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Chapter 2: Problem 46

A steam pipe is wrapped with insulation of inner and outer radii \(r_{i}\) and \(r_{o}\), respectively. At a particular instant the temperature distribution in the insulation is known to be of the form $$ T(r)=C_{1} \ln \left(\frac{r}{r_{o}}\right)+C_{2} $$ Are conditions steady-state or transient? How do the heat flux and heat rate vary with radius?

Short Answer

Expert verified
The conditions in the insulation are steady-state as there is no time-dependency in the given temperature distribution. The heat flux varies with radius as \(q(r) = -k \frac{C_1}{r}\), while the heat rate is constant and does not vary with the radius, given by \(Q(r) = -2\pi k C_1 L\).

Step by step solution

01

Determine steady-state or transient

Since there is no time-dependency in the given temperature distribution, we conclude that the conditions are steady-state.
02

Calculate heat flux using Fourier's Law

Fourier's Law of thermal conduction states that \( q = -k \frac{dT}{dr} \), where \(q\) is the heat flux, \(k\) is the thermal conductivity, and \(\frac{dT}{dr}\) is the temperature gradient with respect to the radial distance. Given the temperature distribution \(T(r) = C_1 \ln(\frac{r}{r_o}) + C_2\), let's differentiate it with respect to \(r\): \[\frac{dT}{dr} = \frac{d}{dr}\left(C_1 \ln\left(\frac{r}{r_o}\right) + C_2\right) = \frac{C_1}{r}\] Now, we can plug this into Fourier's Law to find the heat flux, \(q\): \[q(r) = -k \frac{dT}{dr} = -k \frac{C_1}{r}\]
03

Calculate heat rate

The heat rate (\(Q\)) is the product of the heat flux (\(q\)) and the area through which the heat is being conducted. For a cylindrical surface, the area is given by \(A = 2 \pi r L\), where \(L\) is the length of the pipe. Heat rate can be calculated as: \[Q(r) = q(r) \cdot A = -k \frac{C_1}{r} \cdot 2 \pi r L\] Simplifying this expression, we obtain the heat rate as a function of the radius: \[Q(r) = -2 \pi k C_1 L\] Since the heat rate is constant with respect to the radius, it does not vary with radius. #Summary of results# The conditions in the insulation are steady-state. The heat flux varies with radius as \(q(r) = -k \frac{C_1}{r}\), while the heat rate is constant and does not vary with the radius, given by \(Q(r) = -2\pi k C_1 L\).

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Key Concepts

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

Steady-State Conditions
Understanding steady-state conditions is crucial when analyzing heat transfer in insulation. Steady-state refers to a situation where the temperature field within a material does not change with time.

In the case of the insulated steam pipe exercise, the absence of time variable in the temperature distribution equation, \(T(r) = C_1 \ln(\frac{r}{r_o}) + C_2\), indicates that the conditions are steady-state. This means that the temperature at any given point in the insulation remains constant over time, implying that the rates of heat input and output through any part of the insulation are balanced.

For students, recognizing these conditions is critical as it simplifies the analysis, eliminating the need to deal with the complexities of time-dependent (transient) heat transfer.
Fourier's Law of Thermal Conduction
Fourier's Law of thermal conduction is the foundational principle for understanding heat transfer in materials. It states that the rate of heat transfer through a material is proportional to the negative gradient of the temperature and the area through which heat is flowing.

Mathematically, Fourier's Law is expressed as \( q = -k \frac{dT}{dr} \), where \( q \) is the heat flux, \( k \) is the thermal conductivity of the material, and \( \frac{dT}{dr} \) is the temperature gradient. For cylindrical coordinates, as in our steam pipe scenario, the temperature gradient is determined with respect to the radial position, \( r \).

Fourier's Law highlights the direct link between temperature differences within a material and the resultant heat flow, which is a key concept for solving problems related to thermal conduction.
Heat Flux Calculation
Heat flux is a measure of the rate of heat transfer per unit area. To calculate heat flux using Fourier's Law, one must first determine the temperature gradient. In the exercise, the given temperature distribution enables us to differentiate and find \( \frac{dT}{dr} \), which is \( \frac{C_1}{r} \).

Applying Fourier's Law, we obtain the heat flux \( q(r) \) as \( -k \frac{C_1}{r} \), showing that the heat flux is inversely proportional to the radial position. It reveals how the heat flux varies with distance from the center of the pipe: as the radius increases, the heat flux decreases. This inverse relationship is a valuable insight for students in understanding how thermal energy moves through materials.
Heat Rate Calculation
Heat rate is a term used to describe the total amount of heat transferred per unit time. It is found by multiplying the heat flux by the area over which the transfer takes place.

In the context of the insulated steam pipe, the heat rate \( Q(r) \) can be calculated using the area of a cylindrical surface, \( A = 2 \pi r L \), where \( L \) is the length of the pipe. This results in a heat rate that is independent of the radius, expressed as \( Q(r) = -2 \pi k C_1 L \). The outcome shows a constant heat rate, providing an important lesson for students: even though the heat flux decreases with increasing radius, the heat rate across the cylindrical surface of the insulation does not change.
Thermal Conductivity
Thermal conductivity, denoted by \( k \), is a material property that quantifies its ability to conduct heat. It appears in the calculations for both heat flux and heat rate, signifying its importance in the heat transfer process.

In the discussed exercise, thermal conductivity plays a pivotal role in determining the rate at which heat moves through the insulation. The negative sign in Fourier's law and the subsequent calculations underscores the fact that heat flows from higher to lower temperatures.

The value of \( k \) can vary widely among different materials, and thus affects the efficiency of thermal insulation. For students, understanding the role of thermal conductivity is essential when choosing materials for heat insulation purposes.

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Most popular questions from this chapter

The temperature distribution across a wall \(0.3 \mathrm{~m}\) thick at a certain instant of time is \(T(x)=a+b x+c x^{2}\), where \(T\) is in degrees Celsius and \(x\) is in meters, \(a=200^{\circ} \mathrm{C}\), \(b=-200^{\circ} \mathrm{C} / \mathrm{m}\), and \(c=30^{\circ} \mathrm{C} / \mathrm{m}^{2}\). The wall has a thermal conductivity of \(1 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). (a) On a unit surface area basis, determine the rate of heat transfer into and out of the wall and the rate of change of energy stored by the wall. (b) If the cold surface is exposed to a fluid at \(100^{\circ} \mathrm{C}\), what is the convection coefficient?

To determine the effect of the temperature dependence of the thermal conductivity on the temperature distribution in a solid, consider a material for which this dependence may be represented as $$ k=k_{o}+a T $$ where \(k_{o}\) is a positive constant and \(a\) is a coefficient that may be positive or negative. Sketch the steady-state temperature distribution associated with heat transfer in a plane wall for three cases corresponding to \(a>0\), \(a=0\), and \(a<0\).

A chemically reacting mixture is stored in a thin-walled spherical container of radius \(r_{1}=200 \mathrm{~mm}\), and the exothermic reaction generates heat at a uniform, but temperaturedependent volumetric rate of \(\dot{q}=\dot{q}_{o} \exp \left(-A / T_{o}\right)\), where \(\dot{q}_{o}=5000 \mathrm{~W} / \mathrm{m}^{3}, A=75 \mathrm{~K}\), and \(T_{o}\) is the mixture temperature in kelvins. The vessel is enclosed by an insulating material of outer radius \(r_{2}\), thermal conductivity \(k\), and emissivity \(\varepsilon\). The outer surface of the insulation experiences convection heat transfer and net radiation exchange with the adjoining air and large surroundings, respectively. (a) Write the steady-state form of the heat diffusion equation for the insulation. Verify that this equation is satisfied by the temperature distribution $$ T(r)=T_{s, 1}-\left(T_{s, 1}-T_{s, 2}\right)\left[\frac{1-\left(r_{1} / r\right)}{1-\left(r_{1} / r_{2}\right)}\right] $$ Sketch the temperature distribution, \(T(r)\), labeling key features. (b) Applying Fourier's law, show that the rate of heat transfer by conduction through the insulation may be expressed as $$ q_{r}=\frac{4 \pi k\left(T_{s, 1}-T_{s, 2}\right)}{\left(1 / r_{1}\right)-\left(1 / r_{2}\right)} $$ Applying an energy balance to a control surface about the container, obtain an alternative expression for \(q_{r}\), expressing your result in terms of \(\dot{q}\) and \(r_{1}\). (c) Applying an energy balance to a control surface placed around the outer surface of the insulation, obtain an expression from which \(T_{s, 2}\) may be determined as a function of \(\dot{q}, r_{1}, h, T_{\infty}, \varepsilon\), and \(T_{\text {sur }}\) (d) The process engineer wishes to maintain a reactor temperature of \(T_{o}=T\left(r_{1}\right)=95^{\circ} \mathrm{C}\) under conditions for which \(k=0.05 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, r_{2}=208 \mathrm{~mm}, h=5\) \(\mathrm{W} / \mathrm{m}^{2} \cdot \mathrm{K}, \varepsilon=0.9, T_{\infty}=25^{\circ} \mathrm{C}\), and \(T_{\text {sur }}=35^{\circ} \mathrm{C}\). What is the actual reactor temperature and the outer surface temperature \(T_{s, 2}\) of the insulation? (e) Compute and plot the variation of \(T_{s, 2}\) with \(r_{2}\) for \(201 \leq r_{2} \leq 210 \mathrm{~mm}\). The engineer is concerned about potential burn injuries to personnel who may come into contact with the exposed surface of the insulation. Is increasing the insulation thickness a practical solution to maintaining \(T_{s, 2} \leq 45^{\circ} \mathrm{C}\) ? What other parameter could be varied to reduce \(T_{s, 2}\) ?

Passage of an electric current through a long conducting rod of radius \(r_{i}\) and thermal conductivity \(k_{r}\) results in uniform volumetric heating at a rate of \(\dot{q}\). The conducting rod is wrapped in an electrically nonconducting cladding material of outer radius \(r_{o}\) and thermal conductivity \(k_{c}\), and convection cooling is provided by an adjoining fluid. For steady-state conditions, write appropriate forms of the heat equations for the rod and cladding. Express appropriate boundary conditions for the solution of these equations.

Two-dimensional, steady-state conduction occurs in a hollow cylindrical solid of thermal conductivity \(k=16 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), outer radius \(r_{o}=1 \mathrm{~m}\) and overall length \(2 z_{o}=5 \mathrm{~m}\), where the origin of the coordinate system is located at the midpoint of the center line. The inner surface of the cylinder is insulated, and the temperature distribution within the cylinder has the form \(T(r, z)=a+b r^{2}+c \ln r+d z^{2}\), where \(a=\) \(-20^{\circ} \mathrm{C}, \quad b=150^{\circ} \mathrm{C} / \mathrm{m}^{2}, c=-12^{\circ} \mathrm{C}, d=-300^{\circ} \mathrm{C} / \mathrm{m}^{2}\) and \(r\) and \(z\) are in meters. (a) Determine the inner radius \(r_{i}\) of the cylinder. (b) Obtain an expression for the volumetric rate of heat generation, \(\dot{q}\left(\mathrm{~W} / \mathrm{m}^{3}\right)\). (c) Determine the axial distribution of the heat flux at the outer surface, \(q_{r}^{\prime \prime}\left(r_{o}, z\right)\). What is the heat rate at the outer surface? Is it into or out of the cylinder? (d) Determine the radial distribution of the heat flux at the end faces of the cylinder, \(q_{r}^{\prime \prime}\left(r,+z_{o}\right)\) and \(q_{r}^{\prime \prime}\left(r,-z_{o}\right)\). What are the corresponding heat rates? Are they into or out of the cylinder? (e) Verify that your results are consistent with an overall energy balance on the cylinder.
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