Why is the exit pressure atmospheric in Torricelli's theorem?

[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?

 

[russ_watters]

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.

 

[tomas123]

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

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 = P_atm 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.

 

[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?

 

[russ_watters]

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.