Question

A sealed vertical cylinder of radius R and height h = 0.616 m is initially filled halfway with water, and the upper half is filled with air. The air is initially at standard atmospheric pressure, p0 = 1.01·105 Pa. A small valve at the bottom of the cylinder is opened, and water flows out of the cylinder until the reduced pressure of the air in the upper part of the cylinder prevents any further water from escaping. By what distance is the depth of the water lowered? (Assume that the temperature of water and air do not change and that no air leaks into the cylinder.)

Answer 1

Using Bernoulli's Equation

pressure of water is hpg = (0.308-h)(1000)(9.8)

The air in the cylinder is sealed, thus, if we simply equate (0.308-h)*1000*9.8 + P0 = P0, we will get h=0. Does this make no sense? No. it simply says that all water will leak out if the cylinder is not sealed.

For a sealed packet of air, pV = constant. Let A be the circular area of the cylinder, thus: p0(A(0.308))=p(A(0.308+h)) and we get: p=p0(0.308)/(0.308+h)

So now we equate: (0.308-h)*1000*9.8 + p = p0
and (0.308-h)*1000*9.8 + p0(0.308)/(0.308+h) = p0
and (0.308-h)(0.308+h)*1000*9.8 + p0*0.308 = p0(0.308+h)
(0.308-h)(0.308+h)*1000*9.8 = p0(h)
0.308^2 - h^2 = p0/9800 * h =(1.01*10^5/9800)*h=10.306h

h^2 + 10.306h-0.094864=0
h= 9.197*10-3 m

Well is the answer believable? I think yes, since if you sub this value into p=p0(0.308)/(0.308+h), we get 98071.54, and about 9.8*1000*0.308 = 3018.4 adding them you get back about p0. That is if the pV assumption is true.