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Leno4ka [110]
4 years ago
7

A Pitot-static probe is used to measure the speed of an aircraft flying at 3000 m. If the differential pressure reading is 3200

N/m2, determine the speed of the aircraft in m/s. The density of air at an altitude of 3000 m is 0.909 kg/m3.

Engineering
1 answer:
coldgirl [10]4 years ago
4 0

Answer:

Speed of aircraft ; (V_1) = 83.9 m/s

Explanation:

The height at which aircraft is flying = 3000 m

The differential pressure = 3200 N/m²

From the table i attached, the density of air at 3000 m altitude is; ρ = 0.909 kg/m3

Now, we will solve this question under the assumption that the air flow is steady, incompressible and irrotational with negligible frictional and wind effects.

Thus, let's apply the Bernoulli equation :

P1/ρg + (V_1)²/2g + z1 = P2/ρg + (V_2)²/2g + z2

Now, neglecting head difference due to high altitude i.e ( z1=z2 ) and V2 =0 at stagnation point.

We'll obtain ;

P1/ρg + (V_1)²/2g = P2/ρg

Let's make V_1 the subject;

(V_1)² = 2(P1 - P2)/ρ

(V_1) = √(2(P1 - P2)/ρ)

P1 - P2 is the differential pressure and has a value of 3200 N/m² from the question

Thus,

(V_1) = √(2 x 3200)/0.909)

(V_1) = 83.9 m/s

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A wing component on an aircraft is fabricated from an aluminum alloy that has a plane strain fracture toughness of 27 . It has b
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Answer:

The stress level at which fracture will occur for a critical internal crack length of 6.2mm is 135.78MPa

Explanation:

Given data;

Let,

critical stress required for initiating crack propagation Cc = 112MPa

plain strain fracture toughness = 27.0MPa

surface length of the crack = a

dimensionless parameter = Y.

Half length of the internal crack, a = length of surface crack/2 = 8.8/2 = 4.4mm = 4.4*10-³m

Also for 6.2mm length of surface crack;

Half length of the internal crack = length of surface crack/2 = 6.2/2 = 3.1mm = 3.1*10-³m

The dimensionless parameter

Cc = Kic/(Y*√pia*a)

Y = Kic/(Cc*√pia*a)

Y = 27/(112*√pia*4.4*10-³)

Y = 2.05

Now,

Cc = Kic/(Y*√pia*a)

Cc = 27/(2.05*√pia*3.1*10-³)

Cc = 135.78MPa

The stress level at which fracture will occur for a critical internal crack length of 6.2mm is 135.78MPa

For more understanding, I have provided an attachment to the solution.

4 0
4 years ago
A sample of semiconductor has a cross sectional area of 1cm^2 and a thickness of 0.1cm.
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Answer:

a. 3.17*10¹⁹ b. 3.17*10¹⁴

Explanation:

a. Area A = 1cm²

Thickness h = 0.1cm

Energy of photon E = hf=hc/λ

Where f = frequency; c=speed of light; λ=wavelength = 6300Α=6300*10⁻¹⁰;

Planck's constant h = 6.626*10⁻³⁴ joule-seconds; speed of light c = 3*10⁸

Therefore E = (6.626*10⁻³⁴)*3*10⁸/6300*10⁻¹⁰=3.155*10¹⁹J

1Watt of light releases 3.17*10¹⁸ photons per second.

Volume of sample = Area * Thickness = 1*0.1=0.1cm³

Therefore, number of electron hole pairs that are generated per unit volume per unit time = 3.17*10¹⁸/0.1 = 3.17*10¹⁹photon/cm³-s

b. Steady state excess carrier concentration = 3.17*10¹⁹*10μs

=3.17*10¹⁹*10*10⁻⁶

=3.17*10¹⁴/cm³

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3 years ago
A well-insulated rigid tank contains 5 kg of a saturated liquid–vapor mixture of water at 200 kPa. Initially, three-quarters of
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Answer:

S_{gen} = 18.519\,\frac{kJ}{K}

Explanation:

Given that rigid tank is a closed system, the following model is constructed after the First Law of Thermodynamics:

W_{heater} + U_{sys,1} - U_{sys,2} = 0

W_{heater} = U_{sys,2} - U_{sys,1}

W_{heater} = m\cdot (u_{2}-u_{1})

The entropy generation inside the rigid tank is determined by appropriate application of the Second Law of Thermodynamics:

S_{sys,1} - S_{sys,2} + S_{gen} = 0

S_{gen} = S_{sys,2} - S_{sys,1}

S_{gen} = m\cdot (s_{2}-s_{1})

The properties of the steam are obtained from steam tables:

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u = 1010.7\,\frac{kJ}{kg}

s = 2.9294\,\frac{kJ}{kg\cdot K}

x = 0.25

Final State

P = 869.567\,kPa

T = 173.88\,^{\textdegree} C

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u = 2578.6\,\frac{kJ}{kg}

s = 6.6332\,\frac{kJ}{kg\cdot K}

x = 1.00

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S_{gen} = (5\,kg)\cdot \left(6.6332\,\frac{kJ}{kg\cdot K} - 2.9294\,\frac{kJ}{kg\cdot K} \right)

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