/*! 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

A steam pipe of \(0.12-\mathrm{m}\) outside diameter is insulated with a layer of calcium silicate. (a) If the insulation is \(20 \mathrm{~mm}\) thick and its inner and outer surfaces are maintained at \(T_{s, 1}=800 \mathrm{~K}\) and \(T_{s, 2}=490 \mathrm{~K}\), respectively, what is the heat loss per unit length \(\left(q^{\prime}\right)\) of the pipe? (b) We wish to explore the effect of insulation thickness on the heat loss \(q^{\prime}\) and outer surface temperature \(T_{s, 2}\), with the inner surface temperature fixed at \(T_{s, 1}=\) \(800 \mathrm{~K}\). The outer surface is exposed to an airflow \(\left(T_{\infty}=25^{\circ} \mathrm{C}\right)\) that maintains a convection coefficient of \(h=25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and to large surroundings for which \(T_{\text {sur }}=T_{\infty}=25^{\circ} \mathrm{C}\). The surface emissivity of calcium silicate is approximately \(0.8\). Compute and plot the temperature distribution in the insulation as a function of the dimensionless radial coordinate, \(\left(r-r_{1}\right) /\left(r_{2}-r_{1}\right)\), where \(r_{1}=0.06 \mathrm{~m}\) and \(r_{2}\) is a variable \(\left(0.06

Copper tubing is joined to a solar collector plate of thickness \(t\), and the working fluid maintains the temperature of the plate above the tubes at \(T_{o}\). There is a uniform net radiation heat flux \(q_{\text {rad }}^{\prime \prime}\) to the top surface of the plate, while the bottom surface is well insulated. The top surface is also exposed to a fluid at \(T_{\infty}\) that provides for a uniform convection coefficient \(h\). (a) Derive the differential equation that governs the temperature distribution \(T(x)\) in the plate. (b) Obtain a solution to the differential equation for appropriate boundary conditions.

A device used to measure the surface temperature of an object to within a spatial resolution of approximately \(50 \mathrm{~nm}\) is shown in the schematic. It consists of an extremely sharp-tipped stylus and an extremely small cantilever that is scanned across the surface. The probe tip is of circular cross section and is fabricated of polycrystalline silicon dioxide. The ambient temperature is measured at the pivoted end of the cantilever as \(T_{\infty}=\) \(25^{\circ} \mathrm{C}\), and the device is equipped with a sensor to measure the temperature at the upper end of the sharp tip, \(T_{\text {sen. }}\). The thermal resistance between the sensing probe and the pivoted end is \(R_{t}=5 \times 10^{6} \mathrm{~K} / \mathrm{W}\). (a) Determine the thermal resistance between the surface temperature and the sensing temperature. (b) If the sensing temperature is \(T_{\text {sen }}=28.5^{\circ} \mathrm{C}\), determine the surface temperature. Hint: Although nanoscale heat transfer effects may be important, assume that the conduction occurring in the air adjacent to the probe tip can be described by Fourier's law and the thermal conductivity found in Table A. \(4 .\)

A composite wall separates combustion gases at \(2600^{\circ} \mathrm{C}\) from a liquid coolant at \(100^{\circ} \mathrm{C}\), with gas- and liquid-side convection coefficients of 50 and 1000 \(\mathrm{W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The wall is composed of a \(10-\mathrm{mm}\)-thick layer of beryllium oxide on the gas side and a 20 -mm-thick slab of stainless steel (AISI 304) on the liquid side. The contact resistance between the oxide and the steel is \(0.05 \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\). What is the heat loss per unit surface area of the composite? Sketch the temperature distribution from the gas to the liquid.

A spherical tank of \(3-\mathrm{m}\) diameter contains a liquifiedpetroleum gas at \(-60^{\circ} \mathrm{C}\). Insulation with a thermal conductivity of \(0.06 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and thickness \(250 \mathrm{~mm}\) is applied to the tank to reduce the heat gain. (a) Determine the radial position in the insulation layer at which the temperature is \(0^{\circ} \mathrm{C}\) when the ambient air temperature is \(20^{\circ} \mathrm{C}\) and the convection coefficient on the outer surface is \(6 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (b) If the insulation is pervious to moisture from the atmospheric air, what conclusions can you reach about the formation of ice in the insulation? What effect will ice formation have on heat gain to the LP gas? How could this situation be avoided?
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