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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

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.

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 \(150 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\).

In a material processing experiment conducted aboard the space shuttle, a coated niobium sphere of \(10-\mathrm{mm}\) diameter is removed from a furnace at \(900^{\circ} \mathrm{C}\) and cooled to a temperature of \(300^{\circ} \mathrm{C}\). Although properties of the niobium vary over this temperature range, constant values may be assumed to a reasonable approximation, with \(\rho=8600 \mathrm{~kg} / \mathrm{m}^{3}, c=290 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), and \(k=\) \(63 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). (a) If cooling is implemented in a large evacuated chamber whose walls are at \(25^{\circ} \mathrm{C}\), determine the time required to reach the final temperature if the coating is polished and has an emissivity of \(\varepsilon=0.1\). How long would it take if the coating is oxidized and \(\varepsilon=0.6\) ? (b) To reduce the time required for cooling, consideration is given to immersion of the sphere in an inert gas stream for which \(T_{\infty}=25^{\circ} \mathrm{C}\) and \(h=\) \(200 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Neglecting radiation, what is the time required for cooling? (c) Considering the effect of both radiation and convection, what is the time required for cooling if \(h=200 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(\varepsilon=0.6\) ? Explore the effect on the cooling time of independently varying \(h\) and \(\varepsilon\).

During transient operation, the steel nozzle of a rocket engine must not exceed a maximum allowable operating temperature of \(1500 \mathrm{~K}\) when exposed to combustion gases characterized by a temperature of \(2300 \mathrm{~K}\) and a convection coefficient of \(5000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). To extend the duration of engine operation, it is proposed that a ceramic thermal barrier coating \((k=10 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), \(\alpha=6 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}\) ) be applied to the interior surface of the nozzle. (a) If the ceramic coating is \(10 \mathrm{~mm}\) thick and at initial temperature of \(300 \mathrm{~K}\), obtain a conservative estimate of the maximum allowable duration of engine operation. The nozzle radius is much larger than the combined wall and coating thickness. (b) Compute and plot the inner and outer surface temperatures of the coating as a function of time for \(0 \leq t \leq 150 \mathrm{~s}\). Repeat the calculations for a coating thickness of \(40 \mathrm{~mm}\).

A long rod of \(60-\mathrm{mm}\) diameter and thermophysical properties \(\rho=8000 \mathrm{~kg} / \mathrm{m}^{3}, \quad c=500 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), and \(k=50 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) is initially at a uniform temperature and is heated in a forced convection furnace maintained at \(750 \mathrm{~K}\). The convection coefficient is estimated to be \(1000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) What is the centerline temperature of the rod when the surface temperature is \(550 \mathrm{~K}\) ? (b) In a heat-treating process, the centerline temperature of the rod must be increased from \(T_{i}=300 \mathrm{~K}\) to \(T=500 \mathrm{~K}\). Compute and plot the centerline temperature histories for \(h=100,500\), and \(1000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). In each case the calculation may be terminated when \(T=500 \mathrm{~K}\).
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