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Darcy law
- 1. 1 Darcy’s Law In 1856, Darcy investigated the flow of water through sand filters for water purification purposes. His experimental apparatus is shown below: Sand Pack A L Water Water Water h1 h2 h1-h2 Datum Input manometer Output manometer Water in Water out at a rate q Flow direction 1 2 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. Single-phase flow (only water) 2. Homogeneous porous medium (sand) 3. Vertical flow. 4. Non-reactive fluid (water) 5. 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.:
- 2. 2 q h h ∝ − 1 2 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 L ∝ 1 Thus we write: ( ) q C A L h h = − 1 2 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 d ∝ = grain size 2 2 It is inversely proportional to tortuosity; i.e.: C ∝ = 1 tortuosity 1 τ
- 3. 3 Flow Direction 2 1 A L Datum P2, D2 P1, D1 and inversely proportional to the specific surface; i.e.: C Ss ∝ = 1 specific surface 1 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 A L h h = − γ µ 1 2 Since: q vA = Thus we can write: ( ) v q A K h h L = = − γ µ 1 2 Now, let us consider the more realistic flow; i.e. the tilted flow for the same sand pack:
- 4. 4 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: h D p h D p 1 1 1 2 2 2 = − = − γ γ & Thus: ( ) ( ) h h D p D p D D p p 1 2 1 1 2 2 1 2 1 2 1 − = − − − = − − − γ γ γ which can be written in a difference form as: h h D p 1 2 1 − = − ∆ ∆ γ Substituting (2) into (1) and rearranging yields: v K D L p L K p L D L = − = − − γ µ γ µ γ ∆ ∆ ∆ ∆ 1 The differential form of Darcy’s equation for single-phase flow is written as follows: v K p L D L = − − µ ∂ ∂ γ ∂ ∂ For multi-phase flow, we write Darcy’s equation as follows: v K k p L D L r = − − µ ∂ ∂ γ ∂ ∂ More conveniently, the compact differential form of Darcy’s equation is written as follows: ( ) r v K k p D r = − ∇ − ∇ µ γ Where K is the absolute permeability tensor which must be determined experimentally. It is written as follows:
- 5. 5 K k k k k k k k k k xx xy xz yx yy yz zx zy zz = Substitute (5) into (4), we obtain: v v v k k k k k k k k k k p x D x p y D y p z D z x y z r xx xy xz yx yy yz zx zy zz = − − − − µ ∂ ∂ γ ∂ ∂ ∂ ∂ γ ∂ ∂ ∂ ∂ γ ∂ ∂ Solve for velocity components yields: v k k p x D x k p y D y k p z D z x r xx xy xz = − − + − + − µ ∂ ∂ γ ∂ ∂ ∂ ∂ γ ∂ ∂ ∂ ∂ γ ∂ ∂ v k k p x D x k p y D y k p z D z y r yx yy yz = − − + − + − µ ∂ ∂ γ ∂ ∂ ∂ ∂ γ ∂ ∂ ∂ ∂ γ ∂ ∂ v k k p x D x k p y D y k p z D z z r zx zy zz = − − + − + − µ ∂ ∂ γ ∂ ∂ ∂ ∂ γ ∂ ∂ ∂ ∂ γ ∂ ∂ In most practical problems, it is necessary to assume that K is a diagonal tensor; i.e.: K k k k xx yy zz = 0 0 0 0 0 0 and thus velocity components of equation (6) become: v k k p x D x x xx r = − − µ ∂ ∂ γ ∂ ∂ v k k p y D y y yy r = − − µ ∂ ∂ γ ∂ ∂ v k k p z D z z zz r = − − µ ∂ ∂ γ ∂ ∂
- 6. 6 Unit Analysis Since: q KA p L K q A L p = ⇒ = µ µ ∆ ∆ ∆ ∆ When q in cc/sec, µ in cp, )L in cm, A in cm2 , and )p in atm, then K is in the units of Darcy, where: ( )( )( ) ( ) ( )( ) ( ) ( ) ( ) ( ) 1 Darcy 1 cm 1 cp 1 cm s 1 cm 1 atm 1 cm 1 cp 1 atm 1 s 3 2 2 = = Since 1 cp = 1 100 poise = 1 100 dyne s cm2 and 1 atm = 1,013,250 dyne cm2 thus we have: ( ) ( ) 1 Darcy md = 1 cm 1 dyne s cm 100 cm 1,013,250 dyne s cm 2 2 2 2 = = 1000 1 101325 000 , , or 1 10 101325 986923 10 9 14 md = cm cm 2 2 − − = . . x Since: q KA p L = µ ∆ ∆ Thus: ( )( )( ) ( )( ) ( )( )( ) ( ) bbl day md ft psi cp ft md ft psi cp x bbl lb ft lb day ft x = = = − = ≈ − − 2 9 2 2 2 9 10 101325 1 254 12 1 56146 144 1 47 880 1 86 400 1127106 663 10 0001127107 . . . , , . .