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August 28, 2013, 03:12 
Inviscid Turbulent flow

#1 
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Vino
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I have a strange question. Is there any possibility that an inviscid flow can be a turbulent flow and do we use any inviscid turbulent model for our practical application?


August 28, 2013, 04:15 
Inviscid Turbulent flow

#2 
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rizzu
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Yes, If the fluid has no viscosity, the viscous force will be small compared to inertial force. So the value of the Reynold’s number will be high. So we will have turbulent flow. But one thing practically, there is no invisid fluid, as every fluid has some viscosity. But for our calculation purpose if the fluid has low viscosity, we assume the viscosity is zero.


August 28, 2013, 04:51 

#3 
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Filippo Maria Denaro
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inviscid turbulence is a theoretical situation: for zero viscosity (infinite Reynolds number) the turbulence features an energy cascade extending up to an infinite wavenumbers since the Kolomogorov scale is zero. No dissipative part of the energy spectrum exists.
In such case, no DNS is realizable and LES can be performed with some care. See this LES: http://wwwpersonal.engin.umd.umich....nn_Dhanak2.pdf 

August 28, 2013, 12:19 

#4  
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duri
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Quote:
There is no such inviscid tubulent model, what ever dissipation a cfd solver gives is purely numerical. Since high reynolds number flow can be approximated to inviscid flow except near boundary layer, it cannot be used for many cases except were viscous effects dominates. 

August 28, 2013, 12:54 

#5  
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cfdnewbie
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Turbulence is generated through a vortex stretching mechanism, which only required threedimensionality. Viscosity is NOT needed to produce a scale cascade, only to limit the cascade to a finite bandwidth. 

August 28, 2013, 12:58 

#6  
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Filippo Maria Denaro
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Vorticity equation exists too for inviscid flows and stretching term does not vanish for Euler equation. The contribution of the viscosity is in the appearance of the end of the inertiale energy cascade due to the dissipation. 

August 28, 2013, 13:44 

#7 
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While it is true that turbulence is generated by vortex stretching, which is present even in inviscid flows, the question still presents itself: where does the vorticity arise. Since the original poster stated that his interest is in practical applications, this question is not a trivial one. In any application with solid walls, the absence of viscosity not only eliminates a source of vorticity production, but also removes a strong local damping function of turbulence. Substituting numerical dissipation for physical viscosity has always seemed a little off to me, and I'm still skeptical of results from such simulations. Of course, I had a colleague once who said that all turbulence models are just rough approximations, and maybe an inviscid approach is just a little rougher than most.


August 28, 2013, 13:57 

#8  
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Filippo Maria Denaro
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August 29, 2013, 09:14 

#9  
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Vino
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My doubt again is, as our friend said above, is there any possibility that an inviscid flow can generate turbulence? even when we stir inviscid flow with spoon for example, we can not generate vorticity, because of slip condition.( i agree that the situation is different for practical idealization which was my original question ) 

August 29, 2013, 09:48 

#10 
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Filippo Maria Denaro
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Ok, let's stop for a moment thinking about vorticity.
Consider the 1D inviscid Burgers equation du/dt + d/dx (u^2/2) =0 This model produces what is called "Burgulence", see http://arxiv.org/pdf/nlin/0012033.pdf Now consider a very smooth initial condition, for example u(x,0) = u0*sin(x), you will simply see that the Fourier component exp(ikx) when inserted into the non linear term will produce exp(2ikx) and so on for time passing... This is the generation of a scale cascade adn is a model to understand the born of turbulence. The issue is that even a very smooth initial field can produce an infinite number of wavelenghts. Of course, if the initial is exactly zero you can not generate nothing but if you have some "small fluctuations" superimposed this generates a cascade. in 2 and 3D cases, the scenario is the same, the non linearity in the momentum equation generates the scale cascade that can be seen by the vorticity in the stretching action. Viscosity will only dissipate vortices at the Kolmogorov scale (well, the dissipation starts at the Taylor microscale and continues up to the Kolmogorov scale) In conclusion, any practical application in which velocity gradients are present and interact by means of the non linear convection, produce scale cascade despite of the absence of viscosity. 

August 29, 2013, 16:07 

#11 
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Vino
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Thank you very much. Now i got some better clarity regarding this.


May 8, 2019, 11:35 
There is an asymptotic theory ...

#12 
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Helmut Z. Baumert
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Folks, there is an asymptotic approach for Re => \infty based on early ideas of Lord Kelvin, Kolmogorov, Hans Herrmann and my self: the KKHBtheory of highRe turbulence. It is based on the idea that Kelvins 'vortex atoms' (these are vortex dipols) form an interacting dense ensemble where collisions of likewise rotating vortices (+,+) or (,) lead to dissipation and collisions of (+,) or (,+) lead to reorientation of the dipol motions. The whole concept is closed and gives for the flow at a plance wall the Karman constant as 1/SQRT(2*Pi) ~ 0.399 where the intl. standard value is 0.40. The title of the theory is "Universal equations and constants of turbulent motion" and published in Physica Scripta. If you have questions: baumert@iamaris.de


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