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 $\left(L_{2}=0.1\; m , A_{2}=0.02\; m ^{2}\right.$ $K_{2}=109 \;Wm ^{-1} K ^{-1}$ are soldered end to end as shown in Figure. 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.

$\text { Given, } L_{1}=L_{2}=L=0.1 m , A_{1}=A_{2}=A=0.02 m ^{2}$

$K_{1}=79 W m ^{-1} K ^{-1}, K_{2}=109 W m ^{-1} K ^{-1}$

$T_{1}=373 K ,$ 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.

$\text { So, } H =H_{1}=\bar{H}_{2}$

$=\frac{K_{1} A_{1}\left(T_{1}-T_{0}\right)}{L_{1}}=\frac{K_{2} A_{2}\left(T_{0}-T_{2}\right)}{L_{2}}$

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 1s

$H =\frac{K_{1} A\left(T_{1}-T_{0}\right)}{L}=\frac{K_{2} A\left(T_{0}-T_{2}\right)}{L}$

$=\left(\frac{K_{1} K_{2}}{K_{1}+K_{2}}\right) \frac{A\left(T_{1}-T_{0}\right)}{L}=\frac{A\left(T_{1}-T_{2}\right)}{L\left(\frac{1}{K_{1}}+\frac{1}{K_{2}}\right)}$

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 $K^{\prime},$ 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) $T_{ o }=\frac{\left(K_{1} T_{1}+K_{2} T_{2}\right)}{\left(K_{1}+K_{2}\right)}$

$=\frac{\left(79 W m ^{-1} K ^{-1}\right)(373 K )+\left(109 Wm ^{-1} K ^{-1}\right)(273 K )}{79 W m ^{-1} K ^{-1}+109 W m ^{-1} K ^{-1}}$

$=315 K$

(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) $H^{\prime}=H=\frac{K^{\prime} A\left(T_{1}-T_{2}\right)}{2 L}$

$=\frac{\left(91.6 W m ^{-1} K ^{-1}\right) \times\left(0.02 m ^{2}\right) \times(373 K -273 K )}{2 \times(0.1 m )}$

$=916.1\, W$