Post on 06-Aug-2015
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1 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
HYDROLOGY
[11-1] to [11-10]Groundwater Hydrology
Introduction
Mohammad N. Almasri
2 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
What is Groundwater?
Groundwater is the water that occurs in the tiny spaces (called pores or voids) between the underground soil particles or in the cracks, much like sponge holds water
The substantial quantities of groundwater are found in aquifers. These aquifers are the source of water for wells and springs
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3 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
Groundwater Occurrence
Void space
Soil particles
4 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
Vertical Distribution of Subsurface Water
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5 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
Saturated Zone
All pores are filled up with water
Extends from the upper surface of saturation down to underlying impermeable rock
Generally, the water table forms the upper surface and it is the level at which water stands in a well penetrating the aquifer
Soil Water
Vadose Water
Capillary Water
Groundwater(Phreatic Water)
Land Surface
Water TableU
nsat
urat
ed Z
one
(Zon
e of
Aer
atio
n)Sa
tura
ted
Zone
Rooting Depth
6 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
Soil Zone
NOT all the pores are fully filled out with water
Extends from the ground surface down through the major root zone
The water content vary from near saturation to nearly air-dry conditions
Important in supplying moisture to roots
Soil Water
Vadose Water
Capillary Water
Groundwater(Phreatic Water)
Land Surface
Water Table
Uns
atur
ated
Zon
e(Z
one
of A
erat
ion)
Satu
rate
d Zo
ne
Rooting Depth
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7 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
Capillary Zone
Water at the water table is subject to an upward attraction due to surface tension
Water will rise until the balance occurs between the upward forces and the weight of water
The water is under tension and thus the pressure will be negative
8 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
Intermediate Vadose Zone
Extends from the lower edge of the soil-water zone to the upper limit of the capillary zone
The thickness may vary from zero to more than 100 m under deep water table conditions
The zone serves primarily as a region connecting the zone near ground surface with that near the water table through which water moving vertically downward must pass
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9 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
What is an Aquifer?
Aquifer. A formation that contains sufficient saturated permeable material to yield significant quantities of water to wells and springs. This implies an ability to store and transmit water
The terms groundwater reservoir, groundwater basins, and water-bearing formation are frequently used to refer to aquifers
10 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
Confining Beds and Covers
Aquifers are bounded by confining beds which are relatively impermeable
Aquiclude: is a saturated but relatively impermeableformation that does not yield appreciable quantities of water to wells. Clay
Aquifuge: is a relatively impermeable formation neithercontaining nor transmitting water. Granite
Aquitard: is a saturated but poorly permeable formation, does not yield water freely to wells but that may transmit appreciable water to or from adjacent aquifers and may constitute an important groundwater storage zone. Sandy clay
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11 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
Confined Aquifers
Groundwater is confined under pressure by overlying a relatively impermeable formation
In a well penetrating such an aquifer, the water level will rise above the bottom of the confining bed. This defines the elevation of the piezometric surface at that point
The piezometric surface of a confined aquifer is an imaginary surface
Should the piezometric surface lie above ground surface, a flowing well result
12 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
Confined Aquifers
Water enters a confined aquifer in an area where the confining bed ends (called recharge area) or by leakagethrough the confining bed
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13 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
Unconfined Aquifers
This aquifer does not have a confining bed (cover) above it
The aquifer can be directly recharged by rainfall or irrigation return flow
Water table is the elevation in wells that tap the aquifer
14 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
Unconfined Aquifers
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15 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
HYDROLOGY
[11-2]Groundwater HydrologyAquifer General Properties
Mohammad N. Almasri
16 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
Permeability
The permeability of a soil defines its ability to transmit a fluid
This is a property of the medium only and is independent of fluid properties
It has units of L2
1 Darcy = 10-8 cm2
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17 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
Hydraulic Conductivity
Hydraulic conductivity measures the ability of the soil to transmit water
The hydraulic conductivity is a function of properties of both the porous medium and the fluid passing through it
The units are (LT-1)
18 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
Hydraulic Conductivity and Permeability
µ is the dynamic viscosity (M/L–T) ρ is the density of the fluid (M/L3)k is permeability (L2)K is hydraulic conductivity (L/T)
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19 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
0.1mm0.01mm
Low Permeability
High Permeability
Permeability
20 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
Soil Structure and Permeability
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21 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
Example
hydraulic conductivity of a silty sand was measured and found to be 1.36×10-5 cm/s at 25°C. What is the intrinsic permeability in cm2?
28253
m10237.1101036.181.9997
1089.0Kg
k −−− ×=×××××
=ρµ
=
We need to find µ (dynamic viscosity) and ρ (density) at 25°C
22 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
Transmissivity
Transmissivity = (Hydraulic conductivity) × (aquifer thickness)
Confined aquifers T = K × b where b is saturated thickness below the confining bed
Unconfined aquifer T = K × h where h is water table elevation
Transmissivity is the rate at which water is transmitted through a unit width of aquifer under a unit hydraulic gradient
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23 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
Transmissivity
Apparently, transmissivity varies in unconfined aquifers while it is constant in confined aquifers
T = K × bT = K × h2T = K × h1
24 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
Storage Coefficient in Aquifers
Storage coefficient: indicates the amount of water released (pumped) from or stored (injected) in the aquifer due to the unit decline/increase in the potentiometric head
Mathematically;
Vreleased = S × A × ∆h
S: storage coefficient [storativity] (L0)A: aquifer surface area (L2)∆h: drop in potentiometric head (L)
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25 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
Storage Coefficient in Aquifers
In the first case, no soil media exists and thus all the released water is related to the decline in head (one-to-one relationship)In the second case, soil particles are large and all the pores contribute water (relationship with n)In the third case, soil particles are very fine. Thus, portion of water is held and not released (relationship with S)
[1]
[2]
∆h
Vreleased
A
[1]
∆h
Vreleased
[2]
∆h
Vreleased
[1]
[2]
Vreleased = 1 × A × ∆hVreleased = n × A × ∆hVreleased = S × A × ∆h
[1][2][3]
26 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
Storage Coefficient in Unconfined Aquifers
In unconfined aquifers, water comes from:
The gravity drainage of pores in the aquifer through which the water table is falling
Expansion of the water
Compression of the grains
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27 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
Storage Coefficient in Unconfined Aquifers
For an unconfined aquifer, the storage coefficient is computed using the following formula:
The above equation accounts for water coming due to compressibility of water and porous medium as well as drainage by gravity
28 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
Storage Coefficient in Unconfined Aquifers
The storativity (storage coefficient) under unconfined conditions is referred to as the specific yield (Sy)
Specific yield is the water released from the medium by gravity drainage
The specific yield is expressed as the ratio of the volume of water yielded by gravity drainage to the total volume of the soil
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29 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
Storage Coefficient in Unconfined Aquifers
Specific retention (Sr) is the ratio of the volume of water the aquifer retains against the force of gravity to the total aquifer volume
30 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
Specific Yield and Specific Retention
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31 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
Confined versus Unconfined
An unconfined aquifer releases much more waterfrom storage than a confined aquifer
32 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
HYDROLOGY
[11-3]Groundwater Hydrology
Springs - Overview
Mohammad N. Almasri
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33 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University33
Springs
A spring is a concentrated discharge of groundwater appearing at the ground surface as a current of flowing water
34 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University34
Types of Springs
DepressionThe ground surface intersects the water table
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35 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University35
Types of Springs
ContactCreated by a permeable water-bearing formationoverlying a less permeableformation that intersects the ground surface
36 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University36
Fracture artesianResulting from releases of water under pressurefrom confined aquifers through an opening in the confining bedthat intersectswith ground surface
Types of Springs
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37 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University37
Types of Springs
TubularOccurring in tubular channels or fractures of impervious rock
38 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University38
Classification of Springs
Mean dischargeMagnitude> 10 m3/sFirst
1 – 10 m3/sSecond0.1 – 1 m3/sThird10 – 100 l/sFourth
1 – 10 l/sFifth0.1 – 1 l/sSixth10 – 100 ml/sSeventh
< 10 ml/sEighth
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39 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
Emerging of SpringsVasey’s Paradise
40 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
HYDROLOGY
[11-4]Groundwater Hydrology
Darcy’s Law and Groundwater Movement
Mohammad N. Almasri
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41 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
Hydraulic Head
42 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
Hydraulic Head
Total head at a point is the summation of the pressure head and elevation head
Total head also equals the distance between ground surface and the datum minus the depth to water in the well
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43 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
Hydraulic Head
Groundwater moves in the direction of decreasing total head, which may or may not be in the direction of decreasing pressure head
44 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
Hydraulic HeadExample
In an aquifer, the ground surface is at 1,000 m above sea level, the depth to the water table is 25 m, and the water table height above the measurement point is 50 m. Calculate:
1. The total hydraulic head at the point of measurement
2. The pressure head, and
3. The elevation head
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45 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
Hydraulic HeadExample Solution
Hydraulic head = distance from the water table to the mean sea level = 1,000 – 25 = 975 m
Pressure head = distance from the water table to the point of measurement = 50 m
Elevation head = ground surface elevation – depth to water table – pressure head = 1,000 – 25 – 50 = 925 m
46 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
Hydraulic Gradient
l
h1
h2
Q
A Water table
Datum
The hydraulic gradient is the change in head over a distance in a given direction
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47 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
Hydraulic GradientSimple Example
Water table elevation was measured at two locations with a distance of 1,000 ft. If the measured elevations were 100 and 99 ft, then what is the direction of the groundwater flow and what is the hydraulic gradient?
48 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
Hydraulic GradientSimple Example
Groundwater flows in the direction of decreasing head
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49 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
Hydraulic Head and Gradient Confined Aquifers
50 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
Hydraulic Head and Gradient Unconfined Aquifers
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51 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
Example
Three piezometers monitor water levels in a confined aquifer.
Piezometer A is located 3,000 ft due south of piezometer B
Piezometer C is located 2,000 ft due west of piezometer B
The surface elevations of A, B, and Care 480, 610 and 545 ft, respectively
The depth to water in A is 40 ft, in Bis 140 ft, and in C is 85 ft
Determine the direction of groundwater flow through the triangle ABC and calculate the hydraulic gradient
52 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
Solution
Compute potentiometric head at each observation well
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53 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
Solution
Show schematically the direction of groundwater flow
54 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
Solution
Compute the hydraulic gradient between each two wells in x and y
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55 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
Solution
Compute the direction of the overall gradient (resultant)
56 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
Darcy’s Law
Q: Total flow (L3/T)
q: (Q/A) Darcy flux (L/T)
A: Cross-sectional area of the flow (L2)
K: Hydraulic conductivity (L/T)
dh: Head difference (L)
dl: Increment distance (L)
dh/dl: Hydraulic gradient (L/L)
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57 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
Velocity
Since the flow occurs across the pores that can transmit water and part of the pore space is occupied by stagnant water, then velocity equals the specific discharge divided by the effective porosity
v: velocity (L/T)ne: is the effective porosity
58 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
Darcy’s LawSolved Example – Unconfined
Find q, Q, v, T?
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59 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
Darcy’s LawSolved Example – Unconfined
We have K = 10 m/day, h1 = 20 m, h2 = 19 m, L = 1,000 mThenq = - (10) [(20-19)/(1,000)] = - 0.01 m/day
60 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
Darcy’s LawSolved Example – Unconfined
q is negative and in the opposite direction of x!!
This is correct since the flow is in the direction of decreasing head or the direction of hydraulic gradient or from high to low head
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61 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
Darcy’s LawSolved Example – Unconfined
Q = q A
What is the area of the flow?
The area is the saturated thickness. This thickness varies from point to point
Take the average height [A = 0.5 (h1 + h2) W]where W is the width of the aquifer (assume unit width if
not given; W = 1 m)
Then A = 0.5(20 + 19) (1) = 19.5 m2
62 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
Darcy’s LawSolved Example – Unconfined
Then
Q = 0.01 x 19.5 = 0.195 m3/dayFor the velocityWe know that v = q/ne = 0.01 /0.2 = 0.05 m/day
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63 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
Darcy’s LawSolved Example – Unconfined
We know that transmissivity (T) is:Hydraulic conductivity x aquifer thickness= 10 X 19.5 = 195 m2/day
64 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
Darcy’s LawSolved Example – Confined
Find q, Q, v, T?
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65 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
Darcy’s LawSolved Example – Confined
q = - (10) [(20-19)/(1,000)] = - 0.01 m/dayQ = q A
What is the area perpendicular to the flow?The area is the thickness of the aquifer which is b.
This thickness is constant as far as the aquifer is totally saturated.
A = b W where W is the width of the aquifer (assume unit width if not given; W = 1 m)
Then A = 10 x 1 = 10 m2
66 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
Darcy’s LawSolved Example – Confined
Then
Q = 0.01 x 10 = 0.1 m3/dayFor the velocityWe know that v = q/ne = 0.01 /0.2 = 0.05 m/day
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67 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
Darcy’s LawSolved Example – Confined
We know that transmissivity (T) is:Hydraulic conductivity x aquifer thickness= 10 X 10 = 100 m2/day
68 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
HYDROLOGY
[11-5]Groundwater Hydrology
Flow Nets
Mohammad N. Almasri
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69 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
Flow Nets
Flow nets are nets of equipotential lines (lines with constant head values) for an aquifer with flow linesperpendicular to them
70 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
Head Contour Maps
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71 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
Head Contour Maps and Groundwater Flow Direction
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72 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
Flow NetsHow to Construct a Flow Net?
Install piezometers in the aquifer of interest
Start a monitoring activity of water table elevations for unconfined aquifers or potentiometric heads for confined ones
Piezometers should be well distributed
Write down for each (x, y) the h
Construct the head contours (preferably using a computer software)
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73 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
Flow NetsHow to Construct a Flow Net?
Use Surfer to construct the contour map4 6 8 10 12 14 16 18 20 22
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74 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
Flow NetsThe Surfer Software
Get a free software demo at:http://www.goldensoftware.com/demo.shtml
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75 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
How to Construct a Flow Net?
76 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
Show the Groundwater Flow Directions?
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77 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
Groundwater Flow Direction
78 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
Why to delineate the groundwater flow directions?
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79 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
Flow NetsExample
Unconfined aquifer and K = 10 m/day
I. What is the total flow across the rectangular (150 m × 200 m)?
II. What is the discharge at A
A
80 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
Flow NetsExample
I. Q = K A dh/dx = K W b dh/dx = 10x200x[(19+18.5)/2]x[(19-18.5)/150]= 125 m3/day
II. QA = KWAbA(dh/dx)A = 10x1x [(13+12.5)/2]x [(13-12.5)/100] = 0.6375 m3/day-m width
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81 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
HYDROLOGY
[11-6]Groundwater Hydrology
Determining hydraulic conductivity, heterogeneity and anisotropy
Mohammad N. Almasri
82 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
Darcy’s LawFlow through Multiple – Layer Systems
Q2
Q1
h2
h1
h
You need to find out the total travel time across layers 1 and 2
Assume continuity of flow and head
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83 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
Darcy’s LawFlow through Multiple-Layer Systems
Continuity of flow (Q1 = Q2)
Continuity of head (head at interface is
constant)
84 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
Homogeneous Aquifers
In a homogeneous aquifer: Same properties at all locations
The values of the hydraulic conductivity would be about the same wherever present
The grain sizes and porosity are variable only within small limits
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85 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
Heterogeneity
In heterogeneous aquifers, hydraulic properties change spatially
In reality, aquifers are always heterogeneous
Values of hydraulic conductivity may vary by orders of magnitude
86 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
Homogeneous Versus HeterogeneousSymbolically
Consider two locations (x1, y1) and (x2, y2), thena homogeneous aquifer implies:
Kx at (x1, y1) = Kx at (x2, y2)
and
Ky at (x1, y1) = Ky at (x2, y2)
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87 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
Anisotropy
When hydraulic properties such as hydraulic conductivity, vary with direction at a given location, then the particular property is considered anisotropic
Otherwise, it is considered to be isotropic
88 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
Anisotropy
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89 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
Isotropic Versus Anisotropic Symbolically
Consider two locations (x1, y1) and (x2, y2), thenAn isotropic aquifer implies:
Kx at (x1, y1) = Ky at (x1, y1)
and
Kx at (x2, y2) = Ky at (x2, y2)
90 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
Guess!
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91 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
Equivalent Hydraulic Conductivity
In layered heterogeneous aquifers, it can be shown that there are equivalent horizontal and vertical hydraulic conductivities such that the entire system acts like a single homogeneous anisotropic layer
92 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
Equivalent Hydraulic ConductivityAveraging
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93 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
Equivalent Hydraulic ConductivityVertical
For vertical flow, Darcy flux q is constant (continuity in flow) and the total head loss is the summation of head loss across all the layers
94 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
Equivalent Hydraulic ConductivityHorizontal
For horizontal flow, total flow equals the summation of flow rates in all the layers and the head loss over a horizontal distance is constant across all the layers
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95 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
Equivalent Hydraulic ConductivityHorizontal and Vertical
Always, Kx > Kz for all possible sets of values of K1, K2,….., Kn.
Ratios of Kx/Kz usually fall in the range of 2 to 10 for alluvium, but values up to 100 or more occur where clay layers are present
96 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
Equivalent Hydraulic ConductivityExample
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97 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
Equivalent Hydraulic ConductivityExample
Kx =
Ky =
98 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
Example(Past Exam)
Find:The headloss in each layer of this aquifer between the observation wells
If the headloss in each layer between the wells were to be equal, what would be the length of each layer
50
99 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
Example(Past Exam)
Find the equivalent hydraulic conductivity. Since the flow is perpendicular to the layering of the aquifers then:
.dm85.14
50400
10100,1
300500
400100,1500
Kd
Kd
Kd
dddK
3
3
2
2
1
1
321 =++
++=
++
++=
We know that q = K×I thus the flux across the entire system equals:
.dm043.0
400100,15006.604.6685.14 =
++−
×
Yet, we know that this flux is constant across all the aquifers which means that:
043.0dh
Kdh
Kdh
Kq3
33
2
22
1
11 =
∆=
∆=
∆=
Substituting gives: ∆h1 = 0.716 m, ∆h2 = 4.73 m, and ∆h3 = 0.344 m.
100 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
Example(Past Exam)
Total headloss across all the aquifers = 66.4 – 60.6 = 5.8 m
Equal headloss in each aquifer yields a 1.93 m headloss
Again, we have a constant flux which is 0.043 m/d. Thus
.m346,1043.093.130
qhKdor
dhKq 1
111
11 =×=
∆=
∆=
The same for the other two aquifers yields d2 = 448 m and d3 = 2,244 m
51
101 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
HYDROLOGY
[11-7]Groundwater Hydrology
Applications
Mohammad N. Almasri
102 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
Applications Confined Aquifers
Steady state analysis
Homogeneous
Flow area (A) is constant(b×W)
Total flow (Q) is constantin the aquifer
If Q and A are constant then gradient is constant
52
103 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
Applications Confined Aquifers
104 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
Applications Unconfined Aquifers
Steady state analysis
Flow area (A) is NOTconstant (h×W)
Total flow (Q) is constant in the aquifer
If Q is constant but Ais not, then gradient is NOT constant
53
105 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
Applications Unconfined Aquifers
106 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
Applications Unconfined Aquifers with Recharge
The change in flow equals the recharge (w)
This implies that:
Water Divide
54
107 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
Applications Unconfined Aquifers with Recharge
At the water divide, flow equals zero
When departing from the water divide, there is a flow at both side
The flow increases with distance from the water divide
Q1=Wdx1
dx1
Q2=Wdx2dx2
108 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
Applications Unconfined Aquifers with Recharge
Upon integration
We get the head distribution across the aquifer:
55
109 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
Applications Unconfined Aquifers with Recharge
Applying Darcy’s law, we get the flow per unit width
There is a water divideat which the flow is zero and the water table is at the maximum value
110 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
Example [1]Unconfined Aquifers with Recharge
Two rivers are located 1,000 m apart and fully penetrate an unconfined aquifer of K=0.5 m/day. The mean annual rainfall and evaporation are 15 cm/yr and 10 cm/yr, respectively. The water elevations in rivers 1 and 2 are 20and 18 m, respectively
a) Determine the location and height of the water divide
b) Determine the flow across the aquifer at river locations per m width of each river
56
111 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
Example [1]Unconfined Aquifers with Recharge
K=0.5 m/d
w=15-10=5 cm/yr
20 m18 m
1,000 m
112 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
Example [1]Unconfined Aquifers with Recharge
Location of water divide = 361.2 m
Height of water divide = 20.9 m
57
113 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
Example [1]Unconfined Aquifers with Recharge
Left river x = 0 and Q = -0.0495 m2/d
Right river x = 1,000 and Q = 0.08745 m2/d
114 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
Example [1]Unconfined Aquifers with Recharge
w=15-10=5 cm/yr
20 m18 m
1,000 m361.2 m
0.08745 m2/d-0.0495 m2/d
hmax=20.9 m
58
115 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
Example [2]
The figure shows an unconfined aquifer with a hydraulic conductivity K (m/year) and an effective porosity ne (-)
The aquifer is bounded by two rivers with a distance L (m). If the travel time between the two rivers is t1 where hB >hA, what would be the travel time t2 if the distance is doubled?
River A
River B
hAhB
L
116 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
Example [2]
hKLn
nLhK
LvL
velocitydistance t timeTravel
2e
e
1 ∆=
∆===
When distance is doubled
hKLn4
hK)L2(n t timeTravel
2e
2e
2 ∆=
∆=
From the two equations
12 t4t =
59
117 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
Example [3]
Consider flow in an unconfined aquifer. The aquifer is recharged at a rate ε and its hydraulic conductivity is 20 m/day. Answer the following:
Estimate the critical recharge rate ε1 in m/day for which the spring in the figure starts flowing
118 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
Example [3]
First of all, find the analytical expression that provides the value of recharge in terms of the remainder of the parameters. Thus:
x)xL(K
xL
)hh(hh22
212
1 −ε
+−
−= )xL(xL
)LhxhxhLh(K 222
21
21
−+−+−
=ε
In order for the spring to start flowing, the water table elevation at the spring location should hit the ground surface. Thus, h = 60 m, x = 1,000 m and the remainder of the parameters are as in the figure and question
Plug in and ε1 > 0.007375 m/d
60
119 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
Example [3]
If the head in River 2 became 50 m, then what would be the critical recharge rate ε2 in m/day for which the spring starts flowing
Again, from the above equation, we get ε2 > 0.0055 m/d
120 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
Example [3]
If the recharge was set to zero, then what would be the head in River 1 for which the spring starts flowing given that the head in River 2 is 25 m
For the water table elevation with no recharge, the analytical expression is given below:
xL
)hh(hh22
212
1−
−=
xL)Lhxh)(xL(
h22
21 −
+−−=
Thus h1 = 65.907 m
61
121 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
Example [4]
The figure depicts a cross-sectional view of an unconfined aquifer of length 1,350 m
122 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
Example [4]
The figure depicts the distributions of water table elevation and flow per unit width across the aquifer. Negative flows indicate a flow in the negative direction of x (flow toward River 1). For this aquifer, answer the following:
-6
-3
0
3
6
9
12
15
18
21
24
27
0 150 300 450 600 750 900 1050 1200 1350
x (m) from River 1
h (m
) and
Q' (
m2/
day
h (m)Q' (m2/day)
62
123 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
Example [4]
Find the distance d at which a water divide exists
Water divide occurs at Q’ = 0. Therefore, from the provided figure, the maximum water table is when Q intersects the x-axis or at d = 640 m
124 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
Example [4]
What is the recharge rate in m/day?
Recharge (w) equals the total baseflow to the rivers. Thus, from the figure, at x = 0, Q’ = 6 m2/day and when x = 1,350 m then Q’ = 7 m2/day
Therefore, total baseflow = 6 + 7 = 13 m2/day and recharge is w × 1,350 = 13 and w = 0.00963 m/d
63
125 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
Example [4]
What is the approximate hydraulic conductivity?
From Darcy’s law, we know that:
At x = 0, then h = 15 m, and the gradient in the vicinity of river 1 computed at some distance is:
Plug this in the above equation, -6 = K × 15 × 0.04 K = 10 m/day
xhKh'Q
∆∆
−=
04.050
1517=
−
126 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
Example [5]
A rectangular unconfined aquifer has a length of 2,500 mand a width of 1,000 m. The area receives an annual uniformly distributed recharge of R
The aquifer has particle and bulk densities of 2,600 and 1,800 kg/m3, respectively
The aquifer is bounded by two rivers; River [1] at x=0 (left) and River [2] at x=2,500. River stages are 10 and 8 m for River [1] and River [2], respectively and aquifer hydraulic conductivity is 18 m/day
Water divide is at 1,205 m. The hydraulic gradient at River [1] is 0.02 m/m. For this scenario, answer the following questions:
64
127 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
Example [5]
Compute the recharge (R)
From Darcy's law we have Q = K A I thus the flow at x = 0 is 18×(10×1,000) ×(0.02) = 3,600 m3/day.
From the mass balance we know that flow at x = 0 comes from recharge to the left of the water divide or R×As = 3,600 where As is the surface area perpendicular to the recharge. Thus:
daym003.0
000,1205,1600,3R =×
=
128 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
Example [5]
What is the maximum head in the aquifer?
m5.18205,1)205,1500,2(18003.0205,1
500,281010
d)dL(KRd
L)hh(
hh
222
22
212
1max
=×−×+×−
−
=−+−
−=
65
129 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
Example [5]
What is the approximate travel time between a location at x=1,500 m and River [2] assuming that the effective porosity equals the total porosity
130 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
HYDROLOGY
[11-8]Groundwater Hydrology
Well Hydraulics – Unsteady State Analysis
Mohammad N. Almasri
66
131 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
Introduction
What is well hydraulics?Concentrates on understanding the processes in effect when one or more wells are pumping from an aquifer. This for instance considers the analysis of drawdown due to pumping with time and distance
Importance of well hydraulicsGroundwater withdrawal from aquifers are important to meet the water demand. Therefore, we need to understand well hydraulics to design a pumping strategy that is sufficient to furnish the adequateamounts of water
132 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
Basic Assumptions
The potentiometric surface of the aquifer is horizontal prior to start of pumping
The aquifer is homogeneous and isotropic
All flow is radial toward the well
Groundwater flow is horizontal
The pumping well fully penetrates the aquifer
67
133 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
Steady versus Transient
Steady state implies that the drawdown in head is a function of location only
Transient state implies that the drawdown in head is a function of location and time
Thus
h = f(r) in case of steady state
h = f(r,t) in case of transient state
134 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
Radial Flow to a Well in Confined Aquifers
68
135 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
Unsteady Radial Flow in a Confined Aquifer
Solution to Thies equationWell function
136 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
Propagation of Cone of Depressionwith Time
0.0
2.5
5.0
7.5
10.0
12.5
15.0
17.5
20.0
0 100 200 300 400 500 600 700 800 900 1000
Distance from well (m)s (m
)
t=0.0001 dayst=0.001 dayst=0.01 dayst=0.1 dayst=1 dayst=10 days
69
137 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
Propagation of Cone of Depressionwith Distance from Well
0
5
10
15
20
25
30
0 1 2 3 4 5 6 7
Time (days)
Dra
wdo
wn
(m)
138 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
Unsteady Radial Flow in a Confined Aquifer – Example
A well is pumping from a confined aquifer at a rate of 10,000 m3/day. T and S of the aquifer are 1,000m2/day and 0.0001, respectively.
Find the drawdown at t = 10 days at a location 100 from the well
70
139 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
Unsteady Radial Flow in a Confined Aquifer – Example
Compute u = (100 × 100 × 0.0001)/(4 × 1,000 × 10) = 2.5 × 10-5
Compute the well functionW(u) = -0.5772 – ln (2.5 × 10-5) = 10
Compute the drawdown s = [10,000/(4 × 3.14 × 1,000)] × 10 = 7.95 m
140 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
HYDROLOGY
[11-9]Groundwater Hydrology
Well Hydraulics – Method of Superposition
Mohammad N. Almasri
71
141 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
Method of Superposition
The principle of superposition can be used in multiple well systems and in situations of multiple pumping rates
By applying the superposition method, the outcome from the entire system is characterized by the summation of the outcomes from the different systems comprising the entire system
142 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
Multiple Well System
72
143 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
Multiple Well System
Where the cones of depression of two nearby pumping wells overlap, one well is said to interferewith another
From the principle of superposition, the drawdown at any point in the area of influence caused by the discharge of several wells is equal to the sum of the drawdowns caused by each well individually
144 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
Four wells are used to depressurize a confined aquifer underlying an excavation for a building project
The wells are located 60 m from the center of the excavation at corners of a square
The radius of each well is 0.3 m, the transmissivity of the aquifer is 50 m2/d, and the storage coefficient is 0.001
What discharge from each well is required if the piezometric head is to be lowered 20 m at the center of the excavation within 30 days?
What will be the drawdown at each well?
Multiple Well System – Example
73
145 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
Multiple Well System – Example
146 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
Multiple Well System – Example
The drawdown at the center equals the summation of drawdowns at the center due to each pumping well
s (center, t) = s1(center, t) + s2(center, t) + s3(center, t) + s4(center, t) or
Using Theis solutions (center, t) = Q1 W(u1)/4πT +
Q2 W(u2)/4πT + Q3 W(u3)/4πT + Q4 W(u4)/4πT
74
147 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
Multiple Well System – Example
From Theis solution for confined aquifers under transient conditions of pumping, we have
For r=60 m, t=30 days, and T=50 m2/day we get u=0.0006 and W(u)=6.84.
Total drawdown caused by the four wells = 20 m. Therefore, each well brings about a drawdown of s = 20/4=5 m
Thus, Q = 459.34 m3/day
148 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
HYDROLOGY
[11-10]Groundwater Hydrology
Capture Zone Analysis
Mohammad N. Almasri
75
149 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
What is a Capture Zone?
A capture zone is the area contributing flow to a particular well
-600
-500
-400
-300
-200
-100
0
100
200
300
400
500
600
-200 -100 0 100 200 300 400 500 600 700 800 900 1000
x (m)
y (m
)
Capture zone w
idth
Well
Stag
natio
n po
int
150 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
Capture Zone Analysis
A prior knowledge of the capture zones is essential to understand the potential impacts to a drinking water well in the event of a contamination inside the capture zone. Thus, this analysis is part of the wellhead protection plan
This analysis reveals the area where it is necessary to restrict land use within a capture zone of a water supply well
Capture zone analysis is also important in aquifer remediation using pump-and-treat systems (containment)
76
151 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
Capture Zone AnalysisThe Use in Plume Containment
Extract the contaminated groundwater, do treatment to reduce pollutant levels, and injectwater back into the aquifer or release to surface area
Possible questions in this regard are the following:Optimum number of pumping wellsWell locationOptimum pumping rate for each wellWhere to inject the treated groundwater into the aquifer
152 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
Capture Zone Analysis
77
153 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
Capture Zone Analysis
154 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
Capture Zone AnalysisSteady-State
Stagnation point
B is the aquifer thickness
Q is the pumping rate
i is the gradient
U is Darcy flux
xL is the location of the stagnation point and
yL is the maximum width
78
155 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
Capture Zone AnalysisSteady-State Example
Q variable, K = 5.7 m/d, i = 0.001, B = 10 m
-450-350-250
-150-5050
150
250350450
-200 0 200 400 600 800 1,000 1,200 1,400
x (m)
y (m
) Q=50 m3/dQ=40 m3/dQ=30 m3/d
156 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
Capture Zone Analysis – Example [2]
Shown in the Figure is a contamination plume present in a shallow confined aquifer having a thickness of 10 m, a hydraulic conductivity of 10-4 m/sec, an effective porosity of 0.2, and a storativity of 3×10-5
The hydraulic gradient for the regional flowsystem is 0.002
The maximum allowable drawdown for wells in the aquifer is 7 m
Given this information, design an optimum pumping system
79
157 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
Capture Zone Analysis – Example [2]
Values of B and U are required for the calculation. B is given as 10 m, but U needs to be calculated from the Darcy equation:
U = l0-4×0.002 = 2×10-7 m/sec
Now we are ready to work with the type curves. Superposition of the type curve for one well on the plume (see the figure) provides a Q/BU curve of about 2,500. Using this number and the values of B and U, the single-well pumping rate is: Q = B×U×(Q/BU) = 10×2×10-7×2,500 = 5×10-3 m3/sec
158 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
Capture Zone Analysis – Example [2]
A check is required to determine whether this pumping rate can be supported for the aquifer. The Cooper-Jacob equation provides the drawdown at the well assuming r = 0.2 m and the pumping period is one year. Now, drawdown is 9.85 m
Even without accounting for well loss, the calculated drawdown exceeds the 7 m available
Thus a multiple well system is necessary
80
159 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
Capture Zone Analysis – Example [2]
Superposing the plume on the two-well-type curve provides a Q/BU value of 1,200, which in turn gives a Q for each of the two wells of 10×(2×10-7)×1200 or 2.4×10-3 m3/sec
The optimum distance between wells is Q/(πBU) or 2.4×10-3/[π×10×(2×10-7)] = 382 m
We check the predicted drawdown at each well. Because of the symmetry, the drawdown in each well is the same
160 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
Capture Zone Analysis – Example [2]
The total drawdown at one of the wells includes the contribution of that well pumping plus the second one 382 m away
The calculated drawdown s is 6.57 m, which is less than the available drawdown
However, well loss should be considered, which makes the two-well scheme unacceptable
81
161 [11] Fall 2007 – Groundwater Hydrology – Mohammad N. Almasri, PhD An-Najah National University
Capture Zone Analysis – Example [2]
Moving to a three-well scheme, Q/BU is 800 (see the figure), which translates to a pumping rate of 1.6×10-3 m3/sec for each well
Carrying out the drawdown calculation for three wells located 1.26Q/(πBU) or 320 m apart provides an estimate of 5.7 m for the center well, which is comfortably less than the available drawdown
Thus we have been able to ascertain the need for three wells, located 320 m apart, and each pumped at 1.6×10-3 m3/sec