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So actually I have some confusions about Torricelli's law.

enter image description here

Based on the picture:

  1. Why is water flowing out only affected by air pressure from outside? Why not by $$P+\rho gh$$ from the water inside too? so the net pressure would be $\rho gh$?

  2. Why doesn't $v_1$ depend on the area of the hole? Wouldn't $v_2$ increase if we made area of the hole larger as well?

  3. Pressure in Bernoulli's equation is the water pressure in our case. Why do we put the atmospheric rather (on both sides of the equation) and then cancel them out, shouldn't it be the pressure of the liquid (water)?

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  • $\begingroup$ Within the tank, in the approach to the exit hole, the liquid pressure decreases rapidly from $\rho g h$ to zero (gauge). This occurs over a distance of only about 3 exit diameters from the exit. So, at the exit, there is atmospheric pressure. $\endgroup$ – Chet Miller Dec 7 '19 at 17:13
  • $\begingroup$ You mean like, the effect of the internal pressure vanishes? $\endgroup$ – passepartout Dec 8 '19 at 15:39
  • $\begingroup$ No. The internal pressure decreases in close proximity to the exit hold as the fluid flow within the tank converges and accelerates toward the exit hole. $\endgroup$ – Chet Miller Dec 9 '19 at 3:19
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Torricelli's law follows from Bernoulli's principle for an incompressible fluid $\rho = const$

$$ \frac{\rho u_i^2}{2} + \rho g h_i + p_i = const $$

assuming that the area ratio between the container $A_1$ and the hole $A_2$ is large

$$ A_1 \gg A_2. $$


If you consider a streamline starting from a point 1 at the top part of the container, there is the pressure $p_1$, which corresponds to the ambient pressure and a velocity that results from the surface moving downwards $u_1$ as water is pouring out. At the point 2, the hole in the container, the potential energy has been converted to kinetic energy and the pressure corresponds to the ambient pressure of the surrounding fluid:

$$ \frac{\rho u_1^2}{2} + \rho g h_1 + p_1 = \frac{\rho u_2^2}{2} + \rho g h_2 + p_2 $$

$$ \frac{\rho u_1^2}{2} + \rho g \underbrace{(h_1 - h_2)}_{h} + p_1 = \frac{\rho u_2^2}{2} + p_2 $$

Assuming the change in pressure in the surrounding fluid can be neglected (of course if the outside was water as well then the velocity would clearly be zero as the ambient pressure would change with height as well) $p_1$ and $p_2$ can be assumed equal $p_1 \approx p_2$.

Furthermore we can consider the 1D continuity equation

$$ \dot{m_1} = \rho \dot{V_1} = \rho A_1 u_1 = \dot{m_2} = \rho \dot{V_2} = \rho A_2 u_2, $$

$$ A_1 u_1 = A_2 u_2. $$

Assuming that $A_1$ is significantly larger than $A_2$, $A_1 \gg A_2$ we can neglect the velocity at the top of the container $u_1 \approx 0$ and thus find Torricelli's law

$$ u_2 \approx \sqrt{2gh}. $$


1) As you can see there is no pressure term as you follow a particle from the top of the surface to the exit and on both sides there is assumed the same ambient pressure. If it would be a closed pressurised container you would end up with $u_2 \approx \sqrt{2gh + 2 \frac{(p_1 - p_2)}{\rho} }$ and there would be indeed a contribution from the higher pressure inside the tank.

2) Clearly the derivation of this idealised law is based on the assumption of a small hole but as long as one can still neglect the velocity at the top of the container $u_1 \approx 0$ the outcome would still be the same and not depend on the area. This is a result of mass continuity.

3) As already mentioned we have to consider the ambient pressure for both sides as it reflects an external "force". Consider that at the top of the container there would be a piston pushing the liquid down, so to say increasing the ambient pressure, the liquid would flow out faster, wouldn't it? If also the surrounding fluid was water there would be also a change in pressure with the height of the surrounding fluid and the ambient pressure at point 2 would be equal to $p_2 = p_1 + \rho g (h_1 - h_2) = p_1 + \rho g h$ and there would be no directed macroscopic motion through the hole.

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  • $\begingroup$ 3) So pressure on both sides of the equation is meant to be from the "external force" that acts on the fluid (i.e external pressure, F = PA) and not the pressure of the fluid itself( that reflect the force that the liquid acts upon surroundings?) $\endgroup$ – passepartout Dec 8 '19 at 15:38
  • $\begingroup$ @passepartout I am not sure if I understand you correctly but the pressure exerted by the liquid and the surrounding must be the same. But conceptually the pressure is imposed externally as it must be compatible with the boundaries. $\endgroup$ – 2b-t Dec 9 '19 at 20:20

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