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I don’t want to learn and apply Bernoulli's equation. It is complicated and you need to remember a lot. I instead want to apply energy conservation. But I don't always find it easy.

For example, when we have a column of water and an orifice at the bottom of it from which it pours out and I want to know the velocity. How can I apply energy conservation? I know that at the end, it's $mgh=\frac{1}{2}mv^2$. But I want the logic behind it. All I know is that the potential energy plus the pressure energy of the water in the column, will become kinetic energy at the bottom orifice.

But what is the total potential energy of the whole water? What is the kinetic energy of the amount of the water that pours out? What is the pressure energy at the various points of the water column?

All these complicated calculations. The kinetic energy depends on the amount of the water that goes out from the orifice! The potential energy is not the same at the top or the bottom of the column! Any hint please?

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    $\begingroup$ "I don’t want to learn and apply Bernoulli's equation. It is complicated and you need to remember a lot. I instead want to apply energy conservation." Bernoulli equation is a convenient restatement of the work-energy theorem. It is also quite simple, ignoring it is not very clever. You can find good explanation here: feynmanlectures.caltech.edu/II_40.html#Ch40-S3, the part beginning with "Imagine a bundle of adjacent streamlines". $\endgroup$ – Ján Lalinský Oct 8 '15 at 16:54
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You should get over your fear of Bernoulli because it is useful in a number of different situations. Besides you will find that the conservation of (total) energy will simply revert to the Bernoulli equation (which is a statement of the conservation of mechanical energy) under most practical assumptions (as you have found out). It is not difficult to remember as it follows the simple recipe:

$$accumulation = in - out + production - destruction$$

In relevant variables and at steady state ($accumulation=0$) we have:

$$0=\left.\phi_{m}\left(\frac{p}{\rho}+\frac{1}{2}v^{2}+gz\right)\right|_1-\left.\phi_{m}\left(\frac{p}{\rho}+\frac{1}{2}v^{2}+gz\right)\right|_1+\phi_{w}-\phi_{m}e_{fr}$$

where $\phi_m$ is the mass flow and $\frac{p}{\rho}$, $\frac{1}{2}v^2$ and $gz$ are contributions to (mechanical) energy due to pressure, velocity and gravity. $\phi_w$ and $e_{fr}$ are a source/sink of mechanical energy due to work done on/by the system and friction which is converted to heat.

Your simple example of the column of water with an orifice indeed works both with Bernoulli and by equating the potential and kinetic energy. However, consider now that at the top of the column a piston imparts an overpressure $\Delta p$; how would you include that in your potential and kinetic energy equivalency? I am sure it is possible, but Bernoulli gives a straightforward framework:

$$0=\phi_{m}\left(\frac{p_0+\Delta p}{\rho}+gH\right)-\phi_{m}\left(\frac{p_0}{\rho}+\frac{1}{2}U^{2}\right)$$

Solving for the velocity gives:

$$U=\sqrt{2\left(\frac{\Delta p}{\rho}+gH\right)}$$

If losses due to friction are important than it becomes more complicated but even then Bernoulli gives a good framework to start. There is a reason the Bernoulli equation has been taught for centuries.

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  • $\begingroup$ One point to keep in mind is that although Bernoulli's equation is a re-statement of the principle of energy conservation, its use is limited to streamline motion of fluid only, i.e. it cannot be employed in cases of turbulent motion. $\endgroup$ – SchrodingersCat Oct 8 '15 at 12:22
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Bernoulli equation is application of energy conservation. I prefer to work with energies and energy conservation and I am confident that there will be NO instance where this will fail. If a piston exerted a pressure on the water column of the example, then the potential due to earth energy would be the same, plus there would be an additional pressure energy. But what is the pressure energy equation? Also, the whole potential energy of the water column is mgh or mgh/2 ??? Also: as the water column empties, the velocity of the exit water will become slower. So we need to apply integrals to solve this. Anyone??

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  • $\begingroup$ ^ This is not answer to your question but a reply to my answer. Your additional questions show that your approach using 'energies' does not provide a convenient framework whereas Bernoulli answers your questions directly. $\endgroup$ – nluigi Oct 9 '15 at 10:38

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