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Help regarding 1D compressible flow (Sod Shock tube) |
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July 25, 2016, 17:22 |
Help regarding 1D compressible flow (Sod Shock tube)
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#1 |
Senior Member
Selig
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Hi all,
I am coding up a 1D compressible flow solver via Lax-Friedrichs method (for now), however have run into a when calculating certain quantities, such as velocity. In other words, I believe my implementation is correct, however something is not being properly updated per time step. I have attached my code for proper documentation. Thoughts: My way of updating at each time step is to have U_current = U_new. EDIT: Just for clarification, I am not sure what I am doing wrong as my plots for pressure, velocity, and density are not right Last edited by selig5576; July 26, 2016 at 12:44. |
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July 26, 2016, 11:40 |
Still having issues
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#2 |
Senior Member
Selig
Join Date: Jul 2016
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Hi All,
I am still having issues with my 1D compressible solver. I have narrowed it down to the time step, however I am not sure what I'm doing wrong in the process of updating it. If someone point/help me in the right direction I would heavily appreciate it. Code:
clear; close all; clc; gamma = 1.4; b = 0.5; a = -0.5; nx = 101; dx = (b-a)/(nx-1); x = zeros(nx,1); tMax = 0.15; dt = 0.004; nt = round(tMax/dt)+1; t = zeros(nt,1); r = zeros(nx,nt); p = zeros(nx,nt); u = zeros(nx,nt); F1 = zeros(nx,nt); F2 = zeros(nx,nt); F3 = zeros(nx,nt); U1 = zeros(nx,nt); U2 = zeros(nx,nt); U3 = zeros(nx,nt); for j=1:nt for i=1:nx x(i) = a + (i-1)*dx; t(j) = (j-1)*dt; end end for i=1:nx if (x(i) <= 0) r(i,1) = 1; u(i,1) = 0; p(i,1) = 1; elseif (x(i) > 0) r(i,1) = 0.125; u(i,1) = 0; p(i,1) = 0.1; end end U1(1) = 1; U1(nx) = 0.125; U2(1) = 0; U2(nx) = 0; U3(1) = 1; U3(nx) = 0.1; E = p./(gamma-1) + 0.5*r.*u.^2; for j=1:nt for i=1:nx U1(i,j) = r(i,j); U2(i,j) = r(i,j).*u(i,j); U3(i,j) = E(i,j); F1(i,j) = r(i,j).*u(i,j); F2(i,j) = r(i,j).*u(i,j).^2 + p(i,j); F3(i,j) = u(i,j).*(E(i,j) + p(i,j)); end for i=2:nx-1 U1(i,j+1) = 0.5*(U1(i+1,j) + U1(i-1,j)) - dt/(2*dx)*(F1(i+1,j) - F1(i-1,j)); U2(i,j+1) = 0.5*(U2(i+1,j) + U2(i-1,j)) - dt/(2*dx)*(F2(i+1,j) - F2(i-1,j)); U3(i,j+1) = 0.5*(U3(i+1,j) + U3(i-1,j)) - dt/(2*dx)*(F3(i+1,j) - F3(i-1,j)); end r = U1; u = U2./U1; E = U3; p = (gamma-1)*(E-0.5*r.*u.^2); end u s1 = subplot(2,2,1); plot(x,r,'-b'); xlabel('x'); ylabel('r'); title('Density'); s2 = subplot(2,2,2); plot(x,u,'-b'); xlabel('x'); ylabel('u'); title('Velocity'); s3 = subplot(2,2,3); plot(x,p,'-b'); xlabel('x'); ylabel('p'); title('Pressure'); |
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July 26, 2016, 12:29 |
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#3 |
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I am not completely familiar with the language you are using, but it looks to me like you are overwriting values during your update that should not be overwritten until you complete the entire sweep. You construct u(i,j+1) and then use that value for u(i,j) in the next pass through the j-loop.
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July 26, 2016, 12:39 |
Thanks
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#4 |
Senior Member
Selig
Join Date: Jul 2016
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Hi agd,
I am using Matlab for my language. In terms of Code:
for j=1:nt for i=1:nx U1(i,j) = r(i,j); U2(i,j) = r(i,j).*u(i,j); U3(i,j) = E(i,j); F1(i,j) = r(i,j).*u(i,j); F2(i,j) = r(i,j).*u(i,j).^2 + p(i,j); F3(i,j) = u(i,j).*(E(i,j) + p(i,j)); end for i=2:nx-1 U1(i,j+1) = 0.5*(U1(i+1,j) + U1(i-1,j)) - dt/(2*dx)*(F1(i+1,j) - F1(i-1,j)); U2(i,j+1) = 0.5*(U2(i+1,j) + U2(i-1,j)) - dt/(2*dx)*(F2(i+1,j) - F2(i-1,j)); U3(i,j+1) = 0.5*(U3(i+1,j) + U3(i-1,j)) - dt/(2*dx)*(F3(i+1,j) - F3(i-1,j)); end r = U1; u = U2./U1; E = U3; p = (gamma-1)*(E-0.5*r.*u.^2); end P.S. I'm new to CFD |
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July 26, 2016, 14:00 |
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#5 |
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OK - I see what your code is doing now. I'll look at it a little more and see if anything looks wrong. One thing you can try for debugging is to turn off the boundary conditions and see if your code will maintain a uniform set of conditions. Sometimes that can quickly point to a problem.
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July 26, 2016, 14:49 |
Further debugging
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#6 |
Senior Member
Selig
Join Date: Jul 2016
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Hi agd (and all),
So I took the boundary conditions out (commented it out) and nothing happens. I've looked at the arrays and I see a lot of "Nans." That being said, I removed: Code:
r = U1; u = U2./U1; E = U3; p = (gamma-1)*(E-0.5*r.*u.^2); Below is the hyperbolic system (compressible euler) and some physical definitions I use for solving for total energy and pressure: The scheme I'm using is the Lax-Friedrichs, i.e. |
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July 27, 2016, 10:53 |
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#7 |
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The conservation equations as you have written them are not correct. Is that simply an error in how you wrote them in your post, or did you code them up that way?
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July 27, 2016, 11:27 |
Equations
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#8 |
Senior Member
Selig
Join Date: Jul 2016
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Hi,
I coded them up two different ways to see if what you said was an issue, however the issue does not change. Code:
clear; close all; clc; gamma = 1.4; b = 0.5; a = -0.5; nx = 101; dx = (b-a)/(nx-1); tMax = 0.15; dt = 0.004; nt = round(tMax/dt)+1; x = zeros(nx,nt); t = zeros(nt,nt); r = zeros(nx,nt); p = zeros(nx,nt); u = zeros(nx,nt); F1 = zeros(nx,nt); F2 = zeros(nx,nt); F3 = zeros(nx,nt); U1 = zeros(nx,nt); U2 = zeros(nx,nt); U3 = zeros(nx,nt); for j=1:nt for i=1:nx x(i,j) = a + (i-1)*dx; t(i,j) = (j-1)*dt; end end for i=1:nx if (x(i) <= 0) r(i,1) = 1.0; u(i,1) = 0.0; p(i,1) = 1.0; elseif (x(i) > 0) r(i,1) = 0.125; u(i,1) = 0.0; p(i,1) = 0.1; end end E = p./((gamma-1).*r) + 0.5*r.*u.^2; U1(1) = 1; U1(nx) = 0.125; U2(1) = 0; U2(nx) = 0; U3(1) = 1; U3(nx) = 0.1; for j=1:nt for i=1:nx U1(i,j) = r(i,j); U2(i,j) = r(i,j)*u(i,j); U3(i,j) = r(i,j).*E(i,j); F1(i,j) = r(i,j)*u(i,j); F2(i,j) = r(i,j)*u(i,j).^2 + p(i,j); F3(i,j) = u(i,j)*(r(i,j).*E(i,j) + p(i,j)); end for i=2:nx-1 U1(i,j+1) = 0.5*(U1(i+1,j) + U1(i-1,j)) - (dt/(2*dx))*(F1(i+1,j) - F1(i-1,j)); U2(i,j+1) = 0.5*(U2(i+1,j) + U2(i-1,j)) - (dt/(2*dx))*(F2(i+1,j) - F2(i-1,j)); U3(i,j+1) = 0.5*(U3(i+1,j) + U3(i-1,j)) - (dt/(2*dx))*(F3(i+1,j) - F3(i-1,j)); end r = U1; u = U2./r; E = U3./r; p = (gamma-1).*r.*(E-0.5*r.*u.^2); end subplot(2,2,1); plot(x,r(:,26),'-b'); xlabel('x'); ylabel('r'); title('Density'); subplot(2,2,2); plot(x,u(:,26),'-b'); xlabel('x'); ylabel('u'); title('Velocity'); subplot(2,2,3); plot(x,p(:,26),'-r'); xlabel('x'); ylabel('p'); title('Pressure'); Link: http://bender.astro.sunysb.edu/hydro...ible/Euler.pdf The link I provided above is to illustrate the equations I'm solving not the method. The part in which really stumps me (and maybe you can have wisdom on that) is why I keep getting NaNs in my arrays. I spent a couple hours debugging with no luck. I believe if I did not get "NaNs" in my arrays I would have a complete solution as it looks very close. |
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July 27, 2016, 11:36 |
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#9 |
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Filippo Maria Denaro
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the RHS of your system must have the minus sign...
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July 27, 2016, 11:45 |
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#10 |
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Filippo Maria Denaro
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for i=1:nx U1(i,j) = r(i,j); U2(i,j) = r(i,j)*u(i,j); U3(i,j) = r(i,j).*E(i,j); F1(i,j) = r(i,j)*u(i,j); F2(i,j) = r(i,j)*u(i,j).^2 + p(i,j); F3(i,j) = u(i,j)*(r(i,j).*E(i,j) + p(i,j)); end
why do you mix the operators "*" and ".*" in the for cycle? they are different operations in matlab |
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July 27, 2016, 11:48 |
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#11 |
Senior Member
Selig
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Hi,
Thanks for pointing that out. That was a mistake in terms of not putting the - sign, however in the code I already took that into account. In terms of wrongly using .* and * I have fixed that and is below. Code:
clear; close all; clc; gamma = 1.4; b = 0.5; a = -0.5; nx = 101; dx = (b-a)/(nx-1); tMax = 0.15; dt = 0.004; nt = round(tMax/dt)+1; x = zeros(nx,nt); t = zeros(nt,nt); r = zeros(nx,nt); p = zeros(nx,nt); u = zeros(nx,nt); F1 = zeros(nx,nt); F2 = zeros(nx,nt); F3 = zeros(nx,nt); U1 = zeros(nx,nt); U2 = zeros(nx,nt); U3 = zeros(nx,nt); for j=1:nt for i=1:nx x(i,j) = a + (i-1)*dx; t(i,j) = (j-1)*dt; end end for i=1:nx if (x(i) <= 0) r(i,1) = 1.0; u(i,1) = 0.0; p(i,1) = 1.0; elseif (x(i) > 0) r(i,1) = 0.125; u(i,1) = 0.0; p(i,1) = 0.1; end end E = p./((gamma-1)*r) + 0.5*u.^2; U1(1) = 1; U1(nx) = 0.125; U2(1) = 0; U2(nx) = 0; U3(1) = 1; U3(nx) = 0.1; for j=1:nt for i=1:nx U1(i,j) = r(i,j); U2(i,j) = r(i,j).*u(i,j); U3(i,j) = r(i,j).*E(i,j); F1(i,j) = r(i,j).*u(i,j); F2(i,j) = r(i,j).*u(i,j).^2 + p(i,j); F3(i,j) = u(i,j).*(r(i,j).*E(i,j) + p(i,j)); end for i=2:nx-1 U1(i,j+1) = 0.5*(U1(i+1,j) + U1(i-1,j)) - (dt/(2*dx))*(F1(i+1,j) - F1(i-1,j)); U2(i,j+1) = 0.5*(U2(i+1,j) + U2(i-1,j)) - (dt/(2*dx))*(F2(i+1,j) - F2(i-1,j)); U3(i,j+1) = 0.5*(U3(i+1,j) + U3(i-1,j)) - (dt/(2*dx))*(F3(i+1,j) - F3(i-1,j)); end r = U1; u = U2./r; E = U3./r; p = (gamma-1)*r.*(E-0.5*u.^2); end subplot(2,2,1); plot(x,r(:,26),'-b'); xlabel('x'); ylabel('r'); title('Density'); subplot(2,2,2); plot(x,u(:,26),'-b'); xlabel('x'); ylabel('u'); title('Velocity'); subplot(2,2,3); plot(x,p(:,26),'-r'); xlabel('x'); ylabel('p'); title('Pressure'); |
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July 28, 2016, 11:45 |
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#12 |
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Stupid question - are you sure that you are obeying the stability limit for the Lax-Friedrichs scheme?
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July 28, 2016, 11:58 |
CFL condition
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#13 |
Senior Member
Selig
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Hi agd,
Yes I am. I've made sure there is no instability aside the downfalls of the method. |
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July 28, 2016, 12:04 |
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#14 | |
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Filippo Maria Denaro
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Quote:
that's never a stupid question he wrote an arbitrary dt and the cfl does not appear explicitly computed: b = 0.5; a = -0.5; nx = 101; dx = (b-a)/(nx-1); x = zeros(nx,1); tMax = 0.15; dt = 0.004; |
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July 28, 2016, 12:06 |
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#15 |
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Filippo Maria Denaro
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remember to estimate that dt/dx = 0.4 and the velocity should be considered as convective and sound velocity
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August 2, 2016, 10:22 |
Thanks
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#16 |
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Selig
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Hi agd and FMDenaro,
I have managed to get my 1d compressible flow solver to work. There were some things I did different: 1. I used the Local Lax Friedrichs which provided a more robust method. There were no small oscillations 2. The indexing I was using was running me into trouble, i.e. the time index 3. I adjusted my CFL condition Thank you for the help |
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January 20, 2020, 17:12 |
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#17 |
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ΑΤΤΙΚΗΣ
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Hello!!Could you post the right code after your changes?
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August 25, 2022, 05:11 |
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#18 |
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javeria
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hello i need lax friedrich matlab codes for 1D shock tube problem. Is there any one to help me?
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August 25, 2022, 05:12 |
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#19 |
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javeria
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I need your codes
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August 25, 2022, 05:16 |
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#20 |
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javeria
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hello selig can you post your codes after correction i need your codes
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