09 Contaminants Transport Mechanisms - Advection

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    Geoenvironmental EngineeringKaniraj Shenbaga

    University Malaysia Sarawak

    http://www.unimas.my/
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    By the end of this course, students should be

    able to apply the processes affecting migration

    and fate of contaminants and evaluate the rate

    of contaminant transport in ground water

    using analytical solutions.

    Course Learning Objective 2

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    Migration and Fate ofContaminant

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    1. Transportprocesses, which cause migration of

    the contaminants from one point to another.

    2. Chemical and mass transferprocesses, which

    affect the ultimate chemical form and

    concentration of the contaminants.3. Biologicalprocesses which cause the

    biodegradation of the contaminants.

    Processes Affecting Contaminants

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    Transport of Contaminants(Nonreactive)

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    1. Advection

    Also called as convectionor advective transport

    2. Diffusion

    Also called as molecular diffusion

    3. DispersionAlso called as mechanical dispersionor

    hydrodynamic dispersion

    Transport Processes

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    Transport of ContaminantsAdvection

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    Advection is the movement ofcontaminant by groundwater flow.

    The contaminant is transported at the

    same velocity of groundwater flow.

    Advection

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    Apparent velocity of flow is given by Darcys Law. It is the

    velocity of flow through the soil cross-section including the

    area of solids also.

    =

    = apparent velocity; = coefficient of permeability

    =

    = hydraulic gradient

    Apparent Velocity of Flow, v

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    Darcys Law for Apparent Velocity

    http://www.env.gov.bc.ca/wsd/plan_protect_sustain/groundwater/gwbc/C02_origin.html

    Cross-sectional area Awith both solids and voids

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    Seepage velocity of flow is the velocity of flow through

    the soil pores excluding the area of solids.

    =

    = seepage velocity =apparent velocity

    =porosity of soil

    Seepage Velocity, vs

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    Under steady-state one-dimensional flow conditions,

    contaminant mass flux due to advection Fvis:

    Fv = vsC

    vs= seepage velocity (length per unit time)

    C= contaminant concentration (mass per unit volume)

    Fv is expressed in mass per unit area per unit time.

    Mass of contaminant transported through areaAin

    time t= FvxAx t

    Contaminant Mass Flux, Fv

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    Chloride dissolved in water at concentration of 650 ppm is

    being transported vertically down from a land fill through asandy silt vadose zone into the groundwater. Coefficient of

    permeability = 2.7 x 108cm/s, soil porosity = 0.16, and

    hydraulic gradient of flow = 0.08. Determine the flux of

    chloride into the ground water due to advection and theamount of chloride (in grams) transported into

    groundwater per day through a flow area of 320 m2.

    Exercise 7

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    Exercise 7 Answer

    GWL

    GLLandfill

    Sandy siltk= 2.7 x 10-8cm/sn= 0.16i = 0.08

    Area = 320 m2

    Flow

    Calculate v

    Calculate vs

    Calculate Fv

    Calculate FvxAx t

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    Exercise 7 Answer

    C= 650 ppm = 650 mg/l

    vs= - [2.7 x 108 x ( 0.08)]/0.16 = 1.35 x 108cm/s

    Fv= vsx C= 8.75 x 109mg/cm2/s

    Chloride transported through 320 m2/day = FvxAx t= 2.42 g

    GWL

    GL Landfill

    Sandy siltk= 2.7 x 10-8cm/sn= 0.16i = - 0.08

    Area = 320 m2

    Flow

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    Advective transport (variation of concentration with time

    and distance) under uniform one-directional flow conditionsis expressed as:

    =

    = concentration

    = time

    = direction of flow

    Advective Transport Equation

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    Groundwater flow will transport a

    contaminant by advection.

    In time t, a contaminant of concentration Co

    will be transported by distancex.

    x= vst

    Advective Transport Equation

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    A non-reactive chemical spill contaminates the water in

    well A. Wells B and C are located respectively at 1 km

    and 1.7 km in opposite directions from A. The RLs of

    water level in wells A, B and C are 151.7 m, 153.6 m and

    148.4 m, respectively. The aquifer is silty sand. The

    coefficient of permeability and porosity are 3.1x104 m/s

    and 0.21, respectively. How many years will it take for

    the contaminant plume to reach wells B and C?

    Exercise 8

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    Exercise 8 Answer

    GL

    GWL RL 153.6 m

    Well B

    WellAWell C

    RL 151.7 mRL 148.4 m

    1 km 1.7 km

    Sandy siltk= 3.1 x 10-4m/sn= 0.21

    Calculate v

    Calculate vs Calculate time for travel = distance/vs

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    i= (148.4 157.7)/1700 = - 0.00194

    vs= - [3.1 x 10-4x (-0.00194)]/0.21 = 2.865 x 10-6m/s

    Distance of travel by the contaminant from wellAto

    well C= 1,700 m

    Time for contaminant plume to reach well C

    = 1700/(2.865 x 10-6) seconds = 6,866 days = 18.81 years

    The contaminant will not reach well Bas it is

    upstream of wellA.

    Exercise 8 Answer

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