Steady and unsteady flow in a tank
Gambit
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Create a brick with size 1x1x1 m centered around the origin
Create a cylinder with height 0.5 m and radius 0.05 m. The height is in the z-direction.
Move the cylinder on top of the brick (the center of the cylinder is the z-axis)
Unite both volumes (there will be no face anymore between brick and cylinder)
Create a brick with size 0.1 x 0.1 x 0.5 m centered around the origin.
Move the brick on top of the first brick but translate it also over a distance of 0.35 m along the x-axis.
Now we will take advantage of the symmetry of the geometry (we assume that the flow will be
symmetrical with respect to the xz-plane).
Create a face in the xz-plane centered around the origin with dimension 3x3 m.
Split the two volumes with this face. First retain the face. With the second split operation this face does not
have to be retained anymore.
Delete the volumes with negative y-coordinates with the lower geometry.
Attach a size function to the upper wall of the brick (connected to the cylinder) with as source the edge of
the cylinder at the connection with the brick. The initial size is 0.01 m and the growth factor is 1.4. Size
limit is large (e.g. 1).
Mesh this face with a triangular mesh.
Mesh the side wall of the cylinder with a structured grid (mapped) with an interval count = 10. This latter
value will be used for the edges which interval count is not yet fixed by the mapping scheme.
Mesh the upper wall of the cylinder with a triangular mesh with an interval count = 10.
Mesh the volume (brick+cylinder) with the cooper meshing schemeand with an interval count = 20. The
face meshes on the top of the cylinder and the brick will be projected throuhout the grid. The grid will
consist of triangular prisms.
Mesh the second volume (small brick on top of the other one) with a structure mesh with an interval count
equal to 5 on the edges of the small faces and an interval count equal to 10 along the long edges..
In order to apply the boundary conditions, switch off the mesh. Use the second button in the bottom row of
the lower right buttons and select mesh off. Then press the apply button.
Specify a velocity inlet at the top of the cylinder.
Specify a pressure outlet at the top of the small brick.
Specify a symmetry boundary conditions for the two faces in the symmetry plane.
Specify an interface boundary condition at the lower face of the small brick.
Do this also for the upper face of the large brick. Don't put both interfaces together in one boundary entity.
Select all other faces and give them a wall boundary condition.
Define the volumes as fluids in one entitiy (with specify continuum types).
Export the mesh. There should be no warnings that a default boundary condition or a default continuum
type is created.
Save the file and quit gambit.
Fluent
We assume that the tank (the complete domain) is completely filled with water. The final velocity at the inlet
will be 5 m/s. The pressure level is of no importance. We will use a gauge pressure (relative to the operating
pressure) of zero Pascal at the outlet. We will first do a steady calculation, afterwards an unsteady calculation
with a rising velocity at the inlet as a function of time. Finally an unsteady calculation will be performed with a
sudden change of the velocity at the inlet, which can be seen as an approximation of a piston that is impacting
on the water in the cylinder. In the latter calculation the compressibility of the water should be taken into
account.
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Start the 3D version of Fluent
Read the mesh file.
Select the SST turbulence model.
In the materials panel, select water-liquid from the fluent database and copy it to the solver.
In the boundary conditions panel, select water-liquid as material for the fluid zone.
Specify an inlet velocity equal to 5 m/s, 5 % turbulence intensity. The hydraulic diameter is the diameter of
the cylinder.
Create an interface from the two interfaces in 'Grid interfaces…'.
Switch on the second order accuracy for the momentum equation and the turbulence transport equations.
Monitor (plot and print) the mass flow rate at the outlet (Solve/monitor/surface…)
Initialize and calculate the flowfield.
Check the mass flow rate at inlet and outlet with Report/Fluxes
Create a surface named 'x=0'. Surface/Iso-surface/surface of constant grid… x-coordinate with iso-value
equal to zero.
12) Display the velocity vectors in the symmetry plane and the plane 'x=0'. Scale the velocity vectors (factor
20) and search for the vortices When displaying the vectors, also display the feature edges of the grid.
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13) Switch on the second order implit unsteady flow solver.
14) Read the profile file veltime.prof (define/profile). The velocity will be zero at t=0 s, 5 m/s at t=0.1 s and
will be kept at 5 m/s afterwards.
15) Specify the profile as boundary condition for the velocity magnitude at the inlet.
16) Define an animation sequence (Solve/animate) with the same vector plot as used above (scaling of vectors
now equal to 5) shown every time step.
17) Initialize and calculate the flowfield, e.g with time step equal to 0.05 s and 30 time steps.
18) Playback the animation sequence.
19) Initialze and calculate only the first time step. Analyze the velocity and pressure field in the symmetry
plane. Remark that the fluid is already leaving the outflow because the fluid is considered to be
incompressible. The acoustic waves are neglegted and are assumed to travel forth and back much faster
than the convective phenomena (tranport of a vortex) that we have seen in the animation.
20) Switch on the density based solver. Use the explicit formulation (i.e. fluxes are treated explicitly and a
multistage stepping scheme will be used) and the explicit unsteady formulation (i.e. no dual time stepping).
21) The material should be compressible now. We have to use the ideal gas law, otherwise the compressible
solver will not work in explicit mode. Of course the ideal gas law is not valid for water, but if we try to
keep the temperature constant (here we choose 300 K), then we have that =(p) instead of =(p,T). So,
- use for Cp a large number, e.g. 1e+7 J/kgK.
- use for the molecular weight M = 1.186 kg/kmol.
- set as operating pressure 2.1e+9 Pa.
The result is the following:
- R = Runiversal / M = 8314/1.186 = 7010 J/kgK.
-  = Cp/Cv = Cp/(Cp-R) = 1 since Cp is much larger than R. Since p = cte *  for isentropic changes and
since p =  R T (ideal gas law) the result with =1 will be that T is constant.
- Speed of sound c = sqrt( R T) = 1450 m/s correspondig to the speed of sound in water.
-  = p/(R T)=(poperating + pauge)/(R T)=(2.1e+9 + small)/(7010*300)=998.6 kg/m 3
So, even with pressure changes in the order of 100 bar = 1e+7 Pa the density will almost be constant.
22) Keep the velocity inlet boundary condition but with a constant velocity of 5 m/s instead of the profile.
Starting from a velocity field at rest, this will result in a sudden rise of the velocity at the inlet at t=0 s.
23) Switch off the SST model. We will do a laminar calculation. The fluid is at rest in the beginning. We will
compute the pressure waves before the onset of vortex transport in the tank.
24) In Solve/Animate/Playback, delete the existing animations.
25) Specify two animation sequences for the velocity field: We will show it every 10 iterations (not time steps
!!). Define for the first sequence the same vector plot as before but use only the symmetry plane (in draw
grid, select only the symmetry plane) and define a pressure contour plot for the second sequence also in the
symmetry plane only. Use 99 levels. Use filled, set auto range off, and use limits (-200000 and 400000).
26) Initialize and calculate the flowfield. Watch at both the animation window and the monitor of the mass
flow rate at the outlet. Now there is a time delay between the onset of the inflow and the onset of the
outflow.
27) Playback the animation sequences. Check the physical time and compare it with the previous calculation.
How long does it take for the acoustic wave to reach the outlet ? Verify with what you expect ! The
distance between inlet and outlet is about 1.35 m. The speed of sound in the water is 1450 m/s.
28) Check the density and temperature contours to verify the assumptions above.