# Bernoulli's equation - fluid mechanics question

1. Oct 13, 2014

### jakethe third

1. A sphere 1 ft in diameter is moving horizontally at a depth of 12 ft below a water surface where the water temperature is 50F. Vmax = 1.5Vo, where Vo is the free stream velocity and occurs at the maximum sphere width. At what speed in still water will cavitation first occur?
Given: speed where cavitation will occur is free stream velocity. Absolute pressure at the center of the sphere is yh. y is the specific weight of water. Vmax acts at the top of the sphere 11.5 ft below water surface.

2. Relevant equations
P + YwZ1 + P(V^2)/2 = P + YwZ2 + P(V^2)/2
Yw is specific weight of water and lower case p is density of water.

3. The attempt at a solution
Cavitation of water at 50F is 25.63 lb/ft^2. YwZ2 = 0 because it is at the bottom of the datum. YwZ1 is 62.4*0.5 because it is 0.5 ft above datum (half of the 1 ft diameter is 0.5 ft).
P2 = 62.4*12ft, P1 = 25.63 for cavitation to occur.

I know how to use the Bernoulli equation but I'm not sure how to set this problem up. I don't have a picture, sorry. Any ideas?
The right answer is around 48 ft/s

2. Oct 18, 2014