/*! 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 78 Spheres of \(40-\mathrm{mm}\) di... [FREE SOLUTION] | 91影视

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Spheres of \(40-\mathrm{mm}\) diameter heated to a uniform temperature of \(400^{\circ} \mathrm{C}\) are suddenly removed from the oven and placed in a forced-air bath operating at \(25^{\circ} \mathrm{C}\) with a convection coefficient of \(300 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) on the sphere surfaces. The thermophysical properties of the sphere material are \(\rho=3000 \mathrm{~kg} / \mathrm{m}^{3}, c=850 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), and \(k=15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). (a) How long must the spheres remain in the air bath for \(80 \%\) of the thermal energy to be removed? (b) The spheres are then placed in a packing carton that prevents further heat transfer to the environment. What uniform temperature will the spheres eventually reach?

Short Answer

Expert verified
Since the Biot number is greater than 0.1, we cannot use the lumped capacitance method for this problem, and a different approach must be used to solve the question.

Step by step solution

01

Calculate the Biot number of the sphere

First, we need to check if we can use the lumped capacitance method. This method is valid only when the Biot number (Bi) is less than 0.1. The Biot number is given by: \( Bi = \frac{hL_c}{k} \) Where: - h is the convection coefficient (300 W/m虏K) - Lc is the characteristic length (We will consider this as the radius of the sphere, which is 0.02 m) - k is the thermal conductivity of the sphere material (15 W/mK)
02

Calculate the time constant

Next, we need to determine the time constant of the sphere to find the time it takes for 80% of the thermal energy to be removed. The time constant (蟿) is given by: \( 蟿 = \frac{蟻 c L_c}{h} \) Where: - 蟻 is the density of the sphere material (3000 kg/m鲁) - c is the specific heat of the sphere material (850 J/kgK)
03

Calculate the time needed for 80% of energy to be removed

Now, we can use the lumped capacitance method to calculate the time (t) needed for 80% of the thermal energy to be removed. This is given by: \( \frac{T - T_\infty}{T_i - T_\infty} = e^{-t/蟿} \) Where: - T is the temperature of the sphere at time t - T岬 is the initial temperature (400掳C) - T鈭 is the surrounding air temperature (25掳C) - 饾湉 is the time constant calculated in Step 2 We are required to find the time for 80% of the energy to be removed, which corresponds to the temperature reaching 20% of the initial value. The equation becomes: \( 0.2 = e^{-t/蟿} \) Now, we can solve for t.
04

Find the final temperature in the packing carton

The spheres will eventually reach a uniform temperature within the packing carton, where further heat transfer is prevented. To find this final temperature, we first need to find the amount of energy lost while in the air bath. The total energy change within the sphere is given by: \( Q = m c \Delta T \) Where: - m is the mass of the sphere - 螖T is the temperature difference between the initial temperature and the temperature after 80% energy removal Then, we can calculate the final temperature (T鈧) within the packing carton by distributing this energy uniformly: \( T_p = T_i - \frac{Q}{m c} \) Now we have the step-by-step solution for this problem. #Solution#
05

Calculate the Biot number of the sphere

Using the given values, let's calculate the Biot number (Bi): \( Bi = \frac{hL_c}{k} = \frac{300 \times 0.02}{15} = 0.4 \) As Bi > 0.1, we cannot use the lumped capacitance method in this problem.

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Key Concepts

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

Biot Number
The Biot number (Bi) plays a crucial role in heat transfer analysis, particularly when dealing with transient conduction problems. It is a dimensionless quantity that is used to determine the thermal resistance inside a body in relation to the thermal resistance across the surrounding fluid boundary layer.

The Biot number is calculated using the formula:
\[\begin{equation}Bi = \frac{hL_c}{k}\end{equation}\]Where:
  • h represents the convective heat transfer coefficient or convection coefficient (with units of W/m虏K),
  • L_c is a characteristic length of the object, typically, for a sphere, it is the radius,
  • k is the thermal conductivity of the material from which the object is made (with units of W/mK).
If the Biot number is less than 0.1, the temperature within the object can be assumed to be uniform. This is because the internal resistance to heat conduction within the object is much smaller compared to the resistance to convection at the surface. As a result, the lumped capacitance method can be applied for a quick and simplified analysis. However, if the Biot number is larger than 0.1, which is the case in our exercise (since it is calculated to be 0.4), this method cannot be used, and a more detailed analysis accounting for temperature variations within the object is necessary.
Lumped Capacitance Method
The lumped capacitance method simplifies the complex process of transient heat transfer. It essentially assumes that temperature gradients within an object are negligible, which in turn implies that the entire object is at a uniform temperature during the transient heat transfer process.

This method is particularly useful and accurate when an object meets the criterion of having a Biot number less than 0.1. For the lumped capacitance method to apply, the formula for the time constant, \(\tau\), is used:
\[\begin{equation}\tau = \frac{\rho c L_c}{h}\end{equation}\]In this equation:
  • \rho denotes the density of the material,
  • c is the specific heat capacity, and
  • h again stands for the convective heat transfer coefficient.
The time \(t\) required for the temperature change can then be calculated using the transcendental equation involving the natural exponential function:\[\begin{equation}\frac{T - T_\infty}{T_i - T_\infty} = e^{-t/\tau}\end{equation}\]Which can be rearranged to solve for \(t\), when a certain temperature decline is specified. Although in our exercise the Biot number exceeds 0.1 and this method cannot be used, it serves as an essential tool for thermal engineers when the criteria are met.
Convection Coefficient
The convection coefficient, often denoted as \(h\), is a measure of the convective heat transfer between a solid surface and a fluid (such as air or water) that is in motion. It quantifies the amount of heat transferred per unit area per unit temperature difference between the surface and the fluid.

In the given exercise, the convection coefficient is \(300 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), which indicates the efficiency of the forced-air bath to remove heat from the surface of the spheres by convection.

It features prominently in Newton's law of cooling, which describes convective cooling:\[\begin{equation}q = hA(T_s - T_f)\end{equation}\]Here:
  • q is the heat transfer rate,
  • A is the surface area,
  • T_s is the temperature of the solid surface, and
  • T_f is the temperature of the fluid.
The accuracy of heat transfer analysis heavily depends on the correct determination of the convection coefficient, which itself can be influenced by several factors including fluid velocity, fluid properties, and the nature of the surface. In our case, knowing the convection coefficient allows us to firstly acknowledge that lumped capacitance method cannot be applied due to a high Biot number, and secondly, aids in the more complex calculation that must be performed to solve the heat transfer problem.

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Most popular questions from this chapter

Consider the system of Problem \(5.1\) where the temperature of the plate is spacewise isothermal during the transient process. (a) Obtain an expression for the temperature of the plate as a function of time \(T(t)\) in terms of \(q_{o}^{\prime \prime}, T_{\infty}, h\), \(L\), and the plate properties \(\rho\) and \(c\). (b) Determine the thermal time constant and the steady-state temperature for a 12 -mm-thick plate of pure copper when \(T_{\infty}=27^{\circ} \mathrm{C}, h=50 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), and \(q_{o}^{\prime \prime}=5000 \mathrm{~W} / \mathrm{m}^{2}\). Estimate the time required to reach steady-state conditions. (c) For the conditions of part (b), as well as for \(h=100\) and \(200 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), compute and plot the corresponding temperature histories of the plate for \(0 \leq t \leq 2500 \mathrm{~s}\).

Stainless steel (AISI 304) ball bearings, which have uniformly been heated to \(850^{\circ} \mathrm{C}\), are hardened by quenching them in an oil bath that is maintained at \(40^{\circ} \mathrm{C}\). The ball diameter is \(20 \mathrm{~mm}\), and the convection coefficient associated with the oil bath is \(1000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) If quenching is to occur until the surface temperature of the balls reaches \(100^{\circ} \mathrm{C}\), how long must the balls be kept in the oil? What is the center temperature at the conclusion of the cooling period? (b) If 10,000 balls are to be quenched per hour, what is the rate at which energy must be removed by the oil bath cooling system in order to maintain its temperature at \(40^{\circ} \mathrm{C}\) ?

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.

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 \(10,000 \leq t \leq 20,000 \mathrm{~s} .\)

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 \(0 \leq t \leq 150 \mathrm{~s}\).

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