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how can I determine the resultant temperature on one side of a wall after a specified time when the wall has a series of different insulation layers and when the temperature on the other side is known, and is constant?

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  • $\begingroup$ What exactly are the boundary conditions on the two sides of the wall, and what is the initial condition? $\endgroup$ – Chet Miller Dec 16 '18 at 12:57
  • $\begingroup$ The answer depends on the accuracy you're seeking. Strictly speaking, temperature is discontinuous across an interface, although usually, but not always, this fact is completely neglected. $\endgroup$ – AccidentalBismuthTransform Dec 16 '18 at 14:17
  • $\begingroup$ cecs.wright.edu/~sthomas/htchapter03.pdf $\endgroup$ – Gert Dec 16 '18 at 16:06
  • $\begingroup$ The standard numerical approach is the finite difference method, discussed in a heat transfer context by Incropera and DeWitt. An alternate method appears here ("1-D Heat Transfer in Multilayer Materials Using a Finite Volume Approach"). Özışık discusses the analytical solution in Chap. 8 of Heat Conduction here. The search for more convenient analytical solutions is an area of active research. $\endgroup$ – Chemomechanics Dec 17 '18 at 20:18
  • $\begingroup$ If this was for example a home it would depend on the furnace setting, no furnace and the house eventually be the same temperature as the outside. An interesting question to as his what is the rate of heat loss so that you can figure out how often your furnace needs to run. $\endgroup$ – PhysicsDave Dec 18 '18 at 0:47
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The relevant partial differential equation is the heat equation:

$$\frac{\partial T}{\partial t}=\alpha(\vec{x})\nabla^2 T$$

where $\alpha$ is the thermal diffusivity of whatever material a layer is made of, and so will depend on position for a wall made of different materials. The spatial variation in temperature along the wall is likely negligible, so we can treat this as a 1-dimensional problem:

$$\frac{\partial T}{\partial t}=\alpha(x)\frac{\partial^2T}{\partial x^2}$$

where the $x$-direction points perpendicular to the wall. It is likely that this must be solved numerically at this point, given the variation of $\alpha$ within the wall.

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  • $\begingroup$ "so will depend on position for a wall made of different materials." What needs to be calculated is the overall thermal resistance of the composite wall. $\endgroup$ – Gert Dec 16 '18 at 16:08
  • $\begingroup$ Also, because this is transient problem, heat capacities must come into play at some point. $\endgroup$ – Gert Dec 16 '18 at 16:11
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    $\begingroup$ @Gert Heat capacity is included in the definition of thermal diffusivity, which is equivalent to thermal conductivity divided by density and specific heat capacity. $\endgroup$ – probably_someone Dec 16 '18 at 16:17
  • $\begingroup$ Yep, momentarily forgot that. But I'm sure this problem does not need a numerical solution, not with this simple geometry. $\endgroup$ – Gert Dec 16 '18 at 16:20
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To show just how tedious this problem is, mathematically, let's take the simpler case of a homogeneous wall (no layers):

Wall

Because heat only flows in the $x$ direction, we can treat it as a quasi-one dimensional problem.

As @probably_someone stated, the relevant partial differential equation (PDE) is Fourier's Equation:

$$\frac{\partial T}{\partial t}=\alpha\nabla^2 T$$

Or in partial derivatives shorthand: $$T_t=\alpha T_{xx}$$

The problem states that for all times, $T(t,0)=T_0$ and for that reason I make a small substitution:

$$U(t,x)=T(t,x)-T_0$$

So that:

$$U(t,0)=T(t,0)-T_0=0$$

This is our first boundary condition.

The PDE becomes:

$$U_t=\alpha U_{xx}$$

For a second boundary condition we'll assume that no heat flows from the $x=L$ boundary because air is a good insulator. This means:

$$\dot{Q}(t,L)=0 \Rightarrow T_t(t,L)=U_t(t,L)=0$$

Finally we need an initial condition ($t=0$) and we'll assume $T(0,x)=T_1$, so that:

$$U(0,x)=T_1-T_0$$

For PDEs of this type we use the Ansatz:

$$U(t,x)=\Theta (t)X(x)$$

where $\Theta (t)$ is a function dependent on $t$ only and $X(x)$ a function dependent on $x$ only. Now insert $U(t,x)$ into the PDE:

$$X(x)\Theta'(t)=\alpha \Theta(t)X''(x)$$

Now divide by $\Theta X$ (abbreviated):

$$\frac{\Theta'}{\Theta}=\alpha \frac{X''}{X}$$

Introducing a separation constant (a Real number):

$$\frac{1}{\alpha}\frac{\Theta'}{\Theta}= \frac{X''}{X}=-m^2$$

The original PDE can now be separated into two ODEs:

$$\frac{X''}{X}=-m^2$$

Or: $$X''+m^2X=0$$ and: $$\frac{\Theta'}{\Theta}=-\alpha m^2$$

Starting with the first, it has a general solution of: $$X(x)=A\sin mx +B\cos mx$$

Using our first boundary condition: $$U(t,0)=0 \Rightarrow X(0)=0$$

$$0=A\sin 0 +B\cos 0 \Rightarrow B=0$$

So that $X(x)=A \sin mx$

$$U_t(t,L)=0 \Rightarrow X'(L)=0$$

Or: $A\cos mL=0 \Rightarrow \cos mL=0$

Cosines become zero for $\pi/2, 3\pi/2, 5\pi/2, etc$

So:

$$mL=\frac{n\pi}{2}\:\mathrm{for}\:\ n=1,3,5,7,...$$

So we have: $m=\frac{n\pi}{2L}$

Finally: $X(x)=A\sin \Big(\frac{n\pi x}{2L}\Big)$

The eigenvalues $m$ can now be used in the first ODE:

$$\frac{\Theta'}{\Theta}=-\alpha \Big(\frac{n\pi}{2L}\Big)^2$$

Solved we get:

$$\Theta(t)=Ce^{-\alpha \Big(\frac{n\pi}{2L}\Big)^2 t}$$

with $C$ an integration constant.

Going back to the Ansatz: $U=\Theta X$, then:

$$U_n(t,x)=D_ne^{-\alpha \Big(\frac{n\pi}{2L}\Big)^2 t}\sin \Big(\frac{n\pi x}{2L}\Big)$$

For $n=1,3,5,7,...$ The integration constants $A$ and $C$ have been lumped together.

Each $n$ presents a solution but the Superposition Principle demands we take the sum of all these solution to have the complete solution:

$$U(t,x)=\displaystyle\sum_{n=1}^{+\infty} U_n(t,x)$$ $$U(t,x)=\displaystyle\sum_{n=1}^{+\infty}D_ne^{-\alpha \Big(\frac{n\pi}{2L}\Big)^2 t}\sin \Big(\frac{n\pi x}{2L}\Big)$$

We now need to determine the coefficients $D_n$, using Fourier. The initial condition was set as:

$$U(0,x)=T_1-T_0$$

Inserting this into our solution, the exponential terms all become $1$, so:

$$T_1-T_0=\displaystyle\sum_{n=1}^{+\infty}D_n\sin \Big(\frac{n\pi x}{2L}\Big)$$

$$D_n=\frac{2}{L}\int_0^{L}(T_1-T_0)\sin\Big(\frac{n\pi x}{2L}\Big)dx$$

$$D_n=\frac{2}{L}(T_1-T_0)\Big(\frac{2L}{n\pi }\Big)\Big[1-\cos\Big(\frac{n\pi }{2}\Big)\Big]$$

$$D_n=4(T_1-T_0)\Big(\frac{1}{n\pi }\Big)$$

Finally we need to add $T_0$ to $U(t,x)$ to obtain $T(t,x)$. Then put Humpty Dumpty back together and presto, done!

From $T(t,x)$ can then be determined the time evolution at the boundary $L$:

$$\boxed{T(t,L)=T_0+\frac{4}{\pi}(T_1-T_0)\displaystyle\sum_{n=1}^{+\infty}\Big[\frac{1}{n} e^{-\alpha \Big(\frac{n\pi}{2L}\Big)^2 t}\sin \Big(\frac{n\pi }{2}\Big)\Big]}$$ For $n=1,3,5,7,...$

Although this is a complete solution to the problem for a simple homogeneous wall, its very user unfriendly.

Furthermore, in this 'simple' case $\alpha$ is a constant but in the case of a composite wall, $\alpha=f(x)$, complicating matters enormously.

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