Talk:Bernoulli's principle/Archive 3

Archive 2 has been created with a link at above right. It is an exact copy of the talk page as it was before this edit. Archive 3, when needed in the future, should be a new subpage (same as creating an article) titled "Talk:Bernoulli's principle/Archive 3" and the link added to the template on this page's code. For further information on archiving see Wikipedia:How to archive a talk page. See also User:5Q5 for the used archiving procedure. Thank you. Crowsnest (talk) 19:46, 30 March 2008 (UTC)

Classic Bernoulli equation

1/2 m v^2 + p + m g y = energy = constant

Replace m g y, the work done by gravity, with w to represent work done by a wing.

1/2 m v^2 + p + w = energy

Restate with initial, work done, and final states:

1/2 m v0^2 + p0 + w = 1/2 m v1^2 + p1 = 1/2 m (v0+vc)^2 + p0

vc = +/- sqrt(2 w / m)

Where m = mass, v0 = initial velocity, p0 = initial pressure (ambient), w = work done, v1 = work done velocity, p1 = work done pressure, and vc = change in velocity of affected air when it's pressure transitions back to p0, it's initial pressure (ambient). vc is negative only when |drag| > |lift| and wing is the frame of reference.

During the interaction when work is done, a non-Bernoulli like change in total energy occurs, an increase in velocity^2 without an equal decrease in pressure (typical wing) or vice versa (typical propeller). Jeffareid (talk) 00:20, 16 September 2008 (UTC)

The form 1/2 v² + Ψ + p/ρ = constant (along a streamline) is a more general statement of Bernoulli's equation for incompressible flow, see e.g. Feynman, Leyton & Sands (1964) Lectures on physics, vol. 2, p. 40-6. Here Ψ is not work, but a force potential of a conservative force field, like e.g. gravity. Mechanical work, according to the work-energy theorem, equals the change in kinetic energy. So if you want to call something work (on the air, not the wing) it is the change in –( ρ Ψ + p ), going from position 0 to position 1. The conclusions regarding drag and lift on the wing are incorrect: for instance for the flow field -- calculated with potential flow theory and the Kutta condition -- around a 2D airfoil, the drag will always be zero (due to the d'Alembert paradox). Loosely speaking: drag is produced in regions of the 2D flow where the Bernoulli equation is not valid (boundary layers and wake).-- Crowsnest (talk) 07:51, 16 September 2008 (UTC)

I was trying to generalize this to a more generic, real world 3d environment. If this doesn't work, at least I tried to come up with a explanation that power consumed during flight doesn't just dissappear into thin air, it changes the energy of the air. Jeffareid (talk) 09:44, 16 September 2008 (UTC)

Sure the energy has to stay somewhere, e.g. heat, sound, turbulence, infra-red radiation. But most often you cannot use Bernoulli's principle to calculate the losses of mechanical energy. Bernoulli's principle is a very handy and powerfull tool, but only within its domain of validity, as so many other tools ;-) Crowsnest (talk) 17:38, 16 September 2008 (UTC)

Note that in a closed system, such as a model flying in a sealed container, that the downforce applied by the wing to the air eventually ends up as a downforce applied by the air to the container, from downwash momentum and/or pressure differential within the container. The earth is also a closed system; eventually the downforce from the wing is exerted by the air onto the surface of the earth. The point here is that work peformed on the air will result in a change in pressure and/or velocity of the air or in velocity after the affected air's pressure returns to ambient. Ignoring temperature changes, and including the speed of turbulent air as part of the air's total kinetic energy, then why would the proposed modified equation not reasonably predict vc - the change in velocity of air after it's pressure returns to ambient? Change what I called "vc" to Ve, and it's essentially the same as what is described here propeller analysis. Jeffareid (talk) 03:33, 17 September 2008 (UTC)

You do not have enough information. In your proposal (and looking at the NASA web site), you have one (Bernoulli) equation and two unknowns: the change in velocity from V₀ tot V_e and the pressure drop over the propeller or rotor disc (or thrust). You either need to know V_e, as on the NASA page, or you need to know the thrust to be able to estimate one from the other. -- Crowsnest (talk) 06:43, 17 September 2008 (UTC)

"w" was stated as the other known value in the example, work performed on the air. The normal relationship between work and a change in velocity is w = 1/2 m vc^2. Why should it be different in this case? Jeffareid (talk) 10:03, 17 September 2008 (UTC)

It is not different, you only made a mistake in its calculation. Work by the propeller or rotor on a small fluid parcel of volume ${\displaystyle \Omega }$  first at the far-field front position with velocity ${\displaystyle v_{0}}$  and finally at the "exit" position with velocity ${\displaystyle v_{e}}$  is: ${\displaystyle {\scriptstyle {\frac {1}{2}}}\rho (v_{e}^{2}-v_{0}^{2})\Omega }$ , which is independent of Bernoulli's principle. Still containing the unknown exit velocity ${\displaystyle v_{e}}$ , which you have to determine somehow. Further, I do not think your ideas on this are going to change the article, since it is original research.

In pipe and open channel (e.g. river) flows, the change (reduction) in the constant of the Bernoulli equation ${\displaystyle {\scriptstyle {\frac {1}{2}}}v^{2}+\Psi +p/\rho ={\text{constant}}}$  - due to irreversible work in a flow expansion - is often expressed in the form of a Borda-Carnot hydraulic head loss of the form: ${\displaystyle {\scriptstyle {\frac {1}{2}}}\mu (v_{e}^{2}-v_{0}^{2})}$ . With ${\displaystyle \mu }$  an empirical loss coefficient (less than or equal to one). The propeller equation on the (before mentioned) NASA site looks like this, in its equation for the thrust (with ${\displaystyle \mu =1}$ ). -- Crowsnest (talk) 23:58, 17 September 2008 (UTC)

OK, was just wondering if there was a way to modify Bernoulli to include work (other than gravity) performed on the air, so it would more accurately decribe air flows in a non idealistic situation. Jeffareid (talk) 03:28, 18 September 2008 (UTC)

I do not know of such an extension for propellers and rotors. Perhaps somebody else does. -- Crowsnest (talk) 08:24, 18 September 2008 (UTC)

I meant wings as well. Jeffareid (talk) 10:05, 22 September 2008 (UTC)

Such modifications have been made utilising empirical methods. Doing a quick search: sorta like this one or This one (better). I don't think you will find a neat analytical solution, because of fun issues like cavitation for fluids & complex boundary conditions in the boundary & log layers. So outside the scope of the article. User A1 (talk) 13:21, 18 September 2008 (UTC)

Thanks, but I was hoping for a modified version of Bernoulli equation for wings in a non-idealized case, one that would include net work done on the air. Similar to a propeller, this work occurs within the pressure jump zone for a wing. Jeffareid (talk) 21:46, 18 September 2008 (UTC)

It doesn't seem there will be anymore activity on this section, so it might as well be archived. Jeffareid (talk) 09:55, 22 September 2008 (UTC)

Air brush - the ones with the paint cup below the air flow.

Flit guns - A bit of nostalgia for baby boomers. A hand pump sprayer used to disperse insecticide. Scroll down to the hand operated spayer sections here equipment and techniques Jeffareid (talk) 10:03, 22 September 2008 (UTC)

Isn't it often used to explain the curling of a football (soccer) ball? —Preceding unsigned comment added by 85.211.99.52 (talk) 16:41, 5 February 2009 (UTC) http://en.wikipedia.org/wiki/Curl_(football) —Preceding unsigned comment added by 85.211.99.52 (talk) 16:44, 5 February 2009 (UTC)

Yes. See Magnus effect in sport. This effect is employed in many ball sports. Dolphin51 (talk) 21:56, 5 February 2009 (UTC)

Mikael Häggström introduced the term "Bernoulli pressure" in this edit, stating it is equal to the constant in the Bernoulli equation for incompressible flow (written in "pressure form"). However, I can not find references for this. I do find references which call minus the dynamic pressure, –½ρv², the Bernoulli pressure. If references cannot be provided, this has to go. -- Crowsnest (talk) 15:10, 12 October 2008 (UTC)

Given the following assumption is it reasonable to consider the air moving across an airfoil a "streamline". If a fluid is flowing horizontally and along a section of a streamline, where the speed increases it can only be because the fluid on that section has moved from a region of higher pressure to a region of lower pressure; I'm trying to reconcile this with a kinetic theory understanding of gas. Viscosity effects seem to make more sense than pressure effects.--OMCV (talk) 05:37, 13 December 2008 (UTC)

Yes, it is reasonable to consider a two-dimensional pattern of streamlines around an airfoil. Streamlines are a powerful and accurate tool for visualising steady fluid flows; and the flow around an airfoil at an angle of attack less than the stalling angle can be assumed to be steady flow.

The statement "where the speed increases it can only be because the fluid has moved from a region of higher pressure to a region of lower pressure" is simply an application of Newton's Second Law. If a solid body accelerates from a low speed to a high speed it can only be because a force has acted on that body for the duration of the period of acceleration. It is no different for a fluid - if a small parcel of fluid accelerates it can only be because a force has acted on that parcel of fluid (and if the fluid is flowing horizontally, gravity forces can be ignored). The force responsible for the acceleration is therefore pressure. The parcel of fluid has moved from a region of higher pressure to one of lower pressure, and that is the force responsible for the acceleration. Dolphin51 (talk) 08:59, 13 December 2008 (UTC)

So I still don't get it completely. When I imagine an airfoil I see a bump at the front of the upper surface which would certainly experience more pressure perpendicular to its curve, I assume this region of high pressure is what induces a low pressure region farther down the "streamline" past the bumps high point. It seems reasonable that these two forces would counteract each other. Further more you say. If a solid body accelerates from a low speed to a high speed it can only be because a force has acted on that body for the duration of the period of acceleration. I agree, but I'm not convinced that force is pressure why can't it be viscosity. The air on the top of the airfoil is actually slowed in a wind tunnel or dragged forward in actual flight. Only in the airfoil's frame of reference does the fluid above the airfoil "speed up". That is in a theoretical fluid without viscosity would the Bernoulli's principle fail. So I don't think thats true but I'm still not sure that "pressure" is the best force to consider when explaining an airfoil. Eager to hear what anyone has to say.--OMCV (talk) 15:26, 13 December 2008 (UTC)

No. The air "on the top of the airfoil" is moving faster than the air "in front" of the airfoil. As an airfoil approaches supersonic speed, the air "on the top of the airfoil" reaches sonic velocity before the air vehicle does. 24.96.135.4 (talk) 23:13, 13 December 2008 (UTC)

The force responsible for the acceleration is therefore pressure. or mechanical interaction with a solid (or the boundary layer near the solid). Form drag involves work being done on the air, the total energy of the air is changed, not just exchanged between pressure and kinetic energy. I don't see this aspect being addressed. Jeffareid (talk) 09:10, 30 January 2009 (UTC)

The reason the air flows faster over the top surface of an airfoil when the airfoil is generating lift is perhaps one of the most non-intuitive concepts in aerodynamics. There are various popular approaches to explaining why this happens. (One of the most popular is the "equal-transit time theory". It is very simple and easy to comprehend, but it is untrue.) The approach I find the best is to comprehend the Kutta condition. The Kutta condition can be seen in operation in various applications other than airfoils - for example, the small perspex wind shield on a motor cycle protects the rider's face from the worst of the oncoming wind, even though the top of the wind shield is below the level of the rider's face.

Trying to grasp the basics of airfoils by contemplating viscosity is unnecessary, and probably will make the task more difficult in the long term. There is a fundamental theorem of fluid dynamics called "Prandtl's hypothesis" (unfortunately no Wikipedia article yet) which says that, in external flows like the flow of the atmosphere around the exterior of a body like an airfoil, the influence of viscosity is zero almost everywhere except close to the boundary of the fluid - that is zero almost everywhere except close to the surface of the body. The pressures experienced by each spot on the surface of the body are the same as the pressure in the nearby free-stream - viscosity causes a boundary layer but this boundary layer has only a very small effect on the pressure distribution around the body. When contemplating the lift on an airfoil, Prandtl's hypothesis gives us confidence to ignore the effect of viscosity for all but the most complex and advanced problems. So forget about viscosity for a while, and focus on the Kutta condition. Dolphin51 (talk) 00:08, 14 December 2008 (UTC)

If a fluid is flowing horizontally and along a section of a streamline, where the speed increases it can only be because the fluid on that section has moved from a region of higher pressure to a region of lower pressure;. Air flow changes also occur because of mechanical interaction via molecular collisions (or lack there of from the void introduced by a solid) between fluid and a solid. This how work is peformed on the air when generating lift and/or form drag. It's why the pressure and heat from a space vehicle re-entry is so high, the leading surfaces of the vehicle collides with the air molecules, speeding them up, resulting in a great deal of pressure and heat. In addition for a real world wing, there's a lot of turbulence in the flow around a wing. Unlike slow speed wind tunnel experiements, the streamlines transition to turbulent flow once the pressure starts increasing and the flow starts decelerating. Kutta condition - a thin trailing edge is not required to generate lift. These lifting body, flat top, curved bottom, airfoils do not have thin trailing edges but glide or fly just fine. M2-F2_glider.jpg M2-F3_rocket_powered.jpg. Jeffareid (talk) 23:15, 23 January 2009 (UTC)

• The reason the air flows faster over the top surface of an airfoil when the airfoil is generating lift is perhaps one of the most non-intuitive concepts in aerodynamics. Perhaps faster over the top isn't intuititve, but the low pressure zone above a wing is easily explained: In that case a low pressure region is generated on the upper surface of the wing which draws the air above the wing downwards towards what would otherwise be a void after the wing had passed. Wing Jeffareid (talk) 10:10, 10 June 2009 (UTC)

The article now starts with forms of Bernoulli's equation including the force potential. While correct and more general, this makes the article less accessible: this is not a simple form, see Wikipedia:Make technical articles accessible. Any suggestions for improvement? -- Crowsnest (talk) 07:40, 22 May 2009 (UTC)

I agree that technical articles must be accessible to a general audience, so they should begin with a suitably general description. I am not familiar with the expression force potential, as applicable to Bernoulli's equation. What do you mean by that expression? Dolphin51 (talk) 11:38, 22 May 2009 (UTC)

A conservative force can be described as the gradient of a potential. In this case, for the Bernoulli equation, it is assumed that the body force per unit volume in the Euler equations can be expressed as ${\displaystyle {\vec {F}}=-\rho \,\nabla \Psi .}$  For example, in case of Earth's gravity: the potential is ${\displaystyle \Psi =g\,z}$  and the gravitational force per unit of volume is ${\displaystyle {\vec {F}}=-\rho g{\vec {e}}_{z},}$  with ${\displaystyle {\vec {e}}_{z}}$  the unit vector in the upward vertical direction. The potential then appears in the Bernoulli equation: for instance in an incompressible flow ${\displaystyle {\tfrac {1}{2}}v^{2}+\Psi +p/\rho =}$ constant. These forms of Bernoulli's equation are more general and can be applied in astrophysics and magnetohydrodynamics. See also Batchelor, Introduction to fluid dynamics, §3.5. -- Crowsnest (talk) 16:11, 22 May 2009 (UTC)

The relative air flow parallel to the top surface of an aircraft wing or helicopter rotor blade is faster than along the bottom surface. ...

This statement doesn't clarify that relative means relative to the wing. Also the flow isn't only parallel, there is significant air flow perpendicular to the surfaces of an air foil, a dynamic situation where the wing movement with an angle of attact involves a component of movement perpendicular to the surfaces (air flows towards the receding surfaces). If the only the air flow parallel to surfaces is considered, then relative to the air, the flow above a wing is slower than the flow below.

In the case of air foils such as the M2 series m2f2.jpg, in the immediate vicinity of the "wing", relative to the air, the vertical (downwards) air flow is greater below than above because of the inverted wedge design (since it was a re-entry prototype, a higher drag component was wanted).

In the case of a rotor or prop, relative to the air, the air flow speed just above and below the rotor and prop is about the same. The main change across the "disk" is a jump in pressure (which violates Bernoulli), and due to the aft flow pressure being higher than ambient, the flow continues to accelerate as it's pressure returns to ambient. The fastest flow is well aft of a prop or rotor. NASA_prop_anl.htm

The statement as is could also imply that wings are designed to speed up parallel airflow relative to the wing rather than to deflect that air flow downwards, which could be misleading.

I would prefer something along the lines of the NASA article. Even in the case of wings, Bernoulli is violated in the imediate vicinity of the wing because of mechanical interaction, but does apply to the flows away from the wing where no work is done.

Besides, Newton's third law states that forces only exist in pairs, so the air's upwards force on the wing coexists with the wing's downward force on the air, which results in a downward acceleration of air.

Who added this statement? A link to wing would seem more appropriate than this single statement.

Jeffareid (talk) 22:55, 12 June 2009 (UTC)

I may have been the author of the sentence in its original form. In May 2008 the first sentence said:
The air flowing past the top of the wing of an airplane, or the rotor blades of a helicopter, is moving very much faster than the air flowing past the under-side of the wing or rotor blade.
Notice the word relative is not present. I don't know who added relative. It appears to be redundant. relative air flow has no particular meaning.

It would be reasonable to write about The air flowing relative to the wing of an airplane but that doesn't distinguish between air flowing past the top of the wing, and air flowing past the bottom of the wing. The purpose of the paragraph is to focus on air flowing past the top surface because that is where the air is accelerated to very high speed compared with the speed of the wing through the air. Air flowing past the bottom surface is not similarly accelerated.

I suggest the word relative should be deleted. Dolphin51 (talk) 08:48, 13 June 2009 (UTC)

The other issue is the word parallel. If you remove relative then the air itself could be used as a frame of reference, and in this case, the air flow parallel to the surfaces is faster at the bottom than at the top (relative to the air, most of the air flow is downwards). Also parallel flows could be mis-interpreted to be horizontal flows, which are related to drag and not lift. Jeffareid (talk) 04:56, 14 June 2009 (UTC)

I'm not sure what you mean exactly, but I think I disagree. It is true that we could select an inertial reference frame attached to the atmosphere. In this frame, the wing is traveling at speed but the general atmosphere is stationary. The air adjacent to the upper surface of the wing has a high speed but the air adjacent to the lower surface has a lower speed.

Perhaps you mean selecting the inertial reference frame in which the air flow past the upper surface of the wing is stationary. The air flow past the lower surface is moving in the same direction as the wing, not in the opposite direction. I agree that such a reference frame could be chosen, but what advantage does it offer? I can see no advantage.

When considering an airfoil moving through a fluid, generating lift and drag, there are only two inertial reference frames worth considering - the frame in which the airfoil is stationary, and the frame in which the freestream is stationary. The former has the advantage that the drag force is stationary, it does no work on the air and so does not cause a change in the internal energy of the air. In the latter case, stagnation pressure is equal to atmospheric pressure; but the pressure against the leading edge of the airfoil is higher than stagnation pressure and this appears to contradict Bernoulli's equation. The reason for the apparent contradiction is that the increase in pressure from stagnation pressure up to the pressure against the airfoil leading edge is due to the work done on the air by the airfoil. (The force exerted on the air by the airfoil is the force that is equal and opposite to the drag force exerted on the airfoil by the air.) Dolphin51 (talk) 05:38, 14 June 2009 (UTC)

In this frame, the wing is traveling at speed but the general atmosphere is stationary. The air adjacent to the upper surface of the wing has a high speed but the air adjacent to the lower surface has a lower speed. The parallel component of air speed adjacent to the upper surface is moving slower than the parallel component of air speed adjacent to the lower surface. The total air speed (including flow perpendicular and parallel to surfaces) would be faster above on most (but not all) wings.

The reason for the apparent contradiction is that the increase in pressure from stagnation pressure up to the pressure against the airfoil leading edge is due to the work done on the air by the airfoil. There is also a decrease in pressure on the trailing surfaces because work is also done by those aft surfaces. In this case because of what would otherwise be a void introduced by the receding surfaces, air is drawn in the direction of those receding surfaces via a reduction in pressure.

The force exerted on the air by the airfoil is the force that is equal and opposite to the drag force exerted on the airfoil by the air. This issue applies to both lift and drag. Work is done in the direction of lift as well, also violating Bernoulli. Mechanical interactions between a solid and fluid or gas normally don't comply with Bernoulli because work is done in those interactions. Away from that interface, where no work is done, then Bernoulli applies. The kinetic energy component of work done on the air is the same if the air is pushed or pulled. The pressure is increased when pushing and decreased when pulling. Since the total amount of mechanical work done includes both kinetic energy and pressure changes, and since the pressure is reduced by pulling, it's more efficient to pull air than it is to push it. Jeffareid (talk) 08:10, 14 June 2009 (UTC)

Jeffareid has written:

The parallel component of air speed adjacent to the upper surface is moving slower than the parallel component of air speed adjacent to the lower surface.
Parallel to what? Parallel to the vector representing the relative velocity of airfoil and the freestream; or parallel to the surface of the airfoil?

work is also done by those aft surfaces. Work is not done by surfaces. Work is done by forces. For example, the drag force on an airfoil does work when the airfoil undergoes a displacement.

Work is done in the direction of lift as well. Work is not done in any particular direction. Work is a scalar - it has magnitude but not direction. If the chosen reference frame is the one in which the freestream has zero velocity then lift does no work because it is perpendicular to the airfoil's velocity vector. In all other reference frames lift does work on the airfoil. For example, if the chosen reference frame is the one attached to the ground below an aircraft, when the aircraft is turning the horizontal component of lift does work on the aircraft causing its kinetic energy to increase when turning downwind, and to decrease when turning upwind. Dolphin51 (talk) 11:18, 14 June 2009 (UTC)

Parallel to what? to the surfaces as written in the main article. Using the air as a frame of reference (observer in a hot air balloon watching an aircraft pass by) the parallel to surface component of the speed of the air (relative to the observer) above a wing is slower than the speed below the wing.

Work is not done by surfaces. Work is done by forces. Pressure from the air exerts a force onto the surfaces of the wing, coexistant with those surfaces exerting equal and opposing forces to the air. Again using the air as a frame of reference, when those surfaces move with respect to the air, work is done (force times distance). Generally the lower surface is pushing the air and the force at the lower surface is increased, while the upper surface is pulling the air and the force at the upper surface is decreased. The change in kinetic energy of the air can be tied to the speed of the air just after the wing passes by. The total energy change would include pressure change as well. The exit velocities of the affected air are the velocities of affected air at the time and place the affected air's pressure returns to ambient. The integral average of these exit velocities also represents the change in total energy (pressure isn't a factor because it's returned to ambient).

Work is not done in any particular direction. Agreed, my point was that the production of lift as well as drag both consume energy and involve work being peformed on the air. The generation of both lift and drag conflict with Bernoulli at the point of application of forces between the surfaces of a wing and the air because work is done. Although lift is perpendicular to the direction of travel of the wing, it's not perpendicular to the surfaces of a wing, so lift involves work (change in kinetic energy ?= lift forces times the vertical (lift) components of distance of wing surfaces? Total work done would need to include pressure changes.). Jeffareid (talk) 11:56, 14 June 2009 (UTC)

I don't see anything here that I agree with. Dolphin51 (talk) 23:00, 14 June 2009 (UTC)

Let's start with the first one. Relative to the freestream air, most of the acceleration of the air affected by a wing is downwards, not forwards since lift is usually much greater than drag. If only the forward (approximately the same as parallel to wing surface) speed is considered, than the forward component of speed (realtive to the free stream) of air above the wing is less than the air below the wing. At some points along the upper surface of a wing, the air may experience a backwards component of speed. The diagrams on this web site might help explain my point: avweb_lift.htm. In the animated diagram, the air above is horizontally displaced less than the air below, so the magnitude of backwards speed above is less than the magnitude of forward speed below.

For the work done by surfaces, an anlogy might be a piston in an open ended cylinder. If the piston moves in the cylinder, it's surfaces perform work on the air via mechanical interaction, pushing the air (increasing pressure) on one side of the piston, and pulling the air (reducing pressure) on the other side of the piston, violating Bernoulli because work is done. The result of the work done is a change in the kinetic energy and pressure of the affected air.

It's my understanding that the Navier-Stokes_equations do an adequate job of modeling the interaction between solid and fluids or gases, but Bernoulli equations aren't as good because they don't take into account the work done by the solid to the fluid or gas, and only apply away from the solid+fluid mechanical interface, where no work is done (turbulence is still an issue). Jeffareid (talk) 00:40, 15 June 2009 (UTC)

Hi Jeff. I have visited the Avweb site you supplied. It is one of many sources aimed at student pilots and newcomers to aviation who are looking for an intuitive approach to explaining the phenomenon of fluid-dynamic lift. Roger Long talks about thought-experiments involving mashed potatoes, and he encourages readers to disregard Bernoulli's principle: The next thing you need to forget is Bernoulli. Roger favours the Newtonian explanation of lift, and deprecates the Bernoullian explanation. Long is not a good source of information for a Wikipedia article on Bernoulli's principle.

I have watched the animated images on the Avweb site. I don't follow your explanations, and I don't see that the images support your claim that If the only the air flow parallel to surfaces is considered, then relative to the air, the flow above a wing is slower than the flow below. Even if there is some sense to be made out of this claim, I don't see that it is relevant to a simple, clear, scientific explanation of Bernoulli's principle.

This thread doesn't appear to me to have the potential to contribute to the quality of the main article. Dolphin51 (talk) 03:41, 16 June 2009 (UTC)

Regardless of my explanations, there's still a problem with this statement: The relative air flow parallel to the top surface of an aircraft wing or helicopter rotor blade is faster than along the bottom surface. ... , and removing relative isn't going to fix it. If the wing is the frame of reference, then the parallel speed above is faster than below. If the air is the frame of reference the parallel speed above is slower than below. Removing the parallel to surfaces phrase would elminate the apparent paradox caused by the choice of frame of reference, since the change in speed would include the components perpendicular to the surfaces, which is the primary direction of flow when using the air as a frame of refernce, in the normal case where lift and perpendicular accelerations and flows are much greater than drag and parallel accelerations and flows.

This still doesn't directly predict a relationship between pressure and speed, because how the change in speed and pressure occurred isn't being explained in the main article. Without knowing the cause for the change in speed, then faster moving air doesn't imply lower pressure. The exhaust from a rotor, propeller, or turbine is both higher speed and higher pressure because work is peformed and energy increased. Bernoulli principle relies on a conversion between speed and pressure where no net mechanical work is done. Bernoulli can be an approximation when the amount of work done is small, such as an efficient wing. This should be explained when using a wing as pratical example of Bernoulli principle. Jeffareid (talk) 06:12, 16 June 2009 (UTC)

If the air is the frame of reference the parallel speed above is slower than below. - my quote. As evidence of this consider the flows relative to air and the direction of travel with respect to the air: the perpendicular flows are related to lift and the parallel flows are related to drag. Clearly the parallel flow below a wing is forward, contributing to drag. Relatve to the wing, the flow above is faster backwards than below, so relative to the air, the flow speed above a wing is less forwards than the flow below, and many diagrams show this flow to be backwards. If the magnitude of the backwards flow speed above a wing were greater than the magntiude of the forwards speed below, then you'd have a conflict as the wing would be generating thrust as opposed to drag. Perhaps this drag analogy can help explain why I have issues with The relative air flow parallel to the top surface of an aircraft wing or helicopter rotor blade is faster than along the bottom surface. ... if the reader assumes the freestream air as the frame of reference. For me the problem with this statment is the term parallel Jeffareid (talk) 03:14, 17 June 2009 (UTC)

• I see that both relative and parallel were removed from the statement. Please note I did not make these changes (appears to be some one named A1 or Dolphin51, I don't understand how to read the history properly). The last part that starts with Besides, Newton ... could be better worded, but it's understandable as is. Sorry for all the nit-picky fuss. Jeffareid (talk) 10:51, 17 June 2009 (UTC)

I removed the words in two consecutive edits - one on 16 June and the other on 17 June. Dolphin51 (talk) 11:35, 17 June 2009 (UTC)

The principle also makes it possible for sail-powered craft to travel faster than the wind that propels them (if friction can be sufficiently reduced). If the wind passing in front of the sail is fast enough to experience a significant reduction in pressure, the sail is pulled forward, in addition to being pushed from behind.

The main reason sailcraft can move faster than the wind is because the "apparent crosswind", the component of wind perpendicular to the sailcrafts direction of travel, is independent of the sailcrafts speed, and equal to true wind x sin(angle between true wind and sail craft direction). A glider in a updraft is an even better example. Instead of a force from the ground opposing the force from deceleration of the cross wind, gravity opposes the force from deceleration of the updraft. High end gliders can achive 60 to 1 glide ratios, which translates into 60 mph of forward speed in level flight with 1 mph of updraft. For the Nimbus 4 (Schempp-Hirth_Nimbus-4) it's more like 68 mph (59 knots) versus 1.133 mph updraft. Any vehicle with a glide ratio greater than 1 can move "faster than the wind". I don't see sailcraft as a special case other than "faster than the wind" sailcraft require a low drag factor from the ground interface.

Jeffareid (talk) 23:26, 12 June 2009 (UTC)

Agree that the reason some sail craft can go faster than the wind is because the crosswind persists even when the craft is going at the same speed as the wind. As long as there's a wind coming in from the side, there will be a force driving the craft forward - the craft's speed is limited by the friction forces, not by the wind speed.

What's unclear to me is what this has to do with the Bernoulli principle. I'd vote to remove this paragraph. I guess I could do it myself, but I'd like to see some consensus before doing so. Mr swordfish (talk) 15:46, 23 June 2009 (UTC)

Bernoulli principle is at least partially involved in any interaction between a solid and fluid or gas. What isn't clear is if the tendency of air to follow Bernoulli's principle in most situations is required for the lift to drag ratio to be high enough to allow a sail craft move faster than the true wind.

Agree that the reason some sail craft can go faster than the wind is because the crosswind persists even when the craft is going at the same speed as the wind. As long as there's a wind coming in from the side, there will be a force driving the craft forward - the craft's speed is limited by the friction forces, not by the wind speed. There are two requirements for a sailcraft to operate at faster than true wind speed: 1. The effective lift to drag ratio must be > 1 to allow the apparent headwind to exceed the apparent crosswind. 2. The upwind component of diverted apparent wind relative to the sailcraft must be faster than the downwind component of the sailcrafts velocity in order to slow down the true wind, since slowing down the true wind is the energy source for any wind powered device. Note that an iceboats component of speed in the direction of the true wind can exceed the true wind speed (even upwind), it's faster to tach downwind than to simply go with the wind. A sailcraft takes advantage of the speed differential between the true wind and the ground, which is independent of the sailcraft's speed. A glider can't do this (glide upwards faster than the updraft) because there is no speed differential between an updraft and gravity. A glider can take advantage of differences in wind speed separated by a thin shear boundary. This is called dynamic soaring dynamic_soaring_333mph.htm. Jeffareid (talk) 12:05, 25 June 2009 (UTC)

Take the classic case where a small diameter pipe transitions into a larger diameter pipe. Why doesn't the fast moving fluid form a virtual pipe and continue flowing along at its fast pace without interacting with the perhaps non-moving fluid beyond the "streamline" of that fast moving fluid? The pressure of that fluid is lower, so the surrounding fluid would be drawn towards the fast moving streamline, but with no viscosity how would the streamlines interact with each other? Jeffareid (talk) 01:18, 13 June 2009 (UTC)

What you describe is a valid solution for inviscid flow. It's just not *the* solution that you expect. The flow along the constant (smaller) diameter tube all has constant velocity. The fluid in the outer annulus downstream of the enlargement is all at rest. The static pressure is the same everywhere in the flow. There is infinite shear at the interface (just as there is at the interface between the original flow and the wall of the smaller diameter pipe). Although it is a valid solution for an inviscid fluid, I suppose that strictly, it could not be called irrotational, but it would be irrotational everywhere except at the infinitesimal boundary between the two regions of fluid.

There are infinitely many valid flows inside that geometry. But the *usual* (or expected) solution is the one that we normally think of in which the fluid slows down smoothly to a slower constant velocity downstream of the expansion. As long as the Bernoulli constant is constant along all streamlines, all is well. In the unusual solution you proposed, the Bernoulli constant (i.e. the stagnation pressure) is different in the two portions of the fluid.

What would a "real" inviscid fluid do? It would simply continue doing whatever you set the initial conditions to be. If you started it off with the flow going fast in the smaller section and slower in the larger section of tube, it would merrily continue doing that forever. If you start it off moving fast down the center and stagnant around the periphery, it would merrily continue doing *that* forever. (Of course, I'm neglecting stability analyses at the shear interface.)

Fun little puzzle. Kimaaron (talk) 03:55, 9 September 2009 (UTC)

Is the existance of air's conformance to Bernoulli principle a major requirement for lift, or does it just make it more efficient? Does the wing+air interface really conform to Bernoulli, or does Bernoulli only approximate the wing+air interface because only a small amount of (Bernoulli violating) work is done by the interaction? Most agree that the air to air interaction away from the wing+air interface conforms to Bernoulli, but I don't find any consistency in articles describing the wing+air interface. I don't see any references that claim wing+air interaces are fully compliant with Bernoulli (as opposed to being an approximation).

What is the significant difference between the rotor of a helicopter in a hover versus a prop? I'm referring to the NASA article again: Downstream of the disk the pressure eventually returns to free stream conditions. But at the exit, the velocity is greater than free stream because the propeller does work on the airflow. We can apply Bernoulli'sequation to the air in front of the propeller and to the air behind the propeller. But we cannot apply Bernoulli's equation across the propeller disk because the work performed by the engine (by the propeller) violates an assumption used to derive the equation. propanal.htm.

Other than the magnitude of the pressure differentials, the amount of induced wash and the smaller amount of work, what is the significant difference bewteen a wing, a rotor, or a propeller, such that a wing is fully compliant with Bernoulli, but not a propeller or a rotor of a hovering helicopter?

The point of all of this is I think that something should be added to explain that Bernoulli principle allows for efficient production of lift because acclerating air via a reduction in pressure reduces the total work done on that air. It might also be noted that using a longer wing to achieve the same amount of lift by diverting more air at a lower rate of acceleration also reduces the work done on the air (twice the air at 1/2 the acceleration equals same lift, but half the kinetic energy). Jeffareid (talk) 12:35, 25 June 2009 (UTC)

Good questions. I'll take a stab:

Is the existance of air's conformance to Bernoulli principle a major requirement for lift, or does it just make it more efficient? It's not a requirement - one of the assumptions made when deriving Bernoulli is that the flow be steady (i.e. not vary with time) and one can certainly envision non-steady flows that generate lift. As for making it "more efficient" I'm not sure what that means - if you mean that steady flow (i.e. flow that conforms to Bernoulli) is more efficient I suspect that this is true but I couldn't prove it in all cases. Most designs seem to attempt to minimize the amount of non-steady flow, if that helps. It's certainly easier to solve the differential equations if the flow is time independent.

Does the wing+air interface really conform to Bernoulli, or does Bernoulli only approximate the wing+air interface because only a small amount of (Bernoulli violating) work is done by the interaction? Bernoulli Principle is only applicable to steady flows. There's a turbulent non-steady region at the air-wing interface, so NO on the first half. As for the second question, most wing design attempts to minimize the turbulant region - if it's small enough, one would be able to ignore it for practical purposes.

What is the significant difference between the rotor of a helicopter in a hover versus a prop? One is tilted 90 degreees. Um...that's about it. Ok, that's not really your question, you were wondering about the conservation of energy thing and whether Bernoulli applies to rotors and props. Here's my take: One way to derive Bernoulli is as an application of conservation of energy. If you rely on that derivation you assume that energy in the airflow is constant, and you can't apply Bernoulli where energy is being extracted from the flow (e.g. a sailboat) or energy is being added (e.g. a propeller).

But, there's a way to derive Bernoulli that doesn't depend on conservation of energy. (both derivations are in the article) This non-COE derivation depends only on integrating newtons 2nd law along a streamline. Since COE is not assumed, one can apply the results to situations where energy is not constant.

what is the significant difference bewteen a wing, a rotor, or a propeller, such that a wing is fully compliant with Bernoulli, but not a propeller or a rotor of a hovering helicopter? For a plane in steady level flight, the lift force is perpendicular to the motion of the aircraft. Therefore, F·d=0 and the lift force does no work, therefore it is reasonable to assume that the energy in the flow is constant. For a rotor or a propeller (or a sailboat) the motion of the foil is at least partially in the direction of the lift force, F·d is non-zero and work is added or extracted from the flow. If you believe Bernoulli Principle is dependent on Conservation of Energy, you can't apply it to a prop or a rotor or a sailboat.

This is why Weltner et al says that deriving Bernoulli from Conservation of Energy sows "confusion". I think NASA fell under the spell of this confusion in this case. However, there may be solid reasons why one cannot apply Bernoulli across the prop/rotor - in particular Bernoulli only applies along a streamline and one has to be careful in comparing pressures in different streamlines.

Mr swordfish (talk) 14:42, 25 June 2009 (UTC)

Thanks for the clear cut response, this is what I've been looking for.

For a plane in steady level flight, the lift force is perpendicular to the motion of the aircraft. Therefore, F·d=0 and the lift force does no work, therefore it is reasonable to assume that the energy in the flow is constant. Although lift is perpendicular to the direction of an aircraft, it's not perpendicular to the wings because of the wings effective angle of attack. In the case of a flat board wing, d = wing chord x sin(angle of attack), and is the same for both the upper and lower surfaces. One way to visualize this is to imangine looking through a tall vertical slit as the wing passes by in a air based frame of reference (say you're in a hovering balloon). As the wing passes through the view of the slit you see the upper and/or lower surfaces of the wing moving downward as it passes by the view of the slit. d would be net movement of these surfaces as viewed through the slit. The upper surface's downwards movement reduces pressure and pulls air downwards (and a bit forwards), the lower surface's downwards movement increases pressure and pushes air downwards (and a bit forwards) and the normal (vertical) component of these combined interactions = F_lift.

I'm not sure how the change in pressure of the air is taken into account. Work done is F·d, regardless of the change in pressure, but if the work done also results in a reduction of pressure, how is this taken into account? Jeffareid (talk) 17:03, 25 June 2009 (UTC)

Although lift is perpendicular to the direction of an aircraft, it's not perpendicular to the wings because of the wings effective angle of attack. For computing the work done, what's important is the direction of the force, not necessarily the wing. However, this does bring up an interesting point - we know that the pressure vectors point in different directions at different points along the wing surface, and this would imply that locally the force is not perpendicular to the motion - so at the local level energy is being added or removed from the stream as F·d is non-zero. If you believe Bernoulli's Principle (BP) depends on the energy being constant, it doesn't make sense to use BP here. Fortunately, BP does not depend on the energy being constant, as the first derivation given in this article shows.

Work done is F·d, regardless of the change in pressure, but if the work done also results in a reduction of pressure, how is this taken into account? If you derive Bernoulli's Principle without assuming that energy is constant you get the same familiar equation. So regardless of whether energy is being added or removed, the relation between speed and pressure remains the same. Bernoulli's principle is basically just F=ma integrated - you don't need to "correct" F=ma when energy is being added; you don't need correct Bernoulli either.

Earlier this week this article claimed that Bernoulli is "equivalent" to conservation of energy. That's a common misconception that stems from one method of deriving the equation. It's not true. Perhaps we should say so clearly in the article.

Thanks for the answers. My only other concern is that a few people think that Bernoulli implies that lift can occur without downwash (downwards acceleration of air), but that probably belongs in the lift or wing articles. Jeffareid (talk) 16:27, 26 June 2009 (UTC)

• So regardless of whether energy is being added or removed, the relation between speed and pressure remains the same. Except that the relationship between speed and pressure changes during the time when energy is added or removed. In the case of a propeller or hovering rotor, the pressure increases, but the speed doesn't decrease (it stays about the same or increases slightly), changing the relationship between speed and pressure. After an aircraft passes through a volume of air, and at the locations and times where the affected air's pressure returns to ambient, the velocity is non zero, so the relationship between speed and pressure have been changed because of the work done by the aircraft to the air. This is explained in that NASA prop analysis article as exit velocity. I'm not sure that adding a density factor covers this case. Jeffareid (talk) 09:12, 27 June 2009 (UTC)

I just re-read this section again, and the inclusion of the helicopter rotor is distracting. This is supposed to be an example, not a compendium of all airfoils. I think the section will be much clearer if we delete the references to helicopter rotors and just talk about wings. e.g.

• Bernoulli's Principle can be used to calculate the lift force on an airfoil if you know the behavior of the fluid flow in the vicinity of the foil. In particular, if the air flowing past the top surface of an aircraft wing is moving faster than the air flowing past the bottom surface then Bernoulli's principle implies that the pressure on the surfaces of the wing will be lower above than below. This pressure difference results in an upwards lift force.

Comments? Mr swordfish (talk) 14:13, 20 July 2009 (UTC)

I agree that mention of aircraft wings and helicopter rotor blades are merely examples of airfoils. Propellers could be mentioned too, but in the case of a propeller it is not sufficient to talk about the top surface and the bottom surface.

If mention of helicopter rotor blades is removed the second sentence should begin For example, if the air flowing ...' rather than In particular, if the air flowing ...

At the same time, I think the final sentence should also be deleted. It says See section below on misunderstandings about the generation of lift, and also Coandă effect. This sentence is entirely unrelated to the final paragraph, and generally unrelated to the subject of Real World Applications of Bernoulli's principle. Dolphin51 (talk) 23:19, 20 July 2009 (UTC)

• Bernoulli's Principle can be used to calculate the lift force on an airfoil - How is this done in a real world situation when there aren't any finite width streamlines? There's just some speed gradient versus distance from an airfoil surface. Are there any significantly thick sections of air flows that approximate streamlines? Jeffareid (talk) 07:43, 26 July 2009 (UTC)

In practice, the usual calculational "trick" is to compute a surface integral of the vector field representing velocity, using Bernoulli's law to relate the pressure to the speed, and assuming that the pressure is normal to the surface. Agree that this is a simplification of the "real world" physics, but it's good enough for engineers (and good enough fore Wikipedia). Mr swordfish (talk) 15:15, 27 July 2009 (UTC)

What type of airfoil program uses this Bernoulli based process? I had the impression that programs such as XFoil use the methods described here: Computational_fluid_dynamics and I find no reference to Bernoulli in that article. Jeffareid (talk) 09:08, 28 July 2009 (UTC)

• Airfoils and Bernoulli. Bernolli principle explains the interaction within a gas or fluid, but not necessarily the interaction between a solid and a gas or fluid, in this case it seems to depend on the nature of the interaction between solid and a gas or fluid. For some airfoils (like the m2-f2), the faster moving air occurs below the air foil (if the air itself is used as a frame of reference). Jeffareid (talk) 07:43, 26 July 2009 (UTC)

Bernoulli's Principle can be used to calculate the lift force on an airfoil - The only citation for this is a reference to a book which infers a relationship between streamlines and prediciton of lift, but doesn't include any actual algorithms demonstrating calculating the lift force on an airfoil. Are there any citable references (verifiablity) for any community accepted air foil program where Bernoulli's Principle is used to calculate the lift force on an airfoil? Airfoil programs used to calculate lift (and drag) are covered in Computational_fluid_dynamics, and there is no reference to Bernoulli Princple there, only the more complicated Navier–Stokes_equations , with some simplifications, especially for turbulence, but not simplified to the point of Bernoulli Principle. Jeffareid (talk) 23:35, 29 July 2009 (UTC)

• In fluid dynamics, the Euler equations govern inviscid flow. They correspond to the Navier–Stokes equations with zero viscosity and heat conduction terms. The Euler equation is a differential relationship between pressure and velocity. To obtain a direct relationship, you can integrate the equation between two points on a streamline. The answer to this derivation is, wait for it...: Bernoulli's Equation! While there's no closed form solution to the general Navier–Stokes equations, Bernoulli's Equation IS the solution for steady, incompressible, inviscid flow. (Note that, despite its name, the equation expression of Bernoulli's principle on this page was actually derived by Euler.) So can I make a general plea? Take an aerodynamics class: if you're confused by Bernoulli, there's no way you're ready for Navier–Stokes... —Preceding unsigned comment added by 69.1.23.134 (talk) 01:27, 3 October 2009 (UTC)

My point was about the statement, Bernoulli's Principle can be used to calculate the lift force on an airfoil, because it's not true. I'm not aware of any airfoil program that uses Bernoulli's Principle alone, which the statement implies, as opposed in conjuction with other principles to calculate lift, all of those use some version of Navier–Stokes_equations. And in case anyone mentions this, figuring out the pressures via some complex process and then integrating pressure differentials across a wing surface is not Bernoulli. Also in the real world, a wing increases the total mechanical energy of the air during its interaction of the air, which violates Bernoulli. Jeffareid (talk) 00:28, 15 February 2010 (UTC)

If you're not willing to take an aero class, you might at least read this technical paper on the NASA VU-FOIL program http://www.grc.nasa.gov/WWW/K-12/airplane/FoilTheory.pdf ( It's a precursor to the FoilSimU applet on their website http://www.grc.nasa.gov/WWW/K-12/airplane/foilsimu.html) It reads, in part: "Knowing the local velocity, the local pressure can be calulated from Bernoulli's equation along a streamline." You can tell whether a program uses Bernoulli or Navier–Stokes by seeing whether it has inputs for viscosity and heat transfer: no viscosity and heat transfer terms=Bernoulli. ```` —Preceding unsigned comment added by 214.4.238.6 (talk) 17:22, 2 March 2010 (UTC)

That NASA VU-FOIL program is a training aid for students and based on ideal flow, not real world flow. Jeffareid (talk) 16:19, 8 May 2010 (UTC)

Alright, I have taken years of "Aero" classes, built airplanes that I have flown, and been a pilot since 1962. The statement Bernoulli's Principle can be used to calculate the lift force on an airfoil is misleading and should NOT be the lead sentence in the Wiki. I have flown airplanes with the airfoil "bump" on the bottom surface. According to that lead sentence, not only would flight have been impossible, but the faster I went, the deeper the ruts I would make in the ground. In addition, Bernoulli and Mach tuck are mutually exclusive. Newtonian principles however ARE compatible with the Mach tuck I have experienced in the real world. The lead sentence should be: "Bernoulli's Principle can be used to partially calculate the lift force on an airfoil". (W David Doiron, 4 Juil 10) —Preceding unsigned comment added by 68.98.112.120 (talk) 21:02, 4 July 2010 (UTC)

Bernoulli's principle says nothing about the bump having to be on the top. Over the years this article, and also Lift (force), have attracted many editors who insist that Bernoulli's principle is not relevant to the lift on an airfoil - when we explore the views of these editors we discover that they think Bernoulli's principle and the Equal-transit time theory are one and the same! (Everyone agrees that the Equal-transit time theory is so excessively simplified that it is incorrect.) Bernoulli's principle is a valid scientific statement that has stood the test of time since 1738. Many of us have flown inverted, or flown aircraft with upside-down airfoils, but this does NOT show that there is anything suspicious about Bernoulli's principle.

You have misquoted the statement in the article. What it actually says is Bernoulli's Principle can be used to calculate the lift force on an airfoil if you know the behavior of the fluid flow in the vicinity of the foil. When the flow field about an airfoil is known, including the flow velocities about the airfoil, Bernoulli's principle allows the air pressure to be computed for any point outside the boundary layer. I suggest you familiarize yourself with Kutta condition and the Kutta-Joukowski theorem. Dolphin (t) 03:27, 5 July 2010 (UTC)

Very nice answer. Do you think we should post a version of your first paragraph at the top of this page? It might cut down on the neverending archives of the same issue...69.1.23.134 (talk) 01:12, 22 August 2010 (UTC)

Kutta-Joukowski theorem is about idealized 2d flows, how does that help a person calculate lift in the real world where you have a 3d wing and air? It seems to me if a person knows the behavior of fluid flow in the vinicity of the foil, they already would have determined the lift during the same process used to the determine the behavior. Another issue is that there are no flow velocities, but instead a gradient of velocities that vary with distance from the boundary layer. How does a person use Bernoulli to determine lift with a gradient range of velocities? Rcgldr (talk) 20:27, 7 March 2011 (UTC)

The article states that Bernoulli's principle can be used to calculate the lift force on an airfoil. It doesn't say the Kutta-Joukowski theorem can be used to calculate lift on a 3D wing. The Kutta-Joukowski theorem was developed a century ago when a small number of scientists were striving to develop a rigorous understanding of how lift was generated. Even today, this theorem is a very useful tool in helping us to understand why lift is generated in some circumstances but not in others. Look at some of the poorly-informed edits about lift and you can see that there are still a lot of people who would benefit from an understanding of the Kutta-Joukowski theorem.

Bernoulli's principle addresses pressures and velocities. If we are looking at a flow situation in which we know the velocity gradient but not the velocity then Bernoulli's principle is not applicable and there is no unique solution for the pressure anywhere in the flow field. Dolphin (t) 23:20, 7 March 2011 (UTC)

First, I need help properly completing the aero thermo fundamentals section of centrifugal compressors. Please help! my poor math and BSME leave me with the understanding that all the physics of turbomachinery flow lays in a glass cube. I look in one way and see Euler. Turn the box i see newton, turn again i see navier-stokes...there is still 3 more sides and i have a headache.

Second, question: isn't viscosity required for a wings to fly?

Third, question: why am I not seeing the "Coanda Effect" and the "Kutta Condition" discussed at the same time with respect to lift?

Fourth, question: is the vorticity generated by the viscous flow over a wing not also part of the story?

Fifth, last comment: if you build a small box to the width of a deck of cards and you drill a tiny hole side to side. then glue the lead edge of a playing card to a thin wire you will have a small wind tunnel with a wing/playing card that pivots about its lead edge. if used carefully in an experiment with pennies taped to the card, you can proved without a doubt that St Bernoulli does not lift 747s! Mkoronowski (talk) 06:31, 20 April 2011 (UTC)

Your post seems to indicate that you accept "Euler" but not "Bernoulli". Could you identify what you think Euler's equation expresses differently from Bernoulli's Principle? Also, you cannot possibly prove any fluid dynamic principle "without a doubt" using a wind tunnel the same size as the test object. (Your entire test area would be in the edge effect.) 98.174.165.66 (talk) 02:30, 22 April 2011 (UTC)

I'd like to ask something about Bernoulli's principle. Does Bernoulli's principle apply to air? In the context, it clearly states that the principle describes fluid or liquid, and not gases, like air. But Bernoulli's principle does apply to air. Does that mean air is a fluid or liquid? Is air not a gas? Should Bernoulli's principle be changed? I am concered because I am studying his principles and am very much confused.

You can reach me at remifree@yahoo.ca or at staarre.nasa@gmail.com . --99.255.86.150 (talk) 22:59, 18 January 2010 (UTC)

Fluid covers both gases and liquids, however this page is for discussion of the article, not the topic in general User A1 (talk) 01:13, 19 January 2010 (UTC)

Bernoulli's principle applies to fluids. Liquids and gases are fluids, so Bernoulli's principle applies to liquids and gases, including air. Dolphin51 (talk) 02:56, 19 January 2010 (UTC)

In addition to the venturi effect, carburetors rely on a dynamic vortice effect that occurs at the end of a open ended tube positioned perpendicular to an air flow. The vortice is related to the air flow that is diverted away from the opening at the leading edge of the tube. If that tube was wall mounted so in effect it was a static port, there would be much less fuel drawn into the flow. This dynamic vortice effect of an open ended tube positioned perpendicular to an air flow is used in some type of sprayers, such as some perfume sprayers and the classic "flit gun" insecticide sprayer, that still work even if the static pressure of the flow across the open end of the tube is higher than ambient.

This can be demonstrated with a straw in water and using a blow dryer to blow air over the open end of the straw, which will draw up water. (Use a leaf blower (ear plugs and a secure container are recommended) and it will become a sprayer). Insert the open end of the straw into a large spool of thread to approximate a static port and the effect is greatly reduced. Use a real static port (you can buy these for < $20 USA), and the effect is even less.

Jeffareid (talk) 23:15, 15 February 2010 (UTC)

This whole article is very misleading. Nowhere does it clearly state that Bernoulli's "dynamic pressure" isn't a real pressure exerted within the fluid, and that it's just an imaginary concept used to represent the kinetic energy of a moving flow with respect to its surroundings.

There's even a picture of a manometer displaying a "pressure differential", implying that the static pressure in the thinner tube of the venturi is lower than in the thicker tube (in a lossless situation, which is what the article is all about, the pressure is equal, and the dynamic pressure can't be measured in this way). Does the article propose the existence of perpetual motion machines? This manometer is poised to do work from an example of a "lossless by definition" system.

This is my take on this:

Imagine a venturi (composed of a thick pipe, constriction, thin pipe, expansion, and back to the thick pipe) passing a fluid (not water, lossless, incompressible etc) in a vacuum. Since no pressure is required to move the fluid through a lossless system (the fluid just moves under its own momentum), the entrance and exit pressures can be zero. Does the fluid in the thin section experience a negative pressure? Does it cavitate as it leaves the constriction and enters the thin pipe? No, if it's going to cavitate, it would only do so as it enters the expansion (or exit as a jet).

Bernoulli's principle is 'just' the result of an elaborate illusion, based on the assumed need for the fluid to conserve energy as it flows. It doesn't - the fluid in the thinner pipe already "has" more kinetic energy than the same volume of fluid in the thicker pipe - it doesn't need to "get" this from elsewhere. Imagine a bubble of fluid (in air or a vacuum) going through the system - in this case energy will be conserved, in effect this bubble of fluid has to "slow down" to maintain an equal kinetic energy. Or imagine filling this system - as the flow enters the thin section of pipe, real pressure and energy is required to accelerate it at the constriction (that's how jets work). Energy isn't required to propel it further along the narrow pipe, but energy is being added to the system while any section of the pipe is being filled (in effect charging it). Pressure is raised at and near the walls of the constriction, as flow is bent and accelerated (radially inward) towards the thinner pipe, in turn this pressure is removed as the flow is squeezed (accelerated) into the thin tube, where the pressure inside the fluid (static pressure) returns to whatever it was in the thick pipe. There is no net flow of energy along the direction of the tube though, because all points in the flow (with respect to the tube) are moving at a constant velocity and already have their energy due to the charging process mentioned above. (Additionally, "pressure energy" can't be added to or removed from an incompressible fluid.) The same but opposite situation occurs in the divergent flow at the expansion end where a low pressure is formed at the walls of the expansion (except in a vacuum there might not be enough absolute pressure at the exit to support the expansion of the divergent flow and it might cavitate, or exit as a jet needing continual application of energy to the system to make up for the higher exit velocity). This lower pressure on the downwind side is what powers carburettors etc (it has nothing to do with "dynamic pressure" directly because that's not a real pressure and exerts no real force - a pitot tube wouldn't work if this were true).

Dynamic pressure can be converted to real pressure when the flow hits something (eg when it impinges on a flat plate, the "stagnation point" at the middle). But this is no more relevant in a practical sense than saying a car has a "dynamic" pressure as it travels towards a wall which gets converted to a "static" pressure when it hits it. Unless you look at the molecular motion.

It's that old problem of reference frames, like gravity and centrifugal force. The intuitively obvious (but supposedly more complex conceptually) concept of how things behave within the frame of the actual rotating or moving substance becomes much more involved when converting to the supposedly simpler fixed (inertial) reference frame.

There's nothing wrong with Bernoulli's principle and "dynamic" pressure, as long as it's clear that it is an imaginary concept and has nothing to do with anything measurable within the moving fluid itself. --Adx (talk) 02:46, 20 June 2010 (UTC)

Thanks for your comments. I'm sure you are aware that editors are not at liberty to simply add their personal views to articles in Wikipedia. Editors are required to identify the source of their information by adding references and in-line citations. Are you aware of an authoritative source (book, journal article, website etc) that has published the notion that dynamic pressure is an imaginary concept? If so, I would be grateful if you would let us know by replying to this thread. Also, are you aware of the Wikipedia articles Dynamic pressure and Impact pressure? Dolphin (t) 13:07, 20 June 2010 (UTC)

This is all a bit philosophical. Almost all mathematical concepts (including pressure) are "imaginary". User A1 (talk) 14:20, 20 June 2010 (UTC)

Yes I'm aware everything in the article has to be verifiable, but I'm glad I didn't edit anything, because I got it spectacularly (but not completely) wrong. Somewhere in the "invention" of the lossless venturi in a vacuum I got carried away and ended up sure enough to present my thinking out loud as fact (spurred on my something I read on the web years ago). It was too late lastnight for damage control so I left it. Fortunately it reads like it was written by a nut, so has got the appropriate response.

Some things I got wrong:

• I didn't mean to imply venturis don't work in real situations, but I did suggest that pressure doesn't actually vary (eg, they work by tearing off fluid at the interface to the port). But high speed flow (thin tube) 'wedging' into low speed fluid has to increase pressure as it does so (how else would it slow down?). At the other end, 'sucking' (accelerating) a high speed flow out of a low speed or even stationary fluid can only be done through a lowering in pressure. So I got that very wrong.
• As I clumsily suggested (by saying "may cavitate"), a venturi in a vacuum is perfectly capable of operating with flow at negative pressure due to the tensile strength of the fluid (which means that it's not really a negative pressure). This real static pressure drop (which I was proposing didn't exist) could be verified by viewing the size of air bubbles travelling through a water venturi, or just noticing that cavitation can occur in venturis.
• A pitot tube measures the pressure difference between the stagnation pressure (which is effectively the inlet pressure for a large area ratio venturi), and the lower static pressure.

Some things I (may?) have got right:

• Flow separation can occur at the exit divergence of a venturi, not that that's relevant.
• Dynamic pressure is a (or really "the") imaginary concept, it can't be measured within the flow, it is a theoretical expression of the energy of the flow if its velocity is brought to zero. In other words it is the number you add to make up for the (real) lower static pressure. Plenty of references around for that. Static pressure is the only 'real' (directly measurable) pressure in the system.
• The pressure gradients in a venturi system are caused by the constriction and expansion (and fact it is a pipe so flow is being forced to follow this exact path), and have nothing to do with a "moving flow" itself (as is often taught).
• In the reference frame of the fixed venturi, there is no "conservation of energy", as the system has already been charged with the extra energy it needs to support the faster flow, as I described. It is impossible for energy to be conserved within a unit of flow itself within this system (it's incompressible for a start, and when accelerated into the thin tube it must contain more energy than it did before it entered). The fixed system however, including venturi walls, has the pressure gradients needed to do the work necessary to add and remove kinetic energy from the flow.
• The article doesn't really make these points that clearly. The article on dynamic pressure is slightly clearer but basically just 'spouts a bunch of technical mumbo jumbo'. An abstract mathematical treatment of a simplified imaginary model conveys very little information to most people.

Maybe this will be useful to help other people avoid the same spectacular trip and fall I made. There are a lot of people that believe (or were never quite sure what to make of it, like me) that lower pressure is somehow a property of moving fluid. Other than that, sorry for wasting peoples' time. If I wanted to help I could draw some diagrams trying to explain what's going on, but I think I've already done enough damage... --Adx (talk) 23:49, 20 June 2010 (UTC)

Things not to change in the article: (aka The article doesn't really make these points that clearly ... because they're not accurate.)

• Dynamic pressure is no more imaginary than total pressure.
• Pressure gradients in a venturi system DO vary due to flow. (If there were no flow, there would be no exact path to be followed.)
• There is "conservation of energy" everywhere. There is no "venturi exception" to this law.

69.1.23.134 (talk) 01:03, 18 July 2010 (UTC)

• Dynamic pressure is obviously more imaginary than static pressure - actually the article does make this clear, when it refers to static pressure as "actual" pressure (implying the dynamic pressure isn't). Within the frame of reference of the flow (eg a bubble), which is what people naturally think of when considering the "flow" itself, there is no such thing as dynamic pressure.
• I didn't mean to imply pressure gradients don't vary due to flow (in my second contribution that is). What I was trying to say is that lower pressure isn't a "property of" flow itself (as many people learn), in the same way that higher pressure in an axial flow turbine shouldn't be considered to be a natural property of "fast moving gas". The pressure gradients are driven by the system as a whole (venturi walls, flow, and heat plus compressor blades in the case of the turbine).
• Maybe I confused things with my reference frame stuff. Energy is NOT conserved within the flow (nor does it need to be). 1 gram of water in the thin pipe has more energy than 1 gram of water in the thick pipe. Neither experiences a pressure gradient, therefore no work is being done. The work is done at the ends of the thin pipe (constriction and expansion), driven by the lower pressure within the thin pipe (adding energy at one end and taking it back at the other). As I demonstrated, initially the system needs to be charged with energy before it will work. Of course there is conservation of energy as a whole, but the concept of a "dynamic pressure head" necessary to reconcile this initial charge of energy with a 'static' analysis IS completely imaginary, in a physical sense.

The article may be technically correct, but in my view it remains quite misleading, because it perpetuates an incorrect explanation of what is going on, and what Bernoulli's principle is ultimately for. (Or so I believe, correct me if you can.) --Adx (talk) 14:35, 24 July 2010 (UTC)

This has shifted too many times for me. What exactly do you still think the article has got wrong? --BozMo talk 20:21, 24 July 2010 (UTC)

It has shifted once. I incorrectly stated that the pressure drop in the thinner tube of the venturi isn't a real measurable pressure (in part because of the confusion within the article), I fairly quickly corrected that. The article perpetuates the myth that dynamic pressure is a real property of the fluid (directly measurable), based on some assumption that energy must be conserved along flowlines, which is incorrect. It is merely a tool (used to represent the initial conditions of the system I suppose). I'm not saying that Bernoulli's principle is incorrect. --Adx (talk) 04:46, 31 July 2010 (UTC)

I agree with Adx insofar as dynamic pressure is entirely dependent on our choice of reference frame. Dynamic pressure is the fluidic analogue of kinetic energy; and the kinetic energy of a particle or rigid body is dependent on choice of reference frame. Because of the dependence on choice of reference frame dynamic pressure (and kinetic energy) cannot be measured directly using a gauge of any sort. Static pressure is measured with a gauge; and it varies with depth in a fluid but not with choice of reference frame. Dynamic pressure is different - it cannot be measured with a gauge, it does not vary with depth in a fluid, but it does depend on choice of reference frame. Dolphin (t) 10:03, 31 July 2010 (UTC)

how the velocity of the fluid increases if the pressure decreases? —Preceding unsigned comment added by 219.64.167.181 (talk) 13:45, 15 November 2010 (UTC)

Hello. This page is for discussions on how to improve the article on Bernoulli's principle, see the top of the page. For questions about it: you can ask them on the Reference desk. -- Crowsnest (talk) 09:38, 16 November 2010 (UTC)

Apologies if I am answering this late(it may be irrelevant now, or pretty obvious, as it is), you will find the answer in bernaulli's equation only. Rearranging bernaulli's equation(incompressible flow) for a constant datum(z) you will find that (p1-p2) is directly proportional to (v2^2-v1^2). So if v2>v1(velocity increases) p1>p2(pressure has decreased). — Preceding unsigned comment added by Shri chillies (talk • contribs) 12:09, 19 July 2011 (UTC)

Hi,

There is a picture showing (water) condensation over an airplane wing claiming this is due to the pressure drop. But this must be wrong, a pressure drop would help evaporation not condensation. Instead I would say it is because of a temperature drop over the wing (due to lower pressure) making water condensate.

Your comments on this?

Ravn-hawk (talk) 13:45, 14 January 2011 (UTC)

You are correct. The reduction in pressure causes a reduction in the partial pressure of the water vapour in the air. If the reduction in pressure was achieved isothermally the relative humidity would decrease and there would be no tendency for cloud to form. However, the reduction in pressure is achieved adiabatically so there is a reduction in temperature. The result is that the saturation pressure of water reduces as the temperature falls and so the relative humidity of the air adjacent to the top of the wing increases, and if it reaches 100% cloud begins to form. I will adjust the photo caption.

For additional information, have a look at the article wingtip vortices. The string of condensation that sometimes forms in the core of wingtip vortices forms by the same mechanism. Dolphin (t) 03:47, 15 January 2011 (UTC)

I feel that the derivation of the equation is way too complex. I studied it for some time and the best way to figure out where it comes from, for me, was the conservation of energy. I could edit the article, but that would mean removing a whole section. So I guess I'm asking for permission. (I'm currently an engineering student and a complete wikiNooB) Athreean (talk) 15:17, 29 October 2011 (UTC)

At the moment, both a derivation using Newton's 2nd law and another using conservation of energy are given. The formula's for the derivation using Newton's 2nd law are very simple and compact, so I do not see your point in that respect. The associated text could be clearer and more precise, though.

The part using energy principles is complicated, and confusing (incorrect?) regarding the inclusion of potential energy into the work done. An example of a better and slightly simpler derivation based on energy principles is e.g. Feynman, Leighton & Sands (1964) The Feynman lectures on physics, vol. 2, section 40-3.

In your edits, please also take the Wikipedia guidelines on verifiability in consideration, see WP:V and WP:NOR. -- Crowsnest (talk) 18:02, 29 October 2011 (UTC)

Thank you for the help, what I mean is the following:

Bernoulli states that the energy in the system will remain constant, and energy can be split into Kinetic and potential energy which gives us

E=C

Ek+Ep=C

mv²/2 + Ep = C

since m=w/g you get

wv²/2g + Ep = C

the weight of the fluid is assumed to stay constant so the w gets removed

and pressure head is the same as potential energy

v²/2g + P/ρg = C

this equation does not take into account the difference in height, so you add it to get

v²/2g + P/ρg + z = C

This seems easier for me to understand, but I can be completely wrong

Just thought I would share it

(Sorry for the quick explanation and bad english, but I'm pressed for time)

Athreean (talk) 09:08, 30 October 2011 (UTC)

For incompressible flow, pressure is not a potential energy. It is just a Lagrange multiplier enforcing the constraint of zero-divergence of the flow velocity field ∇•u=0. The kinetic energy density is ½ρ|u|² and the potential energy density is ½ρgz, with the total energy density the sum of these two. The pressure comes in through the work performed on a fluid parcel, by the first law of thermodynamics. -- Crowsnest (talk) 14:22, 30 October 2011 (UTC)

The principle that the energy in a system will remain constant is the principle of conservation of energy. This principle is universal and is a much more fundamental principle than Bernoulli's principle. An important element of the principle of conservation of energy is the principle of conservation of mechanical energy. (Mechanical energy is the sum of kinetic energy plus gravitational potential energy.) This principle only applies in situations where the only forces are conservative forces, such as gravity and springs. Friction is not a conservative force so this principle is an idealised situation because friction acts in almost all dynamic situations.

Bernoulli's principle is the fluidic analogue of the principle of conservation of mechanical energy. The fluidic analogue of kinetic energy is dynamic pressure. The fluidic analogue of gravitational potential energy is static pressure, or static pressure plus extra potential energy relative to a datum ρgh. The fluidic analogue of mechanical energy is total pressure. So in a fluid flow field where friction forces (viscous forces) are non-existent or can be ignored, the principle of conservation of mechanical energy can be expressed as:

total pressure = static pressure + dynamic pressure + ρgh

If the variation of h is zero, or so small it can be ignored, this equation reduces to:

static pressure + dynamic pressure = total pressure

This not the only correct explanation of Bernoulli's principle - there are many different ways to explain it, and all of them are equally correct. Whichever one appeals to you the most is the one that you should bear in mind, but that doesn't mean your explanation is the one that should appear in Wikipedia. The important considerations that govern how Wikipedia explains Bernoulli's principle are that it should be technically correct and rigorous; and it should be supported by citation of a reliable published source to allow independent verification. Happy editing! Dolphin (t) 09:36, 30 October 2011 (UTC)

Thank you for taking the time to respond, I understand it more clearly now Athreean (talk) 14:03, 30 October 2011 (UTC)

Link

http://cereview.info/book/fluid-mechanics-and-hydraulics/bernoullis-energy-theorem-and-equations

About the link

The article pointed by the link is a condensed discussion on how to do calculations involving Bernoulli's equation. It is intended for students and practicing civil engineers who needs a summary from lengthy principles of fluid flow. The highlight of the article are the equations of continuous flow which are common in the practice of Hydraulics, and the graphical representation of the energy of flow. The article was written by me and I strongly believe it will be a good source of additional information to those who seek for more about this topic. — Preceding unsigned comment added by RTFVerterra (talk • contribs) 23:48, 16 March 2012 (UTC)

I think it should be pointed out that the first equation in section 'Compressible flow in thermodynamics' applies only for the case of steady flow. See Landau Lifshitz, Fluid Mechanics, chapter on Bernoulli equation. For example, in a case of an unsteady potential flow, an additional term will have to be added which is equal to the time derivative of the velocity potential, similarly to what mentioned in the section 'Unsteady potential flow.' — Preceding unsigned comment added by 68.59.127.239 (talk) 02:24, 10 April 2012 (UTC)

The background to the above comment is as follows. IP 68.59.127.239 added the qualification for steady flow. See the diff. I removed these words, summarising my edit as Surely the first law of thermodynamics is applicable in all situations, not only in steady flow. See my diff. Dolphin (t) 02:41, 10 April 2012 (UTC)

See Landau Lifshitz, Fluid Mechanics, Second Edition, page 17 (potential flow). Exactly the situation of unsteady compressible potential flow is discussed and the Bernoulli equation is obtained as the first integral of equations of motion, formula 9.3. There is an extra term, time derivative of velocity potential, which is equal to zero in the case of steady flow, and equation (9.3) becomes equation (9.4) which is identical to the first equation in section 'Compressible flow in thermodynamics'. The book can be read for free online http://www.scribd.com/doc/8314363/Fluid-Mechanics-L-D-Landau-E-M-Lifschitz and it is one of the most authoritative references in fluid mechanics. —Preceding unsigned comment added by 68.59.127.239 (talk) 03:03, 10 April 2012‎

The article contains detailed derivations of the equations - see Bernoulli's principle#Derivations of Bernoulli equation. If there is a qualification that an equation is only applicable in steady flow, it should first be introduced in the derivation of the equation. Dolphin (t) 03:24, 10 April 2012 (UTC)

The derivation in this article is only applicable to the stationary flow. For unsteady flows there must be added extra terms in each conservation equation which account for the direct time change. Please check conservation equation equations in integral form in any fluid/gaz dynamic book, e.g. Anderson.

Also, quoting from this article derivation section: 'Conservation of energy is applied in a similar manner: It is assumed that the change in energy of the volume of the streamtube bounded by A1 and A2 is due entirely to energy entering or leaving through one or the other of these two boundaries.' Then the article directly says, quoting: 'and assuming the flow is steady so that the net change in the energy is zero,...'

They clearly say that they assume that the change is only due to convective nature, and no unsteady phenomena, such as radiation, is assumed. Even the conservation of mass equation is only written for a case of steady flow, they assume that there are no sinks or sources of mass (e.g. injection) and mass change in the control volume is due to convection. This is all very basic , first-year fluid mechanics student level material. Also, I hope it is already mentioned somewhere in the article that all these equations apply only for the case of idealized fluid, when we can neglect effects of viscosity and heat conduction. Obviously, a professional in the field, such as myself will immediately know that this is all for the case of the steady flow, but people new to the field may get confused, and assume that the first equation in section 'Compressible flow in thermodynamics' applies also to unsteady flows which is wrong. — Preceding unsigned comment added by 68.59.127.239 (talk) 14:50, 10 April 2012 (UTC)

I agree. You are right, in that most of the Bernoulli equations given here are only valid for (quasi) steady flows. Unsteady flows, like the Küssner effect require a different treatment (Küssner solved that by using a frame of reference relative to the air, instead of to the wing as is customary for steady flows). The parameter to determine the unsteadiness of the flow is the Keulegan-Carpenter number K_C = vT / L, with v the flow speed relative to the body, L a characteristic length (e.g. chord for a wing) and T a characteristic time of the velocity changes (e.g. the time needed to make a turn with a plane). Note that for many situations in aerodynamics v is large, compared to situations in hydrodynamics. And a resulting large value of K_C means the flow is quasi-steady and can be treated with equations valid for steady flow. So unsteadiness becomes often earlier important in hydrodynamics (e.g. wave forces on offshore oil platforms) than in aerodynamics. Landau & Lifshitz Fluid Mechanics is of course a very reliable source, being a classic text on its subject. -- Crowsnest (talk) 16:05, 10 April 2012 (UTC)

I have created a new section for this article over in my user space. Comments and revisions cheerfully accepted. If well received I'll move it to the main article in a few days.

http://en.wikipedia.org/wiki/User:Mr_swordfish/Bernoulli_principle#False_Demonstrations_of_Bernoulli.27s_Principle Mr. Swordfish (talk) 19:06, 29 May 2012 (UTC)

The edits are substantially complete. Please review this "beta" version. Pending comments and discussion, I'll incorporate it into the main article in a few days. Mr. Swordfish (talk) 20:10, 19 June 2012 (UTC)

I have made some comments - see my diff. Dolphin (t) 23:10, 19 June 2012 (UTC)

I think it would be appropriate to call the incompressible flow equation Bernulli's equation and name the article that ^[1]. Bernulli's principle could get a short stub or be part of the article as one of the consequences of Bernulli's equation, which is more fundamental than Bernulli's principle, even though Bernulli's principle might be known by more people from high school. --80.219.201.54 (talk) 20:12, 10 February 2013 (UTC)

A new section appeared today. I've quickly made some obvious but superficial edits, but the section has the following problems:

• It is unsourced. We need citations if the section is going to remain. I've added a ref improve tag to encourage this.
• It presents the "pressure term" as the potential energy due to the pressure. I don't have a cite at hand at the moment, but my understanding is that this is an oversimplification. I'll try to chase down a reference in coming days.
• It doesn't really provide a definition, despite being titled "definition".
• It describes BP as primarily (solely?) a consequence of the law of conservation of energy, but it is equally valid to derive it directly form Newton's laws. The article should be clear that both derivations are correct. Interestingly, BP was derived almost 70 years before Thomas Young defined the concept of energy.
• It doesn't meaningfully add to the article, although it "cuts to the chase" more quickly.
• It mixes styles (i.e. its style is inconsistent with the rest of the article). For instance, it uses y for height when the rest of the article uses z.

None of this is fatal, of course. I think the section can be improved to meet wiki standards, which is the reason I didn't simply revert it, although I'd go along with that if other editors reach consensus. I'm curious to hear other's thoughts. Mr. Swordfish (talk) 17:30, 3 July 2012 (UTC)

I agree with Mr Swordfish's comments. The new section ends with the statement that when v (speed) is zero the equation reduces to delta v equals pgx delta y. This is the barostatic equation but the new section makes no comment about the significance of the equation. Dolphin (t) 12:39, 4 July 2012 (UTC)

I've chased down a reference for item 2:

Sometimes people who use equation 3.5 are tempted to interpret it as an application of the principle of conservation of energy. That is, they try to interpret Bernoulli’s equation as equivalent to the law of the roller coaster (figure 1.9) in the sense that the parcel loses speed when it climbs up a pressure gradient and gains speed when it slides down a pressure gradient. This is plausible at the level of dimensional analysis, since ½ ρ v2 is in fact the kinetic energy per unit volume, and pressure has the same dimensions as energy per unit volume. Alas, this interpretation is not correct. There is more to physics than dimensional analysis. The parcel of air is unlike a roller coaster in that it changes size and shape as it flows up and down the pressure gradient. Furthermore, the pressure is not numerically equal to the potential energy per unit volume. Actually, for nonmoving air, the pressure is numerically equal to about 40% of the energy per unit volume. http://www.av8n.com/how/htm/airfoils.html#sec-bernoulli

Based on this, the present Definition section is misleading at best, and possibly incorrect. Seeing as how there are no citations and that what's there contradicts a published source, I'm going to remove the new section. If we want to revive it, let's edit it in a sandbox and get it to the point where it meets wiki standards before making it live. Mr. Swordfish (talk) 18:01, 6 July 2012 (UTC)

The recently-added section Definition was unsourced so I have no objection to it being removed. The reference for item 2 is from the av8n website - this has the advantage that it is immediately visible to anyone with access to the www, but at the end of the day it is still popular science and not necessarily well-explained or reliable. For example, it makes the bold statement Alas, this interpretation is not correct. And what is stated to support this bold statement? - There is more to physics than dimensional analysis. It is a pity we can't slap a Citation needed tag on statements like this one.

In Physics by Robert Resnick and David Halliday, section 18-4 is titled Bernoulli's Equation. It states We will find it convenient to derive [Bernoulli's equation] from the work-energy theorem, for it is essentially a statement of the work-energy theorem for fluid flow.

Elsewhere on this av8n website it states that Equation 3.5 (a common form of Bernoulli's equation) is a statement that enthalpy is constant. Now, where does enthalpy come from? It is derived from the First Law of Thermodynamics. And what is the relationship between the First Law and the Law of Conservation of Energy? The First Law is a specialised version of the Law of Conservation of Energy suitable for use in thermodynamics.

The av8n website attempts to demolish the idea that static pressure is a measure of potential energy. It does so by pointing out correctly that pressure is not a measure of gravitational potential energy, and that pressure is only about 40% of the internal energy of air. No-one is saying static pressure is a measure of gravitational potential energy. In the absence of viscous forces, pressure forces in a fluid are conservative forces so the work done to oppose pressure forces can be described as a form of energy; the speed of a fluid escaping from a vessel of fluid is directly proportional to the pressure in that vessel. (This is Torricelli's law.) So the pressure of a parcel of fluid is a measure of the potential speed, or potential kinetic energy, of that parcel of fluid.

Bernoulli's equation (and Euler's differential equations) can be presented in a variety of ways. One way is to present each term as a pressure. In this form, each term is a measure of work or energy - kinetic energy, gravitational potential energy and flow potential energy, all adding to total mechanical energy. However, this is not the only way. The av8n website doesn't explain what it means, or what is intended, when it says Bernoulli's equation is not an energy equation so we are left wondering. I think it is merely saying that it is possible to write Bernoulli's equation in ways that don't look obviously like an energy equation. Dolphin (t) 08:14, 7 July 2012 (UTC)

The new section is back. It now has has one source, but it still has the same deficiencies as noted above. In addition, the opening sentence is a statement of conservation of mass, not Bernoulli's principle. One would expect that a section titled "Definition" would start with the definition not begin talking about a different topic entirely. I do not think this addition improves the article. Other opinions? Mr. Swordfish (talk) 16:23, 10 July 2012 (UTC)

I agree it doesn't improve the quality of the article. A principle, such as B's principle, is not subject to a "definition". It may be subject to a quotation or explanation or derivation or experimental observation; but not a definition. I don't see anything in the new section that doesn't already exist elsewhere in the article. In addition, the new section uses math symbols that are different to the symbols already in use. Dolphin (t) 08:47, 11 July 2012 (UTC)

It's now been over a week and the author of the new section has made no effort to correct the remaining shortcomings nor has he or she engaged on the talk page. Since the consensus here is that the new section does not improve the article, I'm going to remove it. Mr. Swordfish (talk) 18:16, 16 July 2012 (UTC)

I support Mr Swordfish's removal of the section. It didn't raise the quality of the article, but we are always willing to discuss new material here on the Talk page. If User:JSquish wishes to add a definition of Bernoulli's principle, of the kind he has added in the past, the best way forward is to discuss it here. Dolphin (t) 00:37, 17 July 2012 (UTC)

FWIW also support removal for reasons given above. --BozMo talk 18:57, 17 July 2012 (UTC)