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October 16, 2020, 11:39 
turbulent flow in steady state

#1 
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dilaw meda
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hi ,i have 2 questions:
1 when performing a numerical study for a turbulent jet using Rans , why can we do hypothesis that the flow is simulated in staeady stade when we know that turbulence is insteady ? 2 numerical models are used because of the closure problem that appears when Reynolds avreagin ? but why we do reynolds avreaging in the first place ? cordially 

October 16, 2020, 13:21 

#2 
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Today, we still typically don't have enough compute power, nor an economically valid reason, to perform a full 3D, unsteady, simulation of a turbulent flow.
So we have to use a method to, somehow, reduce the dynamical complexity of the NavierStokes equations (i.e., reduce the grid and time step size required to simulate them). There are several such methods, among which the most used ones are spatial filtering,that leads to LES, and Reynolds averaging, that leads to RANSURANS. URANS is too complex to discuss here in brief, so let's just say that RANS has the advantage of greatly reducing the complexity because it is based on a time average and you can also exploit several symmetries in your simulation. That is, a RANS field requires the smallest amount possible of grid points for a given flow, and it is steady, so you don't need time integration. Of course, all of these advantages have a cost. RANS, URANS, LES and others remove some information from the simulated flow and, as the flow dynamics is nonlinear, removing that part is not ideally possible. We typically take care of this with models (which I would more likely call physical rather than numerical), but with alternate fortunes. And note that there is no perfect model out there awaiting, it is just a cost we need to take. 

October 16, 2020, 15:05 

#3  
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Filippo Maria Denaro
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Quote:
Turbulence is unsteady and 3D of course but the key to using RANS is about the physical behavior of the mean. If the flow is "statistically" steady (the mean does not depend on time) you can use RANS. A very rude example: the function sin(t) defines an unsteady behavior but the mean is a constant (here is zero). 

October 16, 2020, 15:51 

#4  
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dilaw meda
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Quote:


October 16, 2020, 16:30 

#5 
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Filippo Maria Denaro
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October 16, 2020, 16:42 

#6  
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dilaw meda
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Quote:
I did the numerical study usigne Rans models, and the results of SST kw was in good agreement with experiments (max difference was 11% for velocity and for 7% temperature). I know that i will be asked why did i choose ''steady state'' in fluent and i have no answer for that, this is why i am searching informations in this amazing forum wich helped me a lot several times. 

October 17, 2020, 10:09 

#7  
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Quote:
Now, if your boundary conditions (which include the geometry of the boundary itself) for the solved scales are steady, guess what, there are great chances (and justification to assume) that your case is steady. There is certainly an exception with flows around bluff bodies. But one could argue that that sort of instability actually starts at scales that shouldn't be present in the simulation in the first place. 

October 26, 2020, 07:28 

#8 
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Turbulence is unsteady and 3D of course but the key to using RANS is about the physical behavior of the mean.
snaptube vidmate word to pdf Last edited by blancot; October 27, 2020 at 14:05. 

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flow, rans, reynolds, steady state, turbulent 
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