Consider a spherical shell of inner radius \(r_{1}\) and outer radius \(r_{2}\) whose thermal conductivity varies linearly in a specified temperature range as \(k(T)=k_{0}(1+\beta T)\) where \(k_{0}\) and \(\beta\) are two specified constants. The inner surface of the shell is maintained at a constant temperature of \(T_{1}\) while the outer surface is maintained at \(T_{2}\). Assuming steady one- dimensional heat transfer, obtain a relation for \((a)\) the heat transfer rate through the shell and ( \(b\) ) the temperature distribution \(T(r)\) in the shell.

Short Answer

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Question: Find the heat transfer rate through a spherical shell and the temperature distribution within the shell, given a temperature-dependent thermal conductivity and constant temperatures at the inner and outer surfaces. Answer: The heat transfer rate through the spherical shell is given by: \(\frac{dQ}{dt}=-k_0(4\pi) \left[\frac{r_2^3 - r_1^3}{3}\right] \left[T_2 - T_1 + \frac{1}{2}\beta(T_2^2 - T_1^2)\right]\) The temperature distribution within the shell, T(r), is obtained by integrating and solving the equation: \(\int_{T_1}^T \frac{1}{1+\beta T'} \, dT' = -k_0(4\pi) \int_{r_1}^r \frac{1}{r'^2}\, dr'\)

Step by step solution

01

Set up the equation of heat conduction in a spherical coordinate system

We have steady-state, one-dimensional heat transfer in a radial direction. Fourier's law of heat conduction is given by: \(\frac{dQ}{dt}=-kA\dfrac{dT}{dx}\) For a spherical coordinate system, the heat transfer rate is given by: \(\frac{dQ}{dt}=-k(4\pi r^2) \dfrac{dT}{dr}\) where \((-k)\) is the temperature-dependent thermal conductivity, \((4\pi r^2)\) is the surface area at radius \(r\), and \((\dfrac{dT}{dr})\) is the temperature gradient.
02

Substitute the given thermal conductivity and integrate

Substitute the given temperature-dependent thermal conductivity in the equation: \(\frac{dQ}{dt}=-k_0(1+\beta T)(4\pi r^2) \dfrac{dT}{dr}\) Integrate the above equation with respect to \(T\) and \(r\), keeping in mind that \(r\) changes from \(r_1\) to \(r_2\) and \(T\) changes from \(T_1\) to \(T_2\). The heat transfer rate, \(\frac{dQ}{dt}\), is constant, so we can move it outside the integral. We get: \(\frac{dQ}{dt}=-k_0(4\pi)\int_{r_1}^{r_2} r^2 dr \int_{T_1}^{T_2} (1+\beta T) dT\) After integrating and simplifying, we get: \(\frac{dQ}{dt}=-k_0(4\pi) \left[\frac{r_2^3 - r_1^3}{3}\right] \left[T_2 - T_1 + \frac{1}{2}\beta(T_2^2 - T_1^2)\right]\)
03

Derive the temperature distribution within the shell

Rearrange the equation derived in Step 2 to find the temperature distribution \(T(r)\): \(\frac{dQ}{dt}= -k_0(4\pi r^2) (1+\beta T) \dfrac{dT}{dr}\) Rearrange this equation to separate \(T\) and \(r\) terms: \(\frac{1}{1+\beta T} \, dT = -\frac{k_0(4\pi)}{r^2}\, dr\) Integrate both sides with respect to \(T\) and \(r\) to obtain the temperature distribution within the shell: \(\int_{T_1}^T \frac{1}{1+\beta T'} \, dT' = -k_0(4\pi) \int_{r_1}^r \frac{1}{r'^2}\, dr'\) After integrating and solving, for T(r), we obtain the temperature distribution within the shell. In summary, we have derived the relation for the heat transfer rate \((\frac{dQ}{dt})\) and the temperature distribution \(T(r)\) within the spherical shell.

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Key Concepts

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

Fourier's Law of Heat Conduction
The cornerstone of understanding heat transfer through materials is Fourier's law of heat conduction. It describes the rate at which heat energy is transmitted through a medium as a result of a temperature gradient. The law is concisely expressed by the formula:
\[ \frac{dQ}{dt} = -kA \frac{dT}{dx} \]
where \( \frac{dQ}{dt} \) is the heat transfer rate (the amount of heat transferred per unit time), \( k \) is the thermal conductivity of the material, \( A \) is the cross-sectional area through which heat is being conducted, and \( \frac{dT}{dx} \) represents the temperature gradient in the direction of heat flow. It's important to note that the minus sign indicates heat flows from high to low temperature regions.
In the context of a spherical shell with temperature-dependent conductivity, Fourier's law helps us determine the heat transfer rate through the material from the inner surface, at a higher temperature, to the outer surface, at a lower temperature. The concept is crucial for predicting and controlling thermal behavior in various engineering applications.
Spherical Coordinate System
A spherical coordinate system is an ideal choice when dealing with problems that have radial symmetry, such as the heat transfer through spherical shells. Compared to Cartesian coordinates, spherical coordinates simplify the mathematics involved in spherical geometries. In this system, any point in three-dimensional space is described by three values: the radial distance from a central point (\( r \)), the polar angle (\( \theta \)), and the azimuthal angle (\( \phi \)).
For our heat transfer problem, we're primarily interested in the radial distance \( r \), as we're assuming steady-state, one-dimensional heat flow radially outward or inward. By applying this to Fourier's law, we get the heat transfer rate as a function of the radius:
\[ \frac{dQ}{dt}= -k(4\pi r^2) \frac{dT}{dr} \]
The surface area of a sphere at any radius \( r \) is \( 4\pi r^2 \), and as we move through the material's thickness, different surface areas impact the heat flow, which is incorporated into the calculations.
Temperature-Dependent Thermal Conductivity
In many real-world scenarios, a material's ability to conduct heat, known as thermal conductivity (\( k \)), is not constant but varies with temperature. This dependency can be represented by an equation or a relationship. For the spherical shell problem, we have a linear relationship:
\[ k(T) = k_{0}(1 + \beta T) \]
where \( k_{0} \) is the thermal conductivity at a reference temperature, and \( \beta \) is a material-specific constant that describes how thermal conductivity changes with temperature (\( T \)).
When dealing with temperature-dependent conductivity, it becomes necessary to adjust Fourier's law accordingly. This results in a more complex differential equation that governs the heat transfer, requiring integration while taking into account the continuous change in \( k \) as a function of temperature.
Steady-State Heat Transfer
Steady-state heat transfer implies that the temperature distribution in the material does not change with time, despite the continuous flow of heat. Essentially, the system has reached an equilibrium where the amount of heat entering a given volume is equal to the amount of heat leaving it.
In our spherical shell problem, we assume steady-state conditions for simplicity, which allows us to perform an integration over the shell's thickness without worrying about temporal changes. As a result, we can find a relation for the heat transfer rate through the shell as well as the temperature distribution across it.
Under steady-state conditions, the derivation leads us to a constant heat transfer rate across the entire shell. This allows us to understand and calculate the thermal behavior within the shell and predict the temperature at any given radius, which is critical for design and safety assessments in engineering applications.

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Most popular questions from this chapter

A large steel plate having a thickness of \(L=4\) in, thermal conductivity of \(k=7.2 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft} \cdot{ }^{\circ} \mathrm{F}\), and an emissivity of \(\varepsilon=0.7\) is lying on the ground. The exposed surface of the plate at \(x=L\) is known to exchange heat by convection with the ambient air at \(T_{\infty}=90^{\circ} \mathrm{F}\) with an average heat transfer coefficient of \(h=12 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{2} \cdot{ }^{\circ} \mathrm{F}\) as well as by radiation with the open sky with an equivalent sky temperature of \(T_{\text {sky }}=480 \mathrm{R}\). Also, the temperature of the upper surface of the plate is measured to be \(80^{\circ} \mathrm{F}\). Assuming steady onedimensional heat transfer, \((a)\) express the differential equation and the boundary conditions for heat conduction through the plate, \((b)\) obtain a relation for the variation of temperature in the plate by solving the differential equation, and \((c)\) determine the value of the lower surface temperature of the plate at \(x=0\).

Does heat generation in a solid violate the first law of thermodynamics, which states that energy cannot be created or destroyed? Explain.

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 a large plane wall of thickness \(I_{\text {, }}\) thermal conductivity \(k\), and surface area \(A\). The left surface of the wall is exposed to the ambient air at \(T_{\infty}\) with a heat transfer coefficient of \(h\) while the right surface is insulated. The variation of temperature in the wall for steady one-dimensional heat conduction with no heat generation is (a) \(T(x)=\frac{h(L-x)}{k} T_{\infty}\) (b) \(T(x)=\frac{k}{h(x+0.5 L)} T_{\infty}\) (c) \(T(x)=\left(1-\frac{x h}{k}\right) T_{\infty}\) (d) \(T(x)=(L-x) T_{\infty}\) (e) \(T(x)=T_{\infty}\)

A cylindrical nuclear fuel rod of \(1 \mathrm{~cm}\) in diameter is encased in a concentric tube of \(2 \mathrm{~cm}\) in diameter, where cooling water flows through the annular region between the fuel rod \((k=30 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) and the concentric tube. Heat is generated uniformly in the rod at a rate of \(50 \mathrm{MW} / \mathrm{m}^{3}\). The convection heat transfer coefficient for the concentric tube surface is \(2000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). If the surface temperature of the concentric tube is \(40^{\circ} \mathrm{C}\), determine the average temperature of the cooling water. Can one use the given information to determine the surface temperature of the fuel rod? Explain.

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