For a spherical shell with inner and outer ra | vedclass

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Question

(C) Consider the spherical shell to be made of many thin concentric spherical layers of thickness $dr$ and radius $r$.For a thin layer of radius $r$ a

Vedclass · Accepted Answer

$\frac{{4\pi k({T_1} - {T_2})}}{{\left[ {\frac{1}{{{r_1}}} - \frac{1}{{{r_2}}}} \right]}}$

Vedclass · Answer

(C) Consider the spherical shell to be made of many thin concentric spherical layers of thickness $dr$ and radius $r$.For a thin layer of radius $r$ and thickness $dr$,the temperature gradient is $-dT/dr$.The surface area of this thin layer is $A = 4\pi r^2$.The heat flow rate through this layer is given by $\frac{dQ}{dt} = -kA \frac{dT}{dr} = -k(4\pi r^2) \frac{dT}{dr}$.Rearranging the terms,we get $dT = -\frac{dQ/dt}{4\pi k} \frac{dr}{r^2}$.Integrating from $r_1$ to $r_2$ with temperatures $T_1$ to $T_2$:$\int_{T_1}^{T_2} dT = -\frac{dQ/dt}{4\pi k} \int_{r_1}^{r_2} \frac{dr}{r^2}$.$T_2 - T_1 = -\frac{dQ/dt}{4\pi k} \left[ -\frac{1}{r} \right]_{r_1}^{r_2} = \frac{dQ/dt}{4\pi k} \left[ \frac{1}{r_2} - \frac{1}{r_1} \right]$.$T_1 - T_2 = \frac{dQ/dt}{4\pi k} \left[ \frac{1}{r_1} - \frac{1}{r_2} \right]$.Therefore,the heat flow rate is $\frac{dQ}{dt} = \frac{4\pi k(T_1 - T_2)}{\left[ \frac{1}{r_1} - \frac{1}{r_2} \right]}$.

For a spherical shell with inner and outer radii $r_1$ and $r_2$ respectively,the formula for the heat flow rate is = ......... where $T_1$ and $T_2$ are the temperatures of the inner and outer surfaces of the shell $(T_1 > T_2)$. The thermal conductivity of the material of the shell is $k$.

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