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Question:

  • A rod of length $1$ m, initially heated to $100^{\circ}$ C before one end is inserted into the ice that maintains that end’s temperature at $0^{\circ}$ C.
  • If we denote $\theta(x,t)$ to be the temperature at distance $x$ from the end with temperature at $0^o$C at time $t$, write down this initial condition.
  • Solve the Fourier series for $\theta(x, t)$. What does the model predict for the long term temperature distribution in the rod $?$.

Recall that initial condition refers to $$\theta(x,0) = f(x)$$ where $f$ gives the initial temperature distribution along the rod. In this case, we know that $f$ satisfies $$f(x) = \begin{cases} 0 & \text{ if } x = 0,\\ 100 & \text{ if } 0<x\leq 1 \end{cases} $$

I know that this is related to the homogeneous heat diffusion equation $$\frac{\partial\theta}{\partial t} = \frac{1}{\alpha} \frac{\partial^2 \theta}{\partial x^2}$$ where $L = 1$. I also know that boundary conditions are $\theta(0,t) = \frac{\partial \theta}{\partial x}(1,t) =0$. However, I am not sure whether $\frac{\partial\theta}{\partial x}(x,1) = 0$.

Also, its Fourier series is $$\theta(x,t) = \sum_{r=1}^\infty B_r \sin \left( r\pi x \right) \exp\left( -\alpha \left( \frac{r\pi}{2} \right)^2 t \right)$$ where $$B_r = 2 \int_0^1 f(x)\sin(r\pi x)dx.$$ I am wondering whether my attempt above answer the question or not.

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  • $\begingroup$ With what you've written I think the summation index should be $r$, not $n$, correct? $\endgroup$
    – K.defaoite
    Oct 19, 2020 at 10:54
  • $\begingroup$ @K.defaoite Yes, you are right. I edited my post. $\endgroup$
    – Idonknow
    Oct 19, 2020 at 10:57
  • $\begingroup$ Why do you say that the boundary conditions are $\theta(0,t)=\theta(1,t)=0$? This wouldn't satisfy the initial condition that you say above, right? If I understood well for your problem the initial datum would be something like $\theta(x,0)=0$ if $x=0$ and $\theta(x,0)=100$ if $0<x\leq 1$. Also the boundary conditions would be just $\theta(0,t)=0$ but on the other side we don't know the temperature, what we know is that the rod finishes upon $1$m and hence the heat cannot escape, that means $\partial\theta/\partial x (1,t) = 0$. $\endgroup$
    – Víctor
    Oct 19, 2020 at 12:37
  • $\begingroup$ @vctrnf Edited my post based on your comments. Thanks. $\endgroup$
    – Idonknow
    Oct 19, 2020 at 12:42

1 Answer 1

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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ Lets $\ds{\theta\pars{x,t} = \sum_{n = 0}^{\infty}\on{a}_{n}\pars{t}\sin\pars{k_{n}x}}$ where $\ds{k_{n} = \pars{2n + 1}\,{\pi \over 2}}$ which already satisfy $\ds{\theta\pars{0,t} = \left.\partiald{\theta\pars{x,t}}{x}\right\vert_{\ x\ =\ 1} = 0,\ \forall\ t}$.


The above expression for $\ds{\theta\pars{x,y}}$ must satisfy the above differential equation. Namely, \begin{align} &\sum_{n = 0}^{\infty}\dot{\on{a}}_{n}\pars{t}\sin\pars{k_{n}x} = -\,{1 \over \alpha} \sum_{n = 0}^{\infty}\dot{\on{a}}_{n}\pars{t}k_{n}^{2}\sin\pars{k_{n}x} \end{align} With $\ds{\int_{0}^{1}\sin\pars{k_{m}x}\sin\pars{k_{n}x} \,\dd x = {1 \over 2}\,\delta_{mn}}$, I got $$ \dot{\on{a}}\pars{t} + {k_{n}^{2} \over \alpha}\on{a}\pars{t} = 0 \implies \on{a}_{n}\pars{t} = \on{a}_{n}\pars{0}\exp\pars{-\,{k_{n}^{2} \over \alpha}\,t} $$ The general solution is reduced to: \begin{align} &\theta\pars{x,t} = \sum_{n = 0}^{\infty} \on{a}_{n}\pars{0}\exp\pars{-\,{k_{n}^{2} \over \alpha}\,t}\sin\pars{k_{n}x} \\[5mm] &\ \mbox{and}\quad 100 = \theta\pars{x,0} = \sum_{n = 0}^{\infty} \on{a}_{n}\pars{0}\sin\pars{k_{n}x} \\[5mm] & \ \implies 100\ \underbrace{\int_{0}^{1}2\sin\pars{k_{n}x}\,\dd x} _{\ds{4/\pi \over 2n + 1}} = \on{a}_{n}\pars{0} \\[5mm] &\ \implies \on{a}_{n}\pars{0} = {400/\pi \over 2n + 1} = {200 \over k_{n}} \\[5mm] &\ \implies \bbx{\theta\pars{x,t} = 200 \sum_{n = 0}^{\infty} {\sin\pars{k_{n}x} \over k_{n}} \exp\pars{-\,{k_{n}^{2} \over \alpha}\,t}} \\ & \end{align}
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