# 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.

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.

Below are the diagrams for the two cases and their the electrical analogs. The pumps are added as more direct analogs to the batteries: you'll need pumps to maintain the pressure in the vessels as the water leaves the left vessels and flows to the right vessel.

It is worth to note that, although it may appear to be logical, the analogy between the flow of fluids in the pipes and electricity is very limited and should not be extended beyond the simplest cases.

• Thanks for helping me understand! I don't undersand why there isn't a different "pressure drop" in the 2 cases since resistance is different. I would expect the pressure at C (picture 1) to be higher than the pressure at C' (Picture 2), since resistance is lower. That would match electricity behaviour, wouldn't it? – Gabriele Scarlatti May 17 '18 at 14:05
• If you want to draw a parallel with electricity, each of the vessels would be replaced by a battery. The voltage at terminals of an ideal battery is determined by the battery's voltage only. Similarly, the pressure at the bottom of each vessel is determined by the height of the corresponding vessel only. – V.F. May 17 '18 at 14:41
• Isn't the flow of molecules from the left vessel to the right one always? That means tha we have a potential difference between the terminals, if we add a pump that push water from the collector to the high vessel we get a complete circuit, and the difference in potential is kept constant. – Gabriele Scarlatti May 17 '18 at 14:49
• Assuming the tank heights are equal and the diameters of the pipes at C and C' are equal, the pressures at C and C' are equal by Bernoulli's equation. The driving force for the flow is provided by the height difference of the liquid surface between the tank on the left and the one on the right. – Mephistopheles May 17 '18 at 14:51
• @Mephistopheles, I don't know but it seems that something it's wrong in how I think of it. If pressure at C and C' is given by the right vessel and we call them P, then if we increase resistance from A to C ( smaller pipe), It seems logical that will be reached a point where flow will be reversed since pressure form the right vessel is decreased – Gabriele Scarlatti May 17 '18 at 14:58