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Lelechka [254]
3 years ago
14

A 40 mph wind is blowing past your house and speeds up as it flows up and over the roof. If the elevation effects are negligible

, determine (a) the pressure at the point on the roof where the speed is 60 mph if the pressure in the free stream blowing toward your house is 14.7 psi. Would this effect tend to push the roof down against the house, or would it tend to lift the roof
Engineering
1 answer:
Finger [1]3 years ago
7 0

The pressure at the point on the roof where the speed is 60mph is 14.66 psi

<u>Explanation:</u>

The question follows Bernoulli's equation

P1 + 1/2 ρ V1² = P2 + 1/2 ρ V2²

V1 = 40 X 5280 / 3600 = 58.7 ft/s

V2 = 60 X 5280 / 3600 = 88 ft/s

P2 = P1 + 1/2 ( 0.00238 ) [(58.7)² - (88)²]

P2 - P1 = -5.12lb/ft²

The negative pressure created tends to lift the roof.

We know,

P1 = 14.7 psi = 2116.8 lb/ft²

P2 = 2116.8 - 5.12 lb/ft²

P2 = 2111.68 lb/ft²

P2 = 14.66 psi

Therefore, the pressure at the point on the roof where the speed is 60mph is 14.66 psi

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0=0-W+m(h_1-h_2)\\h_2=h_1-\frac{W}{m}\\\\h_2=3658.8-\frac{2626}{2}\\\\=2345.8kJ/kg

Use the isentropic relations:

s_1=s_{2s}\\s_1=s_{f2}+x_{2s}s_{fg2}\\7.1693=6492+x_{2s}(7.4996)\\x_{2s}=87

Calculate the enthalpy at isentropic state 2s.

h_{2s}=h_{f2}+x_{2s}.h_{fg2}\\=191.81+0.87(2392.1)\\=2272.937kJ/kg

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\eta_{turbine}=\frac{h_1-h_2}{h_1-h_{2s}}\\\\=\frac{3658.8-2345.8}{3658.8-2272.937}=0.947=94.7%

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h_2=h_{f2}+x_2(h_{fg2})\\2345.8=191.81+x_2(2392.1)\\x_2=0.9

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S=\frac{Q}{T}+m(s_2-s_1)

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S=m(s_2-s_1)\\=2\frac{kg}{s}(7.398-7.1693)kJ/kg\\=0.4574kW/k

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A cylindrical specimen of this alloy 12 mm in diameter and 188 mm long is to be pulled in tension. Assume a value of 0.34 for Po
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This question is incomplete, the missing image in uploaded along this answer below.

Answer:

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