Radial Flow at a wellRemoval 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

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 [m2] k = K/gwhere is the dynamic viscosity [kg/m.sec]and K has units m/sec

Water Production Well

Casing, screen, centrifugal pumpOuter casing >12” above grade with cement grout, lower casing with grout seal, pump in screen, sand gravel pack, developed

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 2r . height.

Grundfos submersible pump

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

Radial Flow- Confined AquiferSolve for T

• If we solve

for T=Kb we get:

We will go through Example 8-4.

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

Radial Flow- Unconfined AquiferSolve for K

• If we solve

for K we get:

We will go through Example 8-5.

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.

Bouwer and Rice Slug Test

rc = radius of casingy0 = vertical difference between water level inside well and water level outside at t = 0yt = vertical difference between water level inside well and water table outside (drawdown) at time tRe = effective radial distance over which y is dissipated; varies with well geometry

rw = radial distance to undisturbed portion of aquifer from centerline (includes thickness of gravel pack)Le = length of screened, perforated, or otherwise open section of well, andt = time

begins middle of page 540 4th edition

Example from Figure 8-23b

A screened, cased well penetrates a confined aquifer. The casing radius is rc = 5 cm and the screen is Le = 1 m long. A gravel pack

2.5 cm thick surrounds the well so rw = 7.5 cm. A slug of water is injected that raises the water level by y0 = 0.28 m.

The change in water level yt with time is as listed in the above table. TODO: Given that Re is 10 cm, calculate K for the aquifer.

t (sec) yt(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

First we estimate the 1/t ln(y0/yt) Data for y vs. t are plotted on semi-log paper as shown. The straight line from y0 = 0.28 m to yt = 0.001 m covers 2.4 log cycles. The time increment between the two points is 24 seconds. To convert the log10 cycles to natural log (ln) cycles, a conversion factor of 2.3 is used. Thus,

1/t ln(y0/yt) = 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

Log10 to ln

• Consider the number 10. • Log10 (10) = 1 because 101 = 10• ln (10) = 2.3• ln (10) / Log10 (10) = 2.3/1

Reading log marks if not labeledAlso, the meaning of “one cycle”

“half a cycle” etc.

One cycle

Half acycle

The SolutionUsing this value (0.23) for 1/t ln(y0/yt) 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/src = 5 cm Le = 1 m Gravel pack = 2.5 cm thickSo R = rw = 2.5 + 5 = 7.5cmy0 = 0.28 m. Re is 10 cm1/t ln(y0/yt) = 0.23 from the previous slides.

More Examples

• As usual we will do examples and similar homework problems.