Thermofluids problem (Viscous Flow)

[Bob Harrison]

Homework Statement

A sphere is moving in water at a depth where the absolute pressure is 124kPa. The maximum velocity on a sphere occurs from the forward stagnation point and is 1.5 times the free stream velocity. Calculate the speed of the sphere at which cavitation will occur.

Can anyone help me with this problem? I'm not sure how to obtain the solution.

 

[nilesh_pat]

The speed at which cavitation will occur is determined by the pressure at the stagnation point. This is the lowest pressure experienced by the sphere as it moves through the water. The pressure at the stagnation point is equal to the ambient pressure (in this case, 124 kPa) minus the dynamic pressure due to the motion of the sphere. The dynamic pressure is equal to one half of the density of the fluid times the square of the velocity of the sphere. Therefore, the equation for the pressure at the stagnation point is:Pstag = Pamb - (1/2)ρV2where Pstag is the pressure at the stagnation point, Pamb is the ambient pressure, ρ is the density of the fluid, and V is the velocity of the sphere. Since we know the ambient pressure and the maximum velocity on a sphere occurs from the forward stagnation point (1.5 times the free stream velocity), we can solve for the velocity of the sphere at which cavitation will occur:Vcav = (2(Pamb - Pcav))/ρwhere Pcav is the pressure at which cavitation will occur (usually taken as the vapor pressure of the fluid). For this problem, we have Pamb = 124 kPa and Pcav = 0 kPa, so the velocity of the sphere at which cavitation will occur is:Vcav = (2(124 kPa - 0 kPa))/ρwhere ρ is the density of the fluid (in this case, water). Plugging in the values, we get:Vcav = (2(124 kPa))/1000 kg/m3Vcav = 248 m/s