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In the Bernoulli equation, if $h$ equals zero, it reduces to $$P_1+\frac12\rho v_1^2 = P_2+\frac12\rho v_2^2$$

The equation does not have an intuitive meaning other than the fact that it is a bare mathematical truth.

I have tried to interpret it in the following manner: The sum of kinetic energy per unit volume and pressure exerted by flowing fluid in a horizontal pipe is constant in the direction of flow. Yet, I still can't understand how (pressure) + (kinetic energy per unit volume) is a conserved quantity.

Please give your views on the same.

Gaurav
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  • It's not clear to me what exactly you're asking - generally "Please give your views" or the like is not really a question. Could you perhaps clarify what exactly you want to know? – David Z Oct 23 '14 at 16:54
  • I'm just asking an intuitive explanation of the given equation. – Gaurav Oct 23 '14 at 17:50
  • Look here. – Mike Dunlavey Oct 24 '14 at 11:29
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    Here is a very clear and concise derivation, which may give you some "intuitive meaning": http://www.feynmanlectures.caltech.edu/II_40.html#Ch40-S3-p1 Contrary to popular interpretation, the presence of the pressure term does not mean pressurized liquid has energy density equal to pressure; the term enters as work of the surrounding fluid, not as internal energy. – Ján Lalinský Jul 13 '15 at 20:31
  • Thanks a lot @JánLalinský ! You could also perhaps reproduce relevant portions from the page in an answer. That way the explanation would be more visible to other users. – Gaurav Jul 14 '15 at 14:05
  • I could, but that would take lots of time. Perhaps you can do it, if you found Feynman's derivation useful. – Ján Lalinský Jul 14 '15 at 17:27
  • @JánLalinský If I were to do that, wouldn't all the rewarded rep go to me instead of you ? :) – Gaurav Jul 14 '15 at 17:30
  • Yes! I would save some typing too. – Ján Lalinský Jul 14 '15 at 17:33

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The fact that the Bernoulli equation contains the kinetic and potential energy density (energies per unit volume) terms should strongly suggest to you that the conserved quantity is actually energy. Seeing the pressure as an energy term is a little bit trickier, but really is easiest to see from your units: $$ \left[\,{p}\,\right]=\frac{\rm N}{\rm A}=\frac{\rm N\cdot m}{\rm A\cdot m}=\frac{\rm J}{\rm V} $$ which is the same unit as energy density.
NB: Torque and energy have the same units, but these are definitely not the same thing, so you cannot necessarily equate things based on their units. In this particular case, the two items (pressure & kinetic energy density) are equated, but not due to units (instead the physics), meaning that pressure can be viewed as an energy density.

A few instances that show pressure as proportional to an energy density are:

And probably a few other places that I am neglecting at the moment.

Further, in the Eulerian framework, the energy conservation equation takes the form $$ \frac{\partial E}{\partial t}+\nabla\cdot\left(\left[E+p\right]\mathbf u\right)=0 $$ where $E=\rho e+\frac12\rho\left(\mathbf u\cdot\mathbf u\right)$ is the total energy per unit volume (total energy density), so the pressure must be an energy term to be added to the energy density (as one cannot simply add unlike quantities).

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Kyle Kanos
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  • Thanks for the time ! Could you elaborate on the "little bit trickier" part please ? Is there any scope of understanding pressure as energy per unit volume beyond dimensional analysis ? – Gaurav Oct 23 '14 at 17:57
  • Of course there's more to it, the units are probably the easiest way to view it directly. The equation of state for an ideal gas, $p\propto\rho e$, also directly shows the energy density relation – Kyle Kanos Oct 23 '14 at 18:04
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    This answer is wrong. The idea that pressure of liquid gives density of some kind of energy (in the sense of mechanics and thermodynamics) stored is not generally valid. This is easy to see in the case of incompressible liquid; however large the pressure, the liquid stores no energy after pressure increase, for there is no volume change of any liquid element and no work done. Generally, equality of units is not sufficient to interpret one quantity as the other. In special cases, internal energy may be proportional to pressure, like for ideal gas. Ideal gas is quite a special case. – Ján Lalinský Oct 23 '14 at 19:30
  • @JánLalinský: I agree, it's not generally valid. However, in terms of hydrodynamics, which is what the OP questions, the relation is usually valid for a large variety of applications. – Kyle Kanos Oct 23 '14 at 19:33
  • Another, more mathematical way to see this is to evaluate the equation of continuous transfer of mechanical energy of an incompressible fluid: $\partial_t(\epsilon) + \nabla \cdot [(\epsilon + p)\mathbf v] = 0$, where $\epsilon=\frac{1}{2}\rho v^2$ is density of kinetic energy. The pressure contributes to energy flux density, but not to energy density itself. – Ján Lalinský Oct 23 '14 at 19:35
  • @JánLalinský: The only thing that can accelerate a fluid is pressure, providing force that works through a distance, producing kinetic energy in the fluid. I think that's what Kyle is saying.
    • – Mike Dunlavey Oct 23 '14 at 20:01
  • @JánLalinský: Every source I have ever seen shows $\epsilon=\rho e+\frac12\rho v^2$, which is the total energy density (internal plus kinetic). From the ideal gas law, we can use $\epsilon=p(\gamma-1)^{-1}+\frac12\rho v^2$. – Kyle Kanos Oct 23 '14 at 20:09
  • My example was about incompressible fluid where $e$ never changes (so does not depend on pressure) and can be omitted. In your example internal energy density depends on pressure, but it is not equal to it. In general, it will not even be linear in pressure. – Ján Lalinský Oct 23 '14 at 20:19
  • @JánLalinský: I don't think one can drop a term from a definition because it's derivative is zero in particular cases (especially one that is not often satisfied). For most applications, the ideal gas law applies so $e\propto p$. – Kyle Kanos Oct 23 '14 at 20:27
  • @KyleKanos you are missing the point. I offered an example - incompressible fluid in adiabatic flow - where $e$ does not matter and can be omitted. This model is as important part of "most applications" as the model of ideal gas. Generally, the only thing that can be said about internal energy is that it is a function of both pressure and temperature. Generally, pressure is not sufficient to calculate internal energy. – Ján Lalinský Oct 23 '14 at 20:52
  • @JánLalinský: I see your point, but I believe it's irrelevant. OP wants to know for the specific case of the Bernoulli equation how the pressure and kinetic energy density are related. I think my answer shows that pressure can be thought of as an energy density term in many cases, this being one of them. – Kyle Kanos Oct 23 '14 at 20:56
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    @KyleKanos unfortunately no, your answer only confuses him because Bernoulli equation in the form he wrote it does not apply to compressible fluids you talk about here. – Ján Lalinský Oct 23 '14 at 21:04
  • @JánLalinský: you are mistaken. Nothing I wrote requires compressibility, it's all from dimensional analysis. Even the Euler equation is valid for both incomp. and comp. fluids. – Kyle Kanos Oct 23 '14 at 21:22
  • @KyleKanos your last statements are all true and explain why your answer to the original question is off the mark. The original question is about the simple Bernoulli equation, which is valid only for incompressible flows. For such flows, internal energy density $e$ of any fluid element does not change and can be omitted entirely from the discussion. In compressible flows internal energy does play role, but there the Bernoulli equation does not hold. – Ján Lalinský Oct 23 '14 at 22:03
  • @JánLalinský: my answer says as much as I've written here. Only for the ideal gas equation of state do I invoke the internal energy. Otherwise my answer relies on dimensional analysis. – Kyle Kanos Oct 23 '14 at 22:19
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    @JánLalinský: This is my favoriate explanation. It assumes incompressibility by assuming $\rho$ is constant. – Mike Dunlavey Oct 24 '14 at 11:17
  • @Mike Dunlavey: Nice link Mike ! What strikes me is the surprising nature of physics (and of course, mathematics). The original equation needs no such deep interpretations, it is completely clear and intuitive. But as soon as you reduce it to the given special case, voila !, you need a completely different level of intuition ! – Gaurav Oct 24 '14 at 13:17
  • @KyleKanos: I'm not actually familiar with compressible fluids. But I got a new insight nonetheless. – Gaurav Oct 24 '14 at 13:18
  • @Simha: I honestly don't see where I've invoked compressibility or incompressibility in my answer. Reading it through as many times as I have, I still maintain that my answer & comments argue the relation entirely based on dimensional grounds. – Kyle Kanos Oct 24 '14 at 13:44
  • @KyleKanos: I thought that the Euler equation had terms specific to compressible fluids at the time I made the comment. However, in the comments I later see that you have clarified on the same. Forgive me for the misunderstanding. – Gaurav Oct 24 '14 at 14:09