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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

Consider a large 3 -cm-thick stainless steel plate \((k=\) \(15.1 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) in which heat is generated uniformly at a rate of \(5 \times 10^{5} \mathrm{~W} / \mathrm{m}^{3}\). Both sides of the plate are exposed to an environment at \(30^{\circ} \mathrm{C}\) with a heat transfer coefficient of \(60 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Explain where in the plate the highest and the lowest temperatures will occur, and determine their values.

The variation of temperature in a plane wall is determined to be \(T(x)=52 x+25\) where \(x\) is in \(\mathrm{m}\) and \(T\) is in \({ }^{\circ} \mathrm{C}\). If the temperature at one surface is \(38^{\circ} \mathrm{C}\), the thickness of the wall is (a) \(0.10 \mathrm{~m}\) (b) \(0.20 \mathrm{~m}\) (c) \(0.25 \mathrm{~m}\) (d) \(0.40 \mathrm{~m}\) (e) \(0.50 \mathrm{~m}\)

The outer surface of an engine is situated in a place where oil leakage can occur. Some oils have autoignition temperatures of approximately above \(250^{\circ} \mathrm{C}\). When oil comes in contact with a hot engine surface that has a higher temperature than its autoignition temperature, the oil can ignite spontaneously. Treating the engine housing as a plane wall, the inner surface \((x=0)\) is subjected to \(6 \mathrm{~kW} / \mathrm{m}^{2}\) of heat. The engine housing \((k=13.5 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) has a thickness of \(1 \mathrm{~cm}\), and the outer surface \((x=L)\) is exposed to an environment where the ambient air is \(35^{\circ} \mathrm{C}\) with a convection heat transfer coefficient of \(20 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). To prevent fire hazard in the event the leaked oil comes in contact with the hot engine surface, the temperature of the engine surface should be kept below \(200^{\circ} \mathrm{C}\). Determine the variation of temperature in the engine housing and the temperatures of the inner and outer surfaces. Is the outer surface temperature of the engine below the safe temperature?

A 2-kW resistance heater wire whose thermal conductivity is \(k=10.4 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft} \cdot \mathrm{R}\) has a radius of \(r_{o}=0.06\) in and a length of \(L=15\) in, and is used for space heating. Assuming constant thermal conductivity and one-dimensional heat transfer, express the mathematical formulation (the differential equation and the boundary conditions) of this heat conduction problem during steady operation. Do not solve.

A large plane wall has a thickness \(L=50 \mathrm{~cm}\) and thermal conductivity \(k=25 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). On the left surface \((x=0)\), it is subjected to a uniform heat flux \(\dot{q}_{0}\) while the surface temperature \(T_{0}\) is constant. On the right surface, it experiences convection and radiation heat transfer while the surface temperature is \(T_{L}=225^{\circ} \mathrm{C}\) and the surrounding temperature is \(25^{\circ} \mathrm{C}\). The emissivity and the convection heat transfer coefficient on the right surface are \(0.7\) and \(15 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), respectively. Show that the variation of temperature in the wall can be expressed as \(T(x)=\left(\dot{q}_{0} / k\right)(L-x)+T_{L}\), where \(\dot{q}_{0}=5130 \mathrm{~W} / \mathrm{m}^{2}\), and determine the temperature of the left surface of the wall at \(x=0\).

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