Consider water flowing through a restriction with a pressure drop of 4 bar. Inlet pressure is 64 bar and the outlet pressure will be 60 bar. A cavitation number, that characterises the potential of the flow to cavitate is give by (p2-pv ) / (0.5 x rho x u x u) where p2 is the recovered pressure downstream of the restriction which will be 60 bar = 60,000 Pa PV is the pressure at the vena contracta at the operating temperature, 30 deg C. This pressure is equal to vapor pressure of the water rho density of water : 998 kg/m3 u : Velocity at the restriction , lets say it is 25 m/s What should be the Pv here ? still goes by the operating temperature and is regardless of the pressure ? http://upload.wikimedia.org/wikipedia/commons/0/08/Phase_diagram_of_water.svg According to this figure the boiling point of water at 60 bar is 275 deg C. 45098 mm hg is the water vapor pressure at 275 deg c Can someone help me understand to calculating cavitation number here ?