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Svetradugi [14.3K]
3 years ago
5

An array of electronic chips is mounted within a sealedrectangular enclosure, and colling is implemented by attaching analuminum

heat sink (k=180 W/m*K). The base of the heat sinkhas dimensions of w1=w2=100mm, while the 6 fins are of thicknesst=10mm and pitch S=18mm. The fin length is Lf=50 mm, and thebase of the heat sink has a thickness of Lb=10mm.
If cooling is implemented by water flow through the heat sink,with u[infinity]=3 m/s and T[infinity]=17 C, what is the base temperatureTb of the heat sink when the power dissipation by the chips isPelec=1800W? The average convection coefficient for surfacesof the fins and the exposed base may be estimated by assumingparallel flow over a flat plate. Properties of the water maybe approximated as k=0.62 W/m*K, ν=7.73E-7 m2/s, andPr=5.2.

Engineering
1 answer:
Licemer1 [7]3 years ago
6 0

Answer:

Base temperature is 46.23 °C

Explanation:

I've attached explanations

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A TV USE 75 WATTS WHILE IN USED ASSMING THAT ITIS USED 4 HOURS EVERY DAY HOW MUCH ENERGY IN 4 IN KWH WOULD THE TV CONSUME ANNUAL
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Read 2 more answers
A 35kg block of mass is subjected to forces F1=100N and F2=75N at agive angle thetha= 20° and 35° respectively.find the distance
Talja [164]

Answer:

21 m

Explanation:

Since F₁ = 100 N and acts at an angle of 20° to the horizontal, it has horizontal component F₁' = 100cos20° = 93.97 N and vertical component F₁" = 100sin20° = 34.2 N.

Also, F₂ = 75 N and acts at an angle of -35° to the horizontal, it has horizontal component F₂' = 75cos(-35°) = 75cos35° = 61.44 N and vertical component F₂" = 75sin(-35°) = -75sin35° = -43.02 N

The resultant horizontal force F₃' = F₁' + F₂' = 93.97 N + 61.44 N = 155.41 N

The resultant vertical force F₃" = F₁" + F₂" = 34.2 N - 43.02 N = -8.82 N

If f is the frictional force on the block, the net horizontal force on the block is F = F₃' - f.

Since f = μN where μ = coefficient of kinetic friction = 0.4 and N = normal force on the block.

For the block to be in contact with the surface, the vertical forces on the block must balance.

Since the normal force, N must equal the resultant vertical force F₃" and the weight, W = mg of the object for a zero net vertical force,

N = mg + F₃" (since both the weight and the resultant vertical force act downwards)

N = mg + F₃"

Since m = mass of block = 35 kg and g = acceleration due to gravity = 9.8 m/s² and F₃" = 8.82 N

So,

N = mg + F₃"

N = 35 kg × 9.8 m/s² + 8.82 N

N = 343 N + 8.82 N

N = 351.82 N

So, the net horizontal force F = F₃' - f.

F = 155.41 N - 0.4 × 351.82 N

F = 155.41 N - 140.728 N

F = 14.682 N

Since F = ma, where a = acceleration of block,

a = F/m = 14.682 N/35 kg = 0.42 m/s²

To find the distance the block moved, x we use the equation

x = ut + 1/2at² where u = initial speed of block = 0 m/s, t = time = 10 s and a = acceleration of block = 0.42 m/s²

Substituting the values of the variables into the equation, we have

x = ut + 1/2at²

x = 0 m/s × 10 s + 1/2 × 0.42 m/s² × (10 s)²

x = 0 m + 1/2 × 0.42 m/s² × 100 s²

x = 0.21 m/s² × 100 s²

x = 21 m

So, the distance moved by the block is 21 m.

4 0
3 years ago
"Write a statement that outputs variable numItems. End with a newline. Program will be tested with different input values."
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Answer:

The solution code is written in Java.

System.out.println(numItems);

Explanation:

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