Question

A system consists of a cylinder surrounded by a cylindrical shell. A cylinder is a radius R and is made of material of thermal conductivity K, whereas a cylindrical shell has an inner radius R and outer radius 2R and is made of material of thermal conductivity twice as that of cylinder. Assuming the system is in steady state and negligible heat loss across the cylindrical surface, find the effective thermal conductivity of the system, if the two ends of the combined system are maintained at two different temperatures.
A) $3K \\$
B) $\dfrac{2}{3}K \\$
C) $\dfrac{{7K}}{4} \\$
D) $\dfrac{{5K}}{4} \\ $

Answer 1

Hint: Whenever heat energy is transferred through conduction, due to the inherent property of the material, it offers obstruction to the flow of heat energy through it. This obstruction is called thermal energy.
The formula for the thermal resistance of a cylindrical surface is –
$
  R = \dfrac{L}{{K.A}} \\
  where \\
 $
L = length of the cylinder
K = thermal conductivity
A = area of cross-section
The net-effective resistance for a parallel combination of individual bodies is given by the formula –
$\dfrac{1}{{{R_{eq}}}} = \dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}}$

Complete step by step answer:
Here is the diagram of the cylinder covered by the cylindrical shell.

The ends of the cylinders are maintained at the temperature ${T_1}$ and ${T_2}$. Since, the heat flows through both the cylinders together at the same time, we need to use the condition of parallel combination.
The thermal resistance offered by the cylinder –
$\Rightarrow$ ${R_1} = \dfrac{{{L_1}}}{{{K_1}{A_1}}} = \dfrac{L}{{K\left( {\pi {R^2}} \right)}}$
The thermal resistance offered by the cylindrical shell –
$\Rightarrow$ ${R_2} = \dfrac{{{L_2}}}{{{K_2}{A_2}}} = \dfrac{L}{{K\left( {\pi {{\left( {2R} \right)}^2} - \pi {R^2}} \right)}} = \dfrac{L}{{K\left( {4\pi {R^2} - \pi {R^2}} \right)}} = \dfrac{L}{{K(3\pi {R^2})}}$
Since, the heat flows through both the cylinders together at the same time, the net effective thermal resistance is equal to the parallel combination of the cylinders:
$\Rightarrow$ $\dfrac{1}{{{R_{eq}}}} = \dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}}$
On solving, we get:
$\Rightarrow$ ${R_{eq}} = \dfrac{{{R_1}{R_2}}}{{{R_1} + {R_2}}}$
Substituting the values of the resistances, we get:
\[
\Rightarrow {R_{eq}} = \dfrac{{{R_1}{R_2}}}{{{R_1} + {R_2}}} \\
\Rightarrow {R_{eq}} = \dfrac{{\dfrac{L}{{{K_1}\left( {\pi {R^2}} \right)}}.\dfrac{L}{{{K_2}\left( {3\pi {R^2}} \right)}}}}{{\dfrac{L}{{{K_1}\left( {\pi {R^2}} \right)}} + \dfrac{L}{{{K_2}\left( {3\pi {R^2}} \right)}}}} \\
\Rightarrow {R_{eq}} = \dfrac{{{L^2}}}{{3L{K_2}\pi {R^2} + L{K_1}\pi {R^2}}} \\
 \]
Substituting
$\Rightarrow$ ${R_{eq}} = \dfrac{L}{{{K_{eq}}A}} = \dfrac{L}{{{K_{eq}}{{\left( {2\pi R} \right)}^2}}}$
Equating the values of ${R_{eq}}$, we get –
$
\Rightarrow \dfrac{L}{{4{K_{eq}}\pi {R^2}}} = \dfrac{{{L^2}}}{{3L{K_2}\pi {R^2} + L{K_1}\pi {R^2}}} \\
\Rightarrow \dfrac{L}{{4{K_{eq}}\pi {R^2}}} = \dfrac{{{L^2}}}{{L\pi {R^2}\left( {3{K_2} + {K_1}} \right)}} \\
rearranging, \\
\Rightarrow 4{K_{eq}} = {K_1} + 3{K_2} \\
 $
As per the problem,
$
\Rightarrow {K_1} = K \\
\Rightarrow {K_{_2}} = 2K \\
 $
Substituting these values in the above equation, we get –
$
\Rightarrow 4{K_{eq}} = {K_1} + 3{K_2} \\
\Rightarrow 4{K_{eq}} = K + 3\left( {2K} \right) \\
\Rightarrow 4{K_{eq}} = 7K \\
\Rightarrow {K_{eq}} = \dfrac{7}{4}K \\
$
Hence, the correct option is Option C.

Note: The concept of thermal resistance is like that of electrical resistance. So, when in doubt, you can draw up the analogies between the two of them and compare the analogous quantities so that you have a better understanding of solving the problem.