Problem with understanding deriving Torricelli's law

[Lotto]

TL;DR Summary

I know two ways of deriving Torricelli's law. And both confuse me.

We can derive it by using Bernoulli's equation $p_0+h_0\rho g+\frac12 \rho {v_0}^2=p_1+h_1\rho g+\frac12 \rho {v_1}^2$, where $v_0=0$ is a velocity of a water surface and $h_0$ is its height. $h_1$ is a height of the outlet above the bottom of a tank the water is in. I am confused about the assumption that both pressures are the same, that $p_0=p_1$ is the atmospheric pressure.

I understand that at the water level, the pressure is equal to the atmospheric one, but why is the pressure at the outlet also atmospheric? Why it isn't a hystrostatic pressure plus the atmospheric ($p_1=(h_0-h_1)\rho g +p_0$)? Because there is a certain amount of water above the outlet.

The second way of deriving the law is that we can imagine it as if we "teleported" a certain amount of water from the water level and gave it a kinetic energy at the outlet. So then $mgh_1=mgh_0+\frac 12 mv^2$. Why can I imagine it this way? I intuitivelly feel that the water will have such energy, but I don't understand that explanation.

 

[Dullard]

The outlet pressure is atmospheric (by definition) for discharge into the atmosphere. You're probably just going to have to make peace with that until it makes sense. What else could it be?

Total energy is conserved. Kinetic energy (velocity) and potential energy (pressure) are fungible.

 

[erobz]

I understand that at the water level, the pressure is equal to the atmospheric one, but why is the pressure at the outlet also atmospheric? Why it isn't a hystrostatic pressure plus the atmospheric ($p_1=(h_0-h_1)\rho g +p_0$)? Because there is a certain amount of water above the outlet.

The outlet jet is surrounded only by atmosphere. I think in reality there is very small transitional zone where the pressure changes from what it is inside the pipe to outside- practically a step change though because the pressure change is communicated at the speed of sound in the jet fluid, and that is usually significantly faster than the speed of the fluid jet itself.

 

[Lotto]

The outlet jet is surrounded only by atmosphere. I think in reality there is very small transitional zone where the pressure changes from what it is inside the pipe to outside- practically a step change though because the pressure change is communicated at the speed of sound in the jet fluid, and that is usually significantly faster than the speed of the fluid jet itself.

But when I have something like this, from one side, there is an air with its atmospheric pressure, but from the other side, there is a water with its hydrostatic pressure. Or not? Why?

 

[kuruman]

Consider mass of water $dm$ that is instantaneously filling the space of the hole that has area $A$. On one side you have hydrostatic pressure $p_h$ and other side you have atmospheric pressure $p_a$. The pressure difference is $\Delta p=(p_h-p_a)=\rho gh$ which means that there is a net force $F_{\text{net}}=\Delta p~A$ on $dm$ which accelerates it out of the hole. How could the water move out of the hole without a pressure difference and hence a force to drive it out?

 

[erobz]

But when I have something like this, from one side, there is an air with its atmospheric pressure, but from the other side, there is a water with its hydrostatic pressure. Or not? Why?

The streamlines are converging in this red circle. So you don't really have $P = \rho g h$ everywhere along the black dotted line, the velocity of a particle along a streamline is increasing as it approaches the outlet, the pressure is dropping. Also the hole in the wall has some diameter, and the pressure varies over its surface. Inside the wall the pressure is still higher than atmospheric and over some very short distance outside the wall it drops to atmospheric. I hope it's clear that Bernoulli's is a derived simplification of reality.

 

[Lotto]

View attachment 336906

The streamlines are converging in this red circle. So you don't really have $P = \rho g h$ everywhere along the black dotted line, the velocity of a particle along a streamline is increasing as it approaches the outlet, the pressure is dropping. Also the hole in the wall has some diameter, and the pressure varies over its surface. Inside the wall the pressure is still higher than atmospheric. I hope its clear that Bernoulli's is a derived simplification of reality.

That streamlines are interesting. Does it mean that the water is leaving the tank by flowing from the water level? Because I thought that the water leaving it is water at the height $h_1$, not $h_0$.

 

[erobz]

That streamlines are interesting. Does it mean that the water is leaving the tank by flowing from the water level? Because I thought that the water leaving it is water at the height $h_1$, not $h_0$.

If water is moving from the outlet, water is moving in from elsewhere to takes its place, and that water, is being replaced, etc... The criterion that $V$ at the surface is $0$ is an approximation. For steady flow (without viscosity - another basically unphysical assumption), Bernoulli's is valid along a streamline, but what's happening on that streamline (the convenient one we force into existence to make some calculations) is not really characterizing what all the fluid particles on all the streamlines are actually doing.

 

[Lotto]

If water is moving from the outlet, water is moving in from elsewhere to takes its place, and that water, is being replaced, etc... The criterion that $V$ at the surface is $0$ is an approximation. For steady flow (without viscosity - another basically unphysical assumption), Bernoulli's is valid along a streamline, but what's happening on that streamline (the convenient one we force into existence to make some calculations) is not really characterizing what all the fluid particles on all the streamlines are actually doing.

Okay, but I don't understand the second derivation: $mg(h_0-h_1)=\frac 12 mv^2$. That equation says that water of a mass $m$ comes from the water level to the outlet, that results in gain of a kinetic energy. Why can we do this calculation? What if the water comes from the height $h_1$?

This second derivation confuses me a lot.

 

[erobz]

Okay, but I don't understand the second derivation: $mg(h_0-h_1)=\frac 12 mv^2$. That equation says that water of a mass $m$ comes from the water level to the outlet, that results in gain of a kinetic energy. Why can we do this calculation? What if the water comes from the height $h_1$?

This second derivation confuses me a lot.

Where is $h_1$, the tank surface?

 

[Lotto]

Where is $h_1$, the tank surface?

That is the distance of the outlet from the bottom. $h_0$ is the distance of the water level from the bottom.

 

[erobz]

Ok, so $h_o$ is the tank water surface. An assumption of Bernoulli's is steady flow. If you are imagining a valve closed at $t= 0$ and a fluid particle sitting at $h_1$ (just inside the outlet), initially at rest, being accelerated through the outlet as the valve is opened you have overextended the model. The assumption of steady flow (fluid particle at a fixed location/passing through a fixed locations velocity must not be varying in time) has been violated. In order to uses Bernoulli's as it is, the valve has been opened for some time, and the flow field is no longer evolving in time.

i.e. a fluid particle at $h_1$ must already possess the required kinetic energy to satisfy continuity at $t = 0$, when Bernoulli's is applied. Likewise, particles on the surface have some kinetic energy too. Its only if $A_{tank} \gg A_{outlet}$ that this can be neglected in comparison.

Have a look at this example that better explains what I'm saying in the context of worked problem.

https://ocw.mit.edu/courses/2-25-ad...edabdfd4b95c1792_MIT2_25F13_Unstea_Bernou.pdf

 

[Lotto]

Ok, so $h_o$ is the tank water surface. An assumption of Bernoulli's is steady flow. If you are imagining a valve closed at $t= 0$ and a fluid particle sitting at $h_1$ (just inside the outlet), initially at rest, being accelerated through the outlet as the valve is opened you have overextended the model. The assumption of steady flow (fluid particle at a fixed location/passing through a fixed locations velocity must not be varying in time) has been violated. In order to uses Bernoulli's as it is, the valve has been opened for some time, and the flow field is no longer evolving in time.

i.e. a fluid particle at $h_1$ must already possess the required kinetic energy to satisfy continuity at $t = 0$, when Bernoulli's is applied. Likewise, particles on the surface have some kinetic energy too. Its only if $A_{tank} \gg A_{outlet}$ that this can be neglected in comparison.

And if there was no atmospheric pressure above the water surface, if the tank was closed, what would the formula for the velocity look like?

I would calculate it from

$h_0\rho g=p_a + h_1\rho g + \frac 12 \rho v^2$, so

$v=\sqrt{2g(h_0-h_1)-\frac{2p_a}{\rho}}$.

Is it correct?

 

[erobz]

And if there was no atmospheric pressure above the water surface, if the tank was closed, what would the formula for the velocity look like?

I would calculate it from

$h_0\rho g=p_a + h_1\rho g + \frac 12 \rho v^2$, so

$v=\sqrt{2g(h_0-h_1)-\frac{2p_a}{\rho}}$.

Is it correct?

Is the scenario one in which the tank is completely filled with water initially at atmospheric pressure? If so, rapidly a relative vacuum is going to form above the surface and flow would come to a stop.

 

[Lnewqban]

That is the distance of the outlet from the bottom. $h_0$ is the distance of the water level from the bottom.

Taking the bottom of the tank as reference may lead to confusion.
The velocity at which the fluid is horizontally leaving the tank is independent from the depth.
All it matters is how much column of fluid exists above each orifice.