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Chapter 2: Problem 49

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.

Short Answer

Expert verified
The inner radius of the cylinder is \(r_i = 0.2\) meters. The expression for the volumetric rate of heat generation is \(\dot{q} = 9600 - \frac{192}{r^2} \, \mathrm{W}/\mathrm{m}^3\). The axial distribution of the heat flux at the outer surface is \(q_r^{\prime\prime}(r_o, z) = -4896 \, \mathrm{W}/\mathrm{m}^2\), and the heat rate is going out of the cylinder. The radial distribution of the heat flux at the end faces of the cylinder is \(q_r^{\prime\prime}(r, +z_o) = q_r^{\prime\prime}(r, -z_o) = 0 \, \mathrm{W}/\mathrm{m}^2\). The overall energy balance on the cylinder is consistent, as the total heat transfer out of the cylinder equals the total heat generated within the volume of the cylinder.

Step by step solution

01

Differentiate the temperature distribution with respect to r

To find the temperature gradient with respect to r, take the first derivative of the temperature distribution (\(T(r, z)=a+br^2+c \ln r+dz^2\)) with respect to r: $$\frac{dT}{dr} = 2br + \frac{c}{r}$$
02

Find the inner radius where the temperature gradient is zero

The inner radius (\(r_i\)) will have zero temperature gradient, as the inner surface is insulated. Therefore, set \(\frac{dT}{dr}\) to zero and solve for \(r_i\): $$ 2b r_i + \frac{c}{r_i} = 0 \Rightarrow r_i^2 = -\frac{c}{2b}$$ Given the values for \(b\) and \(c\), we can calculate \(r_i\): $$ r_i = \sqrt{-\frac{-12^{\circ}\mathrm{C}}{2(150^{\circ}\mathrm{C}/\mathrm{m}^2)}} = \sqrt{0.04} = 0.2\mathrm{m}$$ So, the inner radius \(r_i\) of the cylinder is 0.2 meters. (b) Obtain an expression for the volumetric rate of heat generation.
03

Calculate the Laplacian of the Temperature

First, find the second derivatives of the temperature distribution with respect to r and z: $$\frac{d^2T}{dr^2} = 2b - \frac{c}{r^2}$$ $$\frac{d^2T}{dz^2} = 2d$$ The Laplacian of the temperature distribution is the sum of these second derivatives: $$\nabla^2 T =\frac{d^2T}{dr^2} + \frac{d^2T}{dz^2} = 2b - \frac{c}{r^2} + 2d$$
04

Calculate the Volumetric Rate of Heat Generation

Now, use the heat conduction equation \(\nabla^2 T = \frac{\dot{q}}{k}\) to find the volumetric rate of heat generation: $$\dot{q} = k \nabla^2 T = 16(2b - \frac{c}{r^2} + 2d) \mathrm{W}/\mathrm{m}^3$$ Substitute the given values for \(b\), \(c\), and \(d\): $$\dot{q} = 16(2(150^{\circ}\mathrm{C}/\mathrm{m}^2) - \frac{-12^{\circ}\mathrm{C}}{r^2} - 2(300^{\circ}\mathrm{C}/\mathrm{m}^2)) \mathrm{W}/\mathrm{m}^3 = 9600 - \frac{192}{r^2} \, \mathrm{W}/\mathrm{m}^3$$ Thus, the expression for the volumetric rate of heat generation is \(\dot{q} = 9600 - \frac{192}{r^2} \, \mathrm{W}/\mathrm{m}^3\). (c) Determine the axial distribution of the heat flux at the outer surface.
05

Calculate the Radial Heat Flux at the Outer Surface

To find the radial heat flux at the outer surface at \(r=r_o\), use Fourier's law of heat conduction, \(q_r^{\prime\prime}(r_o, z) = -k\frac{dT}{dr}(r_o, z)\): $$q_r^{\prime\prime}(r_o, z) = -16(2b r_o + \frac{c}{r_o})\mathrm{W}/\mathrm{m}^2$$ Substitute the given values for \(b\), \(c\), and \(r_o\): $$q_r^{\prime\prime}(r_o, z) = -16(2(150^{\circ}\mathrm{C}/\mathrm{m}^2)(1\mathrm{m}) - \frac{-12^{\circ}\mathrm{C}}{1\mathrm{m}}) \mathrm{W}/\mathrm{m}^2 = -4896 \, \mathrm{W}/\mathrm{m}^2$$ Since the radial heat flux is negative, the heat rate at the outer surface is going out of the cylinder. (d) Determine the radial distribution of the heat flux at the end faces of the cylinder.
06

Calculate the radial heat flux at the end faces of the cylinder

The temperature gradient at the end faces of the cylinder has no r-component as the heat flux is parallel to the end faces, meaning that there is no radial heat flux at \(z = \pm z_o\) for any \(r\). So, \(q_r^{\prime\prime}(r, +z_o) = q_r^{\prime\prime}(r, -z_o) = 0 \, \mathrm{W}/\mathrm{m}^2\). (e) Verify that your results are consistent with an overall energy balance on the cylinder.
07

Confirm energy balance

According to the energy balance, the total heat rate entering the cylinder must equal the total heat rate leaving the cylinder. As there is no heat rate at the end faces and the heat rate at the inner surface is zero due to insulation, the only outward heat transfer is at the outer surface. Thus, the energy balance for the cylinder is \(-\int\limits_{-z_o}^{z_o} q_r^{\prime\prime}(r_o, z) dz = \int\limits_{V} \dot{q} dV\), where \(V\) is the volume of the annular space between \(r_i\) and \(r_o\). The integral on the left-hand side of the equation represents the total heat transfer out of the cylinder, which can be calculated as: $$-\int\limits_{-z_o}^{z_o} (-4896)\, dz = 4896(2z_o) = 4896 \cdot 5 \, \mathrm{W}$$ The integral on the right-hand side of the equation represents the total heat generated within the volume of the cylinder: $$\int\limits_{V} \dot{q} dV = \int\limits_{-z_o}^{z_o} \int\limits_{r_i}^{r_o} (9600 - \frac{192}{r^2}) rdzdr = 2z_o \int\limits_{r_i}^{r_o} (9600r - \frac{192}{r}) dr$$ Calculate this integral with the known values: $$5 \int\limits_{0.2}^{1} (9600r - \frac{192}{r}) dr = 5 \left[ 4800r^2 - 192 \ln r \right]_{0.2}^{1} = 4896 \cdot 5 \, \mathrm{W}$$ As both sides are equal, the energy balance is consistent and our results are verified.

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Key Concepts

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

Steady-State Conduction
In the realm of heat conduction, the term "steady-state conduction" refers to the condition where the temperature within an object does not change over time. Imagine a scenario where heat is continuously supplied to one end of an object, and removed at the other end, causing no change in temperature distribution with respect to time.

For steady-state conduction to occur, the amount of heat entering a system must equal the amount of heat leaving it, meaning there's a balance between energy input and output. The heat no longer accumulates within the material because it's evenly distributed, maintaining a constant temperature profile.

In many engineering applications, achieving a steady state is crucial because it allows for more predictability in how heat will behave within a material.
Thermal Conductivity
Thermal conductivity is a property of a material that indicates its ability to conduct heat. It is a crucial factor in determining how quickly heat will pass through a material. With a high thermal conductivity, heat moves through the material rapidly, whereas low thermal conductivity suggests that the material is more of an insulator.

In the given problem, the material at hand has a thermal conductivity of 16 W/m路K, meaning it can transfer 16 watts of heat per meter per degree of temperature difference across its surface.
  • This property is intrinsic to the material itself, not affected by the shape or size of the object.
  • It has a major impact on the thermal management of systems, such as in building insulation and heat sinks.
Volumetric Heat Generation
Volumetric heat generation refers to the amount of heat produced within a unit volume of a material. This can occur due to internal sources like chemical reactions or electromagnetic energy being converted into heat.

In our problem, we derived an expression for the volumetric rate of heat generation by using the Laplacian and the heat conduction equation. This represents how much heat is produced internally within the cylindrical system.

  • The volumetric heat generation was found to be dependent on the radius due to its expression being \( 9600 - \frac{192}{r^2}\) . Thus, it changes as we move from the inner radius \(r_i\) to the outer radius \(r_o\).
  • This detail is important in understanding and calculating how energy is transformed inside the cylinder.
Energy Balance
Energy balance in heat conduction ensures that the sum of all energies entering, leaving, or being generated within a system remains consistent. It's like checking a ledger to make sure the inputs and outputs line up distinctly.

In the provided exercise, the energy balance was verified by showing that the total heat rate leaving the system matched the total internal volumetric heat generation.
  • We calculated the heat flow across various segments of the cylinder, such as the outer surface and end faces, ensuring the integral of heat generation across the volume matched this outflow.
  • Ensuring energy balance is fundamental in engineering to confirm the accuracy and reliability of heat transfer designs.

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Most popular questions from this chapter

A pan is used to boil water by placing it on a stove, from which heat is transferred at a fixed rate \(q_{\sigma}\). There are two stages to the process. In Stage 1, the water is taken from its initial (room) temperature \(T_{i}\) to the boiling point, as heat is transferred from the pan by natural convection. During this stage, a constant value of the convection coefficient \(h\) may be assumed, while the bulk temperature of the water increases with time, \(T_{\infty}=T_{\infty}(t)\). In Stage 2, the water has come to a boil, and its temperature remains at a fixed value, \(T_{\infty}=T_{b}\), as heating continues. Consider a pan bottom of thickness \(L\) and diameter \(D\), with a coordinate system corresponding to \(x=0\) and \(x=L\) for the surfaces in contact with the stove and water, respectively. (a) Write the form of the heat equation and the boundary/ initial conditions that determine the variation of temperature with position and time, \(T(x, t)\), in the pan bottom during Stage 1. Express your result in terms of the parameters \(q_{o}, D, L, h\), and \(T_{\infty}\), as well as appropriate properties of the pan material. (b) During Stage 2, the surface of the pan in contact with the water is at a fixed temperature, \(T(L, t)=\) \(T_{L}>T_{b}\). Write the form of the heat equation and boundary conditions that determine the temperature distribution \(T(x)\) in the pan bottom. Express your result in terms of the parameters \(q_{o}, D, L\), and \(T_{L}\), as well as appropriate properties of the pan material.

A plane wall has constant properties, no internal heat generation, and is initially at a uniform temperature \(T_{i \cdot}\) Suddenly, the surface at \(x=L\) is heated by a fluid at \(T_{\infty}\) having a convection coefficient \(h\). At the same instant, the electrical heater is energized, providing a constant heat flux \(q_{o}^{\prime \prime}\) at \(x=0\). (a) On \(T-x\) coordinates, sketch the temperature distributions for the following conditions: initial condition \((t \leq 0)\), steady-state condition \((t \rightarrow \infty)\), and for two intermediate times. (b) On \(q_{x}^{\prime \prime}-x\) coordinates, sketch the heat flux corresponding to the four temperature distributions of part (a). (c) On \(q_{x}^{n}-t\) coordinates, sketch the heat flux at the locations \(x=0\) and \(x=L\). That is, show qualitatively how \(q_{x}^{\prime \prime}(0, t)\) and \(q_{x}^{\prime \prime}(L, t)\) vary with time. (d) Derive an expression for the steady-state temperature at the heater surface, \(T(0, \infty)\), in terms of \(q_{o}^{\prime \prime}\), \(T_{\infty}, k, h\), and \(L\).

The steady-state temperature distribution in a onedimensional wall of thermal conductivity \(50 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and thickness \(50 \mathrm{~mm}\) is observed to be \(T\left({ }^{\circ} \mathrm{C}\right)=a+b x^{2}\), where \(a=200^{\circ} \mathrm{C}, b=-2000^{\circ} \mathrm{C} / \mathrm{m}^{2}\), and \(x\) is in meters. (a) What is the heat generation rate \(\dot{q}\) in the wall? (b) Determine the heat fluxes at the two wall faces. In what manner are these heat fluxes related to the heat generation rate?

A method for determining the thermal conductivity \(k\) and the specific heat \(c_{p}\) of a material is illustrated in the sketch. Initially the two identical samples of diameter \(D=60 \mathrm{~mm}\) and thickness \(L=10 \mathrm{~mm}\) and the thin heater are at a uniform temperature of \(T_{i}=23.00^{\circ} \mathrm{C}\), while surrounded by an insulating powder. Suddenly the heater is energized to provide a uniform heat flux \(q_{o}^{\prime \prime}\) on each of the sample interfaces, and the heat flux is maintained constant for a period of time, \(\Delta t_{o}\). A short time after sudden heating is initiated, the temperature at this interface \(T_{o}\) is related to the heat flux as $$ T_{o}(t)-T_{i}=2 q_{o}^{\prime \prime}\left(\frac{t}{\pi \rho c_{p} k}\right)^{1 / 2} $$ For a particular test run, the electrical heater dissipates \(15.0 \mathrm{~W}\) for a period of \(\Delta t_{o}=120 \mathrm{~s}\), and the temperature at the interface is \(T_{o}(30 \mathrm{~s})=24.57^{\circ} \mathrm{C}\) after \(30 \mathrm{~s}\) of heating. A long time after the heater is deenergized, \(t \geqslant \Delta t_{0}\), the samples reach the uniform temperature of \(T_{o}(\infty)=33.50^{\circ} \mathrm{C}\). The density of the sample materials, determined by measurement of volume and mass, is \(\rho=3965 \mathrm{~kg} / \mathrm{m}^{3}\). Determine the specific heat and thermal conductivity of the test material. By looking at values of the thermophysical properties in Table A.1 or A.2, identify the test sample material.

A composite one-dimensional plane wall is of overall thickness \(2 L\). Material A spans the domain \(-L \leq x<0\) and experiences an exothermic chemical reaction leading to a uniform volumetric generation rate of \(\dot{q}_{\mathrm{A}}\). Material B spans the domain \(0 \leq x \leq L\) and undergoes an endothermic chemical reaction corresponding to a uniform volumetric generation rate of \(\dot{q}_{\mathrm{B}}=-\dot{q}_{\mathrm{A}}\). The surfaces at \(x=\pm L\) are insulated. Sketch the steady-state temperature and heat flux distributions \(T(x)\) and \(q_{\mathrm{x}}^{\prime \prime}(x)\), respectively, over the domain \(-L \leq x \leq L\) for \(k_{\mathrm{A}}=k_{\mathrm{B}}, k_{\mathrm{A}}=0.5 k_{\mathrm{B}}\), and \(k_{\mathrm{A}}=2 k_{\mathrm{B}}\). Point out the important features of the distributions you have drawn. If \(\dot{q}_{\mathrm{B}}=-2 \dot{q}_{\mathrm{A}}\), can you sketch the steady-state temperature distribution?
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