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K-Epsilon Turbulent Model - Turbulence Damping for paper pulp
Posted 15 nov 2011, 11:08 GMT-5 Fluid & Heat, Computational Fluid Dynamics (CFD) Version 4.2, Version 4.2a 9 Replies
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Hi folks.
I am modelling paper pulp flow in pipes.
I have a lot of experimental data that I have to simulate and validate.
I am using single phase turbulent k-epsilon model to do so. However, I get pressure drops higher than
expected.
I suspect that the turbulence damping that occurs for these flows is not being implemented.
Does anyone have suggestions or experience with this matter?
I could use a few pointers....
Best Regards,
Rui
I am modelling paper pulp flow in pipes.
I have a lot of experimental data that I have to simulate and validate.
I am using single phase turbulent k-epsilon model to do so. However, I get pressure drops higher than
expected.
I suspect that the turbulence damping that occurs for these flows is not being implemented.
Does anyone have suggestions or experience with this matter?
I could use a few pointers....
Best Regards,
Rui
9 Replies Last Post 17 nov 2011, 03:03 GMT-5
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Posted:
1 decade ago
Hi,
I am not sure what you mean with 'turbulence damping for these flows'; The starting point for these analysis is always the question: what is your Reynolds number?
At any rate a coarse mesh will cause large-than-expected pressure drops, however if the pipes have several change of directions it may be a good idea to use the k-omega model.
Cheers
I am not sure what you mean with 'turbulence damping for these flows'; The starting point for these analysis is always the question: what is your Reynolds number?
At any rate a coarse mesh will cause large-than-expected pressure drops, however if the pipes have several change of directions it may be a good idea to use the k-omega model.
Cheers
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Posted:
1 decade ago
I guess I should have provided more details :)
It's a straight 4 meters pipe, with a paper pulp solution flowing at 6 m/s with constant temperature.
I believe the behavior of this solution is Non-Newtonian.
And typically, the pressure drop is lower than for water, for example, for the same pipe length.
This is due to Turbulence Damping, because it occurs a damp in the turbulence production in the wall.
The mesh is quite fine with boundary layer.
Hope this helps.
Cheers.
It's a straight 4 meters pipe, with a paper pulp solution flowing at 6 m/s with constant temperature.
I believe the behavior of this solution is Non-Newtonian.
And typically, the pressure drop is lower than for water, for example, for the same pipe length.
This is due to Turbulence Damping, because it occurs a damp in the turbulence production in the wall.
The mesh is quite fine with boundary layer.
Hope this helps.
Cheers.
Hi,
I am not sure what you mean with 'turbulence damping for these flows'; The starting point for these analysis is always the question: what is your Reynolds number?
At any rate a coarse mesh will cause large-than-expected pressure drops, however if the pipes have several change of directions it may be a good idea to use the k-omega model.
Cheers
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Posted:
1 decade ago
Hi,
All right, non-Newtonian flow with turbulence? That sounds tough enough, can you provide the Re number for the averaged density and the max and min value?
Cheers
All right, non-Newtonian flow with turbulence? That sounds tough enough, can you provide the Re number for the averaged density and the max and min value?
Cheers
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Posted:
1 decade ago
if the pipes have several change of directions it may be a good idea to use the k-omega model
Hi Amir,
If you have 2 minutes, could you please indicate a reference where it's advised to use the k-w model instead of k-e for flow in complex geometries?
Thanks!
Francois
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Posted:
1 decade ago
Hi,
There is quite a large volume of literature about the failure of the k-epsilon and how Menter developed the k-omega for bounded flows. You can start with two papers by Menter:
Menter, "Zonal Two Equation k-ω Turbulence Models for Aerodynamic Flows", AIAA Paper 93-2906, 1993
Menter "Two-Equation Eddy-Viscosity Turbulence Models for Engineering Applications", AIAA Journal, vol. 32, no 8. pp. 1598-1605, 1994
Cheers
There is quite a large volume of literature about the failure of the k-epsilon and how Menter developed the k-omega for bounded flows. You can start with two papers by Menter:
Menter, "Zonal Two Equation k-ω Turbulence Models for Aerodynamic Flows", AIAA Paper 93-2906, 1993
Menter "Two-Equation Eddy-Viscosity Turbulence Models for Engineering Applications", AIAA Journal, vol. 32, no 8. pp. 1598-1605, 1994
Cheers
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Posted:
1 decade ago
Hello Amir,
I apologize for the delay but I have been away from my desk.
The Re number for my flow is 45 730 for a pipe diameter with 0.0762 m and 2 meters long. The velocity with 6 m/s.
The density is 1004.77 and the viscosity is a function of the velocity:
ETA=0.0132*V^(-0.1524)
Most of the times I can't get it to converge, and when it does, the pressure drop is 3x larger...
Hope you can give some more insight...
Thank you.
Cheers.
I apologize for the delay but I have been away from my desk.
The Re number for my flow is 45 730 for a pipe diameter with 0.0762 m and 2 meters long. The velocity with 6 m/s.
The density is 1004.77 and the viscosity is a function of the velocity:
ETA=0.0132*V^(-0.1524)
Most of the times I can't get it to converge, and when it does, the pressure drop is 3x larger...
Hope you can give some more insight...
Thank you.
Cheers.
Hi,
All right, non-Newtonian flow with turbulence? That sounds tough enough, can you provide the Re number for the averaged density and the max and min value?
Cheers
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Posted:
1 decade ago
Hi,
What are the units of your density? Kg/m^3 I think. How about the viscosity? Is it kinematic or dynamic? Did you try to average it to have an idea of the average Re?
It seems the viscosity goes to infinity as V approaches zero, do you have an idea of the actual limits on your viscosity?
Do you have an idea of the length scales of the flow in your pipe?
Cheers
What are the units of your density? Kg/m^3 I think. How about the viscosity? Is it kinematic or dynamic? Did you try to average it to have an idea of the average Re?
It seems the viscosity goes to infinity as V approaches zero, do you have an idea of the actual limits on your viscosity?
Do you have an idea of the length scales of the flow in your pipe?
Cheers
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Posted:
1 decade ago
Yes, those the units for the density.
The viscosity is dynamic.
In principle, the paper pulp behaves like a shear thinning fluid and our experiments tell us that the
range for the dynamic viscosity is [0.0405 - 0.0101]Pa.s, and for higher velocities the dynamic viscosity decreases.
The length scales are : It=0.01*Re^(-1/8) and Lt=0.0005.
Hope this helps.
Rui
The viscosity is dynamic.
In principle, the paper pulp behaves like a shear thinning fluid and our experiments tell us that the
range for the dynamic viscosity is [0.0405 - 0.0101]Pa.s, and for higher velocities the dynamic viscosity decreases.
The length scales are : It=0.01*Re^(-1/8) and Lt=0.0005.
Hope this helps.
Rui
Hi,
What are the units of your density? Kg/m^3 I think. How about the viscosity? Is it kinematic or dynamic? Did you try to average it to have an idea of the average Re?
It seems the viscosity goes to infinity as V approaches zero, do you have an idea of the actual limits on your viscosity?
Do you have an idea of the length scales of the flow in your pipe?
Cheers
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Posted:
1 decade ago
Hi,
Crunching the numbers you gave I get:
Re=rho*V*D/eta=rho*D*V^1.1524/0.0132 where D is a characteristic length and can be thought to be the diameter of the pipe
for a critical value of Re=2000 then:
Vcritical=(2000*0.0132/(rho*D))^(1/1.1524)=0.399m/s
It is quite crude but it tells us that for a characteristic length equal to the diameter of the pipe with a velocity-dependent viscosity your flow becomes laminar as the velocity reaches 0.399 m/s.
So I would say that you need to mix laminar and turbulent in one go, honestly I do not know how that can be done in Comsol.
Cheers
Crunching the numbers you gave I get:
Re=rho*V*D/eta=rho*D*V^1.1524/0.0132 where D is a characteristic length and can be thought to be the diameter of the pipe
for a critical value of Re=2000 then:
Vcritical=(2000*0.0132/(rho*D))^(1/1.1524)=0.399m/s
It is quite crude but it tells us that for a characteristic length equal to the diameter of the pipe with a velocity-dependent viscosity your flow becomes laminar as the velocity reaches 0.399 m/s.
So I would say that you need to mix laminar and turbulent in one go, honestly I do not know how that can be done in Comsol.
Cheers