Question

The three rods shown in figure (28−E7) have identical geometrical dimensions. Heat flows from the hot end at a rate of 40 W in the arrangement (a). Find the rates of heat flow when the rods are joined as in arrangement (b) and in (c). Thermal condcutivities of aluminium and copper are 200 W m−1°C−1 and 400 W m−1°C−1 respectively.
Figure

Answer 1

Explanation:

Given, rods have identical geometrical dimensions that is length and area are equal

Rate of heat flow from hot end = 40W

Thermal conductivity of aluminum = $200 \mathrm{Wm}^{-1}$ $^{\circ} \mathrm{C}^{-1}$

Thermal conductivity of copper = $400 \mathrm{Wm}^{-1}$ $^{\circ} \mathrm{C}^{-1}$

(a) As the rods are connected in series so total thermal resistance is given as

     $\mathrm{R}=\mathrm{R}_{1}+\mathrm{R}_{2}+\mathrm{R}_{3}$

         $=\frac{l}{A}\left[\frac{2}{200}+\frac{1}{400}\right]$

     $R=\frac{l}{A} \frac{1}{80}$

     $\frac{Q}{t}=\frac{\left(\theta_{1}-\theta_{2}\right)}{R}$

     $\frac{A}{l}=\frac{1}{200}$

(b) $\mathrm{R}=\mathrm{R}_{1}+\mathrm{R}_{2}$

  $=\mathrm{R}_{1}+\left[\frac{1}{R_{c u}}+\frac{1}{R_{A I}}\right]^{-1}$

         $=\frac{1}{A} \frac{4}{600}$

By using $\frac{A}{l}=\frac{1}{200}$

       $\frac{Q}{t}=\frac{\left(\theta_{1}-\theta_{2}\right)}{R}$

       $\frac{Q}{t}=75 \mathrm{W}$

The rate of heat flow is $75 \mathrm{W}$ in (b) arrangement  

(c) $\frac{1}{R}=\frac{1}{R_{1}}+\frac{1}{R_{2}}+\frac{1}{R_{3}}$

       $R=\frac{l}{A} \frac{1}{800}$

Again using $\frac{A}{l}=\frac{1}{200}$

        $\frac{Q}{t}=\frac{\left(\theta_{1}-\theta_{2}\right)}{R}$

         $\frac{Q}{t}=400 \mathrm{W}$

The rate of heat flow is $400 \mathrm{W}$ in (c) arrangement