A composite block is made of slabs A, B, C, D and E: Heat Conduction


Problem: A composite block is made of slabs A, B, C, D and E of different thermal conductivities (given in terms of constant K) and sizes (given in terms of length, L) as shown in the figure. All slabs are of same width. Heat Q flows only from left to right through the blocks. Then in steady state,  (IIT JEE 2011)

A Composite Block is Made of Slabs
  1. heat flow through A and E slabs are same.
  2. heat flow through slab E is minimum.
  3. temperature difference across slab E is minimum.
  4. heat flow through C = heat flow through B + heat flow through D.

Answer: The options (A), (C) and (D) are correct.

Solution: Let $T_1$, $T_2$, $T_3$ and $T_4$ be the temperatures at slab interfaces as shown in the figure.

A Composite Block is Made of Slabs

The rate of heat flow across a slab, with thermal conductivity $\kappa$, width $w$, thickness $l$, length $x$, and temperature difference $\Delta T$, is given by \begin{align} \frac{\mathrm{d}Q}{\mathrm{d}t}=\kappa lw \frac{\Delta T}{x}.\nonumber \end{align} Thus, the heat flow rates through the given slabs are \begin{align} \label{jaa:eqn:1} &\frac{\mathrm{d}Q_A}{\mathrm{d}t}=\frac{(2K)(4Lw)(T_1-T_2)}{L}=8Kw(T_1-T_2),\\ \label{jaa:eqn:2} &\frac{\mathrm{d}Q_B}{\mathrm{d}t}=\frac{(3K)(Lw)(T_2-T_3)}{4L}=\frac{3}{4} Kw(T_2-T_3),\\ \label{jaa:eqn:3} &\frac{\mathrm{d}Q_C}{\mathrm{d}t}=\frac{(4K)(2Lw)(T_2-T_3)}{4L}=2Kw(T_2-T_3),\\ \label{jaa:eqn:4} &\frac{\mathrm{d}Q_D}{\mathrm{d}t}=\frac{(5K)(Lw)(T_2-T_3)}{4L}=\frac{5}{4}Kw(T_2-T_3),\\ \label{jaa:eqn:5} &\frac{\mathrm{d}Q_E}{\mathrm{d}t}=\frac{(6K)(4Lw)(T_3-T_4)}{L}=24Kw(T_3-T_4). \end{align} In the steady state, heat flow into the system is equal to the heat flow out of the system i.e., $\mathrm{d}Q_A/\mathrm{d}t=\mathrm{d}Q_E/\mathrm{d}t$. Use above equations to get \begin{align} \label{jaa:eqn:6} T_1-T_2=3(T_3-T_4). \end{align} Similarly, the steady state condition for the interface at temperature $T_2$ is \begin{align} \frac{\mathrm{d}Q_A}{\mathrm{d}t}=\frac{\mathrm{d}Q_B}{\mathrm{d}t} +\frac{\mathrm{d}Q_C}{\mathrm{d}t} +\frac{\mathrm{d}Q_D}{\mathrm{d}t},\nonumber \end{align} which gives \begin{align} \label{jaa:eqn:7} 2(T_1-T_2)=T_2-T_3. \end{align} Above equations gives, \begin{align} T_3-T_4=\frac{1}{3}(T_1-T_2)=\frac{1}{6}(T_2-T_3).\nonumber \end{align} Also, above equations give \begin{align} \frac{\mathrm{d}Q_C}{\mathrm{d}t}=\frac{\mathrm{d}Q_B}{\mathrm{d}t}+\frac{\mathrm{d}Q_D}{\mathrm{d}t}. \end{align}

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