# Demystifying the Often Misunderstood Bernoulli’s Equation

Table of Contents

## Introduction

Bernoulli’s equation is one of the most useful equations in the field of fluid mechanics, owing to its simplicity and broad applicability. That applicability does have limits, however, which has led to it being one of the most misunderstood and also misused equations in fluid mechanics. The goal of this post is to present Bernoulli’s equation, its derivation based on energy conservation, and interpretations based on both momentum and energy conservation, and its limitations. The bulk of this article will be written at a level appropriate for an undergraduate STEM student who is familiar with calculus and differential equations.

## The control volume and streamtubes

In fluid mechanics, it is helpful to utilize the concept of a control volume (CV): a fixed volume in space through which fluid flows. For a given *steady* flow field, if we imagine a line drawn by a particle moving through the flow, this line will trace out a streamline. (Note: strictly speaking, this is called a pathline, but for a steady flow, pathlines and streamlines are identical.) Given that the fluid velocity is everywhere tangent to the streamline, no flow can cross such lines. Therefore, it is convenient to draw a curve through the flow field and follow the streamlines along that curve some distance downstream. This procedure traces out a streamtube which is commonly and conveniently employed as a control volume since the flow will only enter and exit the streamtube through the inlet and outlet curves. Additionally, we can draw the inlet and outlet curves such that the local flow is normal to the inlet and outlet surfaces, simplifying the mathematics considerably.

The above image is a typical streamtube created by drawing a curve, ##C_1##, and following the streamlines to some end curve, ##C_2##.

## Conservation of energy and Bernoulli’s equation

We typically concern ourselves with the conservation of three quantities in a control volume: mass, momentum, and energy. It is beyond the scope of this article to discuss all of the forms that these laws may take, but for now, we will apply the Reynolds transport theorem to a steady flow through a streamtube as it pertains to energy.

Conservation of energy in a steady flow states that the rate of external work done on the CV plus shaft work added to the CV plus the rate of heat conducted into the CV plus the rate of internal heat generation equals the integrated net outward energy flux across the control surface (CS). If we *assume the flow is inviscid, shaft work is zero, the heat conduction is zero, and the heat generation is zero*, then we have

[tex]

0 = \oint\limits_{CS}\rho\left(e+\dfrac{U^2}{2} + \dfrac{p}{\rho} + gz\right)\vec{U}\cdot\hat{n}\;dA,

[/tex]

where ##\rho## is density, ##e## is internal energy, ##\vec{U}## is the velocity, ##p## is the static (thermodynamic) pressure, ##g## is the acceleration due to gravity, ##z## is the height above the reference surface for gravitational potential energy, ##\hat{n}## is the outward surface normal vector of the CS, and ##dA## is a differential area on the CS. This is simply a mathematical statement that the net flux of mechanical energy out of our control volume is zero under these conditions.

So, if we note that we have defined the streamtube inlet and outlet as being everywhere normal to the velocity vector and denote the inlet surface bounded by ##C_1## as ##S_1## and the outlet surface bounded by ##C_2## as ##S_2##, then we have

[tex]

0 = -\oint\limits_{S_1}\rho_1\left(e_1+\dfrac{1}{2} U_1^2 + \dfrac{p_1}{\rho_1} + gz_1\right)U_1\;dA + \oint\limits_{S_2}\rho_2\left(e_2+\dfrac{1}{2} U_2^2 + \dfrac{p_2}{\rho_2} + gz_2\right)U_2\;dA,

[/tex]

If we then take the limit of this streamtube as ##A_1## and ##A_2## (the areas of ##S_1## and ##S_2## respectively) go to zero such that the integrands are constant over the integral (i.e. as the streamtube approaches being simply a streamline), we find the relationship

[tex]

0 = -\left(e_1+\dfrac{1}{2} U_1^2 + \dfrac{p_1}{\rho_1} + gz_1\right)\rho_1 U_1 A_1 + \left(e_2+\dfrac{1}{2} U_2^2 + \dfrac{p_2}{\rho_2} + gz_2\right)\rho_2 U_2 A_2.

[/tex]

Next, we note that conservation of mass implies that the mass flow rate into the CV equals the mass flow rate out of the CV, so ##\rho_1 U_1 A_1 = \rho_2 U_2 A_2##. This gives us

[tex]

0 = -\left(e_1+\dfrac{1}{2} U_1^2 + \dfrac{p_1}{\rho_1} + gz_1\right) + \left(e_2+\dfrac{1}{2} U_2^2 + \dfrac{p_2}{\rho_2} + gz_2\right).

[/tex]

Finally, we make one final assumption that the *flow is incompressible*. This implies that the density is constant and that the volume of a fluid element does not change, which, when combined with the lack of heat transfer, implies the the internal energy is constant. Therefore, we are left with

[tex]

\dfrac{1}{2} U_1^2 + \dfrac{p_1}{\rho} + gz_1 = \dfrac{1}{2} U_2^2 + \dfrac{p_2}{\rho} + gz_2.

[/tex]

This is one form of the Bernoulli equation. In effect, it amounts to a statement of conservation of energy. It is perhaps more intuitive to write the version that has been multiplied by density,

[tex]

\dfrac{1}{2} \rho U_1^2 + p_1 + \rho gz_1 = \dfrac{1}{2} \rho U_2^2 + p_2 + \rho gz_2.

[/tex]

In this form, it should be very clear to see that the ##\rho U^2/2## term represents kinetic energy per unit volume (often called dynamic pressure), the ##\rho g z## term represents the gravitational potential energy, and the static pressure, ##p##, represents the remaining pool of total energy, in this case, stored by moving molecules in the form of pressure.

## Limitations of the Bernoulli equation

The limitations of the Bernoulli equation arise due to the assumption used to derive it. In fact, this is the reason for the choice to derive it based on energy conservation rather than momentum, as the assumptions are more obvious. These limitations are

- Steady flow
- Incompressible flow
- Inviscid flow
- No shaft work, heat conduction, or heat generation
- Flow along a streamline*

The final assumption has an asterisk because this is a “weak” requirement. If we rewrite the Bernoulli equation,

[tex]

\dfrac{1}{2} \rho U^2 + p + \rho gz = B = const,

[/tex]

where we will call ##B## the Bernoulli constant. We can then note that ##B## is constant under the preceding assumptions along a streamline. However, if an entire region of streamlines originates from the same reservoir with the same ##B##, then ##B## is constant throughout the entire region and Bernoulli’s equation can be applied more broadly. As it turns out, this is often the case in practical situations and the final requirement above can be dropped.

## Static, dynamic, and stagnation pressure

Often, changes in elevation (##z##) are negligible in a given flow situation. In that case, Bernoulli’s equation can be expressed as

[tex]

p_0 = p + \dfrac{1}{2}\rho U^2,

[/tex]

where ##p_0## is called the total or stagnation pressure and replaces the Bernoulli constant under these conditions. It represents the pressure that would be achieved if the flow was brought isentropically (adiabatically and reversibly) to zero velocity, thus the “stagnation” nomenclature. At any point in a flow, it is the static pressure ##p## that is actually “felt” by any submerged surface. Stagnation pressure is only felt at a stagnation point, and dynamic pressure is never felt directly.

## Interpretation as a force balance

As a sanity check, we can perform a thought experiment based on a force balance (momentum conservation). If the velocity increases from point 1 to point 2, then that acceleration must be the result of a force in the same direction. Bernoulli’s equation tells us that the velocity increase corresponds to a decrease in pressure. This means that a fluid element passing through this pressure gradient will experience a higher pressure on its side in the direction of point 1 than on its side in the direction of point 2, and will have a net force accelerating it in the 2 direction. This is consistent with the simple thought experiment. In fact, this is the basis of the equation’s derivation based on Euler’s equation.

## Conclusion

The Bernoulli equation is a powerful and deceptively simple relationship. The simplicity comes with a cost, however, in terms of a relatively strict set of limitations. With a bit of care in its use, as outlined above, it can be applied to a broad range of fluid dynamics problems, either directly or with slight modification, ranging from pipe flows to airplanes.

PhD – Fluid mechanics, nonlinear dynamical systems, hydrodynamic stability.

Swimming, grilling, beer, movies, watching most sports, playing a few, and a dash of video games

I guess fluid mechanics is a bit like quantum mechanics in being highly counter-intuitive, although for different reasons :smile:

Perhaps even more confusing than QM in some sense, given that there is a Clay Millennial bounty for the Navier-Stokes theorem but none for a QM problem.Wait, I was wrong there… there

isa QM problem in the Clay Millennium list.. http://www.claymath.org/millennium-problems/yang–mills-and-mass-gapIn figure (a), assume that the wing has negligible density compared with the fluid. It is constrained from rotating or moving left / right. But it is totally free to move up or down (no restoring force in Y direction).

In that case, would it drift downwards over time? And if we now add a constraint against this drift, how much force would we measure?

I just realized that the animation in your

Liftarticle (i.e. the one I pasted in my previous post) is actually showing inviscid flow, not viscid flow. So that answers my question. I had wrongly assumed that this is viscid flow.It's inviscid, but it's including an important element of viscous flow. Specifically, inviscid theory alone would predict streamlines more like the top image shown here (which would generate zero lift):View attachment 235557

Viscosity is what enforces that the flow comes smoothly off the sharp trailing edge of the airfoil, modifying the flow from the top image to the bottom image. Frequently, when analyzing lift and flow about an airfoil, it is modeled as inviscid, but with this condition of smooth flow off the trailing edge enforced to give you a physically reasonable flow.

What I really would like to know is the Bernoulli affect the main reason why airplanes can fly? If it is, then I wonder how airplanes are able to fly upside down.It's complicated, so I wouldn't say it's the main effect, but it is always in effect. The simple answer to your question is that the strength of the effect depends on the angle the wing makes with the wind, and that angle is different for an inverted plane than an upright plane.

I just realized that the animation in your

Liftarticle (i.e. the one I pasted in my previous post) is actually showing inviscid flow, not viscid flow. So that answers my question. I had wrongly assumed that this is viscid flow.I guess it makes kind of sense, because the incoming and outgoing

anglesare nearly equal for each and every streamline, hence no net momentum change so no lift.Edit: Wait, that's obviously not correct. It means there is a net upward lift. Sorry.

It may be instructive to see the streamlines side by side for exactly the same airfoil, with and without viscosity.

If that flow was inviscid, the flow on bottom would wrap around the sharp trailing edge and leave the shape somewhere back around on the top surface.

That lift article is just perfect for the (fairly) uninitiated.

Quick question. You said there:

Why does the air move faster over the top, and for that matter, how do we know how much downwash is produced?It turns out, these two questions are related, and also why lift is incorrectly explained so often. It also turns out that the answer to this question is extraordinarily complicated. The simplest answer is that inviscid theory predicts one stagnation point at the front where the air impacts the tip of the airfoil, and one at the back at some arbitrary point that would result in no lift and no flow deflection.A rough hand sketch of this non-viscid case would be nice. Or even better, the following simulation with viscosity set to zero — how would that look?

All those formulas. I don't have a Pizza hut Delivery to my name so they could be mean anything. What I really would like to know is the Bernoulli affect the main reason why airplanes can fly? If it is, then I wonder how airplanes are able to fly upside down.Again, as linked before, there's a totally separate article for that.

https://www.physicsforums.com/insights/airplane-wing-work-primer-lift/

Greg Bernhardt submitted a new blog post

Demystifying the Often Misunderstood Bernoulli's Equation

View attachment 235151

Continue reading the Original Blog Post.All those formulas. I don't have a Pizza hut Delivery to my name so they could be mean anything. What I really would like to know is the Bernoulli affect the main reason why airplanes can fly? If it is, then I wonder how airplanes are able to fly upside down.

So he decided to go for the low-hanging fruit…

Heisenberg quit fluids to do QM. Draw your own conclusions. :wink:

I guess fluid mechanics is a bit like quantum mechanics in being highly counter-intuitive, although for different reasons :smile:

Perhaps even more confusing than QM in some sense, given that there is a Clay Millennial bounty for the Navier-Stokes theorem but none for a QM problem.

A couple of things here:

First, I didn't create the title. I don't know who is in charge of that, but I submitted the article and it magically appeared with a title. I agree it does give off the air of being written at a lower level than it is.

Second, I did try to start it out by saying it was written for someone already familiar with some basic undergraduate-level math and science. There are a surprising number of people at this level with a lot of misconceptions. It's certainly not the majority of them, but a sizable chunk.

Third, I agree that, when this article was finished and I looked back over it, it turned out to be at a higher level than my original intention. This came from going back over it and filling in enough holes to try to make it "reasonably complete" but without digging into, for example, the derivation of the energy equation. I don't honestly see any way around this at this point if I do it from an energy approach, and there are myriad sources online who discuss the same issue from the point of view of a force balance (conservation of momentum).

Finally, I also agree that there is a need for a similar article (or maybe an expansion of this one is some way) that does address this at a lower level, but I am just not sure how to implement it without an appeal to authority rather than letting the math/physics speak for themselves.

(a)makes the principle come alive using more intuition (graphics) and less calculusPart of the issue is that fluid dynamics is particularly counterintuitive for many people who don't have any training in the topic. I am not how to appeal to their intuition when it often serves them wrong in the first place. I was the same way when I was first learning and it all felt backward.(b)debunks specific misunderstandings that people have about airplane wings,https://www.physicsforums.com/insights/airplane-wing-work-primer-lift/Having said that, I'm not sure what fraction of the target audience (meaning people who actually need to be demystified) would be able to take profit from this particular level of exposition. I'm not an expert (which perhaps entitles me to represent the mystified ?:) ones). Is it the case that people with enough math and physics competence to understand the article are in fact likely to have held wrong ideas before reading it? For example, believing in some of the wrong notions of aircraft flight that we often find on YouTube etc? And when I say wrong notions, I mean that other people, the debunkers, say that they are wrong — and not that I'm personally qualified to point out the errors in precise terms.

What I'd like to suggest is, maybe you could write a companion article that

(a)makes the principle come alive using more intuition (graphics) and less calculus and(b)debunks specific misunderstandings that people have about airplane wings, balls floating in jets and so on. To sum up, something to address misconceptions among laymen, as much as misconceptions among specialists.It's a difficult balance between being concise and being complete. A "concise" answer is likely to have holes or thin spots that create an opportunity to be or be perceived as wrong or, well, incomplete, whereas a "complete" answer may lose part of the target audience who is looking for a fast answer to a small question.The title and intro look a bit like a hook to bring people in for a quick answer to a particular misconception or two, but really is a pretty complete explanation….frankly, a quick list of two or three misconceptions would be too short for an article. But I'll provide a concise example of what I think is the most common misconception we get about Bernoulli's principle itself (not the applications such as lift):

Many people don't grasp the full implications of the steady flow and streamline assumptions. We often get people who ask, "but what if we add an orifice…?" or "what if I use a smaller pipe?" Now, not only is it no longer the same streamline or steady flow, but it isn't even the same system!

There's an active thread in biology right now where someone asks; shouldn't the volumetric flow rate through a small straw be the same as through a large straw because a lower cross sectional area results in higher velocity? We often get this very misconception with regard to blood circulation. As described, this is called "continuity" (not specifically mentioned in the insight), but it doesn't apply across two different systems. Continuity says that the mass flow rate in two parts of the same system must be the same unless mass is accumulating or draining. But of course the mass flow rate in two different systems doesn't need to be the same.

Hi,

Congratulations! There has long been a need to demystify Bernoulli's principle, and this article surely fills a significant part of that vacuum.

Having said that, I'm not sure what fraction of the target audience (meaning people who actually need to be demystified) would be able to take profit from this particular level of exposition. I'm not an expert (which perhaps entitles me to represent the mystified ?:) ones). Is it the case that people with enough math and physics competence to understand the article are in fact likely to have held wrong ideas before reading it? For example, believing in some of the wrong notions of aircraft flight that we often find on YouTube etc? And when I say wrong notions, I mean that other people, the debunkers, say that they are wrong — and not that I'm personally qualified to point out the errors in precise terms.

What I'd like to suggest is, maybe you could write a companion article that

(a)makes the principle come alive using more intuition (graphics) and less calculus and(b)debunks specific misunderstandings that people have about airplane wings, balls floating in jets and so on. To sum up, something to address misconceptions among laymen, as much as misconceptions among specialists.Of course, it may turn out that it is impossible to do justice at all to these things without the kind of mathematical support that you have already used in the article. I don't know.

Thanks!

Edit: There's a lot of stuff out there that tries to straighten out the subtle business of Coanda Effect versus Bernoulli effect, but personally I'm still not clear about this kind of thing. This could be one thing that you could perhaps clarify, should you decide to go ahead with the companion article idea.

Wow!! This is$$frac{4.5~hr}{the~past~3~wk}=1.5frac{hr}{the~past~wk}.$$You oughta get more sleep than that. :smile:Now you are catching me in more poorly-worded statements that may or may not be related to sleep deprivation. :sleep:

4.5 hours of sleep over the past 3 weeksWow!! This is$$frac{4.5~hr}{past~3~wk}=1.5frac{hr}{past~wk}.$$You oughta get more sleep than that. :smile:

It is mentioned in the text that "Given that the fluid velocity is everywhere normal to the streamline, no flow can cross such lines." Surely that's a typo. The accompanying drawing correctly shows the local velocity vectors

tangentto the streamlines.Good catch.Let's just say I am averaging about 4.5 hours of sleep over the past 3 weeks but thought this article addressed a misconception that is so frequent that I thought it worthwhile to attempt to write it now anyway. If anyone finds other blatant typos, please let me know.

It is mentioned in the text that "Given that the fluid velocity is everywhere normal to the streamline, no flow can cross such lines." Surely that's a typo. The accompanying drawing correctly shows the local velocity vectors

tangentto the streamlines.