On the filling of a glass of water The Reynolds number is about 10 4 - PowerPoint PPT Presentation

On the filling of a glass of water The Reynolds number is about 10 4 − 10 5 . The viscous time scale is about 10 4 s . R. Nguyen van yen (FU Berlin) Boundary layers and dissipation October 16, 2012 2 / 28

Outline The vanishing viscosity limit 1 The Prandtl 1904 theory 2 Numerical method 3 Results 4 R. Nguyen van yen (FU Berlin) Boundary layers and dissipation October 16, 2012 4 / 28

The vanishing viscosity limit The role of vorticity In an incompressible flow ( ρ = 1), � u 2 d E d t = d � ω 2 = − 2 ν Z 2 = − ν (1) d t where ω z = ∂ x u y − ∂ y u x , (2) . . . To dissipate energy, vorticity needs to be created and/or amplified, in such a way that Z ∼ ν − 1 . Examples: ω ∼ ν − 1 / 2 over O (1) area, ω ∼ ν − 1 over O ( ν ) area. R. Nguyen van yen (FU Berlin) Boundary layers and dissipation October 16, 2012 5 / 28

The vanishing viscosity limit The 2d Euler and Navier-Stokes equation  ∂ t ω + u · ∇ ω = 0   ( E ) : ω = ( ∇ × u ) · e z , ∇ · u = 0 (3)  ω (0) = ω i u | ∂ Ω · n = 0 ,   ∂ t ω + u · ∇ ω = ν ∆ ω   ( NS ) : ω = ( ∇ × u ) · e z , ∇ · u = 0 (4)  ω (0) = ω i u | ∂ Ω = 0 ,  When friction forces are negligible, vorticity is conserved by the flow. This is no longer the case when friction forces are introduced. One can show that there is a Neumann boundary condition for ω : ν∂ y ω = − ∂ x p , (5) where p is the pressure field. This is the source of vorticity we were looking for! R. Nguyen van yen (FU Berlin) Boundary layers and dissipation October 16, 2012 6 / 28

The vanishing viscosity limit Vorticity and pressure boundary conditions Vorticity is produced in reaction to pressure gradient, in the right amount to keep the tangential velocity to zero: ν∂ n ω = − ∂ τ p , (6) This gives a hope of resolving the paradox. Can we get sufficient vorticity when ν → 0? How do we compute the pressure field? ∆ p = − ∇ (( u · ∇ ) u ) inside Ω (7a) ∂ n p | ∂ Ω = ν∂ τ ω | ∂ Ω ( Stokes pressure ) (7b) Note: this boundary condition for ω is nonlocal and nonlinear ! In fact it is possible to reformulate it in a linear way, but it always remain nonlocal. R. Nguyen van yen (FU Berlin) Boundary layers and dissipation October 16, 2012 7 / 28

The vanishing viscosity limit Link with d’Alembert’s paradox For a solid immersed in a perfect fluid (i.e. satisfying the Euler equations), the only contact force at the boundary is the pressure force. The overall force is thus: � � F = f drag = p n dS (8) ∂ Ω ∂ Ω D’Alembert basically showed that in a parallel stationnary flow, the streamwise component F drag = 0. In a viscous flow, there is a tangential component to the drag: f viscousdrag = ν ( ∂ x u y + ∂ y u x ) | ∂ Ω = ν∂ y u x | ∂ Ω = − νω | ∂ Ω (9) This seems to resolve d’Alembert’s paradox. R. Nguyen van yen (FU Berlin) Boundary layers and dissipation October 16, 2012 8 / 28

The Prandtl 1904 theory The Prandtl theory For simplicity we work with Ω = R × ]0 , + ∞ [. 1 2 y , we take the Ansatz : In the limit ν → 0, and letting y 1 = Re 1 2 ω 1 ( x , y 1 , t ) + R ( x , y , t , Re ) ω ( x , y , t ) = ω 0 ( x , y , t ) + Re (10) By injecting that into the NSE, we get that ω 0 should be a solution of (E), and ω 1 of the Prandtl equations:  ∂ t ω 1 + u 1 , x ∂ x ω 1 + u 1 , y ∂ y 1 ω 1 = ∂ 2 y 1 ω 1   � y 1  0 d y ′ 1 ω 1 ( x , y ′  u 1 , x = − 1 , t )  ( P ) : , (11) � y 1 � y ′ 0 d y ′ 0 d y ′′ 1 ∂ x ω 1 ( x , y ′′ u 1 , y = 1 1 , t ) 1     ∂ y 1 ω 1 ( x , 0 , t ) = − ∂ x p 0 ( x , 0 , t )  where p 0 is the pressure computed only from ω 0 . R. Nguyen van yen (FU Berlin) Boundary layers and dissipation October 16, 2012 9 / 28

The Prandtl 1904 theory Shortcoming of Prandtl’s theory 1 2 ω 1 ( x , y 1 , t ) + R ( x , y , t , Re ) ω ( x , y , t ) = ω 0 ( x , y , t ) + Re (12) By applying the Biot-Savart kernel to (11), we see that as long as the Prandtl boundary layer theory is valid, the flow converges to a solution of the Euler equations. In particular, there can be no injection of vorticity into the bulk flow. Enstrophy production requires breakdown of Prandtl scaling. How does this happen? R. Nguyen van yen (FU Berlin) Boundary layers and dissipation October 16, 2012 10 / 28

The Prandtl 1904 theory Mathematics Theorem (Kato ’84) In 2D with smooth initial conditions and smooth domain boundaries, the following assertions are equivalent: ν → 0 u 0 ( t ) in L 2 uniformly in time for t ∈ [0 , T ] , 1 u ν ( t ) − − − → � T { x ∈ Ω | d ( x ,∂ Ω) < c ν } � ∇ u � 2 − 2 ∃ c > 0 , ν � ν → 0 0 − − → 0 For convergence to break down, there has to be energy dissipation in a very thin layer of thickness O ( ν ) along the wall. We study this (hypothetical) breakdown numerically: solve the Navier-Stokes equations in a 2D channel. solve the Euler and Prandtl equations for the same initial data. What happens? How do things scale with Reynolds number? R. Nguyen van yen (FU Berlin) Boundary layers and dissipation October 16, 2012 11 / 28

Numerical method The geometry 2π 2π π π y x R. Nguyen van yen (FU Berlin) Boundary layers and dissipation October 16, 2012 13 / 28

Numerical method The initial data 2π π y x R. Nguyen van yen (FU Berlin) Boundary layers and dissipation October 16, 2012 14 / 28

Numerical method Euler solver Use mirror symmetry around y = 0 to impose boundary condition. Spatial discretization: Fourier 2π pseudo-spectral with hyperdissipation, k max = 682. Time discretization: third order low y storage Runge-Kutta, with x exponential propagator for the 2π viscous term. Play movie R. Nguyen van yen (FU Berlin) Boundary layers and dissipation October 16, 2012 15 / 28

Numerical method Prandtl solver Artificial boundary condition ω 1 = 0 at y 1 = 32. Spatial discretization : second order finite differences. Time discretization : second order semi-implicit Runge-Kutta. Neumann boundary condition for ∂ t ω 1 at y 1 = 0 appears when inverting Helmholtz problem at each timestep: ∂ y 1 ∂ t ω 1 = − ∂ xt p 0 ( x , 0 , t ) (13) Time derivative of pressure approximated from Euler solution and using 2nd order Lagrange polynomials in time. Play movie R. Nguyen van yen (FU Berlin) Boundary layers and dissipation October 16, 2012 16 / 28

Numerical method Navier-Stokes solvers Solver A : Same as Euler solver, but using standard dissipation, and boundary conditions approximated using volume penalization method. 2π y x 2π Solver B : Second order finite volumes in space, Second order time-splitting predictor, R. Nguyen van yen (FU Berlin) Boundary layers and dissipation October 16, 2012 17 / 28 Regular 1024 × 4096 grid on subdomain [0 0 25] × [0 0 5], and use

Results Finite-time singularity in Prandtl’s equations Prandtl equation has well-known finite time singularity (van Dommelen and Shen, 1980): | ∂ x ω 1 | and u 1 , y blows up, ω 1 remains bounded. This is observed in our computations as expected, for t → t D ≃ 55 . 8: −3 8 x 10 40 Nx = 512 Nx = 512 7 Nx = 1024 35 Nx = 1024 Nx = 2048 Nx = 2048 6 30 Nx = 4096 Nx = 4096 5 25 ∞ || ∂ x ω || L ∞ || ω 1 || L 4 20 3 15 2 10 1 5 0 0 0 10 20 30 40 50 60 0 10 20 30 40 R. Nguyen van yen (FU Berlin) Boundary layers and dissipation October 16, 2012 18 / 28 time time

Results Convergence before the singularity According to Kato’s theorem, and since ω 1 remains bounded L 2 uniformly until t D , we expect that u ν − ν → 0 u 0 uniformly on [0 , t D ]. − − → Show convergence! R. Nguyen van yen (FU Berlin) Boundary layers and dissipation October 16, 2012 19 / 28

Results What happens at the singularity? Play movie 70 Re = 5e+06 Re = 1e+07 −7 5 x 10 Re = 2e+07 60 Re = 4e+07 4.5 50 4 maximum vorticity 3.5 40 3 −dE/dT/Z 2.5 30 NSE/Finite Volumes NSE/Penalization 2 data3 NSE/Finite Volumes 20 NSE/Penalization 1.5 data6 NSE/Finite Volumes 1 NSE/Penalization 10 data9 NSE/Finite Volumes 0.5 NSE/Penalization data12 0 0 0 10 20 30 40 50 0 10 20 30 40 50 60 70 time time R. Nguyen van yen (FU Berlin) Boundary layers and dissipation October 16, 2012 20 / 28

Results What happens after the singularity? Dissipative structures detach from the wall and are advected into the bulk flow 1 . These structures are likely to play a role in any kind of wall-bounded turbulent flows. The scaling Re − 1 for the wall-normal extension of the collapsed boundary layer is roughly consistent with the von K´ arm´ an theory, although it is not clear why or how. 1 Romain Nguyen van yen, Marie Farge, and Kai Schneider. “Energy Dissipating Structures Produced by Walls in Two-Dimensional Flows at Vanishing Viscosity”. In: Phys. Rev. Lett. 106.18 (2011), p. 184502. R. Nguyen van yen (FU Berlin) Boundary layers and dissipation October 16, 2012 21 / 28