/*! 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 32 A vacuum system, as used in sput... [FREE SOLUTION] | 91Ó°ÊÓ

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A vacuum system, as used in sputiering electrically conducting thin films on microcircuits, is comprised of a buseplate maintained by an electrical heater at \(300 \mathrm{~K}\) and a shroud within the enclosure maintained at \(77 \mathrm{~K}\) by a liquid-nitrogen coolant loop. The circular baseplate, insulated on the lower side, is \(0.3 \mathrm{~m}\) in diameter and has an emissivity of \(0.25\). (a) How much clectrical power must be provided to the baseplate heater? (b) At what rate must liquid nitrogen be supplied to the shiroud if its heat of vaporization is \(125 \mathrm{~kJ} / \mathrm{kg}\) ? (c) To reduce the liquid-nitrogen consumption, it is proposed to bond a thin sheet of aluminura foil \((\varepsilon=0.09)\) to the baseplate. Will this have the desfred effect?

Short Answer

Expert verified
(a) 28.74 W; (b) 0.00023 kg/s; (c) Yes, it reduces consumption.

Step by step solution

01

Calculate the Radiative Heat Exchange

The heat exchange by radiation between the baseplate and the shroud is given by:\[ Q = \frac{\sigma A (T_b^4 - T_s^4)}{ \frac{1}{\varepsilon_b} + \frac{1}{\varepsilon_s} - 1} \] Where: - \(T_b = 300 \text{ K}\) is the baseplate temperature,- \(T_s = 77 \text{ K}\) is the shroud temperature,- \(\sigma = 5.67 \times 10^{-8} \text{ W/m}^2\text{K}^4\) is the Stefan-Boltzmann constant,- \(A = \pi d^2 / 4\) is the area of the baseplate with \(d = 0.3 \text{ m}\),- \(\varepsilon_b = 0.25\) is the emissivity of the baseplate,- \(\varepsilon_s \approx 1\) considering the shroud is a good blackbody radiator.Substituting these values yields:\[ A = \pi (0.3^2) / 4 = 0.0707 \text{ m}^2 \]\[ \frac{1}{0.25} + 1 - 1 = 4 \]Thus, \[ Q = \frac{5.67 \times 10^{-8} \times 0.0707 \times (300^4 - 77^4)}{4} = 28.74 \text{ W} \].The electrical power required is therefore approximately 28.74 Watts.
02

Calculate the Rate of Liquid Nitrogen Consumption

Given the heat of vaporization \(L_v = 125 \text{ kJ/kg}\), the rate at which liquid nitrogen must be supplied is given by:\[ \dot{m} = \frac{Q}{L_v} \]Where \(Q = 28.74 \text{ W} = 28.74 \text{ J/s}\). Converting \(Q\) to kJ/s:\[ Q = 0.02874 \text{ kJ/s}\]Thus,\[ \dot{m} = \frac{0.02874 \text{ kJ/s}}{125 \text{ kJ/kg}} = 2.3 \times 10^{-4} \text{ kg/s} \].Hence, the flow rate of liquid nitrogen is approximately \(2.3 \times 10^{-4} \text{ kg/s}\).
03

Evaluate the Effect of Adding Aluminum Foil

By attaching aluminum foil with emissivity \(\varepsilon = 0.09\), recalculate the radiative heat exchange:\[ Q_{new} = \frac{\sigma A (T_b^4 - T_s^4)}{ \frac{1}{\varepsilon_{foil}} + \frac{1}{\varepsilon_s} - 1} \]With \(\varepsilon_{foil} = 0.09\),\[ \frac{1}{0.09} + 1 - 1 = 11.11 \]\[ Q_{new} = \frac{5.67 \times 10^{-8} \times 0.0707 \times (300^4 - 77^4)}{11.11} = 10.34 \text{ W} \]The heat exchange is reduced, so the aluminum foil addition will lower the liquid nitrogen consumption.

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Key Concepts

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

Radiative Heat Exchange
Radiative heat exchange is a fundamental concept in thermal systems where heat is transferred through electromagnetic radiation. This process does not require any medium, unlike conduction or convection, and occurs across surfaces at different temperatures. In scenarios where components like a baseplate and a shroud exist in an enclosure, radiative heat exchange is crucial to understanding the thermal dynamics.
The amount of radiative heat transferred is primarily determined by temperature differences and the surface properties, such as emissivity, of the involved bodies. In the given exercise, the baseplate exchanges heat with the shroud maintained at different temperatures, which is essential for calculating the required electrical power to maintain the baseplate temperature. The calculation involves significant equations, integrating surface area and temperature differences using the Stefan-Boltzmann Law.
Emissivity
Emissivity is a measure of a material's ability to emit infrared energy. It is central to determining how efficiently a surface radiates energy compared to a perfect black body, which has an emissivity of 1. In the exercise, the baseplate's emissivity is given as 0.25. This implies that it is not a very efficient radiator of heat. The importance of emissivity lies in its influence on the transfer of radiative heat.
  • A higher emissivity means more radiative heat loss.
  • A lower emissivity results in reduced heat exchange.
Adding materials like aluminum foil, with a lower emissivity (0.09 in the exercise), can considerably reduce heat loss by radiation. This is because less energy is radiated away, consequently requiring less input power to sustain the baseplate temperature. The additional layer effectively cuts down on energy transfer, maximizing efficiency.
Thermal Systems
Thermal systems involve various components for the regulation, transfer, and conservation of heat energy. In the context of the exercise, the vacuum system incorporating a baseplate and a shroud serves as an intriguing thermal system application. Thermal systems include components that manage temperature gradients through design considerations like insulation, heating elements, and reflective barriers.
Understanding the interplay of different elements allows for optimizing the thermal performance. For example, in this exercise, the insulated baseplate and the cooling provision of a shroud define the system boundaries. Cooling systems such as the liquid-nitrogen loop are essential for maintaining desired operational temperatures. Evaluating these systems often involves balancing heat input/output to achieve stability at the required conditions.
Stefan-Boltzmann Law
The Stefan-Boltzmann Law is fundamental in understanding radiative heat transfer in thermal systems. It provides the scientific basis for calculating the power radiated by a body in relation to its absolute temperature, given by:
\[ Q = rac{ ext{σ A} (T_b^4 - T_s^4)}{ rac{1}{ ext{ε}_b} + rac{1}{ ext{ε}_s} - 1} \]
In the formula:
  • σ signifies the Stefan-Boltzmann constant, essential for quantifying thermal radiation.
  • A represents the surface area engaged in radiative exchange.
  • \(T_b\) and \(T_s\) are the absolute temperatures of the baseplate and shroud, respectively.
  • ε values (emissivities) reflect how real surfaces deviate from perfect black bodies.
This equation enables precise calculation of the heat radiated between two objects, considering both temperatures and surface properties. By manipulating factors like emissivity or temperature in radiative calculations, systems can be optimized for energy efficiency, as demonstrated when applying the Stefan-Boltzmann Law to potential design improvements in the exercise.

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Most popular questions from this chapter

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 \(35^{\circ} \mathrm{C}\) ?

A rectangular forced air heating duct is suspended from the ceiling of a basement whose air and walls are at a termperiture of \(T_{\mathrm{w}}=T_{\mathrm{er}}=5^{\circ} \mathrm{C}\). The duct is \(15 \mathrm{~m}\) long. and its cross-section is \(350 \mathrm{~mm} \times 200 \mathrm{~mm}\). (a) For an uninsulated duct whose average surfice temperature is \(50^{\circ} \mathrm{C}\), estimate the rate of heat loss from the duct. The surface emissivity and convection coefficient are approximately \(0.5\) and \(4 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). respectively. (b) If heated air enters the duct at \(58^{\circ} \mathrm{C}\) and a velocity of \(4 \mathrm{~m} / \mathrm{s}\) and the heat loss corresponds to the result of part (a), what is the outlet temperature? The density and specific heat of the air may be assumed to be \(\rho=1.10 \mathrm{~kg} / \mathrm{m}^{\prime}\) and \(c_{p}=1008 \mathrm{~J} / \mathrm{kg}\) * \(\mathrm{K}\), respectively.

An instrumentation package has a spherical outer surface of diameter \(D=100 \mathrm{~nm}\) and emissivity \(c=0.25\). The package is placed in a large space simulation chamber whose walls are maintained at \(77 \mathrm{~K}\). If operation of the electronic components is restricted to the lemperature range \(40 \leq 7 \leq 85^{\circ} \mathrm{C}\), what is the range of acceptable power dissipation for the package? Display your results graphically, showing also the effect of variations in the emissivity by censidering values of \(0.20\) and \(0.30\).

The free convection heat transfer cocfficient on a thin hot vertical plate suspended in still air can be determined from observations of the change in plate temperature with time as it cuoks. Ascuming the plate is iscthermal and radiation exchange with its surroundings is negligible, evaluate the convection cocfficient at the instant of time when the plate temperature is \(225^{\circ} \mathrm{C}\) and the change in plate temperature with time \((d T / d)\) is \(-0.022 \mathrm{~K} / \mathrm{s}\). The ambient air temperature is \(25^{\prime} \mathrm{C}\) and the plate measures \(0.3 \times 0.3 \mathrm{~m}\) with a mass of \(3.75 \mathrm{~kg}\) and a specific heat of \(2770 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\).

An aluminum plate 4 mun thick is moxinted in a horizontal position, and its botlom surface is well insulated. \(\mathrm{A}\) special, thin costing is applied to the top surface such that it atsorts \(80 \%\) of any incident solar radiation, while having an cmissivity of \(0.25\). The density \(\rho\) and specific heat \(c\) of aluminum are known to be \(2700 \mathrm{~kg} / \mathrm{m}^{3}\) and \(900 \mathrm{~J} / \mathrm{kg}+\mathrm{K}_{\mathrm{q}}\) respectively. (a) Consider conditions for which the plate is at a temperature of \(25^{\circ} \mathrm{C}\) and its top surface is suddenly exposed to ambient air at \(T_{=}=20^{\circ} \mathrm{C}\) and to solar radiation that provides an incident flux of \(900 \mathrm{~W} / \mathrm{m}^{2}\). The convection heat transfer cuefficient between the surfixce and the uir is \(h=20 \mathrm{~W} / \mathrm{m}^{2}+\mathrm{K}^{2}\). What is the initial rate of change of the plate temperature? (b) What will be the equitibrium temperature of the plate when sleady-state conditions are reached? (c) The surface radialive properties depend on the specific nature of the appliced couting, Compute and plot the steady-state temperature as at functions of the crissivity for \(0.05 \leq \varepsilon \leq 1\). with all other oondations remaining as prescrited. Repeat your calculations for values of \(\alpha_{3}=0.5\) and \(1.0\), and plot the testalts with those obtained fot \(\sigma_{3}=0.8\). If the intent is to maximize the plate temperature, what is the most desirable combination of the plate emissivity and its absorptivity to molar radiation?

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