# Using Bernoulli's Equation

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1. Mar 5, 2016

### Alettix

Hello!
I have encountered some trouble with choosing the right reference points when using Bernoulli's equation and I would be glad if you could help me sort it out with this made up example. :)

1. The problem statement, all variables and given/known data

There is a large, open, cylindrical water tank with a cross section area of $A_1 = 2 m^2$. 1 meter down the water surface ($\Delta h = 1 m$) is a small pipe with cross section $A_2 = 0.0003 m^2$. Determine the speed of the water flowing from the pipe.

2. Relevant equations
Bernoulli's equation: $\frac{p_1}{\rho} + gh_1 + v_1^2/2 = \frac{p_2}{\rho} + gh_2 + v_2^2/2$ (1)
Continuity equation: $A_av_a =A_bv_b$ (2)

3. The attempt at a solution
We can assume that the flow is laminar, the fluid incompressible and the viscosity very small, therefor we can apply Eq.(1). We choose to place reference point 1 at the surface of the water in the tank, and point 2 at the small pipes opening. Because $A_1>>A_2$ we can assume $v_1 \approx 0$. We set $h_2 = 0$ and therefore $h_1 = 1$. We do also know that $p_1 = p_0$, where $p_0$ is the atmospheric pressure.
Solving for $v_2$ we have:
$v_2 = \sqrt{\frac{2(p_0-p_2)}{\rho} + 2g\Delta h}$ (3)

Now to my question: Where exactly should I put reference point number 2? If I put it just outside the opening of the smal pipe we have $p_2=p_0$ and consequently $v_2 =\sqrt{2g\Delta h}$. However, if I put the point just inside the opening the pressure will be: $p_2 = p_0 + \rho g\Delta h$ which yields $v_2 = 0$.

I think something here must be wrong. It isn't reasonable that the water just inside the small opening should be completely still, but outside it flow with a speed as large as in case of free fall. This result seem especially crazy if we consider that the continuity equation must hold in the small pipe. So where is the fault in my reasoning? Should point 2 be placed just outside or inside the opening of the pipe? Why?

PS: I really wanted to draw a figure, but there is something very wrong with paint on my computer right now.

2. Mar 5, 2016

### Staff: Mentor

Either point is OK, and both points give the same answer. Just inside the opening, the velocity is also equal to the velocity just outside, and the pressure is equal to the pressure just outside. Your mistake is assuming that $p_2 = p_0 + \rho g\Delta h$ just inside the opening. The system is not in static equilibrium, so this equation cannot be applied.

Chet

3. Mar 7, 2016

### Alettix

Thank you very much Sir! This really clarified it!