Question

A rod of length $\ell$ and cross-sectional area A has a variable conductivity given by $K = \alpha + \beta T$, where $\alpha, \beta$ are a positive constants and T is temperature (in Kelvin) of the isothermal surface. Two ends of the rod are maintained at temperatures $T_1$ and $T_2$ ($T_1 > T_2$). Heat current flowing through the rod will be

• A. $\frac{A(T_2 - T_1)}{\ell} \left[\beta + \frac{\alpha}{2}(T_2 + T_1)\right]$
• B. $\frac{A(T_2 - T_1)}{\ell} \left[\beta - \frac{\alpha}{2}(T_2 + T_1)\right]$
• C. $\frac{A(T_2 - T_1)}{\ell} \left[\alpha - \frac{\beta}{2}(T_2 + T_1)\right]$
• D. $\frac{A(T_2 - T_1)}{\ell} \left[\alpha + \frac{\beta}{2}(T_2 + T_1)\right]$

Answer 1

Answer: (D)

$\frac{A(T_2-T_1)}{\ell} \left[\alpha + \frac{\beta}{2}(T_2+T_1)\right]$

Answer 2

For one‐dimensional steady state conduction, Fourier’s law gives

$i = -KA\,\frac{dT}{dx}$,

where the conductivity is

$K = \alpha + \beta T$.

Since the system is in a steady state, $i$ is constant. Writing:

$\frac{dT}{dx} = -\frac{i}{A(\alpha+\beta T)}$,

we separate variables and integrate from $x=0$ to $x=\ell$ and $T=T_1$ to $T=T_2$:

$\int_{0}^{\ell} dx = -\frac{A}{i} \int_{T_1}^{T_2} (\alpha + \beta T)\,dT$.

This gives:

$\ell = -\frac{A}{i} \left[ \alpha (T_2-T_1)+ \frac{\beta}{2}(T_2^2-T_1^2) \right]$.

Solving for $i$:

$i = -\frac{A}{\ell}\left[ \alpha (T_2-T_1)+\frac{\beta}{2}(T_2^2-T_1^2) \right]$.

Now note that:

$T_2^2-T_1^2 = (T_2-T_1)(T_2+T_1)$.

Thus,

$i = -\frac{A}{\ell}(T_2-T_1) \left[\alpha + \frac{\beta}{2}(T_2+T_1)\right]$.

Since $T_1>T_2$, the factor $T_2-T_1$ is negative. Therefore, the heat current (directed from $T_1$ to $T_2$) is given by

$i = \frac{A(T_2-T_1)}{\ell} \left[\alpha + \frac{\beta}{2}(T_2+T_1)\right]$.