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Darcy’s Law (1856)

Darcy’s Law (1856). Henry Darcy (1803 – 1858) First Attempt to describes the flow of a fluid through a porous medium. Units. Hydraulic Conductivity (K)……[L/T] Specific Storage (S) ……. [1/L] Hydraulic Head Elevation (h) ……[L] Pressure head (P/ ρ g) …….[L]

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Darcy’s Law (1856)

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  1. Darcy’s Law (1856) Henry Darcy (1803 – 1858) First Attempt to describes the flow of a fluid through a porous medium

  2. Units • Hydraulic Conductivity (K)……[L/T] • Specific Storage (S) ……. [1/L] • Hydraulic Head • Elevation (h) ……[L] • Pressure head (P/ρg) …….[L] • Velocity head (v2/2g) ……[L] negligible

  3. Darcy’s Law (1856) H h1 Water flow h2 q H1 H2 L Area, A “Water flows from high to low total head” z1 Total head, H = h + z z2 Hydraulic gradient, i = H/L

  4. DARCY’S LAW q = Darcy flux or hydraulic flux (L/T) K = Saturated hydraulic conductivity (L/T) dH = difference in total head (L) dl = distance increment (L) dH/dl = I = hydraulic gradient (unitless)

  5. DARCY’S LAW • Darcy’s law is valid for laminar flow only • Reynold’s number, Re < 10 • Darcy’s law becomes nonlinear with turbulent flow (q=-Kin) Darcy’ Law

  6. DARCY’S LAW • Flow velocities can be high near the well screen in pumping wells or in fractures causing a higher Re. • Although the Darcy’s law considers flow across the total cross-sectional area, the actual flow occurs in the void space. Darcy’ Law

  7. DARCY’S LAW Q = total flow (L3/T) A = total cross-sectional area (L2) Q = qA solids water

  8. DARCY’S LAW Actual water velocity is higher than q and called the pore water velocity, v. v = q/n n = porosity solids water

  9. Example Consider measurements taken at two monitoring wells in an unconfined aquifer. The wells are located 200 m apart. The average K of the aquifer is 3.5 m/day. The observed values are 23.1 m and 24.2 m. Find the total flow across the aquifer and the transmissivity.

  10. water table h2=24.2 K=3.5 m/day h1=23.1 Flow Unconfined x =200 x

  11. Example q = -K dh/dx Using consistent units of meters and days, h1 = 23.1, h2=24.2, K=3.5 q = -K . (h2-h1)/(x2-x1) = -3.5 . (24.2-23.1)/(200-0) = - 0.01925 m/day

  12. Since q is negative, flow is opposite to the direction of x. This is correct because flow occurs from high potential to low potential. Q = qA = 0.01925 x 0.5(24.2+23.1)x1 = 0.455 m3/m-day T = KD D = 0.5(24.2+23.1) = 23.65 m T = 3.5x23.65 = 82.8 m2/day

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