What is the cavitation number at the higher pressures ?

[Dutta]

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 , let's 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 ?

 

[LightHero]

The cavitation number is given by: (p2 - pv) / (0.5 x ρ x u x u)where p2 = 60,000 Pa (outlet pressure) pv = 45098 mm hg (water vapor pressure at 275 deg C) ρ = 998 kg/m3 (density of water) u = 25 m/s (velocity at the restriction) Therefore, the cavitation number is: (60000 - 45098) / (0.5 x 998 x 25 x 25) = 0.0129