Question

Find the rate of heat flow through a cross section of the rod shown in figure (28-E10) (θ2 > θ1). Thermal conductivity of the material of the rod is K.
Figure

Answer 1

The rate of heat flow through a cross section of the rod is given by $q=\frac{K \pi\left(\theta_{2}-\theta_{1}\right)\left(r_{2} r_{1}\right)}{L}$

Explanation:

Step 1:

Let the thermal conductivity of the material of the rod is K.

From attached figure by similar triangles

       $\frac{x}{L}=\frac{y-r_{1}}{\left(r_{2}-r_{1}\right)}$

       $y-r_{1}=\frac{x\left(r_{2}-r_{1}\right)}{L}$

       $y=r_{1}+\frac{x\left(r_{2}-r_{1}\right)}{L}$

       $\text { Let }\left(r_{2}-r_{1}\right) / L=a$

       $y=r_{1}+a x$

Heat resistance is given as

       $d R=\frac{d x}{K A}$

       $d R=\frac{d x}{K \pi r^{2}}$

Where A = area of circle  

Step 2:

Integrating both sides

      $\int_{0}^{R} d R=\int_{0}^{L} \frac{d x}{K \pi\left(r_{1}+a x\right)^{2}}$

$\text { Let } r_{1}+a x=t$

     $a d x=d t$

     $R=\frac{\int_{r_{1}}^{r_{2}+a L} \frac{d t}{t^{2}}}{K \pi}$

    $R=\frac{1}{a K \pi}\left(-\frac{1}{t_{r_{1}}}^{r_{2}+a L}\right)$

    $R=\frac{1}{a K \pi}\left(\frac{1}{r_{1}}-\frac{1}{r_{2}}\right)$

Step 3:

Rate of heat flow is given by,

    $q=\frac{\frac{\left(\theta_{2}-\theta_{1}\right)\left(r_{1} r_{2}\right)}{L\left(r_{2}-r_{1}\right)}}{K \pi\left(r_{2}-r_{1}\right)}$

    $q=\frac{K \pi\left(\theta_{2}-\theta_{1}\right)\left(r_{2} r_{1}\right)}{L}$

Answer 2

Answer:

ans.....

Explanation: