is the one exposed to the source of the highest temperature.

But in cylindrical condition

$$R_ {avg1} = R_i + \dfrac {x}{2}$$

$$R_ {avg2} = R_i + \dfrac {3x}{2}$$

$$A_1 \cong2 \pi R_ {avg1} L$$

$$A_2 \cong2 \pi R_ {avg2} L$$

$$A_2> A_1$$

$$e = R_i + x-R_i = R_i + 2x- (R_i + x) = x$$

then if the material with the highest conductivity is internal

$$k_1> k_2$$

if the material with the highest conductivity is in the inner layer

$$Q_A = \dfrac {T_i-T_e}{x} \dfrac {k_1 A_1 k_2 A_2}{k_1A_1 + k_2A_2}$$

if the material with the highest conductivity is in the outer layer

$$Q_B = \dfrac {T_i-T_e}{x} \dfrac {k_1 A_1 k_2 A_2}{k_1A_2 + k_2A_1}$$

If we make the ratio of both equations

$$\dfrac {Q_A}{Q_B} = \dfrac{k_1A_2 + k_2A_1}{k_1A_1 + k_2A_2}$$

Replacing what their surfaces are worth

$$\dfrac {Q_A}{Q_B} = \dfrac {k_1 (R_i + \dfrac {3x}{2}) + k_2 (R_i + \dfrac {x}{2})}{k_1 (R_i + \dfrac {x}{2}) + k_2 (R_i + \dfrac {3x}{2})}$$

Distributing and ordering

$$\dfrac {Q_A}{Q_B} = \dfrac {(k_1 + k_2) R_i + \dfrac {x}{2} (k_1 + k_2 + 2k_1)}{(k_1 + k_2) R_i + \dfrac {x}{ 2} (k_1 + k_2 + 2k_2)}$$

So if $$k_1> k_2$$ is satisfied

It follows that $$\dfrac {Q_A}{Q_B}> 1$$

Then $$Q_A> Q_B$$ there is a better transfer rate if the smallest surface area is the one with the highest conductivity and is in contact with the hottest fluid