Flujo en medio poroso
Flujo subterráneo
REYES ROQUE, Esteban REYES ROQUE, Esteban PedroPedro
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Flujo Flujo subterránesubterráne
oo
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Water table
Potentiometric surface – elevation to which water will rise
Artesian well – water under pressure will rise up in well – flowing artesian well if water comes out at surfaceRecharge area
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Hydraulic head Elevation of the water column in each well
(for the same aquifer) Water pressure + elevation Water pressure comes from a higher
potential energy
P
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LH1-H2
RechargeArea – water in
DischargeArea –
Water out Using elevations of
water table or pot. surface
Groundwater flows to lower hydraulic head (not always to lower elevation)
Change in head (H1-H2)
over a given distance (L)
Just like stream gradient
Hydraulic gradient
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High P
Lower P
Hydraulic gradient
Flujo en medio poroso Groundwater
movement Movement depends on permeability – we will use the
term Hydraulic conductivity (K) The ease in which a porous medium can transmit
water Units are length/time Velocity = Hydraulic gradient I = (H1-H2)/L)
multiplied by the hydraulic conductivity (permeability)
V = K (H/L) Need to know the area of the aquifer in order to the
get the discharge
Q = KIA Darcy’s Law
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Darcy’s Law Darcy’s law provides an accurate description of
the flow of ground water in almost all hydrogeologic environments.
Henri Darcy established empirically that the flux of water through a permeable formation is proportional to the distance between top and bottom of the soil column. The constant of proportionality is called the hydraulic conductivity (K).
V = Q/A, v –∆h, and v 1/∆L
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Experimento de Experimento de DarcyDarcy
Ley de DarcyLey de Darcy
Observó que cantidad de agua que fluía a través de muestra de arena, por unidad de tiempo, era proporcional a diferencia de carga hidráulica entre entrada y salida de muestra e inversamente proporcional a longitud de muestra
En 1856 estableció Ley General de flujo de fluidos en medios porosos
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Q = Akdh/dlDonde:
Q = caudal que pasa a través de sección transversal A
A = área de sección transversal (m2); k = constante de proporcionalidad, equivalente a
permeabilidad o conductividad hidráulica (m/d). dh/dl = gradiente hidráulico (adim.) Q/A = representa la descarga por unidad de área de
sección transversal y se denomina velocidad aparente (v).
Por tanto, Ley de Darcy, llamada también ley de resistencia lineal, será:
v =- ki
Ley de DarcyLey de Darcy
Flujo en medio poroso Darcy experiment
Screen
²
² ²
²
h
h
1
2
L A
Screen
Q
Q
Sand
Darcy's experiment.
h = z + p
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V= - K dh/dlQ = - KA dh/dl
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Aplicación de Ley de Darcy
Q/A=K(h/l) Q = Caudal
A = Area sección transv.
K = conductiv. hidráulicah = diferencia de cargal = Distancia
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’ = /
= K (h/L)
Q = APoros
Sólidos
Flujo agua
AreaTotal
Velocidad aparente () y real
(’) Velocidad aparente () es calculada de descarga específica,
Q/ADescarga específica: razón del flujo y área de sección
transversal.Velocidad real (’): razón de velocidad aparente y porosidad.
Flujo en medio poroso Ejemplo
Calcular velocidad aparente (Darcy) del flujo subterráneo de un acuífero, con gradiente hidráulico de 0,002 y K = 6.9 x 10-4 m/s
m/s10 x 1.4
m/m)m/s)(0.00210x(LΔh
K
6-
4
96.v
’ = velocidad real = / = (velocidad aparente/porosidad)
’ = (1.4 x 10-6 m/s)/0.30 = 4.7 x 10-6 m/s
Tiempo = distancia/’ =d 148
s 86,400day
m/s 10 x 4.7m 60
6-
Flujo en medio poroso Conditions
In General, Darcy’s Law holds for:
1. Saturated flow and unsaturated flow
2. Steady-state and transient flow
3. Flow in aquifers and aquitards
4. Flow in homogeneous and heteogeneous systems
5. Flow in isotropic or anisotropic media
6. Flow in rocks and granular media
Flujo en medio poroso Darcy Velocity
V is the specific discharge (Darcy velocity). (–) indicates that V occurs in the direction of the
decreasing head. Specific discharge has units of velocity. The specific discharge is a macroscopic
concept, and is easily measured. It should be noted that Darcy’s velocity is different ….
..from the microscopic velocities associated with the actual paths if individual particles of water as they wind their way through the grains of sand.
The microscopic velocities are real, but are probably impossible to measure.
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Darcy & Seepage Velocity Darcy velocity is a fictitious velocity since it
assumes that flow occurs across the entire cross-section of the soil sample.
Flow actually takes place only through interconnected pore channels.
From the Continuity Eqn:Q = A vD = AV Vs
o Where:Q = flow rateA = cross-sectional area of materialAV = area of voids
Vs = seepage velocity
vD = Darcy velocity
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Darcy & Seepage Velocity
Therefore: VS = VD (A/AV)
Multiplying both sides by the length of the medium (L)
VS = VD (AL/AVL) = VD (VT/VV)
Where:VT = total volume
VV = void volume
By Definition, Vv/VT = n, the soil porosity
Thus VS = VD/n
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Example of Darcy’s Law
A confined aquifer has a source of recharge.
K for the aquifer is 50 m/day, and 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 is 5 km.
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Calculate:a) the rate of flow through the aquifer(b) the time of travel from the head of the
aquifer to a point 4 km downstream *assume no dispersion or diffusion
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The solution
Cross-Sectional area =
30(5)(1000) = 15 x 104 m2
Hydraulic gradient = (55-50)/1000 = 5 x 10-3
Rate of Flow for K = 50 m/day Q = (50 m/day) (75 x 101 m2) = 37,500 m3/day
Darcy Velocity:
V = Q/A = (37,500m3/day)/(15x 104
m2) = 0.25m/day
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And Seepage Velocity:
Vs = V/n = (0.25)/(0.2) =
1.25 m/day (about 4.1 ft/day)
Time to travel 4 km downstream: T = 4(1000m)/(1.25m/day) = 3200 days or 8.77 years
This example shows that water moves very slowly underground.
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Limitations of the Darcian Approach1. For Reynold’s Number, Re > 10 where the
flow is turbulent, as in the immediate vicinity of pumped wells.
2. Where water flows through extremely fine-grained materials (colloidal clay)
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Darcy’s Law: 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 and 110 ft in the channel.
A pervious formation averaging 30 ft thick and with K of 0.25 ft/hr joins them.
Determine the rate of seepage or flow from the river to the channel.
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Confined Aquifer
Confining Layer
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Example 2
Consider a 1-ft length of river (and 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 = [6 (30) (120 – 110)]/2000 = 0.9 ft3/day/ft length = 0.9 ft2/day
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V = av. pore velocity = 20 m3 1
3 day
= 6 mday
q = n V = 0.15 * 6 = 0.9 mday
= K hx
= 0.520
K
hx
= 0.520
K = 36 mday
A tracer travels 3 days and 8 hours between 2 wells that are 20 m apart. Water elevation difference between the wells is 0.5 m, porosity n = 0.15. Estimate q, V, and K
Problem
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Constant head Permeameter
Apply Darcy’s Law to find K:
V/t = Q = KA(h/L)or: K =
(VL)/(Ath) Where:
V = volume flowing in time tA = cross-sectional area of the sampleL = length of sampleh = constant head
t = time of flow
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Darcy’s Law
Darcy’s Law can be used to compute flow rate in almost any aquifer system where heads and areas are known from wells.
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Groundwater Flow
Equations
qqx inx in qqx outx outdzdz
dydydxdx
ContinuumContinuum
qqzz
qqyy
Mass (t)Mass (t)
Flujo en medio poroso Assumptio
ns-Examples
xx
((KKxx
hhxx
))
yy((KK
yy
hhyy
))
zz((KK
zz
hhzz
)) == SSss
hh tt
three dimensional, anisotropic, three dimensional, anisotropic, heterogeneous, transient.heterogeneous, transient.
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three-dimensional. homogeneous (K everywhere same). isotropic. transient.
KK ((22
hh
xx22
22hh
yy22
22hh
zz22 )) SSsshh tt
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One-dimensional Homogeneous Steady-state since
KK xxdd 22 hhdxdx 22
00
hh
tt== 00
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Ecuación de Ecuación de LAPLACELAPLACE
0
zv
yv
xv zyx
xh
kvx
02
2
2
2
2
2
zh
yh
xh
th
TS
zh
yh
xh
2
2
2
2
2
2
Flujo permanente
Flujo no permanente
Flujo en medio poroso
)r/rln()hh(mk
Q12
122 )r/rln()hh(k
Q12
2
1
2
2
Acuífero libreAcuífero confinado
Régimen permanente
Flujo radial hacia Flujo radial hacia pozospozos
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Flujo radial hacia pozo en acuífero Flujo radial hacia pozo en acuífero librelibre
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Flujo Radial en acuífero confinado
r0
QGround surface
rer 2
r1
h0
h2
h1
h
Imprevious strata
Confinedaquifer
Cone ofdepression of
Observation bores
Original piezometric surface
Pumpedbore
s
piezometricsurface
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Cuando se bombea un pozo, se extrae un diferencial de volumen de agua de su interior, provocando el descenso de nivel, formándose cono de depresión.
Descenso se denomina abatimiento (s). Tiempo requerido para estabilización de abatimiento
(régimen permanente) depende de: S (Coeficiente de almacenamiento), T (Transmisividad), CL (condiciones límite) y Q (caudal bombeo).
Monitoreo de desarrollo y forma final del cono en pozos de observación, alrededor del pozo de bombeo permite determinar propiedades del acuífero (T & S), a través de Pruebas de Bombeo.
Hidráulica de pozos
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r1
r2
Pozo observación 1Pozo observación 2
Pozo bombeoQ
Q
s1s2
Acuífero confinado
Hidráulica de pozos
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Flujo radial: cono de
depresiónBombeo
Descenso de tabla de agua cerca del pozo: cono
de depresión
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What happens when this well is heavily pumped?
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Solución de Theis para
acuíferos confinados
)u(WTQ
s
4
u
u
)u( ...!.
u!.
u!.
uuuln,du
ue
W443322
57720432
Flujo en medio poroso
Solución de Jacob para acuíferos
confinados
]uln,[TQ
s
577204
u
u
)u( uln,duue
W 57720
SrTt,
logTQ,
s 2
2521830
Si uSi u 0,010,01
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Steady State One Dimensional Flow
Confined Aquiferso Simple application of Darcy’s Law
Unconfined Aquiferso Dupuit Assumptions
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Darcy’s LawQ= -KiA buto A varieso i varies
Solutiono W.T.
• Height above impermeable layer represents ‘A’• Slope of W.T. represents ‘i’
Dupuit assumptions:o All flow lines are horizontal.o Hydraulic gradient is equal to the slope of the water
table.o Consequences:
• Vertical component of flow is ignored• Problem reduced to one-dimensional• Calculation of Q simplified (need long shallow flow system)
h1L
dzdx
xh2
Q?
Unconfined Aquiferz
Dupuit Solution to Unconfined Flow
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Dupuit Formulation –
Darcy’s Law
2h
1h
L
0K.z.dzdx Q
:gIntegratin
.zdx
dzK.- = q'or K.i.A - = Q
L2
)h(hK=Qor
2
)h(hK- = QL
22
21
21
22
z
h1L
dzdx
x
h2
Q?
z
h1L
dzdx
x
h2
Q?h1L
dzdx
x
h2
Q?
evident aregradient and b' average' where
L2
)h).(h1
h + 2
(h-KQ as Same
:Note
12
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Dupuit Equation with Infiltration
z
h1L
dzdx
x
h2
Q?
z
h1L
dzdx
x
h2
Q?h1L
dzdx
x
h2
Q?
R
x2
LR
L2
)h(hK= Q and
x)xL(K
R
L
x)h(hhh
22
21
2/122
212
1
Flujo en medio poroso Example:
A canal is constructed parallel to a river 1500 ft away. Both fully penetrate a sand aquifer with K = 1.2 ft/d. The area is subject to rainfall of 1.8 ft/y and evaporation of 1.3 ft/y. The elevation of the water in the river and canal are 31 ft and 27 ft respectively. o What is the Q between the canal/river w/o infiltration?o With infiltraton
• Where is the water divide?• What is the maximum water table elevation?• What is the daily is discharge into the river and canal?
Answers:o W/o infiltration Q =.093 ft3/day/ft o With infiltration
• 816 ft from the canal, 684 ft from the river• 38.3 ft• 1.14 ft3/day/ft into the canal, 0.96 ft3/day/ft into the rive
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Steady Flow in a
Confined Aquifer Apply
Darcy’s Law: Q = -K I A
Notes:o Horizontal flowo i is the change in the elevation of the
potentiometric surface (change in head) over distance
o A is the aquifer height (b) times the width into the page.
Equations:
o Q = - K (h2 - h1)(b/x) or Q = - K b dh/dx
o Or h2 = h1 - (Q x) / (K b)
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Steady Flow in an Unconfined
Aquifer: Dupuit Formula
h1 h2L
x = 0 x = L
x
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Dupuit Equation
q’ = -K h dh/dx
q’ = 1/2 K (h12 - h2
2)/L
Dupuit Assumptions1. Hydraulic gradient = slope of water table.2. For small hydraulic gradients, flow lines
are horizontal and equipotential lines are vertical.
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How much water has to be pumped
to keep excavation dry?
h1 h2
LIf If KK = 1.2 m/day, = 1.2 m/day, hh11 = 17 m, = 17 m, hh22 = 12 m and = 12 m and LL = = 400 m; how much water would you have to 400 m; how much water would you have to pump per meter of trench to keep the trench pump per meter of trench to keep the trench dry?).dry?).
Impermeable Base
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q’ = 1/2 K (hq’ = 1/2 K (h1122 - h - h22
22)/L)/L = 1/2 1.2 (17= 1/2 1.2 (1722 - 12 - 1222)/400)/400 = 0.2175 m= 0.2175 m33 / day / m of / day / m of trenchtrench = 217.5 L/day/m of trench.= 217.5 L/day/m of trench.
Need to know K.q’ = -K h dh/dxq’ = 1/2 K (h1
2 - h22)/L
q’ is the flow of water per unit width into the excavation.
Use Dupuit Formula
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W = 0.0002 m/day
17m12m
4525 m
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Equations of Ground Water Flow
Description of ground water flow is based on:
1. Darcy’s Law2. Continuity Equation - describes
conservation of fluid mass during flow through a
porous medium; results in a partial differential equation of flow.
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Ground water movement
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An unconfined aquifer
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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, vertical3) Slopes of the free surface and hydraulic gradient are equal
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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(-K/2) d2h2/dx2 = 0
Which implies that,d2h2/dx2 = 0
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Derivation of the Dupuit
EquationIntegration yields
h2 = ax + bWhere 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/dxFrom Darcy’s law: hdh/dx = -q/K
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Derivation of the Dupuit
EquationSo, by substitution
h2 = h02 – 2qx/K
Setting h = hL2 = h0
2 – 2qL/K
Rearrangement givesq = K/2L (h0
2- hL2) 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.
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Schematic
q
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Dupuit Equation with Recharge W
Flujo en medio poroso Dupuit
Example: 2 rivers 1000 m apartK is 0.5 m/dayaverage rainfall is 15 cm/yrevaporation is 10 cm/yrwater elevation in river 1 is 20 mwater elevation in river 2 is 18 m
Determine the daily discharge per meter width into each
River.
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Derivation of the Dupuit Equation
with Recharge WDupuit 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
Flujo en medio poroso Dupuit Equation
with Recharge W
For discharge into River 1, set x = 0 mq = K/2L (h0
2- hL2) + 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 directionFrom the x direction. Therefore,
q = 0.05 m2 /day into river 1
Flujo en medio poroso Dupuit Equation
with Recharge W
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