Question

A spherical container, with an inner radius \(r_{1}=1 \mathrm{~m}\) and an outer radius \(r_{2}=1.05 \mathrm{~m}\), has its inner surface subjected to a uniform heat flux of \(\dot{q}_{1}=7 \mathrm{~kW} / \mathrm{m}^{2}\). The outer surface of the container has a temperature \(T_{2}=25^{\circ} \mathrm{C}\), and the container wall thermal conductivity is \(k=1.5 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). Show that the variation of temperature in the container wall can be expressed as \(T(r)=\left(\dot{q}_{1} r_{1}^{2} / k\right)\left(1 / r-1 / r_{2}\right)+T_{2}\) and determine the temperature of the inner surface of the container at \(r=r_{1}\).

Short answer

Question: Determine the temperature at the inner surface of a spherical container, given the following information: the inner radius, r1 = 1 m; outer radius, r2 = 1.05 m; heat flux at the inner surface, qₒ = 7,000 W/m²; thermal conductivity of the wall material, k = 1.5 W/m·K; and the temperature at the outer surface, T₂ = 25°C. Answer: T(r1) = 138.89°C

Step-by-step solution

Write down the heat conduction equation in spherical coordinates

The heat conduction equation in spherical coordinates is given by: $$ \frac{d}{dr}\left(k\frac{dT}{dr}\right) = 0 $$ In this problem, the thermal conductivity \(k\) is constant, so we can rewrite the equation as: $$ \frac{d}{dr}\left(k \frac{dT}{dr}\right) = \frac{d}{dr}\left(k \frac{dT_{r}}{dr}\right) = 0 $$

Integrate the equation once

Integrate the equation with respect to \(r\): $$ \int \frac{d}{dr}\left(k \frac{dT_{r}}{dr}\right) dr = \int 0 dr $$ $$ k \frac{dT_{r}}{dr} = C_1 $$ Where \(C_1\) is the integration constant.

Integrate the equation again

Integrate the equation with respect to \(r\) once more: $$ \int \frac{1}{k} \left(k \frac{dT_{r}}{dr}\right) dr = \int \frac{C_1}{k} dr $$ $$ T_{r}(r) = \frac{C_1}{k} r + C_2 $$ Where \(C_2\) is another integration constant.

Apply the boundary conditions

Now we apply the boundary conditions to determine the values of \(C_1\) and \(C_2\): 1. At \(r = r_{1}\), the heat flux is given by \(\dot{q}_{1} = -k \frac{dT_{r}}{dr}\). Therefore: $$ \dot{q}_{1} = -k \frac{d}{dr}\left(\frac{C_1}{k} r\right) \big|_{r = r_1} $$ Solve for \(C_1\): $$ C_1 = - \dot{q}_{1} r_1 $$ 2. At \(r = r_{2}\), the temperature is given by \(T_{r}(r_{2}) = T_{2}\). Therefore: $$ T_{2} = -\frac{\dot{q}_{1} r_1}{k}r_2 + C_2 $$ Solve for \(C_2\), using the expression for \(C_1\) found above: $$ C_2 = T_2 + \frac{\dot{q}_{1} r_1 r_2}{k} $$

Write down the temperature variation equation with the determined constants

Substitute \(C_1\) and \(C_2\) back into the temperature equation: $$ T(r) = \left(\frac{\dot{q}_{1} r_1^2}{k}\right)\left(\frac{1}{r} - \frac{1}{r_2}\right) + T_2 $$

Find the temperature at the inner surface, r = r1

Finally, substitute \(r = r_1\) into the temperature equation to find the temperature at the inner surface: $$ T(r_1) = \left(\frac{\dot{q}_{1} r_1^2}{k}\right)\left(\frac{1}{r_1} - \frac{1}{r_2}\right) + T_2 $$ Insert the given values for \(r_1, r_2, \dot{q}_1, k,\) and \(T_2\): $$ T(r_1) = \left(\frac{7\,000\,W/m^2 \cdot 1^2\,m^2}{1.5\,W/m\cdot K}\right)\left(\frac{1}{1\,m} - \frac{1}{1.05\,m}\right) + 25^{\circ}C $$ Calculate the temperature: $$ T(r_1) = 138.89^{\circ}C $$ So the temperature at the inner surface of the container is \(138.89^{\circ}C\).

Key concepts

Heat Flux

Understanding heat flux is crucial when studying heat conduction in spherical coordinates. Heat flux, represented by \( \dot{q} \), refers to the rate of heat energy transfer per unit area. In a spherical container, heat flux might be uniform or it can vary depending on the radial position. For instance, in our exercise, the inner surface of the spherical container has a specified heat flux of \( \dot{q}_{1}=7 \text{ kW/m}^{2} \). This value indicates the amount of thermal energy flowing through each square meter of the inner surface every second.

The heat flux informs us about the thermal energy transfer occurring at the surface. It is directly proportional to the thermal gradient but inversely proportional to the surface area through which the heat travels. Therefore, as the radius increases in a spherical geometry, the area increases, and the heat flux spreads over a larger surface area, resulting in a decrease in flux with distance from the center.

Thermal Conductivity

Thermal conductivity, denoted by \( k \), is a fundamental material property that quantifies a material's ability to conduct heat. It typically appears in the heat conduction equation as a constant for a homogeneous material, but it can change with temperature or composition if the material is not uniform.

In the provided exercise, the thermal conductivity of the container wall is a constant \( k=1.5 \text{ W/m}\cdot\text{K} \), which makes it easier to integrate the heat conduction equation in spherical coordinates. The higher the thermal conductivity, the more efficiently a material transfers heat. This constant is crucial in determining how quickly heat will spread through the material and hence significantly affects the temperature distribution within the object.

Boundary Conditions

Boundary conditions are essential for solving physical problems as they provide the information needed to define the behavior of a system at its limits. In thermal problems, they often specify temperatures or heat fluxes at the boundaries. In our exercise, two boundary conditions are given. The first is the heat flux on the inner surface of \( r=r_{1} \) with a value of \( \dot{q}_{1}=7\text{ kW/m}^{2} \), and the second is the temperature on the outer surface of \( r=r_{2} \) which is \( T_{2}=25^{\text{°C}} \).

These boundary conditions allow us to solve for the constants of integration \( C_{1} \) and \( C_{2} \) in the general solution of the heat conduction equation. Hence, boundary conditions are used to tailor the general solution to the specifics of a physical situation and are critical to finding the exact temperature distribution across the material.

Temperature Distribution

Temperature distribution describes how temperature varies within a medium. In the context of our exercise, the temperature distribution in the container wall is spherical and depends on the radial position, \( r \). The derived temperature function, \( T(r) \), reflects the balance of energy inputs and outputs within the medium, considering the thermal conductivity, applied heat flux, and boundary conditions.

For this spherical case, the temperature decreases inversely with the radius from the inner surface to the outer surface due to the nature of spherical heat spread. The mathematical expression for temperature distribution, derived in the exercise, was \( T(r)=\left(\dot{q}_{1} r_{1}^{2} / k\right)\left(1 / r-1 /r_{2}\right)+T_{2} \), showcasing that temperature at a point inside the material depends on the geometry (i.e., the radius), the material properties (i.e., thermal conductivity), and the boundary conditions.