1
$\begingroup$

I am struggling a bit with the concept of siphon. Firstly, if we consider a pipe filled with water, the inlet is in a tank of water at the altitude $z_I$, and the outlet is in the air at $z_O$.

Here is a sketch of the problem:

enter image description here

Where $S$ is the interface water-air in the tank, so $z_S$ is the altitude of the interface, $T$ is a point in the highest area of the pipe and $H$ is the distance between the interface and the top of the column water above $I$.

1) I am trying to understand first, the condition where the velocity at the outlet is positive, that is to say, the siphon works, the draining of the tank is on. I used the Bernoulli theorem between the inlet $I$ and the outlet $O$:

$P_I + \frac{1}{2} \rho v_I^2 + \rho g z_I = P_O + \frac{1}{2}\rho v_O^2 + \rho g z_O$

To evaluate $P_I$, I made the hypothesis that at $t = 0$ s, the water is at rest in the pipe and the tank. Then I used the hydrostatic law and said that the pressure at $I$ is the atmospheric pressure + pressure due to the weight of the water column above $I$:

$P_I = P_{atm} + \rho g (z_S + H - z_I) $

Then the simplification of the Bernoulli theorem lead to:

$V_O^2 = 2g[z_S + H - z_O]$ and $ V_O > 0 \Leftrightarrow z_S + H > z_O$

However, I saw on the internet, and especially in a french video, that the condition is more $z_S > z_O$ ... and I don't understand why? And why I am wrong?

2) Then, I saw in another french video, that there is a condition on the point $T$. With the Bernoulli theorem between $T$ and $O$, he finds the pressure $P_T = P_O - \rho g(z_T - z_O)$ and then says that $P_T$ must be positive to not have a depression in the pipe, and to have the siphon works. I don't understand that. Why $P_T$ should not just be greater than $P_O = P_{atm}$ ? Is he right about that?

3) Finally, if the pipe is at the beginning empty (or full of air), why the water does not get sucked by the pressure difference $P_I > P_O = P_{atm}$, and so make automatically the siphon works?

If someone could help me with all these questions, thank you in advance.

$\endgroup$

1 Answer 1

0
$\begingroup$
  1. For the pressure at $z_I$ you take the height of the water column above point $I$ through the pipe. However, the pressure at the top of the pipe is not atmospheric. The water column above point $I$ is $z_S-z_I$. There is no lateral variation in the pressure.

  2. Well, $p_T<p_{atm}$. You find that from the equation you give (which is just Bernoulli applied between $T$ and $O$). There is no physical meaning of $p_T<0$ here. It would just mean that water flows out at $O$, but does not flow upward at $I$ anymore.

  3. Here you would have to take into account that $\rho_{air}\ll\rho_{water}$. You cannot really apply Bernoulli in this situation, but maybe you can see where your equations start to fail?

$\endgroup$
10
  • $\begingroup$ Firstly, thanks for your answer. 1) Do you mean that the pressure at $I$ is $P_I = P_{atm} + \rho g (z_S - z_I)$, and the $\rho g H$ doesn't count ? Doesn't weigh on $I$, right? 2) it is pretty straightforward that $P_T < P_{atm}$ indeed, but because of that, why this do not stop the flow? I mean, fluid flows are going from $P+$ to $P-$, and here $P+$ is $P_{atm}$ and $P-$ is $P_T$. It is because $P_I$ is still > $P_O$ so the flow is started and it is enough to continue? 3) $\rho_{air} \ne \rho_{water}$, so there is no streamline and we can't apply Bernoulli, but in terms of aspiration... $\endgroup$
    – Jejouze
    Feb 23 at 16:06
  • $\begingroup$ ... why it doesn't work? I mean physically, what's happening? Finally, another question, why exactly we can assume that $V_I = 0$? First, at $t = 0$ s, the fluid is at rest so ok, but when the flow begins, why can we still assume that $V_I = 0$? I was thinking about the fact that the surface of the tank is much greater than the surface of the outlet, so we could say $V_O >> V_S$ and consider $V_S = 0$, but here it is more about the section of the inlet $I$ which is the same of the outlet and not the section of the tank I think... So I don't know. Could you help me again with that? $\endgroup$
    – Jejouze
    Feb 23 at 16:14
  • $\begingroup$ 1) Yes. 2) Hydrostatic pressure in a container doesn't cause flow in itself (i.e, what from the bottom of a bucket does not flow upward). The driving force is the pressure difference between entrance and exit. Pressure gradient after subtraction of the hydrostatic pressure counts. 3) You're driving pressure gradient is interrupted. $\endgroup$
    – Bernhard
    Feb 24 at 15:44
  • $\begingroup$ Nowhere you really need to assume $v_I=0$. You Assume $v_I=v_O$. And that is why they cancel out of the equation. $\endgroup$
    – Bernhard
    Feb 24 at 15:45
  • $\begingroup$ Thanks for your answers! But if $V_I = V_O$ and then they cancel out of the equation, how can I find the condition where $V_O > 0$? Finally, how can you say that $P_T < 0$ means only that water flows out at $O$ but does not flow upward at $I$ anymore? What is the concrete thing behind this? $\endgroup$
    – Jejouze
    Mar 1 at 16:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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