Applied Mathematics Minicourse

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If we dig into the ground we will eventually
reach water. This may happen just a few feet into
the ground it might require drilling several
hundred feet. However, eventually we reach the
saturated zone, where all the spaces between the
soil particles or all the cracks in the rock are
filled with water. This water is called ground
water.
Question 1 What factors affect the depth to the
water? Question 2 Once we reach ground water,
how deep will this extend?
It depends on GEOLOGY/GEOGRAPHY
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Question 3 For a given location, say the center
of the Adel Mathematics Parking Lot, is this
depth always the same? If there is variation,
explain why this is.
If the depth stayed constant, this would imply
that the water table is in a steady state water
flowing from each spot being replaced by water
flowing down from higher elevations. This implies
an infinite source of water! The water table
will be higher after snow melt in the spring, and
for a while after each rain shower.
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Question 4 Why is the water moving, and what
factors affect the speed at which it moves?
The water is moving due to the effects of
gravity. Its speed will depend on the slope it
runs at as with a mountain spring on the
surface. And also on the medium it is running
through. Water will meet more resistance in some
rocks than in others.
Question 5 In which direction does the water go?
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• Consider the problem of a ball rolling on a
  surface. Assume that we know the exact topography
  of the surface. Can we predict the motion of the
  ball?
• Consider the following contour map, and assume
  that balls are placed at positions X and Y.

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Steepest descent

• Physics dictates that the ball will follow the
  path of steepest descent. (Something to do with
  minimizing potential energy)
• This means that the flow lines will be always
  perpendicular to the contour lines. (Of course,
  not only to the contour lines drawn in, but all
  lines of equal altitude.)

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In which direction will the water flow?

• Is there anything different about the flow of
  water rather than of a ball?

No.
But we will have less of an idea about the
contour lines under the ground than those on the
surface. We may have some rough geologic map to
help us. Perhaps we will also have some data
points.
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Pollutant in the water

• Suppose some chemical pollutant has leaked into
  the water supply perhaps directly above point X
  or Y in our contour map. Is it reasonable to
  think of this having
• Reached the water table directly below where it
  left the surface?
• Having followed the path we previously indicated
  from X and Y?

Maybe. This is a reasonable first assumption.
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Darcys Law

• The flow rate q of water in the aquifer (volume
  per unit time) is proportional to the hydraulic
  gradient i.

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Darcys Law

• The hydraulic gradient is the slope. So a 5 foot
  decline over a distance of 2000 feet gives a
  hydraulic gradient of 0.0025.
• The constant of proportionality called the
  hydraulic conductivity -depends on the geology.
  This is usually denoted K.
• Darcys Law is therefore qKi

What are the units for K?
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Examples
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An aquifer is 40 feet thick and 500 feet wide and
has a hydraulic conductivity of 25 feet per
day. Two test wells are drilled 450 feet apart
and have head values of 120 and 105 feet
respectively. What is the total flow rate in the
aquifer?
Recall Darcys Law qKi
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Our previous analysis was based on
What if we have more information? Say
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What if we have several data points?
What if, for each point x along the aquifer we
have head values specified by the function h(x)?
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Interstitial Velocity

• Darcys Law gives the amount of ground water
  flowing through a unit-square cross-section of
  the aquifer.
• The water is moving at a speed known as the
  interstitial velocity. This speed would be the
  same as q if water can flow through every piece
  of the cross-section. But much of it will be
  blocked by geologic particles. The speed
  therefore depends on the porosity ? of the
  medium.

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Interstitial velocity is given by Vq/?
?1
?0
?0.2
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Examples
What are the units for the porosity values?
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Applying Vq/?
Given two data values, we assume the aquifer has
a constant slope and in the above example (the
units are feet) arrive at i(80-72)/20000.004
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Applying Vq/?

• Given the geologic values of K50 feet/day and
  ?0.25 we compute
• qKi(50)(0.004)0.2
• and
• V0.2/0.250.8 feet/day
• As we were considering a portion of the aquifer
  2000 feet in length, it would take 2000/0.8
  2500 days for water to flow from one end to the
  other.

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Example An aquifer has cross-sectional area of
10,000 square feet. It has hydraulic conductivity
0.05 feet/day and porosity 0.35. Wells yield the
following data
Suppose 2 pounds of contaminant leaks into the
aquifer each day. What is the concentration of
pollutant in the aquifer in parts-per-million
(ppm)?
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Our previous analysis was based on
What if we have more information? Say
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General Case

• Use information to find V over each section of
  the aquifer. Then compute total time taken.

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In fact, given
could we use something other than line segments
between the data points?
A polynomial? Polynomials patched together?
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Fitting a polynomial to data

• We can fit a polynomial of degree n-1 through n
  data points.
• A line through 2 points, a parabola through 3
  etc.
• As an example consider fitting the cubic
  polynomial f(x)abxcx2dx3 through the 4 points
  (0,80), (500,74),(1000,73), (2000,72). This
  requires solving 4 linear equations in 4
  variables to find the values of a, b, c and d.

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• More precisely, we may solve (using Mathematica?)
• 80ab(0)c(0)2d(0)3
• 74ab(500)c(800)2d(500)3
• 73ab(1000)c(1000)2d(1000)3
• 72ab(2000)c(2000)2d(2000)3
• and then graph the resulting cubic (again with
  Mathematica?) to make double check the work and
• to look for potential problems.

Given many data values we may choose not to fit
a single polynomial through all the points but
rather to use 2 or more different polynomials
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Cubic Splines
The most common way of fitting data with
nonlinear polynomials is with a cubic spline.
Here a cubic curve is used between every adjacent
pair of data points. The choice of a cubic curve
(rather than a parabola) is made to have enough
freedom to match derivatives on either side of
the data points. This assures a smooth curve
across all the data points.
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Back to Darcys Law

• With line segments replaced with nonlinear curves
  the constant slope i in Darcys law is replaced
  by the derivative of the height-against-distance
  function h(x).
• This gives the interstitial velocity at point x,
  V(x)Kh(x)/? and the time taken to be

This can be computed using Mathematica.
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Modeling Problems

• Direction of flow (Contour map)
• Accuracy of available data values
• Accuracy of values used
• Accuracy of chosen interpolation method (line
  segments, splines, polynomials)
• Diffusion of contaminant
• Adsorption of contaminant