Bernoullis versus Continuity

Today in my AP Physics class we were discussing fluids and certain theorys. We were solving a problem in class and came onto a discrepency. We think there is something that we are missing but, we couldnt figure it out during class.

The problem was as so: You are given a tube with a constant Area throughout the length of the tube. The tube was then hung and therefore would have the openings at two different heights. Both ends would be open to Atmospheric Pressure.

In the equation of continiuity it is given that A1V1=A2V2 the Velocity remains constant throughout.
However by Bernoulli’s Principle, the raised end of the tube will be affected RHOGH…G as gravity and H as Height and RHO as the density.

If you then attempt to solve the problem, you run into a problem, Bernoulli’s Principle and the Equation of Continiuty do not equal.

We think we are missing something fairly obvious, but like I said, we couldnt figure it out.

If you need more to answer my question, post your questions back at me.

Can you give us the exact question? What are you trying to find or prove? I can’t help you find what’s wrong with your attempted solution when I don’t know what the problem is.

I agree, this seems like a pretty special problem. Generally, what Bernoulli’s is trying to tell you is that if you have a tube with equal pressures at each end, you cannot have flow in the pipe (kind of makes sense, right?). So your constant velocity is zero. Remember, Bernoulli’s relates the substantial derivative of the velocity (DV/Dt) with the pressure gradient, so with both equaling zero you can’t say much.

There’s the resolution to your dilemma. The upper end is at a lower pressure than the lower one. In a compressible fluid, continuity of velocity is not maintained when the pressure changes along its path. Mass flow might still be constant, but at lower pressure, the velocity must be faster to get the same number of molecules to pass in the same time.

Alan beat me to it.

Its funny how we tend to forget that we are all semi-floating in a sea of air

and that we would weigh more in a vacuum

Ahh, i think that answers what i had wanted. Ill bring it up to my teacher tomorrow and then if it still isnt the wanted solution, ill be back with more questions.

We are reviewing it again today, and we still run into the same problem. It’s just not making sense to us.

We assumed it as an incompressible fluid though. And at this point, we cannot figure it out. We have solved numerous problems that give us contrasting answers.

Both ends would be open to Atmospheric Pressure.

Don’t get me wrong, but the atmosphere I live in is compressible. Air has a bulk modulus. For that matter so does water, but most people refer to it incorrectly as a “incompressible”.

Unless the “atmosphere” that the tube is opened to at both ends is stated otherwise, I would start by assuming it to be air and therefore, compressible. (I hate assuming, but in this case I see no alternative)

What billbo911 is hinting at, is that incompressible fluid does not mean constant pressure. It only means that you are allowed to use Bernoulli’s equation. Bernoulli’s has three parts and they are all pressure terms:

static pressure + dynamic pressure + head pressure = constant along a streamline

p + .5densityvelocity^2 + densitygheight = C

You can use this equation to relate two different points on a streamline (for example, the ends of a pipe). Because this is a fixed relation, you cannot just “decide” the values for them. In Corey’s problem all three terms are being predetermined, which wont give you an answer. Alan gave a very visual description of how you can’t say the velocity gradient is zero AND the pressure gradient is zero AND there is a height change. If you say there is constant velocity and pressure, solving makes height zero (horizontal).

It would really be helpful if you drew a picture, which is where all good solutions begin.

HA! thats what my father use to say when he tried to explain something to me, and I didnt get it:

“Do I have to draw you a picture?” :^)

Am I just missing something here? What is the “Find” part of the problem? So you have a tube hanging vertically in air, whoopdee do. What are we trying to prove or what are we looking for in the problem?

I give you some things to think on.

First Bernoulli’s Eqn is applicable when you assume the fluid is invisid (or the viscosity equals zero).

I assume the continuty equation you are using is for an imcompressible fluid. I know air is compressible but at small velocities it can act fairly imcompressible. We mostly use the imcompressible version of the contuity equation. Basically it says if you push a fluid into a cube that fluid in that cube must react to offset it. For compressible fluid this is not entirely true. Stick to imcompressible though, I think VA=VA pretty much assumes it.

For my uses I’ve been using the Bernoulli equation in the form

(1/2)Rho*v^2+P=Constant

Rho=density (air)
v= velocity
P=pressure

Using this equation twice and subtracting will get rid of the constant.

delta§=(1/2)Rho((v1)^2-(v2)^2))

This will give you the pressure difference, delta§, difference when the velocities, v1 and v2, over the tube are different (imagine blowing over a cup with a ping pong ball in it)

This is where I think maybe the equation of continuity you use, VA=VA, comes into play.

Think about having a small tube that expands into a large tube. When the air goes from the small area to the large area the velocity goes down.

A typical problem we do for the bernouillis is a prarie dog (I am from Kansas) tunnel where one hole is lower and the other is at the top of the hill. As the air flows from the lower hole to higher hole it speeds up. This creates a pressure difference at the two holes and let air flow through the tunnel. Basically keeping the prarie dogs from suffocating.

Anyways, using continuity you can find the two velocities. Then you can use Bernouillis equation to determine pressure difference, delta§.

Now the fun part is finding the velocity through the tube.

Here is the equation I use for a pressure driven flow in a tube

v=((delta§)(radius^2))/8u*L

v=average velocity
delta§=change in pressure (from Bernouillis)
radius=radius of the tube
u=viscosity of the fluid (at 20C it is .01813 milliPascals*seconds)
L=length of tube

That is how I’ve been doing it lately. If this was fun for you look into to Chemical engineering or go out and buy Transport Phenomena by Robert Bird. It really is a fun class.

I hope this helps I wasn’t sure of the exact question so I tried to answer as many as possible.

Seems like a very old topic, but very interesting :slight_smile:
You can use both the Bernoulli and equation of continuity in this case too.
However, you cannot dictate both the pressure and the area of cross-section of the flow. Given one, the other chooses its own value.
Here, as height changes (and atm pressure remains const) the velocity of the fluid increases. To account for this, c/s area of the flow decreases according to the continuity equation. The fluid does not use the whole of the tube cross-sectional area for its flow, which is the mistake you seem to have made.


Ganesh Mani
Mechanical Engineering
IIT Bombay

Waiting for replies :slight_smile:


Ganesh Mani
Mechanical Engineering
IIT Bombay