Dear Vishal,

We are assuming wedge as an approximation to simplify the domain.

Radius is R

{because R*0.999 is approximately equals R only right ??};)

--
KANNAN

You are right :P... was just curious why it was thought about to construct the geometry in this manner :). The geometry works fine. Thanks for your tips.

Hi Kanan,

I am now modeling a concentric pipe geometry. I have difficulty in understanding which faces I should generate the mesh. I would like to give boundary conditions on both the surface of inner pipe and on the outer pipe. Here is the sample blockmesh dict file that I have generated.

// length of pore = 36 nm and diameter: 10.2nm , theta = 2 deg (consider this as outer pipe)
// length of DNA = 36 nm and diameter of DNA: 2.2nm , theta = 2 deg (consider this as inner pipe)
//r_DNA*cos(theta/2) = 1.1*cos(0.0349/2 radians) = 1.0998
// r_DNA*sin(theta/2) = 1.1*sin(0.0349/2 radians) = 0.0192
//r_pore*cos(theta/2) = 5.1*cos(0.0349/2 radians) = 5.0992
//r_pore*sin(theta/2) = 5.1*sin(0.0349/2 radians) = 0.0890
vertices
(
(0 0 0) // vertex 0
(36 0 0) // vertex 1
(36 1.0998 -0.0192) // vertex 2 (L,r_DNA*cos(theta/2),-r_DNA*sin(theta/2))
(0 1.0998 -0.0192) // vertex 3 (0,rcos(theta/2),-rsin(theta/2))
(36 5.0992 -0.0890) // vertex 4 (L,r_pore*cos(theta/2),-r_pore*sin(theta/2))
(0 5.0992 -0.0890) // vertex 5 (0,r_pore*cos(theta/2),-r_pore*sin(theta/2))


(0 0 0.0192) // vertex 6 (0, 0,r_DNA*sin(theta/2))
(36 0 0.0192) // vertex 7 (L, 0,r_DNA*sin(theta/2))
(36 1.0998 0.0192) // vertex 8 (0, r_DNA*cos(theta/2),r_DNA*sin(theta/2))
(0 1.0998 0.0192) // vertex 9 (L, r_DNA*cos(theta/2),r_DNA*sin(theta/2))
(0 0 0.0890) // vertex 10 (0, 0,r_pore*sin(theta/2))
(36 0 0.0890) // vertex 11 (L, 0,r_pore*sin(theta/2))
(36 5.0992 0.0890) // vertex 12 (L,r_pore*cos(theta/2),r_pore*sin(theta/2))
(0 5.0992 0.0890) // vertex 13 (0,r_pore*cos(theta/2),r_pore*sin(theta/2))
);

Please let me know if the vertices are correct. If so, how to generate the mesh for the same.

Thanks

Regards
Vishal

hi thanks every one for posting the reply. i have a question i would like to simulate a steady state simulate in ico foam so i changed to
ddtScheme
{
type steadystate;
}

Then i got this error

FOAM Warning :
From function gaussConvectionScheme
in file finiteVolume/convectionSchemes/gaussConvectionScheme/gaussConvectionScheme.H at line 123
Reading "/home/yathuru/task1steady/system/fvSchemes.divSchemes.div(phi,U)" at line 31
Unbounded 'Gauss' div scheme used in steady-state solver, use 'bounded Gauss' to ensure boundedness.
To remove this warning switch off 'boundedGauss' in "/opt/OpenFOAM-2.3.0/etc/controlDict".


can some one help me.. i m lost.

Hi,

icoFoam is a transient solver. If you need steady state, simpleFoam is probably what you need.

Regards,
Jose

Dear Naresh,

It's just a warning I guess you can run your case with that.

OR, you can remove the warning as follows:

-
Best Regards!

Thank you so much for your reply santos and tushar. I knew i can do it in simple foam but just out of curiosity i wanted to try it in ico foam. I ran the simulation in simple foam its working well then i changed the case to turbulent simulation its is also working well now. :) however i have some issues with the convergence. i will try to figure it out.:) if not i will get back again .

Dear Naresh,

Please refer my previous post again. I said that you can run your icofoam solver using the steady state command. Anyways I am writing these again:

Best Luck!

Thanks once again Tushar. I m afraid if i can post this question here.

could you please give me some tips for which solver to use for this case below:

Geometry : a room with a box inside. the room has a inlet and outlet.

I want to simulate the flow through this room with some velocity and temperature at the inlet, the roof is maintained at a low temperature (20 C) and the box inside is maintained at a temperature (35C).


could you please suggest a solver. :)

Dear Naresh,

I will prefer you to go for "BuoyantBoussinesqPisoFoam", this will do the job. Wish you Good Luck for your work.

-
Best Regards!

Thank you once again for your continuous support. U have no idea how much this tipp mean to me. thanks.:)

Welcome :)

-
Best Luck!