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Annette [7]
3 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]3 years ago
3 0

Answer:

Explanation:

Find attached the solution

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With the help of diagrams explain the differences in the functions of electrostatic
ololo11 [35]

Answer:

Good Luck!

Explanation:

5 0
3 years ago
Aviation ppl onlyyyy ayoooo, thoughts on the A-10 Thunderbolt II?
Vaselesa [24]

Answer:

It’s cool I’d say 8/10 I guess

Explanation:

8 0
3 years ago
A trapezoidal section has a 5.0-ft bed width, 2.5-ft depth, and 1:1 side slope. Evaluate its geometric elements (Area, water dep
aleksley [76]

Answer:

a) 18.75 ft^2

b) 2.5 ft

c) 12.07 ft

d) 10 ft

e) 1.553 ft

f) 1.875 ft

Explanation:

Given data :

5.0-ft bed width, ( b )

2.5-ft depth ( y )

1 : 1 side slope

Evaluate

a) Area of trapezoidal section

A = by + my^2

we assume m = 1

A = [5 + (1 * 2.5 ) ] *2.5

   = ( 7.5 ) 2.5  = 18.75 ft^2

b) Calculate water depth

water depth = 2.5 ft

c) Calculate wetted perimeter

P = b + 2y √ 1 + m^2

  = 5 + (2.5*2) √ 1 + 1 ^2  =   12.07 ft

d) calculate top width

    T = b + 2my

       = 5 + 2 ( 1 * 2.5 ) = 10 ft

e) calculate hydraulic radius

R = A / P = 18.75 / 12.07

               = 1.553 ft

f) calculate hydraulic depth

 D = A / T = 18.75 / 10 = 1.875 ft

4 0
3 years ago
The maximum gage pressure is known to be 10 MPa in a spherical steel pressure vessel having a 200 mm outer diameter and a(n) 2 m
VikaD [51]

Answer:

factor of safety = 0.8

Explanation:

given data

maximum gauge pressure in steel p = 10 MPa

outer diameter do = 200  mm

wall thickness t = 2 mm

ultimate stress of the steel σU = 400 MPa

solution

We check the type of shell that is

\frac{t}{do} =\frac{6}{200}

so we can see that that is    

so that it mean vessel is a thin shell now we can we get maximum pressure by the maximum tensile stress is

σ = \frac{p\times do}{2t}     .......................1

put here value and we get

σ =  \frac{10\times 200}{2\times 2}  

σ =  500 MPa  

so factor of safety will be express as

factor of safety = \frac{\sigma U}{\sigma}     ..........2

factor of safety = \frac{400}{500}  

factor of safety = 0.8

6 0
4 years ago
A First Stage in a turbine receives steam at 10 MPa, 800 C with an exit pressure of 800 KPa. Assume the stage is adiabatic and r
9966 [12]

Answer

given,

P₁  = 10 MPa                 T₁ = 800

P₂ = 800 KPa               T₂ = ?

Using formula

\dfrac{T_2}{T_1} = (\dfrac{P_2}{P_1})^{\dfrac{n-1}{n}}

For steam   n = 1.33                    

\dfrac{T_2}{800+273} = (\dfrac{800\times 10^{3}}{10\times 10^{6}})^{\dfrac{1.33-1}{1.33}}

T₂ = 573.368 K                    

T₂ = 573.368 - 273 = 300.368 °C

W = \dfrac{P_1V_1-P_2V_2}{n-1}

          =\dfrac{mR(T_1-T_2)}{n-1}

          =\dfrac{1\times 0.287\times (800 - 300.368)}{1.33-1}

      W = 434.53 kJ/kg

7 0
3 years ago
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