- Capture of the pressure discontinuity by enriching the pressure space in the elements crossed by the interface, with condensation of the associated degrees of freedom before system assembly.
- Filtered level-set method to capture the liquid-air interface with redistancing by solving a Hamilton-Jacobi equation.
# 2D and 3D Jurin's rises
# 2D and 3D Jurin's Rises
Capillarity is the sole driving force. Liquid rises until the equilibrium between capillary and gravitational forces is achieved.
Gravity is neglected in this simulation. 4,000 time increments are performed with a time step of 2,00x10^-4 s. The liquid viscosity is 1.0 mPa.s, the air viscosity is 1.0 µPa.s, the liquid/air surface tension is 37.07 mN/m, the liquid/fiber energy is 30.03 mN/m, and the air/fiber surface energy is 60.9 mN/m. The normal velocity is imposed to be vanishing using an augmented Lagrangian technique at each node of the fiber boundaries. Additionally, the normal velocity is set to vanish (Dirichlet condition) at the left and right boundaries of the computational domain. Finally, a normal stress of 1 Pa is considered as Neumann's condition at the bottom boundary.
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@@ -31,3 +31,13 @@ The mesh consists of 141,000 triangles and 71,200 nodes. The simulation ran for
Additionally, the 'volume' of liquid is plotted against time. It evolves as t^0.4:
The 3D computational domain is obtained by extrusion of the 2D domain described below along the *z* axis (from *z = 0* to *z = 0.5*. The mesh consists of 4.9 million tetrahedra and 844,000 nodes (mesh size of 10^-2). The simulation parameters are the same of those used in the 2D case. The simulation involves 3,000 time increments completed in 28.75 hours using 64 cores.
In addition to visualizing the progression of liquid volume through viscosity, visualizing the pressure, velocity (vectors), and the zero-isosurface of the level-set function (shown as a green surface) in the animations below, the liquid volume is plotted over time. The dependency appears to be almost in square-root of the time (t^0.49 determined by least squares).
