# Darcy's Law

advertisement ```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
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 = Km/rg
where m is the dynamic
viscosity [kg/m.sec]
and K has units m/sec
Water Production Well
Casing, screen,
centrifugal pump
Outer casing &gt;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 2pr . height.
Grundfos submersible pump
Radial Flow- Confined Aquifer
• If the aquifer is confined, the cylinder has height h = b,
[m.m.(m/sec).m/m
and Darcy’s Equation for flow
Area xK x dh/dr
can be integrated for a solution for r and h:
]
To solve we integrate (get the total area), i.e. we add up
many thin cylinders
Separate the variables r and h
sum terms in r = constants x sum terms in h
d ln r = 1/r dr
so:
Solve the above
for Q
d h = dh
Radial Flow- Confined Aquifer
Solve 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 Aquifer
Solve 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
begins middle of page 540 4th edition
rc = radius of casing
y0 = vertical difference between
water level inside well and water
level outside at t = 0
yt = vertical difference between
water level inside well and water
table outside (drawdown) at time t
Re = 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, and
t = time
Example from Figure 8-23b
t (sec)
yt(m)
1
2
3
4
6
9
13
19
20
40
0.24
0.19
0.16
0.13
0.07
0.03
0.013
0.005
0.002
0.001
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.
First we estimate the 1/t ln(y0/yt)
0.3
0.2
4/10ths of a log cycle, read as a
proportion of the length of one log
cycle, NOT on the log scale
second log cycle
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 log cycles
to natural log (ln) cycles,
a conversion factor of 2.3
is used. Thus,
10
first log cycle
1/t ln(y0/yt) = 2.3 x 2.4/24
= 0.23.
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 labeled
Also, the meaning of “one cycle”
“half a cycle” etc.
One cycle
Half a
cycle
The Solution
Using 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,
rc = 5 cm
Le = 1 m
-3
K = 8.27 x 10 cm/s Gravel pack = 2.5 cm thick
So R = rw = 2.5 + 5 = 7.5cm
y0 = 0.28 m.
Re is 10 cm
1/t ln(y0/yt) = 0.23 from the previous
slides.
More Examples
• As usual we will do examples and similar
homework problems.
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