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Prof.Dr.Ir. Sunjoto Dip.HE, DEA-Hydrology of Groundwater-Post Graduate Program JTSL-FT-UGM 2
I. INTRODUCTION
1. Etymology
Hydrogeology (eng) Geohydrologie (fr) Geohidrologi (id)
Geohydrology (eng) Hydrogeologie (fr) Hidrogeologi (id)
2. Hydrology
a. Water cycle
Fig. 1.1. Hydrological cycle
THE WATER CYCLE
Water storage
in ice and snow
Water storage in oceans
Evaporation
Groundwater discharge
Infiltration
Precipitation
Sublimation
Water storage in the atmosphere
Evapotranspiration
Spring Fresh water storage
Groundwater storage
Surface runoff
Snowmelt runoff to stream
SUN
Condensation
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Prof.Dr.Ir. Sunjoto Dip.HE, DEA-Hydrology of Groundwater-Post Graduate Program JTSL-FT-UGM 3
b. Water Balance
Water balance on the ground surface is:
Fig 1.2. Water balance on the ground surface
Fig 1.3. Water balance of the storage
Acccording to Lee R. (1980): P + Ev annual 5 .105 km3/y, = the depth 973 mm
and needs 28 ceturies to evaporate by atmospheric destilation.
I OS
I - O = S
I : InflowO : OutflowS : Storage
P E
I
RP E = R + I
P : PrecipitationE : Evapotranspiration
R : RunoffI : Infiltration
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Prof.Dr.Ir. Sunjoto Dip.HE, DEA-Hydrology of Groundwater-Post Graduate Program JTSL-FT-UGM 5
Table 1.4. Water distribution in the earth (Baumgartner and Reichel, 1975)Items Volume Percentage
Solid 2.782 .107 Km3 2.010 %
Liquid 1.356 .109
Km3
97.989 % Oceans 1.348 .109 Km3 97.390 % Continent; groundwater 8.062 .106 Km3 0.583 % Continent; surface water 2.250 .105 Km3 0.016 %
Vapor 1.300 .104 Km3 0.001 %Total (all forms) 1.384 .109 Km3 100.000 %
Saline water 1.348 .109 Km3 97.938 % Fresh water 3.602 .107 Km3 2.202 %
Table 1.5. Fresh water distribution in the earth (Baumgartner and Reichel, 1975)Items Volume Percentage
Solid 2.782 .107 Km3 77.23 %Liquid 8.187 .106 Km3 22.73 %
Groundwater 7.996 .106 Km3 22.20 % Soil moisture 6.123 .104 Km3 0.17 %
Lakes 1.261 .105
Km3
0.35 % Rivers, organic 3.602 .103 Km3 0.01 %Vapor 1.300 .104 Km3 0.04 %Total (all forms) 3.602 .107 Km3 100.00 %
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Prof.Dr.Ir. Sunjoto Dip.HE, DEA-Hydrology of Groundwater-Post Graduate Program JTSL-FT-UGM 7
d. Management of Groundwater
1). Advantages and Disadvantages of Groundwater
Table 1.7. Conjunctive use of Surface and Groundwater Resources (after Clendenenin Todd, 1980)
Advantages Disadvantages
1. Greater water conservation2. Smaller surface storage3. Smaller surface distribution system4. Smaller drainage system5. Reduced canal lining6. Greater flood control7. Ready integration with existing
development8. Stage development facilitated9. Smaller evapotranspiration losses10. Greater control over flow11. Improvement of power load12. Less danger than dam failure13. Reduction in weed seed distribution14. Better timing of water distribution15. Almost good quality of water resources
1. Less hydroelectric power2. Greater power consumption3. Decreased pumping efficiency4. Greater water salination5. More complex project operation6. More difficult cost allocation
7. Artificial recharge is required8. Danger of land subsidence
Table 1.8. Advantages and Disadvantages of subsurface and Surface Reservoir s(after US Bureau of Reclamation)
Subsurface Reservoirs Surface Reservoirs
Advantages1. Many large-capacity site available2. Slight to no evaporation loss
3. Require little land area4. Slight to no danger of catastrophic
structural failure5. Uniform water temperature6. High biological purity7. Safe from immediate radio active
fallout
Disadvantages1. Few new site available2. High evaporation loss even in humid
climate
3. Require large land area4. Ever-present danger of catastrophic
failure5. Fluctuating water temperature6. Easily contaminated7. Easily contaminated radio active
fallout
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Prof.Dr.Ir. Sunjoto Dip.HE, DEA-Hydrology of Groundwater-Post Graduate Program JTSL-FT-UGM 8
8. Serve as conveyance systems-canalsor pipeline across land of othersunnecessary
Disadvantages1. Water must be pumped
2. Storage and conveyance use only3. Water maybe mineralized
4. Minor flood control value5. Limited flow at any point6. Power head usually not available7. Difficult and costly to evaluate,
investigate and manage8. Recharge opportunity usually
dependent of surplus of surface flows9. Recharge water maybe require
expensive treatment10. Continues expensive maintenance of
recharge area or wells
8. Water must be conveyed
Advantages1. Water maybe available by gravity flow2. Multiple use3. Water generally of relatively low
mineral content4. Maximum flood control value5. Large flows6. Power head available7. Relatively to evaluate, investigate and
manage8. Recharge dependent o annual
precipitation
9. No treatment require recharge ofrecharge water
10. Little maintenance required offacilities
Table 1.9. Attributes of Groundwater ( http://www.tn.gov.in/dtp/rainwater.htm) There is more ground water than surface water
Ground water is less expensive and economic resource.
Ground water is sustainable and reliable source of water supply.
Ground water is relatively less vulnerable to pollutionGround water is usually of high bacteriological purity.
Ground water is free of pathogenic organisms.
Ground water needs little treatment before use.
Ground water has no turbidity and color.
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Prof.Dr.Ir. Sunjoto Dip.HE, DEA-Hydrology of Groundwater-Post Graduate Program JTSL-FT-UGM 9
Ground water has distinct health advantage as art alternative for lower
sanitary quality surface water.
Ground water is usually universally available.
Ground water resource can be instantly developed and used.
There is no conveyance losses in ground water based supplies.
Ground water has low vulnerability to drought.
Ground water is key to life in arid and semi-arid regions.
Ground water is source of dry weather flow in rivers and streams.
e. Data collection
1). Topographic data
2). Geologic data
3). Hydrologic data
(a). Surface inflow and outflow
(b). Imported and exported water
(c). Precipitation
(d). Consumptive use
(e). Changes in surface storage
(f). Changes in soil moisture
(g). Changes in groundwater storage
(h). Subsurface inflow and outflow
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Prof.Dr.Ir. Sunjoto Dip.HE, DEA-Hydrology of Groundwater-Post Graduate Program JTSL-FT-UGM 10
3. History
Dug well
Fig. 1.5. A crude dug well in Shinyanga Region of Tanzania. (after DHV Con. Eng.,in Todd, 1980)
The simplest dug well is crude dug well where the people go down to draw awater directly. Then brick or masonry casing dug well which were build beforecentury. The dug well with casing equipped by bucket, rope and wheel to drawwater.
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Prof.Dr.Ir. Sunjoto Dip.HE, DEA-Hydrology of Groundwater-Post Graduate Program JTSL-FT-UGM 11
Fig. 1.6. Sketch of crude dug well cross section
Fig. 1.7. A modern domestic dug well with rock curb, concrete seal and handpump. (after Todd, 1980)
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Prof.Dr.Ir. Sunjoto Dip.HE, DEA-Hydrology of Groundwater-Post Graduate Program JTSL-FT-UGM 12
Fig 1.8. Communal well equipped by recharge systems.
Fig 1.9. Traditional step well in India it is called baollis or vavadi were built from8th to 15th century (Source: Nainshree G. Sukhmani A. Design of WaterConservation System Through Rain Water Harvesting; An Excel Sheet Approach)
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Prof.Dr.Ir. Sunjoto Dip.HE, DEA-Hydrology of Groundwater-Post Graduate Program JTSL-FT-UGM 15
Crush Bore Well (Cable tool)
Crush Bore Well is a well which is build to provide drinking water by crush or
impact of a sharp cylindrical metal using cable tool to rise on the certain height
and then be released and fall down to the ground and create a hole which reachground water table. In Egypt this system was implemented since 3000 BC, in
Rome near the first century and in a small town in south French Artois, which
well had a hydraulic pressure and it created an artesian well due to the water
squirt out from the well (Fig.1.13.).
Fig. 1.13. Schematic cross section illustrating unconfined and confined aquifer(after Todd, 1980)
Rotary Bore Well
Rotary bore well was implemented since 1890 in USA to draw gas and oil and the
hole reach 2,000 meter depth. Nowadays, the rotary bore well reach 7,000
meter depth.
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Prof.Dr.Ir. Sunjoto Dip.HE, DEA-Hydrology of Groundwater-Post Graduate Program JTSL-FT-UGM 16
Springs
Spring is an outflow of ground water to the ground surface due to hydraulic
head or gravitational force (Fig. 1.14). This technique had been implanted since
before century like in Greek or Roman Kingdom. Spring water as a drinkingwater is usually be conveyed by network of pipes or canals to the town. Like in
Trowulan as capital of Majapahit Kingdom it was implemented since 12nd century
that on the site of spring was built a temple is now called Tikus Temple.
Nowadays from this temple still flowing water even though with small discharge
and this building installed by inflow-outflow and overflow system and
conveyance pipes.
Fig. 1.14. Diagrams that illustrating types of gravity springs. (a). Depressionspring. (b). Contact springs. (c). Fracture artesian spring. (d). Solution tabular spring(after Bryan, in Todd, 1980)
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Prof.Dr.Ir. Sunjoto Dip.HE, DEA-Hydrology of Groundwater-Post Graduate Program JTSL-FT-UGM 17
Above:Fig 1.15. Kaptering or springwater catcher of MajapahitKingdom in Java was build in 12century recently its called TikusTemple
Left:Fig 1.16. Water pipes system withdiameter about 60 cm, convey thewater to the pond and housing ofthe Kingdom(Photo: Prof. Hardjoso P.)
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Prof.Dr.Ir. Sunjoto Dip.HE, DEA-Hydrology of Groundwater-Post Graduate Program JTSL-FT-UGM 19
Fig 1.19. Water pond with brick structure which is called Segaran Pond, about 6hectares area where the water flow from the spring of Tikus Temple.
Fig 1.20. Ancient dug well cased by bricks in the housing of the Kingdom(Photo: Prof. Hardjoso P.)
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Prof.Dr.Ir. Sunjoto Dip.HE, DEA-Hydrology of Groundwater-Post Graduate Program JTSL-FT-UGM 20
4. Qualitative Theory
a. Early Greek Philosophers
Homer, Thales (624-546 BC) and Plato (428-347 BC) hypothesized that
springs were formed by sea water conducted through subterranean channels
below the mountains, then purified and raised to the surface.
b. Aristoteles (384-322 BC
Water is every day carried up and is dissolved into vapor and rises to the upper
region, where it is condensed again by the cold and so returns to the earth.
c. Marcus Vitruvius (15 BC)
Theory of the hydrologic cycle, in which precipitation falling in the mountains
infiltrated the Earth's surface and led to streams and springs in the lowlands.
d. Early Roman Philosophers
Lucius Annaeus Seneca (1 BC AD 65) and Pliny clarify theory of Aristoteles is
precipitation fall down in the mountain, a part of water infiltrate to the ground
as a storage water and then flow out as springs. e. Bernard Palissy (1509-1589)
He described more clearly about hydrological cycle from evaporation in the sea
till water come back again to the sea in his book: Des eaux et fontaines .
f. Johannes Kepler (1571-1630)
The earth as a big monster whose suck water from the sea, be digested and
flow out as fresh water in springs.
g. Athanasius Kircher (1602-1680)
Interaction with magma heat which causes heated water to rise through
fissures and tidal and surface wind pressure on the ocean surface which forces
ocean water into undersea.
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Prof.Dr.Ir. Sunjoto Dip.HE, DEA-Hydrology of Groundwater-Post Graduate Program JTSL-FT-UGM 21
5. Quantitative Theory
a. Pierre Perrault (1608-1690)
He observed rainfall and stream flow in the Seine River basin, confirmingPalissy's hunch and thus began the study of modern scientific hydrology. He
said that the depth of precipitation in the Seine river, France was 520 mm/y
b. Edme Mariotte (1620-1684)
In his book Des mouvements des eaux Seine River: Discharge Q = 200.000
ft 3/min, local flow is 1/6 part, evaporation is 1/3 part and infiltration is 1/3
part.
c. Edmund Halley (16561742)
He developed the equation of balance : I O = S
d. Daniel Bernoulli (1700-1782)
He stated that, in a steady flow, the sum of all forms of mechanical energy in a
fluid along a streamline is the same at all points on that streamline.
e. Jean Leonard Marie Poiseuille (1797-1869).
The original derivation of the relations governing the laminar flow of water
through a capillary tube was made by him in the early of 19 th century.
f. Reynold (1883)
The Reynolds number NR is a dimensionless number that gives a measure of
the ratio of inertial forces V2/L to viscous forces V/L2 and consequentlyquantifies the relative importance of these two types of forces for given flow
conditions.
http://en.wikipedia.org/wiki/Edme_Mariottehttp://en.wikipedia.org/wiki/Streamlines,_streaklines,_and_pathlineshttp://en.wikipedia.org/wiki/Dimensionless_numberhttp://en.wikipedia.org/wiki/Ratiohttp://en.wikipedia.org/wiki/Inertial_forcehttp://en.wikipedia.org/wiki/Viscoushttp://en.wikipedia.org/wiki/Viscoushttp://en.wikipedia.org/wiki/Inertial_forcehttp://en.wikipedia.org/wiki/Ratiohttp://en.wikipedia.org/wiki/Dimensionless_numberhttp://en.wikipedia.org/wiki/Streamlines,_streaklines,_and_pathlineshttp://en.wikipedia.org/wiki/Edme_Mariotte8/10/2019 Bahan Kuliah Aat
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Prof.Dr.Ir. Sunjoto Dip.HE, DEA-Hydrology of Groundwater-Post Graduate Program JTSL-FT-UGM 22
g. Henry Philibert Gaspard Darcy (June 10, 1803 January 3, 1858)
On his books Les fontaines publiques de Dijon (1856), he developed
mathematical equation for flow in porous media.
h. Badon Gabon (1888) and Herzberg (1901)
They developed equilibrium theory of fresh water and saline water in the
circular island with porous soil.
i. Jules Dupuit (1863)
In his book: Estudes Thoriques et Pratiques sur le mouvement des Eaux dans
les canaux dcouverts et travers les terrains permables , Dupuit developed
the formulas for groundwater flow from trench to trench with definite
distance, radial flow in unconfined and confined aquifer with definite distance.
j. Adolph Thiem (1870)
a German engineer who developed equation for the flow toward well and
infiltration galleries.
k. Gunther Thiem (1907)
In 1906, he continued Dupuit principle and his father research he developed
steady stage equation for the circular flow, using two test wells and drawdown
data, and the formula is nowaday called Dupuit-Thiem.
l. Lugeon (1930)
o Lugeon developed the double packer bore hole inflow test made at constant
head. Lugeon is a measure of transmissivity in rocks, determined by pressurized
injection of water through a bore hole driven through the rock.
m. Theis (1936)
The Theis equation was developed to determine transmissivity storage
coefficient by drawdown measuring at any given radius from the well.
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Prof.Dr.Ir. Sunjoto Dip.HE, DEA-Hydrology of Groundwater-Post Graduate Program JTSL-FT-UGM 23
n. Expansion of Theis
Cooper-Jacob simplified the Theis formula by negligible after the first two
terms, etc
o. Forchheimer (1930)
He developed the flow equation in borehole using new parameter is shape
facto r and neglected data of observation well.
p. Expansion of Forchheimer
Development of formulas of shape factors by Samsioe (1931), Dahler (1936),
Taylor (1948), Hvorslev (1951), Aravin (1965), Wilkinson (1968), Al-Dahir &
Morgenstern (1969), Luthian & Kirkham (1949), Kirkham & van Bavel (1948),
Raymond & Azzouz (1969), Smiles & Young (1965) and Sunjoto (1988-2008).
q. Taylor (1940)
Certain guiding principles are necessary such as the requirement that the
formation of the flownet is only proper when it is composed of curvilinear
squares.
r. Sunjoto (1988)
Base on Forchheimer (1930) principle, Sunjoto (1988) developed an unsteady
state radial flow equation for well which was derived by integration solution.
6. Interest of Research
Russian Groundwater in ice region Dutch Groundwater in sand dunes Japanese Hot groundwater Indonesian Recharge Systems
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Prof.Dr.Ir. Sunjoto Dip.HE, DEA-Hydrology of Groundwater-Post Graduate Program JTSL-FT-UGM 24
7. Dimension and Unit
a. Georgy System (mks)
Table 1.8. Dimension and Unit
Description Dimension Unitmass length time
m l t
gram meter second
Force
Energy
Power
Pressure
mlt-2
ml2t -2
ml2t -3
ml-1t -2
N (Newton) = kgm.s-2
J (Joule) = N.m
W (Watt) = N.m.s -1
N.m-2
b. Metric prefixes
Table 1.9. Metric preficesPrefix Symbol Factor Prefix Symbol Factor
tera T 10 12 centi c 10 -2
giga G 109 milli m 10-3
mega M 106 micro 10-6
kilo k 103 nano n 10-9
hecto h 10 2 pico p 10-12
deca da 101 femto f 10 -15
deci d 10-1 atto a 10 -18
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Prof.Dr.Ir. Sunjoto Dip.HE, DEA-Hydrology of Groundwater-Post Graduate Program JTSL-FT-UGM 25
c. Conversion of unit
Table 1.10. ConversionDescription Unit mks NoteForce
Energy
Power
1 kg
1 kg.m
1 kg.ms-1
g.N
g.J
g.W
1 N = 105 dynes
g = 9.78 m.s-2 = 32.3 ft.s -2
1 HP = 75.g.W = 734 W
d. Metric-English equivalents
Table 1.11. Metric-English equivqlent1). Length
1 cm = 0.3937 in
1 m = 3.281 ft
1 km = 0.6214 mi
2). Area
1 cm2 = 0.1550 in2
1 m2 = 10.76 ft 2
1 ha = 2.471 acre
1 km2 = 0.3861 mi2
3). Volume
1 cm3 = 0.06102 in3
1 l = 0.2642 gal = 0.03531 ft 3
1m3 = 264.2 gal = 35.31 ft 3
= 8.106 .10-4 acre.ft
4). Mass1 g = 2.205 .10-3 lb (mass)
1 kg = 2.205 lb (mass)
= 9.842 .10-4 long ton
5). Velocity
1 m/s = 3.281 ft/s
= 2.237 mi/hr
1 km/hr = 0.9113 ft/s
= 0.6214 mi/hr
6). Temperatureo C = K 273.15
= (o F 32)/1.8
7). Pressure
1 Pa = 9.8692 .10-6 atm
= 10-5 bar
= 10-2 millibar
= 10 dyne/cm2
= 3.346 .10-4 ft H 2O (4o C)
= 2.953 .10-4 in Hg ( 0o C)= 0.0075 mm Hg
= 0.1020 kg (force)/m 2
= 0.02089 lb (force)/ft 2
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Prof.Dr.Ir. Sunjoto Dip.HE, DEA-Hydrology of Groundwater-Post Graduate Program JTSL-FT-UGM 26
8). Flow rate
1 l/s = 15.85 gpm
= 0.02282 mgd = 0.03531 cfs
1 m3
/s = 1.585 .104
gpm= 22.82 mgd = 35.31 cfs
1 m3/d = 0.1834 gpm
= 2.642 .10-4 mgd = 4.087 .10-4 cfs
9). Force
1 N = 105 dyne
= 0.1020 kg (force)
= 0.2248 lb (force)
10). Power
1 W = 9.478 .10-4 BTU/s
= 0.2388 cal/s
= 0.7376 ft.lb (force)/s
11). Water quality
1 mg/l = 1 ppm = 0.0584 grain/gal
12). Hydraulic conductivity
1 m/d = 24.54 gpd/ft 2
= 1.198 darcy (water 20o C)
1 cm/s = 2.121 .104 gpd/ft 2
= 1035 darcy (water 20 o C)
13). Viscosity
1 Pa.
s = 103
centistoke= 10 poise= 0.02089 lb (force) .s/ft 2
1 m2/s = 106 centistoke = 10.76 ft 2/s
14). Gravitational acceleration, g
9.807 m/s 2 = 32.2 ft/s 2 (std., free fall)
15). Heat
1 J/m2
= 8.806 .10-5
BTU/ft2
= 2.390 .10-5 cal/cm2
1 J/kg = 4.299 .10 -4 BTU/lb (mass)
= 2.388 .10-4 cal/g
16). Density of water,
1000 kgmass/m3 = 1.94 slugs/ft 3
(when 50o F/10o C)
17). Specific weight of water,
9.807 .103 N/m3 = 62.4 lb/ft 3 (50oF/10 oC)
18). Dynamic viscosity of water,
1.30 .10-3 Pa.s=2.73 .10-5lb.s/ft 2(50o/10oC)
10-3 Pa.s = 2.05 .10-5 lb.s/ft 2 (68o F/20 o C)
19). Kinematic viscosity of water,
1.30.10-6m2/s=1.41 .10=5 ft 2/s(50 o F/10oC)
10-6 m2/s = 1.06 .10-5 ft 2/s (68 o F/20 o C)
20). Atmospheric pressure, p (std)
1.013 .105 Pa = 14.70 psia
21). Energy
1 J = 9.478 .10-4 BTU
= 0.2388 cal
= 0.7376 ft.lb (force)= 2.788 .10-7 kw.hr
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Prof.Dr.Ir. Sunjoto Dip.HE, DEA-Hydrology of Groundwater-Post Graduate Program JTSL-FT-UGM 27
e. Legends
1). Density
Symbol :
Dimension : ml-3 Unit : kgmass.m-3 or slug.ft -3
Detail:
1 slug = 14.60 kgmass
1 feet = 0.305 m
1 slug.ft -3 = 514.580 kgmass.m-3
In practical use: pure water = 1,000 kgmass.m-3 = 1.94 slug.ft-3
sea water = 1,026 kgmass.m-3 = 1.99 slug.ft-3
Table 1.12. Density of pure water in kg mass.m-3 dependent temperature t o Ct t t t 0
2 4
6
8
999.8679
999.9267 1000.0000
999.9081
999.8762
10
12 14
16
18
999.7277
999.5247 999.2712
998.9701
998.6232
20
22 24
26
28
998.2323
997.7993 997.3256
996.8128
996.2623
30
32 34
36
38
995.6756
995.0542 994.3991
993.7110
992.9936
2). Specific weight
Symbol : = .g Dimension : ml-2t -2
Unit : N.m-3 atau lbs.ft -3
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Prof.Dr.Ir. Sunjoto Dip.HE, DEA-Hydrology of Groundwater-Post Graduate Program JTSL-FT-UGM 28
3). Specific Gravity
Symbol : s s = / w = / w
Dimension : -
Unit : -
4). Viscosity
(a). Dynamic viscosity
Symbol :
Dimension : ml-1t -1
Unit : N.s.m-2
1 N.s.m-2 = 10 poise; 478 poise = 1 lbs.ft -2
Table 1.13. Dynamic viscosity of water in 10-2 poisses dependent temperature t o Ct t t t
0
2
4 6
8
1.7921
1.6728
1.5674 1.4728
1.3860
10
12
14 16
18
1.3077
1.2363
1.1709 1.1111
1.0559
20
22
24 26
28
1.0050
0.9579
0.9142 0.8737
0.8360
30
32
34 36
38
0.8007
0.7679
0.7371 0.7085
0.6814
(b). Cinematic viscocity
Symbol :
Dimension : l2t -1
Unit : m2s-1 or stokes
1 m2s-1 = 10-4 stokes
1 ft 2s-1 = 929 stokes
= /
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Prof.Dr.Ir. Sunjoto Dip.HE, DEA-Hydrology of Groundwater-Post Graduate Program JTSL-FT-UGM 29
5). Surface Tension
Symbol :
Dimension : mt-2
Unit : N.m-1 water/air = 0.074 N.m-1
Table 1.14. Relationship of , and of watert = 10o C; p = atm t = 60o F; p = atm
Water Air Unit Water Air Unit
1000
1.3 .10-2
1.3 .10-6
1.37
1.8 .10-4
1.3 .10-5
kgmass.m-3
poise
m2s-1
1.94
2.3 .10-5
1.2 .10-5
2.37 .10-3
3.7 .10-7
1.6 .10-4
slug.ft -3
lbs.s.ft -2
ft 2s-1
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II. GENERAL DESCRIPTION
1. Terminology
a. Aquifer
The origin of aqua is water and ferre is contain.
b. Aquiclude
The origin of claudere is to shut.
c. Aquifuge
The origin of fugere is to expel.
d. AquitardThe origin of tard is late.
2. Vertical Distribution
Fig. 2.1. Diagram of zones in permeable soil
Ground surface
Soil water zone
Intermediatevadoze
zone
Capillary zone
Saturated zone
ZONE OFAERATION
ZONE OFSATURATION
VADOZEWATER
GROUND /PHREATICWATER
Groundwater table
Impermeable
P
e
r
m
e
a
b
l
e
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Prof.Dr.Ir. Sunjoto Dip.HE, DEA-Subsurface Hydrology-Post Graduate Program JTSL-FT-UGM=2012 32
Table 2.1. Capillary rise in samples of unconsolidated materials (after Lohman in Todd,1980)
Soils Type Grain size (mm) Height of capillary (cm)
Fine gravel
Very coarse sand
Coarse sand
Medium sand
Fine sand
Silt
Silt
5 - 2
2 - 1
1 0.5
0.5 0.2
0.2 0.1
0.1 0.05
0.05 0.002
2.50
6.50
1.50
24.60
42.80
105.50
200.00
Table 2.3. Capillary rice of some soils type (Murthy, 1977)Soils Type Size of particles (mm) Capillary rise (cm)
Sand, coarse
Sand, medium
Sand, fine
Silt
Clay, coarse
Clay, colloid
2.00 - 0,60
0.60 0.20
0.20 0.06
0.06 0.002
0.002 0.0002
< 0.0002
1.50 5
5 15
15 - 50
50 - 1,500
1,500 15,000
>15,000
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Prof.Dr.Ir. Sunjoto Dip.HE, DEA-Subsurface Hydrology-Post Graduate Program JTSL-FT-UGM=2012 33
b. Zone of Saturation
1). Specific retention (Sr)
Sr = Wr / V
Wr : the rest water volume after drainage
V : total volume of soil
2). Specific yield (Sy)
S y = W y / V
W y : volume of water which be drained
= Sr + S y
c. Solid Liquid and Air System
Solid phase : geometricly difficult be soluble
Liquid phase : solution organic & unorganic
Air phase : vapor
Fig. 2.3. Diagram of solid, water and air relationship
V
Vv
Va
Vw
Vs
Wa
Ww
Ws
1
air
water
solid
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Prof.Dr.Ir. Sunjoto Dip.HE, DEA-Subsurface Hydrology-Post Graduate Program JTSL-FT-UGM=2012 34
1). Void ratio (e)
The ratio of the volume of voids (Vv) to the volume of solids (Vs), is defined as void
ratio, and:
=
2). Porosity (n)
The ratio of the volume of voids (Vv) to the total volume (V), is defined as porosity,
so:
= 100%
3). Degree of saturation (S)
The ratio of volume of water (Vw) to the volume of voids (Vv) sis defined as degree
of saturation so:
= 100%
4). Water content (w)
The ratio of weight of water (Ww) in the voids to the weight of solids so:
= 100%
5). Unit Weight
a). Unit weight of water ( w)
The ratio of weight of water to the volume of water in the same temperature ( w)
and (o) is designated as unit weight of water at 4 o C.
= 1 3 = 1 3 = 1 3 = 1000 3
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Prof.Dr.Ir. Sunjoto Dip.HE, DEA-Subsurface Hydrology-Post Graduate Program JTSL-FT-UGM=2012 36
3. Type Aquifer
gs
gwt
gwtH
e. Suspended aquifer
Note:gs : ground surfaceps : piezometric surfacegwt : groundwater table
gwt : groundwater table ofperched waterD : thickness of aquiferH : depth of groundwaterK : coefficient of permeability
Note:Compare to Todd (1980) page 44 about leaky aquifer, which the elevation ofgwt is higher than ps.
Fig. 2.4. Types of aquifers
gs
gwt
K=0
gs
gwt
KD H
ps
D=H K
psK 1
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Prof.Dr.Ir. Sunjoto Dip.HE, DEA-Subsurface Hydrology-Post Graduate Program JTSL-FT-UGM=2012 38
The equation becomes:
= =
+
= + = + = + = (3.4)The essential point of above equation is that the flow through the soils is also
proportional to the first power of the hydraulic gradient i as propounded by Posseuilles
Law.And the discharge is by Darcys equation is:
=
(3.5)
where,
Q : dischargeK : coefficient of permeabilityA : section area of aquiferdh : difference water elevationdl : length of aquifer
2). Similar equations Fouriers Law on heat transfer {Jean Baptiste Joseph Fourier (1768 1830)}:
= (3.6) where,
H : rate of heat flowk : thermal conductivityA : cross section areadT : temperature differencedx : thickness
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Prof.Dr.Ir. Sunjoto Dip.HE, DEA-Subsurface Hydrology-Post Graduate Program JTSL-FT-UGM=2012 39
Ohms Law on electrical current flow { George Simon Ohm (1787 - 1854)}:
= (3.7) where,
I : currentC : coefficient of conductivitya : sectional area of conductordv : drop in voltagedl : length of conductor
3). Validity of Darcy Law
= (3.8)
It can be written in other equation as:
=
(3.9)
where,NR : Reynolds Number
D : diameter of pipe : density of water : flow velocity : viscosity of fluid : unit weight of fluidg : acceleration of gravity
Experiments show that Darcys law is v alid for NR < 1 and does not depart seriously up
to NR = 10, and this value represents an upper limit to the validity of Darcys law (Todd,1980).
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Prof.Dr.Ir. Sunjoto Dip.HE, DEA-Subsurface Hydrology-Post Graduate Program JTSL-FT-UGM=2012 41
= = 2
8 =
2
32 . (3.28)
3). Temperature
The coefficient of permeability K is product of k which is dependent on
temperature and a function of the void ratio e, and t he value of k is expressed :
= 1,16 22 . = (3.29) Where, C is constant which is independent of temperature and the expression ofK may now be as below and K varies asw/ .
=
. . ( ) .
(3.30)
4). Structure and stratification
Fig 3.1. Diagram of soil layers structure
a). Flow in the Horizontal Direction
Q = V.A = V. Z = K.i.Z
Q = (V1.Z1 + V2.Z2 + + Vn-1.Zn-1 + Vn.Zn)
Q = (K1.i.Z1 + K2.i.Z2 + + Kn-1.i.Zn-1 + Kn.i.Zn)
= ( + + + ) (3.31)
K 1
K 2
K n-1
K n
Z
Z2
Zn-1
Zn
Z
K v
K h
V1.i.K 1
V2.i.K 2
Vn-1.i.K n-i
Vn.i.K n
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Prof.Dr.Ir. Sunjoto Dip.HE, DEA-Subsurface Hydrology-Post Graduate Program JTSL-FT-UGM=2012 42
b). Flow in the Vertical Direction
The hydraulic gradient is h/Z and:
=
=
1 1 =
2 2 =
If h1, h 2 hn are the loss of heads in each of the layers, therefore:
H = h1 + h2 + hnor,
H = Z1h1 + Z2H2+ ..ZnHn
Substitution:
=
+
+ +
(3.32)
b. Method of Determination
1). Laboratory Method
a). Constant head permeability method
The coefficient of permeability K is computed:
=
(3.33)
= (3.34)
b). Falling head permeability method
The coefficient of permeability K can be determined on the basis of drop in
head (h o h 1 ) and the elapse time ( t 1 t o ).
=
=
. .
(3.35)
= ( ) (3.38) when A = a the equation be:
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Prof.Dr.Ir. Sunjoto Dip.HE, DEA-Subsurface Hydrology-Post Graduate Program JTSL-FT-UGM=2012 43
=( ) (3.39)
where:
K : coefficient of permeability
L : length of sampleA : cross section area of samplea : cross section area stand pipeho h1 : head of water in observation well 1 and 2 respectivelyt o t 1 : duration of flow in observation well 1 and 2 respectively
c). Computation from consolidation test data
In the case of materials of very low permeability with K less than 10 -6 cm/s
consolidation test apparatus with permeability attachment may be used. The
coefficient of permeability K of sample can be computed from equation:
=. . (3.40)
where,K : coefficient of permeabilityL : length of sampleA : cross section area of sampleQ : discharge in certain time th : average headt : duration of flow
d). Computation from grain size distribution
On the basis of Poiseuilles Law the coefficient of permeability can be
computed:
=
2 (3.41)
According to Allen Hazen (1911) in Murthy (1977) the empirical equation can
be computed as:
= 102 (3.42) where,
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K : coefficient of permeability (cm/s)C : a factor (100
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Fig 3.2. Bore hole in some conditions
(1). Murthy (1977)
The coefficient of permeability is calculated by making use of formula:
=0.18
where:Q : discharge (L 3/T)K : coefficient of permeability (L/T)H : hydraulic head (L) Fig. 3.2.
Note:Compare to Forchheimer (1930) that Q= FKH and to Harza (1935), Taylor (1948) and
Hvorslev (1951) that F = 5,5 r. And Sunjoto (2002) developed the formula for the same
condition that F = 2r.
(2). Forchheimer (1930)
Forchheimer (1930) proposed to find a coefficient of permeability (K) by bore hole with
certain diameter and depth.
=( ) (3.49)
Q Q Q & h p Q & h p
hw
hw
hw
hw
(1). H=h w (2). H=h w (3). H=h w+ hp (4). H=h w+ hp
H b
Hg
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where:K : coefficient of permeability (L/T)R : radius of well (L)F : shape factor (L) (F = 4 R, Forchheimer, 1930)t 1 t 2 : time of the measurement respectively (T)
h1 h2 : height of water of the measurement respectively (L)As : cross section area of well (L 2 , As = R2)
c). Partial permeable casing bore hole test
(1). Suharyadi (1984)
There are two conditions of hydraulic head (Fig. 3.3) as:
The hole is submerged in groundwater:
H = difference of groundwater table to the water elevation test
The hole above the groundwater table:
H = Depth of water test on the hole minus half of permeable hole length
Fig. 3.3. Hydraulic head dimension on bore hole test according to Suharyadi
(1984)
Q
Q
(2). The hole test above groundwater table
L L
2R 2R
gwtHw
Hw
(1). The hole test below groundwater table
gwt
(H=H w) H=H c+ / 2L
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Prof.Dr.Ir. Sunjoto Dip.HE, DEA-Subsurface Hydrology-Post Graduate Program JTSL-FT-UGM=2012 47
The coefficient of permeability can be computed by:
=2.302 = 2 (3.44)
where,
K : coefficient of permeabilityL : length of permeable partH : Hydraulic head (L R)R : radius of casing
d). Uncasing bore hole test
(1). Pecker test
Suharyadi (1984)
=. = (3.50)
=
+ (3.51)
Fig. 3.6. Hydraulic head dimension on packer test (after Suharyadi, 1984)
Q and H 2 Q and H 2 Q and H 2 Q and H 2
(b). One pecker testwhich zone test is
above groundwater table
(c). Two peckers testwhich zone testis submerged
(d). Two peckers testwhich zone test is
above groundwater table
L
2R
L L
2R 2R 2R
(a). One pecker testwhich zone testis submerged
gwt
gwt
gwt
gwt
H1
H1
1/2L
H1
H1
1/2LL
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Prof.Dr.Ir. Sunjoto Dip.HE, DEA-Subsurface Hydrology-Post Graduate Program JTSL-FT-UGM=2012 48
(2). Boast and Kirkham (in Todd, 1980)
= . (3.52)
Fig. 3.7. Diagram of auger hole and dimensions for determining coefficient ofpermeability (after Boast and Kirkham, in Todd, 1980)
(3). Sunjoto (1988)
= 1
2
(8.53 )
where:H : depth of hollow well (L)F : shape factor (L)K : coefficient of permeability (L/T)Q : inflow discharge (L3/T), dan Q = C I AC : runoff coefficient of roof ( )I : precipitation intensity (L/T)
A : roof area (L2
)
Note:
When steady flow condition (8.53) become F =Q/KH The solution of this equation by trial and error.
Lw y
2r w H
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Table 3.1. Value of C after Boast and Kirkham (in Todd, 1980)Lw/r w
y/Lw
(H-L w)/Lw for Impermeable Layer H-Lw (H-L w)/Lw for InfinitelyImpermeable Layer
0 0.05 0.1 0.2 0.5 1 2 5 5 2 1 0.5
1 1.00 447 423 404 375 323 286 264 255 254 252 241 213 1660.75 469 450 434 408 360 324 303 292 291 289 278 248 1980.50 555 537 522 497 449 411 386 380 379 377 359 324 264
2 1.00 186 176 167 154 134 123 118 116 115 115 113 106 910.75 196 187 180 168 149 138 133 131 131 130 128 121 1060.50 234 225 218 207 188 175 169 167 167 166 164 156 139
5 1.00 51.9 48.6 46.2 42.8 38.7 36.9 36.1 35.8 35.5 34.6 32.40.75 54.8 52.0 49.9 46.8 42.8 41.0 40.2 40.0 39.6 38.6 36.30.50 66.1 63.4 61.3 58.1 53.9 51.9 51.0 50.7 40.3 49.2 466
10 1.00 18.1 16.9 16.1 15.1 14.1 13.6 13.4 13.4 13.3 13.1 12.60.75 19.1 18.1 17.4 16.5 15.5 15.0 14.8 14.8 14.7 14.5 14.00.50 23.3 22.3 21.5 20.6 19.5 19.0 18.8 18.7 18.6 18.4 17.8
20 1.00 59.1 55.3 53.0 50.6 48.1 47.0 46.6 46.4 46.2 45.8 44.60.75 62.7 59.4 57.3 55.0 52.5 51.5 51.0 50.8 50.7 50.2 48.90.50 76.7 73.4 71.2 68.8 66.0 64.8 64.3 64.1 63.9 63.4 61.9
50 1.00 1.25 1.28 1.14 1.11 1.07 1.05 1.04 1.03 1.020.75 1.33 1.27 1.23 1.20 1.16 1.14 1.13 1.12 1.110.50 1.64 1.57 1.54 1.50 1.46 1.44 1.43 1.42 1.39
100 1.00 0.37 0.35 0.34 0.34 0.33 0.32 0.32 0.32 0.310.75 0.40 0.38 0.37 0.36 0.35 0.35 0.35 0.34 0.340.50 0.49 0.47 0.46 0.45 0.44 0.44 0.44 0.43 0.43
Table 3.2. Coefficient of Permeability of some Soils (Casagrande and Fadum)
K (cm/sec) Soils type DrainageCondition
Recommended method ofdetermining K
101 - 102 Clean gravels Good Pumping Test
101 Clean sand Good Constant head or Pumping test
10-1 10-4 Clean sand and gravel
mixtures
Good Constant head, Falling head
or Pumping test10-5 Very fine sand Poor Falling head
10-6 Silt Poor Falling head
10-7 10-9 Clay soils Practicallyimpervious
Consolidation test
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3). Lugeon Test
Maurice Lugeon(July 10, 1870 - October 23, 1953) was a Swiss geologist, and the
pioneer of nape tectonics. He was a pupil of Eugne Renevier. The Lugeon test,
extensively used in Europe, is a special case of double packer bore hole inflow test made
at constant head.
Lugeonis a measure of transmissivity in rocks, determined by pressurized injection
of water through a bore hole driven through the rock.
o One Lugeon (LU) is equal to one liter of water per minute injected into 1 meter
length of borehole at an injection pressure of 10 bars.o 1 Lugeon Unit = a water take of 1 liter per meter per minute at a pressure of 10
bars.
o Lugeon value : water take (liter/m/min) x 10 bars/test pressure (in bars)
The Lugeon unit is not strictly a measure of hydraulic conductivity but it is a good
approximation for grouting purposes and 1 Lugeon is approximately equivalent to 1x10 -5
cm/s or 1x10 -7 m/s.
The three successive test runs, each of 5 minutes duration enable a rough
assessment of the water behavior.
Analysis:
This test will be analyzed by principle of:
Forhheimer (1930) for steady flow condition
= = Sunjoto (2010) for the shape factor of each condition (Fig.3.8.)
http://en.wikipedia.org/wiki/July_10http://en.wikipedia.org/wiki/1870http://en.wikipedia.org/wiki/October_23http://en.wikipedia.org/wiki/1953http://en.wikipedia.org/wiki/Swisshttp://en.wikipedia.org/w/index.php?title=Nappe_tectonics&action=edithttp://en.wikipedia.org/wiki/Eug%C3%A8ne_Renevierhttp://en.wikipedia.org/w/index.php?title=Lugeon&action=edithttp://en.wikipedia.org/w/index.php?title=Lugeon&action=edithttp://en.wikipedia.org/wiki/Eug%C3%A8ne_Renevierhttp://en.wikipedia.org/w/index.php?title=Nappe_tectonics&action=edithttp://en.wikipedia.org/wiki/Swisshttp://en.wikipedia.org/wiki/1953http://en.wikipedia.org/wiki/October_23http://en.wikipedia.org/wiki/1870http://en.wikipedia.org/wiki/July_108/10/2019 Bahan Kuliah Aat
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Fig. 3.8. Schematic of condition of well and packers location.
Data:o Radius of hole : R = 0.05 m (according to Suharyadi that diameter
of hole = 3 or 4 inches)o Hydraulic head : H = 10 bar = 102 mo Discharge : Q = 1 l/min = 1.66667 .10-05 m3/so Length of hole : L = 1 mo The three successive test runs, each of 5 minutes duration, in constant
dischargeo Hole diameter usually used:
Drill bit : 73 mm Drill hole : 76 mm Casing : 85 or 87 mm
To compute the value of Shape Factor, Sunjoto (2010) proposed formula for three
conditions of well as: Condition of (a) well Fig. 3.8.(a):
=2
2( + 2 ) + 2 2 + 1 (3.53)
Q in 10 bar Q in 10 bar in 10 bar in 10 bar
(2). Condition of well b. (3). Condition of well b. (4). Condition of well c.
L
2R
L L L
2R 2R 2R
(1). Condition of well a.
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Prof.Dr.Ir. Sunjoto Dip.HE, DEA-Subsurface Hydrology-Post Graduate Program JTSL-FT-UGM=2012 52
Condition of well (b) Fig. 3.8.(b):
=2
( + 2 )+
2
+ 1
(3.54)
Condition of well (c) Fig. 3.8.(c):
=2
( + 2 )2 + 2 2 + 1 (3,55)
Hole with diameter 76 mm
Shape Factor of each hole (3.53, 3.54, 3.55):
=2 1
2(1 + 2 0.0376 )0.0376 + 2 10.0376 2 + 1 = 1.33570 m
=2 1
( 1 + 2 0.0376 )0.0376 + 10.0376
2+ 1
= 1.56643 m
=2 1
(1 + 2 0.0376 )2 0.0376 + 12 0.0376 2 + 1 = 1.89308 m
The test will be measured on the constant discharge or in steady flow condition, sothe computation of the coefficient of permeability using Forchheimer formula (1930):
Condition of well a.:
=1.66667 .10 51.33570 102
= 1.22332 .10 7 m/s
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Prof.Dr.Ir. Sunjoto Dip.HE, DEA-Subsurface Hydrology-Post Graduate Program JTSL-FT-UGM=2012 54
IV. PARALLEL FLOW
Dupuit (1863) developed the formulas for groundwater flow from trench to trench with
definite distance, radial flow in unconfined and confined aquifer with definite distance.
The assumption of simplification of this formula are (Castany, 1967):
Steady flow Incompressible water and soil Equipotentiales are plane In accordance with Darcys Law Vertical flow is neglected Homogeneous and isothrope
Distance of flow is constant
1. Free Aquifer
In this case the flow through the permeable layer as unconfined or free aquifer like inFig. 4.1.
Fig. 4.1. Flow through unconfined embankment
a. Discharge Equation
Dupuit (1863) in Castany (1967):
V = K.i i = sin
H1
L
H2
x
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Prof.Dr.Ir. Sunjoto Dip.HE, DEA-Subsurface Hydrology-Post Graduate Program JTSL-FT-UGM=2012 55
= 2 + 2 =
2 + 2
=
1 +
2
Due to the assumption of vertical velocity is neglected so :
1 + 2 = 1 = (4.1) Darcy.s Law (1856)
= = (4.2) so:
= = = = . ( . 1) . =
. .
= . 0 = . 12 . 2 21
=12
( 12 22 )
=
(4.3)
where,H1 : depth of upstreamH2 : depth of downstream
L : length of aquiferh : height of flow line in distance of x
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Prof.Dr.Ir. Sunjoto Dip.HE, DEA-Subsurface Hydrology-Post Graduate Program JTSL-FT-UGM=2012 56
x : distance from upstreamb : length of embankmentK : coefficient of permeability
b. Flow line Equation
Continuity general flow:
2 2 22 + 2 22 + = 0 In this case that h is function of x and N = 0 (precipitation) and equation becomes:
2 2
2 = 0
and the general solution is:
2
= 2 = + Boundary condition:
x = 0 h = H1 ; H12 = B
x = L h = H2 ; h2= AL + B
= 22 12 So the equation of flow line is:
= + (4.4)
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Prof.Dr.Ir. Sunjoto Dip.HE, DEA-Subsurface Hydrology-Post Graduate Program JTSL-FT-UGM=2012 57
2. Confined aquifer
In this case the flow through a permeable layer as laid down under impermeable layer
likes Fig. 4.2.
Fig. 4.2. Flow through confined embankment
a. Discharge
Laplace equation:
= + + ; = . . = = 2 1 = + 1 2 = +
(4.5)
where,D : thickness of aquiferb : width of aquifer
H1
D H2
L
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Prof.Dr.Ir. Sunjoto Dip.HE, DEA-Subsurface Hydrology-Post Graduate Program JTSL-FT-UGM=2012 58
b. Flow line equation
Laplace Equation
2
2 = 0 dx = A = A x + B
Boundary condition:
= 0 = 1 1 = = = 2 2 = + 1
=
2
1
= + (4.6) The examples below are:
Unconfined aquifer embankment Horizontal stratified embankment Embankment with rain Embankment with evaporation Embankment between trench and impermeable zone Confined aquifer Confined aquifer with variation thickness
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Prof.Dr.Ir. Sunjoto Dip.HE, DEA-Subsurface Hydrology-Post Graduate Program JTSL-FT-UGM=2012 59
EXAMPLES
1. Unconfined aquifer embankment
Fig. 4.3. Flow through unconfined embankment
Data:K = 10-6 m/s, H1 = 6 m, H2 = 2 m and L = 40 m.
Compute:a). Flow line equationb). Discharge through the embankmentc). Height of flow line in 20 m from upstream
Answer:
a. Flow line equationGeneral continuity equation:
2
2 2
2 2 +2 2
2 2 +
= 0
Due to the flow only in one direction and no precipitation so the equation onlyh function x and N = 0 and differential equation becomes:
2 2
2 = 0
H1
L
H2
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Prof.Dr.Ir. Sunjoto Dip.HE, DEA-Subsurface Hydrology-Post Graduate Program JTSL-FT-UGM=2012 60
The solution is:2
=
2 =
+
Boundary condition:
x = 0 h = H1 ; H12 = B
x = L h = H2 ; H22 = AL + B
H22 = AL + H12
= 2
2
12
Using data above the equation will be:
2 = + = +
2 = 2262
40 + 62
2 = 0.80 + 36 b. Discharge through the embankment
=
=
=
2
( 2 )
( 2 )= 2
2 12 + 12 = 22 12 So,
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Prof.Dr.Ir. Sunjoto Dip.HE, DEA-Subsurface Hydrology-Post Graduate Program JTSL-FT-UGM=2012 61
=( 2
2 12 )2 For b = 1 meter so:
=106 . 1. (62 22 )2.40 = 4.10 7 3
b. Height of flow line in 20 m from upstream:
2 = 1.25 + 36 2 =
0.80 20 + 36 = 20
h = 4.472 m
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Prof.Dr.Ir. Sunjoto Dip.HE, DEA-Subsurface Hydrology-Post Graduate Program JTSL-FT-UGM=2012 62
2. Horizontal stratified embankment
Fig. 4.4. Flow through unconfined stratified embankment
Data:K1 = 10-4 m/s, K2 = 10-5 m/s, K3 = 10-6 m/s, L1 = 80 m, L2 = 70 m,L3= 40 m, H1 = 10 m, H4 = 4 m.
Compute:a). Discharge through the embankmentb). Height of H 2 and H3 c). Flow line equation equation
Answer:
Q = 7,368 m3/s
H2 = 9,926 m
H3 = 9,470 m
Flow line equation equation for each strata
Layer I : h12 = -0,015 x1 + 100
Layer II : h22 = -0,147 x2 + 98,525
Layer III : h32 = -0,147 x3 + 89,681
L1 L2 L3
H1 H2 H3 H4
K 1 K 2 K 3
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The discharge Q = Q 1 = Q2 = Q3
The flow line for each layer are:
Layer I:
12 = 2
2 121
1 + 12
Layer II:
22 = 3
2 222
2 + 22
Layer III:32 = 4
2 323 3 + 32 The discharge for each layer are:
Layer I:
1 =1 ( 1
2
22 )
2 1 12
22
=2 1
1
Layer II:
2 =2 ( 2
2 32 )2 2 22 32 = 2 22 Layer III:
3 =3 ( 3
2 42 )2 3 32
42 =
2 33
The defference is:
12 42 = 2 11 + 22 + 33
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=( 1
2 42 )2 1
1+ 22 + 33
= (10 2 + 4 2 )2
80104 + 70105 + 40106
= 1.213 10 6 3 Te flow line equation needs the values of H 2 and H3 are:
10 2 22 = 2 1.213 10 6 801 1 0 4 2 = 9.915 9.915 2 2
2 = 2 1.213 10 6 70
1 1 0 5 3 = 9.018 The flow line for each layer are:
Layer I:
12 =
9.915 2 10 280 1 + 10 2 12 = 0.212 1 + 100 Layer II:
22 =
9.018 2 9.915 270 2 + 9.915 2 22 = 0.243 2 + 98.307 Layer III:
32 =
429.018 240 3 + 9.018 2 32 = 1.633 3 + 81.324
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3. Embankment with rain
Fig. 4.5. Flow through unconfined embankment with precipitation
Data:H1 = 8 m, H2 = 2 m, L = 50 m, K = 10-6 m/s, N = 4,8 10-8 m/s.
Compute:a). Flow line equationb). Highest elevation of flow line (when horizontal)c). Discharge through the embankment
Answer:a. Flow line equation
2 2 22 + = 0 2 22 = 2 2 = 2 + 2 = 2
2 =
2 2
2+
+
2 =
2
+
+
Boundary condition:
= 0 = 1 = 12
H1
L
H2
N
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= = 2 22 = 2 + +
=
2
+
22 =
12 1
= + 22 = 12
Flow line equation becomes:
2 = 2 + + 22 12 + 12 2 = 12 ( 2212 ) + ( )
Substitution of the data:
2 =4.80 10 8
106 . (50 ) + 22 8250 + 8 2 2 = 4,80. 10 2 . 2 + 1,20. + 64
b. Highest flow line.Location of highest flow line
2
= 0 2
=
4.80 . 10
2 .2 + 1.20 = 0
9.60 . 10 2 . + 1.20 = 0 = 1.209.60 . 10 2 12.50 So x = 12.50 m from upstream/left ward
The heigt of water table when in horizontal condition or when x = 12.50 m
2=
4.80 . 10
2. 12.50
2+ 1.20 .12.50 + 64 = 71.50
h = 8,46 m
c. Discharge through the embankment.
= = = 12 2
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= 12 12 ( 12 22 ) + ( ) Boundary condition:
= = ( 12 = 22 )2 = 2 =
106 . 1. (82 22 )2 5 0 = 4.80 . 10 8 . 1. 502 = . = = =
( 12 = 22 )
2+
2
=106 . 1. (82 22 )2 5 0 + 4.80 . 10 8 . 1. 502
= , . = d. The maximum elevation of flow line when H 1 = H 2 = H
2 = 2 + ( ) The extreem point is:
2
= 0 2
= + ( ) = 0 2
=
2
+ . 2 2
2=
2
+ 2
4
= 2 + 24
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4. Embankment with evaporation
Fig 4.6. Flow through unconfined embankment with evaporation
Data:H = H1 = H2 = 5 m, K = 0,25 10-3 m/s, E = 0,12 10-6 m/s
Compute:Lowest elevation of flow line
Answer:
Flow line equation,
Darcys Law:
= Continuity equation:
= =
Boundary condition:
x = 0 h = ho
x = L, h = H
H1
L
H2
E
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= 2 = 2
2 = 2 Substitution of data:
25 2
2 =
0.12 . 10 60.25 . 10
3 150
2 ho2 = 24.20 ho = 3.77m
So te lowest point is 3.77 m.
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6. Unconfined aquifer
Fig. 4.8. Flow through confined embankment
Data:
H1 = 15 m, H2 = 10 m, D = 7 m, L = 100 m, K = 10-5 m/s
Compute:a. Flow line equationb. Discharge through the embankment
Answer:a. Flow line equation
Laplace equation,
2
2 = 0 = = + Boundary condition:
= 0 =
1
1
=
= = 2 2 = + 1 = 2 1
H1
D H2
L
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Flow line equation becomes:
=10 15100 + 15 = 0.015 + 15
b. Discharge
= = 2 1 = 1 2 = 1. 10 5 .1.7 15 10100 = 3.50 . 10 6 3
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7. Unconfined aquifer with variable thickness
Fig. 4.9. Flow through confined embankment
Data:
H1 = 8 m, H2 = 6 m, D1 = 2 m, D2 = 3 m, L = 100 m, K 1.10-7 m/s
Compute:a. Discharge through the embankmentb. Flow line equation
Answer:
a. Discharge through the embankment
= = 2 +
32100 = 2 + 0.01 =
(2 + 0.01 ) =
= ( 2 + 0.01 ) = (2 + 0.01 ) Solution by integration:
== .1.107 . (2 + 0.01 ) +
H1
D2 H2
L
D1
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Boundary condition:
= 0 1 = 1 = 8 = .1.107 2 + = 100
2=
2
= 6 =
.1.10
7 3 +
1 2:2 = .1.107 32
= . . b. Flow line equation
Substitute q:8 = .1.107 . 2 + = 11.42
Substitute C:
= .1.107 . (2 + 0.01 ) + So the flow line equation will be:
= . . .( + . ) + .
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V. RADIAL FLOW
Assumptions for the equations are (Dupuit-Thiem):
The soils surrounding the well is assumed homogeneous
The flow towards the well is assumed as steady, laminar, radial and horizontal
The horizontal velocity is independent of depth
The ground water table is assumed as horizontal in all direction
The hydraulic gradient at any point on the drawdown is equal to the slope of
the tangent at the point. According to Castany G. (1967) that value is sinus at
the point.
1. Unconfined aquifer
a. Dupuit (1863)
Fig. 5.1. Circular unconfined aquifer
Let h be the depth of water at radial distance r . The area of the vertical cylindrical
surface of radius r and depth h through which water flow is:
A = 2rh (5.1)
r w r
R
hw
h H
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The hydraulic gradient is:
= (5.2)
Discharge of inf low when the water levels in the well remain stationary (Darcys Law)
V = Ki (5.3)
Q = KiA (5.4)
Substituting for Eqn (1) and (2) for (3), the rate inflow across the cylindricalsurface is:
=
2
(5.5)
The equation for discharge outflow from pumping is:
= ( ) (5.8) The equation for permeability of soil is:
=( ) (5.8 )
where,H : depth of water outside of aquifer layerhw : depth of water at face of pumping wellR : radius of outside of aquifer layer
rw : radius of pumped well
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b. Dupuit-Thiem
1). According to UNESCO (1967),
G. Thiem (1906) based on Dupuit and Darcy principle developed a formula
of pumping and the formula is called Dupuit-Thiem.Let h be the depth of water at radial distance r (Fig. 5.2.). The area of thevertical cylindrical surface of radius r and depth h through which waterflow is:
Fig. 5.2. Pumping in unconfined aquifer
Area of cylinder of piezometric h and radius r: A = 2rh
The hydraulic gradient is: =
Darcys Law: V = Ki and Q = KiA
Substituting, so the rate inflow across the cylindrical surface is:
= 2 (5.9) Rearranging the terms, so:
r 1 r
r 2
h1 hh2
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=2
The equation for permeability of soil is:
= (5.12 ) The equation for discharge outflow from pumping is (Fig, 5.2):
Dupuit-Thiem Formula for the full penetration well in free aquifer:
=
(5.12)
where,Q : discharge of pumpingK : coefficient of permeabilityD : thickness of aquifer layerr1 r 2 : distance from well to observation well 1 and 2 respectivelyh1 h2 : head of water in observation well 1 and 2 respectively
2). According to Castany (1967)G. Thiem (1906) based on Dupuit principle developed a formula of pumpingin unconfined aquifer and the formula is called Dupuit-Thiem (Fig. 5.3.).
Darcys law:
= 2 (5.13)
dr/dh = tg (5.14)
= 2 . tg (5.15)
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Fig. 5.3. Pumping in unconfined aquifer
tg= 1 2r 2 r1 (5.16) For first permanent regime:
= 2 1 1 . tg (5.17) For second permanent regime:
= 2 1 11 . tg
1 (5.18)
Dupuit-Thiem equation for the full penetration well in free aquifer:
= ( + )( ) (5.20)=
( + )(
)
(5.20 )
where:
Q : discharge of pumpingK : coefficient of permeabilityr1 r 2 : distance from well to observation well 1 and 2 respectively 1 2 : drawdown in observation well 1 and 2 respectively
r 1 r 2
h1
h2
2 1
w
hw
r w
R i
H
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3). According to Murthy V.N.S. (1977)Murthy developed the formula for unconfined aquifer by other parameters
and can be found as (Fig.5.3.):
=( ) (5.21)
=( ) (5.21 )
If we write hw = (H - w) where w is the depth of maximum drawdown in the
test well or pumped well so (Castany, 1967): = = ( ) (5.22) =
(
)
(5.22 )
where:
Q : discharge of pumpingK : coefficient of permeabilityRi : radius of influencerw : radius of pumped wellH : depth of water before pumping w : maximum drawdown (on well)
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2. Confined aquifera. Dupuit (1863)
Fig. 5.4. Circular unconfined aquifer
=
= . = 2 = 2 ] = 2 ]
Dupuit (1863) formula for full penetration well on confined aquifer:
= (5.23) =
( ) (5.23 ) where,
Q : discharge of pumpingK : coefficient of permeabilityD : thickness of aquiferR : radius of influencerw : radius of pumped wellH : depth of water outside of aquifer layerhw : depth of water at face of pumping well
hw
H
D
r w R
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b. Dupuit-Thiem (1906)
1). According to UNESCO (1967)
Fig. 5.5. Circular unconfined embankment
= Dupuit-Thiem formula for full penetration well on confined aquifer:
= (5.24) =
(
)
(5.24 )
where,Q : discharge of pumpingK : coefficient of permeabilityD : thickness of aquiferr1 r 2 : distance from well to observation well 1 and 2 respectivelyh1 h2 : head of water in observation well 1 and 2 respectively
h1 h2 D
rr 2
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2). According to Castany (1967)
Fig. 5.6. Circular unconfined aquifer
Dupuit-Thiem equation for the full penetration well in confined aquifer:
= ( ) (5.25) =
(
)
(5.25 )
where:Q : discharge of pumpingK : coefficient of permeabilityD : thickness of aquifer layerr1 r 2 : distance from well to observation well 1 and 2 respectively 1 2 : drawdown in observation well 1 and 2 respectively
r 1
r 2
h2
2 1
h1
D
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3. Alternate equations of the Dupuit-Thiem principle for radial flow are:
1). Pumping in circular aquifer
a). Unconfined aquifer:
o Without observation well and with piezometric head data:
=( ) (5.8 )
o Without observation well and with drawdown data:
=
(
)
5.22 )
b). Confined aquifer:
o Without observation well and with piezometric head data:
=( ) (5.23 )
2). Pumping in unlimited aquifer
a). Unconfined aquifer:
o Without observation well and with piezometric head data:
=( ) (5.21 )
o Without observation well and with drawdown data:
= ( ) (5.22 )
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o With one observation well and with piezometric head data:
=
(5.12 )
o With one observation well and with drawdown data:
= ( ) (5.22 ) =
( + )( ) (5.20 ) o With two observation wells data and piezometric head data:
= (5.12 ) o With two observation wells and drawdown data:
=( +
)(
)
(5.20 )
b). Confined aquifer:
o Without observation well and with piezometric head data:
=( ) = . (5.24 )
o With one observation well and with piezometric head data:
=( ) (5.24 )
o With one observation well and with drawdown data:
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Case 2 : (hw < D)
=( ) (5.27)
= ( ) (5.27 )
4. Correction to flow line
Fig. 5.10. Pumping in unconfined aquifer
a. Castany (1967) implemented Dupuit (1868) equation:
For the lateral flow:
= 2( + ) 22
(
) = [
(
) ] (5.28)
For the free aquifer and parallel flow:
= 2( + ) 2
Real curve
Theoretic curve
h h+h
H
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( ) = [ ( ) ] (5.28 ) b. Ehrenberger (1928)
= , ( ) (5.29) c. Vodgeo Institut (1954)
= , ( ) , (5.30) d. Iokutaro Kano (1939)
= (5.31) 0,324 < C < 1,60
e. Vibert (1949)
= , + (5.32)
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5. Radius of depletion
According to many researchers, the radius of depletion depends on the depressioncone because the drawdown of pumping:
a. W.Sichardt (in Castany, 1967)
= ( ) (5.33) where,Ri : radius of depletion (m)H h : drawdown (m)K : permeability (m/s)
b. H.Cambefort (in Castany, 1967)
= (5.34) where,Ri : radius of depletion (m)H : drawdown (m)Ki : permeability (m/s)
c. I. Choultse (in Castany, 1967)
= (5.35) where,
me : porosity of soil
T : duration of pumping (s or h)H : drawdown (m)K : permeability (m/s or m/h)Ri : radius of depletion (m)
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d. I.P. Koussakine (in Castany, 1967)
=
(5.36)
where,
K : permeability (m/s)T : duration of pumping (hour)
e. Dupuit1). Lateral flow :
1). Dupuit (in Castany, 1967)
= (5.37) 2). Castany (1967)
=
(5.38)
2). Radial flow (in Castany, 1967):
Using Darcys Law, Castany (1967) proposed an equation:
=( 2 2 ) + (5.39)
Sunjoto tried to improve above formula as:
=( ) +
= ( )
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= .( ) (5.40 )
where,Ri : radius of depletion (m)r : radius of observation well location (m)Q : discharge (m 3/h)H : drawdown (m)K : permeability (m/h)h : height of water on observation well (m)
f. Some authors (in Castany, 1967)
=
(5.41)
where,
Ri : radius of influence (L)Q : rate of pumping (L/T 3)I : precipitation intensity (debit/L 2/T)
g. Kozen (in Bogomolov et Silin-Bektchoutine (1955)
= (5.42)
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h. G.V. Bogomolov (in Castany, 1967)
Table 5.1. Coefficient of permeability and Radius of depletionAquifer material Granulometric
fraction
(mm)
Coefficient ofPermeability
(m/day)
Welldischarge
(m3
/hour)
Radius ofDepletion
(m)Clay sand 0,01-0,05 0,500-1,000 0,100-0,300 65Fine sand 0,01-0,05 1,500-5,000 0,200-0,400 65Clay sand in smallgrains
0,10-0,25 10,00-15,00 0,500-0,800 75
Sand in small grains 0,10-0,25 20,00-25,00 0,800-1,700 75Clay sand in mediumgrains
0,25-0,50 20,00-25,00 1,600-10,00 100
Sand in medium grains 0,25-0,50 35,00-50,00 15,00-20,00 100Clay sand in big grains 0,50-1,00 35,00-40,00 20,00-25,00 100Sand in big grains 0,50-1,00 60,00-75,00 40,00-50,00 125Gravels - 100,0-125,0 75,00-100,0 150
Note: drawdown 5-6 meter
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VI. FRESH AND SALINE WATER BALANCE
1. Basic equation
Badon Ghyben (1888) and Herzberg (1901),
Fig. 6.1. Schematic of cross section circular homogenous, isotropic and porous island.
=
(6.3)
Normal condition:
Sea water s = 1.025 tmass/m3 = 1,025 kgmass/m3 } so: = Fresh water f = 1.00 tmass/m3 = 1,000 kgmass/m3
hf
hs
A
h
precipitation
ground surface
groundwater surface
sea level
fresh water boundary area of salinewater and fresh water
saline water
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2. Shape of the Fresh-Salt Water interface
Fig. 6.2. Flow pattern of fresh water in an unconfined coastal aquifer
The exact shape of the interface is (Glover in Todd, 1927):
2 =2
+
2
(6.4)
The corresponding shape for the water table is given by:
= 2( + )1 2 (6.5)
The width xo of the submarine zone through which fresh water dischargesinto the sea can be obtained for z=0,
= 2 (6.6)
Sea
Saline water
Fresh water
Ground surface
Water table
Interface
xo
zo
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The depth of the interface beneath the shoreline z o, occurs where x = 0 sothat:
=
(6.7)
3. Upconing
Upconing is phenomenon that occurs when an aquifer contains an underlying of
saline water and is pumped by a well penetrating only the upper freshwater
portion of the aquifer, a local rise of the interface bellow the well occurs.
Fig. 6.2. Diagram of upconing of underlying saline water to a pumping well(after Schmorak and Mercado ini Todd, 1980)
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4. Drawdown versus Built upa. Theory of Dupuit-Thiem
Fig.6.3. Schematic of pumping
Discharge (Dupuit- Thiem) base on Darcys Law :
=
(6.11)
Problem: Solution of this equation needed minimum two dependent unknown (h2 & r2)so this formula is difficult for predicting computation.
From the above legends and schematic (Fig. 6.3) so the Power:
= ( + ) (6.12)
pump axis level
gsH
S
Q
gwl
r 1
r 2
h1 h
= ( +
)
=( )
Drawdown due to pumping
where,P : power (kN.m/s = kW)Q : discharge (m 3/s)
: specific weight of water(9.81 kN/m3)
H : gap of groundwater level to pump axis (m)S : drawdown m
: pump efficiency K : coefficient of permeability (m/s)h1 : piezometric of observation well 1h2 : piezometric of observation well 2
r1 : radius of observation well 1r2 : radius of observation well 2
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b. Theory of Forhheimer (1930)
Fig.6.4. Theory of Forchheimer (1936)
According to Forchheimer (1930) discharge (Q) on the hole with casing is hydraulichead (H) multiplied by coefficient of permeability (K) multiplied by shape factor (F),and for the hole with casing F = 4 R.
On his auger test with Q = 0, or water was poured instantly and then be measured therelationship between duration (t) and height of water on hole (h), he derivedmathematically the equation to compute coefficient of permeability:
=( ) (6.13)
where,K : coefficient of permeabilityR : radius of holeF : shape factor (F=4R)h1 : depth of water in the beginningh2 : depth of water in the end
t 1 : time in the beginningt 2 : time in the end
=
=( )
t2
t1 h1
h2
2R
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c. Theory of Sunjoto (1988)
Fig.6.5. Theory of recharge well and anti-drawdown (Sunjoto, 1988)
1). Discharge
Base on the steady flow condition theory of Forchheimer (1930), Sunjoto (1988)developed the equation of discharge through the hole with continue discharge flow tothe hole which was derived mathematically by integration and the result is unsteadyflow condition:
Forchheimer (1936) formula:
= (6.14) Sunjoto (1988) formula:
= = (6.15) This formula (6.14) when duration T is infinite so the equation will become Q = FKH
(see Fig. 6.5)
H
T
Q/FK
= 0
Built up due torecharging
Q
K
H
=
Relationship between H an T
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EXAMPLE:Pumping system with discharge Q = 0.1667 m3/s, gab between pumping axis to thegroundwater level H = 6.50 m, coefficient of permeability K = 0.00047 m/s, length ofscreen casing or perforated pipe L = 18 m and diameter of casing is 45 cm, freshwater: f = 1,000 kg/m3 or f = 9.81 kN/m3 and saline water: s = 1,025 kg/m3 or s =
10.552 kN/m3. Tip of the well in -28 m and the pumps are installed on the sandycostal which beneath of the pump in -160.00 m laid the boundary of fresh and saline
water.Compute:Power needed and how is the pumping system related to salt water intrusion.
Fig.6.6. Pumping data
Shape factor installed:
=2 18 + 2 0.225 2
18 + 2 0.2252 0.225 + 182 0.225 2 + 1 = 25.95
K=4.70*10 -4
S
5.00 m
Q=0.1667 m 3/s
6.50 m
23.00 m
18.00 m
+1.5
-5.00
-28.00
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The drawdown of 1 pump installed:
= =0.1667
25.95 0.00047= 13.667
To decrease of drawdown value S is by increasing value of F value, in this case beinstalled 4 wells with same dimension and each well equipped by P = 4.30 KW.
The drawdown of 4 pumps installed:
=0.1667
4 25.95 0.00047= .
The pumps are installed on the sandy costal which beneath of them laid down theboundary of fresh and saline water in 200,00 m.
Upconing: According to Sunjoto Eq.(6.9) is:
=3.41
1,025 1,0001,000 = 136,40 Power needed:
P = 0.1667 m3/s x 9.81 kN/m
3 x (6.50+3.41) m/ 0.60 = 27 kN.m/s = 27 kW
Conclusion:
The level of boundary will move upward to 200 + 136.40 = 63.60 m and due to thetip of the well level is 28 m so the saline water will not flow into tip of pipe so thereis not sea water intrusion.
Recommendation:
To avoid saline water intrusion to the pump so the shape factor F d should beincreased by enlarging the diameter of well or/and adding the length of porous well.
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5. Saline water pumping
Since the last three decades, the cultivation of fish in coastal area speedy increasedue to the demand of fish consumption increases. The fishpond in fresh water andbrackish water had been developed largely in Indonesia and then the fish cultivation
in seawater is now it s beginning to be developed. A seawater fishpond in sandycoastal area which was equipped by geo-membrane had been developed in YogyakartaSpecial Province with 7.20 ha area, 60 cm depth. One third of water should bereplaced by seawater. The needed pumping system for hydraulic head H = 7.50 mand coefficient of permeability K = 0.00047 m/s and saline water: s = 1,025 kg/m3 or
s = 10.552 kN/m3. This fishpond was installed 4 types of pumping system and onesystem still under design. The problem is that the discharge of pumping only less thanhalf of the design discharge even though the power was doubled.
Volume of pond:
Vp = 72,000 m2
x 0.60 m = 43,200 m3
Daily seawater volume needed:Vn = 33 % x 43,200 m3 = 14,400 m3
Daily seawater discharge needed:Qn = 14,400/24/3,600 = 0.1667 m 3/s 10 m3/mnt
Power needed (without drawdown occurs):Pn = Q H / kNm/sPn = 0.1667 m3/s x 10.552 kN/m 3x 7,50 m/ 0.60 = 21.99 kN.m/s = 21.99 kW
Analysis:According to Forchheimer (1930) that radial flow in porous media, discharge (Q) isequal to shape factor (F) multiplied by coefficient of permeability (K) multiplied byhydraulic head (h).
= (6.18) Pumping power is discharge multiplied by specific weight of water multiplied byhydraulic head divided by efficiency of pump system.
=
(6.19)
According to Sunjoto (2008), when drawdown of pumping is equal to hydraulic headthe equation becomes:
=2
(6.20)
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where,Q : discharge (m 3/s)F : shape factor of well (m)K : coefficient of permeability (m/s)H : hydraulic head of pumping (m)
P : power (kN.m/s): specific weight of water (kN/m 3) : pump efficiency
Due to there is not data of saline water and fresh water boundary so it wasdecided that the value of drawdown should be big enough to achieve the high upconingand it will get get saline water discharge, In this case the drawdown was decidedequal to hydraulic gradient and shape factor needed can be computed by (6.20):
=0,1667 2 1.025
0,60 2.135,85 0,00047 = 47,29
1. Lying pipesThis pumping system consists of four pipes of 20 cm diameter non-perforated andthe tip of pipes was covered by screen filter. The pipes were lied down about 1 munder the ground (sand) surface and always sink under low sea water surface toachieve the discharge water free from predators. The installed shape factors is(Sunjoto, 2002):
F = 2 R (6.21)where,
F : shape fator of pipe (m)R : radius of pipe (m)
Computed by (5), the installed shape factor for the 4 pipes is (6.21):Fi = 4 x 2 x x 0,10 = 2,51 m
This system was not installed the pump due to the current of the sea is bigenough to destroy the lied pipes.
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Fig.6.7. Lying pipes
2. Cubical Water IntakeThis system consist of hollow 6 m sides cubical concrete structure and the base ofcube without concrete slab lied down on the costal sand and sink always underlowest sea level. The aim of this system is keeping of 2 pumps from fast currentand high wave. Inside of the cube was installed two cylinder concrete of 60 cmdiameter where the tip of suction pumps take a water. So the shape factor of thisinstall system is (6.21):
Fi = 2 x 2 x x 0,30 = 3,77 mThis system was installed 2 pumps of 1x3.00 KW and 1x4.00 KW
Fig.6.8. Cubical Water Intake
3. Impermeable Deep wellThis system consists of 2 steel non perforated pipes of 45 cm diameter with length60 m and the installed shape factor can be computed by (6.21):
F = 2 x 2 x x R = 2 x 2 x x 0,225 = 2,827 mThis system was installed 2 pumps of 16.00 KW
6.00
Indian Ocean
Q
4 0,20 mQ
Indian Ocean
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Fig.6.9. Deep Well
4. Perforated swallow pipesThis system consists of 6 meter perforated pipes 30 cm diameter was installed in
costal sandy area and according to Sunjoto (2002) the shape factor is:
=2 + 2 2
+ 22 + 2 2 + 1 (6.22)
where,F : shape factor of pipe (m)
R : radius of pipe (m)L : porous length (m)
So shape factor (6.22) :
=2 6 + 2 0,15 2
6 + 2 0,152 0,15 + 62 0,15 2 + 1
= 10,326
This system was installed 1 pumps of 1x3.00 KW
Q
60 m
Indian Ocean
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Fig.6.10. Swallow Porous Pipes
Analysisa. Installations
Acctually there were 4 types of pumping systems were built in this project butthe Lying Pipes was broken down by the current and the wave of the ocean and thepump was not installed so its rest 3 pumping systems operate with the conditions:
1). Total installed powerP = 0 + (3,00 + 4,50) + (16.00 + 16.00) + 3,00 = 42,50 KW
Design power was 21,99 KW2). Total installed shape factor:
F = 0 + 3,770 + 2,827 + 10,326 = 16,923 mNeeded shape factor is 47,29 m.
3). Total real discharge:
Q = Q1 + Q2 + Q3 + Q4
Q = 0 + (0,18 + 0,27) + (1,80 + 1,80) + 0,18 = 4,23 m3/mnt
Design discharge was 10 m3/mnt.
b. Shape factor point of view1). Cubic Water IntakeWhen this system without 60 cm cylinder concrete, it will get bigger shape
factor as: = 4 (6.23) = 4 6 6 = 24 , = 3,77
6,00
Q
Indian Ocean
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To get shape factor F = 47,29 m you can build Cubical Water Intake PumpingSystem with dimension:
When Cylinder form so the radius is:
R = 47,29 / 2 = 7,50 m.
When Rectangular form the sides are (6.23):
= 4 = 47,29 = 11,83 12 To provide the discharge of the project 10 m 3/mnt it can be built only one Cubic
Water Intake Pumping System with dimension radius 7.50 m for the Cylinder form orRectangular form with the sides 12 m, equiped by 5 x 4,50 KW pumps.
2). Deep wellTo provide the discharge of the project 10 m 3/mnt it can be built only 3 Deep
Wells equiped by 16 m perforated pipes and the shape factor (6.22):
= 2 2 1 6 + 2 0,225 2
16 + 2 0,2252 0,225 + 162 0,225 2 + 1
= 2 23,726 = 47,452
To provide the discharge demand of the project 10 m 3/mnt it can be built only 2Deep Wells with 16 m perforated pipe each, equiped by 2 x 12 KW pumps.
3). Perforated swallow wellTo provide the discharge of the project 10 m 3/mnt it can be built only 5
Perforated Swallow Well Systems due to total shape factor is 5 x 10,326 = 51.63 m >47,26 m with 5 x 4.50 Kw Pumps.
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c. Horizontal perforated pipes (Imron Rosyadi, 2004)
According to Imron Rosyadi (2004) in his Master Thesis that the best solutionis 3 m diameter concrete cylinder with height of 13 m shoud be sunk 8 m on the sandand equiped 5 perforated pipes 4 m length and 10 cm diameter (Fig.6.11)
Fig.6.11. Horizontal perforated pipes
1). Shape factor of concrete cylinder is (6.21):
F1 = 2 x x 1,50 = 9,42 m
Indian Ocean
13.00
Q
3.0
4.00
10 cm
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2). Shape factor of perforated pipes is (6.22):
2 = 5 2 4 + 2 0,05 2
4 + 2 0 , 0 52 0,05 + 42 0,05 2 + 1
= 5 5,769 = 28,845
Total shape factor of concrete cylinder and horizontal perforated pipes is:F = F1 + F2 = 9,42 + 28, 845 = 38,265 m < 47,29 m.
Conclusion:The all designs never considerated shape factor of tip of well therefore the powerwas doubled but the discharge was only less than half of the designed value.
5. Horizontal Perforated Pipe (HPP)
HPP is perforated pipe which are installed horizontally to get bigger discharge or
recharge of the well. For discharge well the hydraulic head is the drawdown of
pumping and for recharge well the hydraulic head is the difference of groundwater
elevation on the well before and after pumping.
Gambar 6.12. Cross section of horizontal perforated pipes
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Data:Coefficient of permeability K = 10-3 m/sLength of HPP L = 4 mRadius of HPP r = 0.15 mRadius of well R = 2 m
Number of pipe n = 8 pcsDiameter of pipe pore f = 0.003 mPores distance 0.15 mAxis of HPP elevation: 9.50 mGroundwater elevation above HPP: -6.50 mGround surface elevation: 0.00 mSome of the methods of computation are:
a. Mikel & Klaers Methode (1956)
= (6.24) where,Q : discharge (m 3/s)n : number of pipeL : length of pipe (m)W : flow velocity (m/s)
=
=
= 10 3 33 = 0.001
Discharge of 8 pore pipes:
= 8 4 2 0,001 = 0.402285 3 b. Spiridonoff & Hantush s Method (1964)
= (6.25) where,
Q : discharge (m 3/s)Sv : specific yield aquifer of sand and gravel (S v = 20 %)Af : total area of pore hole of each pipe (m 2)h : distance between axis of pipe to groundwater level (m)D : diameter of pipe (m)
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Total area of pore holes of each pipe:
= 14 2 = 14 0.003 2 162 = 0.114557 2 Distance between axis of HPP to gr