Static pressure intuition-Why does the local static pressure change as the local flow rate changes? From Bernoulli ' theorem, we know that the local static pressure changes as the local flow rate does. But why should the static pressure change anyway, intuitively ? What really makes that happen ? Consider incompressible and isentropic flows.
 A: As Munson puts it in his book "Fundamentals of Fluid Mechanics", the work done on a particle is equal to the change of its kinetic energy. It's the same principle used in Newtonian mechanics in high school! In the absence of non-conservative forces(say friction), energy is conserved. It's just that in the analysis of fluids you go from a microscopic analysis of each particle to a more macroscopic picture of parts of a fluid.  
So, applying the same logic, say you have a cylindrical, horizontal pipe with frictionless walls:  
[Image from Khan academy]  
Along the streamline that runs across the middle of the pipe, the equation that is one step away from Bernoulli's formula reads(gravitational potential energy is the same along the middle streamline):
$\frac{dP}{dx}=-\frac{1}{2}\rho\frac{d(v^2)}{dx}=-\rho v\frac{dv}{dx}$ where $x$ is the parameter that runs accross the streamline.
Since $u>0$ (it's the length of $\vec{u}$, the velocity vector), the above equation tells us that the fluid moves in the direction in which pressure decreases. This is a restatement of Newton's second law in "fluid language"!
Thus, on the left cross-section, if the fluid has pressure $P_1$ and velocity $u_1$ and on the right cross-section it has $P_2$ and $u_2$, intuition(and the continuity equation) tells us that $u_2>u_1$ and so we expect $P_1>P_2$ since(from the above logic) the net force that must accelerate the fluid from $u_1$ to $u_2$ must have a direction to the right.  
So, this is how the local pressure and local fluid velocity are connected. It's a "simple" relation of cause and effect, the same relation found in Newton's mathematical statement of his second law.  
To answer your question more clearly, if the velocity on the right cross-section changes, then also the pressure there must change in order to have the right net force that drives the fluid from $u_1$ to (the new) $u_2$.
A: The movement of a flow (of a gas or a fluid) over a surface, which has the same temperature as the flow standing still, causes a decrease in pressure on the surface because in both cases the temperature stays the same (which means the average velocity of the molecules or atoms stays the same), and adding a velocity to the particles (of the gas or fluid) parallel to the surface means that because the average velocities of the particles stay the same, the velocities perpendicular to the plane are decreased, causing a decrease in the pressure. If the velocities of the particles have a very high value parallel to the plate, then there are almost no components of the velocities of the particles perpendicular to the surface left to impart pressure to the plate (for the temperature to stay the same the velocities of the particles must stay the same, so in the moving flow, the particles have a greater parallel velocity and for the temperature to stay the same a smaller component perpendicular to the surface). Almost all the molecules or atoms have a velocity (because the temperature stays the same as the non-moving flow) almost parallel to the plate (of course there is always a little velocity component normal to the surface, which gets smaller as the flow velocity increases).  
