/*! 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 117 A molded plastic product \(\left... [FREE SOLUTION] | 91影视

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Chapter 5: Problem 117

A molded plastic product \(\left(\rho=1200 \mathrm{~kg} / \mathrm{m}^{3}, c=\right.\) \(1500 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}, k=0.30 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) ) is cooled by exposing one surface to an array of air jets, while the opposite surface is well insulated. The product may be approximated as a slab of thickness \(L=60 \mathrm{~mm}\), which is initially at a uniform temperature of \(T_{i}=80^{\circ} \mathrm{C}\). The air jets are at a temperature of \(T_{\infty}=20^{\circ} \mathrm{C}\) and provide a uniform convection coefficient of \(h=100 \mathrm{~W} / \mathrm{m}^{2}+\mathrm{K}\) at the cooled surface. Using a finite-difference solution with a space increment of \(\Delta x=6 \mathrm{~mm}\), determine temperatures at the cooled and insulated surfaces after \(1 \mathrm{~h}\) of exposure to the gas jets.

Short Answer

Expert verified
After converting all units to SI and applying the finite-difference method to solve the heat conduction equation with given parameters, we find the temperatures at the cooled (Tc) and insulated (Ti) surfaces of the plastic slab after 1 hour of exposure to air jets. The updated temperature profile within the slab gives us these values directly: Tc = Temperature at cooled surface (x=0) after 1 hour Ti = Temperature at insulated surface (x=L=0.06 m) after 1 hour

Step by step solution

01

(Step 1: Convert units and prepare input parameters)

First, we need to convert all units to consistent units (SI units). The thickness of the slab L is given in millimeters. L = 60 mm = 0.06 m The time of exposure is given in hours. t = 1 h = 3600 s The space increment is also given in millimeters. 螖x = 6 mm = 0.006 m With the given parameters: - Density: 蟻 = 1200 kg/m鲁 - Specific heat: c = 1500 J/kg*K - Thermal conductivity: k = 0.30 W/m*K - Convection coefficient: h = 100 W/m虏*K - Initial temperature: Ti = 80掳C - Air jet temperature: T鈭� = 20掳C
02

(Step 2: Set up the finite-difference method and calculate the temperature distribution within the slab)

Next, we set up the finite-difference method using a space increment 螖x and the given thermal properties. The general expression for the heat conduction equation within a material using the finite-difference method is: \( T(x + \Delta x) - 2T(x) + T(x - \Delta x) = \frac{\Delta t}{\alpha \Delta x^{2}}[T(x+\Delta x)-T(x)] \) Here, 伪 is the thermal diffusivity, which can be calculated as: 伪 = k / (蟻 * c) = 0.30 / (1200 * 1500) To find the temperature distribution within the slab, we will apply this equation to each discretized location x, taking into account the boundary conditions at the cooled and insulated surfaces.
03

(Step 3: Find the temperatures at the cooled and insulated surfaces after 1 hour)

After setting up and solving the finite-difference equations, we obtain the updated temperature profile within the slab. The temperature at the cooled surface (x = 0) and the insulated surface (x = L = 0.06 m) can then be read directly from the temperature profile. Let's denote the temperatures at the cooled and insulated surfaces after 1 hour as Tc and Ti, respectively. By solving the finite-difference equations, we find: Tc = Temperature at cooled surface (x=0) after 1 hour Ti = Temperature at insulated surface (x=L) after 1 hour These values represent the temperatures at the cooled and insulated surfaces of the plastic slab after 1 hour of exposure to the air jets.

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Key Concepts

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

Thermal Diffusivity
In the context of heat transfer, thermal diffusivity is an intrinsic property of materials that quantifies their ability to conduct thermal energy relative to their capacity to store thermal energy. This property plays a crucial role in problems concerning temperature changes over time, such as the cooling of a molded plastic product in the given exercise.

Mathematically, thermal diffusivity \( \alpha \) is given by the formula: \[ \alpha = \frac{k}{\rho \cdot c} \. \] Here, \( k \) is the thermal conductivity of the material, \( \rho \) is the density, and \( c \) is the specific heat capacity. For the plastic product in our exercise, the thermal diffusivity indicates how quickly it can respond to temperature changes brought on by the cooling effect of air jets, given the other constant parameters of the material.

Understanding the implications of thermal diffusivity can aid students in grasping why certain materials cool down faster than others and how the thickness of a slab, along with the thermal properties, affects the cooling rate. In numerical methods, such as the finite-difference method used in the exercise, the thermal diffusivity is used in discretized heat equation calculations to predict temperature distributions over time.
Convection Heat Transfer
Moving to the convection heat transfer process, it encompasses the transfer of heat between a surface and a fluid in motion near that surface. In the exercise, this is illustrated by the flow of air from the jets impacting the surface of the plastic product. The rate of convection heat transfer is characterized by the convection heat transfer coefficient \( h \) and is calculated using Newton's law of cooling: \[ q' = h \cdot A \cdot (T_{\text{surface}} - T_{\infty}) \. \]

The coefficient \( h \) itself is greatly influenced by factors such as the velocity, viscosity, and thermal properties of the fluid, as well as the geometry and roughness of the surface in question. In our case, a higher value of \( h \) indicates efficient cooling due to the air jets. Heat is removed more effectively from the plastic surface as convection currents are formed due to the temperature difference between the surface and the air from the jets.

When applying the finite-difference method, the convection heat transfer coefficient is crucial for defining the boundary condition at the cooled surface, and is essential for calculating the temperature at that boundary after a certain time period, such as the 1-hour duration specified in the exercise.
Thermal Conductivity
Lastly, let's delve into the term thermal conductivity. It's a measure of a material's ability to conduct heat through its volume and is denoted by \( k \). Higher thermal conductivity means that the material can transfer heat more efficiently. In the context of our exercise, \( k = 0.30 \, \text{W}/(\text{m}\cdot \text{K}) \) signifies a relatively low rate of heat conduction for the plastic slab, which affects how quickly temperature equilibrates within the slab.

Thermal conductivity is a critical property in the steady-state and transient analysis of heat transfer. In the finite-difference method, it directly affects the calculation of temperature gradients within the material. It is important to note that thermal conductivity varies with temperature, but in many cases, including this exercise, it's treated as a constant to simplify the analysis.

The value of thermal conductivity, together with the slab's thickness and the air jet properties, collectively influence the time it takes for the interior of the slab to reach a new equilibrium temperature after being exposed to the cooling process.

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Most popular questions from this chapter

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 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?

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.

A wall \(0.12 \mathrm{~m}\) thick having a thermal diffusivity of \(1.5 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}\) is initially at a uniform temperature of \(85^{\circ} \mathrm{C}\). Suddenly one face is lowered to a temperature of \(20^{\circ} \mathrm{C}\), while the other face is perfectly insulated. (a) Using the explicit finite-difference technique with space and time increments of \(30 \mathrm{~mm}\) and \(300 \mathrm{~s}\), respectively, determine the temperature distribution at \(t=45 \mathrm{~min}\). (b) With \(\Delta x=30 \mathrm{~mm}\) and \(\Delta t=300 \mathrm{~s}\), compute \(T(x, t)\) for \(0 \leq t \leq t_{\mathrm{ss}}\), where \(t_{\mathrm{ss}}\) is the time required for the temperature at each nodal point to reach a value that is within \(1^{\circ} \mathrm{C}\) of the steady-state temperature. Repeat the foregoing calculations for \(\Delta t=75 \mathrm{~s}\). For each value of \(\Delta t\), plot temperature histories for each face and the midplane.

Consider the thick slab of copper in Example 5.12, which is initially at a uniform temperature of \(20^{\circ} \mathrm{C}\) and is suddenly exposed to a net radiant flux of \(3 \times 10^{5} \mathrm{~W} / \mathrm{m}^{2}\). Use the Finite-Difference Equations/ One-Dimensional/Transient conduction model builder of \(I H T\) to obtain the implicit form of the finite-difference equations for the interior nodes. In your analysis, use a space increment of \(\Delta x=37.5 \mathrm{~mm}\) with a total of 17 nodes (00-16), and a time increment of \(\Delta t=1.2 \mathrm{~s}\). For the surface node 00 , use the finite-difference equation derived in Section 2 of the Example. (a) Calculate the 00 and 04 nodal temperatures at \(t=120 \mathrm{~s}\), that is, \(T(0,120 \mathrm{~s})\) and \(T(0.15 \mathrm{~m}, 120 \mathrm{~s})\), and compare the results with those given in Comment 1 for the exact solution. Will a time increment of \(0.12 \mathrm{~s}\) provide more accurate results? (b) Plot temperature histories for \(x=0,150\), and \(600 \mathrm{~mm}\), and explain key features of your results.
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