Asymptotic Analysis of the Lattice Boltzmann Method for Generalized - PowerPoint PPT Presentation

✬ ✩ Asymptotic Analysis of the Lattice Boltzmann Method for Generalized Newtonian Fluid Flows Wen-An Yong Tsinghua University (joint with) Zaibao Yang Sino-German Symposium on Advanced Numerical Methods for Compressible Fluid Mechanics and Related Problems Beijing, May 2014 ✫ ✪

✬ ✩ Outline (7 parts) • Generalized Newtonian Fluids • Lattice Boltzmann Method • Asymptotic Expansion • Theorem • Corollaries • Observation • Conclusions ✫ ✪

✬ ✩ 1. Generalized Newtonian Fluids Newtonian fluids: air, water (in ordinary conditions) non-Newtonian fluids: shampoo, toothpaste, silly putty, blood, polymers, foams, ceramics, cement slurry, paint, food like oil, vinegar, ketchup, mayonnaise, yogurt, etc. The mechanical behaviors of non-Newtonian fluids are quite different from those of the Newtonian ones. For example, the stress tensor of a non-Newtonian fluid depends on the rate-of-deformation tensor NOT in a simple linear fashion. ✫ ✪

✬ ✩ Generalized Newtonian fluids are among the most common non-Newtonian fluids. The macroscopic motion of a generalized Newtonian fluid is governed by the Navier-Stokes equations ∇ · v = 0 , ( mass ) v t + v · ∇ v + ∇ p = ∇ · T + external force , ( momentum ) . Here v = v ( x, t ) : velocity, p = p ( x, t ) : pressure, T = µ (˙ γ ) S : stress tensor, ✫ ✪

✬ ✩ where the viscosity µ = µ (˙ γ ) is not a constant (Newtonian), but a √ tr ( S 2 ) / 2 with S the non-negative function of the shear rate ˙ γ = rate-of-strain tensor or rate-of-deformation tensor S = ∇ v + ( ∇ v ) t , where the superscript t indicates the transpose. Ref.: R. B. Bird & R. C. Amstrong & O. Hassager, Dynamics of polymeric liquids, Vol. 1: fluid mechanics (2nd Ed.), John Wiley & Sons, Inc., York, 1987 (Chapter 4). ✫ ✪

✬ ✩ Examples: The power-law model: γ n − 1 . µ (˙ γ ) = µ p ˙ ( n < 1 : shear-thinning or pseudo-plastic fluid, n > 1 : shear-thickening or dilatant fluid, n = 1 : classical Newtonian fluid.) The Bingham plastic fluid:  µ p + τ 0 if | T | ≥ τ 0  γ ˙ µ (˙ γ ) = ∞ if | T | < τ 0 .  Here µ p is the plastic viscosity, τ 0 is the yield stress, and | T | is the ✫ ✪ magnitude of the stress tensor T .

✬ ✩ 2. Lattice Boltzmann Method (LBM) The general form is f i ( x + c i h, t + δt ) − f i ( x, t ) = Ω i ( f 1 , f 2 , · · · , f N ) for i = 0 , 1 , 2 , · · · , N . Here N is a given integer, f i = f i ( x, t ) is the i -th density distribution function of particles at the space-time point ( x, t ) , c i is the i -th given velocity, h is the lattice spacing, δt is the time step, and Ω i = Ω i ( f 1 , f 2 , · · · , f N ) is the i -th given collision term. ✫ ✪

✬ ✩ Motivated by the diffusive-scaling analysis (De Masi & Esposito & Lebowitz, 1989; Sone, 2002; Inamuro & Yoshino & Ogino, 1997; Junk & Y., 2003; Junk & Klar & Luo, 2005), we take δt = h 2 in what follows, implying that the non-dimensional speed h/δt = 1 /h → ∞ as h → 0 . This indicates that we are using transport (advection) processes (LBM) to approximate the diffusion ones (NS equations). ✫ ✪

✬ ✩ For simplicity, we only consider the D2Q9 lattice: 2D problem, N = 8 and  (0 , 0) for i = 0    c i = (1 , 0) , (0 , 1) , ( − 1 , 0) , (0 , − 1) for i = 1 , 2 , 3 , 4   (1 , 1) , ( − 1 , 1) , ( − 1 , − 1) , (1 , − 1) for i = 5 , 6 , 7 , 8 .  For i ∈ { 0 , 1 , 2 , · · · , 8 } , we define  0 for i = 0    ¯ i = 3 , 4 , 1 , 2 for i = 1 , 2 , 3 , 4   7 , 8 , 5 , 6 for i = 5 , 6 , 7 , 8 .  It is clear that c i = − c ¯ i . In this sense, c i is said to be odd. ✫ ✪

✬ ✩ The first collision term is (Aharonov & Rothman, 1993; Gabbanelli et al. 2005; Psihogios et al. 2007; · · · ) 1 f eq ( ) Ω i = i ( ρ, v ) − f i . τ ( S/h ) Here the relaxation time is taken as γ ) + 1 τ ( S ) = 3 µ (˙ 2 , while the equilibrium distribution is quite standard: i ( ρ, v ) = w i [ ρ + 3 c i · v + 9 2( c i · v ) 2 − 3 f eq = f eq 2 v · v ] ✫ ✪ i

✬ ✩ ( · indicates the inner product) with 8 8 ∑ ∑ ρ = f i , v = c i f i i =0 i =0 and weighted coefficients  16 for i = 0   w i = 1  4 for i = 1 , 2 , 3 , 4 36   1 for i = 5 , 6 , 7 , 8 .  Note that w i = w ¯ i , that is, w i is even. ✫ ✪

✬ ✩ For convenience, we decompose the equilibrium distribution above as f eq = f iL ( ρ, v ) + f iQ ( v, v ) i with f iL ( ρ, v ) := w i ( ρ + 3 c i · v ) , f iQ ( u, v ) := w i [ 9 2 ( c i · u )( c i · v ) − 3 2 u · v ] . Since c i is odd and w i is even, it is easy to see that ∑ ∑ i f iL ( ρ, v ) ≡ ρ, i f iQ ( u, v ) ≡ 0 , ∑ ∑ i c i f iL ( ρ, v ) ≡ v, i c i f iQ ( u, v ) ≡ 0 . ✫ ✪

✬ ✩ The second collision term (M. Yoshino & Y. Hotta & T. Hirozane & M. Endo, 2007; C.H. Wang & J.R. Ho, 2008) Ω i = f eq i ( ρ, v ; S, h ) − f i τ with a constant relaxation time τ . The equilibrium distribution is i ( ρ, v ; S, h ) = f iL ( ρ, v ) + f iQ ( v, v ) + w i hA ( S f eq h ) S : c t i c i with scalar function A ( S ) = 3 2( τ − 1 2) − 9 2 µ (˙ γ ) . Here S : c t i c i is the standard contraction of two symmetric tensors S and c t i c i . ✫ ✪

✬ ✩ 3. Asymptotic Expansion Motivated by G. Strang (1964), we notice that the LB solution f i = f i ( x, t ; h ) depends on the lattice spacing h which is small. Thus, we seek a power-series expansion h n f ( n ) ∑ f i ( x, t ; h ) ∼ ( x, t ) . i n ≥ 0 Different from the Chapman-Enskog expansion! ✫ ✪

✬ ✩ Referring to the expansion above, we introduce 8 8 ρ ( n ) := v ( n ) := S ( n ) := ( ∇ v ( n ) )+( ∇ v ( n ) ) T f ( n ) c i f ( n ) ∑ ∑ , , i i i =0 i =0 and ∑ ∑ ∑ h n ρ ( n ) , h n v ( n ) , h n S ( n ) . ρ h ∼ v h ∼ S h ∼ n ≥ 0 n ≥ 0 n ≥ 0 ✫ ✪

✬ ✩ Take f (0) = w i . It follows clearly from the even/odd properties of w i i and c i that ρ (0) = 1 , v (0) = 0 , S (0) = 0 . By using the Taylor expansion, we can write the the left-hand side of the LBM as a power series n − 1 D i,s f ( n − s ) ∑ ∑ f i ( x + c i h, t + h 2 ) − f i ( x, t ) = h n ( x, t ) i s =1 n ≥ 2 t ( c i · ∇ ) l a differential operator. 1 l ! m ! ∂ m with D i,s = ∑ ✫ ✪ l +2 m = s

✬ ✩ For the first collision, we expand = τ ( S (1) + hS (2) + · · · ) ∼ ∑ n ≥ 0 h n F ( n ) . τ ( S h /h ) It is not difficult to see that F ( n ) is determined with S ( l ) for l = 1 , 2 , · · · , n + 1 . In particular, F (0) = τ ( S (1) ) . Thus, we may rewrite the LBM as n ≥ 2 h n ∑ n − 1 s =1 D i,s f ( n − s ) r ≥ 0 h r F ( r ) ∑ ∑ i n ≥ 0 h n [ f ( n ) − f iL ( ρ ( n ) , v ( n ) )] + ∑ n ≥ 1 h n ∑ p + q = n f iQ ( v ( p ) , v ( q ) ) . − ∑ = i ✫ ✪

✬ ✩ By equating the coefficient of h k in the two sides of the last equation and using v (0) = 0 , we obtain h 0 : f (0) = f iL ( ρ (0) , v (0) ) = f iL (1 , 0) , i h 1 : f (1) = f iL ( ρ (1) , v (1) ) , i h 2 : τ ( S (1) ) D i, 1 f (1) + f (2) = f iL ( ρ (2) , v (2) ) + f iQ ( v (1) , v (1) ) , i i h k : τ ( S (1) ) ∑ k − 1 s =1 D i,s f ( k − s ) + i + F (1) ∑ k − 2 s =1 D i,s f ( k − 1 − s ) + · · · + F ( k − 2) D i, 1 f (1) + f ( k ) i i i f iL ( ρ ( k ) , v ( k ) ) + ∑ p + q = k f iQ ( v ( p ) , v ( q ) ) = for k ≥ 3 . ✫ ✪

✬ ✩ With this hierarchy of equations, the expansion coefficient f ( k ) can be i uniquely determined in terms of ( ρ ( l ) , v ( l ) ) for l = 1 , 2 , · · · , k . Moreover, ( ρ ( l ) , v ( l ) ) can be inductively obtained by solving a hierarchy of quasilinear or linear PDEs. In particular, ρ (1) ≡ 0 and ( ρ (2) , v (1) ) satisfies the Navier-Stokes equations ∇ · v (1) = 0 + ∇ ρ (2) + v (1) · ∇ v (1) ∂v (1) = ∇ · [ 1 τ ( S (1) ) − 1 S (1) ] . ( ) ∂t 3 3 2 Namely, v (1) and ρ (2) are the respective velocity and pressure of the 3 generalized Newtonian fluid, for the relaxation time is taken as γ ) + 1 τ ( S ) = 3 µ (˙ 2 . ✫ ✪

✬ ✩ 4. Theorem Assume ρ (2 k +1) | t =0 = 0 and v (2 k ) | t =0 = 0 for k = 0 , 1 , 2 , · · · , and the viscosity µ = µ (˙ γ ) is smooth and satisfies that ˙ γµ (˙ γ ) is increasing with respect to ˙ γ ≥ 0 . Then, for periodic boundary-value problems, the expansion coefficients possess the following nice property f ( k ) = ( − 1) k f ( k ) ¯ i i for all i and all k . √ √ √ Remark: Thanks to T : T = µ (˙ γ ) S : S = 2˙ γµ (˙ γ ) , the structural condition is equivalent to that the magnitude of the stress tensor T increases with increasing the shear rate ˙ γ . ✫ ✪

✬ ✩ Moreover, ( ρ ( k +2) , v ( k +1) ) with k ≥ 1 can be consecutively obtained by solving the following linear PDEs ∇ · v ( k +1) = G k , + ∇ ρ ( k +2) ∂v ( k +1) + v (1) · ∇ v ( k +1) + v ( k +1) · ∇ v (1) ∂t 3 2 ) S ( k +1) + F ( k ) S (1) ] + ˜ 1 3 ∇ · [( τ ( S (1) ) − 1 = G k , where G k and ˜ G k depend only on ( ρ ( l ) , v ( l ) ) and their derivatives with l ≤ k . This theorem is valid also for the second collision mechanism. ✫ ✪