(By Frederic DEGHETTO, Eric VALETTE)
Description of the task
The newtonian fluid problem
Differential viscoelastic isothermal problem
The EVOL function
Even if we just study newtonian fluid at the ENSEEIHT, an important part of industrial fluids are non newtonian. More other, if certain fluid like oil could be consider as newtonian fluids in the most of cases, theirs a lot of extremal conditions where they must be considareted as non newtonian fluids. That why we considered important to have an ontroduction to this process.
Description of the task : Mixing in a rotating frame of reference
This example discribes the simulation of a mixer. Most of the time , the than is axisymmetric and one wishes to a 2 D 1/2 " swirling " simulation. however if the rotor is not axisymmetric, it is possible to simulate the flow in a rigid refernce frame without going to time-dependent simulation and complex remeshing procedures. A way to circumvent this difficulty is to simulate the flow in a rotatating reference frame, by adding or substrating an angular velocity W . In this case, a zero velocity boundary condition will apply on the rotor. If inertia is taken into account, one must then add the centrifugal and Coriolis forces as body forces. This example describes how to proceed.
Let us consider a 2-D planar mixing flow process, as displayed in Fig 1. A fluid is contained into a long cylindrical vessel, and is set into motion by means of a propeller. the angular velocity W of the propeller is - 2Pi s-1 while the radius of the cylindrical vessel is 10cm.
Fig 1 : Mixing
This description correspond to what
is seen from a fixed observer.
In a rotating frame of reference, the observer sees a fixed propeller, the wall of the vessel being rotating in the opposite direction. Since will be taken into account, centrifugal and Coriolis forces will be added into the momentum equation
The newtonian fluid problem
We consider a Newtonian fluid, the material data of which are :
nu = 2.5 poise
rau = 1g/cm3
The angular velocity W at the wall will be -2Pi s-1. The corresponding tangential velocity equals -WR , where R is the radius of vessel.
In view of the relatively high Reynolds number which is involved, we apply an evolution strategy on the fuid density.
In fig 2, we display the finite element mesh. This mesh contains one single sub domain : The whole mesh. However it has two boundaties : the external boundary ( BS 1), and the internal boundary which coincides with the propeller ( BS 2).
Fig 2 : Finite element mesh.
Prefix >>>>>>>>to give the prefix of the output file ex : rigid
Read a mesh file : rigid.msh
Create a new task : F.E.M. task,
evolution, D-planar geometry, +Rigid rotation : W=-6.2831998 see
Create a sub-task : Generalized Newtonian isothermal flow problem
Title : Mizing process
Domain of the sub-task : Whole mesh (S1)
Material data :
Constant viscosity = 2.5
Density : 1.0
with EVOL on density : f(S)=S see note 2
Inertia terms : will be taken into account
Flow boundary conditions :
BS 1 : (vn,vs) = (0.0, 62.832) see note 3
BS 2 : (vn,vs) = (0.0,0.0)
Initial S = 0.0
Final S = 1.0
Initial S-interval = 0.01
max. S-interval = 0.25
Probes (optional) :
1. Prefix : prob_1, location : ( 0, 5 )
2. Prefix : prob_2, location : ( -5, 5 )
3. Prefix : prob_3, location : ( 5, 5 )
Save & Exit
Mesh file : rigid.msh
Data file : rigid.dat
Result file : rigir.res
Fluent result files : flum,flur
The input files for POLYFLOW are RIGID.MSH ( the mesh) and RIGID.DAT ( the data file generated by POLYDATA). RGID.DAT is the standard input file for POLYFLOW. After execution, a result file RES is generated by POLYFLOW for an eventual restart. Several files for graphic post-processing FLUM.j for the meshing and FLUR.j for the result are also generated, where j stand for the corresponding S-step.
The mesh file like the other result can be visualised with flpost. this software have the same interface that fluent.
We can see on the image under the variation of the pressure and velocity vector versus time :
Result for time 1 :
Result for time 10 :
In this case the Renolds number is equal to Re = (Rau * N *D**2) / Nu
Re = 1e3*1*(2*7.5e-2)**2/2.5e-1 = 90
This Reynolds number explain how we used the EVOL function for the density.