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How to define Boundary conditions for Quadrature Method of Moment

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Old   February 5, 2019, 12:23
Default How to define Boundary conditions for Quadrature Method of Moment
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Hi guys,
I’m working on population balance model and have some issues with the definition of QMOM boundary conditions.

I used to work with descretization method and there BC was quite simple to define. It was either a log normal or a manual distribution but here with method of moment it is confusing. I can't understand how to set up a boundary value for each moment.

I'm using fluent 18.2 with the mixture multiphase method. The only phenomena occur here is growth rate (2.8e-8 m/s) and the particle min and max size are 1e-7m 1e-5m respectively.



Any comment would be highly appreciated.
Amir

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Old   May 17, 2019, 07:46
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Hi Amir,
did you got the solution, that what are the boundary conditions for QMOM?
If yes, please let me know. I am also having the same problem.


Thank you
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Old   June 21, 2019, 03:12
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Quote:
Originally Posted by Amir.11 View Post
Hi guys,
I’m working on population balance model and have some issues with the definition of QMOM boundary conditions.

I used to work with descretization method and there BC was quite simple to define. It was either a log normal or a manual distribution but here with method of moment it is confusing. I can't understand how to set up a boundary value for each moment.

I'm using fluent 18.2 with the mixture multiphase method. The only phenomena occur here is growth rate (2.8e-8 m/s) and the particle min and max size are 1e-7m 1e-5m respectively.



Any comment would be highly appreciated.
Amir

“Hi Amirhossein, You have to set the boundary values at the inlet as follows,
1. At the inlet you have to decide (e.g. 1 mm, 2mm, etc) what diameter dispersed phase is entering/existing into the system. This is very simple, if you know the experimental PSD/BSD/DSD, take the mean value of the curve.
2.Suppose you decide 2mm as a inlet diameter then 0th moment (i.e. total number of bubbles) will be 1/volume of 2mm diameter i.e.238732414 (1/4.18E-9).
3. The 1st moment will be the total length. it is calculated as diameter/total volume (0.002/4.18E-9=477464).
4. The second moment will be the total area which is calculated as 0.002^2/4.18E-9 = 954.92.
5. Similarly third and fourth moments will be 0.002^3/4.18E-9= 1.9098 and 0.002^4/4.18E-9 = 0.0038.
Hope this will help you.
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Old   October 18, 2019, 08:02
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Quote:
Originally Posted by dhiraj.lote@gmail.com View Post
“Hi Amirhossein, You have to set the boundary values at the inlet as follows,
1. At the inlet you have to decide (e.g. 1 mm, 2mm, etc) what diameter dispersed phase is entering/existing into the system. This is very simple, if you know the experimental PSD/BSD/DSD, take the mean value of the curve.
2.Suppose you decide 2mm as a inlet diameter then 0th moment (i.e. total number of bubbles) will be 1/volume of 2mm diameter i.e.238732414 (1/4.18E-9).
3. The 1st moment will be the total length. it is calculated as diameter/total volume (0.002/4.18E-9=477464).
4. The second moment will be the total area which is calculated as 0.002^2/4.18E-9 = 954.92.
5. Similarly third and fourth moments will be 0.002^3/4.18E-9= 1.9098 and 0.002^4/4.18E-9 = 0.0038.
Hope this will help you.
Hi Dhiraj,
It helped a lot thanks!
Just to add a small point to your comment, I think for each moment and for total number definition it is necessary to consider the dispersed phase volume fraction.
For instance, for 0th moment it will be the VF/total volume instead of 1/total volume and so on for higher moments.
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Old   October 18, 2019, 08:20
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Originally Posted by fjois View Post
If yes, please let me know. I am also having the same problem.

It is been a while but if you are still on this issue here is a way that may help to define the initial moments.

Total number of particles can be calculated as:


N0=VF/(Kv ∙ L0^3 )


N0=total number of particles (#/m3 of solution)
VF = Volumic fraction
Kv = Shape factor

Then the initial moments obtain as:
m0 = L0^0 ∙ N0 = N0
m1 = L0^1 ∙ N0 = m0 ∙ N0
m2 = L0^2 ∙ N0 = m1 ∙ N0
m3 = L0^3 ∙ N0 = m2 ∙ N0
.
.
.
mk = L0^k ∙ N0
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