Groundwater 1
Groundwater flows slowly through the voids between grains or the cracks in
solid rock. Much of our knowledge depends on field and laboratory
observations. Here, for example, is an experiment to measure head loss in an
aquifer.
Darcy’s Law
• Henri Darcy established empirically that the energy
lost ∆h in water flowing through a permeable
formation is proportional to the length of the
sediment column ∆L.
• The constant of proportionality K is called the
hydraulic conductivity . The Darcy Velocity VD:
VD = – K (∆h/∆L)
and since Q = VD A ( where A = total area)
Q = – KA (dh/dL)
Darcy’s Experiment
1. Velocities small, V ~ 0, so:
Piezometers before and
after sand. Pipe is full, so
flow rate is constant
2. Head difference doesn’t change with inclination of the sand filter
3. Again, Darcy related reduced flow rate to head loss and length of
column through a constant of proportionality K,
V = Q/A = -K dh / dL
3
Darcy’s Data (One set of 10 experiments)
L
diam.
n
A
Experiment
No.
1
2
3
4
5
6
7
8
9
10
0.58 m
0.35 m
0.38
0.096211 m2
Duration
(min)
25
20
15
18
17
17
11
15
13
10
Q
L/min
3.6
7.65
12
14.28
15.2
21.8
23.41
24.5
27.8
29.4
dp
(m)
1.11
2.36
4
4.9
5.02
7.63
8.13
8.58
9.86
10.89
Ratio
V/dp
3.25
3.24
3
2.91
3.03
2.86
2.88
2.85
2.82
2.7
Calc
K
(m/min)
0.019552
0.019541
0.018085
0.017568
0.018253
0.017224
0.017359
0.017214
0.016997
0.016275
K
cm/s
3.26E-02
3.26E-02
3.01E-02
2.93E-02
3.04E-02
2.87E-02
2.89E-02
2.87E-02
2.83E-02
2.71E-02
1.Darcy collected data with
his apparatus, then …
4
Plotted it. Note the strong coefficient of determination R2 .
5
Darcy’s allows an estimate of:
• The velocity or flow rate moving within the aquifer
• The average time of travel from the head of the aquifer to a
point located downstream
• Very important for prediction of contaminant plume arrival
Confined Aquifer
Darcy & Seepage Velocity
• Darcy velocity VD is a fictitious velocity
since it assumes that flow occurs across
the entire cross-section of the sediment
sample. Flow actually takes place only
through interconnected pore channels
(voids), at the seepage velocity VS.
Av voids
A = total area
Darcy & Seepage Velocities
• From the Continuity Eqn. Q = constant
• “Pipe running full” means “Inputs = Outputs”
•
Q = A VD = AV Vs
– Where:
Q = flow rate
A = total cross-sectional area of
material
AV = area of voids
Vs = seepage velocity
VD = Darcy velocity
Since A > AV , and Q = constant, Vs > VD
Pinch hose, reduce area, water goes faster
Darcy & Seepage Velocity: Porosity
• Q = A VD = AV Vs ,therefore VS = VD ( A/AV)
• Multiplying both sides by the length of the
medium (L) divided by itself, L / L = 1
• VS = VD ( AL / AVL ) = VD ( VolT / VolV )
we get
volumes
• Where:
VolT = total volume
VolV = void volume
• By definition, Volv / VolT = n, the sediment
porosity
So the actual velocity:
VS = VD / n
Turbulence and Reynolds Number
• The path a water molecule takes is called a streamline.
In laminar flow, streamlines do not cross, and the
viscous forces due to hydrogen bonds are important.
• In turbulent flow acceleration and large scale motion
away from a smooth path is important (this is the
familiar inertial force F = ma) and streamlines cross.
• We could take the ratio of inertial to viscous forces.
When this number is “large,” inertial forces are more
important, and flows are turbulent.
• This ratio is known as the Reynolds number Re:
Viscosity
• Viscosity is a fluid’s resistance to flow.
• Dynamic viscosity m, units Pa·s = N·s/m2, or
kg/(m·s) is determined experimentally. If a fluid
with a viscosity of one Pa·s is placed between two
plates, and one plate is pushed sideways with a
shear stress of one Pascal, it moves a distance
equal to the thickness of the layer between the
plates in one second.
• Kinematic viscosity n, is the dynamic viscosity
divided by the density. The SI unit of ν is m2/s.
Reynolds: Inertial/Viscous forces
Recall the ratio of Kinetic/Potential Energy
(KE/PE) is the Froude Number Fr
Fr = V / sqrt( g L) we saw last time.
Limitations of Darcy’s Equation
1. For Reynold’s Number, Re, > 10 or where the flow
is turbulent, as in the immediate vicinity of pumped
wells.
Darcy’s Law works
water
for2.
1.0Where
< Re < 10
flows through extremely fine-grained
Q = – KA (dh/dL)
q = – Ky (dh/dL)
materials (colloidal clay)
Example 1
Q = KA (dh/dL)
The hydraulic conductivity
K is a velocity, length / time
and n = Vol voids/ Vol total
• A confined aquifer has a source of recharge.
• K for the aquifer is 50 m/day, and porosity n is 0.2.
• The piezometric head in two wells 1000 m apart is
55 m and 50 m respectively, from a common datum.
• The average thickness of the aquifer is 30 m, and
the average width of the aquifer is 5 km = 5000m.
A piezometer is a small-diameter observation well used to measure the piezometric head of
groundwater in aquifers.
Piezometric head is measured as a water surface elevation, expressed in units of length.
Example 1 Compute:
Q = KA (dh/dL)
• a) the rate of flow through the aquifer
• (b) the average time of travel from the head of the
aquifer to a point 4 km downstream
Example 1 Solution
Q = KA (dh/dL)
• Cross-Sectional area= 30(5000) = 1.5 x 105 m2
• Hydraulic gradient dh/dL= (55-50)/1000 = 5 x 10-3
• Find Rate of Flow for K = 50 m/day
5
Q = (50 m/day) (1.5 x 10 m2) ( 5 x 10-3)
Q = 37,500 m3/day
• Darcy Velocity: V = Q/A
5
3
• = (37,500m /day) / (1.5 x 10 m2) = 0.25m/day
And
• Seepage Velocity:
Vs = VD/n = (0.25) / (0.2) =
1.25 m/day (about 4.1 ft/day)
• Time to travel 4 km downstream:
T = (4000m) / (1.25m/day) =
3200 days or 8.77 years
• This example shows that water moves
very slowly underground.
Lesson: Groundwater moves very slowly
Example 2
• A channel runs almost parallel to a river, and they are 2000 ft apart.
• The water level in the river is at an elevation of 120 ft . The channel
is at an elevation of 110ft.
• A pervious formation averaging 30 ft thick and with hydraulic
conductivity K of 0.25 ft/hr joins them.
• Determine the flow rate Q of seepage from the river to the channel.
Confining Layer
30 ft
Aquifer
Example 2: Confined Aquifer
• Consider 1-ft (i.e. unit) lengths of the river and
small channel.
Q = KA [(h1 – h2) / L]
• Where:
A = (30 x 1) = 30 ft2
K = (0.25 ft/hr) (24 hr/day) = 6 ft/day
• Therefore,
Q = [6ft/day (30ft2) (120 – 110ft)] / 2000ft
Q = 0.9 ft3/day for each 1-foot length