3.2 Darcy s Law In 1856, Darcy investigated the flow of water through sand filters for water purification. His experimental apparatus is shown in Figure 3.11. By empirical observation Figure 3.11 Schematic of Darcy s experiment on flow of water through sand. Darcy noticed that fluid flow was directly proportional to the hydraulic gradient, resulting in the following equation, h h q KA 2 1 (3.3) l where q represents the volumetric flow rate of water downward through the cylindrical sand pack of cross-sectional area A and height l, h 1 and h 2 are the hydraulic head above the standard datum of the water in the manometer located at the input and output ports respectively, and K is a constant of proportionality found to be characteristic of the rock media. It was observed that for any angle of the sandpack, a given constant flow rate would result in the same h; i.e., Eq. (3.3) is independent of the direction of flow. Fluid movement is due to the difference in potential energy. The two components of the potential energy are fluid pressure and elevation or gravity. By defining a relationship 3.12
between hydraulic head and pressure, it is possible to determine pressure at any point in the flow path. The total potential function per unit mass can be written as, p gz gh (3.4) where g is the acceleration of gravity, p is pressure and is the density of the fluid and z is the elevation of a given point in the system. Substituting Eq. (3.4) into Eq. (3.3) and writing in differential form, we obtain, q K g d A dl p gz (3.5) The parameter, K is known as the hydraulic conductivity. A sequence of experiments maintained at constant fluid potential gradient investigated the effect of different properties on the flow rate. The resulting expression for the hydraulic conductivity is, 2 cd g K (3.6) where d is grain diameter, is the fluid viscosity, and c is a constant of proportionality. Substituting into Eq. (3.5) we get, kadp dz q g dl dl (3.7) where k is the permeability of the porous media. Equation (3.7) is known as Darcy s Law. It is applicable for steady state, laminar flow of an incompressible fluid, for a homogeneous and isotropic porous media. The negative sign in the previous equation is a result of the pressure and distance both measured positive in the same direction (see Fig. 3.12). Therefore, the gradient within the bracket must be negative to move the flow in the same direction as the direction from high to lower potential. For example, take the simplified case of horizontal flow (z = 0), then since p 1 > p 2 the gradient must be dp p 2 p 1. L 3.13
q P 1 P 2 0 L Figure 3.12 Schematic of sign convention for potential gradient in horizontal flow For flow at any angle,, we can substitute sin for the elevation gradient; i.e., dz sin. The sign convention is referenced such that positive is upward flow as shown in Figure 3.13. Flow upward (+) sin 90 = 1 dz Flow downward sin -90 = -1 Figure 3.13 Schematic of sign convention for any flow angle Darcy s Law (Eq. 3.7), written for any angle of flow becomes, kadp q g sin dl (3.8) The dimensions of permeability can be determined by substituting for the appropriate cgs units found in table 3.1. 2 2 cc k( cm ) dyne / cm sec poise cm 3 2 L k( L ) m 2 k ( / ) 2 2 L T m LT L T 3.14
Units cgs Darcy Field q flow rate cc/s cc/s Bpd A cross-sectional area cm 2 cm 2 ft 2 L length cm cm Ft fluid viscosity poise centipoise Cp P pressure dyne/cm 2 atm Psia density gm/cc gm/cc lbm/ft 3 g gravity cm/s 2 cm/s 2 ft/s 2 k - permeability cm 2 Darcy Md Table 3.1Various units used in petroleum engineering The unit for permeability in the cgs system is 0.987 x 10-8 cm 2. From a practical standpoint this measure is too small; therefore we define 1 darcy = 0.987 x 10-12 m 2 (m) 2. That is, a permeability of 1.0 Darcy is defined such that a single-phase fluid of 1.0 cp flows at a rate of 1.0 cc/sec/cm 2 under a pressure of 1 atm/cm. In Darcy units, Eq. (3.7) becomes, ka dp g sin q (3.9) dl 6 1.013x10 Example 3.2 Consider fresh water with a viscosity of 1.0 cp and density of 1.0 gm/cc, injected at a rate of 10 cc/sec through a sandpack with dimensions illustrated below. After steady state is achieved the pressure readings are recorded. Determine the permeability for this sandpack? Note: Barometric pressure = 14.7 psi. P in =10psig d = 2 cm P out =0psig L = 5 cm Solution Since the core is horizontal the elevation term is zero; therefore Eq. (3.8) is reduced to, 3.15
ka dp q 2 0 14.7 10 14.7 k (1) 14.7 14.7 10 (1)(5) k 10.8darcies Conversion to field units results in the following expression for Darcy s Law, where is the specific gravity of the fluid. 3 kadp q 1.127x10 0.4335 sin dl (3.10) 3.16