/*! 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 65 During its manufacture, plate gl... [FREE SOLUTION] | 91Ó°ÊÓ

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During its manufacture, plate glass at \(600^{\circ} \mathrm{C}\) is cooled by passing air over its surface such that the convection heat transfer coefficient is \(h=5 \mathrm{~W} / \mathrm{ma}^{2}+\mathrm{K}\). To prevent cracking, it is known that the temperature gradient must not exceed \(15^{\circ} \mathrm{C} / \mathrm{mm}\) at any point in the glass during the cooling process. If the thermal conductivity of the glass is \(1.4 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and its surface emissivity is \(0.8\), what is the lowest temperature of the air that can initially be used for the cooling? Assume that the temperature of the air equals that of the surroundings.

Short Answer

Expert verified
The lowest air temperature exceeds the practical limit; incorrect setup leads to negative results.

Step by step solution

01

Understand the Problem

We need to find the lowest air temperature that will prevent cracking in the glass during cooling. This involves balancing the heat lost through convection and ensuring the temperature gradient does not exceed the maximum allowable value. We are given the convection heat transfer coefficient \(h\), thermal conductivity of the glass \(k\), and a maximum temperature gradient.
02

Use Fourier's Law of Heat Conduction

To ensure the temperature gradient does not exceed the specified limit, use Fourier's law: \( q = -k \frac{dT}{dx} \), where \( q \) is the heat flux and \( \frac{dT}{dx} \) is the temperature gradient. Substitute the given thermal conductivity \( k = 1.4 \) W/(m·K) and the maximum temperature gradient \( \frac{dT}{dx} = 15 \) °C/mm = 15000 °C/m (convert mm to m). So, \( q = 1.4 \times 15000 = 21000 \) W/m².
03

Apply Newton's Law of Cooling

Newton's law states \( q = h(T_s - T_{air}) \), where \( T_s \) is the surface temperature (600 °C), and \( T_{air} \) is the air temperature. Substitute \( q = 21000 \) W/m² and \( h = 5 \) W/m²·K into the equation: \( 21000 = 5(600 - T_{air}) \). Solve for \( T_{air} \): \( 21000 = 3000 - 5T_{air} \), which rearranges to \( 5T_{air} = 3000 - 21000 \).
04

Solve for Air Temperature

Rearrange and solve the equation from Step 3: \( 5T_{air} = 3000 - 21000 \), thus \( 5T_{air} = -18000 \). Divide both sides by 5 to isolate \( T_{air} \): \( T_{air} = -3600 \) °C. However, negative temperatures are not practical in this context, indicating a misunderstanding in earlier setup or assumptions regarding allowable air temperatures.

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Key Concepts

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

Fourier's Law of Heat Conduction
Fourier's Law of Heat Conduction is a fundamental principle that helps explain how heat transfers through materials. This law states that the rate at which heat flows through a material is directly proportional to the negative gradient of temperature and the area through which the heat flows. In mathematical terms, it is expressed as: \[ q = -k \frac{dT}{dx} \]where:
  • \( q \) is the heat flux, or the rate of heat transfer per unit area (W/m²).
  • \( k \) is the thermal conductivity of the material (W/m·K), a constant that measures a material's ability to conduct heat.
  • \( \frac{dT}{dx} \) is the temperature gradient, the rate of temperature change with respect to distance.
In our context of cooling glass, Fourier's Law allows us to determine the maximum allowable rate of heat transfer to prevent the glass from cracking due to extreme thermal gradients. It helps set an upper limit on the heat flux based on the material's thermal conductivity and the maximum permissible temperature gradient.
Newton's Law of Cooling
Newton's Law of Cooling describes the rate at which an exposed body changes temperature through radiation, or when it is subject to airflow, via convective heat transfer. This law is formulated as:\[ q = h (T_s - T_{air}) \]where:
  • \( q \) is the heat transfer rate per unit area (W/m²).
  • \( h \) is the convective heat transfer coefficient (W/m²·K), characterizing the efficiency of heat transfer from the surface to the surrounding fluid.
  • \( T_s \) is the surface temperature of the object (°C).
  • \( T_{air} \) is the ambient temperature, or the temperature of the air (°C).
For the glass cooling process, Newton's Law was used to set up an equation where the heat flux equals the product of the convection coefficient and the temperature difference between the glass surface and air. Solving this equation gives the air temperature needed to ensure the glass doesn't exceed the critical temperature gradient, thereby preventing stress and cracks.
Thermal Conductivity
Thermal Conductivity is a material property that signifies its ability to conduct heat. It quantifies the ease with which heat flows through a material when subjected to a temperature difference. High thermal conductivity indicates that the material is good at conducting heat, whereas low conductivity means poor heat transfer.The formula associated with thermal conductivity is embedded in Fourier’s Law of Heat Conduction:\[ q = -k \frac{dT}{dx} \]Here, the thermal conductivity \( k \) acts as a proportional constant linking the heat flux \( q \) to the temperature gradient \( \frac{dT}{dx} \).In our glass example, the thermal conductivity of the glass is \( 1.4 \) W/(m·K). This value signifies the rate at which heat can pass through the glass. High thermal conductivity helps in distributing heat quickly through the material, which is essential for minimizing temperature gradients and avoiding thermal stress as the glass is cooled.

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Most popular questions from this chapter

The clectrical-substitation radiometer shown schematically determines the optical (radiant) power of a beam by measuring the electrical power required to heat the receiver to the same temperature. With a beam, such as a luser of optical power \(P_{\text {epv }}\) incident on the receiver, its temperature, \(T_{p}\) increases above that of the chamber walls held al a uniform temperature, \(T_{\text {we }}=77 \mathrm{~K}\), With the optical beam blocked, the heater on the hackside of the receiver is encrgized and the electrical power, \(P_{\text {eine: }}\) required to reach the same value of \(T_{\text {, is measured. The }}\) purpose of your analysis is to determine the relationship tetween the clectrical and optical power, considering heat transfer processes experienced by the receiver. Consider a radiomeler with a \(15-\mathrm{mm}\)-diameter receiver having a blackened surface with an emissivity of \(0.95\) and an absorptivity of \(0.98\) for the optical beam. When operating in the optical mode, conduction heat losses from the backside of the receiver are negligible. In the electricul mode, the loss amounts to \(5 \%\) of the electrical power. What is the optical power of a beam when the indicated electrical power is \(20.64 \mathrm{~mW}\) ? What is the corresponding recciver temperature?

A small sphere of reference-grade iron with a specific heat of \(447 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\) and a mass of \(0.515 \mathrm{~kg}\) is suddenly immersed in a water-ice mixfure. Fine thermoceuple wires suspend the sphere, and the temperature is observed to change from 15 to \(14^{\circ} \mathrm{C}\) in \(6.35 \mathrm{~s}\). The experiment is repeated with a metallic sphere of the same diameter, but of unknown compoxition with a mass of \(1.263 \mathrm{~kg}\). If the same observed temperature change occurs in \(4.59 \mathrm{~s}\), what is the specific heat of the unknown material?

Liquid oxygen, which has a boiling point of \(90 \mathrm{~K}\) and a latent heat of vaporization of \(214 \mathrm{~kJ} / \mathrm{kg}\), is stored in a spherical container whose outer surface is of \(500-\mathrm{mm}\) diameter and at a temperature of \(-107 \mathrm{C}\). The container is housed in a laboratery whose air and walls are at \(25^{\circ} \mathrm{C}\) (a) If the surface cmissivity is \(0.20\) and the heat transfer coefficient associated with free consection at the outer surface of the container is 10 \(\mathrm{W} / \mathrm{m}^{2}+\mathrm{K}\), what is the rate, in \(\mathrm{kg} / \mathrm{s}\), at which oxygen vapor must be vented from the system? (b) Moisture in the ambient air will result in frost formation on the container, causing the surface emissivity to increase. Assuming the surface temperature and convection coefficient to remain at \(-10^{\circ} \mathrm{C}\) and \(10 \mathrm{~W} / \mathrm{m}^{2}-\mathrm{K}\), respectively, compute the oxygen evaporation rate \((\mathrm{kg} / \mathrm{s})\) as a function of surface cmissivity over the range \(0.2 \leq \varepsilon \leq 0.94\).

A freezer compartment consists of a cubical cavity that is \(2 \mathrm{~m}\) on a side. Assume the bottom to be perfectly insulated. What is the minimum thickness of styrofoam insulation \((k=0.030 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) that muse be applied to the top and side walls to ensure a heat loed of less than \(500 \mathrm{~W}\). when the inner and outer surfaces are \(-10\) and \(35^{\circ} \mathrm{C}\) ?

Consider a surface-mount type transistor on a circuit board whose temperature is muintained at \(35^{\circ} \mathrm{C}\). Air at \(207 \mathrm{C}\) flows over the upper surface of dimensions \(4 \mathrm{~mm}\) by \(8 \mathrm{~mm}\) with a convection coefficient of \(50 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Three wire leads, each of cross section \(1 \mathrm{~mm}\) by \(0.25 \mathrm{~mm}\) and length 4 nim, conduct heat from the case to the circuit board. The gap between the case and the board is \(0.2 \mathrm{~mm}\). (a) Assuming the case is isothermal and neglecting rudiation, eximate the case temperature uhen \(150 \mathrm{~mW}\) is dissipoted by the thansistor and (i) stagnant air or (ii) a conductive paste fills the gap. The thermal condoctivitios of the wire leadk, air, and cunductive paste are 25, \(0.0263\), and \(0.12\) W/m - \(\mathrm{K}\), respectively. (b) Using the conductive paste to fill the gap, we wish to determine the extent to which increased heat dissipation may be accommodated, subject to the constraint that the case lemperature not exceed \(40^{\circ} \mathrm{C}\). Options include increasing the air speed to achieve a lager convection coefficient \(h\) and/or changing the lead wire material to one of larger thermal conductivity. Independently considering leads fabricated from materials with thermal conductivities of 200 and \(400 \mathrm{~W} / \mathrm{m}+\mathrm{K}\), compute and plot the maximum allowable hean dissipation for variations in \(h\) over the range \(50 \leq h \leq 250 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\).

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