1. Radial Flow at a well Removal of groundwater faster than it can flow back lowers the water table near the well. The GWT becomes a radially symmetrical funnel shape called the cone of depression. Everything depends on the ability of the aquifer to transmit water between pore spaces Last time we saw hydraulic conductivity K, a velocity, m/sec

2. Transmissivity and Permeability Transmissivity is a term applied to confined aquifers. It is the product of the hydraulic conductivity K and the saturated thickness b of the aquifer. T = K [m/day]. b [ m ] Permeability symbol k, “little k” has units [m 2 ] k = K  /  g where  is the dynamic viscosity [kg/m. sec] and K has units m/sec

3. Water Production Well Casing, screen, centrifugal pump Outer casing >12” above grade with cement grout, lower casing with grout seal, pump in screen, sand gravel pack, developed

4. Radial Flow The drawdown of the GWT during flow from a well varies with distance, the cone of depression. If we want to know the difference in height h of the GWT at different distances r away from the well, we can sum thin cylinders of water around the well. Cylinders have lateral area 2  r. height. Grundfos submersible pump

5. Radial Flow- Confined Aquifer If the aquifer is confined, the cylinder has height h = b, and Darcy’s Equation for flow can be integrated for a solution for r and h: [m. m. (m/sec ). m/m ] Separate the variables r and h Area xK x dh/dr sum terms in r = constants x sum terms in h d ln r = 1/r dr d h = dh so: To solve we integrate (get the total area), i.e. we add up many thin cylinders Solve the above for Q

6. Radial Flow- Confined Aquifer Solve for T If we solve for T=Kb we get: We will go through Example 8-4.

7. Radial Flow- Unconfined Aquifer If the aquifer is unconfined, the cylinder has height h, and Darcy’s Equation for flow can be integrated for a solution for r and h as well. The extra h gives a little different solution: Such expressions can be solved for K

8. Radial Flow- Unconfined Aquifer Solve for K If we solve for K we get: We will go through Example 8-5.

9. Slug Tests These use a single well for the determination of aquifer formation constants Rather than pumping the well for a period of time, a volume of pure water is added to the well and observations of drawdown are noted through time Slug tests are often preferred at hazardous waste sites, since no contaminated water has to be pumped out and disposed of.

10. Bouwer and Rice Slug Test r c = radius of casing y 0 = vertical difference between water level inside well and water level outside at t = 0 y t = vertical difference between water level inside well and water table outside (drawdown) at time t R e = effective radial distance over which y is dissipated; varies with well geometry r w = radial distance to undisturbed portion of aquifer from centerline (includes thickness of gravel pack) L e = length of screened, perforated, or otherwise open section of well, and t = time begins middle of page 540 4 th edition

11. Example from Figure 8-23b A screened, cased well penetrates a confined aquifer. The casing radius is r c = 5 cm and the screen is L e = 1 m long. A gravel pack 2.5 cm thick surrounds the well so r w = 7.5 cm. A slug of water is injected that raises the water level by y 0 = 0.28 m. The change in water level y t with time is as listed in the above table. TODO: Given that R e is 10 cm, calculate K for the aquifer. t (sec) y t (m) 1 0.24 2 0.19 3 0.16 4 0.13 6 0.07 9 0.03 13 0.013 19 0.005 20 0.002 40 0.001

12. First we estimate the 1/t ln(y 0 /y t ) Data for y vs. t are plotted on semi-log paper as shown. The straight line from y 0 = 0.28 m to y t = 0.001 m covers 2.4 log cycles. The time increment between the two points is 24 seconds. To convert the log 10 cycles to natural log (ln) cycles, a conversion factor of 2.3 is used. Thus, 1/t ln(y 0 /y t ) = 2.3 x 2.4/24 = 0.23. 0.2 0.3 first log cycle second log cycle 4/10ths of a log cycle, read as a proportion of the length of one log cycle, NOT on the log scale

13. Log 10 to ln Consider the number 10. Log 10 (10) = 1 because 10 1 = 10 ln (10) = 2.3 ln (10) / Log 10 (10) = 2.3/1

14. Reading log marks if not labeled Also, the meaning of “one cycle” “half a cycle” etc. One cycle Half a cycle

15. The Solution Using this value (0.23) for 1/t ln(y 0 /y t ) in the Bouwer and Rice equation gives: K = [(5 cm) 2 ln(10 cm/7.5 cm)/(2 x 100 cm)](0.23 sec -1 ) and, K = 8.27 x 10 -3 cm/s r c = 5 cm L e = 1 m Gravel pack = 2.5 cm thick So R = r w = 2.5 + 5 = 7.5cm y 0 = 0.28 m. R e is 10 cm 1/t ln(y 0 /y t ) = 0.23 from the previous slides.

16. More Examples As usual we will do examples and similar homework problems.