Water Flow in Saturated Soils Darcy’s Law Hillel, pp. 173 - 177 Water Flow in Saturated Soils Darcy’s Law P1 P2

Non Equilibrium and Flow Flow occurs from locations with high potential energy to locations of lower potential energy in pursuit of equilibrium state. The driving force for flow is called potential (energy) gradient, the difference in potentials between two points in a system separated by a certain distance. Potential Gradient i High potential energy Low potential energy y1 y2 L Dy1= y1- y2 i…… potential gradient y….. potential energy L…... distance between the locations [L]

Hydraulic Potential yh = yz + ym + yp In general, gradients can develop due to differences in: - Pressure - Position in a gravity field - Chemical concentration - Temperature - Position in an electrical field leading to spontaneous flow of mass or energy. We will focus on flow due to differences in hydraulic potential in this section (neglecting solute potential). yh = yz + ym + yp

Definition of Liquid Viscosity Before we discuss flow in soils it is advantageous to introduce some basic concepts related to flow in general. Newton’s Law of Viscosity Early concepts in fluid dynamics are based on perfect fluids that are assumed to be frictionless and incompressible. In a perfect fluid contacting layers can exhibit no tangential forces (shearing stresses) only normal forces (pressures). Perfect fluids do not exist. In the flow of real fluids adjacent layers do transmit tangential stresses (drag), and the existence of intermolecular attraction causes fluid molecules in contact with solid surfaces to adhere to it rather than to slip over it. The flow of a real fluid is associated with the property of viscosity.

Liquid viscosity The nature of viscosity can be visualized considering fluid motion between two parallel plates; one at rest, the other one moving at constant velocity. Under laminar flow conditions water molecules are moving in adjacent parallel layers. The layers transmit tangential stresses (drag) due to attraction between fluid molecules. Motion of fluid between parallel plates The existence of intermolecular attraction causes fluid molecules to adhere on the solid walls.

Newton’s law of viscosity The velocity distribution in the liquid is linear. Maintaining the relative motion of the plates at constant velocity requires the application of a constant tangential force to overcome the frictional resistance in the fluid. This resistance per unit area of the plate is proportional to the velocity of the upper plate and inversely proportional to the distance between the plates. The shearing stress t at any point is proportional to the velocity gradient. The viscosity h is the proportionality factor between t and the velocity gradient Newton’s Law of Viscosity t shearing stress (force F acting on an area A) [M L-1 t-2] dv/dy velocity gradient perpendicular to the stressed area (shear rate) [t-1] h viscosity coefficient of the liquid in [Pa s] [M L-1 t-1] Viscosity is the property of the fluid to resist the rate of shearing and can be visualized as an internal friction.

Fluid flow in cylindrical tubes Fluid flows through a cylindrical tube having a diameter of 2R and length L. We assume that the flow is laminar and caused by a pressure gradient DP=P2-P1. Pressure Force: Frictional Resistance Force: We equate the pressure and frictional resistance forces and solve for t

Flow through cylindrical tubes Now we can introduce Newton's law of viscosity. The resulting ODE can be solved by integration. Since we know that the velocity at y=R is equal to zero we can solve for the integration constant Substituting the integration constant back into our previous result yields the expression for the velocity profile as a function of distance from the tube axis

Poiseuille’s law for flow in cylindrical tubes We know that the velocity is maximum at the center of the tube where y=0, and can calculate vmax . To calculate the Discharge Rate (volume of water flowing through the tube per unit time) we have to integrate the velocity profile over the cross-sectional tube area. This can be done very simple by calculating the volume of a paraboloid of revolution. This relationship is known as Poiseuille’s law. It shows that the volume of flow is proportional to the pressure trop per unit distance and to the fourth power of the tube radius. If we divide this expression by the tube cross section we receive the average flow velocity as:

Example: laminar flow in tubes What is the average (laminar) flow velocity of water at 20oC through a 50m long tube having a diameter of d=0.1m under a pressure difference of 100 Pa ? Viscosity of water at 20 oC: h = 0.001 Pa s

Water Flow in Soils Images of porous media pore space reveal that pores do not resemble uniform and smooth circular tubes that form the basis for Poiseuille’s law. P1 P2 Flow in porous media is generally described by macroscopic or averaging terms that replace microscopic description of individual flow pathways. The first one able to quantitatively describe saturated flow through porous media was HENRY DARCY a French engineer.

Flow of Water in Saturated Soil (Darcy’s Law) Historical Background Henry Darcy, a French engineer, was commissioned by the city of Dijon to find a solution for cleaning the city's water supply that was contaminated by the waste of mustard industry. Darcy, in search of suitable filtering media, conducted experiments with sand-packed filters.

Flow of Water in Saturated Soil (Darcy’s Law) Historical Background Henry Darcy, a French engineer, was commissioned by the city of Dijon to find a solution for cleaning the city's water supply that was contaminated by the waste of mustard industry. Darcy, in search of suitable filtering media, conducted experiments with sand-packed filters The pioneering work of Darcy published in 1856, provided the fundamental law for fluid flow in porous media. Darcy’s Law Water flux density (flux) Jw Volume of water flowing through a unit cross section per unit time. Saturated hydraulic conductivity Ks Proportionality coefficient between water flux density and hydraulic gradient. JW water flux density [L/t] Q discharge rate [L3/t] V volume of water [L3] A cross-sectional area [L2] Ks saturated hydraulic conductivity [L/t] Dyh/Dz hydraulic gradient [L/L]

Flow of Water in Saturated Soil (Darcy’s Law) Darcy, in search of suitable filtering media, conducted experiments with sand-packed filters The pioneering work of Darcy published in 1856, provided the fundamental law for fluid flow in porous media. Darcy’s Law

Coordinates and conventions For the application of Darcy’s law it is convenient to introduce a sign convention for flux and heads when expressed in energy per unit weight [L]. Upward flux is given a positive sign The differences DH and Dz, should be taken at the same order (if taken DH=H1-H2 then Dz=z1-z2) The negative sign in Darcy’s law ensures the algebraic consistency of the equation. + - H1, z1 H2, z2 1 2

Potentials and Heads Potential Energy of Soil Water As previously mentioned the potential energy of soil water can be expressed in terms of chemical potential m (energy/mass), soil water potential y (energy/volume), or soil water head H (energy/ weight). g acceleration of gravity rw density of water For many hydrological applications it is advantageous to express potential as energy on weight basis (length). This results in a simple notation for expressing heads as H=h+z H the hydraulic head h pressure (positive) or matric (negative) head z gravitational head

Darcy’s Law - Vertical Flow Example A constant 20 mm of water is ponded on the surface of a 50 mm long saturated vertical sand column. What is the water flux from the bottom of the column if the saturated hydraulic conductivity is 50 mm/day? Solution (1) Define a convenient reference level and designate it as z=0. (2) Calculate the difference in hydraulic head across the soil length

Darcy’s results

Darcy’s Law - Vertical Flow (3) Calculate the hydraulic gradient i: with units of hydraulic head it is a dimensionless quantity (4) Calculate the flux. Flux is downward Note the energy loss in the soil!

Darcy’s Law - Horizontal Flow Example The sand column from case a is now placed horizontally with 90 mm of water ponded on the left side and 20 mm on the right side. Find: (1) the water flux density, and (2) the volume of water collected at the outlet during 12 hr if the cross-sectional area of the column was 1000 mm2. Solution (1) Set the reference level z=0 to coincide with the axes of the column (2) Mark the column inlet by x=0. Sign convention + -

Darcy’s Law - Horizontal Flow Solution –Continued (3) Calculate the difference in hydraulic head across the soil length (4) Calculate the hydraulic gradient i: (5) Calculate the flux. with units of hydraulic head it is a dimensionless quantity

Darcy’s Law - Horizontal Flow Solution –Continued (6) Calculate the cumulative volume of flow

Saturated Flow - Potential Diagram A constant water pressure of 20 kPa was maintained at the bottom of a 0.5 m vertical saturated soil column, and the water height at the column’s top was also kept constant at 20 mm. Given the soils saturated hydraulic conductivity Ks = 5 mm/hr, find: (1) The direction of flow; draw a system sketch and a potential diagram (2) The water flux density Jw (3) The height of ponded water on top of the column that causes a cessation of flow.

Saturated Flow - Potential Diagram First we convert the pressure at the bottom of the column from potential (kPa) to head (m): Then we assume the bottom of the column as reference level and calculate the head at the top and the bottom as:

Saturated Flow - Potential Diagram With known Ks we now calculate the flux density Jw as: Positive Jw means flow from bottom to top. The flow ceases when the hydraulic head at the top equals the head at the bottom:

Saturated Flow - Potential Diagram

Saturated Hydraulic Conductivity and Flow Through Layered Soils Hillel, pp. 185-190 & 193-195 Saturated Hydraulic Conductivity and Flow Through Layered Soils

Measurement of Saturated Hydraulic Conductivity Saturated hydraulic conductivity is an important medium property used in many model calculations for flow and transport in soils. Constant Head Method A constant pressure head (50 mm) is maintained on the top of a saturated soil column of known cross-sectional area (1000 mm2) and length (50 mm). The outflow on the bottom is collected over a certain period of time (5 hr) and the outflow volume is determined (25000 mm3). With known and determined quantities we can calculate the saturated hydraulic conductivity.

Measurement of Saturated Hydraulic Conductivity First we rearrange Darcy’s law to receive an explicit expression for saturated hydraulic conductivity Ksat: We calculate the flux density from our measurements and the column dimensions: Negative sign because of downward flow We apply Darcy’s law to calculate Ksat: Note that the negative sign of Darcy’s law ensures positive Ks (There is no physical meaning to a negative hydraulic conductivity)

Laboratory Setup

Measurement of Saturated Hydraulic Conductivity Falling Head Method An alternative method for measurement of saturated hydraulic conductivity does not require the maintenance of a of a constant head nor any outflow measurement is called Falling Head Method. Only initial and final depths of water expressed as pressure head in length units need to be recorded as a function of time. The rate of decrease of depth of ponding is equal to the flux density. Darcy’s Law

Measurement of Saturated Hydraulic Conductivity We can equate Darcy’s law with the expression for flux density derived from the rate of decrease of ponding and integrate the resulting expression to derive a relationship for Ksat:

Typical Values of Ks in Soils

Limitations of Darcy’s Law Reynolds Number d effective pore diameter v mean flow velocity r liquid density h liquid viscosity inertial forces viscous forces At high flow velocities inertial forces are no longer negligible – TURBULENT FLOW In very fine textured media (clays) adsorptive surface forces affect flow. The flux density at low gradients is smaller than predicted according to Darcy’s law

Saturated Steady Flow Through Layered Soil Under steady-state flow conditions the flux through both layers is equal. We solve for H2 and obtain two equations: Solving for flux density and introducing a effective saturated hydraulic conductivity yields:

Saturated Steady Flow Through Layered Soil This solution can be generalized to a soil profile having multiple layers. The effective hydraulic conductivity for a soil profile consisting of n layers, each with distinct hydraulic conductivity Ks and thickness L is obtained by setting Jw=Ks-eff (Hn-H1)/Li This solution is valid for flow perpendicular to the layering (harmonic mean). For flow parallel to the layering we use an arithmetic mean weighed by layer thickness

Effective Saturated Hydraulic Conductivity - Example A 1 m long glass tube having a radius of 1 mm was inserted into a 1 m long saturated cylindrical soil column with a diameter of 100 mm, and Ks of 0.01 mm/min. The water head at the top of the column and at the tube’s inlet was 0.25 m while the outlets where at atmospheric pressure. (1) What would be the total flux through the column tube system and what percentage is contributed by the tube? (2) What is the effective Ks of the column-tube system

Effective Saturated Hydraulic Conductivity - Example To apply Poiseuille’s law for average flow velocity within the tube we first have to convert hydraulic head h(m) to hydraulic potential yh (Pa) using the following relationship: Note that the hydraulic head is the sum of water head and tube length. With known hydraulic potential we now can solve Poiseuille’s law:

Effective Saturated Hydraulic Conductivity - Example The tubes flux Qtube is simply the product of average velocity and the tubes cross-sectional area: The flux of the soil column QC is given as:

Effective Saturated Hydraulic Conductivity - Example Then we calculate the total flux as the sum of column and tube flux: The contribution of the tube to the total flux is calculated as: The systems (tube & column) flux density JT is simply the ratio of the total flux QT and the total cross-sectional area AT. The system’s effective saturated conductivity is calculated as:

Effective Saturated Hydraulic Conductivity - Example Total flux density Effective saturated conductivity