/*! 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 54 A plate of thickness \(2 L=25 \m... [FREE SOLUTION] | 91影视

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Chapter 5: Problem 54

A plate of thickness \(2 L=25 \mathrm{~mm}\) at a temperature of \(600^{\circ} \mathrm{C}\) is removed from a hot pressing operation and must be cooled rapidly to achieve the required physical properties. The process engineer plans to use air jets to control the rate of cooling, but she is uncertain whether it is necessary to cool both sides (case 1 ) or only one side (case 2) of the plate. The concern is not just for the time-to-cool, but also for the maximum temperature difference within the plate. If this temperature difference is too large, the plate can experience significant warping. The air supply is at \(25^{\circ} \mathrm{C}\), and the convection coefficient on the surface is \(400 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The thermophysical properties of the plate are \(\rho=3000 \mathrm{~kg} / \mathrm{m}^{3}, c=750 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), and \(k=15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). (a) Using the IHT software, calculate and plot on one graph the temperature histories for cases 1 and 2 for a \(500-s\) cooling period. Compare the times required for the maximum temperature in the plate to reach \(100^{\circ} \mathrm{C}\). Assume no heat loss from the unexposed surface of case 2 . (b) For both cases, calculate and plot on one graph the variation with time of the maximum temperature difference in the plate. Comment on the relative magnitudes of the temperature gradients within the plate as a function of time.

Short Answer

Expert verified
In summary, to analyze the cooling process of a plate with an initial temperature of 600掳C and compare the cooling of both sides versus only one side of the plate: 1. Gather the given information, including plate dimensions, temperatures, convection coefficient, and plate properties. 2. Analyze each case: Case 1 with convection on both sides and Case 2 with convection only on one side of the plate. 3. Set up and input the parameters into IHT software. 4. Using the software, calculate and plot the temperature histories for Cases 1 and 2, showing the temperature variation within the plate for a cooling period of 500 seconds. 5. From the plot, compare the times required for the maximum temperature in the plate to reach 100掳C for both cases. 6. Calculate and plot the maximum temperature difference in the plate for both cases, showing how the differences change as a function of time. 7. Comment on the relative magnitudes of the temperature gradients for both cases, analyzing the suitability of each cooling method in terms of avoiding plate warping. The exact values and plots should be obtained using the IHT software; this answer provides a general approach and methodology for analyzing the problem.

Step by step solution

01

Gather the given information

Plate thickness: \(2L = 25 mm = 0.025 m\) Initial plate temperature: \(T_{initial} = 600 掳C\) Air temperature: \(T_{air} = 25 掳C\) Convection coefficient: \(h = 400 W/m^2路K\) Plate properties: - Density: \(\rho = 3000 kg/m^3\) - Specific heat: \(c = 750 J/kg路K\) - Thermal conductivity: \(k = 15 W/m路K\)
02

Analyze each case

Case 1: There will be heat loss through convection on both sides of the plate. Case 2: There will be heat loss through convection only on one side of the plate.
03

Set up the IHT software

Import the IHT software and input the given parameters: plate thickness, air temperature, convection coefficient, and plate properties.
04

Calculate and plot the temperature histories for cases 1 and 2

Configure the software to analyze both cases (cooling on one side or both, as described in Step 2). Run the IHT software, making sure to use the proper boundary conditions for each case. It should calculate the temperature histories inside the plate for a cooling period of 500 seconds. Plot the temperature histories for both cases on a single graph, showing how the temperature varies within the plate for the given study time.
05

Compare the times for the maximum temperature to reach 100掳C

From the plot obtained in Step 4, read and compare the times required for the maximum temperature in the plate to reach 100掳C for cases 1 and 2.
06

Calculate and plot the maximum temperature difference in the plate for both cases

Using IHT software, configure it to measure and calculate the maximum temperature difference within the plate as a function of time. Plot the calculated maximum temperature differences for both cases on a single graph, showing how the difference within the plate changes for the given study time.
07

Comment on the relative magnitudes of the temperature gradients

Analyze the plot obtained in Step 6. Discuss and compare the respective temperature gradients within the plate for both cases (one-sided cooling and both-sided cooling) as a function of time. Comment on the results, providing insights as to which case is more suitable in terms of cooling time and temperature difference to avoid significant warping of the plate. Note that the exact data values and plots should be obtained using the IHT software. This solution outlines the approach and methodology for tackling the problem using the software.

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Key Concepts

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

Convection Coefficient
The convection coefficient, symbolized by the variable 'h' in thermal calculations, is a measure of how effectively heat is transferred between a surface and a fluid moving over it. In our context involving the cooling of a plate, the convection coefficient is critical because it dictates the rate at which heat is drawn away from the plate by the air jets. A larger convection coefficient indicates a more effective cooling process.

When approaching a problem like this, we assume a uniform convection coefficient across the entire surface to simplify calculations. However, in actuality, this coefficient can vary due to changes in fluid velocity, surface roughness, and other factors. For students or engineers optimizing the cooling process, enhancing the convection coefficient could involve increasing air velocity or changing the nature of the air jets.
Thermal Conductivity
Thermal conductivity, represented by 'k' in heat transfer equations, embodies the material's ability to conduct heat. Within the scope of our plate-cooling scenario, the material's thermal conductivity influences how quickly the heart propagates through the plate's thickness. A high thermal conductivity means that heat from the plate鈥檚 hot center will spread to its cooler surfaces more rapidly.

In practical situations, if a plate has low thermal conductivity, we might observe that the surface cools down quickly while the core remains hot for a longer time, leading to a large temperature gradient, which in turn could induce warping. The value of the thermal conductivity plays an essential role in predicting and controlling the internal temperature distribution of the plate during cooling.
Temperature Gradient
The temperature gradient is defined as the rate of temperature change with respect to distance within an object. In our case, the gradient relates to how the temperature changes across the thickness of the plate. An excessive temperature gradient, particularly in materials that are sensitive to thermal stress, can lead to warping as different parts of the material expand or contract unevenly.

Understanding this concept is critical for engineers who are tasked with preventing damage during the cooling process. The temperature gradient is closely linked to both the material's thermal conductivity and the convection coefficient as these properties determine how heat flows within and away from the plate. When analyzing the cooling process, minimizing the temperature gradient to avoid warping is as important as the overall cooling time.
Cooling Process Optimization
Optimizing the cooling process involves careful balance between cooling the plate quickly and preventing thermal gradients that can cause warping. To optimize cooling, both the convection coefficient and the thermal conductivity must be considered. If the convection coefficient is too high, the surface cools rapidly, creating large temperature differentials. Conversely, if the thermal conductivity is too low, the plate's interior retains heat, again inducing significant gradients.

Moreover, factors like cooling uniformity are essential. For instance, cooling both sides of the plate may lead to a more homogeneous temperature distribution, reducing gradients and the risk of warping, which is an essential consideration in the cooling of large plates. The approach must therefore balance these factors to expedite the cooling process while maintaining the structural integrity of the plate.

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Most popular questions from this chapter

The stability criterion for the explicit method requires that the coefficient of the \(T_{m}^{p}\) term of the one-dimensional, finite-difference equation be zero or positive. Consider the situation for which the temperatures at the two neighboring nodes \(\left(T_{\mathrm{m}-1}^{p}, T_{\mathrm{m}+1}^{p}\right)\) are \(100^{\circ} \mathrm{C}\) while the center node \(\left(T_{m}^{p}\right)\) is at \(50^{\circ} \mathrm{C}\). Show that for values of \(F o>\frac{1}{2}\) the finite-difference equation will predict a value of \(T_{m}^{p+1}\) that violates the second law of thermodynamics.

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