Download Lecture 5

HONR 297
Environmental Models
Chapter 2: Ground Water
2.4: Darcy’s Law
Henry Darcy (1803 – 1858)
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In the mid 19th century, the
city of Dijon, France decided
to improve and enlarge the
city’s public water works.
Henry, Darcy, the the local
Inspecteur Divisionnaire des
Ponts et Chausees (Inspector
General of Bridges and
Roads), was responsible for
this civil engineering program.
One of the challenges faced
by Darcy was how to design
sand filters for the city’s
public fountains.
Henry Darcy Video!
Courtesy Wikipedia Commons:
http://commons.wikimedia.org/wiki/File:Henry_Darcy.jpg
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Henry Darcy
Since there weren’t any
models for the flow of fluid
through a porous medium,
Darcy had to create one.
 To do so, he performed his
own experiments on the
flow of water through
sand.
 His results, including what
we now know as Darcy’s
Law, were published in
1856 as an appendix to his
647-page report on the
project: Les Fontaines
Publiques de la Ville de Dijon.
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Darcy’s Question
One of the basic
questions Darcy
wanted to answer
was:
 How much water could
move at what rate
through what size
filter?
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Experimental Set-up
Force
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Water
Under
Pressure
Geologic
Sample
Water
Under
Negligible
Pressure
Flow
The experimental set-up above, based on Figure 2.7 in our text is similar
that used by Darcy – all results that follow were observed experimentally
by Darcy!
Geologic material (sand, clay, gravel, silt, etc.) is placed in the middle of a
tube.
Water is introduced on one end and pushed through the sample material
by means of forced a piston.
If water passes through the geologic material easily (sand, gravel, etc.) then
minimal force (which means less pressure) needs to be applied.
For material more resistant to flow (clay, silt, etc.), more force (which
means greater pressure) has to be applied – sufficient pressure will bring
the flow rate back up to what it was for the sand or gravel.
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Base Set-up for Experiments
Length = L
Flow rate Q
Force
Pressure = P
CrossSectional
Area is A
A
Pressure = 0
For all experiments that follow, we will use this
set-up as a base case.
 Assume for each experiment that is performed,
the same type of geologic material is placed in the
tube.
 Suppose for the base set-up, with pressure P,
length of material L, and cross-sectional area A,
we get a flow rate Q.
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Experiment 1 – Double Pressure
Length = L
Flow rate ?
More Force
Pressure = 2P
CrossSectional
Area is A
A
Pressure = 0
Suppose we increase the force so that the
pressure on the left end is doubled to 2P,
while all other parameters stay the same.
 Question: How will this affect the flow
rate on the right end?
 Answer: The flow rate will double to 2Q!
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Experiment 2 – Double Length
Length = 2L
Flow rate ?
Force
Pressure
=P
CrossSectional
Area is A
A
Pressure
=0
Suppose we use the original amount of force on the
left end, but double the sample length to 2L, again
leaving all other parameters the same as in the base
case.
 Question: How will this affect the flow rate on the
right end?
 Note that the volume of the material resisting the
flow has doubled!
 Answer: The flow rate will be cut in half to Q/2!
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Experiment 3 – Double Length and
Double Pressure
Length = 2L
Flow rate ?
More Force
Pressure
= 2P
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CrossSectional
Area is A
A
Pressure
=0
Now, let’s start with the set-up for the last experiment, but
increase the force so that the pressure on the left end is doubled.
Question: How will this affect the flow rate on the right end?
Note that the volume of the material resisting the flow has
doubled, but the pressure has increased to compensate for this!
Answer: The flow rate will be the same as the base case – namely
Q!
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Experiment 4 – Double Crosssectional Area
Length = L
Force
CrossSectional
Area is A
A
Pressure = P
Flow
Pressure = 0
CrossSectional
Area is A
A
Now, let’s suppose that we place two copies of the base-case set-up in
parallel.
 Question: How much water will be flowing through this new set-up, as
compared to the base-case set-up?
 Answer: Twice as much water!
 Reason: Since we’ve doubled the cross – sectional area of the water
pathway through the sample material, this doubles the volume of water
that can be pushed through!
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Our Observations so Far …
Double Pressure (2P) => Double Flow
Rate (2Q)
 Double Sample Length (2L) => One Half
Flow Rate (Q/2)
 Double Cross-Sectional Area (2A) =>
Double Flow Rate (2Q)
 In each case, we were changing a
parameter by a factor of two!
 What if we changed our parameters by a
factor of three? How about K?
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Parameter Change Factor
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Triple Pressure (3P)
=> Triple Flow Rate
(3Q)
Triple Sample Length
(3L)
=> One Third Flow
Rate (Q/3)
Triple Cross-Sectional
Area (3A)
=> Triple Flow Rate
(3Q)
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K*P
=> K*Q
K*L
=> Q/K
K*A
=> K*Q
So, how is the flow
rate related to
pressure, length, and
cross-sectional area in
general?
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General Relationships Between
Flow Rate and Parameters
The flow rate is proportional to the pressure.
The flow rate is inversely proportional to the
sample length.
3. The flow rate is proportional to the crosssectional area of the sample.
 Mathematically, we can write this as
Q = K*P*(1/L)*A
(1)
where K is a constant of proportionality that will
depend on the sample material.
 Equation (1) provides an initial mathematical
model to describe the flow rate of through the
sample material!
1.
2.
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A Revised Experimental Set-up!
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Instead of using a piston
at the left end to control
pressure, we can use
columns of water to do Flow
the same thing (this is
what Darcy actually did).
If the left column of
water is higher than the
right column, then the
net pressure will be equal
to the difference in
water column heights,
which we will denote by
h, i.e.
h = h1 – h2
Flow
Height
= h1
Geologic Material,
length L,
cross-section A
Height
= h2
Flow
Reference level for measuring heights,
for example, sea level.
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A Revised Mathematical Model
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Replacing pressure P
with change in height
h, our flow rate
equation (1) becomes
Q = K* h*(1/L)*A.
In this expression, Q is
the total flow rate
through the sample,
measured in units that
represent volume per
unit time, for example
gallons per minute,
cubic feet per day, etc.
Flow
Flow
Height
= h1
Geologic Material,
length L,
cross-section A
Height
= h2
Flow
Reference level for measuring heights,
for example, sea level.
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Hydraulic Conductivity
In order for our flow equation (1) to
make sense, we need to figure out the
units on the constant of proportionality,
K, which we will call the hydraulic
conductivity.
 Units of Q: [Q] = (length^3)/(time)
 Units of h: [ h] = length
 Units of length: [L] = length
 Units of area A: [A] = length^2
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Hydraulic Conductivity
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Units of hydraulic conductivity K:
[K] = [(Q*L)/( h*A)]
= ((length^3)/(time)*length)/(length*length^2)
= length/time
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Thus, the units for hydraulic conductivity are [K] =
length/time!
Examples of hydraulic conductivity (found
experimentally) for various materials are given in our
text on page 26 – Table 2.1.
Notice that the hydraulic conductivity of gravel and
sand is higher than that for silt or clay – does this make
sense with which materials allow water to flow more
freely?
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Hydraulic Gradient
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We need to define one more quantity, after which we
can present Darcy’s Law!
This is related to our Experiment 3 above, in which
we noted that if we double the sample length and
double the pressure, the flow rate stays the same.
The hydraulic gradient, i, is defined by
i = h/L.
The hydraulic gradient describes the change in
pressure over the change in length of the sample –
note that the hydraulic gradient is dimensionless, as
[i] = [ h/L] = length/length.
Replacing h/L with i in equation (1) yields Darcy’s
Law!
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Darcy’s Law
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Q = K iA
◦ where
 Q is the volumetric flow rate
 K is the hydraulic conductivity of the geologic
material
 i is the hydraulic gradient, with i = h/L.
 A is the cross-sectional area
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Hydraulic Head
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Definition:
◦ The height of the water level at any point is called
the (hydraulic) head. The change in hydraulic
head between two points is called the head loss
or change in head and denoted h.
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Example 1
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Imagine an underground sand aquifer that is
40 feet thick, 200 feet wide, and has a
hydraulic conductivity of 25 ft/day. Two test
wells 750 feet apart are drilled into the
aquifer, along the axis of flow, and the
measured head values at these wells were
found to be 120 and 114 feet, respectively.
◦ (a) Draw a really clear diagram of this situation.
◦ (b) Find the total flow rate in a given cross
section of the aquifer.
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Example 1
(a) Sketch diagram of situation
on board.
 (b) Use Darcy’s Law with the
following:
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◦ Cross-sectional area is
A = (40 ft)(200 ft) =
= 8000 ft^2.
◦ Change in head is
h = h1 – h2 =
= 120 ft – 114 ft = 6 ft.
◦ Length is
L = 750 ft
◦ Thus, the hydraulic gradient is
i = h/L = (6 ft)/(750 ft).
◦ Finally, the hydraulic conductivity is
K = 25 ft/day.
Thus, we have from Darcy’s Law:
 Q = K iA
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= K ( h/L) A
=(25 ft/day)(6 ft)/(750 ft)(8000 ft^2)
= 1600 (ft^3)/day
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Estimating Contamination
Concentration
If we know the rate at which a contaminant
is leaking into an aquifer, then we can use
calculations like those in Example 1 to
estimate the concentration of the
contaminant in an aquifer!
 For contaminant concentration calculations,
we assume that once the contaminant enters
the aquifer it moves downstream at the
same rate as the ground water.
 We also assume that the contaminant
“immediately” spreads out and becomes
uniformly mixed in the water.
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Example 2
As an example, suppose 3 lb of contaminant is leaking
into the aquifer from Example 1 each day (say at the
left end).
 Then, in the course of one day, 1600 cubic feet of
water will pass through an imaginary cross-sectional
plane perpendicular to the flow of the groundwater.
 Also, three pounds of contaminant will pass through
this plane each day!
 It follows that the concentration of contaminant in
the water is
C = (3 lb/day)/(1600 (ft^3)/day)
= 3/1600 lb/ft^3
= 0.001875 lb/ft^3.
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Unit Conversion
Often, one needs to convert results in
terms of given units to another set of
units.
 An example of why this may be necessary
is that data from reports may not always
be in the units needed for a calculation or
data comparison.
 Suppose for instance that we want to
know the contaminant concentration
from Example 2 in kg/m^3.
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Example 3
Convert 0.001875 lb/ft^3 into kg/m^3.
 One solution is to use a unit conversion
tool such as this one, found online:
http://www.calculatorsoup.com/calculator
s/conversions/density.php.
 We can also convert “by hand”!
 Use the facts that 1 kg = 2.20462 lb and 1
in = 2.54 cm.
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Example 3
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(0.001875 lb/ft^3)
=(0.001875 lb/ft^3)*(1 kg/(2.20462 lb))
*(1 ft/(12 in))^3*(1 in/(2.54 cm))^3
*(100 cm/(1 m))^3
= 0.0300347 kg/m^3
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Parts per Million (ppm)
Often, concentration of pollutants in ground
water are measured in parts per million,
denoted “ppm”.
 For ground water contamination, we are
dealing with contaminant amounts that are
typically measured in units of mass.
 For example, if we have contamination
concentration given in the following units:
[concentration] = mass/volume,
 First convert the volume to mass units, then
convert the denominator mass units to one
million mass units.
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Example 4
From Example 3, we have a contaminant
concentration of 0.0300347 kg/m^3.
 Use the fact that one cubic meter of water has a
mass of 1000 kg to convert this concentration
measurement to ppm.
 Solution:
0.0300347 kg/m^3
= (0.0300347 kg contaminant)/(m^3 water)
*(1 m^3 water)/(1000 kg water)
*(10^6 kg water)/(1 million kg water)
= (30.0347 kg contaminant)/1million kg water)
= 30.0347 ppm
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Resources
http://commons.wikimedia.org/wiki/File:Henr
y_Darcy.jpg
 https://www.youtube.com/watch?v=9qGd29d
oiII
 Groetsch, Charles W., Inverse Problems,
Mathematical Association of America, 1999.
 http://www.calculatorsoup.com/calculators/c
onversions/density.php
 Charles Hadlock, Mathematical Modeling in the
Environment, Chapter 2, Section 4.
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