A spherical shell has inner radius 3.00 cm and outer radius...

A spherical shell has inner radius $3.00 \mathrm{~cm}$ and outer radius $7.00 \mathrm{~cm} .$ It is made of material with thermal conductivity $k=0.800 \mathrm{~W} / \mathrm{m} \cdot{ }^{\circ} \mathrm{C} .$ The interior is maintained at temperature $5^{\circ} \mathrm{C}$ and the exterior at $40^{\circ} \mathrm{C}$. After an interval of time, the shell reaches a steady state with the temperature at each point within it remaining constant in time. (a) Explain why the rate of energy transfer $P$ must be the same through each spherical surface, of radius $r$, within the shell and must satisfy $$\frac{d T}{d r}=\frac{P}{4 \pi k r^{2}}$$ (b) Next, prove that $$\int_{5}^{40} d T=\frac{P}{4 \pi k} \int_{0.03}^{0.07} r^{-2} d r$$ where $T$ is in degrees Celsius and $r$ is in meters. (c) Find the rate of energy transfer through the shell. (d) Prove that $$\int_{5}^{T} d T=1.84 \int_{0.03}^{r} r^{-2} d r$$ where $T$ is in degrees Celsius and $r$ is in meters. (e) Find the temperature within the shell as a function of radius. (f) Find the temperature at $r=5.00 \mathrm{~cm}$, halfway through the shell.

A spherical shell has inner radius $3.00 \mathrm{~cm}$ and outer radius $7.00 \mathrm{~cm} .$ It is made of material with thermal conductivity $k=0.800 \mathrm{~W} / \mathrm{m} \cdot{ }^{\circ} \mathrm{C} .$ The interior is maintained at temperature $5^{\circ} \mathrm{C}$ and the exterior at $40^{\circ} \mathrm{C}$. After an interval of time, the shell reaches a steady state with the temperature at each point within it remaining constant in time. (a) Explain why the rate of energy transfer $P$ must be the same through each spherical surface, of radius $r$, within the shell and must satisfy
$$\frac{d T}{d r}=\frac{P}{4 \pi k r^{2}}$$
(b) Next, prove that
$$\int_{5}^{40} d T=\frac{P}{4 \pi k} \int_{0.03}^{0.07} r^{-2} d r$$
where $T$ is in degrees Celsius and $r$ is in meters. (c) Find the rate of energy transfer through the shell. (d) Prove that
$$\int_{5}^{T} d T=1.84 \int_{0.03}^{r} r^{-2} d r$$
where $T$ is in degrees Celsius and $r$ is in meters. (e) Find the temperature within the shell as a function of radius.
(f) Find the temperature at $r=5.00 \mathrm{~cm}$, halfway through the shell.

Key Concepts

Steady State Heat Conduction

In steady state heat conduction, the temperature distribution does not change with time, which implies that the heat entering any region is exactly balanced by the heat leaving it. This concept is fundamental when analyzing thermal systems, as it allows for the derivation of differential equations that govern the spatial temperature profile under constant energy flux conditions.

Fourier’s Law of Heat Conduction

Fourier’s Law states that the heat transfer rate through a material is proportional to the negative of the temperature gradient and the area through which the heat is flowing. This law is essential for modeling heat flow in different geometries, linking the thermal conductivity of the material with the spatial temperature differences, and providing a basis for solving conduction problems.

Spherical Coordinate System in Heat Conduction

When analyzing heat conduction in spherical geometries, the symmetry of the problem allows the area available for conduction to vary with the square of the radius. This leads to a modified form of Fourier’s Law where the heat transfer rate is related to the temperature gradient divided by the area factor, 4?r², making it crucial for setting up the appropriate differential equations in spherical symmetric systems.

Integration of Differential Equations

The formulation of the heat conduction problem in differential form typically requires integration to obtain the temperature distribution as a function of the spatial coordinate. Integrating the differential equation derived from Fourier’s Law enables one to relate temperature differences across the material to the heat transfer rate, particularly when dealing with non-uniform cross-sectional areas.

Application of Boundary Conditions

Boundary conditions are used to determine integration constants when solving the differential equation for temperature distribution. In conduction problems, the known temperatures at the inner and outer surfaces of the material provide the necessary constraints to uniquely solve for the heat transfer rate and the spatial variation of temperature, ensuring the solution accurately reflects the physical situation.

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