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Cross Section of Unconfined and Confined Aquifers
Unconfined Aquifer Systems
• Unconfined aquifer: an aquifer where the water table exists under atmospheric pressure as defined by levels in shallow wells
• Water table: the level to which water will rise in a well drilled into the saturated zone
Confined Aquifer Systems
• Confined aquifer: an aquifer that is overlain by a relatively impermeable unit such that the aquifer is under pressure and the water level rises above the confined unit
• Potentiometric surface: in a confined aquifer, the hydrostatic pressure level of water in the aquifer, defined by the water level that occurs in a lined penetrating well
Special Aquifer Systems
• Leaky confined aquifer: represents a stratum that allows water to flow from above through a leaky confining zone into the underlying aquifer
• Perched aquifer: occurs when an unconfined water zone sits on top of a clay lens, separated from the main aquifer below
Ground Water Hydraulics
• Hydraulic conductivity, K, is an indication of an aquifer’s ability to transmit water
– Typical values:
10-2 to 10-3 cm/sec for Sands
10-4 to 10-5 cm/sec for Silts
10-7 to 10-9 cm/sec for Clays
Ground Water Hydraulics
Transmissivity (T) of Confined Aquifer
-The product of K and the saturated thickness of the aquifer T = Kb
- Expressed in m2/day or ft2/day
- Major parameter of concern
- Measured thru a number of tests - pump, slug, tracer
Ground Water HydraulicsIntrinsic permeability (k)
Property of the medium only, independent of fluid properties
Can be related to K by:K = k(g/µ)where: µ = dynamic
viscosity = fluid densityg = gravitational
constant
Storage CoefficientRelates to the water-yielding capacity of an aquifer
S = Vol/ (AsH)
– It is defined as the volume of water that an aquifer releases from or takes into storage per unit surface area per unit change in piezometric head - used extensively in pump tests.• For confined aquifers, S values range between
0.00005 to 0.005• For unconfined aquifers, S values range
between 0.07 and 0.25, roughly equal to the specific yield
Regional Aquifer Flows are Affected by Pump Centers
Streamlines and Equipotential lines
Derivation of the Dupuit Equation - Unconfined Flow
Dupuit Assumptions
For unconfined ground water flow Dupuit developed a theory that allows for a simple solution based off the following assumptions:
1) The water table or free surface is only
slightly inclined
2) Streamlines may be considered horizontal
and equipotential lines, vertical
3) Slopes of the free surface and hydraulic
gradient are equal
Derivation of the Dupuit Equation
Darcy’s law gives one-dimensional flow per unit width as:
q = -Kh dh/dx
At steady state, the rate of change of q with distance is zero, or
d/dx(-Kh dh/dx) = 0
OR (-K/2) d2h2/dx2 = 0
Which implies that,
d2h2/dx2 = 0
Dupuit Equation
Integration of d2h2/dx2 = 0 yieldsh2 = ax + b
Where a and b are constants. Setting the boundary
condition h = ho at x = 0, we can solve for b
b = ho2
Differentiation of h2 = ax + b allows us to solve for a,a = 2h dh/dx
And from Darcy’s law,hdh/dx = -q/K
Dupuit Equation
So, by substitution
h2 = h02 – 2qx/K
Setting h = hL2 = h0
2 – 2qL/KRearrangement gives
q = K/2L (h02- hL
2) Dupuit Equation
Then the general equation for the shape of the parabola is
h2 = h02 – x/L(h0
2- hL2) Dupuit Parabola
However, this example does not consider recharge to the aquifer.
Cross Section of Flow
q
Adding Recharge W - Causes a Mound to Form
Divide
Dupuit Example
Example:
2 rivers 1000 m apart
K is 0.5 m/day
average rainfall is 15 cm/yr
evaporation is 10 cm/yr
water elevation in river 1 is 20 m
water elevation in river 2 is 18 m
Determine the daily discharge per meter width into each
River.
ExampleDupuit equation with recharge becomes
h2 = h02 + (hL
2 - h02) + W(x - L/2)
If W = 0, this equation will reduce to the parabolicEquation found in the previous example, and
q = K/2L (h02- hL
2) + W(x-L/2)Given:
L = 1000 m K = 0.5 m/day
h0 = 20 m
hL= 28 m
W = 5 cm/yr = 1.369 x 10-4 m/day
Example
For discharge into River 1, set x = 0 m
q = K/2L (h02- hL
2) + W(0-L/2)
= [(0.5 m/day)/(2)(1000 m)] (202 m2 – 18 m2 ) +
(1.369 x 10-4 m/day)(-1000 m / 2)
q = – 0.05 m2 /dayThe negative sign indicates that flow is in the opposite direction
From the x direction. Therefore,
q = 0.05 m2 /day into river 1
Example
For discharge into River 2, set x = L = 1000 m:
q = K/2L (h02- hL
2) + W(L-L/2)
= [(0.5 m/day)/(2)(1000 m)] (202 m2 – 18 m2 ) +
(1.369 x 10-4 m/day)(1000 m –(1000 m / 2))
q = 0.087 m2/day into River 2
By setting q = 0 at the divide and solving for xd, the
water divide is located 361.2 m from the edge of
River 1 and is 20.9 m high
Flow Nets - Graphical Flow Tool
Q = KmH / n
n = # head dropsm= # streamtubesK = hyd condH = total head drop
Flow Net in Isotropic Soil
Portion of a flow net is shown below
Stream tube
Curvilinear Squares
Flow Net Theory1. Streamlines and Equip. lines are .2. Streamlines are parallel to no flow
boundaries.3. Grids are curvilinear squares, where
diagonals cross at right angles.4. Each stream tube carries the same
flow.
Seepage Flow under a Dam