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Fluid Structure Interaction - Gravity - Equilibrium state
Posted 2015年11月17日 GMT-5 13:23 Version 5.1 3 Replies
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Hello,
In the simplified model attached, I am trying to have an initial state of equilibrium, in which the pressure from the fluid would counteract the gravity load from the solid around it. Both fluid and solid have the same density so I am not expecting any buoyancy effect.
The solid part is loaded using a body load -rho*g*z. I thought that applying the volume force of rho*g*z should cancel the load. However there is still surface displacement following this approach.
By disabling the load around the fluid, I noticed that the volume force is not being applied on the coordinate z but a relative z within the domain in which 0 is more or less the center of the sphere. The fluid pressure from gravity at the top should be 0 and rho*g*-2000 (thickness of reservoir) at the bottom. Not rho*g*+1000 at the top and -1000 at the bottom. I do not know how to modify the default coordinate on which the volume force applies.
Please let me know if you have any idea how to work around this problem and to properly apply gravity in the case of a fluid embedded in a solid. I haven't found any useful tutorials in the model library.
Thanks,
Helene
In the simplified model attached, I am trying to have an initial state of equilibrium, in which the pressure from the fluid would counteract the gravity load from the solid around it. Both fluid and solid have the same density so I am not expecting any buoyancy effect.
The solid part is loaded using a body load -rho*g*z. I thought that applying the volume force of rho*g*z should cancel the load. However there is still surface displacement following this approach.
By disabling the load around the fluid, I noticed that the volume force is not being applied on the coordinate z but a relative z within the domain in which 0 is more or less the center of the sphere. The fluid pressure from gravity at the top should be 0 and rho*g*-2000 (thickness of reservoir) at the bottom. Not rho*g*+1000 at the top and -1000 at the bottom. I do not know how to modify the default coordinate on which the volume force applies.
Please let me know if you have any idea how to work around this problem and to properly apply gravity in the case of a fluid embedded in a solid. I haven't found any useful tutorials in the model library.
Thanks,
Helene
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3 Replies Last Post 2015年11月18日 GMT-5 16:43
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Posted:
9 years ago
Hi
first of all there are a few BC I do not understand in your model:
1) the body load on the solid and the Volume force have different sign, that's probably a typo, it also applies to the initial strain, why should it be diagonal identical like that ? I would start without your initial Strain case and check its form, or use a stationary case to "preload" the material
2) you fix with prescribed mesh displacement the fluid, that is not really required as you are on the axial symmetry, anyhow I would not prescribe the "z" direction, OK for the "r"
3) Roller condition is like a symmetry condition, since you are in 2D-axi the axis remains as is, therefore the right roller is probably not required but probably that far away it does not influence a lot
4) you have selected the nearly incompressible material, but your nu is 0.25, normally one uses the additional "p" variable for materials with nu>0.49. it makes the calculations longer, but probably does not make anything wrong, you can get a "p" plot also in the solid like that
Then some general comments:
a) your solid is very "large" compared to the fluid, it just makes the model rather big and heavy, but OK it solves on my laptop, not critical
b) you could probably work quicker by using Stockes flow (neglecting the inertia terms), this makes the time dependent solver work quickly as you will not see any sound wave transients
--
Good luck
Ivar
first of all there are a few BC I do not understand in your model:
1) the body load on the solid and the Volume force have different sign, that's probably a typo, it also applies to the initial strain, why should it be diagonal identical like that ? I would start without your initial Strain case and check its form, or use a stationary case to "preload" the material
2) you fix with prescribed mesh displacement the fluid, that is not really required as you are on the axial symmetry, anyhow I would not prescribe the "z" direction, OK for the "r"
3) Roller condition is like a symmetry condition, since you are in 2D-axi the axis remains as is, therefore the right roller is probably not required but probably that far away it does not influence a lot
4) you have selected the nearly incompressible material, but your nu is 0.25, normally one uses the additional "p" variable for materials with nu>0.49. it makes the calculations longer, but probably does not make anything wrong, you can get a "p" plot also in the solid like that
Then some general comments:
a) your solid is very "large" compared to the fluid, it just makes the model rather big and heavy, but OK it solves on my laptop, not critical
b) you could probably work quicker by using Stockes flow (neglecting the inertia terms), this makes the time dependent solver work quickly as you will not see any sound wave transients
--
Good luck
Ivar
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Posted:
9 years ago
Hi Ivar,
Thank you very much for your reply. I have taken your advice into account.
Regarding 1), the initial stress is to reproduce the initial lithostatic crustal stress. It has been validated in several solid mechanics-only models. It should be equivalent to your solution of preloading it by first running the stationnay solution (I will check that).
I am still not understanding how the volume force works.. On the figure attached, you can see the fluid pressure when just subjected to the applied volume force of - rho * g , not any other force, and just solving for the fluid part. I am not understanding the sign and magnitude of it, which are not equivalent to -rho*g*z.
Thanks,
Helene
Thank you very much for your reply. I have taken your advice into account.
Regarding 1), the initial stress is to reproduce the initial lithostatic crustal stress. It has been validated in several solid mechanics-only models. It should be equivalent to your solution of preloading it by first running the stationnay solution (I will check that).
I am still not understanding how the volume force works.. On the figure attached, you can see the fluid pressure when just subjected to the applied volume force of - rho * g , not any other force, and just solving for the fluid part. I am not understanding the sign and magnitude of it, which are not equivalent to -rho*g*z.
Thanks,
Helene
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Posted:
9 years ago
Hi
I was a bit too quick with some of my comments I see:
Volume force is the force applied to "dx*dy*dz[m^3]" hence it's "spf.rho*g_const[N/m^3]". Formally its the product of both that gives you the force, and total force once integrated over the full domain(s).
Now gravity is mostly defined along "-z" hence the traditional minus sign for Volume Forces, to add to the formula above.
Now, for the initial condition it is different.
This depends where you set the gauge pressure p=0[Pa] level. So if your fluid is just above z>=0 then the initial condition (again applied to the element volume "dx*dy*dz") is "-spf.rho*g_const*z".
The means that the pressure would be p(z=0)=0[Pa] and it will be relative with a negative sign above z=0.
If you try to solve fluids alone and set the initial condition correctly your model might not converge fully, but still it should give a correct pressure value.
Then for the initial stress tensor, the diagonal elements should not be all equal from my understanding, the two first OK are isotropic in R & Phi, but the Z component should be greater. Here again one need to define the "zero gauge Z level", so if gravity is along -Z the diagonal elements should be from my understanding: "-g_const*solid.rho*( nu/(1-nu) , nu/(1-nu) , 1)*z" all in [Pa]. with Z=0 at top level of the solid.
--
Good luck
Ivar
I was a bit too quick with some of my comments I see:
Volume force is the force applied to "dx*dy*dz[m^3]" hence it's "spf.rho*g_const[N/m^3]". Formally its the product of both that gives you the force, and total force once integrated over the full domain(s).
Now gravity is mostly defined along "-z" hence the traditional minus sign for Volume Forces, to add to the formula above.
Now, for the initial condition it is different.
This depends where you set the gauge pressure p=0[Pa] level. So if your fluid is just above z>=0 then the initial condition (again applied to the element volume "dx*dy*dz") is "-spf.rho*g_const*z".
The means that the pressure would be p(z=0)=0[Pa] and it will be relative with a negative sign above z=0.
If you try to solve fluids alone and set the initial condition correctly your model might not converge fully, but still it should give a correct pressure value.
Then for the initial stress tensor, the diagonal elements should not be all equal from my understanding, the two first OK are isotropic in R & Phi, but the Z component should be greater. Here again one need to define the "zero gauge Z level", so if gravity is along -Z the diagonal elements should be from my understanding: "-g_const*solid.rho*( nu/(1-nu) , nu/(1-nu) , 1)*z" all in [Pa]. with Z=0 at top level of the solid.
--
Good luck
Ivar