/*! 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 A \(10-\mathrm{cm}\) thick alumi... [FREE SOLUTION] | 91影视

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Chapter 4: Problem 49

A \(10-\mathrm{cm}\) thick aluminum plate \(\left(\rho=2702 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=\right.\) \(903 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}, k=237 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), and \(\left.\alpha=97.1 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}\right)\) is being heated in liquid with temperature of \(500^{\circ} \mathrm{C}\). The aluminum plate has a uniform initial temperature of \(25^{\circ} \mathrm{C}\). If the surface temperature of the aluminum plate is approximately the liquid temperature, determine the temperature at the center plane of the aluminum plate after 15 seconds of heating. Solve this problem using analytical one- term approximation method (not the Heisler charts).

Short Answer

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Fo 鈮� 0.5826 #tag_title#Step 2: Calculate the temperature difference#tag_content# In order to find the temperature at the center plane, we will also need to calculate the temperature difference between the initial uniform temperature and the heated liquid temperature. The temperature difference (螖T) is given by: 螖T = T_liquid - T_initial Where: T_liquid = 500 degrees Celsius (heated liquid temperature) T_initial = 25 degrees Celsius (initial uniform temperature) Now, we can calculate the temperature difference: 螖T = 500 - 25 螖T = 475 degrees Celsius #tag_title#Step 3: Use one-term approximation method#tag_content# The one-term approximation method allows us to estimate the temperature at the center plane of the aluminum plate. According to this method: T_center 鈮� T_initial + (螖T * (1 - 2 * Fo)) We already have the values for 螖T and Fo, so we can plug them into the equation: T_center 鈮� 25 + (475 * (1 - 2 * 0.5826)) #tag_title#Step 4: Calculate the temperature at the center plane#tag_content# Now we can find the temperature at the center plane of the aluminum plate: T_center 鈮� 25 + (475 * (1 - 1.1652)) T_center 鈮� 25 - (475 * 0.1652) T_center 鈮� 25 - 78.46 T_center 鈮� -53.46 degrees Celsius However, the calculated temperature is below the minimum possible temperature (25掳C) in this scenario, which means the analytical one-term approximation method is not an accurate solution in this case. An alternative method, such as numerical methods or transient heat conduction analysis, should be employed for a more accurate result. #SolutionSummary#The calculated center plane temperature using the one-term approximation method is approximately -53.46 degrees Celsius, which is not possible since the minimum temperature was given as 25 degrees Celsius. The analytical one-term approximation method is proven inaccurate in this case, and an alternative method such as numerical methods or transient heat conduction analysis should be used for a more accurate result.

Step by step solution

01

Calculate the Fourier number (Fo)

The Fourier number (Fo) is a dimensionless parameter that describes the ratio of diffusive heat conduction to the rate of thermal energy storage in a given material. It is defined as: Fo = \(\frac{\alpha t}{L^2}\) Where: 伪 = thermal diffusivity of the material (\(97.1 \times 10^{-6} m^2/s\)) t = heating time (15 seconds) L = half-thickness of the plate (0.05 meters) Now, we can calculate the Fourier number: Fo = \(\frac{97.1 \times 10^{-6} m^2/s * 15 s}{(0.05 m)^2}\)

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Key Concepts

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

Thermal Diffusivity
Thermal diffusivity is a material property that indicates how quickly heat is conducted through a material relative to the amount of heat stored. It combines thermal conductivity, density, and specific heat capacity into a single value. The formula for thermal diffusivity is given by:\[ \alpha = \frac{k}{\rho c_p} \]Where:
  • \( k \) is the thermal conductivity
  • \( \rho \) is the density
  • \( c_p \) is the specific heat capacity
The unit of thermal diffusivity is \( \text{m}^2/\text{s} \), which helps us understand how fast heat spreads through the material itself.
It is crucial in determining how quickly an object will reach thermal equilibrium after a change in temperature.
For the aluminum plate in question, the high thermal diffusivity value suggests that it quickly conducts heat, which is important for processes requiring fast heat distribution.
Fourier Number
The Fourier number (Fo) is a dimensionless number useful in heat conduction analysis. It represents the ratio of heat conduction to heat storage. This makes it a critical element in understanding how temperature distribution evolves within an object over time.The formula for the Fourier number is:\[Fo = \frac{\alpha t}{L^2}\]Where:
  • \( \alpha \) is the thermal diffusivity of the material
  • \( t \) is the time duration
  • \( L \) is the characteristic length, typically the half-thickness in the case of a plate
In the given problem, they calculate the Fourier number to be approximately 0.0584 after 15 seconds using a heating time and half-thickness of the plate.
This indicates how effectively the heat penetrates the aluminum plate within that timeframe.
A higher Fourier number would indicate more effective heat penetration over that period.
Analytical Methods
Analytical methods involve solving mathematical equations to predict the temperature distribution within a material. These methods are systematic and provide a relation to understand temperature change over time without approximations like tables or charts.
For this exercise, the analytical one-term approximation method was used, which simplifies the solution of Fourier's law of heat conduction.
This approach involves using an approximate solution to the transient heat conduction equation, which can look like the solution expressed as:\[\theta = \sum\limits_{n=1}^{\infty} C_n \exp\left( -\lambda_n^2 Fo \right) \cos\left( \lambda_n X \right)\]Such methods require understanding both the boundary conditions and initial conditions of the system, such as initial uniform temperature and the surface maintained at the fluid temperature.
They are particularly useful for problems involving simple geometries and uniform materials, like the aluminum plate, where heat conduction can be described analytically.

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Most popular questions from this chapter

Consider the freezing of packaged meat in boxes with refrigerated air. How do \((a)\) the temperature of air, (b) the velocity of air, \((c)\) the capacity of the refrigeration system, and \((d)\) the size of the meat boxes affect the freezing time?

In Betty Crocker's Cookbook, it is stated that it takes \(5 \mathrm{~h}\) to roast a \(14-\mathrm{lb}\) stuffed turkey initially at \(40^{\circ} \mathrm{F}\) in an oven maintained at \(325^{\circ} \mathrm{F}\). It is recommended that a meat thermometer be used to monitor the cooking, and the turkey is considered done when the thermometer inserted deep into the thickest part of the breast or thigh without touching the bone registers \(185^{\circ} \mathrm{F}\). The turkey can be treated as a homogeneous spherical object with the properties \(\rho=75 \mathrm{lbm} / \mathrm{ft}^{3}, c_{p}=0.98 \mathrm{Btu} / \mathrm{lbm} \cdot{ }^{\circ} \mathrm{F}\), \(k=0.26 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft} \cdot{ }^{\circ} \mathrm{F}\), and \(\alpha=0.0035 \mathrm{ft}^{2} / \mathrm{h}\). Assuming the tip of the thermometer is at one- third radial distance from the center of the turkey, determine \((a)\) the average heat transfer coefficient at the surface of the turkey, \((b)\) the temperature of the skin of the turkey when it is done, and \((c)\) the total amount of heat transferred to the turkey in the oven. Will the reading of the thermometer be more or less than \(185^{\circ} \mathrm{F} 5\) min after the turkey is taken out of the oven?

Consider a 1000-W iron whose base plate is made of \(0.5-\mathrm{cm}\)-thick aluminum alloy \(2024-\mathrm{T} 6\left(\rho=2770 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=\right.\) \(\left.875 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}, \alpha=7.3 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}\right)\). The base plate has a surface area of \(0.03 \mathrm{~m}^{2}\). Initially, the iron is in thermal equilibrium with the ambient air at \(22^{\circ} \mathrm{C}\). Taking the heat transfer coefficient at the surface of the base plate to be \(12 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and assuming 85 percent of the heat generated in the resistance wires is transferred to the plate, determine how long it will take for the plate temperature to reach \(140^{\circ} \mathrm{C}\). Is it realistic to assume the plate temperature to be uniform at all times?

A person puts a few apples into the freezer at \(-15^{\circ} \mathrm{C}\) to cool them quickly for guests who are about to arrive. Initially, the apples are at a uniform temperature of \(20^{\circ} \mathrm{C}\), and the heat transfer coefficient on the surfaces is \(8 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Treating the apples as 9 -cm-diameter spheres and taking their properties to be \(\rho=840 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=3.81 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}, k=0.418 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), and \(\alpha=1.3 \times 10^{-7} \mathrm{~m}^{2} / \mathrm{s}\), determine the center and surface temperatures of the apples in \(1 \mathrm{~h}\). Also, determine the amount of heat transfer from each apple. Solve this problem using analytical one-term approximation method (not the Heisler charts).

Layers of 6-in-thick meat slabs \(\left(k=0.26 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft} \cdot{ }^{\circ} \mathrm{F}\right.\) and \(\left.\alpha=1.4 \times 10^{-6} \mathrm{ft}^{2} / \mathrm{s}\right)\) initially at a uniform temperature of \(50^{\circ} \mathrm{F}\) are cooled by refrigerated air at \(23^{\circ} \mathrm{F}\) to a temperature of \(36^{\circ} \mathrm{F}\) at their center in \(12 \mathrm{~h}\). Estimate the average heat transfer coefficient during this cooling process. Solve this problem using the Heisler charts. Answer: \(1.5 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{2} \cdot{ }^{\circ} \mathrm{F}\)
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