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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.

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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.

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Most popular questions from this chapter

Heat is lost steadily through a \(0.5-\mathrm{cm}\) thick \(2 \mathrm{~m} \times 3 \mathrm{~m}\) window glass whose thermal conductivity is \(0.7 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). The inner and outer surface temperatures of the glass are measured to be \(12^{\circ} \mathrm{C}\) to \(9^{\circ} \mathrm{C}\). The rate of heat loss by conduction through the glass is (a) \(420 \mathrm{~W}\) (b) \(5040 \mathrm{~W}\) (c) \(17,600 \mathrm{~W}\) (d) \(1256 \mathrm{~W}\) (e) \(2520 \mathrm{~W}\)

Eggs with a mass of \(0.15 \mathrm{~kg}\) per egg and a specific heat of \(3.32 \mathrm{~kJ} / \mathrm{kg} \cdot{ }^{\circ} \mathrm{C}\) are cooled from \(32^{\circ} \mathrm{C}\) to \(10^{\circ} \mathrm{C}\) at a rate of 200 eggs per minute. The rate of heat removal from the eggs is (a) \(7.3 \mathrm{~kW}\) (b) \(53 \mathrm{~kW}\) (c) \(17 \mathrm{~kW}\) (d) \(438 \mathrm{~kW}\) (e) \(37 \mathrm{~kW}\)

A concrete wall with a surface area of \(20 \mathrm{~m}^{2}\) and a thickness of \(0.30 \mathrm{~m}\) separates conditioned room air from ambient air. The temperature of the inner surface of the wall \(\left(T_{1}\right)\) is maintained at \(25^{\circ} \mathrm{C}\). (a) Determine the heat loss \(\dot{Q}(\mathrm{~W})\) through the concrete wall for three thermal conductivity values of \((0.75,1\), and \(1.25 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) and outer wall surface temperatures of \(T_{2}=-15,-10,-5,0,5,10,15,20,25,30\), and \(38^{\circ} \mathrm{C}\) (a total of 11 data points for each thermal conductivity value). Tabulate the results for all three cases in one table. Also provide a computer generated graph [Heat loss, \(\dot{Q}(\mathrm{~W})\) vs. Outside wall temperature, \(\left.T_{2}\left({ }^{\circ} \mathrm{C}\right)\right]\) for the display of your results. The results for all three cases should be plotted on the same graph. (b) Discuss your results for the three cases.

The rate of heat loss through a unit surface area of a window per unit temperature difference between the indoors and the outdoors is called the \(U\)-factor. The value of the \(U\)-factor ranges from about \(1.25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (or \(0.22 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{2}{ }^{\circ} \mathrm{F}\) ) for low-e coated, argon-filled, quadruple-pane windows to \(6.25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (or \(1.1 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{2}{ }^{\circ} \mathrm{F}\) ) for a single-pane window with aluminum frames. Determine the range for the rate of heat loss through a \(1.2-\mathrm{m} \times 1.8-\mathrm{m}\) window of a house that is maintained at \(20^{\circ} \mathrm{C}\) when the outdoor air temperature is \(-8^{\circ} \mathrm{C}\).

In a power plant, pipes transporting superheated vapor are very common. Superheated vapor is flowing at a rate of \(0.3 \mathrm{~kg} / \mathrm{s}\) inside a pipe with \(5 \mathrm{~cm}\) in diameter and \(10 \mathrm{~m}\) in length. The pipe is located in a power plant at \(20^{\circ} \mathrm{C}\), and has a uniform pipe surface temperature of \(100^{\circ} \mathrm{C}\). If the temperature drop between the inlet and exit of the pipe is \(30^{\circ} \mathrm{C}\), and the specific heat of the vapor is \(2190 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), determine the heat transfer coefficient as a result of convection between the pipe surface and the surrounding.

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