Bernoulli's Principle -- correct derivation

[Wisemetrics]

Homework Statement

I've tried looking around for a proof of Bernoulli's principle online, but in almost all cases I find a proof similar to the one shown here: https://www.toppr.com/guides/physics/mechanical-properties-of-fluids/bernoullis-principle/ In the proof the author states that the change in energy must be equal to the amount of work done on the fluid which is a reasonable statement but they also say that this work must be equal to $W_2-W_1= P_2 A_2 v_2 t - P_1 A_1 v_1 t = \Delta P \Delta v$ . This is the part that I'm not entirely okay with. I'll put my reasoning behind why I think this is incorrect in the section below, but I wanted to know if this argument is actually logically justified and if not what is the more mathematically complete justification of it.

Relevant Equations

$W_2-W_1= P_2 A_2 v_2 t - P_1 A_1 v_1 t = \Delta P \Delta v$

In this scenario I'm assuming that there is a shared velocity of water within the pipes, as well as a shared pressure and that water is non-compressible. If I understand correctly when someone says that pressure at a point is P at some point, it is the same as saying that if I put a small cube with side length A, the amount of force exerted on it from each side will be equal to PA^2 and result in a zero net force overall (assuming the pressure is equal at the nearby points). If we were to look at the problem and assume that the amount of pressure is constant through the first and second portions of the liquid then the amount of work done on the liquid would be 0 in both cases (since the net force is 0) which would make their argument invalid. This leads me to believe that the change in pressure is not negligible and is probably the key to deriving Bernoulli's principle, but this is never explained in the proof. I think some theorem from calculus would have to be involved to make the proof work, but I'm not certain which one.

 

[haruspex]

assume that the amount of pressure is constant through the first and second portions of the liquid then the amount of work done on the liquid would be 0 in both cases

Since the pressure does change along the streamline, the pressure will not, in general, be exactly the same at both ends of a small element. It only tends to the same as the element length tends to zero.
But what is perhaps not clear is that the writer is considering the movement of the fluid in the whole BD section to being along CE. The pressures on this are the pressures at B-C and at D-E.

 

[Chestermiller]

But what is perhaps not clear is that the writer is considering the movement of the fluid in the whole BD section to being along CE. The pressures on this are the pressures at B-C and at D-E.

The time interval $\Delta t$ is supposed to be very small. This set-up is the standard approach used in developing the open system (control volume) version of the first law of thermodynamics. See Smith and van Ness, Introduction to Chemical Engineering Thermodynamics. W1 and W2 the amounts of work being done over the time interval $\Delta t$ in pushing fluid into the control volume (by the material behind acting like a piston) and pushing fluid out of the control volume (against the pressure of the fluid ahead acting like a piston). The system is assumed to be operating at steady state.