AEM Computational Fluid Dynamics Instructor: Dr. M. A. R. Sharif

Transcription

1 AEM Computational Fluid Dynamics Instructor: Dr. M. A. R. Sharif Numerical Solution Techniques for 1-D Parabolic Partial Differential Equations: Transient Flow Problem by Parshant Dhand September 16, 2002

2 ABSTRACT The primary purpose of this work was to become familiar with numerical techniques, namely finite difference schemes, available for solving parabolic partial differential equations (PDEs) such as the governing equation for viscous diffusion. An application of such viscous diffusion is the transient flow problem. This problem is investigated here and the various finite difference schemes are applied to solve this problem for various time steps. These schemes include the FTCS, Crank-Nicholson, DuFort Frankel (explicit) and IMPLICIT finite difference schemes. For the FTCS scheme, stability is examined and indicated the numerical scheme to be conditionally stable. The results indicate that the FTCS scheme is conditionally stable while the Crank-Nicholson and DuFort-Frankel schemes are unconditionally stable. This realization allows for selection of the best scheme for implementation in solving this problem. INTRODUCTION Background The Flow problem is classic example of viscous diffusion. The governing equation for such problem was derived using boundary layer theory to reduce the full Navier- Stokes equations to the single parabolic PDE,. with the necessary initial and boundary conditions, t = 0: u(0) = 0, u(0.04m) = 0; t > 0: u(0) = 40.0, u(0.04m) = U = 0.0m/s. This problem may be described physically as transient viscous-driven flow between two plates of infinite extent and separated by a distance of 0.04m. Initially both plates are at rest. After time,t=0, the upper plate is set in motion in the positive x-direction with a velocity of 40.0m/s. Due to the viscosity of the fluid filling the space between the plates, successive lamina of fluid are set in motion as time elapses. Eventually, the system reaches a quasi-steady state, as the velocity profile becomes more or less constant in time. The governing equation lends nicely to the use of finite difference techniques to solve the problem in the transient domain. Solution Techniques Part 1: FTCS Finite Difference Scheme: The FTCS (Forward Time Centered Space) explicit finite difference scheme for the diffusion equation may be written as

3 Stability may be examined using the Von Neumman stability analysis. This analysis was completed and is included as appendix B. The results indicate that the FTSC explicit scheme is conditionally stable. That is the scheme is stable provided R less then half or 1/2. Part 2: Crank-Nicholson Finite Difference Scheme: The Crank-Nicholson scheme is an implicit and unconditionally stable scheme that may bewritten as,, In this case the coefficients on the LHS are known and the entire RHS is known at each time step. Thus, the scheme may be written in matrix form as a tridiagonal system of equations that can be subsequently solved at each time step using Numerical Recipes1, TRIDAG. Part 3: DuFort-Frankel (Explicit ) Scheme The DuFort-Frankel explicit scheme, also unconditionally stable, may be written as ujn+1 = [2R/(1+2R)](uj+ n + uj- n) + [(1-2R)/(1+2R)] ujn-1 Part 4: The IMPLICIT scheme for this problem may be written as Ruj- n+1 2(1 + R) n+1 + n+1 = - ujn Again, this scheme is unconditionally stable and as is easily seen can also be solved using the tridiagonalmatrix method of Numerical Recipes1, TRIDAG.In all of the cases above, Each of the above numerical schemes was used to solve the viscous diffusion problem for various values of T. Th value of R = for all cases and Time T = 0.18, T = 0.36, T= The FORTRAN code used to implement these schemes is included as appendix B. Code for the Crank Nicolson scheme was giving a plot which was not accurate, so I decided to write code in MAtlab and it gave fine results. RESULTS AND DISCUSSION The results of each case investigated are presented in the distabce covere by the velocity of the fluid, where u is the velocity computed for each spatial grid point

4 Similarly, y is the distance between spatial grid points and the plate separation is m. These plots represent the velocity profiles computed at 3 different time steps and illustrate the transient nature of the solution for each case. Part 1: FTCS Finite Difference Scheme The results of the stability analysis indicate that the FTCS scheme is stable provided R < ½. Thus, it was apparent that the solution for the case, Since in our case R is constant and is equal to 0.217, hence the solution is stable and the results are presented as figure 1. This plot represents the velocity profiles for three time steps from t=0.18, t= 0.36 and t = In this case one would expect the velocity to be zero beyond certain points across the separation distance until the viscous diffusion has reached the lower plate. This can actually be seen by inspection of the raw output data file. In any case, it is obvious that numerical stability must be investigated to ensure a quality solution to the problem at hand and if instability is predicted, an alternate scheme should be considered. Part 2: Crank-Nicholson Finite Difference Scheme Unlike the FTCS scheme, the Crank-Nicholson finite difference scheme has the primary advantage of being unconditionally stable for all possible values of R and hence having a greater range of application than the previous FTCS scheme. Results are very similar in shape to those obtained using FTCS scheme. Part 3: DuFort-Frankel (Explicit) Scheme: The DuFort-Frankel explicit scheme (also unconditionally stable) generates results similar to the FTCS scheme Figure 3 is much like figure 1 as expected since the time steps are the same. Part 4: The IMPLICIT scheme: Finally, the fully implicit scheme was used to solve this problem. Figure 2 represents the results presented in the same manner as previously described. One would expect the results to look similar to the previous cases. The results are more or less similar. Unlike the explicit scheme which has accuracy of O(Dt2), the accuracy is reduced to O(Dt). The little difference is seen in accuracy of the solution because of this. CONCLUSIONS Based on the results of the Von Neumann stability analysis, the FTCS scheme is conditionally stable. That is numerical stability can only be achieved for R < 0.5. In contrast, the Crank-Nicholson and DuFort-Frankel schemes provide unconditional stability and hence allow for solutions corresponding with various values of R. The primary advantages of unconditional stability are improved efficiency and accuracy as larger time steps may be used to converge to a solution in fewer time iterations and hence reduce the truncation error associated with the time domain. The results of this investigation demonstrate the importance of choosing the appropriate scheme for the problem at hand. In

5 addition, these results provide a nice survey of those techniques available for solving parabolic PDEs such as the diffusion equation and the advantages/disadvantages associated with each. One aspect of Numerical Computation that was learned here (definitely the hard way) was that it is always better to write different subroutines and then put them in a project together. However I tried to do that, but could not do that because of the debugging problem. This was first time to write programs in Fortran, hopefully in later stages things will be easy to do.

6 APPENDIX A: Plots for the four schemes: 1. Plot for FTCS Explicit Scheme VELOCITY : FTCS EXPLICIT SCHEME VELOCITY T = 0.18 T = 0.36 T = For T = 0.18 T = 0.36 T = LENGTH

7 2. Plot for FTCS IMPLICIT Scheme VELOCITY RESULTS: IMPLICIT SCHEME VELOCITY T = 0.18 T = 0.36 T = LENGTH For T = 0.18 T = 0.36 T = 1.08

8 3. Plot for FTCS DUFORT-FRANKEL Scheme VELOCITY RESULTS: DUFORT FRANKEL SCHEME Velocity T =1.08 T = 0.36 T = Length For T = 0.18 T = 0.36 T = 1.08

9 4. Plot for CRANK - NICHOLSON Scheme For T = 0.18 T = 0.36 T = 1.08

10 APPENDIX B: COMPUTER CODE FOR SCHEMES. 1. Computer Code in Fortran for FTCS EXPLICIT SCHEME. Homework#1 - FTCS EXPLICIT METHOD Program computes the numerical solution to the Transient Flow Problem. The following initial and bounadry conditions are applied: t=0: u(y=0)=40.0m/s t>0: u(y=0)=0.0; u(y=0.04m)=0.0 parameter(maxn=42,eps=1.0e-3) integer m,mm,count real*8 u_old(maxn),u_new(maxn),y(maxn) real*8 t,tau,h,r,tmax,u_init,nu,sum,error data h,m,u_init,nu,r,tmax /0.001,41,0.0,2.17e-4,0.217,2.5e5/ open(unit=1,file='hw1_explicit.out',status='unknown') tau=r*h**2/nu mm=m-1 error=1.0 count=0 t=0.0 y(1)=0.0 do 2 i=2,m y(i)=y(i-1)+h 2 continue do 3 i=1,mm u_old(1)= continue u_old(m)=0.0 write(1,*)'velocity Results:' write(1,10)t,(u_old(j),j=1,m) do while count=count+1 sum=0.0 t=t+tau do 4 i=2,mm u_new(i)=u_old(i)+r*(u_old(i-1)-2.0*u_old(i)+u_old(i+1))

11 end do 10 format(2x,f10.3,5x,41f8.4) write(1,'(" Number of steps for convergence = ",i4)')count end 2. Computer Code in Fortran for FTCS IMPLICIT SCHEME. Homework#1 - FTCS IMPLICIT METHOD MODULE VARIABLE PARAMETER (NX=45, NY=45) Declaration of constant parameters PARAMETER t = 2, dt = 0.002, y = 0.04, dy = 0.001, nu = , term = nu*dt/(dy*dy) Declaration for the integer variables INTEGER i, j INTEGER tmax = (t/dt) + 1, ymax = (y/dy)+1 Declaration for the real array variables REAL, DIMENSION(NX) :: a, b, c,d, h, g REAL, DIMENSION(NX,NY) :: u END MODULE VARIABLE PROGRAM IMPLICIT USE VARIABLE CALL INITIAL(u) CALL BOUNDARY(u) CALL LASONNEN(u) END PROGRAM Subroutine for boundary conditions SUBROUTINE BOUNDARY(u) USE VARIABLE DO j=1,tmax u(j,1) = 40 u(j,41) = 0

12 END DO END SUBROUTINE END of subroutine SUBROUTINE LASONNEN(u) USE VARIABLE DO j=2,tmax DO i=2,ymax-1 a(i) = term b(i) = -(1+2*term) c(i) = term d(i) = u(j-1,i) END DO CALL TRIDIAGONAL(a,b,c,d,u,j) IF j = 91.OR. 181.OR. 271.OR. 361.OR. 451.OR. 541 CALL OUTPUT END DO END SUBROUTINE Subroutine for initializing arrays SUBROUTINE INITIAL(u) USE VARIABLE DO i=1,ymax u(1,i) = 0 u(1,1)= 40 END DO END SUBROUTINE END of subroutine SUBROUTINE TRIDIAGONAL(a,b,c,d,u,j) h(1) = 0 g(1) = 40 DO i=2,ymax-1 h(i) = c(i)/(b(i)-a(i)*h(i-1)) g(i) = (d(i)-a(i)*g(i-1))/(b(i)-a(i)*h(i-1)) END DO CALL CALCU(g,h,u,j)

13 END SUBROUTINE SUBROUTINE OUTPUT USE VARIABLE OPEN (UNIT=1,FILE='FILEOUT1.DAT') DO i=1, ymax WRITE (1,100)(u(i)) 101 FORMAT (41F10.3) END DO END SUBROUTINE 3. Computer Code in Fortran for DUFORT FRANKEL SCHEME. Homework1 DUFORT FRANKE SCHEME Program computes the numerical solution to the Transient Flow Problem. The following initial and bounadry conditions are applied: t=0: u(y=0)=40.0m/s t>0: u(y=0)=0.0; u(y=0.04m)=0.0 parameter(maxn=30,eps=1.0e-3) integer k,m,mm,count real*8 u_old(1001,maxn),u_new(1001,maxn),y(maxn) real*8 t,tau,h,r,tmax,u_init,nu,sum,error data h,m,u_init,nu,r,tmax /0.001,41,0.0,2.17e-4,0.217,2.5e5/ open(unit=1,file='hw1_dufort.out',status='unknown') tau=r*h**2/nu mm=m-1 error=1.0 count=0 k=1 t=0.0 y(1)=0.0 do 2 i=2,m y(i)=y(i-1)+h

14 2 continue write(1,*)'velocity Results:' write(1,10)t,(u_old(k,j),j=1,m) do while ((error.gt.eps).and.(count.lt.1080)) count=count+1 sum=0.0 t=t+tau u_old(k,1)=40.0 u_old(k,m)=0.0 do 4 i=2,mm if (k.lt.2) then u_new(k,i)=(2.0*r/( *r))*(u_old(k,i+1)+u_old(k,i-1)) else u_new(k,i)=(2.0*r/( *r))*(u_old(k,i+1)+u_old(k,i-1))+ (( *r)/( *r))*u_old(k-1,i) end if end do 10 format(2x,f10.3,2x,41f8.4) write(1,'(" Number of steps for convergence = ",i4)')count end 4. Computer Code in MATLAB for CRANK NICOLSON SCHEME. m=1080;n=41; dt=0.001; dx=0.001; a= ; h=0.5*(a*dt)/(dx^2); u = zeros(41,1080); u(1,1:1080) =40; u(41,1:1080) = 0; u(1:41,1080) = 0; u(1,1) = 40; for j =2:m for i =2:n-1 a1(i)=h; b1(i)=-1-2*h; c1(i)=h; d1(i)=-u(i,j-1)-h*(u(i+1,j-1)-2*u(i,j-1)+u(i-1,j-1)); end s(1)= 0;

15 t(1)=u(1,j-1); for i =2:n-1 s(i)=c1(i)/(b1(i)-a1(i)*s(i-1)); t(i)=(d1(i)-a1(i)*t(i-1))/(b1(i)-a1(i)*s(i-1)); end val(n)=0; for i =n-1:-1:2 val(i)=-s(i)*u(i+1,j-1)+t(i); end for i=2:n-1 u(i,j)=val(i); end end figure(1) for i = 300:300:1080 plot(0:0.001:0.04,u(:,i),'black'); hold on; end APPENDIX C: ERROR VECTOR for Explicit Scheme at T = 0.18 sec and for dt = Exact Solution T = 1.08 sec Numerical Solution y(m) Explicit EXPLICIT ERROR

16 FOR T = 1.08 for dt = Exact Solution Numerical Solution y(m) Explicit Explicit ERROR

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