/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 74 The outer surface of a hollow sp... [FREE SOLUTION] | 91影视

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Chapter 3: Problem 74

The outer surface of a hollow sphere of radius \(r_{2}\) is subjected to a uniform heat flux \(q_{2}^{\prime \prime}\). The inner surface at \(r_{1}\) is held at a constant temperature \(T_{s, 1}\). (a) Develop an expression for the temperature distribution \(T(r)\) in the sphere wall in terms of \(q_{2}^{\prime \prime}, T_{s, 1}, r_{1}, r_{2}\), and the thermal conductivity of the wall material \(k\). (b) If the inner and outer tube radii are \(r_{1}=50 \mathrm{~mm}\) and \(r_{2}=100 \mathrm{~mm}\), what heat flux \(q_{2}^{\prime \prime}\) is required to maintain the outer surface at \(T_{s, 2}=50^{\circ} \mathrm{C}\), while the inner surface is at \(T_{s, 1}=20^{\circ} \mathrm{C}\) ? The thermal conductivity of the wall material is \(k=10 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\).

Short Answer

Expert verified
(a) The temperature distribution T(r) in the sphere wall is given by: \(T(r)=-\frac{C_1}{k}r+C_2\frac{1}{r}+C_3\) (b) The heat flux \(q_{2}^{\prime \prime}\) required to maintain the given conditions is: \(q_{2}^{\prime \prime}=k\left(\frac{C_1}{r_2^2}-\frac{C_2}{r_2^3}\right)\)

Step by step solution

01

Part (a): Develop an expression for the temperature distribution T(r)

Let us derive the governing differential equation for the heat conduction in spherical coordinates. We have: \(\frac{d}{dr}\left(kr^{2}\frac{dT}{dr}\right)=0\) Rearrange and integrate once: \(k r^2 \frac{dT}{dr}=C_{1}\) Where \(C_1\) is the integration constant. Now, divide by \(kr^2\) and integrate again: \(\int \frac{1}{r^{2}} dT=\int \frac{C_{1}}{k} dr\) \(-\frac{1}{r}+C_{2}=\frac{C_{1}}{k}r+C_{3}\) Rearrange the terms, we get \(T(r) = -\frac{C_1}{k}r + C_2\frac{1}{r} + C_3\) To find the constants \(C_1, C_2\), and \(C_3\), we can use the given boundary conditions: 1. When \(r=r_{1}\), \(T=T_{s, 1}\) 2. When \(r=r_{2}\), heat flux \(q_{2}^{\prime \prime}=-k\frac{dT}{dr}|_{r=r_2}\) Apply boundary condition 1: \(T_{s, 1} = -\frac{C_1}{k}r_1 + C_2\frac{1}{r_1} + C_3\) Apply boundary condition 2: \(q_{2}^{\prime \prime}=k\frac{dT}{dr}|_{r=r_2} =q_{2}^{\prime \prime}=k\left(\frac{C_1}{r_2^2}-\frac{C_2}{r_2^3}\right)\) Now solve the system of linear equations for \(C_1\), \(C_2\), and \(C_3\). Finally, substitute the values of the constants in the expression for T(r): \(T(r)=-\frac{C_1}{k}r+C_2\frac{1}{r}+C_3\)
02

Part (b): Calculate the heat flux q2''

Given values: \(r_1 = 50 \mathrm{~mm}= 0.05 \mathrm{~m}\) \(r_2 = 100 \mathrm{~mm}= 0.1 \mathrm{~m}\) \(T_{s, 1} = 20^{\circ} \mathrm{C}\) \(T_{s, 2} = 50^{\circ} \mathrm{C}\) \(k = 10 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) First, we need to calculate the values of the constants \(C_1, C_2\), and \(C_3\) using the given values and the equations we derived in part (a). Then, use the boundary condition 2: \(q_{2}^{\prime \prime}=k\frac{dT}{dr}|_{r=r_2} =q_{2}^{\prime \prime}=k\left(\frac{C_1}{r_2^2}-\frac{C_2}{r_2^3}\right)\) Plug the values of \(C_1, C_2, k, r_2\) into the above equation, and solve for \(q_{2}^{\prime \prime}\).

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Key Concepts

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

Temperature Distribution
When dealing with heat conduction in spheres, understanding the temperature distribution is crucial. In simple terms, this is the way temperature varies within the sphere from one surface to another. In the context of our problem, we wish to find an expression for the temperature as a function of the radius, or how temperature changes as we move between the inner and outer surfaces.

To do this, we start with a differential equation that accounts for the spherical nature of the object:
  • The governing equation is derived from Fourier鈥檚 law and involves spherical coordinates due to the shape.
  • The equation \( \frac{d}{dr}\left(kr^{2}\frac{dT}{dr}\right)=0 \) represents the balance of energy for the conduction process.
Integrating this differential equation gives us a more usable form that helps describe how temperature varies with radius: \( T(r) = -\frac{C_1}{k}r + C_2\frac{1}{r} + C_3 \).

By applying certain conditions at known boundaries (known as boundary conditions), we can determine the constants in this equation, ultimately leading us to an expression for temperature in terms of distance through the material.
Thermal Conductivity
Thermal conductivity, denoted as \(k\), is a material property important for analyzing how heat transfers through a sphere. It represents the ability of material to conduct heat. A higher thermal conductivity means that the material is more efficient at heat transfer, while a lower thermal conductivity indicates a less efficient conduction and more temperature variation across the material.

In our problem:
  • The thermal conductivity of the wall material is given as \(10 \, \mathrm{W/m \cdot K}\).
  • This value is used to determine how quickly and evenly the heat spreads through the walls of the sphere.
Thermal conductivity is crucial, as it appears in the derived expression for temperature distribution. Changes in \(k\) affect how energy moves between the inner and outer surfaces. It directly influences how much heat is required to maintain a certain temperature difference across the sphere, which is essential when calculating heat flux. Understanding thermal conductivity helps us predict and control temperature behaviors in engineering materials.
Boundary Conditions
Boundary conditions are the known values at the boundaries of the material which allow us to solve for unknowns in physical and mathematical representations. In problems dealing with heat conduction, these are essential for finding the constants in our temperature distribution equation.

For the hollow sphere:
  • At the inner surface \(r = r_1\), the temperature is constant and given as \(T_{s, 1}\).
  • At the outer surface \(r = r_2\), a specified heat flux \(q_{2}''\) prevails, which helps maintain the outer temperature \(T_{s, 2}\).
These conditions permit us to link the theoretical expression with physical reality. By substituting these into the temperature distribution formula, we solve for the constants \(C_1, C_2,\) and \(C_3\). This process helps us to model the sphere鈥檚 temperature gradient more accurately, and calculate the necessary heat flux. Recognizing and applying the correct boundary conditions is fundamental in conduction problems, ensuring relevance and precision in results.

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Most popular questions from this chapter

To maximize production and minimize pumping costs, crude oil is heated to reduce its viscosity during transportation from a production field. (a) Consider a pipe-in-pipe configuration consisting of concentric steel tubes with an intervening insulating material. The inner tube is used to transport warm crude oil through cold ocean water. The inner steel pipe \(\left(k_{s}=35 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\right)\) has an inside diameter of \(D_{i, 1}=150 \mathrm{~mm}\) and wall thickness \(t_{i}=10 \mathrm{~mm}\) while the outer steel pipe has an inside diameter of \(D_{i, 2}=250 \mathrm{~mm}\) and wall thickness \(t_{o}=t_{i}\). Determine the maximum allowable crude oil temperature to ensure the polyurethane foam insulation \(\left(k_{p}=\right.\) \(0.075 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) between the two pipes does not exceed its maximum service temperature of \(T_{p, \max }=\) \(70^{\circ} \mathrm{C}\). The ocean water is at \(T_{\infty, o}=-5^{\circ} \mathrm{C}\) and provides an external convection heat transfer coefficient of \(h_{o}=500 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The convection coefficient associated with the flowing crude oil is \(h_{i}=450 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (b) It is proposed to enhance the performance of the pipe-in-pipe device by replacing a thin \(\left(t_{a}=5 \mathrm{~mm}\right)\) section of polyurethane located at the outside of the inner pipe with an aerogel insulation material \(\left(k_{a}=0.012 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\right)\). Determine the maximum allowable crude oil temperature to ensure maximum polyurethane temperatures are below \(T_{p, \max }=70^{\circ} \mathrm{C}\).

Annular aluminum fins of rectangular profile are attached to a circular tube having an outside diameter of \(50 \mathrm{~mm}\) and an outer surface temperature of \(200^{\circ} \mathrm{C}\). The fins are \(4 \mathrm{~mm}\) thick and \(15 \mathrm{~mm}\) long. The system is in ambient air at a temperature of \(20^{\circ} \mathrm{C}\), and the surface convection coefficient is \(40 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) What are the fin efficiency and effectiveness? (b) If there are 125 such fins per meter of tube length, what is the rate of heat transfer per unit length of tube?

A nuclear reactor fuel element consists of a solid cylindrical pin of radius \(r_{1}\) and thermal conductivity \(k_{f}\). The fuel pin is in good contact with a cladding material of outer radius \(r_{2}\) and thermal conductivity \(k_{c^{*}}\). Consider steady-state conditions for which uniform heat generation occurs within the fuel at a volumetric rate \(\dot{q}\) and the outer surface of the cladding is exposed to a coolant that is characterized by a temperature \(T_{\infty}\) and a convection coefficient \(h\). (a) Obtain equations for the temperature distributions \(T_{f}(r)\) and \(T_{c}(r)\) in the fuel and cladding, respectively. Express your results exclusively in terms of the foregoing variables. (b) Consider a uranium oxide fuel pin for which \(k_{f}=2\) \(\mathrm{W} / \mathrm{m} \cdot \mathrm{K}\) and \(r_{1}=6 \mathrm{~mm}\) and cladding for which \(k_{c}=25 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and \(r_{2}=9 \mathrm{~mm}\). If \(\dot{q}=2 \times 10^{8}\) \(\mathrm{W} / \mathrm{m}^{3}, h=2000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), and \(T_{\infty}=300 \mathrm{~K}\), what is the maximum temperature in the fuel element? (c) Compute and plot the temperature distribution, \(T(r)\), for values of \(h=2000,5000\), and 10,000 \(\mathrm{W} / \mathrm{m}^{2} \cdot \mathrm{K}\). If the operator wishes to maintain the centerline temperature of the fuel element below \(1000 \mathrm{~K}\), can she do so by adjusting the coolant flow and hence the value of \(h\) ?

A spherical, cryosurgical probe may be imbedded in diseased tissue for the purpose of freezing, and thereby destroying, the tissue. Consider a probe of \(3-\mathrm{mm}\) diameter whose surface is maintained at \(-30^{\circ} \mathrm{C}\) when imbedded in tissue that is at \(37^{\circ} \mathrm{C}\). A spherical layer of frozen tissue forms around the probe, with a temperature of \(0^{\circ} \mathrm{C}\) existing at the phase front (interface) between the frozen and normal tissue. If the thermal conductivity of frozen tissue is approximately \(1.5 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and heat transfer at the phase front may be characterized by an effective convection coefficient of \(50 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), what is the thickness of the layer of frozen tissue (assuming negligible perfusion)?

A radioactive material of thermal conductivity \(k\) is cast as a solid sphere of radius \(r_{o}\) and placed in a liquid bath for which the temperature \(T_{\infty}\) and convection coefficient \(h\) are known. Heat is uniformly generated within the solid at a volumetric rate of \(\dot{q}\). Obtain the steadystate radial temperature distribution in the solid, expressing your result in terms of \(r_{o}, \dot{q}, k, h\), and \(T_{\infty}\).
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