Physics > Heat Transfer > 1.0 Conduction
Heat Transfer
1.0 Conduction
2.0 Convection
3.0 Radiation
4.0 Kirchhoff's Law
5.0 Wein's displacement law
6.0 Stefan-Boltzman law
7.0 Newton's laws of cooling
1.2 Parallel connection of slabs
If two slabs are connected in parallel connection, then the total thermal current is equal to the sum of the currents through slabs $1$ and $2$.
$$\begin{equation} \begin{aligned} i = {i_1} + {i_2} \\ = \left( {{T_1} - {T_2}} \right)\left( {\frac{1}{{{R_1}}} + \frac{1}{{{R_2}}}} \right) \\ \Rightarrow \frac{1}{R} = \frac{1}{{{R_1}}} + \frac{1}{{{R_2}}} \\\end{aligned} \end{equation} $$
Question 6. Two metal cubes with $3\ cm$ edges of aluminium and copper are arranged as shown in the figure. Find
(a) The total current from one end to the other
(b) The ratio of the thermal current carried by the aluminium cube to that ot copper cube.
( ${{K_{aluminium}} = 237\,W\,{m^{ - 1}}{K^{ - 1}}}$, ${{K_{copper}} = 401\,W\,{m^{ - 1}}{K^{ - 1}}}$).
Solution: (a) Thermal resistance of aluminium cube is $$\begin{equation} \begin{aligned} {R_1} = \frac{l}{{KA}} \\ {R_1} = \frac{{\left( {3.0 \times {{10}^{ - 2}}} \right)}}{{\left( {237} \right){{\left( {3.0 \times {{10}^{ - 2}}} \right)}^2}}} = 0.14\,K/W \\\end{aligned} \end{equation} $$and the thermal resistance of copper cube is $$\begin{equation} \begin{aligned} {R_2} = \frac{l}{{KA}} \\ {R_2} = \frac{{\left( {3.0 \times {{10}^{ - 2}}} \right)}}{{\left( {401} \right){{\left( {3.0 \times {{10}^{ - 2}}} \right)}^2}}} = 0.08\,K/W \\\end{aligned} \end{equation} $$As these two resistances are in parallel, their equivalent resistance will be,$$\begin{equation} \begin{aligned} R = \frac{{{R_1}{R_2}}}{{{R_1} + {R_2}}} \\ = \frac{{\left( {0.14} \right)\left( {0.08} \right)}}{{\left( {0.14} \right) + \left( {0.08} \right)}} \\ = 0.05\,K/W \\\end{aligned} \end{equation} $$Thermal current is $$\begin{equation} \begin{aligned} \frac{temperature\ difference}{Thermal\ resistance} \\ = \frac{{\left( {100 - 20} \right)}}{{0.05}} = 1.6 \times {10^3}\,Watt \\\end{aligned} \end{equation} $$
(b) In parallel, thermal current distributes in inverse of resistance. Hence, $$\frac{{{i_{cu}}}}{{{i_{al}}}} = \frac{{{R_{al}}}}{{{R_{cu}}}} = \frac{{{R_1}}}{{{R_2}}} = \frac{{0.14}}{{0.08}} = 1.75$$
Question 7. A cylinder of radius $R$ made of a material of thermal conductivity ${K_1}$ is surrounded by a cylindrical shell of inner radius $R$ and outer $2R$ made of a material of thermal conductivity of ${K_2}$. The two ends are maintained at different temperature and let there not be any heat loss through the cylindrical surface. What is the effective thermal conductivity?
Solution: In this situation a rod of length $L$ and area of cross section $\pi {R^2}$ and another of same length $L$ and area of cross section $\pi \left[ {{{\left( {2R} \right)}^2} - {R^2}} \right] = 3\pi {R^2}$ will conduct the heat simultaneously, so the heat flow will be
$$\begin{equation} \begin{aligned} \frac{{dQ}}{{dt}} = \frac{{d{Q_1}}}{{dt}} + \frac{{d{Q_2}}}{{dt}} \\ = \frac{{{K_1}\pi {R^2}\left( {{T_1} - {T_2}} \right)}}{L} + \frac{{{K_2}3\pi {R^2}\left( {{T_1} - {T_2}} \right)}}{L} \\\end{aligned} \end{equation} $$Now, if the equivalent conductivity is $K$,$$\frac{{dQ}}{{dt}} = K\frac{{4\pi {R^2}\left( {{T_1} - {T_2}} \right)}}{L}\quad as\;A = \pi {\left( {2R} \right)^2}$$ Then by the above two equations $$\begin{equation} \begin{aligned} 4K = {K_{_1}} + 3{K_2} \\ \Rightarrow K = \frac{{{K_{_1}} + 3{K_2}}}{4} \\\end{aligned} \end{equation} $$
