/*! 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 27 A pan is used to boil water by p... [FREE SOLUTION] | 91影视

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Chapter 2: Problem 27

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.

Short Answer

Expert verified
(a) During Stage 1, the heat equation is given by: \[\frac{\partial T(x, t)}{\partial t} = \alpha \frac{\partial^2 T(x, t)}{\partial x^2}\] The boundary conditions are: \[\frac{\partial T(0, t)}{\partial x} = -\frac{q_{\sigma}}{k}\] \[\frac{\partial T(L, t)}{\partial x} = -\frac{h}{k} [ T(L, t) - T_{\infty}(t) ]\] And the initial condition is: \[T(x, 0) = T_i\] (b) During Stage 2, the heat equation becomes: \[\frac{d^2 T(x)}{d x^2} = 0\] The boundary conditions are: \[\frac{dT(0)}{dx} = -\frac{q_{o}}{k}\] \[T(L) = T_{L}\]

Step by step solution

01

Write the heat equation for Stage 1

The heat equation for a conducting material (like the pan) is given by: \[\frac{\partial T(x, t)}{\partial t} = \alpha \frac{\partial^2 T(x, t)}{\partial x^2}\] where \(T(x, t)\) is the temperature of the pan at position \(x\) and time \(t\), and \(\alpha\) is the thermal diffusivity.
02

Write the boundary/initial conditions for Stage 1

The heat transfer rate from the stove to the pan is fixed at \(q_{\sigma}\). So at \(x = 0\), the boundary condition is: \[\begin{aligned} -k \frac{\partial T(0, t)}{\partial x} &= q_{\sigma} \\\\ \frac{\partial T(0, t)}{\partial x} &= -\frac{q_{\sigma}}{k} \end{aligned}\] At \(x = L\), the boundary condition is given by the heat transfer through convection to the water: \[\begin{aligned} -k \frac{\partial T(L, t)}{\partial x} &= h [ T(L, t) - T_{\infty}(t) ] \\\\ \frac{\partial T(L, t)}{\partial x} &= -\frac{h}{k} [ T(L, t) - T_{\infty}(t) ] \end{aligned}\] The initial condition is: \[T(x, 0) = T_i\] Where \(T_i\) is the initial temperature of the pan bottom. #For part (b)#
03

Write the heat equation for Stage 2

Since there is no temperature change with respect to time during Stage 2, the heat equation becomes: \[\frac{d^2 T(x)}{d x^2} = 0\]
04

Write the boundary conditions for Stage 2

The boundary condition at x = 0 remains the same as in Stage 1: \[\frac{dT(0)}{dx} = -\frac{q_{o}}{k}\] At \(x = L\), the surface temperature of the pan is fixed at \(T_L\): \[T(L) = T_{L}\] These are the required heat equations and boundary conditions for both stages.

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Key Concepts

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

Conduction
When we think about heat transfer, conduction is often the first mechanism that comes to mind. It is the process by which heat energy is transmitted through collisions between neighboring atoms or molecules. In the context of our pan on the stove, conduction is what allows heat to travel from the hot stove through the metal of the pan to the water.

The mathematical model that describes how temperature changes in a solid material, like our pan, as a function of time is known as the heat equation. For one-dimensional heat transfer (which is a good approximation if the pan's thickness is small compared to its diameter), the heat equation during the first stage of heating can be written as:
c\[\begin{equation}\frac{\partial T(x, t)}{\partial t} = \alpha \frac{\partial^2 T(x, t)}{\partial x^2}\end{equation}\]
Here, \( \alpha \) represents the thermal diffusivity of the pan material, a property that indicates how quickly heat spreads through the material.
Convection
Unlike conduction, convection involves the transfer of heat through a fluid (like water or air) that is caused by the fluid's motion. In our exercise, as the pan heats up, the water near the bottom becomes warmer and less dense. This less dense water rises, and the cooler, denser water moves down to replace it, creating a natural circulation pattern that helps distribute heat. This is natural convection.

The boundary condition at the pan-water interface during Stage 1 accounts for this convective heat transfer and is expressed by:
\[\begin{equation}-k \frac{\partial T(L, t)}{\partial x} = h [ T(L, t) - T_{\infty}(t) ]\end{equation}\]
\( h \) is known as the convection heat transfer coefficient and it quantifies how effectively the pan's surface transfers heat to the moving water.
Boundary Conditions
Boundary conditions are crucial for solving the heat equation as they define how the temperature behaves at the surfaces of the material. For our pan, we have two important surfaces: where the pan touches the stove and where it touches the water.

During Stage 1, the boundary condition at the stove-pan interface is dictated by the constant heat transfer rate from the stove, which can be written as:
\[\begin{equation}\frac{\partial T(0, t)}{\partial x} = -\frac{q_{\sigma}}{k}\end{equation}\]
This equation indicates that the heat flux through the pan's bottom is proportional to the temperature gradient at that surface. At the water-pan interface, we have just seen the convective boundary condition. For Stage 2, when the water boils, the boundary condition at the water-pan surface changes to reflect a static temperature because the water temperature doesn't rise above boiling:
\[\begin{equation}T(L) = T_{L}\end{equation}\]
Because of the differing conditions between Stages 1 and 2, solving the heat transfer problem requires different approaches for each stage.
Thermal Diffusivity
Another key term that appears in the heat equation is thermal diffusivity, denoted as \( \alpha \). It is a material-specific property that measures the rate at which heat diffuses through a material. Higher thermal diffusivity means that the material can spread heat more rapidly, while a lower value would indicate that the material holds heat in a particular area for longer.

In the context of heat transfer through the pan's bottom, thermal diffusivity plays a pivotal role in determining how quickly the pan will reach a uniform temperature and how effectively it will transfer that heat to the water. For different materials, the value of thermal diffusivity can vary significantly, affecting both the rate of heating and the energy efficiency of cooking. Hence, choosing a pan with high thermal diffusivity could be preferable for faster and more uniform cooking.

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Most popular questions from this chapter

A spherical shell of inner and outer radii \(r_{i}\) and \(r_{o}\), respectively, contains heat-dissipating components, and at a particular instant the temperature distribution in the shell is known to be of the form $$ T(r)=\frac{C_{1}}{r}+C_{2} $$ Are conditions steady-state or transient? How do the heat flux and heat rate vary with radius?

Steady-state, one-dimensional conduction occurs in a rod of constant thermal conductivity \(k\) and variable crosssectional area \(A_{x}(x)=A_{o} e^{a x}\), where \(A_{o}\) and \(a\) are constants. The lateral surface of the rod is well insulated. (a) Write an expression for the conduction heat rate, \(q_{x}(x)\). Use this expression to determine the temperature distribution \(T(x)\) and qualitatively sketch the distribution for \(T(0)>T(L)\). (b) Now consider conditions for which thermal energy is generated in the rod at a volumetric rate \(\dot{q}=\dot{q}_{s} \exp (-a x)\), where \(\dot{q}_{o}\) is a constant. Obtain an expression for \(q_{x}(x)\) when the left face \((x=0)\) is well insulated.

A plane wall that is insulated on one side \((x=0)\) is initially at a uniform temperature \(T_{i}\), when its exposed surface at \(x=L\) is suddenly raised to a temperature \(T_{s}\). (a) Verify that the following equation satisfies the heat equation and boundary conditions: $$ \frac{T(x, t)-T_{s}}{T_{i}-T_{s}}=C_{1} \exp \left(-\frac{\pi^{2}}{4} \frac{\alpha t}{L^{2}}\right) \cos \left(\frac{\pi}{2} \frac{x}{L}\right) $$ where \(C_{1}\) is a constant and \(\alpha\) is the thermal diffusivity. (b) Obtain expressions for the heat flux at \(x=0\) and \(x=L\). (c) Sketch the temperature distribution \(T(x)\) at \(t=0\), at \(t \rightarrow \infty\), and at an intermediate time. Sketch the variation with time of the heat flux at \(x=L, q_{L}^{\prime \prime}(t)\). (d) What effect does \(\alpha\) have on the thermal response of the material to a change in surface temperature?

An electric cable of radius \(r_{1}\) and thermal conductivity \(k_{c}\) is enclosed by an insulating sleeve whose outer surface is of radius \(r_{2}\) and experiences convection heat transfer and radiation exchange with the adjoining air and large surroundings, respectively. When electric current passes through the cable, thermal energy is generated within the cable at a volumetric rate \(\dot{q}\). (a) Write the steady-state forms of the heat diffusion equation for the insulation and the cable. Verify that these equations are satisfied by the following temperature distributions: Insulation: \(T(r)=T_{s, 2}+\left(T_{s, 1}-T_{s, 2}\right) \frac{\ln \left(r / r_{2}\right)}{\ln \left(r_{1} / r_{2}\right)}\) Cable: \(T(r)=T_{s, 1}+\frac{\dot{q} r_{1}^{2}}{4 k_{c}}\left(1-\frac{r^{2}}{r_{1}^{2}}\right)\) Sketch the temperature distribution, \(T(r)\), in the cable and the sleeve, labeling key features. (b) Applying Fourier's law, show that the rate of conduction heat transfer per unit length through the sleeve may be expressed as $$ q_{r}^{\prime}=\frac{2 \pi k_{s}\left(T_{s, 1}-T_{s, 2}\right)}{\ln \left(r_{2} / r_{1}\right)} $$ Applying an energy balance to a control surface placed around the cable, obtain an alternative expression for \(q_{r}^{\prime}\), 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 sleeve, 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) Consider conditions for which \(250 \mathrm{~A}\) are passing through a cable having an electric resistance per unit length of \(R_{e}^{\prime}=0.005 \Omega / \mathrm{m}\), a radius of \(r_{1}=15 \mathrm{~mm}\), and a thermal conductivity of \(k_{c}=200 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). For \(k_{s}=15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \quad r_{2}=15.5 \mathrm{~mm}, \quad h=25\) \(\mathrm{W} / \mathrm{m}^{2} \cdot \mathrm{K}, \varepsilon=0.9, T_{\text {o }}=25^{\circ} \mathrm{C}\), and \(T_{\text {sur }}=35^{\circ} \mathrm{C}\), evaluate the surface temperatures, \(T_{s, 1}\) and \(T_{s, 2}\), as well as the temperature \(T_{o}\) at the centerline of the cable. (e) With all other conditions remaining the same, compute and plot \(T_{o}, T_{s, 1}\), and \(T_{s, 2}\) as a function of \(r_{2}\) for \(15.5 \leq r_{2} \leq 20 \mathrm{~mm}\).

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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