Pressure variation across streamlines in a free jet

[Soumalya]

"The static pressure at all points within a free jet of a liquid is uniform throughout and is equal to the pressure surrounding the jet "
I am looking for an explanation to the above statement.

If we consider a vertical stream of a liquid flowing through an orifice at the bottom of a liquid filled open tank, we observe the streamlines to be nearly straight at the exit. Applying, Newton's second law of motion across these straight streamlines (radius of curvature is infinite)one can easily show that the pressure at a point on the surface of the jet(often atmospheric pressure) is equal to the pressure at a point within the jet (assuming both points are at the same horizontal plane surface normal to the free jet so that elevation heads are the same for both the points).Thus, for a vertical vertically downward flowing free jet of liquid pressure is uniform throughout and the same as the surrounding or atmospheric pressure.

But again if we consider a horizontal free jet of a liquid, and apply Newton's second law of motion across the straight streamlines at the exit, the difference in pressure between a point on the surface of the jet (atmospheric pressure) and at a point within the jet is not zero as was in the previous case since both points are at different elevations this time. So, application of F=ma normal to the streamlines (straight) implies that pressure should vary hydrostatically within the free jet as is true for straight streamlines (as in open channel flows).

But even for horizontal free jets the pressure at all points within the jet is said to be equal to the pressure surrounding the jet which is atmospheric pressure under normal operations.

Any clarification on the possible misinterpretation of the analogy presented above would be highly appreciated.

Thank You

 

[256bits]

But again if we consider a horizontal free jet of a liquid, and apply Newton's second law of motion across the straight streamlines at the exit, the difference in pressure between a point on the surface of the jet (atmospheric pressure) and at a point within the jet is not zero as was in the previous case since both points are at different elevations this time. So, application of F=ma normal to the streamlines (straight) implies that pressure should vary hydrostatically within the free jet as is true for straight streamlines (as in open channel flows).

The points are at a different elevation, that much is true.

Are you assuming that from point 1 to point 2 there is a gravitational gradient worthy of consideration ( so that tidal forces become significant for small diameters ).
In most cases, with a small diameter jet, and for short distances, all parts of the jet are assumed to fall at the same rate.ie velocity and acceleration.
Not sure how you justify using F=ma for your argument.

( You should, though, take note that even a vertically falling stream will break up into parts and droplets over distance due to the acceleration of gravity )

 

[Soumalya]

The points are at a different elevation, that much is true.

Are you assuming that from point 1 to point 2 there is a gravitational gradient worthy of consideration ( so that tidal forces become significant for small diameters ).
In most cases, with a small diameter jet, and for short distances, all parts of the jet are assumed to fall at the same rate.ie velocity and acceleration.
Not sure how you justify using F=ma for your argument.

( You should, though, take note that even a vertically falling stream will break up into parts and droplets over distance due to the acceleration of gravity )

Thanks for the reply.

From your point of view, what I understand is since from the basic equation of pressure field for a fluid in which there are no shearing stresses present, a mass of fluid undergoing free fall would have no pressure gradients in all three coordinate directions in a rectangular coordinate system.Since in the jet of liquid all fluid particle are falling under the same acceleration i.e, "g" there would be no vertical surface force between adjacent fluid elements and hence no pressure in the vertical direction.

But what I am confused about is while applying the Euler's equation of motion to a fluid element in a direction normal to streamlines,we have between two points in the flow field, points 1 and 2 on different streamlines but in a direction normal to the streamline in which point 1 lies, we have

p1+∫z1z2 (V2/R) dn+ϒz1=p2+ϒz2

For a horizontal free jet the streamlines are almost straight and parallel at the exit, so we can choose a point 1 on the surface of the jet and another point 2 on the middle of the jet,to have,

p₁+ϒz₁=p₂+ϒz₂

since, for straight streamlines R=∞.Thus between the points pressure should vary hydrostatically. This is very similar to the pressure variation with depth in open channel flows with the only difference being that open channel flows are open to atmosphere at one side whereas jets are exposed to atmosphere on all sides. But Euler's equation across streamlines should be valid for both cases whatever might be the flow conditions!

This is precisely where I am most confused.

 

[Chestermiller]

You agree that the pressure at the surface of the "horizontal" jet is atmospheric, correct? You also noted the, if fluid is in free fall vertically, the pressure does not vary vertically (i.e., hydrostaticly) because the fluid is accelerating. When you have a horizontal jet, is the fluid experiencing vertical acceleration or not? Are the streamlines perfectly straight? Is the fluid not experiencing vertical acceleration even if, at the exit of the tube, its velocity is perfectly horizontal?

Chet

 

[Soumalya]

You agree that the pressure at the surface of the "horizontal" jet is atmospheric, correct? You also noted the, if fluid is in free fall vertically, the pressure does not vary vertically (i.e., hydrostaticly) because the fluid is accelerating.

Chet

Yes correct.In fact for a mass of fluid undergoing free fall in vacuum the pressure doesn't vary in any direction and the pressure is uniform and equal to the surrounding pressure throughout the fluid mass.

When you have a horizontal jet, is the fluid experiencing a vertical acceleration or not?Are the streamlines perfectly straight?

In practice we don't have a horizontal jet but a trajectory over some distance for which streamlines would be curved and this is due to the fact that the fluid experiences both a horizontal and vertical acceleration the former being the result of a pressure gradient and the latter due to gravity.

But for simplicity we focus at the fluid stream in the vicinity of the exit where streamlines converge to become "perfectly" straight for a very short extent.Yes even at this section the fluid does experience a vertical acceleration due to gravity that causes the fluid to "dip" alongwith a horizontal motion and thus we have curved streamlines.

When you have a horizontal jet, is the fluid not experiencing vertical acceleration even if, at the exit of the tube, its velocity is perfectly horizontal?

Yes the fluid experiences a vertical acceleration due to gravity that results in the curved trajectory of the fluid.

 

[Chestermiller]

Yes correct.In fact for a mass of fluid undergoing free fall in vacuum the pressure doesn't vary in any direction and the pressure is uniform and equal to the surrounding pressure throughout the fluid mass.

This doesn't only apply in vaccum.

In practice we don't have a horizontal jet but a trajectory over some distance for which streamlines would be curved and this is due to the fact that the fluid experiences both a horizontal and vertical acceleration the former being the result of a pressure gradient and the latter due to gravity.

There is no horizontal pressure gradient in the fluid after it exits the oriface. The fluid jet is surrounded by air on its entire surface, and the pressure of this air is constant at 1 atm. So the fluid pressure at the surface of the jet is 1 atm. The streamline is not perfectly straight (horizontal) even at the exit of the oriface, so the fluid is in free fall vertically. With this situation, there is no significant pressure variation across the jet cross section.

Yes even at this section the fluid does experience a vertical acceleration due to gravity that causes the fluid to "dip" alongwith a horizontal motion and thus we have curved streamlines.

So there is no significant pressure variation across the jet cross section.

Chet

 

[russ_watters]

So to be direct: the contradiction results from using contradictory assumptions: You can't simultaneously assume that gravity acts and that the stream is horizontal.

 

[Soumalya]

There is no horizontal pressure gradient in the fluid after it exits the oriface.
Chet

Applying Euler's equation along a streamline which is mostly vertical( in the tank) I arrived at the conclusion that ∂P/∂s=0 at least for the vertically straight part of the streamlines which extend from the fluid surface in the tank to a little above the orifice where they begin to curve. So this implies the pressure gradient is zero in the tank as well and at all points within the tank the pressure should be atmospheric!

Is this true?

 

[Soumalya]

So to be direct: the contradiction results from using contradictory assumptions: You can't simultaneously assume that gravity acts and that the stream is horizontal.

Well textbooks are to be blamed for incorrect diagrams which shows streamlines to be nearly perfectly horizontal at the orifice!

So if one applies Euler's equation of motion normal to the streamlines in the jet for pressures to be invariant across streamlines i.e,p₁ =p₂we must have,

∫z1 z2 (V2/R) dn=γ(z2-z1)

Is that correct?

 

[Chestermiller]

Soumalya:

Case 1: Suppose you had a perfectly horizontal finite length of fluid "rope" that is not advancing horizontally, but is falling vertically. Assume that surface tension forces are negligible so that the rope remains intact and cylindrical. Is there a vertical pressure gradient within the rope, or not?

Case 2: Same as Case 1, except that the rope has a constant horizontal velocity component. Is there a vertical pressure gradient within the rope, or not?

Chet

 

[Chestermiller]

Well textbooks are to be blamed for incorrect diagrams which shows streamlines to be nearly perfectly horizontal at the orifice!

So if one applies Euler's equation of motion normal to the streamlines in the jet for pressures to be invariant across streamlines i.e,p₁ =p₂we must have,

∫z1 z2 (V2/R) dn=γ(z2-z1)

Is that correct?

That's pretty close. It should be V²/R~g.

Chet

 

[Chestermiller]

Applying Euler's equation along a streamline which is mostly vertical( in the tank) I arrived at the conclusion that ∂P/∂s=0 at least for the vertically straight part of the streamlines which extend from the fluid surface in the tank to a little above the orifice where they begin to curve. So this implies the pressure gradient is zero in the tank as well and at all points within the tank the pressure should be atmospheric!

Is this true?

No. The fluid is not in free fall in a tank.

Chet

 

[russ_watters]

Well textbooks are to be blamed for incorrect diagrams which shows streamlines to be nearly perfectly horizontal at the orifice!

I'd have to see the diagram and description to know if that is an error or not.

 

[Soumalya]

Soumalya:

Case 1: Suppose you had a perfectly horizontal finite length of fluid "rope" that is not advancing horizontally, but is falling vertically. Assume that surface tension forces are negligible so that the rope remains intact and cylindrical. Is there a vertical pressure gradient within the rope, or not?

Case 2: Same as Case 1, except that the rope has a constant horizontal velocity component. Is there a vertical pressure gradient within the rope, or not?

Chet

Case I: No.Since pressure is the same at all points within the fluid rope which is equal to atmospheric pressure. Also one could apply Euler's equation of motion along the streamlines to obtain ∂P/∂s=0,which follows as all the potential energy is converted into kinetic energy of the fluid.

Case II: A constant horizontal velocity means no horizontal acceleration and hence no pressure gradient in horizontal direction. Of course this will not affect the absence of a vertical pressure gradient in anyway.

 

[Soumalya]

No. The fluid is not in free fall in a tank.

Chet

If the fluid is not under a free fall in the tank then why is the expression for fluid velocity obtained applying Bernoulli's equation between a point at the free surface of liquid and the orifice section is identical with the one, if the fluid element were allowed to fall freely over the same elevation as the distance between the free surface and the orifice section?

I would like to know what exactly happens along a streamline from the free surface to the orifice in terms of velocity and pressure variations until the fluid particle leaves the orifice.

 

[Soumalya]

That's pretty close. It should be V²/R~g.

Chet

It's exactly in accordance with our conclusion!

Actually I missed a "ρ" in the left hand side of the equation formulated earlier which was Euler's equation of motion applied normal to streamlines for the free jet. The correct balance for two points 1 and 2 on the surface of the jet and the middle of it respectively should be,

p1+ ρ ∫z1z2 (V2/R) dn + ϒz1=p2 + ϒz2

For the curved streamlines we have V2/R equal to "g" and dn=-dz so that, ρ ∫z1z2 (V2/R) dn = ϒ(z2-z1)

Thus we have, p₁=p₂

 

[Soumalya]

I'd have to see the diagram and description to know if that is an error or not.

This is a diagram from my textbook.

 

[russ_watters]

The part that may be confusing in that diagram (the one on the left) is that the straight lines might look like streamlines, but they aren't; They are the velcity profile at the exit. The diagram doesn't say anything at all about what happens to the water after leaving the nozzle.

 

[Soumalya]

I

The part that may be confusing in that diagram (the one on the left) is that the straight lines might look like streamlines, but they aren't; They are the velcity profile at the exit. The diagram doesn't say anything at all about what happens to the water after leaving the nozzle.

Yes you are correct!

I misjudged the velocity vectors for the streamlines.

What about the diagram to the right?

The flow geometry after the fluid exits through the orifice seems to the quite straight.Is it correctly depicted?

 

[Chestermiller]

If the fluid is not under a free fall in the tank then why is the expression for fluid velocity obtained applying Bernoulli's equation between a point at the free surface of liquid and the orifice section is identical with the one, if the fluid element were allowed to fall freely over the same elevation as the distance between the free surface and the orifice section?

I would like to know what exactly happens along a streamline from the free surface to the orifice in terms of velocity and pressure variations until the fluid particle leaves the orifice.

The pressure distribution throughout most of the tank is practically the same (hydorstatic) as if the oriface were not even present in the tank wall. Only in very close proximity to the oriface (say 3 or 4 oriface diameters away) does the pressure vary from its hydrostatic value to zero gauge pressure at the oriface exit. The streamlines near the oriface are all pointing radially toward the exit, with nearly spherical symmetry. The exit hole basically acts as a 3D potential flow sink.

There was a thread in Physics Forums several months ago which spelled out all the details of the streamline- and pressure patterns. Unfortumately, I haven't kept track of the link.

Chet

 

[russ_watters]

What about the diagram to the right?

The flow geometry after the fluid exits through the orifice seems to the quite straight.Is it correctly depicted?

The second one shows streamlines and is correctly depicted. Don't read into it more than is there: you may be assuming that it is a minor alteration of the first picture, but it isn't. Look again: where's the fluid surface and what is the height?

For all we are told, this could well be a top-down view of a totally different system!

 

[Soumalya]

The pressure distribution throughout most of the tank is practically the same (hydorstatic) as if the oriface were not even present in the tank wall. Only in very close proximity to the oriface (say 3 or 4 oriface diameters away) does the pressure vary from its hydrostatic value to zero gauge pressure at the oriface exit. The streamlines near the oriface are all pointing radially toward the exit, with nearly spherical symmetry. The exit hole basically acts as a 3D potential flow sink.

There was a thread in Physics Forums several months ago which spelled out all the details of the streamline- and pressure patterns. Unfortumately, I haven't kept track of the link.

Chet

If the pressure varies hydrostatically from the free surface of the liquid in the tank to some very close proximity to the orifice it implies a fluid particle at the surface or at any depth for which the pressure distribution is hydrostatic is not accelerating in the vertical direction. Thus,a fluid particle at the surface or within a depth for which pressure variation is hydrostatic must be stationary.Only the fluid particles at close proximity to the orifice accelerates to the velocity at the orifice. Thus we may say for the part of the streamlines upto which the pressure varies hydrostatically fluid particles possesses only elevation and pressure heads and no kinetic heads? Near the orifice where pressure changes from its hydrostatic value to zero gauge pressure the elevation and pressure heads of the fluid particles are entirely converted to kinetic head?

This implies that for the velocity distribution along the streamline extending from the surface to the orifice, the velocities are zero for the part of the streamline within the depth for which pressure variation is hydrostatic and then near the orifice along the streamlines velocity changes from zero to the value at exit?

 

[Soumalya]

The second one shows streamlines and is correctly depicted. Don't read into it more than is there: you may be assuming that it is a minor alteration of the first picture, but it isn't. Look again: where's the fluid surface and what is the height?

For all we are told, this could well be a top-down view of a totally different system!

Correct again!

It looks like a top down view of a similar arrangement as in the first figure with the smooth, well contoured orifice being replaced by a flat plate with a sharp edged hole!

 

[Chestermiller]

If the pressure varies hydrostatically from the free surface of the liquid in the tank to some very close proximity to the orifice it implies a fluid particle at the surface or at any depth for which the pressure distribution is hydrostatic is not accelerating in the vertical direction. Thus,a fluid particle at the surface or within a depth for which pressure variation is hydrostatic must be stationary.Only the fluid particles at close proximity to the orifice accelerates to the velocity at the orifice. Thus we may say for the part of the streamlines upto which the pressure varies hydrostatically fluid particles possesses only elevation and pressure heads and no kinetic heads?

Yes. Pretty much so, neglecting the small small velocity up to that point.

Near the orifice where pressure changes from its hydrostatic value to zero gauge pressure the elevation and pressure heads of the fluid particles are entirely converted to kinetic head?

This implies that for the velocity distribution along the streamline extending from the surface to the orifice, the velocities are zero

Close to zero, not exactly equal to zero.

then near the orifice along the streamlines velocity changes from zero to the value at exit?

Yes. In the thread that I was referring to, we sketched the streamlines. The sketches showed how the streamlines strongly converge in the region very close to the exit hole, and how they look throughout the rest of the tank. See if you can find that thread.

Chet

 

[Chestermiller]

This is the link I was referring to: https://www.physicsforums.com/threads/calculating-velocity-for-water-pressure.794517/#post-4990608

 

[Soumalya]

This is the link I was referring to: https://www.physicsforums.com/threads/calculating-velocity-for-water-pressure.794517/#post-4990608

Thanks sir Chet :)

The thread was very useful in aiding my imagination understanding the flow within the tank.I am going through the rest of my text studying fluid dynamics and it seems I am able to understand a lot of things more clearly than before.

Thanks to you and Russ