# Time taken for water to drain from a tank through a pipe (fluid mech./Bernoulli)

1. May 10, 2012

### tome101

1. The problem statement, all variables and given/known data
A) A tank of water of cross-sectional area A empties through a pipe
of cross-sectional area a, where a A so that the speed of the water flow in the tank can be neglected compared to that in the pipe. The initial height of the water in the tank is H and the pipe extends a vertical distance h below the bottom of the tank. Find an expression for the time taken for the tank to empty. You may assume an inviscid flow. [Take the height of the water in the tank to be z. The pressure at the water surface and at the bottom of the pipe are both atmospheric.]

B) Explain what would happen if a small hole was made:
a) halfway up the side of the tank;
b) halfway along the pipe.

Only a brief calculation/explanation is needed for this part.

2. Relevant equations
$\frac{v^{2}}{2}+gz+\frac{p}{ρ}=const.$

3. The attempt at a solution
Ok, from Bernoulli we can see that at the exit of the pipe:

$gz=0$

and at the entrance to the pipe at the bottom of the tank

$v=0$

As the speed is said to be negligible in the tank.
Since p and ρ are constant, final Bernoulli equation is:

$\frac{v^{2}}{2}=gh$

Rearranging we get:

$v=\sqrt{2gh}$

$\frac{∂h}{∂t}=\sqrt{2gh}$

$∂t=(2gh)^{-1/2}∂h$

And I'm not really sure where to go from here, or if it's correct up until this point. I assume you integrate both sides but i'm not entirely sure what the boundaries would be on the right hand side, and also I'm not sure where the areas come into it? And part b I don't know where to start...

2. Nov 28, 2015

### AslamRandhawa

∂t=(2gh)−1/2∂h
OR
dt=dh/Sqrt(2gh)

t=2*sqrt(h)/(2g)+C
I donot know, How to find the value of 'C' (A Constant value of integration).

Last edited: Nov 28, 2015
3. Nov 28, 2015

### haruspex

You cannot ignore the pressure created by the water still in the tank.