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January 4, 2019, 00:28 
Why transonic flow is highly nonlinear?

#1 
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Hi,
According to J.D.Anderson, Fundamentals of aerodynamics (2017), it is the nonlinear velocity potential equation can be (and should be) applied to both transonic/hypersonic flows, whereas we can apply a simpler version  linearized potential equation to supersonic flow. I can't wrap my mind around this. I can understand hypersonic flow having strong nonlinearity, but why supersonic doesn't? And wireder, why transonic is highly nonlinear? (and this makes me remember that I read somebody said that transonic flow is harder to solve than supersonic due to strong nonlinearity) I mean, based on the flow regime, supersonic "sits" between transonic and hypersonic. Intuitively, if transonic/hypersonic flow is highly nonlinear, supersonic should be as well. But the theory says the other way. So why? Can anyone explain this? Thanks. Last edited by TurbJet; January 4, 2019 at 16:22. 

January 4, 2019, 01:26 

#2  
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Lucky
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The general nonlinear potential flow equation is valid for all regimes. Andersons' point is that when the flow is entirely subsonic (M<1)or entirely supersonic (M>1), the linearized perturbation equation can be used. If you have a transonic case (when M=1 somewhere) the linearized potential equation makes no sense. Just look at the equation! There's a (1M^2) in it.
Hypersonic means highsubsonic where nonlinearities in the equation of state take over and to use the linearized perturbation method no longer makes any sense. That part does not need any explanation. Hypersonic flows is moreorless defined as flows where the linearized perturbation equation cannot be applied. Quote:
Also Anderson is referring to the perturbations in the potential equation when he refers to linear/nonlinear. Your linear/nonlinear refers to what.....? Last edited by LuckyTran; January 5, 2019 at 15:56. 

January 4, 2019, 04:19 

#3 
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Filippo Maria Denaro
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The key is the linearization method. You introduce a small perturbation to linearize the equation but you need to assess that all disregarded terms are small compared to the basis function. In transonic flows you deduce that you cannot disregard the non linear terms.
That has nothing to do with turbulence. The nature of the global unsteady viscous model is in the non linear NS equations system. 

January 4, 2019, 16:49 

#4  
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Quote:
As for the turbulence part, please forget it. I know that was stupid and I don't even know what was I thinking. 

January 5, 2019, 15:56 

#5  
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Lucky
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Right. Linearity/nonlinearity is a property so you need to be specific when you say what is linear. For example y = Ax^2 + B is linear in A and B but nonlinear in x. So y can be linear or nonlinear depending on what it is you are talking about....
Quote:
It's not just a strange inbetween Mach number. The perturbation parameter is 1M. When M is 1, the first order perturbation is 0. You have to actually have to do the algebra to show that the attempt at linearization results in a different equation. Meaning, it's nontrivial to show that the linearization perturbation equation does not work, but the reason why it does not work should not be a surprise. 

January 6, 2019, 18:26 

#6  
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Quote:


January 6, 2019, 18:35 

#7  
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Filippo Maria Denaro
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Quote:
The equation for the strady potential is elliptic for subsonic condition and hyperbolic for supersonic condition. They are both linear and cannot describe the nonlinearity nature in the shock wave. In transonic condition you assume that elliptic, parabolic and hyperbolic features of propagation of waves cohexist. Have a reading here https://ntrs.nasa.gov/archive/nasa/c...0020078395.pdf 

January 7, 2019, 00:21 

#8  
Senior Member

Quote:
But, it's really hard for me to wrap my mind around it. I mean, typically, hyperbolic equation has a variable in time, and so it's convective in nature. However, for this hyperbolic potential equation, both variables are only in space, no one is in time. So how does this equation behave? Is it still convective? Finally, computing resources are very powerful today. Under many circumstances, we can directly solve NS (either by applying DNS/LES/RANS) for aerodynamic problems. So I am wondering is it still necessary/meaningful to study this potential theory/equation? 

January 7, 2019, 03:45 

#9  
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Filippo Maria Denaro
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No, the hyperbolic character appears also in steady flow, it is a mathematical characteristics of a PDE with a solution f(x1,x2, ..xn). In general, I think that the study of potential flows in real applications is less relevant today but has still some meaning for educational purpose. However, is a very fast tool for preliminary results. 

January 7, 2019, 12:36 

#10  
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Lucky
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I wouldn't underestimate its educational purpose either. It is a critical ingredient in boundary layer theory. 

January 7, 2019, 20:23 

#11  
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Quote:


January 7, 2019, 20:26 

#12  
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Quote:
BTW, could you give some examples how this is used in boundary layer theory? I know linearization plays a crucial part in BLT, but I've never seen potential theory coming into play. 

January 8, 2019, 03:05 

#13 
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Filippo Maria Denaro
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Steady model drives to steady solution, no moving in time.
BL theory requires the effects of the viscosity and is no longer in the framework of potential flows. Classically, the potential solution is used as external solution in the BL theory 

January 8, 2019, 15:11 

#14 
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So I am wondering what kind of numerical schemes can I use to solve this kind of steady hyperbolic equations? Can I do central in space for both x/y directions?


January 8, 2019, 15:19 

#15  
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Lucky
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Quote:
Are you talking about solving the potential equation or solving NS? For NS these flows are strongly upwind biased. 1st order upwind works really nicely! A supersonic flow doesn't care what happens downstream. 

January 8, 2019, 16:13 

#16 
Senior Member

I was talking about the potential equation for supersonic flow.


Tags 
hypersonic flow, linear velocity, nonlinear equation, supersonic flow, transonic flow 
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