# Flow work and kinetic energy in fluid flow

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1. Jul 23, 2015

### Soumalya

Considering horizontal flow in a pipe (under the action of pressure forces only) if we focus on a fluid element we have a pressure gradient across the same i.e, a net pressure force will always act over the element that essentially maintains the flow.Now, in a region in the flow field if the specific volume of the fluid is 'v' and the average pressure immediately upstream (in excess of the downstream pressure) of a fluid element (of unit mass) considered in the region is 'p' then the work done by the resultant pressure force to push this element across a cross section is pv which is what is known as the 'flow work' or sometimes 'flow energy'.

Now, considering the fluid element again a net pressure force is always acting on it (as a result of the pressure gradient) and hence it should either accelerate or decelerate during the flow.This acceleration or deceleration is basically due to the net pressure force which is constantly doing work on the fluid element.According to the work energy principle of dynamics,the work done on the fluid element should result in an increase or decrease in the kinetic energy of the element.Thus, the kinetic energy of the fluid element seems to be a manifestation of the flow work being done on the fluid element.

Then my question is, why do we consider 'flow work' or 'flow energy' and kinetic energy of a fluid as separate quantities in the conservation of energy principle for fluid flow i.e, the Bernoulli equation?

The kinetic energy of the fluid seems to be a result of the work done by pressure forces in accordance with the work energy principle.For eliminating confusion one may also consider the flow of incompressible fluids whose specific volume is practically constant during the flow and hence a fluid element could be treated as a rigid body.

2. Jul 23, 2015

### Staff: Mentor

If the pressure causes the acceleration, how can the kinetic energy and "flow energy" be the same thing? Acceleration converts one form of energy into another (like dropping a rock converts potential energy into kinetic energy). Also, consider the non-accelerating, flow field: the kinetic energy is constant. The "flow energy" may or may not be constant depending on if we're considering losses, but it can be anything and isn't tied to the kinetic energy (they are not exchanging energy).

Consider the real-world example of a fan blowing air through a duct. The kinetic energy of an element is constant, but the "flow energy" drops along the duct as energy is lost to friction against the duct walls.

3. Jul 26, 2015

### Soumalya

If the kinetic energy of a fluid element is constant(considering a non-accelerating flow field) then the velocity of the element should also be constant which by Newton's second law of motion implies that there is no net force and hence no net pressure acting on the element.Then flow work or flow energy seems to be absent.

Let's say we have a body initially moving with a constant velocity 'v'.The kinetic energy at this stage is mv2/2.Now let's say we wish to increase it's kinetic energy.The only way it can be done is to increase its velocity.An increase in velocity can be accomplished over a certain time interval the rate of which is acceleration.Therefore, to increase the velocity of the body it must undergo some acceleration which again by Newton's second law of motion requires a force acting on the mass of the body.A force when acting over a distance in the direction of force is said to do work..Indirectly we arrive at the conclusion that work must be done on the body to accelerate it over some distance or time.

"Flow energy" is basically work done by pressure forces on a fluid element and this work must result in a change in kinetic energy of the element.What I mean to say is that change in kinetic energy of the fluid element is the result of work done by pressure forces.Thus, flow work is expended to increase kinetic energy of the fluid as the work- energy principle outlines, "we must expend equal amount of work to increase the energy of a mass by an equal amount"..

Then while considering energy conservation why do we take the work (flow work) and energy (kinetic energy) separately as flow work will quite definitely be converted into a change in kinetic energy or potential energy or both during flow?

Last edited: Jul 26, 2015
4. Jul 26, 2015

### Staff: Mentor

Correct.
Not necessarily correct. In real fluid flow, there are losses, so the flow energy decreases as the fluid travels through the duct/pipe. This can be observed by measuring the pressure in different locations in the duct/pipe.

And even if there are no losses, that can just mean the flow energy is also constant, not that it is absent (just like the kinetic energy).
Correct.
That isn't true. Flow energy can be stored, indefinitely, as pressure. It does not have to be expended. Consider an inflated balloon. The air inside is under pressure and has energy because of it, but it isn't moving.
Correct.

Then while considering energy conservation why do we take the work (flow work) and energy (kinetic energy) separately as flow work will quite definitely be converted into a change in kinetic energy or potential energy or both during flow?

5. Jul 27, 2015

### Soumalya

From the definition of flow work or flow energy it is equal to 'pv' where 'p' is the net pressure acting on a fluid element and 'v' is the volume of the element.
From an elementary dynamic analysis of a fluid particle motion if its moving with constant velocity then the net pressure acting on the particle must be zero i.e, p=0.

Then, pv =0

How is it a constant value then?

What is the correct definition, mathematical formulation and physical interpretation of flow work then?

I assume flow work and flow energies are similar terms implying the same thing.

6. Jul 27, 2015

### Staff: Mentor

I think you are confusing "pressure" and "net pressure" and therefore work and energy. A flow element that has energy has pressure. A flow element that expends work has a change in pressure(and energy): net pressure.

You do see that a stationary element of a fluid can still have a pressure, right?

And the other side of the coin: energy is what you have, work is what you do.

7. Jul 27, 2015

### Jano L.

I am not sure what the term "flow energy" is supposed to mean, but otherwise your description is correct; for incompressible fluid body moving without friction, work of pressure forces acting on it equals increase in its kinetic energy:

$$\Delta W = \Delta K.$$

The equality is numeric; of course, work is a different concept from energy.

When a fluid body of mass $m$ and volume $V$ moves from point having pressure $p_1$ to point having pressure $p_2$ (assumed constant in time), the work done on it by pressure forces is

$$(p_1 - p_2) V.$$

Let the corresponding speeds of fluid at those two points be $v_1$, $v_2$ (again, assumed constant in time). Then increase in kinetic energy can be expressed as

$$\Delta K = \frac{1}{2} m (v_2^2 - v_1^2).$$

Using the work-energy theorem, we obtain

$$p_1 + \frac{1}{2}\rho v_1^2 = p_2 + \frac{1}{2}\rho v_2^2$$

where $\rho = m/V$ is density of the fluid.

I am not sure what you mean by "we take the work and energy separately". The work-energy theorem states that work done on a body equals change in its energy.

8. Jul 27, 2015

### Staff: Mentor

I don't really like the terms "flow work" and "flow energy" (particularly, "flow energy") because they sound too much alike, which appears to be what is confusing you. I'm pretty sure the terms were not used in my fluids/thermo classes. But let's try to use them both together, mathematically, to show how they are different:

Flow energy is pressure (not net pressure) times volume: pv
Flow work is net pressure times volume: (p1-p2)v

Edit: oops, beaten to it.

9. Jul 27, 2015

### Soumalya

Exactly what I was contemplating on! So 'flow work' and 'flow energy' aren't the same thing right?
In the expression for flow work derived using elementary dynamics the 'p' in 'pv' is taken to be the net pressure acting on a fluid element.Thus it represents work done by pressure forces on a fluid element of volume 'v'.This cannot be the expression for energy associated with pressure for a fluid(which might be what is actually termed as 'flow energy' which is perhaps distinct from 'flow work') if 'p' is to represent the net pressure acting on the element.The source of confusion is what you have correctly pointed out.

Then could you lead me to the derivation of the expression for flow energy?
It should look like 'pv' where 'p' would represent the thermodynamic pressure for the fluid and 'v' volume of fluid considered.Every textbook I have gone through derives the expression for flow work and uses the same expression for flow energy replacing 'p' as the actual thermodynamic pressure instead of the net pressure acting on the fluid element as was assumed during the derivation.

10. Jul 27, 2015

### Staff: Mentor

So, here's an example where they are called the same thing:
http://www.ecourses.ou.edu/cgi-bin/ebook.cgi?topic=th&chap_sec=04.1&page=theory

I don't really like that description because what it glosses over is the fact that when the flow element moves to its next position in steady, lossless flow, that work expended does not get converted to kinetic energy. Indeed, it hasn't been converted to anything: it gets used again to move the flow element to the next position and the next position and the next position. I don't like calling something "work" when it hasn't changed/converted any energies....however, I can see how a physicist might prefer it because it is all-inclusive. But as an engineer, I try to ignore things that don't matter.

In some ways, this is a bit like an argument between absolute and gauge pressure. If the tire gauge you use to pump up your car tires read in absolute pressure, it would be more inclusive -- does that make it more correct? I don't care, but what I do care about is that makes it a pain in the a to use.

11. Jul 27, 2015

### Soumalya

You seem to understand exactly what I am trying to convey! In the expression of the Bernoulli equation p/ρ + V2/2 + gz = constant along a streamline which can also be written as pv + V2/2 + gz = constant along a streamline the term 'pv' is the flow energy or 'pressure energy' as stated in my thermodynamic and fluid mechanics textbooks respectively.In my thermodynamics textbook, the 'pv' has been derived as the expression for flow work where 'p' is taken as the net pressure acing on a fluid element that is a pressure difference.So it is basically work done by the net pressure force on unit mass of a fluid if 'v' represents specific volume of fluid.But it seems as Sir Russ pointed out the first term in the Bernoulli equation represents 'pressure energy' or 'flow energy' as in my textbook where 'p' is the thermodynamic pressure of the fluid and hence representing energy associated with pressure.I was confusing it with 'flow work' which is not the same as flow enrgy or pressure energy.

12. Jul 27, 2015

### Staff: Mentor

When I Google "flow energy", I get very little, implying it is not a widely use term. However, because they are related, the only difference/issue is that the link I quoted above is ignoring the fact that there is a pressure pushing on the element from both sides, not just one. So there isn't much else to derive.

But to see it in use, try this link, halfway through section 3:
http://ocw.mit.edu/ans7870/16/16.unified/thermoF03/chapter_6.htm

13. Jul 27, 2015

It sure seems to me that the flow work term is included in elementary energy analyses in the form of the Bernoulli equation through the $p$ terms in the equation. I feel like this was pretty well illustrated up there by @Jano L. where he showed the derivation of the Bernoulli equation directly from the work-energy theorem.

On the other hand, if the confusion here is arising in view of the situation in a real pipe where you need a pressure gradient to maintain a constant flow speed with no acceleration, this is because of the effects of viscous dissipation, where the energy added due to the work done by the pressure gradient is being balanced by the energy dissipated by viscosity. This situation is not captured by Bernoulli's equation without modification because it assumes zero viscosity.

14. Jul 27, 2015

### Soumalya

Exactly looks like something similar in my textbook.This description simply is in disagreement with the work energy principle as we know.How is it possible that work done on something doesn't change the energy content of it?In fact work is a form of energy interaction and it must result in a change in the total energy of the body it is expended on.According to this definition of 'flow work' the work done by a net pressure force maintains the flow without resulting in a change in kinetic energy i.e, a net force is acting on a fluid element and it is still moving with the same constant velocity - a violation of Newton's second law!

15. Jul 27, 2015

### Jano L.

Sometimes people call $pV$ "pressure energy", but that is misleading, because fluid in incompressible flow does not change its internal energy.
There is no additional energy of fluid associated with pressure when dealing with incompressible flow. The pressure term enters purely as work, not as fluid energy. The only energy of the fluid is kinetic energy; only if different heights are compared, potential energy enters as well.

16. Jul 27, 2015

### Soumalya

Very precise very meaningful! Unfortunate that even some of the highly rated texts out there carry controversial illustrations.

17. Jul 27, 2015

### Soumalya

Sir the confusion arised because the 'p' used in the 'pv' term in the Bernoulli equation should actually represent a pressure difference across a fluid element not absolute pressure at a point in fluid as it is basically expression for work done by the net pressure force on a fluid element.Some textbooks have termed it as 'flow energy' or 'pressure energy' which is technically incorrect as Sir Jano L. pointed out and in practice we substitute this pressure difference across a fluid element by the thermodynamic pressure at a point in flow.This messed up things when I was trying to relate the 'pv' term as some kind of energy associated with the fluid as is commonly misinterpreted since its basically work.

18. Jul 27, 2015

### Staff: Mentor

That's what I was referring to in post #12.

19. Jul 27, 2015

But it needn't be a pressure difference in that term. Bernoulli's equation essentially considers the flow in a control volume. Using an absolute pressure at two ends of it gives you all the pressure differential you need for the analysis. One of the drawbacks is therefore ignoring everything that happens between those two points you choose.

If you want something that is valid more universally and in a differential sense, you would have to move to the energy equation typically used in concert with the Navier-Stokes equations.

20. Jul 28, 2015

### Soumalya

Sir Boneh3ad even if we consider an energy balance for a control volume and try to derive the Bernoulli equation from the Steady Flow Energy Equation we have the term specific enthalpy h=u+pv on both sides the latter term on the right hand side of the equation being defined as the work done by pressure forces on a fluid parcel that is about to enter the C.V. by the fluid immediately upstream to push it within the C.V. or the work done on a fluid parcel that is about the leave the C.V. again by the fluid immediately upstream to push the parcel out of the C.V..

http://www.ecourses.ou.edu/cgi-bin/ebook.cgi?topic=th&chap_sec=04.1&page=theory

you could see the way the expression for flow work has been derived.Thus, it makes no sense if 'p' in the term 'pv' represents an absolute pressure because a fluid element considered in a flow always has a pressure differential acting on it i.e, pressure forces acting on it from all sides.So what they have done is to consider 'p' as the net pressure acting on the fluid parcel that does a work by an amount 'pv' to push a fluid parcel of volume 'v' across a section (the section maybe the inlet or exit of the C.V.). Now by that way the steady flow energy equation if reduced for an incompressible fluid would look like,

p1v+V12/2+gz1=p2v+V22/2+gz2

where p1 and p2 should represent the net pressure acting on a fluid parcel at the inlet and exit conditions for a C.V. (by the way the expression for flow work has been derived in the link above).Now, if the flow is perfectly horizontal the potential energy terms drop out and we are left with,

p1v+V12/2=p2v+V22/2

or, p1v-p2v=V22/2-V12/2

where p1v and p2v represents work done on a fluid a parcel of volume 'v' to push it into and out of the C.V. respectively.If p1=p2 (pressure gradient is constant throughout the flow field) we have, V12/2=V22/2 or, V1=V2 which implies the fluid parcel continues to move with the same velocity even if there is a net pressure, p1=p2=p acts on the parcel.This is in violation of the Newton's 2nd Law of motion as a body acted upon by a force must accelerate or decelerate.Then the definition of flow work as stated in the link cannot be correct.

Moreover, if p1v and p2v represent the work done on a fluid parcel at inlet and exit of a C.V. this work must result in a change in the total energy of the fluid parcel (work energy theorem) as it enters or leaves the C.V..Since we are considering only mechanical forms of energy in the Bernoulli equation this work done on the fluid parcel must be converted into kinetic and/or potential energy of the parcel which are already included in the equation.Thus the flow work done on a fluid parcel as it enters or leaves the C.V. cannot be stored like energy rather it must get converted into energy.My question is then why do we consider flow work as a separate term in the Bernoulli equation along with kinetic and potential energies?

The values of kinetic and potential energies for a fluid parcel as recorded when it enters or leaves a C.V. should already include the flow work done on it.

If 'p' is to represent the thermodynamic pressure at a point in flow then I am clueless as to how it could represent work done by pressure forces on a fluid element since a fluid element cannot have pressure acting on it from a single side there always lies a pressure differential across a fluid element considered in flow.

21. Jul 28, 2015

First, a fluid element does not always have a pressure differential on it. It may or may not depending on the flow. It very well may just have the same pressure acting on all sides. Second, there is nothing that would imply that using absolute pressure is invalid here. Let's take Bernoulli's equation as an example (ignoring gravity for convenience):
$$p_1 + \dfrac{1}{2}\rho v_1^2 = p_2 + \dfrac{1}{2}\rho v_2^2.$$
All of those are absolute pressures. The pressure differential comes from the fact that you have two of them in the equation that are not necessarily equal. The source you cited is incredibly inexact in what it is describing. The flow work is basically the work on a given system (in this case a control volume) done by the flow entering and exiting that system. This flow work is $pV$ and is based on the absolute pressure. The flow entering the system does work ($W_1 = -p_1 V_1$) on the system and the flow leaving the system has work done on it ($W_2 = p_2 V_2$), so the total work is $p_2V_2 - p_1 V_1$ and we are back where @Jano L. had us. These are absolute presssures.

If you want a source that is substantially better than the one you linked, try this one:
http://ocw.mit.edu/ans7870/16/16.unified/thermoF03/chapter_6.htm

So in light of the previous, it should be noted that $p$ is absolute pressure, not a pressure differential. In the illustrations, they show a large box because it is illustrative, but in reality, we are talking about the pressure of an infinitesimal fluid element moving past a point and the energy passing through that point as a result. That is the definition of enthalpy, after all. It is the energy carried along by the flow due to internal energy ($u$) and the specific flow work ($p\nu$) where $p$ is absolute pressure.

Therefore, this statement does not represent a constant pressure gradient, but instead a constant pressure, and this...

should end with "because the pressure is constant and there is nothing to accelerate the fluid".

I think this is the crux of the matter. The link provides a definition of work that is mathematically correct but confusingly explained.

That's just it. They don't. The represent the work done on the system due to the added energy of a fluid parcel (or done by the system to expel such a parcel).

So getting down to this, the best way that I have found to visualize the flow work when it comes down to trying to look at the Bernoulli equation is to view it sort of like a potential energy. There is gravitational potential energy that has its own term, and the pressure term is sort of akin to a spring potential energy that is stored up as pressure in the flow and can be converted into kinetic energy. This is an inexact analogy, but it is a decent illustrative example.

22. Jul 28, 2015

### Soumalya

Well that's right!

I was considering the net work done by pressure forces on a fluid parcel all the time.But it isn't the definition flow work as was clarified by the link you provided(even though I had looked at the link before but failed to study it with more insight until you pointed out the precise root of confusion).I was wondering if a fluid moves with constant velocity during a flow then the net force acting on it is always zero and thus the net work should also be zero.But flow work isn't the net work done on a fluid parcel by the fluid upstream and downstream.Rather it is the work required to push a fluid parcel of volume 'v' and pressure 'p' across the boundary of a control volume.So one needs to consider only the work done by the fluid immediately upstream of the fluid parcel to push it into the C.V. and not the work done on the parcel by the fluid downstream in addition(because the downstream fluid is within the C.V. and we need to know about the work that is delivered across the boundary of the C.V.).So we can imagine the fluid upstream as a imaginary piston exerting a force 'F' to the fluid parcel which has a pressure 'p' and thus in absence of any acceleration the force exerted by the piston on the fluid parcel which is 'F' should be equal to the force exerted by the fluid parcel on the piston which is 'pA' where 'A' is the cross sectional area of the imaginary piston.Thus the work done by this imaginary piston to push the parcel within the C.V. boundary is equal to 'FL' where 'L' is the distance through which the piston pushes the parcel before it enters completely within the C.V..This is flow work by definition i.e,

Wflow=FL=pAL=pV

A similar situation can be imagined for the flow work associated with pushing the parcel out of the C.V. for which the expression remains the same but this time work is delivered by the C.V.

I hope I got it right this time!

Just to re-emphasize, net work done on a parcel by pressure forces acting on it is still zero for a flow of constant velocity but the work done by the upstream fluid to push the parcel within the C.V. is not zero which is flow work by definition.

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Last edited: Jul 28, 2015