Bernoulli's equation - fluid mechanics question

[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.

Homework 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.

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

 

[Greg Bernhardt]

Thanks for the post! Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?