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Chapter 2: Problem 93

Consider a long solid cylinder of radius \(r_{o}=4 \mathrm{~cm}\) and thermal conductivity \(k=25 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). Heat is generated in the cylinder uniformly at a rate of \(\dot{e}_{\text {gen }}=35 \mathrm{~W} / \mathrm{cm}^{3}\). The side surface of the cylinder is maintained at a constant temperature of \(T_{s}=80^{\circ} \mathrm{C}\). The variation of temperature in the cylinder is given by $$ T(r)=\frac{\dot{e}_{\text {gen }} r_{o}^{2}}{k}\left[1-\left[1-\left(\frac{r}{r_{o}}\right)^{2}\right]+T_{s}\right. $$ Based on this relation, determine \((a)\) if the heat conduction is steady or transient, \((b)\) if it is one-, two-, or three-dimensional, and \((c)\) the value of heat flux on the side surface of the cylinder at \(r=r_{o^{*}}\)

Short Answer

Expert verified
Is it one-, two-, or three-dimensional? What is the heat flux on the side surface of the cylinder at r = r_o? Answer: The heat conduction is steady and one-dimensional. The heat flux on the side surface of the cylinder at r = r_o is 1120 W/cm虏.

Step by step solution

01

Determine if the heat conduction is steady or transient

In a steady heat conduction process, the temperature does not change over time. To check if it's steady or transient, we need to examine if time (t) is present in the temperature equation given: $$ T(r) = \frac{\dot{e}_{\text{gen}} r_{o}^{2}}{k} \left[1-\left(1-\left(\frac{r}{r_{o}}\right)^{2}\right)\right] + T_{s} $$ No time-dependent term (t) is present in the equation. Therefore, the heat conduction is steady. (a) The heat conduction is steady.
02

Determine if the heat conduction is one-, two-, or three-dimensional

To identify if the heat conduction is one-, two-, or three-dimensional, we need to examine how the temperature varies with different coordinates. The given temperature equation has only the radial coordinate (r) playing a role in temperature variation: $$ T(r) = \frac{\dot{e}_{\text{gen}} r_{o}^{2}}{k} \left[1-\left(1-\left(\frac{r}{r_{o}}\right)^{2}\right)\right] + T_{s} $$ Since only the radial coordinate (r) affects the temperature, it is a one-dimensional heat conduction. (b) The heat conduction is one-dimensional.
03

Calculate the heat flux on the side surface of the cylinder at r = r_{o}

To find the heat flux at the side surface of the cylinder, we need to compute the radial heat flux at \(r = r_{o}\) using Fourier's Law of heat conduction: $$ q_r = -k\frac{dT}{dr} $$ First, we compute the derivative of the temperature function with respect to r: $$ \frac{dT}{dr} = \frac{\dot{e}_{\text{gen}} r_{o}^{2}}{k} \cdot \left(\frac{-2r}{r_{o}^2}\right) = - \frac{2 \dot{e}_{\text{gen}} r_{o} r}{k} $$ Now, substituting the values of \(k\), \(\dot{e}_{\text{gen}}\), and \(r_{o}\), and calculating the heat flux at the side surface \(r = r_{o}\): $$ q_{r_{o}} = -k \cdot \left(- \frac{2 \dot{e}_{\text{gen}} r_{o} r_{o}}{k}\right) = 2 \dot{e}_{\text{gen}} r_{o}^2 = 2 \cdot 35 \mathrm{~W/cm^3} \cdot (4\mathrm{~cm})^2 = 1120 \mathrm{~W/cm^2} $$ (c) The heat flux on the side surface of the cylinder at \(r = r_{o}\) is 1120 W/cm虏.

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Key Concepts

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

Steady State Heat Transfer
In heat conduction, the term "steady state" refers to a condition where the temperature profile within a medium does not change as time progresses. This implies that the heat entering one side is equal to the heat leaving the other side, maintaining a constant temperature distribution over time. In our example, the temperature equation provided does not include a time variable, meaning it's independent of any changes over time. This characteristic is crucial because problems involving steady-state heat transfer often require simpler mathematical treatment, as we do not have to solve complex transient heat equations. Knowing that a system is in steady state helps predict how a material will behave under consistent thermal conditions.
One-Dimensional Heat Conduction
One-dimensional heat conduction occurs when the heat transfer within a material depends on a single spatial variable. In simpler terms, it only needs one direction to describe how heat moves across the object. In the case of the cylinder from the exercise, the temperature changes are linked solely to the radial coordinate \(r\), with no dependency on the axial or circumferential coordinates. This indicates one-dimensional heat conduction. This simplification is vital for making calculation manageable and instincts in situations where temperature changes in other dimensions are negligible or constant. By reducing the problem to a single dimension, you can use simpler mathematical approaches to predict how heat moves within the material.
Heat Flux Calculation
Heat flux quantifies the rate of heat energy transfer through a specific surface area. It is essentially the amount of heat flowing per unit area per unit time, often expressed in \ \(\mathrm{W/cm^2}\ \) or \ \(\mathrm{W/m^2}\ \). In the exercise, we use Fourier鈥檚 Law, which states that the heat flux \(q_r\) through a surface is proportional to the negative of the temperature gradient:
  • \(q_r = -k \frac{dT}{dr} \)
To find the heat flux, we need to determine the derivative of the temperature with respect to \(r\). This derivative captures how quickly temperature changes at a particular distance \(r\) from the center. After calculating this derivative and multiplying with the material's thermal conductivity \(k\), the heat flux at \(r = r_{o}\) is 1120 \(\mathrm{W/cm^2}\), reflecting the high rate of energy transfer happening across the surface of the cylinder. Understanding heat flux is crucial in design and safety, ensuring that materials can withstand the heat energy flow that occurs in various engineering applications.

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Most popular questions from this chapter

Consider a large plane wall of thickness \(L=0.8 \mathrm{ft}\) and thermal conductivity \(k=1.2 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft} \cdot{ }^{\circ} \mathrm{F}\). The wall is covered with a material that has an emissivity of \(\varepsilon=0.80\) and a solar absorptivity of \(\alpha=0.60\). The inner surface of the wall is maintained at \(T_{1}=520 \mathrm{R}\) at all times, while the outer surface is exposed to solar radiation that is incident at a rate of \(\dot{q}_{\text {solar }}=300 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{2}\). The outer surface is also losing heat by radiation to deep space at \(0 \mathrm{~K}\). Determine the temperature of the outer surface of the wall and the rate of heat transfer through the wall when steady operating conditions are reached.

Consider a large plate of thickness \(L\) and thermal conductivity \(k\) in which heat is generated uniformly at a rate of \(\dot{e}_{\text {gen. }}\) One side of the plate is insulated while the other side is exposed to an environment at \(T_{\infty}\) with a heat transfer coefficient of \(h\). \((a)\) Express the differential equation and the boundary conditions for steady one-dimensional heat conduction through the plate, \((b)\) determine the variation of temperature in the plate, and \((c)\) obtain relations for the temperatures on both surfaces and the maximum temperature rise in the plate in terms of given parameters.

A \(1200-W\) iron is left on the iron board with its base exposed to ambient air at \(26^{\circ} \mathrm{C}\). The base plate of the iron has a thickness of \(L=0.5 \mathrm{~cm}\), base area of \(A=150 \mathrm{~cm}^{2}\), and thermal conductivity of \(k=18 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). The inner surface of the base plate is subjected to uniform heat flux generated by the resistance heaters inside. The outer surface of the base plate whose emissivity is \(\varepsilon=0.7\), loses heat by convection to ambient air with an average heat transfer coefficient of \(h=\) \(30 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) as well as by radiation to the surrounding surfaces at an average temperature of \(T_{\text {surr }}=295 \mathrm{~K}\). Disregarding any heat loss through the upper part of the iron, \((a)\) express the differential equation and the boundary conditions for steady one-dimensional heat conduction through the plate, \((b)\) obtain a relation for the temperature of the outer surface of the plate by solving the differential equation, and (c) evaluate the outer surface temperature.

What is the difference between an algebraic equation and a differential equation?

Consider a large plane wall of thickness \(L=0.4 \mathrm{~m}\), thermal conductivity \(k=1.8 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), and surface area \(A=\) \(30 \mathrm{~m}^{2}\). The left side of the wall is maintained at a constant temperature of \(T_{1}=90^{\circ} \mathrm{C}\) while the right side loses heat by convection to the surrounding air at \(T_{\infty}=25^{\circ} \mathrm{C}\) with a heat transfer coefficient of \(h=24 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Assuming constant thermal conductivity and no heat generation in the wall, \((a)\) express the differential equation and the boundary conditions for steady one-dimensional heat conduction through the wall, \((b)\) obtain a relation for the variation of temperature in the wall by solving the differential equation, and \((c)\) evaluate the rate of heat transfer through the wall. Answer: (c) \(7389 \mathrm{~W}\)
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