Bahan Kuliah Aat

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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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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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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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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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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.
http://www.tn.gov.in/dtp/rainwater.htmhttp://www.tn.gov.in/dtp/rainwater.htm


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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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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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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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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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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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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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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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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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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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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_Mariotte


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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.
http://en.wikipedia.org/w/index.php?title=Lugeon&action=edithttp://en.wikipedia.org/w/index.php?title=Lugeon&action=edit


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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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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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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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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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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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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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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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Prof.Dr.Ir. Sunjoto Dip.HE, DEA-Subsurface Hydrology-Post Graduate Program JTSL-FT-UGM=2012 30

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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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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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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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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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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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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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).


40/138


41/138


= = 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


42/138


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:


43/138


=( ) (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,


44/138


K : coefficient of permeability (cm/s)C : a factor (100


45/138


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


46/138


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


47/138


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


48/138


(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


49/138


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


50/138


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_10


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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.


52/138


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


53/138


54/138


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


55/138


= 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


56/138


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)


57/138


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


58/138


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


59/138


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


60/138


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,


61/138


=( 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


62/138


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


63/138


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


64/138


=( 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


65/138


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


66/138


= = 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


67/138


= 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


68/138


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


69/138


= 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.


70/138


71/138


72/138


73/138


6. Unconfined aquifer


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


74/138


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


75/138


7. Unconfined aquifer with variable thickness


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


76/138


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:

= . . .( + . ) + .


77/138


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


78/138


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


79/138


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


80/138


=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)


81/138



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)


=

= . = 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)


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:


=( ) (5.23 )

2). Pumping in unlimited aquifer

a). Unconfined aquifer:


=( ) (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:


=( ) = . (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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89/138


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Case 2 : (hw < D)

=( ) (5.27)

= ( ) (5.27 )

4. Correction to flow line


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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100/138


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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104/138


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


109/138


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


110/138


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


111/138


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

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