Question
Two rods $$A$$ and $$B$$ of different materials are welded together as shown in figure. Their thermal conductivities are $${K_1}$$ and $${K_2}.$$ The thermal conductivity of the composite rod will be :
Two rods $$A$$ and $$B$$ of different materials are welded together as shown in figure. Their thermal conductivities are $${K_1}$$ and $${K_2}.$$ The thermal conductivity of the composite rod will be :
A.
$$\frac{{3\left( {{K_1} + {K_2}} \right)}}{2}$$
B.
$${{K_1} + {K_2}}$$
C.
$$2\left( {{K_1} + {K_2}} \right)$$
D.
$$\frac{{{K_1} + {K_2}}}{2}$$
Answer :
$$\frac{{{K_1} + {K_2}}}{2}$$
Solution :
Heat current $$H = {H_1} + {H_2}$$
$$\eqalign{ & = \frac{{{K_1}\;A\left( {\;{T_1} - {T_2}} \right)}}{d} + \frac{{{K_2}\;A\left( {\;{T_1} - {T_2}} \right)}}{d} \cr & \frac{{{K_{EQ}}2A\left( {\;{T_1} - {T_2}} \right)}}{d} = \frac{{A\left( {{T_1} - {T_2}} \right)}}{d}\left[ {{K_1} + {K_2}} \right] \cr} $$
Hence equivalent thermal conductivities for two rods of equal area is given by
$${K_{EQ}} = \frac{{{k_1} + {k_2}}}{2}$$
Heat current $$H = {H_1} + {H_2}$$
$$\eqalign{ & = \frac{{{K_1}\;A\left( {\;{T_1} - {T_2}} \right)}}{d} + \frac{{{K_2}\;A\left( {\;{T_1} - {T_2}} \right)}}{d} \cr & \frac{{{K_{EQ}}2A\left( {\;{T_1} - {T_2}} \right)}}{d} = \frac{{A\left( {{T_1} - {T_2}} \right)}}{d}\left[ {{K_1} + {K_2}} \right] \cr} $$
Hence equivalent thermal conductivities for two rods of equal area is given by
$${K_{EQ}} = \frac{{{k_1} + {k_2}}}{2}$$
