Consider a solid sphere (capable of withstanding full vacuum) of 3m diameter filled completely with water resting at sea level. It has a 10cm hole at the bottom with a cork on it. If you open the cork, what is the time taken for water to completely drain out.
What happens for higher diameter spheres?
let a be the area of the drainage hole and v be the velocity of water out of that hole, then av is the rate at which the water is leaving the sphere and V be the volume of the sphere then
(1) DV/Dt=av
now when the sphere is filled to hight h the volume is equal to
V=(Pi*(2.25h)*h^3)/3
(2) DV/dh=Pi*(3h)*h
now using (1), (2), and the chain rule
DV/dt=(DV/dh)(Dh/Dt)=Pi*(3h)*h*(Dh/Dt)=av
Dh/Dt=av/(Pi*(3h)*h)
a=Pi*(.1)^2=.01*Pi
Dh/Dt=.01*v/(h*(3h))
v(h)=Sqrt(2*g*h) where g is acceleration due to gravity on earth (9.8m/s^2)
Dh/Dt=.01*Sqrt(2*g*h)/(h*(3h))
100*h*(3h)/Sqrt[2*g*h] dh = Dt
(3) 20*(5h)*sqrt(2*g*h)/g=t+c
when t=0 h=3
thus
c=20*2*Sqrt(2*g*3)/g=40*sqrt(6*g)/g
now the sphere is empty when h=0 thus we can find how long it takes to drain by setting h=0 in (3) and solving for t thus
t=40*sqrt(6*g)/g
or about 31.2984 seconds
now that seems a little quick for that much water to be able to drain so I think I may have made a mistake somewhere, if someone could please point it out I would be greatly appreciate it :D

Posted by Daniel
on 20060714 01:41:54 