/*! 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 108 The stability criterion for the ... [FREE SOLUTION] | 91Ó°ÊÓ

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The stability criterion for the explicit method requires that the coefficient of the \(T_{m}^{p}\) term of the one-dimensional, finite-difference equation be zero or positive. Consider the situation for which the temperatures at the two neighboring nodes \(\left(T_{\mathrm{m}-1}^{p}, T_{\mathrm{m}+1}^{p}\right)\) are \(100^{\circ} \mathrm{C}\) while the center node \(\left(T_{m}^{p}\right)\) is at \(50^{\circ} \mathrm{C}\). Show that for values of \(F o>\frac{1}{2}\) the finite-difference equation will predict a value of \(T_{m}^{p+1}\) that violates the second law of thermodynamics.

Short Answer

Expert verified
When the Fourier number (Fo) is greater than \(\frac{1}{2}\), the explicit finite-difference equation for heat conduction predicts a temperature \(T_{m}^{p+1}\) that violates the second law of thermodynamics, as it will exceed the temperatures of the neighboring nodes. This results in an incorrect prediction of heat flow, as the temperature must fall between the initial temperature and the neighboring nodes' temperatures.

Step by step solution

01

Write the finite-difference equation for heat conduction

The one-dimensional, explicit finite-difference equation for heat conduction is given by: \[ T_{m}^{p+1} = T_{m}^{p} + Fo \left( T_{m+1}^{p} - 2T_{m}^{p} + T_{m-1}^{p} \right) \] Where \(Fo\) is the Fourier number, \(T_{m}^{p}\) is the temperature at the center node at time level p, \(T_{m+1}^{p}\) and \(T_{m-1}^{p}\) are the temperatures at the neighboring nodes at time level p, and \(T_{m}^{p+1}\) is the predicted temperature at the center node for the next time step.
02

Substitute the given temperatures and analyze Fo

Now, we have to substitute the given temperature values into the equation: - \(T_{m}^{p} = 50^{\circ} \mathrm{C}\) - \(T_{m+1}^{p} = T_{m-1}^{p} = 100^{\circ} \mathrm{C}\) We are also given the condition that \(Fo > \frac{1}{2}\). We now substitute these values into the finite-difference equation: \[ T_{m}^{p+1} = 50 + Fo \left( 100 - 2\cdot50 + 100 \right) \]
03

Simplify the equation and analyze the result

Simplify the equation: \[ T_{m}^{p+1} = 50 + Fo \cdot 100 \] Remember that the second law of thermodynamics states that the predicted temperature should be between the initial temperature (50) and the neighboring nodes' temperatures (100): \[50 \leq T_{m}^{p+1} \leq 100\] However, for any \(Fo > \frac{1}{2}\), the predicted temperature \(T_{m}^{p+1}\) will exceed 100: \[ T_{m}^{p+1} = 50 + 100 \cdot \left( \frac{1}{2} \right) = 100\] For \(Fo\) values greater than 1/2, \(T_{m}^{p+1}\) will be greater than 100 and the inequality above will not hold true. Thus, we have shown that if the Fourier number is greater than 1/2, the finite-difference equation used in the explicit method will predict a temperature that violates the second law of thermodynamics.

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Key Concepts

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

Fourier number (Fo)
The Fourier number (Fo) is a dimensionless quantity used in heat conduction analysis to characterize the transient thermal response of a material. It represents the ratio of heat conduction rate to the heat storage rate within a material and is expressed mathematically as:\[ Fo = \frac{\alpha \cdot t}{L^{2}} \]where \(\alpha\) is thermal diffusivity, \(t\) is the time, and \(L\) is the characteristic length. In the context of the explicit finite-difference method, it is a crucial factor that affects the stability and accuracy of the solution. And as we have seen in the original exercise, a value of Fo greater than 1/2 can lead to physically unrealistic results that conflict with the second law of thermodynamics. This demonstrates the importance of selecting appropriate values for \(Fo\) to ensure that the model predictions are both stable and accurate.
Second law of thermodynamics
The second law of thermodynamics is a fundamental principle that dictates the direction of heat transfer, stating that heat flows spontaneously from regions of higher temperature to regions of lower temperature, and not the other way round. It also implies that in a closed system, the total entropy, a measure of disorder or randomness in the system, can never decrease over time.

Within the context of heat conduction and the Fourier number, it is crucial for predicting temperatures to stay true to this law. When the system is in a stable condition, the second law dictates that the node's temperature, such as \(T_{m}^{p+1}\) from our exercise, should not surpass the neighboring nodes' temperatures after heat has been conducted. If the predicted temperature exceeds this limit, it is a clear indication that the model violates the second law, which is not physically possible. This understanding serves as a critical check on any results obtained from thermal analysis and simulation.
Heat conduction
Heat conduction is the process by which thermal energy is transferred from hotter to cooler areas due to the particles' kinetic energy within a material. Conduction occurs in solids, liquids, or gases that are at rest. The heat transfer continues until thermal equilibrium is reached and there is no net movement of thermal energy. In the physics of heat conduction, Fourier's law is one of the fundamental laws which states that the heat transfer rate is proportional to the negative gradient in the temperature and to the area through which the heat is flowing. In mathematical terms:\[ q = -k \cdot A \cdot \frac{dT}{dx} \]where \(q\) is the heat transfer rate, \(k\) is the material's thermal conductivity, \(A\) is the cross-sectional area through which heat is being conducted, and \(dT/dx\) is the temperature gradient. For accurately modeling heat conduction, the explicit finite-difference method may be used, where the temperature at any node is computed based on its own previous temperature and the temperatures of neighboring nodes.
Stability criterion
In numerical analysis, the stability criterion refers to the conditions under which a numerical solution method provides results that are not only correct but also stable over time, without yielding increasing numerical errors or physically unrealistic outcomes. For heat conduction problems solved by the explicit finite-difference method, stability is tied to the Fourier number. A critical limitation, as explored in our exercise, is that for the stability and physical realism of the result, Fo must be constrained to a certain range - in our case, it should not exceed 1/2.

This stability criterion is part of von Neumann stability analysis, which ensures that errors in the numerical solution do not grow uncontrollably over successive time steps. When adhered to, it allows for a stable progression of the temperature field through a material over time, without violating physical laws such as the second law of thermodynamics. Neglecting the stability criterion can lead to nonsensical results, which is why understanding and applying this principle is essential in explicit finite-difference thermal analysis.

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Most popular questions from this chapter

A very thick plate with thermal diffusivity \(5.6 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}\) and thermal conductivity \(20 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) is initially at a uniform temperature of \(325^{\circ} \mathrm{C}\). Suddenly, the surface is exposed to a coolant at \(15^{\circ} \mathrm{C}\) for which the convection heat transfer coefficient is \(100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Using the finite- difference method with a space increment of \(\Delta x=15 \mathrm{~mm}\) and a time increment of \(18 \mathrm{~s}\), determine temperatures at the surface and at a depth of \(45 \mathrm{~mm}\) after \(3 \mathrm{~min}\) have elapsed.

A chip that is of length \(L=5 \mathrm{~mm}\) on a side and thickness \(t=1 \mathrm{~mm}\) is encased in a ceramic substrate, and its exposed surface is convectively cooled by a dielectric liquid for which \(h=150 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(T_{\infty}=20^{\circ} \mathrm{C}\). In the off-mode the chip is in thermal equilibrium with the coolant \(\left(T_{i}=T_{\infty}\right)\). When the chip is energized, however, its temperature increases until a new steady state is established. For purposes of analysis, the energized chip is characterized by uniform volumetric heating with \(\dot{q}=9 \times 10^{6} \mathrm{~W} / \mathrm{m}^{3}\). Assuming an infinite contact resistance between the chip and substrate and negligible conduction resistance within the chip, determine the steady-state chip temperature \(T_{f}\). Following activation of the chip, how long does it take to come within \(1^{\circ} \mathrm{C}\) of this temperature? The chip density and specific heat are \(\rho=2000 \mathrm{~kg} / \mathrm{m}^{3}\) and \(c=700 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), respectively.

In heat treating to harden steel ball bearings \(\left(c=500 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}, \rho=7800 \mathrm{~kg} / \mathrm{m}^{3}, k=50 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\right)\), it is desirable to increase the surface temperature for a short time without significantly warming the interior of the ball. This type of heating can be accomplished by sudden immersion of the ball in a molten salt bath with \(T_{\infty}=1300 \mathrm{~K}\) and \(h=5000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Assume that any location within the ball whose temperature exceeds \(1000 \mathrm{~K}\) will be hardened. Estimate the time required to harden the outer millimeter of a ball of diameter \(20 \mathrm{~mm}\), if its initial temperature is \(300 \mathrm{~K}\).

For each of the following cases, determine an appropriate characteristic length \(L_{c}\) and the corresponding Biot number \(B i\) that is associated with the transient thermal response of the solid object. State whether the lumped capacitance approximation is valid. If temperature information is not provided, evaluate properties at \(T=300 \mathrm{~K}\). (a) A toroidal shape of diameter \(D=50 \mathrm{~mm}\) and cross-sectional area \(A_{c}=5 \mathrm{~mm}^{2}\) is of thermal conductivity \(k=2.3 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). The surface of the torus is exposed to a coolant corresponding to a convection coefficient of \(h=50 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (b) A long, hot AISI 304 stainless steel bar of rectangular cross section has dimensions \(w=3 \mathrm{~mm}\), \(W=5 \mathrm{~mm}\), and \(L=100 \mathrm{~mm}\). The bar is subjected to a coolant that provides a heat transfer coefficient of \(h=15 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) at all exposed surfaces. (c) A long extruded aluminum (Alloy 2024) tube of inner and outer dimensions \(w=20 \mathrm{~mm}\) and \(W=24 \mathrm{~mm}\), respectively, is suddenly submerged in water, resulting in a convection coefficient of \(h=37 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) at the four exterior tube surfaces. The tube is plugged at both ends, trapping stagnant air inside the tube. (d) An \(L=300-m m\)-long solid stainless steel rod of diameter \(D=13 \mathrm{~mm}\) and mass \(M=0.328 \mathrm{~kg}\) is exposed to a convection coefficient of \(h=30 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (e) A solid sphere of diameter \(D=12 \mathrm{~mm}\) and thermal conductivity \(k=120 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) is suspended in a large vacuum oven with internal wall temperatures of \(T_{\text {sur }}=20^{\circ} \mathrm{C}\). The initial sphere temperature is \(T_{i}=100^{\circ} \mathrm{C}\), and its emissivity is \(\varepsilon=0.73\). (f) A long cylindrical rod of diameter \(D=20 \mathrm{~mm}\), density \(\rho=2300 \mathrm{~kg} / \mathrm{m}^{3}\), specific heat \(c_{p}=1750 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), and thermal conductivity \(k=16 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) is suddenly exposed to convective conditions with \(T_{\infty}=20^{\circ} \mathrm{C}\). The rod is initially at a uniform temperature of \(T_{i}=200^{\circ} \mathrm{C}\) and reaches a spatially averaged temperature of \(T=100^{\circ} \mathrm{C}\) at \(t=225 \mathrm{~s}\). (g) Repeat part (f) but now consider a rod diameter of \(D=200 \mathrm{~mm}\).

The heat transfer coefficient for air flowing over a sphere is to be determined by observing the temperature-time history of a sphere fabricated from pure copper. The sphere, which is \(12.7 \mathrm{~mm}\) in diameter, is at \(66^{\circ} \mathrm{C}\) before it is inserted into an airstream having a temperature of \(27^{\circ} \mathrm{C}\). A thermocouple on the outer surface of the sphere indicates \(55^{\circ} \mathrm{C} 69 \mathrm{~s}\) after the sphere is inserted into the airstream. Assume and then justify that the sphere behaves as a spacewise isothermal object and calculate the heat transfer coefficient.

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