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The Reynolds Number is defined as $Re = \frac{\rho v L}{\mu}$, where $\rho$ is the density, $v$ is the fluid velocity, $L$ is the characteristic length, and $\mu$ is the dynamic viscosity. By definition, inviscid flow implies that $\mu=0$. By this formula, this would make the reynolds number infinite and thus turbulent.

Is it possible to have laminar inviscid flow? Must inviscid flow always (necessarily) be turbulent?

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  • $\begingroup$ Could you provide a rough description of the problem you want to solve? $\endgroup$
    – rul30
    Oct 8 '15 at 18:04
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No, inviscid flows are not necessarily turbulent. If there is nothing to "trip" the turbulence, then the flow will remain laminar. Features which could trip the turbulence include vibration, small temperature fluctuations, any geometric imperfections, velocity field imperfections, and other similar things.

For example, potential flow is a type of inviscid flow. Potential flow solutions are laminar solutions to the Navier-Stokes equations.

As another example, if care is taken to avoid vibration and other flow imperfections, it seems you can generate laminar pipe flows at arbitrarily high Reynolds numbers. Laminar pipe flows have been obtained at Reynolds numbers of about 100,000 under these conditions, which is far higher than the typical transition Reynolds number of about 2,000. From what I understand, there is no indication that 100,000 is any hard limit; you probably could get higher with better experimental setups.

How I think about it is this: Viscosity helps damp out flow imperfections, allowing laminar flows to occur with more imperfections. If you have a truly inviscid flow, it needs to be perfect to not trigger instabilities which lead to turbulence. If you were able to obtain an inviscid flow, you should expect turbulence due to the imperfections inherent in reality.

To answer a question you put in a comment in the other answer, yes, I do believe using a turbulence model for inviscid flow is prudent. For RANS, the Reynolds stress still will exist if the flow is inviscid, and for LES the same is true for the residual stress.

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The simple answer

Reynolds number can be more simply defined as the ratio of Inertial forces to Viscous forces.

  • viscous >> inertial = Laminar
  • viscous << inertial = Turbulent

This is why when speed/flow rate increases fluids go towards/become turbulent as you're increasing the inertia of the fluid.

Based on this logic if you had inviscid (no viscous forces) flow then the inertial forces would have to dominate which should lead to Turbulent flow.

The fringe case

If you're asking if it's possible as a 'fringe case' well... then we're getting to the boundary between engineering and science. Inviscid flow in itself is just an assumption to simplify the NS equations.

AFAIK In real life there's nothing that is truly inviscid, just things that are almost inviscid. Thus as soon as you have some incredibly-close-to-zero viscosity value, then you should have some equally incredibly-close-to-zero flow value that allows laminar flow.

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  • $\begingroup$ Indeed, inviscid flow is a simplifying assumption. My concern is whether or not I should use a turbulence model for most practical purposes when simulating inviscid flow (i.e. the velocity is definitely not close to zero). $\endgroup$
    – Paul
    Oct 8 '15 at 1:59
  • $\begingroup$ While I am definitely not a CFD guru I would assume you should given that in your case viscous<<<inertial forces. Which is what gives rise to turbulence. You might be better off asking on cfd-online $\endgroup$
    – m4p85r
    Oct 8 '15 at 4:15
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    $\begingroup$ cfd-online.com/Forums/main/10165-inviscid-flow-turbulnces.html It appears to vary based on the application/what is important in your model. $\endgroup$
    – m4p85r
    Oct 8 '15 at 4:17

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