I'm having trouble understanding Bernoulli's principle, in particular, why an increase in the section area of a hose increases the pressure?
All the answers I read say: " In order for the energy to be conserved" or "when the tube is compressed, the fluid has to speed up so that the same amount of the fluid gets out of the tube in the same interval of time (continuity law)".
I'm not looking for the finality : energy conservation. I'm trying to understand the mechanism, what happens at the molecular level?
Best Answer
Bernoulli's Principle belongs to continuum mechanics, so it's not well suited to make pronouncements about action at the molecular level (but I'll get back to that further down).
Between two points on the same flowline Bernoulli states:
$$P_1+\frac12 \rho v_1^2+\rho gz_1=P_2+\frac12 \rho v_2^2+\rho gz_2$$
The case below is for an inviscid, incompressible fluid, with no potential energy changes ($z=\mathrm{constant}$) and regularly shaped conduits (the cross sections $A_1$ and $A_2$ are well defined):
Then:
$$P_1+\frac12 \rho v_1^2=P_2+\frac12 \rho v_2^2$$
The relationship between the flow speeds is given by incompressible continuity:
$Q_v=A_1v_1=A_2v_2=\mathrm{constant}\implies v_2=\frac{A_1}{A_2}v_1$
So that:
$P_2=P_1+\frac12 \rho (v_1^2-v_2^2)$
$P_2=P_1+\frac12 \rho \Big(1-\frac{A_1^2}{A_2^2}\Big)v_1^2$
Thus: $\boxed{A_2>A_1\implies v_2<v_1\implies P_2>P_1}$
So what you call the 'mechanism' is entirely due to satisfying the continuity requirement. Counter-intuitively perhaps, here a decrease in flow speed $v$ results in an increase in pressure $P$.
At the molecular level, lower bulk fluid pressure is caused by increased average distances between molecules, as these increases decrease Coulombic repulsions between the electron clouds that make up the fluid's molecules. This results in lower numbers of collisions with the counduit's wall and thus lower pressure.