/*! 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 6 Steel balls \(12 \mathrm{~mm}\) ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Steel balls \(12 \mathrm{~mm}\) in diameter are annealed by heating to \(1150 \mathrm{~K}\) and then slowly cooling to \(400 \mathrm{~K}\) in an air environment for which \(T_{\infty}=325 \mathrm{~K}\) and \(h=20 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Assuming the properties of the steel to be \(k=40 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), \(\rho=7800 \mathrm{~kg} / \mathrm{m}^{3}\), and \(c=600 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), estimate the time required for the cooling process. .

Short Answer

Expert verified
The time required for the cooling process of the steel balls can be estimated using the Lumped System Analysis. Following the steps from calculation of Biot number, checking its applicability, calculating Fourier number, and obtaining the temperature ratio, we find that the time required for cooling is given by the formula: \( t = \frac{Fo \times L_c^2}{\alpha} \) By substituting the values of Fourier number, characteristic length, and thermal diffusivity into the equation, we can calculate the time, \(t\), required for the cooling process.

Step by step solution

01

Calculate the Biot number (Bi)

The Biot number is a dimensionless number used to determine whether or not to use the Lumped System Analysis. It is given by the formula: \( Bi = \frac{hL_c}{k} \) where, \(h\) = heat transfer coefficient (\(20 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\)), \(L_c\) = characteristic length (radius of the sphere, \(r = \frac{D}{2} = \frac{0.012}{2} = 0.006 \mathrm{~m}\)), \(k\) = thermal conductivity of steel (\(40 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\)). Now, we can calculate the Biot number: \( Bi = \frac{20 \times 0.006}{40} \)
02

Check if Lumped System Analysis is applicable

Lumped System Analysis is applicable if the Biot number is less than or equal to 0.1. From step 1, the Biot number is: \( Bi = \frac{20 \times 0.006}{40} = 0.003 \) Since the Biot number is less than 0.1, we can use Lumped System Analysis.
03

Calculate the Fourier number (Fo)

Fourier number is a dimensionless number used in heat transfer analysis. It is given by the formula: \( Fo = \frac{\alpha t}{L_c^2} \) where, \(\alpha = \frac{k}{\rho c} = \frac{40 \mathrm{W/m.K}}{7800 \mathrm{kg/m^3} \times 600 \mathrm{J/kg.K}} \) = thermal diffusivity, \(t\) = time required for cooling, \(L_c\) = characteristic length (\(0.006 \mathrm{~m}\)). We can rewrite the above formula for \(t\) as: \( t = \frac{Fo \times L_c^2}{\alpha} \)
04

Calculate the temperature ratio

The temperature ratio is given by the formula: \( \frac{T - T_{\infty}}{T_i - T_{\infty}} \) where, \(T_i\) = initial temperature (\(1150 \mathrm{~K}\)), \(T_{\infty}\) = air temperature (\(325 \mathrm{~K}\)), \(T\) = final temperature (\(400 \mathrm{~K}\)). Now, let's calculate the temperature ratio: \( \frac{400 - 325}{1150 - 325} \)
05

Use Lumped System Analysis to find the Fourier number

According to Lumped System Analysis, we have: \( \frac{T - T_{\infty}}{T_i - T_{\infty}} = e^{-Bi \times Fo} \) Using the temperature ratio from step 4 and Biot number from step 1, we have: \( \frac{400 - 325}{1150 - 325} = e^{-0.003 \times Fo} \) Let's solve for the Fourier number (Fo).
06

Calculate the time required for cooling

Now we have the Fourier number, and we can calculate the time required for cooling using the formula obtained in step 3: \( t = \frac{Fo \times L_c^2}{\alpha} \) Substitute the values of Fourier number, characteristic length, and thermal diffusivity into this equation and solve for time, \(t\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

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 thermal processing and heat transfer analysis. It is a dimensionless quantity that indicates the ratio of internal thermal resistance within a body to the external thermal resistance across the body’s surface. In the context of cooling steel balls during the annealing process, it helps us to decide whether the temperature within the steel balls can be considered uniform at any given moment. A Biot number significantly lower than 1 (typically less than 0.1) suggests that we can apply the Lumped System Analysis. This means that the temperature difference within the object is negligible, and we can treat the object as if it is at a uniform temperature throughout. In the given exercise, the Biot number calculation is provided as:

\[\[\begin{align*} Bi = \frac{h \times L_c}{k}\end{align*}\]\]
With a low Biot number of 0.003, it confirms the appropriateness of using the Lumped System Analysis for the steel balls as they cool down. In real-world applications, this simplifies the analysis and calculations for engineers and students alike.
Lumped System Analysis
Lumped System Analysis simplifies heat transfer problems when the Biot number justifies its use. This approach assumes that the temperature within a solid body varies negligibly with time and position, and that any temperature changes can be described by an average temperature for the entire object. Using this simplification, we are able to apply the classical Newton’s law of cooling to solve for the time it takes for an object to cool down to a certain temperature. The formula derived from this law, \[\[\begin{align*} \frac{T - T_{\infty}}{T_i - T_{\infty}} = e^{-Bi \times Fo}\end{align*}\]\]
is the foundation for estimating the cooling time. This reduces complex partial differential equations to a much more straightforward exponential decay relationship. For students and professionals, Lumped System Analysis represents a potent shorthand when assessing quick thermal equilibrium processes.
Fourier Number
The Fourier number (Fo) is another dimensionless number, specific to transient heat conduction problems, that represents the ratio of heat conduction to heat storage within a body. Unlike the Biot number, which is a spatial comparison (between external and internal resistance), the Fourier number is a temporal comparison of rates. It is defined as:\[\[\begin{align*}Fo = \frac{\alpha \times t}{L_c^2}\end{align*}\]\]
where \( \alpha \) is the thermal diffusivity of the material, \( t \) is the time, and \( L_c \) is the characteristic length. The Fourier number aids in understanding how quickly a material responds to temperature changes. A high Fourier number corresponds to a system that has reached thermal equilibrium, while a low Fourier number indicates that the system is still undergoing temperature changes due to heat transfer. In the given exercise, we use the Lumped System Analysis to deduce the Fourier number as part of calculating the steel balls' cooling time.
Dimensionless Numbers in Heat Transfer
Dimensionless numbers serve as a foundational concept in the study of heat transfer, enabling the comparison and scaling of physical phenomena across different systems. By removing the units, these numbers allow for the analysis of various heat transfer scenarios, independent of the scale of application. The Biot and Fourier numbers are just two examples of these powerful tools. Others include the Reynolds number for fluid flow, the Prandtl number relating momentum to thermal diffusivity, and the Nusselt number for convective heat transfer. These dimensionless numbers also permit the use of similarity and modeling, where solutions to complex problems can be generalized and applied to many different situations. In the steel ball exercise, the understanding and application of both Biot and Fourier numbers demonstrate their utility in simplifying complex heat transfer problems and enabling precise calculations for practical engineering tasks.
Cooling of Steel Balls
In the context of the exercise, understanding how steel balls cool down involves applying several key concepts from heat transfer. After these balls are annealed, they must be cooled down to a specific temperature. The physics behind this process utilizes our knowledge of material properties, such as thermal conductivity and heat capacity, and combines it with dimensionless numbers to predict the cooling duration. By using the Lumped System Analysis, it is possible to calculate the time it will take for the steel balls to cool, based on a known heat transfer coefficient and ambient temperature. This simplification is only valid if the Biot number confirms a uniform temperature distribution within the steel balls. This kind of analysis is essential in industrial applications where the control of cooling rates can affect the mechanical properties and structural integrity of the material. It shows the direct link between our mathematical models and the physical realities they represent.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

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

Derive the explicit finite-difference equation for an interior node for three- dimensional transient conduction. Also determine the stability criterion. Assume constant properties and equal grid spacing in all three directions.

Carbon steel (AISI 1010) shafts of 0.1-m diameter are heat treated in a gas- fired furnace whose gases are at \(1200 \mathrm{~K}\) and provide a convection coefficient of \(100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). If the shafts enter the furnace at \(300 \mathrm{~K}\), how long must they remain in the furnace to achieve a centerline temperature of \(800 \mathrm{~K}\) ?

A plane wall of a furnace is fabricated from plain carbon steel \(\left(k=60 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \rho=7850 \mathrm{~kg} / \mathrm{m}^{3}, c=430 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) and is of thickness \(L=10 \mathrm{~mm}\). To protect it from the corrosive effects of the furnace combustion gases, one surface of the wall is coated with a thin ceramic film that, for a unit surface area, has a thermal resistance of \(R_{t, f}^{\prime \prime}=0.01 \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\). The opposite surface is well insulated from the surroundings. At furnace start-up the wall is at an initial temperature of \(T_{i}=300 \mathrm{~K}\), and combustion gases at \(T_{\infty}=1300 \mathrm{~K}\) enter the furnace, providing a convection coefficient of \(h=25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) at the ceramic film. Assuming the film to have negligible thermal capacitance, how long will it take for the inner surface of the steel to achieve a temperature of \(T_{s, i}=1200 \mathrm{~K}\) ? What is the temperature \(T_{s, o}\) of the exposed surface of the ceramic film at this time?

The density and specific heat of a particular material are known \(\left(\rho=1200 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=1250 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\), but its thermal conductivity is unknown. To determine the thermal conductivity, a long cylindrical specimen of diameter \(D=40 \mathrm{~mm}\) is machined, and a thermocouple is inserted through a small hole drilled along the centerline. The thermal conductivity is determined by performing an experiment in which the specimen is heated to a uniform temperature of \(T_{i}=100^{\circ} \mathrm{C}\) and then cooled by passing air at \(T_{\infty}=25^{\circ} \mathrm{C}\) in cross flow over the cylinder. For the prescribed air velocity, the convection coefficient is \(h=55 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) If a centerline temperature of \(T(0, t)=40^{\circ} \mathrm{C}\) is recorded after \(t=1136 \mathrm{~s}\) of cooling, verify that the material has a thermal conductivity of \(k=0.30 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). (b) For air in cross flow over the cylinder, the prescribed value of \(h=55 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) corresponds to a velocity of \(V=6.8 \mathrm{~m} / \mathrm{s}\). If \(h=C V^{0.618}\), where the constant \(C\) has units of \(\mathrm{W} \cdot \mathrm{s}^{0.618} / \mathrm{m}^{2.618} \cdot \mathrm{K}\), how does the centerline temperature at \(t=1136 \mathrm{~s}\) vary with velocity for \(3 \leq V \leq 20 \mathrm{~m} / \mathrm{s}\) ? Determine the centerline temperature histories for \(0 \leq t \leq 1500 \mathrm{~s}\) and velocities of 3,10 , and \(20 \mathrm{~m} / \mathrm{s}\).

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.