/*! 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 129 A group of 25 power transistors,... [FREE SOLUTION] | 91Ó°ÊÓ

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A group of 25 power transistors, dissipating \(1.5 \mathrm{~W}\) each, are to be cooled by attaching them to a black-anodized square aluminum plate and mounting the plate on the wall of a room at \(30^{\circ} \mathrm{C}\). The emissivity of the transistor and the plate surfaces is \(0.9\). Assuming the heat transfer from the back side of the plate to be negligible and the temperature of the surrounding surfaces to be the same as the air temperature of the room, determine the size of the plate if the average surface temperature of the plate is not to exceed \(50^{\circ} \mathrm{C}\). Answer: \(43 \mathrm{~cm} \times 43 \mathrm{~cm}\)

Short Answer

Expert verified
Answer: The required dimensions of the square aluminum plate are approximately 43 cm × 43 cm.

Step by step solution

01

Calculate the total power dissipated by the transistors

First, we calculate the total power dissipated by the 25 power transistors by multiplying the individual power dissipation by the number of transistors. Total Power Dissipated = Number of transistors × Power per transistor = \(25 \times 1.5 \mathrm{~W} = 37.5 \mathrm{W}\).
02

Calculate the surface temperature of the plate and room in Kelvin

Since we will be working with the Stefan-Boltzmann law, which requires temperatures in Kelvin, We convert the given temperatures to Kelvin. Plate surface temperature, \(T_s = 50^{\circ} \mathrm{C} + 273.15 = 323.15 \mathrm{K}\) Room temperature, \(T_\infty = 30^{\circ} \mathrm{C} + 273.15 = 303.15 \mathrm{K}\)
03

Use the Stefan-Boltzmann law to find the radiative heat transfer rate

The Stefan-Boltzmann law states that the radiative heat transfer rate \(q_r\) is equal to the product of the emissivity \(\epsilon\), the Stefan-Boltzmann constant \(\sigma\), the surface area \(A\), and the temperature difference between the plate and room, raised to the fourth power. The equation can be written as: $$q_r = \epsilon A\sigma \left(T_s^4 - T_\infty^4\right)$$ Since the total power dissipated by the transistors is equal to the heat transfer rate, we can find the required surface area of the plate. The given emissivity is \(0.9\), and the Stefan-Boltzmann constant is \(5.67\times10^{-8} \mathrm{W/m^2.K^4}\).
04

Calculate the required surface area

Now we can calculate the required surface area A using the following equation: $$A = \frac{q_r}{\epsilon\sigma \left(T_s^4 - T_\infty^4\right)}$$ Substitute the values and solve for \(A\): $$A = \frac{37.5}{(0.9) (5.67\times10^{-8})(323.15^4 - 303.15^4)} = 0.0186 \mathrm{m^2}$$
05

Find the dimensions of the square plate

Since the plate is square, its sides are equal, and the area of the square plate can be calculated as \(A = L^2\), where \(L\) is the length of one side of the square. Now we can find the side length of the square plate: $$L = \sqrt{A} = \sqrt{0.0186 \mathrm{m^2}} = 0.4305\, \mathrm{m}$$ To stick to significant figures and have a practical answer, we round this value to: $$L \approx 0.43\, \mathrm{m}$$ So, the required dimensions of the square plate are \(43\, \mathrm{cm} \times 43\, \mathrm{cm}\).

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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 crucial in understanding radiative heat transfer. It reveals how much radiant energy, or power, is emitted by a surface due to its temperature. The basic idea is that every object emits radiation as a function of its temperature. This law is expressed mathematically as:\[ q_r = \epsilon A \sigma (T_s^4 - T_\infty^4) \]Where:
  • \(q_r\) is the radiative heat transfer rate
  • \(\epsilon\) is the emissivity
  • \(A\) is the surface area
  • \(\sigma\) is the Stefan-Boltzmann constant, \(5.67 \times 10^{-8} \, \mathrm{W/m^2.K^4}\)
  • \(T_s\) and \(T_\infty\) are the temperatures of the surface and surroundings in Kelvin
This formula shows that the heat transferred by radiation depends on the temperature difference raised to the fourth power, the surface's emissivity, and the area through which the heat is transferred.
Emissivity
Emissivity is a measure of a surface's ability to emit thermal radiation compared to a perfect emitter, known as a black body. It's a key factor in calculating radiative heat transfer. Emissivity values range between 0 and 1. - A value of 1 means the surface behaves like a black body, emitting the maximum possible radiation. - A value of 0 would imply that the surface emits no radiation at all. In the context of the problem, the emissivity of the black-anodized plate and transistor surfaces is 0.9. This high value indicates that these surfaces are very efficient at radiating energy, which was factored into calculating the heat transfer rate using the Stefan-Boltzmann Law.
Radiative Heat Transfer
Radiative heat transfer is one of the principal modes of heat transfer, alongside conduction and convection. It involves the transfer of heat through electromagnetic waves, primarily in the infrared spectrum. Unlike conduction, which requires direct physical contact, or convection, which involves the movement of fluids, radiation can occur across a vacuum or air, allowing heat to be transferred through space via electromagnetic waves. In this problem, radiative heat transfer was the key method for cooling the plate attached to the transistors. The heat emitted by the transistors is transferred to the room's surroundings, computed with the necessary area and temperature parameters using the Stefan-Boltzmann Law.
Power Dissipation
Power dissipation refers to the process where electronic components, like transistors, release electrical energy in the form of heat. This is an inevitable by-product of operating electronic devices. - Each transistor in the exercise dissipates 1.5 watts of power as heat. - With 25 transistors, the cumulative power dissipation amounts to 37.5 watts. This total heat output had to be efficiently handled to prevent overheating. The calculations to determine the appropriate size for the aluminum plate ensured that the heat could be effectively radiated away, keeping the system within the desired temperature limits.

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Most popular questions from this chapter

Why are finned surfaces frequently used in practice? Why are the finned surfaces referred to as heat sinks in the electronics industry?

Consider a \(1.2\)-m-high and 2-m-wide double-pane window consisting of two 3-mm-thick layers of glass \((k=\) \(0.78 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) separated by a \(3-\mathrm{cm}\)-wide air space. Determine the steady rate of heat transfer through this window and the temperature of its inner surface for a day during which the room is maintained at \(20^{\circ} \mathrm{C}\) while the temperature of the outdoors is \(0^{\circ} \mathrm{C}\). Take the heat transfer coefficients on the inner and outer surfaces of the window to be \(h_{1}=10 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(h_{2}=25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and disregard any heat transfer by radiation. Evaluate air properties at a film temperature of \(10^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\) pressure. Is this a good assumption?

Consider a thin 16-cm-long and 20-cm-wide horizontal plate suspended in air at \(20^{\circ} \mathrm{C}\). The plate is equipped with electric resistance heating elements with a rating of \(20 \mathrm{~W}\). Now the heater is turned on and the plate temperature rises. Determine the temperature of the plate when steady operating conditions are reached. The plate has an emissivity of \(0.90\) and the surrounding surfaces are at \(17^{\circ} \mathrm{C}\). As an initial guess, assume a surface temperature of \(50^{\circ} \mathrm{C}\). Is this a good assumption?

A spherical tank \((k=15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) with an inner diameter of \(3 \mathrm{~m}\) and a wall thickness of \(10 \mathrm{~mm}\) is used for storing hot liquid. The hot liquid inside the tank causes the inner surface temperature to be as high as \(100^{\circ} \mathrm{C}\). To prevent thermal burns on the skin of the people working near the vicinity of the tank, the tank is covered with a \(7-\mathrm{cm}\) thick layer of insulation \((k=0.15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) and the outer surface is painted to give an emissivity of \(0.35\). The tank is located in a surrounding with air at \(16^{\circ} \mathrm{C}\). Determine whether or not the insulation layer is sufficient to keep the outer surface temperature below \(45^{\circ} \mathrm{C}\) to prevent thermal burn hazards. Discuss ways to further decrease the outer surface temperature. Evaluate the air properties at \(30^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\) pressure. Is this a good assumption?

A 0.2-m-long and \(25-\mathrm{mm}\)-thick vertical plate \((k=\) \(1.5 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) separates the hot water from the cold air at \(2^{\circ} \mathrm{C}\). The plate surface exposed to the hot water has a temperature of \(100^{\circ} \mathrm{C}\), and the surface exposed to the cold air has an emissivity of \(0.73\). Determine the temperature of the plate surface exposed to the cold air \(\left(T_{s, c}\right)\). Hint: The \(T_{s, c}\) has to be found iteratively. Start the iteration process with an initial guess of \(51^{\circ} \mathrm{C}\) for the \(T_{s, c}\).

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