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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.

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  • $\begingroup$ 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? $\endgroup$ – David Z Oct 23 '14 at 16:54
  • $\begingroup$ I'm just asking an intuitive explanation of the given equation. $\endgroup$ – Gaurav Oct 23 '14 at 17:50
  • $\begingroup$ Look here. $\endgroup$ – Mike Dunlavey Oct 24 '14 at 11:29
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    $\begingroup$ Here is a very clear and concise derivation, which may give you some "intuitive meaning": 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. $\endgroup$ – Ján Lalinský Jul 13 '15 at 20:31
  • $\begingroup$ 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. $\endgroup$ – Gaurav Jul 14 '15 at 14:05
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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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  • $\begingroup$ 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 ? $\endgroup$ – Gaurav Oct 23 '14 at 17:57
  • $\begingroup$ 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 $\endgroup$ – Kyle Kanos Oct 23 '14 at 18:04
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    $\begingroup$ 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. $\endgroup$ – Ján Lalinský Oct 23 '14 at 19:30
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    $\begingroup$ @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. $\endgroup$ – Ján Lalinský Oct 23 '14 at 21:04
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    $\begingroup$ @JánLalinský: This is my favoriate explanation. It assumes incompressibility by assuming $\rho$ is constant. $\endgroup$ – Mike Dunlavey Oct 24 '14 at 11:17

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