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Chapter 1: Problem 67

Four power transistors, each dissipating \(12 \mathrm{~W}\), are mounted on a thin vertical aluminum plate \(22 \mathrm{~cm} \times 22 \mathrm{~cm}\) in size. The heat generated by the transistors is to be dissipated by both surfaces of the plate to the surrounding air at \(25^{\circ} \mathrm{C}\), which is blown over the plate by a fan. The entire plate can be assumed to be nearly isothermal, and the exposed surface area of the transistor can be taken to be equal to its base area. If the average convection heat transfer coefficient is \(25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine the temperature of the aluminum plate. Disregard any radiation effects.

Short Answer

Expert verified
Answer: The temperature of the aluminum plate is approximately 44.84°C.

Step by step solution

01

Calculate the total heat dissipated by the transistors

There are 4 transistors, each dissipating 12 W of power. Therefore, the total heat generation (Q) can be found by multiplying the power dissipation of each transistor by the number of transistors, which is given by: $$Q = 4 \times 12 \mathrm{~W} = 48 \mathrm{~W}$$
02

Convert the plate dimensions to meters

The dimensions of the aluminum plate are given in centimeters. We need to convert them to meters to use in the convection heat transfer equation. The conversion factor is 1 meter = 100 centimeters. Therefore, the dimensions of the aluminum plate in meters are: $$0.22 \mathrm{~m} \times 0.22 \mathrm{~m}$$
03

Calculate the surface area of the plate

Since heat is dissipated by both surfaces of the plate, we need to calculate the surface area of both sides to use in the convection heat transfer equation. The surface area (A) of one side of the plate is given by: $$A = 0.22 \mathrm{~m} \cdot 0.22 \mathrm{~m} = 0.0484 \mathrm{~m}^2$$ The surface area of both sides of the plate is then: $$2 \times A = 2 \times 0.0484 \mathrm{~m}^2 = 0.0968 \mathrm{~m}^2$$
04

Apply the convection heat transfer equation and solve for the temperature difference

The convection heat transfer equation is given by: $$Q = h \cdot A \cdot \Delta T$$ Where Q is the heat dissipation, h is the convection heat transfer coefficient, A is the surface area, and \(\Delta T\) is the temperature difference between the plate and the surrounding air. We want to find the \(\Delta T\). Rearranging the formula gives us: $$\Delta T = \frac{Q}{h \cdot A}$$ Now, we can plug in the given values: Q = 48 W, h = 25 W/m²·K, and A = 0.0968 m². $$\Delta T = \frac{48 \mathrm{~W}}{(25 (\mathrm{~W} / \mathrm{m}^2 \cdot \mathrm{K})) \times (0.0968 \mathrm{~m}^2)} \approx 19.84 ^{\circ}\mathrm{C}$$
05

Calculate the aluminum plate temperature

Now that we have the temperature difference between the aluminum plate and the surrounding air, we can determine the aluminum plate temperature by adding this temperature difference to the surrounding air temperature. The surrounding air temperature is given as \(25^{\circ} \mathrm{C}\). Therefore, the aluminum plate temperature can be calculated as follows: $$T_\text{plate} = T_\text{air} + \Delta T = 25^{\circ} \mathrm{C} + 19.84^{\circ} \mathrm{C} \approx 44.84^{\circ} \mathrm{C}$$ The temperature of the aluminum plate is approximately \(44.84^{\circ} \mathrm{C}\).

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Most popular questions from this chapter

A flat-plate solar collector is used to heat water by having water flow through tubes attached at the back of the thin solar absorber plate. The absorber plate has a surface area of \(2 \mathrm{~m}^{2}\) with emissivity and absorptivity of \(0.9\). The surface temperature of the absorber is \(35^{\circ} \mathrm{C}\), and solar radiation is incident on the absorber at \(500 \mathrm{~W} / \mathrm{m}^{2}\) with a surrounding temperature of \(0^{\circ} \mathrm{C}\). Convection heat transfer coefficient at the absorber surface is \(5 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), while the ambient temperature is \(25^{\circ} \mathrm{C}\). Net heat rate absorbed by the solar collector heats the water from an inlet temperature \(\left(T_{\text {in }}\right)\) to an outlet temperature \(\left(T_{\text {out }}\right)\). If the water flow rate is \(5 \mathrm{~g} / \mathrm{s}\) with a specific heat of \(4.2 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\), determine the temperature rise of the water.

Consider a house in Atlanta, Georgia, that is maintained at \(22^{\circ} \mathrm{C}\) and has a total of \(20 \mathrm{~m}^{2}\) of window area. The windows are double-door type with wood frames and metal spacers and have a \(U\)-factor of \(2.5 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (see Prob. 1-125 for the definition of \(U\)-factor). The winter average temperature of Atlanta is \(11.3^{\circ} \mathrm{C}\). Determine the average rate of heat loss through the windows in winter.

A 2-kW electric resistance heater in a room is turned on and kept on for 50 minutes. The amount of energy transferred to the room by the heater is (a) \(2 \mathrm{~kJ}\) (b) \(100 \mathrm{~kJ}\) (c) \(6000 \mathrm{~kJ}\) (d) \(7200 \mathrm{~kJ}\) (e) \(12,000 \mathrm{~kJ}\)

Conduct this experiment to determine the combined heat transfer coefficient between an incandescent lightbulb and the surrounding air and surfaces using a \(60-\mathrm{W}\) lightbulb. You will need a thermometer, which can be purchased in a hardware store, and a metal glue. You will also need a piece of string and a ruler to calculate the surface area of the lightbulb. First, measure the air temperature in the room, and then glue the tip of the thermocouple wire of the thermometer to the glass of the lightbulb. Turn the light on and wait until the temperature reading stabilizes. The temperature reading will give the surface temperature of the lightbulb. Assuming 10 percent of the rated power of the bulb is converted to light and is transmitted by the glass, calculate the heat transfer coefficient from Newton's law of cooling.

Consider heat loss through the two walls of a house on a winter night. The walls are identical, except that one of them has a tightly fit glass window. Through which wall will the house lose more heat? Explain.
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