Chapter 6 – Air Valves
Surge Analysis and the Wave Plan Method
Supplementary Material: Example Problems and Solutions
Chapter 6 – Problem 6.1
6.1 Consider a kinetic air valve with the same inflow and outflow orifice diameter of 200 mm and the same inflow and outflow discharge coefficient of 0.515. Determine whether the flow takes place under choked or non-choked conditions and compute the associated air flowrates in both mass and volumetric units for the following scenarios. Use an atmospheric pressure of 101.325 kPa (abs), a polytropic constant of 1.4, and a density of atmospheric air of 1.2 kg/m3 for all scenarios. Approximate air densities at different pressures may be obtained from Figure E.9 or any other reliable source.
a. Water pressure surrounding the air pocket at the active air valve is 50m (gage)
b. Water pressure surrounding the air pocket at the active air valve is 1 atm
c. Water pressure surrounding the air pocket at the active air valve is 10 kPa (gage)
d. Water pressure surrounding the air pocket at the active air valve is 10 kPa (abs)
e. Water pressure surrounding the air pocket at the active air valve is 5 m (abs)
f. Water pressure surrounding the air pocket at the active air valve is 10 m (abs)
Solution:
a. Water pressure surrounding the air pocket at the active air valve is 50m (gage)
Pressure in the air pocket (50m gage or 490.33 kPa gage or 591.655 kPa abs) is higher than the atmospheric pressure (0 m gage or 101.325 kPa abs), implying outflow.
Pa/P = 0.171 and is less than 0.528 (see Eq. 6.5) implying choked flow
Use Eq. 6.2 to compute the flowrate in mass units. In Eq. 6.2, CDo = 0.515, Ao = 0.0314 m2, P = 591655 Pa, ρ = 6 kg/m3 (read from Figure E.9), γ = 1.4
Mass flowrate m = 20.86 kg/s
Volumetric flowrate = m/ρ = 3.47 m3/s
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