SOE3213/4: CFD Lecture 1

Transcription

1 What is CFD SOE3213/4: CFD Lecture 1 3d

2 3d Computational Fluid Dynamics { use of computers to study uid dynamics. Solve the Navier-Stokes Equations (NSE) : r:u = 0 Du Dt = rp + r 2 u + F 4 s for 4 unknowns, coupled, non-linear, and often involve complex geometries.

3 3d 4 stage process 1 Modelling N-S s, Turbulence Combustion, multiphase, shocks : : : 2 Discretisation 3 Solving. M u = q ) p u = M 1 q p 4 Interpretation

4 3d From the user's point of view, CFD modelling is a 3 stage process : 1 Problem denition. geometry, mesh, boundary conditions physical models, constants 2 Numerical solution. dierencing schemes, soln techniques. tolerances actually running the solver 3 Postprocessing.

5 3d Finite Dierence methods { variables represented at discrete points in space. = p j+1 p j 1 2 x { very complex if the grid distorted or complicated. Finite Element (FE) methods have been used, but are not common. Finite Volume (FV) method most common. { used in most commercial codes { Fluent, STAR-CD, CFX etc.

6 3d ow region is divided into small boxes { cells or control volumes, forming mesh. s reexpressed in terms of ow into and out of each cell. This is done by integrating the s over the volume of each cell. The result is a set of dierence s which can be solved numerically as before. The advantage of doing things this way is that the cells can be any shape required - cubes, tetrahedra, distorted cubes, or more complicated structures.

7 Water owing in a river is contaminated by a chemical leak. Determine where the chemical has reached after a given time t. u u 3d δx Α w e W P E Flow uniform throughout the channel : const. speed u at all points.

8 3d Concentration q @x To apply the FV method, we 1 Split the channel into cells V = Ax 2 integrate eqn over the cell. { the 1-d.

9 3d First term : dv = d dt Second term : dv = Adx dv = ZZZV V qdv = d dt (qv Adx = (uqa) e (uqa) w These two terms are now being evaluated on the boundary face between the cells. However we do not know the value of the variables on w and e. Need to interpolate to get these values.

10 Upwind dierencing (uq) w = (uq) W ; (uq) e = (uq) P 3d Central dierencing (uq) w = (uq) W + (uq) P 2 Diusion can be treated similarly

11 Discretised s : d dt (qv ) + (uqa) e (uqa) A w 3d Rearanging a bit and using central dierencing to interpolate, we get dq dt + u 2x (q E q W ) = x (q 2 E 2q P + q W ) Number the cells i = 0 : : : N! : dq i dt = u 2x (q i+1 q i 1) + x 2 (q i+1 2q i + q i 1)

12 3d This problem is parabolic one { solve by timestepping { of time into M timesteps q j. Time derivative dq dt = qj+1 q j t At what timestep are the values on the r.h.s. evaluated? Take values at timestep j,! explicit scheme. Take values at timestep j + 1,! implicit scheme. Explicit scheme : q j+1 i q j i = ut 2x (q i+1 q i 1) j + t x 2 (q i+1 2q i + q i 1) j

13 Write C = ut x ; D = t x 2 C = Courant number { signicant parameter in determining the stability of the scheme. D relates to the diusion. 3d Writing this out as an algorithm : q j+1 i = C 2 + D q j i 1 + (1 2D) qj i + C 2 D q j i+1 A rule for advancing the values of q through one timestep, which could be written into a spreadsheet.

14 3d Implicit scheme : write as a matrix : 0 D U 0 L D U 0 L D C A q j+1 i 1 q j+1 i q j+1 i+1. 1 C A = 0 q j i 1 q j i q j i+1 Inverting this matrix provides the solution without the stability problems. (although the Courant number is still worth calculating if there are problems).. 1 C A

15 3d 3d This scheme could be derived by standard FD techniques. This is because the mesh we are dealing with was regular and uniform. However this is not always the case. FV approach makes it much easier to evaluate the various terms for irregular shaped cells. The + r:uq = r: rq + S q When we integrate the second term here, we nd ZZZ X r:uq dv = uq V A { Gauss' theorem in vector calculus.