# B Torricelli's theorem

1. Mar 14, 2016

### tomas123

Torricelli's theorem relates the speed of a fluid exiting an opening in a reservoir to the height of the opening relative to the top of the reservoir... V=√2gh https://en.wikipedia.org/wiki/Torricelli's_law

As seen in the wiki-link provided, the equation is essentially a Bernoulli's equation problem. My question is why the pressure at the opening where the fluid flows out is set as atmospheric pressure? Whenever you do a typical nozzle problem to find the speed in which a fluid exits a nozzle, you use the pressure on the inside of the nozzle, i.e. not atmospheric pressure. Why is this different?

2. Mar 14, 2016

### Staff: Mentor

Could you give an example of such a problem? I would think it would be typical to use atmospheric pressure as the exit pressure - I don't see how it could be any other pressure.

You aren't referring to a converging-diverging nozzle (like a rocket engine), are you? Totally different.

3. Mar 14, 2016

### tomas123

Assuming Patm is correct to use, for a horisontal pipe with an opening to the atmosphere we would have: V12/2 + P1/ρ = V22/2 + Patm

where 1 = somewhere downstream and 2 = directly outside the pipe

If we assume the pipe has a constant area, the continuity equation gives us V1 = V2, leaving us with P1 = Patm which in most cases isn't true.

I'm obviously misunderstanding something.. It's been a long time since I've done problems like these.

4. Mar 15, 2016

### gleem

If you to suspend small but visible particles in the tank so that you could see the motion of the fluid at the level of the orifice what would you observe?

5. Mar 15, 2016

### Staff: Mentor

For a theoretical pipe with no friction, the pressure does indeed need to be constant along its length. For a real pipe, with friction, there is a gradient ending at atmospheric pressure at the discharge.