Consider a fluid falling down a tube of uniform cross-section \(A\) with an initial velocity \(v_{o}\) at the top of the tube.

Pressure at the top and bottom of the tube is maintained constant and is equal to the pressure outside.

Find it's(i.e that of the particles at the front) velocity after having fallen a height \(H\).

Now, the paradox arises when we consider an ideal fluid falling solely under the influence of gravity and no factors like viscosity or surface tension are considered. The fluid is incompressible.

Argument 1:

Use Bernoulli's theorem to find the new velocity. Clearly, the fluid will move faster as it reaches the bottom.

Argument 2:

Use the equation of continuity. As the cross-sections at the top and the bottom are the same, the velocities must also be the same as the fluid is incompressible.

So which argument, if any, is valid here?

Also, what happens when the fluid is not ideal, but say, just normal water:

In this case, what I believe will happen is that the stream of water will become narrower as it reaches the bottom, just like what happens to water coming out of a tap.

I would like to discuss the following:

1)Is this an example where the scope of applicability of either Bernoulli's theorem or the equation of continuity is exceeded? If so, why and what is their actual scope?

2)While an ideal fluid can't exist, is it, as a purely mathematical construct, consistent with the equations used to define it?

3)What happens in real life? i.e what factors \(will\) come into play? What happens if we assume a fluid that has surface tension but no viscosity(though I don't think that is possible as the same kind of forces between particles underlie both phenomena)? Can surface tension by itself resolve the paradox? If so, is some crude mathematical analysis possible?

4)I do understand that there will be some confusion as to treat this as a physics problem or a mathematics problem. For one thing, I have my doubts as to whether constant pressure can be maintained at the top and bottom (as mentioned). I would definitely like to hear every physical argument, but what I would really like to know is why the math/equations are not working simultaneously, the assumption again being, that a self-consistent model for an ideal fluid exists.

I remember discussing this problem with Raghav a while back and we reached the aforementioned consensus for the case of a real fluid.

I was reluctant to post this, but of late, discussions on physics have seen a lot of activity and thus, I'm hoping that this discussion will also gain some traction.

P.S: I think that this is closely related to D'Alembert's paradox

This might also be applicable.

Thanks in advance :)

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

TopNewestLet us begin with what we already know,

\(Au\) = \(av\)

\( \rho (v^{2}-{v_{0}}^{2}) = 2 \rho gh\) (Pressure must remain same throughout the length, as otherwise, the layers of water wont be in a steady state due to external pressure over powering or getting over powered due to lesser or more internal pressure)

so we easily get 'a' , the cross section at any height as

\( a = \frac {Av_{0}}{{{(v_{0}}^{2} + 2gh)}^{1/2}} \)

but , the important question you raise here is who causes the compression, and i claim that the reason is outside pressure and since outisde pressure must equal pressure inside fluid, so we have the compression explained qualitatively , since we already know that compression causes increase in number of collisions and hence rise in pressure from kinetic gas theory

Now ,after reading this, can you please be a little bit more elaborate, for i feel for sure that this is insufficient explanation of the real thing we are hunting, but i am unable to catch

So please slightly tell me what it is the exact question in your mind so that i can work on it and understand

Log in to reply

Sorry for the late reply...I had gone to sleep.

I also arrived at the same expression using your logic of pressure being constant. But I have a doubt which may seem a bit stupid but I would still like to have it cleared: I understand that the pressure at any given height should be \(constant\) as the shape of the fluid stream does not change in steady state. But I can't see why the pressure has to be the \(same/equal\) everywhere. i.e why can't the pressure be constant at a \(given\) height and yet be different at different heights as equilibrium between fluid pressure and external pressure can be established for any value of pressure.

Also, I guess what I really had in mind is the following question: Given a set of initial conditions, is it always possible to find all associated parameters after some time? I thought I had found a counterexample to this claim but looks like I was wrong. Obviously discussing such a question in general seems rather pointless. I guess what also motivated me was the question of \(complete\) and \(consistent\) (you must have heard of Godel's Incompleteness theorem)

But this was not the only scenario which brought this doubt into my mind:

Consider the following crude diagram:

My question is to find v1, v2 and also the pressure inside both the smaller tubes. The pressure on the left is say P and on the extreme right is say P'.

If it were a real fluid then we could easily convert the question into and equivalent electrical resistance question and proceed, but what happens when it is an ideal fluid? I can't seem to get enough equations. Raghav asked me this question but he doesn't seem to have seen this note yet. We also asked our physics teacher who said that it is an indeterminate scenario and this is what caused me to have the doubt.

Anyways, I guess the main thing is that I wanted to know if anyone else has also seen other such seemingly paradoxical situations. But maybe the ones I have posted do have solutions and I would stand corrected.

Also, I read a bit of the pdf file that you shared a link for. It's taking me a long time to get through all of it and if you have already learnt everything there, then hats-off to you!

I think I might be taking up too much of your time. I had thought that there would be at least 3-4 people as usual and thus this discussion would be over way sooner. So, do take your time to reply. No hurries and thanks for the help.

@Mvs Saketh

Log in to reply

Having , said that internal and external pressure , must be equal. I can further claim that since the external pressure is that of the air in the tube which is connected to outside Air, and since the gradient of pressure with height in Air is \(negligible\) Hence the pressure at all parts of the tube must be equal for the air and hence for the water

The motion of water cannot affect Air pressure

As for your Circuit question, that is interesting and surely you cannot simply use continuity and bernoulli to solve that as you know nothing of pressure here, However, there is one way to get sufficient equations. Assume that water being uniformly distributed all over the cross section must get distributed Ratio wise in the branches

Hence we have

\(Av\frac { A_{ 1 } }{ { A }_{ 1 }+{ A }_{ 2 } } =A_{ 1 }v_{ 1 }\\ \\ Av\frac { A_{ 2 } }{ { A }_{ 1 }+{ A }_{ 2 } } =A_{ 2 }v_{ 2 }\)

\(which \quad btw \quad also \quad satisfies \quad continuity \quad equation\) , now simply using cotinuity at other branch

we can get speed at second branch from

\(Av={ A }'v'\\ \)

Now , you can couple this with bernoullis equation to get pressure at different points,

I am assuming however that this is being done in a gravity free atmosphere, where fluid fills whole Area

Now, if you want to do it in a real atmosphere, then i will require more varriables, Namely, The air you have put in and related stuff

I believe, it is not so much that the equations are inconsistent but rather that external varriables affect the system through pressure and potential energy and hence without their knowledge we can't predict everything,

And no, its fine, i am helping because i have faced other Paradoxical situations and i believe its due to flaws in my concept perhaps and hence i was naturally interested at the question , maybe i will be able to solve it after this

And yes, keeping in account of time, Reply , whenever you want, no hurries

Log in to reply

Log in to reply

And I think we are finally getting to the crux of the matter....I also feel that the water will be divided ratio wise.

But my doubt is why we are not able to derive this condition from just the equations even though all the initial conditions are given (P,P',A,A1,A2,v,v')...Can it be derived? If not, what additional information needs to be specified and why?

Also, if some parameters do need to be mentioned, what is the \(complete\) set of parameters needed to specify the motion of an \(ideal\) fluid in addition to pressure, velocity and area of cross-section?(let's first discuss things in the absence of gravity).

It might be that this is a case where there are indeed infinitely many solutions...this is possible in a math question and not a physics question...And I think this is what I have muddled up...I have been \(assuming\) that there must be a unique solution because I have assumed that in this \(particular\) scenario both a real and an ideal will behave in the same way and using either should give the same answer (an analogy might be how we replace the working fluid of a Carnot engine with an ideal gas, calculate efficiency using ideal gas and then claim that efficiency must be same for all fluids).

So in a way, I guess I am asking in what scenario can I just use an ideal fluid? The obvious and perfectly valid answer from the standpoint of physics is obvious: You can't use an ideal fluid when it does not work!

Log in to reply

Hence, we include other formulas based on symmetry, here there is no viscous drag and hence the flow is circular and hence logically all portions of face must be equivalent and will enter whatever comes into their way .

Also note that my solution to your problem is not entirely correct, and it cannot be solved unless you tell me exactly what the cross sections are at the brancing point (because that is where fluid will be ratio wise distributed)

And, no there are infinitely many solutions because there are infinitely many varriables here , namely, the environment, whether there is gravity? whether there is air in the tube ? Is the fluid filling the whole cross section? Now if you want to fully analyse it, you must take into account all these but it will be a waste of time in the end because not only is it hard, but most real life fluids are viscous and tend to move with terminal velocity.

However, imagine how complex it would be if you allowed flow to be turbulent or density to change

Log in to reply

P.S. It would be complex enough for people to put a million dollar bounty on it. #Navier-Stokes #Millenium prize.

Log in to reply

Log in to reply

And I think we have done almost everything that can be done to analyse it under ideal conditions and I've understood things far more clearly. Thanks a lot for taking the time to go through my arguments and for posting yours.

I owe you one :)

Log in to reply

Log in to reply

@Raghav Vaidyanathan @Mvs Saketh @Ronak Agarwal @Azhaghu Roopesh M @Deepanshu Gupta @Pratik Shastri and anyone else who's interested :)

Log in to reply

I have thought about your first statement about the contradiction between bernoullis theorem and equation of continiuty before as well,

The thing i believe is that you are

assumingthe fluid always remains in contact with the walls, that is not necessary unless you claim there is some adhesive force between them,In reality it will become narrower as it falls down so that its speed can increase and yet \(Av\) remains constant,

This way, there is no need for the speeds in upper and lower portion to remain same (which by the way is physically impossible)

However, if you are asking how there can be radial contraction when there is no surface tension pulling the surface inward, Here is the explanation, when the fluid moves faster, the pressure inside it decreases, hence the outisde air is now able to compress it

But, i have faced other problems in other situations regarding fluid dynamics, where idealisation is probably the culprit

However there is an interesting example which i believe you might have seen is the solution to the diffusion equation when a finite amount of particles are released from a point in an infinite medium, it turns out that the solution is exponential and for any time \(t \geq 0 \), there are some particles at \( \infty\) as well, which is physically impossible and the paradox arises because we assume too much continuity in a discrete situation

Log in to reply

Yea, I mentioned the narrowing of the fluid stream and I believe it to be a correct explanation. My primary interest is to deal with this question only based on the equations for an ideal fluid. I want to know if the concept of ideal fluids has self consistent mathematical equations.

Log in to reply

Interesting, but can you please tell why you think the equations are inconsistent

Log in to reply

And I've not read about solutions to the diffusion equation for a finite number of particles. It sounds interesting and any link would be much appreciated :)

Log in to reply

However, the pressure in the tube is not changeing, And we are assuming that the pressure in fluid adjust so that even as it gains velocity leading to a fall in pressure inside it, it is getting compressed causing a rise in pressure both effects cancel each other.

also here : http://engineering.dartmouth.edu/~d30345d/courses/engs43/Chapter2.pdf (its not for discrete particles really)

Log in to reply

Log in to reply