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Non Life Sciences => Physics, Astronomy & Cosmology => Topic started by: Eternal Student on 22/11/2022 03:50:42

Title: Why does free vortex flow have this velocity field?
Post by: Eternal Student on 22/11/2022 03:50:42

Hi.

   Anyone know of a good explanation / derivation of the free vortex velocity formula?

   V(r) =  k/r      ;     with V(r) = velocity of the fluid at radius r (and it will be tangential);  k = constant ;   r = radial co-ordinate.

   Just to formalise,  using   cylindrical co-ordinates (r, θ, z)   a free vortex around the z-axis  can be defined to be a fluid flow where the velocity field is given by:

          

(https://upload.wikimedia.org/wikipedia/commons/c/c0/Vorticity_Figure_02_a-m.gif)   Animated image from Wikimedia.

That's all some texts do, just define it that way.   However some try and go a bit further and explain why this sort of velocity field would be reasonable from some elementary assumptions about what is happening.   What I'd like to see is a good derivation of that formula.

   I've been Googling the thing for a while now and not found much that is reliable.

This post is already long and the main thing is for me to ask you something - I would like a good derivation from some elementary assumptions.   I'll put one weak argument under a spoiler, you can see it if you want.  The post is not about me telling you stuff.   It's just that if you do find a derivation based on angular momentum, pause for a moment and see if it really is reasonable.   I'm always happy for any discussion.

Spoiler: show

A weak derivation can be presented as follows:
    You've got your fluid swirling around the z-axis to start with (you may have needed to drive it around with something to start this, that doesn't matter).   For the free vortex we consider the situation where the fluid has now been left for a while, so it is not driven and there is no net torque applied to the fluid.   The flow of fluid has adjusted and settled down to a "free vortex" instead of a "forced vortex" fluid flow.

    Most of the fluid moves in circular rings around the z-axis (that's either an assumption or you could physically go and observe it to check it's reasonable).   Then we consider the angular momentum of a particle of that fluid at radial position r from the z-axis   (so let's say we're using cylindrical co-ordinates   (r, θ, z) :    That particle of fluid has mass m and angular momentum  mrv  around the z-axis .   Specifically we can set   L = mr v(r) where v(r) is the (size of) the tangential velocity of the particle  (velocity in the direction).  We can argue that v = v(r) = function of r only because of the symmetry of the situation and that the entire velocity is (mainly) tangential from an earlier assumption.
     Typically, the particle just has velocity tangentially so it never moves towards or away from the z-axis BUT assume that for some reason (in the real world there's lots of slightly imperfect flows and random events), that particle actually does move towards or away from the z-axis a little bit, so that it's now at radial position R ≠ r .  It will have to acquire a new tangential velocity v(R) - if it didn't it will be bashed into by other particles at that radial position until it does acquire the same general velocity v(R).   (You could add an assumption here, we're only interested in smooth flow solutions, so the particle must acquire the velocity v(R) in some smooth way because we can't have any of this turbulent rubbish with particles crashing into each other).
    Then it has a new angular momentum   mR v(R)    with  R ≠r.   For a "free vortex" fluid flow we assume there is nothing driving the fluid around, specifically there is nothing giving an overall torque to the fluid.   So conservation of angular momentum should apply and we have   mR v(R)  =  mr v(r) = L = constant.   Specifically we have  mr v(r) = constant = L,  irrespective of the value of r.     Re-arrange this and you get   v(r) =  L/(mr)  =   k/r    for some constant k.

   -----
It's a weak argument / derivation of the velocity field but still the best I've been able to come up with by blending other arguments together.   It's weak for many reasons but one of the main ones is that the lack of an externally applied torque to the fluid as whole does not prevent an individual particle from getting some torque.  The particle can receive some torque from one ring of fluid provided that ring receives the same (but directionally opposite) amount of torque.   At best, the argument is just that if v(r) = k/r  then  no particle needs any torque when it moves inward or outward.   The conservation of angular momentum does nothing to prevent a particle actually getting some internally sourced torque if you wanted it and thus changing it's angular momentum as it moves to a new radius.

   
Best Wishes.

Title: Re: Why does free vortex flow have this velocity field?
Post by: alancalverd on 22/11/2022 17:47:40

Perhaps begin with linear flow of a viscous fluid. The physics model is of momentum transfer from the driving shear force into successive layers at increasing distances. Then bend your model into a circle! 

Title: Re: Why does free vortex flow have this velocity field?
Post by: Eternal Student on 23/11/2022 05:15:20

Hi.
   That's an outstanding answer and you can have the "best answer" award for that. It's taken me a while to think about and I'm just astounded by your ability to put something into a few words.   Thank you very much.

   I don't have the ability to compress something into a handfull of words but I think you were saying something like this   (which I'll put under a spoiler because it would be better to keep the post short and any other comments coming in).

Spoiler: show

Starting from the results of Couette flow we see that for a viscous fluid with laminated flow along a straight line we should have a shear force between planes  given by:

   

[1]

(With  μ = coefficient of dynamic viscosity,    = velocity of the fluid orthogonal to the r direction and hence actually in the plane,   A = Area of the plane, r = direction orthogonal to the plane,  directional derivative shown with grad operator if you prefer it).

For motion mainly in circular rings we just consider infinitesimal sections as planes (which is the comment you made, "bend your model into a circle!").  The direction r orthogonal to this plane is precisely the radial direction r in the original co-ordinates used to specify the problem.    (Diagram should appear below)


Circular flow.jpg (53.72 kB . 907x421 - viewed 1276 times)

   Then we have:
   =   

[2]

   which we can take as the force, δF, on the infinitesimal plane due to the plane just below it (at radial position R>r in the diagram).
   We can simplify the above expression by assuming either that we are going to work in plane polar co-ordinates and the z-axis is redundant or that the entire vortex is contained within some vessel of height  ΔZ   along the z-axis.   If we do the latter then, the factor ΔZ  just appears as a multiplicative constant in our formulas.
   We are also only interested in the stable or equilibrium flow which the vortex will settle down to.   By symmetry arguments we can see that the velocity v(x) of this final state of the fluid flow will only depend on r  (not on angle θ  or position on the z-axis, where we can assume we're only interested in the flow in the x-y plane through the middle of the vessel in the z-axis  - or that the vortex extends to infinity in the z-axis if you prefer, this is just a simplified and idealized model - the real flow would not be a perfect free vortex near the ends of a finite length container because the walls will interfere and slow down the fluid there).
   Additionally we can see that (in the final steady flow state) the velocity at any position x cannot have any component of velocity radially or along the z-axis.   Since the flow is symmetric any inward flow at say position x₀ is an inward flow at all positions with r = r co-ordinate of x₀.  We cannot sustain such an inward flow, even the most compressible fluid cannot support a region r< r₀ that becomes infintely dense.  Similar arguments can apply to a sustained outward flow or flow in the z-axis direction.  Hence, the final steady flow solution must be such that the velocity field v =  .

  So we have, velocity is perpendicular to the radius    = v = v(r) and ΔZ is a constant (or redundant) and we can make these amendements to equation [2].

[3]

   (where η = a constant = μ or  μΔZ if you keep the z-axis).

  Equation [2] describes the force (in the positive direction) that is due to shear from the fluid at a greater radius.  (We haven't considered shear from fluid at a lower radius yet - because the original "Couette flow" formulation of shear forces describes only the shear force on the top plate due to fluid below it, there is no fluid above the top plate).  This force was acting on a small piece of our fluid at radius r from the z-axis.  So there is a torque about the z-axis.   We will be interested in the torque applied to a complete ring of the fluid.    From equation [3] we have a small element of torque   on a section of the ring at radius r and with angle between θ and θ+dθ.  Hence,
   

[4]

since v=v(r) so nothing in that integrand has any θ dependance.

   Equation [4] describes the torque on a ring of fluid at radius r due only to the fluid at a greater radius.   The critical point is that this has to be independant of r, despite the appearance of an r² term on the RHS of equation [4], which obviously tells us exactly what the derivative dv/dr must be.   However, we'll need to proceed slowly and explain why that expression can't have r dependance.     Unlike the situation with Couette flow where the top plate didn't have any fluid above it, we do have some fluid with a radius less than r.   So that fluid at smaller radius gets some torque from the fluid at a greater radius than itself.  Specifically, that fluid which is interior to our ring at radius r is being pulled by our ring, and in accordance with Newton, our ring at radius r is experiencing the equal but opposite force.  So there are two torques acting on our ring at radius r.
   Let's explain it this way:   Consider a thin ring of the fluid around the z-axis with radius between  r  and r+δr.   It gets a torque (anti-clockwise around the z-axis) from the fluid at a greater radius given by eqn [4]:
   
   The fluid interior to our ring is also getting some anti-clockwise torque from the fluid at a greater radius than itself,  however that interior fluid is not in contact with any fluid at a radius bigger than our ring   (because our ring was a width δr wide).   We can make arguments about viscous forces being well approximated as contact (or microscpically close contact) forces - which is what we'll do because this post is already too long.   It's sufficient for an idealisation of some viscous shear forces.   (LATE EDITING: surplus discussion removed).
    By Newton's third law (with equal and opposite forces),  our ring experiences a clockwise torque from the fluid interior to it.
     

   Hence the net torque on our ring of width δr  is given by        to a first order approximation  (and becomes exact for a fluid where viscous shear forces are treated as pure contact forces and δr → dr ).   If you're happy working with a ring of width δr >0  that is small but not a differential, then skip straight to the paragraph labelled [Jump here].   If you're keen to take δr down to a differential  dr  then you will notice that the net torque just described was itself just a differential and it will vanish as δr → 0.   However, so does the mass of that ring and hence it's inertia.   We have   I = moment of Inertia =  ρ(r) .2πr.δr.r².ΔZ  with ρ(r) = density of the fluid.   (Drop the ΔZ and replace ρ(r) with mass per unit area,  if you're working in plane polar co-ords and have no z-axis).    Thus   (Net Torque) =  I α     =>   angular acceleration, α,  of the ring can be expressed as a ratio of two differentials.
   α(r) =  .  That denominator is finite for all finite values of r.   So we establish that α(r) = 0   <=>  = 0.   Hence, we can find a streamline at radius r (a ring of infinitesimal width if you like) where the flow velocity changes with time while the simplified discussion just continues to talk about rings of a small width and a small net torque on that ring.

[Jump here]
    For the final steady flow solution there can't be any torque on the fluid in this ring.  If there was, then the tangential velocity would be changing here.    Hence, for the steady flow solution, the gradient is 0  (just to be clear it's everywhere 0, there's no radius r where it could be non-zero or else we can find a ring with net torque on it).
   Finally,  = 0         =>       = constant   and  from equation [4] we must then have      for some constant c     =>    v  =  (k/r) + c  for some constants c,k  and  v = k/r is the simplest solution.

    I think that does show that   is the only velocity field that could describe a steady flow free vortex when the fluid has some coefficient of viscosity μ > 0.       Thanks again for your time.

 
Best Wishes.

LATE EDITING:  Made some adjustments, it'll do.

Title: Re: Why does free vortex flow have this velocity field?
Post by: alancalverd on 23/11/2022 11:44:04

Ars longa, vita brevis*, my friend!

As for the constant of integration, clearly as r → ∞, V → 0, so surely c = 0?   Or is the entire universe rotating under some mysterious ∑ c > 0  thanks to an unpaired butterfly in orbit somewhere?  ;)


*Big bum, short of breath - that's me!

Title: Re: Why does free vortex flow have this velocity field?
Post by: Eternal Student on 23/11/2022 14:25:33

Hi.

    I agree the constant c should be 0 for a vortex extending to infinity (which is the simplest idealistic version of a free vortex flow).  Well spotted.

    I was thinking it could be negative for a free vortex (an approximate free vortex) contained in some finite region like a big cylinder (or a big flooded road).  That's then interesting because we can have some inner region with a clockwise flow, a quiet patch and a counter-clockwise flow further out.    I think I've seen effects like this in a You Tube video with flooded roads that had leaves and debris on the surface to trace streamlines naturally.   It's not that this sort of odd motion will usually be established when you start up a vortex and all the water was calm and static to begin, that's not what I think would happen.    It's just that if the outer part of a large body had some rotation clockwise to begin with,  then this can continue for a long time much as a steady flow solution while someone unblocks a drain cover  and starts up a counter-clockwise vortex near the centre of that body of water.   Provided the velocity profile settles down to   v(r) = (k/r)  - c  then this should be a viable un-forced steady flow solution for a fluid with a small viscosity.

Velocity profiles of the form  v(r) = (k/r) + c  with c positive are never likely to be sustained just because the fluid has more kinetic energy in that solution than with c=0.  Viscous forces tend to reduce the kinetic energy and dissipate heat.   However, for a negative value of c,  I think you could get the clockwise and counter-clockwise motion described and in a container of the right size that doesn't need to be a higher total kinetic energy.

   Hmmm....  Wish I could find the YT video of someone unblocking a road drain with exactly the sort of outer counter-clockwise and  inner clockwise rotations being sustained for a very long time.... can't find it now.   

Best Wishes.

Title: Re: Why does free vortex flow have this velocity field?
Post by: alancalverd on 23/11/2022 15:39:36

The old pedant says you can't have a free vortex if it's contained!

V will have to decrease faster than 1/r because the fluid in contact with the pipe isn't moving, so you need to invert your analysis, beginning with the stationary boundary layer and allowing V to increase towards the centre.

Fortunately the classical models for viscous laminar flow in a pipe will give you the answer, with the mass flow now being in the plane of the page instead of across or out of it.

Once you introduce opposing vortices you are moving towards turbulent flow characteristics at the boundary between them, and as soon as you introduce nonzero viscosity and velocities at the boundary, the system necessarily becomes unstable.

Title: Re: Why does free vortex flow have this velocity field?
Post by: Eternal Student on 23/11/2022 21:07:55

Hi.

Quote from: alancalverd on 23/11/2022 15:39:36

The old pedant says you can't have a free vortex if it's contained!

    There seems to be several different definitions of "free"  used to describe a "free vortex flow" that I can find from different websites and sources. ^*
1.    It's not driven.
2.    It's not contained and not driven.
3.    The velocity of the fluid is given by v = (k/r) and tangential.   Make of it what you will. 

*  Shipping costs, tax and handling fees obviously still apply.  You can't avoid that.

Definition 3 seems to be favoured (as mentioned in an earlier post) and you could obviously argue there were no limits imposed on r.

Quote from: alancalverd on 23/11/2022 15:39:36

V will have to decrease faster than 1/r because the fluid in contact with the pipe isn't moving...

   Set  c =  -(k/R) where the pipe has radius R.     Then a flow given by   v(r) =  (k/r) + c  (and in the direction) satisfies the boundary condition  v(R) = 0.      I'm not sure if you'd still call that a free vortex flow but it has most of what you'd want.    E.g. the velocity still falls off only as 1/r,   it's curl free (irrotational) everywhere except at the origin.   
    On the other hand, it doesn't really offer any automatic conservation of angular momentum for a particle that wanders from an outer track to an inner track like the profile  v = (k/r) would   and it just doesn't have the profile  v(r) = (k/r) if you were taking that as a definition of a free vortex flow.    The lack of an automatic conservation of momentum probably makes the flow less stable, in real life as particles do sometimes wander from one radius to another they need torque which they must be obtaining from other parts of the fluid if the vortex is not driven.

    I'm just about done with Fluid dynamics for a few days,  I'm ready to put the book back on the shelf.   I'm happy there's at least a few reasons why the velocity   k/r   is reasonable for a free vortex instead of just accepting the definition of free vortex flow as being one with that velocity field.  Thanks again for your time.

Best Wishes.

Title: Re: Why does free vortex flow have this velocity field?
Post by: paul cotter on 23/11/2022 21:31:54

You could exercise your brain with some convection studies next, Eternal Student. One runs into a host of dimensionless factors in the equations such as Nusselt, Grashof , Prandtl numbers. Don't however ask me to explain any of it!