Question
The temperature of the two outer surfaces of a composite slab, consisting of two materials having coefficients of thermal conductivity $$K$$ and $$2K$$ and thickness $$x$$ and $$4x,$$ respectively, are $${T_2}$$ and $${T_1}\left( {{T_2} > {T_1}} \right).$$ The rate of heat transfer through the slab, in a steady state is $$\left( {\frac{{A\left( {{T_2} - {T_1}} \right)K}}{x}} \right)f,$$ with $$f$$ equal to
The temperature of the two outer surfaces of a composite slab, consisting of two materials having coefficients of thermal conductivity $$K$$ and $$2K$$ and thickness $$x$$ and $$4x,$$ respectively, are $${T_2}$$ and $${T_1}\left( {{T_2} > {T_1}} \right).$$ The rate of heat transfer through the slab, in a steady state is $$\left( {\frac{{A\left( {{T_2} - {T_1}} \right)K}}{x}} \right)f,$$ with $$f$$ equal to
A.
$$\frac{2}{3}$$
B.
$$\frac{1}{2}$$
C.
$$1$$
D.
$$\frac{1}{3}$$
Answer :
$$\frac{1}{3}$$
Solution :
The thermal resistance is given by
$$\eqalign{ & \frac{x}{{KA}} + \frac{{4x}}{{2KA}} = \frac{x}{{KA}} + \frac{{2x}}{{KA}} = \frac{{3x}}{{KA}} \cr & \therefore \frac{{dQ}}{{dt}} = \frac{{\Delta T}}{{\frac{{3x}}{{KA}}}} = \frac{{\left( {{T_2} - {T_1}} \right)KA}}{{3x}} \cr & = \frac{1}{3}\left\{ {\frac{{A\left( {{T_2} - {T_1}} \right)K}}{x}} \right\} \cr & \therefore f = \frac{1}{3} \cr} $$
The thermal resistance is given by
$$\eqalign{ & \frac{x}{{KA}} + \frac{{4x}}{{2KA}} = \frac{x}{{KA}} + \frac{{2x}}{{KA}} = \frac{{3x}}{{KA}} \cr & \therefore \frac{{dQ}}{{dt}} = \frac{{\Delta T}}{{\frac{{3x}}{{KA}}}} = \frac{{\left( {{T_2} - {T_1}} \right)KA}}{{3x}} \cr & = \frac{1}{3}\left\{ {\frac{{A\left( {{T_2} - {T_1}} \right)K}}{x}} \right\} \cr & \therefore f = \frac{1}{3} \cr} $$
