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Chapter 9: Problem 129

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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