/*! 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 26 An instrumentation package has a... [FREE SOLUTION] | 91Ó°ÊÓ

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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 \(0.30\).

Short Answer

Expert verified
Power dissipation range for \(\epsilon = 0.25\) is calculated using temperatures converted to Kelvin. Plot results for emissivity 0.20, 0.25, and 0.30 to visualize power impact.

Step by step solution

01

Understand the Problem

We need to find the range of acceptable power dissipation for a spherical instrument package based on given temperatures and emissivities. The system is inside a chamber with wall temperature at \(77\,\text{K}\). We need to calculate the net radiated power using these values and consider variances in emissivity (0.20, 0.25, 0.30).
02

Determine Surface Area

Calculate the surface area of the sphere using the formula: \(A = 4\pi r^2\), where \(r\) is the radius of the sphere. Since the diameter \(D = 100\,\text{nm}\), we find \(r = D/2 = 50\,\text{nm} = 50 \times 10^{-9}\,\text{m}\). Then, \(A = 4\pi (50 \times 10^{-9})^2\).
03

Apply Stefan-Boltzmann Law

The power radiated by a body is given by \(P = \epsilon \sigma A (T^4 - T_s^4)\), where \(\epsilon\) is the emissivity, \(\sigma = 5.67 \times 10^{-8} \, \text{W/m}^2\, \text{K}^4\) is the Stefan-Boltzmann constant, \(A\) is the surface area, \(T\) is the operational temperature in Kelvin, and \(T_s\) is the surrounding temperature in Kelvin (77 K in this case).
04

Convert Temperature

Convert the operational temperature limits from Celsius to Kelvin: \(40^\circ \text{C} = 313\,\text{K}\) and \(85^\circ \text{C} = 358\,\text{K}\).
05

Calculate Power Dissipation at Emissivity 0.25

Using \(\epsilon = 0.25\), calculate power dissipation at \(T = 313\,\text{K}\) and \(T = 358\,\text{K}\) using the formula from Step 3. Calculate \(P_\text{min} = 0.25 \times 5.67 \times 10^{-8} \times A \times (313^4 - 77^4)\) and \(P_\text{max} = 0.25 \times 5.67 \times 10^{-8} \times A \times (358^4 - 77^4)\).
06

Calculate Power Dissipation for Other Emissivities

Repeat calculations for \(\epsilon = 0.20\) and \(\epsilon = 0.30\) to find the power dissipation across the range of operational temperatures. Use the formula with \(\epsilon = 0.20\) and \(\epsilon = 0.30\) for \(T = 313\,\text{K}\) and \(T = 358\,\text{K}\).
07

Interpret Results and Graph

Compile the results for different emissivities and plot the power dissipation across the temperature range in graphical form. Comparison will show the variation in power dissipation depending on emissivity values.

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Key Concepts

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

Stefan-Boltzmann Law
The Stefan-Boltzmann Law is pivotal in understanding heat transfer by radiation. It states that the power radiated from a black body is directly proportional to the fourth power of its absolute temperature. Mathematically, it is expressed as \[ P = \sigma Ae(T^4 - T_s^4) \]where:
  • \(P\) is the power radiated.
  • \(\sigma\) is the Stefan-Boltzmann constant, \(5.67 \times 10^{-8} \text{W/m}^2 \text{K}^4\).
  • \(A\) is the area of the emitting surface.
  • \(e\) is the emissivity of the material.
  • \(T\) is the temperature of the object in Kelvin.
  • \(T_s\) is the surrounding temperature in Kelvin.
The law highlights that even small changes in temperature can significantly affect the radiative heat transfer because of the fourth power dependence on temperature. In practical applications, this informs us about the amount of radiant heat energy an object will emit when it is at a certain temperature.
Radiative Heat Transfer
Radiative heat transfer involves the exchange of energy through electromagnetic waves. This process does not require a medium, so it can take place in a vacuum.
When considering radiative heat transfer, you need to understand that it depends on the temperatures of the object and its surroundings, as well as the surface properties like emissivity. The primary equation used to model this type of heat transfer is closely related to the Stefan-Boltzmann law, as it helps determine the net radiative heat loss or gain by an object.
In scenarios like the space simulation chamber, the net energy transfer is calculated between the instrument package and the chamber walls. Here, radiation becomes the dominant mode of heat transfer since conduction and convection are minimal. The focus is on calculating how much power the spherical package must dissipate to remain within safe operating temperatures.
Spherical Geometry
Understanding spherical geometry is essential when dealing with objects like a spherical instrument package.
The surface area of a sphere is calculated using the formula \[ A = 4\pi r^2 \] where \(r\) is the radius of the sphere. For example, with a diameter of \100\ nm, the radius is half of that, \(50\ nm\).
Spherical shapes are particularly interesting in thermal considerations because they naturally minimize the surface area for a given volume, which can influence the rate of heat loss through radiation. Such considerations become especially important in equations of radiative heat transfer, where the surface area, \(A\), directly affects the calculation of radiative power loss according to the Stefan-Boltzmann Law.
Thermal Emissivity
Thermal emissivity is a material-specific property that measures the efficacy with which a surface emits thermal radiation. It is a ratio, with values ranging from 0 to 1. A surface with an emissivity of 1 is considered a perfect emitter (black body), while a lower emissivity means less effective radiation.
Emissivity significantly affects radiative heat transfer calculations; higher emissivity increases heat emission, and vice versa. In the context of the space simulation chamber, varying the emissivity of the spherical package (e.g., 0.20, 0.25, 0.30) directly impacts power dissipation as it determines how efficiently the package radiates heat away.
This property is crucial in optimizing thermal management, especially for equipment that operates within a specific temperature range. Understanding and controlling emissivity leads to better designs and operational efficiency by finely tuning the radiative heat exchange.

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Most popular questions from this chapter

An overhead 25-m-long, uninsulated industrial steam pipe of \(100 \mathrm{~mm}\) diameter is routed through a building whose walls and air are at \(25^{\circ} \mathrm{C}\). Pressurized steam maintains a pipe surface temperature of \(150^{7} \mathrm{C}\), and the coeffcient associated with natural convection is \(h=10\) \(W / m^{2} \cdot K\). The surface emissivity is \(\varepsilon=0.8\). (a) What is the rate of heat loss from the steam line? (b) If the steatm is generated in a gas-fired boiler operating at an efficiency of \(\eta f=0.90\) and natural gas is priced at \(C_{\varepsilon}=50.01\) per MJ, what is the annual cost of heat loss from the line?

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