# Fluid flow

1. Feb 5, 2004

### tandoorichicken

A large storage tank is filled with water. Neglecting viscosity, show that the speed of water emerging through a hole in the side a distance h below the surface is [itex] v = \sqrt{2gh} [/tex].

2. Feb 6, 2004

### HallsofIvy

Staff Emeritus
Please show us what you have to work with and what you have tried on this problem.

3. Feb 6, 2004

### ShawnD

Square the whole thing and it starts to look familiar.

$$V^2 = 2gh$$

Remember that gravity applies a downward force on the fluid and that a force on any side of a fluid is applied to ALL sides of the fluid.

4. Feb 6, 2004

### HallsofIvy

Staff Emeritus
Good! A good solid hint without completely solving the problem!

5. Feb 6, 2004

### himanshu121

Write the Bernoulli's Theorem at the Point just inside the tank and just outside the tank.

6. Feb 9, 2004

### tandoorichicken

ummmmm.....
If I use Bernoulli's Equation, one side of it says
$$P+ \rho gh + \frac{1}{2}\rho v^2$$.
What does the other side of the equation look like?

If the other side is 0,
then
$$P = \rho gh$$
$$2\rho gh = -\frac{1}{2}\rho v^2$$
But since h is going below the surface the negative is expected and can be ignored. Then-
$$2\rho gh = \frac{1}{2}\rho v^2$$
The density of water is 1 g/cm^3.
$$2gh = \frac{1}{2} v^2$$
$$4gh = v^2$$
But then I get
$$v = 2\sqrt{gh}$$, not $$v = \sqrt{2gh}$$