Heat Transfer

Analytical solutions for heat conduction

Lecturer: Jakob Hærvig

Why/why not analytical solutions?

Limitations

Limited to simplified geometries:

Infinitely long plane wall
Infinitely long cylinder
Perfect sphere

Limited to simple conditions with thermal symmetry:

Uniform and temperature-independent properties
Only convection on boundary with constant heat transfer coefficient, e.g. no heat generation inside, radiation on boundary etc
Uniform initial temperature

Reasons to apply

Super fast to compute, e.g. doing millions of evaluations
Yield exact results, e.g. excellent for verifying numerical codes

Starting point of analytical solutions

Partial differential equation:

$$ \frac{\partial^2 T(x,t)}{\partial x^2} = \frac{1}{\alpha} \frac{\partial T(x,t)}{\partial t} $$

Boundary conditions:

At $x=0$:

$$ \frac{\partial T(0,t)}{\partial x} = 0 $$

At $x=L$:

$$ - k A \frac{\partial T(L,t)}{\partial x} = h A \left(T(L,t)-T_\infty\right) $$

Initial condition:

At $t=0$:

$$ ~\hspace{20pt} $$

$$ T(x,0) = T_i $$

Introduction of non-dimensional variables

Convert problem into non-dimensional form to reduce number of variables

Non-dimensional temperature:

$$ \theta(x,t) = \frac{T(x,t)-T_\infty}{T_i-T_\infty} $$

Non-dimensional position:

$$ X = \frac{x}{L} $$

Non-dimensional heat transfer coefficient (Biot number):

$$ \text{Bi} = \frac{hL}{k} $$

Non-dimensional time (Fourier number):

$$ \tau = \frac{\alpha t}{L^2} $$

Re-casting problem in non-dimensional form

Starting from original partial differential equation:

$ \frac{\partial^2 T(x,t)}{\partial x^2} = \frac{1}{\alpha} \frac{\partial T(x,t)}{\partial t} $

Then, inserting definitions of non-dimensional numbers:

Inserting $X=x/L$:

$ \frac{\partial^2 T(X,t)}{\partial (X\cdot L)^2} = \frac{1}{\alpha} \frac{\partial T(X,t)}{\partial t} $ $ \Rightarrow $ $ \frac{\partial^2 T(X,t)}{\partial X^2} = \frac{L^2}{\alpha} \frac{\partial T(X,t)}{\partial t} $

Inserting Fourier number $\tau=\alpha t/L^2$:

$ \frac{\partial^2 T(X,\tau)}{\partial X^2} = \frac{\partial T(X,\tau)}{\partial \tau} $

Inserting non-dimensional temperature $\displaystyle \theta=\frac{T-T_\infty}{T_i-T_\infty}$:

$ \frac{\partial^2 \theta(X,\tau)}{\partial X^2} = \frac{\partial \theta(X,\tau)}{\partial \tau} $

Equation now transformed to non-dimensional form

Final formulation in non-dimensional form

Partial differential equation:

$$ \frac{\displaystyle \partial^2 T(x,t)}{\partial x^2} = \frac{1}{\alpha} \frac{\partial T(x,t)}{\partial t} $$

Boundary conditions:

$\displaystyle x=0$: $\displaystyle \quad \frac{\partial T(0,t)}{\partial x} = 0$

$\displaystyle x=L$: $\displaystyle \quad -k \frac{\partial T(L,t)}{\partial x} = h(T(L,t)-T_\infty)$

Initial conditions:

$\displaystyle t=0$: $\displaystyle \quad T(x,0) = T_i$

Original problem:

9 variables: $T$, $x$, $L$, $t$, $k$, $\alpha$, $h$, $T_i$, $T_\infty$

Non-dimensional partial differential equation:

$$ \frac{\partial^2 \theta(X,\tau)}{\partial X^2} = \frac{\partial T(X,\tau)}{\partial \tau} $$

Non-dimensional boundary conditions:

$X=0$: $\displaystyle \quad \frac{\partial \theta(0,\tau)}{\partial X} = 0$

$X=1$: $\displaystyle \quad \frac{\partial \theta(1,\tau)}{\partial X} = -\text{Bi}\theta(1,\tau)$

Non-dimensional initial conditions:

$\tau=0$: $\displaystyle \quad \theta(X,0) = 1$

Non-dimensional transformed problem:

Only 4 variables: $\theta$, $X$, $\text{Bi}$, $\tau$

A look at the solution for a plane wall

Analytical solution:

$\displaystyle \theta = \sum_{n=1}^\infty \frac{4 \text{sin} \lambda_n}{2\lambda_n+\text{sin}(2\lambda_n)}e^{-\lambda_n^2 \tau} \text{cos}\left(\frac{\lambda_n x}{L}\right) $

where $\lambda_n$ are the roots of: $\lambda_n \text{tan}\lambda_n = \text{Bi}$

which we can expand as:

$ \displaystyle \theta = \textcolor{#008000}{\frac{4 \text{sin} \lambda_1}{2\lambda_1+\text{sin}(2\lambda_1)}e^{-\lambda_1^2 \tau} \text{cos}\left(\frac{\lambda_1 x}{L}\right)} + \textcolor{#524fa1}{\frac{4 \text{sin} \lambda_2}{2\lambda_2+\text{sin}(2\lambda_2)}e^{-\lambda_2^2 \tau} \text{cos}\left(\frac{\lambda_2 x}{L}\right)} + \textcolor{#00bbd6}{\frac{4 \text{sin} \lambda_3}{2\lambda_3+\text{sin}(2\lambda_3)}e^{-\lambda_3^2 \tau} \text{cos}\left(\frac{\lambda_3 x}{L}\right)} + ...$

Example: centre temperature in insulation

Find the centre temperature in a 10 cm thick layer of insulation after 30 minutes. Initially, its temperature is 10$^\circ$C and at time $t=0$ s it's suddenly exposed to an ambient temperature of $25^\circ$C on one side. Assume its properties are:

Plate half thickness, $L=0.05$ m
Initial temperature, $T_i=10^\circ$C
Ambient temperature, $T_\infty=25^\circ$C
Heat transfer coefficient, $h=20$ W/(m$^2$K)
Thermal conductivity, $k=0.05$ W/(m K)
Thermal diffusivity, $\alpha=3\cdot 10^{-7}$ m$^2$/s

One-term approximations

Assume all terms for $n\geq 2$ to be zero, e.g. $\textcolor{#524fa1}{\theta_2=0}$, $\textcolor{#00bbd6}{\theta_3=0}$ ...

$$ \theta = \textcolor{#008000}{\frac{4 \text{sin} \lambda_1}{2\lambda_1+\text{sin}(2\lambda_1)}e^{-\lambda_1^2 \tau} \text{cos}\left(\frac{\lambda_1 x}{L}\right)} + \cancel{\textcolor{#524fa1}{\frac{4 \text{sin} \lambda_2}{2\lambda_2+\text{sin}(2\lambda_2)}e^{-\lambda_2^2 \tau} \text{cos}\left(\frac{\lambda_2 x}{L}\right)} + \textcolor{#00bbd6}{\frac{4 \text{sin} \lambda_3}{2\lambda_3+\text{sin}(2\lambda_3)}e^{-\lambda_3^2 \tau} \text{cos}\left(\frac{\lambda_3 x}{L}\right)} + ...} $$

Typically less than 2% error if $\tau=\alpha t / L_c^2 > 0.2$