During a certain movie on cavitation, Dr. Kavi Tae Shun held in his hand a close

Question

During a certain movie on cavitation, Dr. Kavi Tae Shun held in his hand a closed cylinder filled with water and air. He said that the pressure in the air above the water had its pressure reduced below atmospheric. He then proceeded to accelerate the cylinder downward causing the water to cavitate near the bottom of the cylinder. Determine the number of g's i.e., the acceleration due to gravity in "g's", that he would have to accelerate the cylinder downward to get cavitation. Assume the pressure in the air above the water was 40 kPa absolute, and the temperature of the water was 15 degrees Celsius. The cylinder was 0.70 meters long.

Given hint: Use Euler's Equation

Explanation / Answer

A liquid will cavitate as soon as its static pressure (also known simply as "pressure") drops below its vapor pressure. At that point, a gas bubble will appear somewhere in the liquid, and it will grow or shrink to keep the static pressure constant at the liquid's vapor pressure.

I'm not sure I understand the geometry of this cylinder. A diagram would be helpful. It is not possible to get cavitation on the advancing side of the piston. If Dr. Shun is pushing the piston downward, cavitation will not occur below the piston. It may occur above it

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