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ehidna [41]
2 years ago
14

Consider a fully-clamped circular diaphragm poly-Si with a radius of 250 μm and a thickness of 4 μm. Assume that Young’s modulus

is 130 GPa and Poisson’s ratio is 0.3. A uniform pressure of 20 kPa is applied.
(a) What is the stress at the center of the diaphragm?
(b) What is the radial stress at the fully-clamped edge of the diaphragm?
(c) What is the transverse stress at the fully-clamped edge of the diaphragm?
(d) What is the amount of deflection at the center of the diaphragm?

Engineering
1 answer:
Alina [70]2 years ago
4 0

Answer:

Explanation:

find attached the solution to the question

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Laminar flow normally persists on a smooth flat plate until a critical Reynolds number value is reached. However, the flow can b
grandymaker [24]

Answer:

At L = 0.1 m

h⁻_lam = 11.004K   W/m^1.5

h⁻_turb = 7.8848K   W/m^1.8

At L = 1 m

h⁻_lam = 3.48K   W/m^1.5

h⁻_turb = 4.975K   W/m^1.8

Explanation:

Given that;

h_lam(x)= 1.74 W/m^1.5. Kx^-0.5

h_turb(x)= 3.98 W/m^1.8 Kx^-0.2

conditions for plates of length L = 0.1 m and 1 m

Now

Average heat transfer coefficient is expressed as;

h⁻ = 1/L ₀∫^L hxdx

so for Laminar flow

h_lam(x)= 1.74 . Kx^-0.5  W/m^1.5

from the expression

h⁻_lam = 1/L ₀∫^L 1.74 . Kx^-0.5   dx

= 1.74k / L { [x^(-0.5+1)] / [-0.5 + 1 ]}₀^L

= 1.74k/L = [ (x^0.5)/0.5)]⁰^L

= 1.74K × L^0.5 / L × 0.5

h⁻_lam= 3.48KL^-0.5

For turbulent flow

h_turb(x)= 3.98. Kx^-0.2 W/m^1.8

form the expression

1/L ₀∫^L 3.98 . Kx^-0.2   dx

= 3.98k / L { [x^(-0.2+1)] / [-0.2 + 1 ]}₀^L

= (3.98K/L) × (L^0.8 / 0.8)

h⁻_turb = 4.975KL^-0.2

Now at L = 0.1 m

h⁻_lam = 3.48KL^-0.5  =  3.48K(0.1)^-0.5  W/m^1.5

h⁻_lam = 11.004K   W/m^1.5

h⁻_turb = 4.975KL^-0.2 = 4.975K(0.1)^-0.2

h⁻_turb = 7.8848K   W/m^1.8

At L = 1 m

h⁻_lam = 3.48KL^-0.5  =  3.48K(1)^-0.5  W/m^1.5

h⁻_lam = 3.48K   W/m^1.5

h⁻_turb = 4.975KL^-0.2 = 4.975K(1)^-0.2

h⁻_turb = 4.975K   W/m^1.8

Therefore

At L = 0.1 m

h⁻_lam = 11.004K   W/m^1.5

h⁻_turb = 7.8848K   W/m^1.8

At L = 1 m

h⁻_lam = 3.48K   W/m^1.5

h⁻_turb = 4.975K   W/m^1.8

3 0
2 years ago
A hand crank generator is used to power a light bulb. The person turning the crank uses 22 Newtons of input force to turn the cr
ikadub [295]

Answer:

1. The input power is 17.6 Watts

2. The output power is 0.7232 Watts

3.The efficiency of the hand crank generator is approximately 4.109%

Explanation:

The given parameters are;

The force with which the crank generator is turned = 22 Newtons

The rate at which the crank is turned = 8 revolutions per 3 seconds

The distance the hand moves during each revolution = 0.3 meters

The current recorded by the multimeter = 0.08 amps

The voltage recorded by the multimeter = 9.04 volts

1. Power = The rate of doing work = Work done/(Time taken to do the work) = Force × Distance/Time

∴ The power input of the hand crank generator = 22 × 0.3 × 8/3 = 17.6 Watts

2. The power output to the bulb, P, is given by the formula for electrical power as follows;

Power = Current, I × Voltage V

∴ P = 0.08 ×  9.04 = 0.7232 Watts

3. Efficiency in percentage = Output/Input × 100

Therefore, the efficiency of the hand crank generator = )Output power/(Input power)) × 100 = (0.7232/17.6) × 100 ≈ 4.109%

8 0
2 years ago
In a production facility, 1.6-in-thick 2-ft × 2-ft square brass plates (rho = 532.5 lbm/ft3 and cp = 0.091 Btu/lbm·°F) that are
Gnoma [55]

Answer:

106600 btu/s

<u>note: </u>

<u><em> solution is attached due to error in mathematical equation. please find the attachment</em></u>

8 0
2 years ago
Sharon is designing a house in an area that receives a lot of rainfall all year. Which material should she use to stick the wood
kakasveta [241]

Explanation:

She is passionate about architecture, typography, and black & white film ... Since moving to Texas, I've heard a lot of people say, "If you don't like ... Oc, 3.74, 56, 80 ... Not only does the weather have to be clear to pour the concrete, but it ... system that goes within the slab) is complete, any additional rain will

4 0
2 years ago
The average grain diameter of an aluminum alloy is 14 mu m with a strength of 185 MPa. The same alloy with an average grain diam
timama [110]

Answer:

See explanations for completed answers

Explanation:

Given that; The average grain diameter of an aluminum alloy is 14 mu m with a strength of 185 MPa. The same alloy with an average grain diameter of 50 mu m has a strength of 140 MPa. (a) Determine the constants for the Hall-Patch equation for this alloy, (b) How much more should you reduce the grain size if you desired a strength of 220 MPa

See attachment for completed solvings

6 0
2 years ago
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