Initial drag for cylinder in cross-flow

orxan.shibli · May 28, 2012, 15:26

Hi everyone!

I am simulating flow over cylinder at Re=100 with my own code. Setting time step to dt=1s, I monitor initial values of drag coef. and it solves it without problem. But when I set it to say dt=0.001 then initially drag coef. rises to very high values (hundreds) and then it becomes negative even. It is not important whether it will be better as time passes since I am also trying to understand why decreasing time step causes this kind of problem.

Other specifications:

Unsteady
Fully Implicit, 1st order Euler temporal discre.
2nd order upwind spatial discre.
Max. Inner Iterations = 10
2D, laminar
Initially velocity and pressure field is zero

Any idea?

cdegroot · May 28, 2012, 19:46

You are starting the flow from a standstill, so that takes a big pressure drop to accomplish. That is one reason for the high drag. Also, you are only doing 10 inner iterations, which probably means it is not converging within each time step. In that case, the initial results don't mean much anyways.

orxan.shibli · May 29, 2012, 05:44

Dear Chris

OK but this should be true for both time steps. That is why I decreased time step however it became worse unexpectedly. Also I tried to make inner iterations 100 but that only caused to reach high values faster.

cdegroot · May 29, 2012, 10:06

Well, your small timestep is 1000 times smaller than your large time step, so I would guess that the large time step just "steps over" a lot of the initial transient stuff.

Also, imagine in reality having completely still fluid and then instantly accelerating it to whatever value you have specified at the inlet. Not going to happen. It will take some (maybe small) amount of time to accelerate. In your simulation that time to accelerate is your first time step. Thus, I believe a smaller timestep will imply a larger initial pressure drop and also drag forces. In any case, these initial transients will die out eventually.

orxan.shibli · June 3, 2012, 17:43

I found the problem. It was due to time-step dependency of original Rhie and Chow MIM. Hence, it is better to use modified version:

"Discussion on Momentum Interpolation Method for Collocated Grids of Incompressible Flow", Yu et al.

cdegroot · June 4, 2012, 01:10

Quote:

Originally Posted by orxan.shibli

I found the problem. It was due to time-step dependency of original Rhie and Chow MIM. Hence, it is better to use modified version:

"Discussion on Momentum Interpolation Method for Collocated Grids of Incompressible Flow", Yu et al.

Very interesting. Thanks for sharing.

May 28, 2012, 15:26	Initial drag for cylinder in cross-flow	#1
orxan.shibli Member Join Date: Aug 2011 Posts: 30 Rep Power: 14	Hi everyone! I am simulating flow over cylinder at Re=100 with my own code. Setting time step to dt=1s, I monitor initial values of drag coef. and it solves it without problem. But when I set it to say dt=0.001 then initially drag coef. rises to very high values (hundreds) and then it becomes negative even. It is not important whether it will be better as time passes since I am also trying to understand why decreasing time step causes this kind of problem. Other specifications: Unsteady Fully Implicit, 1st order Euler temporal discre. 2nd order upwind spatial discre. Max. Inner Iterations = 10 2D, laminar Initially velocity and pressure field is zero Any idea?

May 29, 2012, 05:44		#3
orxan.shibli Member Join Date: Aug 2011 Posts: 30 Rep Power: 14	Dear Chris OK but this should be true for both time steps. That is why I decreased time step however it became worse unexpectedly. Also I tried to make inner iterations 100 but that only caused to reach high values faster. Last edited by orxan.shibli; June 3, 2012 at 17:45.

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May 28, 2012, 19:46		#2
cdegroot Senior Member Chris DeGroot Join Date: Nov 2011 Location: Canada Posts: 414 Rep Power: 17	You are starting the flow from a standstill, so that takes a big pressure drop to accomplish. That is one reason for the high drag. Also, you are only doing 10 inner iterations, which probably means it is not converging within each time step. In that case, the initial results don't mean much anyways.

May 29, 2012, 10:06		#4
cdegroot Senior Member Chris DeGroot Join Date: Nov 2011 Location: Canada Posts: 414 Rep Power: 17	Well, your small timestep is 1000 times smaller than your large time step, so I would guess that the large time step just "steps over" a lot of the initial transient stuff. Also, imagine in reality having completely still fluid and then instantly accelerating it to whatever value you have specified at the inlet. Not going to happen. It will take some (maybe small) amount of time to accelerate. In your simulation that time to accelerate is your first time step. Thus, I believe a smaller timestep will imply a larger initial pressure drop and also drag forces. In any case, these initial transients will die out eventually.

June 3, 2012, 17:43		#5
orxan.shibli Member Join Date: Aug 2011 Posts: 30 Rep Power: 14	I found the problem. It was due to time-step dependency of original Rhie and Chow MIM. Hence, it is better to use modified version: "Discussion on Momentum Interpolation Method for Collocated Grids of Incompressible Flow", Yu et al.