# Where will pressure be stronger in these 2 pipe scenarios?

I initially started to be interested in fluid pressure and pipes to learn electricity. It seemed the only way I could help my head to get around some electrical concepts.
That's the reason of my previous question.
Anyway I would like to understand how fluid molecules behave in pipes on a deeper level, at least as much as I can, given my poor background.
I drew this 2 pictures, hoping that they will help to illustrate what is a doubt that I currently have.

Picuture 1

Picture 2

I imagined 2 vessels connected by a pipe. The highest vessel is coninuosly alimented with new water. Water in excess flows down on the sides.
Water will try to reach the same height in both vessels, so some water will pour out of the shortest one and will be collected by the water collector.
The 2 figures are identical apart from the shape of the pipe distance d:

• In Picture 1 the pipe distance d starts with a restriction (from R1 to r2), then returns to the same diameter(R1) and then is restricted again(r2).
• In Picture 2 the pipe distance d is restricted (from R1 to r2) from the beginning to the end.

Given that we know all we need to know to calculate the pressure at point A, B and C ( Picture 1 ) and A' and C' ( Picture 2 ). I would like to know the difference in pressure between C and C', if there's any and why it's there. I'd love to hear an answer that explain everything in terms of water molecule's behaviour, if it's reasonable.

Edited the original version to make a clear distinction between pressure, small p, and force of pressure, F and to define resistance in terms of force of pressure rather than pressure.

The pressure at point C is approximately determined by the height in the right vessel. Since the height of the vessels in the two examples are the same, the pressure at point C is going to be the same as well.

What is going to be different is the flow rate.

The flow through a length of pipe could be approximately determined using the Poiseuille's law:

$Q=\frac {\Delta p \pi r^4} {8 \eta L}$

This law relates the flow rate Q to the pressure gradient (or difference) $\Delta p$, viscosity $\eta$, length of the pipe L and the radius of the pipe r.

We can rewrite this equation as follows:

$Q=\frac {\Delta p (\pi r^2) r^2} {8 \eta L}=\frac {\Delta p A r^2} {8 \eta L}=\frac {\Delta F r^2} {8 \eta L}=\frac {\Delta F} {R}$, where A is the cross-section area of the pipe, $\Delta F$ is the difference in the force of pressure and $R=\frac {8 \eta L} {r^2}$ is the resistance to the flow.

You can see that this law is analogous to the Ohm's law, $I=\frac {\Delta V} R$, where the force of pressure corresponds to the voltage and the flow rate corresponds to the current. In fact, for analysis of the flow in the pipes, we can use well familiar techniques developed for electrical circuits.

Looking at the fist case, for simplicity, we can neglect the resistance of the pipe in the sections A, B and C, where the radius $r_1$ (I'll use small r to differentiate radius from resistance) appears to be relatively big, and focus on the resistance of the pipe in the sections $d_1$ and $d_2$, $r_2$, with corresponding resistances, $R_1$ and $R_2$.

Since these two resistances are connected in series, the total resistance would be equal to their sum: $R=R_1+R_2$ or $R=\frac {8 \eta (d_1+d_2)} {r_2^2}$.

In the second case, there is only one narrow section and its resistance is $R'= \frac {8 \eta d} {r_2^2}$.

Since $d>(d_1+d_2)$, $R'>R$ and, since the difference in pressure force $\Delta F=F_1-F_2$ (shown as $\Delta P=P_1-P_2$ on the diagram) is the same for both cases, the flow rate in the first case will be greater then in the second case.