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Chapter 3: Problem 102

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 \(T_{\infty}\).

Short Answer

Expert verified
The steady-state radial temperature distribution in the solid sphere is given by the following expression: \[T(r) = T_{\infty} + \frac{\dot{q}}{2h}r_o-\frac{\dot{q}}{6k}r^2+\frac{\dot{q}}{6k}\left(\frac{3}{2}r_o^2\right)\] where \(T_{\infty}\) is the surrounding liquid temperature, \(r_o\) is the radius of the sphere, \(\dot{q}\) is the volumetric heat generation rate, \(k\) is the thermal conductivity of the solid sphere, and \(h\) is the convection coefficient.

Step by step solution

01

Write down the heat conduction equation with heat generation

Inside the sphere, heat is generated as well as conducted. Therefore, the heat conduction equation is given by: \[\frac{1}{r^2}\frac{d}{dr}\left(r^2\frac{dT}{dr}\right) = -\frac{\dot{q}}{k}\]
02

Solve the partial differential equation (PDE)

To solve this PDE, we need to do the following integration: \[\int\left[\frac{1}{r^2}\frac{d}{dr}\left(r^2\frac{dT}{dr}\right) + \frac{\dot{q}}{k}\right]dr = 0\] First, integrate both sides with respect to r: \[\int \frac{1}{r^2}\frac{d}{dr}\left(r^2\frac{dT}{dr}\right)dr + \int\frac{\dot{q}}{k} dr = \int 0 dr\] Which results in the equation: \[\frac{dT}{dr} + \frac{\dot{q}}{2k}r = C_1\] Now we can integrate again, with respect to r, to find the temperature distribution: \[\int dT + \int \frac{\dot{q}}{2k}r dr = \int C_1 dr\] This yields the equation: \[T(r) - T(0) = -\frac{\dot{q}}{6k}r^2 + C_1r + C_2\] Where T(0) is the constant of integration.
03

Apply boundary conditions

We have two boundary conditions to consider: 1. At the surface of the sphere (r=r_o), heat is convected to the surrounding liquid such that: \[h \cdot \left[T(r_{o}) - T_{\infty}\right] = -k \cdot \left.\frac{dT}{dr}\right|_{r=r_{o}}\] 2. Since it is a solid sphere, the heat transfer is finite at the center (r=0), so: \[\left.\frac{dT}{dr}\right|_{r=0}=0\] By applying these boundary conditions, we can solve for the constants C_1 and C_2.
04

Find constants C_1 and C_2

Using the second boundary condition: \[\frac{dT}{dr}\Big|_{r=0}= C_1 = 0\] Now, using the first boundary condition while substituting C_1=0: \[h \cdot \left[T(r_{o}) - T_{\infty}\right] = -k\cdot \left(-\frac{\dot{q}}{2k}r_{o}\right)\] Which simplifies to: \[T(r_{o}) = T_{\infty} + \frac{\dot{q}}{2h}r_{o}\] Substituting the expression for T(r) and r=r_o: \[-T(0) = -\frac{\dot{q}}{6k}r_o^2 + \frac{\dot{q}}{2}r_o\] Solving for T(0): \[T(0) = \frac{\dot{q}}{6k}\left(3-\frac{1}{2}\right)r_o^2\]
05

Final temperature distribution

Substitute C_1=0 and T(0) back into the temperature distribution expression: \[T(r) = T_{\infty} + \frac{\dot{q}}{2h}r_o-\frac{\dot{q}}{6k}r^2+\frac{\dot{q}}{6k}\left(\frac{3}{2}r_o^2\right)\] This is the steady-state radial temperature distribution in the solid sphere, expressed in terms of r_o, q虈, k, h, and T_鈭�.

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Key Concepts

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

Heat Conduction Equation
The heat conduction equation is a fundamental part of understanding how temperature changes within a solid material due to internal heat generation. In this context, a sphere with internal heat generation requires us to use the differential form of the heat conduction equation:

\[ \frac{1}{r^2}\frac{d}{dr}\left(r^2\frac{dT}{dr}\right) = -\frac{\dot{q}}{k} \]
Here, \( T \) represents the temperature at a radial distance \( r \) from the center. This equation considers the balance between the generated heat within the sphere and the heat being conducted radially outward. The term \( \dot{q} \) denotes the volumetric rate of internal heat generation, while \( k \) symbolizes the thermal conductivity of the sphere material. Understanding this equation is crucial as it lays the groundwork for calculating how temperature varies within the sphere.
Boundary Conditions
Boundary conditions are essential for solving differential equations like the heat conduction equation. They allow us to find specific solutions that fit the physical situation. In this example, two boundary conditions are applied:

  • At the surface of the sphere \((r=r_o)\), the heat is transferred to the surrounding liquid. This process is described by the equation: \[ h \cdot \left[T(r_{o}) - T_{\infty}\right] = -k \cdot \left.\frac{dT}{dr}\right|_{r=r_{o}} \]
  • At the center of the sphere \((r=0)\), the temperature gradient \(\frac{dT}{dr}\) must be zero to maintain finite heat transfer, so we use: \[ \left.\frac{dT}{dr}\right|_{r=0}=0 \]
These conditions ensure that the solution we find respects the physical constraints of heat transfer at both the center and surface of the sphere.
Thermal Conductivity
Thermal conductivity \( k \) is a property of the material that defines how well it conducts heat. In problems involving heat transfer, knowing the thermal conductivity of the material helps determine the rate at which heat is conducted through the material. It appears in the heat conduction equation as it relates to the thermal resistance of the sphere. A higher thermal conductivity implies that the material will conduct heat more efficiently, leading to a temperature distribution that is more spread out over the sphere.

In engineering and physics, accurately determining and applying thermal conductivity values allows us to predict temperature behaviors in materials subject to heat generation and conduction. This knowledge is vital for designing systems where heat management is critical, such as in reactors or electronic devices.
Convection Coefficient
The convection coefficient \( h \) is a measure of the heat transfer between the surface of a solid and a fluid (in this case, a liquid bath). It influences how quickly or slowly heat is lost from the surface of the sphere to the surrounding fluid. In many situations, the convection coefficient depends on the fluid properties, velocity, and surface conditions.

Understanding convection is crucial because it provides the link between the solid sphere and its environment. When solving the heat conduction problem, we must know \( h \) to correctly apply the first boundary condition at the surface. This allows for an accurate solution for the temperature at the outer radius \( r_o \) and influences the overall temperature distribution within the sphere. A higher \( h \) indicates more efficient heat transfer, leading to a lower surface temperature for the same heat generation within the material.

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Most popular questions from this chapter

A wire of diameter \(D=2 \mathrm{~mm}\) and uniform temperature \(T\) has an electrical resistance of \(0.01 \Omega / \mathrm{m}\) and a current flow of \(20 \mathrm{~A}\). (a) What is the rate at which heat is dissipated per unit length of wire? What is the heat dissipation per unit volume within the wire? (b) If the wire is not insulated and is in ambient air and large surroundings for which \(T_{\infty}=T_{\text {sur }}=20^{\circ} \mathrm{C}\), what is the temperature \(T\) of the wire? The wire has an emissivity of \(0.3\), and the coefficient associated with heat transfer by natural convection may be approximated by an expression of the form, \(h=C\left[\left(T-T_{\infty}\right) / D\right]^{1 / 4}, \quad\) where \(C=1.25\) \(\mathrm{W} / \mathrm{m}^{7 / 4} \cdot \mathrm{K}^{5 / 4}\). (c) If the wire is coated with plastic insulation of 2-mm thickness and a thermal conductivity of \(0.25 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), what are the inner and outer surface temperatures of the insulation? The insulation has an emissivity of \(0.9\), and the convection coefficient is given by the expression of part (b). Explore the effect of the insulation thickness on the surface temperatures.

A bonding operation utilizes a laser to provide a constant heat flux, \(q_{o}^{\prime \prime}\), across the top surface of a thin adhesivebacked, plastic film to be affixed to a metal strip as shown in the sketch. The metal strip has a thickness \(d=1.25 \mathrm{~mm}\), and its width is large relative to that of the film. The thermophysical properties of the strip are \(\rho=7850 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=435 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), and \(k=60 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). The thermal resistance of the plastic film of width \(w_{1}=40 \mathrm{~mm}\) is negligible. The upper and lower surfaces of the strip (including the plastic film) experience convection with air at \(25^{\circ} \mathrm{C}\) and a convection coefficient of \(10 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The strip and film are very long in the direction normal to the page. Assume the edges of the metal strip are at the air temperature \(\left(T_{\infty}\right)\). (a) Derive an expression for the temperature distribution in the portion of the steel strip with the plastic film \(\left(-w_{1} / 2 \leq x \leq+w_{1} / 2\right)\). (b) If the heat flux provided by the laser is 10,000 \(\mathrm{W} / \mathrm{m}^{2}\), determine the temperature of the plastic film at the center \((x=0)\) and its edges \(\left(x=\pm w_{1} / 2\right)\). (c) Plot the temperature distribution for the entire strip and point out its special features.

In a test to determine the friction coefficient \(\mu\) associated with a disk brake, one disk and its shaft are rotated at a constant angular velocity \(\omega\), while an equivalent disk/shaft assembly is stationary. Each disk has an outer radius of \(r_{2}=180 \mathrm{~mm}\), a shaft radius of \(r_{1}=20 \mathrm{~mm}\), a thickness of \(t=12 \mathrm{~mm}\), and a thermal conductivity of \(k=15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). A known force \(F\) is applied to the system, and the corresponding torque \(\tau\) required to maintain rotation is measured. The disk contact pressure may be assumed to be uniform (i.e., independent of location on the interface), and the disks may be assumed to be well insulated from the surroundings. (a) Obtain an expression that may be used to evaluate \(\mu\) from known quantities. (b) For the region \(r_{1} \leq r \leq r_{2}\), determine the radial temperature distribution \(T(r)\) in the disk, where \(T\left(r_{1}\right)=T_{1}\) is presumed to be known. (c) Consider test conditions for which \(F=200 \mathrm{~N}\), \(\omega=40 \mathrm{rad} / \mathrm{s}, \tau=8 \mathrm{~N} \cdot \mathrm{m}\), and \(T_{1}=80^{\circ} \mathrm{C}\). Evaluate the friction coefficient and the maximum disk temperature.

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 \(T_{p, \max }=70^{\circ} \mathrm{C}\).

A \(0.20\)-m-diameter, thin-walled steel pipe is used to transport saturated steam at a pressure of 20 bars in a room for which the air temperature is \(25^{\circ} \mathrm{C}\) and the convection heat transfer coefficient at the outer surface of the pipe is \(20 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) What is the heat loss per unit length from the bare pipe (no insulation)? Estimate the heat loss per unit length if a 50 -mm-thick layer of insulation (magnesia, \(85 \%\) is added. The steel and magnesia may each be assumed to have an emissivity of \(0.8\), and the steam-side convection resistance may be neglected. (b) The costs associated with generating the steam and installing the insulation are known to be \(\$ 4 / 10^{9} \mathrm{~J}\) and \(\$ 100 / \mathrm{m}\) of pipe length, respectively. If the steam line is to operate \(7500 \mathrm{~h} / \mathrm{yr}\), how many years are needed to pay back the initial investment in insulation?
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