# Proving Torricelli's Law Using Bernoulli's Principle

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• BrandonInFlorida

#### BrandonInFlorida

There is a standard proof of this kind in which two points are taken - one at the top of the water and one just outside the spout or opening. I guess my question kind of assumes that you've seen something like this.

A key step of the proof is to say that the difference of pressures, perhaps p2 - p1, is zero since they are both at atmospheric pressure.

What I don't understand is this. If we say that the point just outside the spout or opening is instead just inside, the velocity must be virtually identical, but the pressure is no longer atmospheric pressure, so you can't do that step in which the pressure's cancel. I think that in this case, this pressure would be:

p0 + rho * g * h

So, if you do the proof this way, which seems also to be correct, you do not end up with SQRT(2gh). I did it quickly yesterday and I got SQRT(4gh), but at any rate, you would not get SQRT(2gh).

What's wrong with my analysis?

The pressure at the liquid's surface is equal to that of the opening, i.e., given by the air pressure around it, ##P_1=P_2=P_0##. Make the height of the opening ##z_2=0## and that of the surface ##h##. Then you use Bernoulli's Law
$$\frac{\rho}{2} v_2^2 + P_2 + \rho g h= \frac{\rho}{2} v_1^2 + P_1.$$
Now ##P_1## and ##P_2## cancel, and the velocity at the surface can be neglected, leading to
$$\frac{\rho}{2} v_1^2=\rho g h \; \Rightarrow \; v_1=\sqrt{2gh}.$$

There is a standard proof of this kind in which two points are taken - one at the top of the water and one just outside the spout or opening. I guess my question kind of assumes that you've seen something like this.

A key step of the proof is to say that the difference of pressures, perhaps p2 - p1, is zero since they are both at atmospheric pressure.

What I don't understand is this. If we say that the point just outside the spout or opening is instead just inside, the velocity must be virtually identical, but the pressure is no longer atmospheric pressure, so you can't do that step in which the pressure's cancel. I think that in this case, this pressure would be:

p0 + rho * g * h

So, if you do the proof this way, which seems also to be correct, you do not end up with SQRT(2gh). I did it quickly yesterday and I got SQRT(4gh), but at any rate, you would not get SQRT(2gh).

What's wrong with my analysis?
What do you think the fluid flow streamlines look like inside the tank, particularly in the region approaching the exit hole: (a) straight downward or (b) converging toward the exit hole?

The pressure at the liquid's surface is equal to that of the opening, i.e., given by the air pressure around it, ##P_1=P_2=P_0##. Make the height of the opening ##z_2=0## and that of the surface ##h##. Then you use Bernoulli's Law
$$\frac{\rho}{2} v_2^2 + P_2 + \rho g h= \frac{\rho}{2} v_1^2 + P_1.$$
Now ##P_1## and ##P_2## cancel, and the velocity at the surface can be neglected, leading to
$$\frac{\rho}{2} v_1^2=\rho g h \; \Rightarrow \; v_1=\sqrt{2gh}.$$

What do you think the fluid flow streamlines look like inside the tank, particularly in the region approaching the exit hole: (a) straight downward or (b) converging toward the exit hole?
My textbook, which is "Resnick and Halliday" circa 1966, didn't relate Bernoulli's equation to streamlines, however, to answer your question, I'd guess they're converging. It sounds like you're approaching some understanding which is crucial and which I do not have.

I did read your question and gave the standard derivation. Maybe I misunderstood your question. At least I've no clue, how you come to your alternative equation.

I did read your question and gave the standard derivation. Maybe I misunderstood your question. At least I've no clue, how you come to your alternative equation.
No, you didn't read it. In it, I said that with a small change to the traditional proof, I can get a completely different answer for virtually the same situation.

You're assuming kind of a step function. So the pressure just outside of the sprout is atmospheric, but just inside it is atmospheric + hydrostatic (i.e. rho*g*h). That is quite a pressure difference over such a short distance! But nature doesn't work that way, the pressure will gradually lower towards the sprout. So, just inside the sprout the pressure is *almost* atmospheric.

 Bernoulli holds, so the pressure is directly related to the velocity. This means that, since just inside the sprout the velocity is still *almost* the exit velocity, the pressure is also *almost* atmospheric.[/edit]

vanhees71
My textbook, which is "Resnick and Halliday" circa 1966, didn't relate Bernoulli's equation to streamlines, however, to answer your question, I'd guess they're converging. It sounds like you're approaching some understanding which is crucial and which I do not have.
That's correct. Now, since, within the tank, in the region in close proximity to the exit hole, the streamlines are converging toward the hole, what does that tell you about the fluid velocity variation in the region approaching the exit hole?

vanhees71
Very clear. Thanks so much!

You're assuming kind of a step function. So the pressure just outside of the sprout is atmospheric, but just inside it is atmospheric + hydrostatic (i.e. rho*g*h). That is quite a pressure difference over such a short distance! But nature doesn't work that way, the pressure will gradually lower towards the sprout. So, just inside the sprout the pressure is *almost* atmospheric.

 Bernoulli holds, so the pressure is directly related to the velocity. This means that, since just inside the sprout the velocity is still *almost* the exit velocity, the pressure is also *almost* atmospheric.[/edit]
Very clear. Thanks so much!

That's correct. Now, since, within the tank, in the region in close proximity to the exit hole, the streamlines are converging toward the hole, what does that tell you about the fluid velocity variation in the region approaching the exit hole?
Maybe that it's converging towards the exit velocity you'd get if your second point had been outside the spout, but I wish I understood how the streamlines play into this kind of analysis at all.

By the way, thank you very much. I'm a retiree who majored in physics in college, and I'm working my way through all of my college physics books and solving every problem at the end of every chapter. Under these conditions, I have no one to discuss the physics with.

Maybe that it's converging towards the exit velocity you'd get if your second point had been outside the spout, but I wish I understood how the streamlines play into this kind of analysis at all.

By the way, thank you very much. I'm a retiree who majored in physics in college, and I'm working my way through all of my college physics books and solving every problem at the end of every chapter. Under these conditions, I have no one to discuss the physics with.
Because, within the tank, the flow streamlines are converging in the vicinity of the exit hole, this tells us that the fluid velocity in the tank is speeding as it approaches the exit hole. The exit hole is basically acting like a sink for the flow. Since the fluid velocity is speeding up, Bernoulli tells us that the pressure is decreasing toward the exit hole. Most of this action occurs within about 3 or 4 hole diameters of the exit hole. So the pressure within the tank is not exactly hydrostatic throughout the tank. It is close to hydrostatic within most of the tank volume, but decreases to atmospheric in the region immediately approaching the exit hole.

By the way, you have a great name and live win a great locale. Let's go Brandon!

vanhees71
Maybe that it's converging towards the exit velocity you'd get if your second point had been outside the spout, but I wish I understood how the streamlines play into this kind of analysis at all.
I hope you don't think that the streamlines are straight lines converging to a point. They are not. They are curved, but they are getting closer together as the exit hole is approached.

vanhees71
Because, within the tank, the flow streamlines are converging in the vicinity of the exit hole, this tells us that the fluid velocity in the tank is speeding as it approaches the exit hole. The exit hole is basically acting like a sink for the flow. Since the fluid velocity is speeding up, Bernoulli tells us that the pressure is decreasing toward the exit hole. Most of this action occurs within about 3 or 4 hole diameters of the exit hole. So the pressure within the tank is not exactly hydrostatic throughout the tank. It is close to hydrostatic within most of the tank volume, but decreases to atmospheric in the region immediately approaching the exit hole.

By the way, you have a great name and live win a great locale. Let's go Brandon!
I'm with you on both counts and appreciate you taking the time to explain this. There are very few people I can ask the occasional points I don't understand. I have a former college roommate who's a professor of physics, but I hate to ask him for too many explanations. Besides, his specialty is cosmology and I'm not sure how much he remembers about fluid dynamics.

vanhees71 and Chestermiller
I'm with you on both counts and appreciate you taking the time to explain this. There are very few people I can ask the occasional points I don't understand. I have a former college roommate who's a professor of physics, but I hate to ask him for too many explanations. Besides, his specialty is cosmology and I'm not sure how much he remembers about fluid dynamics.
It's a pleasure for us at Physics Forums to help other members improve their understanding of various aspects of Physics. Look forward to helping in the future.

vanhees71
I'm with you on both counts and appreciate you taking the time to explain this. There are very few people I can ask the occasional points I don't understand. I have a former college roommate who's a professor of physics, but I hate to ask him for too many explanations. Besides, his specialty is cosmology and I'm not sure how much he remembers about fluid dynamics.
A cosmologist knows at least about ideal relativistic hydrodynamics, because that's inevitably the matter model used in cosmology :-).

A cosmologist knows at least about ideal relativistic hydrodynamics, because that's inevitably the matter model used in cosmology :-).
We actually skipped the fluid mechanics chapters in our textbook when we were in school in the 70s. He's very good at inflationary cosmology and theories of gravity though.

vanhees71
The neglect of hydrodynamics in the modern physics curriculum is really unfortunate, because if there is a way to gain intuition of the operations used in vector calculus it's to think in terms of fluid dynamics!

The neglect of hydrodynamics in the modern physics curriculum is really unfortunate, because if there is a way to gain intuition of the operations used in vector calculus it's to think in terms of fluid dynamics!
That's an interesting angle. Most of my experience with vector calculus came from electrodynamics. I'll keep it in mind. Thanks.

vanhees71
The neglect of hydrodynamics in the modern physics curriculum is really unfortunate, because if there is a way to gain intuition of the operations used in vector calculus it's to think in terms of fluid dynamics!

In one of the polish old textbooks on general physics for physics students fluid dynamics and electrodynamics are compiled together in a separate volume of the series, with fluid dynamics being presented first. It makes a lot of sense.

vanhees71