The permeability of two-dimentional porous medium of square fibers (cell model)

In the present study we have solved numerically the problem of a creeping two-dimensional viscous incompressible flow through a fibrous porous medium. The medium is modeled by a system of square fibers (rods) arranged across the flow. The viscous flow around the fibers is solved under the Stokes approximation using the cell model. The Happel boundary conditions (zero normal velocity and tangential shear stresses) have been posted on the boundary of the fluid cell. The numerical realization is based on finite differences of the second order accuracy with successive over-relaxation method for solving a system of linear equations. For regular lattice of square cylinders the hydrodynamic microscopic velocity fields were determined also by the direct numerical simulations. The permeability coefficient of the model porous medium is evaluated by macroscopic averaging and Darcy low assumption. It was found that the cell model with the Happel boundary conditions gives higher values of permeability in comparison with direct numerical calculations. A comparison between these results obtained by cell model and other direct numerical calculations, empirical and analytical relationships in a wide range of the solid concentration made it possible to make the following inferences. The permeability of a rarefied system of square cylinders can be described by an explicit expression that was obtained for circular cylinders [11] but with form factor 2 that reflects the difference between the square and circular cross sections of the rods 𝑎2 = 1 2 1 8 (︂ln 1 - 1.476 + 2 )︂, ≤ 0.2. For moderately rarefied porous media the semi-empirical Kozeny - Carman formula is appropriate. In this case corrective factor 1/ 2 gives error of the permeability more than 25 %. To decrease the disagreement of the numerical values with the Kozeny - Carman formula this relation is better to use with the factor 8/9. The specified expression for permeability is 𝑎2 = 1 9 (︂ln 1 - 1 - 2 1 + 2)︂, 0.1