Solution :
Given :
Water have quality x = 0.7 (dryness fraction) at around pressure of 200 kPa
The phase diagram is provided below.
a). The phase is a standard mixture.
b). At pressure, p = 200 kPa, T =
Temperature = 120.21°C
c). Specific volume
d). Specific energy ()
e). Specific enthalpy
At
f). Enthalpy at m = 0.5 kg
= 1022.91 kJ
Answer:
The fluid level difference in the manometer arm = 22.56 ft.
Explanation:
Assumption: The fluid in the manometer is incompressible, that is, its density is constant.
The fluid level difference between the two arms of the manometer gives the gage pressure of the air in the tank.
And P(gage) = ρgh
ρ = density of the manometer fluid = 60 lbm/ft³
g = acceleration due to gravity = 32.2 ft/s²
ρg = 60 × 32.2 = 1932 lbm/ft²s²
ρg = 1932 lbm/ft²s² × 1lbf.s²/32.2lbm.ft = 60 lbf/ft³
h = fluid level difference between the two arms of the manometer = ?
P(gage) = 9.4 psig = 9.4 × 144 = 1353.6 lbf/ft²
1353.6 = ρg × h = 60 lbf/ft³ × h
h = 1353.6/60 = 22.56 ft
A diagrammatic representation of this setup is presented in the attached image.
Hope this helps!
