/*! 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 127 A constant-property, one-dimensi... [FREE SOLUTION] | 91Ó°ÊÓ

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A constant-property, one-dimensional plane wall of width \(2 L\), at an initial uniform temperature \(T_{i}\), is heated convectively (both surfaces) with an ambient fluid at \(T_{\infty}=T_{\infty, 1}, h=h_{1}\). At a later instant in time, \(t=t_{1}\), heating is curtailed, and convective cooling is initiated. Cooling conditions are characterized by \(T_{\infty}=T_{\infty, 2}=T_{i}, h=h_{2}\) (a) Write the heat equation as well as the initial and boundary conditions in their dimensionless form for the heating phase (Phase 1). Express the equations in terms of the dimensionless quantities \(\theta^{*}, x^{*}, B i_{1}\), and \(F o\), where \(B i_{1}\) is expressed in terms of \(h_{1}\). (b) Write the heat equation as well as the initial and boundary conditions in their dimensionless form for the cooling phase (Phase 2). Express the equations in terms of the dimensionless quantities \(\theta^{*}, x^{*}, B i_{2}\), \(\mathrm{Fo}_{1}\), and \(\mathrm{Fo}\) where \(\mathrm{Fo}_{1}\) is the dimensionless time associated with \(t_{1}\), and \(B i_{2}\) is expressed in terms of \(h_{2}\). To be consistent with part (a), express the dimensionless temperature in terms of \(T_{\infty}=T_{\infty, 1^{*}}\) (c) Consider a case for which \(B i_{1}=10, B i_{2}=1\), and \(F o_{1}=0.1\). Using a finite-difference method with \(\Delta x^{*}=0.1\) and \(\Delta F o=0.001\), determine the transient thermal response of the surface \(\left(x^{*}=1\right)\), midplane \(\left(x^{*}=0\right)\), and quarter-plane \(\left(x^{*}=0.5\right)\) of the slab. Plot these three dimensionless temperatures as a function of dimensionless time over the range \(0 \leq F o \leq 0.5\). (d) Determine the minimum dimensionless temperature at the midplane of the wall, and the dimensionless time at which this minimum temperature is achieved.

Short Answer

Expert verified
(a) For Phase 1: Dimensionless heat equation: \(\frac{\partial \theta^{*}}{\partial t} = Fo \frac{\partial^2 \theta^{*}}{\partial (x^{*})^2}\) Initial condition: \(\theta^{*} = 0\) at \(t = 0\) Boundary conditions: \(\theta^{*} = 1 - Bi_1(1-x^{*})\) at \(x^{*} = 0\) and \(\theta^{*} = 1 - Bi_1(x^{*})\) at \(x^{*} = 1\) (b) For Phase 2: Dimensionless heat equation: \(\frac{\partial \theta^{*}}{\partial t} = Fo \frac{\partial^2 \theta^{*}}{\partial (x^{*})^2}\) Initial condition: \(\theta^{*}\) at \(t = Fo_1\) Boundary conditions: \(\theta^{*} = 1 - Bi_2(1-x^{*})\) at \(x^{*} = 0\) and \(\theta^{*} = 1 - Bi_2(x^{*})\) at \(x^{*} = 1\) (c) Use finite-difference method with \(\Delta x^{*} = 0.1\) and \(\Delta Fo = 0.001\) to solve for the transient thermal response at \(x^{*} = 1\), \(x^{*} = 0\), and \(x^{*} = 0.5\) for \(0 \leq Fo \leq 0.5\). (d) From the finite-difference calculations, find the minimum dimensionless temperature \(\theta^{*}_{min}\) at \(x^{*} = 0\) and the dimensionless time \(Fo_{min}\) at which this minimum temperature is achieved.

Step by step solution

01

Write the heat equation for Phase 1

The heat equation for a one-dimensional plane wall without heat generation is given by: \(\frac{\partial T}{\partial t} = \alpha \frac{\partial^2 T}{\partial x^2}\)
02

Define dimensionless quantities for Phase 1

To express the heat equation in dimensionless form, we need to define dimensionless temperature, position, Biot number, and Fourier number as: \(\theta^{*} = \frac{T - T_{\infty,1}}{T_i - T_{\infty,1}}\) \(x^{*} = \frac{x}{L}\) \(Bi_1 = \frac{h_1 L}{k}\) \(Fo = \frac{\alpha t}{L^2}\)
03

Write the dimensionless heat equation for Phase 1

By rewriting the heat equation using the above dimensionless variables, we obtain the dimensionless heat equation for Phase 1: \(\frac{\partial \theta^{*}}{\partial t} = Fo \frac{\partial^2 \theta^{*}}{\partial (x^{*})^2}\)
04

Write the dimensionless initial and boundary conditions for Phase 1

The initial condition is: \(\theta^{*} = 0\) at \(t = 0\) The boundary conditions are: \(\theta^{*} = 1 - Bi_1(1-x^{*})\) at \(x^{*} = 0\) \(\theta^{*} = 1 - Bi_1(x^{*})\) at \(x^{*} = 1\) (B) Find the dimensionless heat equation and initial and boundary conditions for Phase 2
05

Define dimensionless quantities for Phase 2

To express the heat equation in dimensionless form for Phase 2, we define new dimensionless Biot number and Fourier numbers as: \(Bi_2 = \frac{h_2 L}{k}\) \(Fo_1 = \frac{\alpha t_1}{L^2}\)
06

Write the dimensionless heat equation for Phase 2

The dimensionless heat equation for Phase 2 is the same as that for Phase 1: \(\frac{\partial \theta^{*}}{\partial t} = Fo \frac{\partial^2 \theta^{*}}{\partial (x^{*})^2}\)
07

Write the dimensionless initial and boundary conditions for Phase 2

The initial condition is: \(\theta^{*}\) at \(t = Fo_1\) The boundary conditions are: \(\theta^{*} = 1 - Bi_2(1-x^{*})\) at \(x^{*} = 0\) \(\theta^{*} = 1 - Bi_2(x^{*})\) at \(x^{*} = 1\) (C) Determine the transient thermal response of the surface (x=1), midplane (x=0), and quarter-plane (x=0.5) of the slab using finite-difference method
08

Grid and Time Step

Set up the finite-difference grid with: \(\Delta x^{*} = 0.1\) \(\Delta Fo = 0.001\)
09

Set the given parameters

Plug in the given values: \(Bi_1 = 10\) \(Bi_2 = 1\) \(Fo_1 = 0.1\)
10

Perform finite-difference calculations

Calculate the dimensionless temperatures at x=1, x=0, and x=0.5 using finite-difference method over the range of 0 ≤ Fo ≤ 0.5. Perform the calculations in two parts: first for Phase 1 (heating phase) and then for Phase 2 (cooling phase). (D) Determine the minimum dimensionless temperature at the midplane of the wall
11

Calculate minimum dimensionless temperature

From the finite-difference calculations in Step 10, find the minimum dimensionless temperature (\(\theta^{*}_{min}\)) at the midplane: \(\theta^{*}_{min}\) = min(\(\theta^{*}\) at \(x^{*} = 0\))
12

Calculate the dimensionless time at minimum temperature

Find the dimensionless time (\(Fo_{min}\)) at which the minimum dimensionless temperature occurs: \(Fo_{min}\) = \(Fo\) at \(\theta^{*}_{min}\)

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Key Concepts

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

Dimensionless Analysis
Dimensionless analysis is a powerful tool in engineering that scales down complex systems into simpler, manageable components. This technique involves transforming variables into dimensionless numbers, helping us identify the fundamental behavior of a system without being influenced by specific units.
In the context of the heat equation, the transformation of regular variables like temperature, position, time, and material properties into dimensionless forms allows us to generalize the problem. For example, by converting actual temperature into a dimensionless form \(\theta^{*} = \frac{T - T_{\infty}}{T_i - T_{\infty}}\), we can focus on the relative change in temperature rather than its absolute value.
This approach is key for solving the heat equation for both phases (heating and cooling) without redoing the entire analysis from scratch. It enables engineers and scientists to predict heat transfer behaviors effectively across different scenarios by focusing on dimensionless parameters like the Biot and Fourier numbers.
Biot Number
The Biot number \(Bi\) is a dimensionless quantity that provides insight into the heat conduction characteristics within a material. It is defined as the ratio of the conduction resistance within a material to the convection resistance at its surface. Mathematically, it is expressed as\( Bi = \frac{hL}{k}\), where
  • \(h\) is the heat transfer coefficient,
  • \(L\) is a characteristic length,
  • \(k\) is the thermal conductivity of the material.
A Biot number much less than 1 indicates that the thermal resistance is higher in the fluid medium than within the solid. This scenario suggests that the material's temperature is relatively uniform throughout its volume. Conversely, a Biot number much greater than 1 implies significant temperature gradients within the material. In the exercise,

we observed \(Bi_1 = 10\) indicating high resistance to heat flow within the material during the initial heating, and \(Bi_2 = 1\) indicating a more moderate distribution during cooling. Understanding these values assists in designing better cooling or heating processes in applications such as metallurgical treatments and thermal management systems.
Fourier Number
The Fourier number \(Fo\) is another crucial dimensionless parameter in transient heat conduction. It represents the ratio of heat conduction over a material's surface to the stored heat within that material. It is given by the formula \(Fo = \frac{\alpha t}{L^2}\), where:
  • \(\alpha\) is the thermal diffusivity (ratio of \(k\) to product of density and specific heat capacity),
  • \(t\) is the time, and
  • \(L\) is the characteristic length.
A higher Fourier number demonstrates that heat has penetrated more deeply into the material, whereas a lower Fourier number indicates that the heat is still near the surface.
In this exercise, the Fourier number helps define how quickly the heat propagates through the material from the moment of start (at Phase 1) until the maximum dimensionless time \(Fo\). Monitoring its progression helps observe the thermal response and adjust processes to optimize for efficiency and safety in real-world applications.
Transient Thermal Response
Understanding the transient thermal response of a material involves tracking how its temperature changes over time when exposed to different thermal conditions. This is particularly important when dealing with non-steady-state or time-dependent problems, where conditions such as the Biot and Fourier numbers influence temperature distribution.
The finite-difference method used in the exercise is a numerical approach to handle spatial and temporal discretization, allowing for the estimation of temperature at specific points \(x^{*}=0\), \(x^{*}=0.5\), and \(x^{*}=1\) during heating and cooling phases.
  • At \(x^{*}=0\), the response tells us about the midplane's behavior over time.
  • At \(x^{*}=1\), it reflects how the surface adapts to external conditions.
  • At \(x^{*}=0.5\), it provides information halfway through the wall’s thickness.
By plotting these points against dimensionless time, we gain invaluable insights into thermal conductivity and internal resistance of materials, guiding effective engineering solutions for heat treatment and thermal systems.

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Most popular questions from this chapter

Thermal energy storage systems commonly involve a packed bed of solid spheres, through which a hot gas flows if the system is being charged, or a cold gas if it is being discharged. In a charging process, heat transfer from the hot gas increases thermal energy stored within the colder spheres; during discharge, the stored energy decreases as heat is transferred from the warmer spheres to the cooler gas. Consider a packed bed of \(75-\mathrm{mm}\)-diameter aluminum spheres \(\left(\rho=2700 \mathrm{~kg} / \mathrm{m}^{3}, c=950 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}, k=\right.\) \(240 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) ) and a charging process for which gas enters the storage unit at a temperature of \(T_{g, i}=300^{\circ} \mathrm{C}\). If the initial temperature of the spheres is \(T_{i}=25^{\circ} \mathrm{C}\) and the convection coefficient is \(h=75 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), how long does it take a sphere near the inlet of the system to accumulate \(90 \%\) of the maximum possible thermal energy? What is the corresponding temperature at the center of the sphere? Is there any advantage to using copper instead of aluminum?

A sphere \(30 \mathrm{~mm}\) in diameter initially at \(800 \mathrm{~K}\) is quenched in a large bath having a constant temperature of \(320 \mathrm{~K}\) with a convection heat transfer coefficient of \(75 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The thermophysical properties of the sphere material are: \(\rho=400 \mathrm{~kg} / \mathrm{m}^{3}, c=1600 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), and \(k=1.7 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). (a) Show, in a qualitative manner on \(T \dashv\) coordinates, the temperatures at the center and at the surface of the sphere as a function of time. (b) Calculate the time required for the surface of the sphere to reach \(415 \mathrm{~K}\). (c) Determine the heat flux \(\left(\mathrm{W} / \mathrm{m}^{2}\right)\) at the outer surface of the sphere at the time determined in part (b). (d) Determine the energy (J) that has been lost by the sphere during the process of cooling to the surface temperature of \(415 \mathrm{~K}\). (e) At the time determined by part (b), the sphere is quickly removed from the bath and covered with perfect insulation, such that there is no heat loss from the surface of the sphere. What will be the temperature of the sphere after a long period of time has elapsed? (f) Compute and plot the center and surface temperature histories over the period \(0 \leq t \leq 150 \mathrm{~s}\). What effect does an increase in the convection coefficient to \(h=200 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) have on the foregoing temperature histories? For \(h=75\) and \(200 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), compute and plot the surface heat flux as a function of time for \(0 \leq t \leq 150 \mathrm{~s}\).

A spherical vessel used as a reactor for producing pharmaceuticals has a 5 -mm-thick stainless steel wall \((k=17 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) and an inner diameter of \(D_{i}=1.0 \mathrm{~m}\). During production, the vessel is filled with reactants for which \(\rho=1100 \mathrm{~kg} / \mathrm{m}^{3}\) and \(c=2400 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), while exothermic reactions release energy at a volumetric rate of \(\dot{q}=10^{4} \mathrm{~W} / \mathrm{m}^{3}\). As first approximations, the reactants may be assumed to be well stirred and the thermal capacitance of the vessel may be neglected. (a) The exterior surface of the vessel is exposed to ambient air \(\left(T_{\infty}=25^{\circ} \mathrm{C}\right)\) for which a convection coefficient of \(h=6 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) may be assumed. If the initial temperature of the reactants is \(25^{\circ} \mathrm{C}\), what is the temperature of the reactants after \(5 \mathrm{~h}\) of process time? What is the corresponding temperature at the outer surface of the vessel? (b) Explore the effect of varying the convection coefficient on transient thermal conditions within the reactor.

Steel-reinforced concrete pillars are used in the construction of large buildings. Structural failure can occur at high temperatures due to a fire because of softening of the metal core. Consider a 200 -mm-thick composite pillar consisting of a central steel core \((50 \mathrm{~mm}\) thick) sandwiched between two 75 -mm-thick concrete walls. The pillar is at a uniform initial temperature of \(T_{i}=\) \(27^{\circ} \mathrm{C}\) and is suddenly exposed to combustion products at \(T_{\infty}=900^{\circ} \mathrm{C}, h=40 \mathrm{~W} / \mathrm{m}^{2}+\mathrm{K}\) on both exposed surfaces. The surroundings temperature is also \(900^{\circ} \mathrm{C}\). (a) Using an implicit finite difference method with \(\Delta x=10 \mathrm{~mm}\) and \(\Delta t=100 \mathrm{~s}\), determine the temperature of the exposed concrete surface and the center of the steel plate at \(t=10,000 \mathrm{~s}\). Steel properties are: \(k_{s}=55 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \rho_{s}=7850 \mathrm{~kg} / \mathrm{m}^{3}\), and \(c_{s}=450 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\). Concrete properties are: \(k_{c}=1.4 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \rho_{c}=2300 \mathrm{~kg} / \mathrm{m}^{3}, c_{c}=880 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\) and \(\varepsilon=0.90\). Plot the maximum and minimum concrete temperatures along with the maximum and minimum steel temperatures over the duration \(0 \leq t \leq 10,000 \mathrm{~s} .\) (b) Repeat part (a) but account for a thermal contact resistance of \(R_{t, c}^{\prime}=0.20 \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\) at the concretesteel interface. (c) At \(t=10,000 \mathrm{~s}\), the fire is extinguished, and the surroundings and ambient temperatures return to \(T_{\infty}=T_{\text {sur }}=27^{\circ} \mathrm{C}\). Using the same convection heat transfer coefficient and emissivity as in parts (a) and (b), determine the maximum steel temperature and the critical time at which the maximum steel temperature occurs for cases with and without the contact resistance. Plot the concrete surface temperature, the concrete temperature adjacent to the steel, and the steel temperatures over the duration \(10,000 \leq t \leq 20,000 \mathrm{~s} .\)

Consider the fuel element of Example \(5.11\), which operates at a uniform volumetric generation rate of \(\dot{q}_{1}=10^{7} \mathrm{~W} / \mathrm{m}^{3}\) until the generation rate suddenly changes to \(\dot{q}_{2}=2 \times 10^{7} \mathrm{~W} / \mathrm{m}^{3}\). Use the finite-element software \(F E H T\) to obtain the following solutions. (a) Calculate the temperature distribution \(1.5 \mathrm{~s}\) after the change in operating power and compare your results with those tabulated in the example. Hint: First determine the steady-state temperature distribution for \(\dot{q}_{1}\), which represents the initial condition for the transient temperature distribution after the step change in power to \(\dot{q}_{2}\). Next, in the Setup menu, click on Transient: in the Specify/Internal Generation box, change the value to \(\dot{q}_{2}\); and in the Run command, click on Continue (not Calculate). See the Run menu in the FEHT Help section for background information on the Continue option. (b) Use your \(F E H T\) model to plot temperature histories at the midplane and surface for \(0 \leq t \leq 400 \mathrm{~s}\). What are the steady-state temperatures, and approximately how long does it take to reach the new equilibrium condition after the step change in operating power?

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