Chapter 1: Q. 1.62 (page 40)

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 $Δ x$
derive the heat equation,
$\frac{\partial T}{\partial t} = K \frac{\partial^{2} T}{\partial x^{2}}$
where $K = k_{t} / c ρ_{i}$ , $c$ is the specific heat of the material, and $ρ$ 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
$T (x, t) = T_{0} + \frac{A}{\sqrt{t}} e^{- x^{2} / 4 K t},$
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.

Short Answer

Expert verified

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

Step by step solution

Heat Conduction

The trying to follow has been the warmth conduction law:

$\frac{Δ Q}{Δ t} = k_{t} A \frac{d T}{d x}$

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 $Δ t \to d t, Δ Q \to d Q$ , giving us:

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

The temperature is calculated using the the subsequent equation:

$Q = m c Δ T$

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

$Q = c ρ Δ V Δ T$

$Δ V = A Δ x$ is that the volumetric, and $A$ is that the larger surface section, thus:

$\frac{d Q}{d x} = c ρ A d T$

$\begin{aligned} \frac{d^{2} Q}{d x d t} = c ρ A \frac{d T}{d t} \end{aligned}$

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

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

$K = \frac{k_{t}}{c ρ A}$

Equation

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

$T (x, t) = T_{0} + \frac{A}{\sqrt{t}} e^{- x^{2} / 4 K t}$

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

$\begin{matrix} LHS = \frac{\partial}{\partial t} [T_{0} + \frac{A}{\sqrt{t}} e^{- x^{2} / 4 K t}] \end{matrix}$

localid="1650352081120" $LHS = - \frac{1}{2} \frac{A}{t^{3 / 2}} e^{- x^{2} / 4 K t} + \frac{A}{t^{5 / 2}} \frac{x^{2}}{4 K} e^{- x^{2} / 4 K t}$

Calculation (3) has had the relevant RHS:

$\begin{matrix} RHS = - K \frac{\partial}{\partial x} \frac{\partial}{\partial x} [T_{0} + \frac{A}{\sqrt{t}} e^{- x^{2} / 4 K t}] \end{matrix}$

$RHS = - \frac{A}{\sqrt{t}} \frac{1}{2 t} \frac{\partial}{\partial x} [x e^{- x^{2} / 4 K t}]$

$\begin{matrix} RHS = - \frac{A}{\sqrt{t}} \frac{1}{2 t} \frac{\partial}{\partial x} [e^{- x^{2} / 4 K t} - \frac{x^{2}}{2 K t} e^{- x^{2} / 4 K t}] \end{matrix}$

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

Plot the answer

They still must plan that approach.

$\frac{T (x, t) - T_{0}}{A} = \frac{1}{\sqrt{t}} e^{- x^{2} / 4 K t}$

$I$ have used following syntax to print the road as an element of localid="1650354167789" $x$ for varying values of $t$ , the variables of localid="1650354177813" $t$ will still be in terms of both the fixed localid="1650354173027" $K$ , the constants are:

localid="1650354128505" $t_{1} = \frac{0.01}{K} t_{2} = \frac{0.1}{K} t_{3} = \frac{1.0}{K}$