/*! 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 72 A cold air chamber is proposed f... [FREE SOLUTION] | 91影视

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Chapter 5: Problem 72

A cold air chamber is proposed for quenching steel ball bearings of diameter \(D=0.2 \mathrm{~m}\) and initial temperature \(T_{i}=400^{\circ} \mathrm{C} .\) Air in the chamber is maintained at \(-15^{\circ} \mathrm{C}\) by a refrigeration system, and the steel balls pass through the chamber on a conveyor belt. Optimum bearing production requires that \(70 \%\) of the initial thermal energy content of the ball above \(-15^{\circ} \mathrm{C}\) be removed. Radiation effects may be neglected, and the convection heat transfer coefficient within the chamber is \(1000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Estimate the residence time of the balls within the chamber, and recommend a drive velocity of the conveyor. The following properties may be used for the steel: \(k=50 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \alpha=2 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}\), and \(c=450 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\).

Short Answer

Expert verified
The residence time of the steel ball bearings within the cold air chamber is 106.4 seconds, and the recommended drive velocity of the conveyor belt is 0.00188 m/s.

Step by step solution

01

1. Calculate the temperature difference required to remove 70% of the initial thermal energy content from the steel balls.

To do this, we need to find the initial and final temperature by converting them into Kelvin (K). The final temperature after 70% of the thermal energy content removal can be found by using the steel properties such as k, 伪, and c.
02

2. Apply the convection heat transfer equation.

The convection heat transfer equation is given by: \[Q = h \cdot A \cdot \Delta T\] Where Q is the heat transfer, h is the convection heat transfer coefficient, A is the surface area of the steel balls, and 螖T is the temperature difference.
03

3. Calculate the required heat transfer (Q).

To find the required heat transfer for the 70% energy removal, we can use the following equation: \[Q = m \cdot c \cdot \Delta T\] Where m is the mass of the steel balls, c is the specific heat capacity of the steel, and 螖T is the temperature difference found in step 1.
04

4. Determine the residence time within the chamber.

Divide the required heat transfer (Q) by the heat transfer rate (h 路 A 路 螖T) to find the residence time (t) within the chamber. \[t = \frac{Q}{h \cdot A \cdot \Delta T}\]
05

5. Recommend a drive velocity for the conveyor belt.

The residence time found in step 4 can be used to determine the drive velocity (v) of the conveyor belt using the following equation: \[v = \frac{D}{t}\] This will give us the required drive velocity to achieve the optimum bearing production. Now, let's put all the information together and calculate the residence time and drive velocity. 1. Initial temperature, Ti: \(400^\circ C = 673 K\) (convert to K) Final temperature, Tf: \(-15^\circ C = 258 K\) (convert to K) To remove 70% of the initial thermal energy content, the temperature difference required is: \(\Delta T_r = 0.7 (T_i - T_f) = 0.7 (673 - 258) K = 291.5 K\) 2. Convection heat transfer equation: \(Q = h \cdot A \cdot \Delta T\) 3. Calculate the required heat transfer (Q): We find the mass (m) of the steel ball bearing using the formula \(m = \frac{4}{3} \pi r^3 \rho\), where r is the radius and 蟻 is the density of the steel. Given 伪 = 2 脳 10鈦烩伒 m虏/s, the thermal diffusivity, we can find the density by using the formula 蟻 = \(\frac{k}{c \alpha}\). Now calculate the mass: \[\rho = \frac{50 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}}{450 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K} \cdot 2\times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}} = 5555.56 \mathrm{~kg}/\mathrm{m}^3\] Mass (m): \(m = \frac{4}{3} \pi (0.1)鲁 \cdot 5555.56 = 92.31 \mathrm{~kg}\) Therefore, the required heat transfer (Q) is: \[Q = m \cdot c \cdot \Delta T_r = 92.31 \cdot 450 \cdot 291.5 = 1.208 \times 10^7 \mathrm{~J}\] 4. Determine the residence time: First, we need the surface area of the steel ball (A): \(A = 4 \pi r^2 = 4 \pi (0.1)^2 = 0.1257 \mathrm{~m}^2\). Now, using the convection heat transfer equation, we can find the residence time: \[t = \frac{Q}{h \cdot A \cdot \Delta T} = \frac{1.208 \times 10^7 \mathrm{~J}}{1000 \mathrm{~W}/\mathrm{m}^2 \cdot 0.1257 \mathrm{~m}^2 \cdot 291.5 \mathrm{~K} } = 106.4 \mathrm{~s}\] 5. Recommended drive velocity for the conveyor belt: \[v = \frac{D}{t} = \frac{0.2 \mathrm{~m}}{106.4 \mathrm{~s}} = 0.00188 \mathrm{~m/s}\] So, the residence time of the steel ball bearings within the cold air chamber is 106.4 seconds, and the recommended drive velocity of the conveyor belt is 0.00188 m/s.

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Key Concepts

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

Convection Heat Transfer Coefficient
Understanding the convection heat transfer coefficient is critical when assessing how heat is exchanged between a solid surface and a fluid. This coefficient, represented by the symbol 'h', measures the efficiency of heat transfer in a convection process. It is typically measured in watts per square meter per Kelvin \( W/m^2 \bullet K \).
For our exercise, the convection heat transfer coefficient within the cold air chamber is \(1000 W/m^2 \bullet K\). This value indicates the amount of heat transferred per unit area per unit temperature difference between the steel ball bearings and the air. A higher 'h' value would mean a more efficient heat transfer, enabling the steel balls to cool faster. Understanding 'h' is important for engineers and scientists when designing systems that involve cooling processes, such as the quenching of steel bearings in this scenario.
To improve understanding, consider a pot of water boiling on a stove. The water receives heat from the pot's surface. If the stove transfers heat effectively, we can see this as a high 'h' value, leading to a rapid increase in water temperature. In the textbook exercise, a high 'h' value means that the cold air chamber efficiently removes heat from the bearings, critical for optimizing the quenching process.
Thermal Energy Content
The concept of thermal energy content relates to the total internal energy possessed by an object due to the random movement of its molecules. In our provided textbook exercise, we are interested in the thermal energy content of steel ball bearings at a high temperature.
To calculate changes in thermal energy content, we typically use the formula \( Q = mc\triangle T \), where 'Q' represents the heat transfer, 'm' is the mass, 'c' is the specific heat capacity, and \(\triangle T\) is the temperature change. In the context of the exercise, the goal is to remove 70% of the initial thermal energy content of the steel balls above \( -15^\circ C \). By doing this, the exercise simulates a realistic industrial scenario where controlling the temperature and thermal energy of materials is crucial for achieving desired material properties.

Practical Real-World Connection

Imagine you are heating a pizza in an oven. The pizza gains thermal energy until it reaches the desired temperature. Similarly, the steel ball bearings start with a high thermal energy that must be reduced. Understanding how to calculate this energy helps us control processes like hardening steel, cooking food, or even managing the temperature of a room.
Residence Time Calculation
Residence time calculation is the determination of the time required for an object to undergo a specific process. In the context of our steel ball bearing exercise, residence time refers to the duration that the bearings must remain in the cold air chamber to achieve the desired reduction in temperature.
To calculate the residence time, \( 't' \), you divide the required thermal energy to be removed, \( 'Q' \), by the product of the convection heat transfer coefficient, \( 'h' \), the surface area of the object, \( 'A' \), and the temperature difference, \( '\triangle T' \). This gives us the formula \( t = \frac{Q}{h \bullet A \bullet \triangle T} \). Having the residence time helps in setting the conveyor belt's speed to ensure that each ball bearing spends sufficient time in the cooling zone without unnecessarily slowing down production.

Applied Example

Consider a car paint shop where the freshly painted cars need to dry. The residence time would be how long the cars need to stay in the drying oven. By calculating it accurately, the shop can optimize the drying process, just like optimizing the cooling process in the steel ball bearings exercise. These calculations ensure efficiency and quality in manufacturing processes.

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Most popular questions from this chapter

A constant-property, one-dimensional plane wall of width \(2 L\), at an initial uniform temperature \(T_{i}\), is heated convectively (both surfaces) with an ambient fluid at \(T_{\infty}=T_{\infty, 1}, h=h_{1}\). At a later instant in time, \(t=t_{1}\), heating is curtailed, and convective cooling is initiated. Cooling conditions are characterized by \(T_{\infty}=T_{\infty, 2}=T_{i}, h=h_{2}\) (a) Write the heat equation as well as the initial and boundary conditions in their dimensionless form for the heating phase (Phase 1). Express the equations in terms of the dimensionless quantities \(\theta^{*}, x^{*}, B i_{1}\), and \(F o\), where \(B i_{1}\) is expressed in terms of \(h_{1}\). (b) Write the heat equation as well as the initial and boundary conditions in their dimensionless form for the cooling phase (Phase 2). Express the equations in terms of the dimensionless quantities \(\theta^{*}, x^{*}, B i_{2}\), \(\mathrm{Fo}_{1}\), and \(\mathrm{Fo}\) where \(\mathrm{Fo}_{1}\) is the dimensionless time associated with \(t_{1}\), and \(B i_{2}\) is expressed in terms of \(h_{2}\). To be consistent with part (a), express the dimensionless temperature in terms of \(T_{\infty}=T_{\infty, 1^{*}}\) (c) Consider a case for which \(B i_{1}=10, B i_{2}=1\), and \(F o_{1}=0.1\). Using a finite-difference method with \(\Delta x^{*}=0.1\) and \(\Delta F o=0.001\), determine the transient thermal response of the surface \(\left(x^{*}=1\right)\), midplane \(\left(x^{*}=0\right)\), and quarter-plane \(\left(x^{*}=0.5\right)\) of the slab. Plot these three dimensionless temperatures as a function of dimensionless time over the range \(0 \leq F o \leq 0.5\). (d) Determine the minimum dimensionless temperature at the midplane of the wall, and the dimensionless time at which this minimum temperature is achieved.

Circuit boards are treated by heating a stack of them under high pressure, as illustrated in Problem 5.45. The platens at the top and bottom of the stack are maintained at a uniform temperature by a circulating fluid. The purpose of the pressing-heating operation is to cure the epoxy, which bonds the fiberglass sheets, and impart stiffness to the boards. The cure condition is achieved when the epoxy has been maintained at or above \(170^{\circ} \mathrm{C}\) for at least \(5 \mathrm{~min}\). The effective thermophysical properties of the stack or book (boards and metal pressing plates) are \(k=0.613 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and \(\rho c_{p}=2.73 \times 10^{6} \mathrm{~J} / \mathrm{m}^{3} \cdot \mathrm{K}\) (a) If the book is initially at \(15^{\circ} \mathrm{C}\) and, following application of pressure, the platens are suddenly brought to a uniform temperature of \(190^{\circ} \mathrm{C}\), calculate the elapsed time \(t_{e}\) required for the midplane of the book to reach the cure temperature of \(170^{\circ} \mathrm{C}\). (b) If, at this instant of time, \(t=t_{e}\), the platen temperature were reduced suddenly to \(15^{\circ} \mathrm{C}\), how much energy would have to be removed from the book by the coolant circulating in the platen, in order to return the stack to its initial uniform temperature?

Consider the fuel element of Example \(5.11\), which operates at a uniform volumetric generation rate of \(\dot{q}_{1}=10^{7} \mathrm{~W} / \mathrm{m}^{3}\) until the generation rate suddenly changes to \(\dot{q}_{2}=2 \times 10^{7} \mathrm{~W} / \mathrm{m}^{3}\). Use the finite-element software \(F E H T\) to obtain the following solutions. (a) Calculate the temperature distribution \(1.5 \mathrm{~s}\) after the change in operating power and compare your results with those tabulated in the example. Hint: First determine the steady-state temperature distribution for \(\dot{q}_{1}\), which represents the initial condition for the transient temperature distribution after the step change in power to \(\dot{q}_{2}\). Next, in the Setup menu, click on Transient: in the Specify/Internal Generation box, change the value to \(\dot{q}_{2}\); and in the Run command, click on Continue (not Calculate). See the Run menu in the FEHT Help section for background information on the Continue option. (b) Use your \(F E H T\) model to plot temperature histories at the midplane and surface for \(0 \leq t \leq 400 \mathrm{~s}\). What are the steady-state temperatures, and approximately how long does it take to reach the new equilibrium condition after the step change in operating power?

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.

A plate of thickness \(2 L=25 \mathrm{~mm}\) at a temperature of \(600^{\circ} \mathrm{C}\) is removed from a hot pressing operation and must be cooled rapidly to achieve the required physical properties. The process engineer plans to use air jets to control the rate of cooling, but she is uncertain whether it is necessary to cool both sides (case 1 ) or only one side (case 2) of the plate. The concern is not just for the time-to-cool, but also for the maximum temperature difference within the plate. If this temperature difference is too large, the plate can experience significant warping. The air supply is at \(25^{\circ} \mathrm{C}\), and the convection coefficient on the surface is \(400 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The thermophysical properties of the plate are \(\rho=3000 \mathrm{~kg} / \mathrm{m}^{3}, c=750 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), and \(k=15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). (a) Using the IHT software, calculate and plot on one graph the temperature histories for cases 1 and 2 for a \(500-s\) cooling period. Compare the times required for the maximum temperature in the plate to reach \(100^{\circ} \mathrm{C}\). Assume no heat loss from the unexposed surface of case 2 . (b) For both cases, calculate and plot on one graph the variation with time of the maximum temperature difference in the plate. Comment on the relative magnitudes of the temperature gradients within the plate as a function of time.
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