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*(2.25-h)*h^3)/3

(2) DV/dh=Pi*(3-h)*h

now using (1), (2), and the chain rule

DV/dt=(DV/dh)(Dh/Dt)=Pi*(3-h)*h*(Dh/Dt)=-av
Dh/Dt=-av/(Pi*(3-h)*h)
a=Pi*(.1)^2=.01*Pi
Dh/Dt=-.01*v/(h*(3-h))

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*(3-h))
100*h*(3-h)/Sqrt[2*g*h] dh = -Dt
(3) 20*(5-h)*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 2006-07-14 01:41:54