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Afina-wow [57]
2 years ago
5

What is the design wind speed for average-risk building:_____.

Engineering
1 answer:
bagirrra123 [75]2 years ago
3 0

Answer:

A. 115 MPH              B. 110 MPH

Explanation:

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Semmy [17]

Answer:

Anne is a mechanical engineer.

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forsale [732]

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True.

Explanation:

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Why efficiency of gas turbines are lower than combustion engine​
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3 years ago
Read 2 more answers
Consider 4.8 pounds per minute of water vapor at 100 lbf/in2, 500 oF, and a velocity of 100 ft/s entering a nozzle operating at
andriy [413]

Answer:

A) v_2 = 2016.80 ft/s

B) \Delta s = 0.006 Btu/lbm R  

Explanation:

Given data:

P-1 = 100 lbf/in^2

T_1 = 500 degree f

V_1 = 100 ft/s

P_2 = 40 lbf/inc^2

effeciency = 80%

from steady flow enerfy equation

h_1 +\frac{V_1^2}{2} = h_2 + \frac{V_2^2}{2}

where h1 and h2 are inlet and exit enthalpy

for P1 = 100 lbf/in^2 and T1 = 500 degree F

H_1 = 1278.8 Btu/lbm

s_1 = 1.708 Btu/lbm -R

for P1 = 40 lbf/in^2

H_1 = 1193.5 Btu/lbm

s_1 = 1.708 Btu/lbm -R

exit enthalapy h_2

\eta = \frac{h_1 - h'_2}{ h_1 - h_2}

0.80 = \frac{1278.8 - h'_2}{1278.8 -1193.5} = 1197.77 Btu/lbm

from above equation

1278.8 \times 25037 + \frac{100^2}{2} = 1197.77   \times 25037 + \frac{v_2^2}{2}                   [1 Btu/lbm = 25037 ft^2/s^2]

v_2 = 2016.80 ft/s

b) amount of entropy

\Delta s = s_2 - s_1

s_1 = 1.708 Btu/lbm -R

at h_2 = 1197.77 Btu/lbm [\tex]  and [tex]P_2 = 40 lbf/in^2

s_2 is 1.714 Btu/lbm -R

\Delta s = 1.714 - 1.708 = 0.006 Btu/lbm R

6 0
3 years ago
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