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Large steel plates \(1.0\)-cm in thickness are quenched from \(600^{\circ} \mathrm{C}\) to \(100^{\circ} \mathrm{C}\) by submerging them in an oil reservoir held at \(30^{\circ} \mathrm{C}\). The average heat transfer coefficient for both faces of steel plates is \(400 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Average steel properties are \(k=45 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \rho=7800 \mathrm{~kg} / \mathrm{m}^{3}\), and \(c_{p}=470 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\). Calculate the quench time for steel plates.

Short Answer

Expert verified
Answer: The quench time for the steel plates is 1000 seconds.

Step by step solution

01

Calculate the temperature difference

First, we need to find the difference between the initial plate temperature (\(T_1\)) and the oil reservoir temperature (\(T_\infty\)): $$\Delta T = T_1 - T_\infty = 600^{\circ}\mathrm{C} - 30^{\circ}\mathrm{C} = 570^{\circ}\mathrm{C}$$
02

Determine the Biot number

The Biot number (Bi) is a dimensionless number used to determine if we can use the lumped capacitance method. $$Bi = \frac{hL_{c}}{k}$$ Where: \(h\) is the average heat transfer coefficient, \(L_{c}\) is the characteristic length (in this case, half the plate's thickness), and \(k\) is the thermal conductivity of the steel. Calculating the Biot number: $$L_{c} = \frac{1}{2} \times 0.01\, \mathrm{m} = 0.005\,\mathrm{m}$$ $$Bi = \frac{400 \,\mathrm{W} / \mathrm{m}^{2} \cdot \mathrm{K} \times 0.005\, \mathrm{m}}{45\, \mathrm{W} / \mathrm{m} \cdot \mathrm{K}} = 0.0444$$ Since the Biot number is less than 0.1, we can use the lumped capacitance method.
03

Calculate the heat transfer resistance

We can determine the heat transfer resistance (\(R_{h}\)) as follows: $$R_{h} = \frac{1}{hA}$$ Where: \(A\) is the surface area of the steel plate. We have not been given the dimensions of the steel plate, so we cannot determine the surface area. However, we can still find the heat transfer resistance in terms of \(A\): $$R_{h} = \frac{1}{400 \,\mathrm{W} / \mathrm{m}^{2} \cdot \mathrm{K} \times A}$$
04

Calculate the volume and mass of the steel plate

The volume (\(V\)) and mass (\(m\)) of the steel plate can also be determined in terms of the unknown dimensions: $$V = A \times 0.01\, \mathrm{m}$$ $$m = \rho V = 7800\, \mathrm{kg} / \mathrm{m}^{3} \times (A \times 0.01\, \mathrm{m}) = 78A\, \mathrm{kg}$$
05

Calculate the quench time

Using the lumped capacitance method, the quench time (\(t\)) is given by: $$t = \frac{(T_1 - T_\infty)R_{h}mc_{p}}{(T_1 - T_2)}$$ Substituting the known values: $$t = \frac{(600^{\circ}\mathrm{C} - 30^{\circ}\mathrm{C})\left(\frac{1}{400 \,\mathrm{W} / \mathrm{m}^{2} \cdot \mathrm{K} \times A}\right)(78A\,\mathrm{kg})(470 \,\mathrm{J} / \mathrm{kg} \cdot \mathrm{K})}{(600^{\circ}\mathrm{C} - 100^{\circ}\mathrm{C})}$$ The surface area (\(A\)) cancels out, simplifying the expression for quench time: $$t = \frac{570\, \mathrm{K} \times 0.0025 \, \mathrm{m} \times 7800 \, \mathrm{kg} / \mathrm{m}^{3} \times 470 \, \mathrm{J} / \mathrm{kg} \cdot \mathrm{K}}{500\, \mathrm{K}} = 1000\, \mathrm{s}$$ The quench time for the steel plates is 1000 seconds.

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Key Concepts

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

Lumped Capacitance Method
The Lumped Capacitance Method is a simplified way to analyze heat transfer in objects where the temperature change is assumed to be uniform throughout the entire object. This method is particularly useful when an object can be approximated as having negligible thermal gradients because its thermal conductivity is high and its size is small compared to its thermal boundary resistance.
This translates to a low Biot number, typically less than 0.1. When this condition is met, we can assume the entire object changes temperature at a uniform rate. This makes calculations much easier.
  • How does it work? This method assumes all energy leaving or entering the object does so simultaneously through its surface. Essentially, it predicts the object's average temperature over time rather than at specific points.
  • Why is it useful? It's a major time-saver by avoiding complex differential equations and using the average heat transfer coefficients.
  • Limitations: The method may not be appropriate for larger objects or those with lower thermal conductivity, as temperature gradients within the material may be significant.
In our steel plate example, the Lumped Capacitance Method helps us determine how long it will take for the plates to reach a certain temperature during quenching because the Biot number confirms its applicability.
Quenching Process
The quenching process involves rapidly cooling a material, often a metal, from a high temperature to a lower one. This process is vital in altering the microstructure and mechanical properties of the material. The objective is to achieve desired material strength or hardness, while avoiding toughness degradation.
Quenching usually involves immersing the material in a bath of oil, water, or air. This provides a medium with a much lower temperature, ensuring quick and uniform cooling.
  • Main Steps: Quenching consists of three main phases: the initial cooling stage where air rapidly escapes the metal surface, the vapor blanket stage where a blanket of vapor slows cooling, and the boiling phase where actual heat transfer is most intense.
  • Applications: It's commonly used in manufacturing for parts requiring specific toughness and durability, such as automotive and machinery components.
In our example with steel plates starting at 600°C, the quenching process is completed to bring the temperature down to 100°C, ensuring the structure achieves the required mechanical properties.
Biot Number
The Biot Number is a dimensionless number used in heat transfer calculations to compare the internal conductive heat resistance within a body to the external convective heat transfer resistance. Understanding the Biot Number helps determine which heat transfer method, like the Lumped Capacitance Method, can be applied for accurate calculations.
The Biot Number (\( Bi \) ) is defined as:
\[ Bi = \frac{hL_{c}}{k} \] where \( h \) is the heat transfer coefficient, \( L_{c} \) is the characteristic length, and \( k \) is the thermal conductivity of the material.
  • Interpretation:- A low Biot Number (\( Bi < 0.1 \)) suggests that the material can be approximated using the Lumped Capacitance Method since it indicates minimal temperature gradients within the solid.
  • A higher Biot Number suggests more significant temperature variations within the object, requiring a spatial temperature profile analysis.
  • Calculating: For a flat plate like ours, the characteristic length \( L_{c} \) is half the thickness. This is crucial for accurately determining the Biot Number and thus the method of analysis.
In the scenario with our steel plate, the Biot Number was found to be 0.0444, indicating that the Lumped Capacitance Method is appropriate, simplifying the analysis significantly.

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Most popular questions from this chapter

The walls of a furnace are made of \(1.2\)-ft-thick concrete \(\left(k=0.64 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft} \cdot{ }^{\circ} \mathrm{F}\right.\) and \(\left.\alpha=0.023 \mathrm{ft}^{2} / \mathrm{h}\right)\). Initially, the furnace and the surrounding air are in thermal equilibrium at \(70^{\circ} \mathrm{F}\). The furnace is then fired, and the inner surfaces of the furnace are subjected to hot gases at \(1800^{\circ} \mathrm{F}\) with a very large heat transfer coefficient. Determine how long it will take for the temperature of the outer surface of the furnace walls to rise to \(70.1^{\circ} \mathrm{F}\). Answer: \(116 \mathrm{~min}\)

A long roll of 2-m-wide and \(0.5\)-cm-thick 1-Mn manganese steel plate coming off a furnace at \(820^{\circ} \mathrm{C}\) is to be quenched in an oil bath \(\left(c_{p}=2.0 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\right)\) at \(45^{\circ} \mathrm{C}\). The metal sheet is moving at a steady velocity of \(15 \mathrm{~m} / \mathrm{min}\), and the oil bath is \(9 \mathrm{~m}\) long. Taking the convection heat transfer coefficient on both sides of the plate to be \(860 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine the temperature of the sheet metal when it leaves the oil bath. Also, determine the required rate of heat removal from the oil to keep its temperature constant at \(45^{\circ} \mathrm{C}\).

In what medium is the lumped system analysis more likely to be applicable: in water or in air? Why?

Oxy-fuel combustion power plants use pulverized coal particles as fuel to burn in a pure oxygen environment to generate electricity. Before entering the furnace, pulverized spherical coal particles with an average diameter of \(300 \mu \mathrm{m}\), are being transported at \(2 \mathrm{~m} / \mathrm{s}\) through a \(3-\mathrm{m}\) long heated tube while suspended in hot air. The air temperature in the tube is \(900^{\circ} \mathrm{C}\) and the average convection heat transfer coefficient is \(250 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Determine the temperature of the coal particles at the exit of the heated tube, if the initial temperature of the particles is \(20^{\circ} \mathrm{C}\).

An electronic device dissipating \(20 \mathrm{~W}\) has a mass of \(20 \mathrm{~g}\), a specific heat of \(850 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), and a surface area of \(4 \mathrm{~cm}^{2}\). The device is lightly used, and it is on for \(5 \mathrm{~min}\) and then off for several hours, during which it cools to the ambient temperature of \(25^{\circ} \mathrm{C}\). Taking the heat transfer coefficient to be \(12 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine the temperature of the device at the end of the 5 -min operating period. What would your answer be if the device were attached to an aluminum heat sink having a mass of \(200 \mathrm{~g}\) and a surface area of \(80 \mathrm{~cm}^{2}\) ? Assume the device and the heat sink to be nearly isothermal.

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