# Bernoulli's principal and law of conservation of energy

1. Dec 7, 2009

### 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.

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2. Dec 7, 2009

### 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.

3. Dec 7, 2009

### stewartcs

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

4. Dec 7, 2009

5. Dec 7, 2009

### Staff: Mentor

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

6. Dec 7, 2009

### dE_logics

But how?

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

Last edited: Dec 7, 2009
7. Dec 7, 2009

### 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.

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8. Dec 7, 2009

### dE_logics

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

Thanks for the help. :)

9. Dec 7, 2009

### Staff: Mentor

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.

10. Dec 7, 2009

### 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?

11. Dec 7, 2009

### 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.

12. Dec 7, 2009

### rcgldr

Re: Bernoulli's principle and law of conservation of energy

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.

13. Dec 7, 2009

### dE_logics

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 -

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14. Dec 7, 2009

### 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.

15. Dec 7, 2009

### rcgldr

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

16. Dec 7, 2009

### 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.

17. Dec 8, 2009

### dE_logics

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?

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

18. Dec 8, 2009

### dE_logics

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

19. Dec 8, 2009

### rcgldr

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).

20. Dec 8, 2009

### dE_logics

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 -

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?

21. Dec 8, 2009

### rcgldr

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.

22. Dec 8, 2009

### dE_logics

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.

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23. Dec 8, 2009

### dE_logics

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

24. Dec 8, 2009

### 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.

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25. Dec 8, 2009

### rcgldr

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.