# High Residual values for converged steady state computations

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 November 15, 2018, 23:59 High Residual values for converged steady state computations #1 New Member   CFD Join Date: Jul 2016 Posts: 2 Rep Power: 0 An equation of the form given below: du/dt+R=0 reaches a steady state solution computationally, when the residual (R) goes to zero. In my CFD simulations, I have observed that in many cases, the Residuals (R) are as high as 0.0004 - 0.01, but the solution has converged to steady state (based on Cl, Cd values and surface pressure (Cp)). However, as per the above equation, since R is not close to zero, we should still see a changing solution. How does one explain this discrepancy?

November 16, 2018, 02:34
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Filippo Maria Denaro
Join Date: Jul 2010
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Quote:
 Originally Posted by N.Ganesh An equation of the form given below: du/dt+R=0 reaches a steady state solution computationally, when the residual (R) goes to zero. In my CFD simulations, I have observed that in many cases, the Residuals (R) are as high as 0.0004 - 0.01, but the solution has converged to steady state (based on Cl, Cd values and surface pressure (Cp)). However, as per the above equation, since R is not close to zero, we should still see a changing solution. How does one explain this discrepancy?

There is no discrepancy. First, Cl, Cd, Cp are integrals of local values over a body and do not determine a steady state that implies a norm on all the values of the acceleration field. Second, the values means nothing if the problem is not properly scaled (non-dimensionalize) in such a way that all the terms are of O(1). You can get 0.01 max value acceptable if the fields are dimensional and the order of the velocity is not unitary

 November 16, 2018, 07:15 #3 Senior Member   Lucky Join Date: Apr 2011 Location: Orlando, FL USA Posts: 5,680 Rep Power: 66 The solution converges when u is (asymptotically) constant regardless of what du/dt is and regardless of what R is. R need not be zero for u to converge. This is a technicality but important in CFD because you solve not the continuous du/dt = 0 problem but you take your governing equation and discretize it which introduces eventually non-zero du/dt and R. Note that your governing equation is satisfied as long as du/dt + R = 0, which means that any du/dt = -R is a solution to your problem. In floating point arithmetic there is no zero regardless but only small numbers. So to say du/dt is 0 is simply imposible. You can only say du/dt is less than some tolerance (often labeled TOL) for u. This same problem shows up in even more basic arithmetic like solving for 3*x = 2. Try doing this and you will find that x is not exactly the analytical 2/3 but 0.6666 with funny terminating digits. Does this mean that computers cannot do any arithmetic? No, it means they do a particular type of arithmetic. Last edited by LuckyTran; November 16, 2018 at 10:37.