# All conditions for a siphon to work?

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: 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.

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, $$p_T. You find that from the equation you give (which is just Bernoulli applied between $$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?
• 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... Feb 23 at 16:06
• ... 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? Feb 23 at 16:14
• 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. Feb 24 at 15:45
• 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? Mar 1 at 16:23