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Darcy law, governing equation.ppt

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Darcy law Darcy law

Generalization of Darcy’s column h/L = hydraulic gradient Q is proportional to h/L q Generalization of Darcy’s column h/L = hydraulic gradient Q is proportional to h/L q = Q/A Figure from Hornberger et al. (1998)

Darcy’s data for two different sands Figure from Hornberger et al. (1998) Darcy’s data for two different sands Figure from Hornberger et al. (1998)

Darcy’s Law • Henri Darcy established empirically that the flux of water through a Darcy’s Law • Henri Darcy established empirically that the flux of water through a permeable formation is proportional to the distance between top and bottom of the soil column. • The constant of proportionality is called the hydraulic conductivity (K). • V = Q/A, V – ∆h, and V 1/∆L

Darcy’s Law V = – K (∆h/∆L) and since Q = VA (A = Darcy’s Law V = – K (∆h/∆L) and since Q = VA (A = total area) Q = – KA (dh/d. L)

Hydraulic Conductivity • K represents a measure of the ability for flow through porous Hydraulic Conductivity • K represents a measure of the ability for flow through porous media: • Gravels - 0. 1 to 1 cm/sec • Sands - 10 -2 to 10 -3 cm/sec • Silts - 10 -4 to 10 -5 cm/sec • Clays - 10 -7 to 10 -9 cm/sec

Range in hydraulic conductivity, K 13 orders of magnitude Figure from Hornberger et al. Range in hydraulic conductivity, K 13 orders of magnitude Figure from Hornberger et al. (1998)

Conditions • Darcy’s Law holds for: 1. Saturated flow and unsaturated flow 2. Steady-state Conditions • Darcy’s Law holds for: 1. Saturated flow and unsaturated flow 2. Steady-state and transient flow 3. Flow in aquifers and aquitards 4. Flow in homogeneous and heterogeneous systems 5. Flow in isotropic or anisotropic media 6. Flow in rocks and granular media

Darcy Velocity • V is the specific discharge ( Darcy velocity). • (–) indicates Darcy Velocity • V is the specific discharge ( Darcy velocity). • (–) indicates that V occurs in the direction of the decreasing head. • Specific discharge has units of velocity. • The specific discharge is a macroscopic concept, and is easily measured. It should be noted that Darcy’s velocity is different ….

Darcy Velocity • . . . from the microscopic velocities associated with the actual Darcy Velocity • . . . from the microscopic velocities associated with the actual paths if individual particles of water as they wind their way through the grains of sand. • The microscopic velocities are real, but are probably impossible to measure.

Darcy & Seepage Velocity • Darcy velocity is a fictitious velocity since it assumes Darcy & Seepage Velocity • Darcy velocity is a fictitious velocity since it assumes that flow occurs across the entire cross-section of the soil sample. Flow actually takes place only through interconnected pore channels. Av voids A = total area

Darcy & Seepage Velocity • From the Continuity Eqn: • Q = A v. Darcy & Seepage Velocity • From the Continuity Eqn: • Q = A v. D = AV Vs – Where: Q = flow rate A = total cross-sectional area of material AV = area of voids Vs = seepage velocity VD = Darcy velocity

Darcy & Seepage Velocity • Therefore: VS = VD ( A/AV) • Multiplying both Darcy & Seepage Velocity • Therefore: VS = VD ( A/AV) • Multiplying both sides by the length of the medium (L) VS = VD ( AL / AVL ) = VD ( VT / VV ) • Where: VT = total volume VV = void volume • By Definition, Vv / VT = n, the soil porosity • Thus VS = V D / n

q = Q/A q is a vector z q qz 2 z x x q = Q/A q is a vector z q qz 2 z x x qx 1 In general: Kz < Kx, Ky

q = - K grad h q = - K grad h

Vector Form of Darcy’s Law q = - K grad h q = specific Vector Form of Darcy’s Law q = - K grad h q = specific discharge (L/T) K = hydraulic conductivity (L/T) grad h = hydraulic gradient (L/L) h = head (L)

q = - K grad h q is a vector with 3 components h q = - K grad h q is a vector with 3 components h is a scalar K is a tensor with 9 components (three of which are Kx, Ky, Kz)

Scalar 1 component Magnitude Head, concentration, temperature Vector 3 components Magnitude and Specific discharge, Scalar 1 component Magnitude Head, concentration, temperature Vector 3 components Magnitude and Specific discharge, (& direction velocity), mass flux, heat flux Tensor 9 components Magnitude, Hydraulic conductivity, direction and Dispersion coefficient, magnitude thermal conductivity changing with direction

Darcy’s law q = - K grad h q equipotential line grad h Isotropic Darcy’s law q = - K grad h q equipotential line grad h Isotropic Kx = Ky = Kz = K q grad h Anisotropic Kx, Ky, Kz

Equations of Groundwater Flow • Description of ground water flow is based on: Darcy’s Equations of Groundwater Flow • Description of ground water flow is based on: Darcy’s Law Continuity Equation - describes conservation of fluid mass during flow through a porous medium; results in a partial differential equation of flow. • Laplace’s Eqn - most important in math

Derivation of 3 -D GW Flow Equation from Darcy’s Law z y Mass In Derivation of 3 -D GW Flow Equation from Darcy’s Law z y Mass In - Mass Out = Change in Storage

Derivation of 3 -D GW Flow Equation from Darcy’s Law Replace Vx, Vy, and Derivation of 3 -D GW Flow Equation from Darcy’s Law Replace Vx, Vy, and Vz with Darcy using Kx, Ky, and Kz Divide out constant , and assume Kx= Ky= Kz = K

Transient Saturated Flow A change in h will produce change in and n, replaced Transient Saturated Flow A change in h will produce change in and n, replaced with specific storage Ss = g( + n ). Note, is the compressibility of aquifer and B is comp of water, therefore,

Transient Saturated Flow Simplifying by assuming K = constant in all dimensions And assuming Transient Saturated Flow Simplifying by assuming K = constant in all dimensions And assuming that S = Ssb, and that T = Kb yields

Steady State Flow to Well Simplifying by assuming K = constant in all dimensions Steady State Flow to Well Simplifying by assuming K = constant in all dimensions and assuming that Transmissivity T = Kb and Q = flow rate to well at point (x, y) yields

Example of Darcy’s Law • A confined aquifer has a source of recharge. • Example of Darcy’s Law • A confined aquifer has a source of recharge. • K for the aquifer is 50 m/day, and n is 0. 2. • The piezometric head in two wells 1000 m apart is 55 m and 50 m respectively, from a common datum. • The average thickness of the aquifer is 30 m, and the average width of aquifer is 5 km.

Compute: • a) the rate of flow through the aquifer • (b) the average Compute: • a) the rate of flow through the aquifer • (b) the average time of travel from the head of the aquifer to a point 4 km downstream • *assume no dispersion or diffusion

The solution • Cross-Sectional area= 30(5)(1000) = 15 x 104 m 2 • Hydraulic The solution • Cross-Sectional area= 30(5)(1000) = 15 x 104 m 2 • Hydraulic gradient = (55 -50)/1000 = 5 x 10 -3 • Rate of Flow for K = 50 m/day Q = (50 m/day) (75 x 101 m 2) = 37, 500 m 3/day • Darcy Velocity: V = Q/A = (37, 500 m 3/day) / (15 x 104 m 2) = 0. 25 m/day

And • Seepage Velocity: Vs = V/n = (0. 25) / (0. 2) = And • Seepage Velocity: Vs = V/n = (0. 25) / (0. 2) = 1. 25 m/day (about 4. 1 ft/day) • Time to travel 4 km downstream: T = 4(1000 m) / (1. 25 m/day) = 3200 days or 8. 77 years • This example shows that water moves very slowly underground.

Limitations of the Darcian Approach 1. For Reynold’s Number, Re, > 10 or where Limitations of the Darcian Approach 1. For Reynold’s Number, Re, > 10 or where the flow is turbulent, as in the immediate vicinity of pumped wells. 2. Where water flows through extremely fine-grained materials (colloidal clay)

Darcy’s Law: Example 2 • A channel runs almost parallel to a river, and Darcy’s Law: Example 2 • A channel runs almost parallel to a river, and they are 2000 ft apart. • The water level in the river is at an elevation of 120 ft and 110 ft in the channel. • A pervious formation averaging 30 ft thick and with K of 0. 25 ft/hr joins them. • Determine the rate of seepage or flow from the river to the channel.

Confined Aquifer Confining Layer 30 ft Aquifer Confined Aquifer Confining Layer 30 ft Aquifer

Example 2 • Consider a 1 -ft length of river (and channel). Q = Example 2 • Consider a 1 -ft length of river (and channel). Q = KA [(h 1 – h 2) / L] • Where: A = (30 x 1) = 30 ft 2 K = (0. 25 ft/hr) (24 hr/day) = 6 ft/day • Therefore, Q = [6 (30) (120 – 110)] / 2000 = 0. 9 ft 3/day/ft length = 0. 9 ft 2/day

Permeameters Constant Head Falling Head Permeameters Constant Head Falling Head

Constant head Permeameter • Apply Darcy’s Law to find K: V/t = Q = Constant head Permeameter • Apply Darcy’s Law to find K: V/t = Q = KA(h/L) or: K = (VL) / (Ath) • Where: V = volume flowing in time t A = cross-sectional area of the sample L = length of sample h = constant head • t = time of flow