Objective

1. Objective • Review course material • Start with numerical methods • Finite volume

2. Momentum equation –Newton’s second law dimensions of fluid particle Stress components in x direction forces per unit of volume in direction x ……………….. ……………… ……………. total derivative

3. Momentum equation Sum of all forces in x direction Internal source x direction y direction z direction

4. Newtonian fluids Viscous stress are proportional to the rate of deformation (e) Elongation: Shearing deformation: For incompressible flow Viscous stress: 0 viscosity

5. Momentum equations for Newtonian fluids x direction: y direction: z direction:

6. First Methods on Analyzing Turbulent Flow - Reynolds (1895) decomposed the velocity field into a time average motion and a turbulent fluctuation vx’ Vx - Likewise f stands for any scalar: vx, vy, , vz, T, p, where: Time averaged component

7. Time Averaged Momentum Equation Instantaneous velocity Average velocities Reynolds stresses For y and z direction: Total nine

8. Modeling of Reynolds stressesEddy viscosity models Average velocity Boussinesq eddy-viscosity approximation Is proportional to deformation Coefficient of proportionality k = kinetic energy of turbulence Substitute into Reynolds Averaged equations

9. Modeling of Turbulent Viscosity Fluid property – often called laminar viscosity Flow property – turbulent viscosity MVM: Mean velocity models TKEM: Turbulent kinetic energy equation models Additional models: LES: Large Eddy simulation models RSM: Reynolds stress models http://www.cfd-online.com/Wiki/Main_Page

10. Reynolds Averaged Navier Stokes equations Continuity: 1) Momentum: 2) 3) 4) General format:

11. General CFD Equation Values of , ,eff and S

12. - Conservation of ffor the finite volume Divide the whole computation domain into sub-domains Finite Volume Method One dimension: n h P E W dx dx w e s Dx w e l - Finite volume is a fixed space in the flow domain with imaginary boundaries that allow the fluid to flow in and out. - Integral conservation of the quantities such as mass, momentum and energy. f

13. General Transport Equation -3D problem steady-state H N W E P S L Equation for node P in the algebraic format:

14. 1-D example of discretization of general transport equation dxw dxe P Steady state 1dimension (x): E W Dx e w Point W and E represent the cell center of the west and east neighbors of cell P and w, e the neighboring surfaces. Integrating with Gaussian theorem on this control volume gives: To obtain the equations for the value at point P, assumptions are used to convert the surface values to the center values.

15. Steady–state 1D example X direction Convection term Central-difference scheme: P E W dxe dxw ……. Upwind scheme: Dx e w and and Diffusion term: When mesh is uniform: DX = dxe = dxw

16. 1D example After substitution a), b) and c) into I): We started with partial differential equation: same and developed algebraic equation: We can write this equation in general format: Unknowns Equation coefficients

17. 1D example multiple (N) volumes N unknowns i N 1 2 N-1 3 Equation for volume 1 N equations Equation for volume 2 …………………………… Equation matrix: For 1D problem 3-diagonal matrix

18. 3D problem Equation in the general format: H N W E P S L Wright this equation for each discretization volume of your discretization domain A F 60,000 elements 60,000 cells (nodes) N=60,000 = x 60,000 elements 7-diagonal matrix This is the system for only one variable ( ) When we need to solve p, u, v, w, T, k, e, C system of equation is larger

19. Convection term dxw dxe P E W Dx e w – Central difference scheme: - Upwind-scheme: If Vx>0 and and If Vx<0

20. Diffusion term dxw dxe P E W Dx e w

21. Summary: Steady–state 1D I) X direction If Vx > 0, If Vx < 0, Convection term - Upwind-scheme: P E W dxe dxw and a) and Dx e w Diffusion term: b) When mesh is uniform: DX = dxe = dxw Assumption: Source is constant over the control volume Source term: c)

22. 1D example - uniform mesh After substitution a), b) and c) into I): We started with partial differential equation: same and developed algebraic equation: We can write this equation in general format: Unknowns Equation coefficients

23. 1D example multiple (N) volumes N unknowns i N 1 2 N-1 3 Equation for volume 1 N equations Equation for volume 2 …………………………… Equation matrix: For 1D problem 3-diagonal matrix

24. 3D problem Equation in the general format: H N W E P S L Wright this equation for each discretization volume of your discretization domain A F 60,000 elements 60,000 cells (nodes) N=60,000 = x 60,000 elements 7-diagonal matrix This is the system for only one variable ( ) When we need to solve p, u, v, w, T, k, e, C system of equation is larger

25. Iteration method Alternative to use matrix solver tool is to use iterations You can use excel if you are not familiar with matrix solver tools General Iteration Procedure: 1) Express equation in explicit form 2) Guess initial values 3) Substitute initial values and calculate new values 4) Substitute new values and calculate newer values 6) Repeat step 4) until convergence is achieved example Iterations -residual 2 2 2 Difference of value between two iteration 2 2

26. Numerical instabilitydivergency divergence variable solution convergence iteration

27. Navier Stokes Equations Continuity equation This velocities that constitute advection coefficients: F=rV Momentum x Momentum y Momentum z Pressure is in momentum equations which already has one unknown • In order to use linear equation solver we need to solve two problems: • find velocities that constitute in advection coefficients • 2) link pressure field with continuity equation

28. Pressure and velocities in NS equations How to find velocities that constitute in advection coefficients? For the first step use Initial guess And for next iterative steps use the values from previous iteration

29. Pressure and velocities in NS equations How to link pressure field with continuity equation? SIMPLE (Semi-Implicit Method for Pressure-Linked Equations ) algorithm Dx Dx P E W Dx Ae Aw Aw=Ae=Aside We have two additional equations for y and x directions The momentum equations can be solved only when the pressure field is given or is somehow estimated. Use * for estimated pressure and the corresponding velocities

30. SIMPLE algorithm Guess pressure field: P*W, P*P, P*E, P*N , P*S, P*H, P*L 1) For this pressure field solve system of equations: x: ……………….. y: ……………….. z: Solution is: 2)The pressure and velocity correction P = P* + P’ P’ – pressure correction For all nodes E,W,N,S,… V = V* + V’ V’ – velocity correction Substitute P=P* + P’ into momentum equations (simplify equation) and obtain V’=f(P’) V = V* + f(P’) 3) Substitute V = V* + f(P’) into continuity equation solve P’ and then V 4) Solve T , k , e equations

31. SIMPLE algorithm start Guess p* p=p* Step1: solve V* from momentum equations Step2: introduce correction P’ and express V = V* + f(P’) Step3: substitute V into continuity equation solve P’ and then V Step4: Solve T , k , e equations no Converged (residual check) yes end

32. Other methods SIMPLER SIMPLEC variation of SIMPLE PISO COUPLED - use Jacobeans of nonlinear velocity functions to form linear matrix ( and avoid iteration )

33. Relaxation Relaxation with iterative solvers: When the equations are nonlinear it can happen that you get divergency in iterative procedure for solving considered time step divergence variable solution convergence Solution is Under-Relaxation: Y*=f·Y(n)+(1-f)·Y(n-1) Y – considered parameter , n –iteration , f – relaxation factor For our example Y*in iteration101=f·Y(100)+(1-f) ·Y(99) f = [0-1] – under-relaxation -stabilize the iteration f = [1-2] – over-relaxation - speed-up the convergence iteration Value which is should be used for the next iteration Under-Relaxation is often required when you have nonlinear equations!

34. Example of relaxation(example from homework assignment) Example: Advection diffusion equation, 1-D, steady-state, 4 nodes 1) Explicit format: 4 3 1 2 2) Guess initial values: 3) Substitute and calculate: Substitute and calculate: 4) Substitute and calculate: ………………………….