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.

Answer 1

For arrangement (a),


Temperature of the hot end ,T₁ = 100°C

Temperature of the cold end ,T₂ = 0°C
R_s = R₁ + R₂ + R₃
$=\frac{l}{K_{\mathrm{Al}}\mathrm{A}}+\frac{l}{K_{\mathrm{cu}}\mathrm{A}}+\frac{l}{K_{\mathit{Al}}A}$

$$\begin{gathered} =\frac{\mathit{l}}{\mathit{A}}\left(\frac{1}{200}+\frac{1}{400}+\frac{1}{200}\right) \\ =\frac{l}{\mathrm{A}}\left(\frac{5}{400}\right) \\ =\frac{l}{\mathrm{A}}\times \frac{1}{80} \end{gathered}$$
$$\begin{gathered} \frac{\mathrm{d}Q}{dt}=\mathit{}q=\mathrm{Rate}\mathrm{of}\mathrm{flow}\mathrm{of}\mathrm{heat}=\frac{T_1-T_2}{R_{\mathrm{S}}} \\ =\frac{100-0}{{\displaystyle \frac{l}{\mathrm{A}}\times \frac{1}{80}}} \\ \mathrm{Given}: \\ q=40\mathrm{W} \\ 40=\frac{100}{{\displaystyle \frac{l}{\mathrm{A}}}\times {\displaystyle \frac{1}{80}}} \\ \Rightarrow \frac{l}{\mathrm{A}}=200 \\ \Rightarrow \frac{\mathrm{A}}{l}=\frac{1}{200} \end{gathered}$$

For arrangement (b),


$$\begin{gathered} R_{\mathrm{net}}\mathit{}=R_{\mathrm{Al}}+\frac{R_{\mathrm{Cu}}\times R_{\mathrm{Al}}}{{\mathrm{R}}_{\mathrm{Cu}}+{\mathrm{R}}_{\mathrm{Al}}} \\ =\frac{l}{K_{\mathrm{Al}}A}+\frac{{\displaystyle \frac{l}{K_{\mathrm{Cu}}A}}\times {\displaystyle \frac{l}{K_{\mathrm{Al}}l}}}{{\displaystyle \frac{l}{K_{\mathrm{Cu}}A}}+{\displaystyle \frac{l}{K_{\mathit{Al}}\mathrm{A}}}} \\ =\frac{\mathit{l}}{\mathit{A}\mathit{·}{\mathit{K}}_{\mathit{Al}}}\mathit{+}\frac{\mathit{l}}{\mathit{A}\mathit{}\left({\mathit{K}}_{\mathit{Al}}\mathit{+}{\mathit{K}}_{\mathit{Cu}}\right)} \\ =\frac{l}{\mathrm{A}}\left(\frac{1}{200}+\frac{1}{200+400}\right) \\ =\frac{l}{\mathrm{A}}\left(\frac{1}{200}+\frac{1}{600}\right) \\ =\frac{4}{600}.\frac{l}{\mathrm{A}} \end{gathered}$$

Rate of flow of heat is given by
$$\begin{gathered} q=\frac{{\mathrm{T}}_1-{\mathrm{T}}_2}{{\mathrm{R}}_{\mathrm{net}}} \\ =\frac{\left(100-0\right)}{4l}600\mathrm{A} \\ =\frac{100\times 600}{4}\times \frac{1}{200} \\ =75\mathrm{W} \end{gathered}$$

For arrangement (c),



$$\begin{gathered} \frac{\mathit{1}}{{\mathit{R}}_{\mathit{net}}}\mathit{=}\mathit{}\frac{\mathit{1}}{{\mathit{R}}_{\mathit{Al}}}\mathit{+}\frac{\mathit{1}}{{\mathit{R}}_{\mathit{Cu}}}\mathit{+}\frac{\mathit{1}}{{\mathit{R}}_{\mathit{Al}}} \\ =\frac{K_{\mathrm{Al}}A}{l}+\frac{K_{\mathrm{Cu}}A}{l}+\frac{K_{\mathrm{Al}}A}{l} \\ \frac{1}{R_{\mathrm{net}}}=\left(K_{\mathrm{Al}}+K_{\mathrm{Cu}}+K_{\mathrm{Al}}\right)\frac{\mathit{A}}{\mathit{l}} \\ \frac{1}{R_{\mathrm{net}}}=\left(200+400+200\right)\times \frac{1}{200} \\ \Rightarrow R_{\mathrm{net}}\mathit{}=\frac{200}{800} \\ =\frac{1}{4} \end{gathered}$$

$$\begin{gathered} \mathrm{Rate}\mathrm{of}\mathrm{heat}\mathrm{flow}=\frac{\Delta T}{R_{\mathrm{net}}} \\ =\frac{100}{{\displaystyle \frac{1}{4}}} \\ =400\mathrm{W} \end{gathered}$$