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Electronic power devices are mounted to a heat sink having an exposed surface area of \(0.045 \mathrm{~m}^{2}\) and an emissivity of \(0.80\). When the devices dissipate a total power of \(20 \mathrm{~W}\) and the air and surroundings are at \(27^{\circ} \mathrm{C}\), the average sink temperature is \(42^{\circ} \mathrm{C}\). What average temperature will the heat sink reach when the devices dissipate \(30 \mathrm{~W}\) for the same environmental condition?

Short Answer

Expert verified
The average temperature the heat sink will reach when the devices dissipate 30 W for the same environmental condition is approximately \(57.424^{\circ} \mathrm{C}\).

Step by step solution

01

Convert temperatures to Kelvin

To use the Stefan-Boltzmann law, we need to convert the temperatures to Kelvin, since it requires absolute temperatures. So, we convert the given temperatures from Celsius to Kelvin by simply adding 273.15 to them: \(T_{a} = 27^{\circ} \mathrm{C} + 273.15 = 300.15 \mathrm{~K}\) \(T_{s1} = 42^{\circ} \mathrm{C} + 273.15 = 315.15 \mathrm{~K}\)
02

Calculate the heat transfer by radiation at 20W

Now, using the Stefan-Boltzmann law, we calculate the heat transfer by radiation when the devices dissipate 20 W. The formula is: \(q = e * A * \sigma * (T_{s}^4 - T_{a}^4)\) where: - \(q\) is the heat transfer by radiation; - \(e\) is the emissivity of the heat sink; - \(A\) is the exposed surface area; - \(\sigma\) is the Stefan-Boltzmann constant (\(\sigma = 5.67 * 10^{-8} \mathrm{~Wm}^{-2} \mathrm{~K}^{-4}\)); - \(T_{s}\) is the heat sink temperature; - \(T_{a}\) is the air temperature. So, when the devices dissipate 20 W, we have: \(20 = 0.80 * 0.045 * 5.67 * 10^{-8} * (315.15^4 - 300.15^4)\)
03

Solve for T_s when the devices dissipate 30 W

Now, when the devices dissipate 30 W, we need to find the new heat sink temperature, \(T_{s2}\). Therefore, we form a similar equation: \(30 = 0.80 * 0.045 * 5.67 * 10^{-8} * (T_{s2}^4 - 300.15^4)\)
04

Solve for T_s2

We can solve for \(T_{s2}\) by either rearranging the equation to find \(T_{s2}^4\) directly or by using a numerical method like the Newton-Raphson method. We choose to rearrange the equation to find \(T_{s2}^4\): \(T_{s2}^4 = 300.15^4 + \frac{30}{0.80 * 0.045 * 5.67 * 10^{-8}}\) Now solve for \(T_{s2}\): \(T_{s2} = \sqrt[4]{300.15^4 + \frac{30}{0.80 * 0.045 * 5.67 * 10^{-8}}}\)
05

Convert T_s2 back to Celsius

Now that we have found the value of \(T_{s2}\) in Kelvin, we can convert it back to Celsius by simply subtracting 273.15: \[T_{s2} = \sqrt[4]{300.15^4 + \frac{30}{0.80 * 0.045 * 5.67 * 10^{-8}}} - 273.15\] Compute the value of \(T_{s2}\): \(T_{s2} \approx 57.424^{\circ} \mathrm{C}\) So, the average temperature the heat sink will reach when the devices dissipate 30 W for the same environmental condition is approximately \(57.424^{\circ} \mathrm{C}\).

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Key Concepts

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

Understanding the Stefan-Boltzmann Law
The Stefan-Boltzmann law is a principle critical in calculating the radiative heat transfer, particularly in the process of cooling electronic devices. This law states that the power radiated per unit area from a black body is directly proportional to the fourth power of the black body's absolute temperature. Essentially, this law provides us with a mathematical relationship to estimate how much heat energy is radiated from a surface at a certain temperature into its surroundings.

Let's clarify this with an example. If you have an electronic device that generates heat and a heat sink that's meant to dissipate this heat, the Stefan-Boltzmann law will be part of the calculation to determine how effectively the heat sink is radiating energy away. In our exercise, the given power dissipated by the electronic devices initially is 20 W (watts), and we used the Stefan-Boltzmann law to calculate the heat transfer which has been observed at the sink's temperature. The formula used encapsulates the emissivity, surface area, Stefan-Boltzmann constant, and the temperature difference between the heat sink and the air, all raised to the fourth power.

The power of this law lies in how it enables us to predict changes in heat transfer. When the dissipated power increases to 30 W, we can anticipate a higher temperature for the heat sink as it must radiate more energy to reach a state of balance with its environment. By applying the law correctly, one can effectively manage and design heat dissipation systems in electronic circuits, preserving component longevity and performance.
Radiative Heat Transfer in Thermal Management
Radiative heat transfer is one of the primary modes of thermal energy transfer – the other two being conduction and convection. Heat radiation occurs through electromagnetic waves and doesn't require a medium to travel through, which makes it distinct in scenarios like the vacuum of space or, relevant to our context, thermal management in electronics. In devices, radiative heat transfer is mainly concerned with managing the thermal energy that electronic components emit due to their operation.

For electronic components mounted onto a heat sink, like in our problem, radiative heat transfer plays a significant role in reducing the overall temperature by emitting infrared radiation. The efficiency of this process is influenced by factors such as surface temperature and emissivity. High temperatures and emissivity closer to one (the maximum for a perfect black body) improve the efficiency of radiation as a method of heat discharge. Moreover, radiative transfer complements conductive and convective cooling processes in dissipating heat away from delicate electronic parts. An accuracy in calculating radiative heat transfer through the Stefan-Boltzmann law is thus essential for predicting and controlling the operating temperatures of electronic devices.
Thermal Management Solutions in Electronics
Thermal management in electronics is a critical field, dedicated to maintaining components within safe operating temperatures. Heat sinks are a common solution, designed to enhance heat dissipation through increased surface area, often assisted by convective airflow. However, managing the heat often involves a multi-faceted approach including material selection, component layout, and active cooling mechanisms like fans.

In the given exercise, we are concerned with the passive cooling process using a heat sink. Its capability to handle the thermal load is being tested by increasing the power dissipation from 20 W to 30 W. Through the calculated results, which predict an increase in the heat sink's operational temperature, we acquire practical insights into choosing appropriate thermal management strategies. These could range from selecting heat sinks with higher emissivity, improving the airflow around the components for better convection or even supplementing with active cooling systems if the temperatures approach the upper limits of safety. Good thermal management prevents overheating, which in turn avoids performance degradation or failure, ensuring reliability and longevity of electronic systems.

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Most popular questions from this chapter

In considering the following problems involving heat transfer in the natural environment (outdoors), recognize that solar radiation is comprised of long and short wavelength components. If this radiation is incident on a semitransparent medium, such as water or glass, two things will happen to the nonreflected portion of the radiation. The long wavelength component will be absorbed at the surface of the medium, whereas the short wavelength component will be transmitted by the surface. (a) The number of panes in a window can strongly influence the heat loss from a heated room to the outside ambient air. Compare the single- and double-paned units shown by identifying relevant heat transfer processes for each case. (b) In a typical flat-plate solar collector, energy is collected by a working fluid that is circulated through tubes that are in good contact with the back face of an absorber plate. The back face is insulated from the surroundings, and the absorber plate receives solar radiation on its front face, which is typically covered by one or more transparent plates. Identify the relevant heat transfer processes, first for the absorber plate with no cover plate and then for the absorber plate with a single cover plate. (c) The solar energy collector design shown in the schematic has been used for agricultural applications. Air is blown through a long duct whose cross section is in the form of an equilateral triangle. One side of the triangle is comprised of a double-paned, semitransparent cover; the other two sides are constructed from aluminum sheets painted flat black on the inside and covered on the outside with a layer of styrofoam insulation. During sunny periods, air entering the system is heated for delivery to either a greenhouse, grain drying unit, or storage system. Identify all heat transfer processes associated with the cover plates, the absorber plate(s), and the air. (d) Evacuated-tube solar collectors are capable of improved performance relative to flat-plate collectors. The design consists of an inner tube enclosed in an outer tube that is transparent to solar radiation. The annular space between the tubes is evacuated. The outer, opaque surface of the inner tube absorbs solar radiation, and a working fluid is passed through the tube to collect the solar energy. The collector design generally consists of a row of such tubes arranged in front of a reflecting panel. Identify all heat transfer processes relevant to the performance of this device.

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 must be applied to the top and side walls to ensure a heat load of less than \(500 \mathrm{~W}\), when the inner and outer surfaces are \(-10\) and \(35^{\circ} \mathrm{C}\) ?

Single fuel cells such as the one of Example \(1.5\) can be scaled up by arranging them into a fuel cell stack. A stack consists of multiple electrolytic membranes that are sandwiched between electrically conducting bipolar plates. Air and hydrogen are fed to each membrane through fiw channels within each bipolar plate, as shown in the sketch. With this stack arrangement, the individual fuel cells are connected in series, electrically, producing a stack voltage of \(E_{\text {stack }}=N \times E_{c}\), where \(E_{c}\) is the voltage produced across each membrane and \(N\) is the number of membranes in the stack. The electrical current is the same for each membrane. The cell voltage, \(E_{c}\), as well as the cell efficiency, increases with temperature (the air and hydrogen fed to the stack are humidified to allow operation at temperatures greater than in Example 1.5), but the membranes will fail at temperatures exceeding \(T \approx 85^{\circ} \mathrm{C}\). Consider \(L \times w\) membranes, where \(L=w=100 \mathrm{~mm}\), of thickness \(t_{m}=0.43 \mathrm{~mm}\), that each produce \(E_{c}=0.6 \mathrm{~V}\) at \(I=60 \mathrm{~A}\), and \(\dot{E}_{c g}=45 \mathrm{~W}\) of thermal energy when operating at \(T=80^{\circ} \mathrm{C}\). The external surfaces of the stack are exposed to air at \(T_{\infty}=25^{\circ} \mathrm{C}\) and surroundings at \(T_{\text {sur }}=30^{\circ} \mathrm{C}\), with \(\varepsilon=0.88\) and \(h=150 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) Find the electrical power produced by a stack that is \(L_{\text {stack }}=200 \mathrm{~mm}\) long, for bipolar plate thickness in the range \(1 \mathrm{~mm}

Convection ovens operate on the principle of inducing forced convection inside the oven chamber with a fan. A small cake is to be baked in an oven when the convection feature is disabled. For this situation, the free convection coefficient associated with the cake and its pan is \(h_{\mathrm{fr}}=3 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The oven air and wall are at temperatures \(T_{\infty}=T_{\text {sur }}=180^{\circ} \mathrm{C}\). Determine the heat flux delivered to the cake pan and cake batter when they are initially inserted into the oven and are at a temperature of \(T_{i}=24^{\circ} \mathrm{C}\). If the convection feature is activated, the forced convection heat transfer coefficient is \(h_{\mathrm{fo}}=27 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). What is the heat flux at the batter or pan surface when the oven is operated in the convection mode? Assume a value of \(0.97\) for the emissivity of the cake batter and pan.

A rectangular forced air heating duct is suspended from the ceiling of a basement whose air and walls are at a temperature of \(T_{\infty}=T_{\text {sur }}=5^{\circ} \mathrm{C}\). 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 surface 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 \(\rho=1.10 \mathrm{~kg} / \mathrm{m}^{3}\) and \(c_{p}=1008 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), respectively.

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