Groundwater 1

1. Groundwater 1 Groundwater flows slowly through the voids between grains or the cracks in solid rock. Much of our knowledge depends on field and laboratory observations. Here, for example, is an experiment to measure head loss in an aquifer.

2. Darcy’s Law • Henri Darcy established empirically that the energy lost ∆h in water flowing through a permeable formation is proportional to the length of the sediment column ∆L. • The constant of proportionality K is called the hydraulic conductivity . The Darcy Velocity VD: VD = – K (∆h/∆L) and since Q = VD A ( where A = total area) Q = – KA (dh/dL)

3. Darcy’s Experiment 1. Velocities small, V ~ 0, so: Piezometers before and after sand. Pipe is full, so flow rate is constant 2. Head difference doesn’t change with inclination of the sand filter 3. Again, Darcy related reduced flow rate to head loss and length of column through a constant of proportionality K, V = Q/A = -K dh / dL

4. Darcy’s Data (One set of 10 experiments) 1.Darcy collected data with his apparatus, then …

5. Plotted it. Note the strong coefficient of determination R2 .

6. Darcy’s allows an estimate of: • The velocity or flow rate moving within the aquifer • The average time of travel from the head of the aquifer to a point located downstream • Very important for prediction of contaminant plume arrival Confined Aquifer

7. Darcy & Seepage Velocity • Darcy velocity VD is a fictitious velocity since it assumes that flow occurs across the entire cross-section of the sediment sample. Flow actually takes place only through interconnected pore channels (voids), at the seepage velocity VS. Av voids A = total area

8. Darcy & Seepage Velocities • From the Continuity Eqn. Q = constant • “Pipe running full” means “Inputs = Outputs” • Q = A VD = AV Vs • Where: Q = flow rate A = total cross-sectional area of        material AV = area of voids Vs = seepage velocity VD = Darcy velocity Since A > AV , and Q = constant, Vs > VD Pinch hose, reduce area,water goes faster

9. Darcy & Seepage Velocity: Porosity • Q = A VD = AV Vs ,therefore VS = VD ( A/AV) • Multiplying both sides by the length of the medium (L) divided by itself, L / L = 1 • VS = VD ( AL / AVL ) = VD ( VolT / VolV ) we get volumes • Where: VolT = total volume VolV = void volume • By definition, Volv / VolT = n, the sediment porosity So the actual velocity: VS = VD / n

10. Turbulence and Reynolds Number • The path a water molecule takes is called a streamline. In laminar flow, streamlines do not cross, and the viscous forces due to hydrogen bonds are important. • In turbulent flow acceleration and large scale motion away from a smooth path is important (this is the familiar inertial force F = ma) and streamlines cross. • We could take the ratio of inertial to viscous forces. When this number is “large,” inertial forces are more important, and flows are turbulent. • This ratio is known as the Reynolds number Re:

11. Viscosity • Viscosity is a fluid’s resistance to flow. • Dynamic viscosity m, units Pa·s = N·s/m2, or kg/(m·s)is determined experimentally. If a fluid with a viscosity of one Pa·s is placed between two plates, and one plate is pushed sideways with a shear stress of one Pascal, it moves a distance equal to the thickness of the layer between the plates in one second. • Kinematic viscosity n, is the dynamic viscosity divided by the density. The SI unit of ν is m2/s.

12. Reynolds: Inertial/Viscous forces Recall the ratio of Kinetic/Potential Energy (KE/PE) is the Froude Number Fr Fr = V / sqrt( g L) we saw last time.

13. Limitations of Darcy’s Equation 1. For Reynold’s Number, Re, > 10 or where the flow is turbulent, as in the immediate vicinity of pumped wells. Darcy’s Law works for 1.0 < Re < 10 2. Where water flows through extremely fine-grained materials (colloidal clay) Q = – KA (dh/dL) q = – Ky (dh/dL)

14. Example 1 Q = KA (dh/dL) The hydraulic conductivity K is a velocity, length / time and n = Vol voids/ Vol total • A confined aquifer has a source of recharge. • K for the aquifer is 50 m/day, and porosity 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 the aquifer is 5 km = 5000m. A piezometer is a small-diameter observation well used to measure the piezometric head of groundwater in aquifers. Piezometric head is measured as a water surface elevation, expressed in units of length.

15. Example 1 Compute: Q = KA (dh/dL) • 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

16. Example 1 Solution Q = KA (dh/dL) • Cross-Sectional area= 30(5000) = 1.5 x 105 m2 • Hydraulic gradient dh/dL= (55-50)/1000 = 5 x 10-3 • Find Rate of Flow for K = 50 m/day Q= (50 m/day) (1.5 x 105 m2) ( 5 x 10-3) Q = 37,500 m3/day • Darcy Velocity: V = Q/A • = (37,500m3/day) / (1.5 x 105 m2)= 0.25m/day

17. And • Seepage Velocity: Vs = VD/n = (0.25) / (0.2) = 1.25 m/day (about 4.1 ft/day) • Time to travel 4 km downstream: T = (4000m) / (1.25m/day) = 3200 days or 8.77 years • This example shows that water moves very slowly underground. Lesson: Groundwater moves very slowly

18. 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 . The channel is at an elevation of 110ft. • A pervious formation averaging 30 ft thick and with hydraulic conductivity K of 0.25 ft/hr joins them. • Determine the flow rate Q of seepage from the river to the channel. Confining Layer Aquifer 30 ft

19. Example 2: Confined Aquifer • Consider 1-ft (i.e. unit) lengths of the river and small channel. Q = KA [(h1 – h2) / L] • Where: A = (30 x 1) = 30 ft2 K = (0.25 ft/hr) (24 hr/day) = 6 ft/day • Therefore, Q = [6ft/day (30ft2) (120 – 110ft)] / 2000ft Q = 0.9 ft3/day for each 1-foot length