I Proving Torricelli's Law Using Bernoulli's Principle

AI Thread Summary
The discussion centers on proving Torricelli's Law using Bernoulli's Principle, focusing on pressure differences at two points: one at the water's surface and the other just outside the spout. A key point is that when considering the pressure just inside the spout, it is not atmospheric due to hydrostatic pressure, complicating the proof and leading to a different velocity outcome. The conversation also explores how fluid streamlines converge toward the exit hole, indicating that fluid velocity increases as it approaches the exit, which correlates with a decrease in pressure. The participants emphasize the importance of understanding these fluid dynamics concepts, noting that the pressure is not strictly hydrostatic near the exit. Overall, the discussion highlights the nuances of applying Bernoulli's equation in fluid dynamics.
BrandonInFlorida
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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?
 
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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
$$\frac{\rho}{2} v_1^2=\rho g h \; \Rightarrow \; v_1=\sqrt{2gh}.$$
 
BrandonInFlorida said:
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?
 
vanhees71 said:
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
$$\frac{\rho}{2} v_1^2=\rho g h \; \Rightarrow \; v_1=\sqrt{2gh}.$$
You didn't read my question.
 
Chestermiller said:
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.
 
BrandonInFlorida said:
You didn't read my question.
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.
 
vanhees71 said:
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.

[edit] 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]
 
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BrandonInFlorida said:
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?
 
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  • #10
Very clear. Thanks so much!
 
  • #11
Arjan82 said:
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.

[edit] 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!
 
  • #12
Chestermiller said:
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.
 
  • #13
BrandonInFlorida said:
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!
 
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  • #14
BrandonInFlorida said:
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.
 
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Chestermiller said:
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.
 
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BrandonInFlorida said:
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.
 
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  • #17
BrandonInFlorida said:
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 :-).
 
  • #18
vanhees71 said:
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.
 
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  • #19
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!
 
  • #20
vanhees71 said:
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.
 
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  • #21
vanhees71 said:
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.
 
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