A Simple Turbulence Closure Model. Atmospheric Sciences 6150
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1 A Simple Turbulence Closure Model Atmospheric Sciences 6150
2 1 Cartesian Tensor Notation Reynolds decomposition of velocity: V = V + v V = U i + u i Mean velocity: V = Ui + V j + W k =(U, V, W ) U i =(U 1,U 2,U 3 ) Turbulent velocity: v = ui + vj + wk =(u, v, w) u i =(u 1,u 2,u 3 )
3 v = ui + vj + wk =(u, v, w) u i =(u 1,u 2,u 3 ) Gradient operator: Advection operator: = x, y, z =, x k x 1 x 2, x 3 V = U x + V y + W z U k = U 1 + U 2 + U 3 x k x 1 x 2 x 3 The covariance matrix is a tensor of rank 2: uu uv uw vu vv vw u i u j = wu wv ww u 1 u 1 u 1 u 2 u 1 u 3 u 2 u 1 u 2 u 2 u 2 u 3 u 3 u 1 u 3 u 2 u 3 u 3
4 Turbulent kinetic energy, e = q 2 /2, and summation over repeated indices: q 2 = uu + vv + ww q 2 = u i u i = u 1 u 1 + u 2 u 2 + u 3 u 3 A tensor of rank 3: uuu uuv uuw uvu uvv uvw, uwu uwv uww = u 1 u 1 u 1 u 1 u 1 u 2 u 1 u 1 u 3 u 1 u 2 u 1 u 1 u 2 u 2 u 1 u 2 u 3 u 1 u 3 u 1 u 1 u 3 u 2 u 1 u 3 u 3 vuu vuv vuw vvu vvv vvw vwu vwv vww, u i u j u k =(u 1 u j u k, u 2 u j u k, u 3 u j u k ) u 2 u 1 u 1 u 2 u 1 u 2 u 2 u 1 u 3, u 2 u 2 u 1 u 2 u 2 u 2 u 2 u 2 u 3 u 2 u 3 u 1 u 2 u 3 u 2 u 2 u 3 u 3 wuu wuv wuw wvu wvv wvw wwu wwv www, u 3 u 1 u 1 u 3 u 1 u 2 u 3 u 1 u 3 u 3 u 2 u 1 u 3 u 2 u 2 u 3 u 2 u 3 u 3 u 3 u 1 u 3 u 3 u 2 u 3 u 3 u 3
5 Kronecker delta: First moments of velocity (3 unique): 1 i = j δ ij = 0 i = j Second moments of velocity (9, 6 are unique): U i u i u j Third moments of velocity (27,? are unique): u i u j u k
6 2 The Closure Problem The additional term is due to momentum transport by the turbulent velocity fluctuations. The momentum equation for a homogeeous incompressible fluid at high Re is V t Decompose variables into means and deviations: p + V V = + ν 2 V. (1) ρ 0 Substitute into (1) and average: V = U i + u i p = P + p U j t + U U j k = p x k x k ρ 0 u ku j x k + ν 2 U j. (2)
7 One way to close the equations is to assume that Uk u k u j = K m + U j. x j x k This is the eddy viscosity model. However, K m is a property of the flow, not of the fluid (as viscosity is), and is not necessarily a constant (as viscosity is). This type of model is generally poor in the atmosphere. Another way to close the equations is to derive equations for the Reynolds stresses, u k u j, and to make assumptions for the unknown terms in these equations in order to close the set of equations. This is a second-moment (or second-order) closure model.
8 3 Closure Models
9 4 A Simple Turbulence Closure Model We use the eddy viscosity model for the turbulent fluxes. We set the eddy viscosity K m to be proportional to the turbulence velocity scale q times a turbulence length scale l. This allows K m to depend on the turbulence properties, which is a more realistic than using a constant K m. Unknown turbulent fluxes u i u j (momentum) u i θ (any scalar) Modeling assumption q 2 3 δ Ui ij ql 1 + U j x j x i Θ ql 2 x i
10 To close these models for the turbulent fluxes, we require an equation for q 2 u i u i. To do this, we start with the full equation for q 2 : dq 2 dt = u iu i u j x j 2 pu j ρ x j 2u i u j U i x j +2 g i Θ u iθ v 2. The terms on the r.h.s. of this equation represent turbulent transport, pressure transport, shear (mechanical) production, buoyancy production (or loss), and dissipation, respectively. To close this equation we 1. Assume that production and dissipation balance: 2. Use the models above for the fluxes. 0=SP + BP D. 3. Model dissipation using = q3 Λ 1.
11 1 The result is where q 2 =Λ 1 l 1 S ij U i x j l 2 l 1 g i Θ S ij U i x j + U j x i Θ v, x i (3)
12 Let l 1 = A 1 l, l 2 = A 2 l, and Λ 1 = B 1 l, where l is the turbulence length scale. A 1 =0.92, A 2 =0.74, and B 1 = 16.6 are constants determined from experiments. Many prescriptions for the turbulent length scale l exist. The only definite constraint is that l kz near the surface so that K m = ku z under neutral conditions. One commonly used form is l l = 1+l /kz, where the asymptotic length scale l is specified to be about 10 percent of the boundary layer depth. The specification of l is usually not very critical.
13 5 Richardson Number Dependence Equation (3) includes the effects of stratification, so it should exhibit a dependence on Richardson number. To show this, we will first simplify (3) by making the boundary layer approximation: Then only and are nonzero, so Also, U 3 =0, S ij U i x j = x 1 = x 2 =0. S 13 = S 31 = U 1 x 3 S 23 = S 32 = U 2 x U1 U2 +. x 3 x 3 g i Θ v Θ x i Using these simplifications in (3), we obtain q 2 =Λ 1 l 1 U1 x 3 = g Θ v. Θ x 3 2 U2 + x 3 2 l 2 l 1 g Θ Θ v x 3. (4)
14 The condition for q 2 > 0 is therefore 2 2 U1 U2 + l 2 g Θ v > 0. x 3 x 3 l 1 Θ x 3 Write this in terms of a gradient Richardson number Ri: or l 1 l 2 > g Θ v Θ x 3 U1 x U2 x 3 2 Ri Ri < l 1 l 2 = A 1l A 2 l = A 1 A 2 = =1.24.
15 Theoretical and laboratory results suggest that laminar flow becomes turbulent when Ri < 0.25, and that turbulent flow becomes laminar when Ri > 1. These are local criteria. Even if Ri > 1 when estimated using resolved variables, it may be < 1 locally within a grid volume.
16 6 Performance This simple closure model works best when its assumption of a local balance between production and dissipation is most nearly met. This condition is most likely to be valid when and where shear production dominates buoyancy production.
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