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1
Ex 9 – Water
Exercise
9
Water Resources
James S. Reichard
Georgia Southern University
Student Name K h a i T r u o n g Section GL-106 In this lab you will:
examine the geology of groundwater resources and the impact that pollution and human
withdrawals can have on natural hydrologic systems.
Background Reading and Needed Supplies
Prior to doing this exercise you should read Chapter 11 on Water Resources in the textbook.
With respect to supplies, you will need a calculator, ruler, and colored markers.
Part I
-
Groundwater
Exploration
In many areas of the world local governments and individual homeowners obtain freshwater by installing wells in a suitable aquifer. As illustrated in Figure 9.1, installing a well involves first drilling a hole into the subsurface using a specialized truck. Here a drill bit is attached to a rotating pipe. As the bit advances into the earth, the drilling operator usually records the type of material that is forced to come up out of the hole. This results in a written record called a drilling
or well log
. The log lists the different types of earth material according to the depth they were encountered in the hole. By knowing the elevation of the land surface, one can easily determine
the elevation of the contacts between the different types of earth materials. For example, suppose the driller in Figure 9.1 continued drilling and encountered the contact between the sand and clay at a depth of 70 feet. By subtracting 70 feet from the land elevation of 575 feet above sea level (ASL) would tell us that this contact is at 505 feet ASL.
Figure 9.1 – collecting samples as a drill bit advances into the subsurface.
2
Ex 9 – Water
1) You are to map the subsurface extent of a sand and gravel aquifer located near Perrysville, Indiana, using actual water-well logs listed below. The logs correspond to the wells numbered
1 through 6 on the USGS topographic map in Figure 9.2.
Well #1
elev. = 608
Well #2
elev. = 568
Well #3
elev. = 540
Well #4
elev. = 550
Well #5
elev. = 530
Well #6
elev. = 670
0-125 clay
125-150 bedrx
0-21 clay
21-154 gravel
0-4 clay
4-133 gravel
133-134 bedrx
0-12 clay
12-34 bedrx
0-80 bedrx
0-106 clay
106-188 bedrx
It is important to note that in this area, glaciers deposited a thick blanket of till (mostly clay) on top of a pre-existing bedrock surface. During the cyclical glaciations, sand and gravel bodies were also deposited in the till by melt-water streams. These sand and gravel bodies now serve as the primary aquifer for the region. To map the aquifer near Perrysville, you will use the above
well data to construct a cross section on the graph paper in Figure 9.3. Carefully follow the step-
by-step instructions provided below and use PENCIL only.
a) From the well-log data, plot the surface elevation of each well on Figure 9.3. Then, using a
ruler, draw a straight line between the surface and bottom depth (BD) of each well (i.e., where drilling stopped). The lines you drew show where we have data in the subsurface.
b) The topographic map shows that along the cross-section line, the river is about 490 feet
above sea level. Plot this elevation below the point indicated for the river in Figure 9.3.
c) Using a hand-drawn line (NO ruler), show the position of the land surface by connecting the surface elevation points of the river and wells. Note that your profile should reflect the
presence of the floodplain located on both sides of the river (see map in Figure 9.2).
d) From the drilling data, determine the elevation of the contacts between the different earth materials in each well. Note that wells 1, 2, 4, and 6 have only one contact, whereas well 3
has two contacts and well 5 has none (i.e., contains all bedrock).
e) Use short horizontal lines (tick marks) on the graph paper in Figure 9.3 to indicate the
elevation of the contacts you determined above.
f) Using a light pencil, record the type of earth material that lies above and below the contacts
in each well.
g) Map the top of the old bedrock surface by drawing a line (by hand) from the contact in each well that marks the top of the bedrock. Note that all but well 2 were drilled deep enough to reach the bedrock. Also, the bottom depths (BD) of the wells are NOT contacts,
so do NOT connect your bedrock line to these points.
h) Now you need to draw the remaining contact between the gravel and clay. This is trickier as both gravel and clay are not present in every well. In some places you’ll have to end the
contact at the land surface, and in others at the top of the bedrock.
i) When finished, check your interpretation with your instructor. After checking, color the
gravel unit yellow
, the clay gray
, and the bedrock brown
.
3
Ex 9 – Water
Figure 9.2 – USGS topographic map showing the location of water wells near Perrysville, Indiana. Circled wells are used in this exercise.
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4
Ex 9 – Water
Figure 9.3 – Graph paper for constructing a cross section along wells 1 through 6 shown in Figure 9.2.
5
Ex 9 – Water
2) If you were given the task of installing a large municipal water-supply well along your cross section, where would the best place be in terms of the geology? Explain why you chose the
location you did.
Well 5 is near the river and this place will be good for the construction because ; successful well are drilled near the rivers so that the groundwater may still be available even if the river is temporarily dry or when the aquifers dry.
3) Considering the location of your cross section on the map in Figure 9.2, could there be an
even better location for the new well than the one you described above? Explain where.
No, because it does not have the particular gravel 2 sustain the wall.
Part II –
Groundwater Flow
Recall that the water level in unconfined aquifers is referred to as the water table
, but in confined aquifers, which are pressurized, the water level is called the potentiometric surface
. Also recall that groundwater flows in the direction of the hydraulic gradient
, which means that water flows from areas where the hydraulic head (potential energy) is high, to areas where the
head is lower
. As illustrated in Figure 9.4, the direction and magnitude of the hydraulic gradient
in the horizontal direction can be found by measuring the height (elevation) of the water in two different wells. If both the head and distance between two wells are measured using the same units (e.g., feet), then the computed value for hydraulic gradient will be unitless (i.e., have no units).
Figure 9.4 – determining the hydraulic gradient in the horizontal direction by measuring the water level in two wells.
By knowing the hydraulic gradient and the conductivity and porosity of earth materials, one can determine the actual velocity of groundwater using the following version of Darcy’s Law
:
groundwater velocity
conductivity
gradient
porosity
6
Ex 9 – Water
4) The following table lists the depth to water and top elevation of the three wells shown in Figure 9.5. From these data, determine the hydraulic head (elevation) in each of the wells.
well
casing elevation above sea level
depth to water
hydraulic head
(ft. above sea level)
1A
231 ft
12 ft
219
1B
231 ft
45 ft
186
2B
237 ft.
44 ft
193
Figure 9.5 – Cross section showing the depth to water for the wells. Note that the screened interval of the shallow well is open only to the unconfined aquifer, whereas the two deep wells are screened in a confined aquifer.
5) Although hydraulic head is determined by measuring the height of the water column in a well,
the head actually represents the amount of potential energy at the BOTTOM where groundwater enters the screened (open) interval of the casing. To help visual this important concept, take the head values from the table above and write them next to the corresponding well screens in Figure 9.5.
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7
Ex 9 – Water
6) Based on the head values you recorded on Figure 9.5, describe whether groundwater within
the confined aquifer is flowing from well 1b to 2b, or from 2b to 1b. Explain how you know.
Because of the difference in elevation between the two sites, groundwater inside the limited aquifer
flows from 2b to 1b. Well 2b stands at 237 feet above sea level, whereas well 1b is at 231 feet.
7) Calculate the magnitude of the hydraulic gradient within the confined aquifer in Figure 9.5. Your answer will be unitless because both head and distance were measured in the same
units, namely feet (i.e., feet over feet cancel out).
To calculate the magnitude of the hydraulic gradient, we subtract the initial hydraulic head from the final hydraulic head and divide it by the distance between the two points. In this case, the calculation would be (193-186)/325 = 0.021.
8) Suppose that the hydraulic conductivity of the confined aquifer in Figure 9.5 was determined
to be 0.55 ft/day, and the porosity 0.15 (15%). Use Darcy's Law to calculate the horizontal groundwater velocity in ft/day within the confined aquifer.
groundwater velocity
conductivity
gradient
porosity
Conductivity = 0.55, Porosity = 0.021, Gradient= 0.15
=> Velocity = 0.55*0.021/0.15 = 0.077 ft/day
9) Use the familiar relationship that velocity equals distance divided by time to determine the
number of years it would take for groundwater to travel between the two deep wells.
Each well has equal time/years that is covered based on their distances. Therefore, to cover one ell, it is necessary for a similar time frame to b covered.
That is same years. Hence, one well will be approximately 11.55 years.
10) In terms of the vertical leakage of groundwater across the aquitard in Figure 9.5, describe
whether the vertical flow is upward or downward. Again, explain how you know.
Aquitard is a geological formation of poor permeability, but through which seepage is possible and hence it won't yield water freely to wells. Vertical flow is downward since the hydraulic head in the unconfined aquifer is higher than the confined aquifer which is comparatively similar as compared to well 1a.
8
Ex 9 – Water
11) Notice that the separation distance between the unconfined and confined aquifers is 50 feet.
Using this distance and the head values in wells 1a and 1b, compute the hydraulic gradient in
the vertical direction between the two aquifers.
The head of 1A from the figure, H
A
= 219ft
The head of 1B from the figure, H
B
= 186ft
The problem asked to use the distance between unconfined and confined aquifers, L = 50ft
With all the values gathered, we can substitute these on the formula:
hydraulicgradient(
i
)=Δ
H
/L
= (HA
−
HB
)/L
= (219−186)/50
= 0.66
12) Suppose that the hydraulic conductivity of the aquitard in the vertical direction is 0.000030
ft/day and the porosity is 0.25 (25%). Use Darcy's Law to calculate the velocity in ft/day of water flowing across the aquitard.
groundwater velocity
conductivity
gradient
porosity
gradient = 0.00003
conductivity = 0.66
porosity = 0.25
= 0.250.66 x 0.00003
= 7.92 x 10
-5
ft/day
13) As before, us the relationship that velocity equals distance divided by time, determine the
number of years it would take for groundwater to travel across the aquitard.
Separation distance = 50 ft
flow velocity = 7.92 x 10^-5 ft/day
We have to solve for time in years.
Time = separation distance/ flow velocity
= 50/ (7.92
∗
10^−5
∗
365)
=1729.625 years
14) Based on what you learned in this section, explain why confined aquifers are less likely to
become contaminated compared to unconfined aquifers.
It will take a long time whereas in an unconfined aquifer contamination can flow faster to it thus contaminating it quicker.
9
Ex 9 – Water
Part
III – Floridan
Aquifer System
In this section we will examine how heavy pumping withdrawals from the Floridan aquifer system in coastal Georgia have led to saltwater intrusion problems. Figure 9.6 illustrates how this confined aquifer system underlies most of the coastal plain in Georgia, Alabama, and parts of South Carolina as well as the entire state of Florida. Here the aquifer system is composed of as much as 2,000 feet of limestone rock, with the more conductive layers serving as the primary
water supply throughout the region. However, saltwater intrusion is occurring in areas of large groundwater withdrawals. As illustrated in Figure 9.7, saltwater intrusion commonly occurs when
pumping withdrawals create a cone-of-depression that draws saltwater into parts of an aquifer that once contained only freshwater.
Figure 9.6 – (A) Extent of the Floridan aquifer system; (B) areas of major water-level declines due to heavy pumping withdrawals (from USGS Professional Paper 1403).
(A)
(B)
Figure 9.7 – Illustration showing how pumping in a confined aquifer creates a cone-of- depression and allows saltwater to move into the freshwater portion of the aquifer.
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10
Ex 9 – Water
Before examining the saltwater intrusion problems with the Floridan aquifer, we need to explore
how a contour map of a water table (unconfined system) or potentiometric surface (confined system) can be used to construct groundwater flow paths. For example, the map in Figure 9.8A
shows how water-level data from individual wells were used to draw a contour map of the water
table—similar to how the land surface is contoured using elevation data. From this map view you should be able to visualize the corresponding cross-sectional view (9.8B) showing how the water table slopes toward the right. Because groundwater flows in the direction where the hydraulic gradient is the steepest, flow lines are drawn on a map such that they cross the contour lines at right angles (9.8A). As can be seen in Figure 9.9, if the contours lines themselves are curved, then the groundwater flow lines must also be curved. Note how this causes flow lines to diverge or converge, depending on the shape of the contours.
Figure 9.8 – Contour map (A) showing the elevation changes of a sloping water table. The slope of the water table can also be seen in cross-section (B).
A)
B)
Figure 9.9 – Map view showing how groundwater flow lines maintain a right angle relationship to water-level contours.
Exercise
Questions
The following questions in this section will pertain to the contour maps in Figure 9.10 and
9.11. These maps, located along the Georgia and South Carolina coast, show the figuration of the potentiometric surface of the Floridan aquifer prior to major groundwater withdrawals (1900)
and after development (1986). These maps also show various landscape features, such as streams, towns, and county boundaries. It is important to remember that the contours represent
the amount of hydraulic head within the aquifer, NOT the elevation of the landscape.
11
Ex 9 – Water
15) There are 4 points labeled with an "X" on the pre-development map in Figure 9.10. Using a
red
pencil, draw flow lines from each point indicating the path that groundwater would have
taken in the subsurface—put arrows at the end of each line to show flow direction. Remember that flow lines should cross the contour lines at right angles.
16) Find the black dot representing the town of Pembroke in Figure 9.10. Suppose that in 1900 the people of Pembroke discovered that the groundwater in the Floridan aquifer had become
contaminated from some unknown source. Could the source possibly have been from the city of Stilson, located about 15 to the northeast? Explain your answer.
No, the flow of groundwater to Stinson to Pembroke is impossible since the flow is from east to
West.
17a) Using a blue
pencil, draw another flow line from Pembroke to the coast—your arrow should come close to the point on Hilton Head Island.
b) Determine the curved distance, in feet
, between Pembroke and the point on Hilton Head—
here you'll need to convert the map distance to ground distance using the map scale.
The curved distance between Pembroke to the point on Hilton Head is 231889 feet.
c) Based on the head contours from the map and the ground distance you just calculated,
determine the hydraulic gradient in unitless form between Pembroke and Hilton Head.
Hydraulic gradient = Difference between hydraulic heads/map distance between the two pints
h
1 = 70 feet
h
2
= 7 feet
Map distance = 231889 feet
Hydraulic gradient = (70-7)/231889 = 0.00027
d) Calculate the pre-development groundwater velocity (ft/day) between Pembroke and Hilton
Head using Darcy's Law. For the conductivity of the Floridan aquifer use 200 ft/day and for
porosity use 0.20 (20%).
groundwater velocity
conductivity
gradient = 0.00027*200/0.2 = 0.27 ft per day
porosity
e) Using the relationship that velocity equals distance divided by time, determine the number
of years it would have taken for groundwater to travel from Pembroke to Hilton Head.
= 231889/(365*0.27) = 2353 years
Ex 9 – Water
98
Figure 9.10 – Potentiometric surface contours (in feet) of the Floridan aquifer in 1900 (USGS Water Supply Paper 2411).
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Ex 9 – Water
99
18) Using a red
pencil, draw groundwater flow lines from the seven points labeled with an "X" on
the 1986 post-development map (Figure 9.11). Again, put arrows at the end of each flow line
and be sure that your lines cross the head contours at right angles.
19) Could contaminated water from Pembroke ever reach Hilton Island under the 1986
conditions? Explain your answer.
The answer is no, contaminated water from Pembroke could not reach Hilton Island under the 1986 conditions. This is because a hydraulic depression has been created near VANNAM, which means that the water flow lines in the area terminate into this depression. As a result, any contaminated water from Pembroke would not be able to flow towards Hilton Island.
20) Notice that some of the flow lines going into the cone-of-depression actually originate in the offshore marine environment. What do these flow lines from the marine environment indicate
might be happening to the aquifer?
The aquifer is being contaminated with salt sea water.
21a) What is the lowest elevation shown for the potentiometric surface in the cone-of- depression (i.e., drawdown cone) on the 1986 map?
-95 feet
b) What was the elevation of the potentiometric surface in this same area under the pre-
development conditions shown on the 1900 map (Figure 9.10)?
Approximately 38 ft
c) Determine the number of feet that the potentiometric surface has declined within the cone-
of-depression between 1900 and 1986.
= 37+95 = 132 ft
Ex 9 – Water
Figure 9.11 – Potentiometric surface contours (in feet) of the Floridan aquifer in 1986 (USGS Water Supply Paper 2411)
100
15
Ex 9 – Water
22)
The major decline in hydraulic head within the Floridan aquifer has indeed allowed saltwater
to intrude into what previously had been freshwater parts of the aquifer. Saltwater today is found in wells on Hilton Head and Tybee islands.
a)
Using a blue pencil, draw a flow line from the point labeled S on Tybee Island to the center of the cone-of-depression.
b)
Determine the actual length of your flow line from point S.
74650 ft
c)
Calculate the hydraulic gradient between point S and the center of the drawdown cone.
= 75/74650 = 0.001
d)
Use Darcy's Law to compute the groundwater velocity. Assume that the aquifer's hydraulic
conductivity is 200 ft/day and porosity is 0.20.
groundwater velocity
conductivity
gradient
porosity
= 200*0.001/0.2 = 1 ft/day
e)
Using the relationship that velocity equals distance divided by time, determine the number of years it should take for the saltwater to travel from Tybee Island and the water-supply wells in the center of the drawdown cone.
Time = 74650/365 = 204.52 years
23)
The salinity of the Floridan aquifer within much of the drawdown cone will eventually become so high that the groundwater would have to undergo expensive treatment before being used for municipal or industrial purposes. Describe several steps that could be taken to
minimize the saltwater intrusion problem.
1. Reducing Groundwater Pumping: This is the most direct way to reduce saltwater intrusion. By reducing the amount of water being pumped out, the pressure in the aquifer can be maintained, preventing saltwater from being drawn in.
2. Artificial Recharge: This involves adding fresh water to the aquifer to replace the water that has
been pumped out. This can be done through methods such as injection wells or spreading basins.
3. Use of Barrier Wells: These are wells that are drilled near the coast and pumped with fresh water to create a hydraulic barrier to saltwater intrusion.
4. Zoning Regulations: Implementing zoning regulations that limit the amount of water that can be
pumped in areas prone to saltwater intrusion. 5. Public Education: Educating the public about the issue and encouraging water conservation can also help reduce the demand for groundwater.
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