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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.

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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.

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Most popular questions from this chapter

A steel strip of thickness \(\delta=12 \mathrm{~mm}\) is annealed by passing it through a large furnace whose walls are maintained at a temperature \(T_{w}\) corresponding to that of combustion gases flowing through the furnace \(\left(T_{w}=T_{\infty}\right)\). The strip, whose density, specific heat, thermal conductivity, and emissivity are \(\rho=7900 \mathrm{~kg} / \mathrm{m}^{3}\), \(c_{p}=640 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}, k=30 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), and \(\varepsilon=0.7\), respectively, is to be heated from \(300^{\circ} \mathrm{C}\) to \(600^{\circ} \mathrm{C}\). (a) For a uniform convection coefficient of \(h=\) \(100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(T_{w}=T_{\infty}=700^{\circ} \mathrm{C}\), determine the time required to heat the strip. If the strip is moving at \(0.5 \mathrm{~m} / \mathrm{s}\), how long must the furnace be? (b) The annealing process may be accelerated (the strip speed increased) by increasing the environmental temperatures. For the furnace length obtained in part (a), determine the strip speed for \(T_{w}=T_{\infty}=\) \(850^{\circ} \mathrm{C}\) and \(T_{w}=T_{\infty}=1000^{\circ} \mathrm{C}\). For each set of environmental temperatures \(\left(700,850\right.\), and \(\left.1000^{\circ} \mathrm{C}\right)\), plot the strip temperature as a function of time over the range \(25^{\circ} \mathrm{C} \leq T \leq 600^{\circ} \mathrm{C}\). Over this range, also plot the radiation heat transfer coefficient, \(h_{r}\), as a function of time.

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

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.

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 \(T=500 \mathrm{~K}\).

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.

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