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"laminar" incomp. N-S eqs. for "turbulent" flows |
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November 25, 2001, 15:56 |
"laminar" incomp. N-S eqs. for "turbulent" flows
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#1 |
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Dear all,
Can anybody explain why it is found in the literature that the Navior-Stokes equations for incompressible flow of constant viscosity, that is, du/dt + u*grad(u) = -grad(p) + (molec.viscosity+turb.viscosity)*div*grad(u) + f are used to solved for turbulent flows? Regards, wowakai |
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November 25, 2001, 17:08 |
Re: "laminar" incomp. N-S eqs. for "turbulent" flo
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#2 |
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In the derivation of these equations, there is really nothing explicit about the condition of the flow being laminar or turbulent. Therefore, if the numerical solution uses fine enough spatial and temporal resolutions to compute all of the flow structures, these equations can be solved directly for turbulent flow, without resorting to a turbulence model.
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November 25, 2001, 22:19 |
Re: "laminar" incomp. N-S eqs. for "turbulent" flo
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#3 |
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(1)N-S eqns (strictly speaking, the set of eqns comprising continuity and motion)are not for laminar or turbulent flow situations - it is for flow. In laminar situation, viscosity is taken as constant. (2)In order to capture 'turbulence' grossly, the N-S eqns are decomposed into 'time averaged' and 'fluctuating' portions. (3)'Time averaged portion' is the one you have written.It is popularly called Reynolds Averaged N S eqns (RANS). (4)'Fluctuation' is addressed thru Reynolds stresses - how one addresses is the detailing of turbulence eqns. (5)You may like to go thru the book 'Boundary Layer Theory' by Herman Schlichting .
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November 26, 2001, 03:50 |
Re: "laminar" incomp. N-S eqs. for "turbulent" flo
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#4 |
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turb.viscosity is there to take turbulence into acount. It is in fact a modelisation of the Reynolds stress tensor wich come from the Reynolds decomposition of velocity :
let u(x,t) be the velocity vector at position x and at time t for a given experiment. if one realises the same experiment N times, one gets N values of u(x,t) (u_i(x,t), i=1,N). from this, the mean velocity [u] is define by : [u(x,t)] = (u_1(x,t) + u_2(x,t) + .... + u_N(x,t))/N and the fluctuating velocity u' comes from : u'(x,t) = u(x,t) - [u(x,t)] writing the NS equation for [u(x,t)] leads to : d[u(x,t)]/dt + [u(x,t)]*grad([u(x,t)]) + [ u'(x,t)*grad(u'(x,t))] = -grad([p(x,t)]) + (molec.viscosity)*div*grad([u(x,t)]) + f since [u'] = 0. If the flow is incompressible it comes : div(u) = 0 then div [u] = 0 then div(u') = 0 so : [ u'(x,t)*grad(u'(x,t))] = div ([u'(x,t)*u'(x,t)] Using the Boussinesq assumption for turbulence (it seems he makes a lot) : [u'(x,t)*u'(x,t)] = - (turb.viscosity)*div*grad([u(x,t)]) and the NS equation for [u(x,t)] comes as you write it. Rem 1 : [u(x,t)] is NOT define as time average velocity, but one can show that for flows such as d[u(x,t)]/dt = 0 (call statically stationnary flows), the time average velocity tends to the realisation average velocity. This leads to the fact that doing unsteady computations with RANS modeling is not correct. |
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November 27, 2001, 05:48 |
Re: "laminar" incomp. N-S eqs. for "turbulent" flo
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#5 |
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> This leads to the fact that doing unsteady computations with RANS modeling is not correct.
I meant : If one wants to solve oscillating flows with RANS one gets wrong, oscillations have been modelised through the averaging process. At the opposite, RANS models can be used with moving geometries. |
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