Thats my point. Bringing back the RPM is also changing the speed of the fluid as well as the total amount of mass that goes by.
Imagine having 100 air molecules going by at one point in the manifold per second. You increase the RPM, now 200 air molecules are going by per second at twice the speed they did before. That does not cause a decrease in static pressure because you changed the mass flow.
Now on the same 100 molecules per second manifold place a venturi tube and you will get the same 100 air molecules going by per second (trough the venturi) at a greater speed but trough a smaller section. The mass flow remains the same but the speed increases and that causes a decrease in static pressure. It could be wrong, but this is how I always though Bernoulli's principle applied.
I apologise if I'm wrong about this one. I'm here to keep learning.
Ok, lets look at it from an airfoil standpoint. If we have a wing moving through the air at 100 kias, there are X air molecules passing over the wings surfaces per second- we'll call this a mass flow of X kg/s. At X kg/s, there is a suction force exerted on the upper wing surface to the tune of, well say, S psi, because the static pressure is less than ambient static pressure (which we can assume given Bernoulli's principle), which is, we'll say, Apsi. We'll say the static pressure above the wing surface is (A-S)psi, or Ambient minus Suction, measured in psi. Now, we're going to decrease our airspeed to 70 KIAS. At 70 KIAS, there are Y fewer air molecules pass over the wing surfaces per second, so mass flow now equals (X-Y) kg/s. Apsi is the same, its the ambient air pressure. The suction force, because of bernoulli's principle, must decrease, so we'll say our suction force is now (S-T)psi, where T is the amount of suction decrease. The static pressure force acting above the wing surface has now increased to {A-(S-T)}psi, the resultant increase in static pressure causes a net loss of lift.
Now, we'll take it back to the RPM scenario. Our mass flow, X kg/s, is moving through the induction system at a velocity Vm/s when RPM is, we'll say, 2700 (We'll use this scenario: The moment after takeoff in a seminole when transitioning to cruise climb, the throttle has already been retarded to the cruise climb setting, the prop hasn't yet). At X kg/s, the MP gauge is indicating, we'll say, 24". Now, we will decrease the RPM to 2500. The airflow moving through the system must slow down, because there's less intake strokes happening per second than before, air is being drawn in at a slower rate, so our new fluid velocity will be (V-W)m/s. This decreased fluid velocity will contribute to a decrease in overall mass flow, which will now be (X-Y) kg/s. The same thing is happening here as is happening when you slow a wing down, and that is that as the velocity slows down, some of the suction force goes away, as will be evidenced by the manifold pressure gauge...if there is less suction, the manifold pressure will be closer to ambient, and in this case, should read about 25 inches now.