# Rate of flow from a leak in a storage tank

1. Nov 23, 2006

### mirandasatterley

"A large storage tank, open at the top and filled with water, develops a small hole in its side at a point 16.0m below the water level. If the rate of flow from the leak is 2.50x10^-3 m^3/min, determine; A) The speed at which the water leaves the hole and (B) The diameter of the hole.

So I used the equation;
P(air) + 1/2 d(density)v1^2 + pgh(total height of water) = P(air) + 1/2dv2^2 + pgh(from the bottom of the container to the leak)

The equation then simplifies to:
gh(total height) = 1/2V2^2 + gh(from bottom to leak)
V2^2 = 2g(h(total) - h(form bottom to leak))
V2^2 = 2g(h(from leak to top of water))

Since this h is known, I can solve for v.

I Just can't figure out the diameter part.

Any help is appreciated.

2. Nov 23, 2006

### Pyrrhus

You're given the volume flow rate (discharge or flux), Q.

Last edited: Nov 23, 2006
3. Nov 23, 2006

### mirandasatterley

okay so the equation is;
flow rate = A(flow area)V1

And I can find the area, but how do i find just the diameter.

A= 4/3 pi r^2 (if it were a circular hole)

I'm unsure because it doesn't tell us what shape the hole is. Does the shape even matter?

4. Nov 23, 2006

### Pyrrhus

Why 4/3? and yes simply use $A = \frac{\pi d^2}{4}$

5. Nov 23, 2006

Okay, Thanks