A vertical brick duct (tube) is filled with cast iron. The lower end of the duct is maintained at a temperature $T_{1}$ which is greater than the melting point $T_{m}$ of cast iron and the upper end at a temperature $T_{2}$ which is less than the melting point of cast iron. It is given that the conductivity of liquid cast iron is equal to $k$ times the conductivity of solid cast iron. The fraction of the duct filled with molten metal is given as $\frac{l_{1}}{l} = \frac{x k (T_{1} - T_{m})}{k (T_{1} - T_{m}) + (T_{m} - T_{2})}$ . Find $x$ .

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Solution

In steady state, $\frac{d Q}{d t}$ is across the boundary of molten cast iron and solid cast iron.
Hence,
$K_{m o l t e n} A \frac{(T_{1} - T_{m})}{l_{m o l t e n}} = K_{s o l i d} A \frac{(T_{m} - T_{2})}{l_{s o l i d}}$
$∴ \frac{l_{s o l i d}}{l_{m o l t e n}} = \frac{(T_{m} - T_{2})}{k ((T_{1} - T_{m})}$
$∴ \frac{l_{s o l i d}}{l_{m o l t e n}} + 1 = \frac{((T_{m} - T_{2})}{k (T_{1} - T_{m})} + 1$
$∴ \frac{l_{m o l t e n}}{l} = \frac{k (T_{1} - T_{m})}{k ((T_{1} - T_{m}) + (T_{m} - T_{2})}$

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In steady state, dQdt is across the boundary of molten cast iron and solid cast iron.Hence,KmoltenA(T1−Tm)lmolten=KsolidA(Tm−T2)lsolid∴lsolidlmolten=(Tm−T2)k((T1−Tm)∴lsolidlmolten+1=((Tm−T2)k(T1−Tm)+1∴lmoltenl=k(T1−Tm)k((T1−Tm)+(Tm−T2)