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Darcy’s Law In 1856, Darcy investigated the flow of water through sand filters for water purification purposes. His experimental apparatus is shown below: Output manometer Input manometer Water h1-h2 Water in 1 Sand Pack Flow direction A h1 L h2 Datum 2 Water out at a rate q Water Water Where q is the volume flow rate of water downward through the cylindrical sand pack. The sand pack has a length L and a cross-sectional area A. h1 is the height above a datum of water in a manometer located at the input face. h2 is the height above a datum of water in a manometer located at the output face. The following assumptions are implicit in Darcy’s experiment: 1. 2. 3. 4. 5. Single-phase flow (only water) Homogeneous porous medium (sand) Vertical flow. Non-reactive fluid (water) Single geometry. From this experiment, Darcy concluded the following points: The volume flow rate is directly proportional to the difference of water level in the two manometers; i.e.: 1 q ∝ h1 − h2 The volume flow rate is directly proportional to the cross-sectional area of the sand pack; i.e.: q∝A The volume flow rate is inversely proportional to the length of the sand pack; i.e.: q∝ 1 L Thus we write: q=C A (h1 − h2 ) L Where: A is the cross-sectional area of the sand pack L is the length of the sand pack h1 is the height above a datum of water in a manometer located at the input face h2 is the height above a datum of water in a manometer located at the output face C is the proportionality constant which depends on the rock and fluid properties. For the fluid effect, C is directly proportional to the fluid specific weight; i.e. C ∝γ and inversely proportional to the fluid viscosity; i.e.: C∝ 1 µ Thus: C∝ γ µ For the rock effect, C is directly proportional to the square of grain size; i.e.: C ∝ (grain size) = d 2 2 It is inversely proportional to tortuosity; i.e.: C∝ 1 1 = tortuosity τ 2 and inversely proportional to the specific surface; i.e.: C∝ 1 1 = specific surface S s where Ss is given by: Ss = Interstitial surface area bulk volume Combine the above un-measurable rock properties into one property, call it permeability, and denote it by K, we get: q=K γ A (h − h ) µ L 1 2 Since: q = vA Thus we can write: v= q γ (h1 − h2 ) =K A µ L Now, let us consider the more realistic flow; i.e. the tilted flow for the same sand pack: Datum P2, D2 2 P1, D1 w Flo 1 io ect r i D L 3 n A Note that the fluid flows from point 1 to point 2, which means that the pressure at point 1 is higher than the pressure at point 2. Since: h1 = D1 − p1 h2 = D2 − & γ p2 γ Thus: p p 1 h1 − h2 = D1 − 1 − D2 − 2 = ( D1 − D2 ) − ( p1 − p2 ) γ γ γ which can be written in a difference form as: h1 − h2 = ∆D − 1 γ ∆p Substituting (2) into (1) and rearranging yields: v=K γ ∆D 1 ∆p ∆D K ∆p − =− −γ µ L γ L µL L The differential form of Darcy’s equation for single-phase flow is written as follows: v=− ∂D K ∂p −γ ∂L µ ∂L For multi-phase flow, we write Darcy’s equation as follows: k ∂p ∂D v = − K r − γ ∂L µ ∂L More conveniently, the compact differential form of Darcy’s equation is written as follows: k r v = − K r (∇p − γ∇D) µ Where K is the absolute permeability tensor which must be determined experimentally. It is written as follows: 4 k xx K = k yx k zx k xy k yy k zy k xz k yz k zz Substitute (5) into (4), we obtain: k xx v x kr v y = − k yx µ v k zx z k xy k yy k zy ∂p −γ k xz ∂x ∂p k yz − γ ∂y k zz ∂p −γ ∂z ∂D ∂x ∂D ∂y ∂D ∂z Solve for velocity components yields: k ∂p ∂p ∂D ∂D ∂D ∂p v x = − r k xx − γ + k xy − γ + k xz − γ ∂z ∂x ∂y ∂z ∂y µ ∂x k ∂p ∂p ∂D ∂D ∂D ∂p v y = − r k yx − γ + k yy − γ + k yz − γ ∂z ∂x ∂y ∂z ∂y µ ∂x k ∂p ∂p ∂D ∂D ∂D ∂p v z = − r k zx − γ + k zy − γ + k zz − γ ∂z ∂x ∂y ∂z µ ∂x ∂y In most practical problems, it is necessary to assume that K is a diagonal tensor; i.e.: k xx K=0 0 0 k yy 0 0 0 k zz and thus velocity components of equation (6) become: k ∂p ∂D v x = − k xx r − γ ∂x µ ∂x k ∂p ∂D v y = − k yy r − γ ∂y µ ∂y k ∂p ∂D v z = − k zz r − γ ∂z µ ∂z 5 Unit Analysis Since: KA ∆p qµ ∆L ⇒ K= A ∆p µ ∆L 2 When q in cc/sec, µ in cp, )L in cm, A in cm , and )p in atm, then K is in the units of Darcy, where: q= (1 cm )(1 cp)(1 cm) = (1 cm ) (1 cp) 1 Darcy = (s) (1 cm )(1 atm) (1 atm) (1 s) 3 2 2 Since 1 cp = 1 1 dyne s poise = 100 100 cm 2 and 1 atm = 1,013,250 dyne cm 2 thus we have: 1 cm 2 1 dyne s cm 2 1 1 Darcy = 1000 md = = cm 2 2 (100) cm (1,013,250) dyne s 101,325,000 or 1 md = 10 − 9 cm 2 = 986.923x10 −14 cm 2 . 101325 Since: q= KA ∆p µ ∆L Thus: bbl day (md )( ft 2 )( psi ) (md )( ft )( psi ) = = (cp)( ft ) (cp) 2 = 10 − 9 lb 1 1 bbl 144 2 . 2.54 x12 5.6146 ft 101325 1 1 lb − day 47,880 86,400 ft 2 ≈ 0.001127107 6 = 1127106.663x10 − 9