Question

A cylinder of radius R made of a material of thermal conductivity ${{k}_{1}}$ is surrounded by a cylindrical sheet of inner radius R and the outer radius 2R made of material of thermal conductivity ${{k}_{2}}$. The two ends of the combined system are maintained at two different temperatures. There is no loss of heat across the cylindrical surface and the system is in a steady state. Calculate the effective thermal conductivity of the system. Assume that heat flow is always axial.

Answer 1

Hint: In this question, we have to find the thermal conductivity. By using the formula of thermal resistance for two different areas and the cylinders are in parallel, so we add both the thermal conductivities and by solving the equations, we get the desirable answer.

Formula Used:
Thermal conductivity for the given cylinder is
$R=\dfrac{L}{KA}$
Where A is the area.
So $R=\dfrac{L}{K(\pi {{R}^{2}})}$

Complete step by step solution:
We know the thermal resistances for the inner and outer cylinders be

${{R}_{1}}=\dfrac{L}{{{K}_{1}}(\pi {{R}^{2}})} \\ $
$\Rightarrow {{R}_{2}}=\dfrac{L}{{{K}_{2}}(4\pi {{R}^{2}}-\pi {{R}^{2}})} \\ $
$\Rightarrow {{R}_{2}}=\dfrac{L}{{{K}_{2}}(3\pi {{R}^{2}})} \\ $
For an equivalent conductor of length L, radius 2R and thermal conductivity K,
Thermal resistance $R=\dfrac{L}{K(4\pi {{R}^{2}})}$
As the inner and the outer cylinders are effectively in parallel between the same temperature difference, then
$\dfrac{1}{R}=\dfrac{1}{{{R}_{1}}}+\dfrac{1}{{{R}_{2}}} \\ $
$\Rightarrow \dfrac{K(4\pi {{R}^{2}})}{L}=\dfrac{{{K}_{1}}(\pi {{R}^{2}})}{L}+\dfrac{{{K}_{2}}(\pi {{R}^{2}})}{L} \\ $
Solving the above equation, we get
$\therefore K=\dfrac{{{K}_{1}}+3{{K}_{2}}}{4}$

Hence the thermal conductivity of the system is $K=\dfrac{{{K}_{1}}+3{{K}_{2}}}{4}$.

Note: Thermal conductivity is the capability of the given object to conduct or transfer heat. Objects with high thermal conductivity are used in heat sinks and the materials with low thermal conductivity are used as the thermal insulators.