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Old   May 22, 2003, 05:39
Default Two phase gas-solid transient
  #1
Meiring Beyers
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I'm running a two-phase simulation in CFX4.4. Air and solid particles running through a slightly inclined horizontal cylinder. Bouyancy is switched on in order to get a feed of solids material running down the bottom of the cylinder. The flow of solid material is very small (0.0001 m/s along the bottom of the cylinder). I can get a stable simulation with timesteps of the order 0.01seconds but this means that my simulation time to put the inlet solids through the cylinder is excessive i.e. for a 5m cylinder length the solids residence time is 50000seconds requiring 5 million time steps! Does anyone know of a way to speed things up, possibly getting a pseudo steady state simulation going first. The problem is inherently transient since this cylinder is also rotating with internal baffles therefore solids does not really get to a steady state profile near the bottom of the cylinder. Any comments please!
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Old   May 22, 2003, 08:38
Default Re: Two phase gas-solid transient
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Mads Bang
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I have done some transient simulations where I simulated vortex shedding behind a square rod. I found that in order to catch the phenomena I needed to set the time to step sizes to maximum 1/100 of the shedding frequency when I used a first order time scheme. However with a second order time scheme the step sizes could be reduced to less than 1/10 of the shedding frequency. –So maybe you could increase the time step size if you switched time scheme.

Secondly, if you want to obtain a steady solution to a transient problem, I found that it is much easier if you use the pseudo transient time stepping algorithm (which CFX4.4 also has).

Good luck, Mads
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Old   May 22, 2003, 12:53
Default Re: Two phase gas-solid transient
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Jan Rusås
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It would like to comment on the reccomendation about just increasing the time step. It is basically true that the time step can be increased when a second order time step is used. BUT it does actually depend on the Courant number for the calculation. The courant number should be around 1 when a 1st order scheme is used, and I have seen that a number on 5 can be used when using a higher order scheme. (I do not know if 10 also can be used) The problem with the definition of the Courant number is that, usually the maximum expected velocity and the minimum cell size is used. In case that the maximum velocity is not located at the smallest cells, then would the local courant number be well below 1 (or the 5). It should be possible with user fortran to calculate the local Corant number which would give a good indication of if the time step could be further decreased.

It is however also my experience that when the time step is increased then it requires more iterations per time step. I actually once did an investigation on vortex shedding and I found in a certain range (in order to predict the correct Strouhal number) that the number of iterations/timestep should be above a certain level. In that case I could not benefit from increasing the time step, because then should the iterations also increase. So take care if you increase the time step to much, maybe you will get an incorrect solution. Maybe it would be a good suggestion for you to calculate your global Courant number and see where it is, if it is less that 1 or 5 (depending on scheme) you should be able to increase it. Or maybe even better calculate the local Courant number and adjust the time step based on that.

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Old   May 22, 2003, 13:58
Default Re: Two phase gas-solid transient
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Mads Bang
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Hi Jan,

Well, good arguments from you as always!

There is the option to use the "courant number time stepping" and set the desired curant number to 1. Then the false timestep size is automatically ajusted to unity for each timestep. Maybe that will help you accelerate the convergence.
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Old   May 23, 2003, 02:37
Default Re: Two phase gas-solid transient
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Meiring Beyers
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Thanks Jan and Mads Your comments are appreciated, this is precisely my dilemma and presently there does not seem to be a way around it. I can go to somewhat higher timesteps - but then my required iterations increase. It does seem that the problem is governed by the fall velocity of the particles, more than any other of the smaller local velocities. The cylinder wall tangential speed is lower than this due to low rotational speed, particles axial speed along cylinder length is lower than this due to very slight incline (1 deg) thus the Courant number is govered by fall velocity i.e terminal particle velocity. Roughly this translates in a dt<dx/wfall=0.025/0.5=0.05 seconds timestep. I'll let you know on progress!
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