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Old   June 14, 2008, 05:33
Default Directional Loss Model
  #1
Jaloha
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I'm doing a case of SCR(Seletive Catalytic Reduction, a De-NOx device applied in power plants) reactor now. The Catalytic Layer is located in the Reactor, and I want to use the Directional Loss Model to simulate the pressure drop across the layer. Though there is a case("Flow in a Catalytic Converter") in the CFX Tutorials, but an important difference is that in my case I don't know the exact "resistance coefficient"(Kq), while in the Tutorial case a value of 650 [kg m^-4] has been directly given. However, what I know is the design parameter of the "Pressure Drop" across the layer(DeltaP = 435[Pa]) and the "Catalytic Layer Height" (DeltaH = 11.4 [m]). According to the equation stated in the Tutorial, I've simplified it into: DeltaP/DeltaH = -Kq * abs(U)*Ui I defined a CEL expression to represent the Kq value, which goes like this, Kq = DeltaP/(DeltaH * sqrt(Velocity u * Velocity u + Velocity v * Velocity v + Velocity w * Velocity w) * Velocity v ) After running it in CFX-solver, the "Fatal Overflow" error occurred unexpectedly, since I've ever tried to directly set a Kq value, say 100 [kg*m^-4], it can complete normally. I dont know how come it happened. If you have any idea about how to solve this problem, plz let me know! If you are interested in my case, personal contact through Email (wateraction@126.com) or MSN (wateraction@hotmail.com) is encouraged. Thanks very much for your attention!

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Old   June 24, 2008, 11:46
Default Re: Directional Loss Model
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hsr
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If Velocity v is zero (usually the initial guess for the velocity data field) your expression will turn to DeltaP/(DeltaH*0), thus division by zero. This might your problem with the "Fatal Overflow".

You have to make sure, that [ sqrt(Velocity u * Velocity u + Velocity v * Velocity v + Velocity w * Velocity w) * Velocity v ] doesn't evaluate to zero.

You could try to check for zero velocities and "replace" it with a sufficiently small velocity in your Kq expression, like: Kq = DeltaP/(DeltaH * ( Velocity * Velocity v + step((Velocity*Velocity v)/1[m^-2 s^2])*step((Velocity*Velocity v)/-1[m^-2 s^2])*1e-10[m s^-1]))

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Old   June 25, 2008, 01:48
Default Re: Directional Loss Model
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Jaloha
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Thank you very much, hsr, but can you tell me what does step mean?
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Old   June 25, 2008, 02:16
Default Re: Directional Loss Model
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hsr
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step(<var>) checks whether <var> is
: 0 returns 1 = 0 returns 0.5 < 0 returns 0 <var> must be dimensionless, that's why you have to divide the velocities by 1 [m s^-1]

so when you evaluate step(<var>)*step(-(<var>)) it will return 0 for |var| > 0 and 0.25 for var = 0.

It would be nice to use something like "if (<var> > 0) then", but for this you'll have to wait until CFX 12 is released or use user fortran.
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Old   June 26, 2008, 12:38
Default Re: Directional Loss Model
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Jaloha
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Thanks for your explaination´╝îhsr, I've learned a lot. I agree with you that using "if" sentence would be nice since it's simple and easy to understand. While for the Fortran language, I've never learned it though people say it's not hard to get on hand. Really expecting the 12.0 version coming soon.

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