I'll try.
I think this latest discussion is all due to the fact that, to find total pressure drop, either the inlet or the outlet pressure must be known and its sometimes difficult to separate the theoretical pressure drop of an individual element (or sum of several) from that practical reality. P1 - Cv dP^2 = P2
Now let P1 and P2 float. Note that it can be Cv for a valve, k for a pipe, maybe L*(r1^2-r2^2)*Q for a reducer, or whatever for any individual flow element, but the sum of all of them only ever equals dP for the system of elements, until you specify either P1 or P2. Even if flow is known, either P1 or P2 must also be known to solve the hydraulic system equations.
Even with a butterfly or sandwich check valve bolted to the tank, theoretically there is still an infitesimal length of pipe both upstream and downstream of the valve's "orifice".
OK, let's take it another step farther and consider just a hole in the side of a tank.
Now, theoretically and practically, inside the tank itself, as a fluid particle goes from rest away from the tank outlet to some infitesimal velocity, there is the transformation from static to dynamic head, and there is also the exit flow coefficient, Cd, to deal with that drops the hydraulic gradient as a fluid particle drifts from an area under the influence of only the liquid surface static pressure, accelerates and moves into the tank outlet and out into free air or into the inlet just before arriving at a valve's orifice. That loss can be significant, depending on shape and form of the outlet at the tank wall, a nozzle protrusion, [] or O x-section, curved or square, sharp or blunt edge, etc.
But, as soon as you say, "constant level"... you've fixed the upstream pressure and can then calculate downstream pressure, or v/v.
Was that it?
BigInch
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-born in the trenches.