Lift as a consequence of streamline arguments

[arildno]

Hmm..your equation is wrong (there shouldn't be a minus sign in front of the first term on the right-hand side).
Restricting myself to the case of constant density&dynamic viscosity and conservative volume forces, taking the curl of Navier-Stokes yields:
\frac{D\vec{c}}{dt}=(\vec{c}\cdot\nabla)\vec{v}+\nu\nabla^{2}\vec{c},\vec{c}\equiv\nabla\times\vec{v},\frac{D}{dt}\equiv\frac{\partial}{\partial{t}}+\vec{v}\cdot\nabla
In the inviscid case (with volume forces being a gradient field), therefore, we have:
\frac{D\vec{c}}{dt}=(\vec{c}\cdot\nabla)\vec{v}
(Note that this is also exact in the inviscid, barotropic case, \rho=\rho(p))
As for the interpretation of the right-hand side, we note that if two points A,B in the fluid is separated by the vector \vec{AB}=dq\vec{s} (where dq is an infinitesemal quantity), we have:
\vec{v}_{B}-\vec{v}_{A}=((\vec{AB}\cdot\nabla)\vec{v})\mid_{A}=dq(\vec{s}\cdot\nabla\vec{v})\mid_{A}
This is proven as follows:
\vec{v}_{B}=\vec{v}(x_{A}+dx,y_{A}+dy,z_{A}+dz,t)=\vec{v}_{A}+dq(\vec{s}\cdot\nabla)\vec{v}\mid_{A},(dx,dy,dz)=dq\vec{s}
If therefore A, B is fluid-particles at the same instantaneous vortex line (i.e, defined with \vec{c} as the tangent vector at time t), we have, by the vorticity equation:
\vec{v}_{B}-\vec{v}_{A}=dq\frac{D\vec{c}}{dt}\mid_{A}
(at time t)
That is, the vorticity change, as experienced by particle A is only affected by the perceived velocity difference along the vortex line to which A instantaneously belongs.

To proceed further, let \vec{AB}_{t+\bigtriangleup{t}} be the vector joining A, B an instant later.
We have therefore the equality:
\vec{AB}_{t+\bigtriangleup{t}}=\vec{AB}+(\vec{v}_{B}-\vec{v}_{A})\bigtriangleup{t}=dq(\vec{c}+\frac{D\vec{c}}{dt}\bigtriangleup{t})\mid_{A}=dq\vec{c}(t+\bigtriangleup{t})\mid_{A}
That is, particles A and B remain joined to the SAME vortex line at time t+\bigtriangleup{t} as they were on time t.
(At time t+\bigtriangleup{t} the vorticity experienced by particle A is \vec{c}(t+\bigtriangleup{t})

Hence, we may conclude that in the inviscid case, vortex lines are MATERIAL curves, in that they consist of the same particles throughout time.

 

[arildno]

The Clebsch decomposition

That vortex lines are material curves, is neatly illustrated in the special case where a Clebsch decomposition of the velocity field is possible, that is:
\vec{v}=\nabla\Phi+\alpha\nabla\beta\to\vec{c}=\nabla\alpha\times\nabla\beta
In this case, the inertia term can be transformed as:
\frac{D\vec{v}}{dt}=\nabla\frac{\partial\Phi}{\partial{t}}+\nabla\beta\frac{\partial\alpha}{\partial{t}}+\nabla(\alpha\frac{\partial\beta}{\partial{t}})-\nabla\alpha\frac{\partial\beta}{\partial{t}}+\nabla(\frac{1}{2}\vec{v}^{2})+(\vec{v}\cdot\nabla\alpha)\nabla\beta-(\vec{v}\cdot\nabla\beta)\nabla\alpha=\nabla(\frac{\partial\Phi}{\partial{t}}+\alpha\frac{\partial\beta}{\partial{t}}+\frac{1}{2}\vec{v}^{2})+\nabla\beta\frac{D\alpha}{dt}-\nabla\alpha\frac{D\beta}{dt}
Hence, assuming constant density (and volume forces derivable from a potential V), we may eliminate the pressure as an independent variable, through setting:
\frac{D\alpha}{dt}=0,\frac{D\beta}{dt}=0, \nabla\cdot\vec{v}=0
(the last equation being the equation of continuity),
and the pressure fullfills:
\frac{p}{\rho}+\frac{\partial\Phi}{\partial{t}}+\alpha\frac{\partial\beta}{\partial{t}}+\frac{1}{2}\vec{v}^{2}+V=K
(where K is some constant).

That is, \alpha,\beta are conserved quantities for the fluid particle, and the vortex lines are the intersections between \alpha,\beta surfaces, since \vec{c}=\nabla\alpha\times\nabla\beta

 

[Clausius2]

Some clarification?

I will use the remainder of this post to focus on the separation phenomenon; in particular, I would like to post some thought on how this is related to STALLING, i.e., the dramatic lift reduction commonly associated with separation.

Now, separation will typically occur at points on the surface where the wall shear is zero, i.e, close to the separation point, there will be some backflow, so that a SEPARATION BUBBLE/vortex forms between the foil surface, and the streamline roughly denoting the limit betweeen the viscous and inviscid layers of the fluid.

There, is however, a subtlety connected to this which is of crucial importance for our purposes in getting some grip on the stalling phenomenon:

At first, we can say that the centers of curvature for the motion (trajectories)of the fluid in the viscous layer is no longer coincident with with the centres of curvature as determined by the foil surface (that is the case when the Prandtl equations are valid).

Rather, the centres of curvature in the viscous layer has now a complicated spatial distribution, in the form of vortex centres situated in the viscous layer itself.

Let us fix our attention to a single vortex (in the stationary case):
Locally, therefore, the particle trajectories has the approximate form of concentric "circles" about the vortex centre (that is, after all, the archetypal vortex representation), and let us furthermore assume that the local, associated velocity field is dominantly dependent on the radial variable (measured from the vortex centre).

It then follows, from Navier-Stokes, that only the pressure gradient can provide the centripetal acceleration associated with the vortex motion (viscous forces affects the tangential accelerations).
That is, the pressure must increase radially outward from the vortex centre.

Yes, my equation was wrong. Thanks for the clarification. I will post the doubts arised.

Now I want to return at the discussion of stalling. A free vortex has a velocity distribution:

v_{\theta}=\frac{A}{r} where A=\frac{\Gamma}{2\pi}

The N-S equations yields:

\frac{v_{\theta}^2}{r}=\frac{\partial P}{\partial r}

that is,

\frac{A^2}{r^3}=\frac{\partial P}{\partial r}

so that,

P(r)=-\frac{1}{2}\frac{A^2}{r^2}+C

where C is a constant solved when imposing an external pressure r\rightarrow\infty / P\rightarrow P_{\infty}

Therefore:

P(r)=-\frac{1}{2}\frac{A^2}{r^2}+P_{\infty}

It was that what you were referring to, when you said the vortex provokes points of despressurizing?. I have assumed v_r=0 which I don't if that's true.

Regards.

 

[arildno]

I'll get back to this, but I'll give what I roughly had in mind:
1) We know that viscosity tends to oppose velocity gradients; in particular, it is unrealistic to assume that infinite velocity gradients can occur in a real fluid.
It therefore seems permissible to say that viscosity imposes a BOUND upon the magnitude of the velocity gradient.
2) In an inviscid flow which is assumed to follow a curved object smoothly (no separation) ,say, a sphere, consider the following case:
Let the free-stream velocity be U, and the object a sphere with radius R.
Now, we can by potential theory solve this problem easily for an arbitrary R.
But:
Letting U be a constant, and regarding R as tiny, this flow has a huge pressure gradient associated with it!

Now let's look at the real case:
We know there will be some separation, and this will be stronger as the radius is smaller (free-stream velocity kept constant). That is, a REAL fluid is unable to set up the pressure gradient which must provide the centripetal acceleration of the fluid predicted by inviscid theory.

3)But cannot we then say that viscosity sets a BOUND upon the magnitude of the pressure gradient (indirectly by putting a bound on the magnitude of the velocity gradient)?
This idea is rather natural to assume from 1), and approaches certainty when considering 2)

Regarding this idea as our "primary", cannot we say that separation occurs whenever the required centripetal acceleration to follow the object becomes so large as to exceed the bound upon the pressure gradient dictated by viscosity?

4) This was basically my rough idea as to why separation will occur.
It is of interest to note, that in the d'Alembert's paradox case with a wing, the presssure gradient at the trailing edge is, indeed infinite (and so is the velocity there).
Hence, in a real fluid, we may say that it is the same mechanism which prevents the evolution of d'Alembert's paradox which also dictates the onset of separation (the mechanism being that the magnitude of the pressure gradient is always subject to some bound due to viscosity).

EDIT:
The argument in the cited paragraph is connected with the effort of finding a rough estimate of the pressure value at the wing. As I read your response again, was this what you were asking about?
In addition, note that the archetypal (only radially dependent) vortex may be of the form:
\vec{v}=(\frac{A}{r}+Br)\vec{i}_{\theta}
(The last vortex contribution, is of course, the velocity field associated with rigid-body motion, whereas the first is the potential point vortex field)

 

[Clausius2]

A break

I know I am storing threads of yours that must be answered. But before that, do you mind answering me this question, which appears in one past exam that I have?. It rushes me, because I have an exam on Monday.

"Under what conditions is the solution of potential flow compatible with the Euler equations?."

I know potential flow is for irrotational flow, and Euler equations can deal with both irrotational and rotational behaviors. So maybe the condition would be the flow must be irrotational?. But the logics of the question doesn't appear to be satisfied with that answer.

What do you think?

Thanks for bearing me.

 

[arildno]

Kelvin's theorem states that for an inviscid, baratropic fluid, the circulation around any material curve is constant through time.
Hence, if at say t=0 the circulation about EVERY material curve is zero, it follows they will remain so throughout time.
But, now, for example by invoking Stokes' theorem, it follows that the flow is irrotational
EDIT: The above argument, assumes, of course, that the Euler equations are at all times the "correct" equations.

Kelvin's theorem&lift:
It is repeatedly stated that lift is generated by the production of a vortex pair in the fluid, one remaining whizzing about the wing, the other (having opposite circulation) being displaced into infinity, (or dissipated by viscous forces).
This explanation is apparently consistent with:
a) An initial position of rest (prior to the plane starting to accelerate)
In this case, the circulation is certainly zero everywhere about any material curve.
b) In the time-dependent period, since the vortices are counter-spinning, the net circulation about any material curve containing both vortices remain zero.
c) The stationary case (when time has gone to infinity), where we can regard the counterspinning vortex to have gone off to infinity, so that for any material curve of finite extent bounding the wing, there remains a net circulation.

However, the trouble with this scheme, is that it introduces a singularitiy in the fluid domain IN A FINITE PERIOD OF TIME, namely that counter-spinning vortex.
This can simply not be accepted!
(I confess I've accepted this myself at some earlier point as well, but, now, I don't think the counter-vortex argument holds water, due to the finite-time singularity it implies)
The only time-dependent inviscid solution which does not introduce any singularities in the fluid domain during a finite period of time, is the solution generating d'Alembert's paradox (which, in an INfinite period of time establishes a singularity).
Hence, lift PRODUCTION cannot be explained solely within inviscid theory; there is an actual generation of NET circulation going on (in the time-dependent phase); i.e, the effect of viscosity is essential.
Inviscid theory can be shown to be lift-SUSTAINING, not lift-producing.
(It can be regarded as lift-sustaining, since Kelvin's theorem states that a given material curve will maintains whatever circulation it has through time, whenever we may approximate the flow as inviscid).

EDIT: This means that, IMO, the Euler equations cannot be regarded as the "correct" equations to view the lift-situation from (throughout time); rather, it is the Navier-Stokes equations which undergoes a surprising simplification when stationary conditions has been reached (i.e, potential theory may be used to determine the lift)