Capillarity - Permeability - The Network Model of Porous Media - III. Dynamic Properties of Networks with Tube Radius Distribution

Relative permeability and relative electrical resistivity curves are obtained for networks of tubes with a tube radius distribution by means of a network of resistors used as an analog model. These curves are similar to those obtained from real porous media. The effects of pore size distribution and network structure on relative permeability and resistivity are demonstrated. INTRODUCTION Study of the dynamic properties of networks of single size tubes in paper IT' of this series has shown that the network model has flow properties similar to those observed on real porous media. The comparison of properties of single size tube networks with those of real porous media cannot be carried far because very few porous media have uniform size pores. The capillary pressure curve from most porous materials such as soil, sandstone, and sintered glass indicate that these materials have pore size distribution similar to those shown in Figs. 5-1, 9-1, and 10-I.† To properly test the network model, it is therefore necessary to compare the dynamic properties of real porous media with those of networks of tubes in which the tube radii are distributed accord- ing to the distributions in the aforementioned figures. A network in which the tube sizes are distributed cannot be treated in the same manner as a network of single size tubes. In the single size tube network, the relation between tube geometry, flow resistance, and tube volume did not have to be known if the assumption was made that whatever relation did exist was the same for all tubes in a particular network. In a network of distributed tube size, the relation between tube geometry and flow resistance must be known or assumed in order to replace the tubes by equivalent electrical resistors. Relation of Tube Geometry to Flow Resistance The interchangeability of tubes and electrical resistors in the network model is possible because of the analogy between Poiseuille's law and Ohm's law. Poiseuille's law for a cylinder is q= pr4/8µl ?P.........(1) where q is the volumetric rate of flow, µ is the viscosity, ?P is the pressure gradient, r is the tube radius, and 1 is the tube length. Ohm's law is I=1/R?E..........(2) where I is the flux and is equivalent to the volumetric rate in Poiseuille's law, R is the resistance, and ?E is the voltage gradient. Both laws give the flux as a function of the potential drop and the resistance of the medium. By analogy then, 1/R=pr4/8µl ?P...........(3)