Small hole at bottom of cylindrical tank draining water?

[khfrekek92]

Homework Statement

the tank has a radius of 2m, containing an initial water level of 3m. A hole at the very bottom (underneath) of the tank has radius .005m. How long will it take to empty the tank?

Homework Equations

Bernoulli's principle.
A=radius at top of tank
a=radius of drain hole

The Attempt at a Solution

after solving V=sqrt(2gh) [at the hole in the bottom], i set dV/dt=--a*sqrt(2gh), so A(dy/dt)=-a*sqrt(2gh). After integrating both sides and solving for C, I got that (A/g)sqrt(2gh)=-8at+(A/g)sqrt(6g)
Then I plugged everything in with h=0 and got that t=15649.2sec=4.35 hours.

Did I do this right? Did you get the same answer?

Thanks so much in advance! I appreciate it!

 

[1994Bhaskar]

Hey You can't apply sqrt(2gh) when area of hole is given.sqrt(2gh) is when area of hole is very small.You need to apply continuity equation to get velocity outside hole Av₁=av₂, where v₁ is velocity just before water is inside cylinder, & v₂ is velocity of water just after coming out of hole.
you should get velocity v₂=sqrt(2gh/(1-(a²/A²))).

 

[SteamKing]

Even accepting the point from 1994Bhaskar, since the radius of the hole is given as 0.005 m and the radius of the tank is 2 m, I think a simple calculation shows the area of the hole is in fact "very small" in relation to the cross sectional area of the tank. The correction factor for v2 is 0.99999375, using the information given. In other words, v = sqrt (2gh).

 

[khfrekek92]

Yes that is a great point 1994Bhaskar, but my professor said we are allowed to assume that the velocity at the top is negligible compared to the bottom, So I'll just use v=sqrt(2gh). Thanks so much for your help guys! :)