Bernoulli's principal and law of conservation of energy

[dE_logics]

In a pipe of varying diameter in which an ideal fluid is flowing, there's an increment in the kinetic energy of each particle as it reaches a region of lower cross section from a higher cross section in the pipe.

Following the law of conservation of energy, there should be an energy transfer in some way which allows this increment in kinetic energy; the energy stored by virtue of pressure in the fluid is said to provide this energy but in an ideal fluid, there can't be any energy stored though pressure...so considering an ideal fluid in this situation, where does this energy come from?

E.g in this case of an ideal fluid flow -

This additional K.E by virtue of y cannot come though conversion of pressure to K.E since an ideal fluid cannot store energy using pressure.
I've tried by best to explain the situation, but if you still do not understand, please complaint.

 

[MaxwellsDemon]

Is it possible that the mass density of your ideal fluid in the little section of pipe is different than the mass density in the big section? This is assuming that the kinetic energy remains constant in both sections.

 

[stewartcs]

In a pipe of varying diameter in which an ideal fluid is flowing, there's an increment in the kinetic energy of each particle as it reaches a region of lower cross section from a higher cross section in the pipe.

Following the law of conservation of energy, there should be an energy transfer in some way which allows this increment in kinetic energy; the energy stored by virtue of pressure in the fluid is said to provide this energy but in an ideal fluid, there can't be any energy stored though pressure...so considering an ideal fluid in this situation, where does this energy come from?

E.g in this case of an ideal fluid flow -

This additional K.E by virtue of y cannot come though conversion of pressure to K.E since an ideal fluid cannot store energy using pressure.
I've tried by best to explain the situation, but if you still do not understand, please complaint.

Flow energy. The pressure is acting on the area of a fluid element (a force) which causes it to move some distance (d) which results in work. Since the total energy is constant, and, in this case equal to,

$$\frac{P}{\gamma} + \frac{v^2}{2g}$$

the KE must increase since the flow energy has decreased in order to satisfy the Conservation of Energy.

CS

 

[SystemTheory]

DE_Logics,

My knowledge of fluids is mostly from the NASA history reference. This page has Figure 25 and Figure 28 in which it appears the dynamic pressure of an ideal fluid can increase:

http://history.nasa.gov/SP-367/chapt3.htm

However I am reading very rapidly and must return to the analysis later.

Here's the link to the Index page:

http://history.nasa.gov/SP-367/contents.htm

 

[russ_watters]

...but in an ideal fluid, there can't be any energy stored though pressure...

Why do you think that is true? It isn't.

 

[dE_logics]

Why do you think that is true? It isn't.

But how?

DE_Logics,

My knowledge of fluids is mostly from the NASA history reference. This page has Figure 25 and Figure 28 in which it appears the dynamic pressure of an ideal fluid can increase:

http://history.nasa.gov/SP-367/chapt3.htm

However I am reading very rapidly and must return to the analysis later.

Here's the link to the Index page:

http://history.nasa.gov/SP-367/contents.htm

CS

Thanks, I'll surely see to it...that will answer the question.

 

[dE_logics]

Flow energy. The pressure is acting on the area of a fluid element (a force) which causes it to move some distance (d) which results in work. Since the total energy is constant, and, in this case equal to,

$$\frac{P}{\gamma} + \frac{v^2}{2g}$$

the KE must increase since the flow energy has decreased in order to satisfy the Conservation of Energy.

CS

There are my derived conclusions after reading a few articles with the topic "Flow energy" -

Now though these principals, there's an increment in the kinetic energy of each particle as it reaches a lower cross section region...following the law of conservation of energy, there should be an energy transfer in some way which allows this increment in kinetic energy. This energy comes from the “flow energy”.
There should be some work done to make the fluid flow. Suppose in such a pipe -

A force F acts though the piston and into a distance d1 doing work F d1; considering the ideal nature of the fluid inside the pipe, the input work done should be equal to the output work done and so on the other end, a force of f will be acted by the fluid though a distance d2 such that -

F d1 = f d2

With the knowledge of F d1 and d2, one can estimate that f will be larger than F -
Assuming F = 10
d1 = 15
d2 = 5
150 = 5f
f = 30

A higher force application will mean more acceleration and so more kinetic energy in the fluid which's in the smaller cross section area of the pipe...All this results in the the fluid fluid gaining more kinetic energy by virtue of the work done on the fluid.
Notice, to allow the fluid to keep flowing, there should be continuous application of force (to do the work)...if the flow is solely to rely on...for example inertia of the fluid (which contains limited kinetic energy) the fluid flow might stop before all the fluid drains out of the arrangement, reason being the motion will be used to do work on the fluid towards the narrower section of the pipe.

 

[dE_logics]

But I think there's a new doubt in queue which I will post in another thread.

Thanks for the help. :)

 

[russ_watters]

But how?

How what? Have you read through the description of how it was derived? http://en.wikipedia.org/wiki/Bernoulli's_principle

A force F acts though the piston and into a distance d1 doing work F d1; considering the ideal nature of the fluid inside the pipe, the input work done should be equal to the output work done and so on the other end, a force of f will be acted by the fluid though a distance d2 such that...

What force f? There aren't any objects there for the fluid to imparting a force on. And as such...

F d1 = f d2

With the knowledge of F d1 and d2, one can estimate that f will be larger than F -
Assuming F = 10
d1 = 15
d2 = 5
150 = 5f
f = 30

That's all just nonsense.

 

[rcgldr]

I think the OP's issue is how do you decribe how pressure varies within an incompressable fluid, or similarly, how do you describe how force works with an incompressable object?

 

[SystemTheory]

In Bernoulli's equation the total pressure of an ideal fluid is derived from the Conservation of Energy. My understanding is the total pressure and energy are related and remain constant for free horizontal flow of an ideal fluid. No work is done by an external agent. If work is done internally (I'm not sure) no energy is lost, i.e., it is reversible.

This plate at the NASA reference shows clearly how the sum of static pressure p and dynamic pressure q remain constant in the venturi tube, so there is no net change in total pressure and no net change in energy of the moving fluid. I don't think work is done in this apparatus.

http://history.nasa.gov/SP-367/f28.htm

The question of a force pushing a piston on an incompressible fluid is a different system, where work is done by the piston. I did not interpret the original question that way.

 

[rcgldr]

These classis examples of Bernoulli's principle are problematic when you also consider the issue of zero viscosity. Adjacent streamlines can flow at different rates wihout interaction. For example if a flow from a narrow section of pipe exits into a wider section of pipe, that flow will only interact with the fluid directly ahead, and not with the surrounding fluid. Eventually you could end up with a narrow stream of fluid flowing through the wider section of pipe, leaving the surrounding fluid undisturbed.

 

[dE_logics]

How what? Have you read through the description of how it was derived? http://en.wikipedia.org/wiki/Bernoulli's_principle
What force f? There aren't any objects there for the fluid to imparting a force on. And as such... That's all just nonsense.

This has to be related to law of conservation of energy...by this formula it's -

K.E (a form of energy) + potential energy (another form of energy) + P (I don't know how do we consider pressure to be a from of energy) = constant (should be the summation of all energies, but that p is causing the problem.)

Question -- How do we prove that p is energy?

I referred this before deriving that -

 

[SystemTheory]

I agree there are interactions between layers in a real fluid. Also between the fluid and bodies at the boundary layer. However the OP implies that energy cannot be stored as static pressure in an ideal fluid.The NASA plates appear to be informed and convince me otherwise.

 

[rcgldr]

How do we prove that p is energy?

Pressure energy = pressure x volume.

Bernoulli equation for ideal fluid: pressure/density + g h + 1/2 v² = constant

Multiply this by mass and you get:

(pressure x volume) + m g h + 1/2 m v² = total mechanical energy = constant

 

[SystemTheory]

A clear way to see the energy connection is to let height change be zero. The mgh term drops out. The total mechanical energy in an ideal fluid:

$$pV + \frac{1}{2}mv^{2}$$

Fluid is incompressible so the density is constant:

$$m = \rho V$$

Substitute for mass m and divide by the volume V:

$$p + \frac{1}{2} \rho v^{2}$$

Thus Bernoulli's equation is the mechanical energy per unit volume stated as the static pressure and dynamic pressure for horizontal flow of an ideal fluid.

 

[dE_logics]

Pressure energy = pressure x volume.

So...if suppose I have a 5 ton object place on the piston of a piston cylinder apparatus holding an ideal fluid and the piston does not move on placement of that object, the energy possessed by it will be pressure x volume?...completely independent of the displacement of the piston?

Bernoulli equation for ideal fluid: pressure/density + g h + 1/2 v² = constant

This is suppose to be true for streamline flow only right?...and the energy is same everywhere in the whole of the fluid?

 

[dE_logics]

By saying "energy storage through pressure" I mean the fluid is like a spring storing pressure...is this the case?

 

[rcgldr]

Pressure energy = pressure x volume.

So...if suppose I have a 5 ton object place on the piston of a piston cylinder apparatus holding an ideal fluid and the piston does not move on placement of that object, the energy possessed by it will be pressure x volume?...completely independent of the displacement of the piston?

Since it's an imaginary incompressable fluid, there is no displacement, just an increase in pressure. Energy is the ability to perform work, but in the case of an incompessable fluid with no movement, there is no means to harness that energy.

For the classic Bernoulli thought experiement in a lossless environment, the incompressable fluid needs to be flowing in a pipe of varying diameters, somehow contained, such as frictionless pistons at both ends of the pipe. In order to be lossless, there's no friction between the walls of the pipe and the fluid, and none of flow in the fluid causes an increase in temperature. Since the mass flow at any cross section of the pipe is costant, the speed of the fluid is faster when the pipe diameter is smaller. Since there's no work being done, there's an internal conversion between pressure energy and kinetic energy within the fluid flow. But if the fluid isn't flowing the pressure is the same everywhere within the pipe. The instant the fluid starts flowing, then there's an instaneous decrease in pressure in the narrower sections of the pipe. The magnitude of the velocity doesn't matter, only if it's zero or non-zero.

By saying "energy storage through pressure" I mean the fluid is like a spring storing pressure...is this the case?

Yes, in real life, forces and pressures are associated with deformation, compression, and expansion at the point of application of the forces, when physical contact is involved. (A constant gravitational field would allow force and acceleration of an object to occur without deformation).

 

[dE_logics]

Since it's an imaginary incompressable fluid, there is no displacement, just an increase in pressure. Energy is the ability to perform work, but in the case of an incompessable fluid with no movement, there is no means to harness that energy.

For the classic Bernoulli thought experiement in a lossless environment, the incompressable fluid needs to be flowing in a pipe of varying diameters, somehow contained, such as frictionless pistons at both ends of the pipe. In order to be lossless, there's no friction between the walls of the pipe and the fluid, and none of flow in the fluid causes an increase in temperature. Since the mass flow at any cross section of the pipe is costant, the speed of the fluid is faster when the pipe diameter is smaller. Since there's no work being done, there's an internal conversion between pressure energy and kinetic energy within the fluid flow. But if the fluid isn't flowing the pressure is the same everywhere within the pipe. The instant the fluid starts flowing, then there's an instaneous decrease in pressure in the narrower sections of the pipe. The magnitude of the velocity doesn't matter, only if it's zero or non-zero.

Yes, in real life, forces and pressures are associated with deformation, compression, and expansion at the point of application of the forces, when physical contact is involved. (A constant gravitational field would allow force and acceleration of an object to occur without deformation).

So the interconversion of energy to pressure is an assertion governed by the velocity of fluid flow.

We saw that the velocity increases towards the bottle neck so we assumed the bottle neck to be at low pressure which causes the fluid to accelerate as it move towards the narrower region.

Since we forcefully have to follow the law of conservation of energy (else the almighty scientists will reject our theories and make make fun of us), we said that the energy came from the loss of pressure; so everyone's happy with that...but in this derivation -

There are my derived conclusions after reading a few articles with the topic "Flow energy" -

Now though these principals, there's an increment in the kinetic energy of each particle as it reaches a lower cross section region...following the law of conservation of energy, there should be an energy transfer in some way which allows this increment in kinetic energy. This energy comes from the “flow energy”.
There should be some work done to make the fluid flow. Suppose in such a pipe -

A force F acts though the piston and into a distance d1 doing work F d1; considering the ideal nature of the fluid inside the pipe, the input work done should be equal to the output work done and so on the other end, a force of f will be acted by the fluid though a distance d2 such that -

F d1 = f d2

With the knowledge of F d1 and d2, one can estimate that f will be larger than F -
Assuming F = 10
d1 = 15
d2 = 5
150 = 5f
f = 30

A higher force application will mean more acceleration and so more kinetic energy in the fluid which's in the smaller cross section area of the pipe...All this results in the the fluid fluid gaining more kinetic energy by virtue of the work done on the fluid.
Notice, to allow the fluid to keep flowing, there should be continuous application of force (to do the work)...if the flow is solely to rely on...for example inertia of the fluid (which contains limited kinetic energy) the fluid flow might stop before all the fluid drains out of the arrangement, reason being the motion will be used to do work on the fluid towards the narrower section of the pipe.

It can be seen that law of conservation of energy holds true without involvement of any sort of "pressure energy transfer", so, if this derivation is actually true, then energy being transferred from the static pressure to the dynamic pressure will be against the law of conservation of energy...so this conversion should not happen according to this derivation.

Sadly these derivations are wrong (according to russ_watters). So our imaginary conversion of energy from static pressure to dynamic pressure forcefully holds true.

Or is it that we have a better explanation?

 

[rcgldr]

F d1 = f d2 ... With the knowledge of F d1 and d2, one can estimate that f will be larger than F

Since d2 is greater than d1, then f will be smaller then F. Also there's no net work being done on the fluid, (the center of mass of the fluid is not accelerating) so the actual equation should be:

F d1 + f d2 = 0.

static pressure ... dynamic pressure ... energy

pressure is energy per unit volume.

 

[dE_logics]

There are my derived conclusions after reading a few articles with the topic "Flow energy" -

Now though these principals, there's an increment in the kinetic energy of each particle as it reaches a lower cross section region...following the law of conservation of energy, there should be an energy transfer in some way which allows this increment in kinetic energy. This energy comes from the “flow energy”.
There should be some work done to make the fluid flow. Suppose in such a pipe -

A force F acts though the piston and into a distance d1 doing work F d1; considering the ideal nature of the fluid inside the pipe, the input work done should be equal to the output work done and so on the other end, a force of f will be acted by the fluid though a distance d2 such that -

F d1 = f d2

With the knowledge of F d1 and d2, one can estimate that f will be larger than F -
Assuming F = 10
d1 = 15
d2 = 5
150 = 5f
f = 30

A higher force application will mean more acceleration and so more kinetic energy in the fluid which's in the smaller cross section area of the pipe...All this results in the the fluid fluid gaining more kinetic energy by virtue of the work done on the fluid.
Notice, to allow the fluid to keep flowing, there should be continuous application of force (to do the work)...if the flow is solely to rely on...for example inertia of the fluid (which contains limited kinetic energy) the fluid flow might stop before all the fluid drains out of the arrangement, reason being the motion will be used to do work on the fluid towards the narrower section of the pipe.

Actually I know there's another paradox flowing through these derivations.

Take 2 syringes...one without a needle and the other with a needle attached to it.

Both of them have identical pressure application on the piston (ignore the surface tension) till the piston reaches the end of the cylinder...the work done by both of these pistons are identical but (by Bernoulli's principal and practical experience) the K.E of each molecule of the fluid flowing out of the syringe with the needle on it is higher relative to the one without a needle.

 

[dE_logics]

Since d2 is greater than d1, then f will be smaller then F. Also there's no net work being done on the fluid, (the center of mass of the fluid is not accelerating) so the actual equation should be:

F d1 + f d2 = 0.

pressure is energy per unit volume.

Oh no...I forgot that the d2 is a length...not the diameter...thanks for correcting...I'll fix it.

 

[dE_logics]

I've fixed the derivations and hopefully it does not have any more issues.

I have a question about f, is this the force applied by the fluid on it's surrounds?...I really think it's not.

 

[rcgldr]

I have a question about f

Both F and f are the forces applied to the imaginary pistons in the pipe in your example. The pressure near each piston is defined as:

pressure = (force on piston) / (area of the piston surface)

The pistons are the source of the pressure on the fluid.

 

[stewartcs]

This has to be related to law of conservation of energy...by this formula it's -

K.E (a form of energy) + potential energy (another form of energy) + P (I don't know how do we consider pressure to be a from of energy) = constant (should be the summation of all energies, but that p is causing the problem.)

Question -- How do we prove that p is energy?

I referred this before deriving that -

I proved it to you in my first post. Flow energy (i.e. Pressure energy) is simply the pressure acting on a fluid element that cause it to flow (or move) a given distance...which results in work.

So...if suppose I have a 5 ton object place on the piston of a piston cylinder apparatus holding an ideal fluid and the piston does not move on placement of that object, the energy possessed by it will be pressure x volume?...completely independent of the displacement of the piston?

No. There is no displacement involved so no work is being done.

Since it's an imaginary incompressable fluid, there is no displacement, just an increase in pressure. Energy is the ability to perform work, but in the case of an incompessable fluid with no movement, there is no means to harness that energy.

There is a displacement, hence why the energy is pressure x volume. The volume results from the displacement of the fluid element as it flows through the pipe (i.e. area x length of the displacement = volume).

CS

 

[dE_logics]

Both F and f are the forces applied to the imaginary pistons in the pipe in your example. The pressure near each piston is defined as:

pressure = (force on piston) / (area of the piston surface)

The pistons are the source of the pressure on the fluid.

Thanks...that clears things up.

 

[dE_logics]

I proved it to you in my first post. Flow energy (i.e. Pressure energy) is simply the pressure acting on a fluid element that cause it to flow (or move) a given distance...which results in work.

Thanks, but the main issue lies with the static pressure...an ideal fluid it cannot store energy.

The static pressure is acting as medium though which the force applies...in an incompressible fluid it cannot store energy.

There is a displacement, hence why the energy is pressure x volume. The volume results from the displacement of the fluid element as it flows through the pipe (i.e. area x length of the displacement = volume).
CS

We were actually referring to a sealed piston cylinder arrangement.

Ok, so there will be a pressure on the piston which will move a certain volume of the fluid into the pipe...so work is done on the fluid...this work will be simply converted to a form of motion...why (or actually how) will it be stored as pressure in an uncompressable fluid?

 

[russ_watters]

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.

 

[stewartcs]

Thanks, but the main issue lies with the static pressure...an ideal fluid it cannot store energy.

The static pressure is acting as medium though which the force applies...in an incompressible fluid it cannot store energy.
We were actually referring to a sealed piston cylinder arrangement.

Ok, so there will be a pressure on the piston which will move a certain volume of the fluid into the pipe...so work is done on the fluid...this work will be simply converted to a form of motion...why (or actually how) will it be stored as pressure in an uncompressable fluid?

I'm am talking about the static pressure.

The idea you are getting hung up on is that the fluid is storing energy like a spring. Since it is incompressible you are reasoning that it cannot store energy like a spring. Which is correct, an incompressible fluid does not contain elastic potential energy like a spring would. 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.

The flow energy (also known as pressure energy) comes from the static pressure acting on a fluid element causing it to move. This results in work being done by the pressure acting on the fluid element. Energy is the capacity to do work, since work is being done, energy exists.

Now add in the assumptions that are required by Bernoulli and you come up with a total amount of energy that is constant. There must be a trade off between the static pressure and the velocity head (dynamic pressure) such that the total pressure is constant in order to conserve energy.

Hence, as the area of the pipe is reduced the velocity head will increase, and the pressure head will decrease accordingly.

CS

 

[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.