1. Water Supply and Treatment

3. Average Precipitation

4. Water Deficits

5. Hydrologic Cycle

6. Groundwater Problems

7. Groundwater Supplies

8. The amount of water stored in an aquifer is equal to the volume of void spaces between the soil grains, the porosity Porosity = volume of voids/total volume The amount of water that can be extracted from the aquifer is called the specific yield: Specific Yield = Volume of water that will drain freely from a soil Total volume of water in a soil

9. Q = A v v = superficial velocity Q = A v = a v’ a = area available for flow (pores) v’ = actual velocity of flow v’ = Av/a = A v L / a L = v / porosity When water flows through a soil it loses energy just like when it flows through a pipe (friction, etc.) dh/dL h = energy, measured as elevation of the water table in an unconfined aquifer or as the pressure in a confined aquifer L = horizontal distance in direction of flow In an unconfined aquifer, the drop in elevation of the water table with distance is dh/dL. The elevation of the water surface is the potential energy of the water. Water flows from higher elevation to lower elevation, losing energy.

10. Flow through a porous medium like a soil is related to the energy loss using the Darcy equation: Q = K A dh/dL K = coefficient of permeability, m/sec A = cross-sectional area, m 2

11. A soil sample is installed in a permeameter as in the figure below. The sample length is 0.1 m with a cross-sectional area of 0.05 m 2. The pressure on the upstream side is 2.5 m and on the downstream side is 0.5 m. A flow rate of 2.0 m 3 /day is observed. What is the coefficient of permeability, K? dh/dL = (2.5 – 0.5)/ 0.1 = 20 K = Q / (A dh/dL) = 2.0 / (0.05 x 20) = 2 m/day

12. Drawdown This is an unconfined aquifer being pumped. As water flows to the well the area through which it can travel becomes smaller and smaller, therefore the velocity gets larger and larger. This means more loss of energy (the gradient dh/dL increases) and the slope of the water surface increases, resulting in a “cone of depression”

13. Given the cylinder shown at the right, Darcy’s law becomes: Q = K (2 B rw) dh/dL Separate the variables and integrate: Q = [ B K(h 1 2 – h 2 2 )]/ ln(r 1 /r 2 )

14. Example A well is 0.2 m in diameter and pumps from an unconfined aquifer 30 m deep at an equilibrium rate of 1000 m 3 /day. Two observation wells are located at distances of 50 and 100 m and have drawdowns of 0.3 and 0.2 m, respectively. What is the coefficient of permeability and estimated drawdown at the well? K = [Q ln(r 1 /r 2 )]/[ B (h 1 2 – h 2 2 )] = [1000 ln(100/50)]/[3.14(29.8 2 – 29.7 2 ] = 37.1 m/day

15. The radius of the well is 0.2/2 = 0.1 m. Now: Q = [( B K(h 1 2 – h 2 2 )] / [ln(r 1 /r 2 )] = [3.14 x 37.1 x (29.7 2 – h 2 2 )]/[ln(50/0.1)] = 1000 m 3 /day h 2 = 28.8 m Since the aquifer is 30 m deep, the drawdown is: 30 – 28.8 = 1.2 m

16. Multiple Wells