Let’s solve a heat conduction problem.A slab, 0 ≤ x ≤ L, is initially at uniform temperature T0. For times t > 0,the boundary surface at x = 0 is maintained at constant temperature T0, and the boundary surface at x = L is maintained at a prescribed time-dependent temperature T = T0 cos ωt, where ω is a positive constant. Obtain an expression for the temperature distribution T(x, t) in the slab using the Laplace transform technique

## Heat Conduction problem solving

I will solve the below problem:

A slab, 0 ≤ x ≤ L, is initially at uniform temperature T0. For times t > 0, the boundary surface at x = 0 is maintained at constant temperature T0, and the boundary surface at x = L is maintained at a prescribed time-dependent temperature T = T0 cos ωt, where ω is a positive constant. Obtain an expression for the temperature distribution T(x, t) in the slab using the Laplace transform technique

(sol)

$$ \frac{\partial^2 T}{\partial x^2} = \frac{1}{\alpha} \frac{\partial T}{\partial t} \text{ in } 0 <x<L, t>0$$

$$BC1: T(x=0)=T_0, BC2: T(x=L) = T_0 \cos \omega t , IC: T(t=0)=T_0$$

Let $\Theta(x,t) = T(x,t) – T_0$

$$ \frac{\partial^2 \Theta}{\partial x^2} = \frac{1}{\alpha} \frac{\partial \Theta}{\partial t} \text{ in } 0 <x<L, t>0$$

$$BC1: \Theta(x=0)=0, BC2: \Theta(x=L) = T_0( \cos \omega t -1), IC: \Theta(t=0)=0$$

By Laplace transform,

$$\frac{\partial^2 \overline{\Theta}}{\partial x^2} = \frac{s}{\alpha} \overline{\Theta} \text{ in } 0 < x < L$$

$$BC1: \overline{\Theta}(x=0)=0, BC2: \overline{\theta}(x=L) = – T_0 \omega \frac{1}{s} \frac{1}{ s^2 + \omega^2}$$

$$\overline{\Theta}(x,s) = C_1 \cosh{\sqrt{\frac{s}{\alpha}}x}+C_2 \sinh{\sqrt{\frac{s}{\alpha}}x}$$

By BC1, $C_1 = 0$. By BC2,

$$\overline{\Theta}(x,s) = – T_0 \omega \frac{1}{s^2 + \omega^2} \frac{\sinh{\sqrt{\frac{s}{\alpha}}x}}{\sinh{\sqrt{\frac{s}{\alpha}}}L}$$

$$\mathcal{L}^{-1} \left[ \frac{1}{s^2+\omega^2}\right]= \frac{1}{\omega} \sin \omega t$$

$$\mathcal{L}^{-1} \left[ \frac{\sinh{\sqrt{\frac{s}{\alpha}}x}}{\sinh{\sqrt{\frac{s}{\alpha}}}L}\right]=\underbrace{\frac{x}{L} + \frac{2}{\pi} \sum_{n=1}^\infty \frac{(-1)^n}{n} \sin{\lambda_n \frac{x}{\sqrt{\alpha}}} e^{-\lambda_n^2 t}}_{G(x,t):=}$$

$$\Theta(x,t) = -T_0 \int_0^t \sin{\omega \tau} G(x,t-\tau) d\tau$$

$$\T(x,t) = -T_0 \int_0^t \sin{\omega \tau} G(x,t-\tau) d\tau+T_0$$

Reference:

Problem 9.10 in Hahn, D. W., & Özisik, M. N. (2012). Heat conduction. John Wiley & Sons.

HW problem in Advanced heat transfer lecture by Prof. Seol (설재훈 교수님) in GIST, Fall semester 2023

[Heat Conduction Problem Solving 6.3] In a one-dimensional infinite medium, −∞