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Chapter 2: Problem 73

A spherical shell, with thermal conductivity \(k\), has inner and outer radii of \(r_{1}\) and \(r_{2}\), respectively. The inner surface of the shell is subjected to a uniform heat flux of \(\dot{q}_{1}\), while the outer surface of the shell is exposed to convection heat transfer with a coefficient \(h\) and an ambient temperature \(T_{c \infty}\). Determine the variation of temperature in the shell wall and show that the outer surface temperature of the shell can be expressed as \(T\left(r_{2}\right)=\left(\dot{q}_{1} / h\right)\left(r_{1} / r_{2}\right)^{2}+T_{\infty \text { co }}\).

Short Answer

Expert verified
Question: Determine the outer surface temperature of a spherical shell with given heat flux at the inner radius, thermal conductivity, convection heat transfer coefficient, and temperature far from the outer surface. Answer: The outer surface temperature of the spherical shell can be found using the following expression: \(T\left(r_{2}\right)=\left(\frac{\dot{q}_{1}}{h}\right)\left(\frac{r_{1}}{r_{2}}\right)^{2}+T_{c \infty}\), where \(\dot{q}_{1}\) is the heat flux at the inner radius, \(h\) is the convection heat transfer coefficient, \(r_{1}\) and \(r_{2}\) are the inner and outer radii of the spherical shell, and \(T_{c\infty}\) is the temperature far from the outer surface.

Step by step solution

01

Governing Equation

The general form of the Fourier's law expresses the heat conduction in the radial direction of a spherical system is: \(\frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2 k \frac{\partial T(r)}{\partial r}\right) = 0\) Let's assume that the thermal conductivity "k" is constant. Therefore, the governing equation will simplify to: \(\frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2 \frac{\partial T(r)}{\partial r}\right) = 0\)
02

Solve the Governing Equation

To find the temperature profile, we will integrate the simplified governing equation twice with respect to "r". Doing so, we get: First Integration: \(\int_{T(r)}\frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2 \frac{\partial T(r)}{\partial r}\right) dr = \int_{T(r)} 0 dr\) \(\Rightarrow r^2\frac{\partial T(r)}{\partial r} = C_{1}\) Second Integration: \(\int_{T(r)}\frac{\partial T(r)}{\partial r} dr = \int_{T(r)} \frac{C_{1}}{r^2} dr \) \(\Rightarrow T(r) = C_{1} \int\frac{1}{r^2} dr + C_{2}\) Evaluating the integrals, we get: \(T(r) = -\frac{C_{1}}{r} + C_{2}\)
03

Apply Boundaries Conditions

We will use the given boundary conditions to find the constants \(C_{1}\) and \(C_{2}\). First boundary condition, at inner radius \(r_1\), is given by the heat flux \(\dot{q}_1\) (We can relate heat flux to the temperature gradient with \(\dot{q}_1 = -k\frac{dT(r_1)}{dr}\)) \(\dot{q}_1 = -k\cdot (-\frac{C_1}{r_1^2}) \Rightarrow C_1 = -\frac{\dot{q}_1 r_1^2}{k}\) Second boundary condition, at outer radius \(r_2\), is given by the convection heat transfer: \(\dot{q}_{\text{conv}} = h(T(r_2) - T_{c\infty}) = -k\cdot\left(-\frac{C_1}{r_2^2}\right)\) \(\Rightarrow T(r_2) = T_{c\infty} + \frac{k}{h}\cdot\frac{\dot{q}_1 r_1^2}{r_2^2 k}\) \(\Rightarrow T(r_2) = T_{c\infty} + \frac{\dot{q}_1 r_1^2}{h r_2^2}\)
04

Final Temperature Expression

Combining the values of \(C_1\) and \(C_2\) obtained from boundary conditions and the expression of temperature profile \(T(r)\), we finally get the variation of temperature in the shell wall: \(T(r) = -\frac{\left(-\frac{\dot{q}_1 r_1^2}{k}\right)}{r} + T_{c \infty}\) The outer surface temperature of the shell is: \(T\left(r_{2}\right)=\left(\frac{\dot{q}_{1}}{h}\right)\left(\frac{r_{1}}{r_{2}}\right)^{2}+T_{c \infty}\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Fourier's Law
Fourier's Law of thermal conduction is a fundamental principle that describes how heat energy transfers through a material due to temperature gradients. In the context of a spherical shell, this law can be expressed mathematically as:
\[\begin{equation}\frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2 k \frac{\partial T(r)}{\partial r}\right) = 0\end{equation}\]
where \(r\) is the radial position within the sphere, \(k\) is the thermal conductivity of the material, and \(T(r)\) is the temperature at any radius \(r\). Fourier's law in spherical coordinates takes into account that the area through which heat is conducted increases with the square of the radius, hence the term \(r^2\) in the equation.
Conduction in Spherical Systems
Heat conduction in spherical systems, such as our spherical shell, differs from planar conduction due to geometry. When dealing with spheres, the heat transfer is radially outwards or inwards, and the amount of material through which heat must conduct gets larger as we move away from the center. This geometry is accounted for in the governing heat conduction equation, which, for a sphere with constant thermal conductivity, simplifies to:
\[\begin{equation}\frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2 \frac{\partial T(r)}{\partial r}\right) = 0\end{equation}\]
Integrating this equation gives us a general form of the temperature profile within the spherical shell. Understanding how to solve this differential equation is essential for predicting temperature variation in such systems, which is especially crucial in applications involving thermal insulation or energy storage.
Boundary Conditions in Heat Transfer
Boundary conditions play a significant role in solving heat transfer problems as they define the behavior of a system at its limits. For our spherical shell, we have two key boundary conditions:
  • The inner surface has a uniform heat flux, \(\dot{q}_{1}\).
  • The outer surface experiences convection with the ambient, characterized by a convection coefficient \(h\) and an ambient temperature \(T_{c \infty}\).
These conditions allow us to determine the unknown constants from the general solution of the temperature profile. They provide the necessary information to predict how the spherical shell will interact with its environment. Specifically, the first condition helps us find the value of constant \(C_{1}\), while the second condition is used to determine \(C_{2}\) and thereby complete our temperature expression.
Temperature Profile in Heat Transfer
The temperature profile in a material describes how temperature varies with location. In the spherical shell scenario, this profile is determined by integrating the heat conduction equation, considering the respective boundary conditions for the system. Once we find the constants \(C_{1}\) and \(C_{2}\), we can write the temperature profile as:
\[\begin{equation}T(r) = -\frac{C_{1}}{r} + C_{2}\end{equation}\]
In the final step of our solution, the profile shows how temperature changes from the inner radius to the outer radius of the spherical shell. By applying the boundary conditions properly, we can accurately predict and express the outer surface temperature of the shell, which is crucial for designing thermal systems and understanding how they will perform in real-world scenarios.

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Most popular questions from this chapter

Heat is generated in a long wire of radius \(r_{o}\) at a constant rate of \(\dot{e}_{\text {gen }}\) per unit volume. The wire is covered with a plastic insulation layer. Express the heat flux boundary condition at the interface in terms of the heat generated.

Consider a large plane wall of thickness \(L=0.8 \mathrm{ft}\) and thermal conductivity \(k=1.2 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft} \cdot{ }^{\circ} \mathrm{F}\). The wall is covered with a material that has an emissivity of \(\varepsilon=0.80\) and a solar absorptivity of \(\alpha=0.60\). The inner surface of the wall is maintained at \(T_{1}=520 \mathrm{R}\) at all times, while the outer surface is exposed to solar radiation that is incident at a rate of \(\dot{q}_{\text {solar }}=300 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{2}\). The outer surface is also losing heat by radiation to deep space at \(0 \mathrm{~K}\). Determine the temperature of the outer surface of the wall and the rate of heat transfer through the wall when steady operating conditions are reached.

A spherical metal ball of radius \(r_{o}\) is heated in an oven to a temperature of \(T_{i}\) throughout and is then taken out of the oven and dropped into a large body of water at \(T_{\infty}\) where it is cooled by convection with an average convection heat transfer coefficient of \(h\). Assuming constant thermal conductivity and transient one-dimensional heat transfer, express the mathematical formulation (the differential equation and the boundary and initial conditions) of this heat conduction problem. Do not solve.

A spherical container, with an inner radius of \(1 \mathrm{~m}\) and a wall thickness of \(5 \mathrm{~mm}\), has its inner surface subjected to a uniform heat flux of \(7 \mathrm{~kW} / \mathrm{m}^{2}\). The outer surface of the container is maintained at \(20^{\circ} \mathrm{C}\). The container wall is made of a material with a thermal conductivity given as \(k(T)=k_{0}(1+\beta T)\), where \(k_{0}=1.33 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \beta=0.0023 \mathrm{~K}^{-1}\), and \(T\) is in \(\mathrm{K}\). Determine the temperature drop across the container wall thickness.

Consider steady one-dimensional heat conduction through a plane wall, a cylindrical shell, and a spherical shell of uniform thickness with constant thermophysical properties and no thermal energy generation. The geometry in which the variation of temperature in the direction of heat transfer will be linear is (a) plane wall (b) cylindrical shell (c) spherical shell (d) all of them (e) none of them
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