I'm working on a certain problem in fluid mechanics, which isn't really my strongest area. The problem is as follows: Curved pipe is partially submerged in a flowing river so that one end is pointing in the same direction as the velocity of river. Level of water inside of the pipe is 7cm higher than the level of river. Determine the speed of river.

Basically what I argued is that this reduces to Torricelli's law: level of water in pipe is constant so the velocity of the surface is zero (or very nearly so, I assume it would oscillate in real life), therefore water should move on the other end with $v_{2}=\sqrt{2gH}$, but it isn't since $v_{2}=v_{river}$ counteracts the movement.

I've tried to be more rigorous so I took pressures at the submerged end of pipe: $$p_{\text{water in pipe}}=p_{\text{atmosphere}}+\rho g h_{\text{depth}} + \rho g H_{\text{above water}}$$ $$p_{\text{river}}=p_{\text{atmosphere}}+\rho g h_{\text{depth}} + \frac{1}{2} \rho v^{2}$$

Since they have to be in equilibrium, pressures are equal and you get Torricelli's expression above.

Is my reasoning correct? It's somewhat counter-intuitive to me because the water is moving away from the pipe.

  • $\begingroup$ "Ceci n'est pas un pipe" OK, seriously, somehow your result has to be consistent with self-bailing boats, where the boat's motion thru the water pulls water out of the boat. $\endgroup$ Jan 6, 2014 at 18:09
  • $\begingroup$ @CarlWitthoft I believe they use openings that allow for water to go out, but not go in. Edit: Very appropriately, yes, "Ceci n'est pas un pipe" for the boats. $\endgroup$
    – user36875
    Jan 6, 2014 at 18:13
  • $\begingroup$ Looking at a pitot tube-- I can't imagine a static high pressure point at a trailing edge, but I'm far from expert on fluid dynamics. (My jab about self-bailing is that if a hole facing behind a moving boat produced a higher pressure, then water would in fact enter the boat's hull. This doesn't happen. And self-bailers need not have a flap valve) $\endgroup$ Jan 6, 2014 at 18:22
  • $\begingroup$ Naturally, water doesn't go in, but water pressure is higher outside because the level of water outside is always (well, unless you have sunk) higher than inside the ship. I suspect they use some sort of flaps which are held shut by higher water pressure outside and open only when ship accelerates and inertia of water overcomes the force keeping the flap shut. $\endgroup$
    – user36875
    Jan 6, 2014 at 18:34
  • 1
    $\begingroup$ In addition to using the link for "back up", a picture is worth 1000 words. $\endgroup$ Sep 5, 2015 at 14:35

1 Answer 1


If you consider the following graphic:

The second one is your scenario, but let's consider the top one first. According to Bernoullis equation, $$ H=z+\frac{p}{\rho g}+\frac{v^2}{2g}=\mathrm{const}.$$ At the tube inlet: $H_\mathrm{left}=H_\mathrm {right}$, where $z$ is the geodesic height, $\frac{v^2}{2g}$ is the energy height, and $\frac{p}{\rho g}$ is the pressure height.

The geodesic height cancels out on both sides. But the energy height from the flow of the river presses against the tube inlet, which has to be balanced by raising the water level of the tube: $$ \frac{v^2}{2g}=\frac{p}{\rho g}=\Delta h $$ Thus $\Delta h$ is positive.

Now for the second scenario where the tube inlet faces in the direction of flow, the energy height is negative, because it "pulls" on the tube inlet, which consequently causes the water level in the tube to lower. $$ -\frac{v^2}{2g}=\frac{p}{\rho g}=\Delta h $$ Thus $\Delta h$ is negative.

I would conclude, that if the tube is pointing in the same direction as the velocity the water inside the pipe should actually be lower than the water surface.

  • $\begingroup$ The second part above is wrong. There is no "pulling" on the tube inlet. $\Delta h$ depends on the pressure at the pipe outlet, which will typically be higher than the pressure at the same level away from the pipe. How much higher, however, depends on the structure of the wake region that will develop at the blunt end of the pipe. Assuming ideal flow, however, it will be equal to the stagnation pressure, and thus the same as for a pipe facing the flow, meaning the original analysis is approximately correct. $\endgroup$
    – Pirx
    Nov 22, 2016 at 12:50
  • 1
    $\begingroup$ @Pirx Shouldn't the wake region created by the bend in the pipe make the pressure lower when the outlet is facing away from the flow? $\endgroup$
    – JMac
    Apr 5, 2017 at 22:50
  • $\begingroup$ But the problem says the first scenario. Why do you say it is the second one? $\endgroup$ Jul 27, 2018 at 3:50
  • $\begingroup$ @AaronStevens In the OP it says «[…] one end is pointing in the same direction as the velocity of river.» I interpreted this to be equivalent to the second scenario in my answer. $\endgroup$
    – Andrew
    Jul 27, 2018 at 10:02
  • $\begingroup$ "Level of water inside of the pipe is 7cm higher than the level of river. " I see the issue though. Maybe the problem just means parallel to the velocity? $\endgroup$ Jul 27, 2018 at 10:03

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