Let try an experiment.

If water enter through an end $A$ with some velocity say $v_1$,and leaving end $B$ with speed $v_2$ in a UNIFORM cylindrical tube $AB$ (which is completely filled with water).

If we consider 3 cases.

  1. tube is horizontal
  2. tube is vertical with $A$ upward
  3. tube is vertical with $B$ upward

And we go through the experiment and we got our results, which is $v_1~=~v_2$

So for which case this is valid to?

Edit Due to lack of logical answers

If we consider Bernoulli's equation

$$P+\rho gh+\frac{1}{2}\rho v^2= \text{Constant}$$

So in the case of vertical pipe, we consider two points $A$ and $B$ and now applying Bernoulli's equation here as follows.

Let's assume $A$ to be upward, and take $B$ as reference level and applying Bernoulli's equation.

$$P_a+\rho gh+\frac{1}{2}\rho v_1^2 = P_b+\frac{1}{2}\rho v_2^2$$

If both are exposed to the atmosphere, then $P_a = P_b = $P_{\text{atm}}$

Then we get

$$\rho gh = \frac{1}{2}\rho(v_2^2 - v_1^2)$$ which implies that $2gh +v_1^2 = v_2^2$

So, finally this will prove that $v_1$ never equal to $v_2$ in vertical pipe, but if we consider equation of continuity then mass going in is same as mass coming out so according to equation of continuity

$$\Delta m = \rho A_1 v_1\Delta t = \rho A_2 v_2 \Delta t$$ which implies $A_1 v_1 = A_2 v_2$, and in our case $A_1 = A_2$ then according to equation of continuity $v_1 = v_2$.

Thus, according to continuity $v_1 = v_2$ in vertical pipe case and according to Bernoulli's equation $v_1$ never equal to $v_2$.

How can this be possible? Please guys help me out. Please go through the question and then answer it?

  • $\begingroup$ Please use MathJax to format equations and variables. $\endgroup$ Sep 22, 2020 at 7:07

3 Answers 3


If the tube is vertical, the fluid can't be filling the tube, and the area of the stream must be changing, thus allowing a velocity change to be consistent with mass conservation. In the case where A is at the bottom, if the fluid fills the tube, the pressure can't be atmospheric at the bottom. Therefore, you can't assume both the same pressure at the ends of the tube as well as a constant area of the fluid stream.


I think BioPhysicist's answer is correct, and addresses the problem. I just have a few points I want to add to possibly help clear it up (and I got pretty into the paint diagrams I was making).

I outlined the two possible situations I can think of where the tube is vertical, with A over B.

In the first situation, the velocity is non-zero, and the tanks at A and B are both exposed to atmosphere. If the velocity is non-zero though, this means that fluid is going from tank A to tank B. For this to happen the water in tank B must be pushing against the atmosphere. If it were just at atmospheric pressure, the water wouldn't be able to move, because the surface of point B would also be suspended by atmospheric pressure. So basically, even if both sides are exposed to atmosphere, if the water is flowing I don't think it's safe to say both sides of the pipe are actually at atmospheric pressure, the one on the receiving side of the flow should be above atmosphere so that the water can actually get into that vessel and flow out of B (or raise it's surface).

Here is a bad diagram I drew:

Fluid flowing

Or, if the fluid isn't flowing, the situation becomes hydrostatic, and you can clearly see the pressure difference, because if one end is above the other and there is no flow, hydrostatics dictates that $\Delta P = \rho g h$, so $P_A \neq P_B$, I drew another bad diagram:


So basically as Biophysicist said, the assumptions just cannot all be held at once. Hopefully this shows why it wouldn't make a lot of sense for them all to hold.

Here is how it must look if pressure is atmospheric on both sides and the flow is vertical:

Atmospheric on both sides

  • $\begingroup$ So why does we take atmospheric pressure at both end while calculating speed of efflux, one is reasonable to take atmospheric pressure at upper part, but why we take pressure at hole where water tend to fall be atmospheric pressure. This is the origin of my confusion, please help me out by editing your answer. $\endgroup$
    – 5 Dots
    Sep 23, 2020 at 16:12
  • 1
    $\begingroup$ @5Dots You can take pressure as atmospheric at both ends in situations where both ends are actually free to interact with the atmosphere; but in that case you cannot have the area and velocity of the flow remain constant while it is falling vertically. You cannot have a situation where there is flow vertically in the tube, the area and velocity don't change, and there is no pressure difference. It's just not what can happen. I'll add another bad drawing to show what $P_{atm}$ on both sides looks like. $\endgroup$
    – JMac
    Sep 23, 2020 at 16:46
  • $\begingroup$ thanks mate..finally i realise where i got wrong... cheers $\endgroup$
    – 5 Dots
    Sep 24, 2020 at 5:56

If air cannot enter the tube all 3 are valid, with the possible exception of case 2. In case 2 if the height of the cylindrical tube is greater than 10 meters vacuum or vapor cavities could form allowing some water to accelerate downward.


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