Does Fluid Dynamics Explain Pressure Gradients in a Frictionless Tube?

In summary, when a flow is created in a tube with a constant diameter, the pressure of the fluid at the entrance needs to be higher than the pressure at the exit to create a flow.
  • #1
Freixas
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Let's start with a horizontal tube with a constant diameter. I'm not sure if it's important, but let's assume it's frictionless. I will have some fluid flowing in this tube and if it's important, we can make the fluid incompressible, inviscid, irrotational, etc.

To create a flow in the tube, I've been told I need a difference in pressure. I've also been told that the pressure of the fluid exiting the tube will be equal to the atmospheric pressure. To get a flow, the pressure of the fluid at the entrance needs to be higher than the pressure at the exit.

For beginners to fluid dynamics, the term "pressure" is problem. Sometimes it means total pressure and sometimes static pressure. For the exit pressure, I'm pretty sure the rule is for static pressure. When I mentioned needing a pressure difference to get a flow, I'm less sure, but I think we're talking about total pressure.

The continuity equation tells me that if the fluid is incompressible and the tube has a fixed diameter, the velocity of the flow will be equal at every point. Then the Bernoulli equation let's me calculate the static pressure at any point given the velocity. Since the velocities are equal, the static pressures will also be equal all the way along the tube.

Yet people talk about a "pressure gradient" in the tube. And a pressure gradient makes sense if one thinks about what might be happening at the molecular level. Do the pressure gradients exist only very close to the entrance and exit of the tube?

One final puzzle piece. Another text talked about a frictional flow. In this example, it was clear that the author was saying that the difference in static pressure between two points in a tube was because of energy lost due to friction--there was no mention of any other pressure gradient other than the one caused by friction.

Hopefully, someone can quickly straighten me out. I'd appreciate the help.
 
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  • #2
Freixas said:
Yet people talk about a "pressure gradient" in the tube.
You've constrained the problem to create the paradox. It's the fluids equivalent of saying a zero resistance wire can't have a current because there's no voltage drop to drive it. So:
I'm not sure if it's important, but let's assume it's frictionless.
Yes, it's important. Don't assume it's frictionless if you want to analyze the pressure drop due to friction!
 
  • #3
russ_watters said:
Don't assume it's frictionless if you want to analyze the pressure drop due to friction!

Fair enough, but I'm not trying to analyze the pressure drop due to friction.

Let me try to clarify. Wikipedia claims:

The following assumptions must be met for the Bernoulli equation to apply:
  • the flow must be steady, i.e. the flow parameters (velocity, density, etc...) at any point cannot change with time,
  • the flow must be incompressible – even though pressure varies, the density must remain constant along a streamline;
  • friction by viscous forces must be negligible.
Let's say all these conditions are met and the static pressure at the entrance is 1,000 kPa above the exit pressure. Given the shape of the tube and combining the continuity equation with the Bernoulli equation, the static pressure in the tube must be equal at all points in the tube. Is it? If not, why? If it is, what is the static pressure in the tube?
 
  • #4
The situation you're describing with frictionless, incompressible flow through a pipe of constant diameter and 1000kPa pressure difference from inlet to outlet is not at steady state. In reality, the flow velocity will accelerate until mechanical equilibrium is reached (equal pressures). If you're not convinced, do a momentum balance on any volume element in the fluid, and you'll see there's unbalanced force due to the pressure gradient.

In the case of say, frictionless, incompressible steady-state flow through a converging nozzle, the momentum balance is completed by normal force by the rigid nozzle on the fluid. However, a pipe of constant diameter can't exert an axial force on the fluid (discounting viscous shear), so it doesn't explain your scenario. The only option is for the fluid to accelerate until the pressure gradient vanishes.
 
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  • #5
Freixas said:
Fair enough, but I'm not trying to analyze the pressure drop due to friction.

...the static pressure at the entrance is 1,000 kPa above the exit pressure.
You are simultaneously saying you don't want a pressure drop and you do want a pressure drop. You can't have it both ways at the same time.

If you don't have a pressure drop, the pressure at the inlet and outlet must be the same. That's what "pressure drop" means.
 
  • #6
Freixas said:
To get a flow, the pressure of the fluid at the entrance needs to be higher than the pressure at the exit.
I do not see that anyone has commented directly on this. It is not correct.

If the fluid is frictionless, you do not need a pressure difference between the tube ends to have a flow. The flow can be there, can have always been there and can be maintained indefinitely into the future with no pressure drop.
 
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  • #7
right, if you write out the Bernoulli eq. for a horizontal frictionless pipe of constant area the terms are equal term by term. v^2, P, and z are the same on both sides.
 
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  • #8
This kind of issue used to worry me a bit, until I figured out a simple little thought experiment. If you take a ring shaped tube, and if there is no viscosity/friction, then the fluid can flow around forever just like current in a superconducting ring. It takes no energy input or pressure gradient to keep it moving around.

Now if you make one segment of the ring narrower than the rest of the ring, the pressure there will be lower than the rest. We can deduce this as follows.

There will be pressure gradients in the tapering parts because the fluid is accelerating or decelerating there. So as an element of fluid passes through the taper towards the thin segment, it needs to accelerate -- hence the pressure at the end of the taper should be less than at the start. Once we're past the taper, there is no change along the thin part itself, but the pressure is lower than in the thick part: after all, we have just gone through a gradient!

Finally, as the liquid passes through the second taper towards the thick part, it needs to decelerate. This means a back pressure gradient that produces the required deceleration -- and that gradient brings up the pressure again by the time we are in the thick part.

None of this has to do with viscosity, because we assumed there is none.
 
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  • #9
jbriggs444 said:
I do not see that anyone has commented directly on this. It is not correct.

If the fluid is frictionless, you do not need a pressure difference between the tube ends to have a flow.

Thanks. Can I assume you mean static pressure? And could I reword the statement as: In order to initiate a flow in a tube, there must be sufficient static pressure to overcome friction in the tube?
 
  • #10
Freixas said:
Thanks. Can I assume you mean static pressure? And could I reword the statement as: In order to initiate a flow in a tube, there must be sufficient static pressure to overcome friction in the tube?
I am making the simplifying assumption that the entry and exits to the tube have the same area. So the flow velocity is the same. So it does not matter what sort of pressure we are talking about. No friction. No pressure difference.

Why this blather about "initiating" a flow? We are talking about the situation once a flow has been established. You don't overcome friction to initiate a flow. You overcome inertia.
 
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  • #11
jbriggs444 said:
I am making the simplifying assumption that the entry and exits to the tube have the same area.

Your simplifying assumption is also one of the assumptions I gave for the problem, so yes.

jbriggs444 said:
So the flow velocity is the same. So it does not matter what sort of pressure we are talking about. No friction. No pressure difference.

Let's say the tube connects regions that have the same static pressure and the same 0 dynamic pressure. Even assuming the tube is frictionless, why would a flow begin? Where would the energy come from?

So: what is the correct statement to make relative to satic/dynamic/total pressure and the conditions needed to create a flow in a tube, whether it is frictionless or not?
 
  • #12
Swamp Thing said:
Now if you make one segment of the ring narrower than the rest of the ring,

It's a clever model, but doesn't really explain what I want. Your narrow ring section is connected on both ends to regions of equal static and dynamic pressure. In my example, the tube connects regions of different static and equal (0) dynamic pressures.
 
  • #13
russ_watters said:
You are simultaneously saying you don't want a pressure drop and you do want a pressure drop. You can't have it both ways at the same time.

I never said that I did or didn't want a pressure drop. I said that if I had a pressure gradient in the tube, it would seem to violate Bernoulli's equation, which I read to say that I should never have a pressure gradient in a frictionless tube with a constant diameter.

There is a pressure difference between the entrance and exit, yes. The question is: what happens in the tube?
 
  • #14
Freixas said:
There is a pressure difference between the entrance and exit, yes. The question is: what happens in the tube?
It evaporates into a cloud of imagination because it cannot exist.
 
  • #15
Twigg said:
The situation you're describing with frictionless, incompressible flow through a pipe of constant diameter and 1000kPa pressure difference from inlet to outlet is not at steady state.

In reality, the flow velocity will accelerate until mechanical equilibrium is reached (equal pressures). If you're not convinced, do a momentum balance on any volume element in the fluid, and you'll see there's unbalanced force due to the pressure gradient.

While I find your answer intriguing, it's also over my head. I used this definition for a steady flow: the flow parameters (velocity, density, etc...) at any point cannot change with time. I have no idea what might be buried in the "etc.". Do you have a different definition?

Let's consider this: I use the tube to connect two regions of different static pressure and 0 dynamic pressure. The fluids in these regions are incompressible and of equal fluid density. The regions are so large that I could maintain any flow rate through the tube for years without changing the pressures on either side by any significant amount. So I begin the flow and let it run for hours.

Now I measure the static pressures at five equally spaced regions, stretching from end to end. I'll list pressures relative to the exit (gauge) pressure, with the entrance pressure at 1,000 kPa. I'll assume a frictionless tube, but if you like you can imagine a tube with a small amount of friction. Here are some guesses as to what I might find:
  • 1,000, 750, 500, 250, 0. This gives a nice gradient but violates Bernoulli's equation. More important, there seems to be an unaccounted energy loss. Where is the pressure going? If I assume it is being converted to dynamic pressure, then the continuity equation is violated.
  • 1,000, 1,000, 1,000, 1,000, 0: This satisfies Bernoulli's and the statement that exit pressure is equal to gauge pressure. However, it doesn't explain how we got a flow--it looks like the dynamic pressure is 0 through most of the tube.
  • 1,000, 0, 0, 0, 0: Again, this satisfies Bernoulli's and the exit pressure and it allows for a flow since the static pressure is converted into dynamic pressure. The only problem is that 1,000 kPa of static pressure needs to be converted to dynamic pressure right at the entrance. This seems unlikely to be instantaneous. We could allow for a transition zone, but then we're back to the first option above, just with a shorter tube.
 
  • #16
Freixas said:
While I find your answer intriguing, it's also over my head. I used this definition for a steady flow: the flow parameters (velocity, density, etc...) at any point cannot change with time. I have no idea what might be buried in the "etc.".

That's the definition I had in mind too.

To show you what I mean, forget about fluid dynamics for a second and just apply good ol' Newton's laws to the fluid in the pipe. Say the pipe has a 1 meter squared cross section, and say the static pressure on the inlet is 2000 kPa and the static pressure on the outlet is 1000 kPa. Those pressures translate to forces: 2000 kPa * 1m^2 = 2 kN going inwards at the inlet and 1000 kPa * 1m^2 = 1 kN going inwards at the outlet. That means there's a net force of 1 kN in the same direction that the fluid flows through the inlet. The fluid must accelerate in response to the net force, meaning the flow velocity changes over time. That's why I say these conditions can't be a steady state. For steady state in the pipe, the pressures at the inlet and outlet must be equal.

The only way to have a pressure difference and the fluid not accelerate is if there's another force on the fluid to balance out the net force of the pressure difference. In the case of frictionless flow through a converging nozzle (a cone-shaped pipe with inlet area > outlet area), there is a normal force by the pipe acting on the fluid that balances out the net force. In the case of flow with friction through a regular pipe, the force of friction by the pipe on the fluid balances the difference in pressures.

Now if you're talking about initiating a flow where there wasn't a flow before, yes you need a net force to get the fluid going and that shows up as a pressure drop, but that isn't the steady state because the fluid is accelerating, so you wouldn't apply Bernoulli's to it. (If you dive into the thermodynamics of it, Bernoulli's is a statement of energy conservation, but when you're initiating a flow there is work being done on the system so the energy of the system is rising. What you want in that case is the full "energy balance" equation you'd find in any engineering thermodynamics book.)
 
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  • #17
The flow lines in the first tank will be converging towards the input end of the tube, so the fluid flow would have effectively a self-imposed taper. Acceleration would happen in this tapering region just before the point where the tube connects to the reservoir.

Therefore the pressure values in the tube would be something like 100, 99, 98, 97, 95. The drop from 1000 to 100 (occurring in the convergent flow region) accounts for the energy required to accelerate the fluid, while the drop from 100 to 95 is the viscosity within the tube.

As to where the energy ends up, it creates turbulence in the second reservoir which then dissipates as heat.

The final transition from 95 to 0 would happen in the complicated turbulent region in the second tank,
 
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  • #18
Correction : the pressure in the tube would have to be negative, so something like -100, -101, -102, -103, -104, -105.

Then the fluid leaving the tube would decelerate in the turbulent region and get back up to 0 by the time it has slowed down to nearly zero velocity.
 
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  • #19
Freixas said:
Let's say the tube connects regions that have the same static pressure and the same 0 dynamic pressure. Even assuming the tube is frictionless, why would a flow begin?
Who said it would begin to flow? All you have been told is that would continue to flow.
 
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  • #20
Freixas said:
Now I measure the static pressures at five equally spaced regions, stretching from end to end. I'll list pressures relative to the exit (gauge) pressure, with the entrance pressure at 1,000 kPa. I'll assume a frictionless tube, but if you like you can imagine a tube with a small amount of friction. Here are some guesses as to what I might find:

Friction makes all the difference here. If there is friction, however 'small' (see my remark below), then the tube can sustain a pressure drop, if there is no friction the tube simply cannot sustain a pressure drop if the flow is assumed to be in steady state condition.

Furthermore: the only thing that will apply a force on a fluid and thus either accelerate it or change its direction is the static pressure. Nothing else (except friction of course...).

So if you add friction this actually becomes a plausible case and you would get a linear drop of static pressure over the tube of exactly this 1000kPa by your definition (assuming the tube is long enough, because you also have some entrance and exit effects causing a pressure drop, if the tube is very short the pressure drop is not linear. But these entrance and exit effects are still mostly inside the tube). It will by definition be a pressure drop of 1000kPa because the two containers at the inlet and outlet of this are by your definition maintained at 1000kPa and 0kPa respectively. The flow will accelerate, increasing the friction which is dependent on velocity, until steady state is reached. This would not violate Bernoulli, rather it would violate the applicability of Bernoulli in this case. For Bernoulli to be valid you need frictionless flow.

Note that you cannot add 'a little bit of friction'. If you allow for friction then the flow will accelerate long enough until the friction balances the pressure drop exactly. However, the roughness of the internal surface of the tube will determine what the steady state velocity will be inside the tube.

If you however do not allow for friction, then the tube cannot sustain a pressure drop in steady state. If you insist both a pressure difference over the tube and steady state regardless, then you get a very weird theoretical discussion. If there is a pressure difference the flow will accelerate because there is a force imbalance. If the flow is to be steady state, there simply cannot be a pressure difference over the tube.
 
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  • #21
There is a lot of circular discussion here making this whole thing confusing. Let me try to circle around to something a bit more basic: ##\vec{F} = m\vec{a}##.

Any fluid element is still subject to Newton's laws. In general, there are two kinds of forces acting on fluid elements: body forces surface forces. Let's neglect body forces here for a moment (gravity, electromagnetic) by assuming there is no height change. That leaves only surface forces.

The two surface forces are pressure (normal stress), which I will call ##\vec{F}_p##, and viscosity (shear stress), ##\vec{F}_v##. In a flow through a pipe with viscosity ("friction"), you need a pressure drop driving the flow that exactly counteracts the viscous forces that retard flow. For this reason, in the real world with viscosity, a steady state flow through a pipe absolutely requires a pressure drop in the direction of flow. Based on Newton for a steady flow:
[tex]\vec{F} = \vec{F}_p + \vec{F}_v = m\vec{a} = 0,[/tex]
therefore
[tex]\vec{F}_p = -\vec{F}_v.[/tex]

If you throw out viscosity so the only force remaining is pressure, then
[tex]\vec{F}_p = 0.[/tex]
This shouldn't be entirely surprising. If no outside force acts on an object, it will not change velocity, so a frictionless fluid that is already flowing does not need a pressure gradient to make it flow, nor will will spontaneously start to flow without some kind of pressure gradient applied.
 
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  • #22
Arjan82 said:
So if you add friction this actually becomes a plausible case and you would get a linear drop of static pressure over the tube of exactly this 1000kPa by your definition (assuming the tube is long enough, because you also have some entrance and exit effects causing a pressure drop, if the tube is very short the pressure drop is not linear. But these entrance and exit effects are still mostly inside the tube). It will by definition be a pressure drop of 1000kPa because the two containers at the inlet and outlet of this are by your definition maintained at 1000kPa and 0kPa respectively. The flow will accelerate, increasing the friction which is dependent on velocity, until steady state is reached. This would not violate Bernoulli, rather it would violate the applicability of Bernoulli in this case. For Bernoulli to be valid you need frictionless flow.

Someone reading this might conclude that it is only friction that could limit the steady-state flow rate, and therefore, if friction is decreased by several orders of magnitude then the steady-state flow rate could increase by a similar factor. (Assuming the pressures in the two "ideal reservoirs" are held constant while varying the friction).

Would that conclusion be correct? (I believe it would be wrong, but I'd like to make sure).
 
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  • #23
Swamp Thing said:
Someone reading this might conclude that it is only friction that could limit the flow rate, and that, if friction is decreased by several orders of magnitude then the flow rate could increase correspondingly.

Keep in mind that @Arjan82 is tackling the case where there is a fixed pressure difference. If the pressures are driven boundary conditions by some blackbox mechanism, then yes the fluid will continue to accelerate until friction limits it (or it goes into the turbulent regime, or the compressible regime yikes). If you drop the viscosity of the fluid, the flow will speed up until the viscous force balances the fixed pressure drop. The result is like taking the Hagen-Poiseuille equation where the pressure drop ##\Delta p## is a constant and dropping the viscosity ##\mu##: the flow rate ##Q## will rise. It's a weird scenario, because it's just not in line with how familiar hydraulics work (you could make something like this using diaphragm regulators, maybe?).
 
  • #24
Twigg said:
Keep in mind that @Arjan82 is tackling the case where there is a fixed pressure difference.

So am I. We're on the same page as far as driven boundary conditions being constant during the experiment and across all variants of the experiment (like changing the viscosity).

But if we reduce viscosity by orders of magnitude, will the result be a flow rate that is necessarily orders of magnitude higher? I have my doubts.

There are some "discharge rate formulas" for orifices, tubes etc where the viscosity is neglected, but the flow rate is calculated to be a finite and reasonable value. The formulas typically contain the relevant diameters, pressures and density - but not visosity. A similar principle would work in the example we are discussing.

IMHO, in this regime the flow rate would adjust itself to a value such that the dynamic Bernoulli drop would be just enough to drive the required acceleration. Those discharge rate formulas are derived by solving for this condition, IIRC.
 
  • #25
Swamp Thing said:
There are some "discharge rate formulas" for orifices, tubes etc where the viscosity is neglected, but the flow rate is calculated to be a finite and reasonable value.

Are you referring to discharge coefficients for orifice plates and venturi nozzles? If so, I don't follow. Aren't they just constant fudge factors in flow measurement devices that account for real world geometries and surface finish? Also, there's no structure in a pipe to correct for, so I'm a little confused here.

Swamp Thing said:
IMHO, in this regime the flow rate would adjust itself to a value such that the dynamic Bernoulli drop would be just enough to drive the required acceleration. Those discharge rate formulas are derived by solving for this condition, IIRC.
Again, I'm a little lost here. I thought the dynamic pressure can't cause acceleration because it doesn't represent an external force? Also, if we're talking about discharge coefficients, aren't those measured not derived?

If we're talking real-world effects that can happen in a pipe and limit the ultimate flow rate, then I guess at some flow velocity you might see cavitation at the inlet? But that's more a function of what's upstream of the pipe than the pipe itself.
 
  • #26
Thank you Twigg. Perhaps there are some errors in my understanding of what's going on.

Having said that, here's a discussion of the kind I was referring to.
https://www.engineeringcivil.com/flow-through-orifices.html
If you scroll right down to "submerged orifices", there is a formula that doesn't include viscosity but determines the flow purely from dynamic considerations. And if I understand correctly, the "discharge factor" is an empirical correction that is applied to a Bernoulli-based formula that predicts finite flow with zero viscosity.

I'll get back with some more questions..
 
  • #27
No worries, but don't take my word as gospel. Sometimes thinking about real world fluid mechanics feels like putting my brain through a pasta press.

In that discussion you linked I think what they are doing is using Bernoulli's to calculate a theoretical flow, then using a discharge coefficient to derate the flow rate to match reality, what with orifices being non-smooth non-textbook-like flow obstacles. Even if they omitted the coefficient of discharge, the final velocity would be finite. As an aside, for idealized things like "sharp-edged orifice plates" the coefficients are probably found numerically. I did not know about the relationship between the coefficient of discharge and the vena contracta, that is pretty neat and mysterious.

I see the formula you're pointing to, but I don't think this is analogous to the case of the pipe. In the pipe, you have the flow accelerate because the net force corresponding to the pressure drop is unbalanced. In the bucket-with-a-hole you linked, the bucket is exerting a normal force on the fluid that balances the static pressure, like in a converging nozzle. That's why you get a limited flow rate even with no viscosity: the walls of the bucket balance the force.
 
  • #28
Swamp Thing said:
But if we reduce viscosity by orders of magnitude, will the result be a flow rate that is necessarily orders of magnitude higher? I have my doubts.

Because the pressure difference is kept constant at both ends of the tube this implies you have an unlimited power source, because of that a lot is possible. In reality however, and if you assume air, compressibility will become important. Then the amount of airflow is limited by how fast the air can expand.

Also, the relation does not have to be linear. I mean halving the viscosity does not necessarily mean doubling the flow speed. If viscosity becomes very low you also generate a lot of turbulence (high Reynolds number). This generation of turbulence may start to dominate meaning that a further lowering viscosity hardly has an effect anymore. Note that normally we include turbulence losses also into the frictional losses because turbulence essentially dissipates into heat because of viscosity.

Swamp Thing said:
There are some "discharge rate formulas" for orifices, tubes etc where the viscosity is neglected, but the flow rate is calculated to be a finite and reasonable value. The formulas typically contain the relevant diameters, pressures and density - but not visosity. A similar principle would work in the example we are discussing.

I know of loss terms which are considered as a factor of the dynamic pressure. These are applied as extras on top of the pressure drop in a straight tube for things like the start and end of a tube, or an elbow, a T joint, a valve, an expansion/contraction etc. and depend on the exact geometry, gate valve/ball valve/plug valve etc. (typically these loss coefficients are provided by the manufacturer). In this way, you can compute the pressure drop over your plumbing as the pressure drop in the pieces of straight pipe plus a loss for each item (valve, bend, etc).

They are measured values and indeed do not depend on viscosity for their calculation as they are a fraction of the dynamic pressure. This is because the effect of these items is that they generate turbulence. Maybe you know that in the Moody Diagram the friction coefficient becomes independent on Reynolds at some point because of roughness (the higher the roughness, the earlier the independence on Reynolds, see figure below). If you keep flow rate equal, medium (air, water) equal and pipe diameter equal, this essentially means independence on viscosity.

So in this regime mostly turbulence is generated, this turbulence dissipates into heat because of viscosity, but because the scale of the flow features at which the turbulence is generated (the scale of the tube) and the scale of the flow features at which the turbulence dissipates (the Kolmogorov length scale, typically millimeters or less) are so far apart, the exact value of the viscosity doesn't matter anymore.

Moody-chart-min.jpg
 
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  • #29
The discharge coefficients take into account the part of the static pressure that is not converted to dynamic pressure (i.e. a velocity) due to losses (friction / turbulence).

So in the case of a draining pool the flow starts at rest far away of the hole and has a certain velocity when it passes the hole. In between it has converted its static pressure (due to gravity in this case) to dynamic pressure. If all of the static pressure is converted to dynamic pressure you can just apply Bernoulli to find the discharge velocity. However, this will overestimate this velocity because in reality there are losses. This is what the discharge coefficient is for.
 
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  • #30
Thank you, everyone! I have been trying to absorb everything and I thought I'd review my original post and see if I actually have answers.

1) The pressure of a fluid exiting the tube will be equal to the atmospheric pressure.

I got this from Arjan82, so I'm pretty sure it's roughly correct, although it's my paraphrasing.

Follow up points:
  • Is the statement correct or does it need some tweaks?
  • Does pressure mean "static pressure"? (I'm pretty sure it means "static".)
2) To create a flow in the tube, I've been told I need a difference in pressure.

I received an answer from boneh3ad: "... a frictionless fluid ... will [not] spontaneously start to flow without some kind of pressure gradient applied."

Follow-up points:
  • Is this true for viscous fluids as well? (I think it is; viscous or inviscid, you need a net force to start the flow.)
  • Does the term "pressure" here mean "total pressure" or "static pressure"? (I think it's "total pressure"—force is force.)
3) If I have a tube with a uniform cross-section, a steady flow and an incompressible fluid, then I can apply the continuity equation. It tells me that the velocity of the fluid is equal through any cross-section.

I don't think I saw any dissenting opinions.

4) Starting from #3, if I also assume the conditions needed for Bernoulli's equation, then the static pressure in the tube is the same at all points.

This is where the discussion veered into some rather interesting areas. It appears that by having different static pressures at the two ends of the tube and specifying a frictionless flow, I created a theoretical monster. I thought I needed a difference in pressures to get a flow started (#2) and a frictionless flow to apply Bernoulli.

You have to understand that many introductory texts start with the Bernoulli equation, give the conditions for its application and then make claims along the line of "it's not exact, but it's close". I thought I was setting up something along those lines.

----------------------------------------------------------------------------------------------

Let's test my understanding with something simpler. My answers depend on some of the points above, so if I got those wrong, I may get off-base pretty quick.
  • I have a large piston filled with water connected to a normal tube (with friction).
  • I move the piston so as to move a constant 1 L/s into the tube.
Let's see if I can get this correct:
  • I'm pretty sure I get a flow.
  • Can I apply the continuity equation? Water is not highly compressible.
  • If I can, then if I push 1 L/s in, I should get 1 L/s out and through every cross-section.
  • If I know the cross-section, I can calculate the velocity of the flow.
  • Going back to #1, the pressure at the end of the tube is whatever the exit pressure is.
  • If I incorrectly apply Bernoulli to this, the equation tells me that the static pressure is equal at every point; therefore, it has to be equal to the exit pressure.
  • I then correct for friction in the tube and from the viscosity of the water. This tells me the static pressure drop over the tube's length. There may be some other corrections that need to be applied.
  • If I were a real physicist, I would then be able to calculate the force needed by the piston to maintain that 1 L/s flow.
Let's move on to something more complicated:
  • I connect a hose to an air compressor's tank. I have now violated the incompressibility requirement of Bernoulli's.
  • I have a limitless supply of compressed air.
  • This is a guess: the mass flow through any cross-section is equal. It would seem that if x grams of air goes in, x grams has to go out. It can't be more and if it is less, the air would be collecting somewhere and the tube would eventually blow up.
  • Another guess: the fluid density is a gradient from entrance to exit. I'm totally guessing; for all I know, the density drops completely right near the entrance.
  • Based on the above: the velocity of the air at any point in the tube will not be a constant. The air will accelerate as it gets closer to the exit.
  • The pressure drops exactly the amount needed to go from the compressed tank's pressure to atmospheric pressure. Bernoulli's equation has nothing to offer us in this case.
  • The air accelerates until viscosity and tube friction balance the forces accelerating it.
The real scenario I want to analyze is just blowing through a tube, but I'm not sure if lungs compress air (making it more like an air compressor) or not (making it more like the piston). Also, maintaining a constant airflow is difficult with lung power, so it makes it harder to set up a realistic problem. One of the two scenarios above should apply.

It seems that Bernoulli's equation is a bit more useful with the piston than with the air compressor.
 
  • #31
Freixas said:
1) The pressure of a fluid exiting the tube will be equal to the atmospheric pressure.

Correct! :wink:

Freixas said:
  • Is the statement correct or does it need some tweaks?

It is correct as long as you are not interested in the pressures within, say, one diameter of the exit.

Freixas said:
  • Does pressure mean "static pressure"? (I'm pretty sure it means "static".)

Correct!

Freixas said:
2) To create a flow in the tube, I've been told I need a difference in pressure.

Correct!

Freixas said:
  • Is this true for viscous fluids as well? (I think it is; viscous or inviscid, you need a net force to start the flow.)

Exactly as you say, a (static!) pressure difference is the only way you can apply a force on any kind of fluid and a force is the only thing that will make a fluid (or anything for that matter) move (or accelerate, to be more precise).

Freixas said:
  • Does the term "pressure" here mean "total pressure" or "static pressure"? (I think it's "total pressure"—force is force.)

No! It's static pressure! Total pressure is static pressure plus kinetic energy (dynamic pressure is kinetic energy). So you can use total pressure for energy analysis (Bernoulli is an energy conservation statement). But ONLY static pressure will apply a force!

Freixas said:
3) If I have a tube with a uniform cross-section, a steady flow and an incompressible fluid, then I can apply the continuity equation. It tells me that the velocity of the fluid is equal through any cross-section.

You can ALWAYS use the continuity equation (as long as we stay on Earth that is, it fails in near vacuum). Even if the flow is compressible, or time dependent or both. The continuity equation looks a bit different in each case, but it states simply that mass is conserved. Thus the continuity equation is the mathematical equation that enforces mass conservation.

Freixas said:
4) Starting from #3, if I also assume the conditions needed for Bernoulli's equation, then the static pressure in the tube is the same at all points.

Yup, that's why applying Bernoulli in a pipe doesn't make much sense. Only when you are analyzing short sections with a contraction or something like that (so a Venturi or just a contraction in the diameter).

Freixas said:
This is where the discussion veered into some rather interesting areas. It appears that by having different static pressures at the two ends of the tube and specifying a frictionless flow, I created a theoretical monster. I thought I needed a difference in pressures to get a flow started (#2) and a frictionless flow to apply Bernoulli.

The theoretical monster truly arises if you insist on all of these three points:
  • Frictionless flow
  • A pressure difference over the tube
  • A steady state case, thus the flow looks the same now as in a bit of time
I'm pretty sure that the universe collapses if you try this...

On the piston:

Freixas said:
  • I'm pretty sure I get a flow.

You do indeed.

Freixas said:
  • Can I apply the continuity equation? Water is not highly compressible.

As I said, the continuity equation is a mass conservation statement. You can always apply this, even for a compressible flow, although the equation looks a bit different in that case.

Freixas said:
  • If I can, then if I push 1 L/s in, I should get 1 L/s out and through every cross-section.
  • If I know the cross-section, I can calculate the velocity of the flow.
  • Going back to #1, the pressure at the end of the tube is whatever the exit pressure is.

Correct, applying the first two points is essentially applying the continuity equation.

Freixas said:
  • If I incorrectly apply Bernoulli to this, the equation tells me that the static pressure is equal at every point; therefore, it has to be equal to the exit pressure.

Correct.

Freixas said:
  • I then correct for friction in the tube and from the viscosity of the water. This tells me the static pressure drop over the tube's length. There may be some other corrections that need to be applied.

Correct! Note that also here, if in steady state, the force you apply though the piston balances with the friction in the tube (In the case of a syringe it is a bit more complicated because you are also continuously accelerating flow in the reservoir to the velocity in the needle, this also takes a force to do, so to be more exact: the pressure difference over the tube balances with the friction).

Freixas said:
  • If I were a real physicist, I would then be able to calculate the force needed by the piston to maintain that 1 L/s flow.

You are already very close. This last step is not the hardest. Pressure is force divided by area, so if you multiply the force by the area of the piston you have your force again.

About the compressor bit:

Freixas said:
  • I connect a hose to an air compressor's tank. I have now violated the incompressibility requirement of Bernoulli's.

For the bread and butter Bernoulli equation indeed, but there also exists a compressible variant (and even a time dependent variant).

Freixas said:
  • I have a limitless supply of compressed air.
  • This is a guess: the mass flow through any cross-section is equal. It would seem that if x grams of air goes in, x grams has to go out. It can't be more and if it is less, the air would be collecting somewhere and the tube would eventually blow up.

Good guess, this is what the continuity equation states mathematically (if you take the correct version)

Freixas said:
  • Another guess: the fluid density is a gradient from entrance to exit. I'm totally guessing; for all I know, the density drops completely right near the entrance.

Phew, this depends on the details (and I usually only deal with incompressible flow, i.e. water...). In general the density varies but it could be that at some point the density hardly changes any further.

Freixas said:
  • Based on the above: the velocity of the air at any point in the tube will not be a constant. The air will accelerate as it gets closer to the exit.

Correct.

Freixas said:
  • The pressure drops exactly the amount needed to go from the compressed tank's pressure to atmospheric pressure. Bernoulli's equation has nothing to offer us in this case.

Correct!

Freixas said:
  • The air accelerates until viscosity and tube friction balance the forces accelerating it.

Correct! Viscosity and tube friction are two sides of the same coin.

Freixas said:
The real scenario I want to analyze is just blowing through a tube, but I'm not sure if lungs compress air (making it more like an air compressor) or not (making it more like the piston). Also, maintaining a constant airflow is difficult with lung power, so it makes it harder to set up a realistic problem. One of the two scenarios above should apply.

I would guess it is more like a piston (unless your head turns red and blue, then it may be compressible... 😂)

Freixas said:
It seems that Bernoulli's equation is a bit more useful with the piston than with the air compressor.

I think it is not useful for either...
 
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  • #32
Arjan82 said:
Correct!

Thanks, Arjan!
 
  • #33
Freixas said:
Thank you, everyone! I have been trying to absorb everything and I thought I'd review my original post and see if I actually have answers.

Kudos on actually taking the time to think through all of the responses you've received.

Freixas said:
1) The pressure of a fluid exiting the tube will be equal to the atmospheric pressure.

I got this from Arjan82, so I'm pretty sure it's roughly correct, although it's my paraphrasing.

Follow up points:
  • Is the statement correct or does it need some tweaks?

I'd argue it needs some tweaks. This is true only for an incompressible flow (which is, admittedly, what we are talking about here).

Freixas said:
  • Does pressure mean "static pressure"? (I'm pretty sure it means "static".)

In almost all cases, the word "pressure" without any modifier refers to static pressure.

Freixas said:
2) To create a flow in the tube, I've been told I need a difference in pressure.

I received an answer from boneh3ad: "... a frictionless fluid ... will [not] spontaneously start to flow without some kind of pressure gradient applied."

Follow-up points:
  • Is this true for viscous fluids as well? (I think it is; viscous or inviscid, you need a net force to start the flow.)

Yes. No matter what, and object at rest tends to stay at rest unless acted on by an outside force. That force just needs to be higher for a viscous fluid and needs to be maintained in order to sustain the flow, unlike in the inviscid case.

Freixas said:
  • Does the term "pressure" here mean "total pressure" or "static pressure"? (I think it's "total pressure"—force is force.)

Static. Static is always the relevant pressure when talking about the force on a fluid element.

Freixas said:
3) If I have a tube with a uniform cross-section, a steady flow and an incompressible fluid, then I can apply the continuity equation. It tells me that the velocity of the fluid is equal through any cross-section.

I don't think I saw any dissenting opinions.

That's correct whether the fluid is viscous or inviscid. In the viscous, case, the statement needs the additional qualifiers that the velocity in question is the average velocity.

Freixas said:
4) Starting from #3, if I also assume the conditions needed for Bernoulli's equation, then the static pressure in the tube is the same at all points.

This is where the discussion veered into some rather interesting areas. It appears that by having different static pressures at the two ends of the tube and specifying a frictionless flow, I created a theoretical monster. I thought I needed a difference in pressures to get a flow started (#2) and a frictionless flow to apply Bernoulli.

You have to understand that many introductory texts start with the Bernoulli equation, give the conditions for its application and then make claims along the line of "it's not exact, but it's close". I thought I was setting up something along those lines.

All of this is effectively true, but you can apply the Bernoulli equation to viscous flows provided you have some kind of modification to account for the effects of viscosity. Usually this is done by incorporating a "head loss" term that serves as an energy sink on the right-hand side of the equation.

Freixas said:
Let's test my understanding with something simpler. My answers depend on some of the points above, so if I got those wrong, I may get off-base pretty quick.
  • I have a large piston filled with water connected to a normal tube (with friction).
  • I move the piston so as to move a constant 1 L/s into the tube.
Let's see if I can get this correct:
  • I'm pretty sure I get a flow.

Of course.

Freixas said:
  • Can I apply the continuity equation? Water is not highly compressible.

The continuity equation applies regardless of compressibility. It's a statement that mass is conserved. The form of the equation simply changes for compressible and/or unsteady flows.

Freixas said:
  • If I can, then if I push 1 L/s in, I should get 1 L/s out and through every cross-section.
  • If I know the cross-section, I can calculate the velocity of the flow.

Correct as long as you note that you get average velocity this way (which is a meaningful distinction for viscous flows, not really for inviscid).

Freixas said:
  • Going back to #1, the pressure at the end of the tube is whatever the exit pressure is.
  • If I incorrectly apply Bernoulli to this, the equation tells me that the static pressure is equal at every point; therefore, it has to be equal to the exit pressure.

That isn't an incorrect application. It's just a trivial one. The result you get is correct under these conditions.

Freixas said:
  • I then correct for friction in the tube and from the viscosity of the water. This tells me the static pressure drop over the tube's length. There may be some other corrections that need to be applied.
  • If I were a real physicist, I would then be able to calculate the force needed by the piston to maintain that 1 L/s flow.

No issues here.

Freixas said:
Let's move on to something more complicated:
  • I connect a hose to an air compressor's tank. I have now violated the incompressibility requirement of Bernoulli's.
  • I have a limitless supply of compressed air.
  • This is a guess: the mass flow through any cross-section is equal. It would seem that if x grams of air goes in, x grams has to go out. It can't be more and if it is less, the air would be collecting somewhere and the tube would eventually blow up.

This is all correct.

Freixas said:
  • Another guess: the fluid density is a gradient from entrance to exit. I'm totally guessing; for all I know, the density drops completely right near the entrance.
  • Based on the above: the velocity of the air at any point in the tube will not be a constant. The air will accelerate as it gets closer to the exit.
  • The pressure drops exactly the amount needed to go from the compressed tank's pressure to atmospheric pressure. Bernoulli's equation has nothing to offer us in this case.
  • The air accelerates until viscosity and tube friction balance the forces accelerating it.

Compressible flows get... messy. There are a lot of effects that can come into play. In general, yes, the flow will have density changes that go along with pressure, temperature, and velocity changes. How those variables couple depends on the nature of the flow, but the usual assumption is that it is a perfect gas (so you can assume ##p=\rho R T##). Then you've got to worry about whether the flow is adiabatic or, better, isentropic, which has implications on how those variables couple as well.

In general, though, if you have no area change in your tube, then the pressure is overcoming friction, which acts along the whole tube, so all of those variables (##v##, ##\rho##, ##p##, ##T##) will be changing as a function of distance along the tube. You are correct that Bernoulli is effectively useless here. The "simple" way to analyze these flows would be what is called Fanno flow, which models friction similar to how it's done in incompressible flows with a friction factor.

Freixas said:
The real scenario I want to analyze is just blowing through a tube, but I'm not sure if lungs compress air (making it more like an air compressor) or not (making it more like the piston). Also, maintaining a constant airflow is difficult with lung power, so it makes it harder to set up a realistic problem. One of the two scenarios above should apply.

It seems that Bernoulli's equation is a bit more useful with the piston than with the air compressor.

Lungs do compress air, but my understanding is that it's maximally on the order of 1 psi, which means the flow is still going to behave as an incompressible one.
 
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  • #34
There is one puzzling thing that came up in several answers in one form or another. It is not really relevant to my initial question, but I'm curious.

boneh3ad said:
This shouldn't be entirely surprising. If no outside force acts on an object, it will not change velocity, so a frictionless fluid that is already flowing does not need a pressure gradient to make it flow, nor will will spontaneously start to flow without some kind of pressure gradient applied.

A.T. said:
Who said it would begin to flow? All you have been told is that would continue to flow.

Arjan82 said:
If there is a pressure difference the flow will accelerate because there is a force imbalance. If the flow is to be steady state, there simply cannot be a pressure difference over the tube.

The impression I'm left with is that if I have a frictionless fluid:
  • To start the flow, I need a (temporary) pressure difference between the entrance and exit.
  • If static pressure is then equal at entrance and exit, it will continue to flow.
  • If the static pressures remain unequal, the flow will accelerate until horrible things happen.
The puzzle is that that mass I accelerated eventually leaves the tube. With the pressures now equal at both ends, there would appear to be no force left to accelerate any further flow. Wouldn't the flow stop?

I would think that to maintain a flow through the tube, I would need to keep applying a force. After all, I am moving an ever increasing amount of mass the length of the tube. But it has been made clear that if I tried this, the world would end.

Of course, another answer is that my three bullet point assumptions are incorrect.

While it's not essential that I understand what's going on here, it'll probably prove useful at some point. I might find some frictionless fluid and inadvertently cause a catastrophe!
 
  • #35
Freixas said:
The puzzle is that that mass I accelerated eventually leaves the tube. With the pressures now equal at both ends, there would appear to be no force left to accelerate any further flow. Wouldn't the flow stop?

I would think that to maintain a flow through the tube, I would need to keep applying a force. After all, I am moving an ever increasing amount of mass the length of the tube. But it has been made clear that if I tried this, the world would end.

Of course, another answer is that my three bullet point assumptions are incorrect.

While it's not essential that I understand what's going on here, it'll probably prove useful at some point. I might find some frictionless fluid and inadvertently cause a catastrophe!

Think of it like pushing a block on a frictionless surface. You need some temporary force to get it going but once you let go and no longer provide a force, it just keeps going. If you keep pushing, it keeps going faster. That's the same principle as what would happen with a frictionless fluid. If you accelerate said fluid and then stop "pushing" with pressure, there is no force to flow it back down so it won't stop. Because fluids are continuous media, not a collection of discrete objects, once you get it going, the remaining fluid will in some sense be dragged along with the fluid in front of it (that's an inexact analogy, but hopefully helps with visualization).
 
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