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8/10/2019 09 Contaminants Transport Mechanisms - Advection
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Geoenvironmental EngineeringKaniraj Shenbaga
University Malaysia Sarawak
http://www.unimas.my/8/10/2019 09 Contaminants Transport Mechanisms - Advection
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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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