11-1

11. Transport processes

Groundwater transports dissolved matter (solutes). Transport processes are important in;- migration of contaminants from the source area

(radioactive waste, landfill leachate, gasoline spill)- nutrients cycle in catchments - formation of hydrothermal ore depositsAgain, the conservation principle is our primary tool.

Jin Jout

∆MAlso, we will use the approach similar to what we used for Darcy’s law. We start from the microscopic scale. The key concept, this time, is “random motion”.

Storage of solutes

We leave out the cases in which solutes participate in chemical reactions (GLGY 503, 505). Solutes are either dissolved in water or attached on solid surfaces. The latter is called sorption.

Consider saturated sediments in a fixed box V. The mass of solutes in water (aqueous phase) in the box is;

Maq = VnC, where C is concentration (kg/m3). The amount of sorbed solutes per unit mass of solid is commonly expressed as KdC, where Kd is called a distribution coefficient.

11-2The mass of solid contained in the box is Vρb. The mass of sorbed solutes is given by;

Ms = VρbKdCThe total mass storage is;

M = Maq + Ms = VnC +VρbKdC = V(n + ρbKd)CNow, n and ρbKd must have the same unit. Note that n has a unit of [m3

-water/m3-total], and ρb has [kg-solid/m3

-total].∴ Kd has a unit of [m3

-water/kg-solid].

high Kd low Kd

The nature of attraction depends on many factors with respect to the properties of solid (area/mass ratio, organic coating, electrical charge, etc.), of solute species (charge, molecular structure, etc.), and of water (pH, temperature, ionic strength,etc.). This will be explored further in higher level courses.

Kd represents how solute molecules are distributed between aqueous and sorbed phases.

When the solid has a strong tendency to attract the solute, only a small amount of solid is needed to sorb a certain number of molecules. This results in a high value of Kd.

Example: Kd = 0.002 m3/kg ρb = 1600 kg/m3 n = 0.4C = 10 mg/L

You have a REV of 1 m3. How much (kg) of solute is dissolved in water? How much is on the solid surface?

11-3

(1) AdvectionA plane having an area A is oriented normal to the direction of specific discharge (q).

A

q

When solute molecules just move with groundwater, the process is called advection. Advective flux is give by;

Ja = qC

The unit of q really is [m3-water/s/m2

-total]. To have it represent velocity, we divide q by porosity n. The unit of porosity in saturated sediment is [m3

-water/m3-total]. Therefore, the quantity

q/n has a unit of [m-total/s], which is called the average linear velocity (v) of groundwater (see page 5-20).

Only the void spaces (holes in the picture) contribute to advection. Each hole has a different flow velocity, and v represents the average velocity of all holes.

Using v, we can write the advective flux as;

Ja = nvC [11-1]

The unit of q is [m/s]. So, does q represent the actual “speed”of solute migration?

Solute transport mechanisms

The transport flux (J) is defined as the rate of mass transport per unit area (kg s-1 m-2).

11-4

(2) Molecular diffusion

Solutes can spread without flow of water. This phenomenon is called molecular diffusion.

The rigorous theory of diffusion is due to Einstein (1905, Ann. d. Physik). He analyzed a then well-known phenomenon called Brownian motion.

Suppose all molecules are initially in the left half of a hallway. The cross sectional area of the hallway is A. Each second, they have equal probability of going left or right.

LIn the above, we have the net transfer of two molecules in ∆tseconds. This represents the transport flux.

Observe that the flux is proportional to the difference in the initial concentration between the left and the right half.

J ∝ ∆C

Also, if L is longer, each molecule needs to travel a longer distance and the flux will be smaller; J ∝ 1/L

t = 0 t = ∆t

11-5

Based on similar, but more elegant argument, Einstein gave a theoretical explanation to an empirical law known as Fick’s law of diffusion;

xCDJ mxm ∂

∂−=

Dm is called molecular diffusion coefficient in free water. The Dm of most chemical species is well defined and can be found in many handbooks. It is usually about 1 × 10-9 to 2 × 10-9

m2/s. Note that Dm is closely related to the random motion of molecules, and hence is dependent on temperature.

In water-saturated sediments, the measured value of diffusion coefficient is usually smaller than Dm, presumably due to tortuous paths that molecules have to travel through. We call this value effective diffusion coefficient and denote it by D*.

xCnDJ xm ∂

∂−= *

D* is related to Dm by D* = wDm, where w is called a tortuosity factor and takes a range of 0.01-0.5. The value of D* is dependent on sediment types and packing condition. It needs to be determined experimentally.

[11-2]

As we saw earlier, only the area occupied by water contributes to solute transport. Fick’s law in sediments is written as;

11-6(3) Mechanical dispersionRecall Poiseuille’s equation for a thin tube.

u

dxdhgruav µ

ρ8

2

−=

The average flow velocity is uav, but u varies within the tube. Let’s do a simple thought experiment. We release particles from one end at t = t0. At t = t1, we place a cursor at some xand count the number of particles in the cursor.

The # of particles represents concentration. Solutes tend to disperse as they travel along the pipe. This idea, called Taylordispersion, was established by another giant in modern physical science, G.I. Taylor (1953). Note that the centre of mass travels at a constant velocity uav.

cursor

uav

# of

par

ticle

x

∼ t0

t1

t2 t3

centre of mass

11-7

C1

C2

C3C4

q1

23 4

centre of mass

Cav

Now, we move one step up the scale and look at a bundle of tubes. This is close to what we see in laboratory column experiments, except that we are ignoring molecular diffusion between tubes. What is the velocity of the centre of mass?

sampleq1

q2

Cav

time

C1

C2

Move up further. Put two bundles together and sample water at the outlets. Concentration curve is further smoothed out. This is similar to a field experiment having a well interceptingtwo formations; medium sand and fine sand.

‘snap shot’

‘break through’

11-8Dispersion has “hierarchy”. In general, dispersion becomes greater as the scale of the problem becomes larger and more heterogeneity is involved. Experimental results suggest;

xCnDJ Lssx ∂∂

−=

where DsL (m2/s) is longitudinal mechanical dispersion coefficient.

Jsx is the mechanical dispersion flux in longitudinal direction, which is parallel to the direction of groundwater flow. In our analysis, the flow is along x-axis.

Experimental results suggest; DsL = αL|v | [11-4]where v is the average linear velocity and αL (m) is called longitudinal dispersivity. The value of αL is scale-dependent, and must be experimentally determined.

[11-3]

This graph shows the scale-dependence of αL for sand and gravel aquifers.

Original chart by Gelhar et al. (1992. Water Resources Research, 28: 1955-1974). The dashed line was fitted by Xu and Eckstein (1995. Ground Water, 33: 905-908),

11-9The dashed line in the graph is given by

αL = 0.83(logL)2.414 [11-5]where L is the scale parameter.

Example:The migration of contaminants a few hundred meters away from the source. Assume L ≅ 200 m.

αL =

zCnDJ

yCnDJ TVsszTHssy ∂

∂−=

∂∂

−=

When the direction of groundwater flow is reasonably constant with respect to time, we can align x-axis parallel to the average flow direction to study longitudinal dispersion.

Mechanical dispersion also occurs in the direction normal to flow. This is called transverse dispersion.

DsTH = αTH|v| DsTV = αTV|v| [11-6] where αTH is horizontal transverse dispersivity, and αTV is vertical transverse dispersivity. Mechanical dispersion flux in transverse (y, z) directions:

A few previous studies suggest αTH > αTV, but the process of transverse dispersion is not well understood.

11-10

They used to believe that transverse dispersion is caused by the spreading flow paths under the steady flow.

Does this diagram make sense?Fetter, Fig. 10.9

Gelhar et al. 1992. Water Resources Research, 28: 1955-1974.

It is commonly observed that αTH is in the order of 0.1αL or less. Compare this graph to the one on page 11-8.

Recent studies suggest that the transient change in flow direction is responsible for transverse dispersion.

11-11(4) Hydrodynamic dispersionMolecular diffusion (Jm) and mechanical dispersion (Js) are both expressed as gradient law.

xCnD

xCnD

xCnDJJJ LLsmxsxdx ∂

∂−=

∂∂

−∂∂

−=+= *

where DL = DsL + D* is called longitudinal hydrodynamic dispersion coefficient. Similar rules apply to y and z directions:

DTH = DsTH + D* DTV = DsTV + D*

For the example in page 11-9,

αL = 6.2 m

Now suppose q = 0.03 m/day and n = 0.3.

DsL =

If we assume the tortuosity w = 0.1.

D* =

In general, when we have significant groundwater flow, longitudinal hydrodynamic dispersion is dominated by mechanical dispersion. In contrast, molecular diffusion dominates the transport processes in low-flow environments such as in tight clays.

xCnDJ Lssx ∂∂

−=xCnDJ xm ∂

∂−= *

In experimental studies, it is impossible to separate the effects of Jm and Js. We lump them together and call it hydrodynamic dispersion (Jd).

[11-7]

11-12Conservation of mass

Let’s look at the mass conservation in a two-dimensional system. Note that groundwater flows along x-axis.

∆x∆y

Jx

x = x0 x = x1

Jy

bxCnDnvC

xCnDqCJJJ

L

Ldxaxx

∂∂

−=

∂∂

−=+=Flux:

yCnD

yCnD

JJJ

THTH

dyayy

∂∂

−=∂∂

−=

+=

0

Using the Taylor series approximation,

xxCnDnvC

xJx

xJJJ Lxxx

xxxxxxx ∆

∂∂

−∂∂

+=∆∂∂

+= === )(0@0@1@

For x-direction,

xxCv

xCDn

xxCnDnvC

xJJJJ

L

Lxxxxxxoutin

∆∂∂

−∂∂

=

∆∂∂

−∂∂

−=−=− ==

)(

)(

2

2

1@0@

For y-direction,

yyCnD

yyCnD

yJJJJ

TH

THyyyyyyoutin

∆∂∂

=

∆∂∂

−∂∂

−=−=−==

2

2

1@0@ )(

Storage: M = V(n + ρbKd)C CnKnVM db ∆+=∆ )1( ρ

11-13

∆x∆y

Jx

Jy

b

Conservation equation; (Jin - Jout)A = ∆M/∆t

Equating the expressions and taking ∆t → 0,

tCR

xCv

yCD

xCD

tC

nKn

yCnD

xCnv

xCnD

THL

dbTHL

∂∂

=∂∂

−∂∂

+∂∂

∂∂

+=

∂∂

+∂∂

−∂∂

2

2

2

2

2

2

2

2

)1( ρ

R = 1 + ρbKd/n [11-9]

where R is called retardation factor.

Eq. [11-8] is called advection-dispersion equation (ADE). ADE describes the transport of solutes in geological materials. Note that we assumed that everything except for C is constantwith respect to (t, x, y). In particular, v does not vary with time. This means the flow is at steady state, and the transport is at transient state.

tC

nKnV

tM db

∆∆

+=∆∆ )1( ρ

VyCnDV

xCv

xCDn

xbyyCnD

ybxxCv

xCDnAJJ

THL

TH

Loutin

2

2

2

2

2

2

2

2

)(

)(

)()()(

∂∂

+∂∂

−∂∂

=

∆∆∂∂

+

∆∆∂∂

−∂∂

=−

[11-8]

11-14Analysis of the ADE

Let’s look at simple cases of the migration of a “plume”.

v

tC

xCv

∂∂

=∂∂

−

The plume travels with groundwater without dispersion. This is a hypothetical case illustrating the concept of pure advection.

Case 1: DL = DTH = 0. R = 1.

v

The plume undergoes advection and dispersion. The center of mass travels at velocity v. This could apply to the migration of non-reactive tracers such as Cl-.

tC

xCv

yCD

xCD THL ∂

∂=

∂∂

−∂∂

+∂∂

2

2

2

2Case 2: R = 1

tC

xC

Rv

yC

RD

xC

RD THL

∂∂

=∂∂

−∂∂

+∂∂

2

2

2

2

v

Case 3: R >1(Kd > 0)

The center of mass travels at velocity v/R, which is smaller than v. This is why R is called a retardation factor. This could apply to the migration of chlorinated solvents, which are easily sorbed on soil particles.

11-15

Groundwater does not flow. Therefore, mechanical dispersion is non-existent and the transport process is completely driven by molecular diffusion. This could apply to the diffusion of radio-active species in tight clays. With some modification, it could also apply to the diffusion of metal ions in rock matrix during hydrothermal events. Molecular diffusion is an extremely slow process. With D* = 10-10 m2/s, it takes 102

years for the solute “front” to advance 1 m.

tC

yC

xCD

∂∂

=

∂∂

+∂∂

2

2

2

2*

Case 4: v = 0, DL = DTH = D*, R = 1

Solution of the ADE: free-moving plume

Suppose M [kg] of contaminant is spilled in a horizontal, confined aquifer having a thickness of b [m]. We want to predict the location of the contaminant plume and the degree of spreading (dispersion) using the ADE.

xy

bFlow direction

Initial spill

11-16

v

−−

tDvtx

L4)(exp

2

−

tDy

TH4exp

2

initial mass M

−

−−=∴ 2

2

2

2

2exp

21

2)(exp

21

yyxx sy

ssmx

sbMC

ππ

where m is mean, sx and sy are standard deviations.

For Case-2 (no retardation) the solution is given by;

−

−−=

tDy

tDvtx

DDtbMC

THLTL 4exp

4)(exp

4/ 22

π

The physical meaning becomes apparent when we set;

tDstDsvtm THyLx 22 ===

[11-10]

11-17For Case-3 (R > 1), we can simply replace v in Eq. [11-10] by v/R, DL by DL/R, and DT by DTH/R.

−

−−= 2

2

2

2

2exp

21

2)(exp

21

yyxx sy

ssmx

sbMC

ππ

tRDst

RDst

Rvm TH

yL

x22

===

Note that the mean position, or the position of the center of mass, is now at vt/R. The movement of the plume is slowed down or retarded because solute particles spend some time being attached to solid surfaces.

For Case-4 (v = 0), we can simply eliminate v and replace DLand DTH in Eq. [11-10] by D*.

−

−= 2

2

2

2

2exp

21

2exp

21

yyxx sy

ssx

sbMC

ππ

tDss yx*2==

Note that, since sx = sy, molecular diffusion is symmetrical.

For Case-1, the solution takes a different, simpler form;C(x, y, t) = C0(x - vt, y)

where C0(x, y) is the initial distribution of concentration.This represents simple translation of the plume.

[11-11]

11-18

0

2

4

6

8

20 30 40 50 60m

s

measured variable (x)

# of

occ

urre

nce

mean (m) = 38 standard deviation (s) = 6

Normal distribution:

−−= 2

2

2)(exp

21)(

smx

sxf

π

Review of normal distribution

This graph shows an experimental result; the number of occurrence is plotted against the value of x. For example, x = 36 occurred four times. If we have a reason to believe that the results are distributed in “bell-shape”, we can fit a normal probability density function, or normal distribution.

Normal distribution usually arises from random processes. Einstein (1905) showed that the Brownian motion of particles lead to normal distribution.

# of occurrence = (total # of samples) × f(x) × ∆x

11-19

Solution of the ADE : one-dimensional case

In this case, we are interested in a plume originating from a fixed source. We will analyze one-dimensional case, which has a reasonably simple solution.

tC

xC

Rv

xC

RDL

∂∂

=∂∂

−∂∂

2

2

Initial condition : C = 0 at t = 0, for all x > 0Boundary conditions : C = C0 at x = 0 for all t > 0

C = 0 at x = ∞ for all t > 0

t = t2

t = t1

x = vt2/Rx = vt1/R

C/C

0

1

0.5

0

x = 0 x = ∞

C = C0v

+

+

−=

RtDRvtx

Dvx

RtDRvtx

CC

LLL /2/erfcexp

/2/erfc

21

0

Solution:

where erfc(β) is called complementary error function. See page 11-20 for the definition and a numerical table.

[11-12]

11-20

)erf(1)erfc( ββ −=

ηηπ

ββ

d∫ −=0

2 )exp(2)erf(

β erf(β) erfc(β) β erf(β) erfc(β) β erf(β) erfc(β) β erf(β) erfc(β)0.001 0.00113 0.99887 0.54 0.55494 0.44506 1.09 0.87680 0.12320 1.64 0.97962 0.020380.003 0.00339 0.99661 0.55 0.56332 0.43668 1.10 0.88021 0.11979 1.65 0.98038 0.01962

0.01 0.01128 0.98872 0.56 0.57162 0.42838 1.11 0.88353 0.11647 1.66 0.98110 0.018900.02 0.02256 0.97744 0.57 0.57982 0.42018 1.12 0.88679 0.11321 1.67 0.98181 0.018190.03 0.03384 0.96616 0.58 0.58792 0.41208 1.13 0.88997 0.11003 1.68 0.98249 0.017510.04 0.04511 0.95489 0.59 0.59594 0.40406 1.14 0.89308 0.10692 1.69 0.98315 0.016850.05 0.05637 0.94363 0.60 0.60386 0.39614 1.15 0.89612 0.10388 1.70 0.98379 0.016210.06 0.06762 0.93238 0.61 0.61168 0.38832 1.16 0.89910 0.10090 1.71 0.98441 0.015590.07 0.07886 0.92114 0.62 0.61941 0.38059 1.17 0.90200 0.09800 1.72 0.98500 0.015000.08 0.09008 0.90992 0.63 0.62705 0.37295 1.18 0.90484 0.09516 1.73 0.98558 0.014420.09 0.10128 0.89872 0.64 0.63459 0.36541 1.19 0.90761 0.09239 1.74 0.98613 0.013870.10 0.11246 0.88754 0.65 0.64203 0.35797 1.20 0.91031 0.08969 1.75 0.98667 0.013330.11 0.12362 0.87638 0.66 0.64938 0.35062 1.21 0.91296 0.08704 1.76 0.98719 0.012810.12 0.13476 0.86524 0.67 0.65663 0.34337 1.22 0.91553 0.08447 1.77 0.98769 0.012310.13 0.14587 0.85413 0.68 0.66378 0.33622 1.23 0.91805 0.08195 1.78 0.98817 0.011830.14 0.15695 0.84305 0.69 0.67084 0.32916 1.24 0.92051 0.07949 1.79 0.98864 0.011360.15 0.16800 0.83200 0.70 0.67780 0.32220 1.25 0.92290 0.07710 1.80 0.98909 0.010910.16 0.17901 0.82099 0.71 0.68467 0.31533 1.26 0.92524 0.07476 1.81 0.98952 0.010480.17 0.18999 0.81001 0.72 0.69143 0.30857 1.27 0.92751 0.07249 1.82 0.98994 0.010060.18 0.20094 0.79906 0.73 0.69810 0.30190 1.28 0.92973 0.07027 1.83 0.99035 0.009650.19 0.21184 0.78816 0.74 0.70468 0.29532 1.29 0.93190 0.06810 1.84 0.99074 0.009260.20 0.22270 0.77730 0.75 0.71116 0.28884 1.30 0.93401 0.06599 1.85 0.99111 0.008890.21 0.23352 0.76648 0.76 0.71754 0.28246 1.31 0.93606 0.06394 1.86 0.99147 0.008530.22 0.24430 0.75570 0.77 0.72382 0.27618 1.32 0.93807 0.06193 1.87 0.99182 0.008180.23 0.25502 0.74498 0.78 0.73001 0.26999 1.33 0.94002 0.05998 1.88 0.99216 0.007840.24 0.26570 0.73430 0.79 0.73610 0.26390 1.34 0.94191 0.05809 1.89 0.99248 0.007520.25 0.27633 0.72367 0.80 0.74210 0.25790 1.35 0.94376 0.05624 1.90 0.99279 0.007210.26 0.28690 0.71310 0.81 0.74800 0.25200 1.36 0.94556 0.05444 1.91 0.99309 0.006910.27 0.29742 0.70258 0.82 0.75381 0.24619 1.37 0.94731 0.05269 1.92 0.99338 0.006620.28 0.30788 0.69212 0.83 0.75952 0.24048 1.38 0.94902 0.05098 1.93 0.99366 0.006340.29 0.31828 0.68172 0.84 0.76514 0.23486 1.39 0.95067 0.04933 1.94 0.99392 0.006080.30 0.32863 0.67137 0.85 0.77067 0.22933 1.40 0.95229 0.04771 1.95 0.99418 0.005820.31 0.33891 0.66109 0.86 0.77610 0.22390 1.41 0.95385 0.04615 1.96 0.99443 0.005570.32 0.34913 0.65087 0.87 0.78144 0.21856 1.42 0.95538 0.04462 1.97 0.99466 0.005340.33 0.35928 0.64072 0.88 0.78669 0.21331 1.43 0.95686 0.04314 1.98 0.99489 0.005110.34 0.36936 0.63064 0.89 0.79184 0.20816 1.44 0.95830 0.04170 1.99 0.99511 0.004890.35 0.37938 0.62062 0.90 0.79691 0.20309 1.45 0.95970 0.04030 2.00 0.99532 0.004680.36 0.38933 0.61067 0.91 0.80188 0.19812 1.46 0.96105 0.03895 2.01 0.99552 0.004480.37 0.39921 0.60079 0.92 0.80677 0.19323 1.47 0.96237 0.03763 2.02 0.99572 0.004280.38 0.40901 0.59099 0.93 0.81156 0.18844 1.48 0.96365 0.03635 2.03 0.99591 0.004090.39 0.41874 0.58126 0.94 0.81627 0.18373 1.49 0.96490 0.03510 2.04 0.99609 0.003910.40 0.42839 0.57161 0.95 0.82089 0.17911 1.50 0.96611 0.03389 2.05 0.99626 0.003740.41 0.43797 0.56203 0.96 0.82542 0.17458 1.51 0.96728 0.03272 2.06 0.99642 0.003580.42 0.44747 0.55253 0.97 0.82987 0.17013 1.52 0.96841 0.03159 2.07 0.99658 0.003420.43 0.45689 0.54311 0.98 0.83423 0.16577 1.53 0.96952 0.03048 2.08 0.99673 0.003270.44 0.46623 0.53377 0.99 0.83851 0.16149 1.54 0.97059 0.02941 2.09 0.99688 0.003120.45 0.47548 0.52452 1.00 0.84270 0.15730 1.55 0.97162 0.02838 2.10 0.99702 0.002980.46 0.48466 0.51534 1.01 0.84681 0.15319 1.56 0.97263 0.02737 2.12 0.99728 0.002720.47 0.49375 0.50625 1.02 0.85084 0.14916 1.57 0.97360 0.02640 2.14 0.99753 0.002470.48 0.50275 0.49725 1.03 0.85478 0.14522 1.58 0.97455 0.02545 2.16 0.99775 0.002250.49 0.51167 0.48833 1.04 0.85865 0.14135 1.59 0.97546 0.02454 2.18 0.99795 0.002050.50 0.52050 0.47950 1.05 0.86244 0.13756 1.60 0.97635 0.02365 2.20 0.99814 0.001860.51 0.52924 0.47076 1.06 0.86614 0.13386 1.61 0.97721 0.02279 2.25 0.99854 0.001460.52 0.53790 0.46210 1.07 0.86977 0.13023 1.62 0.97804 0.02196 2.30 0.99886 0.001140.53 0.54646 0.45354 1.08 0.87333 0.12667 1.63 0.97884 0.02116 2.40 0.99931 0.00069

Table of error function

)erf(1)erfc( ββ +=−

11-21

Transport processes in the vadose zone

In the forgoing analysis we assumed that the material is saturated so that volumetric water content (θ) is equal to porosity (n). The mass conservation principle is applicable in the vadose zone, when n is replaced by θ.

However, some of the assumptions used to derive the ADE are no longer valid in the vadose zone. In particular, the flow in the vadose zone is highly transient; θ and v varies with time and space. The ADE in the vadose zone takes a much more complicated form than Eq. [11-10].

Three-dimensional form of the ADE

It is straight-forward to extend the mass conservation analysis in page 11-12 to three dimensions. The resulting ADE takes the form;

tC

xC

Rv

zC

RD

yC

RD

xC

RD TVTHL

∂∂

=∂∂

−∂∂

+∂∂

+∂∂

2

2

2

2

2

2

where DTH is horizontal transverse dispersion coefficient and DTV is vertical transverse dispersion coefficient. In 3-D form, Eq. [11-10] becomes:

−−

−−=

RtDz

RtDy

RtDRvtx

RDDDt

MCTVTHL

TVTHL/4/4/4

)/(exp/)(8

222

323

π

[11-13]

11-22

Application of the ADE

Eqs. [11-10] and [11-12] can be used to gain insight into many important environmental and geological problems.

(1) Landfill leachateWhat is C in the well t years after the construction of the landfill?

C = C0

slow diffusionL

(2) Nuclear waste disposalHow long will it take to have C/C0= 0.001 near the water table? How does it compare to the half life of 226Ra (1600 yr)?Note that v = 0 in this problem.

(3) Chemical spillA spill of pesticide generated a plume of dissolved pesticide, which is transported in an aquifer toward a lake. When the plume reaches the lake, what is the concentration of pesticide?

spill v

C = C0

x = Lx = 0

q