An electronic package in the shape of a sphere with an outer diameter of \(100 \mathrm{~mm}\) is placed in a large laboratory room. The surface emissivity of the package can assume three different values \((0.2,0.25\), and \(0.3)\). The walls of the room are maintained at a constant temperature of \(77 \mathrm{~K}\). The electronics in this package can only operate in the surface temperature range of \(40^{\circ} \mathrm{C} \leq T_{s} \leq 85^{\circ} \mathrm{C}\). Determine the range of power dissipation \((\dot{W})\) for the electronic package over this temperature range for the three surface emissivity values \((\varepsilon)\). Plot the results in terms of \(\dot{W}(\mathrm{~W})\) vs. \(T_{s}\left({ }^{\circ} \mathrm{C}\right)\) for the three different values of emissivity over a surface temperature range of 40 to \(85^{\circ} \mathrm{C}\) with temperature increments of \(5^{\circ} \mathrm{C}\) (total of 10 data points for each \(\varepsilon\) value). Provide a computer generated graph for the display of your results and tabulate the data used for the graph. Comment on the results obtained.

Short Answer

Expert verified
## Short Answer Calculate the power dissipation for a spherical electronic package with a diameter of 100 mm and surface temperatures ranging from 40 to 85 °C using emissivity values of 0.2, 0.25, and 0.3, and a room temperature of 77 K. Create a graph and table that compares the power dissipation values for each combination of surface temperature and emissivity. Analyze the results, and comment on how power dissipation varies with changes in surface temperature and emissivity, as well as its implications on the electronic package's performance and efficiency.

Step by step solution

01

Calculate the Surface Area of the Sphere

To calculate the surface area of the sphere, we'll use the following equation: \(A = 4\pi r^2\) where \(r\) is the radius of the sphere. Given that the sphere has a diameter of 100 mm, we have: \(r = \frac{100}{2}\mathrm{~mm} = 50\mathrm{~mm} = 0.05\mathrm{~m}\) Now, we can calculate the surface area: \(A = 4\pi (0.05)^2 = 0.0314\mathrm{~m^2}\)
02

Convert the Surface Temperature Range to Kelvin

To work with the radiation heat transfer equation, we need to convert the surface temperature range from Celsius to Kelvin. To do this, simply add 273.15 to the temperatures in Celsius: \(T_{s, \mathrm{min}} = 40^{\circ} \mathrm{C} + 273.15 = 313.15\ \mathrm{K}\) \(T_{s, \mathrm{max}} = 85^{\circ} \mathrm{C} + 273.15 = 358.15\ \mathrm{K}\)
03

Calculate the Power Dissipation for Different Emissivity Values and Surface Temperatures

Now we need to calculate the power dissipation, \(\dot{W}\), over the temperature range with \(\Delta T_s = 5^{\circ} \mathrm{C}\) increments for the three different emissivity values \((0.2, 0.25 , 0.3),\) using the formula: \(\dot{W} = A \cdot \varepsilon \cdot \sigma \cdot (T_s^4 - T_\infty^4)\) Given that the room's temperature is \(77\ \mathrm{K}\), we will iterate through all the combinations of surface temperatures and emissivity values, calculating the associated power dissipations using the radiation heat transfer equation, and store the results in a table.
04

Plot and Tabulate the Results

Once the power dissipation values are calculated for each combination of surface temperature and emissivity, plot the results on a graph with \(\dot{W}(\mathrm{~W})\) on the y-axis and \(T_{s}\left({ }^{\circ} \mathrm{C}\right)\) on the x-axis. Tabulate the data used for the graph, including the values of power dissipation, emissivity, and surface temperature for each data point.
05

Comment on the Results

Analyze the graph and table, and comment on how the power dissipation values vary with surface temperatures and emissivities. Determine the range of power dissipation for each emissivity value and discuss the implications for the electronic package's performance and efficiency.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Surface Emissivity
Surface emissivity is a critical factor in the study of radiation heat transfer. It is a measure of how effectively a material can emit thermal radiation compared to an ideal black body. An ideal black body has an emissivity value of 1, while non-black bodies have values between 0 and 1. Materials with higher emissivity emit more radiation at a given temperature than those with lower emissivity.

When designing electronic packages, engineers must consider the emissivity of the material to predict how much heat the device can dissipate. For instance, in the given exercise, the emissivity values of 0.2, 0.25, and 0.3 suggest that the electronic package is not an ideal emitter, affecting its ability to transfer heat to its surroundings. As emissivity increases, the rate at which the package can dissipate heat through radiation also increases, which is crucial for maintaining the electronic components within their operational temperature range.
Power Dissipation Calculation
Power dissipation in the context of electronic packages refers to the process of eliminating excess heat generated by the electronic device during operation. Calculating power dissipation helps ensure that the device operates within safe temperature limits to prevent overheating and potential damage.

Power dissipation can be calculated using the formula:
\[\begin{equation}\dot{W} = A \cdot \varepsilon \cdot \sigma \cdot (T_s^4 - T_\infty^4)\end{equation}\] where \(\dot{W}\) is the power dissipation, \(A\) is the surface area of the device, \(\varepsilon\) is the surface emissivity, \(\sigma\) is the Stefan-Boltzmann constant, \(T_s\) is the surface temperature of the device in Kelvin, and \(T_\infty\) is the ambient temperature in Kelvin. The fourth power in the formula signifies that radiative heat transfer is highly sensitive to temperature changes, which means that even small changes in surface temperature can result in significant changes in power dissipation.
Sphere Surface Area
The surface area of a sphere is one of the key parameters in calculating radiation heat transfer for spherical objects like the electronic package in the exercise. The formula to calculate the surface area \(A\) of a sphere is: \[\begin{equation}A = 4\pi r^2\end{equation}\] where \(r\) is the radius of the sphere. For a sphere with a given diameter, the radius is half that value. Surface area directly influences the power dissipation calculation because it represents the area available for heat transfer. Larger surfaces dissipate more heat while smaller ones dissipate less. In the given example, knowing the surface area is crucial for determining how the electronic package manages heat through radiation.
Temperature Conversion to Kelvin
Temperature conversion to Kelvin is a fundamental step in thermal calculations involving radiation. The Kelvin scale is essential because it is an absolute temperature scale, starting at absolute zero, where no thermal motion occurs at the atomic level.

To convert Celsius to Kelvin, the formula is:
\[\begin{equation}T(K) = T(^\text{o}C) + 273.15\end{equation}\]In the exercise, the surface temperatures given in Celsius must be converted to Kelvin to apply the radiation heat transfer equation correctly. The accuracy of such conversions directly impacts the reliability of power dissipation calculations and the overall analysis of the thermal management of electronic packages.
Electronic Package Thermal Management
Thermal management of an electronic package is a critical aspect of its design and operation. It involves maintaining the temperature of electronic devices within safe operating limits to prevent thermal-related failures and extend their lifespan. Effective thermal management strategies can include passive methods, such as conduction, convection, and radiation, as well as active systems involving cooling devices like fans or heat sinks.

In the context of the problem at hand, thermal management is focused on ensuring that the power dissipation via radiation keeps the electronic package's surface temperature within a specific range. Engineers must consider various factors, such as ambient temperature, material properties, and surface characteristics, to manage the heat produced by the electronics. The relationship between surface emissivity and power dissipation is especially crucial because it determines how effectively the package can radiate heat away, influencing the device's performance and safety.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Can all three modes of heat transfer occur simultaneously (in parallel) in a medium?

A 40-cm-long, 0.4-cm-diameter electric resistance wire submerged in water is used to determine the convection heat transfer coefficient in water during boiling at \(1 \mathrm{~atm}\) pressure. The surface temperature of the wire is measured to be \(114^{\circ} \mathrm{C}\) when a wattmeter indicates the electric power consumption to be \(7.6 \mathrm{~kW}\). The heat transfer coefficient is (a) \(108 \mathrm{~kW} / \mathrm{m}^{2} \cdot \mathrm{K}\) (b) \(13.3 \mathrm{~kW} / \mathrm{m}^{2} \cdot \mathrm{K}\) (c) \(68.1 \mathrm{~kW} / \mathrm{m}^{2} \cdot \mathrm{K}\) (d) \(0.76 \mathrm{~kW} / \mathrm{m}^{2} \cdot \mathrm{K}\) (e) \(256 \mathrm{~kW} / \mathrm{m}^{2} \cdot \mathrm{K}\)

An engineer who is working on the heat transfer analysis of a house in English units needs the convection heat transfer coefficient on the outer surface of the house. But the only value he can find from his handbooks is \(22 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), which is in SI units. The engineer does not have a direct conversion factor between the two unit systems for the convection heat transfer coefficient. Using the conversion factors between \(\mathrm{W}\) and \(\mathrm{Btu} / \mathrm{h}, \mathrm{m}\) and \(\mathrm{ft}\), and \({ }^{\circ} \mathrm{C}\) and \({ }^{\circ} \mathrm{F}\), express the given convection heat transfer coefficient in Btu/ \(\mathrm{h} \cdot \mathrm{ft}^{2}{ }^{\circ} \mathrm{F}\). Answer: \(3.87 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{2}{ }^{\circ} \mathrm{F}\)

An AISI 316 stainless steel spherical container is used for storing chemicals undergoing exothermic reaction that provides a uniform heat flux of \(60 \mathrm{~kW} / \mathrm{m}^{2}\) to the container's inner surface. The container has an inner diameter of \(1 \mathrm{~m}\) and a wall thickness of \(5 \mathrm{~cm}\). For safety reason to prevent thermal burn on individuals working around the container, it is necessary to keep the container's outer surface temperature below \(50^{\circ} \mathrm{C}\). If the ambient temperature is \(23^{\circ} \mathrm{C}\), determine the necessary convection heat transfer coefficient to keep the container's outer surface temperature below \(50^{\circ} \mathrm{C}\). Is the necessary convection heat transfer coefficient feasible with free convection of air? If not, discuss other option to prevent the container's outer surface temperature from causing thermal burn.

Solve this system of three equations with three unknowns using EES: $$ \begin{array}{r} x^{2} y-z=1.5 \\ x-3 y^{0.5}+x z=-2 \\ x+y-z=4.2 \end{array} $$

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free