Doubt in Bernoulli's theorem proof When proving Bernoulli's theorem, we say that the total work done by external forces on a mass of fluid going from A to B (at different heights) is the work of gravitational field plus the work of pressure forces at the endpoints. Then we can say that this is equal to the variation of kinetic energy of the mass of fluid in point B and in point A. But why do we consider the work done by pressure forces only at the endpoints? Don't they do work on the mass of fluid even when it's travelling in the tube, and the only work done there is only the gravitational one? If they don't, then what happens when we have an horizontal tube?
 A: Sorry for my poor english. My native language is french.
When you write a kinetic energy balance, you actually should have to take into account the work of external forces and of the inner forces.

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*For the external forces, you have the pressure forces at the ends and gravity. The forces on the side faces of the tube do not work because the fluid is assumed to be without viscosity.

*For internal forces: by assumption, no viscosity. Only the pressure forces remain. But since the fluid is assumed to be incompressible, the work of these internal pressure forces is zero since the volume of a fluid particle does not change.

For a perfect and incompressible fluid, there is no conversion of mechanical energy into internal energy and the two can be treated separately. It is possible to obtain Bernoulli's theorem by a balance of momentum rather than a balance of mechanical energy. As we can get the Kinetic Energy Theorem from Newton's law for a point mass.
A: If one end of the fluid has a force $F_1$ acting on it from one end, and the other end has a force $F_2$ acting on it, then the net force is $F_1-F_2$, and the net work done by the fluid force is $W=(F_1-F_2)\Delta x$, where $\Delta x$ is the displacement of the section of fluid along the fluid.
If you want to look at the work per volume, this is where pressure arises:
$$\frac WV=\frac{(F_1-F_2)\Delta x}{V}=\frac{F_1-F_2}{A}=P_1-P_2$$
where $A$ is the cross-sectional area of the fluid.$^*$
So, the difference in pressure does account for the total work done.
If you are still unconvinced, imagine a simpler example where you are just pushing a block on a frictionless surface. Of course, you could break the block into small subsections, and say that your force does work on the first subsection, which pushes on the second subsection and does work, etc., But at the end of the say you can just look at your pushing force; there is no need to think about all of the smaller internal forces.

$^*$This can be generalized to non-constant cross-sections, but the idea is still the same.
