/*! 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 96 A cylindrical shell of inner and... [FREE SOLUTION] | 91Ó°ÊÓ

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A cylindrical shell of inner and outer radii, \(r_{i}\) and \(r_{o}\), respectively, is filled with a heat-generating material that provides a uniform volumetric generation rate \(\left(\mathrm{W} / \mathrm{m}^{3}\right)\) of \(\dot{q}\). The inner surface is insulated, while the outer surface of the shell is exposed to a fluid at \(T_{\infty}\) and a convection coefficient \(h\). (a) Obtain an expression for the steady-state temperature distribution \(T(r)\) in the shell, expressing your result in terms of \(r_{i}, r_{o}, \dot{q}, h, T_{\infty}\), and the thermal conductivity \(k\) of the shell material. (b) Determine an expression for the heat rate, \(q^{\prime}\left(r_{o}\right)\), at the outer radius of the shell in terms of \(\dot{q}\) and shell dimensions.

Short Answer

Expert verified
The steady-state temperature distribution in the shell is given by: \[ T(r) = - \frac{\dot{q}}{6k} r^3 + \frac{\dot{q}}{6k} r_i^2 r + T_\infty - \frac{\dot{q}}{6k}(r_i^2-r_o^2)\] The heat rate at the outer radius is given by: \[ q'(r_o) = \pi\dot{q}r_o\left(\frac{1}{3} r_i^2 - \frac{1}{2} r_o^2\right) \]

Step by step solution

01

Write down the heat equation for a cylindrical coordinate system with heat generation

The equation governing the temperature distribution in a cylindrical coordinate system, in radial direction only and considering the heat generation, is given by: \[\frac{1}{r}\frac{d}{dr}\left(r\frac{dT}{dr}\right) + \frac{\dot{q}}{k} = 0\]
02

Solve the heat equation for T(r)

Let's integrate the equation twice to find the temperature distribution T(r) within the shell. The first integration gives: \[ \int \left( r \frac{dT}{dr} \right) dr = -\int \frac{\dot{q}}{k} r dr + C_1\] \[ r \frac{dT}{dr} = - \frac{\dot{q}}{2k} r^2 + C_1\] Now, we'll integrate it again to get T(r): \[ \int \frac{dT}{dr} dr = - \int \frac{\dot{q}}{2k} r dr + \int C_1 dr\] \[ T(r) = - \frac{\dot{q}}{6k} r^3 + C_1 r + C_2\]
03

Apply the boundary conditions

We have two boundary conditions for this problem: 1. The inner surface is insulated, which means there is no radial heat flux, and so \( \frac{dT}{dr}(r_i) = 0 \). 2. The outer surface of the shell is exposed to a fluid at temperature \(T_\infty\) with a convection coefficient h. Applying the first boundary condition gives us the value of \(C_1\): \[ 0 = - \frac{\dot{q}}{6k} r_i^3 + C_1 r_i\] Hence, \[ C_1 = \frac{\dot{q}}{6k} r_i^2\] For the second boundary condition, we need to use Newton's Law of Cooling at the outer surface: \[ q''(r_o) = -k \frac{dT}{dr}(r_o) = h\left(T(r_o) - T_\infty\right)\]
04

Temperature distribution within the shell

Now we'll substitute \(C_1\) into the temperature distribution formula and simplify: \[ T(r) = - \frac{\dot{q}}{6k} r^3 + \frac{\dot{q}}{6k} r_i^2 r + C_2\] To solve for \(C_2\), we can use Newton's Law of Cooling at the outer surface. Rewrite the temperature distribution formula for \(r = r_o\): \[ T(r_o) - T_\infty = - \frac{\dot{q}}{6k} r_o^3 + \frac{\dot{q}}{6k} r_i^2 r_o + C_2 - T_\infty\] Substitute the Newton's Law of Cooling expression into the equation: \[ \frac{k \frac{dT}{dr}(r_o)}{h} = - \frac{\dot{q}}{6k} r_o^3 + \frac{\dot{q}}{6k} r_i^2 r_o + C_2 - T_\infty\] Differentiate T(r) with respect to r, and evaluate it at \(r=r_o\): \[ \frac{dT}{dr}(r_o) = - \frac{\dot{q}}{2k} r_o^2 + \frac{\dot{q}}{6k} r_i^2\] Substitute the expression for \(\frac{dT}{dr}(r_o)\) in the equation, and solve for \(C_2\): \[ C_2 = T_\infty - \frac{\dot{q}}{6k}(r_i^2-r_o^2)\] Now we have the temperature distribution T(r) within the shell: \[ T(r) = - \frac{\dot{q}}{6k} r^3 + \frac{\dot{q}}{6k} r_i^2 r + T_\infty - \frac{\dot{q}}{6k}(r_i^2-r_o^2)\]
05

Step 5. Calculate heat rate at the outer radius

The heat rate \(q'(r_o)\) can be calculated using the following equation: \[ q'(r_o) = -2\pi r_o k \frac{dT}{dr}(r_o)\] Substitute our previously derived expression for \(\frac{dT}{dr}(r_o)\): \[ q'(r_o) = -2\pi r_o k \left( - \frac{\dot{q}}{2k} r_o^2 + \frac{\dot{q}}{6k} r_i^2 \right)\] Simplify to get the final expression for the heat rate at the outer radius: \[ q'(r_o) = \pi\dot{q}r_o\left(\frac{1}{3} r_i^2 - \frac{1}{2} r_o^2\right) \]

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Key Concepts

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

Steady-State Temperature Distribution
Understanding the steady-state temperature distribution in cylindrical coordinates is crucial when dealing with problems involving long-term heat conduction in systems like pipes, rods, and other cylindrical objects. The term 'steady-state' implies that the temperature at any given point does not change with time. It reaches an equilibrium where the amount of heat entering a particular section is equal to the heat leaving that section.

In a cylindrical shell with uniform volumetric heat generation, as described in the exercise, we're looking for a temperature profile, denoted as \(T(r)\), which should remain constant over time. The inner surface insulation prevents heat from flowing through it, which imposes a boundary condition for our mathematical model. We further apply Newton's Law of Cooling at the shell's outer surface, which interacts with the surrounding fluid.

To make this concept more accessible, think of the cylindrical shell as a circular rod uniformly producing heat along its length. The inner surface acts like a perfect insulator—no heat escapes from it—while the outer surface loses heat to the surrounding environment. Our goal is to understand how the temperature varies from the insulated inner surface to the convective outer surface based on the supplied thermal parameters.
Volumetric Heat Generation
Volumetric heat generation is a common phenomenon in materials that produce heat due to internal mechanisms such as chemical reactions, radioactive decay, or electrical resistance. It refers to the amount of heat produced per unit volume of the material. In the provided exercise, this is depicted by the symbol \(\frac{W}{m^3}\) or \(\dot{q}\).

Let's visualize this concept using a real-world example. Take an electric heater's coil: as electricity passes through the coil, it produces heat. The volumetric heat generation would be the heat produced throughout the volume of the coil. In the case of the cylindrical shell, the heat is assumed to be generated uniformly throughout its volume. This assumption simplifies our calculations, as we can treat the heat generation rate as a constant when solving the heat conduction equation.

In practice, the rate of heat generation can be influenced by various factors such as material properties, the geometry of the material, and the specific mechanisms of heat production. In cases where heat generation is not uniform, the problem becomes more complex and may require a more sophisticated approach to solve.
Radial Heat Conduction Equation
The radial heat conduction equation in cylindrical coordinates is a mathematical expression that describes how heat is transferred radially in a cylindrical object. For problems involving radial symmetry and steady-state conditions, this equation acts as the governing equation for heat conduction.

In essence, the radial heat conduction equation balances the heat flowing through a cylindrical shell with the heat generated inside it. It is derived by applying the law of conservation of energy to a differential volume element in radial coordinates. The equation considers not just the heat flowing in and out of the cylindrical shell but also accounts for any heat being generated or absorbed within the material.

In the exercise, the equation \(\frac{1}{r}\frac{d}{dr}\left(r\frac{dT}{dr}\right) + \frac{\dot{q}}{k} = 0\) is the starting point. Through integration, it yields the temperature distribution function \(T(r)\). Here, \(\dot{q}\) represents the volumetric heat generation, and \(k\) is the material's thermal conductivity—a measure of a material's ability to conduct heat. Finally, the equation is solved subject to the boundary conditions described in the exercise, completing the model that describes radial heat flow in the shell.

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Most popular questions from this chapter

An air heater consists of a steel tube \((k=20 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\), with inner and outer radii of \(r_{1}=13 \mathrm{~mm}\) and \(r_{2}=16\) \(\mathrm{mm}\), respectively, and eight integrally machined longitudinal fins, each of thickness \(t=3 \mathrm{~mm}\). The fins extend to a concentric tube, which is of radius \(r_{3}=\) \(40 \mathrm{~mm}\) and insulated on its outer surface. Water at a temperature \(T_{\infty, i}=90^{\circ} \mathrm{C}\) flows through the inner tube, while air at \(T_{\infty, o}=25^{\circ} \mathrm{C}\) flows through the annular region formed by the larger concentric tube. (a) Sketch the equivalent thermal circuit of the heater and relate each thermal resistance to appropriate system parameters. (b) If \(h_{i}=5000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(h_{o}=200 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), what is the heat rate per unit length? (c) Assess the effect of increasing the number of fins \(N\) and/or the fin thickness \(t\) on the heat rate, subject to the constraint that \(N t<50 \mathrm{~mm}\).

A thin flat plate of length \(L\), thickness \(t\), and width \(W \geqslant L\) is thermally joined to two large heat sinks that are maintained at a temperature \(T_{o}\). The bottom of the plate is well insulated, while the net heat flux to the top surface of the plate is known to have a uniform value of \(q_{o}^{\prime \prime}\) (a) Derive the differential equation that determines the steady-state temperature distribution \(T(x)\) in the plate. (b) Solve the foregoing equation for the temperature distribution, and obtain an expression for the rate of heat transfer from the plate to the heat sinks.

The wall of a spherical tank of \(1-m\) diameter contains an exothermic chemical reaction and is at \(200^{\circ} \mathrm{C}\) when the ambient air temperature is \(25^{\circ} \mathrm{C}\). What thickness of urethane foam is required to reduce the exterior temperature to \(40^{\circ} \mathrm{C}\), assuming the convection coefficient is \(20 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) for both situations? What is the percentage reduction in heat rate achieved by using the insulation?

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)?

An air heater may be fabricated by coiling Nichrome wire and passing air in cross flow over the wire. Consider a heater fabricated from wire of diameter \(D=\) \(1 \mathrm{~mm}\), electrical resistivity \(\rho_{e}=10^{-6} \Omega \cdot \mathrm{m}\), thermal conductivity \(k=25 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), and emissivity \(\varepsilon=0.20\). The heater is designed to deliver air at a temperature of \(T_{\infty}=50^{\circ} \mathrm{C}\) under flow conditions that provide a convection coefficient of \(h=250 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) for the wire. The temperature of the housing that encloses the wire and through which the air flows is \(T_{\text {sur }}=50^{\circ} \mathrm{C}\). If the maximum allowable temperature of the wire is \(T_{\max }=1200^{\circ} \mathrm{C}\), what is the maximum allowable electric current \(I\) ? If the maximum available voltage is \(\Delta E=110 \mathrm{~V}\), what is the corresponding length \(L\) of wire that may be used in the heater and the power rating of the heater? Hint: In your solution, assume negligible temperature variations within the wire, but after obtaining the desired results, assess the validity of this assumption.

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