# how many BC's should be assigned physically in navier stokes runs?

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 August 9, 2013, 17:26 how many BC's should be assigned physically in navier stokes runs? #1 Senior Member     Ehsan Join Date: Oct 2012 Location: Iran Posts: 2,209 Rep Power: 18 how many BC's on inlet and outlet have to be assigned in navier stokes equations?and what are they? There is a table for euler and navier stokes but isn't clear enough for inlet and outlet. __________________ Injustice Anywhere is a Threat for Justice Everywhere.Martin Luther King. To Be or Not To Be,Thats the Question! The Only Stupid Question Is the One that Goes Unasked.

 August 10, 2013, 09:36 #2 Senior Member   Jonas T. Holdeman, Jr. Join Date: Mar 2009 Location: Knoxville, Tennessee Posts: 108 Rep Power: 10 It can be as few as none with the pressure-free modified-Hermite finite element method.

 August 10, 2013, 12:42 #3 Senior Member     Ehsan Join Date: Oct 2012 Location: Iran Posts: 2,209 Rep Power: 18 I'm not familiar with this method dear Jonas.I want to know about ordinary cases with navier-stokes equations. I know we need p,T and direction of U for inlet and only p for outlet.is it true? then whats the difference between navier-stokes and Euler equations?

August 10, 2013, 14:39
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Filippo Maria Denaro
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Quote:
 Originally Posted by immortality I'm not familiar with this method dear Jonas.I want to know about ordinary cases with navier-stokes equations. I know we need p,T and direction of U for inlet and only p for outlet.is it true? then whats the difference between navier-stokes and Euler equations?

August 10, 2013, 17:17
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Jonas T. Holdeman, Jr.
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Quote:
 Originally Posted by immortality I'm not familiar with this method dear Jonas.I want to know about ordinary cases with navier-stokes equations. I know we need p,T and direction of U for inlet and only p for outlet.is it true??
(1) It is well-established that the NS equation is composite, and can be decomposed into a pressure-free governing equation for the velocity, and a pressure equation which depends on the velocity. Just how one can implement this with continuous, truly divergence-free finite elements is less well known. (2) If a (2D) divergence-free velocity field exists, then there necessarily must exist a stream function such that the velocity is the curl of that stream function. It would seem then that the velocity degrees-of-freedom go hand in hand with the stream function (there may be div-free finite elements in primitive variables where the stream function is implicit, but these are rather complicated). (3) The stream function and solenoidal velocities can be unified (a stream function - velocity method?) using a modified Hermite element for the stream function (the modification consists of identifying the derivatives v=dPsi/dy, u=-dPsi/dx). The velocity element is the curl of this stream function element. If one uses such an element with the pressure-free form of the NSE, one finds it necessary to impose BCs on the stream function. For problems like duct flow, the BC on the stream function determines the direction and magnitude of the flow in the duct. Thus, an additional BC is necessary, but it is imposed on no-flow boundaries and not imposed on the inlet or outlet, which was the point of your question. Having found the velocity, the pressure gradient is recovered from the "pressure" equation. The pressure requires an additional integration constant, which may be specified anywhere, including a boundary. This is why no inlet/outlet BCs are necessary for solving certain NSE problems.

In principle, one could form exactly div-free functions from non div-free functions by subtracting off the gradient of an irrotational function. The problem is that one cannot do this with a FE and still preserve the element BCs. So it is common to satisfy the div-free condition only weakly. Here it gets complicated with inf-sup conditions and such. This function to be subtracted out satisfies an equation just like the pressure equation, but with different boundary conditions. But here it gets confusing because people use the symbol p to represent this projection function, and worse yet, they call it the pressure. And, they apply physical arguments about the pressure to this function.

So you are really asking about conditions on this projection function. I don't know if a firm mathematical answer exists. The practical answers are based on lore and are part of the mystique of CFD.

Including a coupled energy equation does not change the argument above, nor does inclusion of magnetic effects with conducting fluids. And there are some interesting possibilities using a fully FE method with moving domains.

While this applies to the Euler equation, that subject is treated differently because of an additional simplification.

 August 10, 2013, 17:33 #6 Senior Member     Ehsan Join Date: Oct 2012 Location: Iran Posts: 2,209 Rep Power: 18 thanks Jonas for this fundamental argument you did.when you speak about FE do you use it in contradiction of finite volume method(FV)? __________________ Injustice Anywhere is a Threat for Justice Everywhere.Martin Luther King. To Be or Not To Be,Thats the Question! The Only Stupid Question Is the One that Goes Unasked.

 August 10, 2013, 23:29 #7 Senior Member   Jonas T. Holdeman, Jr. Join Date: Mar 2009 Location: Knoxville, Tennessee Posts: 108 Rep Power: 10 My interest is in FEM for incompressible flow by the method described using modified Hermite finite elements. I have also applied the method to magnetohydrodynamics and am presently extending the work to moving fluid domains. I have placed a couple of educational codes using the method on the CFD Wiki.

August 18, 2013, 10:29
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Quote:
 Originally Posted by immortality then whats the difference between navier-stokes and Euler equations?
Navier-Stokes equations are elliptic and have 2nd order spatial derivative,while Euler equations are hyperbolic and have only 1rst order spatial derivative.
These features are very important especially for outlet boundaries treatment where non reflective BC is repquired especially for Euler equations.

 August 25, 2013, 09:01 #9 New Member   JAY Join Date: Jul 2013 Posts: 10 Rep Power: 5 I have some problem with my boundary condition that how i utilize the neumann condition(∂φ/∂x = 0) at outlet. Can i select pressure outlet at outlet? my inlet condition is : flat velocity profile =40m/s Re=1.916*10^5 turbulent intensity=4.5% And Inlet height of duct=5mm I select velocity inlet at inlet because my flow is incompresible. And axis boundary condition applied along the centerline. and no slip condition at wall. pls help me.

August 25, 2013, 09:09
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Ehsan
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Quote:
 Originally Posted by JAY ADHVARYU I have some problem with my boundary condition that how i utilize the neumann condition(∂φ/∂x = 0) at outlet. Can i select pressure outlet at outlet? my inlet condition is : flat velocity profile =40m/s Re=1.916*10^5 turbulent intensity=4.5% And Inlet height of duct=5mm I select velocity inlet at inlet because my flow is incompresible. And axis boundary condition applied along the centerline. and no slip condition at wall. pls help me.
it seems you use Fluent?if yes ask in Fluent Forum.
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August 25, 2013, 09:14
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Ehsan
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 Originally Posted by leflix Navier-Stokes equations are elliptic and have 2nd order spatial derivative,while Euler equations are hyperbolic and have only 1rst order spatial derivative. These features are very important especially for outlet boundaries treatment where non reflective BC is repquired especially for Euler equations.
Hi Reflix.
I have to say one thing,Navier-Stokes in steady-state mode is elliptic but in unsteady form,is parabolic.and why you say its necessarily 2nd order in space?we can descritize space in 1st order methods like upwind.
and in related to Euler,I think it depends on Mach number,if Mach<.7 its elliptic,if M>1.4 its hyperbolic and in .7<M<1.4 its a combination of elliptic and hyperbolic.
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August 25, 2013, 09:42
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Filippo Maria Denaro
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Quote:
 Originally Posted by immortality Hi Reflix. I have to say one thing,Navier-Stokes in steady-state mode is elliptic but in unsteady form,is parabolic.and why you say its necessarily 2nd order in space?we can descritize space in 1st order methods like upwind. and in related to Euler,I think it depends on Mach number,if Mach<.7 its elliptic,if M>1.4 its hyperbolic and in .7
Actually, the time-dependent NS equations are incompetley parabolic, the continuity equation is always hyperbolic ...
Note also that the time-dependent Euler equations are always hyperbolics, whatever the Mach numeber is, since the characteristic lines (or surface) are always real.

August 25, 2013, 13:01
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Lefteris
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Hi immortality!

Quote:
 Originally Posted by immortality Hi Reflix. I have to say one thing,Navier-Stokes in steady-state mode is elliptic but in unsteady form,is parabolic.and why you say its necessarily 2nd order in space?we can descritize space in 1st order methods like upwind.
Because the viscous term has a second order spatial derivative.

Quote:
 and in related to Euler,I think it depends on Mach number,if Mach<.7 its elliptic,if M>1.4 its hyperbolic and in .7
Where did you find that?
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August 30, 2013, 20:08
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Hi Immortality

Quote:
 Originally Posted by immortality Hi Reflix. why you say its necessarily 2nd order in space?we can descritize space in 1st order methods like upwind.
I do not mean order of the scheme but order of thespatial derivative.

NS equation is second order derivative for the diffusive term:

term with d²u/dx², d²u/dy² etc...

Quote:
 Originally Posted by immortality in related to Euler,I think it depends on Mach number,if Mach<.7 its elliptic,if M>1.4 its hyperbolic and in .7
As far as I know Euler is always hyperbolic because you don't have diffusive term.
I agree with Filippo

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