Consider this picture with pump on left hand side with pressure P1 (or some very big container with water, so big that its volume is not changing) and right hand side is open to air (pressure PA). According to Bernoulli equation (P1-PA)=1/2rhov^2, where rho is density and v is maximal outlet velocity. So maybe stupid question, but what is P2 and v2 on the picture? Is P2=(P1-PA) and v2=v ? Then, what happens when we push piston against water in horizontal part (where was P2)? PA should remain the same so P2 is reduced and then also v is reduced, which implies the flow rate Q is also reduced ? If so, is our piston in fact pressure reduction valve ?


1 Answer 1


or some very big container with water, so big that its volume is not changing

Yes, let's do that: imagine the left-hand side of the pipe to be connected with an infinitely large reservoir.

The fluid is considered both inviscid and incompressible.

Let's also make the height difference $h_1$ between the surface of the reservoir and the centre line of the (horizontal) pipe, such that with Pascal's Law we have:

$$p_1=p_A+\rho g h_1$$

Now we imagine a flow line that connects a point on the surface of the reservoir with a point right on the centre of the pipe and at its entrance. We can apply Bernoulli and because the flow speed $v_1$ at the surface of the reservoir is $\text{zero}$, so that with Torricelli's Law (basically a simple application of Bernoulli's principle) for inviscid fluids we get for the entrance speed $v_{ent}$:


Because the fluid is incompressible and the pipe's cross-section is constant, then per the continuum equation the flow speed is constant along the flowlines of the entire length of pipe. I'll call that velocity $v_2(=v_{ent})$.

Now imagine a flow line going from the centre of the entrance of the pipe to the top of the manometer tube (where you write $h_2$ and $p_2$). $p_2$ (top of the manometer tube) is of course $p_A$ and the flow velocity is $\text{zero}$ at that point.

We apply Bernoulli between these points:

$$p_1+\frac12 \rho v_{2}^2=p_A+\rho g h_2$$ $$p_A+\rho g h_1+\frac12 \rho v_{2}^2=p_A+\rho g h_2$$ $$\rho g h_1+\frac12 \rho v_{2}^2=\rho g h_2$$ So that: $$\boxed{h_2=h_1+\frac{1}{2g}v_{2}^2}$$

Due to the nature and spirit of the question I've answered it using 'typical' approximations, used to approach such problems.

If one inserts the value of $v_2$ into the last equation, one obtains $h_2=2h_1$, which doesn't sound realistic.

There are two approximations in particular which deserve some scrutiny and both derive from the inviscidity assumption.

Firstly, when a viscous fluid enters a pipe the way it does in this problem, then the sudden contraction of flow causes pressure loss ('head loss', for engineers). This reduces $v_2$ somewhat.

Secondly, flow of a viscous fluid through a pipe, whether laminar or turbulent also causes pressure loss. So by the time flow reaches the point $2$, there is further loss of pressure.

If the sum of both pressure losses is written as $-\Delta p$, then the equation for $h_2$ could be rewritten as:

$$h_2=2h_1-\frac{\Delta p}{\rho g}$$

But it would be impossible to estimate this pressure loss without additional data such as fluid viscosity, pipe diameter, pipe length, position of point $2$ and pipe relative roughness.

  • $\begingroup$ Not sure what you mean there? $\endgroup$
    – Gert
    Sep 4, 2021 at 20:22
  • $\begingroup$ OK, I agree with your answer until We apply Bernoulli. When pressure on the left hand side is P1 and we apply Torricelli's law, then Vent is like you said. I agree with that, but should not be then across pipe line the pressure just atmosferic one , which mean h2 is zero ? Where I have mistake ? Because if p1 is add to 1/2*rho*v^2 than the energy is somehow bigger than energy of water in our reservoir, or not ? $\endgroup$
    – NiobTitan
    Sep 4, 2021 at 20:33
  • $\begingroup$ No, the pressure at the bottom of the mano is $> p_A$ for sure. But $h_2$ is overestimated due to the neglecting of the head loss at the entrance of the pipe and the head loss due to viscous flow (see Darcy-Hasselbach equation e.g.). The problem doesn't make a lot of sense if we oversimplify it... $\endgroup$
    – Gert
    Sep 4, 2021 at 20:46
  • $\begingroup$ Then I think, you don't have it right, sorry. Then the P2 (at the bottom of mano) should be as I said in OP (P1-PA), which means h1*rho*g. Or something between PA and P1, i don't know, that's the point of the OP. Maybe it depends on which type of tube is opened first, if is it outlet or mano tube at the bottom... However I'm quite sure (of course we are neglecting losses) that h2 cannot big higher than h1... $\endgroup$
    – NiobTitan
    Sep 4, 2021 at 21:02

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