# Continuity equation and Bernoulli equation on inclined pipes

I came across the following problem in Fluid Mechanics. We are asked to find the velocity in 3.

The pipe diameter from 2 to 3 is constant. I calculated the velocity in 2 using $\sqrt{2g\times50}$ and then from the continuity equation I deduced that the velocity in 3 had to be equal to the velocity in 2 (same diameter implies same cross sectional area, plus water is incompressible).

However the textbook solution is different. They arrive to the following form of the ideal Bernoulli equation:

$$p_1 + \rho g 350 = p_3 + \frac{1}{2}\rho v_3^2$$

They then assume $p_1 = p_3$ and find $v_3 = \sqrt{2g\times350}$, a very different value. By my previous reasoning, it seems to me that this would violate the continuity equation. I also do not understand the assumption that $p_1 = p_3$, since point 3 is inside the pipe. Is the text book right? If so, what is the flaw with my reasoning regarding the continuity equation?

• See what I have added to my answer. For this system, the fluid will be cavitating, and most of the column between points 2 and 3 will be filled with vacuum. – Chet Miller May 8 '18 at 12:07

In this problem, the two points that are used in the application of the Bernoulli equation are points 1 and 3. As @V.F. correctly points out, the pressures at points 1 and 3 are both atmospheric, so these pressures cancel. The fluid velocity at point 1 is essentially zero, and the elevation z at point 1 is 350 m above the datum, situated at point 3.

For point 3, the elevation z is 0, and the velocity is $v_3$. This all leads to the equation presented in the OP. And, of course, the continuity equation is not violated, since $v_2=v_3$.

Of course, the velocity at point 2 is not equal to $\sqrt{2g\times 50}$ because the pressure at point 2 is not equal to the pressure at point

Before completing this solution, it is of interest to check the pressure at point 2 to ascertain whether the fluid is cavitating. Applying Bernoulli to points 2 and 3, we have: $$p_2+\rho g(300)+\frac{1}{2}\rho v_2^2=p_3+\frac{1}{2}\rho v_3^2$$Solving for $p_2$ yields $$p_2=-300\rho g$$This is well below the value of $-10\rho g$ which would be required to cause cavitation. Since the fluid will be cavitating, the tube between points 2 and 3 will not be running full, and there will be a vacuum of nearly $10 \rho g$ in the empty section above the column of fluid in the tube (situated above point 3). The pressure at point 2 will be $p_2=-10\rho g$. Then, applying the Bernoulli equation to points 1 and 2, we would have: $$0+50 \rho g+0=-10\rho g+0+\frac{1}{2}\rho v_2^2$$So, $$v_2=\sqrt{120 \rho g}$$We can now calculate the elevation of the fluid level in the partially filled column between points 2 and 3. Applying the Bernoulli equation in conjunction to the continuity equation to the column gives: $$p_2+\rho g h+\frac{1}{2}\rho v_2^2=p_3+\frac{1}{2}\rho v_3^2$$or$$-10\rho g+\rho g h=0$$Therefore, as one might expect, the height of the column will be 10 meters. Because of the cavitation, the other 290 meters will essentially be vacuum (more precisely, at the absolute vapor pressure of water at the system temperature).

• You are making no assumptions here about the diameter of the pipe. If we just ballpark it from the picture, it is going to be more than 20m. Is it realistic to have a column 10m high and 20m wide? Will cavitation occur regardless of the pipe diameter? Is it possible that the water will just flow on the bottom of the pipe, as the textbook solution possibly implies? – V.F. May 8 '18 at 15:36
• I think the diagram is just schematic. Lengths are often not shown to scale in geophysical situations, particularly diameters of pipes. – Chet Miller May 8 '18 at 16:04
• The diagram aside, assuming pipe diameter 20m, could you answer my three questions above? – V.F. May 8 '18 at 16:07
• In my judgment, if the diameter were 20 meters, the pipe would not be running full over most of the distance between points 2 and 3. So certainly, under these conditions, cavitation would not be occurring. It is definitely possible then that the water would just be flowing on the bottom of the pipe. – Chet Miller May 8 '18 at 16:36
• @ChesteMiller if the water is flowing on the bottom of the pipe, the pressure in the pipe still will be close to 0, so the cavitation could occur in the water. We just can't assume that at some depth the water will fill in the pipe cross-section fully as it does at point 1. And then, in general, $v_3$ should be greater than $v_2$, as V.F. suggests in his answer. – Ján Lalinský May 8 '18 at 16:43

Since the pipe at the bottom is open, the pressure at the bottom, $p_3$, will be atmospheric. This explains the statement $p_1=p_3$.

The water in the pipe, lacking support at the bottom, will submit to gravity and accelerate. As a result, $v_3$ will be greater than $v_2$.

So, for the continuity equation $v_2A_2=v_3A_3$ to hold, $A_3$ will have to be smaller than $A_2$, i.e., the pipe at the bottom won't be filled.

What's left in the Bernoulli equation reflects the transformation of the potential energy of the water at the top to its kinetic energy at the bottom.

• This is not correct. For a constant pipe cross sectional area, $v_2=v_3$. So what? That has nothing to do with the solution to the problem. – Chet Miller May 8 '18 at 2:13
• ChesterMiller In the Bernoulli equation for points 2 and 3, $p_2+\rho gh_2+\frac{\rho v_2^2}{2} = p_3+\rho gh_3+\frac{\rho v_3^2}{2}$, pressure and potential energy terms are greater in point 2, therefore kinetic energy term has to be greater in point 3, therefore $v_3$ must be greater and $v_2$. – V.F. May 8 '18 at 11:40
• @ChesterMiller To simplify the problem, we could remove 50m column at the top, in which case $p_2$ would be obviously equal $p_3$, both at atmospheric pressure level. Would $v_3$ be equal $v_2$ then? – V.F. May 8 '18 at 12:22