Chapter 3 – Permeability 3.17 3.3 Porosity-Permeability Relationships To this point we have independently developed the fundamental properties of porosity and permeability. Environmental and depositional factors influencing porosity also influence permeability, and often there is a relationship between the two. The relationship varies with formation and rock type, and reflects the variety of pore geometry present. Typically, increased permeability is accompanied by increased porosity. Figure 3.14 illustrates the various trends for different rock types. For example, a permeability of 10 md can have a porosity range from 6 to 31%, depending on the rock type and its pore geometry. Constant permeability accompanied by increased porosity indicates the presence of more numerous but smaller pores. Figure 3.14 Permeability and porosity trends for various rock types [CoreLab,1983] For clastic rocks, the k- trend is influenced by the grain size as shown in Figure 3.15. Post depositional processes in sands including compaction and cementation will result in a shift to the left of the permeability-porosity trend line. Dolomitization of limestones tends to shift the permeability- porosity trend lines to the right.Chapter 3 – Permeability 3.18 Figure 3.15 Influence of grain size on the relationship between porosity and permeability [Tiab & Donaldson, 1996] The inter-relationship of rock properties has lead to numerous correlations to estimate permeability. Several of the more notable are as follows. Darcy’s Law (1856) uses empirical observations to obtain permeability as previously shown. Slichter in 1899 performed theoretical analysis of fluid flow through packed spheres of uniform size and introduced packing as a factor influencing permeability. skdk22.10 (3.11) where d is the sphere diameter (cm) and ks is a packing constant and function of porosity ( = 26% & hexagonal packing  ks = 84.4;  = 45% & cubic packing  ks = 13.7). One of the more well-known correlations was developed by Kozeny (1927) and later modified by Carmen(1939). It is based on fundamental flow principles by considering the porous media as a bundle of capillary tubes with the spaces between filled with a non-porous cementing material. Figure 3.16 is a schematic representation of the capillary tube model.Chapter 3 – Permeability 3.19 Figure 3.16 Capillary tube model We can define the porosity for the model shown in Figure 3.16 as, 2rnt (3.12) where r is the radius of the capillary tube and nt is the number of tubes per unit area (A). Also, the permeability can be derived from combining Poiseuille’s Equation for flow through a conduit with Darcy’s Law for flow in porous media, 84rnkt (3.13) Combining Eqs (3.12) and (3.13) leads to an expression relating k and . 82rk  (3.14) Example 3.3 For the cubic packing arrangement shown in the diagram below, determine the porosity and permeability. rChapter 3 – Permeability 3.20 Solution The number of tubes per unit area is: 2)4/(4 rtubes. Substituting into Eq. (3.12) results in an estimate for porosity. 42*241 rr The permeability from Eq. (3.14) is r2/32. To relate the capillary radius, r, to the porous media, we must first define Spv, the specific surface area per unit pore volume. In the case of cylindrical pore shape, Spv = 2/r. Similar expressions can be derived for Sbv, specific surface area per unit bulk volume and Sgv, specific surface area per unit grain volume. pvSgvSpvSbvS1* (3.15) Substitution of Spv for pore radius in Eq. (3.14) results in the Carmen-Kozeny equation for porous media. 2pvSzkk (3.16) The Kozeny constant, kz, is a shape factor to account for variability in cross-sectional shape and length. It can be separated into two components, kz = ko* , where is known as the tortuosity and describes the variability in length between the capillary tube, La, and unit length, L. 2LaL (3.17) ko is a shape factor to account for various cross-sectional shapes; e.g., ko = 2 for circular, = 1.78 for square. If we go back to our example of circular tubes and substitute ko = 2,  = 1 (tubes and unit length are equal and parallel), and Spv = 2/r into Eq. (3.16), we obtain Eq. (3.14).Chapter 3 – Permeability 3.21 Example 3.4 Measurements from capillary pressure, adsorption and statistical techniques are available to obtain the specific surface area per unit pore volume. A given sample measurement resulted in a reading of 182 mm-1. Tortuosity is measured from electrical resistivity and was determined to be 3.6. Porosity of the sample is 27.7% Assuming circular, capillary tubes to represent the porous media what is the permeability of the sample? Solution For a circular cross-sectional area of a capillary tube, the pore radius is 1011.018222 mmpvSr Substituting into the Carmen-Kozeny equation (Eq. 3.16) results in, 2610164.1)4)(6.3(22)011.0)(277(.mmxk Converting to darcies, Darcyscmxdarcymmcmmmxk 18.12810987.01*2101*2610164.1  If porosity and pore radius are substituted into Eq. (3.12) we can solve for n, 27292)011.0(277.2 mmrtn that is 729 tubes per unit area. Generalized Capillary tube model The above derivation illustrates the simple capillary tube model. We can modify this model to be more complex by considering the tube length to be greater than the unit length of the sample. The results are expressions for porosity and permeability which include tortuosity. 2rtn (3.18) 82rk  (3.19)Chapter 3 – Permeability 3.22 Furthermore, let’s introduce the concept of hydraulic radius, the ratio of the volume open to flow to the wetted surface area or, pvSgvShr111 (3.20) thus for a bundle of circular capillary tubes, Eq (3.19) reduces to, okhrk2 (3.21) where ko = 2 for circular tubes. As another example, consider spherical particles with diameter, dp. In this case, Sgv = 6/dp, substitute into Eq. (3.20) and then the result into Eq. (3.21) provides, 