0
$\begingroup$

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.

$\endgroup$

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

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – CuriousOne, Community, Gert
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 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
1
$\begingroup$

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 = $(p_4-p_3)Q$

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.