Bernoulli's Equation Applied to Two Free Surfaces

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  • #1
tjm2444
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Consider two cylindrical containers of the same diameter. Fill them with a liquid to different levels. We all know that the liquid can be "siphoned" from one container to the other using a thin hose, which we will consider to be thin and of uniform diameter.

Bernoulli's Equation can be used to derive the flow rate (which is essentially the same as Torricelli's Theorem).

However, consider applying Bernoulli's Equation to the free surface of each container.

P1 + pgh1 + 1/2 p v1^2 = P2 + pgh2 + 1/2 p v2^2

Because the cylinders have the same diameter, v1 = v2.

Because a free surface is at atmospheric pressure, P1 = P2.

Bernoulli's Equation then reduces to h1 = h2, which makes no sense.

Does anyone have any insight to why Bernoulli's Equation does not work in this case?
 

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  • #2
Studiot
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Does anyone have any insight to why Bernoulli's Equation does not work in this case?

It would help if you drew a diagram and identified your symbols.

The three points of interest in the flow are

1) The free surface in the first vessel. Here the velocity is zero. Call it A

2) The knuckle of the connecting pipe. Call it B This can be calculated from Bernoulli.

3) The discharge point into the second vessel. Call it C The velocity here is the same as at B
 
  • #3
tjm2444
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It would help if you drew a diagram and identified your symbols.
The three points of interest in the flow are

I understand that, but my question is why Bernoulli's Equation give nonsensical results when applied to the liquid surface on each side.
 
  • #4
Studiot
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Diagram?

What is v1 and v2? Or rather where are they taken?
 
  • #5
tjm2444
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Diagram?

What is v1 and v2? Or rather where are they taken?

In the diagram attached, Bernoulli's Equation applied to points 1 and 2 gives h = 0, which makes no sense.
 

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  • #6
Studiot
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That is not a syphon, that is two vessels connected by a pipe at the bottom.

Look at my diagram and think about the value of velocity at A
 

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  • #7
tjm2444
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That is not a syphon, that is two vessels connected by a pipe at the bottom.

Look at my diagram and think about the value of velocity at A

I'm familiar with your analysis. My question is why does applying Bernoulli's Equation at points 1 and 2 in my diagram give h = 0. Applying Bernoulli's Equation in this way yields the same result whether it's a siphon, or if the vessels are connected by a pipe at the bottom.
 
  • #8
Studiot
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Why do you think your v1 = your v2 ?
 
  • #9
tjm2444
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Why do you think your v1 = your v2 ?

Because the cylinder's have the same cross sectional area, as I said in the original post.
 
  • #10
Studiot
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I know you said that but you are not correctly applying Bernoulli's equation.

You will understand much more quickly if you stop accusing the equation of being wrong and look at your own work.

At the start of the process (vessel 1 full, vessel 2 empty) v1 = 0, v2 can be calculated by Bernoulli and h1 is not equal to h2.

Since the vessels are of limited capacity the velocity changes through the discharge.

At the end of the process v1 = v2 = 0 and h1 = h2 as you have already observed.
 
  • #11
tjm2444
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At the start of the process (vessel 1 full, vessel 2 empty) v1 = 0, v2 can be calculated by Bernoulli and h1 is not equal to h2.

OK. Why can't we start the process with initial conditions as they are in my diagram, and why does the application of Bernoulli's Equation to points 1 and 2 yield the non-sensical result that h = 0 in that case? I'm not accusing Bernoulli's Equation of being wrong, I'm simply trying to understand its range of application.
 
  • #12
Studiot
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Why can't we start the process with initial conditions as they are in my diagram, and why does the application of Bernoulli's Equation yield the non-sensical result that h = 0 in that case?

Of course you can, but you don't have the information to fill in the Bernoulli equation part of the way through the process so you cannot solve it.

What you have written down is not correct for your diagram.
 
  • #13
tjm2444
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Of course you can, but you don't have the information to fill in the Bernoulli equation part of the way through the process so you cannot solve it.

What you have written down is not correct for your diagram.

OK, let's work through this:

v1 = v2 because the cylinders are identical.

P1 = P2 because both points are at the atmosphere.

So we get h = 0.


What did I do wrong?
 
  • #14
Studiot
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OK, let's work through this:

You have shown a picture.

Is this your initial state, ie is there initially some water in each vessel and a tap in your pipe?

So do you start the process by opening the tap?
 
  • #15
tjm2444
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You have shown a picture.

Is this your initial state, ie is there initially some water in each vessel and a tap in your pipe?

So do you start the process by opening the tap?


Sure. The initial state is as shown, and the process starts by opening a tap.
 
  • #16
Studiot
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So at the instant of opening the tap, can you describe the situation, that is put values into Bernoulli's equation?
 
  • #17
tjm2444
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So at the instant of opening the tap, can you describe the situation, that is put values into Bernoulli's equation?

It doesn't matter how the cylinders are connected or whether or not there are valves. We can start with one empty and one full, and then apply Bernoulli's equation at the time the situation is as depicted in my diagram.
 
  • #18
Studiot
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It doesn't matter how the cylinders are connected or whether or not there are valves.

If you know the answers, why are you asking?

Of course it matters or I wouldn't have introduced a valve.

At the instant of opening the valve the velocity at both ends of the tube is zero.

As a result, Bernoulli will give you the result that the difference in fluid levels equals the static pressure difference between the fluids in each vessel.

You have three variables to insert into each side of Bernoulli. You can reduce this to two (or even one) if you can make one (or two) of the height, static pressure or velocity equal zero, or some other known value such as atmospheric pressure.

The problem arises because as soon as flow starts the static pressure drops and you therefore don't know either.

You keep repeating that the velocities are equal because the vessels are the same size. The velocity is quite independent of the vessel size.
 
  • #19
tjm2444
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If you know the answers, why are you asking?
You keep repeating that the velocities are equal because the vessels are the same size. The velocity is quite independent of the vessel size.

Points 1 and 2 are at the surface in each container. If the containers have the same diameter, then v1 = v2.
 
  • #20
Studiot
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To solve this you obtain an expression for v using Bernoulli.

If h is the head difference between the higher free surface and the lower free surface, producing the flow then by Bernoulli

h = v2/2g

You can estimate the time to equalise levels by integrating the equation

-Adh = Qdt.

Where A is the area of the upper free surface and t is time

You can obtain an expression for Q in terms of t from the velocity and connecting pipe diameter.

To complete the task you may need to consider the reverse pressure build up in the second vessel, due to increasing head
 
  • #21
tjm2444
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To solve this you obtain an expression for v using Bernoulli.

If h is the head difference between the higher free surface and the lower free surface, producing the flow then by Bernoulli

h = v2/2g

When you apply Bernoulli's Equation to two points -- the free surface on each side -- it
yields h = 0. Clearly this is a misapplication of Bernoulli's Equation, but why? What is wrong with using those two points?

Thanks for trying to help.
 
  • #22
Studiot
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One last time it doesn't yield h = o.

Read again my last post.
 
  • #23
tjm2444
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One last time it doesn't yield h = o.

One last time, yes it does.

See image.
 

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  • #24
boneh3ad
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Here's the thing: it should predict zero velocity. By applying Bernoulli's equation at the two free surfaces, you are basically asking the equation to predict any motion that would occur directly between two surfaces at the exact same hydrostatic conditions. Of course it will predict no flow. If you ignore the atmospheric pressure, there is no head between the two surfaces.

The problem is that your fluid motion is occurring at the bottom of the tank. What you need to do is relate the pressure at one end of the connecting pipe to the height of the water in that tank and then do the same with the other tank. Then you have a pressure differential and can determine the flow rate. The surfaces will certainly move at the same speed since one is losing liquid at the same rate the other is gaining it and the tanks are identical. Keep in mind that the flow rate will change depending on the height, so a differential equation would need to be solved to get the time history of the heights of the surfaces (or the flow rates).
 
  • #25
Studiot
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Not once have you asked me how I arrived at my equations.

You also said that you understood them and they are standard stuff.

It is clear that you have a fundamental misunderstanding about the Bernoulli equation and its application.

I have drawn just the full vessel at two times.

At t1 just as the emptying process begins and again at t2 when the top parcel of fluid has travelled some distance down the vessel.

At t1 it is important to realise that the velocity is zero. Thus v1 = 0

I have filled in the other conditions for the Bernoulli equation.

At t2 the same parcel of fluid has now lost some potential energy but gained some kinetic energy. The pressure energy has not altered and is still atmospheric. Consequently it is now travelling with velocity v2

Finally I have substituted into the Bernoulli equation which leads to the conclusion that the loss of head (h1 - h2) supplies the energy to drive the fluid at velocity v2.

Can you see this ?
 

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  • #26
tjm2444
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Not once have you asked me how I arrived at my equations.

I don't need to ask because I understand how you arrive at each of your equations. The v = sqrt (2gh) result is standard, and I've derived it many times on my own.

Can you see this ?

Yes, and now you're applying Bernoulli's equation to the same free surface at two different times, with the result that the free surface is moving with a speed v = sqrt (2gh), which is wrong.

I'm happy to be engaged by you, and I thank you for it, but clearly we're going around in circles here.
 
  • #27
Studiot
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with the result that the free surface is moving with a speed v = sqrt (2gh), which is wrong.

Please explain.
 
  • #28
boneh3ad
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Of course that still won't give you the answer you want. It doesn't take into account the water moving through the pipe and the influence of the pressure on the other side. What you really need is to know the pressure at each end of the connecting pipe (in this case, hydrostatic pressure) and then the size of the pipe so you can tell what the flow rate through the pipe will be.

You will end up with
[tex]\rho g h(t) = \Delta p[/tex]
From there you need to know the flow rate through the pipe, since the rate of volume removal/addition to your tanks is what will actually determine the movement of your surfaces. For a laminar flow, that is given by the Hagen-Poiseuille equation (which is almost certainly going to be a fine estimate for you).
[tex]Q = \dfrac{\Delta p \pi d^4}{128 \mu L} = \dfrac{\rho g h \pi d^4}{128 \mu L}[/tex]
That will give you the volumetric flow rate through the connecting pipe. To change that into a velocity of the free surfaces, you just need to use that change in volume and divide by the area of the surfaces. If you say the surfaces have diamter D, that gives:
[tex]V = \dfrac{\rho g h d^4}{32 \mu L D^2}[/tex]
This works for an instant in time. If you wanted to get the time history, a differential equation would be needed.

Of course, you could also use Bernoulli here if you felt so inclined. Consider the two points to be one at the top of the left tank and one at the pipe exit of the right tank. You would have:
[tex]p_1 + \rho g y_1 + \dfrac{1}{2}\rho v_1^2 = p_2 + \rho g y_3 + \dfrac{1}{2}\rho v_2^2[/tex]
On the left side, [itex]p_1[/itex] and [itex]v_1[/itex] are assumed zero and on the right side, [itex]y_3[/itex] is the height of the pipe and has been set as the zero reference and [itex]p_2=\rho g y_2[/itex], the hydrostatic pressure.
[tex]\rho g y_1 = \rho g y_2 + \dfrac{1}{2}\rho v_2^2[/tex]
So solving for [itex]v_2[/itex]
[tex]v_2 = \sqrt{2 g h}[/tex]
This is the velocity of the water moving through the pipe. The flow rate is then based on the pipe diameter, d
[tex]Q = \sqrt{\dfrac{gh}{2}}\pi d^2[/tex]
Dividing by the area of your surface gives
[tex]V = \sqrt{2gh}\dfrac{d^2}{D^2}[/tex]
Once again, this applies only at a given instant and really only applies if [tex]D\gg d[/tex] since it is based on the assumption that the surfaces are not moving. Otherwise the equation gets a bit more complicated and you are better off just going with the Hagen-Poiseuille route, which doesn't assume invsicid flow anyway (not a particularly valid assumption in a pipe).
 
  • #29
tjm2444
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Please explain.

Your notes suggest that you think that a free surface that has descended a height h is moving at a speed √2gh.
 
  • #30
tjm2444
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First, thank you for understanding my question, bone3ead!

Most texts say that Bernoulli's Equation applies to any two points along a streamline. Is there no streamline that can be traced from the free surface on one side to that on the other? And if there is (which I believe is the case) then why does Bernoulli's Equation fail in this case?


The problem is that your fluid motion is occurring at the bottom of the tank.

The motion is occurring everywhere. The left surface is moving down and the right up.
 
  • #31
tjm2444
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and [itex]p_2=\rho g y_2[/itex], the hydrostatic pressure.
[tex]\rho g y_1 = \rho g y_2 + \dfrac{1}{2}\rho v_2^2[/tex]
So solving for [itex]v_2[/itex]
[tex]v_2 = \sqrt{2 g h}[/tex]
This is the velocity of the water moving through the pipe.

But aren't you saying here that the pressure in the pipe is simply the hydrostatic pressure? Is the water moving at point 2? If so, then the pressure at 2 is lower than the hydrostatic pressure.
 
  • #32
256bits
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One last time, yes it does.

See image.

You are using 2 seperate datums, which is why this is not the way to apply Bernoulli.

P1 + pgh1 + 1/2 p v1^2 = P2 + pgh2 + 1/2 p v2^2

Using a datum level as the surface of liquid 2:

v1 = -v2
P2 = 0
h2 =0

the equation reduces to
P1 + pgh1 + 1/2 p v1^2 = P2 + pgh2 + 1/2 p v2^2
P1 + pgh1 + 1/2 p v1^2 = 0 + 0 + 1/2 p v2^2

And 1/2 p v1^2 = 1/2 p v2^2

So P1 = pgh1
 
  • #33
boneh3ad
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First, thank you for understanding my question, bone3ead!

Most texts say that Bernoulli's Equation applies to any two points along a streamline. Is there no streamline that can be traced from the free surface on one side to that on the other? And if there is (which I believe is the case) then why does Bernoulli's Equation fail in this case?

The motion is occurring everywhere. The left surface is moving down and the right up.

I am sure you could find a streamline somewhere that would connect two points such as those you are using, but in reality, if the tank is much larger than the pipe connecting the two, most of the motion will be in the region of the pipe. At any rate, the entire problem is unsteady, so Bernoulli's equation doesn't really apply. There is an unsteady analog to Bernoulli's equation, but then you get significantly more complicated. There are much easier ways to analyze the flow than in that case.

The solution is to say that the surfaces aren't changing much (i.e. D >> d from my previous post) and approximate the situation as steady. Again, it is only an approximation though.

But aren't you saying here that the pressure in the pipe is simply the hydrostatic pressure? Is the water moving at point 2? If so, then the pressure at 2 is lower than the hydrostatic pressure.

The static pressure at one opening is the hydrostatic pressure from that tank and the static pressure at the other opening is the hydrostatic pressure from the height of the water in that tank.

The pressure at point 2 is zero (gauge). There is no pressure other than atmospheric at that point. Same with point 1. The potential energy then that leads to the motion between tanks is gravitational in nature and the flow rate between the two (and thus velocity of the surfaces) is tied to the height difference. However, it is between the inlet and outlet of the pipe that you really need to apply Bernoulli's equation to get your estimate. Picking two points at each surface loses any information about what occurs between them because both points are essentially in equilibrium.
 
  • #34
Studiot
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So P1 = pgh1

That's an interesting assertion for a free surface of a liquid.

Would you like to explain it?
 
  • #35
256bits
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That's an interesting assertion for a free surface of a liquid.

Would you like to explain it?

P1 is the pressure of the liquid in container 1 at the same level as the surface of container 2.
P1 is NOT the pressure of the free surface of container 1 which is causing the confusion here.

The energy level at the two free surfaces 1 and 2 are of course not going to be equal.
By how much - by the value P1 = pgh1.
 
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