Question
An iron bar $\left(L_1=0.1 m , A_1=\right. \left.0.02 m ^2, K_1=79 W m ^{-1} K ^{-1}\right)$ and a brass bar $\left(L_2=0.1 m , A_2=0.02 m ^2\right.$, $K_2=109 W m ^{-1} K ^{-1}$ ) are soldered end to end as shown in Fig. $10.16.$ The free ends of the iron bar and brass bar are maintained at $373 K$ and $273 K$ respectively. Obtain expressions for and hence compute $(i)$ the temperature of the junction of the two bars, $(ii)$ the equivalent thermal conductivity of the compound bar, and $(iii)$ the heat current through the compound bar.
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Answer

Given, $L_1=L_2=L=0.1 m , A_1=A_2=A=0.02 m ^2$
$K_1=79 W m ^{-1} K ^{-1}, K_2=109 W m ^{-1} K ^{-1}$, $T_1=373 K$, and $T_2=273 K$.
Under steady state condition, the heat current $\left(H_1\right)$ through iron bar is equal to the heat current $\left( H _2\right)$ through brass bar.
$\text { So, }H =H_1=H_2$
$ =\frac{K_1 A_1\left(T_1-T_0\right)}{L_1}=\frac{K_2 A_2\left(T_0-T_2\right)}{L_2}$
For $A_1=A_2=A$ and $L_1=L_2=L$, this equation leads to
$K_1\left(T_1-T_0\right)=K_2\left(T_0-T_2\right)$
Thus, the junction temperature $T_0$ of the two bars is
$T_0=\frac{\left(K_1 T_1+K_2 T_2\right)}{\left(K_1+K_2\right)}$
Using this equation, the heat current $H$ through either bar is
$H =\frac{K_1 A\left(T_1-T_0\right)}{L}=\frac{K_2 A\left(T_0-T_2\right)}{L}$
$ =\left(\frac{K_1 K_2}{K_1+K_2}\right) \frac{A\left(T_1-T_0\right)}{L}=\frac{A\left(T_1-T_2\right)}{L\left(\frac{1}{K_1}+\frac{1}{K_2}\right)}$
Using these equations, the heat current $H^{\prime}$ through the compound bar of length $L_1+L_2=2 L$ and the equivalent thermal conductivity $K$, of the compound bar are given by
$H^{\prime} =\frac{K^{\prime} A\left(T_1-T_2\right)}{2 L}=H$
$K^{\prime} =\frac{2 K_1 K_2}{K_1+K_2}$
$(i) T_0=\frac{\left(K_1 T_1+K_2 T_2\right)}{\left(K_1+K_2\right)}$
$=\frac{\left(79 W m ^{-1} K ^{-1}\right)(373 K )+\left(109 W m ^{-1} K ^{-1}\right)(273 K )}{79 W m ^{-1} K ^{-1}+109 W m ^{-1} K ^{-1}}$
$=315 K$
$(ii)\ K^{\prime} =\frac{2 K_1 K_2}{K_1+K_2}$
$ =\frac{2 \times\left(79 W m ^{-1} K ^{-1}\right) \times\left(109 W m ^{-1} K ^{-1}\right)}{79 W m ^{-1} K ^{-1}+109 W m ^{-1} K ^{-1}}$
$ =91.6 W m ^{-1} K ^{-1}$
$(iii)\ H^{\prime}=H=\frac{K^{\prime} A\left(T_1-T_2\right)}{2 L}$
$=\frac{\left(91.6 W m ^{-1} K ^{-1}\right) \times\left(0.02 m ^2\right) \times(373 K -273 K )}{2 \times(0.1 m )}$
$=916.1 W$

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