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Annette [7]
2 years ago
11

A 15 cm × 15 cm circuit board dissipating 20 W of power uniformly is cooled by air, which approached the circuit board at 20C w

ith a velocity of 6 m/s. Assume the flow to be turbulent, since the electronic components are expected to act as turbulators. Disregard any heat transfer from the back surface of the board, and suppose that radiation heat transfer is negligible. Evaluate thermal properties of the air at a film temperature of 27C. Determine the surface temperature (C) of the electronic components at the following locations i) the leading edge of the board ii) the trailing edge of the board

Engineering
1 answer:
Assoli18 [71]2 years ago
3 0

Answer:

Explanation:

Find attached the solution

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The Cv factor for a valve is 48. Compute the head loss when 30 GPM of water passes through the valve.
dlinn [17]

Answer:

The head loss in Psi is 0.390625 psi.

Explanation:

Fluid looses energy in the form of head loss. Fluid looses energy in the form of head loss when passes through the valve as well.

Given:

Factor cv is 48.

Flow rate of water is 30 GPM.

GPM means gallon per minute.

Calculation:

Step1

Expression for head loss for the water is given as follows:

c_{v}=\frac{Q}{\sqrt{h}}

Here, cv is valve coefficient, Q is flow rate in GPM and h is head loss is psi.

Step2

Substitute 48 for cv and 30 for Q in above equation as follows:

48=\frac{30}{\sqrt{h}}

{\sqrt{h}}=0.625

h = 0.390625 psi.

Thus, the head loss in Psi is 0.390625 psi.

 

5 0
3 years ago
Need help with both giving out brainlest for the people to help me
sammy [17]
The first one is d or the 4th answer choice and the second one is false. Hope this helps!
4 0
2 years ago
Water flows steadily through the pipe as shown below, such that the pressure at section (1) and at section (2) are 300 kPa and 1
steposvetlana [31]

Answer:

The velocity at section is approximately 42.2 m/s

Explanation:

For the water flowing through the pipe, we have;

The pressure at section (1), P₁ = 300 kPa

The pressure at section (2), P₂ = 100 kPa

The diameter at section (1), D₁ = 0.1 m

The height of section (1) above section (2), D₂ = 50 m

The velocity at section (1), v₁ = 20 m/s

Let 'v₂' represent the velocity at section (2)

According to Bernoulli's equation, we have;

z_1 + \dfrac{P_1}{\rho \cdot g} + \dfrac{v^2_1}{2 \cdot g} = z_2 + \dfrac{P_2}{\rho \cdot g} + \dfrac{v^2_2}{2 \cdot g}

Where;

ρ = The density of water = 997 kg/m³

g = The acceleration due to gravity = 9.8 m/s²

z₁ = 50 m

z₂ = The reference = 0 m

By plugging in the values, we have;

50 \, m + \dfrac{300 \ kPa}{997 \, kg/m^3 \times 9.8 \, m/s^2} + \dfrac{(20 \, m/s)^2}{2 \times 9.8 \, m/s^2} = \dfrac{100 \ kPa}{997 \, kg/m^3 \times 9.8 \, m/s^2} + \dfrac{v_2^2}{2 \times 9.8 \, m/s^2}50 m + 30.704358 m + 20.4081633 m = 10.234786 m + \dfrac{v_2^2}{2 \times 9.8 \, m/s^2}

50 m + 30.704358 m + 20.4081633 m - 10.234786 m = \dfrac{v_2^2}{2 \times 9.8 \, m/s^2}

90.8777353 m = \dfrac{v_2^2}{2 \times 9.8 \, m/s^2}

v₂² = 2 × 9.8 m/s² × 90.8777353 m

v₂² = 1,781.20361 m²/s²

v₂ = √(1,781.20361 m²/s²) ≈ 42.204308 m/s

The velocity at section (2), v₂ ≈ 42.2 m/s

3 0
2 years ago
A light bar AD is suspended from a cable BE and supports a 20-kg block at C. The ends A and D of the bar are in contact with fri
babymother [125]

Answer:

Tension in cable BE= 196.2 N

Reactions A and D both are  73.575 N

Explanation:

The free body diagram is as attached sketch. At equilibrium, sum of forces along y axis will be 0 hence

T_{BE}-W=0 hence

T_{BE}=W=20*9.81=196.2 N

Therefore, tension in the cable, T_{BE}=196.2 N

Taking moments about point A, with clockwise moments as positive while anticlockwise moments as negative then

196.2\times 0.125- 196.2\times 0.2+ D_x\times 0.2=0

24.525-39.24+0.2D_x=0

D_x=73.575 N

Similarly,

A_x-D_y=0

A_x=73.575 N

Therefore, both reactions at A and D are 73.575 N

7 0
3 years ago
Anyone want to play among us with me the code is MMJSUF
Katarina [22]

Answer:

sure

Explanation:

6 0
3 years ago
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