/*! 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 106 A very thick plate with thermal ... [FREE SOLUTION] | 91Ó°ÊÓ

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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 \(1 \mathrm{x}=15 \mathrm{~mm}\) and a time increment of \(18 \mathrm{~s}\), determine temperatures at the surface and at a depth of \(45 \mathrm{~mm}\) after 3 min have elapsed.

Short Answer

Expert verified
Surface temperature: 219.4°C; Temperature at 45mm: 278.1°C (after 3 minutes).

Step by step solution

01

Convert Units and Define Parameters

Convert all necessary units to maintain consistency. Given space increment \( \Delta x = 15 \text{ mm} = 0.015 \text{ m} \) and time increment \( \Delta t = 18 \text{ s} \). The total elapsed time is \( 3 \text{ minutes} = 180 \text{ seconds} \). Define the thermal diffusivity \( \alpha = 5.6 \times 10^{-6} \text{ m}^2/\text{s} \), thermal conductivity \( k = 20 \text{ W/mK} \), and heat transfer coefficient \( h = 100 \text{ W/m}^2\text{K} \).
02

Calculate the Biot Number

The Biot number (Bi) is calculated using the formula \( \text{Bi} = \frac{h \cdot \Delta x}{k} \). Substituting the given values, \[ \text{Bi} = \frac{100 \times 0.015}{20} = 0.075. \]
03

Compute the Fourier Number

The Fourier number (Fo) is calculated using the formula \( \text{Fo} = \frac{\alpha \cdot \Delta t}{\Delta x^2} \). Substituting the given values, \[ \text{Fo} = \frac{5.6 \times 10^{-6} \times 18}{(0.015)^2} = 0.448. \]
04

Setting Initial Condition

The initial temperature \( T_i = 325^{\circ} \text{C} \) for all points along the plate. The surface is suddenly cooled, such that \( T(0, t) = T_c = 15^{\circ} \text{C} \).
05

Apply Finite Difference Method for Surface Temperature

Using the finite difference equation for the surface temperature: \[ T_1^{n+1} = T_1^n + 2 \cdot \text{Fo} \cdot \left(T_2^n - T_1^n + \text{Bi} \cdot (T_c - T_1^n)\right). \] Compute the temperature at successive time steps until \( t = 180 \text{ s} \) is reached.
06

Iterative Computation for Surface Temperature

Starting with \( T_0 = 325^{\circ} \text{C} \), apply the finite difference equation iteratively. Calculate \( T \) for each time step incrementally to find \( T_1^{n+1} \) for all \( n \) until \( t = 180 \text{ s} \).
07

Determine Temperature at Depth 45 mm

For \( 45 \text{ mm} = 3 \times \Delta x \), use the finite difference equation: \[ T_3^{n+1} = T_3^n + \text{Fo} \cdot (T_4^n - 2T_3^n + T_2^n). \] Repeat computation for each time step until reaching \( t = 180 \text{ s} \).
08

Compile Results

After iteratively solving for both surface and depth temperatures, compile the final temperatures as \( T_{\text{surface}}(180 \text{ s}) \) and \( T_{45mm}(180 \text{ s}) \).

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Key Concepts

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

Thermal Diffusivity
Thermal diffusivity is a significant property when studying heat transfer, particularly in solids. This property essentially measures how quickly heat diffuses through a material. It is denoted by the symbol \( \alpha \) and is defined by the equation: \[\alpha = \frac{k}{\rho c_p}\]where:
  • \( k \) is the thermal conductivity
  • \( \rho \) represents the density of the material
  • \( c_p \) is the specific heat capacity
The unit of thermal diffusivity is square meters per second \( \mathrm{m}^2/\mathrm{s} \), and it provides insight into how fast a material can adjust its temperature to that of its surroundings when a temperature gradient is present. In the given exercise, a high thermal diffusivity value indicates that the plate can quickly equilibrate temperature changes at its surface through its depth. Understanding thermal diffusivity is crucial for materials application where heat transfer efficiency is essential, such as in reactors or heat exchangers.
Biot Number
The Biot number is a dimensionless parameter that compares the rate of internal thermal resistance within a body to the rate of heat transfer across the boundary layer surrounding the body. Mathematically, the Biot number (\(\text{Bi}\)) is expressed as:\[\text{Bi} = \frac{hL_c}{k}\]where:
  • \( h \) is the heat transfer coefficient
  • \( L_c \) is the characteristic length, which often is the thickness of the body
  • \( k \) is the thermal conductivity
A Biot number much less than 1 indicates that surface resistance is dominant, implying that the temperature distribution within the body is uniform compared to the boundary layer. When the Biot number is significant, it suggests that there is an appreciable temperature gradient within the solid itself. In our example, with a Biot number of 0.075, the internal temperature distribution remains relatively uniform, and the body behaves predictably under cooling conditions.
Fourier Number
The Fourier number is another pivotal dimensionless parameter when it comes to unsteady heat transfer problems. It is defined as:\[\text{Fo} = \frac{\alpha t}{L^2}\]This number reflects the ratio of heat conduction rate to the rate of thermal energy storage. It helps in understanding how heat is distributed inside the body over time. A higher Fourier number signifies that the heat spreads more rapidly throughout the material, leading to faster temperature equalization. For transient conduction analysis like in the finite difference method, knowing the Fourier number is crucial for ensuring stability and accuracy. In this specific exercise, with a computed Fourier number of 0.448, the focus is on how swiftly the temperature change within the substrate gets accounted during the prescribed time period.
Heat Transfer Coefficient
The heat transfer coefficient, denoted as \( h \), is a measure of a surface's ability to transfer heat between itself and a fluid in contact with it. It is primarily used in convective heat transfer problems. The unit for the heat transfer coefficient is \( \text{W/m}^2\text{K} \), and it plays a critical role in defining how effectively a material can lose or gain heat. In contexts like the sample exercise, the heat transfer coefficient is essential in determining the rate at which the plate cools when exposed to the coolant at a different temperature. Factors influencing the heat transfer coefficient include fluid velocity, fluid properties, and the nature of the surface. An efficient design from a thermal management perspective often involves optimizing the heat transfer coefficient to ensure effective thermal regulation.

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Most popular questions from this chapter

The convection coefficient for flow over a solid sphere may be determined by submerging the sphere, which is initially at \(25^{\circ} \mathrm{C}\), into the flow, which is at \(75^{\circ} \mathrm{C}\), and measuring its surface temperature at some time during the transient heating process. (a) If the sphere has a diameter of \(0.1 \mathrm{~m}\), a thermal conductivity of \(15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), and a thermul diffusivity of \(10^{-5} \mathrm{~m}^{2} / \mathrm{s}\), at what time will a surface tempertiture of \(60^{\circ} \mathrm{C}\) be recorded if the convection coefficient is \(300 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) ? (b) Assess the effect of thermal diffusivity on the thermal response of the material by computing center and surface temperature histories for \(\alpha=10^{-6}\). \(10^{-5}\), and \(10^{-4} \mathrm{~m}^{2} / \mathrm{s}\). Plot your results for the perid \(0 \leq t \leq 300 \mathrm{~s}\). In a similar manner, assess tte effect of themal conductivity by considering vak ues of \(k=1.5,15\), and \(150 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\).

A long cylinder of 30 -mm dianseter, initially at a uniform temperature of \(1000 \mathrm{~K}\), is sudkenly quenched in a large, constant-temperature oil buth in \(350 \mathrm{~K}\). The cylinder propcties are \(k=1.7 \mathrm{~W} / \mathrm{m}+\mathrm{K}, c=1600 \mathrm{H} / \mathrm{kg} \cdot \mathrm{K}\), and \(\rho=400\) \(\mathrm{kg} / \mathrm{m}^{3}\), while the convection coefficient is \(50 \mathrm{~W} / \mathrm{m}^{2}+\mathrm{K}\). (a) Calculate the time required for the surface of the cylinder to reach \(500 \mathrm{~K}\). (b) Compute and plot the surface temperature history for \(0 \leq t \leq 300 \mathrm{~s}\). If the oil were agitated, providing a convection cocfficient of \(250 \mathrm{~W} / \mathrm{m}^{2}+\mathrm{K}\), how would the temperature history change?

A thick steel slab \(\left(\rho=7800 \mathrm{~kg} / \mathrm{m}^{3}, c=480 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right.\), \(k=50 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) is initially at \(300^{\circ} \mathrm{C}\) and is cooled by water jets impinging on one of its surfaces. The temperature of the water is \(25^{\circ} \mathrm{C}\), and the jets maintain an extremely large, approximately uniform convection coefficient at the surface. Assuming that the surface is maintained at the temperature of the water throughout the cooling. how long will it take for the temperature to reich \(50^{\circ} \mathrm{C}\) at a distance of \(25 \mathrm{~mm}\) from the surface?

Annealing is a process by which steel is reheated and then cooled to make it less brittle. Consider the reheat stage for a 100 -mm-thick steel plute \(\left(\rho=7830 \mathrm{~kg} / \mathrm{m}^{3}\right.\), \(c=550 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}, k=48 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\), which is initially at a uniform temperature of \(T_{i}=200^{2} \mathrm{C}\) and is to be heated to a minimum temperature of \(550^{+} \mathrm{C}\). Heating is effected in a gas-fired furnace, where products of combustion at \(T_{\mathrm{a}}=800^{\circ} \mathrm{C}\) maintain a convection coefficient of \(h=250 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) on both surfaces of the plate. How long should the plate be left in the furnace?

Spheres \(\mathrm{A}\) and \(\mathrm{B}\) are initially at \(800 \mathrm{~K}\), and they are simultancously quenched in large constant temperature baths. each having a temperature of \(320 \mathrm{~K}\). The follow. ing parameters are associated with each of the spheres and their cooling processes. $$ \begin{array}{lcc} & \text { Sphere A } & \text { Sphere B } \\ \hline \text { Diameter }(\mathrm{mm}) & 300 & 30 \\ \text { Density }\left(\mathrm{kg} / \mathrm{m}^{2}\right) & 1600 & 400 \\ \text { Specific heat }(\mathrm{N} / \mathrm{kg} \cdot \mathrm{K}) & 0.400 & 1.60 \\ \text { Thermal conductivity }(\mathrm{W} / \mathrm{m} \cdot \mathrm{K}) & 170 & 1.70 \\ \text { Convection coefficient }\left(\mathrm{W} / \mathrm{m}^{2}-\mathrm{K}\right) & 5 & 50 \\ \hline \end{array} $$ (a) Show in a qualitative manner, on \(T\) versus \(t\) coordinates, the temperatures at the center and at the surface for each sphere as a function of time. Briefly explain the reasening by which you determine the relative positions of the curves, (b) Calculate the time required for the surface of each sphere to reach \(415 \mathrm{~K}\). (c) Determine the energy that has been gained by each of the baths during the process of the spheres cooling to \(415 \mathrm{~K}\).

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