# Why the transient state initial velocity profile is not identical to the final steady

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 October 14, 2019, 06:27 Why the transient state initial velocity profile is not identical to the final steady #1 New Member   Ahmed Join Date: Sep 2018 Location: Daejeon Posts: 8 Rep Power: 7 I am studying flow in a 3D reactor which is rotating due to rotating disk present inside the reactor. Firstly, I simulated it with a steady-state RSM model and a converged solution has been obtained. Later on, I studied it with a Laminar transient state and initialized the flow field with the steady-state solution of RSM. Now, when the transient state simulation is completed, I extracted velocity profiles at different time steps starting from the 0-time step (0 sec). To my understanding, the velocity profile of the 0-time step should be identical to the velocity profile obtained from RSM steady-state solution. But there is a slight difference in the velocity near the wall of a rotating disk. I am wondering why is that? I just checked the Solution Methods of both cases. In RSM, I used the SIMPLE scheme with the second order. While, in the Laminar transient state case, I used the PISO scheme with 1st order Momentum. So, are these changes resulting in different velocity values? I much appreciate if anyone could share his thoughts over it. Thanks in advance.

 October 14, 2019, 12:51 #2 Senior Member   Lucky Join Date: Apr 2011 Location: Orlando, FL USA Posts: 5,679 Rep Power: 66 RSM the turbulence model? If you run one with a turbulence model and one with a laminar model, of course the results will be different. Turbulent flows are different than laminar flows.

October 15, 2019, 02:22
#3
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Ahmed
Join Date: Sep 2018
Location: Daejeon
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Quote:
 Originally Posted by LuckyTran RSM the turbulence model? If you run one with a turbulence model and one with a laminar model, of course the results will be different. Turbulent flows are different than laminar flows.
Yes, the turbulence model.

You are right. Solver formulations are different from each other and there is no issue in getting different final results as I already got.
But my question is "the initial Laminar transient solution should not be identical to the final steady RANS solution?" Since Laminar transient solver did not start solving it as I extracted result at 0 time-step.
That is why I was hoping for the same results. In fact, they are same but differ in just two points near the wall of the rotating disk.
Please correct me, if my understanding is wrong.

 October 15, 2019, 09:04 #4 Senior Member   Lucky Join Date: Apr 2011 Location: Orlando, FL USA Posts: 5,679 Rep Power: 66 Oh. My mistake, you mentioned a bunch of irrelevant details and fed me a red herring. They should be identical, they could be identical. But it depends on what steps you used to "initialize" and "extracted." If it's the same mesh, then you shouldn't have to extract or initialize and there would've been a 1-to-1 correspondence.

October 16, 2019, 04:53
#5
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Ahmed
Join Date: Sep 2018
Location: Daejeon
Posts: 8
Rep Power: 7
Quote:
 Originally Posted by LuckyTran Oh. My mistake, you mentioned a bunch of irrelevant details and fed me a red herring. They should be identical, they could be identical. But it depends on what steps you used to "initialize" and "extracted." If it's the same mesh, then you shouldn't have to extract or initialize and there would've been a 1-to-1 correspondence.
Okay, I accept my mistake of sharing irrelevant details which confused you.

However, the velocity profile is identical everywhere except near the rotating disk. Mesh is the same in both cases. I don't know why is that. Is it due to the friction of the rotating wall which hinders the fluid to flow near the wall?
I initialized the solution in this way: File > Interpolate > write.