Bernoulli's principal and law of conservation of energy

[rcgldr]

A good rule of thumb I learned was the incompressible flow assumption is typically valid below 220 mph.

From what I recall, in the of aircraft wings, compression and expansion factors are about 5% at 1/3rd mach depending on wing loading, which is only a bit faster than 220 mph. In the case of wings, no one really uses Bernouulli equations to calculate lift and drag, instead polar generation progams (like Xfoil) are based on some simplified implementation of Navier Stokes equations.

Venturi tubes are a better real world example of Bernoulli principle.

 

[dE_logics]

I've been thinking about how best to explain this issue and I think I now have it. The key is in understanding what a "simplifying assumption" is and how the concept is applied. A simplifying assumption is used to make calculations easier and is something that is assumed to be true but really isn't. Now how can such a thing be acceptable? How can you assume something to be true if it isn't? The answer is that as long as the error caused by the simplifying assumption is small enough not to significantly affect your conclusions, it is acceptable to use it.

So now to the incompressible flow assumption itself. We all know that air is compressible and any change, no matter how small will compress it. But when you are dealing with low speed flow, the pressure changes and the resulting density changes are very small when compared to atmospheric pressure and the density of air at atmospheric pressure. So the density changes are ignored: The density is assumed to be constant, ie., the fluid is incompressible. And that's all the incompressible flow assumption is about: it does not mean we are assuming there is no internal energy due to the pressure and density of the air.

Now how do we know that the density change is small enough to ignore? Well it depends on your needs. If you need your calculations to be 99% accurate, the maximum speed at which the assumption is valid is higher than if you need the calculation to be 99.9% accurate. A good rule of thumb I learned was the incompressible flow assumption is typically valid below 220 mph. I actually don't know what the error is at 220 mph, but now it's your turn to do some work here:

Step 1: Calculate the change in static pressure required for airflow out of a tank at 220 mph.

Step 2: Use the ideal gas law to determine the fractional change in the density of air with that static pressure change. That fraction is your percent error caused by the assumption.

We are talking mostly liquids here and that too an ideal liquid (eveything is hypothetical and following the law explicitly), according to Bernoulli's equation, an ideal liquid holds energy by the pressure it possesses...my question is how.

Since the fluid is ideal, there are no approximations.

However, the energy in the "p" term in the Bernoulli equation doesn't come from elastic potential energy of the fluid, is comes from the static pressure acting on the fluid element causing it to move, thus doing work.

So actually the pressure is not doing work...the piston is doing work to make the fluid gain K.E...the pressure is just acting like a medium to transfer the work.

For e.g -

Here, the rigid black colored rod is pushing the red colored piston...the work is done by the object pushing the piston not the rod...similarly the blue colored fluid is pushing the second piston (towards the RHS) and since it's uncompressable and poses (in this context) similar properties to the black colored rod, it will not do work, it won't even act as a significant pressure buffer.


On working on the derivations further (the latest PDF that I attached), I got a few new issues...a very general mechanical issue actually.

If a second piston is attached to the other end of the pipe...it will experience a force f governed by the formula...Since there's a force f there should also be acceleration i.e velocity of the fluid should increase as the force F continuously applies.
But is there any acceleration?

 

[dE_logics]

Also I tried some syringe calculations.

There should be some work done on the whole fluid inside the syringe when I push the piston.

I computed what maximum height can the fluid reach doing this same amount of work...the work done should be equal to the increase in P.E of the whole fluid.


Area of the cylinder - As
Force application on the piston = F
distance for which the piston is pushed = d
Density of fluid = ds
Work done = Fd
Heigh of elevation if the whole fluid is lifted to a height equal to the work done - h
Total mass in the syringe - As * ds * d
The mass will contain the potential energy Fd, thus
mgh = Fd
(As * ds * d) * g * h = Fd
diameter of syringe = 1.3/100 m
As = pi r^2 = 1.327322896e-4 m^2
ds = 1 kg/m^2
d = 5.2/100
F = 0.2 kg equivalent on Earth = 1.961N
g = 10
(As * ds * d) * g * h = Fd
(1.327322896e-4 * 1 * (5.2/100)) * 10 * h = 1.961*(5.2/100)
Solve for h, I get a very small value, but actually with the needle on the syringe, the height elevation is like .8 m...the thinner the needle more the elevation with the same amount of work done.

Sorry...I do not get a small value...I get a very large and insane value.

 

[rcgldr]

Is there any acceleration?

That ideal fluid has no friction with the pipe it flows through. If it's not accelerating then no work is done. If the piston at the right end doesn't have an external force being applied, then the ideal fluid moving at constant velocity has zero pressure, and there's no force being applied by either piston. It would be similar to a closed pipe with the fluid inside, moving at constant velocity in space with no external forces.

If you're goal is to turn this into some kind of Bernoulli example, then both pistons need to exert an inwards force to create a pressure on the fluid. If the fluid is flowing and the ends of the pipe and pistons have differing cross sectional areas, then the forces will need to be equal to the fluid pressure times the cross sectional area of the piston surface.

 

[dE_logics]

That ideal fluid has no friction with the pipe it flows through. If it's not accelerating then no work is done. If the piston at the right end doesn't have an external force being applied, then the ideal fluid moving at constant velocity has zero pressure, and there's no force being applied by either piston. It would be similar to a closed pipe with the fluid inside, moving at constant velocity in space with no external forces.

If you're goal is to turn this into some kind of Bernoulli example, then both pistons need to exert an inwards force to create a pressure on the fluid. If the fluid is flowing and the ends of the pipe and pistons have differing cross sectional areas, then the forces will need to be equal to the fluid pressure times the cross sectional area of the piston surface.

The acceleration here can be taken in 2 ways -

1) The acceleration that occurs when the fluid reaches a narrower cross section form a larger one.

2) An acceleration occurs even when the diameter of the pipe does not change (I'm referring to this one).

So there is an acceleration even in the second case?...practically I don't see so.

The question here is the P in the Bernoulli's equation, I need an explanation as to how it stores energy...and why P is directly taken as energy. By working on this example, I tried to see the same; i.e. if the static pressure is converting to K.E, but actually the work done on the piston is increasing the K.E of the fluid, not the pressure.

 

[SystemTheory]

This discussion may yield some insight, although NASA needs a better online equation editor! The questions ring a bell in my memory of thermodynamics with the term "isentopic flow."

The Molecular Scale Derivation makes some sense to me.

http://www.grc.nasa.gov/WWW/K-12/airplane/bern.html

Obviously acceleration (F = ma) and constant flow are two different aspects. In Bernoulli's Equation acceleration occurs in a portion of the fluid with a drop in static pressure within the accelerated portion, such that the energy per unit volume remains the same. Please try to think of pressure not as energy, but as energy per unit volume.

I also think there can be internal (non-zero) pressure in the moving ideal fluid. There would be no difference in pressure from inlet to outlet for ideal fluid flow in a uniform pipe (no pressure drop). This is like having no voltage drop in a perfect conductor.

 

[dE_logics]

In the molecular scale division, yes, it does make a bit of sense...when randomness exhibits some sort of order, we get an increase in velocity.

But the randomness in an ideal fluid has absolutely no energy...so how can it convert to K.E.? So the drop is static pressure should really not make a difference in that total energy of the fluid.

There is pressure in a moving ideal fluid, but the energy possessed by it cause of the same reason (static pressure) is negligible.

 

[SystemTheory]

But the randomness in an ideal fluid has absolutely no energy...

This plate shows the SI units of energy as Newton-meter {N-m}. The pressure is Newton per square meter {N/m^2}. The volume is meter cubed {m^3}. The dimensions of pressure times volume are energy.

http://physics.nist.gov/cuu/Units/units.html

The randomness of an ideal fluid does have static pressure, which one can measure, and this static pressure has the units of energy per unit volume. Therefore a given volume of ideal fluid with static pressure has an associated energy state.

 

[SystemTheory]

Another way to look at Bernoulli's equation is that because the fluid is ideal it can exchange energy between two forms, one potential, one kinetic, and there is no energy loss in the process. I can visualize this clearly about midway through this reference:

Derivation of Bernoulli's equations via Newton's Second Law:

http://home.earthlink.net/~mmc1919/venturi_discuss_math.html

This is a good reference for the novice and advanced interpratation of Bernoulli's equation, so check it out. The animation page is the logical starting point:

http://home.earthlink.net/~mmc1919/venturi.html

 

[dE_logics]

This plate shows the SI units of energy as Newton-meter {N-m}. The pressure is Newton per square meter {N/m^2}. The volume is meter cubed {m^3}. The dimensions of pressure times volume are energy.

http://physics.nist.gov/cuu/Units/units.html

The randomness of an ideal fluid does have static pressure, which one can measure, and this static pressure has the units of energy per unit volume. Therefore a given volume of ideal fluid with static pressure has an associated energy state.

This really makes sense. But how do we extract this energy?


Moreover it can happen that 2 quantities are having the same unit, but used in a different way and have different meanings.

For e.g (here) ρ is used as a substitution for m...but ρ and m are different things having different units.

Another e.g...I would like to mimic the unit of speed -

distance(d) = 10m
t = 5s
then speed = d/t with the unit in m/s

if I suppose multiply the d in the formula with 2 more variables (x, y) so as to depict the volume, then divide it by 2 variables (n,k) which also have the unit m so as to divide this volume into sub parts -

(d*x*y)/(t*n*k)

It's unit will be m/s but the results will not match with the actually speed...thus it can be said to be a quantity different from speed...actually it will be the rate at which the sections get compressed.

Yet again, the unit of pressure and K.E are different, but K.E is called as dynamic pressure. So I think unit cannot be always said to be depicting a certain quantity.

Anyway, I'm not sure about this.

In the initial pipe configuration, the static pressure on the right side is higher than the static pressure on the left. This does not mean the fluid is flowing from a lower pressure towards a higher pressure, since the total pressure is the same at every point.

Fluids flow is by difference in static pressure.

It was clear that some force had to be acting on the water to increase its speed. Earlier, Robert Boyle (1627-1691) had demonstrated that pressure and volume have an inverse relationship, and Bernoulli seems to have applied Boyle’s findings to the present situation. Clearly the volume of water flowing through the narrower pipe at
any given moment was less than that flowing through the wider one. This suggested, according to Boyle’s law, that the pressure in the wider pipe
must be greater.

In such a case, for a fluid to accelerate, there has to be a difference in pressure.

Another perspective of taking Bernoulli's principle is assuming that the components of the equation is pressure, not energy, but insertion of the potential energy cause of gravity make things a havoc.

We have use P as the pressure not P*volume; thus P at it's raw form cannot be energy.

 

[rcgldr]

But the randomness in an ideal fluid has absolutely no energy...so how can it convert to K.E.?

I'm at a loss for what you are trying to get at in this thread regrading Bernoullie princple and an ideal fluid. As mentioned earlier, an 'ideal fluid' is going to end up violating some common laws of physics. For example, the speed of sound in an incompressable fluid or solid is infinite (faster than the speed of light). In an ideal inviscid fluid, there is no drag on an object moving through that fluid. The fluid just 'slips' around the object without imparting any net directional force.

http://home.earthlink.net/~mmc1919/venturi_discuss_math.html

In the initial pipe configuration, the static pressure on the right side is higher than the static pressure on the left.

This isn't possible with an ideal fluid. All pressure changes are instantaneous. There are only two states. Either the idal fluid isn't moving and pressure is the same at all points, or the ideal fluid is flowing and the pressure is lower in the smaller diameter sections of pipe. The transition between these two states is instaneous.

 

[dE_logics]

I'm at a loss for what you are trying to get at in this thread regrading Bernoullie princple and an ideal fluid. As mentioned earlier, an 'ideal fluid' is going to end up violating some common laws of physics. For example, the speed of sound in an incompressable fluid or solid is infinite (faster than the speed of light). In an ideal inviscid fluid, there is no drag on an object moving through that fluid. The fluid just 'slips' around the object without imparting any net directional force.

So Bernoulli's principal violates the laws of physics?...cause it uses an ideal fluid in it's model?


I'm simply trying to understand (notice, the topic has been deviated from the 1st post) how can an ideal fluid store energy by static pressure...since Bernoulli's equation states it.

 

[rcgldr]

So Bernoulli's principal violates the laws of physics?...cause it uses an ideal fluid in it's model?

It's the ideal fuild that violates the laws of phyiscs, not Bernoulli's princple. Bernoulli's principle establishes an approximate relationship between pressure and speed, in an environment where no work is done. Note that something had to create the initial pressure difference in the first place, typically violating Bernoulli in the process (work was done), but once the pressure difference exists, then a gas or fluild will flow from higher pressure zones to lower pressure zones, and approximately in compliance with Bernoulli's equation.

 

[dE_logics]

It's the ideal fuild that violates the laws of phyiscs, not Bernoulli's princple.

Ok, why not just remove modern physics from the picture...this is the classical physics forum...so here an ideal fluid will not violate any physics laws.

in an environment where no work is done.

I absolutely did not put that in the picture.


So, suppose the fluid gains an inertia, and then the force stops acting; then the fluid will flow out of such a pipe -

without any more force application?

That means, if there is a continuous force application as in this case -

There will be a constant acceleration?

 

[SystemTheory]

dE_logics,

Frequently I write out symbolic equations with the units in brackets (dimensional analysis) to convince myself of the proper analysis and help keep track of terms. This is best done in SI units which are coherent and don't inject numbers for unit conversion factors into the symbolic equation. If you work Bernoulli's equation forward and backward to the energy equation this way you'll see how the fluid stores energy per unit volume, and that this property is called "static pressure" and "dynamic pressure."

In the first case (no applied force), if the pipe were infinitely long with diameter A1 any small volumetric element of the fluid would flow under its own inertia per Newton's First law. If the volume slice is constant, the density is constant, so the mass is constant. Write Bernoulli's equation for any point in this A1 section and it applies to all points. Use subscript A1.

Now let the pipe neck down from A1 to A2. Write Bernoulli's equation here using subscript A2 and equate this on the right side of an equal sign with the prior relation for section A1. The point here is that (1) flow energy is conserved; and therefore (2) energy per unit volume is also conserved; which is more conveniently measured as (3) the sum of the static and dynamic pressures per Bernoulli's equation.

In the second case (applied force) the fluid accelerates but it may be possible to treat it as a steady flow (not accelerating) just prior to the transition from A1 to A2, then apply Bernoulli's equation just as above. This is a simplifying assumption but may be reasonable in many cases.

 

[dE_logics]

If you work Bernoulli's equation forward and backward to the energy equation this way you'll see how the fluid stores energy per unit volume, and that this property is called "static pressure" and "dynamic pressure."

But I'm questioning Bernoulli's equation here, there's the issue.

In the first case (no applied force), if the pipe were infinitely long

I hope by saying infinity long, you're only assuming that the fluid does not flow out of the pipe...since if the pipe is infinitely long and filled with the fluid, the fluid will have infinite inertia...it will be like a case of a constantly applied force.

In the second case (applied force) the fluid accelerates but it may be possible to treat it as a steady flow (not accelerating) just prior to the transition from A1 to A2

Just prior?...you mean the fluid in the pipe with area A₂ and A₁ accelerates (I do not mean the transaction from A₁ to A₂)?...well is should accelerate.

 

[russ_watters]

So Bernoulli's principal violates the laws of physics?...cause it uses an ideal fluid in it's model?

Bernoulli's principle is a law of physics.

What you need to get your head around is the concept of a "simplifying assumption", which I described in my post # 29. At the time, you responded with this:

Since the fluid is ideal, there are no approximations.

No, a calling the fluid "ideal" is an approximation.

Please understand: every physics problem ever done in the real world is answered wrong. The goal isn't to get an exact answer, it is simply to get an answer close enough. Simplifying assumptions allow you to discard the little nuances to a problem that don't have a significant impact on the outcome of the problem. For many Bernoulli's principle problems, the compressibility of the fluid is one of those little nuances that doesn't have a significant effect on the answer to the problem.

I'm simply trying to understand (notice, the topic has been deviated from the 1st post) how can an ideal fluid store energy by static pressure...since Bernoulli's equation states it.

Again, from post #29: the incompressible flow assumption doesn't say the fluid isn't storing energy via static pressure, it is just saying the fluid's density change is negligible.

de_logistics, you are making a mountain out of a molehill here. It isn't a complicated issue.

 

[russ_watters]

So, suppose the fluid gains an inertia, and then the force stops acting; then the fluid will flow out of such a pipe -

without any more force application?

That means, if there is a continuous force application as in this case -

There will be a constant acceleration?

Bernoulli's principle deals with steady flow only. By mixing together different issues, you are confusing yourself.

 

[dE_logics]

I absolutely do not understand what is the significance of the assumption you are talking about regarding the main question.

In the first place, how do you come to the topic of assumption when when I'm questioning the theory? An assumption absolutely does not have to do anything here...we're taking an imaginary case and I'm not comparing it to any real life scenario.

Your answer answers a question "Why am I not practically getting this observation?" not "How can we say static pressure stores energy?" and the latter is my question.

Yeah, I did a few experiments with the syringe but it was a different topic.

There is a question here with the following dependencies -

1) Bernoulli's principal assumes an ideal fluid and the fact that pressure can store energy...how can an ideal fluid store energy in terms of static pressure?

This is the only problem with Bernoulli's equation I'm having...if we just remove the law of conservation of energy part and just take the equation to be the summation of a set of values towards the left and RHS...I'm fine with it.

In the whole of the above question, where am I comparing the thing to a real life scenario so as to give the assumption a significance?

Since the fluid is ideal, there are no approximations.

I meant we are taking a hypothetical case of an ideal fluid...the the theory with the ideal fluid should be absolute.

Your theory of approximation absolutely does not have to do anything with the question here.

When we are manipulating circuits we assume the wires to be a superconductor and then analyze the theory. The assumption does not have a significance when we're talking everything in paper...I absolution do not get the heads or tails of what you mean!

 

[dE_logics]

Bernoulli's principle deals with steady flow only. By mixing together different issues, you are confusing yourself.

I'm not mixing the 2 I'm asking which 1 of the 2 does bernoulli's principal assume?The simple question here is that suppose the fluid inside the first tube has inertia twards the right (the tube is not infinitely long, i.e it has a limited amount of fluid in it) will all the fluid easily pass out of the thinner cross section following Bernoulli's equation?

 

[SystemTheory]

The simple question here is that suppose the fluid inside the first tube has inertia twards the right (the tube is not infinitely long, i.e it has a limited amount of fluid in it) will all the fluid easily pass out of the thinner cross section following Bernoulli's equation?

Yes, assuming the fluid is ideal and there is zero gravity. Of course a real fluid is not ideal.

I'm questioning the theory (of Bernoulli's equation)

How can static pressure of an ideal fluid store energy? The microscopic particles in an ideal fluid have kinetic energy. No energy or momentum is lost when the particles collide within an ideal fluid. The kinetic theory of ideal gases is a good place to study this conceptual model.

When a force acts at one end of an ideal fluid, it sees the total mass of the fluid, so the acceleration in many cases is relatively small. The piston accelerates quite slowly. The pressure is transmitted instantaneously in an ideal fluid and therefore Bernoulli's law is a reasonable approximation anyway. This is like the voltage spike on an inductor with zero starting current flow, a long pipe full of fluid is like an inductor in a circuit analogy when applying the pressure-voltage analogue.

 

[russ_watters]

I absolutely do not understand what is the significance of the assumption you are talking about regarding the main question.

In the first place, how do you come to the topic of assumption when when I'm questioning the theory? An assumption absolutely does not have to do anything here...we're taking an imaginary case and I'm not comparing it to any real life scenario.

When you said this in the OP:

...considering an ideal fluid in this situation...

...you were invoking the incompressible flow simplifying assumption (as well as inviscid, steady, lossless, and no heat transfer assumptions). How could you not know that? Didn't you know that there is a version of Bernoulli's equation that does not assume incompressible flow? Did you not read the wiki on the subject?

This is why I said in post #29 that the entire key to this thread is that you don't understand the purpose and scope of simplifying assumptions...

1) Bernoulli's principal* assumes an ideal fluid and the fact that pressure can store energy...how can an ideal fluid store energy in terms of static pressure?

This is the only problem with Bernoulli's equation I'm having...

I answered this directly, already: the incompressible flow form of Bernoulli's equation only assumes no change in density. You are equating the incompressible flow assumption with an assumption that static pressure can't store energy. It does not assume that static pressure can't store energy. You can see that in the equation!

*And agin, Bernoulli's principle DOES NOT assume an ideal fluid. A particular form of Bernoulli's equation might, though.

I'm questioning the theory (of Bernoulli's equation)

You need to stop doing that. You need to stop with making these diagrams and inventing situations to try to prove physical theories wrong. You don't understand Bernoulli's principle anywhere near well enough to challenge it and you're meandering around with lines of thought that don't have anything to do with Bernoulli's principle. This is why you are having so much trouble here (and why people are getting frustrated with you): You are not really trying to learn.

Since you clearly didn't read it, here's the opening two paragraphs from the wiki on the subject. Read them!

In fluid dynamics, Bernoulli's principle states that for an inviscid flow, an increase in the speed of the fluid occurs simultaneously with a decrease in pressure or a decrease in the fluid's potential energy.[1][2] Bernoulli's principle is named after the Dutch-Swiss mathematician Daniel Bernoulli who published his principle in his book Hydrodynamica in 1738.[3]

Bernoulli's principle can be applied to various types of fluid flow, resulting in what is loosely denoted as Bernoulli's equation. In fact, there are different forms of the Bernoulli equation for different types of flow. The simple form of Bernoulli's principle is valid for incompressible flows (e.g. most liquid flows) and also for compressible flows (e.g. gases) moving at low Mach numbers. More advanced forms may in some cases be applied to compressible flows at higher Mach numbers (see the derivations of the Bernoulli equation).

The simple question here is that suppose the fluid inside the first tube has inertia twards the right (the tube is not infinitely long,

The phrase "inertia towards the right" is gibberish. I don't think you understand what the word "inertia" means...

...perhaps the word you are looking for is momentum?

i.e it has a limited amount of fluid in it)

Bernoulli's principle doesn't work in empty pipes, only full pipes.

...will all the fluid easily pass out of the thinner cross section following Bernoulli's equation?

What does "easily" mean?

That line of questioning is incoherent.

 

[russ_watters]

When we are manipulating circuits we assume the wires to be a superconductor and then analyze the theory.

That's a great example of exactly what I'm talking about: it is a simplifying assumption.

The assumption does not have a significance when we're talking everything in paper...

Oh, it absolutely does! If the resistance of the wire is high enough, it will affect the accuracy/usefulness of the calculations. For example, if the wires are 10 miles long (ie, power lines), the resistance can't be ignored. If the circuit is a flashlight, they can.

Understanding what simplifying assumptions are available, which ones to use and when you can use them is a critical skill for "doing" science.

 

[dE_logics]

Yes, assuming the fluid is ideal and there is zero gravity. Of course a real fluid is not ideal.
How can static pressure of an ideal fluid store energy? The microscopic particles in an ideal fluid have kinetic energy. No energy or momentum is lost when the particles collide within an ideal fluid. The kinetic theory of ideal gases is a good place to study this conceptual model.

When a force acts at one end of an ideal fluid, it sees the total mass of the fluid, so the acceleration in many cases is relatively small. The piston accelerates quite slowly. The pressure is transmitted instantaneously in an ideal fluid and therefore Bernoulli's law is a reasonable approximation anyway. This is like the voltage spike on an inductor with zero starting current flow, a long pipe full of fluid is like an inductor in a circuit analogy when applying the pressure-voltage analogue.

Ok...I'm sort of satisfied with the fact that it has energy in terms of K.E. of each molecule...but you know the real doubt still persists.

There are 2 things that manifest high K.E of molecules -

1) High temperature

2) High pressure.

Now, depending on the type of fluid it might be that at high pressure the amount of energy stored in the fluid is high or low...but for an ideal fluid it's 0, although the K.E of the molecules is very high.

So the K.E is not a direct measure of the energy stored in the fluid...I think. Cause it can also happen that under certain amount of pressure the molecules do not gain certain amount of K.E as compared to other fluids.

It does not assume that static pressure can't store energy.

Okay...I see.

So we assume an ideal fluid just to put ρ instead of m...but in reality since fluids can store energy in terms of pressure and that is quiet significant, we do not assume P as 0...or when it comes to P, we do not assume the fluid as ideal.

I conclude that at certain places the unmodified Bernoulli's equation assumes an ideal fluid, and at some places a real one.I think that solves the issue...thanks a lot, it could not have been cleared without bringing in this perspective of the assumption.However there's still a small doubt left -

So, suppose the fluid gains an inertia, and then the force stops acting; then the fluid will flow out of such a pipe -

without any more force application?

That means, if there is a continuous force application as in this case -

There will be a constant acceleration?

 

[russ_watters]

However there's still a small doubt left -

You are misusing the word "inertia", so the question is meaningless (perhaps you mean "momentum"?). Please rephrase.

Also, Bernoulli's principle and equation don't deal with partially empty pipes, nor does it deal with unsteady (accelerating) flow. So your question doesn't seem to me to have anything to do with Bernoulli's principle.

 

[rcgldr]

So, suppose the fluid gains an inertia, and then the force stops acting; then the fluid will flow out of such a pipe

Except that once there is no inwards force from the pistons, the pressure drops to zero. Each piston need to exert some exact amount of force to maintain a constant mass flow rate within the pipe. As you pointed out earlier, this occurs when the there is no net work done on the fluid.

 

[dE_logics]

You are misusing the word "inertia", so the question is meaningless (perhaps you mean "momentum"?). Please rephrase.

Also, Bernoulli's principle and equation don't deal with partially empty pipes, nor does it deal with unsteady (accelerating) flow. So your question doesn't seem to me to have anything to do with Bernoulli's principle.

I meant momentum.

What actually do you mean by partially empty?...vertically (i.e the whole cross section of the pipe is not filled) or horizontally (the whole pipe is not filled along it's length, but filled along it's cross section)?

Except that once there is no inwards force from the pistons, the pressure drops to zero. Each piston need to exert some exact amount of force to maintain a constant mass flow rate within the pipe. As you pointed out earlier, this occurs when the there is no net work done on the fluid.

So, the answer is no, there has to be a constant force application to make all the fluid flow out...but it's not a function of how much force...even the slightest force will be good enough to make all the fluid flow out.

 

[Phrak]

1. If I take a piston and change the pressure of an ideal, incompressible, nonviscous fluid then no work is done on the fluid.

2. Changing the pressure on this fluid does no work, whether it is in motion or not.

Any logical or physical errors so far?

 

[dE_logics]

1. If I take a piston and change the pressure of an ideal, incompressible, nonviscous fluid then no work is done on the fluid.

2. Changing the pressure on this fluid does no work, whether it is in motion or not.

Any logical or physical errors so far?

That problem has been solved...this is an extended issue.

Maybe I'll start a new thread on this.

 

[russ_watters]

What actually do you mean by partially empty?...vertically (i.e the whole cross section of the pipe is not filled) or horizontally (the whole pipe is not filled along it's length, but filled along it's cross section)?

If the fluid is air, the air will spread out to fill the pipe due to its own pressure. This will happen whether you are applying a force to it or not. The far end can't just stay a vacuum.

If the fluid is water and there is air at the end of the pipe, the water will spill out and the air will move backwards to fill the pipe.

Also, this "force" has to come from somewhere. If you have a piston, for example, and you stop moving the piston, the air or water will have to stop too, otherwise you'll generate a vacuum where it meets the piston.

So, the answer is no, there has to be a constant force application to make all the fluid flow out...but it's not a function of how much force...even the slightest force will be good enough to make all the fluid flow out.

Ehh, water in an unfilled pipe with air on each end will gain momentum that will carry it for a while if you ram it with a piston (then somehow remove the piston and let air in that end). But again, these things have bear no relationship to the scenarios described by Bernoulli's principle and your description is not detailed enough for us to know what your scenarios are really trying to describe. You can't just draw an arrow and say there is a force. You need to be more specific about what is causing the force and how it can just go away.

I really have no idea what you are trying to describe with that first diagram. For the second diagram, the system will accelerate and quickly reach an equilibrium with steady flow, based on the applied pressure equalling the total pressure required to move the fluid at a certain velocity through the small pipe.