Suppose we have the following scenario:

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A very simple case where we have a pipe going into a large reservoir (filled with water) and a pump pushing the water outwards (from left to right).

What first comes to mind is to apply Bernoulli's equation which in the case of there being a pump is expressed as: $P_1 + P_p + \frac{1}{2} \rho v_{1}^2 + \rho g h_1 = P_2 + \frac{1}{2} \rho v_{2}^2 + \rho g h_2$

Here $P_p$ is the pressure difference caused by the pump. We can simplify the previous equation as just $P_p = \frac{1}{2} \rho v_{2}^2$ as there is no height difference, no pressure difference ($P_1 = P_2 = P_{ATM}$) and the initial velocity at the reservoir is $0$.

So far so good until I try to find out the force exerted by the pump, in this case I would say that $F_p = P_p A_p = \frac{1}{2} \rho A_p v_{2}^2$ and by applying conservation of momentum the force exerted by the water is $|F_w| = \dot{m}v_2 = \rho A_2 v_{2}^2 = |F_p|$ which implies that $A_p = 2 A_2$.

This is the part that doesn't make sense, it looks like the area where the pressure of the pump is applied must be twice of that of the exit area as if I'm "forced" to place the pump in a specific region. Perhaps what is going on is that the pump exerts a force elsewhere (where the area is $2A_2$ which would be in the slipstream in the reservoir, kind of like an axial fan) but then there wouldn't be a pressure gradient across the actual location of the pump. Clearly, I'm missing something.


  • 1
    $\begingroup$ The silent downvote, how charming $\endgroup$
    – rr1303
    Sep 10, 2022 at 15:21
  • $\begingroup$ Those make me crazy as well $\endgroup$
    – basics
    Sep 10, 2022 at 17:01
  • $\begingroup$ The blades of the pump supply the missing force. The liquid exerts a reaction force on the pump blades. This is very similar to what happens with a convergent nozzle. The walls of he nozzle exert a backward force on the flowing fluid. $\endgroup$ Sep 12, 2022 at 11:46
  • $\begingroup$ @ChetMiller Interesting, now if the power is $P = \frac{dW}{dt} = \frac{1}{2} \dot{m} v_2^{2}$ and the pump power is $P = F_p v_p = \dot{m} v_2 v_p$, applying conservation of energy we get that $v_p = \frac{v_2}{2}$ therefore the blades spin half as fast as the water that traverses it? $\endgroup$
    – rr1303
    Sep 12, 2022 at 15:02
  • $\begingroup$ @ChetMiller I must have got that wrong, $v_p = \frac{v_2}{2}$ has nothing to do with the speed of the blades, it's really just the average speed of the water which is where the work is being done. That said, if it is the walls (and nozzle) what contribute to the missing force then does that mean that the real force the pump "feels" is $F_p = \frac{1}{2} \rho A_p v_2^{2} = \frac{1}{2} \rho A_2 v_2^{2}$? $\endgroup$
    – rr1303
    Sep 12, 2022 at 21:19

2 Answers 2


Your determination of the pressure rise across the pump is correct: $$P_p=\Delta P=\frac{1}{2}\rho v_2^2$$The horizontal momentum balance on the pump alone reads as follows:$$-\Delta P_pA_2+F_p=\rho v_2A_2(v_2-v_2)=0$$or$$F_p=\frac{1}{2}\rho v_2^2A_2$$assuming that the inlet and outlet pipe areas to the pump are both $A_2$. This horizontal force that the pump exerts on the water flowing through it includes the force of the water on the pump housing plus the force of the water on the pump blades. This is opposite in sign to the net horizontal force the water exerts on the pump (which is in the negative x-direction).

  • $\begingroup$ Thanks! For completeness, the macroscopic force acting on the pump and pipe (all as one single mass) should be $F = \dot m v_2 = \rho A_2 v_2^{2}$ whereas the internal force of the pump is half of that (in this case). One could therefore say that using pipes in specific configurations allows for force amplification (or reduction) kind of like a gearbox? $\endgroup$
    – rr1303
    Sep 13, 2022 at 16:15
  • $\begingroup$ Oh really? At the very entrance to the pipe and pump (all as a single mass), do you think that the pressure and velocity are $p_{atm}$ and 0, respectively, or are they $p_{atm}-\rho\frac{v_2^2}{2}$ and $v_2$? $\endgroup$ Sep 14, 2022 at 0:17
  • $\begingroup$ Ah, you’re right, now it makes sense. You’ve just earned yourself the green check mark :) $\endgroup$
    – rr1303
    Sep 14, 2022 at 8:34

I don't think your model is consistent. Neglecting all the effects of viscosity and assuming water as an incompressible fluid with constant density, in the straight horizontal pipe with constant area, the only effect of the pump it's a pressure increase across it, while velocity is uniform for mass balance. If it's so, $P_2>P_1= P_{atm}$, and I guess that the streamlines of the water jet are not straight but "expands" forming a qualitatively conical shape, at the end 2 of the pipe, because of the pressure difference $P_2 >P_{atm}$.

This is a very qualitative description of the problem, while for a more detailed description of the flow field and the pressure field I guess that you need some more detailed model (likely, with numerical solution) or measurements.

  • $\begingroup$ Thanks for the answer, I thought of the expanding (or contracting) idea at the end of the pipe but we would still have $A_{exit} = \frac{A_p}{2}$ which means that if $A_p$ is very big or very small we should see some radical change of area as the water exists the pipe, something which I doubt resembles reality. $\endgroup$
    – rr1303
    Sep 10, 2022 at 15:36
  • $\begingroup$ You could always have a try $\endgroup$
    – basics
    Sep 10, 2022 at 17:02

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