# Heat transfer through 2 slabs of same material properties, but different thickness

Let'se say we have two walls, wall A and wall B. Both walls are made from the same material, but they differ in thickness. Wall A is L cm, whereas Wall B is L + ΔL cm.

Wall A and Wall B make up parts of two separate compartments with identical dimensions, and in both compartments there is fire, causing the temperature to increase at the same pace in both rooms.

Now, my question is the following: How do I go about proving that the temperature at L cm in Wall A is lower than the temperature at L cm in wall B at time t, given that steady state has not been reached.

My intuition tells me that the temperature at L cm in Wall A will be lower than the temperature at L cm in Wall B since heat will be lost due to convection. Could the solution really be this simple?

• So the fires provide a different $q$ to allow for the temperature to increase at the same pace while having different insulation? – JMac Jun 20 '19 at 13:11
• Is $dL$ a differential increase in length? If so, that means the increase in length approaches zero in the limit. I think there needs to be a finite difference in length between the walls to have an effect on heat transfer through the walls. – Bob D Jun 20 '19 at 13:12
• @JMac Sorry about the confusion. The temperature is regulated according to a standardized time-temperature curve. I used the example of fire to simplify, but you're right of course.. – user1904218 Jun 20 '19 at 13:17
• @BobD Not a differential, it's a finite difference in length. Have clarified. – user1904218 Jun 20 '19 at 13:18
• Suppose the convective heat transfer coefficient were infinite, so the outside temperatures of the walls are the same (and constant). In the case of the thicker wall, the extra layer of thickness $\Delta L$ will act as a layer of insulation for the part of the wall between 0 and L. So the temperature at L will be higher. – Chet Miller Jun 20 '19 at 14:27

## 1 Answer

Consider a 1-D system of thickness $$L$$ (a slab) with conduction coefficient $$k$$, density $$\rho$$, and specific heat capacity $$\tilde{C}_p$$. It is subjected to a sudden temperature change from $$T_o$$ to $$T_\infty$$ on one side (the left side). The convection coefficient on that side is $$h$$. The resulting formulation and boundary conditions are developed in this example PowerPoint posting.

$$\rho \tilde{C}_p\frac{\partial T}{\partial t} = k \frac{\partial^2 T}{\partial x^2}$$

We can make this dimensionless and allow an unrealized temperature to be defined as $$\theta = (T(z,t) - T_o)/(T_\infty - T_o)$$ where $$T_o$$ is the temperature at time $$t = 0$$ throughout the entire slab. In dimensionless form, the temperature profile through the slab as a function of time and position is determined by the equivalent of Fourier's second law. A defining time constant of the system is found from the Fourier number

$$N_{Fo} = \frac{k t}{\rho \tilde{C}_p L^2}$$

The Fourier number can be viewed essentially as the dimensionless time needed to reach a specific temperature $$T$$ at a specific position $$x$$ in the slab.

You have two slabs A and B with $$L_B > L_A$$. For all else the same, $$N_{Fo,A} < N_{Fo,B}$$. This means, at the same relative position $$Z = z/L$$, slab A will reach temperature $$T$$ sooner than slab B. Alternatively, at the same time $$t$$ for the same distance $$x$$ from the surface exposed to $$T_\infty$$, slab A will be at a temperature $$T_A$$ closer to $$T_\infty$$ while slab B will be at a temperature $$T_B$$ closer to $$T_o$$. The opposite is true at the position closer to the side of the slab that is held at $$T_o$$.

The exact solution requires separation of variables and is shown in the PowerPoint link. A simpler analysis is to show the case for unrealized temperature as a function of dimensionless position in a slab with the left wall held at $$T_\infty$$ and the right wall at $$T_o$$. Such a sketch is shown below. At the instant $$T_\infty$$ is applied on the left wall, the profile is abrupt at that wall and $$T = T_o$$ throughout. The left is at an unrealized temperature $$\theta = 1$$ and the right is at $$\theta = 0$$. As time progresses, the temperature profile develops roughly as shown. At steady state (infinite time), the profile is a straight line between $$\theta = 1$$ on the left and $$\theta = 0$$ on the right.

Consider that we are half way in slab A. This is the solid vertical line in the picture. Now consider slab B with $$L_B = 2L_A$$. The same absolute distance into slab B from the left side is marked as B1 in the picture. We see for the same time, the unrealized temperature in slab B is always higher at that absolute position than at the same absolute distance from the left wall in slab A. Now consider a position that is the same relative distance from the right side in slab B. This is B2. We see for the same time, the unrealized temperature in slab B is always lower at that absolute position than at the same absolute distance from the right wall in slab A.

• Thank you. It seems a bit counterintuitive to me that A will reach a higher temperature at a distance x from the surface than B. It also goes against the answer Chet Miller provided. – user1904218 Jun 25 '19 at 7:39
• The difference is the starting point to reference the temperature. I'll add a picture and revise shortly. – Jeffrey J Weimer Jun 25 '19 at 12:27
• @user1904218 The revision should address your question and provide links for details on the mathematics. – Jeffrey J Weimer Jun 26 '19 at 20:40
• Thank you very much! Will review the solution shortly. – user1904218 Jun 27 '19 at 7:59