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Chapter 5: Problem 74

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 \(40^{\circ} \mathrm{C}\) ?

Short Answer

Expert verified
The ball bearings must be kept in the oil bath for approximately 1078 seconds for their surface temperature to reach 100掳C. The center temperature of the ball bearings at the conclusion of the cooling period is also approximately 100掳C. To quench 10,000 ball bearings per hour, the oil bath cooling system must remove energy at a rate of 3.46 kW to maintain its temperature at 40掳C.

Step by step solution

01

Relevant Equations and Constants

We will be using the following equations: 1. Lumped System Analysis: \[Q = - hA_s (T - T_\infty)\] 2. Energy equation for cooling: \[mc\frac{dT}{dt} = -hA_s(T - T_\infty)\] 3. Lumped capacitance time constant: \[\tau = \frac{V\rho c}{hA_s}\] where: - Q is the heat transfer rate - h is the convection coefficient - A_s is the surface area of the ball - T is the temperature of the ball - T_\infty is the temperature of the oil bath - m is the mass of the ball - c is the specific heat of the stainless steel - V is the volume of the ball - \(\rho\) is the density of the stainless steel - \(\tau\) is the time constant Constants: - Stainless steel specific heat (c) = 500 J/(kg路K) - Stainless steel density (\(\rho\)) = 7900 kg/m鲁 - Ball diameter (D) = 0.02 m - Convection coefficient (h) = 1000 W/(m虏路K) (a)
02

Surface Area and Volume of the Ball

We can calculate the surface area and volume of the ball: Surface area A_s = \[4 \pi r^2\] Volume V = \[\frac{4}{3} \pi r^3\] where: - r is the radius of the ball Using the given diameter of 0.02 m, the radius can be found as r = D / 2 = 0.01 m.Putting the value of r into both equations, we get: Surface area A_s = 4 * \(\pi\) * (0.01 m虏) =\(1.256 \times 10^{-3}\) m虏 Volume V = (4 / 3) * \(\pi\) * (0.01 m鲁) = \(4.189 \times 10^{-6}\) m鲁
03

Lumped Capacitance Time Constant

Calculate the time constant, \(\tau\), using the lumped capacitance equation: \[\tau = \frac{V\rho c}{hA_s}\] \(\tau\) = \[\frac{(4.189 \times 10^{-6} \text{m}鲁)(7900 \text{kg/m}^3)(500 \mathrm{J/(kg路K)})}{(1000 \mathrm{W/(m虏路K)})(1.256 \times 10^{-3} \text{m}虏)}\] \(\tau\) = 837 s
04

Calculate the Time to Reach Final Temperature

Using the time constant equation, we can solve for the time to reach the temperature 100掳C: \[t = \tau \ln\frac{T_i - T_\infty}{T_f - T_\infty}\] where: - t is the time - T_i is the initial temperature of the ball - T_f is the final temperature of the ball Substituting the given values (T_i = 850掳C, T_f = 100掳C, T_\infty = 40掳C, \(\tau\) = 837 s) into the equation, we get: \[t = (837\text{s})\ln\frac{850 - 40}{100 - 40}\] \[t = 1078 \text{s}\]
05

Find the Center Temperature

Now we will use lumped capacitance method to find the center temperature, T_center: \[T_center = T_\infty + (T_i - T_\infty)e^{-t/\tau}\] T_center = 40掳C + (850掳C - 40掳C) * \[\mathrm{e}^{-1078/837}\] T_center 鈮� 100掳C (b)
06

Calculate the Rate of Energy Removal of the Cooling System

First, find the energy removed from one ball: Q_single = mc螖T where 螖T = (T_i - T_f) The mass of single ball, m = 蟻V = (7900 kg/m鲁)(4.189 脳 10鈦烩伓 m鲁) = 0.0331 kg Q_single = (0.0331 kg)(500 J/(kg路K))(850掳C - 100掳C) = 1245 J If 10,000 balls are to be quenched per hour, we can find the rate at which energy must be removed, Q_rate: Q_rate = (10,000 balls/h)(1245 J/ball)(1 h/3600 s) = 3.46 kW

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Key Concepts

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

Lumped System Analysis
Lumped System Analysis is a simplified approach for solving heat transfer problems. It's particularly useful when temperature variations within a solid are negligible. This assumption allows the body to be treated as a "lump" with uniform temperature throughout.
For this analysis to be valid, the Biot number, a dimensionless parameter, must be much less than 1. The Biot number is defined as \[Bi = \frac{hL_c}{k}\] where \(h\) is the convection coefficient, \(L_c\) is the characteristic length (which is the volume-to-surface area ratio for spheres), and \(k\) is the thermal conductivity of the material.
Once the lumped capacitance model is applicable, the heat transfer can be described using a simple exponential relation. The temperature of the body decreases exponentially over time as expressed by:\[t = \tau \ln\frac{T_i - T_\infty}{T_f - T_\infty}\]This equation helps us determine the time required for a material to cool to a desired temperature, such as in the quenching process.
Convection Coefficient
The convection coefficient, denoted as \(h\), plays a critical role in heat transfer between a solid and a fluid. It's measured in watts per square meter per Kelvin (\(W/m^2 \cdot K\)) and quantifies the heat exchange rate per unit area per unit temperature difference.
This parameter depends on various factors such as the nature of the fluid, the flow regime, and the surface roughness of the solid. In this problem, a high convection coefficient of \(1000 \mathrm{~W/m}^2 \cdot \mathrm{K}\) indicates efficient heat transfer between the stainless steel ball bearings and the oil bath.
A high convection coefficient accelerates the temperature equilibration process, thereby shortening the time required to reach a specific temperature. Understanding and calculating this parameter is essential in designing efficient thermal systems, such as in the quenching and cooling processes.
Quenching Process
The quenching process is a rapid cooling technique used to alter the microstructure of metal components like the AISI 304 stainless steel ball bearings in this exercise.
This process involved immersing the balls in an oil bath, which is significantly cooler than the initial temperature of the balls. The primary goal of quenching is to achieve specific mechanical properties, such as increased hardness or strength, by controlling the cooling rate.
Several factors impact quenching effectiveness:
  • Initial and Final Temperatures: A higher initial temperature and a significant temperature drop (e.g., from \(850^{\circ}C\) to \(100^{\circ}C\)) facilitate the required microstructural transformation.
  • Cooling Medium: Oil baths are commonly used due to their ability to provide a uniform quench and reduce thermal shock.
  • Heat Transfer Characteristics: The high convection coefficient helps rapidly lower the surface temperature, ensuring that the transition progresses uniformly through the material.
This understanding aids in optimizing the cooling process, guaranteeing that the metal attains desired properties without distortion or cracking.
Energy Removal Rate
The energy removal rate is crucial for maintaining the temperature of the quenching medium, ensuring consistent cooling conditions for each batch of parts. In this exercise, it refers to the rate at which energy must be extracted from the oil bath.
For a scenario of quenching \(10,000\) balls per hour, the total energy removed is determined by calculating the energy lost by each steel ball, and then scaling up to the entire batch.
To calculate, we use:
  • Specific Heat Capacity: Determines the amount of energy required to change the temperature of the stainless steel.
  • Mass and Temperature Change: Mass is derived from the ball's volume and density, and the temperature change is the difference between the initial and final states.
Balls lose a calculated amount of energy (e.g., \(1245\) Joules per ball). Multiplying this by the number of balls quenched per hour and dividing by \(3600\) seconds per hour gives the required energy removal rate in kilowatts. This ensures that the cooling system effectively handles the thermal load, maintaining a stable bath temperature.

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Most popular questions from this chapter

Common transmission failures result from the glazing of clutch surfaces by deposition of oil oxidation and decomposition products. Both the oxidation and decomposition processes depend on temperature histories of the surfaces. Because it is difficult to measure these surface temperatures during operation, it is useful to develop models to predict clutch-interface thermal behavior. The relative velocity between mating clutch plates, from the initial engagement to the zero-sliding (lock-up) condition, generates heat that is transferred to the plates. The relative velocity decreases at a constant rate during this period, producing a heat flux that is initially very large and decreases linearly with time, until lock-up occurs. Accordingly, \(q_{f}^{\prime \prime}=q_{o}^{\prime \prime}=\left[1-\left(t / t_{\mathrm{lu}}\right)\right]\), where \(q_{o}^{\prime \prime}=1.6 \times 10^{7} \mathrm{~W} / \mathrm{m}^{2}\) and \(t_{1 \mathrm{u}}=100 \mathrm{~ms}\) is the lock-up time. The plates have an initial uniform temperature of \(T_{i}=40^{\circ} \mathrm{C}\), when the prescribed frictional heat flux is suddenly applied to the surfaces. The reaction plate is fabricated from steel, while the composite plate has a thinner steel center section bonded to low- conductivity friction material layers. The thermophysical properties are \(\rho_{s}=\) \(7800 \mathrm{~kg} / \mathrm{m}^{3}, c_{\mathrm{s}}=500 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), and \(k_{s}=40 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) for the steel and \(\rho_{\mathrm{im}}=1150 \mathrm{~kg} / \mathrm{m}^{3}, c_{\mathrm{fm}}=1650 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), and \(k_{\mathrm{fm}}=4 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) for the friction material. (a) On \(T-t\) coordinates, sketch the temperature history at the midplane of the reaction plate, at the interface between the clutch pair, and at the midplane of the composite plate. Identify key features. (b) Perform an energy balance on the clutch pair over the time interval \(\Delta t=t_{\mathrm{lu}}\) to determine the steadystate temperature resulting from clutch engagement. Assume negligible heat transfer from the plates to the surroundings. (c) Compute and plot the three temperature histories of interest using the finite-element method of FEHT or the finite-difference method of IHT (with \(\Delta x=0.1 \mathrm{~mm}\) and \(\Delta t=1 \mathrm{~ms}\) ). Calculate and plot the frictional heat fluxes to the reaction and composite plates, \(q_{\mathrm{rp}}^{\prime \prime}\) and \(q_{\mathrm{cp}}^{\prime \prime}\), respectively, as a function of time. Comment on features of the temperature and heat flux histories. Validate your model by comparing predictions with the results from part (b). Note: Use of both \(F E H T\) and \(I H T\) requires creation of a look-up data table for prescribing the heat flux as a function of time.

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

Plasma spray-coating processes are often used to provide surface protection for materials exposed to hostile environments, which induce degradation through factors such as wear, corrosion, or outright thermal failure. Ceramic coatings are commonly used for this purpose. By injecting ceramic powder through the nozzle (anode) of a plasma torch, the particles are entrained by the plasma jet, within which they are then accelerated and heated. During their time-in-fbht, the ceramic particles must be heated to their melting point and experience complete conversion to the liquid state. The coating is formed as the molten droplets impinge (splat) on the substrate material and experience rapid solidification. Consider conditions for which spherical alumina \(\left(\mathrm{Al}_{2} \mathrm{O}_{3}\right.\) ) particles of diameter \(D_{p}=50 \mu \mathrm{m}\), density \(\rho_{p}=\) \(3970 \mathrm{~kg} / \mathrm{m}^{3}\), thermal conductivity \(k_{p}=10.5 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), and specific heat \(c_{p}=1560 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\) are injected into an arc plasma, which is at \(T_{\infty}=10,000 \mathrm{~K}\) and provides a coefficient of \(h=30,000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) for convective heating of the particles. The melting point and latent heat of fusion of alumina are \(T_{\text {mp }}=2318 \mathrm{~K}\) and \(h_{s f}=3577 \mathrm{~kJ} / \mathrm{kg}\), respectively. (a) Neglecting radiation, obtain an expression for the time-in-flight, \(t_{i-f}\), required to heat a particle from its initial temperature \(T_{i}\) to its melting point \(T_{\text {mp }}\), and, once at the melting point, for the particle to experience complete melting. Evaluate \(t_{i-f}\) for \(T_{i}=300 \mathrm{~K}\) and the prescribed heating conditions. (b) Assuming alumina to have an emissivity of \(\varepsilon_{p}=0.4\) and the particles to exchange radiation with large surroundings at \(T_{\text {sur }}=300 \mathrm{~K}\), assess the validity of neglecting radiation.

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 thermal diffusivity of \(10^{-5} \mathrm{~m}^{2} / \mathrm{s}\), at what time will a surface temperature 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 period \(0 \leq t \leq 300 \mathrm{~s}\). In a similar manner, assess the effect of thermal conductivity by considering values of \(k=1.5,15\), and \(150 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\).

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?
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