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May 9, 2013, 05:16 |
initial residual and final residual
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#1 |
New Member
Maosong Cheng
Join Date: Aug 2012
Posts: 19
Rep Power: 14 |
Hi FOAMERs,
what is the definition of the initial residual and final residual, which can be used to decide convergence ? |
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May 15, 2013, 06:20 |
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#2 |
Senior Member
Anne Gerdes
Join Date: Aug 2010
Location: Hamburg
Posts: 168
Rep Power: 16 |
The initial residual is evaluated based on the current values of the field before solving an equation for a particular field . The final residual is evaluated after the solution of the equation is performed.
The intial residual is more important to decide whether a computation converges, or not. |
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May 15, 2013, 09:23 |
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#3 |
Member
Timo K.
Join Date: Feb 2010
Location: University of Stuttgart
Posts: 66
Rep Power: 16 |
Hi Anne,
can you give some rule of thumbs for values, which should be achieved for the initial/final residual and how many iterations should be used for pEqn per cycle e.g. for a transient simulation with a standard PCG solver. Best Timo |
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May 15, 2013, 09:29 |
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#4 |
Senior Member
Anne Gerdes
Join Date: Aug 2010
Location: Hamburg
Posts: 168
Rep Power: 16 |
Initial residual: e^-4 or e^-5 is a good result. Final residual has to be smaller (e^-8).
I cannot answer your 2nd question. |
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August 17, 2016, 16:45 |
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#5 | |
New Member
Manm
Join Date: Apr 2015
Posts: 16
Rep Power: 11 |
Quote:
Hi, based on this definition, I was curious as to why the final residual in one time step NOT equal to initial residual in NEXT time step? |
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August 17, 2016, 16:55 |
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#6 |
Member
Bruno Blais
Join Date: Sep 2013
Location: Canada
Posts: 64
Rep Power: 13 |
Assuming you are solving in steady state, it is because the equations are non-linear.
If you solve the momentum equation for U, at the next iteration, P and phi will have been changed. Therefore, the initial residual will be different from the final residual of the last iteration. If you were to solve a linear equation, non-coupled with any other variables, in steady-state, then you would solve it in a single iteration... and therefore your "next initial" residual would be your last final one... |
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August 17, 2016, 18:13 |
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#7 | |
New Member
Manm
Join Date: Apr 2015
Posts: 16
Rep Power: 11 |
Quote:
If I understand it right, Say : U =f(p,phi,t) where f is a complex function If t0 =0: During 1 st iteration, we start with a p0, phi0 and dt time step and get U1 =f(p0, phi0,dt) as well as p1, phi1 values. The corresponding final error/ residual is calculated from U1 and true U expected. Before 2nd iteration starts, a (say) U12 is calculated based on new available values: U12 =f(p1,phi1,dt) and this is used to calculate the initial error/ residual before the 2nd time step starts. and then we go on to calculate U2 =f(p1, phi1, dt+dt) and corresponding p2, phi2 and final error/ residual at the end of 2nd time step and so on.... |
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August 18, 2016, 09:31 |
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#8 | |
Member
Bruno Blais
Join Date: Sep 2013
Location: Canada
Posts: 64
Rep Power: 13 |
Yup, That is exactly so.
OpenFOAM is a segregated solver, that is U, V, W, phi and P are solved segregated. This is why such a thing happen. There are block coupled solvers in foam-extend, but those are still linear solver and don't implement a non-linear newton method or something like that. Quote:
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