# 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.

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:

$$\frac{dP}{dx}=-\frac{1}{2}\rho\frac{d(v^2)}{dx}=-\rho v\frac{dv}{dx}$$ 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.
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) $$u_2$$.