Processes which move pollutants and other compounds through the air, surface water, or subsurface environment or through engineered systems (for example, treatment reactors) are of particular interest to environmental engineers and scientists. Pollutant transport acts to move pollutants from the location at which they are generated, resulting in impacts which can be distant from the pollution source. On the other hand some pollutants, such as sewage sludge, can be degraded in the environment if they are sufficiently dilute. For these pollutants, slow transport---slow dilution---can result in excessively high pollutant concentrations, with resulting increased adverse impacts.
In this section, we discuss the processes which distribute pollutants in the environment. The goals of this discussion are twofold: to provide an understanding of the processes which cause pollutant transport, and to present and apply the mathematical formulas used to calculate pollutant fluxes.
Transport processes in the environment may be divided into two categories: advection and diffusion. Advection refers to transport with the mean fluid flow. For example, if the wind is blowing toward the east, advection will carry any pollutants present in the atmosphere toward the east. Similarly, if a bag of dye is emptied into the center of a river, advection will carry the resulting spot of dye downstream. In contrast, diffusion refers to the transport of compounds through the action of random motions. Diffusion works to eliminate sharp discontinuities in concentration and results in smoother, flatter concentration profiles. Advective and diffusive processes can usually be considered independently. In the example of a spot of dye in a river, while advection moves the center of mass of the dye downstream, diffusion spreads out the concentrated spot of dye to a larger, less concentrated region.
We used the term mass flux (
, with units of mass/time)
earlier when we calculated the rates at which mass was transported
into and out of a control volume in a mass balances. Because mass
balance calculations are always made with reference to a specific
control volume, it was clear that this value referred to the rate at
which mass was transported across the boundary of the control
volume. In our calculations of advective and diffusive fluxes, however, we
will not restrict ourselves to a specific, well-defined control
volume. Instead, we will calculate the flux density across an
imaginary plane oriented perpendicular to the direction of mass
transfer. The resulting mass flux density is defined as the rate of
mass transferred across the plane per unit time per unit area,
with units of (mass)/
. We will
use the symbol J to represent the flux density.
J represents the mass flux density, expressed as the rate per unit area at which mass is transported across an imaginary plane. J has units of [M]/[LT].
The total mass flux across a boundary (
) can be calculated from
the flux density simply by multiplying J by the area of the
boundary:
In the following sections, we will consider the flux density which results from advection and from diffusion. The symbol J will be used to represent the flux density in each case, whether the flux is a result of advection, diffusion, or a combination of both processes.
The advective flux refers to the movement of a compound along with flowing air or water. The advective flux density depends simply on concentration and flow velocity.
The fluid velocity, v, is a vector quantity---it has both magnitude and direction, and the flux J refers to the movement of pollutant mass in the same direction as the fluid flow. In this course, we will generally define our coordinate system so that the x-axis is oriented in the direction of fluid flow. In this case, the flux J will reflect a flux in the x-direction, and we will generally ignore the fact that J is really a vector.
.
. Calculation of the advective flux
density.
Calculate the average flux density J of phosphorus
downstream of the sewage pipe of example 2.1. The
cross-sectional area of the river is 30 m
.
and 
as phosphorus. The
average river velocity is 
. Using
the definition of flux density (equation 32), we find:
Diffusion results from random motions of two types: the random motion of molecules in a fluid, and the random eddies which arise in turbulent flow. Diffusion from the random molecular motion is termed molecular diffusion; diffusion which results from turbulent eddies is called turbulent diffusion or eddy diffusion. We will compare these two types of diffusion below. First, however, we consider why diffusion occurs.
In this section, we will derive Fick's Law, the equation which is used to calculate the diffusive flux density, by analyzing the results of random motion of a hypothetical box of molecules. The purpose of this derivation is to provide a qualitative and intuitive understanding of the reason that diffusion occurs, and the derivation itself is useful only for that purpose. In problems where it is necessary to actually calculate the diffusive flux, we will normally start at the end of this derivation---that is with the Fick's Law equation (equation 40). (This derivation is based closely on one presented in Mixing in Inland and Coastal Waters, H. B. Fischer, E. J. List, J. Imberger, and N. H. Brooks, Academic Press, New York, 1979.)
Consider a box which is initially divided into two parts, as shown in
Figure 8. Each side of the box has a height
and depth of 1 unit, and a width of length 
. Initially, the
left portion of the box is filled with 10 molecules of gas x and
the right side is filled with 20 molecules of gas y, as shown in the
top half of Figure 8. We now remove the
divider, and observe what happens.
As we know, molecules are never stationary. All of the molecules in
our box are constantly moving around, and at any moment they have some
probability of crossing the imaginary line at the center of the box.
We will check the box and count the molecules on each side every
seconds; we will call the probability that a molecule crosses
the central line during the period between observations k. Let's
say for now that 
. So the first time we check the box, after
a period 
, 20% of the molecules that were originally on the
left will have moved to the right, and 20% of the molecules that were
originally on the right will have moved to the left. When we count
the molecules on each side, then, we will find the situation shown in
the bottom of Figure 8, with 8 ``x''
molecules remaining on the left and 2 on the right, and 16 ``y''
molecules remaining on the right, 4 having moved to the left. If we
assume that the mass of each molecule is 1 unit, we can calculate
concentrations in units of mass/volume, and find that the
concentration differences between the two boxes have been reduced from
10 to 6 for molecule ``x'', and from 20 to 16 for molecule ``y.''
This is a fundamental result of diffusion---that concentration
differences are reduced. We also observe that the movement of
molecule ``x'' is essentially independent from that of ``y'', so the
diffusion of each molecule can be considered as a separate problem.
That is, we don't have to worry about molecule ``x'' when we are
calculating the diffusion of molecule ``y'', or vice versa.
So diffusion moves mass from regions of higher concentration to
regions of lower concentration, and if left to continue indefinitely,
it would result in equal concentrations on both sides of the box.
We now need to find the flux density J which diffusion
causes. For this calculation, we will again use the situation shown
in Figure 8, with a probability of any
molecule crossing the central boundary during a period 
\
equal to k. Since each molecule can be considered independently, we
will analyze the movement of a single molecule type, say
molecule ``y.''
Let 
be the total mass of molecule ``y'' in the left half of the
box, and 
equal the mass in the right half. Since our box has
unit height and depth, the area perpendicular to the direction of
diffusion is one square unit. Thus, the flux density---the flux per
unit area---is just equal to the rate of mass transfer across the
boundary. The amount of mass transferred from the left to right in a
single time step is equal to 
, while the amount transferred
from right to left during the same period is 
. Thus, the
rate of mass flux from left to right across the boundary is equal to
divided by 
, or
Since it is more convenient to work with concentrations than with total mass values, we need to convert this equation to concentration units. The concentration in each half of the box is given by
Thus, substituting 
for the mass in each half of the box,
the flux density is equal to
Finally, we note that as 
, 
, and therefore
(Note that the negative sign in this equation is simply a result of
the convention that flux is positive when it flows from left to right,
but the derivative is positive when concentration increases
toward the right.) This equation states that the flux of mass across
an imaginary boundary is proportional to the concentration gradient at
the boundary. Since the resulting flux cannot depend on arbitrary
values of 
or 
, the product 
must be constant. This product is the value we call
the Diffusion Coefficient, D. Thus, we obtain Fick's
Law:
The units of the diffusion coefficient are clear from an analysis of
the units of equation 40 or from the units of the
parameters in equation 39; the diffusion
coefficient has the same units as 
. Since
k is a probability, and thus has no units, the units of D must
be (length)
/(time). Diffusion coefficients are commonly reported
in cm
/s.
Before we go on, note the form of equation 40:
This form of equation will also appear later when we discuss Darcy's Law, which governs the rate at which groundwater flows through soil pores. The same equation also governs heat transfer, if we replace the concentration gradient with a temperature gradient.
The molecules-in-a-box analysis used above is essentially an analysis
of molecular diffusion. Purely molecular diffusion is relatively slow.
Typical values of the diffusion coefficient D are approximately
10
--10
cm
/s for gases, and much lower, around
cm
/s for liquids. The difference in diffusion
coefficient between gases and liquids is understandable if we consider
that gas molecules are free to move much greater distances before
being stopped by ``bumping'' into another molecule. The diffusion
coefficient also varies with temperature and the molecular weight of
the diffusing molecule. This is because the average speed of the
random molecular motions is dependent on the kinetic energy of the
molecules. As heat is added to a material and temperature increases,
the thermal energy is converted to random kinetic energy of the
molecules, and the molecules move faster. This results in an increase
in the diffusion coefficient with increasing temperature. However, if
we compare molecules of differing molecular weight, we find that
at a given temperature a heavier molecule moves more slowly, and thus
the diffusion coefficient decreases with increasing molecular weight.
.
. Molecular Diffusion
Gasoline-contaminated groundwater has been
transported under a house from a nearby gas station. Two meters below
the dirt floor of the house's basement, the concentration of
hydrocarbon vapors in the airspace within the soil is
g/cm
. Estimate
the flux density of gasoline vapor transported into the basement by
molecular diffusion. The diffusion coefficient for gasoline vapor in
the air space within the soil is equal to 
cm
/s. Assume
that the basement is well-ventilated, so that the concentration of
gasoline in the basement is very small in comparison to the
concentration in the soil.
The calculated flux density of 
can
be used to calculate the total mass flux of gasoline into the basement
using equation 31. If the basement floor has an area
of 100 m
, then the total flux into the basement is
In turbulent diffusion, mass is transferred through the mixing of turbulent eddies within the fluid. This is fundamentally different from the processes which determine molecular diffusion---in turbulent diffusion, it is the random motion of the fluid that does the mixing, while in molecular diffusion it is the random motion of the pollutant molecules that is important. Earlier in this section, we used an example in which a spot of dye was dropped into the center of a river. If we follow the center of the dye spot down the river, we would see that spreading out of the spot by molecular diffusion would occur (slowly) as a result of the random motion of dye molecules across the edges of the spot. Turbulent diffusion would occur (much more quickly), as a result of eddies in the river mixing clean water from outside the spot with dye-colored water within the spot.
To indicate this difference in causation, the diffusion coefficient for turbulent diffusion is often referred to as the eddy diffusion coefficient. The value of the eddy diffusion coefficient depends on the properties of the fluid flow, rather than on the properties of the pollutant molecule we are interested in. Most important is the flow velocity---turbulence is only present at flow velocities above a critical level, and the degree of turbulence is correlated with velocity. (More precisely, the presence or absence of turbulence depends on the Reynolds Number, a non-dimensional number which depends on velocity, width of the river or pipe, and the viscosity of the fluid.) In addition, the degree of turbulence depends on the material over which the flow is occurring, so that flow over bumpy surfaces will be more turbulent than flow over a smooth surface, and the increased turbulence will cause more rapid mixing.
Finally, the value of the eddy diffusion coefficient depends to some extent on the size scale of the problem we are considering. This is best illustrated by an example. Figure 9 shows three examples of the mixing of an isolated puff of pollutant (or spot of dye) in a turbulent flow. In Figure 9 a, the size of the pollutant puff is large compared to the size of the turbulent eddies. The result is that turbulent diffusion is slow. The opposite extreme is shown in Figure 9 b, where the size of the puff is very small compared to the turbulent eddies. In this case, the entire puff is moved along with the fluid eddies. The result is not diffusion at all---we would normally call this advection, since the puff is being ``blown'' along intact. The third example (Figure 9 c) shows the intermediate situation. Here, the size of the puff is comparable to the size of the turbulent eddies, and the puff is rapidly stretched out and mixed with the surrounding fluid. In this case, the eddy diffusion coefficient would be rather large. In the real world, of course, turbulent eddies of all sizes are present simultaneously. Therefore, any given case of turbulent diffusion will be a mixture of the three situations shown in Figure 9, and only a single eddy diffusion coefficient would be used.
Fick's Law applies to turbulent diffusion just as it does to molecular diffusion. Thus, flux density calculations are the same for both processes; only the magnitude of the diffusion coefficient is different.
The final diffusion-like process that we will consider is similar to turbulence, in that it is a result of variations in the movement of the water (or air) which carries our pollutant. In mechanical dispersion, these variations are the result of (a) variations in the flow pathways taken by different fluid parcels that originate in the nearby locations near one another, or (b) variations in the speed at which fluid travels in different regions.
Dispersion in groundwater flow provides a good example of the first
process. Figure 10 shows a magnified
depiction of the pores within a soil sample, through which groundwater
flows. (Note that, as shown in
Figure 10, groundwater movement is not a
result of underground rivers or creeks, but rather is caused by the
flow of water through the pores of the soil, sand, or other material
underground.)
Because transport through the soil is limited to the pores
between soil particles, each fluid particle takes a convoluted path
through the soil and, as it is transported horizontally with the mean
flow, it is displaced vertically a distance that depends on the exact
flow path it took. The great variety of possible flow paths results
in a random displacement in the directions perpendicular to the mean
flow path. Thus, a spot of dye introduced into the groundwater flow
between points B and C in Figure 10
would be spread out, or dispersed into the region between points
B
and C
as it flows through the soil.
The second type of mechanical dispersion results from differences in flow speed, and is only important in non-steady state problems, such as the accidental, sudden release of a pollutant into a flowing stream. Anywhere that a flowing fluid contacts a stationary object, the speed at which the fluid moves will be slower near the object. For example, the speed of water flowing down a river is fastest in the center of a river, and can be very slow near the edges. Thus, if a line of dye were somehow laid across the river at one point, it would be stretched out as it flowed down the river, with the center part of the line moving faster than the edges. This type of dispersion spreads things out in the longitudinal direction---in the direction of flow---in contrast to diffusion and dispersion in groundwater, which spread things out in the direction perpendicular to the direction of mean flow.
In this section, we will analyze the forces which determine the movement of a particle suspended in a fluid. Suspensions of particles must be analyzed in a variety of applications, from the cleaning of particulate pollutants from coal-fired power plants and the settling out of suspended particles in wastewater treatment plants, to the dense suspension of particles in fresh cement. In each of these examples, it is useful to understand the movement of particles within the air or water fluid.
The movement of a particle in a fluid can be determined by a balance of the viscous drag forces resisting the particle movement with gravitational or other forces which cause the movement. The solution to this balance of forces in the specific problem of particle settling under gravity is known as Stoke's Law. This law is derived and applied in this section.
Consider the settling particle shown in
Figure 11. The particle is settling in a
fluid, which may be air, water, or any other fluid. The
particle moves in response to the gravitational force, 
, which is
equal to the mass of the particle times the gravitational constant,
. An upward buoyancy force is also present, and is equal to
the mass of the displaced fluid times g. Let us assume that the
particle is a sphere of radius 
and density 
\
(
), and that
the fluid density is 
. Then the volume of the particle is
, and we have
The only remaining force to determine is the drag force, 
.
The drag force is the result of frictional resistance to the flow of
fluid past the surface of the particle. This resistance depends on
the speed at which the particle is falling through the fluid, the size
of the particle, and the viscosity, or resistance to shear of
the fluid. (Viscosity is essentially what one would qualitatively
call the ``thickness'' of the fluid---honey has a high viscosity,
while water has a relatively low viscosity and the viscosity of
air is much lower yet.) Over a wide range of
conditions, the friction force can be correlated with the
Reynolds number, which is discussed in CE361. However, for most
situations we are interested in, we can use Stoke's Law, which states
that
where 
is the fluid viscosity (units of 
) and
is the velocity of the particle relative to the fluid.
Using this relationship, we now have an equation for all of the forces acting on the particle. The net downward force acting on the particle is equal to:
The particle will respond to this force according to Newton's second
law, which states that 
. Thus,
This differential equation can be solved to determine the time-varying velocity of a particle which is initially at rest. The solution indicates that, in almost all cases of environmental interest, the period of time required before the particle reaches its final settling velocity is very short. For this reason, we will consider here only the steady-state situation in which the velocity is constant, and thus the right-hand side of equation 49 is equal to zero.
In this case, 
. Setting 
equal to zero and
noting that 
is equal to the settling velocity 
at steady
state, we can rearrange equation 48 to obtain
This is the fundamental equation which is used to calculate terminal
settling velocities of particles in both air and water. Because it is based on the Stoke's Law for the drag force,
this equation is also often referred to as Stoke's Law, and the
settling velocity is often call the Stokes velocity. A fundamental
result of this equation is that the settling velocity increases as the
square of the particle diameter, so that larger particles settle
much faster than do smaller particles.
.
. Settling in water.
A tank called a
grit chamber is to be used to remove suspended sand particles from
water by allowing them to settle out. The main purpose of this
chamber is to prevent the sand from wearing out pipes and pumps within
the wastewater treatment plant. To design the tank in a
sufficient size, we must first determine the settling velocity of the
sand particles. If the sand particles are spheres with diameter
m and density 
, what would be their
settling velocity? (Note that the viscosity of water is
and the density of water is
.)
(Note that 
.)
The settling velocity is relatively slow. For very small
particles, the drag force is significant in comparison to the
gravitational force acting on their small mass.
.
. Settling in air.
Calculate the
gravitational settling velocity for two atmospheric particles having
diameters of 0.1 
and 100. 
, respectively, and with
density 
. Based on the result, determine whether
removal of these particles in a settling chamber of 1 m depth would be a realistic
proposition. The viscosity of air is
. The density of air is
extremely small, and thus the buoyancy force may be neglected.
particle, we
obtain
and, for the 100. 
particle,
The 0.1 micron particle would take about 900 hours to settle a distance
of 1 m. However, because the settling velocity is
proportional to the square of the particle size, the 100 
m particle
settles 10
times faster, and would settle a distance of 1 m in
only 3 s. It would clearly be realistic to construct a chamber with a
residence time of >3 s. However, a much larger chamber would be
required to reach a residence time of 9 hr, and such a chamber would
likely not be economical.
Note that the 100
particle in this example settles with a
velocity that is much greater than that of the 100
particle
settling through water in example 3.3. This is
due to the much lower viscosity of air relative to water.
.
. Calculation of the minimum diameter
removed by a settling chamber.
A settling chamber is used to remove sand particles from the
sewage flow through a wastewater treatment plant. The chamber is
2 m deep and the residence time (retention time) of water in the
chamber is 4.4 hr. What is the minimum size particle which would be
completely removed by settling to the bottom of the chamber during
passage through the chamber? The density of sand particles is
2.65 g/cm
, and the viscosity of water is
0.01185 g cm
s
. Assume that any particle that settles to
the bottom of the chamber is removed.
Plugging in our equation for settling velocity,
Thus, the minimum size particle removed entirely has a diameter of
13
.
Analyses similar to the one we used to calculate the gravitational settling velocity can be used in a number of other applications. Essentially, in any situation where a particle moves in response to an applied force, the applied force can be balanced against the drag force to determine the particle's terminal velocity. In air pollution control, devices called electrostatic precipitators are used to remove smaller particles by applying an electric force to the particles, which causes them to move out of the airstream and onto charged collection plates. Other air pollution control devices, called cyclones, are based on the use of inertial forces to remove particles. To determine the particle removal efficiency in these devices, it is necessary to balance the drag force against a centrifugal force.
Groundwater is water which fills the cracks and pore spaces of underground soil and rock. It constitutes the world's largest freshwater resource, containing more than twice as much freshwater as the world's glaciers and about 40 times the freshwater in all the world's lakes and rivers. Groundwater is used extensively as a water source for agricultural and industrial use, and about half of the U.S. population uses groundwater resources for drinking water. However, groundwater flow is extremely slow in comparison to surface water flow speeds. In combination with the large volume of groundwater reservoirs, this results in slow pollutant transport but very long residence times. For this reason, contaminated groundwater moves quite slowly, but once groundwater is contaminated it can be very difficult to clean up. In this section, we will present and apply the equations that govern the rate at which groundwater moves through the subsurface environment. Before we begin, however, some definitions are in order.
The hydraulic gradient is the gradient of the height of the water
table. For example, if two wells are drilled a distance of 
m
apart and the height of the standing water within each well (called
the head of water at that location) is
measured, the hydraulic gradient between the two wells is given by
Just like any fluid above or below the ground, groundwater flows from regions of higher head to regions of lower head.
Porosity is defined as the fraction of the total volume of soil
or rock that is empty pore space capable of containing
water or air. Thus, porosity, with the symbol 
, is defined as
Since the units cancel, porosity is a unitless value. Typical values of porosity range from 15--20% for sandstone, sand, and gravel, to 45% for clay.
We are now ready to consider the rate at which water flows through the subsurface environment. The force which drives this flow is proportional to the hydraulic gradient. Based on the drag force we used for particles (equation 45), we might expect that the drag force is proportional to the velocity of groundwater flow. Thus, since the driving force must be equal to the drag force, we would expect the velocity of groundwater flow to be proportional to the hydraulic gradient. Thus, we would expect
where Q is the volumetric flow rate of groundwater, and A is the cross-sectional area through which the flow is occurring. This result was observed in the nineteenth century by a French civil engineer named D'Arcy. The equation he used to describe this relationship is termed Darcy's Law:
The constant K in equation 58 is called the
hydraulic conductivity. The hydraulic conductivity depends on
properties of the soil or rock medium and on properties of the fluid.
For example, it depends on the smoothness or roughness of the pore
surfaces and on how wide or narrow or how straight or tortuous the
pores are. It also depends on the viscosity of the fluid (just as the
drag force in Stoke's Law depends on the fluid viscosity).
Representative values of hydraulic conductivity K are .2--.5 cm/s
for gravel (which has big pores that are easy for water to flow
through); 
--
cm/s for sand; and
cm/s for clay (which is composed of very fine
particles and thus has very small, tight pores that are difficult to
force water through).
Darcy's Law relates the volumetric flow rate of groundwater, Q\ (units of volume water per unit time) to the cross-sectional area A\ of the soil or rock through which the flow occurs. The ratio of these two quantities gives the Darcy velocity:
However, this velocity does not reflect the true speed at which
groundwater moves through the subsurface environment,
because the cross-sectional area of the pores through which
groundwater flows is smaller than the total soil or rock
cross-sectional area. The ratio of pore cross-sectional area to
total cross-sectional area is equal to the porosity, 
. Thus,
the true velocity is given by
.
. Transport time of groundwater between two
wells
An underground storage tank has been discovered to be
leaking diesel fuel into groundwater. A drinking water well
is located 200 m from the fuel spill. To ensure the safety of the
drinking water supply, a monitoring well is drilled halfway between
the drinking water well and the fuel spill. The difference in
hydraulic head between the drinking water well and the monitoring well
is 40 cm (with the head in the monitoring well higher). If the
porosity is 39 percent and hydraulic conductivity is
45 m/day, how long after it reaches the monitoring well would the
contaminated water reach the drinking water well?
. The hydraulic gradient is equal
to 
. The Darcy velocity is given by
(equation 59)
The true velocity is equal to this value divided by the porosity (equation 61):
Thus, the period for flow from the monitoring well to the drinking
water well is 
days. This
result is typical of groundwater flow speeds---groundwater transport
is usually very slow.
.
.
The concentration of of a
pollutant along a quiescent water-containing tube is shown in the
figure below. The diffusion coefficient for this pollutant in water
is equal to 
cm
/s.
, 1.5, 2.5, 3.5, and
4.5?
answer: (a) At 
, 
.
(b) At 
, 
.
.
.
The tube in
problem 1 is connected to a source of
flowing water, an water is passed through the tube at a rate of
100 cm
/s. If the pollutant concentration in the water is constant
at 2 mg/l, what is
answer: (a) 
.
(b) 
.
.
.
A settling chamber is used to remove
particles from an air flow. If the settling chamber height is 3 m and
the residence time of air in the settling chamber is 10 s, what is the
minimum size particle that the chamber would remove? Assume that the
particles settle according to Stoke's Law, and that, to be removed, a
particle must settle the full 3 m during the period it is in the
chamber. The particle density is 1 g/cm
, air viscosity is
, and the air density is small
enough that the buoyance force may be ignored. Express your answer as
particle diameter in 
.
answer: 96 
.
.
A wastewater treatment plant uses a
grit chamber to settle out sand particles. Water flows at a rate of
100 gallons per minute through the chamber, which is 2 m in height and
has a volume of 100 m
. Calculate the minimum size sand particle
that would be removed in the chamber. To be removed, the particles
must settle the depth of 2 m during the period the wastewater spends
in the chamber (
). (Sand density is 2.65 g/cm
,
water density is 1.0 g/cm
, and the viscosity of water is
.
answer: 12.9 
.
.
Two groundwater wells are located 100 m apart
in permeable sand and gravel. The water level in well one
is 50 m below the surface, and in well two the water level is 75 m
below the surface. The hydraulic conductivity is
1 m/day, and porosity is 0.60. What is:
.
answer: (a) 0.25 m/day; (b) 0.42 m/day; (c) 238 days.
.
.
The hydraulic gradient of groundwater in a
certain location is 2 ft/100 ft. Groundwater in this location flows
through sand, with a high hydraulic conductivity equal to 40 m/day and
a porosity of 0.5. An oil spill has caused the pollution of the
groundwater in a small region beneath an industrial site. How long
would it take the polluted water from that location to reach a
drinking water well located 100 m downgradient?
answer: 62.5 days
