/*! 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 137 A thin circular disk is subjecte... [FREE SOLUTION] | 91Ó°ÊÓ

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A thin circular disk is subjected to induction heating from a coil, the effect of which is to provide a uniform heat generation within a ring section as shown. (a) Derive the transient, finite-difference equation for node \(m\), which is within the region subjected to induction heating. (b) On \(T r\) coordinates sketch, in qualitative manner, the steady-state temperature distribution, identifying important features.

Short Answer

Expert verified
The transient, finite-difference equation for node \(m\) is: \( T_m^{n+1} = T_m^n + \alpha \Delta t \frac{T_{m+1}^n - 2T_m^n + T_{m-1}^n}{\Delta r^2} + Q \Delta t \). In the steady-state, the temperature distribution follows a curve that starts at the highest temperature at \(r = 0\) and gradually decreases as we move radially outward. Important features in the distribution include the location of maximum temperature and the parabolic shape of the curve.

Step by step solution

01

(a) Deriving the transient finite-difference equation for node \(m\)

To derive the transient finite-difference equation for node \(m\), we will first write the general 1D transient heat conduction equation, which is given by: \( \frac{\partial T}{\partial t} = \alpha \frac{\partial^2 T}{\partial r^2} + Q \) where \(T\) is the temperature, \(t\) is the time, \(r\) is the radial distance from the disk center, \(\alpha\) is the thermal diffusivity, and \(Q\) is the uniform heat generation per unit volume. Now, let's discretize the equation using the finite-difference method. We will replace all the derivatives with approximations using finite differences: \( \frac{T_m^{n+1} - T_m^n}{\Delta t} = \alpha \frac{T_{m+1}^n - 2T_m^n + T_{m-1}^n}{\Delta r^2} + Q \) We can rewrite this equation in terms of the unknown \(T_m^{n+1}\) as follows: \( T_m^{n+1} = T_m^n + \alpha \Delta t \frac{T_{m+1}^n - 2T_m^n + T_{m-1}^n}{\Delta r^2} + Q \Delta t \) This is the transient, finite-difference equation for node \(m\).
02

(b) Sketching the steady-state temperature distribution

In the steady-state, there is no more change in the temperature with respect to time, i.e., \( \frac{\partial T}{\partial t} = 0 \). So, the 1D transient heat conduction equation becomes: \( 0 = \alpha \frac{\partial^2 T}{\partial r^2} + Q \) To sketch the temperature distribution, we should integrate this equation twice: 1. Integrating once with respect to \(r\): \( 0 = \alpha \frac{\partial T}{\partial r} + Qr + C_1 \) 2. Integrating again with respect to \(r\): \( 0 = \alpha T + \frac{Qr^2}{2} + C_1r + C_2 \) Now, we can sketch the temperature distribution in a qualitative manner using \(T r\) coordinates. We should note that: - The temperature is expected to be highest at the center of the disk, where the heating effect is the strongest. - The temperature decreases as we move radially outward from the center of the disk since the heating effect becomes weaker. We can draw the curve by starting at the highest temperature at \(r = 0\) and gradually decreasing the temperature as we move radially outward. We should also note important features such as the location of maximum temperature and the specific shape of the curve (e.g., parabolic). Consequently, we have analyzed the given exercise and successfully derived the transient, finite-difference equation for node \(m\) and sketched the steady-state temperature distribution for the thin circular disk subjected to induction heating from a coil.

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Key Concepts

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

Heat Conduction
Heat conduction is a fundamental concept in thermodynamics and heat transfer that describes the process of transfer of heat within a body due to the temperature gradient. This kind of heat movement occurs within materials or between different materials that are in direct contact. In physical terms, heat moves from regions of higher temperature to regions of lower temperature, driven by the thermal energy difference.

During this process, thermal energy is passed from molecule to molecule without any net motion of the material as a whole. It's crucial to understand that conduction is possible in solids, liquids, and gases, but it is most efficient in solids because the molecules are closer together and can transfer energy more easily.

The mathematical description of heat conduction is governed by Fourier's law, which states that the time rate of heat transfer through a material is proportional to the negative gradient of the temperature and the area at right angles to that gradient, through which the heat flows. The general equation for one-dimensional heat conduction is represented by:
\[ q = -k \frac{dT}{dx} \]
where \( q \) is the heat flux, \( k \) is the thermal conductivity, and \( dT/dx \) is the temperature gradient. This principle is crucial for solving problems related to thermal systems, like the transient, finite-difference equation for a disk under induction heating, which involves analyzing how the temperature at a point changes over time due to the balance of heat conduction and internal heat generation.
Induction Heating
Induction heating is a non-contact heating process used to heat conductive materials by subjecting them to an alternating electromagnetic field. It is primarily used in processes where metals need to be melted, hardened, or bonded without direct contact with the heat source.

Here's how it works: An induction coil powered by an AC electric current creates a rapidly alternating magnetic field. When a conductive material is placed within this magnetic field, eddy currents are induced within the material. These eddy currents, in turn, generate heat due to the electrical resistance of the material. This is the fundamental principle behind the uniform heat generation within a ring section of a circular disc as described in the example exercise.

The efficiency of induction heating depends on several factors, such as the properties of the material being heated (like electrical conductivity and magnetic permeability), the frequency of the alternating current in the coil, and the strength of the magnetic field. Induction heating allows for precise control over the heating zone and quick heating rates, which are desirable features in many industrial applications. Furthermore, the mathematical representation involving the uniform heat generation per unit volume, \( Q \), is intrinsic to modeling the transient temperature changes in the material under study.
Steady-State Temperature Distribution
Steady-state temperature distribution refers to a condition where the temperature within a material does not change with time. This state is achieved when the amount of heat entering a section of the material is equal to the amount of heat leaving it, resulting in an equilibrium condition.

In the context of heat transfer, reaching a steady-state means that the system has become thermally stable and there is no longer an internal net flow of thermal energy. This concept is essential when analyzing systems designed to operate for extended periods, as it allows prediction of component temperatures and assessment of whether materials will withstand thermal stresses over time.

For instance, the steady-state distribution in a circular disk subjected to induction heating shows a temperature gradient from the center, where heat generation is at its maximum, to the edges, where it dissipates to the surroundings. Mathematically, the steady-state condition for radial heat conduction in a disk can be determined from the equation:\[ 0 = \alpha \frac{\partial^2 T}{\partial r^2} + Q \]
By integrating the above equation, we can achieve an expression for temperature as a function of radial position. This theoretical understanding aids in designing efficient thermal systems, such as heat exchangers, insulation, and electronic devices, ensuring they can sustain steady-state operation without overheating or failure.

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Most popular questions from this chapter

Consider the fuel element of Example 5.11. Initially, the element is at a uniform temperature of \(250^{\circ} \mathrm{C}\) with no heat generation. Suddenly, the element is inserted into the reactor core, causing a uniform volumetric heat generation rate of \(\dot{q}=10^{8} \mathrm{~W} / \mathrm{m}^{3}\). The surfaces are convectively cooled with \(T_{\infty}=250^{\circ} \mathrm{C}\) and \(h=1100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Using the explicit method with a space increment of \(2 \mathrm{~mm}\), determine the temperature distribution \(1.5 \mathrm{~s}\) after the element is inserted into the core.

In a manufacturing process, long rods of different diameters are at a uniform temperature of \(400^{\circ} \mathrm{C}\) in a curing oven, from which they are removed and cooled by forced convection in air at \(25^{\circ} \mathrm{C}\). One of the line operators has observed that it takes \(280 \mathrm{~s}\) for a \(40-\mathrm{mm}\) diameter rod to cool to a safe-to-handle temperature of \(60^{\circ} \mathrm{C}\). For an equivalent convection coefficient, how long will it take for an 80 -mm-diameter rod to cool to the same temperature? The thermophysical properties of the rod are \(\rho=2500 \mathrm{~kg} / \mathrm{m}^{3}, c=900 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), and \(k=15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). Comment on your result. Did you anticipate this outcome?

A metal sphere of diameter \(D\), which is at a uniform temperature \(T_{i}\), is suddenly removed from a furnace and suspended from a fine wire in a large room with air at a uniform temperature \(T_{\infty}\) and the surrounding walls at a temperature \(T_{\text {sur }}\) (a) Neglecting heat transfer by radiation, obtain an expression for the time required to cool the sphere to some temperature \(T\). (b) Neglecting heat transfer by convection, obtain an expression for the time required to cool the sphere to the temperature \(T\). (c) How would you go about determining the time required for the sphere to cool to the temperature \(T\) if both convection and radiation are of the same order of magnitude? (d) Consider an anodized aluminum sphere ( \(\varepsilon=0.75\) ) \(50 \mathrm{~mm}\) in diameter, which is at an initial temperature of \(T_{i}=800 \mathrm{~K}\). Both the air and surroundings are at \(300 \mathrm{~K}\), and the convection coefficient is \(10 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). For the conditions of parts (a), (b), and (c), determine the time required for the sphere to cool to \(400 \mathrm{~K}\). Plot the corresponding temperature histories. Repeat the calculations for a polished aluminum sphere ( \(\varepsilon=0.1)\).

Small spherical particles of diameter \(D=50 \mu \mathrm{m}\) contain a fluorescent material that, when irradiated with white light, emits at a wavelength corresponding to the material's temperature. Hence the color of the particle varies with its temperature. Because the small particles are neutrally buoyant in liquid water, a researcher wishes to use them to measure instantaneous local water temperatures in a turbulent flow by observing their emitted color. If the particles are characterized by a density, specific heat, and thermal conductivity of \(\rho=999 \mathrm{~kg} / \mathrm{m}^{3}\), \(k=1.2 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), and \(c_{p}=1200 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), respectively, determine the time constant of the particles. Hint: Since the particles travel with the flow, heat transfer between the particle and the fluid occurs by conduction. Assume lumped capacitance behavior.

Thermal stress testing is a common procedure used to assess the reliability of an electronic package. Typically, thermal stresses are induced in soldered or wired connections to reveal mechanisms that could cause failure and must therefore be corrected before the product is released. As an example of the procedure, consider an array of silicon chips \(\left(\rho_{\text {ch }}=2300 \mathrm{~kg} / \mathrm{m}^{3}\right.\), \(\left.c_{\text {ch }}=710 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) joined to an alumina substrate \(\left(\rho_{\mathrm{sb}}=4000 \mathrm{~kg} / \mathrm{m}^{3}, c_{\mathrm{sb}}=770 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) by solder balls \(\left(\rho_{\mathrm{sd}}=11,000 \mathrm{~kg} / \mathrm{m}^{3}, c_{\mathrm{sd}}=130 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\). Each chip of width \(L_{\mathrm{ch}}\) and thickness \(t_{\mathrm{ch}}\) is joined to a unit substrate section of width \(L_{\mathrm{sb}}\) and thickness \(t_{\mathrm{sb}}\) by solder balls of diameter \(D\). A thermal stress test begins by subjecting the multichip module, which is initially at room temperature, to a hot fluid stream and subsequently cooling the module by exposing it to a cold fluid stream. The process is repeated for a prescribed number of cycles to assess the integrity of the soldered connections. (a) As a first approximation, assume that there is negligible heat transfer between the components (chip/solder/substrate) of the module and that the thermal response of each component may be determined from a lumped capacitance analysis involving the same convection coefficient \(h\). Assuming no reduction in surface area due to contact between a solder ball and the chip or substrate, obtain expressions for the thermal time constant of each component. Heat transfer is to all surfaces of a chip, but to only the top surface of the substrate. Evaluate the three time constants for \(L_{\mathrm{ch}}=15 \mathrm{~mm}\), \(t_{\mathrm{ch}}=2 \mathrm{~mm}, L_{\mathrm{sb}}=25 \mathrm{~mm}, t_{\mathrm{sb}}=10 \mathrm{~mm}, D=2 \mathrm{~mm}\), and a value of \(h=50 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), which is characteristic of an airstream. Compute and plot the temperature histories of the three components for the heating portion of a cycle, with \(T_{i}=20^{\circ} \mathrm{C}\) and \(T_{\infty}=80^{\circ} \mathrm{C}\). At what time does each component experience \(99 \%\) of its maximum possible temperature rise, that is, \(\left(T-T_{i}\right) /\left(T_{\infty}-T_{i}\right)=0.99\) ? If the maximum stress on a solder ball corresponds to the maximum difference between its temperature and that of the chip or substrate, when will this maximum occur? (b) To reduce the time required to complete a stress test, a dielectric liquid could be used in lieu of air to provide a larger convection coefficient of \(h=200 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). What is the corresponding savings in time for each component to achieve \(99 \%\) of its maximum possible temperature rise?

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