/*! 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 49 Annealing is a process by which ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Annealing is a process by which steel is reheated and then cooled to make it less brittle. Consider the reheat stage for a \(100-\mathrm{mm}\)-thick steel plate \(\left(\rho=7830 \mathrm{~kg} / \mathrm{m}^{3}\right.\), \(c=550 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}, k=48 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\), which is initially at a uniform temperature of \(T_{i}=200^{\circ} \mathrm{C}\) and is to be heated to a minimum temperature of \(550^{\circ} \mathrm{C}\). Heating is effected in a gas-fired furnace, where products of combustion at \(T_{\infty}=800^{\circ} \mathrm{C}\) maintain a convection coefficient of \(h=250 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) on both surfaces of the plate. How long should the plate be left in the furnace?

Short Answer

Expert verified
The steel plate should be left in the furnace for approximately 521.90 minutes to reach a minimum temperature of 550°C.

Step by step solution

01

Determine the governing equation for temperature distribution

We need to find the governing equation for the temperature distribution within the steel plate while being heated in the furnace. One-dimensional unsteady conduction equation for a solid semi-infinite medium with constant temperature boundary conditions is given by: \[ T(x,t) = T_i + (T_\infty - T_i)\left(1 - \operatorname{erf} \frac{x}{2\sqrt{\alpha t}} \right) \] where \(T(x, t)\) is the temperature distribution at a distance \(x\) and time \(t\), \(T_i\) is the initial temperature, \(T_\infty\) is the furnace temperature, \(\alpha\) is the thermal diffusivity (which can be obtained from \(\alpha = \frac{k}{\rho c}\)), and \(\operatorname{erf}\) is the error function.
02

Determine the centerline temperature

For our analysis, we only need to consider the centerline temperature of the steel plate because once we reach the desired minimum temperature at the centerline, we can assume the rest has reached the required temperature as well. For the center of the plate (\(x = \frac{100\mathrm{~mm}}{2} = 50\mathrm{~mm}\)), the formula becomes: \[ T_c(t) = T_i + (T_\infty - T_i)\left(1 - \operatorname{erf} \frac{x}{2\sqrt{\alpha t}} \right) \] Where \(T_c(t)\) is the centerline temperature.
03

Solve for the heating time

To find the time required to heat the centerline temperature to \(550^{\circ} \mathrm{C}\), set \(T_c(t) = 550^{\circ} \mathrm{C}\) and solve for \(t\). \[ 550 = 200 + (800 - 200)\left(1 - \operatorname{erf} \frac{50}{2\sqrt{\alpha t}} \right) \] \[ \frac{350}{600} = 1 - \operatorname{erf} \frac{50}{2\sqrt{\alpha t}} \] First, we need to calculate the thermal diffusivity \(\alpha\): \[\alpha = \frac{k}{\rho c} = \frac{48}{7830\times 550} = 0.00000123542 \mathrm{~m}^{2}/\mathrm{s} \] Now, find \(t\) by solving the error function: \[ \operatorname{erf}^{-1}\left(1-\frac{350}{600}\right) = \frac{50}{2\sqrt{\alpha t}}\] \[ t = \frac{50^2}{4\cdot \alpha \cdot \operatorname{erf}^{-1} \left(\frac{250}{600}\right)^2} \] Now substitute the values into the equation: \[ t = \frac{50^2}{4\cdot 0.00000123542 \cdot \operatorname{erf}^{-1} \left(\frac{250}{600}\right)^2} \] Using a calculator, we can determine the inverse error function: \(\operatorname{erf}^{-1}(0.4167) \approx 0.2346\). Now calculate the time: \[ t = \frac{50^2}{4\cdot 0.00000123542 \cdot (0.2346)^2} \approx 31314.05 \mathrm{s} \]
04

Convert time to minutes

Lastly, convert the time from seconds to minutes: \[ t = 31314.05 \mathrm{~s} \cdot \frac{1\mathrm{~min}}{60\mathrm{~s}} \approx 521.90\mathrm{~min} \] The steel plate should be left in the furnace for approximately 521.90 minutes to reach a minimum temperature of 550°C.

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.

Thermal Diffusivity
Understanding thermal diffusivity is essential when analyzing heat conduction processes. It represents how quickly heat spreads through a material, indicating the rate at which a material attains thermal equilibrium when subjected to a temperature change. Thermal diffusivity, denoted as \( \alpha \), is calculated using the formula:\[ \alpha = \frac{k}{\rho c} \]where:
  • \( k \) is the thermal conductivity of the material, measured in watts per meter-kelvin (W/m·K).
  • \( \rho \) is the density of the material, measured in kilograms per cubic meter (kg/m³).
  • \( c \) is the specific heat capacity of the material, measured in joules per kilogram-kelvin (J/kg·K).
Thermal diffusivity combines these three fundamental properties to give insight into how a material behaves under thermal conditions. A material with high thermal diffusivity will quickly adjust to temperature changes, while a low thermal diffusivity material will take longer to reach thermal equilibrium. In our example, the steel plate has a calculated thermal diffusivity of \( 0.00000123542 \text{ m}^2/ ext{s} \), contributing to how heat travels through it during the annealing process.
Error Function
The error function is a special mathematical function often encountered in thermal conduction problems involving diffusion or probability. Denoted as \( \operatorname{erf}(x) \), it describes the probability of a random variable falling within a certain range in a normal distribution, but in thermal physics, it helps model heat distribution in materials.In problems of heat conduction like ours, the error function is used to express solutions of the diffusion equation with boundary conditions. The error function allows the expression of complex temperature profiles in a simplified form. The solution formula for temperature distribution in the given problem is:\[ T(x,t) = T_i + (T_\infty - T_i)\left(1 - \operatorname{erf} \left(\frac{x}{2\sqrt{\alpha t}}\right) \right) \]This equation demonstrates how the temperature at any point \( x \) in the steel evolves over time \( t \), with \( \operatorname{erf} \) accounting for the diffusive effects. The inverse error function \( \operatorname{erf}^{-1} \) comes into play when solving for time \( t \) given a specific target temperature. It’s a critical step in handling these kinds of problems, providing insight into when the material reaches a desired thermal state.
Unsteady Heat Transfer
Unsteady heat transfer, also known as transient heat transfer, focuses on how temperature varies with time and position in a system. Unlike steady-state heat transfer, where temperatures remain constant over time, unsteady heat transfer considers scenarios where temperatures change until thermal equilibrium is achieved.In the annealing process of the steel plate, the heating stage is modeled using the unsteady heat transfer approach because the temperature inside the steel isn't uniform initially but changes as heat penetrates the plate. The goal is to reach a uniform minimum temperature of \( 550^{\circ} \text{C} \) at the centerline.The mathematical representation of this process involves the time-dependent heat conduction equation. For our example:\[ T_c(t) = T_i + (T_\infty - T_i)\left(1 - \operatorname{erf} \frac{x}{2\sqrt{\alpha t}} \right) \]This equation models how the centerline temperature \( T_c(t) \) changes over time. It includes variables such as the initial temperature, boundary conditions, and thermal properties, precisely the thermal diffusivity \( \alpha \), which influence the rate of heat transfer throughout the plate.Unsteady heat transfer provides the necessary framework for calculating the time it takes to achieve uniform heating within a material, enabling effective design and control of thermal processes in industrial applications.

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

Thin film coatings characterized by high resistance to abrasion and fracture may be formed by using microscale composite particles in a plasma spraying process. A spherical particle typically consists of a ceramic core, such as tungsten carbide (WC), and a metallic shell, such as cobalt \((\mathrm{Co})\). The ceramic provides the thin film coating with its desired hardness at elevated temperatures, while the metal serves to coalesce the particles on the coated surface and to inhibit crack formation. In the plasma spraying process, the particles are injected into a plasma gas jet that heats them to a temperature above the melting point of the metallic casing and melts the casing before the particles impact the surface. Consider spherical particles comprised of a WC core of diameter \(D_{i}=16 \mu \mathrm{m}\), which is encased in a Co shell of outer diameter \(D_{o}=20 \mu \mathrm{m}\). If the particles flow in a plasma gas at \(T_{\infty}=10,000 \mathrm{~K}\) and the coefficient associated with convection from the gas to the particles is \(h=20,000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), how long does it take to heat the particles from an initial temperature of \(T_{i}=300 \mathrm{~K}\) to the melting point of cobalt, \(T_{\mathrm{mp}}=1770 \mathrm{~K}\) ? The density and specific heat of WC (the core of the particle) are \(\rho_{c}=16,000 \mathrm{~kg} / \mathrm{m}^{3}\) and \(c_{c}=300 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), while the corresponding values for Co (the outer shell) are \(\rho_{s}=8900 \mathrm{~kg} / \mathrm{m}^{3}\) and \(c_{s}=750 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\). Once having reached the melting point, how much additional time is required to completely melt the cobalt if its latent heat of fusion is \(h_{s f}=2.59 \times 10^{5} \mathrm{~J} / \mathrm{kg}\) ? You may use the lumped capacitance method of analysis and neglect radiation exchange between the particle and its surroundings.

Joints of high quality can be formed by friction welding. Consider the friction welding of two 40 -mm-diameter Inconel rods. The bottom rod is stationary, while the top rod is forced into a back-and-forth linear motion characterized by an instantaneous horizontal displacement, \(d(t)=a \cos (\omega t)\) where \(a=2 \mathrm{~mm}\) and \(\omega=1000 \mathrm{rad} / \mathrm{s}\). The coefficient of sliding friction between the two pieces is \(\mu=0.3\). Determine the compressive force that must be applied to heat the joint to the Inconel melting point within \(t=3 \mathrm{~s}\), starting from an initial temperature of \(20^{\circ} \mathrm{C}\). Hint: The frequency of the motion and resulting heat rate are very high. The temperature response can be approximated as if the heating rate were constant in time, equal to its average value.

In a thin-slab, continuous casting process, molten steel leaves a mold with a thin solid shell, and the molten material solidifies as the slab is quenched by water jets en route to a section of rollers. Once fully solidified, the slab continues to cool as it is brought to an acceptable handling temperature. It is this portion of the process that is of interest. Consider a 200-mm-thick solid slab of steel \(\left(\rho=7800 \mathrm{~kg} / \mathrm{m}^{3}, \quad c=700 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}, \quad k=30 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\right)\), initially at a uniform temperature of \(T_{i}=1400^{\circ} \mathrm{C}\). The slab is cooled at its top and bottom surfaces by water jets \(\left(T_{\infty}=50^{\circ} \mathrm{C}\right)\), which maintain an approximately uniform convection coefficient of \(h=5000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) at both surfaces. Using a finite-difference solution with a space increment of \(\Delta x=1 \mathrm{~mm}\), determine the time required to cool the surface of the slab to \(200^{\circ} \mathrm{C}\). What is the corresponding temperature at the midplane of the slab? If the slab moves at a speed of \(V=15 \mathrm{~mm} / \mathrm{s}\), what is the required length of the cooling section?

In Section \(5.2\) we noted that the value of the Biot number significantly influences the nature of the temperature distribution in a solid during a transient conduction process. Reinforce your understanding of this important concept by using the IHT model for one-dimensional transient conduction to determine radial temperature distributions in a 30 -mm-diameter, stainless steel rod \(\left(k=15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \rho=8000 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=475 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\), as it is cooled from an initial uniform temperature of \(325^{\circ} \mathrm{C}\) by a fluid at \(25^{\circ} \mathrm{C}\). For the following values of the convection coefficient and the designated times, determine the radial temperature distribution: \(h=100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) \((t=0,100,500 \mathrm{~s}) ; h=1000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}(t=0,10,50 \mathrm{~s})\); \(h=5000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}(t=0,1,5,25 \mathrm{~s})\). Prepare a separate graph for each convection coefficient, on which temperature is plotted as a function of dimensionless radius at the designated times.

The heat transfer coefficient for air flowing over a sphere is to be determined by observing the temperature-time history of a sphere fabricated from pure copper. The sphere, which is \(12.7 \mathrm{~mm}\) in diameter, is at \(66^{\circ} \mathrm{C}\) before it is inserted into an airstream having a temperature of \(27^{\circ} \mathrm{C}\). A thermocouple on the outer surface of the sphere indicates \(55^{\circ} \mathrm{C} 69 \mathrm{~s}\) after the sphere is inserted into the airstream. Assume and then justify that the sphere behaves as a spacewise isothermal object and calculate the heat transfer coefficient.

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.