A. I believe, lmk if I’m right
Answer:
the pressure at a closed valve attached to the tank 10 ft above its bottom is 37.88 psi
Explanation:
Given that;
depth 1 = 71 ft
depth 2 = 10 ft
pressure p = 17 psi = 2448 lb/ft²
depth h = 71 ft - 10 ft = 61 ft
we know that;
p = P_air + yh
where y is the specific weight of ethyl alcohol ( 49.3 lb/ft³ )
so we substitute;
p = 2448 + ( 49.3 × 61 )
= 2448 + 3007.3
= 5455.3 lb/ft³
= 37.88 psi
Therefore, the pressure at a closed valve attached to the tank 10 ft above its bottom is 37.88 psi
Answer:
89375 N
Explanation:
Rearrange the formula for normal stress for F:


Convert given values to base units:
275 MPa =
Pa
325
= 0.000325 
Substituting in given values:
F =
N
Answer:

The entropy change of the carbon dioxide is -0.1104 kJ/kg.K
Explanation:
We are given that carbon dioxide undergoes a process in a closed system.
We are asked to find the entropy change of the carbon dioxide with the assumption that the specific heats are constant.
The entropy change of the carbon dioxide is given by

Where Cp is the specific heat constant
Cp = 0.846 kJ/kg.K
R is the universal gas constant
R = 0.1889 kJ/kg.K
T₁ and T₂ is the initial and final temperature of carbon dioxide.
P₁ and P₂ is the initial and final pressure of carbon dioxide.




Therefore, the entropy change of the carbon dioxide is -0.1104 kJ/kg.K
Answer:
LOS = A
Explanation:
Given all the parameters the level of service as seen from the attached graph
is LOS = A
<u>To determine the LOS from the attached graph </u>
<em>calculate the trial value of Vp</em>
Vp = V / PHF
= (100 + 150) / 0.95 = 263 pc/h
since the trial value of Vp = ( 0 to 600 ) pc/h . hence E.T = 1.7 , ER = 1
next we will calculate the flow rate
flow rate = 1 / [ ( 1 + PT(ET - 1 ) + PR ( ER - 1 ) ]
Fhr = 1 / 1.035 = 0.966 ≈ 1
next calculate the real value of Vp
Vp = V / ( PHF * N * Fhr * Fp )
= ( 100 + 150 ) / ( 0.95 * 2 * 1 * 1 )
Vp ≈ 126 pc/h/In
Next calculate the density
D = Vp / S = 126 / ( 45 * 1.61 ) = 1.74 pc/km/In