flujo subterraneo.ppt

Flujo en medio poroso

Flujo subterráneo

REYES ROQUE, Esteban REYES ROQUE, Esteban PedroPedro


Flujo Flujo subterránesubterráne

oo


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


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


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


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


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


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


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

Flujo en medio poroso Darcy’s Law

V= - K dh/dlQ = - KA dh/dl


Aplicación de Ley de Darcy

Q/A=K(h/l) Q = Caudal

A = Area sección transv.

K = conductiv. hidráulicah = diferencia de cargal = Distancia


’ = /

= 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.


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


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


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.


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


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


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.


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)


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.


Confined Aquifer

Confining Layer


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


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


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


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.


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.


three-dimensional. homogeneous (K everywhere same). isotropic. transient.

KK ((22

hh

xx22

22hh

yy22

22hh

zz22 )) SSsshh tt


One-dimensional Homogeneous Steady-state since

KK xxdd 22 hhdxdx 22

00

hh

tt== 00


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


)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


Flujo radial hacia pozo en acuífero Flujo radial hacia pozo en acuífero librelibre


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


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


r1

r2

Pozo observación 1Pozo observación 2

Pozo bombeoQ

Q

s1s2

Acuífero confinado

Hidráulica de pozos


Flujo radial: cono de

depresiónBombeo

Descenso de tabla de agua cerca del pozo: cono

de depresión


What happens when this well is heavily pumped?


Solución de Theis para

acuíferos confinados

)u(WTQ

s

4

u

u

)u( ...!.

u!.

u!.

uuuln,du

ue

W443322

57720432


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


Steady State One Dimensional Flow

Confined Aquiferso Simple application of Darcy’s Law

Unconfined Aquiferso Dupuit Assumptions


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


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


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


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)


Steady Flow in an Unconfined

Aquifer: Dupuit Formula

h1 h2L

x = 0 x = L

x


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.


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


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


W = 0.0002 m/day

17m12m

4525 m


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.


Ground water movement


An unconfined aquifer


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


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



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



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.


Schematic

q


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.


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

flujo subterraneo.ppt

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