This example models a rotating disk in a tank. The model geometry is shown in Figure 6-13. Because the geometry is rotationally symmetric, it is possible to model it as a 2D cross section. However, the velocities in the angular direction differ from zero, so the model must include all three velocity components, even though the geometry is in 2D.Figure 6-13: The original 3D geometry can be reduced to 2D because the geometry is rotationally symmetric.
The flow is described by the Navier-Stokes equations: ρ ∂u T + ρ ( u ⋅ ∇ )u = ∇ ⋅ [ – pI + η ( ∇u + ( ∇u ) ) ] + F ∂t ∇⋅u = 0
C H A P T E R 6 : M I X E R S A N D S T I R R E D VE S S E L S
In these equations, u denotes thevelocity (m/s), ρ the density (kg/m3), η the dynamic viscosity (Pa·s), and p the pressure (Pa). For a stationary, axisymmetric flow the equations reduce to (Ref. 1): ∂u v ∂u ∂p 1 ∂ ∂u u- ∂ u ρ ⎛ u ------ – ----- + w ------⎞ + ------ = η -- ----- ⎛ r ------⎞ – ---- + --------- + F r ⎝ ∂r r ∂z ⎠ ∂r r ∂r ⎝ ∂r ⎠ r 2 ∂z 2 ∂v uv ∂v 1 ∂ ∂v v- ∂ v ρ ⎛ u ----- + ------ + w ----- ⎞ = η -- ----- ⎛ r ----- ⎞ –---- + -------- + F ϕ ⎝ ∂r r ∂z⎠ r ∂r ⎝ ∂r⎠ r 2 ∂z 2 ∂w 1 ∂ ∂w ∂p ∂ w ⎛ ∂w ρ ⎝ u ------- + w -------⎞ + ----- = η -- ----- ⎛ r -------⎞ + ---------- + F z ∂r ∂z ⎠ ∂z r ∂r ⎝ ∂r ⎠ ∂z 2 Here u is the radial velocity, v the rotational velocity, and w the axial velocity (m/s). In the model you set the volumetric force components Fr, F ϕ , and Fz to zero. The swirling flow is 2D even though the modelincludes all three velocity components.
2 2 2 2
Figure 6-14 below shows the boundary conditions.
Sliding wall Symmetry
No slip Axial symmetry
Figure 6-14: Boundary conditions. On the stirrer, use the Sliding wall boundary condition to specify the velocities. The velocity components in the plane are zero, and that in the angular direction is equal to theangular velocity, ω, times the radius, r: w w = rω (6-11)
SWIRL FLOW AROUND A ROTATING DISK
At the boundaries representing the cylinder surface a No slip condition applies, stating that all velocity components equal zero: u = ( 0, 0, 0 ) (6-12)
At the boundary corresponding to the rotation axis, use the Axial symmetry boundary condition allowing flow in the tangential directionof the boundary but not in the normal direction. This is obtained by setting the radial velocity to zero: u = 0 (6-13)
On the top boundary, which is a free surface, use the Symmetry condition to allow for flow in the axial and rotational directions only. The boundary condition is mathematically similar to the axial symmetry condition.
Because there is no outflow boundary inthis model, where the pressure would be specified, you need to lock the pressure to some reference pressure in a point. In this model, set the pressure to zero in the top right corner.
The parametric solver provides the solution for four different angular velocities. Figure 6-15 shows the results for the smallest angular velocity, ω = 0.25π rad/s.
C H A P T E R 6 : M I X E RS A N D S T I R R E D VE S S E L S
Figure 6-15: Results for angular velocity ω= 0.25π rad/s. The surface plot shows the magnitude of the velocity field and the white lines are streamlines of the velocity field. The shape of the two recirculation zones, which are visualized with streamlines, changes as the angular velocity increases. Figure 6-16 shows the streamlines of the velocity field forhigher angular velocities.
Figure 6-16: Results for angular velocities ω = 0.5π, 2π, and 4π rad/s. The surface plot shows the magnitude of the velocity and the white lines are streamlines of the velocity field.
SWIRL FLOW AROUND A ROTATING DISK
Figures 6-17 and 6-18 show isocontours of the rotational velocity component together with surface plots of the velocity magnitude...