Question

A spherical annular thermal resistor of thermal conductivity $K$ and inner and outer radii $2R$ and $5R$ respectively, is in steady state with its inner surface maintained at ${100}^{\circ }C$ and its outer surface maintained at ${10}^{\circ }C$ (heat flow is radially outwards). The temperature of the concentric spherical interface of radius $3R$ in steady state is $T$.

Answer 1

Given: A spherical annular thermal resistor of thermal conductivity $K$ and inner and outer radii $2R$ and $5R$ respectively, is in steady state with its inner surface maintained at ${100}^{\circ }C$ and its outer surface maintained at ${10}^{\circ }C$ (heat flow is radially outwards). The temperature of the concentric spherical interface of radius $3R$ in steady state is $T$.

To find the temperature of the concentric spherical interface of radius $3R$ in steady state

Solution:

Cut a sphere of width $dr$ and radius

$\frac{dQ}{dt}=H=\frac{KAdT}{dr}$

This can be understood,

$H=\frac{KA{\pi r}^{2}dT}{dr}$

$\frac{Hdr}{r^2}=KdT$

$H[\frac{1}{R_1}-\frac{1}{R}]=K\left(\Delta T\right)$

$H[\frac{1}{R_1}-\frac{1}{R_2}]=K(T_1-T_2)$.....(1)

Now in fig(ii):

There are two layers and those layers are in series.

Heat current will be same across layers from formula (1)

For layer A heat current $=\frac{K(100-T)(2R\times 3R)}{R}$......(2)

For layer B heat current $=\frac{K(T-10)(3R\times 5R)}{2R}$.....(3)

eqn(2) and eqn(3) should be equal

$\frac{K(100-T)\times {6R}^2}{R}=\frac{K(T-10)15R^2}{2R}$

$(100-T)6\times 2=15(T-10)$

$1200-12T=15T-150$

$1350=27T$

$T=50°C$

is the temperature of the concentric spherical interface of radius $3R$ in steady state