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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

4 6 8 10 12 14 16 18 20 22

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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

4 6 8 10 12 14 16 18 20 22

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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