I have a doubt on the use of Bernoulli equation for pumps. Consider the situation in the picture.

enter image description here I marked different points: $1$ on the surface of first tank, $2$ in the exit from first tank, $3$ just before the pump, $4$ just after the pump and $5$ entering the second tank.

Now consider Bernoulli equation in the "normal form" (ignoring the pump)

$$p_a+\frac{1}{2} \rho v_a^2 +\rho g h_a=p_b +\frac{1}{2} \rho v_b^2 +\rho g h_b\tag{1}$$

And in the form for the presence of pump delivering power $\mathscr{P}$

$$(p_a+\frac{1}{2} \rho v_a^2 +\rho g h_a) Q +\mathscr{P}=(p_b +\frac{1}{2} \rho v_b^2 +\rho g h_b) Q \tag{2}$$

$a$ and $b$ are two generic points among the ones listed above.

My question now is: can I use $(2)$ between any point before the pump and any point after the pump, regardless the height, velocity and pressure in such points?

I have this doubt because usually one takes point $1$ and $5$ and uses $(2)$ - and I'm ok with that- but, if the answer to previous question is yes, I could also choose to use $(2)$ between $1$ and $4$ or $2$ and $5$ or $2$ and $3$ and so on and that sound strange because the quantity $p+\frac{1}{2} \rho v^2 +\rho g h$ should be the same before and after the pump, indipendently from the particular point chosen. In other words I should be able to use $(1)$, normal Bernoulli equation, between $1$ and $2$, which is not very realistic, since the fluid in $2$ will probably move with a velocity that is influenced by the pump.

That is, even if $2$ is before the pump, the velocity there is different from the situation with no pump. And that's what I cannot understand here. How is that possible? And can I use $(1)$ between $1$ and $2$?

Any suggestion is highly appreciated.


closed as off-topic by CuriousOne, ACuriousMind, user36790, honeste_vivere, Gert Jun 26 '16 at 22:38

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  • 2
    $\begingroup$ Without the pump, the velocities at points 1 and 2 are both zero. With the pump, the velocities are both nonzero. So I see no contradiction here. $\endgroup$ – knzhou Jun 24 '16 at 21:59
  • $\begingroup$ @knzhou Thanks for the reply! If I understand what you are saying, I can use eq. $(1)$ between $1$ and $2$ both with and without the pump (in eq $(1)$ the power $\mathscr{P}$ does not appear), the only things that change are the velocities $v_1$ and $v_2$. Is that possibly correct? $\endgroup$ – Sørën Jun 24 '16 at 22:11
  • 1
    $\begingroup$ Yes. (I mean, as long as Bernoulli's equation itself is still true. If the pump is too powerful and the flow too fast, it'll break down, no matter how or where you apply it.) $\endgroup$ – knzhou Jun 24 '16 at 22:13

You need to get $p_4-p_3$. Taking the datum of elevation z as that of points 3 and 4, we have

$$p_{atm}+(10)\rho g=p_2+\frac{1}{2}\rho v^2$$ $$p_3+\frac{1}{2}\rho v^2=p_2+\frac{1}{2}\rho v^2$$ $$p_{atm}+(120)\rho g=p_5+\frac{1}{2}\rho v^2+(120-h)\rho g$$where h is the depth of point 5 below the surface of the tank on the right. $$p_4+\frac{1}{2}\rho v^2=p_5+\frac{1}{2}\rho v^2+(120-h)\rho g$$ If we combine these equations, we obtain: $$p_4-p_3=(120-10)\rho g$$

Power = $(p_4-p_3)Q$


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