Q. 1.62 - An Introduction to Thermal Physics

Question

Consider a uniform rod of material whose temperature varies only along its length, in the $x$ direction. By considering the heat flowing from both directions into a small segment of length $\mathrm{\Delta }x$

derive the heat equation,

$\frac{\mathrm{\partial }T}{\mathrm{\partial }t}=K\frac{{\mathrm{\partial }}^{2}T}{\mathrm{\partial }{x}^2}$

where $K=k_t/c{\rho }_i$, $c$ is the specific heat of the material, and $\rho$ is its density. (Assume that the only motion of energy is heat conduction within the rod; no energy enters or leaves along the sides.) Assuming that $K$ is independent of temperature, show that a solution of the heat equation is

$T(x,t)=T_0+\frac{A}{\sqrt{t}}{e}^{-x^2/4Kt},$

where $T_0$ is a constant background temperature and $A$ is any constant. Sketch (or use a computer to plot) this solution as a function of $x$, for several values of $t$. Interpret this solution physically, and discuss in some detail how energy spreads through the rod as time passes.

Step-by-Step Solution

Answer

The energy spreads through the rod as time passes $\frac{dT}{dt}=K\frac{d^{2}T}{d{x}^2}$

Step 1: Heat Conduction

The trying to follow has been the warmth conduction law:

$\frac{\mathrm{\Delta }Q}{\mathrm{\Delta }t}=k_{t}A\frac{dT}{dx}$

The warmth capacity is stated with $k_t$ Examine the diagram follows, then evaluate the derivation of all this problem to relation to $x$ and let $\mathrm{\Delta }t\to dt,\mathrm{\Delta }Q\to dQ$ , giving us:

$\frac{d^{2}Q}{dxdt}=k_{t}A\frac{d^{2}T}{d{x}^2}$

The temperature is calculated using the the subsequent  equation:

$Q=mc\mathrm{\Delta }T$

The heft is determined by multiplying the length of the slice by both the packing density, as follows:

$Q=c\rho \mathrm{\Delta }V\mathrm{\Delta }T$

 $\mathrm{\Delta }V=A\mathrm{\Delta }x$is that the volumetric, and $A$ is that the larger surface section, thus:

$\frac{dQ}{dx}=c\rho AdT$

$\begin{array}{r}\frac{d^{2}Q}{dxdt}=c\rho A\frac{dT}{dt}\end{array}$

$\begin{array}{r}c\rho A\frac{dT}{dt}=-k_{t}A\frac{d^{2}T}{d{x}^2}\\ \end{array}$

$\frac{dT}{dt}=K\frac{d^{2}T}{d{x}^2}$

$K=\frac{k_t}{c\rho A}$

                     

Step 2: Equation

Perhaps we must always show that such regression relation is simply the differential equation's response.

$T(x,t)=T_0+\frac{A}{\sqrt{t}}{e}^{-x^2/4Kt}$

Calculation (3)'s LHS was even as continues to follow:

$\begin{array}{c}\mathrm{LHS}=\frac{\mathrm{\partial }}{\mathrm{\partial }t}\left[T_0+\frac{A}{\sqrt{t}}{e}^{-x^2/4Kt}\right]\\ \end{array}$

$\mathrm{LHS}=-\frac{1}{2}\frac{A}{t^{3/2}}{e}^{-x^2/4Kt}+\frac{A}{t^{5/2}}\frac{x^2}{4K}{e}^{-x^2/4Kt}$

Calculation (3) has had the relevant RHS:

$\begin{array}{c}\text{ RHS }=-K\frac{\mathrm{\partial }}{\mathrm{\partial }x}\frac{\mathrm{\partial }}{\mathrm{\partial }x}\left[T_0+\frac{A}{\sqrt{t}}{e}^{-x^2/4Kt}\right]\end{array}$

$\text{ RHS }=-\frac{A}{\sqrt{t}}\frac{1}{2t}\frac{\mathrm{\partial }}{\mathrm{\partial }x}\left[x{e}^{-x^2/4Kt}\right]$

$\begin{array}{c}\mathrm{RHS}=-\frac{A}{\sqrt{t}}\frac{1}{2t}\frac{\mathrm{\partial }}{\mathrm{\partial }x}\left[e^{-x^2/4Kt}-\frac{x^2}{2Kt}{e}^{-x^2/4Kt}\right]\\ \end{array}$

$\mathrm{RHS}=-\frac{1}{2}\frac{A}{t^{3/2}}{e}^{-x^2/4Kt}+\frac{A}{t^{5/2}}\frac{x^2}{4K}{e}^{-x^2/4Kt}$

Step 3: Plot the answer

They still must plan that approach.

$\frac{T(x,t)-T_0}{A}=\frac{1}{\sqrt{t}}{e}^{-x^2/4Kt}$

$I$ have used following syntax to print the road as an element of $x$ for varying values of $t$, the variables of $t$ will still be in terms of both the fixed $K$, the constants are:

$t_1=\frac{0.01}{K}{t}_2=\frac{0.1}{K}{t}_3=\frac{1.0}{K}$

Step 4: Graph