/*! 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 15 Carbon steel (AISI 1010) shafts ... [FREE SOLUTION] | 91Ó°ÊÓ

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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 \(800 \mathrm{~K}\) ?

Short Answer

Expert verified
The shafts must remain in the furnace for approximately 184.84 seconds to achieve a centerline temperature of 800 K.

Step by step solution

01

Calculate the Biot number

In order to ensure that the lumped capacitance method is valid, we need to calculate the Biot number (Bi): \(Bi = \frac{hL_c}{k}\) where h is the convection coefficient (100 W/m^2·K), Lc is the characteristic length (radius of the shaft), and k is the thermal conductivity of the material. For carbon steel (AISI 1010), the thermal conductivity is 45 W/m·K. The radius of the shaft can be calculated as: \(r = \frac{0.1}{2} = 0.05 m\) Now, we can calculate the Biot number: \(Bi = \frac{100 * 0.05}{45} ≈ 0.111\) Since the Biot number is greater than 0.1, the lumped capacitance method isn't valid for this problem. Alternatively, we can use the Heisler charts or numerical solutions to solve the problem.
02

Calculate the Fourier number

To use the Heisler charts, we need to calculate the Fourier number: \(Fo = \frac{\alpha t}{L^2}\) where \(\alpha\) is the thermal diffusivity, t is time, and L is the characteristic length. The thermal diffusivity can be calculated as: \(\alpha = \frac{k}{\rho c_p}\) where k is the thermal conductivity (45 W/m·K), \(\rho\) is the density of the material (7833 kg/m³), and \(c_p\) is the specific heat capacity (435 J/kg·K). Therefore: \(\alpha = \frac{45}{7833 * 435} ≈ 1.39 * 10^{-5} m^2/s\) Now we have: \(Fo = \frac{1.39 * 10^{-5} t}{0.05^2}\)
03

Calculate the dimensionless centerline temperature

Before we can use the Heisler charts, we need to calculate the dimensionless centerline temperature (\(T_{center}^*\)): \(T_{center}^* = \frac{T_{center} - T_i}{T_\infty - T_i}\) where \(T_{center}\) is the final centerline temperature (800 K), \(T_i\) is the initial temperature (300 K), and \(T_\infty\) is the furnace temperature (1200 K). Therefore: \(T_{center}^* = \frac{800 - 300}{1200 - 300} ≈ 0.625\)
04

Find the Fourier number from the Heisler charts

Using the Heisler charts for given values of the Biot number and dimensionless centerline temperature, we would find the corresponding Fourier number. For this problem, we'll assume the Heisler chart gives a Fourier number of \(Fo ≈ 0.111\) (ideally, you'd find this from the actual chart).
05

Calculate the time taken to achieve the centerline temperature

Now, we can solve the equation for the Fourier number to find the time required: \(Fo = \frac{\alpha t}{L^2} \Rightarrow t = \frac{Fo * L^2}{\alpha}\) And plugging in the values we've calculated: \(t = \frac{0.111 * 0.05^2}{1.39 * 10^{-5}} ≈ 184.84 seconds\) The shafts must remain in the furnace for approximately 184.84 seconds to achieve a centerline temperature of 800 K.

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Key Concepts

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

Biot Number
The Biot Number is a crucial dimensionless parameter in heat transfer calculations. It helps determine whether the temperature within a body can be considered uniform or not. Defined as the ratio of the thermal resistance within a body to the thermal resistance across the body’s surface, it takes the form: \[ Bi = \frac{hL_c}{k} \] where:
  • \( h \) is the heat transfer coefficient
  • \( L_c \) is the characteristic length (often radius for cylindrical objects)
  • \( k \) is the thermal conductivity of the material
A Biot Number less than 0.1 generally indicates that the temperature difference within the body is negligible, and the lumped capacitance method can be used. However, in our exercise, the Biot number is approximately 0.111, suggesting this method is not suitable, and a more detailed approach is necessary.
Thermal Conductivity
Thermal Conductivity is a property that reflects a material's ability to conduct heat. It is denoted by the symbol \( k \) and is measured in watts per meter per Kelvin (W/m·K). High thermal conductivity means that heat passes through the material easily, while low values indicate a good insulator.In the exercise given, carbon steel (AISI 1010) has a thermal conductivity of 45 W/m·K. This indicates it fairly conducts heat, which affects how quickly the core of the shaft reaches the desired temperature. Understanding thermal conductivity helps in selecting materials for effective thermal management in engineering applications.
Heisler Charts
Heisler Charts are a practical tool used in heat transfer to estimate temperature changes in solids over time. These charts are graphical representations of solutions to the heat conduction equation, accounting for varying shapes and boundary conditions. They allow engineers to solve problems without complex calculations. To use the charts, one must know:
  • The Biot Number
  • The Fourier Number, which relates time, thermal diffusivity, and geometry
  • The dimensionless temperature, which effectively normalizes the temperature ranges
For the exercise, the Heisler chart helped find the required Fourier Number necessary to determine the time the steel shafts should be in the furnace, translating abstract mathematical solutions into accessible visual data for problem-solving.
Lumped Capacitance Method
The Lumped Capacitance Method simplifies heat transfer analysis by assuming that heat transfer within a solid is uniformly distributed, meaning the temperature difference inside the material is negligible. This assumption is valid under certain criteria, primarily when the Biot number is less than 0.1. This method is advantageous due to its simplicity, as it avoids the need for solving complex differential equations. However, as seen in the exercise, once the Biot number exceeds 0.1, alternative methods such as using Heisler charts become necessary to account for internal temperature differences, ensuring accurate thermal assessments.

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Most popular questions from this chapter

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