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frosja888 [35]
3 years ago
5

Four power transistors, each dissipating 12 W, are mounted on a thin vertical aluminum plate 22 cm 3 22 cm in size. The heat gen

erated by the transistors is to be dissipated by both surfaces of the plate to the surrounding air at 25°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 W/m2·K, determine the temperature of the aluminum plate. Disregard any radiation effects.
Physics
1 answer:
Sveta_85 [38]3 years ago
6 0

Answer:

The temperature of the Aluminium plate 44.84⁰C

Explanation:

Number of transistors = 4

Since the heat dissipated by each transistor is 12W

Total heat dissipated, Q = 4 * 12 = 48 W

Q = 48 W

Cross sectional Area of the Aluminium plate, A = 2(l * b)

l = Length of the aluminium plate = 22 cm = 0.22 m

b = width of the aluminium plate = 22 cm = 0.22 m

A =2( 0.22 * 0.22 )

A = 0.0968 m²

From the heat balance equation, Q = hAΔT

h = 25 W/m²·K

A = 0.0968 m²

ΔT = T - T(air)

T(air) = 25°C

ΔT = T - 25°C

Q = 25 * 0.0968 * (  T - 25)

Q = 2.42 (T - 25)

Substitute Q = 48 into the equation above

48 =  2.42 (T - 25)

T - 25 = 19.84

T = 25 + 19.84

T = 44.84 ⁰C

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Complete Question

You are performing a double slit experiment very similar to the one from DL by shining a laser on two nattow slits spaced 7.5 * 10^{-3} meters apart. However, by placing a piece of crystal in one of the slits, you are able to make it so that the rays of light that travel through the two slits are Ï out of phase with each other (that is to say, Ao,- ). If you observe that on a screen placed 4 meters from the two slits that the distance between the bright spot closest to center of the pattern is 1.5 cm, what is the wavelength of the laser?

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