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So I'm not OK with how some people derive this equation.

These people consider a pipe whose endings have cross-sectional areas and heights which are different. They then use the conservation of energy principle by saying $dW = dK + dU$ (Where $W$ is work, $K$ is kinetic energy, and $U$ is potential energy).

For this they consider that the work done on the system would be due to external pressure forces exerted on the whole system of water along the pipe. And here comes the part where I disagree: they use this Work to calculate the change in Potential and Kinetic energy for just a small slab of water within the whole system. This is completely invalid isn't it? I mean you would have to consider the entire system, I think.

My way of interpreting the derivation is if you consider just one slab the whole time. Is this a valid way of thinking?

Thanks!

edit: In fact, in one video I saw, the person just says "the middle chunk of water stays the same the whole time, so we can just ignore it".

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2 Answers 2

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You have to consider the assumptions that go into deriving Bernoulli's equation in that manner; that the fluid is incompressible, non-viscous, and experiences non-turbulent flow.

If these are the assumptions under which you've built your model for fluid flow, when you apply some sort of pressure to one end that results in a work on the fluid, because the fluid can't compress at all to absorb the energy nor can it interact with other parts of the fluid to have some resistive drag that causes some of the fluid to behave different then other sections of the fluid, every part of the fluid responds similarly.

Treating one small part of the fluid and its behavior should therefore be the same as any other part of the fluid. Indeed, it's as if the whole fluid is reacting to the force applied to it at once time. This is why Bernoulli's Equation tells us that energy is conserved per unit volume of the fluid, regardless of where it is.

In general, a more rigorous derivation is needed for more complicated fluid models, but that one suffices for the basic dynamics of fluid flow.

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  • $\begingroup$ I can see how applying pressure at one side results in work done on the whole fluid. What I can't see is how the magnitude of the work calculated is used directly to know the change in kinetic energy of just one small part. I mean if I consider a system of two blocks which is pulled by some tension for some distance, I won't directly use that magnitude for work to calculate the kinetic energy in just one block. $\endgroup$
    – DLV
    Commented May 23, 2014 at 3:12
  • $\begingroup$ Bernoulli's Equation deals with external interactions on a fluid in the form of pressures. A pressure P is defined as $$ P = \frac{F}{A} = \frac{F d}{A d} = \frac{W}{V} $$ This is why Bernoulli's Equation is conserved per unit volume. $\endgroup$
    – user28754
    Commented May 23, 2014 at 3:23
  • $\begingroup$ Yeah I think I can see that being true, but at first (in the derivation) it's not about W/V, it's just the work-energy theorem using P*A as a Force. And it doesn't make sense to me how the exact magnitude of the work done on the whole fluid will be equal to dK+dU for just one small infinitesimally small slab of water. What should I think about in order to make this clear? $\endgroup$
    – DLV
    Commented May 23, 2014 at 3:45
  • $\begingroup$ This happens when you take the original conservation of energy and attempt to apply it to a fluid. You are quickly presented with the issue of how to deal with the "mass" of a fluid. Fluids aren't typically characterized by a mass per se but instead a mass per characteristic volume otherwise known as a density. Your next guess might be to maybe talk about a total mass over a volume. $$ \frac{1}{2}\frac{m}{V}v^{2} + \frac{m}{V}gh + \frac{W}{V} = \frac{E}{V} $$ $\frac{m}{V}$ is the density in terms of a unit volume, leading us to a work per unit volume adding to an energy per unit volume. $\endgroup$
    – user28754
    Commented May 23, 2014 at 3:55
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You cannot derive the classic Bernoulli Equation from conservation of energy, because, contrary to popular opinion, it is actually not an expression of conservation of energy at all. It is more accurately construed as an integrated expression of the conservation of linear momentum, $F=ma$.

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