Dam and Reservoir Engineering

8/12/2019 Dam and Reservoir Engineering

http://slidepdf.com/reader/full/dam-and-reservoir-engineering 1/798

T ARIQ . 2008. D AM AND R ESERVOIR E NGINEERING 1-1

Ch-1: Introducti on

Chapter - 1

INTRODUCTION

1.1 GENERAL

Dam : Dam is a barrier built across a river to hold back river water for safe retention andstorage of water or control the water flow. Dams allow to divert the river flow into a pipeline,a canal or channel (Fig 1.1). Dams results in substantially raising water levels in the riverover a large area, thus create a storage space. Dams may be of temporary or permanentnature. Dams may be built by constructing an embankment across the river at some suitablelocation. Natural processes as landslide and rock falling into the river may obstruct the riverflows for some time and create a dam like condition. The earthquake of 2005 resulted in adebris embankment of more than 200 m width and 70 m height across Karli/Tang Nullah near

Hattian Balla in AJK (Fig. 1.2); and after ascertaining the stability of the debris fill the waterimpoundment is being converted into a tourist point. However dams are built by humans toobtain some economic benefits. The water body created behind a constructed embankment ordam is called a man made lake or reservoir. Wildlife (Beaver) may also create ponds or smalldams for their habitat purposes.

Figure 1.1a: Water reservoir created by Tarbela Dam.




Ch-1: Introducti on

Figure 1.1b: Tarbela Dam aerial view (Source: Earth-Google).

L en gt h o f L ak e = 2 00 0 M t r Aver age Width = 350 Mtr Aver age Depth = 50 Mtr

X-SECTIONKARLI NULLAH LAKE

2.2 KM

202’ 189’ 171’ 149’ 137’ 122’ 110’ 95’ 77’ 57’44’

100 M100 M

100 M100 M

100 M100 M

100 M100 M

100 M100 M100 MBED OF NULLA H

150 M60 m

4’

INLETDISCHARGE

30’

Figure 1.2 : Natural dam across Kalri Nullah AJK formed by land slide due to earthquake.

Reservoir: Reservoir is defined the as a man-made lake or fresh water body created orenlarged by the building of embankment, dams, barriers, or excavation and on which manexerts major control over the storage and use of the water (Golze 1977, P-619). Theembankment may be constructed on one or more or all four sides of the reservoir.

Need:

(1) River supply usually does not match with the demand at all times/months. Damsstorage reservoir is created to match releases with the water demand.

(2) Dam created to substantially raise water level and thus working head for

hydropower production or to direct water into off taking canals (e.g. irrigationcanal, feeder to on off-channel dam).




Ch-1: Introducti on

Purposes

Dams and reservoirs are built to raise water level for storage and safe retention oflarge quantity of water. Water is subsequently released to achieve various purposes. Damsmay be constructed to meet one or more purposes as (USBR 2001, P:1-3):

1. Irrigation

2. Hydropower development

3. Domestic, municipal, industrial water supply (Hub dam, Simly dam)

4. Stock watering

5. Flood control

6. Recreation (picnic, camping, fishing, swimming, kayaking, white waterrafting)

7. Fish and wildlife protection and development, and improvement of riverecology

8. River water quality / pollution control and management

9. Stream flow regulation for various purposes

10. Navigation

Multipurpose dams :

Most dams are multi-purpose, serving more than one purpose. Mostly these additional purposes are achieved as byproduct outcome, e.g., hydropower, recreation, etc. Formultipurpose dams, the storage is allocated and prioritized for different purposes and costallocation (Fig. 1.4).

1.2 DAM AND RESERVOIR DEVELOPMENT STRATEGY

Reservoir design can be considered in a broader sense. It is really selected with suchimprovements or remedial work as may be considered necessary to assure safe andsatisfactory performance of its intended purpose. Development of a reservoir must assurestructural integrity and adequacy of the reservoirs. The reservoir site is evaluated in terms ofgeology, rim stability against slides, water tightness and water holding capability, seismicity,

bank storage, evaporation, sedimentation, land use and mineral resources, right-of-way and property ownership, relocation of the populace, utilities, and transportation facilities,historical-cultural and religious monuments etc.

The water stored behind the dam exerts a lrge water pressure on the dam. A dam must be ableto withstand such high pressures. In addition dam must be safe against failure due toovertopping, foundation thrust failures, destruction of dam body due to internal erosion andmaterial failure, foundation uplift, and retain storage contents – practically no loss of water

due to seepage.




Ch-1: Introducti on

Figure 1.3 : Upper Reservoir of Taum Sauk 450 MW pumped power plant (ReynoldsCounty, Missouri, on the East Fork of the Black River) made of ridge top 6562 ft long84 ft high CFRD dike with 10 ft parapet wall. The reservoir dike constructed in1960’s failed on Dec 14, 2005 due to internal leakage and slope failure. Plantremained out of use as of Jan 2007. [http://www.ferc.gov/industries/hydropower/safety/projects/taum-sauk/consult-rpt/sec-2-summ.pdf].




Ch-1: Introducti on

Natural or man-made water bodies, albeit large ones, has high aesthetical appeal andthus attract huge number of visitors for recreation. The reservoir design must include

provisions of recreation facilities as parking area, picnic area, camping area, hiking and biking trails, nature walk trails, horse trails, rock climbing, enjoying surrounding scenery,water sports, motel, public services, restrooms, emergency services, indoor shelter areas,

project guided tours, etc. These should be evaluated in terms of need vs luxury and security

concerns for the structure and public.

Reservoir area requires clearing of brush/shrubs/trees from below maximum reservoirlevels for safe use of reservoir surface. Such clearing may be done by cutting/pulling or by

protected fires. In flat side reservoirs large surface area is exposed or reservoir lowering.Suitable alternatives may be evaluated to make economic use of this area for short timeactivities, as farming, sand mining etc.

1.3 CLASSIFICATION OF DAMS

1.3.1 Classification of Dams According To Location

On-Channel : Dam is constructed across the main water feeding river. Examples Tarbela,Mangla, Simly, Hub dam. To increase the water availability water from other riversmay be diverted to the dam through feeder channels e.g. Kurram Tangi dam.

Off-Channel : Dam is constructed on a channel having much smaller flow. Major storagewater is transferred from a different nearby river. This is done due to non-availabilityof suitable/economic dam site on the major flow river. Example Akhori dam,Replacement dams for Mangla and Tarbela.

Irrigation storage

Flood storageFlood surcharge

Free board

H y

d r o p o w e r p

l a n t

Normal conservation level Max spillwaycrest level

Dam crest

Figure 1.4: Multipurpose dam.

Dead storage

P o w e r

t u n n e l

/

i r r i g a t i o n o u

t l e t

Dead storage level

River profile




Ch-1: Introducti on

1.3.2 Classification of Dams According to Release Pattern

Storage dam : Water is stored and later released through an outlet for consumptive or non-consumptive purposes as per requirements.

Recharging dam. There is no outlet provided to release water and all incoming water isretained. The water infiltrates through the foundation and/or dam body. The main

purpose of the dam is to induce recharge to ground water system in the area. Smallrelease in d/s channel to allow seepage in the channel bed.

Delay action dam / retarding dam. These dams are used to retard the peak flow of flashfloods. There may or may not be any control over the outflow. For no control over theoutflow the outflow rate varies as function of storage volume / water depth in thedam. The flood peak is thus considerably attenuated. The outlet capacity is set thatmaximum outflow discharge do not exceed the safe capacity of the downstream river

during highest flood. The reservoir empties fully after the flood. For control onoutflow by gates (detention dam) , the flow is released in such a pattern to retain thewater for long time but there is enough storage available to store next flood event.These dams are usually meant to reduce flood damages as well as to induce maximumrecharge in the area. One type of such dam is a porous dam built of a porousembankment, e.g. stone gabions.

Tailings dam These dams are constructed away from any river along a topographic slope byconstructing small dikes on three or all four sides to store slurry / waste of mineralmining and processing facilities. The water evaporates or is evacuated and the solidcontents dry up filling up the storage capacity.

Diversion dam These are hydraulic structures with a main purpose to raise water level todivert flow into the off taking channels / canals/ hydropower pressure tunnels and

penstock. These are preferably called as barrage or canal head works. The storagecreated by these is minimal. E.g. Patrind Weir.

Coffer dam: These are small temporary dams built across the river on upstream anddownstream side of the main dam in order to keep the flow away and the workingarea dry. The u/s coffer dam causes the flow through the diversion system and d/s

coffer dam prevents the flooding of the working from backwater effects. Aftercompletion of the main dam the u/s coffer is usually abandoned and drowns in thereservoir while d/s coffer dam is dismantled and removed.

1.3.3 Classification of dams according to Hydraulic Design

Non-Overflow dam : Flow is not allowed over the embankment crest for reasons of damsafety. (earth, rock) dams.

Overflow dam The dam body is made of strong material as concrete and flow is allowedover the dam crest Concrete dams




Ch-1: Introducti on

1.32.4 Classification of dams according to Size

Dams may be classified as small, medium or large as under:

Small . USBR defined small dam as one having maximum height < 15 m (50 ft).

Medium: Intermediate sizes 40-70 ftLarge: ICOLD defined large dam as: a dam that follows one or more of following

conditions. (Thomas 1976 P-0)

• Dam height > 15 m (50 ft) measured from lowest portion of the general foundationarea to the crest

• A dam height 10-15 m but it compiles with at least one of the following condition:

a. crest of dam longer than 500 m

b. capacity of the resulting reservoir more than 1 million mc. maximum flood discharge more than 2000 m

3

3

d. dam has specially difficult foundation problems

/s (70,000 cfs)

e. dam is of unusual design

Unique: Dams exceeding 100 m are considered as unique. Every aspect of its design andconstruction must be treated as a problem specifically related to that particularsite.

1.3.5 Classification of Dams According to Filling and Emptying ModeThe storage of a dam may be filled and emptied in short time (one season) or long

time (several seasons). The dams are defined as:

Seasonal: Seasonal dams are filled and then emptied within the same water year (Septemberto August). Example Tarbela dam. Thus water level in the dam varies from maximum(normal conservation level) to minimum (dead storage level) in most years. Suchdams have annual releases usually equal or little more than the minimum annual flow.For very wet or very dry years the reservoir may not reach the extreme levels. Theseasonal dams spread the water stored in wet months over to dry months in the sameyear.

Carry over: Filling and emptying of a carry-over dam reservoir continues over more thanone year (e.g. 4 to 5 years). Example. Hub Dam, Kurram Tangi Dam. Thus waterstored in wet years may be released during subsequent dry years The annual releasesare usually more than minimum annual flow but equal to long term average annualflow. Applicable where wide variations in annual flows. Carry over dams spreadstorage during wet years/months over to dry years and months.




Ch-1: Introducti on

1.3.6 Classification according to location of service area

Local : The service area of the dam is limited to a single contiguous localized geographic arealocated very near the dam. Far located areas and geographic regions do not benefit.E.g. Kurram Tangi, Simly, Khanpur dams.

Regional: The service area of the dam extends to many widely apart geographic regionslocated any distance from the dam. Thus all near and far located areas and geographicregions get the benefit. The water supply to all areas is possible through a network ofriver and canal systems. Exampleas are Tarbela, Diamir-Basha, Kalabagh, Mangladams.

1.3.7 Classification according to type of material

A dam can be made of earth, rock, concrete or wood. Dams are classified according tothe materials used as under: (Navak P: 11-18, 33)

A. Embankment Dams (Figs. 1.6, 1.7)

1. Earthfill Dam : These are constructed of selected soils (0.001 ≤ d ≤ 100 mm)compacted uniformly and intensively in relatively thin layers (20 to 60 ± cm) and atcontrolled optimum moisture content. Compacted natural soils form more than 50%of the fill Material. Dams may be designed as: Homogeneous, Zoned or withimpermeable core (Figs. 1.5-1.7). Zoned part is made of relatively finer material thatreduces seepage flow, e.g. clay. The fill material is placed as rolled, hydraulic fill orsemi-hydraulic fill.

Figure 1.5 : Earthfill dam. Left-homogeneous, right-zoned dam.

2. Rockfill dam : Over 50% of fill material be of class ‘rock’ usually a graded rockfill(0.1 ≤ d ≤ 1000 mm) is filled in bulk or compacted in thin layers by heavy plant.Some impervious membranes/materials are placed in the interior or on u/s face of theembankment to stop/reduce seepage through the dam embankment. Dams section may

be homogeneous, zoned, with impermeable core, or with asphalt or cement concreteface. Zoned part is made of relatively finer material that reduces seepage flow, e.g.clay. Core is made of clay, concrete, asphalt concrete etc.

3. Earthfill-rockfill or Earth-rock dams These dams are made of mix of large proportions of earthfill and rockfill materials.




Ch-1: Introducti on

B. Concrete Dams

Concrete dams are formed of cement-concrete placed in the dam body (Figs. 1.8, 1.9).Concrete dam section designed such that the loading produces compression stress only andno tension is induced any where. The reinforcement is minimum mainly as temperature

control. Concrete is placed in two ways: Reinforced concrete dam (RC dam) or Rollercompacted concrete (RCC) dams. The variations of concrete dam include:

1. Concrete gravity dam,

2. Concrete arch dam and arch-gravity dam

3. Multiple arch dam

4. Double curvature or dome/cupola dam

5. Buttress dam (head as diamond, roundhead, massive, decked etc)

6. Hollow gravity dam

7. Brick or rock masonry gravity dam

Rubble/random/stone masonry to fill dam section. Concrete / mass concrete as bulk materialin dam section with steeper side slope. RCC section to take loadings, thus decrease section.

1 Gravity dam: Stability due to its mass. Dam straight or slightly curved u/s in plan (noarch action). The u/s face is vertical or nearly vertical, d/s sloping.

2. Buttress dam: It consists of continuous u/s face supported at regular intervals by d/s

buttress (massive buttress /diamond head, round head) with each section separate. Ambursen / flat slab buttress / decked buttress.

3. Arch dam: Arch dam has considerable u/s plan curvature. U/s and d/s faces arenearly straight / vertical. Water loads are transferred onto the abutments or valleysides by arch action. Arch dam is structurally more efficient than concrete gravitydams (requires only 10-20% concrete). However abutment strength and geologicstability is critical to the structural integrity and safety of the dam. Multiple archdams.

4. Cupola/Dome/Double curvature dam: . U/s & d/s faces curved in plan and profilesection, curved in plan as well/ as arch (Part of a dome or shell structure).

5. Hollow gravity section made hollow to reduce uplift pressure at d/s side and smallertotal construction materials. (between gravity and buttress dams)

C. Timber/steel dam

The bulk of the dam is made of timber braces with timber board facings. Such dams weremostly constructed by early gold miners in California USA for obtaining river water forseparating gold dust and getting water power; such dams are not practically used any

longer. The face of earthfill or rockfill dams may be also fitted with timber board forseepage control.




Ch-1: Introducti on

Figure 1.6: Earthfill embankment dams.

Figure 1.7: Rockfill embankment dams.




Ch-1: Introducti on

Figure 1.8: Concrete dams.

Figure 1.9 : Future Concrete dams.




Ch-1: Introducti on

1.4 PLANNING AND DESIGN OF DAM

1.4.1 Stages

Any dam project is carried out at following stages

• Initial screening based on river profile and topographic maps.• Reconnaissance plan-uses only any available data

• Pre-feasibility plan-little exploration and additional field data

• Feasibility plan-Extensive exploration and additional field data

• Design stage: – point tests/surveys to finalize design

At each succeeding stage, the plan is firmed up with more precise details, dimensions andanalysis; More data at each successive stage. The design stage ends up with drawings

appropriate for construction activities. Still further details/revision continues well during theconstruction of the dam as new information is gathered or some already available informationis found to be incorrect and not valid.

1.4.2 Data Required

Large amount of data is required for planning/designing of dams (Golze P. 47-50, USBR1949 P.5-10). These include as:

• Location & vicinity map

• Topographic maps/aerial photographs of dam site

• Elevation surveys/triangulation + bench mark

• Transportation map (road, rail, air)

• Geological / rock formations data of dam site

• Seismic/tectonic activity map

• Climatic data (P, T, ET, wind, sunshine)

• Stream flow data (daily average flows)

• Sediment data

• Demographic/land ownership/housing data for the reservoir area

• River environment/ecology (u/s, at site, d/s) (fish, w/life, birds, flora, fauna,vegetation)

• Project water requirement

• Power requirements & national grid / transmission lines

• Flood data (instantaneous peak flow rates, time to peak, base time, flood duration,flood volumes, flow hydrograph, etc) of all or major floods




Ch-1: Introducti on

• Water rights

• River hydrographic data (bed levels, flood levels, cross section, bank/valleylevels)

• Groundwater table data in the vicinity, u/s and d/s area

• Public recreation need

• Land evaluation

• Public/Private buildings

• Availability of construction materials

• River stage-discharge data (u/s, tail water)

• Geo-political economic data

1.4.3 The Planning/Design Team

Dam planning/design multi-task activity; various tasks are as:

1. Site selection, 2. topographic surveys, 3. water availability assessment, 4. sizing andlayout, 5. geologic surveys and construction materials investigations, 6. geologic evaluationof foundation, rim, abutment and pond area, 7.dam section design, 8. dam seepage andstability analysis, 9. Diversion arrangements details (diversion tunnel, coffer dam), 10. floodsand spillways, 11. hydropower works, 12. irrigation outlets and irrigation system design, 13.Reservoir sedimentation, 14. Reservoir operation studies, 15. Material quantities and costing,16. Environmental studies, 17. Land acquisition and replacement, etc.

Thus planning and design of dam is a multi-disciplinary task and require teamwork offollowing disciplines:

1. Project Manager

2. Water resources engineer

3. Layout planner

4. Surveyors (topographic and elevation)

5. Hydrology + meteorology

6. Engineering geologist, Geophysist/Siesmologist

7. Geotechnical and Geophysical exploration specialist / Drillers

8. Geo-technical / foundation design engineers

9. Hydraulic engineer

10. Structural engineer (for structural design of outlets, spillway. Powerhouse, energydissipation)

11. Mechanical engineer (for design of controls, gates, valves, hoists, )




Ch-1: Introducti on

12. Hydropower engineer

13. Electrical engineer

14. Infrastructure/road/municipal engineer / Civil engineer

15. Instrumentation and telecommunication engineer16. Environmental engineer, Environmental scientists (fish, wild life, flora, fauna, etc)

17. Economists

18. Construction planner / manager

19. Quantity Surveyor / Costing engineer

20. Irrigation engineer

21. Irrigation agronomist

22. Soil expert

1.5 DAM SITE SELECTION

The purpose of a dam is to retain and store large quantities of water in a safe way.Many considerations are analyzed. Dams can be built anywhere if you can spend enoughmoney. However preferred site have following characteristics which lead to lower projectcosts. Thus alternate dam sites/axis location are evaluated for most cost effective choice.

1. Small river channel width with steep side gorge: short dam crest length, leads tolarge storage for small dam length

2. A wide and gently sloping valley upstream of the dam site (for storage dams) andnarrow and steeply sloping valley for hydropower dams.

3. River channel and valley has very flat slopes u/s of dam site (leads to largestorage for small dam heights).

4. Deep reservoir possible – require less area and lesser land costs, less surfaceevaporation

5. Enough water flow/yield available to meet requirements/demand

6. High sediment load tributaries are excluded

7. Geology favorable for foundation (foundation can be designed at any site, but itincreases costs), competent hard rock is most suitable.

8. Abutments are water tight, and reservoir rim allow minimum percolation andseepage losses.

9. Small river sediment rate (longer dam life) Depend on river morphology andcatchment characteristics. Gomal Zam has 10 times sediment load than KurramTangi dam, thus large dead storage space is adopted.




Ch-1: Introducti on

10. Land use of reservoir area is minimal – lower economic values means lowercompensations.

11. Reservoir area not very sensitive to environment (wild life parks, endangeredspecies, historical places, monuments etc).

12. No seismic and tectonic activities or active faults in and near the site.

13. Socio-political stability (no unstable gestures) (Gomal-Zam, Mirani, KurramTangi dams), Diamer-Basha vs Kalabagh dams.

14. Reservoir and dam area less populated

15. Site have adequate stream flow record

16. Site is easily accessible; approach road is present or can be developed easily.

17. Construction material available nearby easily

18. Site near load center (demand area) for water+ power

19. No mineral resources in reservoir area (present or future)

20. Site allows a deep reservoir & small surface area (less land costs and smallevaporation losses).

21. Existing infrastructure, e.g. highway, least affected. E.g. KKH and Bhasha-Diamere dam.

1.6 DAM COMPONENTS

Elements of a typical dam include (Figs. 1.10 and 1.11):

1.6.1 Main Dam

This is the main structure built across the river. The height of a dam depends upondesired storage capacity and the site conditions. The crest length of he dam depend upontopography at the dam site. The dam may be built of many different materials. The storedwater is released from the dam as per requirements.

1.6.2 Flanks/Abutment:

The rock mass on right & left banks of the river constitute abutments. Dam is joinedwith and supported by the abutments. In addition outlet tunnels, diversion tunnel, spillway arealso placed in the flanks. The geology of the abutments has to be strong enough to enable

placing various structural components without any risk. In addition abutments need to be ofcompetent rock without any structural defects and lowest permeability

1.6.3 Saddle Dam:

The reservoir is usually formed by the main dam on one side and low/high hills on allother sides of the reservoir. In most cases the elevation of the hills along the rim of the dam is

much higher than the reservoir maximum water level. In some other cases elevations ofsurrounding hills along a part of the rim/periphery of the reservoir is not high enough over a




Ch-1: Introducti on

small section to completely contain the stored water and a saddle (low level place) is formed.Water can flow out through the saddle. A small embankment is then constructed at thislow/saddle point to seal off the reservoir rim and is called as saddle dam. Example: Sukiandam and Jari dam for Mangla Dam project.

1.6.4 Diversion Channel/Tunnel

These channel or tunnel are constructed prior to dam construction such that river flowis passed around and away from the dam site through the diversion tunnels and that than damsite remain dry and accessible to construction at all time. The capacity of diversion structureis set such that most probable floods likely to occur during the construction period can be

passed over without danger of overtopping of cofferdam and inundation of construction area. Necessary arrangements are made at d/s end for energy dissipation. These tunnels may beabandoned (plugged – Simly dam) after project completion or converted to irrigation / power/ desilting tunnels. Diversion tunnel may not be provided (Mirani dam) and u/s coffer dam.

Figure 1.10 : Dam components (http://www.dnr.state.wi.us/ORG/WATER/WM/dsfm/dams/gallery.html )

http://rds.yahoo.com/_ylt=A9iby4JGKYZGbDIBQxOjzbkF/SIG=12i1i1rgq/EXP=1183283910/**http%3A/www.dnr.state.wi.us/ORG/WATER/WM/dsfm/dams/gallery.html







Ch-1: Introducti on

Figure 1.11: Dam layout showing main dam, saddle dam, u/s and d/s coffer dam, spillwayand stilling basin, diversion tunnel(s), power tunnel, power house and irrigation canal.

550

550

550

500450

500

450

400500

400

PH450

N




Ch-1: Introducti on

1.6.5 Cofferdam

These are small temporary dams built u/s and d/s of the dam site to make theconstruction area dry and workable. The u/s cofferdam causes water to flow through thediversion tunnel and the d/s cofferdam prevents backwater level to inundate the constructionarea. Coffer dam may be dovetailed in u/s part of dam (Mangla) or abandoned. Material usedearth, rock, concrete etc. Arrangemnet are required for control of seepage across the cofferdam.

1.6.6 Spillway

This is a water release/conveyance structure to pass the large flood volumes safelyacross the dam without danger of overtopping of the dam crest. There would be one or morespillways usually at different levels (Service, additional, emergency). The lower spillway isused to release often occurring flood and regular inflows and is called as service spillway. Ithas usually more elaborate arrangements and may be free flowing or gated. The auxiliary oremergency spillway is set at or above normal conservation level and has fewer arrangementsand is usually free flowing. This is used only during flood events of extra-ordinary nature.Fuse plug, rubber dam etc may be used to delay water release and possible additional storageat the reservoir.

The spillway may be a integral part of the main dam (mostly for concrete dams) or be aseparate structure in the dam abutments.

1.6.7 Outlet Works

(a) Intake Structure / Tower: This is a structure to admit and control flow of water into theirrigation/power outlets. It would be a tower or inlet flush with reservoir side walls. Gatesmay be provided at u/s, intermediate or d/s end of the outlet tunnel. Necessary provision ismade to keep the intake operation for long after sedimentation by having multiple water entrylevels particularly for domestic supply purposes. Multi level inlet openings may be used.

(b) Irrigation/Power Outlet Tunnel: This is a large water conveyance structure to releasewater to irrigation network and/or powerhouse turbines. The outlet is in the form of a tunneldug or formed through the abutment / flank for earth / rockfill dams or through the dam body

for a concrete dam. At the u/s end an intake is provided along with gates, trash rack. Thetunnel design must eliminate risk of cavitation and/or aeration. Gates may be placed at u/s,d/s or intermediate location. The power tunnel is transitioned into surge chamber,

penstock/scroll case etc. Energy dissipation structure may be provided at d/s end, if needed.Irrigation outlet may release into a canal or into the river if demand site is at distance fromthe dam. The intake level of the tunnel is kept below or at the dead storage level. Air vent is

provided to minimize cavitation. Water cushon for vortex control are also provided.

(c) Low Level Outlet: A low outlet tunnel may be provided to flush sediments, draw waterfrom below dead storage level under very drought condition, emptying of reservoir in

emergencies, draw water during repair of outlet tunnel/gates, etc. The intake level is keptmuch lower than the intake level main irrigation tunnel. May discharge into stilling basin for




Ch-1: Introducti on

spillways/outlet works or as a separate energy dissipation structure provided. [Similar tounder sluices in a barrage.](d) Gates/Valves/(e) Trash Rack, air duct for cavitation control

1.6.8 Drainage System

Dams are designed to store water with least seepage through the dam embankmentand the foundation but seepage do occur. The drainage/seepage water also causes tremendousuplift pressure particularly at d/s half of the dam base. Features are included in the damdesign to minimize seepage through the foundation and through the dam embankment anduplift pressure. Cutoff wall, sheet piles, slurry trench, etc.

• Grout Certain: An impermeable zone is created under the dam.

• Grout Blanket: Impermeable area is created u/s of dam.

• Pressure relief / Drainage Wells: Wells are installed at d/s area to pick andremove seepage water to reduce uplift pressure in thefoundation area.

• Drainage gallery A horizontal/inclined gallery is formed in the body of the dam(specially in concrete dams) where water from drainage wellsdischarge into and is ultimately flow out of the dam body. Italso intercepts leaks through dam body.

• Horizontal Blanket Drain: To intercept seepage lines at base of dam on d/s side.

• Chimney Drain: Vertical or inclined drainage filter layer (usually d/s of theimpermeable clay core) to intercept seepage flow.

• Toe Drain: A drain is provided at toe of dam (homogeneous coarse fill) tointercept seepage flow inside the dam body.

• D/S Trench: Trench provided at d/s of dam to intercept seepage flow lines.

• Impermeable blanket to lengthen the seepage path and lower hydraulicgradient and seepage rate.

1.6.8 Preliminary Works

Civil works, infrastructures, buildings required to be provided before start ofconstruction of main dam work. These include offices, staff housing, community buildings,water supply, approach road, client/consultant/contractor camp, labor camp, securityarrangements, rest house, rail sidings, air strip, hele-pad, etc.

1.6.9 Hydropower Development

(a) Powerhouse: Building to house turbine, generators, mechanical workshop, valves, draft

tube, office, control room, visitor area, up transformer, etc for hydropower generation.(b) Penstock: This is a large diameter pressure pipe used to deliver water to turbines.




Ch-1: Introducti on

(c) Surge chamber. To contain water hammer surge on plant load rejection / sudden shut-down.

(d) Switchyard: This is an area to install electrical equipment to change low to high tension power supply for further transmission.

1.6.10 Slope protection/Riprap

Stone is placed on u/s & d/s dam slopes for protection against damage due to waveaction, rain water, burrowing animals. Parapet wall may be used to protect dam top againstsudden waves generated by strong winds, tsunami, etc.

1.6.11 Dam Instrumentation

Various gages/instruments are installed in the dam body, foundations, spillway tomonitor settlement, movement, stresses, pore water/uplift pressure, earthquake.

1.6.12 Stilling Basin

To dissipate excess energy of diversion tunnel, low level outlet, irrigation tunnel,spillway, etc.

1.6.13 Gallery/Shafts

These are provided in the dam body for access to interior of concrete dam body.These are horizontal, vertical (with round stair ways), sloping.

1.6.14: Operational buildings

These are buildings required for operation of the dam and works. These includeOffice buildings, Rest House, Security buildings, Staff residences and other community

buildings, gate control room.

1.6.15: Temporary works:

These are installations required for temporary use and are removed after projectcompletion. These include contractors camp, material processing, handling and stock area,machine room, casting yard, steel fabrication, labor camp, etc.

1.7 MERITS AND DEMERITS OF DAMS

1.7.1 Embankment Dam

a Merits (Novak P-14)

1. Suitable to type of sites in wide valleys and relatively steep sided gorges alike.

2. Adoptable to a broad range of foundation conditions-from competent rock to softand compressible or relatively pervious soil foundation.

3. Use of natural materials at smaller cost thus no need to import or transport largequantities of processed materials or cement to the side.

4. Subject to the design criteria, embankment dams are extremely flexible toaccommodate different fill materials (rock, earth) if suitably zoned internally.




Ch-1: Introducti on

5. Construction process highly mechanized and continuous (less human handling asform work, curing time)

6. If properly designed, dam can safely accommodate appreciable degree ofsettlement-deformation without risk of serious cracking and possible failure.

Embankment dams withstand earthquake better. However the foundation of thesedams, if deep and of unconsolidated origin, is more liable to settlement and failure

by earthquake (liquification).

b Demerits

• Inherent greater susceptibility to damage or destruction due to over topping(require adequate flood relief and separate spillway).

• Vulnerable to concealed leakage and internal piping/erosion in dam or foundation.

c. limitations• Spillway outlet are separate from main dam.

1.7.2 Concrete/Masonry Dams

a Concrete Dam Merits (Novak P-17)

1. Concrete dams, except arch and cupola, are suitable to site topography of wide ornarrow valley alike, provided that a competent rock foundation is present atmoderate depths (< 5 m) (arch best for narrow section)

2. Concrete dams are not sensitive to overtopping under extreme flood conditions.3. All concrete dams can accommodate a crest spillway, if necessary, over the entire

length, provided that steps are taken to control d/s erosion and possibleundermining of the dam. Thus cost of separate spillway is avoided.

4. Outlet pipe works, valves and ancillary works are readily and safely housed inchambers or galleries within the dam.

5. Has high inherent ability to withstand seismic disturbances.

6. Cupola dam is extremely strong and efficient structure for a narrow valley with

competent abutments.

b Demerits

1. Concrete dams require sound and stable rock foundations.

2. These require processed natural materials of suitable quality and quantity foraggregate and importation to site and storage of bulk cement and other materials.

3. Traditional mass concrete construction is slow, labor intensive and discontinuous,and require adequate skill for formwork, concreting etc.

4. Cost per unit of concrete dam much higher than embankment fill. Smallerquantities seldom counter balance for dams of given height.




Ch-1: Introducti on

1.8 DAM FOCUS POINTS (Novak P 10-11)

Dams have following focus points and thus differ from other major civil engineeringstructures.

1. Every dam, large or small, is quite unique; foundation geology, materialcharacteristics, catchment yield and flood hydrology are each site specific.

2. Dams are required to function at or close to their design loadings for extended periods.

3. Dams do not have a structural life span, components must be designed for longlife). Dams may have notional life for accounting/economic purposes, or afunctional life span dictated by the reservoir sedimentation.

4. Majority of dams are of earth fill made from a range of natural soils, and are least

consistent of construction materials.5. Dam engineering draws together a range of disciplines to a quite unique degree

(hydrology, hydraulics, geology, geotech, structure etc).

6. FIRST PLAN: All type of dams may be constructed at the site, thus planalternative design until discarded due to technical, financial or environmentalreasons

7. Dam engineering is critically dependent upon the application of informedengineering judgment.

1.9: ELEVATION-AREA-VOLUME RELATIONSHIPThe elevation-volume-area relationship for a reservoir/dam describes the variations of

volume and surface area with elevation/height. This relationship is determined from elevationcontour map of the reservoir area. The elevation is determined by topographic survey at gridor random locations (grid spacing varies with level of investigation from 200 m for pre-feasibility study to 50 m or less for feasibility study). Wide contours indicate a gently slopingflat valley area and closed spaced contours indicate steeply sloping cliff sides. Contours aredrawn at an interval of 5 to 10 ft (Fig. 1.12). Surface area is measured for each contour. Theincremental volume between two consecutive contours is determined as ∆V = (A 1+A 2

Vol. =

)/2 x

∆h { ∆h is contour interval}. Total volume at any elevation is obtained by adding successiveincremental area as V = ∑ ∆V. Table 1.1 below show calculations for elevation-volume-arearelationship. The reservoir surface area and volume is related as (H = Elevation – datum):

∫ H

0

dH Area and Area = dV/dH, (1.1)

The data points are plotted with volume or area on x-axis and elevation on y-axis (volume on primary x-axis, and area on secondary x-axis) (Fig. 1.13).Equations may be developed (usually a power function) to find elevation for a given storageor area as

El = A (Vol) B + datum and El = C (Area) D + datum




Ch-1: Introducti on

Table 1.1 : Elevation-Area-Volume Relationship for a Dam.

Map Scale:1 inch = 5000 ft

1 sq in = 5000 2 = 25,000,000 sq ft1 sq in = 5000 2 / 43,560 = 573.92 Acres

Selected datum (ft amsl) = 1800Elevation Height above

datumMap area Plan Area Incremental

volumeTotal storage

capacity(ft amsl) (ft) (sq. in) (Acres) (AF) Acre Feet ThAF

1820 20 0.00 0 0 0 01850 50 0.49 281 4,993 5,043 51900 100 1.88 1,079 34,005 39,048 391950 150 4.11 2,359 85,945 124,993 1252000 200 7.17 4,115 161,846 286,838 2872050 250 11.03 6,330 261,134 547,972 5482100 300 15.69 9,005 383,379 931,352 9312150 350 21.14 12,133 528,438 1,459,789 1,460

Example . For Kurram Tangi dam the elevation-storage-area relation are described as:(volume in AF, elevation is ft amsl, and area is in acres and 1805 is datum) (Figs. 1.14 to1.17).

El = 2.6905 x (Vol) 0.3432

El = 2.5821 x (Area)

+ 18050.5226

For some cases more than one equation may be needed to describe the data for differentranges. Inverse equations may be derived to find volume or area corresponding to anyelevation, e.g. for Kurram Tangi dam elevation-area-volume dam is described as (Volume inAF, Elevation in ft amsl, Area in acres and Datum = 1805 ft amsl..

+ 1805

Vol.= 0.05595 (Elevation - Datum) 2.913

Area = 0.163 (Elevation ft - Datum)

Equation form of the elevation-area-volume relationship may be useful for various purposes,e.g. reservoir simulations, flood routing for spillway design and diversion tunnel design.

1.9132

1.10 DAM HEIGHT

The height of any dam above the lowest level in the river channel is determined from(i) the gross storage (live storage + dead storage) capacity of the dam, (ii) the space requiredto pass maximum design flood over the spillway (called flood surcharge), (iii) the waveheight generated from extreme winds, (iv) the wave runup over the upstream sloping face dueto wind gusts and (v) the free board. The reservoir level corresponding to normal reservoirstorage is called as normal conservation level NCL and is determined from the elevation-volume relationship of the dam. Referring to Figs 1.13, the normal conservation level isdetermined as 2076.2 for gross storage capacity of 0.716 MAF. The wave height and waverunup is determined from reservoir area, depth and prevailing wind speeds in the vicinity of




Ch-1: Introducti on

the dam. Free board of 5 to 10 ft are customary depending upon the reservoir importance andother factors.

For Gross storage = 0.716 MAF (Live storage = 0.55 as determined from mass curve /reservoir operation studies, and dead storage = 0.166 MAF as determined from sedimentationanalysis), the required dam height is worked as:

Minimum River bed level at dam site = 1805.0 ft amsl Normal conservation level for 0.716 MAF = 2.6905x(716000) 0.3432

Maximum reservoir depth = 2076.2-1805.0 = 271.2 ft+1805 = 2076.2 ft amsl

Flood surcharge (from PMF routing) = 6.5 ftWave height e.g. = 3.5 ftWave runup e.g. = 4.7 ftFree board e.g. = 10 ft

Total dam height = 271.5 + 6.5 + 3.5 + 4.7 + 10.0 = 295.9 ftDam crest level = 1805.0 + 295.9 = 2100.90 ft (say 2101 ft amsl)

190

2100 ft

2050 ft

2000 ft

1950 ft

Kurram TangiDam

2150 ft

Fi ure 1.12 : To o ra hic surface contours of Kurram Tan i Dam.




Ch-1: Introducti on

KURRAM TANGI DAM: Area-Elevation-Capacity-Curves

0

5

39

125

287

548

931

1,460

0.05

0.28

1.08

2.36

4.12

6.33

9.00

12.13

1800182518501875

190019251950197520002025205020752100212521502175

0 200 400 600 800 1,000 1,200 1,400 1,600

Capacity (Th.Acre-ft)

E l e v a

t i o n

( f t )

1800182518501875

190019251950197520002025205020752100212521502175

012345678910111213141516

Area (Thousand Acres)

E l e v a

t i o n

( f t )

Reservoir Capacity

Reservoir Surface Area

Figure 1.13: Kurram Tangi Dam: Elevation-Volume-Surface Area Curves.

KTD: Elevation vs Reservoir Surface Area Curve

52

281

1,079

2,359

4,115

6,330

9,005

12,133

y = 2.582141x 0.522649

R 2 = 0.999916

0

25

50

75

100

125

150

175

200

225

250

275

300

325

350375

0 1,000 2,000 3,000 4,000 5,000 6,000 7,000 8,000 9,000 10,000 11,000 12,000 13,000 14,000

Surface Area (Acres)

E l e v

a t i o n

F t + 1 8 0 0

Figure 1.14 : Elevation-Surface Area curve fit to data.




Ch-1: Introducti on

KTD: Elevation vs Res er voir Capacity Cur ve

20

50

100

150

200

250

300

350

0

20

40

60

80

100

120

140

160

180

200

220

240

260

280300

320

340

360

380

0 200 400 600 800 1,000 1,200 1,400 1,600Volum e (ThAF)

E l e v a

t i o n

F t + 1 8 0 0

Figure 1.15: Kurram Tangi Dam: Elevation-volume curve fit to data.

Figure 1.16: Kurram Tangi Dam. Surface area vs. elevation curve.

KTD Elevation v s Area Curve

52 281

1,079

2,359

4,115

6,330

9,005

12,133

y = 0.162962x 1.913170

R2 = 0.999916

-

1,000

2,000

3,000

4,000

5,000

6,000

7,000

8,000

9,000

10,000

11,000

12,000

13,000

0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360 380

Elevation (1800 +ft)

S u r f a c e

A r e a

( A c r e s

)




Ch-1: Introducti on

KTD Elevation v s Capacity Curve

0 5 39125

287

548

931

1,460

0

200

400

600

800

1,000

1,200

1,400

1,600

0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360 380Elevation (1800+ft)

V o

l u m e

( T h A F )

Figure 1.17: Kurram Tangi Dam: Volume vs. elevation curve.

1.11 DAM LAYOUT

Dam embankmentOnce the site of a dam is selected, the layout of dam embankment is carried out. The

outline of dam is done on a contour map of potential dam location. Following steps are taken(Fig. 1.18).

Data: Let dam crest level = 2100 ft, u/s face slope = 1:3.5, d/s face slope = 1:3.0; contourinterval = 50 ft, river bed level = 1805 ft

Earthfill-Rockfill dam:

Crest:

1. Locate the centerline of dam crest by connecting two points on 2100 ft contour linealong right and left abutments such that the dam has smallest crest length. Thegeologic makeup of the foundations and abutments is also considered. Measure thecrest length.

2. Mark the crest width (e.g. 30 ft) parallel to the selected centerline.

3. Mark chainage along the dam crest with 0+00 at one of abutments, e.g. rightabutment. Determine the dam crest length.




Ch-1: Introducti on

U/s face:

4. Determine the horizontal distance corresponding to 50 ft vertical height for u/s face (= 50 x 3.5 = 175 ft). [3.5 :1 is slope of u/s face]

5. Mark a line A-A’ on u/s face parallel to crest edge spaced 175 ft apart between 2 nd

6. Mark lines B-B’, C-C’, D-D’, E-E’ 175 ft apart between other contour lines of 2000,1950, 1900, 1850 ft, respectively.

contour line of 2050 ft.

7. Mark location of point F of lowest elevation in the river channel.

8. Connect points A-B-C-D-E-F-E’-D’-C’-B’-A’ with a smooth line and connect theoutline with crest edge on u/s face. This defines the dam outline along u/s slopingface.

D/s face :

9. Determine the horizontal distance corresponding to 50 ft vertical height for d/s face (=50 x 3.0 = 150 ft). [3:1 is slope of d/s face]

10. Mark a line G-G’ on d/s face parallel to crest edge spaced 150 ft apart between 2 nd

11. Mark lines H-H’, I-I’, J-J’, K-K’ 150 ft apart between other contour lines of 2000,1950, 1900, 1850 ft, respectively.

contour line of 2050 ft.

12. Locate point L of lowest elevation in river channel on d/s side.

13. Connect points G-H-I-J-K-L-K’-J’-I’-H’-G’ with smooth line and connect this withcrest edge on d/s side. This defines the dam outline along d/s sloping face.

Crest length, Longitudinal Section and Cross section

14. Draw longitudinal section (L-section) along centerline of dam crest. This will providevalley profile between the river left and right abutments (Fig. 1.19).

15. Draw dam cross section at maximum depth (section F-L at Ch 7+45 in Fig. 1.19), andalso at other chainage, e.g. at every 200 ft apart (Fig. 1.19).

The layout of concrete gravity dam is similar to earthfill dams with the exception thatu/s and d/s face slopes are very small (u/s ~ 1H:10V, d/s ~ 0.7H:1V)

Concrete gravity dam:

Dam appurtenants

The layout of dam appurtenants (spillway, outlet, diversion tunnel, power house, etc)are determined such that space requirements of all dam components is adequately met. Manytrial may be needed to finalize the layout of dam embankment and dam appurtenants.

Figs 1.20 to 1.23 describe the alternate layouts for Kurram Tangi dam for damembankment and dam appurtenants. Figs 1.24 to 1.29 show layout of major dams in Pakistan.




Ch-1: Introducti on

Figure 1.18: Topographic surface contours at a dam and layout of dam outline.

2 0 5 0

2 0 0 0

1 9 5 0 1

9 0 0 1

8 5 0

2 1 0 0

2 1 0 0 2

0 5 0

2 0 0 0 1

9 5 0 1

9 0 0 1

8 5 0

Dam Crest;El = 2100 ft

R I V E R

DOWNSTREAMSLOPING FACE

UPSTREAMSLOPING FACE

A

B

C

D

E

A’

B’

C’

D’

E’

G

H

I

G’

H’

I’

J J’

K K’

L

F

SLOPE: u/s = 3.5H:1V; d/s = 3.0H:1V; SCALE = 1:5000.

2+00 4+00 6+00 8+00 10+00 12+00 14+00

Crestlength =1650 ft

175 ft

175 ft

175 ft

175 ft

175 ft

30 ft150 ft

150 ft

150 ft

150 ft

150 ft




Ch-1: Introducti on

Dam Crest; El = 2100 Ft, Length = 1650 ft

Chainage (ft) 2 + 0 0

4 + 0 0

6 + 0 0

8 + 0 0

1 0 + 0 0

1 2 + 0 0

1 4 + 0 0

0 + 0 0

1 6 + 0 0

1800

1900

2000

2100

E l e v a t i o n

( f t )

(a) Longitudinal section

Dam crest: El = 2100 ft, width = 30 ft Normal conservation level = 2081.6 ft

U/s slope =1V:3.5H

D/s slope =1V:3.0H

River level = 1805 ft885 ft1032 ft

(b): Dam maximum cross section at F-L Ch 7+45 ft.

295 ft

1947 ft

Dam crest: El = 2100 ft

River level = 1805 ft

675 ft787 ft

(c): Dam X-section at Ch 4+00 ft.

225 ft

1492 ft

Valley El = 1875-1950 ft

225 ftEl = 1875 ftEl = 1875 ft


River level = 1805 ft

765 ft578 ft

(d): Dam X-section at Ch 12+00 ft.

255 ft

1373 ft

El = 1845

El = 1935 ft165 ft


420 ft368 ft

(e): Dam X-section at Ch 14+00 ft.

140 ft

818 ft

El = 1960 ftEl = 1995 ft105 ft

Figure 1.19 : Longitudinal and cross section of dam of Fig. 1.18. Scale:




Ch-1: Introducti on

Figure 1.20 : Contour map of dam area of Kurram Tangi Dam.




Ch-1: Introducti on

Figure 1.21 : Dam embankment layout of Kurram Tangi Dam.




Ch-1: Introducti on

Figure 1.22 : Layout plan of concrete face rockfill dam (CFRD) embankment andappurtenances for Kurram Tangi Dam.




Ch-1: Introducti on

Figure 1.23 : Layout plan of concrete gravity dam embankment and appurtenances forKurram Tangi Dam.




Ch-1: Introducti on

1.12 DAM ENVIRONMENTAL IMPACTS

Construction of dams significantly alters the river flow regime. The flow in floodseason is considerably reduced while the flow in other months is increased. The changed flow

pattern affects the ecology and echo system of the river d/s reaches. The dam constructionaffects the migration of cold-water fish for their annual spawning voyage to u/s cold-waterregions. However the dam reservoir provide an excellent place for supervised fishdevelopment. The river may have cropped area which is seasonally flooded by the river floodflows (sailaba area). Construction of dam may lower the flood flows thus the sailaba areaneed to be irrigated by alternative means. Affected area adjacent to the dam may be providedsupplemental canal or tubewell irrigation facilities. Waterlogging and high watertable mayappear in some places above or below the dam site.

The sediment carried by the flood water get trapped in the dam and thus a small amountof sediments enters the d/s reach of the rivers. The imbalance in the sediment flow combinedwith educed flood flows causes a aggradations of the river bed. This slowly lead to raising ofthe flood levels in the affected river reach requiring a constant raising of flood dikes andspurs. The sediment reduction due to dams lead to erosion/degradation of the river delta at theentrance to the ocean. Thus erosion of coastal areas is negatively affected by the constructionof dams.

It is required that environmental impacts of dam may be evaluated independently andnecessary mitigation measures may be taken to mitigate and minimize the adverseenvironmental impacts.

1.13 RESETTLEMENT

The construction of dam requires large land area to be occupied by dam embankment,spillway channel, outlet canals, hydropower plant, offices, approach roads, housing facilities,etc. In addition the reservoir occupies very large surface area in many square kilometers. Thearea to be occupied by a dam and reservoir has to be possessed before the construction of thedam. The affected area may be under mix of private and public ownership. The area may be

partly or wholly used for various productive purposes as cropping, grazing, rock quarrying,

public entertainment, parks, residential, commercial or industrial purposes, etc. Most of damsites are usually remote to present urban and industrial centers; thus a significant part of theaffected area may be barren and unproductive.

Construction of dam will deprive the current occupants of the area from productive benefits. Nevertheless some inhabitants occupying the river banks and nearby villages will beneeded to be moved out of the area and resettled. The affected persons will not only loosetheir residential houses but most often their means of livelihood (agriculture, small tomedium business etc.) In addition the dam and reservoir may inundate some places of social-religion nature. Some transportation corridors (rail lines, highway, and other roads) may getsubmerged. Thus dam project must include a plan to resettle the affected persons to new




Ch-1: Introducti on

places, restoring their economic livelihood, etc which is socio-politically acceptable to theaffected population groups. The affected persons may be provided compensation in the formof cash, kind (equivalent housing and business units in some nearby areas). It is alsoimportant to ensure the social and cultural harmony and adjustment of the people moving to

new locations.The transportation corridors have to be moved to new locations above and away from

the dam and reservoirs. The religious and social/cultural monuments and places must be planned to be protected by flood dikes, by moving to higher and safer levels, etc. Else theaffected persons will react very strongly to the dam project, jeopardizing the whole project.Monuments of lesser importance may not be protected due to the large numbers. Varioussocio-cultural-political groups must be approached, contacted and satisfied to come withsuitable resettlement plans, which is acceptable to both the affected persons and the damowners.

Fig. Dam failure.




Ch-1: Introducti on

Figure 1.24: Layout and cross section of Mangla Dam.




Ch-1: Introducti on

Figure 1.25: Layout plan and cross section of Tarbella Dam.




Ch-1: Introducti on

Figure 1.26: Layout plan and cross section of Hub Dam.




Ch-1: Introducti on

Figure 1.27: Layout plan and cross section of Khanpur Dam.




Ch-1: Introducti on

Figure 1.28: Layout plan and cross section of Simly Dam.




Ch-1: Introducti on

Figure 1.29: Layout plan and cross section of Bolan Dam.

References:

To be completed.



T ARIQ . 2008 . D AM AND R ESERVOIR E NGINEERING 2-1 Chapter-2 Dam Hydrology and Sedimentation

Chapter - 2

DAM HYDROLOGY AND SEDIMENTATION

2.1 PURPOSES

Hydrologic analysis is very important study for any dam and reservoir project.

Hydrological study is required to establish:

i. Water availability/yield (average, dependable, probable etc); flow duration curve

(FDC) for run-of-the-river hydropower projects

ii. Water demand (in coordination with irrigation team)

iii. To determine storage volume required to meet the demand (live, dead, gross)

iv. Flood analysis for purposes of river diversion

v. Capacity of diversion tunnels and height of coffer dam (with hydraulic team)

vi. PMP/PMF/ Project design flood

vii. Flood surcharge for spillway design flood vis-à-vis spillway capacity (with

hydraulic team)

viii. Wave height/wave run up (+ hydraulic team)

ix. Reservoir Sedimentation

x. Reservoir operation study /reservoir simulations

xi. Reservoir rule curves

2.2 TERMINOLOGY (Punmia p-219)

Water yield . This is the amount of water that can be supplied from the reservoir in a specific

interval of time (usually one year).

Safe/firm yield . The maximum of water that can be guaranteed during a critical dry period.

Secondary yield . Quantity available in excess of safe yield during periods of wet years / high

floods.

Average yield . Arithmetic average of firm + secondary yield over a long period of time.






at dam site may be extended/ synthesized by known hydrologic/ statistical procedures. The

accuracy of the measured flow data is also important. Whenever possible the accuracy of the

measured flow data must be checked for symmetric/non symmetric errors, measurement

errors and stationarity/homogeneity/trend in the data. The dam may be used to store the flows

of the river over which dam is constructed or else flows of other rivers may be diverted to the

dam site for supplementing the storage. For an off-channel storage dam all river flows may

not be diverted to the dam due to limited diversion feeder channel capacity.

The length of flow record is preferably 100+ years. For most dam sites minimum flow

data of 20 to 30 years is needed to undertake meaningful hydrological analysis. Following

methods may be used to determine the river flows in order of preference

1- Historic stream flow is data available at the dam site for sufficient long period :

Data of measured flows (hourly, average daily record as cfs or m 3

2- Flow data at dam site (Q

/s) are available for

long time. Use the data directly to determine river yield.

d ) is available for short period but flow data of same

river at a u/s or d/s distant location (Q L

Develop correlation between flow at the two sites Q

) is available for long period.

d = F(Q L

3- Historic flow record is available on the same river for long time period but for a

location at some distance u/s or d/s from the dam site (Q

).and then extend flow

record for dam site using the flow record of the u/s or d/s site and the derivedcorrelation.

S

A suitable catchment yield/runoff model or snow melt relationship is derived from Qs

and P record. The flow from the area intervening the measurement and dam site is

synthesized from the derived yield/runoff model for the intervening area, Q

). There is no

measurement record for the dam site. Historic rainfall (P) record is also

available for the catchment area.

I =

F(A I ,P I) The flow at dam site is determined from the flow record of the distant site

and the synthesized flow of the intervening area as: Q d = Q S ± Q I . The relationship

should preferably be verified from available record, if any. This procedure was used

for Kurram Tangi Dam.




4 Short flow data at dam site and long rainfall data for the catchment area.

Develop P-Q relationship of the catchment area using the data for the period of flow

record. Extend the flow record for other period using the derived P-Q relationship and

the historic P data for the remaining period. This is done for monthly or annual basis.In some cases rainfall record for dam catchment area is available for short time only

but long time rainfall record is available at a nearby site. Then correlate the rainfall

for the two sites on the basis of short time concurrent data and then use the rainfall

data of the nearby site to synthesis rainfall record for the dam site on the basis of

correlation. This method was used for generating long term flow synthesis for Mirani

Dam (NESPAK 1992).

5- Short flow record at dam site but a long flow record at a nearby river (Q N

Develop a correlation between flow at the two sites for the period of overlapping

record as Q

)having similar hydrologic conditions (rainfall, catchment hydrologic

characteristics, etc).

d = F(Q N

6. No flow data for the dam site river but satisfactory flow record for a nearby

basin (Q

). Then extend flow record for the dam site using the flow record

of the second river site and the derived correlation.

B

Develop P-Q relationship for the site having rainfall and flow record as Q

) of similar or different hydrologic characteristics in the region.

Precipitation data is available for the two sites/basins.

B =

F(A B ,P B). Due to similarity of hydrologic conditions in the two areas, the runoff

generation is expected to be similar. Thus use the derived P-Q relation using the

rainfall record for the dam site as Q d = F(A d ,P d

7 No flow record at dam site or nearby location. Rainfall data available at dam site

or a nearby location:

). In case the hydrologic conditions are

not same, then the underlying factors F in P-Q relationship may be modified in

consideration of hydrology characteristics of the two basins and the rainfall to account

for differing hydrologic conditions at the dam site. Small Dam

Determine flow at dam site using local or regional P-Q models as Q= F(P) to convert

monthly rainfall data to flow data. e.g. for small dam Jammergal Dam average

monthly flows Q (mm) were determined by Small Dams Organization (SDO, 1992)




using average monthly rainfall P (mm) for a nearby station (Jhelum city 20 miles from

dam site) as Q = 0.045 (P – 20) 0.35 . A subsequent study based on the measured flow

and rainfall data at the dam site for the period 1991-1999 (Tariq 2000 and Tariq 2004)

showed that the regional model could produce better results if site-specific rainfall

data is used and using the same model or a modified models as Q = 0.046 (P - 10) 0.35 .

The rainfall- monthly runoff (Q, mm) at Gandiali dam was found from monthly

rainfall (P, mm) as Q = 0.00815 P + 0.001938 P 2

2.3.2 Stochastic Data Generation from Short Data:

on the basis of 1961-79 data, and the

long term monthly flow data was synthesized by the equation (Nespak 1988).

Stochastic principle may be used to generate long time data on the basis of short-term

data statistics (mean, variance, skewness, kurtosis). Various models used to extend datainclude Auto-correlation (AR) models, Moving Average (MA) models, ARMA model,

ARIMA models, Seasonal/non-seasonal flow models (e.g. Thomas-Fierring). The generated

data have the same statistical properties as the original short term data. Seasonal models will

provide monthly flows, and Non-seasonal models will provide annual flows.

2.3.3 Flows Diverted From Other River

For a dam built on the active river (on channel), all river flows (small or large flows,

base flow, flood flow) will enter into the dam and hence is fully available for storage and

usage. Thus 10-d, monthly, yearly average flows volumes are meaningful.

When canals/feeder channels carry water from a river to an off-channel storage dam,

or from another river to the nearby dam, the amount of diverted water depends on the

feeder/canal capacity as well as on the instantaneous river flows. Then Q diversion equals

minimum of Q river , and Q canal capacity

For Kaitu river the total flow in 1985 at Spinwam is 208 ThAF out of which 65 ThAF

is reserved for local and d/s uses and 143 ThAF is available for diversion. But only 122 ThAF

could be diverted to KT dam for a diversion channel of 1500 cfs capacity (1500 cfs = 1086

ThAF/annum) due to capacity limit of the diversion channel and/or river flow.

(Fig. 2.2). The diversion flow data is required on

continuous basis or with very short measurement interval e.g. 1 hour, particularly if discharge

variations in the feeder river are very rapid. Using average daily flow data will result in an

over estimate. Diverted flow statistics are subsequently determined.




Kaitu River Flows 1985

0

200

400

600

800

1000

1200

1400

1600

1800

2000

1 - J u n

8 - J u n

1 5

- J u n

2 2

- J u n

2 9

- J u n

6 - J u

l

1 3

- J u

l

2 0

- J u

l

2 7

- J u

l

3 - A u g

1 0

- A u g

1 7

- A u g

2 4

- A u g

3 1

- A u g

7 - S e p

1 4

- S e p

2 1

- S e p

2 8

- S e p

F l o w

( C f s )

Total Flow available for diversion

Actual diversion

Figure 2.2: Flow diversion from Kaitu River to KT dam.

2.3.4 Data Processing

Average flows for Storage Reservoir

Data is processed to determine river inflow volumes on 10-daily, monthly, and annual basis. Determine average, standard deviations, skewness, minimum, and maximum flows on

10-daily, monthly and yearly basis, average annual flow, average monthly flow, etc (Table

2.1, Figs. 2.3 to 2.5).

Flow duration curve for hydropower projects

The flows for the run-of-the-river hydropower projects are shown in the form of flow

duration curve (FDC) which describes the exceedence probability for selected flow discharge.

The number of days (N i) when flow exceeded selected flow (Q i) is determined for each year

of record; this can be done by using spreadsheet function as: N i = COUNTIF(Range of data

cells,">Q i"). This is r epeated for all discharge ranges with discharge increment of ΔQ. The

percent exceedence is given as: P(Q ≥Q i) (%) = N i/ΣN i

×100. A curve is drawn between Q (on

y-axis) and P(on x-axis). This procedure is explained in Tables 2.2 and 2.3 and the resulting

flow duration curve is shown in Fig. 2.6.




Table 2.1: Flows into Kurram Tangi Dam (Thousand acre feet).

Year Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec Total

1971 29 22 17 31 25 35 93 89 25 18 21 36 441

1972 22 22 48 112 168 74 72 66 62 28 29 46 7481973 39 40 84 125 83 41 274 127 59 49 27 32 9791974 34 31 33 55 130 29 104 88 51 28 21 34 6381975 32 30 50 83 102 80 50 166 96 29 26 38 7811976 36 41 56 135 110 47 67 96 55 52 29 41 7651977 37 30 24 93 56 30 98 55 35 37 28 36 5591978 31 26 178 80 52 50 74 148 32 40 39 35 7861979 34 50 85 136 111 55 85 205 57 53 41 42 9551980 42 59 124 99 57 142 98 74 51 57 41 52 8971981 73 43 145 145 131 56 134 91 56 52 41 48 1017

1982 73 43 134 139 135 57 127 89 56 52 41 47 9931983 48 36 100 273 300 105 76 170 84 60 47 53 13501984 51 44 50 70 39 38 151 137 94 40 44 49 8081985 49 35 34 91 26 26 54 92 34 35 25 40 5411986 40 32 56 87 92 49 86 123 35 37 28 49 7151987 40 30 153 74 132 68 84 52 46 41 27 34 7811988 34 29 108 89 49 59 94 129 60 32 22 45 7491989 42 24 63 91 70 43 97 96 36 34 23 34 6541990 36 36 95 145 101 62 156 120 56 45 37 37 9271991 44 45 107 206 259 115 120 111 74 50 40 47 1219

1992 50 54 83 216 246 140 117 142 85 59 47 55 12951993 54 40 153 152 122 91 149 81 76 53 39 46 10561994 48 51 73 103 88 36 122 63 68 67 41 52 8131995 51 35 79 189 127 59 85 74 42 57 31 44 8741996 45 40 71 70 110 147 83 129 39 69 29 33 8651997 34 30 50 161 180 125 55 65 30 87 48 47 9101998 52 86 215 265 160 60 97 94 111 64 40 48 12941999 64 82 60 33 30 17 71 100 58 39 36 33 6212000 41 36 32 22 24 40 30 41 42 24 14 26 3702001 27 17 30 38 17 56 81 61 40 15 13 18 414

Average 43 39 84 116 107 66 99 102 56 45 33 41 833st dev 12 15 49 64 70 36 44 39 21 16 10 9 250






Kurram Tangi Dam: 1971-2001 10-daily Synthesised Inflow (Th. AF)

0

20

40

60

80

100

120

140

J a n 0 1 - 1

0 ,

7 1

J a n 0 1 - 1

0 ,

7 3

J a n 0 1 - 1

0 ,

7 5

J a n 0 1 - 1

0 ,

7 7

J a n 0 1 - 1

0 ,

7 9

J a n 0 1 - 1

0 ,

8 1

J a n 0 1 - 1

0 ,

8 3

J a n 0 1 - 1

0 ,

8 5

J a n 0 1 - 1

0 ,

8 7

J a n 0 1 - 1

0 ,

8 9

J a n 0 1 - 1

0 ,

9 1

J a n 0 1 - 1

0 ,

9 3

J a n 0 1 - 1

0 ,

9 5

J a n 0 1 - 1

0 ,

9 7

J a n 0 1 - 1

0 ,

9 9

J a n 0 1 - 1

0 ,

0 1

Month,10-Day period and Year

1 0 - d a y

K T D i n f l o w

( T h

. A F )

Figure 2.5: Historic 10-day inflows to Kurram Tangi Dam site.

Golen Gol Hydro Power Project

Flow Duration Curve (1993-2006)

05

101520253035404550556065707580859095

100

0 5 10 15 20 25 30 35 40 45 50 55 60 65 70 75 80 85 90 95 100

Exceedence Time (%)

D i s c

h a r g e

( m 3 / s

e c

)

Av 93-06199319952004

Figure 2.6: Flow duration curve for Golen Gol Hydropower Project.




Table 2.2: Flow duration analysis: Golen Gol River 1995.

GOLEN GOL: Historic daily discharge (m 3 Flow durationanalysis /s) for 1995

Day Jan. Feb. Mar. Apr. May Jun. Jul. Aug. Sep. Oct. Nov. Dec. Qm3 No./s

%Exced

1 5.6 6.0 5.0 4.6 5.6 11.4 46.4 58.7 46.5 23.4 15.1 5.8 4.56 365 100

2 5.7 6.0 5.1 4.6 6.1 13.2 48.6 54.9 45.5 23.1 13.8 5.9 5 332 913 5.9 5.9 5.2 4.7 6.3 12.8 49.4 58.0 43.0 22.8 12.5 6.0 10 180 494 6.1 5.8 5.2 4.7 6.5 12.0 49.3 59.7 43.1 22.5 11.3 6.0 15 150 415 6.2 5.8 5.4 4.7 6.7 13.1 50.9 57.4 42.5 22.0 10.4 6.1 20 128 356 6.3 5.7 5.4 4.8 7.1 16.9 56.1 60.6 41.7 21.3 9.4 6.2 25 112 317 6.5 5.6 5.5 4.8 7.6 19.5 61.7 60.8 41.7 21.1 8.5 6.2 30 103 288 6.5 5.6 5.5 4.9 8.1 23.6 59.9 58.8 39.5 20.7 8.3 6.1 35 96 269 6.4 5.5 5.4 4.9 8.6 30.1 59.3 55.9 38.8 20.6 8.2 6.0 40 89 24

10 6.4 5.5 5.3 5.0 9.8 33.2 61.8 56.3 37.6 20.5 8.0 6.1 45 67 1811 6.3 5.4 5.3 5.0 10.6 38.7 63.6 56.7 37.1 20.3 7.9 6.1 50 55 1512 6.3 5.4 5.1 5.0 11.8 41.9 63.1 59.7 36.2 20.0 7.8 6.0 55 47 1313 6.3 5.4 5.0 5.0 13.1 43.7 61.1 61.9 35.6 19.8 7.5 6.0 60 26 714 6.2 5.3 5.0 5.0 13.0 42.7 59.9 61.2 34.0 19.6 7.3 5.9 65 5 115 6.2 5.3 4.8 5.1 13.2 42.9 60.9 61.0 33.2 19.5 7.2 5.9 70 4 116 6.2 5.3 4.9 5.0 13.6 44.0 61.4 58.5 32.6 19.7 7.1 5.9 75 3 117 6.3 5.3 4.8 5.1 13.9 46.0 59.3 57.6 31.7 19.4 6.9 5.9 80 3 118 6.2 5.4 4.8 5.1 13.1 45.9 60.6 57.7 30.6 19.5 6.8 5.9 85 3 119 6.2 5.4 4.7 5.1 12.4 43.1 61.3 57.9 29.4 19.3 6.6 5.8 90 3 120 6.2 5.3 4.7 5.1 12.2 43.2 61.6 58.7 28.0 19.0 6.5 5.8 95 1 021 6.3 5.4 4.7 5.2 12.1 43.6 62.5 55.2 27.0 18.9 6.5 5.8 100 0 0

22 6.3 5.4 4.7 5.4 12.0 41.9 62.5 53.7 26.6 18.8 6.5 5.7

23 6.2 5.3 4.7 5.4 11.7 42.5 65.2 52.7 26.2 18.7 6.3 5.7

24 6.3 5.3 4.7 5.4 11.8 42.5 92.6 51.2 25.9 18.6 6.3 5.7

25 6.2 5.2 4.7 5.4 11.9 42.8 96.8 53.6 25.5 18.4 6.2 5.7

26 6.3 5.1 4.8 5.6 11.6 44.6 93.8 52.1 25.4 18.3 6.1 5.727 6.3 5.0 4.7 5.7 11.4 44.2 72.8 55.1 25.2 17.9 6.0 5.7

28 6.3 5.0 4.7 5.7 11.3 43.8 62.1 53.0 24.6 17.4 6.0 5.7

29 6.2 4.6 5.7 11.2 43.0 63.0 49.6 24.4 17.0 5.9 5.6

30 6.2 4.6 5.5 11.3 42.8 62.9 47.2 24.1 16.8 5.8 5.6

31 6.2 4.6 11.2 45.6 62.6 45.1 16.5 5.56.

Av-1 6.2st 5.7 5.3 4.8 7.2 18.6 54.4 58.1 42.0 21.8 10.5 6.0

Av 2 6.2nd 5.4 4.9 5.0 12.7 43.2 61.3 59.1 32.8 19.6 7.2 5.9

Av 3 6.2rd 5.2 4.7 5.5 11.6 43.4 72.4 51.7 25.5 17.9 6.2 5.7

Av mnth 6.2 5.4 5.0 5.1 10.5 35.3 63.0 56.1 33.4 19.7 8.0 5.9

Table 2.3: Result of flow duration analysis: Golen Gol River 1993-2005.

Flow 1993 1994 1995 1996 1997 1998 1999m No /s % No % No % No % No % No % No %

0 364 100 365 100 365 100 365 100 365 100 365 100 365 1005 331 91 365 100 332 91 364 100 365 100 334 92 365 100

10 223 61 159 44 180 49 188 52 187 51 224 61 250 6815 156 43 122 33 150 41 129 35 132 36 149 41 139 3820 130 36 112 31 128 35 109 30 116 32 93 25 117 3225 96 26 83 23 112 31 86 24 85 23 68 19 105 2930 86 24 79 22 103 28 76 21 61 17 23 6 92 25

35 64 18 72 20 96 26 63 17 54 15 9 2 69 19




40 45 12 68 19 89 24 46 13 27 7 1 0 7 245 32 9 48 13 67 18 25 7 14 4 0 0 0 050 15 4 43 12 55 15 3 1 7 2 0 0 0 055 2 1 30 8 47 13 0 0 2 1 0 0 0 060 0 0 15 4 26 7 0 0 2 1 0 0 0 065 0 0 5 1 5 1 0 0 2 1 0 0 0 0

70 0 0 4 1 4 1 0 0 2 1 0 0 0 075 0 0 2 1 3 1 0 0 2 1 0 0 0 080 0 0 1 0 3 1 0 0 2 1 0 0 0 085 0 0 1 0 3 1 0 0 1 0 0 0 0 090 0 0 1 0 3 1 0 0 0 0 0 0 0 095 0 0 1 0 1 0 0 0 0 0 0 0 0 0

100 0 0 0 0 0 0 0 0 0 0 0 0 0 0Flow 2000 2001 2002 2003 2004 2005 Totalm No /s % No % No % No % No % No % No %

0 366 100 365 100 365 100 364 100 365 100 365 100 4744 1005 366 100 365 100 365 100 298 82 363 99 361 99 4574 96

10 161 44 190 52 167 46 155 43 161 44 228 62 2473 5215 115 31 72 20 107 29 105 29 101 28 128 35 1605 3420 69 19 7 2 53 15 80 22 56 15 98 27 1168 2525 3 1 1 0 19 5 55 15 16 4 74 20 803 1730 0 0 0 0 11 3 31 9 0 0 57 16 619 1335 0 0 0 0 1 0 0 0 0 0 48 13 476 1040 0 0 0 0 0 0 0 0 0 0 38 10 321 745 0 0 0 0 0 0 0 0 0 0 16 4 202 450 0 0 0 0 0 0 0 0 0 0 8 2 131 355 0 0 0 0 0 0 0 0 0 0 3 1 84 260 0 0 0 0 0 0 0 0 0 0 2 1 45 165 0 0 0 0 0 0 0 0 0 0 1 0 13 070 0 0 0 0 0 0 0 0 0 0 0 0 10 075 0 0 0 0 0 0 0 0 0 0 0 0 7 0

80 0 0 0 0 0 0 0 0 0 0 0 0 6 085 0 0 0 0 0 0 0 0 0 0 0 0 5 090 0 0 0 0 0 0 0 0 0 0 0 0 4 095 0 0 0 0 0 0 0 0 0 0 0 0 2 0

100 0 0 0 0 0 0 0 0 0 0 0 0 0 0

2.3.5 Dependable Yield

Most often the water requirements for agriculture and other purposes are very

enormous and all river flows could be used for economic gains. However it is needed that

scale of demand be selected such that the projected or design demand must be met adequately

in most years. Else the scale of infrastructure development will remain underutilized for large

proportion of time. The annual dependable yield is determined from the historic or

synthesized data on annual/ seasonal basis. Since actual flows of river can vary considerably

over the days and may not be utilized without a storage dam of suitable capacity. The

dependable yield sets the maximum usable potential of water from the river system. Thus the

project demand is tailored to match the dependable yield. Following procedure is followed to

determine the dependable yield.




• Arrange annual flow volume data of N years in descending order

• Assign serial number n (n = 1 to N)

• Dependability (p%) of ‘n’the discharge event = n/(N+1) * 100

• For pre-selected dependability P%, find m th value where m = (N+1) * P/100. Read the

mth

• This procedure is valid for a seasonal storage only where volume stored in one season

is released in next irrigation season within one water cycle of one year.

flow value and is P% dependable yield of the river.

• For a large size carry over dam dependable flow equals the average flow over a

couple of years since storage reservoir will considerably alter the outflow volumes.

Example 2.1:

The annual synthesized inflow of Kurram and Kaitu Rivers into the Kurram Tangi

dam is given in Table 2.4 (average annual inflow to the dam is 833 Th.AF). Determine the

dependable yield with and without dam.

Solution:

The dependability of different flows is determined in Table 2.4 and shown in Figure

2.7. From the Table and Figure it is seen that 50, 60, 70, 80 90 and 95% dependable yield of

the river without the dam is as 810, 775, 745, 630, 460, and 400 ThAF per annum,

respectively. Construction of the dam will increase the dependable yield at 90% level from

460 ThAF to as 795 ThAF per annum with dam of 600 ThAF live storage capacity and as 785

ThAF with dam of 550 ThAF live storage capacity (Table 2.6 and Figs. 2.18 for

dependability analysis with dam). The dependability of 795 ThAF without dam is about 52%

only.




Table 2.4: Dependable flow of Kurram and Kaitu Rivers into Kurram Tangi Dam.

A: Historic flow data (ThAF) B: Data arranged in descending order

Year Kurram R Kaitu River Total Year Flow (ThAF) Order No (n) DependabilityP (%)

1971 362 79 441 1983 1,350 1 3.131972 663 86 748 1992 1,295 2 6.25

1973 814 165 979 1998 1,294 3 9.381974 508 130 638 1991 1,219 4 12.50

1975 644 137 781 1993 1,056 5 15.631976 617 147 765 1981 1,017 6 18.751977 481 78 559 1982 993 7 21.88

1978 661 125 786 1973 979 8 25.001979 753 203 955 1979 955 9 28.131980 713 183 897 1990 927 10 31.25

1981 761 256 1,017 1997 910 11 34.38

1982 664 329 993 1980 897 12 37.501983 1,083 267 1,350 1995 874 13 40.63

1984 625 183 808 1996 865 14 43.75

1985 419 122 541 1994 813 15 46.881986 589 125 715 1984 808 16 50.00

1987 598 183 781 1978 786 17 53.13

1988 573 177 749 1987 781 18 56.251989 490 164 654 1975 781 19 59.381990 608 319 927 1976 765 20 62.50

1991 872 346 1,219 1988 749 21 65.631992 882 413 1,295 1972 748 22 68.751993 749 307 1,056 1986 715 23 71.88

1994 577 236 813 1989 654 24 75.001995 601 273 874 1974 638 25 78.131996 670 194 865 1999 621 26 81.25

1997 605 305 910 1977 559 27 84.381998 865 429 1,294 1985 541 28 87.50

1999 426 195 621 1971 441 29 90.632000 279 91 370 2001 414 30 93.75

2001 306 108 414 2000 370 31 96.88

Average 628 205 833 P(%) = n/32*100




Inflows to Kurram Tangi Dam from Kurram and Kaitu Rivers

05

101520253035404550556065707580859095

100

300 400 500 600 700 800 900 1,000 1,100 1,200 1,300 1,400

Annual Deprndable Flow (ThAF)

D e p e n

d a

b i l i t y / e x c e e

d a n c e (

% )

Figure 2.7: Dependable flow/yield of Kurram River without dam at KT dam site.

2.4 RESERVOIR LIVE STORAGE CAPACITY

The inflows into the dam (Q) and releases (R) from the dam to meet the planned

demand (D) vary over time (Fig. 2.8); during some months Q < D and for other period Q > D.

Water goes into storage when Q > D and is later released from storage when Q < D. Ripple

Mass Curve of inflows vs demand determines the required live storage capacity for a dam to

meet the target demand adequately over extended service period.

2.4.1 Ripple Mass Curve Analysis

Ripple mass curve compares the cumulative inflows vs. the cumulative demand over a period of time. A hypothetical reservoir of capacity S is considered to be full at some time

after which any deficit (deficit = demand D – flows Q) is met out of the storage. The storage

S required to avert any deficit during the year must be equal to maximum deficit in the year.

The required storage S is determined for each year. The storage S for a reservoir is selected to

ensure averting of deficits in most or all years depending upon the purpose of the storage.

Following procedure is followed.




• Arrange flow (Historic or synthesized data) and demand data for each time period

(10-day or month). The flow data may vary over the years but the demand remains

almost same for all years. The demand may vary over the months (as for irrigation

purposes) or remain constant (as for hydropower development).

• Select apex point P where the reservoir is most likely to be filled up at this point (flow

condition changes from Q > D to Q < D) and start depleting subsequently. Apex point

may be determined by comparing inflows and demand over the years and select a

common time point such that at apex Q ≈ D and later Q < D and extra demand is met

out of storage creating a storage deficit (Fig. 2.8)

• Apex point may also be determined by drawing a tangent line to accumulated inflow

line (dΣQ/dt) and selecting a point where (dΣQ/dt) is largest and start decreasing

subsequently (Fig 2.9).

• Determine accumulated flow ΣQ and demand ΣD. Plot accumulated flow discharge

against time (Figure 2.9).

• Superimpose accumulated demand ΣD starting at point P (Fig. 2.10).

KT Dam: River inflows vs. Demand

0

10

20

30

40

50

60

J a n

0 1 - 1

0 , 7

1

M a r

0 1 - 1

0

M a y

0 1 - 1

0

J u l 0 1 - 1

0

S e p

0 1 - 1

0

N o v

0 1 - 1

0

J a n

1 - 1

0 , 7

2

M a r

0 1 - 1

0

M a y

0 1 - 1

0

J u l 0 1 - 1

0

S e p

0 1 - 1

0

N o v

0 1 - 1

0

J a n

1 - 1

0 , 7

3

I n f l o w

( T h A F )

Flows

Demand

P P

Figure 2.8: Inflows vs Demand for apex point P.




KT Dam: Commulative Inflow

0

200

400

600

800

1000

1200

1400

J a n

0 1 - 1

0 , 7

1

F e

b 0 1 - 1

0

M a r

0 1 - 1

0

A p r

0 1 - 1

0

M a y

0 1 - 1

0

J u n

0 1 - 1

0

J u l 0 1 - 1

0

A u g

0 1 - 1

0

S e p

0 1 - 1

0

O c t

0 1 - 1

0

N o v

0 1 - 1

0

D e c

0 1 - 1

0

J a n

1 - 1

0 , 7

2

F e

b 0 1 - 1

0

M a r

0 1 - 1

0

A p r

0 1 - 1

0

M a y

0 1 - 1

0

J u n

0 1 - 1

0

J u l 0 1 - 1

0

A u g

0 1 - 1

0

S e p

0 1 - 1

0

O c t

0 1 - 1

0

N o v

0 1 - 1

0

D e c

0 1 - 1

0

J a n

1 - 1

0 , 7

3

I n f l o w

( T h A F )

P

P

Figure 2.9: Cumulative inflows for apex point P.

KT Dam: Commulative Inflow

0

200

400

600

800

1000

1200

J u l 0 1 - 1

0

A u g

0 1 - 1

0

S e p

0 1 - 1

0

O c t

0 1 - 1

0

N o v

0 1 - 1

0

D e c

0 1 - 1

0

J a n

1 - 1

0 , 7

2

F e

b 0 1 - 1

0

M a r

0 1 - 1

0

A p r

0 1 - 1

0

M a y

0 1 - 1

0

J u n

0 1 - 1

0

J u l 0 1 - 1

0

A u g

0 1 - 1

0

S e p

0 1 - 1

0

O c t

0 1 - 1

0

N o v

0 1 - 1

0

D e c

0 1 - 1

0

I n f l o w

( T h A F )

P

P

D1

D2

D3

S1 S2 S3

Figure 2.10: Cumulative flow and cumulative demand curves and required storage.




• For small demand the ΣD curve will meet the ΣQ curve before next apex point P. This

ensures that reservoir will become full at this time of the year. For large demand the

cumulative demand curve may meet the cumulative flow curve after more than 1 year

(Fig. 2.10).

• Determine supply deficit for each year as the maximum difference between supply

ΣQ and demand ΣD curve s. This gives required storage for each year corresponding

to the demand. Thus for Fig. 2.10 the required storage to ensure meeting of demand in

the year is S1, S2, and S3 for demands D1, D2, and D3, respectively.

• For large demand, the reservoir may not become full at end of each water year

(example demand D3 in Fig. 2.10). This represents condition for a carry over dam.

• In case demand varies during the year, use appropriate data to determine accumulative

demand and deficit for each flow period (Fig. 2.11).

• Analysis is started from a time when reservoir is most likely to be full (e.g. by 1 st

• Determine maximum deficit and the required storage S for each year of analysis.

Sept.) each year depending upon average flow pattern of the particular river.

• The calculated storage requirements represent live storage for the particular purpose

e.g. irrigation.

• Determine the required reservoir capacity that will ensure supplies for some selected

level by the probability procedure of Section 2.4.3 (also Table 2.6).

2.4.2 Ripple Mass Curve Analysis Using Spreadsheet

The reservoir storage space may be determined conveniently by using a spreadsheet (e.g.

Excel) as under (Table 2.5).

1. Arrange data in columns (time, Q, D) for all years. The flow and demand may be

available on 10-daily basis or on monthly basis.

2. Start the analysis at latest apex point P (e.g. 1 st

3. Determine storage deficit SD for subsequent periods as:

Sept.) when dam may be considered

full every year.

SD t = MAX [{(D t-Q t)+SD t-1

4. Draw graph between time and storage deficit. (Fig. 2.11)

}, 0]




5. Seasonal dams become full and the storage deficit becomes zero on or before the next

apex time of 1 st

6. Determine largest value of the storage deficit SD for each water year of the analysis period. This is the required live storage for that year (Fig. 2.12).

Sep. The carryover dams become full after more than one year and

the storage deficit remain non-zero for few years in a row (Fig. 2.11).

7. Determine the required reservoir capacity that will ensure supplies for some selected

probability level by the probability procedure.

8. The deficit for Kurram Tangi Dam is shown in Figs. 2.13 for different annual

demands 785, 700, 600, 500 ThAF (+ 20 ThAF as annual evaporation losses from

reservoir surface), and varies considerably. Thus required storage capacity depends on

the target annual demand.




Table 2.5 KTD: Mass curve of inflows Q, demand D, Storage deficit SD and Maximum

storage deficit in the water year SD mx

#

(All values in ThAF)

Period Q D SD SD mx Annual CWR = 785, System losses = 20,Total annual Demand = 805

1 Jan 01-10,71 9.8 10.392 Jan 11-20 9.8 11.563 Jan 21-31 9.7 14.494 Feb 01-10 7.8 15.075 Feb 11-20 7.8 19.276 Feb 21-28 6.7 24.817 Mar 01-10 7.8 17.238 Mar 11-20 4.5 23.949 Mar 21-31 5.2 28.65

10 Apr 01-10 5.6 31.74

11 Apr 11-20 7.1 28.8512 Apr 21-30 18.4 18.3513 May 01-10 6.9 24.2214 May 11-20 8.6 26.7415 May 21-31 9.5 28.8716 Jun 01-10 14.5 17.1817 Jun 11-20 14.2 19.3618 Jun 21-30 5.8 23.9319 Jul 01-10 24.6 25.7220 Jul 11-20 15.7 20.3521 Jul 21-31 52.4 19.49

22 Aug 01-10 43.4 14.7523 Aug 11-20 24.5 17.2624 Aug 21-31 21.2 27.12 025 Sep 01-10 13.7 31.25 1826 Sep 11-20 8.1 30.88 4027 Sep 21-30 3.2 31.42 6828 Oct 01-10 5.0 31.01 9529 Oct 11-20 7.0 30.77 11830 Oct 21-31 5.7 31.75 14431 Nov 01-10 6.4 34.62 17232 Nov 11-20 6.6 33.83 20033 Nov 21-30 7.7 18.93 21134 Dec 01-10 11.1 6.94 20735 Dec 11-20 11.5 7.24 20236 Dec 21-31 13.0 6.79 19637 Jan 1-10,72 6.0 10.39 201 27738 Jan 11-20 6.7 11.56 20539 Jan 21-31 9.0 14.49 21140 Feb 01-10 7.2 15.07 21941 Feb 11-20 8.1 19.27 23042 Feb 21-28 6.6 24.81 24843 Mar 01-10 7.9 17.23 25844 Mar 11-20 15.3 23.94 26645 Mar 21-31 25.0 28.65 270

# Period Q D SD SD mx 46 Apr 01-10 27.1 31.74 27447 Apr 11-20 26.4 28.85 27748 Apr 21-30 58.2 18.35 23749 May 01-10 97.7 24.22 16450 May 11-20 38.7 26.74 15251 May 21-31 31.3 28.87 14952 Jun 01-10 18.5 17.18 14853 Jun 11-20 12.3 19.36 15554 Jun 21-30 43.6 23.93 13555 Jul 01-10 32.6 25.72 12856 Jul 11-20 11.9 20.35 13757 Jul 21-31 27.4 19.49 129

58 Aug 01-10 23.3 14.75 12059 Aug 11-20 24.0 17.26 11460 Aug 21-31 18.5 27.12 12261 Sep 01-10 4.6 31.25 14962 Sep 11-20 41.8 30.88 13863 Sep 21-30 15.3 30.88 15464 Oct 01-10 8.9 30.47 17565 Oct 11-20 8.4 30.23 19766 Oct 21-31 10.6 31.15 21867 Nov 01-10 7.5 34.00 24468 Nov 11-20 6.8 33.21 270

69 Nov 21-30 14.8 18.65 27470 Dec 01-10 17.1 6.81 26471 Dec 11-20 14.0 7.11 25772 Dec 21-31 15.0 6.67 24973 Jan 1-10,73 13.9 10.20 245 27474 Jan 11-20 12.8 11.35 24475 Jan 21-31 12.2 14.21 24576 Feb 01-10 11.4 14.79 24977 Feb 11-20 9.9 18.90 25878 Feb 21-28 18.4 24.40 26479 Mar 01-10 33.1 16.91 248

80 Mar 11-20 19.8 23.50 25181 Mar 21-31 31.2 28.12 24882 Apr 01-10 47.8 31.16 23283 Apr 11-20 48.6 28.33 21184 Apr 21-30 28.8 18.02 20185 May 01-10 26.4 23.79 19886 May 11-20 31.7 26.29 19387 May 21-31 25.0 28.38 19688 Jun 01-10 14.5 16.92 19989 Jun 11-20 9.4 19.08 20890 Jun 21-30 16.8 23.56 21591 Jul 01-10 101.5 25.35 13992 Jul 11-20 65.2 20.04 94




# Period Q D SD SD mx 93 Jul 21-31 107.2 19.16 694 Aug 01-10 65.9 14.54 095 Aug 11-20 27.6 17.00 096 Aug 21-31 33.3 26.66 0

97 Sep 01-10 19.5 30.72 1198 Sep 11-20 10.7 30.34 3199 Sep 21-30 29.2 30.88 33

100 Oct 01-10 19.9 30.47 43101 Oct 11-20 16.3 30.23 57102 Oct 21-31 12.9 31.15 75103 Nov 01-10 9.3 34.00 100104 Nov 11-20 8.4 33.21 125105 Nov 21-30 9.4 18.65 134106 Dec 01-10 10.4 6.81 130107 Dec 11-20 9.9 7.11 128108

Dec 21-31 11.36.67 123

109 Jan 1-10,74 9.7 10.20 124 221110 Jan 11-20 9.8 11.35 125111 Jan 21-31 14.7 14.21 125112 Feb 01-10 10.6 14.79 129113 Feb 11-20 10.6 18.90 137114 Feb 21-28 9.8 24.40 152115 Mar 01-10 10.4 16.91 158116 Mar 11-20 7.3 23.50 174117 Mar 21-31 15.1 28.12 187118 Apr 01-10 28.8 31.16 190119 Apr 11-20 16.6 28.33 202120 Apr 21-30 9.7 18.02 210121 May 01-10 13.1 23.79 221122 May 11-20 81.1 26.29 166123 May 21-31 36.0 28.38 158124 Jun 01-10 9.1 16.92 166125 Jun 11-20 8.2 19.08 177126 Jun 21-30 11.2 23.56 189127 Jul 01-10 18.3 25.35 196128 Jul 11-20 33.5 20.04 183129 Jul 21-31 52.2 19.16 150130 Aug 01-10 52.1 14.54 112

131 Aug 11-20 16.6 17.00 112132 Aug 21-31 19.8 26.66 119133 Sep 01-10 17.3 30.72 133134 Sep 11-20 7.2 30.34 156135 Sep 21-30 26.3 30.88 160136 Oct 01-10 11.2 30.47 180137 Oct 11-20 8.2 30.23 202138 Oct 21-31 8.4 31.15 225139 Nov 01-10 6.4 34.00 252140 Nov 11-20 6.6 33.21 279141 Nov 21-30 7.7 18.65 290

142 Dec 01-10 11.1 6.81 285

# Period Q D SD SD mx 143 Dec 11-20 10.7 7.11 282144 Dec 21-31 12.6 6.67 276145 Jan 1-10,75 11.6 10.20 274 335146 Jan 11-20 9.8 11.35 276

147 Jan 21-31 10.2 14.21 280148 Feb 01-10 10.9 14.79 284149 Feb 11-20 12.1 18.90 291150 Feb 21-28 6.9 24.40 308151 Mar 01-10 11.9 16.91 313152 Mar 11-20 11.3 23.50 325153 Mar 21-31 26.6 28.12 327154 Apr 01-10 28.6 31.16 330155 Apr 11-20 22.5 28.33 335156 Apr 21-30 32.3 18.02 321157 May 01-10 37.6 23.79 307158

May 11-20 43.426.29 290

159 May 21-31 20.7 28.38 298160 Jun 01-10 16.1 16.92 299161 Jun 11-20 35.3 19.08 282162 Jun 21-30 28.8 23.56 277163 Jul 01-10 14.5 25.35 288164 Jul 11-20 19.0 20.04 289165 Jul 21-31 16.1 19.16 292166 Aug 01-10 29.1 14.54 277167 Aug 11-20 68.3 17.00 226168 Aug 21-31 69.0 26.66 184169 Sep 01-10 42.0 30.72 173170 Sep 11-20 14.3 30.34 189171 Sep 21-30 39.4 30.88 180172 Oct 01-10 9.7 30.47 201173 Oct 11-20 10.6 30.23 221174 Oct 21-31 8.5 31.15 243175 Nov 01-10 7.8 34.00 269176 Nov 11-20 7.8 33.21 295177 Nov 21-30 10.0 18.65 303178 Dec 01-10 13.2 6.81 297179 Dec 11-20 11.9 7.11 292180 Dec 21-31 13.1 6.67 286

181 Jan 1-10,76 10.2 10.20 286 315182 Jan 11-20 10.3 11.35 287183 Jan 21-31 15.8 14.21 285184 Feb 01-10 13.0 14.79 287185 Feb 11-20 15.7 18.90 290186 Feb 21-28 11.9 24.40 303187 Mar 01-10 12.9 16.91 307188 Mar 11-20 21.2 23.50 309189 Mar 21-31 22.0 28.12 315190 Apr 01-10 38.8 31.16 307191 Apr 11-20 29.1 28.33 307

192 Apr 21-30 67.4 18.02 257




# Period Q D SD SD mx 193 May 01-10 38.3 23.79 243194 May 11-20 33.5 26.29 236195 May 21-31 37.8 28.38 226196 Jun 01-10 21.5 16.92 222

197 Jun 11-20 16.1 19.08 225198 Jun 21-30 9.5 23.56 239199 Jul 01-10 6.7 25.35 257200 Jul 11-20 24.6 20.04 253201 Jul 21-31 35.3 19.16 237202 Aug 01-10 8.2 14.54 243203 Aug 11-20 63.1 17.00 197204 Aug 21-31 24.8 26.66 199205 Sep 01-10 21.6 30.72 208206 Sep 11-20 25.6 30.34 213207 Sep 21-30 8.1 30.88 235208

Oct 01-10 26.630.47 239

209 Oct 11-20 12.2 30.23 257210 Oct 21-31 13.4 31.15 275211 Nov 01-10 9.6 34.00 299212 Nov 11-20 8.5 33.21 324213 Nov 21-30 10.8 18.65 332214 Dec 01-10 12.1 6.81 327215 Dec 11-20 15.2 7.11 319216 Dec 21-31 13.3 6.67 312217 Jan 1-10,77 11.8 10.20 310 434218 Jan 11-20 11.8 11.35 310219 Jan 21-31 13.3 14.21 311220 Feb 01-10 11.2 14.79 314221 Feb 11-20 11.5 18.90 322222 Feb 21-28 7.4 24.40 339223 Mar 01-10 9.0 16.91 347224 Mar 11-20 6.7 23.50 363225 Mar 21-31 8.2 28.12 383226 Apr 01-10 41.3 31.16 373227 Apr 11-20 33.6 28.33 368228 Apr 21-30 18.1 18.02 368229 May 01-10 16.9 23.79 375230 May 11-20 17.6 26.29 383

231 May 21-31 21.0 28.38 391232 Jun 01-10 9.7 16.92 398233 Jun 11-20 8.1 19.08 409234 Jun 21-30 12.5 23.56 420235 Jul 01-10 11.8 25.35 434236 Jul 11-20 50.6 20.04 403237 Jul 21-31 36.0 19.16 386238 Aug 01-10 32.4 14.54 368239 Aug 11-20 6.5 17.00 379240 Aug 21-31 15.8 26.66 390241 Sep 01-10 22.6 30.72 398

# Period Q D SD SD mx 242 Sep 11-20 4.6 30.34 424243 Sep 21-30 7.6 30.88 447244 Oct 01-10 12.1 30.47 465245 Oct 11-20 16.5 30.23 479

246 Oct 21-31 8.8 31.15 501247 Nov 01-10 6.9 34.00 528248 Nov 11-20 9.5 33.21 552249 Nov 21-30 11.6 18.65 559250 Dec 01-10 12.0 6.81 554251 Dec 11-20 11.4 7.11 550252 Dec 21-31 12.9 6.67 544253 Jan 1-10,78 10.7 10.20 543 580254 Jan 11-20 10.2 11.35 544255 Jan 21-31 10.3 14.21 548256 Feb 01-10 9.0 14.79 554257

Feb 11-20 9.418.90 563

258 Feb 21-28 8.0 24.40 580259 Mar 01-10 16.8 16.91 580260 Mar 11-20 139.6 23.50 464261 Mar 21-31 21.8 28.12 470262 Apr 01-10 20.6 31.16 481263 Apr 11-20 29.8 28.33 479264 Apr 21-30 29.4 18.02 468265 May 01-10 23.5 23.79 468266 May 11-20 15.0 26.29 479267 May 21-31 13.8 28.38 494268 Jun 01-10 9.0 16.92 502269 Jun 11-20 21.2 19.08 500270 Jun 21-30 20.2 23.56 503271 Jul 01-10 32.3 25.35 496272 Jul 11-20 20.3 20.04 496273 Jul 21-31 21.1 19.16 494274 Aug 01-10 31.1 14.54 477275 Aug 11-20 69.2 17.00 425276 Aug 21-31 47.6 26.66 404277 Sep 01-10 5.9 30.72 429278 Sep 11-20 8.1 30.34 451279 Sep 21-30 17.9 30.88 464

280 Oct 01-10 21.5 30.47 473281 Oct 11-20 9.6 30.23 494282 Oct 21-31 9.1 31.15 516283 Nov 01-10 16.7 34.00 533284 Nov 11-20 11.0 33.21 555285 Nov 21-30 11.2 18.65 563286 Dec 01-10 11.7 6.81 558287 Dec 11-20 11.0 7.11 554288 Dec 21-31 12.4 6.67 548289 Jan 1-10,79 10.7 10.20 548 565




KT Dam: Ripple Mass Curve Analysis, 1984-1992

137

208

372

427 439

502

613

467

105

0

100

200

300

400

500

600

700

800

S e p

0 1 - 1

0

J a n

1 - 1

0 , 8

4

M a y

0 1 - 1

0

S e p

0 1 - 1

0

J a n

1 - 1

0 , 8

5

M a y

0 1 - 1

0

S e p

0 1 - 1

0

J a n

1 - 1

0 , 8

6

M a y

0 1 - 1

0

S e p

0 1 - 1

0

J a n

1 - 1

0 , 8

7

M a y

0 1 - 1

0

S e p

0 1 - 1

0

J a n

1 - 1

0 , 8

8

M a y

0 1 - 1

0

S e p

0 1 - 1

0

J a n

1 - 1

0 , 8

9

M a y

0 1 - 1

0

S e p

0 1 - 1

0

J a n

1 - 1

0 , 9

0

M a y

0 1 - 1

0

S e p

0 1 - 1

0

J a n

1 - 1

0 , 9

1

M a y

0 1 - 1

0

S e p

0 1 - 1

0

J a n

1 - 1

0 , 9

2

M a y

0 1 - 1

0

S e p

0 1 - 1

0

S t o r a g e

D e

f i c i t ( T h A F )

0

1000

2000

3000

4000

5000

6000

7000

8000

C o m m u

l a t i v e

I n f l o w s a n

d D e m a n

d ( T h A F )Max defict in the water

CommulativeInflows

CommulativeDemand

Figure 2.11: Deficits for Kurram Tangi dam (1984-1992).

KTD: Annual Storage Deficit

2 7 7

2 7 4

2 2 1

3 3 5

3 1 5

4 3 4

5 8 0

5 6 5

4 0 5

3 1 0

1 3 2

1 3 2

1 3 7

2 0 8

3 7 2

4 2 7

4 3 9

5 0 2

6 1 3

4 6 7

1 0 5

8 8 1

0 7

9 3

1 4 6 1 6

9

1 0 8

1 6 6

4 6 5

0

100

200

300

400

500

600

700

J a n 1 - 1

0 , 7

1

J a n 1 - 1

0 , 7

2

J a n 1 - 1

0 , 7

3

J a n 1 - 1

0 , 7

4

J a n 1 - 1

0 , 7

5

J a n 1 - 1

0 , 7

6

J a n 1 - 1

0 , 7

7

J a n 1 - 1

0 , 7

8

J a n 1 - 1

0 , 7

9

J a n 1 - 1

0 , 8

0

J a n 1 - 1

0 , 8

1

J a n 1 - 1

0 , 8

2

J a n 1 - 1

0 , 8

3

J a n 1 - 1

0 , 8

4

J a n 1 - 1

0 , 8

5

J a n 1 - 1

0 , 8

6

J a n 1 - 1

0 , 8

7

J a n 1 - 1

0 , 8

8

J a n 1 - 1

0 , 8

9

J a n 1 - 1

0 , 9

0

J a n 1 - 1

0 , 9

1

J a n 1 - 1

0 , 9

2

J a n 1 - 1

0 , 9

3

J a n 1 - 1

0 , 9

4

J a n 1 - 1

0 , 9

5

J a n 1 - 1

0 , 9

6

J a n 1 - 1

0 , 9

7

J a n 1 - 1

0 , 9

8

J a n 1 - 1

0 , 9

9

J a n 1 - 1

0 , 0

0

A n n u a l S t o r a g e

D e f i c i t ( T

h A F )

Figure 2.12: KTD-Annual max deficit for annual demand of 785+20=805 ThAF.




KTD: Deficit Curve, Demand = 785 + 20 = 805 ThAF

2 7 7

2 7 4

2 2 1

3 3 5

3 1 5

4 3 4

5 8 0

5 6 5

4

0 5

3 1 0

1 3 2

1 3 2

1 3 7

2 0 8

3 7 2

4 2 7

4 3 9

5 0 2

6 1 3

4 6 7

1 0 5

8 8 1

0 7

9 3

1 4 6 1

6 9

1 0 8

1 6 6

4 6 5

0

100

200

300

400

500

600

700

J a n

1 -

1 0

, 7 1

J a n

1 -

1 0

, 7 2

J a n

1 -

1 0

, 7 3

J a n

1 -

1 0

, 7 4

J a n

1 -

1 0

, 7 5

J a n

1 -

1 0

, 7 6

J a n

1 -

1 0

, 7 7

J a n

1 -

1 0

, 7 8

J a n

1 -

1 0

, 7 9

J a n

1 -

1 0

, 8 0

J a n

1 -

1 0

, 8 1

J a n

1 -

1 0

, 8 2

J a n

1 -

1 0

, 8 3

J a n

1 -

1 0

, 8 4

J a n

1 -

1 0

, 8 5

J a n

1 -

1 0

, 8 6

J a n

1 -

1 0

, 8 7

J a n

1 -

1 0

, 8 8

J a n

1 -

1 0

, 8 9

J a n

1 -

1 0

, 9 0

J a n

1 -

1 0

, 9 1

J a n

1 -

1 0

, 9 2

J a n

1 -

1 0

, 9 3

J a n

1 -

1 0

, 9 4

J a n

1 -

1 0

, 9 5

J a n

1 -

1 0

, 9 6

J a n

1 -

1 0

, 9 7

J a n

1 -

1 0

, 9 8

J a n

1 -

1 0

, 9 9

J a n

1 -

1 0

, 0 0

S t o

r a g e

D e f i c i t T

h A F


2 2 2

1 6 4

1 7 0

2 1 8

1 3 7

1 7 2

2 7 5

1 9 8

9 4 1

0 1

1 0 5

1 0 6

8 3

1 4 9

2 5 8

2 6 3

1 9 4

1 7 3

2 2 5

1 0 8

8 1

6 9 8

3

6 9

1 1 6

1 2 8

8 0

1 3 9

3 5 9

0

100

200

300

400

500

600

700

J a n

1 -

1 0

, 7 1

J a n

1 -

1 0

, 7 2

J a n

1 -

1 0

, 7 3

J a n

1 -

1 0

, 7 4

J a n

1 -

1 0

, 7 5

J a n

1 -

1 0

, 7 6

J a n

1 -

1 0

, 7 7

J a n

1 -

1 0

, 7 8

J a n

1 -

1 0

, 7 9

J a n

1 -

1 0

, 8 0

J a n

1 -

1 0

, 8 1

J a n

1 -

1 0

, 8 2

J a n

1 -

1 0

, 8 3

J a n

1 -

1 0

, 8 4

J a n

1 -

1 0

, 8 5

J a n

1 -

1 0

, 8 6

J a n

1 -

1 0

, 8 7

J a n

1 -

1 0

, 8 8

J a n

1 -

1 0

, 8 9

J a n

1 -

1 0

, 9 0

J a n

1 -

1 0

, 9 1

J a n

1 -

1 0

, 9 2

J a n

1 -

1 0

, 9 3

J a n

1 -

1 0

, 9 4

J a n

1 -

1 0

, 9 5

J a n

1 -

1 0

, 9 6

J a n

1 -

1 0

, 9 7

J a n

1 -

1 0

, 9 8

J a n

1 -

1 0

, 9 9

J a n

1 -

1 0

, 0 0

S t o

r a g e

D e f i c i t

T h A F


1 6 6

9 5 1

0 0

1 1 4

8 4

8 6

1 3 1

1 0 0

6 0

6 4

6 8

6 8

3 9

7 3

1 2 7

1 1 3

1 0 2

9 9

1 1 8

7 4

4 9

4 5 5

3

3 8

8 1

7 4

4 6

1 0 4

2 1 2

0

100

200

300

400

500

600

700

J a n

1 -

1 0

, 7 1

J a n

1 -

1 0

, 7 2

J a n

1 -

1 0

, 7 3

J a n

1 -

1 0

, 7 4

J a n

1 -

1 0

, 7 5

J a n

1 -

1 0

, 7 6

J a n

1 -

1 0

, 7 7

J a n

1 -

1 0

, 7 8

J a n

1 -

1 0

, 7 9

J a n

1 -

1 0

, 8 0

J a n

1 -

1 0

, 8 1

J a n

1 -

1 0

, 8 2

J a n

1 -

1 0

, 8 3

J a n

1 -

1 0

, 8 4

J a n

1 -

1 0

, 8 5

J a n

1 -

1 0

, 8 6

J a n

1 -

1 0

, 8 7

J a n

1 -

1 0

, 8 8

J a n

1 -

1 0

, 8 9

J a n

1 -

1 0

, 9 0

J a n

1 -

1 0

, 9 1

J a n

1 -

1 0

, 9 2

J a n

1 -

1 0

, 9 3

J a n

1 -

1 0

, 9 4

J a n

1 -

1 0

, 9 5

J a n

1 -

1 0

, 9 6

J a n

1 -

1 0

, 9 7

J a n

1 -

1 0

, 9 8

J a n

1 -

1 0

, 9 9

J a n

1 -

1 0

, 0 0

S t o

r a g e

D e f i c i t

T h A F


1 1 3

6 7

4 1

7 7

6 2

4 7

8 3

6 6

3 1

2 8

3 1

3 2

1 9

5 0

8 3

7 7

6 5 7

2 8 4

4 3

2 7

2 1 2

8

1 7

4 7

3 9

3 1

7 1

9 9

0

100

200

300

400

500

600

700

J a n

1 -

1 0

, 7 1

J a n

1 -

1 0

, 7 2

J a n

1 -

1 0

, 7 3

J a n

1 -

1 0

, 7 4

J a n

1 -

1 0

, 7 5

J a n

1 -

1 0

, 7 6

J a n

1 -

1 0

, 7 7

J a n

1 -

1 0

, 7 8

J a n

1 -

1 0

, 7 9

J a n

1 -

1 0

, 8 0

J a n

1 -

1 0

, 8 1

J a n

1 -

1 0

, 8 2

J a n

1 -

1 0

, 8 3

J a n

1 -

1 0

, 8 4

J a n

1 -

1 0

, 8 5

J a n

1 -

1 0

, 8 6

J a n

1 -

1 0

, 8 7

J a n

1 -

1 0

, 8 8

J a n

1 -

1 0

, 8 9

J a n

1 -

1 0

, 9 0

J a n

1 -

1 0

, 9 1

J a n

1 -

1 0

, 9 2

J a n

1 -

1 0

, 9 3

J a n

1 -

1 0

, 9 4

J a n

1 -

1 0

, 9 5

J a n

1 -

1 0

, 9 6

J a n

1 -

1 0

, 9 7

J a n

1 -

1 0

, 9 8

J a n

1 -

1 0

, 9 9

J a n

1 -

1 0

, 0 0

S t o

r a g e

D e f i c i t

T h A F

Figure 2.13: KTD- Deficit for annual irrigation demand of 785, 700, 600 and 500 ThAF(1971-2000).




2.4.3: Required Storage Capacity

Storage may be provided to meet the maximum deficit determined during the period

of analysis. This is true when 100% dependable supplies are required for the purposes, e.g.

domestic water supply. For other cases, as for irrigation, providing large enough storage tosatisfy all deficits may be too costly and that some shortage could not be averted at all and

thus may be accepted during few years. In that case storage is provided for selected

probability level in concordance with the scope of water delivery, e.g. 80 to 90% for

irrigation, 50 to 80% for hydropower, etc. Following procedure is followed to determine the

storage required to avert deficits for selected probability levels (Table 2.6).

• Determine the yearly maximum deficit for N years from Ripple curve analysis for

known inflows and selected annual demand. The storage required to meet all deficit inany year equals the maximum deficit of that year. Thus if in the beginning of any year

the storage is available equal to or more than the maximum deficit in that year, all the

deficit in the year will be met out of the storage.

• Arrange yearly required storage (i.e. live storage) data of N years in ascending order.

• Assign serial number n (n = 1 to N)

• Determine the dependability (P %), i.e. percent time (in years) when shortages areaverted when live storage value equals the n th

• Draw a graph between live storage capacity (x-axis) versus % dependability (y-axis)

(Fig. 2.14).

. deficit: P = n/(N+1) * 100.

• For pre-selected dependability P (%), read out the required storage from the graph or

find m where m = (N+1)*P/100 (round up to next integer value). Read the m th

The computations for the ripple mass curve are given in Table 2.5 for Kurram Tangi Dam

for target demand of 805 ThAF (irrigation = 785, evaporation losses less direct rainfall = 20

ThAF). The yearly deficits for annual demand of 785, 700, 600 and 500 ThAF are shown in

Figure 2.13 and given in Table 2.6. The probability of averting shortages for various live

storage capacities is worked out in Table 2.6 and shown in Figure 2.14. It is seen that annual

demand has a significant effect on the required live storage capacity for same level of

flow

value and is P% dependable storage requirements for the dam.




shortage averting. Reservoir operation simulations may be carried out to further evaluate the

live storage requirements.

Table 2.6: Storage requirements at Kurram Tangi dam to avert seasonal shortage. Average

annual flow = 833 ThAF.A: Yearly storage required (ThAF) to

avert seasonal shortageB: Probability of averting shortage forvarious live storage capacity: Data ofyearly storage required arranged in

ascending order

Year

Max storage required for annualdemand of: Annual demand:

RankNo.m

ProbP %785 700 600 500 785 700 600 500

1971 250 150 115 85 88 69 38 17 1 3.21972 277 222 166 113 93 69 39 19 2 6.51973 274 164 95 67 105 80 45 21 3 9.71974 221 170 100 41 107 81 46 27 4 12.91975 335 218 114 77 108 83 49 28 5 16.11976 315 137 84 62 132 83 53 28 6 19.41977 434 172 86 47 132 94 60 31 7 22.61978 580 275 131 83 137 101 64 31 8 25.81979 565 198 100 66 146 105 68 31 9 29.01980 405 94 60 31 166 106 68 32 10 32.31981 310 101 64 28 169 108 73 39 11 35.51982 132 105 68 31 208 116 74 41 12 38.71983 132 106 68 32 221 128 74 43 13 41.91984 137 83 39 19 250 137 81 47 14 45.21985 208 149 73 50 274 139 84 47 15 48.41986 372 258 127 83 277 149 86 50 16 51.61987 427 263 113 77 310 150 95 62 17 54.81988 439 194 102 65 315 164 99 65 18 58.11989 502 173 99 72 335 170 100 66 19 61.31990 613 225 118 84 372 172 100 67 20 64.51991 467 108 74 43 405 173 102 71 21 67.71992 105 81 49 27 427 194 104 72 22 71.01993 88 69 45 21 434 198 113 77 23 74.21994 107 83 53 28 439 218 114 77 24 77.41995 93 69 38 17 465 222 115 83 25 80.6

1996 146 116 81 47 467 225 118 83 26 83.91997 169 128 74 39 502 258 127 84 27 87.11998 108 80 46 31 565 263 131 85 28 90.31999 166 139 104 71 580 275 166 99 29 93.52000 465 359 212 99 613 359 212 113 30 96.8




Kurram Tangi Dam: Live storage vs. Shortage Probability

0

5

1015

20

25

30

35

40

45

50

55

60

65

70

75

80

85

90

95

100

0 50 100 150 200 250 300 350 400 450 500 550 600 650 700Required Live Storage (ThAF)

P e r c e n t t i m e

( y e a r s ) s h o r t a g e a v e r t e d

Annual demand = 785 ThAFAnnual demand = 700 ThAFAnnual demand = 600 ThAFAnnual demand = 500 ThAF

Figure 2.14 : Shortages averted for different live storage capacity and annual demand.

2.4 RESERVOIR TOTAL CAPACITY

The reservoir total capacity is made up of live storage capacity required to regulate

the river flows for the intended purposes (store during high flows and release during low river

flows in relation to target/design demand). Rivers carry large amounts of silt. Under

unobstructed flow conditions the sediment is carried away with water. When a dam/reservoir

obstructs the natural flow regime, a very large part of the sediment (80 to 99%) gets

deposited in the reservoir. The sediment deposition will soon reduce the storage/regulating

capacity of the dam. To ensure that reservoir live/usable capacity remains available for long

time, additional capacity is provided to store the sediment load corresponding to a long time

period. The required reservoir capacity is determined as under:

• Determine required live storage capacity from Ripple mass curve analysis described

in previous section.

• Dead storage volume is selected in view of annual sediment inflow volumes such that

dead storage space is filled up in not less than 50 to100 years.

Example: The annual sediment inflow for Kurram Tangi Dam. is 6.2 MST (million

short tons). For sediment wet specific weight of 62.3 lb/cft the sediment volume equal




to = 6.2 x 10 6

• Flood storage space (for a dam with part objective of flood control) is determined by

knowing flood volume which has to be temporarily stored in the dam and then

released, e.g. a 100 year frequency flood may be stored temporarily, with a total flood

volume of say 120 ThAF.

x 2000 /(62.3 x 43560) = 4570 Acre feet/year. Sediment volume inflow

in 50 years = 4570 * 50 = 228,500 AF. Sediment inflow in 100 years = 4570 * 100 =

457,000 AF. Considering 10 to 15% sediment outflow/flushing/setting in live storage

space, sediment volume in 50 years = 194,225 AF and in 100 years = 388,450 AF.

Thus dead storage space may be taken as say 300,000 AF (to hold sediments for 50 –

70 years).

• Total gross storage = live storage + dead storage + flood storage.

• The gross/dead/live storage may be adjusted on basis of dam height, geotechnical,

environmental issues and cost and economic return basis. e.g. KTD 1.2 MAF requires

a dam of height 345 ft + 15 = 360 ft and a 0.9 MAF dam is of height 285 ft+ 15 = 300

ft. Average irrigation shortages for 0.9 MAF dam is = 5% and for 1.2 MAF dam

shortage is 3%, showing a possible small incremental benefits due to larger dam.

• Additional height of dam is provided in excess of gross storage for: i. Flood

surcharge, ii. Wave height for sustained winds, iii. Wave runup over the dam face for

very high winds, iv. Free board for unforeseen emergencies, etc.

2.5 FLOOD ANALYSIS

Dams are required to handle various flood events adequately without any threat to the

safety of the dam. These floods may occur during the construction stage (diversion floods) or

after completion of the construction (spillway floods). Thus it is very essential that magnitude

of various floods, that the dam may face, must be ascertained with accuracy. The floods arecharacterized by the (1) return period/frequency of occurrence, (2) the peak flow rate, (3)

time to the peak flow, (4) duration of the flood, (5) the volume of the flood. All these

parameters affect the design of various components of the dam. Flood analysis describes the

peak flow discharge and complete hydrograph of the selected flood. Flood analysis can be

based on frequency analysis of historic flood data or by catchment modeling (using projected

rain and rainfall-runoff relationship).




2.5.1 FREQUENCY ANALYSIS OF HISTORIC DATA

Frequency analysis of historic flood data is performed to estimate the flood events of

large magnitude / return period. The floods are based on the historic data of the flood at the

dam site during last many years. During flood event the discharge must be observed andrecorded frequently preferably at half to one-hour interval. In case actual flood data is not

available, the floods may be synthesized from u/s, d/s locations, from adjacent rivers, from

historic rainfall data combined with catchment characteristics, or other accepted methods.

Frequency analysis is carried out to determine peak flood discharge corresponding to various

return periods of large values 100 to 1000 years by using Gumbel method, Extreme Value

Type-I (EV-I) method, Log-Pearson Type-III (LP-3) method, and other methods.

The probability of exceedence, p x>X of rainfall P (or discharge Q) is given as p x>X =1/T and probability of non-exceedence (or safe probability) is given as: p x<X = 1 - P x ≥ X(x) =1 - 1/T. For T = 5 years, p x>X = 1/5 = 0.20 and p x<X = 1 - 1/5 = 0.80. The probability of non-exceedence in N consecutive years is as: (1 - 1/T) N. The risk / probability of failure (p F)[exceedence of flow than design discharge is considered as failure, however the structure mayor may not sustain any damages] of system is given by the probability of exceedence ofrainfall larger than P for at least once in N years as: p F = 1 - (1 - 1/T) N. For T = 5, the p F forvarious periods is as: N = 5 years, p F = 0.67; N = 10 years, p F = 0.89; N = 20 years, p F

The reliability of estimated precipitation (P) corresponding to return period T (i.e. P

=0.989.

T)is good to very good for T ≤ n, fair for n ≤ T ≤ 1.5 n and poor for T ≥ 2 n where n is thenumber of years of record.

Following procedure is adopted to obtain depth-frequency and/or depth-duration-frequency curves for the area of interest.

2.5.1.1 Gumbel method

Collect data of daily instantaneous peak flow for the dam site for sufficient long time periodsof 20 to 30 years (Table 2.7); longer the time period, better the results. The analysismay be based on data in selected critical months (e.g. annual flood, winter flood,summer flood etc).

Record the subjective values of peak flow (Q) e.g. annual maximum flood (annual series) orall flow above some selected threshold values (partial duration series). For annualseries there will be same number of selected values as the number of years of record.

Arrange the flow data in descending order of magnitude (largest value first and smallest valuelast). Assign an order number (m) to each value; order number 1 is assigned to largestevent (Table 2.8).




Determine frequency or return period for any flood storm by Weibull plotting positionmethod as:

m1+n

=T (8.1)

whereT = return period (in years)n = total number of observations in the recordm = order number of any event; 1 ≤ m ≤ n.

The return period so computed indicate that the flood of peak flow Q m shall beequaled or exceeded at least once in T number of years

Draw a graph between flood frequency T and flood peak flow Q on a Gumbel's probability paper with return period or frequency (T) on transformed scale and peak flow rate Q

on normal scale. A straight line is usually drawn through the graph points.

.

[NOTE: The Gumbel's probability paper has x-axis transformed as double log(natural log to base e, ln) scale as: v = - ln [ ln T/(T-1)]. The T data is transformedinto variate v and then graph is drawn between the reduced variate v and flood peakflow] At T = 1.582, v = 0.0.

Draw a smooth curve, usually a straight line, through the points. This is the frequency curvefor the flood peak flow (Fig. 2.15).

Read out rainfall values corresponding to desired frequency levels / return periods.

Table 2.7: Kurram River at Thal: Annual Instantaneous Peak Flows Source: SWHP / WAPDA

Year DateAnnual

Peak (cfs)Daily Average

Flow (cfs) Year DateAnnual Peak

(cfs)Daily Average

Flow (cfs)1971 4-Jul-1971 37,500 5,530 1986 18-Jul-1986 18,900 4,9101972 19-Jul-1972 30,400 3,700 1987 26-Jul-1987 62,900 6,5601973 26-Jul-1973 30,400 8,330 1988 23-Jul-1988 39,800 11,9001974 5-Jul-1974 37,400 1,440 1989 17-Jul-1989 15,180 4,5031975 13-Aug-1975 16,900 4,780 1990 18-Sep-1990 29,830 3,307

1976 18-Aug-1976 13,300 4,390 1991 13-Jul-1991 73,200 29,4501977 26-Jul-1977 4,550 1,420 1992 22-Jul-1992 37,680 8,0271978 6-Jul-1978 63,900 21,700 1993 23-Jul-1993 66,300 8,5081979 9,700 1994 26-Jun-1994 18,965 13,6001980 27-Jun-1980 4,420 3,490 1995 24-Apr-1995 22,725 13,5901981 29-Mar-1981 12,500 5,940 1996 22-May-1996 15,644 5,6101982 1-Jul-1982 18,000 3,160 1997 28-Aug-1997 1,911 1,3231983 19-Aug-1983 24,400 4,170 1998 24-Apr-1998 2,860 1,9181984 2-Aug-1984 49,600 6,230 1999 31-Jul-1999 9,911 1,5081985 20-Aug-1985 19,500 2,460 2000 27-Dec-2000 549 544




Table 2.8: Frequency analysis of Annual Instantaneous Peak Flows of Kurram River at Thal. by Gumbel method

Ranking in descendingorder

FrequencyDistribution

Ranking in descendingorder

FrequencyDistribution

Year Flow

Cfs

Rank

m T years variate v

Year Flow

Cfs

Rank

m T years variate v1991 89,890 1 34.00 3.51 1994 23,289 18 1.89 0.281993 81,416 2 17.00 2.80 1986 23,209 19 1.79 0.201978 78,469 3 11.33 2.38 1982 22,104 20 1.70 0.121987 77,241 4 8.50 2.08 1975 20,753 21 1.62 0.041984 60,909 5 6.80 1.84 1996 19,211 22 1.55 -0.041988 48,874 6 5.67 1.64 1989 18,641 23 1.48 -0.121970 47,401 7 4.86 1.47 1976 16,332 24 1.42 -0.201992 46,271 8 4.25 1.32 1981 15,350 25 1.36 -0.281971 46,050 9 3.78 1.18 1999 12,171 26 1.31 -0.371974 45,927 10 3.40 1.05 1979 11,912 27 1.26 -0.46

1969 42,120 11 3.09 0.94 1968 7,822 28 1.21 -0.551972 37,331 12 2.83 0.83 1977 5,587 29 1.17 -0.651973 37,331 13 2.62 0.73 1980 5,428 30 1.13 -0.761990 36,631 14 2.43 0.63 1998 3,512 31 1.10 -0.891983 29,963 15 2.27 0.54 1997 2,347 32 1.06 -1.041995 27,906 16 2.13 0.45 2000 674 33 1.03 -1.261985 23,946 17 2.00 0.37

KURRAM RIVER AT KT DAM SITE: Flood Frequency Analysis

0

10,000

20,000

30,000

40,000

50,000

60,000

70,000

80,000

90,000

100,000

110,000

120,000

130,000

-2.0 -1.5 -1.0 -0.5 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0

Reduced variate: v = - Ln{Ln[T/(T-1)]}

A n n u a

l P e a

k F l o w

( C f s )

2 5 10 20 30 40 50Return Period (Years) ---> 75 100

Log Pearson Type III

Gumble distribution

Figure 2.15: Flood frequency analysis for Kurram Tangi Dam.




The rainfall depths for various frequency storms (i.e. P

2.5.1.2 Extreme Value-I (EV-I) method

T

) can also be determined fromthe historic data using Extreme Value type-I (EV-I) distribution given as:

0.5772 -

1 - T T -

6 += P T lnlnπ

σ µ ( )

where μ and σ is the mean and standard deviation of the population, respectively, and T is thereturn period or frequency (in years). The above eqation holds only for large data (N ∞) .For small data the mean ( μ) and standard deviation ( σ) of the population is replaced by

sample average (x av ) and sample standard deviation (s N-1 6π ) and the parameters and

0.5772 are replaced by S N and Y N

Table 2.9 : Gumble frequency factors Y

from Table 2.9 given below. The Gumble method is usedfor data with standard deviation of log transformed values to be less than 1.13.

N and S N

N

(Source: Subramanya P 246) Y SN NN Y SN NN Y SN N

10 0.4952 0.9496 41 0.5442 1.1436 72 0.5552 1.187311 0.4996 0.9676 42 0.5448 1.1458 73 0.5555 1.188112 0.5035 0.9833 43 0.5453 1.1480 74 0.5557 1.189013 0.5070 0.9971 44 0.5458 1.1499 75 0.5559 1.189814 0.5100 1.0095 45 0.5463 1.1519 76 0.5561 1.1906

15 0.5128 1.0206 46 0.5468 1.1538 77 0.5563 1.1915

16 0.5157 1.0316 47 0.5473 1.1557 78 0.5565 1.1923

17 0.5181 1.0411 48 0.5477 1.1574 79 0.5567 1.1930

18 0.5202 1.0493 49 0.5481 1.1590 80 0.5569 1.193819 0.5220 1.0565 50 0.5485 1.1607 81 0.5570 1.194520 0.5236 1.0628 51 0.5489 1.1623 82 0.5572 1.195321 0.5252 1.0696 52 0.5493 1.1638 83 0.5574 1.195922 0.5268 1.0754 53 0.5497 1.1658 84 0.5576 1.196723 0.5283 1.0811 54 0.5501 1.1667 85 0.5578 1.197324 0.5296 1.0864 55 0.5504 1.1681 86 0.5580 1.198025 0.5309 1.0915 56 0.5508 1.1696 87 0.5581 1.198726 0.5320 1.0961 57 0.5511 1.1708 88 0.5583 1.199427 0.5332 1.1004 58 0.5515 1.1721 89 0.5585 1.200128 0.5343 1.1047 59 0.5518 1.1734 90 0.5586 1.200729 0.5353 1.1086 60 0.5521 1.1747 91 0.5587 1.201330 0.5362 1.1124 61 0.5524 1.1759 92 0.5589 1.202031 0.5371 1.1159 62 0.5527 1.1770 93 0.5591 1.202632 0.5380 1.1193 63 0.5530 1.1782 94 0.5592 1.203233 0.5388 1.1226 64 0.5533 1.1793 95 0.5593 1.203834 0.5396 1.1255 65 0.5535 1.1803 96 0.5595 1.204435 0.5402 1.1285 66 0.5538 1.1814 97 0.5596 1.204936 0.5410 1.1313 67 0.5540 1.1824 98 0.5598 1.205537 0.5418 1.1339 68 0.5543 1.1834 99 0.5599 1.206038 0.5424 1.1363 69 0.5545 1.1844 100 0.5600 1.206539 0.5430 1.1388 70 0.5548 1.1854

40 0.5436 1.1413 71 0.5550 1.1863 Infinity 0.5772 1.2825






A=-0.52+0.30|C s| (if |C s | > 0.90), and B = 0.94 – 0.26 |C s|, if |C s| ≤ 1.50 Or B = 0.55 if |C s | >

1.50. Then K T is obtained using factor C w

Table 2.11: Frequency Factor K

.

T = F(C s

Coefficient ofSkewness, Cs

,T) for Log-Pearson Type-III Distribution

Recurrence Interval T in Years2 10 25 50 100 200 1000

3.0 -0.396 1.180 2.278 3.152 4.051 4.970 7.2502.5 -0.360 1.250 2.262 3.048 3.845 4.652 6.6002.2 -0.330 1.284 2.240 2.970 3.705 4.444 6.2002.0 -0.307 1.302 2.219 2.912 3.605 4.298 5.9101.8 -0.282 1.318 2.193 2.848 3.499 4.147 5.6601.6 -0.254 1.329 2.163 2.780 3.388 3.990 5.3901.4 -0.225 1.337 2.128 2.706 3.271 3.828 5.1101.2 -0.195 1.340 2.087 2.626 3.149 3.661 4.8201.0 -0.164 1.340 2.043 2.542 3.022 3.489 4.540

0.9 -0.148 1.339 2.018 2.498 2.957 3.401 4.3950.8 -0.132 1.336 1.998 2.453 2.891 3.312 4.2500.7 -0.116 1.333 1.967 2.407 2.824 3.223 4.1050.6 -0.099 1.328 1.939 2.359 2.755 3.132 3.9600.5 -0.083 1.323 1.910 2.311 2.686 3.041 3.8150.4 -0.066 1.317 1.880 2.261 2.615 2.949 3.6700.3 -0.050 1.309 1.849 2.211 2.544 2.856 3.5250.2 -0.033 1.301 1.818 2.159 2.472 2.763 3.3800.1 -0.017 1.292 1.785 2.107 2.400 2.670 3.2350.0 0.000 1.282 1.751 2.054 2.326 2.576 3.090-0.1 0.017 1.270 1.716 2.000 2.252 2.482 2.950-0.2 0.033 1.258 1.680 1.945 2.178 2.388 2.810-0.3 0.050 1.245 1.643 1.890 2.104 2.294 2.675-0.4 0.066 1.231 1.606 1.834 2.029 2.201 2.540-0.5 0.083 1.216 1.567 1.777 1.955 2.108 2.400-0.6 0.099 1.200 1.528 1.720 1.880 2.016 2.275-0.7 0.116 1.183 1.488 1.663 1.806 1.926 2.150-0.8 0.132 1.166 1.448 1.606 1.733 1.837 2.035-0.9 0.148 1.147 1.407 1.549 1.660 1.749 1.910-1.0 0.164 1.128 1.366 1.492 1.588 1.664 1.880-1.4 0.225 1.041 1.198 1.270 1.318 1.351 1.465-1.8 0.282 0.945 1.035 1.069 1.087 1.097 1.130-2.2 0.330 0.844 0.888 0.900 0.905 0.907 0.910

-3.0 0.396 0.660 0.666 0.666 0.667 0.667 0.668The K T

( ) ( ) ( ) 5432232 3/1163/11 k zk k z k z z k z z K T ++−−−+−+=

may also be determined as:

where k = C s

22

2

001308.0189269.0432788.11010328.0802853.0515517.2

www

www z

+++++−=

/6, and z is standard normal variable, given as:




and ( )[ ] 2/121ln pw = for 0 < p ≤ 0.5 [p = 1/T]. For p > 0.5, use 1-p instead of p and z is given

a negative sign.

2.5.2 FLOOD ANALYSIS BASED ON CATCHMENT MODELING

2.5.2.1 General procedure

The steps are:

1. Determine the expected rain P T

2. Use a catchment loss model to determine runoff depth Q

for desired return period T by performing frequency

analysis of historic rainfall data. Use the rainfall data of critical months (annual,

summer, winter, month). Some areas of Pakistan have winter snowfalls and summer

rainfall. Winter snowfall will not produce any large runoff, thus rainfall in summer

months produce runoff and used for frequency analysis. Adjust the rainfall for aerialreduction factor.

T

3. Use a historic or synthesized unit hydrograph to convert runoff depth to discharge

hydrograph Q

from the catchment area

considering the watershed hydrologic characteristics.

t = U t*Q T

4. Computer model HEC-HMS may be used to do the above steps.

and determine the peak flow rate.

The time to peak may be determined from analysis of maximum historic floods. The

shape of the flood and flood volume are also derived from historic record. A unit hydrograph

may be derived from the historic data and used further to synthesize floods of various

frequencies. Table 2.7 and Fig. 2.15 describe /show a flood frequency curve and Table 2.12

and Fig. 2.16 describe computation of design flood hydrograph.

2.5.2.2 Design Flood Hydrograph




Table 2.12 : Derivation of T year flood from historic maximum flood.Hours Max

historic

floodcfs

30 yearfrequency

floodcfs

Hours Maxhistoric

floodcfs

30 yearfrequency

floodcfs

Hours Maxhistoric

floodcfs

30 yearfrequency

floodcfs

Historic maximum peak = 79,885 cfs, 30-year flood peak = 86,500 cfs, Factor= 1.0830 3,823 4,116 9 50,470 54,333 18 17,192 18,5081 3,857 4,152 10 66,860 71,978 19 14,368 15,4672 4,030 4,339 11 79,885 86,000 20 11,666 12,5593 4,587 4,938 12 63,877 68,766 21 9,382 10,1004 5,641 6,073 13 42,462 45,713 22 7,378 7,9435 7,140 7,687 14 31,805 34,240 23 6,111 6,5796 13,059 14,059 15 27,062 29,133 24 4,210 4,5327 22,811 24,557 16 23,332 25,118 25 4,155 4,473

8 34,551 37,195 17 19,894 21,416 26 4,030 4,339

Hydrograph of 30 year frequency Flood at Kurram Tangi Dam

0

5,000

10,000

15,000

20,000

25,000

30,000

35,000

40,000

45,000

50,000

55,00060,000

65,000

70,000

75,000

80,000

85,000

90,000

0 2 4 6 8 10 12 14 16 18 20 22 24 26 28

Time (Hours)

D i s c h a r g e

( c f s )

Figure 2.16 : A 30-year frequency flood for Kurram Tangi Dam.




2.6 DIVERSION FLOODS

The construction of the dam embankment takes many years to complete. There has to

be arrangement that allows flows including flood flows to be diverted away from the

construction area during the whole construction period. Diversion tunnels/channels are

provided to pass the diversion flows. The selection of the capacity and size of the diversion

tunnel/channel has to be made in view of the design diversion flood that the system has to

pass. The magnitude/frequency of diversion flood taken as (10 x T c) , where T c

• Q

- is the number

of years of construction period.

T

• Both peak flow rate and flood volume (i.e. flood hydrograph Q

selected from frequency analysis of historic annual maximum flood flow.

t

• Determine time gap between successive flood events.

distribution) are

important for routing

• Frequency analysis by log-Pearson III or Gumble extreme value method. Small

extrapolation Ok

• Determine confidence limits on Q T

• If historic flood data is not available, synthesize Q

and select appropriate value.

T from historic P T

• If measurements not at dam site, transpose flood to dam site from u/s or d/s location

by area method. Q

values using

appropriate catchment runoff models (e.g. HEC-HMS).

2 = Q 1 * (A 2/A 1 )

• If no record of flow is available, regional flow values may be used.

0.5

2.7 PROBABLE MAXIMUM FLOOD, PMF

For large dams the spillway is usually designed to cater for probable maximum flood

(PMF). The PMF is derived from the knowledge of probable maximum precipitation (PMP)

over the catchment area, the catchment hydrologic characteristics to convert rainfall into

runoff and transformation of excess rainfall into storm runoff by using unit hydrograph. The

unit hydrograph derived from the project historic flood record may be used to determine

PMF. PMF for KT dam is shown in Fig. 2.17.

• PMF results from occurrence of PMP on the catchment area.




• PMP determined by meteorologists from historic data of extreme rainfall events,

humidity, dew point, precipitable water in the air column, after due maximization to

account possible worst conditions.

• PMP averaged over catchment area.

• PMP as rainfall over short/long time periods.

• PMP converted to PMF by using appropriate catchment runoff models considering

worst hydrologic conditions over the catchment area. Models as HEC-HMS may be

used.

• PMP peak flow rate, flow volumes and time distribution of flow important for

routing.

• PMP is routed through the spillway with designed/selected elevation-outflow

relationship.

• Maximum rise of water surface above the normal conservation level provide the flood

surcharge for fixing the free board of the dam.

Kurram Tangi Dam : Probable Maximum Flood Inflow HydrographSCS CN = 92, Lag time = 9 hrs, PMP-68

0

10,000

20,000

30,000

40,000

50,000

60,000

70,000

80,000

90,000

100,000

110,000

120,000

130,000

140,000

150,000

160,000

170,000

180,000

190,000

200,000

0 6 12 18 24 30 36 42 48 54 60 66 72 78 84 90 96 102Time (hours)

D i s c h a r g e ( c f s )

Figure 2.17: PMF for KT dam.




Selection of Spillway Design Flood

Selection of design flood for spillways is a very crucial matter, and many countries havedevised standards for this purpose; however no such standards exist for Pakistan. ICOLD(1992) noted that the spillway design discharge has direct bearing on structure safety on one

side and project costs on the other side. Ideally this should be based on engineering andeconomic considerations relevant to the site and its environment. In most cases two distinctdischarges are set: (1) the design flood: the flood that must be discharged through thestructure under normal conditions with a safety of margin provided by the free board; this isusually taken as flood of selected recurrence probability / return period; and (2) the safetycheck flood, which is the discharge which can be passed by the crest structure, thewaterway and energy dissipater on the verge of failure but to exhibit marginally safeperformance.

ICOLD (1992) narrated the factors to be considered for selection of spillway design flood and

include analysis of (i) downstream economic hazard – loss of economic values, (ii)downstream life hazard – loss of human life, (iii) type of dam – its susceptibility to breachand (major) damage in case of structure overtopping, (iv) spillway type – susceptibility tomalfunction, (v) structure height, (vi) storage volume at maximum water depth, (vii)consequences of dam failure -- in terms of being very vital for area population or mere someeconomic costs for replacement of works.

Different standard practices in dam engineering for selection of spillway design flood in allover the world.

1. Indian Standard for design of spillway capacity recommends 100 year or standardproject flood for design of intermediate structures (Engineering Hydrology by KSubramanya P 257).

2. ASCE recommend standard project flood for structure less than 30 m height andcapacity of head pond less than 62 MCM (Introduction to hydrology by Warren andWiseman P 587).

3. US Army Corp of Engineering recommend 100 year or upto 50% PMF forintermediate structure where loss of property and damage to structure is minor(capacity less than 60 MCM and height less than 30 m (ICOLD Bulletin 82 P 177)

4. In China for structure of this category return period of 200 yr is recommended (Ref:Selection of design floods in South Asia by Jian Liu; [[email protected]]).

5. Australian National Committee on Large Dam (ACOLD) recommends 100-1000years flood for design of structure where loss of structure and downstream loss of lifeand property is very high (ICOLD Bulletin 82 P 176).

6. In United Kingdom recommendations for selection of design flood for overflowsection are 0.3 PMF to 1000 year (General) and 0.2 PMF to 150 year (if overtoppingis tolerable). (ICOLD Bulletin 82 P 197).




7. Ohio Dept of Natural Resource recommends 40-100 % PMF for design of spillwaywhere height of dam is more than 60 ft and capacity greater than 5000 acre-ft[[email protected]].

8. When the spillway design flood is selected as of 100-1000 year frequency, then two

flood estimates should be developed. 1) Design flood – for which the structure isdesigned and suitable free board is provided above this capacity. 2) Safety checkflood. – a higher flood, e.g. PMF for which the safety of the structure must beestablished with less than acceptable safety criteria and/or free board.

The pertinent detail of risk factors for the e.g. Patrind HPP project is as under:

1) D/s Economic hazard: Low – minimal (underdeveloped to occasional structures oragriculture). The houses and small agriculture is located well above the maximumhistoric flood levels. The d/s bridges across the Kunhar river are well above the floodlevels (these have either stood against the 1992 flood levels or have beenconstructed a new after 1992 flood). There is almost negligible chance of loss ofproperty downstream of the weir.

2) D/s Loss of life: Low – Most dwelling are located well above maximum flood levelsand due to large peak time, people are likely to move quickly above the probableflood levels.

3) Type of dam: Concrete dams are predominantly considered safe against structuraldamages, have factor of safety of more than 2 against overturning, sliding failure andmaterial stresses. Some damage may be incurred to bridge deck in case flood

reaches the weir structure top.

4) Spillway type: Spillway will be mostly in operational condition before the onset of anyexceptionally high flood to pass the usual rainy season flood flows. The seven baysare 12 m wide each and its blockage susceptibility due to debris flow is consideredas very low.

5) Structure height Storage volume: The structure height is only 26 m with a storagecapacity of only 6.42 MCM and is thus classified as small to intermediate structure.Due to very small storage volume the enhancement of any flood flows will be limitedto 2 to 4% only.

6) Dam failure consequences: The weir structure is planned to divert water forhydropower production only. Thus in case of any dam breach there will be economicloss in terms of loss of hydropower production only and such the weir structure canbe refurbished after necessary repairs/improvements. Local or other population willnot be directly affected.

7) Summary: The weir structure is defined as small to intermediate on capacity basiswith low risk on economic and loss of life basis.




2.8 LEVEL POOL RESERVOIR ROUTING

Reservoir routing is required to determine the water surface elevations for known

inflow hydrograph and designed elevation-outflow relationship of structures as diversion

tunnel or spillway. The inflow for diversion tunnel is the design flood flow against which protection is planned during the construction. The inflow for the spillway is the design

project flood (usually PMF or a 10,000 year frequency flood) for the spillway. For large

storage reservoirs the incoming flood water is temporarily stored in the flood surcharge space

of the dam. This results in attenuation of the outflowing flood peak to a smaller value. Thus

the spillway design discharge is taken as equal to maximum outflow rates.

Alternate procedures can be used to perform flood routing through the reservoir.

These include level pool reservoir routing, hydrologic routing e.g. HEC-HMS, hydraulicrouting e.g. HEC-RAS. The basic data needed for reservoir routing id incommimng flood

hydrograph, reservoir elevation-storage-outflow relation, reservoir cross section data, the

spillway geometric data and coefficients describing head losses, and head-discharge relation.

Level pool reservoir routing is discussed below.

The flow entering into a large reservoir quickly mingles with the storage water. The

water surface profile attains some grade at entry into the reservoir but then quickly levels off

to a uniform elevation (actually with a very small gradient but approaching to uniform level).

This happens due to a very large flow cross section of the reservoirs. Thus the outflow is

function of water surface elevation of the reservoir. The reservoir elevation in turn depends

on the storage volume. Note that the water surface elevation and storage volume are related

by the elevation-volume-area relationship of the reservoir. Thus outflow = F (reservoir

elevation) e.g. O = K 1 + K 2 x (El – datum) K3

• Initial condition: outflow = inflow (for diversion structure and for spillway

discharging freely before the design flood; Else spillway outflow = zero.)

. Following steps are taken for reservoir routing.

• Inflow will add to storage and raise water level. Increased water level result in greater

outflows

• Incremental storage over a time interval equals the difference of inflow and outflow

volume and will add to storage

• Analysis at ∆t interval with ∆t → 0




• Let V 0 , El 0 , d 0 , I 0 , O 0

• Let O = F(d). e.g. O = 13.56 d

represent the initial condition for the storage volume (V),

water surface elevation (El), depth (d) over the invert of diversion tunnel (or depth

over the normal conservation level for spillway routing), inflows (I), outflows (O) at

starting time, i.e. t = 0.

2.5 (O in cfs, d = depth ft over invert). Also find d =

F’(O), e.g. d = 0.07375 O 0.4

• Let elevation-volume relation is V = F(El) e.g. V = 0.11284*(El–1800)

.

2.790 and El =

1800 + 2.18878 * V 0.3583

• Set O

(El – ft amsl, V - AF)

0 = I 0

• Determine d

.

0 corresponding to O 0

• Find initial elevation El

from the relevant hydraulic equation d = F’(O).

0 = tunnel invert level EL T + d 0

• Determine initial reservoir storage volume corresponding to El

0

• For t = 1, set O

.

1 = O 0

• For all next time steps determine the flow depth over the invert d

.

t+1 = El t - EL

• Determine O

T

t+1 = F (d t+1

• Determine the inflow and outflow volumes during the time interval t to t+1 as: V

)

IN =

(I t + I t+1 )/2 * ∆t and V OUT = (O t + O t+1

• Determine change in storage volume during the time interval as: ∆V = V

)/2 * ∆t

IN - V

• Determine new storage volume as V

OUT

t+1 = V t

• Find new reservoir surface elevation as:El

+ ∆V

t+1 = F(V

• Table 2.13 may be used for systematic calculations:

t+1)

Table 2.13: Reservoir level pool flood routing.Time t

(hr)InflowI (cfs)

Depthd (ft)

OutflowO (cfs)

V(AF)

IN V(AF)

OUT ∆V(AF)

VolumeV (AF)

WS Elevation(ft)

0 2500 8.06 2500 1955.81 1833.061 2850 8.06 2500 221.07 206.61 14.97 1970.78 1833.162 3550 8.16 2579 264.46 209.88 54.58 2025.36 1833.493 4675 8.49 2848 339.88 224.26 115.62 2140.98 1834.1645




Example: (see Table 2.13 above)

Given: ∆t = 1 hr and I 0 = 2500c I 1 = 2850 I 2 = 3550, I 3

• Let invert level of tunnel = El

= 4675, cfs etc.

T

• Outflow O = 13.56 * d

= 1825 ft.2.5

Also d = 0.07375 O0.4

• V = 0.11284 * (El – 1800)

(O in cfs, d in ft above tunnel invertlevel).

2.790 and El = 1800 + 2.18878 * V 0.3583

At t = 0

(El - ft, V - AF)

O 0 = I 0

d = (O/13.56)

= 2500 cfs.1/2.5 , and d 0 = (2500/1356) 1/2.5

El

= 8.06 ft

0

V

= 1825 + 8.06 = 1833.06 ft.

IN = 0.0, V

OUT

V

= 0.0, ∆V = 0.0

0 = 0.11284 * (1833.06 – 1800) 2.79

At t = 1

= 1955.81 AF.

I1

d

= 2850 cfs

1 = d 0

O

= 8.06 ft

1 = O 0

V

= 2500 cfs

IN

V

= (2500+2850)/2*3600/43560 = 221.07 AF

OUT

∆V = 221.07 – 206.61 = 14.97 AF

= (2500+2500)/2*3600/43560 = 206.61 AF

V 1

El

= 1955.81 + 14.97 = 1970.78 AF

1 = 2.1887 * 1970.78 0.3583

At t = 2

+ 1800 = 1833.16 ft

I2

d

= 3550 cfs

2

O

= 1833.16 – 1825 = 8.16 ft

2 = 13.56 * d 22.5 = 13.56 * 8.16 2.5

V

= 2579 cfs

IN

V

= (2850+3550)/2*3600/43560 = 264.46 AF

OUT

∆V = 264.46 – 209.88 = 54.58 AF

= (2500+2579)/2*3600/43560 = 209.88 AF

V 2

El

= 1970.78 + 54.58 = 2025.36 AF

2 = 2.1887 * 2025.36 0.3583

At t = 3

+ 1800 = 1833.49 ft

I3

d

= 4675 cfs

3 = 1833.49 – 1825 = 8.49 ft




O 3 = 13.56 * 8.49 2.5

V

= 2848 cfs

IN

V

= (3550+4675)/2*3600/43560 = 339.88 AF

OUT

∆V = 339.88 – 224.26 = 115.62 AF

= (2579+2848)/2*3600/43560 = 224.26 AF

V 3

El

= 2025.36 + 115.62 = 2140.98 AF

3 = 2.1887 * 2140.98 0.3583

Example:

+ 1800 = 1834.16 ft

The 30-year flood data of Table 2.12 is routed through 2 Nos. diversion tunnels (one

of 14 ft dia and other of 20 ft dia) with invert at 1825 ft. The elevation-volume and depth-

outflow relationship is given as: El = 2.18878*Vol 0.358 , Vol = 0.11284*(El-1800) 2.79 , O =

18.2351 d 2 + 374.738 d – 1000.58 for d < 17 ft and O = - 5.356 d 2

The Table 2.14 and Fig. 2.18 reveal that the water surface and the outflow rise quickly

due to sudden influx of large amount of flood water. The outflow reaches at its maximum

after 8 hours and stays so for next 10 hrs. The water surface level drops steadily and reaches

the pre-flood conditions after 15 hours. The second flood is not likely to occur within this

time period. Therefore the diversion capacity provided is adequate. The results for PMF

routing of Kurram Tangi dam are shown in Fig. 2.19.

+ 649.652 d – 3744.7 for d

>= 17 ft. The routing computations are given in Table 2.14 and results are shown in Fig. 2.18.

The maximum inflow is 86,000 cfs, maximum outflow is 34,650 cfs, and maximum water

level is 1880.23 ft amsl. This defines the height of coffer dam plus usual other allowances.

Reservoir routing can also be performed conveniently by using HEC-HMS model.

The model requires description of inflow hydrograph, the elevation-storage relationship data,

the elevation-outflow relationship data, the description of clock time for start and end of

simulation. The model output (graph and tabular) provides reservoir elevations, inflows,

outflows, storage at different times (Figs. 2.20 and 2.21).




Table 2.14:Routing of diversion flood through the diversion tunnelsFlow in cfs, voume in acre feet, elevation in ft.

Time hr Inflow In Vol WS El Depth Outflow Out Vol Storage V

0 4116 0 1834.38 9.38 4116 0 21841 4152 342 1834.40 9.40 4129 341 21852 4339 351 1834.40 9.40 4133 341 21943 4938 383 1834.45 9.45 4171 343 22344 6073 455 1834.68 9.68 4334 351 23385 7687 569 1835.25 10.25 4753 375 25316 14059 899 1836.26 11.26 5532 425 30057 24557 1596 1838.56 13.56 7434 536 40648 37195 2552 1842.97 17.97 11405 778 58389 54333 3782 1848.92 23.92 16469 1152 8468

10 71978 5219 1855.89 30.89 21749 1579 12108

11 86000 6528 1863.53 38.53 26724 2003 1663312 68766 6395 1871.19 46.19 30859 2379 2064913 45713 4731 1876.92 51.92 33400 2655 2272414 34240 3304 1879.61 54.61 34426 2803 2322515 29133 2619 1880.23 55.23 34650 2854 2299016 25118 2242 1879.94 54.94 34546 2859 2237217 21416 1923 1879.16 54.16 34264 2843 2145218 18508 1650 1877.98 52.98 33817 2813 2028819 15467 1404 1876.44 51.44 33204 2769 1892320 12559 1158 1874.55 49.55 32409 2711 1736921 10100 936 1872.30 47.30 31390 2636 1566922 7943 746 1869.68 44.68 30113 2541 1387423 6579 600 1866.71 41.71 28542 2424 1205024 4532 459 1863.42 38.42 26659 2281 1022825 4473 372 1859.81 34.81 24403 2110 849026 4339 364 1855.95 30.95 21786 1909 694627 4339 359 1852.06 27.06 18936 1683 562128 4339 359 1848.26 23.26 15936 1441 453929 4339 359 1844.70 19.70 12934 1193 370530 4339 359 1841.57 16.57 10211 956 310731 4339 359 1839.03 14.03 7843 746 271932 4339 359 1837.21 12.21 6291 584 249433 4339 359 1836.07 11.07 5383 482 237034 4339 359 1835.42 10.42 4883 424 230535 4339 359 1835.07 10.07 4618 393 227136 4339 359 1834.88 9.88 4480 376 225337 4339 359 1834.78 9.78 4410 367 224438 4339 359 1834.73 9.73 4375 363 224039 4339 359 1834.71 9.71 4357 361 223840 4339 359 1834.70 9.70 4348 360 2237

El = 2.18878*Vol 0.358 Vol = 0.11284*(El-1800) 2.79 O = 18.2351 d 2 + 374.738 d – 1000.58 for d < 17 ft and O = - 5.356 d 2 + 649.652 d – 3744.7 for d >= 17




KTD Flood diversion

0

10000

20000

30000

40000

50000

60000

70000

80000

90000

100000

0 5 10 15 20 25 30 35 40

Time (hrs)

I n ,

o u

t F l o w

( c f s )

1800

1810

1820

1830

1840

1850

1860

1870

1880

1890

1900

R e s e r v o

i r w a t e r s u r f a c e e l e v a

t i o n

F t

Inflows

Outflow

WS Elev

Figure 2.18: Diversion flood routing.

KTD PMF Routing for Spillway Design

0

20000

40000

60000

80000

100000

120000

140000

160000

180000

200000

0 5 10 15 20 25 30 35 40 45 50Time (hrs)

F l o w

( c f s )

2100

2101

2102

2103

2104

2105

2106

2107

2108

2109

2110

R e s e r v o

i r w a t e r e

l e v a

t i o n

( f t )

InflowsOutflow

Res Water Elev

Figure 2.19: Routing of PMF for spillway design.




Figure: 2.20 : Summary of diversion flood routing by HEC-HMS.

Figure 2.21 : Detail routing output of HEC-HMS model.

Water Surface Elevations (ft)Storage

Inflow hydrograph

Outflow hydrograph




2.9 RESERVOIR OPERATION

Reservoirs are created to store water during periods of large inflows and release the stored

water subsequently. Reservoir regulation requires that reservoir water levels at different time

of the year may be known. For hydropower generation, the determination of net head isdependent on the reservoir water levels. Reservoir operation study is carried out to determine

temporal changes in storage volume, reservoir water levels, actual supplies vis-à-vis target

releases, shortages, spills, evaporation losses, over life (or part life) of dam for given inflows.

In addition energy generation, and alternate modes of reservoir filling or emptying are

studied. Reservoir operation is also done to establish required live storage capacity for

various inflow and demand patterns. Reservoirs considerably improve the dependable flow of

the river.

• Reservoir operation is based on long period (30-50 years) historic data of inflows

into the dam, evaporation, and rainfall. Data may be on the basis of 10-daily or

monthly periods.

• Data may be actual historic or synthesized. Assumed data may be used for

scenario testing only.

• Reservoir operation is carried to meet target water demand and for selected

gross/dead storage values and site specific elevation – volume – area function

• Reservoir operation is essentially an accounting procedure on volume basis.

Different variables used in reservoir operation calculations be expressed as:

I = Inflows,

D = demand,

R = actual releases,

S = shortage if any; S = D - R,El = reservoir water surface elevation,

LS = reservoir live storage

DS = reservoir dead storage

GV = reservoir gross storage volume = LS + DS,

V = reservoir current volume

RV = Maximum volume of water releasable in the current simulation interval,

A = reservoir surface area,




E = pan evaporation (mm/period),

Ev = evaporation volume from the lake surface,

P = rainfall (mm/period),

Pv = direct precipitation/rain volume over the lake surface.

Reservoir current gross storage volume V is further described as:

V 1

V

= volume at beginning of simulation interval,

2

V

= volume at end of simulation interval after accounting releases only,

a

V

= average volume during the simulation interval = (V1+V2)/2,

3

V = volume at end of simulation interval after accounting all releases, evaporation,

rainfall, and spillages.

= volume after accounting releases, evaporation, and direct rainfall but excluding

spillage and

Simulations are carried over small time periods, e.g. 10-day or one month, and are started

with assumed starting live storage volume V 0

1. Given data is Inflow I, rainfall data P, pan evaporation data E.

2. Determine period starting volume: for first time period V 1 = V 0 and for other periods

V 1 = V t-1 (i.e. volume V

3. Net water available for release: RV = V

at end of last period)

1

4. Release demand = D

– DS + I

5. Irrigation release R: R = D

6. Shortage: S = R – D

If D ≤ RV Else R = RV

7. % Shortage = S/D * 100

8. V 2 = V 1

9. Average volume: V

+I - R

a = (V 1+V 2

10. Average reservoir El = Function of (V

)/2

a

11. Average surface area: A = Function of (average E1and El -V-A relationship)

and El -V- A relationship)

12. Lake evaporation: Ev = E * lake factor ( ≈ 0.7) * average lake surface area A *

simulation time period * units conversion factor.

13. Rain volume falling over the reservoir surface are: Pv = P * lake surface area (use

appropriate conversion units)

14. V 3 = V 2

15. Check for any spillage: If V

- Ev + Pv

3 > gross storage GV, spill = V 3

16. Reservoir volume at end of simulation interval: V = V

– GV, Else spill = 0

3 – spill




17. Reservoir water surface elevation: El = Function of (V, and El-V-A relationship)

Example:

The reservoir operation for Kurram Tangi dam is given in Table 2.15 for the period

1971-73. The summary of results for 31 years period 1971-2001 (live storage capacity = 600

ThAF) are given in Table 2.16. The results are shown for reservoir water level, reservoir

volume, and deficits/shortage from target demand are shown in Figs. 2.22 to 2.24.The

dependability of water delivery for various reservoir live storage capacities of 50, 100, 200,

300, 400, 500, 600, 700 and 800 ThAF are shown in Fig. 2.25 and summarized in Table 2.17.

It is seen that larger size reservoirs considerably improves the dependability of water

delivery. However the incremental benefits of larger reservoirs decreases. This explains that

the selection of the reservoir live storage capacity should be evaluated from both availabilityand economic point of view.

Reservoir operation simulations also provide the reservoir water levels, and therefore

the net head, for the potential hydropower plant at the toe of the dam. Thus the reservoir

water levels and the flow releases may be used to develop head duration, flow duration and

power production duration curves and the total power production under future dam operation

conditions.




Table 2.15: Reservoir operation. Volume in ThAF. (Starting V = 700)

Time I P V RV1 D R S V V2 Aa PE Pv V Spill3 V ElJan 1-10,71 9.8 0.00 700.0 409.8 9.79 9.79 0.0 700.0 700.0 7471 0.34 0.00 699.7 0.0 699.7 2072.3

Jan 11-20 9.8 0.00 699.7 409.5 10.94 10.94 0.0 698.6 699.1 7464 0.37 0.00 698.2 0.0 698.2 2072.1

Jan 21-31 9.7 4.57 698.2 407.9 14.11 14.11 0.0 693.8 696.0 7441 0.40 0.09 693.5 0.0 693.5 2071.5

Feb 01-10 7.8 6.03 693.5 401.3 14.31 14.31 0.0 687.0 690.3 7399 0.43 0.12 686.7 0.0 686.7 2070.5

Feb 11-20 7.8 0.00 686.7 394.5 18.50 18.50 0.0 676.0 681.3 7333 0.51 0.00 675.5 0.0 675.5 2068.9

Feb 21-28 6.7 16.76 675.5 382.1 23.67 23.67 0.0 658.5 667.0 7227 0.57 0.34 658.2 0.0 658.2 2066.5

Mar 01-10 7.8 2.22 658.2 366.0 16.47 16.47 0.0 649.6 653.9 7130 0.75 0.04 648.9 0.0 648.9 2065.1

Mar 11-20 4.5 0.20 648.9 353.4 22.99 22.99 0.0 630.4 639.6 7023 0.87 0.00 629.5 0.0 629.5 2062.3

Mar 21-31 5.2 1.71 629.5 334.7 27.46 27.46 0.0 607.2 618.4 6862 0.98 0.03 606.3 0.0 606.3 2058.8

Apr 01-10 5.6 0.00 606.3 311.8 30.29 30.29 0.0 581.5 593.9 6674 0.89 0.00 580.7 0.0 580.7 2054.9

Apr 11-20 7.1 17.21 580.7 287.8 27.38 27.38 0.0 560.4 570.5 6493 1.02 0.31 559.7 0.0 559.7 2051.6

Apr 21-30 18.4 28.96 559.7 278.1 17.38 17.38 0.0 560.7 560.2 6413 1.16 0.52 560.1 0.0 560.1 2051.7

May 01-10 6.9 0.00 560.1 267.0 22.76 22.76 0.0 544.3 552.2 6349 1.13 0.00 543.1 0.0 543.1 2048.9

May 11-20 8.6 14.16 543.1 251.8 24.65 24.65 0.0 527.1 535.1 6214 1.30 0.25 526.1 0.0 526.1 2046.1

May 21-31 9.5 25.02 526.1 235.6 26.47 26.47 0.0 509.1 517.6 6074 1.46 0.42 508.1 0.0 508.1 2043.1

Jun 01-10 14.5 0.00 508.1 222.6 15.37 15.37 0.0 507.2 507.6 5994 1.38 0.00 505.8 0.0 505.8 2042.7

Jun 11-20 14.2 8.76 505.8 220.0 17.41 17.41 0.0 502.6 504.2 5966 1.42 0.15 501.4 0.0 501.4 2042.0

Jun 21-30 5.8 14.67 501.4 207.2 22.47 22.47 0.0 484.7 493.0 5875 1.35 0.24 483.6 0.0 483.6 2038.9

Jul 01-10 24.6 23.56 483.6 208.2 23.93 23.93 0.0 484.3 483.9 5800 1.26 0.38 483.4 0.0 483.4 2038.8

Jul 11-20 15.7 46.74 483.4 199.1 18.31 18.31 0.0 480.8 482.1 5785 1.20 0.75 480.4 0.0 480.4 2038.3

Jul 21-31 52.4 70.36 480.4 232.8 17.98 17.98 0.0 514.8 497.6 5912 1.16 1.16 514.8 0.0 514.8 2044.2

Aug 01-10 43.4 36.64 514.8 258.2 13.05 13.05 0.0 545.1 529.9 6173 1.18 0.63 544.6 0.0 544.6 2049.2

Aug 11-20 24.5 0.00 544.6 269.1 15.60 15.60 0.0 553.5 549.0 6325 1.09 0.00 552.4 0.0 552.4 2050.4

Aug 21-31 21.2 0.00 552.4 273.6 25.41 25.41 0.0 548.2 550.3 6335 0.99 0.00 547.2 0.0 547.2 2049.6

Sep 01-10 13.7 0.00 547.2 260.9 29.08 29.08 0.0 531.9 539.5 6249 1.07 0.00 530.8 0.0 530.8 2046.9

Sep 11-20 8.1 0.00 530.8 238.9 29.60 29.60 0.0 509.3 520.1 6094 0.93 0.00 508.4 0.0 508.4 2043.1Sep 21-30 3.2 0.00 508.4 211.6 29.48 29.48 0.0 482.1 495.3 5893 0.79 0.00 481.3 0.0 481.3 2038.5

Oct 01-10 5.0 0.00 481.3 186.3 29.15 29.15 0.0 457.2 469.2 5679 0.84 0.00 456.3 0.0 456.3 2034.0

Oct 11-20 7.0 0.00 456.3 163.3 29.02 29.02 0.0 434.3 445.3 5479 0.73 0.00 433.6 0.0 433.6 2029.8

Oct 21-31 5.7 0.00 433.6 139.3 30.04 30.04 0.0 409.3 421.4 5276 0.64 0.00 408.6 0.0 408.6 2025.0

Nov 01-10 6.4 0.00 408.6 115.1 32.91 32.91 0.0 382.1 395.4 5050 0.42 0.00 381.7 0.0 381.7 2019.6

Nov 11-20 6.6 0.00 381.7 88.3 32.23 32.23 0.0 356.1 368.9 4816 0.33 0.00 355.8 0.0 355.8 2014.2

Nov 21-30 7.7 0.00 355.8 63.5 18.31 18.31 0.0 345.2 350.5 4649 0.26 0.00 344.9 0.0 344.9 2011.9

Dec 01-10 11.1 0.00 344.9 56.0 6.87 6.87 0.0 349.2 347.0 4618 0.29 0.00 348.9 0.0 348.9 2012.7

Dec 11-20 11.5 0.00 348.9 60.4 6.93 6.93 0.0 353.4 351.2 4655 0.27 0.00 353.2 0.0 353.2 2013.7

Dec 21-31 13.0 2.16 353.2 66.2 6.55 6.55 0.0 359.6 356.4 4703 0.25 0.03 359.4 0.0 359.4 2015.0Jan 1-10,72 6.0 0.00 359.4 65.4 9.79 9.79 0.0 355.7 357.5 4713 0.23 0.00 355.4 0.0 355.4 2014.1Jan 11-20 6.7 3.11 355.4 62.1 10.94 10.94 0.0 351.2 353.3 4675 0.25 0.04 351.0 0.0 351.0 2013.2Jan 21-31 9.0 32.89 351.0 60.0 14.11 14.11 0.0 345.9 348.4 4631 0.27 0.42 346.0 0.0 346.0 2012.1Feb 01-10 7.2 0.00 346.0 53.2 14.31 14.31 0.0 338.9 342.5 4576 0.22 0.00 338.7 0.0 338.7 2010.5Feb 11-20 8.1 15.37 338.7 46.9 18.50 18.50 0.0 328.4 333.5 4494 0.25 0.19 328.3 0.0 328.3 2008.2Feb 21-28 6.6 0.00 328.3 34.9 23.67 23.67 0.0 311.2 319.7 4366 0.28 0.00 310.9 0.0 310.9 2004.2Mar 01-10 7.9 9.33 310.9 18.8 16.47 16.47 0.0 302.4 306.6 4242 0.24 0.11 302.2 0.0 302.2 2002.2Mar 11-20 15.3 18.03 302.2 17.5 22.99 17.55 5.4 300.0 301.1 4190 0.28 0.21 299.9 0.0 299.9 2001.6Mar 21-31 25.0 16.45 299.9 25.0 27.46 24.95 2.5 300.0 300.0 4179 0.32 0.19 299.9 0.0 299.9 2001.6

Apr 01-10 27.1 12.76 299.9 26.9 30.29 26.95 3.3 300.0 299.9 4178 0.44 0.15 299.7 0.0 299.7 2001.6 Apr 11-20 26.4 8.83 299.7 26.1 27.38 26.08 1.3 300.0 299.9 4178 0.52 0.10 299.6 0.0 299.6 2001.6 Apr 21-30 58.2 26.92 299.6 57.8 17.38 17.38 0.0 340.4 320.0 4368 0.62 0.33 340.1 0.0 340.1 2010.8May 01-10 97.7 23.37 340.1 137.8 22.76 22.76 0.0 415.0 377.6 4893 0.89 0.32 414.5 0.0 414.5 2026.1

May 11-20 38.7 1.93 414.5 153.2 24.65 24.65 0.0 428.5 421.5 5276 1.13 0.03 427.4 0.0 427.4 2028.6May 21-31 31.3 0.00 427.4 158.7 26.47 26.47 0.0 432.2 429.8 5347 1.31 0.00 430.9 0.0 430.9 2029.3




Time I P V RV1 D R S V V2 Aa PE Pv V Spill3 V El

Jun 01-10 18.5 3.56 430.9 149.4 15.37 15.37 0.0 434.1 432.5 5370 1.22 0.05 432.9 0.0 432.9 2029.7Jun 11-20 12.3 0.00 432.9 145.2 17.41 17.41 0.0 427.8 430.3 5352 1.26 0.00 426.5 0.0 426.5 2028.5Jun 21-30 43.6 39.94 426.5 170.1 22.47 22.47 0.0 447.6 437.1 5409 1.23 0.60 447.0 0.0 447.0 2032.3Jul 01-10 32.6 29.46 447.0 179.6 23.93 23.93 0.0 455.7 451.4 5530 1.36 0.45 454.8 0.0 454.8 2033.7Jul 11-20 11.9 0.00 454.8 166.7 18.31 18.31 0.0 448.4 451.6 5532 1.30 0.00 447.1 0.0 447.1 2032.3Jul 21-31 27.4 28.89 447.1 174.4 17.98 17.98 0.0 456.5 451.8 5533 1.23 0.45 455.7 0.0 455.7 2033.9

Aug 01-10 23.3 5.69 455.7 178.9 13.05 13.05 0.0 465.9 460.8 5609 1.25 0.09 464.7 0.0 464.7 2035.5 Aug 11-20 24.0 7.30 464.7 188.7 15.60 15.60 0.0 473.1 468.9 5676 1.15 0.12 472.1 0.0 472.1 2036.8 Aug 21-31 18.5 19.69 472.1 190.6 25.41 25.41 0.0 465.2 468.6 5674 1.03 0.31 464.4 0.0 464.4 2035.5Sep 01-10 4.6 0.00 464.4 169.0 29.08 29.08 0.0 440.0 452.2 5537 1.03 0.00 438.9 0.0 438.9 2030.8Sep 11-20 41.8 47.69 438.9 180.7 29.60 29.60 0.0 451.1 445.0 5476 0.91 0.73 451.0 0.0 451.0 2033.0Sep 21-30 15.3 0.00 451.0 166.2 29.48 29.48 0.0 436.7 443.9 5466 0.80 0.00 435.9 0.0 435.9 2030.2Oct 01-10 8.9 0.00 435.9 144.8 29.15 29.15 0.0 415.7 425.8 5313 0.73 0.00 415.0 0.0 415.0 2026.2Oct 11-20 8.4 0.00 415.0 123.4 29.02 29.02 0.0 394.4 404.7 5131 0.64 0.00 393.7 0.0 393.7 2022.1Oct 21-31 10.6 1.91 393.7 104.4 30.04 30.04 0.0 374.3 384.0 4950 0.56 0.03 373.8 0.0 373.8 2018.0Nov 01-10 7.5 0.00 373.8 81.3 32.91 32.91 0.0 348.4 361.1 4746 0.46 0.00 348.0 0.0 348.0 2012.5Nov 11-20 6.8 0.00 348.0 54.8 32.23 32.23 0.0 322.6 335.3 4510 0.36 0.00 322.2 0.0 322.2 2006.8

Nov 21-30 14.8 22.48 322.2 37.0 18.31 18.31 0.0 318.7 320.5 4372 0.28 0.27 318.7 0.0 318.7 2006.0Dec 01-10 17.1 33.21 318.7 35.8 6.87 6.87 0.0 329.0 323.8 4404 0.22 0.41 329.2 0.0 329.2 2008.4Dec 11-20 14.0 13.27 329.2 43.2 6.93 6.93 0.0 336.2 332.7 4486 0.21 0.17 336.2 0.0 336.2 2010.0Dec 21-31 15.0 15.94 336.2 51.2 6.55 6.55 0.0 344.7 340.4 4557 0.20 0.20 344.7 0.0 344.7 2011.8Jan 1-10,73 13.9 0.00 344.7 58.6 9.79 9.79 0.0 348.8 346.7 4615 0.22 0.00 348.5 0.0 348.5 2012.7Jan 11-20 12.8 9.97 348.5 61.4 10.94 10.94 0.0 350.4 349.5 4640 0.24 0.13 350.3 0.0 350.3 2013.0Jan 21-31 12.2 0.00 350.3 62.5 14.11 14.11 0.0 348.4 349.4 4639 0.26 0.00 348.2 0.0 348.2 2012.6Feb 01-10 11.4 17.21 348.2 59.5 14.31 14.31 0.0 345.2 346.7 4615 0.26 0.22 345.2 0.0 345.2 2011.9Feb 11-20 9.9 0.00 345.2 55.1 18.50 18.50 0.0 336.6 340.9 4562 0.30 0.00 336.3 0.0 336.3 2010.0Feb 21-28 18.4 45.97 336.3 54.7 23.67 23.67 0.0 331.0 333.7 4495 0.34 0.58 331.3 0.0 331.3 2008.9Mar 01-10 33.1 55.37 331.3 64.4 16.47 16.47 0.0 347.9 339.6 4550 0.51 0.70 348.1 0.0 348.1 2012.6Mar 11-20 19.8 0.00 348.1 67.8 22.99 22.99 0.0 344.8 346.5 4613 0.61 0.00 344.2 0.0 344.2 2011.7

Mar 21-31 31.2 12.45 344.2 75.5 27.46 27.46 0.0 348.0 346.1 4610 0.70 0.16 347.5 0.0 347.5 2012.4 Apr 01-10 47.8 8.26 347.5 95.3 30.29 30.29 0.0 365.0 356.2 4701 0.56 0.11 364.5 0.0 364.5 2016.1 Apr 11-20 48.6 7.11 364.5 113.1 27.38 27.38 0.0 385.7 375.1 4871 0.68 0.10 385.2 0.0 385.2 2020.3 Apr 21-30 28.8 0.00 385.2 113.9 17.38 17.38 0.0 396.6 390.9 5010 0.80 0.00 395.8 0.0 395.8 2022.5May 01-10 26.4 0.00 395.8 122.1 22.76 22.76 0.0 399.4 397.6 5069 0.92 0.00 398.5 0.0 398.5 2023.0May 11-20 31.7 31.12 398.5 130.2 24.65 24.65 0.0 405.5 402.0 5108 1.09 0.44 404.9 0.0 404.9 2024.3May 21-31 25.0 0.00 404.9 129.8 26.47 26.47 0.0 403.4 404.1 5126 1.26 0.00 402.1 0.0 402.1 2023.7Jun 01-10 14.5 0.00 402.1 116.6 15.37 15.37 0.0 401.2 401.7 5105 1.34 0.00 399.9 0.0 399.9 2023.3Jun 11-20 9.4 0.00 399.9 109.3 17.41 17.41 0.0 391.9 395.9 5054 1.37 0.00 390.5 0.0 390.5 2021.4Jun 21-30 16.8 20.96 390.5 107.3 22.47 22.47 0.0 384.8 387.7 4982 1.31 0.29 383.8 0.0 383.8 2020.1Jul 01-10 101. 28.51 383.8 185.3 23.93 23.93 0.0 461.4 422.6 5286 0.98 0.42 460.8 0.0 460.8 2034.8Jul 11-20 65.2 35.69 460.8 226.1 18.31 18.31 0.0 507.8 484.3 5803 1.02 0.58 507.3 0.0 507.3 2043.0Jul 21-31 107. 139.0 507.3 314.5 17.98 17.98 0.0 596.5 551.9 6347 1.06 2.46 597.9 0.0 597.9 2057.6

Aug 01-10 65.9 21.97 597.9 363.8 13.05 13.05 0.0 650.8 624.3 6907 1.28 0.42 649.9 0.0 649.9 2065.3 Aug 11-20 27.6 30.35 649.9 377.5 15.60 15.60 0.0 661.9 655.9 7145 1.20 0.60 661.3 0.0 661.3 2066.9 Aug 21-31 33.3 50.29 661.3 394.6 25.41 25.41 0.0 669.2 665.3 7214 1.09 1.01 669.1 0.0 669.1 2068.0Sep 01-10 19.5 0.00 669.1 388.6 29.08 29.08 0.0 659.6 664.3 7207 1.16 0.00 658.4 0.0 658.4 2066.5Sep 11-20 10.7 0.00 658.4 369.1 29.60 29.60 0.0 639.5 648.9 7093 1.02 0.00 638.4 0.0 638.4 2063.6Sep 21-30 29.2 14.54 638.4 367.6 29.48 29.48 0.0 638.1 638.3 7013 0.88 0.28 637.5 0.0 637.5 2063.5




Table 2.16: Summary of reservoir operation study for Kurram Tangi Dam..

Dam capacity Th.AF = 900 Demand vs. supply summary results

Dead storage (Th AF) = 300Target cropping intensity (%) and

demand (ThAF)=135% and757

Conservation level ft = 2097.8 No. of years when any shortage occur = 5

Dead storage level ft = 2001.6 Average % shortage during shortageyears = 13.8

Starting storage ThAF = 700 Average % shortage during ALL years = 2.2

Year DaminflowThAF

Irrigationdemand and

supplies

Shortage duringthe year(yes = 1,No = 0)

IrrigationShortage

(ThAF) and% shortageon annual

basis

No. of 10-dayshort irrigationsupply periods

and %shortage

during theshortageperiods

Avragesurface

area(Acres)

AnnualLake

Evaporation

(ThAF)

Directrainover

reservoir area

(ThAF)

AverageReservoir volume

(ThAF) andelevation (ft

amsl) during theyear

Annualspillagefrom the

Dam(ThAF)

Demand Supply ThAF %short

Nos. % Vol El

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

1971 441 757 757 0 0 0 0 0 6,136 30 5 524 2,045 01972 748 757 744 1 13 2 4 12 4,915 25 6 381 2,019 01973 979 757 757 0 0 0 0 0 5,698 27 9 479 2,036 01974 638 757 757 0 0 0 0 0 5,897 30 7 494 2,040 01975 781 757 757 0 0 0 0 0 5,220 25 7 416 2,026 01976 765 757 757 0 0 0 0 0 5,427 23 7 439 2,031 01977 559 757 636 1 121 16 8 52 4,555 18 6 339 2,010 01978 786 757 738 1 19 3 2 43 4,927 19 8 383 2,019 01979 955 757 757 0 0 0 0 0 5,659 22 9 472 2,036 01980 897 757 757 0 0 0 0 0 7,044 29 10 645 2,064 01981 1,017 757 757 0 0 0 0 0 8,259 23 14 812 2,087 1101982 993 757 757 0 0 0 0 0 8,592 32 9 855 2,092 2151983 1,350 757 757 0 0 0 0 0 8,654 28 16 861 2,093 5301984 808 757 757 0 0 0 0 0 8,470 28 4 840 2,090 471985 541 757 757 0 0 0 0 0 7,697 29 6 729 2,076 01986 715 757 757 0 0 0 0 0 6,563 27 7 579 2,054 01987 781 757 757 0 0 0 0 0 6,637 28 4 590 2,056 01988 749 757 757 0 0 0 0 0 6,235 24 5 537 2,048 01989 654 757 757 0 0 0 0 0 5,685 20 5 469 2,036 01990 927 757 757 0 0 0 0 0 5,904 25 7 501 2,041 01991 1,219 757 757 0 0 0 0 0 7,890 31 10 764 2,080 1511992 1,295 757 757 0 0 0 0 0 8,717 30 11 870 2,094 4951993 1,056 757 757 0 0 0 0 0 8,724 33 13 873 2,095 3091994 813 757 757 0 0 0 0 0 8,572 31 3 855 2,092 91995 874 757 757 0 0 0 0 0 8,582 30 6 854 2,092 1461996 865 757 757 0 0 0 0 0 8,340 29 5 821 2,088 881997 910 757 757 0 0 0 0 0 8,408 28 6 831 2,089 921998 1,294 757 757 0 0 0 0 0 8,771 26 10 878 2,095 4941999 621 757 757 0 0 0 0 0 8,159 23 10 794 2,085 152000 370 757 738 1 19 3 2 45 5,967 19 3 504 2,040 02001 414 757 405 1 352 46 24 61 4,239 16 2 306 2,003 0

Average 833 757 740 0.161 17 2.2 40 62.3 6,921 26 7 635 2,060 87




Reservoir Elevation

20002010202020302040205020602070

2080209021002110

J a n

1 - 1

0 , 7

1

J a n

1 - 1

0 , 7

2

J a n

1 - 1

0 , 7

3

J a n

1 - 1

0 , 7

4

J a n

1 - 1

0 , 7

5

J a n

1 - 1

0 , 7

6

J a n

1 - 1

0 , 7

7

J a n

1 - 1

0 , 7

8

J a n

1 - 1

0 , 7

9

J a n

1 - 1

0 , 8

0

J a n

1 - 1

0 , 8

1

J a n

1 - 1

0 , 8

2

J a n

1 - 1

0 , 8

3

J a n

1 - 1

0 , 8

4

J a n

1 - 1

0 , 8

5

J a n

1 - 1

0 , 8

6

J a n

1 - 1

0 , 8

7

J a n

1 - 1

0 , 8

8

J a n

1 - 1

0 , 8

9

J a n

1 - 1

0 , 9

0

J a n

1 - 1

0 , 9

1

J a n

1 - 1

0 , 9

2

J a n

1 - 1

0 , 9

3

J a n

1 - 1

0 , 9

4

J a n

1 - 1

0 , 9

5

J a n

1 - 1

0 , 9

6

J a n

1 - 1

0 , 9

7

J a n

1 - 1

0 , 9

8

J a n

1 - 1

0 , 9

9

J a n

1 - 1

0 , 0

0

J a n

1 - 1

0 , 0

1

R e s e r v o

i r E l e v a

t i o n

( F t ) a m s

l

Figure 2.22: KT dam reservoir water levels.

Reservoir Volume

250300350400450500550600650700750800850900950

J a n

1 - 1

0 , 7 1

J a n

1 - 1

0 , 7 2

J a n

1 - 1

0 , 7 3

J a n

1 - 1

0 , 7 4

J a n

1 - 1

0 , 7 5

J a n

1 - 1

0 , 7 6

J a n

1 - 1

0 , 7 7

J a n

1 - 1

0 , 7 8

J a n

1 - 1

0 , 7 9

J a n

1 - 1

0 , 8 0

J a n

1 - 1

0 , 8 1

J a n

1 - 1

0 , 8 2

J a n

1 - 1

0 , 8 3

J a n

1 - 1

0 , 8 4

J a n

1 - 1

0 , 8 5

J a n

1 - 1

0 , 8 6

J a n

1 - 1

0 , 8 7

J a n

1 - 1

0 , 8 8

J a n

1 - 1

0 , 8 9

J a n

1 - 1

0 , 9 0

J a n

1 - 1

0 , 9 1

J a n

1 - 1

0 , 9 2

J a n

1 - 1

0 , 9 3

J a n

1 - 1

0 , 9 4

J a n

1 - 1

0 , 9 5

J a n

1 - 1

0 , 9 6

J a n

1 - 1

0 , 9 7

J a n

1 - 1

0 , 9 8

J a n

1 - 1

0 , 9 9

J a n

1 - 1

0 , 0 0

J a n

1 - 1

0 , 0 1

R e s e r v o

i r V o

l u m e

( T h A F )

Figure 2.23: KT dam: reservoir volume.

C: 10-day irrigation shortage %

0102030405060

708090

100

J a n

1 - 1

0 , 7

1

J a n

1 - 1

0 , 7

2

J a n

1 - 1

0 , 7

3

J a n

1 - 1

0 , 7

4

J a n

1 - 1

0 , 7

5

J a n

1 - 1

0 , 7

6

J a n

1 - 1

0 , 7

7

J a n

1 - 1

0 , 7

8

J a n

1 - 1

0 , 7

9

J a n

1 - 1

0 , 8

0

J a n

1 - 1

0 , 8

1

J a n

1 - 1

0 , 8

2

J a n

1 - 1

0 , 8

3

J a n

1 - 1

0 , 8

4

J a n

1 - 1

0 , 8

5

J a n

1 - 1

0 , 8

6

J a n

1 - 1

0 , 8

7

J a n

1 - 1

0 , 8

8

J a n

1 - 1

0 , 8

9

J a n

1 - 1

0 , 9

0

J a n

1 - 1

0 , 9

1

J a n

1 - 1

0 , 9

2

J a n

1 - 1

0 , 9

3

J a n

1 - 1

0 , 9

4

J a n

1 - 1

0 , 9

5

J a n

1 - 1

0 , 9

6

J a n

1 - 1

0 , 9

7

J a n

1 - 1

0 , 9

8

J a n

1 - 1

0 , 9

9

J a n

1 - 1

0 , 0

0

J a n

1 - 1

0 , 0

1

1 0 - d a y

I r r i g a t i o n s

h o r t a g e

%

Figure 2.24: KT dam: Shortage to target demand.




Table 2.17: Water availability with various size dams.

Y e a r

Water availability with live storage(ThAF) of

Water availability arranged in descending order for livestorage (ThAF) of

50 100 200 300 400 500 600 700 800 Sr. No

Dependability p (%)

50 100 200 300 400 500 600 700 800

71 437 509 608 705 757 757 757 757 757 1 3.13 733 757 757 757 757 757 757 757 757

72 597 647 699 699 744 744 744 744 744 2 6.25 729 757 757 757 757 757 757 757 757

73 679 729 757 757 757 757 757 757 757 3 9.38 725 757 757 757 757 757 757 757 757

74 553 602 688 757 757 757 757 757 757 4 12.50 720 757 757 757 757 757 757 757 757

75 638 688 716 742 757 757 757 757 757 5 15.63 709 757 757 757 757 757 757 757 757

76 671 720 757 757 757 757 757 757 757 6 18.75 707 757 757 757 757 757 757 757 757

77 532 552 588 588 636 636 636 636 636 7 21.88 697 746 757 757 757 757 757 757 757

78 636 686 738 738 738 738 738 738 738 8 25.00 688 738 757 757 757 757 757 757 757

79 697 746 757 757 757 757 757 757 757 9 28.13 683 735 757 757 757 757 757 757 757

80 688 738 757 757 757 757 757 757 757 10 31.25 683 733 757 757 757 757 757 757 75781 683 733 757 757 757 757 757 757 757 11 34.38 682 733 757 757 757 757 757 757 757

82 683 733 757 757 757 757 757 757 757 12 37.50 679 731 757 757 757 757 757 757 757

83 729 757 757 757 757 757 757 757 757 13 40.63 678 729 757 757 757 757 757 757 757

84 667 735 757 757 757 757 757 757 757 14 43.75 674 724 757 757 757 757 757 757 757

85 538 541 640 737 757 757 757 757 757 15 46.88 671 724 757 757 757 757 757 757 757

86 613 662 691 691 757 757 757 757 757 16 50.00 670 720 757 757 757 757 757 757 757

87 647 697 757 757 757 757 757 757 757 17 53.13 667 720 757 757 757 757 757 757 757

88 642 692 745 745 754 757 757 757 757 18 56.25 647 697 746 757 757 757 757 757 757

89 620 654 654 654 654 740 757 757 757 19 59.38 642 692 745 757 757 757 757 757 757

90 670 720 746 746 746 746 757 757 757 20 62.50 638 688 738 746 757 757 757 757 757

91 709 757 757 757 757 757 757 757 757 21 65.63 636 686 738 745 757 757 757 757 757

92 725 757 757 757 757 757 757 757 757 22 68.75 620 662 716 742 757 757 757 757 757

93 707 757 757 757 757 757 757 757 757 23 71.88 613 654 699 738 757 757 757 757 757

94 720 757 757 757 757 757 757 757 757 24 75.00 597 647 691 737 754 757 757 757 757

95 674 724 757 757 757 757 757 757 757 25 78.13 571 639 688 705 746 746 757 757 757

96 682 731 757 757 757 757 757 757 757 26 81.25 553 602 654 699 744 744 757 757 757

97 678 724 757 757 757 757 757 757 757 27 84.38 538 552 640 691 738 740 744 757 757

98 733 757 757 757 757 757 757 757 757 28 87.50 532 541 608 654 654 738 738 744 744

99 571 639 738 757 757 757 757 757 757 29 90.63 437 509 588 588 636 641 738 738 73800 367 367 367 446 544 641 738 757 757 30 93.75 399 405 405 446 544 636 636 636 636

01 399 405 405 405 405 405 405 482 579 31 96.88 367 367 367 405 405 405 405 482 579

Average annual irrigation releases No. of years with any deficit supply

63231

67525

70612

71910

7308

7367

7405

7434

7464




Dependability of Kurram Tangi Dam for various Live StorageCapacity (ThAF)

05

101520

253035404550556065707580859095

100

300 400 500 600 700 800 900 1,000 1,100 1,200 1,300 1,400

Annual Available flow (ThAF)

D e p e n

d a

b i l i t y / e x c e e

d a n c e

( % )

No damDam=50

Dam=100

Dam=200

Dam=300

Dam=400

Dam=500

Dam=600

Dam=700

Dam=800

Figure 2.25: Improvement of water availability for different live storage capacities of the dam as

determined by reservoir operation.

2.10 RESERVOIR MAXIMUM MINIMUM RULE CURVEReservoir inflow varies over the years. The releases are required to be tailored to the

inflows and the storage available in the reservoir. Unplanned releases could lead to situation

where full demand is met during part of a crop season and very large shortage result in

remaining crop season. Crop are very sensitive to water shortages during later part of the

growing season. Thus it is required that any anticipated shortage should be spread equally for

the whole growing season. This requires that rules be framed to regulate releases (maximum

and minimum) during different periods/months.

These rules are defined for seasonal dams in terms of maximum and minimum

reservoir level during different months. These rules are developed/optimized from reservoir

operation studies. The minimum rules are to ensure temporal equity of water releases and

maximum rules are set to ensure reservoir filling and flood handling particularly during later

part of reservoir filling.

For a carryover dam these rules are defined differently. The reservoir releases for a

crop season are adjusted in view of the reservoir levels before the start of the season. The




releases are curtailed by a factor corresponding to various reservoir levels, e.g. for KT dam

the releases are reduced by 10, 15 and 20% for reservoir elevation of 2035, 2025 and 2015 ft

amsl. Thus shortages are reduced from maximum of 80% in 10-day period (Fig. 2.24) to only

10 to 20% (Fig. 2.26).

0

10

20

3040

50

60

70

80

90

100

J a n

0 1 - 1

0 , 7

1

J a n

1 - 1

0 , 7

2

J a n

1 - 1

0 , 7

3

J a n

1 - 1

0 , 7

4

J a n

1 - 1

0 , 7

5

J a n

1 - 1

0 , 7

6

J a n

1 - 1

0 , 7

7

J a n

1 - 1

0 , 7

8

J a n

1 - 1

0 , 7

9

J a n

1 - 1

0 , 8

0

J a n

1 - 1

0 , 8

1

J a n

1 - 1

0 , 8

2

J a n

1 - 1

0 , 8

3

J a n

1 - 1

0 , 8

4

J a n

1 - 1

0 , 8

5

J a n

1 - 1

0 , 8

6

J a n

1 - 1

0 , 8

7

J a n

1 - 1

0 , 8

8

J a n

1 - 1

0 , 8

9

J a n

1 - 1

0 , 9

0

J a n

1 - 1

0 , 9

1

J a n

1 - 1

0 , 9

2

J a n

1 - 1

0 , 9

3

J a n

1 - 1

0 , 9

4

J a n

1 - 1

0 , 9

5

J a n

1 - 1

0 , 9

6

J a n

1 - 1

0 , 9

7

J a n

1 - 1

0 , 9

8

J a n

1 - 1

0 , 9

9

J a n

1 - 1

0 , 0

0

1 0 - d a y I r r i g a

t i o n s

h o r t a g e

%

Figure 2.26. Effect of reservoir release control on seasonal shortages.

2.11 WAVE HEIGHT (USBR P-271)

Free board to prevent overtopping of embankment by abnormal and severe waves actions of

rare occurrences created due to unusual sustained winds of high velocity.

• Wave considered coincident with design inflow flood.

• Wave height depends on: 1. Wind velocity, 2. Duration of the wind, 3. The fetch, 4.

Depth of the water, 5. Width of the reservoir

• Wave height at approaching dam may be altered due to increasing water depth or

decrease of reservoir width.

• Wave run-up affected by slope angle, surface texture, angle of wave train.

• Wind energy lost in raising the water up the sloping face of the dam.






2.12 RESERVOIR SEDIMENTATION

(Ref: USBR DESIGN OF SMALL DAMS, Appendix-H, Pages 767-796; GOLZE;

HANDBOOK OF DAM ENGINEERING Pages 142-146)

Rivers carries a lot of sediments especially in the monsoon season. The amount of

sediments depend on many factors, some are as under.

• Climate, particular annual rainfall and resulting runoff.

• Vegetal cover, which is dependent on rainfall

• Surface geology and soil cover

• Land and river slopes

• Land use

2.13 EFFECTS OF RESERVOIRS ON SEDIMENT ACCUMULATIONS:

Whenever a dam is built across a river, retention of water borne sediments can be

expected. Accumulation of the sediments in the dam over time can occupy storage space that

was provided to carryout the basic functions of the dam and reservoir. To extend useful life

of the dam, the planned storage of the dam is increased by an amount necessary for the

storage of the sediments entering the dam during its planned life.

Factors that determine the rate and total volume of sediment accumulation in a

reservoir are:

• Average sediment load carried by a river (expressed as dry weight, tons)

• Periods of abnormally high or low runoff which may result in wide variations

from the average sediment load.

• Ratio between reservoir capacity and the annual volume of inflow, a measure

of trap efficiency.

• Grain size of sediments, sediment size gradation, sediment unit weight

• Method of reservoir operation.




2.13 ESTIMATING SEDIMENT ACCUMULATION IN RESERVOIRS:

Two methods are used: (1). Sampling of suspended sediments; establishment of

sediment-discharge relation, computation of total annual sediment load in the river, and

estimation of trap efficiency. (2). Results of surveys of sediment accumulations in existingreservoirs, expressed in tons or acre feet per sq. mile per year. The average annual sediment

load may be set accordingly.

2.14.1 Sediment Estimation Based On Sediment Sampling.

The suspended and bed load sediments are measured by sampling for the known flow.

Measurements are made often throughout the year. A general sediment rating curve is

developed for each period or year showing sediment load versus river flow. The sediment

rating curve is used to compute the total sediment load for an average year from known flow

record or by use of flow duration curve. The sediment measurements and estimation is done

for long time period. The bed load may be estimated by using various formulas (e.g.

Schoklitsch formula, Meyer-Peter Muller formula) in combination with flow duration curve.

Else the bed load may be taken as 10 to 20% of the suspended load. The size gradation of the

suspended and bed load is based on measured samples showing percentage of clay, silt and

sand grains.

The average sediment load of the Kurram River at Thal is estimated by SWHP of

WAPDA as 3.77 Million Short Tons per year (MST/yr) (Table 2.18, Figs. 2.27, 2.28). About

86% of the sediments were observed for monsoon period of 4 months (April-September). The

annual sediment yield may be expressed as 1.48 Ac.ft. per sq. mile of drainage area. The

sediment concentration is 0.383% (by weight) or 3,830 ppm. The maximum observed

concentration was 142,000 ppm and minimum observed concentration was 2 ppm. The

computed maximum concentration was 47,800 ppm. At average discharge, the suspended

sediment consists of about 6% sand, 70% silt and 24% clay (actually varies during and over

the years). SWHP estimated the unit weight of fresh deposits of this sediment as about 58.3

lbs per cft.

If sediments measurements are not available at the dam site, then sediment is

estimated from sediment measurements at any u/s or d/s location on the basis of sediment

load per unit area or per unit flow volume (assuming that sediment generation potential being

same for the two sites). The sediment load may be estimated at required site in proportion to

average annual flow at the two sites. Thus S 2 = S 1 x Q 2/Q 1 . The sediment load of Kurram




River was measured at Thal, which is located 10 miles u/s of the dam site. Also there were no

measurements of sediment for the Kaitu River. The sediment load of the Kurram River flows

between Thal and Kurram Tangi Dam site and of the diversions of Kaitu River through

Kaitu-Kurram Feeder was estimated on the basis of annual flow volumes. The sediment load

of Kaitu River was considered same as for Kurram River. The bed load is considered as 10%

of the suspended load. Thus total sediment inflow in Kurram Tangi Dam is determined on

area basis as under.

Kurram River suspended sediments at Thal = 3.77 MST/yrKurram River suspended sediments at KT dam site = 3.77 * (628/574) = 4.12 MST/yrKurram River bed load @ 15% = 0.62 MST/yrTotal sediment load of Kurram River at dam site = 4.74 MST/yrKaitu-Kurram Feeder suspended load = 3.77 * (180/574) = 1.19 MST/yr

Kaitu-Kurram Feeder bed load = 0 (a silt ejector is to be installed at feeder head)Total sediment load entering Kurram Tangi Dam = 5.93 MST/yr(Suspended = 4.12 + 1.19 = 5.31, bed load = 0.62 MST/year)

Gomal zam dam has much higher sediment load (~ 12 MST/year) in spite of smaller

catchment area due to nature of geologic formation and land cover.

Table 2.18: Historic sediment load of Kurram River at Thal.Year Yearly

sedimentload

(m.s.t).

Monsoonload Apr-

Sept(m.s.t.)

Monsoonload as %of Whole

Year

Year Yearlysediment

load(m.s.t).

Monsoonload Apr-

Sept(m.s.t.)

Monsoonload as %of Whole

Year

1968 2.55 2.47 97 1984 4.74 4.62 961969 1.43 1.27 89 1985 0.99 0.89 901970 1.4 1.38 98 1986 4.72 4.08 761970 3.99 3.96 99 1987 3.5 4.66 761972 7.05 6.17 87 1988 4.87 3.93 811973 10.4 9.38 90 1989 1.39 1.14 821974 4.49 1.31 53 1990 5.52 5.3 961975 4.65 4.62 99 1991 10.7 0.6 991976 5.13 4.22 82 1992 4.87 4.7 971977 1.01 0.75 74 1993 6.45 6.35 981978 4.05 4.03 99 1994 5.04 4.93 98

1979 1.01 0.94 93 1995 4.56 4.15 841980 3.17 4.48 78 1996 4.27 1.68 741981 1.74 1.02 57 1997 0.82 0.7 861982 1.56 1.38 88 1998 4.49 4.26 911983 10 9.94 99 Average 3.77 3.24 86




Kurram River at Thal: Historic Sediment and Flow Data

0

100

200

300

400

500

600

700

800

900

1000

1100

1200

1 9 7 1

1 9 7 3

1 9 7 5

1 9 7 7

1 9 7 9

1 9 8 1

1 9 8 3

1 9 8 5

1 9 8 7

1 9 8 9

1 9 9 1

1 9 9 3

1 9 9 5

1 9 9 7

A n n u a l F

l o w ( T h A F )

0

1

2

3

4

5

6

7

8

9

10

11

12

S e d i m e n t L o a d

M S T / y r

Sediment LoadAnnual FlowMonsoon Flow

Figure 2.27 : Historic sediment data of Kurram River at Thal.

Kurram River at Thal: Sediment -Flow Relation

0

1

2

34

5

6

7

8

9

10

11

0 200 400 600 800 1000 1200Annual Flow (ThAF)

S e d i m e n t L o a d

( M S T )

Figure 2.28: Kurram River at Thal: Flow vs. sediment load.




PATRIND HYDROPOWER PROJECTSUSPENDED SEDIMENT RATING

1

10

100

1000

10000

100000

1000000

10000000

1 10 100 1000

Flow Discharge (Cumecs)

S e

d i m e n

t l o a

d ( T o n s

)

Figure 2.29: Sediment load for Kunhar River at Ghari Habibullah.

PATRIND HYDROPOWER PROJECTGRADATION CURVE FOR SUSPENDED SEDIMENT

AT GARHI HABIB ULLAH

05

101520253035404550556065707580859095

100

0.001 0.01 0.1 1

Sediment Size (mm)

% p a s s

i n g

/ F i n e r

16-10-1984 10/10/198413-10-1984 17-9-844/9/1984 13-10-198418-9-1984 19-9-198419-1-1984 11/2/19842/11/1983

Figure 2.30: Sediment size gradation for Kunhar River at Ghari Habibullah.




The particle size of the all suspended sediments are considered same as found at Thal (6%

sand, 70% silt and 24% clay) while the bed load is considered as 10% silt and 90% sand. The

particle size of total sediment is weighted and estimated as under:

Suspended Bedload Weighted

Sand (> 62 µ) 6% 90% 15% (=0.06*5.31+0.9*0.62)/5.93

Silt (4 to 62 µ) 70% 10% 64%

Clay (< 4 µ) 24% 0% 21%

2.14.2 Sediment Estimation Based On Survey Of Existing Reservoirs.

The sediment load may also be estimated from a record of sediment accumulation in

the existing reservoirs. The volume of sediment accumulated over a number of years is

measured by topographic and hydrographic surveys. The depth of reservoir bottom at any

point below the water surface is measured by sounding while traversing the reservoir on a

boat. For this purpose section pillars are placed along the reservoir r im to form a straight line

during hydrographic survey. In addition samples of deposits are taken to measure dry weight

and size gradation. Comparison of reservoir bottom level with earlier or first survey provides

the depth of sediments over the intervening period. However considerations must be given to

the following.

• The observed record must be from a drainage area that is equivalent in size to the

project area or an adjustment of the unit yield must be made.

• The hydrologic characteristics of the drainage areas being compared should be

similar, particularly the annual rainfall.

• The physical characteristics of the two areas must essentially be the same including

topography, geology and soils, vegetal cover, and land use.

• The trap efficiency, or the capacity-inflow ratio of the two reservoirs must be same.

• The existing record of sediment accumulation must be at least 10 years long. If only a

short record is available for comparison, a regional hydrologic investigation must be

made to determine whether the sediment record was obtained in a period of high,

average or low runoff/flows.






The capacity V of the reservoir will decrease with time due to sediment deposition.

Therefore for same annual inflow Q the trap efficiency will decrease due to decrease of V/Q.

The sediment deposition in the reservoir is computed over small intervals, e.g. 5 year, during

which V may be considered constant.

For Kurram Tangi Dam, V/Q = 900,000/808,000 = 1.11, and T E = 97.5%. Thus

97.5% of the all incoming sediments will be trapped in the reservoir, causing a continuous

decline in the reservoir capacity over the years. As the reservoir capacity become smaller

with lower V/Q ratio, then trap efficiency will continuously decrease each year (Table 2.19).

The trapping efficiency varied from 97.5% for 1 st

2.16 UNIT WEIGHT OF THE DEPOSITED SEDIMENT

year to 86.3% after 100 years.

The sediment estimate of weight per time (e.g. million short tons per year) is

converted into volume by knowing the sediment unit weight (e.g. lb/cft). The various sizes of

particle are classified as clay (less than 0.004 mm), silt (0.004 to 0.0625 mm) and sand

(0.0625 to 2.0 mm). The unit weight of the fresh deposited sediments is dependent by the

proportion of the sediments of clay, silt and sand sizes. The sediments deposited in the

reservoir get compacted to a denser form over many years due to its self-weight. There are

several factors influencing the weight of the deposited and compacted sediment as (1) manner

in which reservoir is to be operated, (2) texture and size of particles, (3) compaction or

consolidation rate, (4) other factors (density currents, thalweg slope of stream, vegetation in

head water areas).

The reservoir operation is most influential factor. Sediments exposed for long periods

due to reservoir drawdown are considerably consolidated. Reservoir operation are classified

of following types:

Type I: Sediment remain always submerged or nearly submerged (small drawdown)

e.g. Bunji hydropower project dam

Type II: Normally moderate to considerable reservoir drawdown, e.g. Tarbela dam

Type III: Reservoir normally empty

Type IV: Riverbed sediments




Table 2.19: Sediment unit weight and deposition with time.K = 2.916 Average.sediment/yr= 5.926 MST Av. Flow Q = 833 ThAF

T

(Years)

Sediment unit

weight lb/cft

V/Q

ratio Trapefficiency

%

Time Sediment Load (MST)

Cum.

SedimentdepositThAF

Final

Storagecapacity VW CWT InflowT Traped Cum. Mass

0 67.34 67.34 916

10 70.26 69.31 1.10 97.50 59.3 57.8 57.8 38 878

20 71.13 70.07 1.05 97.50 59.3 57.8 115.6 76 84030 71.65 70.53 1.01 97.50 59.3 57.8 173.3 113 803

40 72.01 70.87 0.96 97.33 59.3 57.7 231.0 150 766

50 72.29 71.13 0.92 97.12 59.3 57.6 288.6 186 73060 72.53 71.35 0.88 96.90 59.3 57.4 346.0 223 693

70 72.72 71.53 0.83 96.66 59.3 57.3 403.3 259 657

80 72.89 71.69 0.79 96.42 59.3 57.1 460.4 295 621

90 73.04 71.84 0.75 96.16 59.3 57.0 517.4 331 585100 73.17 71.96 0.70 95.89 59.3 56.8 574.2 366 550

110 73.29 72.08 0.66 95.60 59.3 56.7 630.9 402 514

120 73.40 72.19 0.62 95.30 59.3 56.5 687.3 437 479130 73.50 72.29 0.57 94.98 59.3 56.3 743.6 472 444

140 73.60 72.38 0.53 94.63 59.3 56.1 799.7 507 409

150 73.69 72.46 0.49 94.25 59.3 55.9 855.6 542 374160 73.77 72.54 0.45 93.85 59.3 55.6 911.2 577 339

170 73.84 72.62 0.41 93.40 59.3 55.4 966.5 611 305

180 73.92 72.69 0.37 92.92 59.3 55.1 1021.6 645 271190 73.98 72.75 0.32 92.37 59.3 54.7 1076.3 679 237

200 74.05 72.82 0.28 91.76 59.3 54.4 1130.7 713 203210 74.11 72.88 0.24 91.06 59.3 54.0 1184.7 746 170

220 74.17 72.94 0.20 90.24 59.3 53.5 1238.2 779 137

230 74.23 72.99 0.16 89.25 59.3 52.9 1291.0 812 104

240 74.28 73.04 0.12 88.01 59.3 52.2 1343.2 844 72

250 74.33 73.09 0.09 86.32 59.3 51.2 1394.3 876 40

The initial deposited weight of the sediment deposit (W i

W

) is determined on the basis of

proportion (P) and unit weight (W) of different size particle (c = clay, m = silt (medium), s =

sand) as:

i = W c × P c + W m × P m + W s × P

where

s

Pc, Pm, Ps = proportion of clay, silt and sand particles in the sediments, respectively.

Wc, Wm, Ws = unit weight of clay, silt, sand respect ively according to type of dam.




The unit weight of different particles is given as function of dam type on operation basis as

(lbs/cft):

Table 2.20 : Unit weight (lb/cft) of sediments constituents.Reservoir type clay W silt Wc sand Wm s

I 26 70 97II 35 71 97III 40 72 97IV 60 73 97

Example: The Kurram Tangi Dam reservoir was classified for reservoir operation as of Type-

II category. The particle weight for category-II reservoir operation is taken as W = 35, 71 and

97 lb/cft for clay, silt and sand respectively. Thus initial sediment weight is determined as

under. (Proportion of clay, silt and sand as 35, 64 and 15%, respectively).

W i

The sediment volume is determined from the total sediment load and the unit

sediment weight as sediment volume = sediment weight (lbs) ÷ unit sediment weight (lb/cft).

The average volume of fresh sediment deposits for KT Dam is estimated as:

= 35 × 0.21 + 71 × 0.64 + 97 × 0.15 = 67.34 lbs/cft

(5,930,000 ST × 2000 lb/ST) ÷ (67.34 lb/cft × 43,560 ft 3

The unit weight of the deposited sediment will increase each year due to compaction

of the sediment it remains in the reservoir according to the equation

/acre-feet) ≈ 4,040 Acre-Feet/ year.

W T = W i + K Log 10

where W

T

T = unit weight after T years of compaction, W i

Table 2.21: K factor for sediment constituents.

= the initial unit weight, and K = a

constant depending on the type of the reservoir and size analysis of the sediment. The factor

K is as under.

Reservoir type Clay K Silt K c Sand K m s

I 16.0 5.7 0.0

II 8.4 1.8 0.0

III 0.0 0.0 0.0

The overall K factor is determined as: K = K c × P c + K m × P m + K s × P s

K = 8.4 × 0.21 + 1.8 × 0.64 + 0.0 × 0.15 = 2.916

. e.g. The factor K

for KT category-II dam (K = 8.4, 1.8, and 0.0 for clay, silt and sand, respectively) is

determined (for sediment with 21, 64 and 15% of clay, silt and sand respectively) as:




The sediment unit weight after, say 50 years, will be as: 67.34 + 2.916 Log 50 = 72.29 lb/cft.

Sediments are deposited in the reservoir in each of T years of operation and each

years deposits will have different compaction time and sediment unit weight. The average

sediment unit weight after T years (of cumulative all sediments deposited from 1st

CW

through Tyears) is determined as:

T = W i

For KT Dam the sediment unit weight W

+ 0.4343 K [T/(T-1) Ln T – 1]

T

The sediments deposited over a period ∆T and the reservoir final volume is then

determined as: S

after, say 50 years, will be as: 67.34 + 0.4343 *

2.916 [50/49 Ln 50 – 1} = 71.13 lb/cft. The cumulative sediment unit weight for different

years T is calculated for KT dam in Table 2.19.

T = S A ∆T T E / CW T and V T = V I - S

where S

T

T = sediment deposition at time T, S A = average annual sediment load (MST), V T =

reservoir volume at time T, V I

2.17 SEDIMENT DISTRIBUTION WITHIN A RESERVOIR

= reservoir volume at time T- ∆T. The sediment deposition and

reservoir volume for KT dam are given in Table 2.19.

A sediment particle is affected in settlement by horizontal force due to water

movement and vertical force due to gravity. A particle will remain in suspension as long as

turbulent forces equal or exceed the force of gravity. When the flow enters the reservoir the

increased cross sectional area results in decrease of velocity and turbulence until it become

ineffective in transporting the sediment and the particles get deposited. The sediment

deposition is not confined to lower storage increments (dead storage space), rather it get

deposited at all levels in the reservoir below the normal conservation level.

2.17.1 Reservoir shape factor

The distribution of sediment deposition at different levels depends upon the shape

classification of the reservoir on the basis of slope (m) of reservoir storage versus depth graph

(on a log-log paper). The slope m also be determined as under (Volume V in acre feet and

depth d in ft):

m = (Log V 2 – Log V 1) / (Log d 2 – Log d 1

The reservoirs are classified according to shape as under.

)




Table 2.22: Reservoir classification according to shape.

Reservoir type Classification Slope mI Lake 3.5 – 4.5II Flood plain – Foothill 2.5 – 3.5

III Hill 1.5 – 2.5IV Gorge 1.0 – 1.5

KT Dam: Depth-volume Curve

y = 2.18878x 0.35826

R2 = 0.9996610

100

1000

1000 10000 100000 1000000Volume (AF)

H e

i g h t ( f t ) a

b o v e

d a

t u m

o f 1 8 0 0

f t

Figure 2.32 : Shape factor m = dV/dh for reservoir classification.

For KT dam m was calculated as m = (Log 100,000 – Log 1000) / (Log 136.35 – Log 26.31)

= 2.80 (Fig. 2.32): and thus dam was termed as of Flood plain – Foothill type II dam.

The sediment depth wise distribution in the reservoir may be determined by following

two methods. (1) By using Reservoir storage design curves, and (2). By using Reservoir area

design curves.

2.17.2 Sediment Depth Wise Distribution By Reservoir Storage Design Curves

The silt deposition at various depths is given for different category of dams in terms

of % sediment deposition versus % reservoir depth (see Figure H-7 of USBR Design of SmallDams, page-781). The silt deposition at any depth is given as under (but not exceeding

original storage volume at the depth) as:

Sediment volume = Total sediment * % sediment at the given depth

The % sediment deposition at different depths for various shape category dams is given in

Table 2.23 and shown in Figure 2.33.




Table 2.23 : Depth wise sediment deposition according to reservoir shape.:

Percent relativedepth (%d)

Percent sediment deposition below %d depth for dam shape

I II III IV

0 0 0 0 00.05 0.1 1.51 1.51 21.5

0.1 0.2 4.02 5.1 35.5

0.2 1.2 11.45 18.3 53.1

0.3 4.5 21.5 35.2 66.8

0.4 11.5 33.5 53.8 77.8

0.5 20.8 46.3 70.8 86.5

0.6 33.2 59.5 84.5 92.1

0.7 48.5 72.4 92.5 95.80.8 66.5 84.1 97.5 98.1

0.9 86.2 93.6 99.5 99.6

0.95 94.5 97.5 99.8 99.9

1 100 100 100 100

Depth wise relative sediment distribution

05

101520253035404550556065707580859095

100

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1Depth (fraction)

% s e d i m

e n t d e p o s i t i o n

Type I dam

Typr II dam

Type III dam

Type IV dam

Figure 2.33 : Depth wise relative sediment deposition for different shape category dams.

(USBR 2001. pp-H-7)




For KT dam the total sediment volume for 50 years is 186,000 acre feet. The normal

conservation level NCL and dead storage depth are as 295 and 196.6 ft at elevation of 2100.0

ft 2001.6 ft amsl respectively. Thus the dead storage depth is 66.7% of total depth. The

sediment deposition below dead storage depth (at 66.7% depth) is about 68.1% for shape

category II dam. Thus sediment deposited below dead storage level = 0.681 * 186,000 =

126,000 acre feet against 300,000 acre of original dead storage capacity. The balance of

186,000 – 126,000 = 60,000 acre feet of sediment (31.9% of total sediments) are deposited

above the dead storage level in the live storage space and thus infringe into the live storage

space and reduce usable storage of the dam. Thus distribution of sediment deposit can also be

worked for other or all depth/levels. Thus after 50 years the dead storage will reduce to

300,000 – 126,000 = 174,000 AF and live storage will reduce to 618,000 – 60,000 = 558,000

AF. The storage (live, dead and gross) space for KT dam available after various years of

reservoir operation is worked out in Table 2.24 and variations are shown in the Figure 2.34.

The reservoir live storage will decrease from present value of 616 ThAF to 556, 499, 374,

204 and 40 ThAF after 50, 100, 150, 200 and 250 years of reservoir operation if present rate

of sediment continues unaltered. The reservoir elevation-volume relationship will vary

considerably over the years. The change in the elevation-volume relationship over the years is

shown in Figure 2.35.




Table 2.24 Sediment deposition computations.

T Years Sediments de osited ThAF Stora e Ca acit Th.AFTotal Live Dead Total Live Dead

0 % De osition in dead 68 916 616 30010 38 12 26 878 604 274

20 76 24 51 840 592 249

30 113 36 77 803 580 223

40 150 48 102 766 568 198

50 186 60 127 730 556 173

60 223 71 151 693 545 149

70 259 83 176 657 533 12480 295 95 200 621 521 100

90 331 106 225 585 510 75

100 366 117 249 550 499 51

110 402 129 273 514 487 27

120 437 140 297 479 476 3

130 472 172 300 444 444 0

140 507 207 300 409 409 0

150 542 242 300 374 374 0160 577 277 300 339 339 0

170 611 311 300 305 305 0

180 645 345 300 271 271 0

190 679 379 300 237 237 0

200 713 413 300 203 203 0

210 746 446 300 170 170 0

220 779 479 300 137 137 0

230 812 512 300 104 104 0240 844 544 300 72 72 0

250 876 576 300 40 40 0




Sediment Deposition in Kurram Tangi Dam

0

100

200

300

400

500

600

700

800

900

1000

0 25 50 75 100 125 150 175 200 225 250

Years after operation

A v a i l a b

l e S t o r a g e c a p a c i t y

( T h A F

Total storage ThAF

Live storage ThAF

Dead storage ThAF

Figure 2.34: Reservoir live and dead storage capacity after silt deposition.

KURRAM TANGI DAM: STORAGE CAPACITY VS. RESERVOIR SEDIMENTATION

1820184018601880190019201940196019802000202020402060208021002120

0 50 100 150 200 250 300 350 400 450 500 550 600 650 700 750 800 850 900 950 1000Storage ThAF

E l e v a t i o n

F t a m s l

Origional

T=25 years

T=50 years

T=75 years

T=100 years

T=125 years

T=150 years

T=175 years

T=200 years

Dead Storage Level = 2001.6 ft

Figure 2.35: Reservoir storage capacity curves after sedimentation.




2.18 DELTA DEPOSITION

The sediment laden river water when enters the reservoir experiences sudden

enlargement of flow area and corresponding decrease of flow velocity. This causes all of the

coarse sediments of suspended load and bed load to settle out of floatation/bed rolling. Thisresults in the formation of a delta at the mouth of the reservoir. Delta is defined by top set

slope, fore set slope, bottom set slope, and density currents. The pivot point is intersection of

top set and fore set slopes. The delta advances with time becoming closer to the dam body.

As reservoir draws down, the inflow causes advancement of the delta. The delta is weak

deposit of sediments without any binding force between the particles. Any seismic activity,

particularly at low reservoir levels, may result in immediate liquification of the delta which

will move towards the dam. The dam may fail on account of the thrust of the moving deltadeposits. Thus formation of delta is monitored strictly. The dam is operated in a way which

may not excessively expose the delta.

Pivot point

Topset slope

Foreset slope

Bottomset slope

Density currents

Normal pond level

Original river bed level

Origional thalweg slope

Figure 2.36: Delta formation at head of reservoir.




References

KTDC 2004. KURRAM TANGI DAM FEASIBILITY REPORT (Draft). Prepared by

Kurram Tangi Dam Consultants for Wapda, Lahore.

Morris, Gregory L. and Jiahua Fan. 1998. RESERVOIR SEDIMENTATION HANDBOOK

(Design and management of dams, reservoirs and watersheds for sustainable use).

McGraw Hill. New Yor. (Cewre library Accession # 6007).

NESPAK. 1988. Gandiali Dam Planning Report. National Engineering Services (NESPAK).

Lahore.

NESPAK. 1992. Mirani Dam Multipurpose Project: Project Planning Report. National

Engineering Services (NESPAK). Lahore. pp 2.3.

SDO. 1992. Project Completion Report of Jammergal Dam. Small Dams Organization,

Irrigation and Power Dept., Govt. of Punjab. Rawalpindi.

Tariq, Ata-ur-Rehman. 2004. Hydrologic Assessment of Small Dams in Potohar Area: Case

Study of Jammargal Dam. Paper presented at 69 th

Tariq, Muhammad. 2000. Hydrologic Assessment of Small Dams in Potohar Area: Case

Study of Jammargal Dam. M.Sc. Thesis. Centre of Excellence in Water Resources

Engineering, University of Engineering and Technology, Lahore.

annual session of Pakistan

Engineering Congress (April 6-8, 2004), Lahore. Paper No. 647. Proceedings pp:623-

645.

USBR. 1992. Flood Hydrology Manual. U.S. Department of the Interior, Bureau of

Reclamation, Denver.







Figure 2.37 : Reservoir bed profile by lead line or sounding disk method (Source:

http://www.usace.army.mil/publications/eng-manuals/em1110-2-1003/c-8.pdf)

Freeboard Design

An important aspect of dam design is the nature of the freeboard - the specified verticaldistance between the reservoir water level and the top of the dam. This is primarily to providea margin of safety against flood events but is also essential to prevent splashing or occasionalovertopping of a dam embankment by extreme waves. Factors which should be considered indetermining the size of the freeboard include:

meteorological factors - wind characteristics, fetch;hydrological/hydraulic factors - flood, reservoir and wave characteristics, spillway

and outlet flow rating curves;downstream flood risks;

structural factors - type, slope, surface, settlement of embankment, spillway andoutlet characteristics etc;

geophysical hazards.Source: http://www.worldbank.org/html/fpd/em/hydro/fd.stm



Tariq. 2008. DAM AND RESERVOIR ENGINEERING 3- 1 Ch. 3: Dam Geological and Geotechnical Studies

Chapter-3

GEOLOGICAL AND GEOTECHNICAL STUDIESFOR DAM DESIGN 1

3.1 PURPOSES

Geology of the dam refers to the study and investigation of foundation materials, overwhich the dam will be placed. Detailed geologic is directed to establish ground conditions interms of strength, durability, susceptibility to weathering, seepage flows, geologic structure,stratigraphy, faulting, foliation/folding, jointing for the dam site itself and the sites for otherstructures such as reservoir, spillways, diversion tunnel, outlet works, power house, etc.

General objectives of the geologic investigations are:

• To determine whether the dam foundation has sufficient strength and durability tosupport the type of dam proposed.

• To establish whether foundation is watertight.

• To set extent of need of any foundation improvement/treatment as curtaingrouting or blanket grouting.

• To evaluate dam foundation against probable settlement and deformation.

• To determine seepage pattern, seepage quantity, and pore water pressures in

foundations.• To establish containment integrity of reservoir basin (is storage area watertight?),

presence of cavernous rock openings (in limestone) leading to leaks of water fromthe reservoirs to underground caverns and adjacent basins.

• To check stability of reservoir rim/ week joints against landslide along reservoirrim (leading to wave of water overtopping of dam crest when dam is full.

• To find sources/locations, nature, suitability of construction materials in nearby places which will be needed to build the dam.

Dams are usually founded over rocks. Most rocks have adequate strength but theirweakness is in the orientation and dip of discontinuities relative to loading from the dam, aswell as in fill material and depth of weathering in such discontinuities. (Fig. 3.1)

1 Sources:Wahlstrom (1974); USBR (1967); USBR (2001); Novak et al. (1985)http : // homepages . ihug . com . au/~richardw/ page 19.html, page25.html, page26.htmlwww.dur.ac.uk/~des0www4/cal/dams/geol/topo.htm

http://www.dur.ac.uk/~des0www4/cal/dams/geol/topo.htm





Figure 3.1: Foundation geology along the dam axis.

The geological services are required for the engineering of a large dam in the followingareas: (1) The safety of the dam on its foundations; (2) The water tightness of the reservoir

basin; (3) The availability of natural materials for its construction.

The engineering geologist is a key member of an engineering team, since he willensure the feasibility of the project, continuing through the design stage and terminating onlywhen construction has either proved that geological conditions revealed are in conformitywith the premises adopted in design, or he has made possible proper evaluation of anyconditions not foreseen in the earlier stages.




The safety, viability and cost of a dam are all dependent upon geology. Most rockshave adequate strength but their weakness is in the orientation and dip of discontinuitiesrelative to the loading from the dam, as well as the infilling material in, and depth of,weathering in such discontinuities. It is necessary to investigate both the regional geology and

the specific local geology to ensure a global picture is developed.

3.2 FOUNDATION MATERIALS CHARACTERISTICS

3.2.1 Terminology

Bedrock is a general term that includes any of the generally indurated or crystallinematerials that make up the Earth's crust. Individual stratigraphic units or units significant toengineering geology within bedrock may include poorly or nonindurated materials such as

beds, lenses, or intercalations. These may be weak rock units or interbeds consisting of clay,silt, and sand (such as the generally soft and friable St. Peter Sandstone, Sugary limestone forTarbela), or clay beds and bentonite partings in siliceous shales of the Morrison Formation.

Surficial Deposits are the relatively younger materials occurring at or near the Earth'ssurface overlying bedrock. They occur as two major classes: (1) transported depositsgenerally derived from bedrock materials by water, wind, ice, gravity, and man's interventionand (2) residual deposits formed in place as a result of weathering processes. Surficialdeposits may be stratified or unstratified such as soil profiles, basin fill, alluvial or fluvialdeposits, landslides, or talus. The material may be partially indurated or cemented by

silicates, oxides, carbonates, or other chemicals (caliche or hardpan). This term is often usedinterchangeably with the imprecisely defined word “overburden.” “Overburden” is a miningterm meaning, among other things, material overlying a useful material that has to beremoved. “Surficial deposit” is the preferred term.

Soil may be defined in engineering applications as generally uninduratedaccumulations of solid particles produced by the physical and/or chemical disintegration of

bedrock and which may or may not contain organic matter. Surficial deposits, such ascolluvium, alluvium, or residual soil, normally are described

Rock as an engineering material is defined as lithified or indurated crystalline ornoncrystalline materials. Rock is encountered in masses and as large fragments which haveconsequences to design and construction differing from those of soil.

Foundation materials are classified according to size, shape and looseness

• Unconsolidated : These include loose materials in the form of single grains-clay, silt,sand, gravel, cobbles, boulders, conglomerates, etc. Grain can be separated andmoved apart easily. None or small bending/cementing forces. These materials may becohesive as clays or non-cohesive as sands.

•

Consolidated : Basic grains are attached together in a strong manner, thus cannot beseparated easily. Such materials are generally termed as rocks.




3.2.2 Size :

Earth crust soil particles vary considerably in size. Particles are classified according tosize following soil classification system as: ASTM, Unified soil classification, USDA, UKsystems etc. Basic particle classes are: clay, silt, sand, gravel, cobbles and main classes may

further be subdivided into fine, medium or coarse particles. The clay, silt and sand whencombined in different proportions are classed into an array of granular soils (Table 3.1)

Terminology for Soils (Source: Engineering Geology Field Manual)

Definitions for soil classification and description are in accordance with USBR 3900Standard Definitions of Terms and Symbols Relating to Soil Mechanics: The sizes aresummarized in Table 3.2.

Cobbles and boulders —particles retained on a 3-inch (75-mm) U.S. Standard sieve. Thefollowing terminology distinguishes between cobbles and boulders:

• Cobbles —particles of rock that will pass a 12-in (300-mm) square opening and beretained on a 3-in (75-mm) sieve. Dia 3-12 inches.

• Boulders —particles of rock that will not pass a 12-in (300-mm) square opening.

Table 3.1: Particle size classification. Size in mmClass ASTM D422

Novak p.42UK

BS1377Wahlstrom1974 p-43

USDA Unified EGFM

Clay colloids < 0.001Clay < 0.005 < 0.002 < 0.002 < 0.002 < 0.075

Silt 0.005-0.075 0.002-0.06 0.002-0.06 0.002-

< 0.074

Fine 0.002-

0.002- Medium 0.006-0.02 0.006-0.02

Coarse 0.02-0.06 0.02-0.06

Sand 0.075-4.75 0.06-2 0.06-2 0.05-1 0.074-5

Fine 0.075-0.425 0.06-0.2 0.06-0.2 0.05-0.25 0.074-

0.075- Medium 0.425-2 0.2-0.6 0.2-0.6 0.25-0.5 0.42-2 0.425-2

Coarse 2-4.75 0.6-2 0.6-2 0.5-2 2-5 2-4.75

Gravel 4.75-76.2 2-60 2-60 2-76

Fine 2-6 2-6 5mm-

4.75-3/4”

Medium 6-20 6-20

Coarse 20-60 20-60 ¾”-3” ¾”-3”

Cobbles > 76.2 60-600 > 60 75-300

Boulders > 600 >300




Table 3.2: Comparative Particle sizes.

Descriptive term Size Familiar example within the size rangeBoulder 300 mm or more Larger than a volleyballCobble 75 to 300 mm Orange -grapefruit- VolleyballCoarse gravel 20 to 75 mm Grapes to orangeFine gravel # 4 sieve (5 mm) to 20 mm Pea to grapesCoarse sand # 10 to # 4 sieve Sidewalk saltMedium sand # 40 to # 10 sieve Openings in aluminum window screenFine sand # 200 to # 40 sieve Grains barely visible-table salt-sugar

Gravel —particles of rock that will pass a 3-in (75-mm) sieve and is retained on a No. 4(4.75-mm) sieve. Gravel is further subdivided as follows:

• Coarse gravel —passes a 3-in (75-mm) sieve and is retained on 3/4-in (19-mm) sieve.

• Fine gravel —passes a ¾-in (19-mm) sieve and is retained on No. 4 (4.75-mm) sieve.

Sand —particles of rock that will pass a No. 4 (4.75-mm) sieve and is retained on a No. 200(0.075-mm or 75-micrometer [µm]) sieve. Sand is further subdivided as follows:

• Coarse sand —passes No. 4 (4.75-mm) sieve and is retained on No. 10 (2.00-mm)sieve.

• Medium sand —passes No. 10 (2.00-mm) sieve and is retained on No. 40 (425-µm)sieve.

• Fine sand —passes No. 40 (425-µm) sieve and is retained on No. 200 (0.075-mm or75-µm) sieve.

Clay —passes a No. 200 (0.075-mm or 75-µm) sieve. Soil has plasticity within a range ofwater contents and has considerable strength when air-dry. For classification, clay is afine-grained soil, or the fine-grained portion of a soil, with a plasticity index greaterthan 4 and the plot of plasticity index versus liquid limit falls on or above the "A"-line(Figure 3.23, later in this chapter).

Silt —passes a No. 200 (0.075-mm or 75-µm) sieve. Soil is non-plastic or very slightly plastic

and that exhibits little or no strength when air-dry is a silt. For classification, a silt is afine-grained soil, or the fine grained portion of a soil, with a plasticity index less than 4or the plot of plasticity index versus liquid limit falls below the "A"-line (Figure 3.23).

Organic clay —clay with sufficient organic content to influence the soil properties is anorganic clay. For classification, an organic clay is a soil that would be classified as aclay except that its liquid limit value after oven-drying is less than 75 percent of itsliquid limit value before oven-drying.

Organic silt —silt with sufficient organic content to influence the soil properties. Forclassification, an organic silt is a soil that would be classified as a silt except that its




liquid limit value after oven-drying is less than 75 percent of its liquid limit value before oven-drying.

Peat —material composed primarily of vegetable tissues in various stages of decomposition,usually with an organic odor, a dark brown to black color, a spongy consistency, and a

texture ranging from fibrous to amorphous. Classification procedures are not applied to peat.

3.2.3 Shapes

Gravels and cobbles shape may be as: Rounded, Sub rounded, Sub angular, Angular.The particles shape may be as Bulky (equi-dimensional), Platy, Flaky, or Fibrous.

3.3 ROCK FEATURES FOR CLASSIFICATION

Rocks are aggregates of minerals. The chemical composition and molecular structure

determine the strength of the rock. Rocks and rock minerals are classified according to:Hardness : determined by scratching the rock surface and compared with standard scale The

10 minerals scale: Tale-mica - 1, Gypsum – 2, Calcite – 3, Diamond – 10), Fingernail –2, Copper coin 3½ (3-4), knife blade 5, window glass 5.5. The rock hardness /strength descriptors are as under.

Alphanumeric descriptor Description / CriteriaH1 Extremely hard Core, fragment, or exposure cannot be scratched with knife or sharp

pick; can only be chipped with repeated heavy hammer blows.H2 Very hard Cannot be scratched with knife or sharp pick. Core or fragment breaks with

repeated heavy hammer blows.H3 Hard Can be scratched with knife or sharp pick with difficulty (heavy pressure).

Heavy hammer blow required to break specimen.H4 Moderately hard Can be scratched with knife or sharp pick with light or moderate

pressure. Core or fragment breaks with moderate hammer blowH5 Moderately soft Can be grooved 1/16 inch (2 mm) deep by knife or sharp pick with

moderate or heavy pressure. Core or fragment breaks with light hammer blow orheavy manual pressure.

H6 Soft Can be grooved or gouged easily by knife or sharp pick with light pressure, can

be scratched with fingernail. Breaks with light to moderate manual pressure.H7 Very soft Can be readily indented, grooved or gouged with fingernail, or carved with a

knife. Breaks with light manual pressure

Cleavage (yes/No). Yes if smooth surface on breaking. cleavage can be along one or more parallel planes.

Fracture -breakage in planes other than cleavage plane: conchoidal –concentric curvedsurface, irregular - rough surface, splintery - wood like appearance

Luster appearance of mineral surface due to quality and intensity of light reflected

Color of the rock block.




Streak -color of fine powder of mineral obtained on rubbing

Uni-axial compression strength : weak – less than 35 MPA, strong – 35 – 115 MPA, verystrong - > 115 MPA. (1 MPA = 145 psi ????)

Pre-failure deformation : elastic or viscous

Gross homogeneity : Massive or layered

Formation continuity : Solid – joint spacing > 2 m, blocky – joint spacing 1 – 2 m, broken / fragmented - < 1m.

3.4 ROCK FORMING MINERALS

Only a dozen out of 2000 minerals are found in most rocks.Quartz - silicon dioxide (example milky quarts, rock crystal quarts)Feldspar - potassium aluminum silicate or No-Ca-Al-silicate

Mica - Complex K – Al-silicateAmphibole - Ca-Mg-Fe-silicatePyroxene - Ca-fe-SilicateOlivine - Mg-Fe-silicateCalcite - Ca-CODolomite - Ca-Mg-Co

3

Clay minerals - Hydrous-Al-silicate3

Limonite - Hydrous ferric oxideHematite - Ferric oxide

3.5 ROCK ORIGIN

3.5.1 Igneous rocks

Igneous rocks are primary rocks. These are formed on cooling/solidification of moltenlava (magma). If lava cools within the earth body, it forms an intrusive igneous (or plutonic)rock. If the lava reaches the earth surface through some channels and then cools, it formsextrusive igneous (or volcanic) rocks. Intrusive-dike-introduced at an angle to bedding plane.Intrusive-sill-introduced on parallel to bedding plane. Cooling of magma results in thesystematic arrangement of ions into orderly patterns. The silicate minerals resulting from

crystallization form in a predictable order. Texture refers to size and arrangement of mineralgrains. Cooling rate, dissolved gases affect crystal size.

Igneous rocks-may be coarse grained-individual crystals visible to naked eye. Finegrained, Glossy-rock non crystalline. Examples- Granite, Rhyolite (Extrusive equivalent ofgranite), Basalt, Gabbro (Intrusive equivalent of basalt). Shrinkage cracks often formed oncooling to give a columnar structure. Intrusive rocks may become exposed subsequently dueto weathering / erosion of upper layers. Pyroclastic rocks are those composed of fragmentsejected during a volcanic eruption.

Figure 3.2.1 describes the classification of igneous rocks and Figs 3.2.2 to 3.2.5 showsome igneous rocks.




Figure 3.2.1: Classification of Igneous rocks. 2

Fig 3.2.2 : Fine grained and coarse grained igneous rocks.

2 Figs 3.2.1- 3.2.16 are taken from lecture materials of Prof. J. David Rogers for GE50: Geology for Engineers[http://web.umr.edu/~rogersda/umrcourses/ge50_....] and GE341: Engineering Geology and Gotechnics,University of Missouri, Missouri School of Mines, Rolla. [http://www.geoengineer.org/learnbyhy-geology.html]






Classtic rocks are made out of fragmental sediments of gravel, sand, silt and clays and havediscrete fragments and particles. Nonclastic rocks have pattern of interlocking crystals.Chemical rocks form from dissolution and precipitation of chemical portions of rocks.Organic rocks form from sediments of organic origin.

Examples: Classtic Conglomerate, Breccia, Sandstone, Siltstone, Mudstone, Shale

Pyroclastic Fine ash, Tuff, coarse (cinder) agglomerate

Chemical CaCO 3 (limestone), Ca-Mg-CO 3 (Dolomite), CaSO 4 (Gypsum), NaCl (Rock salt), CaCO 3

Organic animal remain (Coral rocks, Chalk), Carbon - plant remains(Coal)

–

Figure 3.2.6 describes the classification of sedimentary rocks and Figs. 3.2.7 to 3.2.9 give

examples of these rocks.Stratification is in approx parallel bands may be flat, tilted or folded. Bedding may be thin(few inches) or thick (few feet). Bedding planes parallel to stratification. Cross bedding alsodevelop at right angle to the stratification.

Figure 3.2.6: Identification of sedimentary rocks.




Figure 3.2.7: Alternate sequences of sandstone and shale in Grand Canyon.

Fig 3.2.8: Quartz sandstone and Conglomerate.




Fig 3.2.9 . Breccia, Coquina

3.5.3 Metamorphic Rocks

Metamorphic rocks are formed by partial to complete re-crystallization of pre-existingrocks (igneous, sedimentary, other metamorphic rocks) due to high temperature and/or

pressure, differential stresses. Differential stress causes mechanical rotation and elongation ofconstituent minerals and clasts. These rocks usually become deformed with complex highlycontrolled fabrics. Layering develop due to original rock layering or generated duringmetamorphism is clearly displayed in most rocks but not present in some massive rocks.Layering is called foliation or schistocity. Most metamorphic rocks have the same overallchemical composition as the parent rock from which they formed. Mineral makeupdetermines, to a large extent, the degree to which each metamorphic agent will cause change.Foliation – any planar arrangement of mineral grains or structural features within a rock.Examples of foliation: Parallel alignment of platy and/or elongated minerals.

Schistosity: Platy minerals are discernible with the unaided eye and exhibit a planar orlayered structure. Rocks having this texture are referred to as schist. Gneissic: During highergrades of metamorphism, ion migration results in the segregation of minerals. Gneissic rocksexhibit a distinctive banded appearance.

Nonfoliated rocks: Marble is a crystalline rock formed by the metamorphosis of limestone.Quartzite is formed from a parent rock of quartz-rich sandstone. Quartz grains are fused

together.

Examples. Gneiss, Schist, Slate, Phyllite, Quartzite (from sandstone), Marble (from lime typesedimentary rock). Fig. 3.2.10 gives classification of metamorphic rocks and Figs 3.2.11 to3.2.16 give examples of metamorphic rocks.




Figure 3.2.10: Classification of metamorphic rocks.

Figure 3.2.11 : Foliation resulting from directed stress.




Figure 3.2.12: Garnet-mica-schist

Figure 3.2.13: Slaty cleavage planes; Slope creep and rock toppling.

Figure 3.2.14: Deformed and folded gneiss; Gneissic texture






3.5.4 Rock symbols

The rocks are shown on the map by standard symbols as shown in Figure 3.2.17.

Figure 3.2.17: Symbols representing various rock types. (Source: Wahlstrom 1974, p-200)

3.5.5 Rock Formations in Pakistan

Various rock formations encountered at different locations are shown in Figs. 3.2.18to 3.2.20 for illustrative purposes.




Figure 3.2.18 : Massive limestone rock with many discontinuities (M-2 Salt range). Also seeare blast drill hole marks.

Figure 3.2.19 : Shale formation cut slopes affected by whethering and erosion (M-2 Saltrange).




Figure 3.2.20 : Massive limestone and sandstone inter bedding (M-2 Salt range).

Figure 3.2.21 : Shale and sandstone inter bedding (M-2 Salt range).




Figure 3.2.22 : Shale and sandstone inter bedding (M-2 Salt range).

Figure 3.2.23 : Fractured sandstone bedding (M-2 Salt range).




Figure 3.2.24 : Massive limestone rock on abutments of Dharabi dam (Dist. Chakwal).




Figure 3.2.25: Rock features along Mastuj River (a Try of Chitral R).

Figure 3.2.26: Rock fracturing Golen Gol, Chitral..






Figure 3.2.29 : Fractured rock cut at Simly dam left abutment..

Figure 3.2.30 : Shale and sandstone layering at right abutment of emergency spillway,Mangla dam raising project.




Figure 3.2.31: Tanpura-I dam: Fractured friable sandstone layers over shale.




3.6 ROCK RESHAPING

Many features result in reshaping of rock masses as under.

3.6.1 Disintegration and Decomposition of rocks

Rocks are subjected to many physical and chemical processes that alter rockformations and properties to varying degrees. Most changes result in reduction in size andconsequently their strength. Rocks are subject to weathering an account of exposure toenvironment and on contract with hot aqueous solutions (hydrothermal solutions). The rate ofweathering depend on rock composition the climate, extent of contact of surface solutionsinto deep locations (along fractures) warm and hot regions has greatest weathering. Physicalweathering produces soil and gravel, which are finally eroded by water, wind, and glaciers.Chemical weathering results in clay minerals. Magnesium, calcium and iron rich rocks breakmore easily by chemical weathering.

Rock masses also break on cooling and shrinking which also causes deep fracturesand joints. Rocks become folded etc due to seismic thrusts, lava eruption and earthquakereasons. These features weaken the rock mass. Some depth of loose materials or alluviummay also cover rock surface and is termed as overburden.

3.6.2 Bedding and Folding

Most rocks have flat-lying beds or layers and are visible on vertical exposed edges.However, subsequent slope failure, lava eruptions, seismic movements lift these straight bedsand become folded and tilted or inclined (Figs. 3.3, 3.4).

Folds are produced by a complex process of dislocation involving bending, shearingor slipping on a large to small scale and/or recrystalization. Anticlines-upfolds, synclines-downfolds. Dome is upfold which dips away in all directions.

3.6.3 Fractures in rocks

Discontinuity .— A collective term used for all structural breaks in geologic materials whichusually have zero to low tensile strength. Discontinuities also may be healed. Discontinuitiescomprise fractures (including joints), planes of weakness, shears/faults, and shear/fault zones.Depositional or erosional contacts between various geologic units may be considereddiscontinuities.

Bedding planes - The planes marking the termination of one sedimentary deposit and the beginning of another; they usually constitute a weakness along which the rock tendsto break.

Foliation - In rocks that have been subjected to heat and deforming pressures during regionalmetamorphism, some new materials such as muscovite and biotite mica, talc andchlorite may be formed by recrystallisation. These new minerals are arranged in

parallel layers of flat or elongated crystals - the property of foliation.




Figure 3.3: Rock layering (Source: Wahlstrom 1974. p-88)

Fractures are discontinues/breaks in geologic materials resulting from failure of rock understress (tensional, compressional or shear) The presence of faults may be recognizedfrom such physical features as; Offset of beds, dykes or veins; Slickensides; Gouge;

Brecciation or crushing; Topographic features like escarpments, linear trenches or sagvalleys.

Joint is a crack which transects a rock with little or slight displacement of adjacent sides parallel to the plane but a slight seperation normal to the joint surface. A series of joints with similar orientation forms a joint set (Figs. 3.5, 3.6).

Shear .— A structural break where differential movement has occurred along a surface or zoneof failure; characterized by polished surfaces, striations, slickensides, gouge, breccia,mylonite, or any combination of these. Often direction of movement, amount of

displacement, and continuity may not be known because of limited exposures orobservations.

Fault

.— fractures along which there has been notable displacement of one side of fracturerelative to other. A shear with significant continuity which can be correlated betweenobservation locations; foundation areas, or regions; or is a segment of a fault or faultzone reported in the literature. The designation of a fault or fault zone is a site-specificdetermination. (A and B in Fig. 3.5)




Figure 3.4: Cross section of several kinds of folds. (no scale). Source: Wahlstrom 1974 p.96.Shear/fault zone

Faults are generally accomplished by joints but joints are not necessarily associatedwith faults. Fractures are of greater concern for engineering proper ties of rocks.Fractures greatly reduce strength of rock and promote valley slope failure. Fracturesalso provide channel ways for movement of water (leading to mechanical andchemical weathering and alternation by solutions of deep seated origin. Fault isdislocation along a fracture. Fault zone movement along a number of parallel, sub-

parallel or intersecting surfaces.

.— A band of parallel or subparallel fault or shear planes. The zone mayconsist of gouge, breccia, or many fault or shear planes with fractured and crushedrock between the shears or faults, or any combination. In the literature, many faultzones are simply referred to as faults.

Stirke of fault is compass bearing of a horizontal line in the plane of fault.

Dip

is inclination of fault from the horizontal (measured at right angle to strike). Many faultsand fault zone contain crushed materials and/or secondary minerals deposited fromgroundwater circulation along the faults.




Figure 3.5: Some aspects of faults. (Wahlstrom 1974, p.97). A. Tensional “gash joints” andtight compressed “shear joints” have developed along a fault. B. Faulting has caused“drag folding” in adjacent sedimentary rocks. C. Fault is accompanied by extensivedevelopment of joints in wall rocks. D. “Sheeted” fault zone. Movement along jointsis accompanied by development of parallel joints. E. Fault zone contains closelyspaced joints. F. “Gash joints” and “shear joints” have developed within and adjacentto fault zone.

Gouge is the mix having large proportion of grains of clay size or slightly larger.

Crush breccia contain angular rock fragments usually embed in gouge.

Primary/Secondary Joints: Primary joints formed at formation of rock due to shrinkage/contraction/ volume change. Fractures can also develop during folding/lifting process.




Figure 3.6: Faults fill materials. (Wahstrom 1974, p.98). A. “Braided” slip surfaces intersectfault filling, consisting of angular “crush breccia”. B. Fault filled with “gouge” and“crush conglomerate”. C. Open space in fault contains a partial filling of mineralsdeposited from solutions that moved along the fault. D. Fault contain a vein of gangueand ore minerals. Wall rocks are altered by the solutions that deposited the faultfilling.

Figure 3.7: Shear and tension joints. (Wahlstrom 1974, p.99) A. Smooth surfaced,intersecting shear joints. B. Rough surfaced tension joints. C. Tension joints that havelocalized alteration of wall rock. D. Closely spaced shear joints associated withdisplacements along bedding plane in shale.






3.6.4 Weathering

Weathering considerably alters the properties of rocks. Weathering is caused byweather, water, chemical and physical processes Weathering is classed as under:

Weathering category Weathering signs

Fresh Rock No visible signs of weathering

Slightly WeatheredPenetrative weathering developed in open discontinuity surfaces but only slightweathering of rock material.

Moderately Weathered Weathering extends throughout the rock mass, but the rock is not friable.

Highly WeatheredWeathering extends throughout the rock mass, but the rock material is partlyfriable.

Completely WeatheredRock is wholly decomposed, and in a friable condition but rock texture andstructure are preserved.

SoilA soil material with the original texture, structure and mineralogy of the rockcompletely destroyed.

Figure 3.9: Rock matrix failure below river channels in soluble rocks.




3.6.5 Rock Classification

Tunnels are frequently constructed as part of outlet works, power tunnels, riverdiversions etc. Tunnels are excavated through the rock matrix of abutments. The followingterms are used for the classification of rocks for tunneling purposes:

Intact rock contains neither joints nor hairline cracks. If it breaks, it breaks across soundrock. On account of damage to the rock due to blasting, spalls may drop off the roofseveral hours or days after blasting. This is known as spalling condition. Hard, intactrock may also be encountered in the popping condition (rock burst) involving thespontaneous and violent detachment of rock slabs from sides or roof.

Stratified rock consists of individual strata with little or no resistance against separationalong the boundaries between strata. The strata may or may not be weakened bytransverse joints. In such rock, the spalling condition is quite common.

Moderately jointed rock contains joints and hairline cracks, but the blocks between jointsare locally grown together or so intimately interlocked that vertical walls do not requirelateral support. In rocks of this type, both the spalling and the popping condition may beencountered.

Blocky and seamy rock consists of chemically intact or almost intact rock fragments whichare entirely separated from each other and imperfectly interlocked. In such rock,vertical walls may require support.

Crushed but chemically intact rock has the character of a crusher run. If most or all of the

fragments are as small as fine sand and no recementation has taken place, crushed rock below the water table exhibits the properties of a water-bearing sand.

Squeezing rock slowly advances into the tunnel without perceptible volume increase.Movement is the result of overstressing and plastic failure of the rock mass and not dueto swelling.

Swelling rock advances into the tunnel chiefly on account of expansion. The capacity toswell is generally limited to those rocks which contain smectite, a montmorillonitegroup of clay minerals, with a high swelling capacity.

Although the terms are defined, no distinct boundaries exist between rock categories. Widevariations in the physical properties of rocks classified by these terms and rock loading areoften the case.

3.7 ENGINEERING PROPERTIES OF ROCKS

The strength of loose aggregates depends on degree of cohesion, confinement andwater content and varies as 5 to 20 psi. The strength of rocks (crushing strength) is a functionof mineralogy, cementation, origin and rock fabric (fabric expressed by grain size, grainshape, grain distribution etc planer arrangement of elements of fabric, stratification, foliation

etc) and degree of fraction.




Engineering properties are considered in terms of ability of rocks to take load(compressive and shear) and ability to pass water flow.

The rock density sp. Gravity

Basalt, gabbro, schist 2.9 – 3.2

Granite, slate, marble, limestone 2.5 – 2.8

In general strongest rocks are denser and weak rocks are most porous. Crushing strength ofrocks is given as under: (Wahlstrom p.48-49)

Rock type strength (thousand psi)

Clastic sedimentary rocksCalcarious mudstones 8-28Dolomite 9-51

Limestone 0.7-29Sandstone 1.5-34Shale 1-33Siltstone 4-45

Igneous/metamorphic rocksBasalt 26-40Gneiss 22-36Granite 6-42Marble 7-34Quantzite 30-53Schist 1.1-20Slate 14-47

Presence of mica lowers crushing strength. Weathering of rocks greatly reduces its strength.Porosity of strong rocks may be less than 1%. Fractures, joints, solution channels mayincrease porosity to as much as 15% (in limestone with cavernous spaces).

3.8 GEOLOGICAL REQUIREMENTS OF DAMS

Most dams can be built on all type of foundation conditions from strength point ofview. But this would require treatment and foundation improvements to make up structuraldeficiencies of the foundation material. This means more costs usually making the projectexpensive, less attractive and may be unacceptable. Thus there are preferred foundationconditions favoring the type of dam and are described below. It may be pointed thatstructurally sound foundations may require treatment for the purposes of reducing andcontrolling seepage through dam foundations.

Earthfi ll Dams

• Can be built over all types of foundation (alluvium, rock, soil, etc).




Rockfi ll Dams

• Moderately hard rocks are preferred but can be built over deep alluvium. Sites withlarge amount of clays are not very suitable.

•

Uncertain or variable foundation, which is unreliable for concrete dam can also beused.

Concrete Gravity Dam

• Hard rock at or near surface. Any soft overlying materials need to be removed.

• Depth of soft material over the rock not to exceed 7-10 m to avoid excessiveexcavation.

Concrete Buttress Dam

• Buttress dam is suitable if rock of bearing strength of 2-3 MPa

Arch Dam

• The dead gravity weight of an arch dam is supported by the foundation rock but itutilizes the strength of an arch to transfer the water loads onto the abutments.Therefore the strength of the rock mass at the abutments and immediately downvalleyof the dam must be unquestionable.

• Its modulus of elasticity must be high enough to ensure deformation under thrust fromarch do not induce excessive stress in the arch.

• Competent foundation and abutments of strong/hard rock.

• Thin arch dam: Rock strength of valley sides between 5.5 to 8 Mpa.

• Thick arch-dams: rock bearing strength more than 3.5 Mpa.

• Multiple Arch Dam: Foundation rock reliable to bear 2-3 MPa or more withoutsettlement. [1 psi = 6.895 KPa, 1 MPa = 145 psi]

3.9 DAM SITE INVESTIGATIONS

3.9.1 Objectives

Geologic information of dam site is required for design of a safe dam structure. Over40% of dam failures on record have occurred due to dam foundation defects. Investigationsmust be able to answer following:

• Depth of overburden that must be removed to reach an acceptable foundation.

• Rock types that make up foundation and affect of weathering on rock quality.

• Engineering properties of foundation rock types (strength, deformability,durability), settlement, elastic properties.




• Geologic structure of foundation in terms of joints, fractures, faults, folding,defect pattern, orientation, spacing, extent and openness/aperture. Characteristicsof infill materials, presence of solution cavities and void continuity.

• Permeability of rock foundation due to defects as joints, faults, bedding open.

• Sources and location of adequate supplies of construction materials such as clay,sand, gravel and rock fill, preferably as close as possible to the dam site;

• Check if the rock excavated to provide a spillway for the dam be acceptable foruse as rock fill in the construction of the dam embankment;

• Check if the spillway requires concrete lining and an energy dissipation structureat its downstream end or is the spillway rock sufficiently erosion resistant thatthese can be omitted (esp for small dams).

In order to be able to answer the above questions an experienced engineeringgeologist must explore the dam site. Most dam site investigations will employ severaldifferent methods, the exact mix of methods and the timing when each is carried out issomething which is tailored to suit the particular geological problems of each individual damsite. These geological site investigations allow the engineering geologist to construct a"geological model " of the site which is then used by the dam designers as a basis on whichthey can design a safe and economic dam structure appropriate to the geology of that

particular site.

It is important to realize that even the most comprehensive site investigation programcannot hope to reveal all the significant geological features of the site. It is therefore ofcritical importance that the actual geological conditions revealed during construction becompared with the geological model of the site derived from the site investigations. It is quitecommon for unexpected geological conditions

3.9.2 Dam Site Investigations/Explorations Include

to be revealed during construction whichrequire changes to be made to the original design. A record of the site geology "as found"during construction is also of great value if problems develop later during the operation andmaintenance phase

• Surface explorations: geological mapping of surface rock outcrops

• Geophysical surveys: seismic refraction, electrical resistivity surveys

• Sub-surface explorations which include:

excavation of pits and trenches using bulldozers, backhoe, etc.

excavation of shafts, addits (tunnels) gallery

diamond core drilling to obtain undisturbed core samples at various depths(few to hundreds of meters deep)

Rotary drilling to obtain disturbed samples




• In-situ loading compressibility tests

• Water pressure intake tests

• Grout intake tests

The test sites include: dam axis, reservoir periphery, spillway, tunnels, power house site, etc.

3.9.3 Surface Explorations

The rock and loose fill material features are obtained from study of exposed surface,rock outcrops, vertical cuts along streams, road sides, building foundation excavations.Features directly observable/visible are evaluated. Loose fill material is noted as to grain size,size gradation, shape and type of fill (alluvium, instu weathering, aeolian, glacial deposits) interms of quantity, quality, volume, depth, aerial extent etc. Rocks are noted for type, bedding,layering, layer thickness, fracturing, dip and strike.

Surface exploration is a cheap and quick method for preliminary reconnaissance and pre-feasibility level studies only. These must be added with other exploration methods forfeasibility and detail design level studies. Surface explorations with limited subsurfaceexplorations are useful to locate construction materials as gravel and sand for concreting,

boulders/cobles for riprap etc..

3.9.4 Geophysical Surveys

These are methods to “read” the foundation profile from the surface withoutexcavating the profile itself by traversing the profile with an electromagnetic signal. These

are non-destructive, cheap and quick methods. The main purpose of geophysical surveys areto (1) - determine depth of bed rock and rock layers, (2) - locate buried bedrock channels, (3)- determine depth of the rock weathering, (4) - major layering conditions, and (5) - delineate

boundaries of different materials.

A signal in the form of aseismic impact (e.g. hammer blow)or electric current or acoustic signalis sent into the earth crust. The wavetraverses through the soil/rock layersrefracted back to soil surface and ismonitored at a distant location bygeophones or electrodes. Thestrength and velocity of wave travelis interpreted in terms of rock type,layering depth, rock integrity. Fieldexperience in the form of bore holelogs at neighboring locations are used for accurate interpretation of results. Geophysicalsurveys require specialized equipment and experienced persons. Geophysical surveys aremostly used to fill/detail out rock description between borehole locations.

Figure 3.10. Geophysical profile testing.




3.9.5 Sub Surface Explorations

Sub surface exploration includes excavating the foundation profile and studying thecontents. Various methods used are:

Test Pit of small plan area is dug to shallow depth. It allows inspection, sampling and insitutests.

Trench is a long continuous test pit to selected depths. Mostly limited to upper weatheredzone.

Tunnels-/adits is a horizontal opening made to explore area under slope, abutments, section5’x7’ rectangular or hoarse shoe. It is slow and expensive. Show rock features. May be usedlater as drainage adit.

Figure 3.11: Top: Exploratory adits on left and right abutments of Monar dam. Bottom:Entrance to the adit. (http://www.corestore.org/DeanieMonar.htm)

Rotary drilling or Auger borings provide undisturbed samples. Augur borings of 4-12 inch

dia and up to 20 ft depth are done manually. Deeper holes are possible with mechanical




augers only including helical augurs for 3-16 inch dia, disk augurs for upto 42 inches dia, and bucket augurs upto 48 inch dia holes. Augur borings provide profile details. Large size holesare preferred for foundation investigations.

Diamond core drilling

3.9.6 Geologic maps

: These are carried by large drilling rigs using diamond drill bit. Water

is circulated through the drill stem to extract the rock cuttings/grindings to the surface. Drillholes may be vertical or inclined. Log is prepared of the nature of rock materials encountered.The penetration speed also provides rock hardness and strength. Undisturbed core samplesare retrieved by wire line without removing drill bit. Bore holes are located along dam axis,and dam periphery. Bore holes are drilled down to level of hard un-weathered rock.Undisturbed samples provide RQD description. Samples are sealed and sent to lab for furtherstrength tests.

The results of geologic investigations are presented in the form of geologic maps.These include map of dam and reservoir area, along dam axis, along right and left abutments,along spillway and other places of interest. Preliminary maps are first prepared from surficalinformation and are improved further as more and more information is obtained by detailedfield investigations. Such maps for are shown in Figs. 3.12 (a to g) for Kurram Tangi dam.

Figure 3.12a : Surface geology map of KTD site.




Figure 3.12b : Alternate dam axis.

Figure 3.12c: Geological section along river channel.






Figure 3.12f: Geological section along right abutment / outlet tunnel.

Figure 3.12g: Geological section along spillway.




3.10 ROCK FOUNDATIONS (Sherard 1963 p-255)

Rocks are generally stronger than alluvium overlying the bed rock. Some rock massesof soft secondary rocks (shale, clay stone, siltstone, mudstone or marl) are weak rocks withtheir strength varying over a wide range. Density of rock may vary from 88 to 155 lbs/cft andwater content 2 to 40%. The rocks may be well cemented to being rock due to higher pressureonly (compaction shale). Shale foundations must be given the most careful and conservativeconsiderations even for low dams.

Characteristics which can reduce the strength of a rock are:

1. Continuous mass of clay or other weak material, even for thickness of a fractionof an inch.

2.

Closely spaced cracks system often associated with severe twisting movementsand faults.

3. Basic constituents consisting of highly plastic and very fine clay with no sand sizeso that individual rock fragments have a greasy surface texture and very littlefrictional resistance when sliding in respect to each other.

4. Horizontal bedding is likely to be more dangerous for two reasons. (a) weakseams are more likely be continuous, (b) it is more likely that high pore water

pressure will be transmitted horizontally.

Test on individual rock specimens, core borings alone may not provide complete picture.Large rock samples, large scale field insitu tests, test pits and shafts should be used. Calyx

borings (large diameter holes drilled with smooth walls) plus a large core is very useful forthis.

3.11 LOGGING OF EXPLORATIONS

The description of formation type and other characteristics is called bore log. This showsresults from rotary drill with adequate arrangements for obtaining samples. The standardsymbols are used to graphically describe the details. The information of multiple explorationlogs are presented in the form of sample log, x-section, or fence diagram. Sample log isshown in Fig. 3.13.

• Test holes should be numbered for identification.

• Hole number prefixed by 1-2 letters as: DH-drill hole, AH augur hole (hand), AP-Augur hole (power), TP-test pit (open), T-trench, etc.

• A log is a written record of data concerning materials and conditions encounteredin each test hole.

• Each log be factual, accurate, clear and complete




• Log describes depth elevation, graphic log, samples, classification and physicalcondition, percolation tests, type and size of hole, core recovery, RQD (rockquality designation = proportions of samples which are 4 inch or longer in length).

• Each log also record of hole No., location, project, ground level, boredip/inclination, total depth, test dates, ground coordinates, etc. type of equipmentused.

Figure 3.13: A typical log of a borehole. (Wahlstrom 1974. p.201)

3.12 FOUNDATION FAILURE (Wahlstrom p-165)

Dam foundations should be designed with generous margins of safety. The dam andthe reservoir behind dam create dead weight loads and water pressures that did not exist

previously. Thus behavior of dam, the materials in the foundations and the abutments, and inthe reservoir site require constant monitoring during and after dam construction and reservoirfilling for short and long term responses to loads (creep response: elongation due to continuedloading for long time). Tarbela dam is known to have numerous earth tremors on account ofrelease of stresses on bedding plans which are imposed by dam and water weight.

Dam body exerts large unit pressure over the foundation. Concrete dams, because ofsmall contact area, have greatest in concentrated loads pressures. Concrete dams are rigid

body and small foundation settlements can induce excessive stresses in the dam body.Earthfill/ rockfill dams are non-rigid and can relatively easily adjust internally to load, and

pressure exerted on foundation are approximately equal to the weight of overlying prism ofmaterials of different height.




Figure 3.14: Geologic conditions promoting foundation failure. Plastic mechanism for shearfailure. A: vertical directed load. B: Load is directed asymmetrically. (Wahlstrom1974, p.181)

Horizontal foundation rock/earth fill dams fail because of seepage/uplift problems

rather for shearing dislocations owing to the load of the dam. On contrary concrete dam fail by shear dislocations. Shear surfaces develop under the dam as a symmetric wedge.Foundations with faults, folds, low strength beds, inclined planes, fault zone or other weaksurface can lead to failure until treated.

Elastic properties of some selected rocks (P-185)Rock Elasticity (10 6

Igneous Granite: 1.5-11.9; Andesite: 4.7-6.9; Baslat 5.9-12.4 psi)

Metamorphic Quartzite: 1.2-6.4; Gneiss: 3.5-15.1Sedimentary Gypsum: 0.17-1.1, Shale: 0.3-9.9; Limestone: 0.4-14.1; Sandstone 0.6-8.0;

Siltstone: 1.0-9.3

3.13 IMPROVEMENTS OF FOUNDATION AND RESERVOIR AREA

FOUNDATION IMPROVEMENT

In spite of geological and geophysical investigations, the many important details become known only at the time of construction and may require modifications of design andconstruction details. Unpredicted delays are not appropriate. It be born in mind that neveragain in the life time of dam it will be possible to examine in detail and take the appropriatenecessary steps to correct adverse conditions that are revealed in future. The foundations ofmain dam through river valley and abutments are required to withstand dead/dynamic loads




with minimum seepage. Treatment of foundations and abutment is done to improve theircompetency and to reduce or eliminate subsurface seepage.

Figure 3.15: Geologic conditions promoting failure of foundation of concrete dams.(Wahlstrom 1974. p-182-183). A. Brittle fractured sandstone rests on a weak shalelayer dipping upstream. B. Horizontally layered limestone rest on a weak shale layer

which extends downstream to a steep slope in the valley floor. C. Fracture crystallinerocks lie above a flat fault containing sheared, gougy materials of very low strength.D. Intersecting strong conjugate joints have attitudes that promote easy mass sheatdislocation. E. Sedimentary rocks dipping downstream are intersected by a faultdipping upstream and containing materials of low strength. F. Folsed rocks containingthin, weak layers of shale present a potential for foundation failures.

3.13.1 Stripping

The top few meters of rock surface are most affected by weathering and development

of fractures, faults etc inclined slippage planes. The fractured part of the rock foundation isstripped off to reach a otherwise hard competent rock. This may be good enough for small




dams but need other treatments for large dams. Foundations with overlying alluvium arestripped to expose the hard bed rock surface for concrete dams; this may not be needed forearthfill or rockfill dams.

3.13.2 Replacement of Weak Layers

Weak inclined layers/beds of shale and other materials do not provide goodfoundation. These layers if few and thin, may be excavated and space filled up by strongconcrete. Thus upper 5-10 m depth may be treated. However, the weak cleavage planes

between adjacent beds/layers may not get treated fully leaving potential planes of failures.

3.13.3 Grouting

Grouting is a process of filling the discontinuities and void spaces of undergroundchannels in rock with a sealant. Grout is a liquid, either a uniform chemical substance or anaqueous suspension of solids, that is injected into rocks or unconsolidated materials throughspecially drilled bore holes to improve bulk physical properties and/or to reduce or eliminateseepage flow paths of water beneath the dam structure. Grout include: 1- Portland cement

based slurries, 2- chemical grouting solutions, and 3- organic resins, epoxy/polymers. Clay,sand, bentonite or chemicals may be added to cement slurries to increase/decrease settingtime.

Three kinds of grouting programs are identified.

1. Comparatively shallow systematic ‘blanket’ or consolidation grouting over critical portions of the foundation.

2. Deep curtain grouting from a gallery or concrete grout cap along a specified locatione.g. dam axis, to produce deep impermeable barrier to subsurface groundwaterseepage.

3. Off-pattern special purpose grouting to improve strength and/or overcome problemscreated by groundwater circulation.

Grouting help to close channel ways, and thus reduce seepage considerably. Smallinaccuracies may allow some seepage past the curtain, however, thus seepage can be takencare of by installing drainage wells.

Grout curtain is located under the clay core in earth fill-rock fill dams, under theconcrete face in CFRF dam, under heel of a concrete dam, below gallery in concrete dams.Depth of grout curtain is upto solid/firm/competent rock mass through weaker or fractured

beds. If complete description of foundation rock is not available, or is highly varied, thengrout depth is taken equal to height of dam at that location. Grouting for deep holes is usuallydone over short intervals upward or downward. Packers may be used to isolate the groutingsection.




Figure 3.16 : Earthfill and rockfill dam foundations in unconsolidated deposits. Wahlstrom p.236-237). A. Rockfill dam, impervious membrane (asphaltic concrete) extends to agrout cap on bed rock. B . Cut-off trench extends to bed rock. C . Cutoff trench

penetrates impervious layer in unconsolidated valley fill. D. Cutoff extends to layersof impervious material in unconsolidated valley fill. Grout holes extend throughlimestone layer in bed rock. E . A cutoff is provided by sheet piling driven into animpervious layer in valley fill. F . Flow beneath dam is reduced by a layer ofimpervious material placed upstream from the dam.

IMPROVEMENT OF ABUTMENTS AND RESERVOIR AREA

The weakness of abutments in terms of rock surface disintegration and falling overmay be improved by (1) Dental work – the open rock joints are cleaned and filled withconcrete or some other filler; (2) Scaling – the loose rock masses over the abutments andreservoir rim are removed to exposed lower lying hard and stable surface, thus future dangeris reduced.

3.14 GROUTING

3.14.1 Curtain Grouting

This is grouting done along a single selected axis (e.g. below core, below concreteslab of rockfill dam, below heel of concrete dam , or grout cap etc) to form a seepage curtain

below the dam. Thus every space on selected axis becomes treated (Figs. 3.17, 3.185).

• Curtain grouting before dam construction for E/F, R/F dams.






• Depth of grouting: For completely described geology = 1/3 of dam height + 50 ft.For unknown or highly variable geology = dam height.

• For unknown geological conditions, depth as may be needed to plug off seepage.

•

Grout spacing commonly 10 ft apart.• A pre-determined closure pattern followed (Fig. 3.20).

• Holes may be vertical or inclined

• The depth of shallower holes is controlled by the experience in the precedingholes

3.14.2 Blanket Grouting

• Blanket grout holes are shallower 20-30 ft.

• Intended to remedy flows in the foundation such as deep fractured rock. Over alarger area.

• Holes normal to foundation

• Blanket grouting completed before construction of dam.

Figure 3.18: Schematic layout of Curtin and blanket grouting of rock foundation of anearthfill dam. A: Plan, B: Section showing formation depth for curtain grout holes.(Wahlstrom 1974. p-244)

3.14.3 Pattern Grouting

• Dam geology unknown before construction. Location/specifications of curtain & blanket grouting not precisely stated. Grouting ‘take’ can be in excess of estimates




• Need to complete grouting ‘as required’. Grouting is an art and not a science

• Pattern grouting included in plans to for prior estimate of footage and grantsvalue.

•

Actual number of holes & specifications determined by the area and crosssectional configuration of the excavation of dam foundation.

Fig. 3.19: Schematic location of pattern blanket and curtain grout holes in foundation of aconcrete gravity dam. Curtain grout holes are drilled from a gallery within the dam.A: Plan, B: Preferred pattern, C: Alternate plan for grouting from the gallery.

(Wahlstrom 1974. p-245).




Figure 3.20: Conventional closure pattern for curtain drilling and grouting. Numbers indicate

sequence of drilling and grouting. (Wahlstrom 1974. p-248).

Figure 3.21: Curtin grouting holes with depth attitudes determined by subsurface geologicconditions. (Wahlstrom 1974. p-249).




Figure 3.22: Grout curtain used on the abutment. (USBR, 2001, p.213).

3.14.4 Grouting pressure

Grout entry into the formation depends on grout consistency, void opening size, and grout pressure. Thick or thin mixtures are used. Grout liquid injected below the rock layers tend totravel outward and also force upward the overlying layers. The lifting of grout under pressurein a horizontal channel way is function of grout pressure and rock density (for granite density== 2.6) as under. (Pressure in psi, Height of rock that can be displaced.): 10 psi 8.8 ft; 100

psi 89 ft, 200 psi 178 ft, 300 psi 276 ft, 500 psi 444 ft. This necessitates thatshallower holes be grouted at smaller pressure, thus will have smaller lateral spacing.

3.15 ROCK SLOPE STABILITY

Fractured rock masses with inclined planes of weakness tend to be unstable and present a threat of collapse by slope failure. This is especially so rock friction is likely to

change by exposure to water from a reservoir. Grouting and or rock bolts or steel tensionedcables are useful.

• Rock bolts tend to close fissures and increase shearing strength along fractures byincreasing frictional resistance.

• These be firmly anchored in solid rock. Expanding anchoring device (rawl bolt),quick set high strength resins are used.

• Rock bolts may be installed at right angle to plane of facture.




Figure 3.23: Rock slab stability improvement by rock tension bolts. (Wahlstrom 1974. p-264).

Figure 3.24: Reinforcement of rock mass by tensioned rock bolts or steel cables. (Wahlstrom1974. p-265) A: Unstable slope in inclined sedimentary rocks. B: Block of crystallinerock above faults inclined toward the valley floor. C: Crystalline rock massintersected by vertical shear. D: A shattered zone along a fault zone. E: Closelyfractured zone in a foundation. F: Rock bolts or heavy steel cables to reduce hazard of

possible dislocation along a horizontal shale layer when reservoir is filled.




3.16 EARTHQUAKE HAZARDS

Earthwork or seismic tremors are important geophysical phenomena. Earthquakes arecaused by volcanic exceptions, tectonic plates movement and release of stresses embedded in

plates in the earth profile along active seismic faults, e.g. St. Andrews fault California, USA.

Earthquakes cause a sudden acceleration of earth surface resulting in shaking, jolting of allstructures resulting on the earth surface. Extreme tremors can cause considerable damage to

property and life. A geo-physist can analyze the earthquake hazard in the area.

Shock waves associated in the earthquakes are P or longitudinal, S or shear and Lwaves. P waves moves with maximum in velocity, S waves at about 60% of P waves. P and Swaves are body waves that travel through rocks below the earth surface. L waves arerelatively slow surface waves of long periods and capable of causing swaying of buildings orwave motion in water bodies at great distances from the point of origin. Most damage fromearthquake is caused by L waves rather than P & S waves.

Earthquake are characterized by location and depth of epicenters, intensity andmagnitude. Intensity is given by Mercalli scale and magnitude by Richter scale. Richtermagnitude is determined from amplitude of ground vibrations. Richter scale size ofearthquake effects at a specific location. The distance of epicenter and nature of groundformation causes large changes on intensity and impact of the earthquake at a specificlocation. Subsurface geology being much more important. Mercalli scale 1 to 12 (1 not felt

by people, 12 damage extreme or total). Richter scale M 1 to 8. M = Log 10 A/A 0 , where A =maximum amplitude of horizontal ground motion at a distance of 100 km and A 0

Concrete dams resting on solid rock are usually not extensively damaged byearthquake. Rock and earthfill dams are dislocated more than concrete dams by shock waves,generally are flexible to adjust to micro adjustments without failure (not on weak foundation)(or not dislocated by extension of fault into the dam). Although visible damage may notdevelop in a dam, small scale displacements in foundation rocks may alter groundwater flow

pattern beneath the dam, disrupt or reopen cracks filled and sealed by grouts.

= amplitude

at 0.001 m.

Earthquakes may promote extensive land sliding in slopes, leading to water waves, of

considerable height. These water waves present a serious threat of destructive over flow of adam and dam failure. Waves may also be generated in the reservoir water body due to surface jolting (called Tsunami). The earthquake hazard require that seismic loads must beadequately accounted for dam and foundation design. Thus active or potential fault zone inthe vicinity of the dam site must be evaluated in the light of historic observed earthquake inthe area. For high risk area, the dam structure must be designed that can withstand shockwaves without damage. If historic record do not point to any earthquake hazard, even then

provision of measures tending to nullify the effects of natural disasters of all kind should beconsidered as a social responsibility.




3.17 FOUNDATION PREPERATIONS

The foundations of a dam must be able to withstand without unacceptabledeformation the loads imposed upon it by the structure, both immediately after filling thereservoir and in the long term. With time, deterioration by saturation and percolation of water

can occur, whilst soft rocks and clays usually exhibit lower residual strengths under sustainedloading (creep) than under rapid testing. It is the 10-20m of rock immediately below the damthat is of greatest importance.

Terzaghi's advice might well apply to foundation testing - "...because of unavoidableuncertainties involved in the fundamental assumptions of the theories and the numericalvalues of the soil constants, simplicity is of much greater importance than accuracy." TheEngineer must use all the available resources, concentrating on the zones of foundation thatappear weak and that will be subject to stresses once loaded.

Introduction

If it is economically feasible, all material under the base of a proposed dam whichcould cause excessive settlement and leakage should be removed. If this cannot be done, thedam design should be modified to take account of such material. Sometimes it may benecessary to remove material to considerable depths in isolated areas of the foundation. Thisis known as dental work. The general overall removal of material is termed stripping,whereas the removal of loose masses of rocks on the abutments is termed scaling. Theengineering geologist has to determine the expected depth of weathered or unsound rock or

overburden that must be removed in advance of construction. Further it has to be ascertainedabout the vertical side angles of all cut-slopes (short or long periods) required for powerhouse, spillway, chute, stilling basin, plunge pool, etc. Also required is the compressivestrength, shear strength and water tightness of various rock formations.

Foundation program

A planned program of foundation excavation should be initiated with the view that thevolume of excavation and configuration of the excavation will approximate reasonably to the

plans and specifications established. It is the responsibility of the construction engineer toensure slopes for excavations will be permanently stable or will not fail during construction.In earth materials slopes of 1.5:1 to 2:1 are excavated in permanent cuts and slopes of 1:1 areestablished in temporary cuts, except where unusual conditions are anticipated. In bedrockthat is not closely fractured or does not contain inclined planes of potential slippage, such as

bedding planes in weak rocks, slopes are excavated at angles up to the vertical.

Problematic foundation materials

In foundations in unconsolidated material excavation of natural deposits may reveal

inadequate localized or widespread foundation materials that require special treatment or total




removal. Unacceptable or inadequate materials rich in organic substances such as topsoil,swamp muck or peat, loose deposits of sand or silt, talus accumulations and plastic, active,sensitive, or swelling clays. Poor foundation conditions in rocks are associated with closefracturing, weathering or hydrothermal alteration, or poorly indurated sedimentary rocks.

Excavation in bedrock

The objective of excavation is the preparation of a clean surface that will provideoptimum contact with the dam materials, whether earth or concrete is to be placed on thatsurface. Therefore excavations in bedrock should extend into firm, fresh rock. Any closelyfractured zones extending downward, especially if containing soft altered materials such asclay gouge or other products of weathering, should be removed if feasible.

Prolonged exposure of both earth and rock foundations to the atmosphere or to waterfrequently results in deterioration by hydration, dehydration, frost action, shrinkage, andexpansion with changes in temperature. It is good practice to protect reactive surfaces thatwill be exposed for long periods of time with bituminous materials. Alternatively, originalcover is not removed until final cleanup and just prior to placement of the dam.

Construction on unconsolidated deposits

At an ideal site, excavations in unconsolidated deposits should extend to solid bedrock for the full width of the dam, whether it is constructed of concrete or earth/rock fill.

However, there are many locations where the depth of the valley fill is so great that damsmust be constructed in part or entirely on unconsolidated deposits (Tarbela dam). Where thisis the case steps must be taken to improve the engineering properties of the foundationmaterials and to reduce subsurface seepage to allowable levels.

Except for low dams of small gross weight, concrete dams are not built onunconsolidated deposits because of their generally low bearing strength. Larger damsconstructed in whole or in part on unconsolidated deposits should without exception, be earthor rockfill dams with the capacity to adjust to settlement in the foundation materials.

Preparation of foundation for concrete dam

The extent of the work that will be necessary in the foundations for a concrete damwill be determined by two main factors, their strength to sustain the loads that will beimposed by dam and the reservoir water, and the effect of water entering the foundationsunder pressure from the reservoir.

Generally the quality of foundations for a gravity dam will improve with depth ofexcavation however the abutments for an arch dam often do not improve with distanceexcavated into the sides of the valley. Deterioration of clay could endanger the dam and/orlead to collapse of abutments downstream from the dam.




Frequently the course of the river has been determined by geological faults orweaknesses; proving of the river bed is therefore of first importance in the investigationstage. The depth to be excavated will depend upon the nature of the infilling material, theshape of the excavated zone, and the depth of cutoff necessary to ensure an acceptable

hydraulic gradient after the reservoir is filled.

Concrete dams may be constructed on foundations other than massive rock, i.e.shales, glacial deposits or even sand for river works. Each case must be examined relative to

permeability, settlement, and load-carrying capacity (vertical and horizontal).

The final preparation of the foundation should be undertaken just prior to the placement of concrete. It should include the removal of loose rock and all debris, rougheningof smooth rock surfaces, washing down of all surfaces, and the removal of excess water from

pools to leave a clean damp surface to receive the concrete.

3.18 GRANULAR MATERIALS IN FOUNDATIONS & FOR CONSTRUCTION(Golze p-151)

Dams are occasionally built over granular foundations. In addition granular materialsare required for constructing earthfill dams and also impermeable core of rock fill dam. Adeep understanding of engineering properties of granular soils is important.

Granular materials are resultant of weathering of rocks over geological time scale.Most frequently the weathering products are transported to other sites by water, ice, wind.

Various landforms of granular deposits are:

1. Fluvial Soils : Soils whose properties have been affected in-situ by action of water towhich they have been subjected. Common characteristics are roundness of individual grainsconsiderable sorting action, soil is stratified with lenses, strata may be thick or thin, and smallrange of grain sizes in each stratum. Further divided as:

a) Torrential outwash. When the steep channels debounch onto plains the sedimentsare deposited as an alluvial fans, as small deposits of steeply sloping coarse rockfragments to gently sloping plains of fine grained alluvium. Coarse materials

deposited first. Sands and gravel sub-rounded to sub-angular. These are goodsource of sand and gravels for construction purposes.

b) Valley fill materials: Flood plain deposits of generally finer materials., morestratified generally flat, stream shape explain nature of deposits, braided silt,sand, gravel, meandering fine grain soils, flood plain as common source ofconstruction material (sand, gravel) for concrete and dam shell material. Miningfrom d/s areas is unfeasible until a positive cutoff is provided. It can also affecttail water levels d/s of mining area ultimately affecting out let and spillway stilling

basin design. River bank terraces from an early stage of valley development can provide good amounts of sand and gravel of excellent quality.




c) Lake beds: These are formed out of sedimentation of fine grain silt and clay in stillwater. Stratification is usually weak. Materials are usually impervious,compressible and of low shear strength. Cracking clays be present. Can be usedfor imperious core of earthfill dams.

2) Glacial Deposits : These are formed due to grinding action of glaciers during itsadvancing and retreats. These deposits are heterogeneous and difficult to exploreeconomically. These contain wide range of particle sizes, and particles are typically sub-rounded to sub-angular with flat sides.

a) Moraine Deposits: Deposits formed from ice movement, heterogeneous mix ofcobbles, gravel, sand and some non-plastic fines. Flat to slightly undulatingsurface. Large accumulation at tip.

b) Glacial outwash. These are deposits from glacial melt water.

3) Aelian Deposits : These are soils formed from blown wind and two types are dunesands and loess. These are very rich in quartz and in fine to medium range sand with nocohesive strength, high permeability and moderate compressibility. Loess has ability ofstanding in vertical faces due to presence of small amount of clay which binds the soil grainstogether. These fall in ML or ML-CL boundary group

4) Residual Soils. These are in-place weathered rock soil particles. Undefined dividingline between parent rock and residual soil. Individual grains are angular and soft. Workingand handling reduces grain size, thus difficult to predict performance. Talus and land slides

are examples.3.19 SOIL CLASSIFICATION

Soil is aggregate of uncemented mineral grains. Soils are classified on the basis of percentage of various size grains in terms of clay, silt, sand, gravels to describe its potentialto produce crops. (See figure of soil classification triangle). Engineers are interested in soil

physical properties as unit weight, permeability, shear strength, compressibility andinteraction with water. Unified soil classification system is widely accepted by engineers.This classification system is developed jointly by US Bureau of Reclamation, Corps ofEngineers and Prof. Casagrande. The system is usable for both visual - manual examinationas well as laboratory testing. The system is based on:

• percentage of various soil fractioning

• shape of grain size distribution curve

• plasticity and compressibility characteristics of very fine grains

The system has established 15 distinctive soil groups (with a 2 letter symbol and a name) asgiven in Table 3.3. The identification is based on:

1. Visual method: simple manual tests and visual observations to estimate size anddistribution of coarse grains and plasticity of fine grains.




2. Laboratory method: Use lab tests for size gradation, moisture content (soilconsistency) other basic soil properties. Help in precise delineation of soil group.

Figure 3.25: Typical soil gradation chart.




C: Gap graded soils.

Figure 3.26 : Examples of soil gradation (USBR p-129).




Table 3.3: Grading of soils.Major Divisions Group

symbolTypical Names Field identification procedure Information required for

describing soils1 2 3 4 5 6

C o a r s e g r a i n e d s o i l s

M o r e t h a n

5 0 % m a t e r i a l l a r g e r t h a n # 2 0 0 s i e v e s i z e

G r a v e l s : M o r e

t h a n h a l f o f

c o a r s e f r a c t i o n l a r g e r t h a n # 4

C l e a n g r a v e l s

( l i t t l e o r n o

f i n e s )

GW

Well graded gravels,gravel-sand mixtures,little or no fines

Wide range in grain sizes andsubstantial amounts of all intermediate

particle sizes

For undisturbed soils addinformation onstratification, degree ofcompaction, cementation,moisture conditions anddrainage characteristics.

Give typical name, indicateapproximate % of sand,gravel, max size,angularity, surfacecondition, hardness ofcoarse grains, local orgeologic name, and otherfeatures and symbol.

Example: Silty sand,gravely, about 20% hard,angular gravel particles ½

in max size, rounded andsubangular, sand grainscoarse to fine; about 15%non plastic fines with lowdry strength; wellcompacted and moist in

place, alluvial sand (SM).

GPPoorly graded gravels,gravel-sand mixtures,little or no fines

Predominantly one size or a range ofsizes with some intermediate sizes

missing

G r a v e l s w

i t h

f i n e s

( a p p r e c i a b l e

a m o u n t o f

f i n e s )

GM

Silty gravels, gravel-sand-silt mixtures

Non plastic fines or fines with low plasticity (procedure as for MC below)

GCClayey gravels, gravel-sand-clay mixtures

Plastic fines(procedure as for CL below)

S a n d s : M o r e t h a n

h a l f o f c o a r s e

f r a c t i o n s m a l l e r t h a n # 4 s i e v e

C l e a n s a n d s

( l i t t l e o r n o

f i n e s )

SW

Well graded sands,gravely sands, little orno fines

Wide range in grain sizes andsubstantial amounts of all intermediate

particle sizes

SPPoorly graded sands,gravely sands, little orno fines

Predominantly one size or a range ofsizes with some intermediate sizes

missing

S a n d s w

i t h f i n e s

( a p p r e c i a b l e

a m o u n t o f f i n e s )

SM

Silty sands, sand-siltmixtures

Non plastic fines or fines with low plasticity (procedure as for MC below)

SCClayey sands, sand clatmixtures

Plastic fines(procedure as for CL below)

F i n e g r a i n e d s o i l s :

M o r e

t h a n 5 0 % m a t e r i a l s m a l l e r t h a n # 2 0 0

s i e v e s i z e

Identification procedureDry

strengthDialatancy Toughness

Silts and ClayLiquid limit less

than 50%ML

Inorganic silts and veryfine sands, rock flour,silty or clayey finesands or clayey siltswith slight plasticity

None toslight

Quick toslow

None Give typical names.Indicate degree andcharacter of plasticity,amount and maximumsize of coarse grains,color in wet condition,odor if any, local orgeologic name, and other

pertinent descriptiveinformation; and symbol..

For undisturbed soils addinformation on structure,stratification, consistencyin undisturbed andremolded states, moistureand drainage conditions.

Example: Clayey silt loam:, brown, slightly plastic,small percentage of finesand, numerous vertical

root holes, firm and dry in place, loess (ML).

CLInorganic clays of lowto medium plasticity,gravely clays, sandyclays, silty clays, leanclays

Mediumto high

None tovery slow

medium

OLOrganic silts andorganic silty clays oflow plasticity

Slight tomedium

Slow Slight

Silts and claysLiquid limit more

than 50%MH

Inorganic silts,micaceous ordiatomaceous, finesandy or silty soils,elastic silts

Slight tomedium

Slow tonone

Slight tomedium

CHInorganic clay of high

plasticity, fat clays.High to

very high None High

OHOrganic clays ofmedium to high

plasticity, organic silts

Mediumto high

None tovery slow

Slight tomedium

Highly organic soilsPt

Peat and other highlyorganic soils.

Readily identified by color, odor,spongy feel and frequently by fibrous

texture.

Table 3.3 continued on next page




(Table 3.3 Continued)Major Divisions Group

symbolLaboratory classificationcriteria

1 2 6 7

C o a r s e g r a i n e d s o i l s

M o r e

t h a n 5 0 % m a t e r i a

l l a r g e r t h a n # 2 0 0 s i e v e s i z e

G r a v e l s : M o r e

t h a n h a l f o f c o a r s e f r a c t i o n

l a r g e r t h a n # 4 s i e v e

C l e a n g r a v e l s ( l

i t t l e o r

n o f i n e s )

GW

D e t e r m

i n e % o f g r a v e l a n d s a n d

f r o m g r a i n s i z e c u r v e .

D e p e n d i n g o n p e r c e n t a g e o f f i n e s

( f r a c t i o n

s m a l l e r t h a n

# 2 0 0 s i e v e s i z e ) c o a r s e g r a i n e d s o

i l s a r e c l a s s i f i e d a s

f o l l o w s .

L e s s

t h a n 5 %

G W

, G P , S W

, S P

M o r e

t h a n 1 2 %

G M

, G C

, S M

, S C

5 t o 1 2 %

B o r d e r l i n e c a s e

( u s e d u a l s y m b o l s )

Uniformity coefficient Cu > 6,

Coefficient of curvature Cc between 1 and 6

GP Not meeting all gradation requirements for GW

G r a v e l s w

i t h f i n e s

( a p p r e c i a b l e a m o u n t

o f f i n e s )

GM

Atterberg limits below A line or PIless than 4

Above ‘A’ line with PI between 4 and 7 are

borderline cases requiringuse of dual symbols.

GCAtterberg limits above A line or PI

greater than 7

S a n d s : M o r e

t h a n h a l f o f c o a r s e f r a c t i o n

s m a l l e r t h a n # 4 s i e v e

C l e a n s a n d s

( l i t t l e o r

n o f i n e s )

SW

Uniformity coefficient Cu > 4,Coefficient of curvature Cc between one and 3

SP Not meeting all gradation requirements for GW

S a n d s w

i t h f i n e s

( a p p r e c i a b l e a m o u n t

o f f i n e s )

SM

Atterberg limits below A line or PIless than 4

Limits plotting in hatchedzone with PI between4 and 7 are borderlinecases requiring use ofdual symbols.

SCAtterberg limits above A line or PI

greater than 7

F i n e g r a i n e d s o i l s :

M o r e

t h a n 5 0 % m a t e r i a l s m a l l e r t h a n # 2 0 0 s i e v e s i z e

Fig. 3.27: Soil plasticity chart for laboratory classification of fine grained soils.

Silts and ClayLiquid limit less

than 50%

ML

CL

OL

Silts and claysLiquid limit

more than 50% MH

CH

OH

Highly organic soils Pt

Table 3.3 continued on next page




(Table 3.3 Continued)Procedures and Criteria for Visual Classification of Fine-Grained SoilsSelect a representative sample of the material for examination and remove particles larger than the No. 40 sieve (mediumsand and larger) until a specimen equivalent to about a handful of representative material is available. Use this specimen for

performing the identification tests. Identification Criteria for Fine-Grained Soils.— The tests for identifying properties of fines are dry strength, dilatency,toughness, and plasticity. 1. Dr y strength.— Select from the specimen enough material to mold into a ball about 1 in (25 mm) in diameter. Mold or

work the material until it has the consistency of putty, adding water if necessary. From the molded material, make atleast three test specimens. Each test specimen should be a ball of material about ½ in (12 mm) in diameter. Allow thetest specimens to dry in air or sun, or dry by artificial means, as long as the temperature does not exceed 60 degreesCentigrade (EC). In most cases, it will be necessary to prepare specimens and allow them to dry over night. If the testspecimen contains natural dry lumps, those that are about ½ in (12 mm) in diameter may be used in place of molded

balls. (The process of molding and drying usually produces higher strengths than are found in natural dry lumps of soil).Test the strength of the dry balls or lumps by crushing them between the fingers and note the strength as none, low,medium, high, or very high according to the criteria below. If natural dry lumps are used, do not use the results of anyof the lumps that are found to contain particles of coarse sand.

None : The dry specimen crumbles with mere pressure of handling.

Criteria for describing dry strength

Low : The dry specimen crumbles with some finger pressure. Medium : The dry specimen breaks into pieces or crumbles with considerable finger pressure. High: The dry specimen cannot be broken with finger pressure. Specimen will break into pieces between thumb and a

hard surface.Very High : The dry specimen cannot be broken between thumb and a hard surface. The presence of high-strength, water-

soluble cementing materials, such as calcium carbonate, may cause exceptionally high dry strengths. The presence ofcalcium carbonate can usually be detected from the intensity of the reaction with dilute hydrocloric acid (HCl). Criteriafor reaction with HCl are presented in a subsequent paragraph.

2. Di latancy.— Select enough material from the specimen to mold into a ball about ½ in (12 mm) in diameter. Mold thematerial, adding water if necessary, until it has a soft, but not sticky, consistency. Smooth the soil ball in the palm ofone hand with the blade of a knife or spatula. Shake horizontally (the soil ball), striking the side of the hand vigorouslyagainst the other hand several times. Note the reaction of the water appearing on the surface of the soil. Squeeze thesample by closing the hand or pinching the soil between the fingers and note reaction as none, slow, or rapid accordingto the criteria below. The reaction criteria are the speeds with which water appears while shaking and disappears whilesqueezing.

None No visible change in the specimen.Criteria for describing dilatancy

Slow Water slowly appears on the surface of the specimen during shaking and does not disappear or disappears slowlyupon squeezing.

Rapid Water quickly appears on the surface of the specimen during shaking and disappears upon squeezing. 3. Toughness.— Following completion of the dilatancy test, the specimen is shaped into an elongated pat and rolled by hand

on a smooth surface or between the palms into a thread about c in (3 mm) diameter. (If the sample is too wet to rolleasily, spread the sample out into a thin layer and allow some water loss by evaporation). Fold the sample threads andreroll repeatedly until the thread crumbles at a diameter of about c in (3 mm) when the soil is near the plastic limit. Notethe time required to reroll the thread to reach the plastic limit. Note the pressure required to roll the thread near the

plastic limit. Also, note the strength of the thread. After the thread crumbles, the pieces should be lumped together andkneaded until the lump crumbles. Note the toughness of the material during kneading. Describe the toughness of thethread and lump as low, medium, or high according to the criteria below.

Low Only slight pressure is required to roll the thread near the plastic limit. The thread and the lump are weak and soft.Criteria for describing toughness

Medium Medium pressure is required to roll the thread to near the plastic limit. The thread and the lump have mediumstiffness.

High Considerable pressure is required to roll the thread to near the plastic limit. The thread and the lump have very highstiffness.

4. Plasticity.— On the basis of observations made during the toughness test, describe the plasticity of the material accordingto the criteria given below.

Nonplastic A 3-mm thread cannot be rolled at any water content.Criteria for describing plasticity

Low The thread can barely be rolled, and the lump cannot be formed when drier than the plastic limit. Medium The thread is easy to roll, and not much time is required to reach the plastic limit. The thread cannot be rerolled

after reaching the plastic limit. The lump crumbles when drier than the plastic limit.High It takes considerable time rolling and kneading to reach the plastic limit. The thread can be rolled several times after

reaching the plastic limit. The lump can be formed without crumbling when drier than the plastic limit. Based on the drystrength, dilatency, toughness, and plasticity tests, decide on whether the soil is an organic or an inorganic fine grained soil.




Figure 3.27 : Plasticity chart for laboratory classification of fine grained soils.

Materials described as (gravel > #4, sands #4-#200, silt+clay <#200)

Coarse-grained, soils (gravel G, sand S modified by grading uniformity W-well graded, P- poor graded). Sub-divisions areGW well graded gravel, GP poor graded gravelGM silty gravel (with 5-10% fines of little or no plasticity)GC clayey gravel sand (with 5-10% fines of slight to medium/ plasticity).SW well graded sand SP poor graded sandSM silty sand with fines of little/no plasticity

SC clayey sand with fines of slight to medium plasticity

Fine grained (silt+ clays) < ≠ 200 sieve. Main groups are: M – silt, C – clay, O – organicsoils modified with liquid limit as: ; L – low liquid limit, H – high liquid limit. Subdivisions are:CL lean clay ML siltOL organic clay (on or above A-line) or organic silt (below A-line)CH fat clay MH elastic siltOH organic clay (on or above A-line) or organic silt (below A-line)

Boundary soil . Soils falling on boundary of fine and coarse grained and are given dualsymbol. This is used when % of fines is 5 to 12% or 45 to 55%.






Figure 3.28 : USDA soil classification system (USBR 2001. p-121).

i.

Nature and thickness of top soils – to avoid mixing up during excavations.ii. Relative elevation of borrow area: to determine grade of haul roads.

iii. Natural water content, its seasonal fluctuations and need for draining or irrigating it inthe embankment:

iv. Quantity of oversized cobbles which would have to be removed before compaction.

v. Influence of excavation on appearance and operation of dam – these sites should not pose underlying pervious areas. No borrow close to dam.

The fill materials are evaluated for:

- Permeability - Density

- Stability - Shear strength

- Compression and shrinkage - Proctor compaction

- Piping and washing out of fines, - Economics of utilization

Various tests conducted on potential embankment fill materials include following.

Water Contents

These tests may be done in-situ or in a lab.




Permeability

Soil less than 1 ft/year (10 -6 cm/s) are considered as impervious. Pervious soils mayhave K of 100 ft/year (10 -4

The permeability of soil may change when compacted in embankments. A perviousgravelly or sandy soil with some fines can become quite impervious after compaction,especially when course grains are well graded. To get imperviousness, adequate compactionand proper gradation (minimum of fines and max of coarse grains) are required.

cm/s) or more. Both pervious and impervious materials are neededfor dam construction in varying amounts.

Stability

Ability of compacted embankment to resist shear force reflect stability. This dependon factors as grain size, gradation, grain shape and drainability. Larger well graded grains,

better stability and shear resistance. Finer soils develop pore water pressure which reducesshear resistance.

Compression and Shrinkage

Soil high in clay content compress under land of upper layers and develop high porewater pressure. This is not desireable. Such materials can also shrink and crack on dryingduring construction shut downs and extended low reservoir periods. Upper layers should bereworked to improve.

Piping and Washing of Fines

Washing of fines from u/s & d/s slope be protected by providing graded filter andriprap.

3.22 FIELD TEST

i. Field Permeability.

(In-situ test.) Tests are carried out by using pump out, method below water level and pump in method above water level. Test over whole depth of the bore hole or packers may beused to test over partial length. Test for each identified layer (both horizontal or vertical K).For large deposits pumping drawdown test is conducted.

ii. In-place Density Test.

Sand density test. (test hole filled with sand of known ρ ). weight sample. Small diahole for fine soils. Large cubic excavation/hole for gravely soil. If clean hole be dug, volume

by measurements of hole size. If broken hole, volume by sand displacement.

3.23 LAB TESTS

I. Fill materials tests

1. Particle gradation:

a. dia < ≠ 200 sieve hydrometer analysis








3.25 CONCRETE AGGREGATES

Large quantities of concrete are required for dam construction even for earth/rock filldams. Concrete require coarse and fine aggregates (gravel and sand). Economy ofconstruction dictates that enough quantities of aggregate are available within reasonable

haulage distance from the site. Highest quality aggregate may not always be the best choice.These materials may be available is ready to use size/quality or some rock processing(crushing/sorting/grading) may be needed. Cost of additional processing be compared withcost of obtaining aggregates from amore distant source.

Aggregates should be tested in Lab for crushing strength, specific gravity (high values

→ materials hard and tough), chemical content (sodium and magnesium sulfate soundnesstest), workability (shape and size of particles and proportion of small fractions.

Angular grains require more cement sand and water. Flat and elongated particles

liable to fear the matrix during finishing operations. Sands required of uniform grading fromcoarse to fine. Coarse sand require more cement, Fine sand require more water

Aggregates be tested against contaminants as

% by weight Max. for Fine CoarseCoal 1 1Clay lumps 1 0.25Shale 1 1Materials passing 200 sieve 3 1

Other substances (alkali, Mica, coated 2 1grains, flaky Particles, loamSum of all contaminants 5 3

3.26 DESCRIPTION OF THE PHYSICAL PROPERTIES OF SOIL

Descriptive information for classification and reporting soil properties such asangularity, shape, color, moisture conditions, and consistency are presented in the following

paragraphs.

AngularityAngularity is a descriptor for coarse-grained materials only. The angularity of the

sand (coarse sizes only), gravel, cobbles, and boulders, are described as angular, sub-angular,sub-rounded, or rounded as indicated by the criteria below. A range of angularity may bestated, such as: sub-rounded to rounded.

Criteria for describing angularity of coarse-grained particles

Angular Particles have sharp edges and relatively planar sides with unpolishedsurfaces.

Sub-angula r Particles are similar to angular description but have rounded edges.




Sub-rounded Particles have nearly planar sides but wellrounded corners and edges.

Rounded Particles have smoothly curved sides and no edges.

Shape

Describe the shape of the gravel, cobbles, and boulders as “flat, elongated” or “flatand elongated” if they meet the criteria given below. Indicate the fraction of the particles thathave the shape, such as: one-third of gravel particles are flat. If the material is to be processedor used as aggregate for concrete, note any unusually shaped particles.

Criteria for describing particle shape

The particle shape is described as follows, where length, width, and thickness refer to thegreatest, intermediate, and least dimensions of a particle, respectively.

Flat Particles with width/thickness >3.

Elongated Particles with length/width >3.

Flat and elongated Particles meet criteria for both flat and elongated.

Color

Color is an especially important property in identifying organic soils and is oftenimportant in identifying other types of soils. Within a given locality, color may also be usefulin identifying materials of similar geologic units. Color should be described for moistsamples. Note if color represents a dry condition. If the sample contains layers or patches ofvarying colors, this should be noted, and representative colors should be described. TheMunsel Color System may be used for consistent color descriptions.

Odor

Describe the odor if organic or unusual. Soils containing a significant amount oforganic material usually have a distinctive odor of decaying vegetation. This is especiallyapparent in fresh samples, but if the samples are dried, the odor often may be revived byheating a moistened sample. If the odor is unusual, such as that of a petroleum product orother chemical, the material should be described and identified if known. The material may

be hazardous, and combustion or exposure should be considered.

Moisture Conditions

Describe the moisture condition as dry, moist, or wet, as indicated by the criteria below.

Criteria for describing moisture condition

Dry Absence of moisture, dusty, dry to the touch.

Moist Damp but no visible water.

Wet Visible free water, usually soil is below water table.




Consistency

Describe consistency (degree of firmness) for intact finegrained soils as very soft,soft, firm, hard, or very hard, as indicated by the criteria given below. This observation isinappropriate for soils with significant amounts of gravel. Pocket penetrometer or torvane

testing may supplement this data.

Criteria for describing consistency of in-place or undisturbed fine-grained soils

Very soft Thumb will penetrate soil more than 1 in (25 mm).

Soft Thumb will penetrate soil about 1 in (25 mm).

Firm Thumb will indent soil about 1/4 in (5 mm).

Hard Thumb will not indent soil but readily indented with thumbnail.

Very hard Thumbnail will not indent soil.

Cementation

Describe the cementation of intact soils as weak, moderate, or strong, as indicated bythe criteria below.

Criteria for describing cementation

Weak Crumbles or breaks with handling or litt le finger pressure.

Moderate Crumbles or breaks with considerable finger pressure.

Strong Will not crumble or break with finger pressure.

Structure (Fabric)

Describe the structure of the soil according to criteria described below. Thedescriptors presented are for soils only; they are not synonymous with descriptors for rock.

Criteria for describing structure

Stratified Alternating layers of varying material or color; note thickness.

Laminated 1

Fissured

Alternating layers of varying material or color with layers less than 6 mmthick; note thickness.

1

Slickensided

Breaks along definite planes with little resistance to fracturing.1

Blocky

Fracture planes appear polished or glossy, sometimes striated.1

Lenses Inclusion of small pockets of different soils, such as small lenses of sandscattered through a mass of clay; note thickness.

Cohesive soil that can be broken down into small angular lumps which resistfurther breakdown.

Homogeneous Same color and textural or structural appearance throughout.




(Note: Do not use for coarse-grained soils with the exception of fine sands which can belaminated.)

Additional Descriptive Information

Additional descriptive information may include unusual conditions, geologicalinterpretation or other classification methods, such as:

Presence of roots or root holes or other organic material or debris;

Degree of difficulty in drilling or auguring hole or excavating a pit;

or Raveling or caving of the trench, hole, pit, or exposure;

or Presence of mica or other predominant minerals.




Geology of Kurram Tangi Dam

• The strata flanks is made of sand stone (soft to medium/hard to very hard),conglomerate with pebbles of limestone, sand stone, and shale (thirty bedded,

blocky and sparely jointed).

• Valley is 150 wide, 20 to 30 ft deep consisting of mix of cobbles, gravel, fine tocoarse sand, etc. Gravel of quartzite, dionite, limestone and sand stone.

• Sand stone well exposed on valley wells.

• Foundations are stable (4000-10700 pm crushing strength) with modulous ofdeformation as 755000 to 14,600,000 psi

• Permeability vary considerably. 2-3 ingeous at KTV-1, 114-193 at KTV-2.

• The nearby seismic faults are sughar fault, Kala Bagh fault, Kurdal fault, Bhittanifault.

• Small to medium seismic risk.




REFERENCES

Novak et al. HYDRAULIC STRUCTURES pp-25, 85

USBR. 1967. “Engineering Geology Field Manual” @www.usbr.gov/prnts/geology/fieldman.htm.

USBR 1990a. “Procedure for Determining Unified Soil Classification (Laboratory Method),” Earth Manual , Part II, 3rd edition, Bureau of Reclamation, U.S. Department of theInterior, USBR-5000.

USBR. 1990b. “Procedure for Determining Unified Soil Classification (Visual Method),” Earth Manual , Part II, 3rd edition. Bureau of Reclamation, U.S. Department of theInterior, USBR-5005.

USBR. 2001. DESIGN OF SMALL DAMS. Oxford & IBH Publishing Co. New Delhi. Chapter 5 pp:107-204, chapter 6 pp 211-220

Wahlstrom, E. E. 1974. DAMS, DAM FOUNDATION AND RESERVOIRS SITES ElsevierScientific publishing company, Amsterdam.

Web resources:

http : // homepages . ihug . com . au/~richardw/ page 19.html, page25.html, page26.html

www.dur.ac.uk/~des0www4/cal/dams/geol/topo.htm

Other sites






Tariq 2008 DAM AND RESERVOIR ENGINEERING 4-1 Ch 4: Earthfill and Rockfill Embankment Dams

Chapter 4

EARTHFILL AND ROCKFILL EMBANKMENT DAMS

4.1 DEFINITION

International Commission on Large Dams (ICOLD) defined embankment dam as “anydam constructed of excavated materials placed without addition of binding material otherthan those inherent in the natural material. The materials are usually obtained at or near thedam site”. An Earthfill Dam is an embankment dam, constructed primarily of compactedearth materials, either homogeneous or zoned, and containing more than 50% of earthgranular materials. Contrary a Rockfill Dam is an embankment dam constructed of naturalrock materials, usually broken down to smaller fragments. Rockfill dam with all voids filled

by finer materials by hydraulic sluicing is usually regarded as earth-fill dam. An embankment

dam where large quantities of both granular materials (earth) and rock fragments are used iscalled as Earthfill-Rockfill Dam .

Example of embankment dam, Stratos Dam, Greece (http://www.geoengineer.org)

I: EARTHFILL DAMS

4.2 GENERAL DESIGN CRITERIA:

Embankment dams are built to meet the following design criteria (Golze 1977 P-291, Novak 19** P-59):

1. Stability: The foundation, abutments and embankments must be stable for all loading/stress conditions during construction, and operation. Some distress can be toleratedduring construction.




2. Control of Seepage: Seepage through embankment, foundation and abutments must be small and not exert excessive uplift on the structure, create high exit gradients, piping not permitted.

3. Overtopping and Free Board: Top of dam must be high enough to allow for

settlement of dam and foundation and to provide sufficient free board to preventwaves at maximum pond level (during maximum flood, e.g. spillway design flood)from overtopping the dam.

4. Maximum Flood Evacuation: Spillway and outlet capacity be large enough to prevent overtopping of the dam (Spillway only, no other outlets) even when few (atleast one) spillway gate become stuck/inoperative.

5. Upstream Slope Protection: Slope of embankment and outlet works be stable underall operational conditions (first filling, quick drawdown, steady pond etc). Cuts into

rock masses for placing spillway must be stable under earthquake conditions.6. Outlet and Ancillary Works: Care must be taken to ensure that outlet or other

facilities constructed through the dam do not permit their perimeter with risk of soilmigration and piping. Same care is needed at embankment joints with abutments.

7. Stability against uplift under structures: Seepage under the various structures asspillway, chute, stilling basins, power house, exert lot of uplift pressure, thus thesestructure must be safe for this condition.

4.3 PLACEMENT OF FILL MATERIALS

Huge quantities of fill material of varying gradation are placed to form theembankment. The embankment materials of a dam may be placed as a rolled fill or hydraulicfill.

Rolled fill. The embankment material of requisite grading is transported to site by haulingmachinery, placed at specific locations in layers, rolled out by earth movingmachinery into layers of suitable thickness, watered and compacted by plain or sheep-foot rollers to requisite density.

Hydraulic fill. The material containing all grades and sizes are thoroughly blended, mixed

with water, transported to site in suspension by pumps and pipes and discharged at thedam edge in inward direction. The material gets deposited by sedimentation. Thus thecoarser particles get deposited near the edges and finer particles reach to the middlesection. The fill is usually not further compacted.

Semi-Hydraulic fill. The material in suspension is transported by hauling units and dumpedat the edge of the embankment. It is then washed in its final position by water jets.

Drainage of hydraulic fill. The excess water reaching inner part of dam percolateshorizontally to outer more pervious shell. Remainder water rises upward to the

surface, allowing the center of dam to consolidate and subside. The downward




movement of the core eventfully develops arching in the core and prevents its furtherconsolidation.

4.4 TYPE OF EARTHFILL DAMS

Earthfill dams can be of types as Homogeneous, Zoned and Diaphragm dam.

4.4.1 Homogeneous Dams

The dam embankment is made of a single type of material (Fig. 4.1). These includefine-grained particles or coarse-grained materials. The materials are compacted mechanicallyto form a watertight fill. The fill material is required to possess following properties

• 1

• It must be capable of being placed and consolidated to form a homogeneous masswithout any potential of piping as paths of percolation through the fill or along itscontact with the foundation and abutments.

:It must be sufficiently impervious to provide an adequate barrier and preventexcessive loss of water through the dam, the acceptable level being determined fromthe safety of the structure and the value of the lost water.

• The fill material should develop maximum practical shear strength under compactionand maintain most of it after the filling of the reservoir.

• It must not consolidate, soften or liquefy upon saturation.

Due to relatively finer materials, the slopes must be able to avoid sloughing. The u/s slope isrelatively flat to ensure safety against sloughing under rapid drawdown conditions after

prolonged high-level storage. The d/s slope must be protected to resist sloughing whensaturated to a high level by rainfall.

For a completely homogeneous embankment, the seepage will eventually emerge onthe d/s slope regardless of its flatness and the impermeability of the soil if reservoir level ismaintained for a sufficiently long time. The surface to the height of 1/3 rd

1 (@ www.dur.ac.uk/~des0www4/cal/dams/emba.htm/embaf1.htm)

of depth of thereservoir will be eventually affected. The exit of seepage may induce sloughing of the damtoe and consequently the dam embankment. Thus measures are included to intercept the

H/3

H

See a e

Figure 4.1 : Seepage through an earthfill homogeneous dam.

Phreatic/Seepage line






4.4.2 Zoned Embankment Dam

A zoned embankment dam is constructed of materials of more than two types. Thezoned dam has a central zone of impermeable materials flanked by zones of materialsconsiderably more pervious called shell or shoulders. The inner zone is usually called a core.The shell materials enclose, support and protect the impervious core. The u/s shell providesstability against rapid drawdown and d/s shall acts as drain to control the line of seepage. Thesection as a whole show progressive increase in permeability from the center outwardstowards each slope. The core is flanked by one or more zones of graded filter.

The central impervious zone consists of clay and outer shell consists of sand, gravel,cobbles or rock or mixture of these materials. If rock is used in shell, it is then called asearthfill-rockfill dam (Tarbela, Mangla dams). The dam is considered as zoned dam only ifthe horizontal width of the impervious zone at any elevation equals or exceeds the height ofthe dam above that elevation, and is not less than 10 feet (Fig. 4.6). The maximum width ofthe core is controlled by stability and seepage criteria and the availability of the material. Theouter shall due to coarse nature and good drainage, may have relatively steeper outer slope,limited only by the strength of the foundation, the stability of the embankment itself andmaintenance/construction considerations. For better stability of a section, longer haulage ofmaterials may be preferred. Graded filters are provided on u/s and d/s sides of the core which

Rockfill toe or horizontal drainage blanket

Fine rock orsand/gravel fillGraded gravel or

crushed rock Toe drain

Figure 4.5 : Toe drain for use with rockfill toe or horizontal drainage blanket.

Dam d/s slope

HU/S

ShellCORE D/S

ShellSee a e

Filter

Figure 4.6: A zoned earthfill dam.




acts as chimney drain. The d/s graded filter is connected with d/s horizontal drainage blanketand toe drain for seepage outflow.

When a variety of soil materials are available, the choice of an earthfill dam shouldalways be a zoned embankment type because of its inherent advantage in reduced cost of

construction. The necessary arrangements are required to collect and dispose off any seepagethat does cross the impervious central zone.

4.4.3 Diaphragm Dam

This dam is similar to a zoned embankment dam with the exception that a thindiaphragm of impervious material is provided to form a water barrier (Fig. 4.7). The bulk ofthe embankment is constructed of pervious material (sand, gravel or rock). The position ofthe diaphragm may vary from a blanket on the u/s face to a central vertical core. Thediaphragm may be made of earth/clay, Portland cement concrete, asphalt concrete or other

material. If the diaphragm material is earth, the horizontal thickness of the diaphragm at anyelevation is less than 10 feet or the height of the embankment above the correspondingelevation of the dam (W ≤ h and W ≤ 10 ft). In some cases the diaphragm may be inclined.

Necessary arrangement for drainage of seepage flow is required. Graded filters are providedon u/s and d/s sides of the core, which acts as chimney drain. The d/s graded filter isconnected with d/s horizontal drainage blanket and toe drain for seepage outflow.

The core may be vertical oriented or inclined. It can be placed near the u/s face, in thecenter, or near the d/s face. The u/s and d/s faces of earthfill dam are protected by suitableriprap.

A thin core dam becomes more economical for reasons as:

• Unit cost of placing impervious materials may be more than the unit cost of placing pervious materials.

• The amount of embankment volume can be reduced in a thin core dam more effectively.

• The construction time available and weather conditions may not permit the use of animpervious core of large thickness.

HU/S

Shell C O R E

D/SShell

See a e

Filter

Figure 4.7: A diaphragm earthfill dam with central vertical core.




The minimum thickness of core depends on a number of factors on:

1. the tolerable seepage loss;

2. minimum width which will allow proper construction (machinery considerations);

3. type of materials chosen for the core and shoulders;4. design of proposed filter layers;

5. past experience of similar projects.

Vertical Core

The core is inclined vertical (Fig. 4.7) and is usually located in alignment with thecrest of the dam.

Advantages of vertical core

• Higher pressure exists on the contact between core and the foundation, and will providemore protection against the possibility of leakage along the contact.

• Vertical core tends to be slightly thicker for a given quantity of impervious soil than thethickness of the sloping core.

Criteria

• Cores with width of 30 to 50% dam height prove satisfactory under diverse conditions.

• Core with width of 15 to 20% (thin) if constructed adequately is satisfactory under most

condition.• Core with less than 10% used only if large leaks through the core would not cause dam

failure.

Inclined Core

The inclined core is oriented at an angle with the base of the dam. The core is locatedcloser to the u/s face of the dam with top of core aligned with the dam crest (Fig. 4.8).

H

U/S

Shell C O R E

D/SShell

Filter

Figure 4.8: A diaphragm earthfill dam with inclined core.




Advantages

1. Core can be constructed after completion of d/s portion of embankment. Especiallyuseful for short dry weather condition. Suitable to allow construction of core from finegrained soils.

2. Foundation grouting can be continued while dam embankment is being placed (thussmaller construction period).

3. Filter zones can be thin (smaller slanting width for same horizontal width) and areeasier to install.

Disadvantages

1. Location of core for deep foundation conditions cannot be determined in advance; thusdifficult to locate grout curtain.

2. Additional grouting, if required after dam completion, cannot be undertaken.

Location of Impervious Core/Diaphragm

The core is preferably located in the center of the dam embankment due to followingadvantages.

1. The core is equally supported and is more stable during a sudden drawdown (ifconstructed from earth).

2. Settlement of dam induces compressive stresses in the core, tending to make it morecompact.

3. There is less core volume.

4. Foundation grouting if required can be done post construction of the dam from the crest.

The choice of impermeable zone depends on stability of the core material. If it is strong toresist cracking under load, a location near u/s is often the most economical. However, if corematerial is weak, a central location is better.

[www.ferc/industries/hydropower/safety/eng-guide/chap4.pdf (embankment dam) and…/chap3.pdf (gravity dam)]

4.5 CONTROL OF SEEPAGE THROUGH EMBANKMENT

The seepage through the dam embankment is controlled by two steps: (1) minimizethe seepage rate and volumes and (2) streamline the any seepage to exit from the dam withoutany damage to the embankment (safe seepage exit gradients).

1. Minimize the seepage: All the fill materials will allow some seepage through theembankment. The impermeability of the core minimizes the seepage rate. Thus

permeability and the thickness of the core will ultimately set the seepage rate through theembankment. Thus thick cores having minimum permeability materials will result in

smaller seepage rates.




2. Contain and streamline the seepage: For a sustained high-level reservoir, the seepageflow occurs through the dam section. The seepage emerges at the d/s face of homogeneousand zoned dams. The seepage flow if unchecked can lead to severe piping, and sloughingof the d/s slope and may ultimately lead to failure of the dam. Following arrangements are

used to contain and streamline the exit of seepage flow from the dam body.4.5.1 Rockfill Toe and Toe Drain

The d/s toe of a homogeneous embankment is constituted of rockfill material with agraded filter between the earthfill and rockfill pervious material. The seepage line willconverge towards the rockfill and is then exits safely across the d/s slope keeping the d/sslope dry and safe. A graded filter is provided between the embankment fill material and therockfill toe to prevent migration of embankment materials into the rockfill toe. Frequently a

perforated toe drain of rockfill grade material (Fig. 4.5) is constructed near and below theouter end of the toe to collect the seepage flow. A perforated pipe is embedded in a trenchfilled with fine rock fill. The toe drain collects the seepage discharging from the embankmentand the foundation and lead it to an outfall into the river channel below.

Toe drains may be made of vitrified clay or concrete, perforated corrugated metal orPVC pipe. Drains are placed in trenches below the ground surface to ensure effectiveinterception of seepage flow. Minimum depth below GS = 4 ft, maximum as required tomaintain uniform gradient. Bottom width of trench is 3-4 ft, pipe dia- 6 ″ to 24 ″ depending ongradient, reach length, seepage rate. Drain pipe is surrounded by geotextile filter to preventclogging. Material surrounding drain must satisfy filter criteria. The fill materials in the

trench and surrounding the drain pipe include: Graded sand, Sand and gravel or selected finerock, and Graded gravel or crushed rock

4.5.2 Drainage Blanket

Blanket drains are provided under the base of embankment fill material and extend d/sof impervious zone, impervious diaphragm or 1/4 to 1/3 base of the dam (Figs. 4.3 and 4.9).The blanket drain will intercept the seepage line. Drainage blanket may contain one or morelayers of coarse filter grade materials of filter criteria to match with the materials on two sidesof the filter. The thickness of the blanket should be enough to carry the seepage flow to the

toe drain at the end of the blanket. The blanket drain may not provide full protection againstseepage over a stratification layer which moves horizontally over the layer and ultimatelyreaches the downstream face.

Blanket drain: It may be a continuous layer along whole length of dam or may be intermittentand connected with chimney drain. The length of the dram should reach to d/s edge of core oru/s water depth for a uniform dam. In some cases it may extend under the core. Large lengthof drain decreases the seepage flow path and increase seepage. The thickness varies 3 ft andabove. Material is of filter criteria. A toe drain or a drainage gallery is also provided at outerend to collect the seepage.




4.5.3 Chimney Drain

This is a vertical or inclined drain (made of graded filter) provided inside the dam

body (Figs 4.4, and 4.6 to 4.8). These are usually placed d/s of the impervious core and may be vertical or inclined (30 o ≤ θ ≤ 120 o

The chimney drain can be equally useful for a homogeneous dam with a toe drain.The dam fill placement and compaction in layers form a pseudo-layered condition whereseepage flow entering in one layer will continue in the same layer and will ultimately appearon d/s slope facing leading to slope failure. The chimney drain will intercept the seepagefrom these layers and lead safely to the toe drain (Fig. 4.4)

). The chimney drain may be composed of one or morezones to match the gradation of the adjacent materials. The chimney drain intercepts theseepage flow that crosses the core. It may be single graded or double graded depending upongradation of fill materials on the two sides of the chimney drain. Chimney drain is used inconjunction with horizontal drainage blanket. Chimney drain is connected to blanket drain at

bottom or into a floor channel of drainage gallery.

4.6 FILTER CRITERIA

The filter material is placed in toe, blanket or chimney drains and its materials mustmatch with the gradation of the adjacent materials to ensure stability of the filter and adjacentmaterials. The filter must have large flow capacity to transmit intercepted seepage flow out ofthe dam body. Following criteria follows. D refers to the size of filter material (having largersize) and d refers to the size of base (adjacent) materials having smaller size. (Sherard P-83,USBR 2001, p-235).

Standard sieve set is used to determine the particle size gradation of fill and filtermaterials. The sieve sizes are as under.

# mm # mm # mm # mm # mm3 6.4 10 2.0 25 0.71 60 0.25 200 0.0744 4.8 12 1.7 30 0.59 70 0.21 270 0.0535 4.0 16 1.19 35 0.50 100 0.149 300 0.0506 3.4 18 1.00 40 0.42 140 0.105 325 0.044

8 2.38 20 0.84 50 0.297 170 0.088 400 0.037

Dam foundation material(fine to coarse grained)

EARTHFILL Dam d/s slope

Fine graded filter

Coarse graded filter

Figure 4.9: D/s horizontal drainage blanket and toe drain.

Toe drain




The fill and filter material are characterized by Uniformity Coefficient C U = d 60 /d 10 andCoefficient of Curvature as: C C = d 30

2/[d 60 xd 10 ] and Self-Filtering Critera C SF =d50

2/[d 60 xd 10

1. D

]. The filter criteria is as under:

15 /d 15

2. D

= 5 to 40

15 /d 85

3. D

≤ 5 [This is to prevent migration of fines.]

85

4. Gradation curve of filter material be parallel to gradation curve of base material(similar C

/drain opening ≥ 2 [for toe drain]

U

5. If base material contains gravel, then filter is designed on the basis of gradation curveof the portion of the material finer than 1 ″ sieve.

as for base material).

6. Filter should contain not more than 5% of fines passing # 200 sieves and the fines, if

any, should be cohesionless.

7. Self filtering is achieved if d 15 coarser ≤ 5 d

An alternate filter criteria for transition zone is described as under.

85 finer

* D 15 /d 15

* D

> 4-5 For sufficient permeability

15 /d 85

* D

< 4-5 To prevent migration of fines

50 /d 50

* D

< 25 To prevent migration of fines

60 /D 10

The filter may have one zone/layer or more than one zone between the adjacent fillmaterials e.g. clay core and rockfill. Single or double filter layers between fine and course fillare selected to ensure filter criteria on both sides of the filter layer. For single zone/layer filterof Fig. 4.10(a) the filter F1 must comply both for the gradation of clay core on one side andthe gradation of rockfill on the other side. Considering the filter criteria between clay coreand filter F1 the D will refer to gradation of filter F1 and d will refer to gradation of claycore. Considering the filter criteria between filter F1 and the rockfill the D will refer togradation of rockfill and d will refer to gradation of filter F1. For double zone/layer filters of

Fig. 4.10(b) the filter F2 abutting the coarser fill material (rockfill) will be coarse than filterF1 abutting the finer fill material (clay core). The filter F1 must comply both for thegradation of clay core on one side and the gradation of filter F2 on the other side. Similarlythe filter F2 must comply both for gradation of filter F1 on one side and the rockfill gradationon the other side. In exceptional cases three layers/zones of filter may become necessary tofully meet the filter criterion between fine and coarse fill materials.

< 20 For well graded filter to prevent segregation of filter

Dimensions of Filter Layer

Filter zone width and thickness is selected from point of view of its carrying capacity afterfew years (when some settlement, particle rearrangement had occurred and some fines mayhave settled) and its constructability.




• Minimum thickness is one which can be constructed without danger of gaps orareas of segregated materials.

• Horizontal filter layers can be thin, as 6 ″ for sand and 12 ″ for gravel but thickerlayers are preferred.

• Chimney drains or transition zones min horizontal width of the filter zone should be 8-10 ft, 10-12 ft preferable to enable placement, handling and somecompaction.

• For cost reasons or limited filter materials, 3-5 ft wide zones may be used, butrequire more supervision and hand labor for good construction.

Example ( USBR 2000, p-236)

Given: d 15 = 0.006 mm, d 85

D

= 0.10 mm, pipe openings = ½ inches.

15

D

lower = 5 * 0.006 = 0.03 mm (1) [criteria 1]

15

D

upper = 40 * 0.006 = 0.24 mm (2) [criteria 1]

15

From eq 2 and 3 select smaller size, D

≤ 5 * 0.10 = 0.50 mm (3) [criteria 2]

15 upper = 0.24 mm; Average D 15

Draw filter gradation line parallel to base material gradation curve and read D = 0.14 mm.

85

D

= 2.4 mm.(4)

85

As D

≥ 2 * 0.5” ≥ 1” (5) [criteria 3]

85 from eq (4) is smaller than from eq (5), thus a single filter layer will not work. Adoptabove criteria for 1 st layer F1 [D 15 = 0.14 mm, D 85 = 2.4 mm] and Work for 2 nd

D

layer F2.

15

D


15 upper = 40 * 0.14 = 5.6 mm (7) [criteria 1]

Clay core Earthfill F i l t e r

F 1

F i l t e r

F 2

Figure 4.10.1 Single or double zone/layer filter between clay core and gravel / coarse fill.


F 1

(a) single filter zone

(b) double filter zones




D 15 ≤ 5 * 0.24 = 12 mm (8) [criteria 2]

Figure 4.10.2 : Mangla dam raising project showing core, and u/s and d/s double filter layers.

Core F1 F2F1F2

U/s fillD/s fill

F1F2 CORED/s fill






• Foundation may be of rock, coarse grained material (sand, gravel), or fine grainedmaterial (silt and clay)

• Infinite variations in the combinations (materials), structural arrangements and physical characteristics of the constituent materials.

• Roughly stratified.

• For hard foundation minimum treatment include stripping of foundation area toremove sand, topsoil, and other unsuitable materials.

• A key trench is provided to improve bonding of impervious zone of embankment tothe foundation.

4.8 ROCK FOUNDATIONS

Most rock foundations have adequate physical strength. However weathering near thesurface make is weaker and prone to excessive seepage flows. Some treatments may be doneto improve strength and/or to reduce seepage potential. The treatments includes: (see detail inchapter 3 on Geology): 1. Stripping, 2. Strengthening of weak zones, 3. Grouting to make itwater tight. Rock foundations are very well suited for earthfill and rockfill dams.

4.9 SAND GRAVEL FOUNDATIONS

4.9.1 Characteristics

• Gravel/sand foundation has enough bearing/shear strength the support small tomedium earthfill and rockfill dams.

• However these foundations are very conducive to seepage and need suitabletreatment for seepage and uplift pressure control.

• These materials usually are laid over impervious geological foundation at somedepth below the surface.

• Usually stratified heterogeneous mixture

• Excessive under seepage could lead to: Large seepage uplift pressures and Damfailure due to piping (if fine sand is present in large quantities).

• Clean sand (fine and uniform) of low density is inherently unstable due to its loosestructure and is liable to collapse under dynamic load as for earthquake.

• Vibrations/shock as for an earthquake tremor causes re-adjustment of grains into adense structure. Pore water pressure increases suddenly (due to slow drainage) andfoundation behaves as liquid and results in sudden liquification.

• Cohesionless sands of low relative density (< 50%) are suspect to failure.

4.9.2 Treatment of Foundation








Figure 4.12a: Sheet pile installation at Taunsa Barrage. (L) – Secondary weir, (R) – Old pileexposed. Note the pile section and the interlocking between the pile sheets.




Figure 4.12b: Taunsa Barrage: Sheet pile interlocking and embedding in concrete.

Figure 4.12c: Taunsa Barrage: Sheet pile installation by vibroinstaller.




Figure 4.12d: Taunsa Barrage: Sheet pile installation by vibroinstaller.




Figure 4.12e: Taunsa Barrage second weir: U/s, mid and d/s sheet pile rows.

4.9.2.4 Cement Bound Curtain Cutoff

• In places piles are cast by mixing cement with foundation material (Fig. 4.13).

• Curtin constructed by successive overlapping individual piles.

• Each pile consist of column of sand intimately mixed with mortar to form a pilelike structure

• Hole drilled, Mortar injected through hollow rotating pipe with mixer head at

bottom.

• May be reinforced.

U/s Mid D/s




4.9.2.5 Concrete Wall

• RCC wall build down tothe bed rock provide

positive cutoff.

• Wall width 5 ft or more

• Dewatering and shoring

bracing/sheeting required.

• High in cost andchallenging in construction

• RCC or PCC

• Depths 150-200 ft in past

4.9.2.6 Slurry Trench

• Trench excavated by drag

lines 5 ± wide or less.Depth to impervious layerrock surface.

• Excavated material stock piled in windrows.

• Trench filled with bentonite mud slurry(slurry density > water

density) which prevent

Plan

Section

Figure 4.13 Cement bound curtain cutoff.

River level

Bed rock levelPictorial

Figure 4.14 Sequential operations in theconstruction of a slurry trench. (SourceUSBR 2001, p-228).




walls from caving in. Slurry weighs more than water.

• Mud slurry level above water table to keep trench sides stabilized.

• Trench bottom cleared with clamshell bucket and air lift pumps

• Bentonite coated excavated slurry material are further blended with 15-20% ofnatural silt.

• Mixture dumped on one end of trench, displacing the slurry until backfillingcomplete.

• Completed slurry trench in a very soft condition for many months afterconstruction, with consistency like a stiff butter.

• Need more care if cobbles, boulders, large blocks in deeper locations.

• The upper weathered/fractured part of the bed rock grouted after completion of theslurry trench.

4.9.2.7 Grouting Alluvial Deposits

• Cement grouts not injected uniformly in alluvial deposits except for coarsematerial

• Chemical grout can be injected in sand, but expensive

• Primary difficulty of keeping hole open with casing, impossibility of using packers, and lack of technique ensuring uniform penetration of grout

• Special techniques developed for grouting in alluvium as packers can not be usedalong with casing.

• Coarse materials grouted successfully

• Usually several rows of grout holes to increase effectiveness.

• Outer rows grouted with cement and cement-clay grouts, inner rows withchemical grout.

• Results of grouting difficult to evaluate.

4.9.2.8 Horizontal U/s Impervious Blanket [Sherard p-312, Fig 6.3.2]

If construction of complete seepage barrier for a dam on pervious foundation is not practicable, then under seepage may be reduced by increasing the width of the base ofimpervious section by a horizontal impervious blanket, which is connected to the dam core.The seepage is reduced due to lengthening of seepage path. This also reduces the d/s porewater pressure and thus increase stability. The u/s impervious blanket is constructed ofimpervious material extending u/s of the dam face toe/ heel and connected with imperviouscore of the dam embankment (Fig. 4.15).




• These may be used in conjunction with partial cutoff located at u/s end or anyother location (e.g. Tarbela, Khanpur dam).

• Blanket is generally used for a stream channel or valley floor of sand and gravel.

• This may also be required for portions of abutments to reduce seepage through theabutments.

• Blanket starts from core of the dam and extends about 400-500 m, upstream.

• Blanket thickness 10% of dam height (minimum 10 feet) at dam face to minimum3 ft at outer end.

• Blanket protected from erosion by 2-3 ft thick riprap over gravel bedding.

• Areas with natural clay blanket if any are cleared of trees/vegetation, defective places repaired, and entire surface rolled to seal root holes.

• No stripping of area us/ of dam to obtain fine construction material for damconstruction (particularly if no +ve cutoff).

• Length of blanket governed by desired reduction in seepage flow.

• Blanket may not eliminate piping in naturally stratified soils as high pressuresmay exist in one or more strata at d/s toe of the dam.

• Tarbella dam has 5700 ft long u/s impervious blanket. Its thickness varies from 42ft at dam u/s toe to 10 ft at the outer end.

Figure 4.15a: U/s horizontal impermeable barrier.

U/s impermeable barrier / blanket

Protective gravel/riprap layer




Figure 4.15b: Mangla dam raising: u/s impervious blanket with top gravel layer for protection in area of Sukhian dike.

Figure 4.15c: Mangla dam raising: Compaction of impervious blanket by sheep foot rollersat optimum moisture content.




4.9.2.9 Horizontal d/s impervious blanket

Likewise u/s blanket, impermeable horizontal blanket may also be provided at d/s ofdam to lengthen seepage path and reduce seepage (Fig. 4.16). However due to its position atd/s of dam it is subjected to excessive uplift pressures. Thus the d/s blanket must be designed

to resist uplift pressure. This is done by providing berm of random fill material to add weightover the impermeable layer. The d/s drainage blanket may be provided above theimpermeable blanket. D/s impervious blanket is not very often used.

4.9.3 Seepage through Foundation

4.9.3.1 Seepage rate

Under seepage through the foundation is determined by Darcy’s law

Q = K I A I = ∆h/L

I = Average hydraulic gradient over the flow length.

L = length of seepage path ≈ = base of impervious bottom or core

∆h = head difference between reservoir water level and the d/s drain water level.

A = 1 x depth of foundation

K = average permeability for all layers (horizontal K)

The Darcy formula is quite valid when depth of the foundation (d) is small incomparison to the flow length L. For other conditions a flow net should be drawn to

Random fill to counteractuplift pressure. Height=H/2

D/s impermeable blanket

Fi ure 4.16 D/s im ervious blanket.

Figure 4.17 Seepage force components. [USBR p-221]

L

d

Seepage exitarea ~ 2-3 d




determine the seepage flow rate. The seepage flow emerges d/s of the dam over a length 2-3 ddepending on the permeability and stratification/ layering of the foundation.

4.9.3.2 Seepage Forces and Piping

• The flow of water through pervious foundation produces seepage force due tofriction of percolating water with the walls of the pores.

• Seepage force proportional to flow velocity.

• Small downward force at entry over large u/s area. This increases submergedweight of soil.

• Under the dam flow velocity increases due to reduced flow area.

• At d/s toe of dam, the seepage force is upward reducing effective weight of thesoil.

• If upward force exceeds soil weight, the soil would be floated out (boilsformation).

• The particle erosion progress backward along the flow line until a continuous pipelike opening is formed (usually irregular and tortuous) to reservoir.

• Piping allows rapid escape of water.

• It can lead to dam failure due to foundation heaving.

• Excessive seepage results in blow out / heave at d/s of dam.

• Piping failure also called as failure by heave or internal/subsurface erosion.

• Magnitude and distribution of seepage forces by flow net analysis (this requiresconsiderable experience to draw flow net).

• Grain size and gradation of the foundation materials has an important bearing onthe piping failure.

• Piping failure takes places after the dam has been in service for some time.

• Piping takes places along minor geological weaknesses.

• Piping can be completely and reliably prevented by controlling the under seepagesuch as (Sherard P-313):

i. Exist velocities are not high.

ii. Water discharges through adequate thickness of progressively coarser soilswhich meets gradation requirements of filter.

• Line of creep- shortest path that a particle of water has to travel in seepage underthe dam.




• Creep ratio i.e. the ratio of length of creep to the pressure head loss (CR = L/ Δh) ,this is inverse of the average hydraulic gradient.

• Weighted creep ratio is for length of line of creep computed as sum of the verticalcomponents of the shortest seepage path plus one-third of the length of thehorizontal seepage path. This is used for stratified soils.

• Criterion for piping potential on the basis of weighted creep ratio as function offoundation soil type.

Creep ratio. Minimum 3 – for gravel/boulder foundation

Minimum 8 – for very fine sands

• This is to be used as guide to judgment but not as a design criteria.

• Valid if no graded filter provided at d/s of dam (graded filter reduces flow length).

• The best plan is to provide drainage blanket of graded filter under d/s section ofthe dam.

• Pressure relief wells placed near d/s toe of dam shall be useful to intercept theseepage and this reduces uplift pressures.

4.9.3.3 Pressure Relief Wells

• Relief wells are to ease out and reduce pressure of the seeping water under thefoundation of an earth dam.

• These has disadvantages as:

a. These decrease the length of average seepage path and cause to increase thequantity of under seepage.

b. These require frequent inspection and maintenance, replacement.

c. The pressure head is lowered to a value nearly equal to the elevation of top (ordischarge level) of the well.

d. Well may discharge into a delivery pipe, a drainage ditch and water is carried

back to river section.e. Wells to be closely spaced (10 to 25 ft) to minimize pressure build up in

between. Spacing based on judgment.

f. Additional wells be installed in between or in d/s row subsequently, if needed.

g. Wells penetrate more than 50% depth of foundation.

h. Screen is placed in center of hole and outer annular spaced filled with suitablegravel pack as per filter criteria. Usually a single pack is used and screenopening is designed to match the selected filter material.




i. Wells are developed to improve efficiency.

j. Pressure relief wells are very helpful to relieve seepage pressure when lower pervious foundation material is overlain by otherwise natural imperviousstratum (no danger of piping, blow out).

k. Depth of wells equal to height of dam (or depth of bed rock) are mostsatisfactory.

l. Pressure relief wells lower uplift pressure but enhance seepage flow rate.

4.9.3.4 Deep toe drain

A toe drain is often provided to collect seepage flow occurring through the drainage blanket. This is generally a shallow ditch filled with gravel/sand material. The toe drain alsointercepts the seepage flow through the dam foundation. Providing a deeper toe drain can

considerably enhance the interception of foundation seepage. A graded filter zone is provided

Bed rock

Drainage ditch

Pressure relief well

Seepage flow

Fi ure 4.17 Pressure relief well.

Well detailsWell head details




between the foundation and toe drain fill to stop migration of fines from the foundation. Thedeep toe drain does not significantly alter the seepage path length and thus seepage rate is notvery much affected.

4.9.4 Sand Gravel Foundation Design

The design criteria require control of seepage flow through the foundation andabutments (no internal erosion, no sloughing in area where seepage emerges). The perviousfoundation may be either exposed or covered at the surface. The pervious foundation may behomogeneous or stratified. Stratification influence foundation treatment method.

4.9.4.1 Case I: Exposed Foundation

The sand gravel foundation is open at the surface. The foundation may be shallow,medium or deep.

A: Shallow Foundation• Provide a positive (complete) cutoff to bedrock.

• Grouting of bedrock, if needed.

• Horizontal drainage blanket not necessary if shallow pervious foundation can actas filter and provide adequate drainage.

• Provide drainage blanket of filter criteria if:

a. embankment is homogeneous or d/s shell is rockfill

b. perviousness of foundation is questionable.

c. Piping potential exist, either from embankment to foundation or fromfoundation to embankment zone (at d/s part).

d. Foundation is stratified.

• If rockfill at d/s portion of dam, provide DB from d/s slope to the imperviouszone/core.

• It positive cutoff not practical due to lack of materials, short construction season,wet climate, high dewatering cost, then other methods of cutoff be used.

B: Intermediate Depth Foundation

• Positive cutoff may be less economical

• Provide other methods of cutoff (sheet pile, slurry trench etc).

• Provide minimum impervious zone/core B 1½:1 u/s slope and 1:1 d/s slope (coreB is described in a later section).

• Provide drainage blanket of filter grade if i) overlying zone is impervious or ii)

overlying zone is rockfill, iii) piping potential is present




• Provide key trench

C: Deep Depth Foundation

• Foundation too deep for a positive cutoff

• Provide u/s impermeable blanket in continuation of impermeable core.

• Minimum core B


• At d/s of embankment, provide adequate thickness of previous or impervious(random fill) materials (berm) (Fig. 4.16) to improve stability against high uplift

pressures.

• Provide filter grade drainage blanket for d/s rock or imp fill against piping hazard.

• Provide toe drains

• For foundations of high K, which cause extensive seepage, ponding and sand boils, then provide drainage trenches, pressure relief wells, extension of d/s toe ofdam or blanket on d/s area.

• For deep stratified layers, provide partial cutoff and u/s blanket.

• Some seepage inadvertent.

4.9.4.2 Case-II: Covered Pervious Foundation

The gravel/sand foundation is covered by some impervious layer. There are three (3)conditions:

A: Top impervious layer thickness 3 ft or less

• Layer usually ineffective as an impervious blanket. Design the foundation asexposed foundation.

• Excavate/remove the foundation material to bottom bed rock layer.

B: Thickness of top impervious layer more than 3 ft, but less than dam height h.

• Provide drainage trenches (of depth as much as to top of lower pervious layer) atd/s of dam or pressure relief wells to relieve uplift pressure.

• May act as u/s impervious blanket depending on thickness, continuity,imperviousness, u/s distance to natural loose deposits

• Need to compact with heavy roller.

• Horizontal drainage blanket also provided if embankment is homogeneous, or permeability of d/s zone questionable – of length reaching to base of imp zone(Z+5’) if d/s slope of core greater that 1




C: Thickness of impervious layer greater than dam reservoir head.

• No major problems for seepage or seepage forces. No treatment is needed forseepage control.

4.10 FINE GRAINED (SILT, CLAY) FOUNDATIONS [USBR p-246]


• Foundation of fine grained soil (silt, clay) are sufficient impermeable and thus nodanger of under seepage and piping

• Main problem is stability against consolidation and shear failure due to low bearing/shear strength

• Characteristics depend on location of water table, and compactness of soil

• State of compactness determined by standard penetration test (soil below watertable) and by density-in-place test (for dry soils above water table)

• Weak soils need to be treated for improving strength (by improving density)

4.10.1.1 Saturated soils

• Determine nature of consolidation as normally consolidated or over consolidated by analyzing the weight to which the soils had been exposed in geologic past.

• Saturated impervious sands (dirty sands - sands having good amounts of fines)also act as fine grained soils

• Ability to resist shear stress (due to embankment weight) may be determined fromsoil group.

• Relative density for cohesionless soils D r = (e max -e)/(e max -e min

• For cohesive soils relative consistency C

) is related tostrength.

r [C r = (LL-W)/(LL-PL)] = (e LL -ew)/(e LL -e PL

• At C

) is also related to strength LL = liquid limit, PL = plastic limit, W =water content.

r = 0 (W = LL), cohesive strength of all remolded soils C LL ≈ 0.2 lb/sq in andshear strength S LL = 0.2 + σ’ tan φs. The φs

• At W = PL, cohesive strength varies considerably; φ is obtained from triaxial teston samples compacted at proctor maximum dry density.

is obtained by slow shear test onsaturated soil (drainage permitted, pore water pressure ≈ zero) tan φ about 0.5

Treatments:

The shear strength can be increased on

i. Remove the soil of low shear strength.




ii. Provide drainage of foundation to permit settlement on drainage and increaseof strength during construction.

iii. To reduce the magnitude of the average shear stress along the potential surfaceof sliding by flattening the slopes of the embankment.

a. Removal of soft foundation is practicable and thin layers of soft soilover lying firm layers are excavated.

b. Vertical drains may be provided to facilitate consolidation. This is practical for low embankments only e.g. under highway (non-hydraulicstructures) not practical for dams.

c. Flattening of slopes lengthen the surface of sliding, decreases averageshear stress along the path and increase factor of safety against sliding(Fig. 4.18).

4.10.1.2Relatively Dry Foundations• These soils exhibit large strength at its present dryness

• The relative density of the material indicates the potential/danger of soil oncompression

• Many soils will undergo quick and sudden volume reduction on wetting/saturatingon reservoir filling (Fig. 4.18).

• Dense soils which will undergo small compaction on loading and wetting may beused as foundation for dams.

• Pre-wetting of soil before loading improves its strength on loading.

• Large compaction and could cause serious rupture/weak section for dam conematerials and consequent dam failure

• Drainage must be assured by an underlying pervious layer or by a verticaldrainage.

h/2

h

Figure 4.18 : Increasing base contact area for stability against shear, sliding.






4.10.3 Fine Grained Foundation Design

Saturated Foundation

• Recommendation according to soil group on USBR P-251

• Add stabilizing fill (u/s + d/s) to embankment designed for a stable foundation(Fig. 4.18).

• Slopes according to strength of foundation material as determined by penetrationtest, consistency and dam height.

• Suitable for small dams only

Dry Foundation

• Design depend on potential to volume change on wetting

• If potential less, design as for saturated soil

• Else pre-densify the soil before construction and later design as for saturateddense soils.

4.11 EMBANKMENT SECTION DESIGN

This defines the crest design, u/s and d/s slopes of impermeable core, the shoulder fill,the slope protection, etc.

4.11.1 Crest Design

Crest width W

• The width W of the crest is governed by height of dam, importance of structure,

width of highway, construction procedure, access required either duringconstruction or as a permanent feature.

• Japanese code W (m) = 3.6 H 1/3

• Special widening may be necessary to provide ahighway or safeguard against freak waves etc. Thiswidening could be done by steepening face slope in

– 3. (Thomas p-384). These are seismic activeareas)

the upper reaches of dam.

• Top crest width should not be less than 30’.

Figure 4.19b

h/2

h

3:1 or flatter

Random fill




• The top width (m) is taken as

W = h/5 + 3 low dams (Punmia P. 365)

W = 0.55 √h + 0.2 h h< 30 m

W = 1.65 (h+1.5)1/3

Crest width of dams in Pakistan are: Hub dam = 28 ft, Mangla dam = 40 ft, Khanpurdam= 35 ft, Tarbella dam = 40 ft, Simly dam = 35 ft, Bolan dam = 40 ft.

h>30 m (USBR 2001)

For ease of construction with power equipment, the crest width should not be lessthan 12 ft. Roadway across the dam set the width.

Sherard p-413: For earthquake area, top of dam is subjected to worst damage and canvibrate with greater amplitude than the base. Thus it is advisable to make dam top thicker byincreasing crest width or using flatter slopes near top. Also that if any crack develops, the

longer seepage path causes less seepage and increases dam safety.

Surface Drainage

Surface drainage of crust be provided by a crown of at least 3”, or by sloping towardsthe upstream floor. For wider crest 2% slope is adequate.

Surfacing

Crest surface should be protected against damage from wave splash, rainfall, wind,frost and traffic wear. A layer of fine rock or gravely material of 4 inches minimum thickness

be provided. If a highway is carried across the dam, then crest width and surfacing mustconform to highway codes.

Safety Requirements

Crest should be made safe by providing metallic or concrete guard rails on bothshoulders of the crest. For minor dam, pillars at 25ft spacing or large boulders placed atintervals along the crest may be provided. Guard rails be at least 2 1/2

Camber

ft from crest edge and bewell supported.

The crest elevation is increased towards center of the dam by an amount equal tofuture consolidation of dam foundation and embankment after completion of the construction.Selection of amount of camber is somewhat arbitrary. It is provided to ensure that someresidual camber will remain after settlement and consolidation. This improves the appearanceof the dam. The camber is provided by increasing the u/s and d/s slopes near the crest of thedam. The camber is not accounted in stability calculations. For non-compressiblefoundations, camber of about 1% of dam height is provided. Several feet (often 8 to 10 ft) ofcamber may be needed for dams constructed on foundations expected to settle.




4.11.2 Embankment Slopes

Embankment slopes are designed to ensure strength, stability and economy ofconstruction: Flat slopes, more cost, more stability/strength; Steeper, lower costs, stability orstrength. Embankments are constructed from infinite conditions of soil materials with varyingsize, gradation, stress-strain relationship, and shear strength (USBR 2001, p-254).

The procedure for designing a cross-section of earthfill dam consists largely ofdesigning to the slopes and characteristics of existing successful dams , making analytical andexperimental studies of unusual conditions and controlling closely select ion and placement of

embankment materials. Except small variation in specific design, radical innovation areavoided. Any fundamental changes in design concepts are adopted gradually as more

practical experience is gathered. This practice is being overly cautious, but probable extent ofloss of property and life in the event of failure of a constructed dam provide ample

justification for these conservative procedures. Whereas design of large dam can be madeincreasingly secure by laboratory test of materials, the design of small dams is heavilyfollowed on the basis of successful structures and past experiences.

Embankment slopes may be continuous or discontinuous. Embankment may have asingle slope over whole height, or multiple slopes may be provided over different sections ofthe depth. The slope discontinuity or change in slope may be with or without a berm.

The u/s and d/s slopes of the embankment and core are selected from generalguidelines, experiences in the light of foundation materials and materials available forconstruction. The seepage analysis and stability of the selected dam section is carried out anddam section may be acceptable if factor of safety for the dam under different construction andoperation conditions are found satisfactory. Alternate dam sections are evaluated for materialneeds/crest and factor of safety and that dam section is adopted which provide higher factorof safety at lowest costs. Stability of the shape is analyzed under static loads as well as under

seismic conditions.

Dam crest design levelConstruction level

Camber ~ 4

Camber

Figure 4.20 Camber for the dam crest.

W

ELEVATIONX-SECTION

1.75:1

2:1




Except where there is surplus of material available from required excavations, themost economical dam is obtained with the minimum volume and therefore most steep slopesconsistent with the dam stability (Sherard p-48). The allowable steepness depends on theinternal zoning and on strength of foundations and the embankment material. Crest length

and pace of construction may also affect the slope selection. Use of excavated material asrandom fill may allow flatter slopes. The random fill material may be placed (Fig. 4.21) at

bottom of u/s face to eliminate slope protection, at d/s face as toe support to improvestability; it may be buried inside the supporting shells, or if it contains coarse materials it may

be used as filter zones to the core. If random fill is impermeable but with poor stability, itmay be buried inside the impervious core. (Also see USBR p-260 fig. 149)

The strength of foundation is also affects the dam face slopes: Weak foundation –average slope 2:1 to 4:1; Strong foundation – steeper slope 1.5:1 to 3:1. The height of damalso affect slope selection. For homogeneous materials dams of fine core: Short height –

steeper slope, Higher dams – flatter slope. The internal zoning permit steeper slopes. For thinclay core slopes are independent of height. For rockfill dam with thin u/s core, the d/s slope isequal to natural angle of repose of rockfill material (1.7:1 – rounded stream gravel, 1.2:1 –angular quarried rock, 1:1 – thin layers of well compacted quarried rock. Central core d/s 1.6

– 1.8. The slopes may be single or multiple slope. Slopes may be continuous or discontinuouswith or without intervening berm.

Slopes are set as following:

• Dams located in narrow rock-walled canyons can be constructed with some what

steeper slopes than otherwise, because of added stability given by the confiningwalls. In narrow valleys broad toe berm or very flat slopes at the toe of dam can

be provided relatively cheaply due to the small quantities of embankment materialrequired.

• U/s slope may vary from 2:1 to as flat as 4:1 for stability, usually it is 2½:1 or 3:1.

• For eliminating slope protection in lower levels (below dead storage) slope may be made flatter

• Berm to act as base for top slope protection

Figure 4.21 Placement of random fill.




• Steeper slopes may be allowed above normal conservation level

• Random fill in lower part to flatten slope

• D/s slope: 2:1 for dams with d/s pervious zone and 2 1/2

• The slope of vertical core as: u/s face – 1.5H:1V to 0.6H:1V; d/s face: 1:1 to0.5H:1V; Inclined core: u/s face – 1:1, d/s face 0.3:1 (reverse slope)

:1 for homogeneous dams.

This provides stability for most soils when drainage is provided to eliminatesaturation of d/s slope.

• The slope depends on materials available, foundation condition, dam height, andvaries widely as: u/s from 2H:1V to 4H;1V. Coarser free draining materials allowsteeper slopes, and finer materials require flatter slope. In general slopes may beas (Pumnia p-366):

Material u/s d/sHomogeneous well graded 2.5:1 2:1

Coarse silt 3:1 2.5:1Silty clay h < 15 m 2.5:1 2:1

h > 15 m 3:1 2.5:1Sand and gravel with clay core 3:1 2.5:1

Concrete core 2.5:1 2:1

• The slope of the dam also depends on the type of the dam and on the nature ofmaterials for construction.

Diaphragm Type

If shoulder material SW- GW or GW. (# 200 < 5%) slopes as for rockfill dam (1.3 to1.7 Horizontal to 1 Vertical)

HomogeneousMaterials No rapid draw down Rapid drawdown

u /s d/s u/s d/sGW GP SW SP Materials not suitable -too perviousGC GM SC SM 2½:1 2:1 3:1 2:1

CL ML 3:1 2½:1 3½:1 2½:1CH MH 3/6:1 2½:1 4:1 2½:1

Zoned embankment

• Impervious core flanked by relatively pervious material.

• Filter transition provided on both sides of the impervious zone to prevent pipingand internal erosion.

• Transition materials partially fill cracks/holes in imp core.

• Transition of rock fines or sand gravel.




• Few feet required but constructed as 8-12 ft to accommodate constructionmaterials/machinery

• Thick transition-design as filter

• Thick transition-less requirements

Impervious Core

• Pervious or impervious foundation with positive cut off - provide minimum core A(top width 10’, base = h, symmetric)

• Exposed pervious foundations or covered pervious foundation (cover < 3 ft). No positive cutoff-minimum core B (Top width – 10’, U/S 1½:1, D/S 1:1)

• Maximum core (Top width – 10’, U/S slope = (x-1/2) : 1, D/S slope = (y – ½) : 1;where x:1 is slope of u/s face, y :1 is slope of d/s face)

• For core greater than maximum core, outer shells become ineffective in stabilizing thedam and embankment may be considered as homogenous for stability analysis.

• Core smaller than minimum core – dam as diaphragm type.

• Impervious cover over foundation more than 3’- select between core A and core Bdepending on extent and effectiveness of the core.

• Top of the core kept 3-5 ft below crest to safeguard against weathering.

• Thickness of impervious cover over foundation more than dam height (d > h): Forsaturated fine grained foundation use Core A + u/s and d/s random stabilizing fill(Fig. 4.18). The slope of stabilizing fill depends on dam height (min 3:1) andconsistency and nature/group of foundation soil (4:1 for SM to 10:1 for CH). Furtherdetails in USBR 2001 p=251, Table 16).

• For core A stability not affected by core material (due to smaller thickness).

Minimum Core A: for dams on impervious foundation or shallow pervious foundation with positive cutoff trench.

Minimum Core B: for dams on deep pervious foundations without positive cutoff.

Maximum Core:

Figure 4.21 Size range of impervious core for zoned embankment. (USBR p-266).

Slope = y:1

Slope = y-½:1

Slope = 1:1

Slope = ½:1

1½:1

x-½:1

x:1

Z

Z




• Outside shell slopes governed by stability of fill material

• Rocks, GW, GP, suitable for shell

• Gravely SW and SP also suitable for shell

• Embankment slopes may be selected according to materials as below.

• USBR p-251 Table 16 defines slopes of stabilizing fill, min 3:1 and max 10:1.

•

Table 4.2: Recommended slopes for small zoned earthfill dams on stable foundations (USBR p-267)

Type Shell material Core material No rapiddrawdown

Rapiddrawdown

U/s D/s U/s D/sMin core A Rock, GW,

GP, SW, SP,gravely

GC, GM, SC, SM,CL, ML, CH, MH

2:1 2:1

Max core Rock, GW,GP, SW, SPgravely

GC, GM 2:1 2:1 2½:1 2¼:1

SC, SM 2¼:1 2¼:1 2½:1 2¼:1

CL, ML 2½:1 2½:1 3:1 2½:1

CH, MH 3:1 3:1 3½:1 3:1

Cross section of some dams in Pakistan are shown in chapter-1.

Tarbela Dam, Pakistan.

Rocks under alluvium and abutments.

Abutments: Metamorphosed sedimentary rocks (sugary limestone, phylite, quartzite, schist)

Alluvium: Boulders/cobbles and gravel choked with sand, depth as much as 600 ft.

4.11.3 Slope Protection

Dam slopes are needed to be protected against action of various destructive forces.

U/S Slopes: Protection is required against destructive waves splashing onto the side slope.Waves generated due to high sustained winds as well as from earthquake action. Also neededto be protected against burrowing animals.

D/S Slope: These need to be protected against erosion by windand rainfall runoff and the borrowing animals. This also needs to

be protected against possibility of seeping of rain water andforming internal erosion (piping and sloughing of inside of

embankment).




I: UPSTREAM SLOPE

U/s slopes are provided protection by: rock riprap, concrete pavement, steel facing, bituminous concrete pavement, pre cast concrete blocks, others as short cement pavement,wood, sacked concrete. Special care is needed against beaching process if water level stays at

one elevation for long times.

Rock Riprap

This refers to placing of fairly large size rock pieces over the slope face. The rockmay be dumped or hand placed. The riprap is placed over a properly graded filter, which may

be a specially placed blanket or may be outer pervious zone of a zoned dam.

Dumped Riprap

The rock fragments/stones are dumped over the slope. The efficiency of dumped rock

riprap depends on following: Quality of the rock, Weight or size of individual stone pieces,Thickness of the riprap, Shape of stones or rock fragments (rounded, angular), Slope of theembankment, Stability and effectiveness of the filter.

Rock for riprap should be hard, dense and durable, resist long exposure to weathering.Igneous, metamorphic rocks, limestone, hard sandstones make excellent riprap. Visualinspection and lab tests, petrographic tests assure quality. Rocks should be free of seams ofshale (low quality rocks).

Individual pieces should be of sufficient weight to resist displacement by waves (mustfor all size dams). The thickness of riprap should be sufficient to accommodate weight andsize of stones necessary to resist wave action (Fig. 4.22). A 3’ minimum thickness is used.Smaller thickness if wave action is less severe. Lesser thickness may be used for upper slopescorresponding to flood control storage (above normal conservation level) due to infrequentexposure of this part to waves. If there is any damage to this section, it can be repaired onreservoir lowering.

Slo e rotection Min 3 ft

Shell material(min 4-6 ft toenable placing)

Filter or bedding layer 1 ftmin (washed gravel fill)

Figure 4.22a: U/s Slope protection.




Figure 4.22b: Dumped rock riprap. (L: placement in progress and R: completed)

Figure 4.22c: Hand placed rock riprap. (USBR p-279)




Figure 4.22d: Mangla dam raising: Protection of d/s slope by hand/machine placed roundedriver-run cobbles. (looking downward)

Figure 4.22e: Mangla dam raising: U/s slope protection by angular rock riprap over filter.






Figure 4.22h: Mangla dam: u/s face protection by angular rock riprap.

Riprap weight(Zipparro eds. Davis’ Handbook of Applied Hydraulics, p-13.58) defined the riprap

weight as:

( ) ( )ba

Cot GK

H W

θ

γ 350

1−

=

W max = 4 W 50 and W min = W 50

W

/8

50

H = Wave height (ft)

= Average stone weight (lbs)

γ = Stone unit weight (lbs/cft) (bulk unit weight after placement) ~ 156 lb/cft

G = Sp. Gravity of stones material (2.3 – 2.7)

θ = angle (degrees) of slope surface with horizontal

K = stability coefficient (K ∼ 4.37)

a, b = empirical coefficient (In general coefficient are as: a = 3, b ∼1)




Novak et al (p-54) defined size of rock armoring necessary for stability under wave action isas: M = 10 3 x H s

3 where M=mass of stone required (kg), and H s

The size of riprap is estimated as : D = [7 W / 5 γ]

=significant wave height(m).

1/3 where D = stone size (ft), W = stone

weight, γ = bulk un i t weight (lbs/cft). The th i ckness must be more than size ofheaviest/largest stones. In no case it should be smaller than 1.5 x D 50

USBR p-277 provided gradation (by weight in lbs) of riprap for slope = 3:1 and angular rocksas:

or 24”

Fetch thickness Max D 50-60 D 40-50 D

< 2.5 miles 30” 2500 lbs 1250 75-1250 75

0-10

> 2.5 ml 36” 4500 2250 1000-2250 100

• Sand and rock dust < 5% by weight

• Rounded size d 0-10

• Rounded rocks require a thicker layer, or slope should be made flatter

is meant to fill the voids in larger rocks.

• For 2:1 slope, 36” minimum thickness be used.

Shape of Rock

Shape of rock fragments influences the ability of riprap to resist displacement bywave action. Angular fragments tend to interlock better than boulders and rounded cobbles.

Thus rounded stones should have more thickness.

Graded Filter

A layer or blanket of graded filter should be provided underneath the riprap if there isdanger of fines from underneath layer to more into the riprap layer by wave action. For azoned dam filter not needed if outer shall is gravel. Blanket of crushed rock or natural gravels3/16” to 3 1/2”

Flexibility

with thickness equal to half of riprap thickness (but not less than 12”) issatisfactory. Follow filter criteria discussed earlier.

Dumped filter should have flexibility to adjust base surface an account of settlementof dam body or local settlement.

Placement

The riprap is dumped from hauling trucks onto the prepared surface. Bulldozers areused to level off-and compact the dumped layer to fill up the voids between larger stones.Smaller stones fit in voids of larger pieces very well. The rock stones must not break duringhandling / placement / compaction. Top surface is uneven, rough and decreases wave riprap.




Riprap materials had been hauled from long distances (200+ miles) due to its satisfactory and proven performance and economy.

Hand Packed rock Riprap

This consists of suitably sized stones carefully laid by hand in a more or less definite pattern with minimum amount of voids and with top surface relatively smooth. Doubled orirregular shapes lay up less satisfactorily than stones of roughly square shape. Stones of flatstratified nature should be placed with principal bedding plane normal to slope. Joints should

be broken as much as possible and voids be avoided carefully by arranging various sizes ofstones and small rock fragments.

The stones of excellent quality should be used. Thickness can be half of dumpedriprap but not less than 12”. Filter blanket be provided underneath the riprap, if required. Dueto tight packing, hand placed riprap is not as flexible, so it cannot adjust to foundation or

local settlement. Thus hand placed riprap should not be used where considerable settlement isexpected. Hand placed riprap could be costly due to extensive labor cost in spite of its smallerthickness.

Concrete Paving

Concrete is placed over the sloping surface to resist wave destruction. It can be used both for rockfill and earthfill dams. Paving thickness depend on dam height, slope steepness.Thickness is 8” for h ∼ 50 for and 12’-18” for high dams. Paving is placed in blocks 6’ x 6’ or

more but monolithic construction gives the best service. A water tight surface will eliminatehydrostatic pressure underneath the pavement. Blocks could be displaced or broken by waveaction and uplift forces under the slab. Concrete can crack requiring frequent maintenance.

For blocks, expansion joints and construction joints should be widely spaced.Reinforcement is (5% area) in both directions and be continuous through the construction

joints. Joints be sealed with plastic fillers and cracks be grouted and sealed properly.Pavement should extend from crest to below the minimum water levels. It should terminate ata berm and against a deep seated curb or header (minimum 18” deep).

The success of concrete pavement is mixed, but successes and problems have beenobserved. Paving is a costly alternate, but may be adopted if enough riprap material is notavailable. Concrete pavement may or may not be held in place firmly by foundation boltsembedded deep inside sloping shell. Concrete paving increase the wave runup and suitablewave breakers, wave deflectors, may be provided to reduced risk of dam overtopping. Failurechance is 30% + due to inherent deficiencies in this type of construction.

Soil Cement

Soil cement is produced by mixing cement with coarse sandy or gravely soil with 10-25% material passing # 200 sieve are ideal (Maximum allowed < 50% of # 200). The cementis 0.7-1.0 barrel of cement per cubic yard of compacted soil cement. 2-4% extra cement may




be added for erosion resistance. For most soils 10-12% cement (% of compacted volume) isconsidered typical. Cement and moisture ratio is determined by lab tests.

Soil cement is placed in 6-8” horizontal layers over the slope (horizontal width as 8 ftnormal thickness 2-3½ ft) and roller compacted in a stair-step horizontal layers. Soil cements

have 500-1000 lb/inch compressive strength at 7-day (10% cement). The edges of the cementlayers are not trimmed to retard wave runup. A reasonably firm foundation is required so thatdeformation following placement of soil-cement is not significant. Normal embankmentconstruction procedures are satisfactory.

Figure 4.23b : Soil cement paving.

II. DOWNSTREAM SLOPE

8-10 ft

2-3.5 ft

Concrete paving with coping wallConcrete paving withwave breakersSoil cement

Figure 4.23a: Paving with soil cement or concrete.




The zoned dams with d/s shell of rock or cobble fill do not need additional slope protection. Slope protection is required for all conditions against erosion by wind and rainfall.If not protected gully can develop.This protection is provided by

placing a layer of rock, cobbles, orsod (grass). However, vegetation

protection can be poor andineffective at places, especially inarid regions. Thus cobbles/rock protection is preferred. The stone is 24” thickness (minimum12”) over a filter bed should be provided. Minimum single layer hand placed cobbles / stones.Berm or a cut slope may be provided at intervals and graded contour drainage channels

provided to collect and dispose the rain runoff from upper portions of dam slope. Drainagechannels discharge into cement lined channels running down the slope and ultimately to safe

disposal point/river bed. A contour drain is also provided along toe of dam. Surface drainageis also provided (as an open gutter) for abutments and valley floor.

4.11.4 Abutments

FLARED ABUTMENT SLOPES

The u/s and d/s slopes of embankment are often flared at abutments to provide flatterslopes for stability and seepage control. The u/s flaring is equivalent to providing u/simpermeable blanket. The flaring design is governed by topography of the site, the length ofconstant desired, for aesthetic value, and ease of construction. For steep side slopes this may

be useful to locate access road across the dam. ABUTMENT SIDE SLOPES

The side slope of impervious abutment are usually discontinuous. The sides aredressed with slope not exceeding 1H:2V to provide a stable contact between the embankmentand abutments. The bottom should be continuous without abrupt level changes. Any hangover should be removed to have good contact.

Remove overhangs

Trim slopes to max 0.5V:1H

Figure 4.24 Abutment shaping.

Cut slope Berm




4.12 ENGINEERING CHARACTERISTICS OF SOILS [Novak et al. 1998, p-36-45]

Soil load – pore water pressur e response

Soils undergo deformation as a result of changes in loading or drainage conditionsdue to alterations in the geometric configuration of the soil particle assembly. The volumechanges and settlement due to external loading takes place slowly through the complex

process of consolidation . Relationships in the form of pore-pressure coefficients are used todescribe immediate response of pore water pressure to applied total stress.Shear strength

The shear strength of a soil is defined as the maximum resistance to shearing stresswhich can be mobilized; when this is exceeded failure occurs usually along identifiable slipsurfaces. The shear strength of any material is described by Mohr-Coulomb failure criterion

based on total stress as: S = c + σ tan φ or based on effective stress as: S = c’ + σ’ tan φ’ σ =

total compressive stress ( σ = σ’ + u), σ’ = effective stress, u = pore water pressure.Laboratory shear tests, e.g. triaxial shear test, are run for the material compacted to the designdensity / moisture content and construction of Mohr circle plot. Coarse soils such as sandsderive their shear strength largely from particle interlock and internal friction, and are calledas cohesionless (c=0) or frictional soils; the shear strength is mostly controlled largely soildensity.y . Most clays soils derive shear strength from both cohesion and internal friction.Following tests are usually carried out. (Sherard p-332)

1. Undrained test: (unconsolidated – undrained test). No drainage and dissipation of pore water pressure. Called as Q-test (quick test). Used for stability analysis fordam during and after construction.

2. Consolidated-undrained test. (sample first consolidated with full pore water pressure dissipation under given consolidation pressure) and then is failed in shearwith no drainage allowed. This is called R test.

3. Drained test – consolidated. Drainage and complete dissipation of pore pressureallowed at all stages (slow test) For parameters in terms of effective stress (c’ andφ’). Called as S test (slow test)

Compressibi l ity and consol idation

When load is applied to a soil, mass volume decreases and settlement may occur dueto (a) elastic deformation of soil particles, (b) compression of the pore fluid, (c) explusion of

pore fluid from the stressed zone with rearrangement of soil particles, with expulsion of porewater being dominant. The consolidation of clays is very slow due to their very low

permeability. Vertical consolidation characteristics are determined in lab in oedometer testsand expressed by:

Coefficient of volume compressibility to determine the magnitude of time dependentconsolidation settlement: m v = Δε v/Δσ’ v

Coefficient of consolidation to establish rates of settlement: c

v = k/m vγw (k=permeability)




Coefficient of secondary consolidation to describe subsequent continuing settlement due tocreep of the soil structure under constant effective stress.

Compaction

Compaction is the process of densification by expulsion of the air from the soil void spaces,and result in closer particle packing, improved strength and reduced settlement. Rollersassisted by vibratory excitation are used for field compaction of embankments. The degree ofcompaction is measured in terms of dry density ρ d

Representati ve engineering properties for soil s:

= ρ/(1+w) where ρ = bulk in -situ density,w is moisture content. Compaction of soil modifies the major engineering characteristics asshear strength, compressibility, volume change due to change in moisture content, and

permeability.

Description Saturatedunitweight γ(kN/m 3

Shear strength (effectivestrength basis)

)

Coefficient ofcompressibility, mv

(x10

-4 m2

Coefficientof horizontal

permeabilityk /kN) h (m/s)

Cohesionc’ (kN/m 2

Friction, φ’(degrees))

Gravels 17-22 0 30-45 0.1 – 1.0 10 -1 – 10 -2

Sand 0 30-45 10 - – 10 -

Silts < 5 20-35 10 - – 10 -

Clay (soft-medium)

15-21 0 20-30 1.0 – 10.0 Intact clays,

< 10-8

, ifweathered,fissured, orwith siltlenses 10 -3 –10 -8

Clays (sensitive,silty)

< 10 < 30

Clays (medium-stiff)

< 50 < 20

4.12 SEEPAGE ANALYSIS

Seepage flow will occur through all types of formation irrespective how small the permeability may be. Seepage occur both through the dam embankment as well as damfoundation. Seepage flow is given by Darcy’s law q = K I per unit flow cross section and asQ = q A = K I A = K ∆hA/L through section of area A.

Flow net method is used for simple flow conditions. Seepage occurs as confined flowthrough the foundation and as unconfined flow through the embankment. A flow net isdrawn with curvilinear squares; different squares may have different area but all have flowlines and potential lines cross at right angle and that all four sides are of equal curved length

(Fig. 4.25). The flow is given as




Q = K ∆H N f / N

K = Permeability (m/s)

d

∆H = Head difference across two ends

N f

N

= No. of flow lines

d

Flow net are drawn on a ‘to-scale’ map of the dam + foundation cross section. Usually the permeability in horizontal orientation is higher than permeability in vertical orientation (an-isotropic case). Experience, skill and practice is required to draw flow net. See figures onnext page.

= No. of potential drops

Flow net provide

1. Seepage flow rate through the section

2. Distribution of water potential (h) and pore water pressure u (u = h – z ; z =elevation and Note: u, h, z has units of length and are taken from a pre-selecteddatum). Also u = P/ρ g = P/ γ, P = water pressure, ρ = water density, γ = sp weight.The water potential h and pore water pressure is given in units of pressure head;this should be multiplied with unit weight of water (γ = 62.4 lbs/cft, 1000 kg mass/m3

) to convert it to pressure units. Pore water pressure distribution is needed forstability analysis.

No. of flow tubes = N f = 4.3

Seepageexit face

≈ h/3

h

Equi-potential drops: 1 2 3 4 5 6 7

8 9 10 12 14 16

Figure 4.25 Seepage flow through the dam embankment by flow net method.

Seepageexit face

≈ h/3

h

Equi-potential drops: 1 2 3 4 5 6 7 8 9 10 12

Figure 4.26 Drawing equi-potential lines through the dam embankment.

∆h

δh=∆h/m

B

H C0.3 L

L




Procedure to draw flow net

1. Draw a to-scale map of dam and foundation

2. Determine the seepage exit area3. The u/s face of the embankment is plane where seepage originates.

4. The d/s exit area is usually up to a height h/3 for a homogeneous dam (h = damheight). For a modified homogeneous dam, the flow will converge towards the toedrain or the horizontal blanket drain. Flow emerges tangentially to the exit face ford/s face or chimney drain or vertical line for rockfill toe or drainage blanket (Fig.4.27).

5. The seepage line is part of parabola and exits tangent to the d/s surface.

6. Establish the seepage line/phreatic surface. The seepage line intersects thereservoir water surface at a distance 0.3 L from the point C (point C is at watersurface at u/s face) where L is the horizontal projection of the u/s face (Fig. 4.26).The actual phreatic line is modified to meet at point C.

7. Determine the head difference ∆h between u/s and d/s exit area ∆h = h u/s – h

8. Select number of potential drops m over ∆h. Determine head drop across one

potential drop δh = ∆h/m. Divide the seepage line into selected equal δh intervals.Thus total number of potential drops N

d/s

d = h/ δh.

a) Seepage exit at d/sface of dam. b: Seepage exit into vertical

face chimney drain

c) Seepage exit intorockfill toe

d) Seepage exit intodrainage blanket

Figure 4.27 Phreatic line for various seepage exit conditions.




9. Draw equipotential lines from these points such that they intersect to the seepageline and lower confining layers at right angle. Towards the u/s face theseequipotential lines will take parallelism to the slope of the u/s face.

10. Draw a flow line in the middle part of the flow area keeping it normal to potential

lines but approximately curvilinear and parallel to seepage line such thatapproximate squares are formed. Extend this flow to the originating surface (i.e.u/s face) and to the exit surface (i.e. d/s face or toe/blanket drain). Draw moreflow lines to the bottom confining layer.

11. The effect of confining surface on the shape of flow lines and equipotential linesdiminishes farther from the confining surface.

12. The flow net is formed of curvilinear squares (equal sides, right angle). Few non-squares will not affect seepage flow rate but may affect internal head distribution.

13. Flow net become more complex if more than one material or anisotropic materialsare present.

14. For large differences in K of the two materials of embankment and foundation theflow lines can cross from foundation into upper embankment or fromembankment into bottom foundation (the material of higher K will attract flowlines from other material). For small differences in K the flow lines originatingfrom any material will continue within the same material and will not cross intothe other material.

15. Flow lines are attracted by chimney drain, drainage blanket and toe drain.

Phreatic Line in earth dams with drainage blanket: Graphical Method (Fig. 4.28)

• L = Horizontal projection B-D of the upstream face length A-D

• Mark point C as CD = 0.3 L

• Taking C as center draw circle of radius CF to point E.

A

B C0.3 L

L

F G HQ xR = x+S

R

P(x,y)

D E

Figure 4.28 Phreatic surface D-P-G for dam with drainage blanket.

y

S

h

T

Directrix




• Draw vertical tangent from E to H (E-H line is directrix )

• G point midway between F and H. This is extremity of seepage line D-P-G

• Draw vertical line at Q (F-Q = x)

• With F as center, Q-H as radius R, cut PQ vertical at P. The distance P-Q = y.(x,y) are the coordinates of the seepage line parabola. Draw other points similar toP. The seepage line meets at C.

• U/S end part of the seepage line is redrawn to meet the water surface at D at rightangle.

Seepage rate

I = dy/dx, A = y x 1, Q = K I A = KS, where S = Focal distance = FH. Also

T hT K Q −+= 22

Figure 4.29 Seepage flow net for dam foundation with partial cutoff.

Figure 4.30 Equipotential contours and flow lines for seepage through dam foundation with sheet pile.




Seepage Through Dam Foundation

Seepage through the dam foundation is also determined by drawing flow net for thefoundation section. The flow net is drawn by procedure similar to for the embankment. Theseepage control measures are also considered while drawing the flow net and determining the

seepage rates.

Example: {Lambe and Whitman 1969, p-273. K = 5 x 10 -4

Top width = 14 ft, u/s and d/s slope = 1.5:1, toe drain = 30 ft, total base = 140 ft, height = 42ft, free board = 2 ft.

ft/sec, Fig. 4.31.

N f = 2.8, N d = 9, = 42 – 2 = 40 ft, Δh = 40/9 = 4.444 ft, saturated flux = 5 x 10 -4 x 2.8/9 x 40= 6.22 x 10 -3 ft3/s/ft, unsaturated flux = 0.58 x 10 -3 ft3/s/ft total seepage flux = 6.80 x 10 -3 ft3

/s/ft.}

Figure 4.31 Seepage flow net for rockfill toe homogeneous e/f dam. (b), (c) by Seep/Wshowing equi potential lines and flow lines respectively.

(a)

(c)

(b)






good material and placed on sound foundation are considered safe against sliding. The safetyagainst shear failure is analyzed for the following conditions: (Sherard p-326)

1. During and after construction for both u/s and d/s faces. Assume pore pressurehigh and not drained; Analysis is based on lab Q – test

2. Full reservoir steady seepage – d/s face; Analysis is based on lab S - test

3. Rapid drawdown – u/s face – pore water undrained and pore pressure high;Analysis is based on lab R- test

4. Seismic loading; Analysis is based on lab R- test

A factor of safety is determined for various situations. The dam section is accepted if thefactor of safety for the selected loading condition is higher than recommended values. Elsethe dam section (i.e. side slope of core and shell and materials) is revised and safety re-

evaluated.

Method of Slices / Sweadish Circle Method

Procedure

• Problem is considered in 2-D space (cross section)

• a continuous potential surface of shear failure (usually called slip surface) passingthrough dam embankment and/or foundation is assumed. Slip surface could be acombination of part of a circle, an arc, line, etc. Sliding surface-circular orcombination of arc and straight lines (Fig. 4.32).

Slip surface as circle Slip surface as lines

Slip surface as arc Slip surface as arc + line

Fi ure 4.32 : Various sha es of sli surface as circle arc lines.






D x b. For marked difference in length of two sides, area of trepezoid as A =(D L + D R

b) Normal component of W force acting on bottom of slice: N = W cos α

)/2 x b.

c) Tangential component of weight: T = W sin α d) Total water potential h acting on the slice bottom is determined from the equi-

potential contour map. The pore water pressure head (units of L) is thendetermined as u = h – z, where z is the elevation of the bottom from selecteddatum. Total pore water pressure head U acting on bottom of slice as: U = u x

b/cos α x 1 = average pore water pressure x area of bottom of slice. Area of bottom of slice = ∆L x 1 = b/cos α x 1. The pore water pressure head U is

converted to force units as U Force units = U Length units x γ

e) The cohesion of the material c or c’ is determined from lab tests of thematerials. The total shearing resistance component due to cohesion C = unitcohesion x area of bottom of slice. Thus C = c ′ b/cos α

w

f) Total shear resistance which can be developed on the bottom of the any sliceat failure: S = C + (N - U) tan φ′ [N-U = α’]

g) Determine sum of tangential force T and shear resistance S on all slices.

h) Safety factor F = ∑S/∑T = ∑[C+(N-U) tan φ′] / ∑ W x Sin α NOTE: T, U and N may be worked as continuous curve across all slices. Normal components

pass through center of rotation and does not cause any driving moment on the slice.Tangential component T causes a driving moment M=T x r, r = radius of slip surface.Resisting forces determined from Columb’s equation.

For homogeneous and uniform cohesive soils a circular arc is considered for slipcircle. The locus of the centre of the critical circle with r u

Z < 0.3 is approximated as:

c = H Cot β(0.6 + 2 tan φ’) and Y c

where Z = H Cot β(0.6 - tan φ’)

c, Y c = coordinate w.r.t. toe of dam (+ve up and left), β = slope angle, H = height, z =depth below ground surface, r u = dimensionless pore pressure ratio, = u w

Method of Sliding block

/γZ.

Same as method of slices, but 2-3 slices only called blocks.

Stability of D/s slope for steady seepage

For steady seepage the d/s slope is liable to shear failure. A slip surface is selected over thed/s slope. The sliding mass is divided into slices or blocks and factor of safety determined byabove procedure. The pore water pressure along the base of sliding slices is determined froman equipotential contour of the d/s slope (Fig. 4.34).




9

1 2

1 4

1 6

1 8

2 0

2 2

2 4

2 6 2 8

3 0

3 0

3 2

3 4

3 6

3 7

3 8

3 9

4 0

4 0

4 2

4 4

4 5

4 6

5 1 5 2

5 3

5 4

Figure 4.34: Equipotential lines for steady seepage in a vertical core earthfill dam.

Stability of U/S slope During Sudden Drawdown

When reservoir is suddenly lowered, the Hydrostatic force acting on u/s force isremoved and weight of water tends to help a sliding failure as no outside pressure tocounteract it. Water in the saturated mass tend to drain towards both U/S face and the

permeable foundation. Permeability of foundation material affects drainage pattern. Iffoundation is permeable then flow is downward, if impermeable flow horizontally outwardtowards outer faces. The U/S face is not an equipotential line but potential varies with heightas h = z = elevation. Water potential within the saturated mass of soil changes according tothe u/s potential.

A slip surface is considered along the u/s face. Determine h, z, and u = h-z on bottomof the slices along the slip surface. The pore water pressure is determined from correspondingequipotential contour map (Figure 4.35). Determine safety factor by procedure above.

No change in water content within the saturated mass of the earthfill.

Hydrostatic force acting on u/s face is removed and potential at face h = z (point height)

Weight of water tend to help sliding failure as no outside pressure to counteract it.

Permeability of foundation material affect drainage pattern. If more permeable, flowdownward, if imp, flow is horizontally outward.

9 1 0 1

2

1 4 1 4

1 6

1 7 1

8

2 0 2 0

2 2 2 2

2 4

2 4

2 5

2 6

2 6 2 7

2 8

2 8

2 9

3 0

3 0

3 2

3 4

3 6

3 7

3 8

3 9 4 0

4 0

4 2

4445 4 6

Figure 4.35: Equipotential lines for sudden drawdown in a vertical core earthfill dam.

U/s + d/s face during and at end of construction

The embankment fill of core is compacted to maximum dry density at optimum moisturecontent. Soil mass changes due to compaction and by its own weight. The pore water




pressure increases on compaction due to decrease of void rat io. The pore water pressure aftercompaction is determined by Hilf’s equation as (Fig. 4.36).

∆−+

∆=

wca

a

V hV P

u

where

u = pore water induced pressure

P a

∆ = embankment compression in % of original total embankment volume

= air pressure in voids of soil mass after initial compaction ≈ atmospheric pressure corrected for the site elevation)

V a

V

= Volume of free air voids as % of original total embankment volume

w

h = Volume of pore water as % of original total embankment volume

c = Henery constant of stability of air in water = 0.02 at 68 o

F

Figure 4.36: Consolidation and water potential in dams.

• Draw graph between effective stress and % compression ∆ (Plot 1)

• For each ∆, determine u from Hilf’s equation. Also determine corresponding σ’from plot 1. Determine σ = σ’ + u.

• Draw u vs. σ (plot 2)

• Find mid height D of each slice bottom

• Calculate total stress σ = γD and find u for each slice from plot 2 corresponding toσ.

• Determine the safety factor as above procedure.

NOTE : DAVIS. HAH P.18-38

In zoned dam, critical circle is located so that a maximum portion of its length passes throughmaterials of lowest shear strength (core or foundation layer). The slip surface can be as partof Toe circle, Slope circle or Midpoint circle.

∆ %

σ’ σ

u

Plot 1 Plot 2




Recommended Factor of Safety (Novak) Design loading F s

During/end of construction 1.25 1.25 (1.3 to 1,5)u/s d/s

+ earth quake 1.0 1.0Reservoir full/partial (steady conditions) - 1.5

+ earth quake - 1.1Rapid drawdown 1.2 -

(Slip circle between highest and lowest water levels)Seismic loading 1.1 1.1Steady seepage + surcharge pool condition - 1.4

Example:

The Fig. 4.38 shows section of an earthfill dam at its maximum depth showing steadyseepage phreatic line, equipotential lines, a trial slip surface A-B-C-D-E. Scale: 1 block = 5x5m. Dam height = 60 m, depth at normal conservation level = 55 m, U/s slope = 2:1, d/s slope= 2:1, Core uniform width = 20 m, core height = 55 m. The material properties are as: Core:c' = 5 KPa, φ' = 30 °, average unit weight γ = 20 KN/m 3, K = 1 x 10 -5 cm/s. Fill: c' = 3 KPa, φ'= 35 °, γ = 18 KN/m 3, K = 5 x 10 -4

Solution:

cm/s. Determine the factor of safety for the d/s face forsteady seepage condition for the shown slip surface by using method of sliding blocks.

The slip area is divided into four sliding blocks 1 to 4. The width, side height are noted foreach block as: width = 20, 20, 40 and 30 m and sections heights as 15, 35, 35, 20 and 0.1: Total weight of each block.W1 = 5x15x18 + 10x15x20 + (15x20)/2x20 = 1350 + 3000 + 3000 = 7350 KNW2 = (35+35)/2x20x18 = 12600 KNW3 = (35+20)/2x40x18 = 19800 KN W4 = (20+0)/2*30*18 = 5400 KN2. Bottom length:L1 = [15 2+25 2]0.5 = 29.15 m; L2 = [20 2+10 2]0.5

L3 = [40 = 22.36m;

2+52]0.5 = 40.31 m; L4 = [30 2+5 2]0.5 = 30.41 m

Mid point circle

Slope circleToe circle

Figure 4.37: Trial slip circle.




9

1 2

1 4

1 6

1 8 2 0

2 2

2 4

2 6 2 8

3 0

3 0

3 2

3 4

3 6

3 7

3 8

3 9

4 0

4 0

4 2

4 4

4 5

4 6

Figure 4.38 : Dam d/s slope stability for constant seepage.

3. Inclination of bottomθ1 = tan -1(25/15) = 59.04° θ2 = tan -1

θ3 = tan(10/20) = 26.57°

-1(5/40) = 7.13° θ4 = tan -1

4. Component of block weight normal to base N:(5/30) = - 9.46°

N1 = 7350 x cos 59.04 = 3781 KN N2 = 12600 x cos 26.57 = 11269 KN N3 = 19800 x cos 7.13 = 19647 KN N4 = 5400 x cos 9.46 = 5327 KNΣN in core = N1 = 3781 KN; ΣN in fill = N1+N2+N3=11269+19647+5327 = 36243 KN 5. Component of block weight normal to base T:T1 = 7350 sin 59.04 = 6303 KN T2 = 12600 x sin 26.57 = 5636 KNT4 = 19800 x sin 7.13 = 2458 KN T4 = 5400 x sin -9.46 = - 888 KNΣT = 6303 + 5636 + 2458 – 888 = 13509 KN = Total shearing force6. Total water pressure head h, elevation, and net pore water pressure at points B, C, D, E,F:

h (m) = 47.5 37.0 33.4 24.9 15z (m) = 45 25 15 10 15u (m) = 2.5 12 18.4 14.9 0u (KPa) = 24.52 117.72 180.50 146.17 0

Average pore water pressure u along bottom of each block:u1 = (24.52+117.72)/2 = 71.12 KPa, u2 = (117.72+180.50)/2 = 149.11 KPa,u3 = (180.50 + 146.17)/2 = 163.33 KPa, u4 = (146.17 + 0)/2 = 73.08 KPa.Total pore water pressure force along block bottom U:U1 = 71.12 KPa x 29.15 m x 1 m= 2073 KN U2 = 149.11 x 22.36 = 3334 KNU3 = 163.33 x 40.31 x 1 = 6584 KN U4 = 73.08 x 30.41 x 1 = 2223 KNΣU in core = U1 = 2073 KN; ΣU in fill = 3334 + 6584 + 2223 = 12141 KN 7. Cohesion along slip surface C:C1 = (3x5 + 5x10 + 5x29.15) x 1 = 211 KN C2 = 3x22.36x1 = 67 KNC3 = 3x40.31 x 1 = 121 KN C4 = 3 x 30.41 x 1 = 91 KNΣC = 211 + 67 + 121 + 91 = 490 KN

A

B

C

DE

F

2

1

3

4




8. Total shearing resistance mobilized:S = 490 + (3781 - 2073) tan 30 + (36243 – 12141) tan 35 = 490 + 986 + 16876 = 18352 KN9. Factor of safety FOS = 18352/13509 = 1.359

STABILITY OF FOUNDTION AGAINST SHEAR

• Foundations of finer loose cohesionless materials or of unconsolidated clays and silts-weak in shear.

• Approximate method

• Assumption earthen material has an equivalent liquid unit weight which would produce same shear stress as the material itself.

P = total horizontal shear down to rigid boundary

)2

45(tan 2

12m

22

2 φ γ −

−=

hhP i

1φ = equivalent angle of friction

1

1m1

ctanh tan

hmγ φ γ

φ +

=

where

φ, c = shear parameters of foundation materials

γm

1

221 )(

h

hhh f d m

γ γ γ

+−=

= mean unit weight of dam and foundation weighted in proportion to depth of each

γd = unit weight of dam material and γf

Max unit shear S

= unit weight of foundation material.

max = 1.4 S av

Now average unit shear = s

and occur at point B which is 0.4 b from C

av

Let S

= P/b

1 = unit shear strength below toe (at A) = c + γf h 2

S

tan φ

2 = unit shear strength at point C = c + γm h 1

Average shear strength – S = (S

tan φ

1 + S 2

Overall factor of safety against shear = S/S

)/2

av

Factor of safety at maximum shear:

. This schould be > 1.5

Let S = Unit shear strength at point of max. shear (point B) = c + γav

γ

h tan φ

av = mean effective unit weight = ( γdh + γf h2) / (h+h 2) and F.S. = S/S max . This should begreater than 1.0

h1

h2 b

0.4b

h PA B C

Rigid boundary




Inter slice ForceSlice Normal force at base= W + (X L – X R ) + N Cos α + S m

Ordinary or Fellenius – No inter slice forcesα - D Sin w

Bishop horizontal - yes, Shear NoJanbu horizontal - No, No shear-but a correctionfactor used to account for interslice shear forceStability may be ascertained by considering Moment or force

equilibriumMoment equilibrium on individual slice or Overall sliding manForce equilibrium ← summation of horizontal forces

Inter slice shear force X = E λ f(x)f(x) = inter slice force function

λ = % (in decimal) of function usedWeight W increased/decreased by X amount

SAFETY AGAINST SETTLEMENT AND DEFORMATION [Novak et al. 1990]

Dams are provided with some free board for safety against overtopping. The dam fillmaterial is however liable to settle and deform resulting in decrease of free board. To assuresafety against future settlement of dam crest, the crest is elevated by the amount of futuresettlement.

The primary consolidation δ1 which develop as excess pore water pressure are

dissipated (during the course of construction of the dam embankment) can be estimated interms of coefficient of compressibility m v , the depth of compressible soil and mean verticaleffective stress increa se Δσ as: δ 1 = m v

Embankment: δ

Δσ. Then:

1e = m ve γ dH2/2, [Δσ = γ d

Foundation: δ

H2/2]

1f = m vf D f Δσ f [Δσ f = I γd

where H = embankment height, D

Ze]

f

The secondary consolidation settlement δ

is depth of compressible foundation, I is influence factor(depending on foundation elasticity and depth-width ratio; for representativeembankment/dam foundation geometries I ~ 0.90-0.99.)

2 can be estimated from the coefficient ofsecondary consolidation C α as: δ 2 = C α z log(t 2 /t 1) where z is the height H or D f asappropriate and times t 2 and t 1 are relative to completion of primary consolidation. Values ofCα

See worked example 2.4 by Noval et al. 1990.

are generally below 0.002 for over consolidated clay fills, rising to 0.005 and 0.5 forsofter normally consolidated clays.

ER EL

XR

XL

W




II: ROCKFILL DAMS

4.21 ROCKFILL DAM (Source: USBR ch-7, Golze ch-7)

Rockfill dams are type of embankment dams where more than 50% rock massof varying size and quality is used for construction of the embankment. The rockfill hasusually very large permeability and thus large amounts of water can seep through the rockfillembankment. Thus a seepage control membrane is used to minimize seepage through theembankment. The alignment of the dam (dam axis) is selected for minimum embankmentvolume and/or membrane exposure. There are two main modes of seepage control membrane.

• Internal membrane (central core). The membrane is located inside theembankment (Figs. 4.41 and 4.42). The membrane materials include earth/clay

core, reinforced cement concrete (RC), roller compacted concrete (RCC), asphaltconcrete, steel or other metals. The clay core may be thin or thick, located in themiddle or near the u/s face, may be vertical or inclined.

• External u/s face. The membrane is located at the u/s face of the rockfill dam (Fig.4.43). The membrane materials include cement concrete, RCC, steel, timber, stone/ rubble or PCC masonry, asphalt concrete.

A and E: RockfillB and D: graded filter / transition

C: Vertical internal seepage controlmembrane (earth core)

D

Figure 4.41 A typical rockfill dam section with vertical internal seepage control membrane

EC

B

A

A and E: RockfillB and D: graded filter / transitionC: Inclined internal seepage control

membrane (earth core)

D

Figure 4.42 . A typical rockfill dam section with inclined internal seepage control membrane

E

C

B

AE1 E2 E3

E1 small size rocksE2 medium size rocksE3 large size rocks




Figure 4.44: Glennies Creek Dam (67 meters high, concrete faced rock fill dam CFRD) onGlennies Creek. [Source: http://members.optusnet.com.au/~richardw2/projects.html ]

A- seepage control and face protection membrane

B- membrane bedding layerC- random fill of low quality

rock

D- rockfill of best qualityrock

C D

Figure 4.43 A typical rockfill dam section with external seepage control membrane

AB

http://members.optusnet.com.au/~richardw2/projects.html







Figure 4.45: Rockfill dam: d/s view.

4.22 CONDITIONS FAVORING CHOICE OF ROCKFILL DAM

• Large quantities of rock usually readily available from structural excavation or anearby quarry site.

• Earth materials are difficult to obtain or require expensive + extensive processing.

• Short construction season (allows simultaneous during unfavorable weather asexcess moisture not an issue for u/s face membrane dams).

• Excessive wet climate restricting placement of earth materials

• The dam is to be raised at a later time

• Rockfill can be placed during rainy season. Grouting foundation can be donesimultaneously with embankment placement.

• Diversion floods are very large and large diversion arrangements very costly. Theflood can flow through/over the dam without failure or with small damage.

• Uplift pressure and erosion due to seepage not a significant design problem




• Short structure base lengths due to steeper slopes

• Short coffer dam (can be placed within the gorge length) and also overtopping can be permitted.

4.23 EMBANKMENT DETAILS4.23.1 Materials

The design of the rockfill dam depends on placement and materials of the seepagecontrol membrane. The typical sections are shown in Figs. 4.41 to 4.43.

U/s Face membrane

A. Face protection

B. Membrane bedding layer of Well graded small size rock/gravel

C. Random fill-low quality Lesser quality – excavation materialsD. Best quality rock Best quality large size this section provide high stability to

dam

Internal membrane

A+E Rockfill

B+D Graded filter/transition (One or more zones)

C Earth core

• Selection depends on availability of rock

• Smaller size rocks close to filter/transition zones

• Larger size rock of highest quality on u/s & d/s slope

• Thin earth core as vertical or inclined

4.23.2 Traditional vs Present Design

Breitenbach 2007 summarized historical record rockfill placement and compactionindicates 4 milestones in rockfill dam construction. The first milestone included the use oflow level hand placed rockfill dumps with timber facing on the upstream slope in the 1850’sfor water storage and gold sluicing operations. The second milestone included a gradualincrease in water storage dam heights to over 300 feet (100 m) high using thick dry and looserockfill dump lift placement by trucks or draglines without compaction into the 1940’s. Thethird milestone included the use of high-pressure water jets and flooding techniques to wetand consolidate the thick loose rockfill dump lifts to achieve up to 85 percent of total damsettlement before reservoir filling from the 1940’s into the 1950’s. The fourth milestoneincluded control of rockfill lift thickness with dozer spreading and roller compaction, inaddition to documentation of rockfill gradation, moisture, and density in large-scale test fillsfrom the 1960’s to the present day.




Figure 4.46: Rockfill embankment traditional construction technique. Top : Rock dump looselift placement in 45 ft (15 m) thickness. Bottom : Rock segregation with boulders at

bottom of rock dump lift (Breitenbach 2007)




Figure 4.47: Present day construction of rockfill embankment. Top: 0.5 m thick rockfill lift placement by haul truck and dozer. Bottom: Fill lift compaction with steel smoothdrum vibratory roller. (Breitenbach 2007)




Figure 4.48: Wetting and compacting 2 ft (0.6 m) thick loose rockfill lift with 20 tonvibratory compactor roller 2005

Figure 4.49: Rockfill bulk density test. (L) - 1 m diameter plate for large scale rockfillsample. (M) - Water replacement test in hand excavated and lined hole, (R) - Bulk

gradation test on excavated rockfill materials. (Breitenbach 2007)

Traditional design (Dumped rockfill)

• Rocks dumped in high lifts 30 to 60 ft.

• No compaction-point to point bearing

• Smaller size rocks sluiced with high pressure water volume 2-4 times rock vol.(dirty rock need more water) Sluicing with 2-4 inch dia nozzles.

• Rock angular corners break easily on wetting and thus denser packing




• Settlement is caused due to wetting and rock mass weight + rock thrust on rollingdown the slope + height of drop

• U/s hand or derrick placed rock zone or rubble masonry required to form supportfor face membrane.

Present design (Compacted rockfill)

• Rock material dumped in thin layers, spreaded by dozers and compacted byvibratory rollers of 5 to 50 ton capacity. Lift varies 1 to 2 m only.

• Has very small post construction settlement.

• Wide range of rock (size, quality) may be used

• Concurrent work in adjacent areas unaffected

4.23.3 U/s and d/s Face Slopes• Slopes depend on type and location of membrane. Slopes evolved from steep (0.5

H: 1V) to flat (1.3-1.7:1)

• Steep slopes used to minimize rock volume and cost

• Steep slopes possible with u/s face membrane

• For past design the steep slopes were stabilized by thick crane-placed dry rubblemasonry (and which provide as support zone for the bedding layer for u/smembrane). No derrick/crane placed rock work required for present design

• Slope flattened to match angle of repose

• Central sloping core: 2:1 to 4:1 both u/s & d/s (flatter slopes for central core)

• U/S face membrane: concrete u/s 1.3-1.7:1, d/s natural angle (1.3 – 1.4 :1),Asphalt concrete face 1.6 – 1.7:1, Steel, u/s 1.3-1.4:1, D/s – 1.3- 1.4

4.23.4 Rock Quality

• Hard, durable and able to withstand disintegration due to weathering.

• Resist excessive breakage due to quarrying (rock blasted at quarry), loading,hauling and placing operations

• Free of unstable minerals

• Individual rocks of uniform size for good rock-to-rock contact.

• Igneous, sedimentary and metamorphic rocks all used successively.

• Each dam site a unique problem, thus General guidelines only.

• Rock quality determined by lab tests and/or in-situ inspections of weathering

marks at the rock quarry site.




• Test embankment to answer i. Use of marginal materials, ii. Performance ofmaterials during compaction operation, iii. Correct compaction equipment, iv.

Number of passes, v. Correct lift for each material

4.23.5 Rock Sources

Rock can be obtained from many sources as: Excavation for foundations, structures,spillway, stilling basin, tunnels, underground power houses etc., Quarry rock near dam site,Talus slopes, etc. Angular rock fragments can be obtained from the from river bed – ifcobbles/boulders

4.23.6 Rock Size

Use Rock of sp gravity = 2.67 – 2.94+, weight not less than 160 lb/cft = 2560 kg/m3

Past Design

Zone B: Mix. of: (between bedding layer and Czone see figure)

40% - quarry chip to 1000 lbs (375 kg) ofcompressive strength > 350 kg/cm 2

30% 1000-3000 lbs (1/2 to 1.5 tonshaving 0.45 to 0.75 m size)

(5000 psi)

30% 3000-14000 lb (1.5 to 6 tons of 0.75 to 1.25 m size)

<3 % quarry dust

Maximum dimension not more than 3 times min dimension

Zone D: Extra large rock

50% > 14000 lbs (4.5 ft)

50% 6000 – 14000 lbs (3.25 to 4.5 ft)

Max dimension not more than 4 times minimum dimension

Zone C: Random quality rock as Mix.Quarry chips to 14000 lbs

Present design with compacted layers

(I) With u/s face membrane (A):

D Good quality rock of 1 cft to 1 cubic yardsize, No slaby rocks (to avoid bridging), wellgraded, minimum finer part.

C Low quality: 3” to 1 cft

B C D

C D

A

B




B Bedding layer: ¼” – 3” to provide smooth uniform bearing surface for the u/s facemembrane

Note: (1) For C D zones fine rock placed nearer to u/s face and coarse rock nearer to d/s.Strongest material is placed in lower part of zone D to improve stability. The internal

friction angle decreases with rock size and confining pressure.

(2) The lift (lift = fill height) should be at least twice the size of the largest rock. TheB zone is dumped in 30 cm (1 ft) lifts, leveled and compacted with crawler orvibratory rollers. The C and D zones are dumped with 1 m and 1-2 m lifts,respectively and compacted with vibratory rollers. The material is thoroughly wettedduring truck dumping time (but not sluiced) before compaction.

(3) For asphalt face, a thin B zone is enough, and compaction is done by smooth drumvibratory rollers up the face.

(II) With Central core:

The u/s and d/s rock shell provide support to the core. Thus strongest and large rocksin d/s rockfill shell/zones. The u/s shell may be formed of lesser quality rock. For both u/sand d/s shells, the smaller size rock is placed nearer and adjacent to the core while larger sizerocks is placed towards the outer faces. The rock material placed on u/s and d/s face to be ofsufficient size and quality to satisfy the riprap requirements. No bedding layer is required

below the riprap due to sufficient porous nature of the rockfill.

General grading of rock material as: 0-10% - 0.6 mm, 0-40% -5 mm, 0-65% - 19 mm,

22-100% - 76 mm, 60-100% - 305 mm, 100 % - 610 mm.

The u/s and d/s shell rockfill is compacted in 1 m lifts with vibratory rollers. Thesluicing is done in such a way that will not clog filters or impermeable materials washedaway.

The filter/transition zones are compacted in 30 cm lifts by crawler or vibratory rollers.The width of filter zone should be enough for placing and compaction. Filters materials

prevent piping of the impervious materials into the rock shells.

The core is compacted in 15 cm layers and compacted by sheep foot rollers +

vibratory or tamping rollers. The top surface is scarify / roughened before the next layer toobtain an effective bond. The core material to have enough plasticity index to allow the coreto deform without cracking.

4.23.7 Rockfill Dam: Overflow and through Flow

• Flow through rock voids

• Unsupported d/s slope liable to erode and collapse

• Provide an anchorage system to support d/s face

• If larger rocks at d/s face, then no support needed






- use or not of marginal materials

- performance of selected materials during compaction operations

- correct type of compaction equipment for each material

- required number of compaction passes for each material- correct lift thickness for each material

- effects of particle crushing

4.24 FOUNDATION

4.24.1 Foundation Requirements

Foundation requirements for rockfill dam is more severe than earthfill dam but lesssevere than concrete gravity dam. Hard erosion resistant bed rock is most suitable. Rockfill

dams are not suitable in soft foundation of sand, silt and clay. Foundation with river gravel +rock fragments is acceptable (A positive cutoff must to bed rock to control seepage).Foundation is treated for minimum dam settlement. Filters to protect migration of fines fromthe foundation into rockfill.

For rock foundation grouting is done to seal-off rock imperfections. If geology of thefoundation is unknown, complete grouting is done, and shallow grouting may be enough ifgeology permits. For cobbles-gravel-sand foundation the under seepage through thefoundation is minimized by providing a positive cutoff. For a deep foundation a partial cutoff(concrete, metal, sheet pile etc) is provided in upper part with cement grouting beneath the

cutoff. Some grouting may be needed into the rock below the positive cutoff. Postconstruction grouting may be required depending on seepage measurements of first few yearsif reservoir can be drawn down to the bottom. Cutoffs Provides leakage control, facilitategrouting operation (as grout cap), provide water tight seal with membrane, and takedownward thrust of membrane4.24.2 Foundation Preparation

• Rockfill dams usually founded on some type of rock

• Rock may be exposed at surface or buried

• Stripping of shallow over burden of sand-gravel-cobbles foundation under thecore (3-5 m deep) to form a key trench. For other areas e.g. under the shells, thefoundations strength may be more than dam fill material, thus no need to strip orexcavate.

• Shallow clay-silt-sand foundation to be stripped for entire dam base (core andshells)

• Over hangs in foundation and abutment be eliminated

• Trimming/excavation not to damage bed rock




• Large depressions below desired bottom contours filled with dental concrete

• Foundation preparation is less severe under rock shells

• Prominent rock projection removed.

4.24.3 Grouting

• Minimize seepage through dam foundation

• Reduce hydrostatic pressure under d/s portion of dam (usually not a problem forclear rocks)

• Eliminate piping through dam foundation

• Blanket and curtain grouting in and adjacent to core foundation area

• Fractured/faulted rocks upper 30 ft blanket grouted to prevent piping of fines from

core into rock crevices

• Single or multiple line of grout under core

• Grouting pressure to avoid fracturing or moving of rocks.

4.25 SEEPAGE MEMBRANE4.25.1 Options

Seepage membrane is required to stop the seepage through the dam embankment.Central core (vertical or sloping/inclined) or u/s membrane are used for this purposes. The

materials for the membrane include reinforced cement concrete (RC), roller compactedconcrete (RCC), steel, timber, stone / rubble or PCC masonry, asphalt concrete for u/s facemembrane and earth/clay, reinforced cement concrete RC, RCC, steel, stone / rubble or PCCmasonry, asphalt concrete for central core dams. Economic and safety analysis is done tochoose type and design of the membrane. The advantages and disadvantages vary accordingto type, materials available, and foundation condition.

Advantages of Internal membranes

- Less total area exposed to water (due to steep slope)

- Shorter grout curtain length (shorter axis length at shortest line of damaxis)

- Potential safety from weathering and external damages

- Core location precisely known (a plus point when additional groutingworks may be needed in future)

Disadvantages of Internal membrane

- Simultaneous construction is must both for membrane and rockfill.

- Inaccessibility to inspection and damage repair




- Small dam base for stability against sliding

- Need flatter dam shell side slope if E/F core

- Filters/transition zone required for earth core

- Adequate construction control required if several filter zones are requireddue to coarse shell.

- Through and over flow not permitted

Advantages of u/s membrane

- Readily available for inspection and repair

- Membrane can be constructed after rockfill section

- Foundation grouting can be performed simultaneously with rockfill dam

- Large portion of dam base for sliding stability

- Membrane works as slope protection

- Dam raising easy

- Flow through dam body permitted during dam construction

4.25.2 Membrane Design Internal Core1: Earth Core

Impervious Central Core of Earth

• Enough quantity of earthfill available for core

• Used when u/s abutments widely apart in comparison to dam axis length

• Or show highly weathered rock to great depth and require adequate grouting/cutoff.

• Or higher elevation of abutment with deep layers of overburden thus trench typeinstallation less economical

• Design same as for earthfill dam, seepage and stability analysis required.

• Material placed in 6” lifts and compacted by tampering rollers

• Core material to have enough plasticity to allow it to deform without cracking ondam deflection.

• Filter zones provided (one or multiple zone of 8-15 ft thick)

• Foundations and abutments opposite to core be treated to prevent piping

• Joints, cracks, fissures in core area be cleaned out and filled with concrete orgrouted; additional future grouting of foundation can be carried from dam crest.

• Vertical side faces/overhangs of abutments trimmed to 1H:2V




• Bottom width 0.5 h to 2.5 h

• U/s and d/s slopes symmetrical (0.3 H:1V → 1.5H:1V), or u/s flatter than d/s face.

• Dam slopes as x+1 H:1V (minimum 2:1) [x = core slope]

• Chimney/blanket drain to drain off seepage flow (from the earth core and othercore / membranes.

• Location is central vertical position

Sloping Earth Cores

• These core located closer to u/s face, almost paralleling the u/s face

• Filter zones on u/s & d/s of core

• Provide more stability against sliding (provide better transfer of water pressure tofoundation and d/s shell)

• Usually thin width, width decreases at top

• Bottom width 30 to 50 ft

• Top width 15 to 20 ft

• Advantages: grouting cutoff can be at same of d/s fill placement

• U/S face: core = 1.4H:1V Dam: u/s face 2:1 or flatter, d/s face = 1.4:1

• Core can be placed after initial settlement of rockfill (less subsequent crackingrisk)

• Section better to pass flood flows as through flow (d/s anchoring needed)

1

1

1

1

> 2

2

1.4

1.9

Z

> Z

Figure 4.51: : Sloping earth core rockfill dams.




• Due to lower contact pressure at foundation → more susceptible to seepage and piping

• Additional grouting, if required difficult as foundation contact area likely underwater.

• Works as u/s earth face rockfill dam with face protected by dumped quarry rock.

Moderate Sloping earth core

U/s 0.5H;1V to 0.9 H:1 V, d/s – 0.5:1

• Moderately sloping core has clear advantage with respect to arching

• U/S dam slope can be made steeper than for extremely sloping core for stabilityreason

2 Other Materials for Central CoreLimited success due to rigid nature

Reinforced Concrete

• 6 ft at bottom to 1 ft at top

• Large deflection (e.g. at one dam 9 ft in 4years & additional 5 ft in next 38 years)

• Core cracking due to lateral movement of shell

• Use of concrete practically discontinued

• Concrete cutoff in foundation and extending partially into impervious earth coreused in modern dam to improve contact and seepage control and sliding safety.

Steel Diaphragm

• Centrally located

• Deterioration of steel due to water contact (oxidation, corrosion, potting, holes)

• Impossibility of repairing

• Limited used in few installation

Bituminous Material

• Used for small height dams

• Thin cores 40-100 cm in thickness

• Transition zone to provide uniform support and for filter, if any leakage/rupture

• Can be vertical or slightly inclined

Earth core

Concretecutoff




Figure 4.52: Top: Cross section of an asphalt core rockfill dam in Norway that wasconstructed with an only one meter thick. Very high quality control is necessary forsuch a thin core. Bottom: Construction.

[Source:http://cee.engr.ucdavis.edu/faculty/boulanger/geo_photo_album/Embankment%20dams/Zoned%20rockfill%20dams/Zoned%20rockfill%20-%20main.html]

http://cee.engr.ucdavis.edu/faculty/boulanger/geo_photo_album/Embankment%20dams/Zoned%20rockfill%20dams/Zoned%20rockfill%20-%20main.html







4.25.3 U/s face membrane

Concrete Faced Rockfill Dam (CFRD )

• RCC slabs placed at face over bedding layer

• Slab thickness and reinforcement requirements by experience, precedent and judgment

• Criteria

- Low permeability

- Sufficient strength to permit large subsided areas beneath the facing

- High resistant to weathering

- Flexible to adjust to small embankment settlements

• Best suited for compacted rockfill dams due to lesser chance of settlement anddeflection.

• Well compacting bedding layer (4+4+8 passes) reduce bridging requirements and provide more uniform support to the face layer.

• Concrete to be dense, durable, weather/chemical resistant

• Slab placed in blocks 20-60 ft square

• Horizontal + vertical expansion joints and construction joints are provided. Gaps

filled with flexible bitumen.

• Metal or rubber water stops (1 or more layers) in joints

• Concrete facing result in smooth surface and increase wave run up (but due tosteeper slopes, net run up may not increase much)

• Coping or parapet walls (5-10 ft) (Fig. 4. (a)) in continuation of face concrete toreduce height of embankment by containing wave run up.

• Concrete placement by slip forming process

• Shortcrete (roller compacted concrete RCC) may also be used

• Facing provided after dam construction (to allow dam settlement)

• Concrete facing anchored to the foundation cutoff wall through continuousreinforcement (Fig. 4. (b))

• May be anchored to flat bottom with dowel anchored footwall which also serve asgrout cap (Fig. 4 (c)).

• Slabs 20-60 ft square slip formed; Contraction joints horizontal

• Minimum temperature reinforcement 5%




• Thickness tapered t = 0.3 + 0.002h (min 1.5’ at base to 1’ at top) [KTD 6.6 ft to 1ft over 315 ft height)

Figure 4.54: concrete face slab construction work by slip forming. Note the reinforcement

and machine control. [@ http://www.dur.ac.uk/~des0www4/cal/dams/emba/embaf23.htm]

Parapet wall

Concreteface slab

Dam crest

Originalgroundsurface Foot wall min

1 m thick

dowelMin 1m

Grout curtain

Cutoff

(a) b cFigure 4.53 : u/s face concrete slab.




Asphaltic Concrete

• provides more flexibility and tolerates larger settlement

• Dam u/s slope 1.7:1 or flatter for easy placement

• Good bedding layer to eliminate uplift pressures and piping if cracks

• If bedding layer B zone not used , provide a 6” thick leveling layer to fill surfacevoids, provide easy travel of paving machinery, and smooth bedding surface forasphalt membrane

• Penetration coat over leveling layer to bind and stabilize it

• Membrane thickness 20 to 25 cm. Asphalt 8.5% by weight of dry aggregates

• Standard road paver used and asphalt placed in 3 layers

• Seal coat on the finished surface (for water proofing) and increased durability

• Placed in 3 to 4 m (10-12’) wide strip – at right angle to dam axis

• Paving placed on upslope pass only

• Rolling operation follow placement

• Smooth wheel rollers, vibratory or tandem type

• Layers compacted to min of 97% density

• Tight joints between adjacent strips• Transverse joints minimum and complete as hot joints

• Cold joints by (a) apply tack coat (b) overlap 10-15 cm (3-6”) joints (c) reheat joint with infrared heating (no open flames) (d) compact joints by rolling afterheating.

• Joints offset by 1-1.5 m (3-4’) from joints of bottom layers

• Formation cutoff allow easy placement

• Membrane must be durable, flexible, impervious, does not creep, and resistweathering

• Membrane material must satisfy: sieve analysis, immersion + unconfinedcompression test, Sustained load test, Permeability, Wave action test

• Special tests may be needed as: Slope flow, Coefficient of expansion, Flexuralstrength and Effect of reservoir ice

• Parapet walls may be used to contain wave action




Steel Face

• Used on few dams

• Performance satisfactory

• Can be rapidly constructed

• Can tolerate greater embankment reverts

• Disadvantage-probability of corrosion

• Cathodic protection on both faces of plates

• Proper maintenance can made facing as permanent

• Dam u/s Slopes 1.3-1.7

Original ground surface

Backfill

Cutoff

Grout curtain

Asphalt membrane

Rockfill embankment

Figure 4.55 Asphalt concrete membrane


Backfill

Cutoff min 1m

Grout curtain

Steel plateRockfill embankment

Figure 4.56 Steel membrane

Anchor dowel




• Steeper slopes construction difficulties

• Plate anchored to embankment by steel anchor rods grouted in bedding material

• Plate raised on a scaffolding, grid, bedding material placed after or during plate

construction• Plate thickness ¼-3/8”

• Jointed by bolts or continuous fillet weld,

• Expansion joints provided at regular interval

• Coping walls can be used to retard over splash

4.26 SEISMIC DESIGN

• Low seismic activity require no additional provisions

• Note: No exact rules for dam design in earthquake regions.

• Fact: Large d/s zone of quarried rock placed in thin layers provide maximumstability

• For Moderate seismic activity areas, provide:

- Large d/s zone of good quality rock

- D/s slope flattened to 1.7:1 in all cases

- For additional conservation u/s slope may also be flattered

- Foundation must be firm rock/blanket grouting

- Free draining cobbles/boulders/rock fragments (if compaction same asrockfill) may be used.

- Provide trench type cutoff

- Provide thicker bedding zone

- Use better quality rock in C zone (routine is random fill of poor qualityrock)

- Limit lift thickness to max of 3’ in zone D

- Use a thicker membrane on U/S.




REFERENCES AND BIBLIOGRAPHY

Breitenbach 2007. History of rockfill dam construction: Parts 1 and 2. @http://www.geoengineer.org/ rockfill1.htm, rockfill2.htm (as on 16 Jun 2007)

http://www.geoengineer.org/






QUESTIONS

1. A homogeneous dam has following data. Total height = 80 m, Free board = 5 m, u/sface slope = 2.5:1, d/s slope face = 2:1, Crest width = 5 m, Foundation thickness = 25m, K of dam fill material = 5 x 10 -6 m/s, K of foundation material = 2 x 10 -7

2. The attached Figure shows section of an earthfill dam at its maximum depth showingsteady seepage phreatic line, equipotential lines, a trial slip surface. Scale: 1 block =5x5 m.

m/s. Tailwater depth = zero. Draw seepage flow net and determine the seepage rates throughthe dam and foundation. Assume seepage from dam do not enter into foundation andvive versa. Also determine uplift pressure at base of the dam.

Dam height = 60 m, depth at normal conservation level = 55 m, U/s slope = 2:1, d/sslope = 2:1, Core uniform width = 20 m, core height = 55 m. The material propertiesare as:

Core: c’ = 12 KPa, φ’ = 29 °, average unit weight γ = 21 KN/m3

, K = 3 x 10-5

Fill: c’ = 5 KPa, φ’ = 33 °, γ = 19 KN/m cm/s.

3, K = 5 x 10 -4

Determine the factor of safety for the d/s face for steady seepage condition for theshown slip surface by using method of slices .

cm/s.

9

1 2

1 4

1 6

1 8 2 0

2 2

2 4

2 6 2 8

3 0

3 0

3 2

3 4

3 6

3 7

3 8

3 9

4 0

4 0

4 2

4 4

4 5

4 6

3. Repeat Q-1 by using SEEP/W program.4. Repeat Q-2 by using SEEP/W and SLOPE/W computer programs.







Figure 4.61 : Dynamic compaction, Townsville Australia.




Figure 4.62: Mangrove Creek concrete faced Rock Fill dam

(http://www.ipenz.org.nz/nzsold/NZSOLD-Newsletter-46-Text.pdf )

Figure 4.63: 125 m high Storglomvatn Rockfill dam with asphalt concrete core

( http://www.ngi.no ) 125 m high RF dam with asphalt concrete core.

http://www.ipenz.org.nz/nzsold/NZSOLD-Newsletter-46-Text.pdf



http://www.ngi.no/

http://www.ngi.no/

http://www.ngi.no/

http://www.ngi.no/





Figure 4.64:

Zoned Rockfill Dams in Norway Dams constructed of soil, of rock, or of acombination of soil and rock are called embankment dams. Embankment dams are most

economical where the materials at the dam site can be used to construct the embankment withlittle or no processing. Small embankment dams can be built of a single type of soil, whichmust hold back the water and provide enough strength for stability of the embankment. Largedams are usually “zoned,” with fine soils (silts or clays) at the center of the dam (the “core”)to hold back the water, and sand, gravel or rockfill in the upstream and downstream parts ofthe dam (the “shells”) to provide the strength needed for stability of the embankment.This

photo shows a cross section through a zoned rockfill dam in Norway. The yellow zonesdownstream from the green core are the filter and the drain. The filter is graded to hold the

particles within the core in place, while allowing seeping water to pass freely. The drains

have high enough permeability to carry the seepage without allowing any significant porewater pressures to develop in the downstream parts of the dam. The rockfill shells are shownin orange. The gray zone between the drains and the shells is a “transition zone,” used tomake economical use of as much of the soil at the site as possible. The steep slopes indicatethat the rockfill of which the shells are constructed has a high angle of internal friction.(http://cee.engr.ucdavis.edu/faculty/boulanger/geo_photo_album/Embankment%20dams/Zoned%20rockfill%20dams/Zoned%20rockfill%20-%20main.html )

http://rds.yahoo.com/_ylt=A9G_bDoYU1ZILu8A.gyjzbkF/SIG=158vq7chi/EXP=1213703320/**http%3A/cee.engr.ucdavis.edu/faculty/boulanger/geo_photo_album/Embankment%2520dams/Zoned%2520rockfill%2520dams/Zoned%2520rockfill%2520-%2520main.html









Figure 4.65: Mohali dam, Lesotho (http://www.trc.org.ls )

Fig 4.66: Kouris EF dam, Cyprus. (http://www.flickr.com/photos/lemessoler/2155179591/ )

http://rds.yahoo.com/_ylt=A9G_bF_uVFZIsF0B1tyjzbkF/SIG=12104cvd7/EXP=1213703790/**http%3A/www.trc.org.ls/news_pages/damcrack.htm



http://www.flickr.com/photos/lemessoler/2155179591/








Fig 4.67: Windamere Dam (69 metres high, earth and rock fill dam) on the CudgegongRiver near Mudgee. Geotechnical problems included excessive grout takes in highly fractured

rock in dam foundation. The spillway was an unlined cutting in andesite about 1 km from thedam site and supplied the entire rock fill requirement for the construction of the damembankment. If a spillway had been built in the weathered sedimentary rocks at the dam sitefull concrete lining would have been required.

Windamere Dam earth and rock fill embankment. The dam foundations are weatheredDevonian conglomerates, sandstones and shales. The spillway is located about 1km awayfrom the dam in mostly unweathered Ordovician andesite. The spillway is an unlined rockcutting that provided all the rock fill required for the construction of the dam embankment.The access road bridge over the spillway cutting is just visible in the upper left of the photo

http://members.optusnet.com.au/~richardw2/img/wind3.jpg







Fig 4.68 : The dam wall of the Upper Yarra Reservoir which is a rolled earthfill and rockfillembankment and is 610 metres long and 90 metres high.

(http://www.flickr.com/photos/splatt/357903508/ )

http://rds.yahoo.com/_ylt=A9G_bF9ANmJIVkMBigyjzbkF/SIG=1222d4b9a/EXP=1214482368/**http%3A/www.flickr.com/photos/splatt/357903508/







Chapter 4


4.1 DEFINITION





I: EARTHFILL DAMS































• 1










H/3

H

See a e






seepage flow before it reaches the d/s slope. Such a dam is then called as modifiedhomogeneous dam (Figs. 4.2 to 4.4). These measures include rockfill toe, horizontal d/sdrainage blanket, and a vertical or inclined chimney drain. These measures do not decreasethe seepage amount but makes the seepage exit safer with no danger of dam toe failure. A toe

drain (Fig. 4.5) is usually used in conjunction with these seepage control measures tointercept the seepage flow and its disposal.

A homogeneous embankment should not be used for storage dam. A homogeneoustype of dam is applicable in localities where readily available soils show little or no variationis permeability and soils of contrasting permeability are available only in minor amounts or atconsiderably greater cost.

Figure 4.3 : Modified homogeneous dam with d/s horizontal drainage blanket.

H

See a e


Figure 4.4: Modified homogeneous dam with a chimney drain and d/s horizontal drainage blanket.

H

See a e


H

See a e

Figure 4.2 : Modified homogeneous dam with rockfill toe and graded filter.












Dam d/s slope

HU/S

ShellCORE D/S

ShellSee a e

Filter








4.4.3 Diaphragm Dam









HU/S

Shell C O R E

D/SShell

See a e

Filter










Vertical Core





Criteria




failure.

Inclined Core


H

U/S

Shell C O R E

D/SShell

Filter





Advantages




Disadvantages
































4.5.3 Chimney Drain






4.6 FILTER CRITERIA



# mm # mm # mm # mm # mm3 6.4 10 2.0 25 0.71 60 0.25 200 0.0744 4.8 12 1.7 30 0.59 70 0.21 270 0.0535 4.0 16 1.19 35 0.50 100 0.149 300 0.0506 3.4 18 1.00 40 0.42 140 0.105 325 0.044

8 2.38 20 0.84 50 0.297 170 0.088 400 0.037



Fine graded filter



Toe drain






2/[d 60 xd 10

1. D


15 /d 15

2. D

= 5 to 40

15 /d 85

3. D


85



U







85 finer

* D 15 /d 15

* D


15 /d 85

* D


50 /d 50

* D


60 /D 10














Given: d 15 = 0.006 mm, d 85

D


15

D


15

D


15


≤ 5 * 0.10 = 0.50 mm (3) [criteria 2]



85

D

= 2.4 mm.(4)

85

As D

≥ 2 * 0.5” ≥ 1” (5) [criteria 3]


D

layer F2.

15

D




F 1

F i l t e r

F 2



F 1






D 15 ≤ 5 * 0.24 = 12 mm (8) [criteria 2]


Core F1 F2F1F2

U/s fillD/s fill

F1F2 CORED/s fill




Figure 4.10.3 : Mangla dam raising project showing core, and double filter layers.

Constriction of chimney filter, Mangla dam raising project.




From eq 7 and 8 select smaller size, D 15 upper = 5.6 mm; Average D 15

Draw filter gradation line parallel to base material gradation curve and read D

= 4.0 mm.

85

D

= 50 mm. (9)

85

Select D

≥ 2 * 0.5” ≥ 1” (10) [criteria 3]

85 from eq 9 and 10 as 50 mm. Then F2 = [D 15 = 4.0 mm, D 85

Dimensions and Permeability of Toe/Blanket/Chimney Drains

= 50 mm.

The dimension and permeability of the drain must be adequate to carry away theanticipated flow with an ample margin of safety for unexpected leaks. For a relativelyimpermeable foundation, then the expected leakage would be low.

A drain should be constructed of material with a coefficient of permeability of at least

10 to 100 times greater than the average embankment material.

Drain material is usually a processed material. Pit run borrow is usually too dirty (i.e.have large fines). Drain materials must have following grading.

Particle size % passing by weight

1½″ 90 – 100¾” 45 – 75# 4 (4.8 mm) 30 - 45# 50 (0.297 mm) 4 - 10

# 100 (0.149 mm) 1 - 3




# 200 (0.074 mm) 0 - 2

Gradation should be such that it will prevent particles of soil from the adjacent location fromentering the filter and clogging it.

4.7 FOUNDATION DESIGN

Foundation includes both valley floor and the abutments. Foundation must ensurefollowing design requirements

1. It provides support for the embankment under all conditions of saturation andloading.

2. It provides sufficient resistance to seepage to prevent excessive loss of water.

• Foundation is not actually designed but treatments are provided in design to ensurethat all essential requirements are met.

• No two foundations exactly alike, each presents its own separate and distinct problems. Foundation improvements be adopted to local conditions.

• 40% dam failures attributed to failure of the foundation.

• Judgment on the basis of foundation exploration and past experiences.


• Infinite variations in the combinations (materials), structural arrangements and

physical characteristics of the constituent materials.

















• Vibrations/shock as for an earthquake tremor causes re-adjustment of grains into adense structure. Pore water pressure increases suddenly (due to slow drainage) and

foundation behaves as liquid and results in sudden liquification.• Cohesionless sands of low relative density (< 50%) are suspect to failure.


The foundation is treated to minimize the seepage through the foundation and reduceuplift pressures for d/s part. Various foundation treatments include positive cut-off, partialcutoff, sheet pile, cement bound curtain, concrete wall, slurry trench, grouting, etc. These aredescribed below.

4.9.2.1 Positive Cut-off Trenches:

Rolled earth/clay is filled and compacted in a trench excavated to the impermeable barrier / underlying hard bed rock (Fig. 4.11a). The compacted clay forms an impermeable barrier to the seepage flow. The cutoff depth varies as 50 to 150 ft with 1:1 or flatter sideslopes. It is located in continuation of the embankment core u/s from centerline of dam crest,

but not beyond where cover of core becomes small. It is made of usually same material as is

HU/S

ShellCORE D/S

Shell

Figure 4.11a: A positive cutoff for earthfill dam.

Gravel sand foundation

Bed rock

OverburdenRiver

bottom




suitable for dam core. Wider trench base is adopted for dams with large depth. For deepertrench smaller base may be used as seepage force at foundation contact decrease withincrease in depth. Grouting of upper part of weathered/fractured bed rock, if required.Generally top width as w = h – d. A minimum bottom width ≈ 20 ft to allow operating

machinery. Trench below water table will require dewatering.4.9.9.2 Partial Cutoff

The cutoff penetrates only partially into the foundation (Fig. 4.11b).

• Suitable if a low K layer of considerable thickness found above the bed rock. Thislayer must be aerially extensive. Thus seepage from upper more pervious layer isintercepted.

• Partial barrier be at least 95% deep to have any appreciable reduction in seepage.

• Partial seepage barrier may be effective at sites where average permeability offoundation decreases with depth.

• For deep foundations the upper part is sealed off against seepage by providing a partial cutoff and lower part may be sealed by providing sheet piling or groutingetc below and in continuation of the partial cutoff.

• In all cases a minimum partial cutoff of 6-10 ft should be provided. This trenchalso provided better understanding of the subsoil conditions.

4.9.2.3 Sheet Piling Cutoff

Steel sheet pile may be driven into soft alluvium.

• Depth to bed rock.

• Used in combination with partial cutoff to seal lower horizons.

HU/S

ShellCORE D/S

Shell

Fi ure 4.11b: A artial cutoff for earthfill dam.

Deep gravel sand foundation

Bed rock

Sheet piling or grouting etc




• Not suitable for cobbles/boulders as these formations cause misalignment/ open joints, interlock liable to tear-off, pile wander off, pile twisting making anineffective barrier.

• Twin steel sections may be used with interior filled with cement grout.

• Not completely water tight

• 80-90% effective if good work

• Poor workmanship, efficiency less than 50%.

• Seepage resistance offered by sheet pile equals 30-40 ft length of soil; field testsshow resistance equivalent of 400-2000 ft. The effectiveness increase with timedue to filling of gap by sediments, encrustation etc.























bottom.


U/s Mid D/s






positive cutoff.





• RCC or PCC








Plan

Section


River level
























































Q = K I A I = ∆h/L











L

d




























































Bed rock

Drainage ditch


Seepage flow











A: Shallow Foundation• Provide a positive (complete) cutoff to bedrock.














• Provide drainage blanket of filter grade if i) overlying zone is impervious or ii)

overlying zone is rockfill, iii) piping potential is present







• Provide u/s impermeable blanket in continuation of impermeable core.

• Minimum core B



pressures.


• Provide toe drains

• For foundations of high K, which cause extensive seepage, ponding and sand boils, then provide drainage trenches, pressure relief wells, extension of d/s toe ofdam or blanket on d/s area.



























• Saturated impervious sands (dirty sands - sands having good amounts of fines)also act as fine grained soils






• At C



• At W = PL, cohesive strength varies considerably; φ is obtained from triaxial teston samples compacted at proctor maximum dry density.

is obtained by slow shear test onsaturated soil (drainage permitted, pore water pressure ≈ zero) tan φ about 0.5

Treatments:










c. Flattening of slopes lengthen the surface of sliding, decreases averageshear stress along the path and increase factor of safety against sliding(Fig. 4.18).

4.10.1.2Relatively Dry Foundations• These soils exhibit large strength at its present dryness







h/2

h





4.10.2 Treatment/Improvements of Fine Grained Foundation

Foundation of dams can be improved by: 1) Pre-consolidation, 2) Densification of

cohesionless soils, and 3) Dynamic compaction

Pre-consolidation

• Useful in compressible soils

• Done by applying artificial surcharge such a soil removed from stripping andscaling of abutments may be piled up

• Allow time for water to drain

• For rapid rate (1-2 months for 50% consolidation) piling of random weight is

useful.

• For slow rate soils, dam weight is used to consolidate the soil. This requires slowconstruction rate and providing drainage. Longer time periods (1-2 years for 50%consolidation) are necessary.

Densification of Cohesionless Soils

This is carried out using shock and vibration. Vibrofloatation is used to improve poorfoundation. This can reduce settlement as much as 50% with substantially increased shearingstrength. Vibrations convert loosely packed soils into a denser soil.

Vibroflat can be used to penetrate the soil and operate below the water table. Bestresults are obtained in coarse sands which can contain little or no silt or clay.

Dynamic Compaction

This is repeated application of very high intensity impacts to the surface. Thisimproves the soil mechanical properties. Compaction is done by dropping a weight, typically10-20 tones from heights of 10-20 meters at regular interval across the surface. Severaltamping/passes may be made at the site. Each imprint is backfilled after tamping. In finersoils increased pore water pressure must be allowed to dissipate between passes, which maytake several weeks.

Stress σ

S t r a i n

ε

Unconsolidated dry sampleUnconsolidated wet sample

Preconsolidated dry sample

Preconsolidated wet sample

Figure 4.19: Consolidation of wet/dry soils.

Sudden consolidationof dry soil on wetting










Dry Foundation






4.11.1 Crest Design

Crest width W

• The width W of the crest is governed by height of dam, importance of structure,

width of highway, construction procedure, access required either duringconstruction or as a permanent feature.


• Special widening may be necessary to provide ahighway or safeguard against freak waves etc. Thiswidening could be done by steepening face slope in




Figure 4.19b

h/2

h

3:1 or flatter

Random fill






W = 0.55 √h + 0.2 h h< 30 m

W = 1.65 (h+1.5)1/3


h>30 m (USBR 2001)


Sherard p-413: For earthquake area, top of dam is subjected to worst damage and canvibrate with greater amplitude than the base. Thus it is advisable to make dam top thicker byincreasing crest width or using flatter slopes near top. Also that if any crack develops, the

longer seepage path causes less seepage and increases dam safety.

Surface Drainage


Surfacing



Safety Requirements


Camber








The procedure for designing a cross-section of earthfill dam consists largely ofdesigning to the slopes and characteristics of existing successful dams , making analytical andexperimental studies of unusual conditions and controlling closely select ion and placement of





The u/s and d/s slopes of the embankment and core are selected from generalguidelines, experiences in the light of foundation materials and materials available forconstruction. The seepage analysis and stability of the selected dam section is carried out anddam section may be acceptable if factor of safety for the dam under different construction andoperation conditions are found satisfactory. Alternate dam sections are evaluated for materialneeds/crest and factor of safety and that dam section is adopted which provide higher factorof safety at lowest costs. Stability of the shape is analyzed under static loads as well as under

seismic conditions.


Camber ~ 4

Camber


W

ELEVATIONX-SECTION

1.75:1

2:1




Except where there is surplus of material available from required excavations, themost economical dam is obtained with the minimum volume and therefore most steep slopesconsistent with the dam stability (Sherard p-48). The allowable steepness depends on theinternal zoning and on strength of foundations and the embankment material. Crest length

and pace of construction may also affect the slope selection. Use of excavated material asrandom fill may allow flatter slopes. The random fill material may be placed (Fig. 4.21) at



The strength of foundation is also affects the dam face slopes: Weak foundation –average slope 2:1 to 4:1; Strong foundation – steeper slope 1.5:1 to 3:1. The height of damalso affect slope selection. For homogeneous materials dams of fine core: Short height –




• Dams located in narrow rock-walled canyons can be constructed with some what

steeper slopes than otherwise, because of added stability given by the confiningwalls. In narrow valleys broad toe berm or very flat slopes at the toe of dam can





















Diaphragm Type



u /s d/s u/s d/sGW GP SW SP Materials not suitable -too perviousGC GM SC SM 2½:1 2:1 3:1 2:1

CL ML 3:1 2½:1 3½:1 2½:1CH MH 3/6:1 2½:1 4:1 2½:1

Zoned embankment











Impervious Core












Maximum Core:


Slope = y:1

Slope = y-½:1

Slope = 1:1

Slope = ½:1

1½:1

x-½:1

x:1

Z

Z









•


Type Shell material Core material No rapiddrawdown

Rapiddrawdown

U/s D/s U/s D/sMin core A Rock, GW,

GP, SW, SP,gravely


2:1 2:1


GC, GM 2:1 2:1 2½:1 2¼:1

SC, SM 2¼:1 2¼:1 2½:1 2¼:1

CL, ML 2½:1 2½:1 3:1 2½:1

CH, MH 3:1 3:1 3½:1 3:1



Rocks under alluvium and abutments.

Abutments: Metamorphosed sedimentary rocks (sugary limestone, phylite, quartzite, schist)




U/S Slopes: Protection is required against destructive waves splashing onto the side slope.Waves generated due to high sustained winds as well as from earthquake action. Also neededto be protected against burrowing animals.

D/S Slope: These need to be protected against erosion by windand rainfall runoff and the borrowing animals. This also needs to

be protected against possibility of seeping of rain water andforming internal erosion (piping and sloughing of inside of

embankment).




I: UPSTREAM SLOPE

U/s slopes are provided protection by: rock riprap, concrete pavement, steel facing, bituminous concrete pavement, pre cast concrete blocks, others as short cement pavement,wood, sacked concrete. Special care is needed against beaching process if water level stays at

one elevation for long times.

Rock Riprap



Dumped Riprap

The rock fragments/stones are dumped over the slope. The efficiency of dumped rock

riprap depends on following: Quality of the rock, Weight or size of individual stone pieces,Thickness of the riprap, Shape of stones or rock fragments (rounded, angular), Slope of theembankment, Stability and effectiveness of the filter.


Individual pieces should be of sufficient weight to resist displacement by waves (mustfor all size dams). The thickness of riprap should be sufficient to accommodate weight andsize of stones necessary to resist wave action (Fig. 4.22). A 3’ minimum thickness is used.Smaller thickness if wave action is less severe. Lesser thickness may be used for upper slopescorresponding to flood control storage (above normal conservation level) due to infrequentexposure of this part to waves. If there is any damage to this section, it can be repaired onreservoir lowering.


Shell material(min 4-6 ft toenable placing)
















Figure 4.22f: Simly dam: u/s slope protection by angular rock riprap.

Figure 4.22g: Tanpura-I dam: u/s slope protection by rounded rock riprap.





Figure: D/s face protection, Dharabi Dam, Potohar, Chakwal.




Riprap weight

(Zipparro eds. Davis’ Handbook of Applied Hydraulics, p-13.58) defined the riprapweight as:

( ) ( )ba

Cot GK H W

θ γ 3501−

=


W

/8

50









3 where M=mass of stone required (kg), and H s

The size of riprap is estimated as : D = [7 W / 5 γ]

=significant wave height(m).

1/3 where D = stone size (ft), W = stone

weight, γ = bulk un i t weight (lbs/cft). The th i ckness must be more than size ofheaviest/largest stones. In no case it should be smaller than 1.5 x D 50

USBR p-277 provided gradation (by weight in lbs) of riprap for slope = 3:1 and angular rocksas:

or 24”


< 2.5 miles 30” 2500 lbs 1250 75-1250 75

0-10

> 2.5 ml 36” 4500 2250 1000-2250 100






Shape of Rock

Shape of rock fragments influences the ability of riprap to resist displacement bywave action. Angular fragments tend to interlock better than boulders and rounded cobbles.Thus rounded stones should have more thickness.




Graded Filter

A layer or blanket of graded filter should be provided underneath the riprap if there isdanger of fines from underneath layer to more into the riprap layer by wave action. For azoned dam filter not needed if outer shall is gravel. Blanket of crushed rock or natural gravels

3/16” to 3 1/2”

Flexibility



Placement

The riprap is dumped from hauling trucks onto the prepared surface. Bulldozers are

used to level off-and compact the dumped layer to fill up the voids between larger stones.Smaller stones fit in voids of larger pieces very well. The rock stones must not break duringhandling / placement / compaction. Top surface is uneven, rough and decreases wave riprap.



This consists of suitably sized stones carefully laid by hand in a more or less definite pattern with minimum amount of voids and with top surface relatively smooth. Doubled or

irregular shapes lay up less satisfactorily than stones of roughly square shape. Stones of flatstratified nature should be placed with principal bedding plane normal to slope. Joints should

be broken as much as possible and voids be avoided carefully by arranging various sizes ofstones and small rock fragments.

The stones of excellent quality should be used. Thickness can be half of dumpedriprap but not less than 12”. Filter blanket be provided underneath the riprap, if required. Dueto tight packing, hand placed riprap is not as flexible, so it cannot adjust to foundation orlocal settlement. Thus hand placed riprap should not be used where considerable settlement isexpected. Hand placed riprap could be costly due to extensive labor cost in spite of its smaller

thickness.

Concrete Paving

Concrete is placed over the sloping surface to resist wave destruction. It can be used both for rockfill and earthfill dams. Paving thickness depend on dam height, slope steepness.Thickness is 8” for h ∼ 50 for and 12’-18” for high dams. Paving is placed in blocks 6’ x 6’ ormore but monolithic construction gives the best service. A water tight surface will eliminatehydrostatic pressure underneath the pavement. Blocks could be displaced or broken by waveaction and uplift forces under the slab. Concrete can crack requiring frequent maintenance.




For blocks, expansion joints and construction joints should be widely spaced.Reinforcement is (5% area) in both directions and be continuous through the construction

joints. Joints be sealed with plastic fillers and cracks be grouted and sealed properly.Pavement should extend from crest to below the minimum water levels. It should terminate at

a berm and against a deep seated curb or header (minimum 18” deep).The success of concrete pavement is mixed, but successes and problems have been

observed. Paving is a costly alternate, but may be adopted if enough riprap material is notavailable. Concrete pavement may or may not be held in place firmly by foundation boltsembedded deep inside sloping shell. Concrete paving increase the wave runup and suitablewave breakers, wave deflectors, may be provided to reduced risk of dam overtopping. Failurechance is 30% + due to inherent deficiencies in this type of construction.

Soil Cement


be added for erosion resistance. For most soils 10-12% cement (% of compacted volume) isconsidered typical. Cement and moisture ratio is determined by lab tests.

Soil cement is placed in 6-8” horizontal layers over the slope (horizontal width as 8 ftnormal thickness 2-3½ ft) and roller compacted in a stair-step horizontal layers. Soil cementshave 500-1000 lb/inch compressive strength at 7-day (10% cement). The edges of the cementlayers are not trimmed to retard wave runup. A reasonably firm foundation is required so thatdeformation following placement of soil-cement is not significant. Normal embankmentconstruction procedures are satisfactory.

8-10 ft

2-3.5 ft








The zoned dams with d/s shell of rock or cobble fill do not need additional slope protection. Slope protection is required for all conditions against erosion by wind and rainfall.If not protected gully can develop.

This protection is provided by placing a layer of rock, cobbles, orsod (grass). However, vegetation


provided to collect and dispose the rain runoff from upper portions of dam slope. Drainage

channels discharge into cement lined channels running down the slope and ultimately to safedisposal point/river bed. A contour drain is also provided along toe of dam. Surface drainageis also provided (as an open gutter) for abutments and valley floor.

4.11.4 Abutments


The u/s and d/s slopes of embankment are often flared at abutments to provide flatterslopes for stability and seepage control. The u/s flaring is equivalent to providing u/simpermeable blanket. The flaring design is governed by topography of the site, the length of

Cut slope Berm




constant desired, for aesthetic value, and ease of construction. For steep side slopes this may be useful to locate access road across the dam.

ABUTMENT SIDE SLOPES





process of consolidation . Relationships in the form of pore-pressure coefficients are used todescribe immediate response of pore water pressure to applied total stress.Shear strength

The shear strength of a soil is defined as the maximum resistance to shearing stresswhich can be mobilized; when this is exceeded failure occurs usually along identifiable slip

surfaces. The shear strength of any material is described by Mohr-Coulomb failure criterion based on total stress as: S = c + σ tan φ or based on effective stress as: S = c’ + σ’ tan φ’ σ =

total compressive stress ( σ = σ’ + u), σ’ = effective stress, u = pore water pressure.Laboratory shear tests, e.g. triaxial shear test, are run for the material compacted to the designdensity / moisture content and construction of Mohr circle plot. Coarse soils such as sandsderive their shear strength largely from particle interlock and internal friction, and are calledas cohesionless (c=0) or frictional soils; the shear strength is mostly controlled largely soildensity.y . Most clays soils derive shear strength from both cohesion and internal friction.Following tests are usually carried out. (Sherard p-332)

Remove overhangs






1. Undrained test: (unconsolidated – undrained test). No drainage and dissipation of pore water pressure. Called as Q-test (quick test). Used for stability analysis fordam during and after construction.

2. Consolidated-undrained test. (sample first consolidated with full pore water

pressure dissipation under given consolidation pressure) and then is failed in shearwith no drainage allowed. This is called R test.








v = k/m vγw


(k=permeability)

Compaction




permeability.



)


(x10

-4 m2


permeabilityk /kN) h (m/s)

Cohesionc’ (kN/m 2

Friction, φ’(degrees))

Gravels 17-22 0 30-45 0.1 – 1.0 10 -1 – 10 -2

Sand 0 30-45 10-

– 10-




Silts < 5 20-35 10 - – 10 -

Clay (soft-medium)

15-21 0 20-30 1.0 – 10.0 Intact clays,< 10 -8, ifweathered,

fissured, orwith siltlenses 10 -3 –10 -8


< 10 < 30


< 50 < 20

4.12 SEEPAGE ANALYSIS

Seepage flow will occur through all types of formation irrespective how small the permeability may be. Seepage occur both through the dam embankment as well as damfoundation. Seepage flow is given by Darcy’s law q = K I per unit flow cross section and asQ = q A = K I A = K ∆hA/L through section of area A.

Flow net method is used for simple flow conditions. Seepage occurs as confined flowthrough the foundation and as unconfined flow through the embankment. A flow net isdrawn with curvilinear squares; different squares may have different area but all have flowlines and potential lines cross at right angle and that all four sides are of equal curved length(Fig. 4.25). The flow is given as

Q = K ∆H N f / N


d


N f

N

= No. of flow lines

d



Flow net provide


2. Distribution of water potential (h) and pore water pressure u (u = h – z ; z =elevation and Note: u, h, z has units of length and are taken from a pre-selecteddatum). Also u = P/ρ g = P/ γ, P = water pressure, ρ = water density, γ = sp weight.The water potential h and pore water pressure is given in units of pressure head;this should be multiplied with unit weight of water (γ = 62.4 lbs/cft, 1000 kg mass





/m3



face chimney drain




Seepageexit face

≈ h/3

h



∆h

δh=∆h/m

B

H C

0.3 LL





1. Draw a to-scale map of dam and foundation

2. Determine the seepage exit area

3. The u/s face of the embankment is plane where seepage originates.4. The d/s exit area is usually up to a height h/3 for a homogeneous dam (h = dam

height). For a modified homogeneous dam, the flow will converge towards the toedrain or the horizontal blanket drain. Flow emerges tangentially to the exit face ford/s face or chimney drain or vertical line for rockfill toe or drainage blanket (Fig.4.27).


6. Establish the seepage line/phreatic surface. The seepage line intersects the

reservoir water surface at a distance 0.3 L from the point C (point C is at watersurface at u/s face) where L is the horizontal projection of the u/s face (Fig. 4.26).The actual phreatic line is modified to meet at point C.




d/s

d


= h/ δh.

10. Draw a flow line in the middle part of the flow area keeping it normal to potentiallines but approximately curvilinear and parallel to seepage line such thatapproximate squares are formed. Extend this flow to the originating surface (i.e.u/s face) and to the exit surface (i.e. d/s face or toe/blanket drain). Draw moreflow lines to the bottom confining layer.

11. The effect of confining surface on the shape of flow lines and equipotential lines

diminishes farther from the confining surface.



14. For large differences in K of the two materials of embankment and foundation theflow lines can cross from foundation into upper embankment or fromembankment into bottom foundation (the material of higher K will attract flowlines from other material). For small differences in K the flow lines originating




from any material will continue within the same material and will not cross intothe other material.









• With F as center, Q-H as radius R, cut PQ vertical at P. The distance P-Q = y.(x,y) are the coordinates of the seepage line parabola. Draw other points similar toP. The seepage line meets at C.


Seepage rate


T hT K Q −+= 22

A

B C 0.3 LL

F G HQxR = x+S

R

P(x,y)

D E


y

S

h

T

Directrix




Seepage Through Dam FoundationSeepage through the dam foundation is also determined by drawing flow net for the

foundation section. The flow net is drawn by procedure similar to for the embankment. Theseepage control measures are also considered while drawing the flow net and determining theseepage rates.



ft/sec, Fig. 4.31.


/s/ft.}


(a)





Blanket drain thickness:

The thickness of the blanket drain required to pass seepage discharge q per unit width is as:

d d d K K H K Lqt 15.1≈= where L is shoulder width at drain level, H is the reservoir

water depth above drain level, K 1 = permeability of dam core material and K d

Seepage Analysis by Computer Software

is permeabilityof drain material.(Novak et al. 1998, p-61)

Computer software are available (e.g SEEP/W) to determine seepage flow through thedam embankment and the foundation. These programs are user friendly and easilyincorporate the seepage control measures, the seepage exit conditions, varying material

properties, etc. The program result provides seepage flux through selected sections, equi potential contours, distribution of potential/head, seepage flux vectors, and seepage flowlines, etc.

Permissible Seepage

Seepage control measures are provided to reduce seepage quantity and the uplift pressures due to seepage flows. The seepage quantity is usually not very large and theseeping water could be used beneficially at some downstream location. The uplift pressuresare usually more critical in the stability of the dam structure. Thus seepage control measures

are sized to achieve acceptable gradients in the flow domain. Cedergen (1967) defined the

(c)

(b)




acceptable average hydraulic gradients exerted by the water seeping through the embankmentand foundations should not exceed the following critical values.

Impervious core 1:4

Impervious blanket 1:15

Alluvial foundation 1:0.066 (15:1)

Rock foundation 1:1

Materials placed around seepage water collection system to comply with filter criteria.

[Source: Cedergren, H. R. 1967. “Seepage, Drainage and Flownets”, 2 nd

Gradient method: 1) determine average hydraulic gradient in soil element, 2) determinemagnitude of seepage force (F = 62.5 I V, V = element volume), 3) determine direction ofseepage force, 4) line of action of seepage force.

edition, John Wiley& Sons Inc., New York] p-115, 16.

4.13 STABILITY ANALYSIS

Dam sections are analysis for safety against failure by shear and sliding. The slidingof dam can occur at base of dam or any height above the base. Most earthfill dams built withgood material and placed on sound foundation are considered safe against sliding. The safetyagainst shear failure is analyzed for the following conditions: (Sherard p-326)

1. During and after construction for both u/s and d/s faces. Assume pore pressurehigh and not drained; Analysis is based on lab Q – test

2. Full reservoir steady seepage – d/s face; Analysis is based on lab S - test



A factor of safety is determined for various situations. The dam section is accepted if thefactor of safety for the selected loading condition is higher than recommended values. Elsethe dam section (i.e. side slope of core and shell and materials) is revised and safety re-evaluated.





Procedure



• Shape and location of the slip surface chosen arbitrarily.

• Material above the selected slip surface is called a trial sliding mass

• The trial sliding mass divided into 8-10 slices as in Fig 4.33. (dam unit thick)

• Width of each slice adjusted so that entire base of a slice is located on a singlematerial and chord length ΔL does not significantly differ from arc length.

• Available shear force from material properties S = c + σ tan φ is determined along base of selected surface

• Actual shear force from loading conditions determined.

• Factor of safety F S

• Procedure is repeated for other potential failure surfaces until a critical surface

obtained with lowest factor of safety.

= Shear strength force available ÷ Shear force applied







• Analysis is based on shear strength derived on the basis of total stress S = c + σ

tan φ, or effective stress as : S = c ′ + σ′ tan φ′, where σ′ = σ - u.

• For each slice of bottom width b, compute forces as:

a) Total weight W of the slice. W = area of slice x slice thickness (unit) x grassunit weight (Soil + water). For same height of the two sides of the slice area =D x b. For marked difference in length of two sides, area of trepezoid as A =(D L + D R


)/2 x b.

c) Tangential component of weight: T = W sin α

d) Total water potential h acting on the slice bottom is determined from the equipotential contour map. The pore water pressure head (units of L) is thendetermined as u = h – z, where z is the elevation of the bottom from selecteddatum. Total pore water pressure head U acting on bottom of slice as: U = u x

b/cos α x 1 = average pore water pressure x area of bottom of slice. Area of

bottom of slice = ∆L x 1 = b/cos α x 1. The pore water pressure head U isconverted to force units as U Force units = U Length units x γw

12

34

56

7

1112

15

b

W5

T5

N5

α

Assumed slip surface (circle) through embankment and foundation

CORE

Figure 4.33: Dam stability analysis by method of slices.

W5

T5

N5

b

∆L

D

α

Slice # 5

Center of slip circle





f) Total shear resistance which can be developed on the bottom of the any slice

at failure: S = C + (N - U) tan φ′ [N-U = α’]


h) Safety factor F = ∑S/∑T = ∑[C+(N-U) tan φ′] / ∑ W x Sin α NOTE: T, U and N may be worked as continuous curve across all slices. Normal components pass through center of rotation and does not cause any driving moment on the slice.Tangential component T causes a driving moment M=T x r, r = radius of slip surface.Resisting forces determined from Columb’s equation.

For homogeneous and uniform cohesive soils a circular arc is considered for slip

circle. The locus of the centre of the critical circle with r u






/γZ.




9 1

2

1 4

1 6

1 8 2 0

2 2

2 4

2 6 2 8

3 0

3 0

3 2

3 4

3 6

3 7

3 8

3 9

4 0

4 0

4 2

4 4

4 5

4 6

5 1 5 2

5 3

5 4




permeable foundation. Permeability of foundation material affects drainage pattern. Iffoundation is permeable then flow is downward, if impermeable flow horizontally outward




towards outer faces. The U/S face is not an equipotential line but potential varies with heightas h = z = elevation. Water potential within the saturated mass of soil changes according tothe u/s potential.

A slip surface is considered along the u/s face. Determine h, z, and u = h-z on bottom

of the slices along the slip surface. The pore water pressure is determined from correspondingequipotential contour map (Figure 4.35). Determine safety factor by procedure above.





9 1 0 1

2

1 4 1 4

1 6

1 7 1

8

2 0 2 0

2 2 2 2

2 4

2 4

2 5

2 6

2 6 2 7

2 8

2 8

2 9

3 0

3 0

3 2

3 4

3 6

3 7

3 8

3 9 4 0

4 0

4 2

4445 4 6





∆−+

∆=

wca

a

V hV P

u

whereu = pore water induced pressure

P a



V a

V


w

h

= Volume of pore water as % of original total embankment volume

c = Henery constant of stability of air in water = 0.02 at 68 oF






• For each ∆, determine u from Hilf’s equation. Also determine corresponding σ’

from plot 1. Determine σ = σ’ + u.







∆ %

σ’ σ

u

Plot 1 Plot 2

Mid point circle











Example:


Solution:



L3 = [40 = 22.36m;

2+52]0.5 = 40.31 m; L4 = [30 2+5 2]0.5

9

1 2

1 4

1 6

1 8 2 0

2 2

2 4

2 6 2 8

3 0

3 0

3 2

3 4

3 6

3 7

3 8

3 9

4 0

4 0

4 2

4 4

4 5

4 6

= 30.41 m


3. Inclination of bottom

A

B

C

DE

F

21

3

4




θ1 = tan -1(25/15) = 59.04° θ2 = tan -1

θ3 = tan(10/20) = 26.57°

-1(5/40) = 7.13° θ4 = tan -1


N1 = 7350 x cos 59.04 = 3781 KN N2 = 12600 x cos 26.57 = 11269 KN

N3 = 19800 x cos 7.13 = 19647 KN N4 = 5400 x cos 9.46 = 5327 KNΣN in core = N1 = 3781 KN; ΣN in fill = N1+N2+N3=11269+19647+5327 = 36243 KN 5. Component of block weight normal to base T:T1 = 7350 sin 59.04 = 6303 KN T2 = 12600 x sin 26.57 = 5636 KNT4 = 19800 x sin 7.13 = 2458 KN T4 = 5400 x sin -9.46 = - 888 KNΣT = 6303 + 5636 + 2458 – 888 = 13509 KN = Total shearing force6. Total water pressure head h, elevation, and net pore water pressure at points B, C, D, E,F:

h (m) = 47.5 37.0 33.4 24.9 15

z (m) = 45 25 15 10 15u (m) = 2.5 12 18.4 14.9 0u (KPa) = 24.52 117.72 180.50 146.17 0

Average pore water pressure u along bottom of each block:u1 = (24.52+117.72)/2 = 71.12 KPa, u2 = (117.72+180.50)/2 = 149.11 KPa,u3 = (180.50 + 146.17)/2 = 163.33 KPa, u4 = (146.17 + 0)/2 = 73.08 KPa.Total pore water pressure force along block bottom U:U1 = 71.12 KPa x 29.15 m x 1 m= 2073 KN U2 = 149.11 x 22.36 = 3334 KNU3 = 163.33 x 40.31 x 1 = 6584 KN U4 = 73.08 x 30.41 x 1 = 2223 KN

ΣU in core = U1 = 2073 KN; ΣU in fill = 3334 + 6584 + 2223 = 12141 KN 7. Cohesion along slip surface C:C1 = (3x5 + 5x10 + 5x29.15) x 1 = 211 KN C2 = 3x22.36x1 = 67 KNC3 = 3x40.31 x 1 = 121 KN C4 = 3 x 30.41 x 1 = 91 KNΣC = 211 + 67 + 121 + 91 = 490 KN8. Total shearing resistance mobilized:S = 490 + (3781 - 2073) tan 30 + (36243 – 12141) tan 35 = 490 + 986 + 16876 = 18352 KN9. Factor of safety FOS = 18352/13509 = 1.359





P = total horizontal shear down to rigid

boundary

h1

h2 b

0.4b

h PA B C

Rigid boundary




)2

45(tan 2

12m

22

2 φ γ −

−=

hhP i


1

1m1

ctanh tanhmγ

φ γ φ +=

where


γm

1

221 )(

h

hhh f d m

γ γ γ

+−=



Max unit shear S


max = 1.4 S av



av

Let S

= P/b


S

tan φ



tan φ

1 + S 2


)/2

av

Factor of safety at maximum shear:. This schould be > 1.5


γ

h tan φ

av = mean effective unit weight = ( γdh + γf h2) / (h+h 2) and F.S. =S/S max

Inter slice Force

. This should be greater than 1.0

Slice Normal force at base= W + (X L – X R ) + N Cos α + S m




Inter slice shear force X = E λ f(x)

ER EL

XR

XL

W




f(x) = inter slice force functionλ = % (in decimal) of function usedWeight W increased/decreased by X amount



The primary consolidation δ1 which develop as excess pore water pressure aredissipated (during the course of construction of the dam embankment) can be estimated interms of coefficient of compressibility m v , the depth of compressible soil and mean verticaleffective stress increa se Δσ as: δ 1 = m v

Embankment: δ

Δσ. Then:

1e = m ve γ dH2/2, [Δσ = γ d

Foundation: δ

H2/2]



Ze]

f









II: ROCKFILL DAMS








D


EC

B

A



D


E

C

B

AE1 E2 E3








rock


C D


AB

























U/s Face membrane

A. Face protection



dam

Internal membrane

A+E Rockfill


C Earth core






































H: 1V) to flat (1.3-1.7:1)







4.23.4 Rock Quality














4.23.5 Rock Sources


4.23.6 Rock Size


Past Design




(5000 psi)


<3 % quarry dust



50% > 14000 lbs (4.5 ft)

50% 6000 – 14000 lbs (3.25 to 4.5 ft)







B C D

C D

A

B













22-100% - 76 mm, 60-100% - 305 mm, 100 % - 610 mm.














• Grid of steel bars anchored by tie back rods extending horizontally into the rockmass (12 to 20 mm rods 30 cm vertical spacing and 1-1.5 m horizontal spacing)

• Alternatively slope stabilization by concrete slabs, asphalt concrete membranes,long flat berm of heavy rock also useful

Figure 4.50: Steel mesh being installed on downstream rock fill face of Windamere Dam as protection against overtopping during diversion. In the background the impermeable brownclay core of the dam can be seen under construction. (Source:http://members.optusnet.com.au/~engineeringgeologist/page11.html)

4.23.8 Test Embankment

• Laboratory tests (abrasion resistance, freeze-thaw characteristics, waterabsorption) used to evaluate suitability of rock.

• Petrographic analysis for minerals identification and rock weathering potential.

• Unconfined or triaxial tests for strength evaluation.

• In-situ examination of rock to check weathering condition.

• Test embankment to evaluate performance of rocks with questionable properties.It is used to determine following issues









4.24 FOUNDATION


















4.24.3 Grouting


































- Dam raising easy























Sloping Earth Cores











1

1

1

1

> 2

2

1.4

1.9

Z

> Z









U/s 0.5H;1V to 0.9 H:1 V, d/s – 0.5:1




Reinforced Concrete






Steel Diaphragm





Bituminous Material





Earth core

Concretecutoff













Figure: Mirani dam u/s face protection by concrete.





• Criteria

- Low permeability








• Horizontal + vertical expansion joints and construction joints are provided. Gapsfilled with flexible bitumen.















Parapet wall

Concreteface slab

Dam crest


1 m thick

dowelMin 1m

Grout curtain

Cutoff





Figure 4.54: concrete face slab construction work by slip forming. Note the reinforcementand machine control. [@ http://www.dur.ac.uk/~des0www4/cal/dams/emba/embaf23.htm]

Asphaltic Concrete




• If bedding layer B zone not used , provide a 6” thick leveling layer to fill surface

voids, provide easy travel of paving machinery, and smooth bedding surface forasphalt membrane













• Tight joints between adjacent strips

• Transverse joints minimum and complete as hot joints








Steel Face• Used on few dams









Backfill

Cutoff

Asphalt membrane

Rockfill embankment




• Steeper slopes construction difficulties• Plate anchored to embankment by steel anchor rods grouted in bedding material

• Plate raised on a scaffolding, grid, bedding material placed after or during plateconstruction

• Plate thickness ¼-3/8”




4.26 SEISMIC DESIGN






Backfill

Cutoff min 1m

Grout curtain



Anchor dowel







- Foundation must be firm rock/blanket grouting- Free draining cobbles/boulders/rock fragments (if compaction same as

rockfill) may be used.




- Limit lift thickness to max of 3’ in zone D- Use a thicker membrane on U/S.












QUESTIONS






, K = 3 x 10-5


3, K = 5 x 10 -4


cm/s.

9

1 2

1 4

1 6

1 8 2 0

2 2

2 4

2 6 2 8

3 0

3 0

3 2

3 4

3 6

3 7

3 8

3 9

4 0

4 0

4 2

4 4

4 5

4 6



















http://www.ngi.no/

http://www.ngi.no/

http://www.ngi.no/

http://www.ngi.no/





Figure 4.64:















































Chapter 4


4.1 DEFINITION





I: EARTHFILL DAMS































• 1










H/3

H

See a e






seepage flow before it reaches the d/s slope. Such a dam is then called as modifiedhomogeneous dam (Figs. 4.2 to 4.4). These measures include rockfill toe, horizontal d/sdrainage blanket, and a vertical or inclined chimney drain. These measures do not decreasethe seepage amount but makes the seepage exit safer with no danger of dam toe failure. A toe

drain (Fig. 4.5) is usually used in conjunction with these seepage control measures tointercept the seepage flow and its disposal.

A homogeneous embankment should not be used for storage dam. A homogeneoustype of dam is applicable in localities where readily available soils show little or no variationis permeability and soils of contrasting permeability are available only in minor amounts or atconsiderably greater cost.

Figure 4.3 : Modified homogeneous dam with d/s horizontal drainage blanket.

H

See a e


Figure 4.4: Modified homogeneous dam with a chimney drain and d/s horizontal drainage blanket.

H

See a e


H

See a e

Figure 4.2 : Modified homogeneous dam with rockfill toe and graded filter.












Dam d/s slope

HU/S

ShellCORE D/S

ShellSee a e

Filter








4.4.3 Diaphragm Dam









HU/S

Shell C O R E

D/SShell

See a e

Filter










Vertical Core





Criteria




failure.

Inclined Core


H

U/S

Shell C O R E

D/SShell

Filter





Advantages




Disadvantages
































4.5.3 Chimney Drain






4.6 FILTER CRITERIA



# mm # mm # mm # mm # mm3 6.4 10 2.0 25 0.71 60 0.25 200 0.0744 4.8 12 1.7 30 0.59 70 0.21 270 0.0535 4.0 16 1.19 35 0.50 100 0.149 300 0.0506 3.4 18 1.00 40 0.42 140 0.105 325 0.044

8 2.38 20 0.84 50 0.297 170 0.088 400 0.037



Fine graded filter



Toe drain






2/[d 60 xd 10

1. D


15 /d 15

2. D

= 5 to 40

15 /d 85

3. D


85



U







85 finer

* D 15 /d 15

* D


15 /d 85

* D


50 /d 50

* D


60 /D 10














Given: d 15 = 0.006 mm, d 85

D


15

D


15

D


15


≤ 5 * 0.10 = 0.50 mm (3) [criteria 2]



85

D

= 2.4 mm.(4)

85

As D

≥ 2 * 0.5” ≥ 1” (5) [criteria 3]


D

layer F2.

15

D




F 1

F i l t e r

F 2



F 1






D 15 ≤ 5 * 0.24 = 12 mm (8) [criteria 2]


Core F1 F2F1F2

U/s fillD/s fill

F1F2 CORED/s fill




Figure 4.10.3 : Mangla dam raising project showing core, and double filter layers.

Constriction of chimney filter, Mangla dam raising project.




From eq 7 and 8 select smaller size, D 15 upper = 5.6 mm; Average D 15

Draw filter gradation line parallel to base material gradation curve and read D

= 4.0 mm.

85

D

= 50 mm. (9)

85

Select D

≥ 2 * 0.5” ≥ 1” (10) [criteria 3]

85 from eq 9 and 10 as 50 mm. Then F2 = [D 15 = 4.0 mm, D 85

Dimensions and Permeability of Toe/Blanket/Chimney Drains

= 50 mm.

The dimension and permeability of the drain must be adequate to carry away theanticipated flow with an ample margin of safety for unexpected leaks. For a relativelyimpermeable foundation, then the expected leakage would be low.

A drain should be constructed of material with a coefficient of permeability of at least

10 to 100 times greater than the average embankment material.

Drain material is usually a processed material. Pit run borrow is usually too dirty (i.e.have large fines). Drain materials must have following grading.

Particle size % passing by weight

1½″ 90 – 100¾” 45 – 75# 4 (4.8 mm) 30 - 45# 50 (0.297 mm) 4 - 10

# 100 (0.149 mm) 1 - 3




# 200 (0.074 mm) 0 - 2

Gradation should be such that it will prevent particles of soil from the adjacent location fromentering the filter and clogging it.

4.7 FOUNDATION DESIGN

Foundation includes both valley floor and the abutments. Foundation must ensurefollowing design requirements

1. It provides support for the embankment under all conditions of saturation andloading.

2. It provides sufficient resistance to seepage to prevent excessive loss of water.

• Foundation is not actually designed but treatments are provided in design to ensurethat all essential requirements are met.

• No two foundations exactly alike, each presents its own separate and distinct problems. Foundation improvements be adopted to local conditions.

• 40% dam failures attributed to failure of the foundation.

• Judgment on the basis of foundation exploration and past experiences.


• Infinite variations in the combinations (materials), structural arrangements and

physical characteristics of the constituent materials.

















• Vibrations/shock as for an earthquake tremor causes re-adjustment of grains into adense structure. Pore water pressure increases suddenly (due to slow drainage) and

foundation behaves as liquid and results in sudden liquification.• Cohesionless sands of low relative density (< 50%) are suspect to failure.


The foundation is treated to minimize the seepage through the foundation and reduceuplift pressures for d/s part. Various foundation treatments include positive cut-off, partialcutoff, sheet pile, cement bound curtain, concrete wall, slurry trench, grouting, etc. These aredescribed below.

4.9.2.1 Positive Cut-off Trenches:

Rolled earth/clay is filled and compacted in a trench excavated to the impermeable barrier / underlying hard bed rock (Fig. 4.11a). The compacted clay forms an impermeable barrier to the seepage flow. The cutoff depth varies as 50 to 150 ft with 1:1 or flatter sideslopes. It is located in continuation of the embankment core u/s from centerline of dam crest,

but not beyond where cover of core becomes small. It is made of usually same material as is

HU/S

ShellCORE D/S

Shell

Figure 4.11a: A positive cutoff for earthfill dam.

Gravel sand foundation

Bed rock

OverburdenRiver

bottom




suitable for dam core. Wider trench base is adopted for dams with large depth. For deepertrench smaller base may be used as seepage force at foundation contact decrease withincrease in depth. Grouting of upper part of weathered/fractured bed rock, if required.Generally top width as w = h – d. A minimum bottom width ≈ 20 ft to allow operating

machinery. Trench below water table will require dewatering.4.9.9.2 Partial Cutoff

The cutoff penetrates only partially into the foundation (Fig. 4.11b).

• Suitable if a low K layer of considerable thickness found above the bed rock. Thislayer must be aerially extensive. Thus seepage from upper more pervious layer isintercepted.

• Partial barrier be at least 95% deep to have any appreciable reduction in seepage.

• Partial seepage barrier may be effective at sites where average permeability offoundation decreases with depth.

• For deep foundations the upper part is sealed off against seepage by providing a partial cutoff and lower part may be sealed by providing sheet piling or groutingetc below and in continuation of the partial cutoff.

• In all cases a minimum partial cutoff of 6-10 ft should be provided. This trenchalso provided better understanding of the subsoil conditions.

4.9.2.3 Sheet Piling Cutoff

Steel sheet pile may be driven into soft alluvium.

• Depth to bed rock.

• Used in combination with partial cutoff to seal lower horizons.

HU/S

ShellCORE D/S

Shell

Fi ure 4.11b: A artial cutoff for earthfill dam.

Deep gravel sand foundation

Bed rock

Sheet piling or grouting etc




• Not suitable for cobbles/boulders as these formations cause misalignment/ open joints, interlock liable to tear-off, pile wander off, pile twisting making anineffective barrier.

• Twin steel sections may be used with interior filled with cement grout.

• Not completely water tight

• 80-90% effective if good work

• Poor workmanship, efficiency less than 50%.

• Seepage resistance offered by sheet pile equals 30-40 ft length of soil; field testsshow resistance equivalent of 400-2000 ft. The effectiveness increase with timedue to filling of gap by sediments, encrustation etc.























bottom.


U/s Mid D/s






positive cutoff.





• RCC or PCC








Plan

Section


River level
























































Q = K I A I = ∆h/L











L

d




























































Bed rock

Drainage ditch


Seepage flow







Sand boilinghttp://research.eerc.berkeley.edu/projects/GEER/GEER_Post%20EQ%20Reports/Peru_2007/Liquefaction.htm#Jahuay%20Reference

Figure 3.4 A large sand boil feature at the southern end of the 400 m long slope failure withshrinkage cracks in the perimeter ejecta material (08/21/07 5:25PM, S13.3950W76.1979).




http://research.eerc.berkeley.edu/projects/GEER/GEER_Post%20EQ%20Reports/Peru_2007/DSC05084.JPG

Figure 3.9 A large sand boil at the base of the Pan American highwayembankment failure. Note the three concrete box culverts running through theembankment that were sheared during the failure (see Figure 3.8). Also notice theshrinkage cracks in the perimeter ejecta material (08/24/07 2:47PM, S13.41307W76.18960).

Grain Size Distribution Curves for Two Sand Boils near the Jahuay SlopeFailure









Sand Boil Specimen USCS %

Fines LL PL PI

Jahuay Boil (LMS-540) SC 43 25 17 8

Pan Am Boil (LMS-504) SP 2.9 NP NP NP

A picture of the sand boil labeled as ?Jahuay Boil? in the grain size distributionplot and table above is shown in Figure 3.4. A picture of the sand boil labeled

as ?Pan Am Boil? is shown to the left Both sand boils were found near theJahuay Slope Failure (Section 3.2)

Sand boil (source Kaplan, 2004)

Injection & grouting

Dynamic compaction




Piping failure. [Indiana DoNR 2007]




ASWCC 2002







Seepage area on downstream

embankment behind tree




Seepage at toe of dam

a. Water flows through dam as a result of piping.




b. Note the eddy or whirlpool in thereservoir, indicating removal

of water by piping





A: Shallow Foundation

• Provide a positive (complete) cutoff to bedrock.

















• Provide drainage blanket of filter grade if i) overlying zone is impervious or ii)overlying zone is rockfill, iii) piping potential is present




• Provide u/s impermeable blanket in continuation of impermeable core.• Minimum core B



pressures.


• Provide toe drains• For foundations of high K, which cause extensive seepage, ponding and sand

boils, then provide drainage trenches, pressure relief wells, extension of d/s toe ofdam or blanket on d/s area.



























• Saturated impervious sands (dirty sands - sands having good amounts of fines)

also act as fine grained soils









• At C



• At W = PL, cohesive strength varies considerably; φ is obtained from triaxial test

on samples compacted at proctor maximum dry density.

is obtained by slow shear test on

saturated soil (drainage permitted, pore water pressure ≈ zero) tan φ about 0.5

Treatments:







c. Flattening of slopes lengthen the surface of sliding, decreases averageshear stress along the path and increase factor of safety against sliding

(Fig. 4.18).

4.10.1.2Relatively Dry Foundations

h/2

h





• These soils exhibit large strength at its present dryness







4.10.2 Treatment/Improvements of Fine Grained Foundation

Foundation of dams can be improved by: 1) Pre-consolidation, 2) Densification of

cohesionless soils, and 3) Dynamic compactionPre-consolidation

• Useful in compressible soils

• Done by applying artificial surcharge such a soil removed from stripping andscaling of abutments may be piled up

• Allow time for water to drain

• For rapid rate (1-2 months for 50% consolidation) piling of random weight is

useful.

Stress σ

S t r a i n

ε

Unconsolidated dry sampleUnconsolidated wet sample

Preconsolidated dry sample

Preconsolidated wet sample

Figure 4.19: Consolidation of wet/dry soils.

Sudden consolidationof dry soil on wetting




• For slow rate soils, dam weight is used to consolidate the soil. This requires slowconstruction rate and providing drainage. Longer time periods (1-2 years for 50%consolidation) are necessary.

Densification of Cohesionless Soils

This is carried out using shock and vibration. Vibrofloatation is used to improve poorfoundation. This can reduce settlement as much as 50% with substantially increased shearingstrength. Vibrations convert loosely packed soils into a denser soil.

Vibroflat can be used to penetrate the soil and operate below the water table. Bestresults are obtained in coarse sands which can contain little or no silt or clay.

Dynamic Compaction

This is repeated application of very high intensity impacts to the surface. This

improves the soil mechanical properties. Compaction is done by dropping a weight, typically10-20 tones from heights of 10-20 meters at regular interval across the surface. Severaltamping/passes may be made at the site. Each imprint is backfilled after tamping. In finersoils increased pore water pressure must be allowed to dissipate between passes, which maytake several weeks.







Dry Foundation




Figure 4.19b

h/2h

3:1 or flatter

Random fill






4.11.1 Crest Design

Crest width W

• The width W of the crest is governed by height of dam, importance of structure,width of highway, construction procedure, access required either duringconstruction or as a permanent feature.


• Special widening may be necessary to provide a

highway or safeguard against freak waves etc. Thiswidening could be done by steepening face slope in






W = 0.55 √h + 0.2 h h< 30 m

W = 1.65 (h+1.5) 1/3


h>30 m (USBR 2001)


Sherard p-413: For earthquake area, top of dam is subjected to worst damage and canvibrate with greater amplitude than the base. Thus it is advisable to make dam top thicker byincreasing crest width or using flatter slopes near top. Also that if any crack develops, thelonger seepage path causes less seepage and increases dam safety.

Surface Drainage


Surfacing






Safety Requirements


Camber





The procedure for designing a cross-section of earthfill dam consists largely ofdesigning to the slopes and characteristics of existing successful dams , making analytical andexperimental studies of unusual conditions and controlling closely selection and placement of


Camber ~ 4

Camber


W

ELEVATIONX-SECTION

1.75:1

2:1








The u/s and d/s slopes of the embankment and core are selected from generalguidelines, experiences in the light of foundation materials and materials available forconstruction. The seepage analysis and stability of the selected dam section is carried out anddam section may be acceptable if factor of safety for the dam under different construction andoperation conditions are found satisfactory. Alternate dam sections are evaluated for materialneeds/crest and factor of safety and that dam section is adopted which provide higher factorof safety at lowest costs. Stability of the shape is analyzed under static loads as well as underseismic conditions.

Except where there is surplus of material available from required excavations, themost economical dam is obtained with the minimum volume and therefore most steep slopesconsistent with the dam stability (Sherard p-48). The allowable steepness depends on the

internal zoning and on strength of foundations and the embankment material. Crest lengthand pace of construction may also affect the slope selection. Use of excavated material asrandom fill may allow flatter slopes. The random fill material may be placed (Fig. 4.21) at



The strength of foundation is also affects the dam face slopes: Weak foundation –average slope 2:1 to 4:1; Strong foundation – steeper slope 1.5:1 to 3:1. The height of dam

also affect slope selection. For homogeneous materials dams of fine core: Short height –








• Dams located in narrow rock-walled canyons can be constructed with some whatsteeper slopes than otherwise, because of added stability given by the confiningwalls. In narrow valleys broad toe berm or very flat slopes at the toe of dam can




















Diaphragm Type



u /s d/s u/s d/sGW GP SW SP Materials not suitable -too perviousGC GM SC SM 2½:1 2:1 3:1 2:1CL ML 3:1 2½:1 3½:1 2½:1CH MH 3/6:1 2½:1 4:1 2½:1

Zoned embankment








Impervious Core




Maximum Core:


Slope = y:1

Slope = y-½:1

Slope = 1:1

Slope = ½:1

1½:1

x-½:1

x:1

Z

Z

















•


Type Shell material Core material No rapid

drawdown

Rapid

drawdownU/s D/s U/s D/s

Min core A Rock, GW,GP, SW, SP,gravely


2:1 2:1


GC, GM 2:1 2:1 2½:1 2¼:1

SC, SM 2¼:1 2¼:1 2½:1 2¼:1

CL, ML 2½:1 2½:1 3:1 2½:1

CH, MH 3:1 3:1 3½:1 3:1






Rocks under alluvium and abutments.Abutments: Metamorphosed sedimentary rocks (sugary limestone, phylite, quartzite, schist)




U/S Slopes: Protection is required against destructive waves splashing onto the side slope.Waves generated due to high sustained winds as well as from earthquake action. Also needed

to be protected against burrowing animals.D/S Slope: These need to be protected against erosion by windand rainfall runoff and the borrowing animals. This also needs to

be protected against possibility of seeping of rain water andforming internal erosion (piping and sloughing of inside ofembankment).

I: UPSTREAM SLOPE

U/s slopes are provided protection by: rock riprap, concrete pavement, steel facing,

bituminous concrete pavement, pre cast concrete blocks, others as short cement pavement,wood, sacked concrete. Special care is needed against beaching process if water level stays atone elevation for long times.

Rock Riprap



Dumped Riprap

The rock fragments/stones are dumped over the slope. The efficiency of dumped rockriprap depends on following: Quality of the rock, Weight or size of individual stone pieces,Thickness of the riprap, Shape of stones or rock fragments (rounded, angular), Slope of theembankment, Stability and effectiveness of the filter.





Individual pieces should be of sufficient weight to resist displacement by waves (mustfor all size dams). The thickness of riprap should be sufficient to accommodate weight andsize of stones necessary to resist wave action (Fig. 4.22). A 3’ minimum thickness is used.Smaller thickness if wave action is less severe. Lesser thickness may be used for upper slopes

corresponding to flood control storage (above normal conservation level) due to infrequentexposure of this part to waves. If there is any damage to this section, it can be repaired onreservoir lowering.


Shell material(min 4-6 ft to

enable placing)













Figure 4.22f: Simly dam: u/s slope protection by angular rock riprap.




Figure 4.22g: Tanpura-I dam: u/s slope protection by rounded rock riprap.





Figure: D/s face protection, Dharabi Dam, Potohar, Chakwal.

Riprap weight

(Zipparro eds. Davis’ Handbook of Applied Hydraulics, p-13.58) defined the riprapweight as:

( ) ( )ba

Cot GK

H W

θ

γ 350

1−

=


W

/8

50









3 where M=mass of stone required (kg), and H s =significant wave height(m).




The size of riprap is estimated as : D = [7 W / 5 γ] 1/3 where D = stone size (ft), W = stoneweight, γ = bulk un i t weight (lbs/cft). The th i ckness must be more than size ofheaviest/largest stones. In no case it should be smaller than 1.5 x D 50

USBR p-277 provided gradation (by weight in lbs) of riprap for slope = 3:1 and angular rocks

as:

or 24”


< 2.5 miles 30” 2500 lbs 1250 75-1250 75

0-10

> 2.5 ml 36” 4500 2250 1000-2250 100






Shape of Rock

Shape of rock fragments influences the ability of riprap to resist displacement bywave action. Angular fragments tend to interlock better than boulders and rounded cobbles.Thus rounded stones should have more thickness.

Graded Filter

A layer or blanket of graded filter should be provided underneath the riprap if there isdanger of fines from underneath layer to more into the riprap layer by wave action. For azoned dam filter not needed if outer shall is gravel. Blanket of crushed rock or natural gravels3/16” to 3 1/2”

Flexibility



Placement

The riprap is dumped from hauling trucks onto the prepared surface. Bulldozers areused to level off-and compact the dumped layer to fill up the voids between larger stones.Smaller stones fit in voids of larger pieces very well. The rock stones must not break duringhandling / placement / compaction. Top surface is uneven, rough and decreases wave riprap.






This consists of suitably sized stones carefully laid by hand in a more or less definite pattern with minimum amount of voids and with top surface relatively smooth. Doubled orirregular shapes lay up less satisfactorily than stones of roughly square shape. Stones of flat

stratified nature should be placed with principal bedding plane normal to slope. Joints should be broken as much as possible and voids be avoided carefully by arranging various sizes ofstones and small rock fragments.

The stones of excellent quality should be used. Thickness can be half of dumpedriprap but not less than 12”. Filter blanket be provided underneath the riprap, if required. Dueto tight packing, hand placed riprap is not as flexible, so it cannot adjust to foundation orlocal settlement. Thus hand placed riprap should not be used where considerable settlement isexpected. Hand placed riprap could be costly due to extensive labor cost in spite of its smallerthickness.

Concrete Paving

Concrete is placed over the sloping surface to resist wave destruction. It can be used both for rockfill and earthfill dams. Paving thickness depend on dam height, slope steepness.Thickness is 8” for h ∼ 50 for and 12’-18” for high dams. Paving is placed in blocks 6’ x 6’ ormore but monolithic construction gives the best service. A water tight surface will eliminatehydrostatic pressure underneath the pavement. Blocks could be displaced or broken by wave

action and uplift forces under the slab. Concrete can crack requiring frequent maintenance.For blocks, expansion joints and construction joints should be widely spaced.

Reinforcement is (5% area) in both directions and be continuous through the construction joints. Joints be sealed with plastic fillers and cracks be grouted and sealed properly.Pavement should extend from crest to below the minimum water levels. It should terminate ata berm and against a deep seated curb or header (minimum 18” deep).

The success of concrete pavement is mixed, but successes and problems have beenobserved. Paving is a costly alternate, but may be adopted if enough riprap material is notavailable. Concrete pavement may or may not be held in place firmly by foundation boltsembedded deep inside sloping shell. Concrete paving increase the wave runup and suitablewave breakers, wave deflectors, may be provided to reduced risk of dam overtopping. Failurechance is 30% + due to inherent deficiencies in this type of construction.

Soil Cement


be added for erosion resistance. For most soils 10-12% cement (% of compacted volume) is

considered typical. Cement and moisture ratio is determined by lab tests.




Soil cement is placed in 6-8” horizontal layers over the slope (horizontal width as 8 ftnormal thickness 2-3½ ft) and roller compacted in a stair-step horizontal layers. Soil cementshave 500-1000 lb/inch compressive strength at 7-day (10% cement). The edges of the cementlayers are not trimmed to retard wave runup. A reasonably firm foundation is required so that

deformation following placement of soil-cement is not significant. Normal embankmentconstruction procedures are satisfactory.


8-10 ft

2-3.5 ft






Riprap protection (ASWCC 2002)

Piping from d/s sink holes (ASWCC 2002)




U/s Slope failure due to wave erosion (ASWCC 2002)




Wave erosion (ASWCC 2002)




Animal burrows (ASWCC 2002)


The zoned dams with d/s shell of rock or cobble fill do not need additional slope protection. Slope protection is required for all conditions against erosion by wind and rainfall.If not protected gully can develop.This protection is provided by

placing a layer of rock, cobbles, orsod (grass). However, vegetation


provided to collect and dispose the rain runoff from upper portions of dam slope. Drainagechannels discharge into cement lined channels running down the slope and ultimately to safedisposal point/river bed. A contour drain is also provided along toe of dam. Surface drainageis also provided (as an open gutter) for abutments and valley floor.

4.11.4 Abutments


Cut slope Berm




The u/s and d/s slopes of embankment are often flared at abutments to provide flatterslopes for stability and seepage control. The u/s flaring is equivalent to providing u/simpermeable blanket. The flaring design is governed by topography of the site, the length ofconstant desired, for aesthetic value, and ease of construction. For steep side slopes this may

be useful to locate access road across the dam. ABUTMENT SIDE SLOPES





process of consolidation . Relationships in the form of pore-pressure coefficients are used todescribe immediate response of pore water pressure to applied total stress.

Shear strengthThe shear strength of a soil is defined as the maximum resistance to shearing stress

which can be mobilized; when this is exceeded failure occurs usually along identifiable slipsurfaces. The shear strength of any material is described by Mohr-Coulomb failure criterion

based on total stress as: S = c + σ tan φ or based on effective stress as: S = c’ + σ’ tan φ’ σ =

total compressive stress ( σ = σ’ + u), σ’ = effective stress, u = pore water pressure.Laboratory shear tests, e.g. triaxial shear test, are run for the material compacted to the designdensity / moisture content and construction of Mohr circle plot. Coarse soils such as sandsderive their shear strength largely from particle interlock and internal friction, and are calledas cohesionless (c=0) or frictional soils; the shear strength is mostly controlled largely soil

Remove overhangs






density.y . Most clays soils derive shear strength from both cohesion and internal friction.Following tests are usually carried out. (Sherard p-332)

1. Undrained test: (unconsolidated – undrained test). No drainage and dissipation of pore water pressure. Called as Q-test (quick test). Used for stability analysis for

dam during and after construction.

2. Consolidated-undrained test. (sample first consolidated with full pore water pressure dissipation under given consolidation pressure) and then is failed in shearwith no drainage allowed. This is called R test.








v = k/m vγw


(k=permeability)

Compaction




permeability.



)


(x10

-4

m2


permeabilityk

/kN)h (m/s)

Cohesion

c’ (kN/m2

Friction, φ’

(degrees))




Gravels 17-22 0 30-45 0.1 – 1.0 10 -1 – 10 -2

Sand 0 30-45 10 - – 10 -

Silts < 5 20-35 10 - – 10 -

Clay (soft-medium)

15-21 0 20-30 1.0 – 10.0 Intact clays,< 10 -8, ifweathered,fissured, orwith siltlenses 10 -3 –10 -8


< 10 < 30


< 50 < 20

4.12 SEEPAGE ANALYSISSeepage flow will occur through all types of formation irrespective how small the

permeability may be. Seepage occur both through the dam embankment as well as damfoundation. Seepage flow is given by Darcy’s law q = K I per unit flow cross section and asQ = q A = K I A = K ∆hA/L through section of area A.

Flow net method is used for simple flow conditions. Seepage occurs as confined flowthrough the foundation and as unconfined flow through the embankment. A flow net isdrawn with curvilinear squares; different squares may have different area but all have flow

lines and potential lines cross at right angle and that all four sides are of equal curved length(Fig. 4.25). The flow is given as

Q = K ∆H N f / N


d


N f

N

= No. of flow lines

d



Flow net provide


2. Distribution of water potential (h) and pore water pressure u (u = h – z ; z =elevation and Note: u, h, z has units of length and are taken from a pre-selected

datum). Also u = P/ρ g = P/ γ, P = water pressure, ρ = water density, γ = sp weight.





The water potential h and pore water pressure is given in units of pressure head;this should be multiplied with unit weight of water (γ = 62.4 lbs/cft, 1000 kg mass/m3



face chimney drain




Seepageexit face

≈ h/3

h



∆h

δh=∆h/m

B

H C0.3 L

L





1. Draw a to-scale map of dam and foundation2. Determine the seepage exit area

3. The u/s face of the embankment is plane where seepage originates.

4. The d/s exit area is usually up to a height h/3 for a homogeneous dam (h = damheight). For a modified homogeneous dam, the flow will converge towards the toedrain or the horizontal blanket drain. Flow emerges tangentially to the exit face ford/s face or chimney drain or vertical line for rockfill toe or drainage blanket (Fig.4.27).


6. Establish the seepage line/phreatic surface. The seepage line intersects thereservoir water surface at a distance 0.3 L from the point C (point C is at watersurface at u/s face) where L is the horizontal projection of the u/s face (Fig. 4.26).The actual phreatic line is modified to meet at point C.




d/s

d


= h/ δh.

10. Draw a flow line in the middle part of the flow area keeping it normal to potentiallines but approximately curvilinear and parallel to seepage line such thatapproximate squares are formed. Extend this flow to the originating surface (i.e.u/s face) and to the exit surface (i.e. d/s face or toe/blanket drain). Draw moreflow lines to the bottom confining layer.

11. The effect of confining surface on the shape of flow lines and equipotential linesdiminishes farther from the confining surface.






14. For large differences in K of the two materials of embankment and foundation theflow lines can cross from foundation into upper embankment or fromembankment into bottom foundation (the material of higher K will attract flowlines from other material). For small differences in K the flow lines originating

from any material will continue within the same material and will not cross intothe other material.









• With F as center, Q-H as radius R, cut PQ vertical at P. The distance P-Q = y.

(x,y) are the coordinates of the seepage line parabola. Draw other points similar toP. The seepage line meets at C.


Seepage rate


T hT K Q −+= 22

A

B C0.3 L

L

F G HQxR = x+S

R

P(x,y)

D E


y

S

h

T

Directrix




Seepage Through Dam Foundation

Seepage through the dam foundation is also determined by drawing flow net for thefoundation section. The flow net is drawn by procedure similar to for the embankment. Theseepage control measures are also considered while drawing the flow net and determining theseepage rates.



ft/sec, Fig. 4.31.


/s/ft.}






Blanket drain thickness:

The thickness of the blanket drain required to pass seepage discharge q per unit width is as:

d d d K K H K Lqt 15.1≈= where L is shoulder width at drain level, H is the reservoir

water depth above drain level, K 1 = permeability of dam core material and K d

Seepage Analysis by Computer Software

is permeabilityof drain material.(Novak et al. 1998, p-61)

Computer software are available (e.g SEEP/W) to determine seepage flow through thedam embankment and the foundation. These programs are user friendly and easilyincorporate the seepage control measures, the seepage exit conditions, varying material

properties, etc. The program result provides seepage flux through selected sections, equi potential contours, distribution of potential/head, seepage flux vectors, and seepage flowlines, etc.

(a)

(c)

(b)




Permissible Seepage

Seepage control measures are provided to reduce seepage quantity and the uplift pressures due to seepage flows. The seepage quantity is usually not very large and theseeping water could be used beneficially at some downstream location. The uplift pressures

are usually more critical in the stability of the dam structure. Thus seepage control measuresare sized to achieve acceptable gradients in the flow domain. Cedergen (1967) defined theacceptable average hydraulic gradients exerted by the water seeping through the embankmentand foundations should not exceed the following critical values.

Impervious core 1:4

Impervious blanket 1:15

Alluvial foundation 1:0.066 (15:1)

Rock foundation 1:1Materials placed around seepage water collection system to comply with filter criteria.

[Source: Cedergren, H. R. 1967. “Seepage, Drainage and Flownets”, 2 nd

Gradient method: 1) determine average hydraulic gradient in soil element, 2) determinemagnitude of seepage force (F = 62.5 I V, V = element volume), 3) determine direction ofseepage force, 4) line of action of seepage force.

edition, John Wiley& Sons Inc., New York] p-115, 16.

Shallow slides or slump on d/s slope (ASWCC 2002)




Shallow slide (ASWCC 2002)

Deep seated slide (ASWCC 2002)







4.13 STABILITY ANALYSIS

Dam sections are analysis for safety against failure by shear and sliding. The slidingof dam can occur at base of dam or any height above the base. Most earthfill dams built withgood material and placed on sound foundation are considered safe against sliding. The safetyagainst shear failure is analyzed for the following conditions: (Sherard p-326)

1. During and after construction for both u/s and d/s faces. Assume pore pressure

high and not drained; Analysis is based on lab Q – test2. Full reservoir steady seepage – d/s face; Analysis is based on lab S - test



A factor of safety is determined for various situations. The dam section is accepted if thefactor of safety for the selected loading condition is higher than recommended values. Elsethe dam section (i.e. side slope of core and shell and materials) is revised and safety re-evaluated.





Procedure



• Shape and location of the slip surface chosen arbitrarily.

• Material above the selected slip surface is called a trial sliding mass

• The trial sliding mass divided into 8-10 slices as in Fig 4.33. (dam unit thick)

• Width of each slice adjusted so that entire base of a slice is located on a singlematerial and chord length ΔL does not significantly differ from arc length.

• Available shear force from material properties S = c + σ tan φ is determined along base of selected surface

• Actual shear force from loading conditions determined.

• Factor of safety F S

• Procedure is repeated for other potential failure surfaces until a critical surface

obtained with lowest factor of safety.

= Shear strength force available ÷ Shear force applied







• Analysis is based on shear strength derived on the basis of total stress S = c + σ

tan φ, or effective stress as : S = c ′ + σ′ tan φ′, where σ′ = σ - u.

• For each slice of bottom width b, compute forces as:

a) Total weight W of the slice. W = area of slice x slice thickness (unit) x grassunit weight (Soil + water). For same height of the two sides of the slice area =D x b. For marked difference in length of two sides, area of trepezoid as A =(D L + D R


)/2 x b.

c) Tangential component of weight: T = W sin α

d) Total water potential h acting on the slice bottom is determined from the equipotential contour map. The pore water pressure head (units of L) is thendetermined as u = h – z, where z is the elevation of the bottom from selecteddatum. Total pore water pressure head U acting on bottom of slice as: U = u x

b/cos α x 1 = average pore water pressure x area of bottom of slice. Area of

bottom of slice = ∆L x 1 = b/cos α x 1. The pore water pressure head U isconverted to force units as U Force units = U Length units x γw

12

34

56

7

1112

15

b

W5

T5

N5

α

Assumed slip surface (circle) through embankment and foundation

CORE

Figure 4.33: Dam stability analysis by method of slices.

W5

T5

N5

b

∆L

D

α

Slice # 5

Center of slip circle





f) Total shear resistance which can be developed on the bottom of the any slice

at failure: S = C + (N - U) tan φ′ [N-U = α’]


h) Safety factor F = ∑S/∑T = ∑[C+(N-U) tan φ′] / ∑ W x Sin α NOTE: T, U and N may be worked as continuous curve across all slices. Normal components pass through center of rotation and does not cause any driving moment on the slice.Tangential component T causes a driving moment M=T x r, r = radius of slip surface.Resisting forces determined from Columb’s equation.

For homogeneous and uniform cohesive soils a circular arc is considered for slip

circle. The locus of the centre of the critical circle with r u






/γZ.




9 1

2

1 4

1 6

1 8 2 0

2 2

2 4

2 6 2 8

3 0

3 0

3 2

3 4

3 6

3 7

3 8

3 9

4 0

4 0

4 2

4 4

4 5

4 6

5 1 5 2

5 3

5 4




permeable foundation. Permeability of foundation material affects drainage pattern. Iffoundation is permeable then flow is downward, if impermeable flow horizontally outward




towards outer faces. The U/S face is not an equipotential line but potential varies with heightas h = z = elevation. Water potential within the saturated mass of soil changes according tothe u/s potential.

A slip surface is considered along the u/s face. Determine h, z, and u = h-z on bottom

of the slices along the slip surface. The pore water pressure is determined from correspondingequipotential contour map (Figure 4.35). Determine safety factor by procedure above.





9 1 0 1

2

1 4 1 4

1 6

1 7 1

8

2 0 2 0

2 2 2 2

2 4

2 4

2 5

2 6

2 6 2 7

2 8

2 8

2 9

3 0

3 0

3 2

3 4

3 6

3 7

3 8

3 9 4 0

4 0

4 2

4445 4 6





∆−+

∆=

wca

a

V hV P

u

whereu = pore water induced pressure

P a



V a

V


w

h

= Volume of pore water as % of original total embankment volume

c = Henery constant of stability of air in water = 0.02 at 68 oF






• For each ∆, determine u from Hilf’s equation. Also determine corresponding σ’

from plot 1. Determine σ = σ’ + u.







∆ %

σ’ σ

u

Plot 1 Plot 2

Mid point circle











Example:


Solution:



L3 = [40 = 22.36m;

2+52]0.5 = 40.31 m; L4 = [30 2+5 2]0.5

9

1 2

1 4

1 6

1 8 2 0

2 2

2 4

2 6 2 8

3 0

3 0

3 2

3 4

3 6

3 7

3 8

3 9

4 0

4 0

4 2

4 4

4 5

4 6

= 30.41 m


3. Inclination of bottom

A

B

C

DE

F

21

3

4




θ1 = tan -1(25/15) = 59.04° θ2 = tan -1

θ3 = tan(10/20) = 26.57°

-1(5/40) = 7.13° θ4 = tan -1


N1 = 7350 x cos 59.04 = 3781 KN N2 = 12600 x cos 26.57 = 11269 KN

N3 = 19800 x cos 7.13 = 19647 KN N4 = 5400 x cos 9.46 = 5327 KNΣN in core = N1 = 3781 KN; ΣN in fill = N1+N2+N3=11269+19647+5327 = 36243 KN 5. Component of block weight normal to base T:T1 = 7350 sin 59.04 = 6303 KN T2 = 12600 x sin 26.57 = 5636 KNT4 = 19800 x sin 7.13 = 2458 KN T4 = 5400 x sin -9.46 = - 888 KNΣT = 6303 + 5636 + 2458 – 888 = 13509 KN = Total shearing force6. Total water pressure head h, elevation, and net pore water pressure at points B, C, D, E,F:

h (m) = 47.5 37.0 33.4 24.9 15

z (m) = 45 25 15 10 15u (m) = 2.5 12 18.4 14.9 0u (KPa) = 24.52 117.72 180.50 146.17 0

Average pore water pressure u along bottom of each block:u1 = (24.52+117.72)/2 = 71.12 KPa, u2 = (117.72+180.50)/2 = 149.11 KPa,u3 = (180.50 + 146.17)/2 = 163.33 KPa, u4 = (146.17 + 0)/2 = 73.08 KPa.Total pore water pressure force along block bottom U:U1 = 71.12 KPa x 29.15 m x 1 m= 2073 KN U2 = 149.11 x 22.36 = 3334 KNU3 = 163.33 x 40.31 x 1 = 6584 KN U4 = 73.08 x 30.41 x 1 = 2223 KN

ΣU in core = U1 = 2073 KN; ΣU in fill = 3334 + 6584 + 2223 = 12141 KN 7. Cohesion along slip surface C:C1 = (3x5 + 5x10 + 5x29.15) x 1 = 211 KN C2 = 3x22.36x1 = 67 KNC3 = 3x40.31 x 1 = 121 KN C4 = 3 x 30.41 x 1 = 91 KNΣC = 211 + 67 + 121 + 91 = 490 KN8. Total shearing resistance mobilized:S = 490 + (3781 - 2073) tan 30 + (36243 – 12141) tan 35 = 490 + 986 + 16876 = 18352 KN9. Factor of safety FOS = 18352/13509 = 1.359





P = total horizontal shear down to rigid

boundary

h1

h2 b

0.4b

h PA B C

Rigid boundary




)2

45(tan 2

12m

22

2 φ γ −

−=

hhP i


1

1m1

ctanh tanhmγ

φ γ φ +=

where


γm

1

221 )(

h

hhh f d m

γ γ γ

+−=



Max unit shear S


max = 1.4 S av



av

Let S

= P/b


S

tan φ



tan φ

1 + S 2


)/2

av

Factor of safety at maximum shear:. This schould be > 1.5


γ

h tan φ

av = mean effective unit weight = ( γdh + γf h2) / (h+h 2) and F.S. =S/S max

Inter slice Force

. This should be greater than 1.0

Slice Normal force at base= W + (X L – X R ) + N Cos α + S m




Inter slice shear force X = E λ f(x)

ER

EL

XR

XL

W




f(x) = inter slice force functionλ = % (in decimal) of function usedWeight W increased/decreased by X amount



The primary consolidation δ1 which develop as excess pore water pressure aredissipated (during the course of construction of the dam embankment) can be estimated interms of coefficient of compressibility m v , the depth of compressible soil and mean verticaleffective stress increa se Δσ as: δ 1 = m v

Embankment: δ

Δσ. Then:

1e = m ve γ dH2/2, [Δσ = γ d

Foundation: δ

H2/2]



Ze]

f









II: ROCKFILL DAMS








D


EC

B

A



D


E

C

B

AE1 E2 E3








rock


C D


AB

























U/s Face membrane

A. Face protection



dam

Internal membrane

A+E Rockfill


C Earth core






































H: 1V) to flat (1.3-1.7:1)







4.23.4 Rock Quality














4.23.5 Rock Sources


4.23.6 Rock Size


Past Design




(5000 psi)


<3 % quarry dust



50% > 14000 lbs (4.5 ft)

50% 6000 – 14000 lbs (3.25 to 4.5 ft)







B C D

C D

A

B













22-100% - 76 mm, 60-100% - 305 mm, 100 % - 610 mm.














• Grid of steel bars anchored by tie back rods extending horizontally into the rockmass (12 to 20 mm rods 30 cm vertical spacing and 1-1.5 m horizontal spacing)

• Alternatively slope stabilization by concrete slabs, asphalt concrete membranes,long flat berm of heavy rock also useful

Figure 4.50: Steel mesh being installed on downstream rock fill face of Windamere Dam as protection against overtopping during diversion. In the background the impermeable brownclay core of the dam can be seen under construction. (Source:http://members.optusnet.com.au/~engineeringgeologist/page11.html)

4.23.8 Test Embankment

• Laboratory tests (abrasion resistance, freeze-thaw characteristics, waterabsorption) used to evaluate suitability of rock.

• Petrographic analysis for minerals identification and rock weathering potential.

• Unconfined or triaxial tests for strength evaluation.

• In-situ examination of rock to check weathering condition.

• Test embankment to evaluate performance of rocks with questionable properties.It is used to determine following issues









4.24 FOUNDATION


















4.24.3 Grouting


































- Dam raising easy























Sloping Earth Cores











1

1

1

1

> 2

2

1.4

1.9

Z

> Z









U/s 0.5H;1V to 0.9 H:1 V, d/s – 0.5:1




Reinforced Concrete






Steel Diaphragm





Bituminous Material





Earth core

Concretecutoff













Figure: Mirani dam u/s face protection by concrete.





• Criteria

- Low permeability








• Horizontal + vertical expansion joints and construction joints are provided. Gapsfilled with flexible bitumen.















Parapet wall

Concreteface slab

Dam crest


1 m thick

dowelMin 1m

Grout curtain

Cutoff





Figure 4.54: concrete face slab construction work by slip forming. Note the reinforcementand machine control. [@ http://www.dur.ac.uk/~des0www4/cal/dams/emba/embaf23.htm]

Asphaltic Concrete




• If bedding layer B zone not used , provide a 6” thick leveling layer to fill surface

voids, provide easy travel of paving machinery, and smooth bedding surface forasphalt membrane













• Tight joints between adjacent strips

• Transverse joints minimum and complete as hot joints








Steel Face• Used on few dams









Backfill

Cutoff

Asphalt membrane

Rockfill embankment




• Steeper slopes construction difficulties• Plate anchored to embankment by steel anchor rods grouted in bedding material

• Plate raised on a scaffolding, grid, bedding material placed after or during plateconstruction

• Plate thickness ¼-3/8”




4.26 SEISMIC DESIGN






Backfill

Cutoff min 1m

Grout curtain



Anchor dowel







- Foundation must be firm rock/blanket grouting- Free draining cobbles/boulders/rock fragments (if compaction same as

rockfill) may be used.




- Limit lift thickness to max of 3’ in zone D- Use a thicker membrane on U/S.






Kaplan, A. 2004. Soil Liquefaction, Mid-America Earthquake Center and Georgia Institute ofTechnology, [http://geosystems.ce.gatech.edu/Faculty/Mayne/papers/Liquefaction%20by%20Alisha%20Kaplan.doc] Visited on 12-4-2010

Indiana DoNR. 2007. Indiana Dam Safety Inspection Manual, Part-4: EmergencyPreparedness. Dept of Natural Resources, Division of Water, Indianapolis, In. USA.[www.in.gov/dhr/water/files/Part-4-Dam_Safety_Manual.pdf] visited on 12-4-2010

ASWCC. 2002. Inspection and Maintenance Manual for Arkansas Dam Owners. ArkansasSoil and Water Conservation Commission, Little Rock, AR, USA.[www.michigan.gov/documents/deq/deq-p2ca-bestpractices-arkansasexample_281013_7.pdf] visited on 12-4-2010.







QUESTIONS






, K = 3 x 10-5


3, K = 5 x 10 -4


cm/s.

9

1 2

1 4

1 6

1 8 2 0

2 2

2 4

2 6 2 8

3 0

3 0

3 2

3 4

3 6

3 7

3 8

3 9

4 0

4 0

4 2

4 4

4 5

4 6



















http://www.ngi.no/

http://www.ngi.no/

http://www.ngi.no/

http://www.ngi.no/





Figure 4.64:





































































Tariq 2008. Dam and Reservoir Engineering 5-1Ch-5: Concrete Dams

Chapter 5CONCRETE DAMS

5.1 INTRODUCTION

The concrete dams get their structural strength by use of Portland cement. Othercementous substances as Pozolan, Fly Ash may be mixed with the cement to derive certain properties. Concrete is used in the form of PC (plain cement concrete), RC (reinforcedconcrete), RCC (roller compacted concrete) etc. Concrete dams may be categorized into three

principle types according to their physical form and features of their design. These areGravity Dams, Arch Dams, and Buttress Dams. Historically stone/brick masonry had also

been used in conjunction with concrete to construct dams.

5.1.1 Concrete Gravity Dam

A concrete dam resists the applied loads primarily by its weight. Gravity dams are

usually straight in plan, but may be slightly curved to take advantage of topography of a site.In cross section gravity dam are roughly triangular. Most of gravity dams are solid and thushave no bending stress in plan. A gravity dam can also be made hollow to decrease concretevolumes as well as uplift pressures, and are called hollow gravity dams. Gravity dams are

particularly suited across gorges with very steep canyon slopes where earth dams might slip.When good foundation is available, gravity dams can be built to any height. The spillway can

be created in the dam by providing an overflow section (Fig. 5.1). Thus gravity dams becomenatural choice for narrow valleys where spillway could not be located due to topography. Thehighest dams in the world are of gravity type. Warsak Dam is example of a concrete gravitydam located across River Kabul.

Figure 5.1a: A concrete gravity dam with overflow spillway section in the middle.




Figure 5.1b: Grand Coulee Dam, Washington. (Photo: USBR @,http://www.pbs.org/wgbh/buildingbig/dam/coulee_dam.html)

5.1.2 Arch Dam

An arch dam is a dam with significantly upstream curvature in plan and transmit amajor part of the applied water load to the canyon walls / abutments by horizontal thrust ofthe arch action. Arch dam has single curvature in horizontal plan only (Fig. 5.3). Arch dammay have curvature both in horizontal plan and the vertical cross section with undercutting atheel and, in most cases, a downstream overhang near the top of dam; such dams are called asCupola or double curvature dam (Fig. 5.4). The overhang is used to place an over-fallspillway.

5.1.3 Buttress Dam

A buttress dam is made of two structural elements: A sloping water supporting deckand A buttress which support the deck. The strength and the stability is provided by the

weight of the water over the deck and buttress concrete body. Buttress dam are furtherclassified according to the type of water supporting deck. Ambersom (slabbed buttress) damhas straight continuous deck supported at upstream edge of the buttress. Multiple arch damconsists of a series of arch segments supported by buttress. A massive head buttress dam isformed by flaring the u/s edge of buttress to span the space between buttress walls (halfwayto either side). Massive head can be diamond head or round head depending on shape of theenlargement section at the u/s face. Spillway may be provided as over-fall across the damface or as separate structure in the abutment (Fig 5.5).




Figure 5.2b: Proposed Diamir-Basha concrete gravity dam (Profile)

Figure 5.2a: Proposed Diamir-Basha concrete gravity dam (cross section)

Ex rock El = 898 m

Ex river bed El ≈ 945 m

Tail WL El = 960 m

Crest El = 1177 m

El = 1150 m

El =1060 m

El = 1150 m

Crest 12 m6.5 m

30 MPa RCC

217.25 m

250.63 m

33.4 m

20 MPa RCC

0.75H:1V

0.22H:1V

0.1H:1V

0.05H:1V

DFL = 1170 m

281 m

El =1000 m

1-m thick enriched groutconcrete on u/s face




5.1.4 Arch Gravity Dam

An arch gravity dam is combination of gravity dam and arch dam where theadvantages of both dams are combined overcoming the limitations and stringent requirementsof arch dam (Figs. 5.6a-f and 5.7).

Figure 5.3: An arch dam with spillway in the left abutment.

Figure 5.4: Typical double curvature arch dam with overflow spillway and plunge-pool




Figure 5.5: Bartlett buttress dam (Verde River, Arizona) with spillway on the right abutment.(Source: U.S. Bureau of Reclamation,@ http://www.pbs.org/wgbh/buildingbig/dam/bartlett_dam.html)

Figure 5.6a: Glen Canyon arch gravity dam on Colorado River. (Total structure height = 710ft, height above river bed = 587 ft, crest length = 1560 ft, crest width = 25 ft, basewidth = 300 ft, Geology = Navajo sandstone, Storage = 27 MAF, Spillway = 2,76,000cfs, irrigation outlets = 15,000 cfs, Power house shown in foreground contain 8 NosFrancis turbine of 155,500 HP each producing 1296,000 KW energy). Note the arch-abutment contacts. [Source earth-google]




Figure 5.6b: Spillway layout of Glen Canyon dam. [Source earth-google]

Figure 5.6c: Panoramic view of Glen Canyon dam (green dam). (Source:

http://www.istockphoto.com/file_closeup.php?id=227013)




Figure 5.6d: D/s canyon view of Glen canyon dam.(http://www.usbr.gov/dataweb/dams/az10307.htm)

Figure 5.6e : Glen canyon dam. Source:

(http://www.flickr.com/photo_zoom.gne?id=388953847&size=l)

http://www.flickr.com/photo_zoom.gne?id=388953847&size=l

http://www.flickr.com/photo_zoom.gne?id=388953847&size=l




Figure 5.7: Hoover arch-gravity dam. [Source earth-google]

5.2 CONSTRUCTION ARRANGEMENT

Concrete dams involve huge concrete volume. (Estimated volume of concrete forDiamir-Basha dam is 22 million m 3

Internal drains are placed d/s of water stops to intercept any flow. Construction keysmay or may not be provided along the joints. Longitudinal joints are parallel to dam axis forall dam height. These joints must necessarily be legged in order to transmit vertical shearingstresses across the section. Key surfaces are placed approximately in the planed principalstress trajectories under full reservoir conditions. [Davis p:11-5 fig 5,6,7,8 Punmia p:311- 13Figs 8.21-23]. Each construction blocks are constructed in 1-2.25 m lifts. Concrete blocks are

poured when adjacent side blocks has gained sufficient strength.

.) Large concrete volume can not be placed once. Thusmost dams are built in blocks separated by transverse joints and longitudinal joints.Transverse joints run from u/s face to d/s face for full height of dam and spaced 20-60 ftapart. Water stops (rubber and metal) are provided to stop flow through the joints.

CONTRACTION AND CONSTRUCTION JOINTS (USACE 1995)

a . To control the formation of cracks in mass concrete, vertical transverse contraction(monolith) joints will generally be spaced uniformly across the axis of the dam about 50 feetapart. Where a powerhouse forms an integral part of a dam and the spacing of the units is inexcess of this dimension, it will be necessary to increase the joint spacing in the intake blockto match the spacing of the joints in the powerhouse. In the spillway section, gate and piersize and other requirements are factors in the determination of the spacing of the contraction

joints. The location and spacing of contraction joints should be governed by the physical

features of the damsite, details of the appurtenant structures, results of temperature studies, placement rates and methods, and the probable concrete mixing plant capacity. Abrupt




discontinuities along the dam profile, material changes,defects in the foundation, and the location of featuressuch as outlet works and penstock will also influence

joint location. In addition, the results of thermal studieswill provide limitations on monolith joint spacing forassurance against cracking from excessive temperature-induced strains. The joints are vertical and normal to theaxis, and they extend continuously through the damsection. The joints are constructed so that bonding doesnot exist between adjacent monoliths to assure freedomof volumetric change of individual monoliths.Reinforcing should not extend through a contraction

joint. At the dam faces, the joints are chamfered aboveminimum pool level for appearance and for minimizing

spalling. The monoliths are numbered, generallysequentially, from the right abutment.

b. Horizontal or nearly horizontal construction joints(lift joints) will be spaced to divide the structure intoconvenient working units and to control construction

procedure for the purpose of regulating temperaturechanges. A typical lift will usually be 5 feet consistingof three 20-inch layers, or 7-1/2 feet consisting of five18-inch layers. Where necessary as a temperaturecontrol measure, lift thickness may be limited to 2-1/2feet in certain areas of the dam. The best lift height foreach project will be determined from concrete

production capabilities and placing methods. EM 1110-2-2000 provides guidance on establishing lift thickness

Construction joints’ key:

Horizontal and vertical keys are provided in theconstruction block joints to ensure an efficient load transfer between the blocks.

WATERSTOPS (USACE 1995)

A double line of waterstops should be provided near the upstream face at all contraction joints. The waterstops should be grouted 18 to 24 inches into the foundation or sealed to thecutoff system and should terminate near the top of the dam. For gated spillway sections, thetops of the waterstops should terminate near the crest of the ogee. A 6- to 8-inch-diameterformed drain will generally be provided between the two waterstops. In the non-overflowmonolith joints, the drains extend from maximum pool elevation and terminate at about thelevel of, and drain into, the gutter in the grouting and drainage gallery. In the spillwaymonolith joints, the drains extend from the gate sill to the gallery. A single line of waterstopsshould be placed around all galleries and other openings crossing monolith joints. EM 1110-2-2102 provides further details and guidance for the selection and use of waterstops and other

joint materials.

Vertical joints

Transverse joints

Longitudinal joints

SECTION

PLANFigure 5.8: Concrete dam joints




Figure 5.9: Block joint, layer green joint, water stop, and internal drain for concrete block.

Figure 5.9b : Waterstop closeup view.

Foundation Grouting and Drainage (USACE 1995)

It is good engineering practice to grout and drain the foundation rock of gravity dams. Awell-planned and executed grouting program should assist in disclosing weaknesses in thefoundation and improving any existing defects. The program should include area grouting forfoundation treatment and curtain grouting near the upstream face for seepage cutoff throughthe foundation. Area grouting is generally done before concrete placement. Curtain grouting

is commonly done after concrete has been placed to a considerable height or even after thestructure has been completed. A line of drainage holes is drilled a few feet downstream from

Water stop

Internal drain

Inspection gallery

Seepage drain




the grout curtain to collect seepage and reduce uplift across the base. Detailed information ontechnical criteria and guidance on foundation grouting is contained in EM 1110-2-3506.

5.3 HEAT OF HYDRATION

Cement concrete releases large heat of hydrations. Procedures used to handle this include

• Water curing – It takes long duration and limit construction lift and pace.

• Use of low cement concrete.

• Use of heat retarding admixtures

• Additions, e.g pozolan

• Pre cooling of concrete by using cold water and ice blocks.

• Internal refrigeration – This is done by inserting metal coils at about 1-2 feet verticaldistance. Coolant or cold water or river water is passed, which cools down theconcrete.

5.4 LOADS

Dams are subjected to many different loads (Fig. 5.10). The determination of all applicableloads is necessary for successful analysis and design of concrete dam.

5.4.1 Water Loads

The water pressure at any depth = p w= γw h. The resultant horizontal force over depthh1 = P wh =γwh1

2/2 and act at height of h 1/3 from base where h 1 is u/s depth of water. For d/ssloping face the total tail water pressure =P twh = γwh2

2/2 and act at h 2 /3 height from basewhere h 2 is d/s depth of tail water. For u/s face batter, the water weight over sloping face isPwv = γw x A 1 . and along d/s face as P twv =γw x A 2 where A 1 and A 2

5.4.2 Self weight load

is profile water areaover u/s and d/s sides respectively.

The structure weight P m act through centroid of the cross section profile area A p, thusPm = γc x A p (γc = concrete unit weight ~ 23.5 N/M 3

5.4.3 Fixtures

, 150-155 lb/cft).

The weight of various fixtures P F as crest gates, other ancillary structure andequipment of significant weight placed/attached at the dam crest act at distance b F

5.4.4 Seepage and uplift load

from thetoe of dam.

Interstitial pore water pressure (u w) develop within a concrete dam and its foundationdue to seepage e.g. due to preferential flow along joint plains, cracks, fine fissures and within

pore structure of concrete and rock foundation. Pressure distribution is locally indeterminate.Formed drains provided in the dam structure near u/s face eliminate seepage pressure withinthe dam concrete structure. The pressure relief drains d/s of cutoff or grout curtain will reduce

pore pressure in the foundation. The pore water pressure in the foundation varies from γw h1 at u/s heel to γw h2 or zero at d/s end and is considered to vary linearly. The mean effectiveuplift head at the line of pressure relief drains h d

( )212 hhK hh d d −+=

can be expressed as:




The empirical coefficient K d is function of geometry of relief drain / pressure relief well(diameter, spacing, and location relative to u/s face). For an efficient drainage system K d

+

=2

21 hhbP wu γ

~0.33. For no relief drains total uplift pressure is as:

KN/m and

++

=21

1223 hh

hhb yu

where h 1 and h 2 is head water and tail water depths and b is base width. The P u acts throughcentroid of the pressure distribution diagram at distance y u from the heel; m. For presence ofrelied drains P u

( ) ( )[ ]2/)(2/ 21 hh Lbhh LP d d d d wu +−++= λ

is as:

5.4.5 Sediment Load

The general accumulation of significant deposits of fine sediments and silt against theface of the dam generate silt pressure P s additional to the water load P wh and is given as: P s =K a γ’s h s

2/2 (kN/m) and act at h s/3 from base where γ’s = γs- γw = silt/sediment submergedunit weight, and γs = silt/sediment unit weight (~ 18-20 kN/m 3), γw = water unit weight, K a = active lateral pressure coefficient given as: K a = (1- sin Φs)/(1+sin Φs), where Φs = angle of

h1

hs

h1/3

h2

P wh

P ewh

A1

A2

P ewv

P wv

P ice P wave

R

P U

P s P m

P emh P emv

P twh

P twv

b m b U

b

Centroid of profile area0.4 h 1

b R

Figure 5.10: Loads acting on a concrete dam.

P F

b F m=dam mass weightw=water weight

s=silt loadv=verticalh=horizontalt=tail watere=earthquakeR=resultantU=upliftF=Fixtures

b Hw

b t




shearing resistance of the sediments (for Φs ~30°, K a ~ 0.33). This yields equivalent fluidunit weight of K a γ’s [≈ 3.0 kN/m2]. The silt depth h s

is complex time dependent function ofsuspended sediment concentration, reservoir characteristics, flow hydrograph, etc.

5.4.6 Ice Load

The ice load becomes important when ice sheets form to a appreciable thickness and persists for long periods. Thus a considerable ice pressure may be generated with horizontalthrust near the crest level. This is a complex function. An acceptable provision for ice load isgiven by USBR (1976) as P ice = 145 kN/m 2

5.4.7 Wind Load (Pumnia p-282)

for ice thickness in excess of 0.6 m. For smallerice thickness less than 0.4 m, or when ice is subject to little restraint as on a sloping surface,ice load may be neglected.

This is usually small force and is usually not accounted in dam stability analysis.Wind pressure act only on that portion of the supper structure which is exposed to windaction above the water level. Wind pressure may be taken as 100-150 kg/m 2

5.4.8 Wave Pressure. (Punmia p-281)

for the u/s areaexposed to wind.

Waves are generated on the reservoir surface because of the wind blowing over it.Wave pressure depends on height of the waves h w , given as: h w = 0.032 (V.F) ½+0.763 -0.271 F ¼ for F < 32 km and h w = 0.032 (V.F) ½ for F > 32 km where F is fetch or straightlength of the water surface (km) and V is wind velocity (km/hr). The pressure intensity due to

wave is given as: p w = 2.4 γw hw (t/m 2) and act at h w/8 above the still water surface.Assuming a linear pressure distribution, total pressure P w is given as: P w = ∫pw dh = 2 γw h w (t/m2) and act at height of 3/8h w

5.4.9 Earthquake Pressure

.

5.4.9.1 E/Q Waves

The earthquake induces various vibration waves in the earth’s crust. These waveimpart acceleration of various magnitudes in the foundation under the dam resulting in itsmovement. In order to avoid destruction/rupture the dam also must move along thefoundation. The acceleration induces an inertia force in the body of dam and sets up stressesin the dam body. The acceleration can take any direction and resultant direction is usually

PU P U

yu yu bu bu Uplift pressure distribution without

relief drains/wellsUplift pressure distribution with

relief drains/wellsFigure 5.11: Seepage uplift pressure distribution without and with relief drains/wells

h1

h2

h1

h2 hd

b

Ld




resolved into horizontal and vertical directions. The dynamic loads generated by seismicdisturbances must be considered in the design of concrete dams situated in “seismic-risk”regions as well as in regions of close proximity to potentially active geologic faultcomplexes. Maximum credible earthquake (MCE) and Operating basis earthquake ( OBE )is established for the dam site after thorough review of local and regional geology inconjunction with historical evidence. For low-risk areas a minimal level of disturbance isspecified for design purposes.

Seismic activity produces complex oscillating patterns of acceleration and groundmovements and these generate transient loads due to inertia of dam and the retained body ofwater. Both horizontal and vertical acceleration are considered in the sense least favorable tostability of the dam.

Under reservoir full condition the most adverse seismic loading will occur when aground shock is associated with (i) horizontal foundation acceleration operating upstream and(ii) vertical foundation acceleration operating downward. Horizontal acceleration willgenerate additional hydrodynamic water load P ewh acting downstream against dam face andalso inertia force due to the mass of the dam P emh acting in downstream sense. Verticalacceleration will effectively reduce the mass (weight) of the dam by an amount P emv

Seismic shock waves have frequency of 1-10 Hz, and consequently oscillate veryrapidly in a transient form (Fig. 5.12). The natural vibrating frequency of structures is as: f

, andhence the stability of the structure.

n =600T/h 2 (Hz) where T is base thickness (m), h is dam height of triangular profile (of concretematerial of elasticity E eff =14 GN/m 2

Seismic loads are approximated by using the simple quasi-static approach of ‘seismiccoefficient analysis’. Alternatively dynamic analysis will provide necessary earthquakeresponse. The simple seismic coefficient analysis approach is conservative approach and usedfor smaller and less vulnerable dams. For other dams sophisticated procedures are necessary.The seismic load is defined primarily by an acceleration coefficient representing ratio ofseismic ground acceleration to gravitational acceleration g, thus α

). Seismic waves are irregular in magnitude, periodicity,and direction and are unlikely to sustain resonance beyond few seconds.

h and αv are horizontal andver tical coefficients with usually αv ~ 0.5 αh and αh = 0.1 to 0.2 are most common; αh

5.4.9.2 Inertia force – mass of dam

= 0.4is used for high risk dams in Japan.

Inertia forces are given as: Horizontal: P emh = ± αh P m and Vertical: P emv = ± αv P m where P m

5.4.9.3 Hydrodynamic force – water reaction

is dam mass weight. The inertia forces are considered to act through the centroid of the dmsection; +ve forces act in upstream and downward sense.

The hydrodynamic force at depth y below the water surface is given as:

p ewh = C e αh γw

where C

y

e is dimensionless pressure factor depending on y/h and slope angle of u/s face φu from the vertical, h is max water depth at the section being studied. C e

is given as:




Figure 5.12 Earthquake acceleration time history.

−+

−=

h y

h y

h y

h yC

C me 22

2 and °

=90

735.0 umC

φ ; values of C e

y/h

are given below.

C e (φu C= 0) e (φu = 15°)0.2 0.35 0.290.4 0.53 0.450.6 0.64 0.550.8 0.71 0.61

1.0 0.73 0.63The pressure variation is elliptical/parabolic curve. The total hydrodynamic load at depth y isas:

h y yC P wheewh .66.0 γ α =

and acts at 4/3π h (~0.4h) above the respective section. The total dynamic pressure and itsmoment arm may be written as:

P ewy = 0.726 p ewy y and : M ehy = 0.299 p ewy y2

The vertical hydrodynamic load P

.

ewv effective above the upstream face batter or flare may beas: P ewv =± αv P wv and act through centroid of area A 1 .




Figure : Earthquake pressure coefficient. (USBR 2001)

5.4.10 Load Combinations

All forces acting on a dam do not act simultaneously. Design of dam is usually basedon most adverse combination of loads.I: Normal (Usual) load combination: Usual loads refer to loads and load conditions, whichare related to the primary function of a structure and can be expected to occur frequentlyduring the service life of the structure. A usual event is a common occurrence and thestructure is expected to perform in the linearly elastic range.

Normal load includes: self weight + normal head water level + minimum tail waterlevel + ice + silt + normal uplift.

II: Unusual load combination: Unusual loads refer to operating loads and load conditionsthat are of infrequent occurrence. Construction and maintenance loads, because risks can becontrolled by specifying the sequence or duration of activities, and/or by monitoring

performance, are also classified as unusual loads. Loads on temporary structures which areused to facilitate project construction, are also classified as unusual. For an unusual eventsome minor nonlinear behavior is acceptable, but any necessary repairs are expected to beminor.

Unusual load includes: self weight + maximum water level (pool at standard projectflood surcharge) + tail water level at flood elevation + ice + silt + normal upliftIII: Extreme load combination: Extreme loads refer to events, which are highly improbableand can be regarded as emergency conditions. Such events may be associated with major

accidents involving impacts or explosions and natural disasters due to earthquakes orflooding which have a frequency of occurrence that greatly exceeds the economic service lifeof the structure. Extreme loads may also result from the combination of unusual loadingevents. The structure is expected to accommodate extreme loads without experiencing acatastrophic failure, although structural damage which partially impairs the operationalfunctions are expected, and major rehabilitation or replacement of the structure might benecessary.

Extreme loads includes: self weight + normal head water level + minimum tail waterlevel + ice + silt + extreme uplift (max water level with drains not functioning) + maximumcredible earthquake MCE loads.




Other combinations of loads can also be analyzed as: Operating basis earthquake (OBE), Poollevel for PMF condition, No tail water, etc. The design is the checked for safety at reservoirfull and reservoir empty conditions particularly for dam interior and around openings

provided in the dam body.




5.5 CONCRETE GRAVITY DAMS5.5.1 Layout ( Golze P- 437 )

The initial layout of dam structure is done on basis of previous experience with

similar dams. Then a standard stability analysis is made to determine the acceptability of thedesign so that stress distribution and stability is satisfactory. Modifications are made byreshaping the structure to improve the design and acceptability. The design is accomplished

by making out successive layout, each one being progressively improved based on the resultof stress and stability analysis of the preceding layout s

5.5.2 Shape

The a/s face of the dam is usually made vertical to concentrate the weight at u/s faceto resist reservoir water loading. The d/s face has usually constant slope from top of the damto the base (usually 0.75H:1V). The base width required to satisfy the stress and stabilityrequirements determine the slope. The thickness of dam at crest is usually set by the roadway

or other access requirements for the non- overflow section. However this should be adequateto withstand all possible loadings including ice pressure, wind pressure and impact of floatingobjects/debris. When additional crest width is used, the d/s face is usually vertical from thed/s edge of crest to an intersection with the sloping d/s face.

A batter may be used on the lower part of the u/s face to increase the base thicknessand thereby improving the sliding safety at the base .The lower edge of u/s batter may also

be used as grout cap.

Spillway may be incorporated in the dam by providing an overflow section but thelayout of the spillway section should be similar to the non-overflow section. The curvesdescribing spillway crest and lower energy dissipater are designed to meet the hydraulic

requirements and slopes are set as tanget to straight segments.Golze P438 fig. 8-4-1

The maximum water surface elevation should not exceed the top of non-overflowsection of dam-A solid concrete parapet wall can be included on the top of the dam to providefreeboard against wind/wave section.

5.6 DAM SAFETY ANALYSIS

Dam safety is required against: i. Rotation and overturning, ii. Translation and sliding,iii. Overstress and material failure. Criteria i and ii must be satisfied with respect to the

profile above all horizontal planes within the dam and the foundation. Overstress criteria

must be satisfied for the dam concrete and for the rock foundation.5.6.1 Safety against Overturning:

This is defined by comparing the moments of the forces resisting overturning(righting moment M R ) to those forces causing overturning (overturning moment M O ) againstthe dam toe. The resisting forces are weight of the dam (W) and weight of the water wedgeover the u/s (W HW ) and d/s (W TW ) batter/slope. The overturning forces are water pressure atthe upstream face (P wh ), the uplift pressure at the base (P u), pressure at u/s face from silt (P sh)/ ice/wind/wave. During an earthquake the dam weight will decrease, but the water pressure(hydrodynamic) will increase. Thus factor of safety against overturning FS O

3/3/3/ 21 hPbPhPhPbW bW bW

M M

FS twhuusshwh

twtw HW HW w

O

RO ×−×+×+××+×+×

=ΣΣ

=

is given as:




where b w , b HW , b tw , b u is distance from toe to centroid of dam section, to centroid of u/swater wedge, to centroid of tail d/s water wedge, and centroid of uplift distribution diagram,respectively and h 1 , h 2 and h s is depth of u/s water depth, d/s water depth and silt deposition.The FS O

5.6.2 Safety against sliding

> 1.25 is acceptable but > 1.5 is desirable under extreme load condition Davis (p-11.14).

The resistance of a gravity dam to sliding is dependent on the development ofsufficient friction resistance and shearing strength along the potential sliding surface, e.g. aconstruction lift joint, weak seam in the foundation or along the dam base. The frictionresistance is created due to interlocking of the grains under load on two sides of the slidingsurface. The shearing resistance develops due to cohesion and the internal friction.

Sliding Factor F SS

The factor F

:

SS is expressed as function of the resistance to simple sliding over the planeconsidered; where resistance is from friction of the contact surfaces only and no shear

strength or cohesion is mobilized. For a horizontal contact plan the F SS

V H

F SS ΣΣ=

is as ratio of the sumof all horizontal load components to sum of all vertical loads:

where ΣH = algebraic sum of all active horizontal forces (water pressure, silt, ice, wave,wind, hydrodynamic water pres sure, horizontal inertia of d am body, etc.), ΣV = alg ebraicsum of all active vertical forces due to weight of dam and water wedge on u/s and d/s faces,vertical inertia of dam body). If the contact plan is inclined at a small angle α, (α is defined+ve if sliding operates in an uphill sense) then factor of safety is as:

α α

tan)/(1tan)/(

V H V H F SS ΣΣ+ −ΣΣ=

The ΣH and ΣV are maximum and minimum values appropriate to the loading condition andΣV is determined allowing for the effects of uplift.

For well constructed mass concrete the FSS should nor exceed 0.75 for normal loading and0.9 for extreme loading. If a plane of low shear resistance is present then the FSS may belimited to 0.5 or less on some limestones, schists, laminated shales and similar low strengthfoundations.

Shear friction factor F

The F

SF

SF

H S

F SF Σ=

is defined as the ratio of total resistance to shear and sliding S which can bemobilized on a plane (from both cohesion c and the frictional component of shear strength tanφ) to the total horizontal load as:

The maximum shear resistance is defined as:

( ) )tan(

tantan1α φ

α φ α +Σ+

−= V

CoscA

S h

where A h is the area of plane of contact or sliding. For a horizontal contact plane (α = 0) thetotal resistance is as:




φ tanV AcS h Σ+= .

If there exists some d/s passive wedge resistance then

F SF = (S + P p

where Pp is wedge resistance given as:

) / ΣH

( ) )tan(

tantan1 0α φ α φ α

++−

= w

AB p W

Cos Ac

P

and W w

is wedge weight and φ is for the base material.

The horizontal shearing resistance is attained by stepping the foundation and by ensuringgood bond between concrete and rock and from successive pours of concrete. For concretedams the resistance to shear on any approx horizontal construction joint above the foundationdepends entirely upon the bond developed between successive concrete pours.

Recommended values of shear friction factor F SF

Sliding plan location:

are as (Noval et al. p-101):

Normalloading

Unusualloading

Extremeloading

1- dam concrete-base interface 3.0 2.0 > 1.0

2- foundation rock 4.0 2.7 1.3

The factor due to combined friction and shearing FS S

H V

H U W cA

FS S ΣΣ=

Σ−Σ+= φ tan)(

is expressed as:

where ΣH = algebraic sum of all active horizontal forces (water pressure, silt, ice, wave,wind, hydrodynamic water pressure, horizontal inertia of dam body, etc.), ΣV = algebraicsum of all active vertical forces due to weight of dam and water wedge on u/s and d/s faces,vertical inertia of dam body), U = total uplift force acting on the base, f’ = coefficient ofmaximum static friction between the two surfaces (e.g. concrete and rock foundation at thedam base, or between concrete and concrete at a lift joint), c = cohesion value of the concreteor rock, A is area of contact at base, φ =coefficient of internal friction (of concrete or rock), b= length of base measured horizontally, q = unit shear resistance of foundation material =c+w tan φ ). The required safety factor is 2-4 for normal loading and 1.25-1.5 for extremeloading.

α α0

B

A

Ww




Table 5.*: Range of shearing resistance parameters. Descript ion Cohesion c

MN/mFrictionTanφ 2

Mass concrete intact 1.5-3.5 1.0-1.5H or construction joint 0.8-2.5 1.0-1.5

Concrete rock interface 1.0-3.0 0.8-1.8Rock mass sound 1.0-3.0 1.0-1.8

inferior < 1.0 < 1.0Sound conditions [competent

parent rock, few significantdiscontinues in mass, nosignificant alteration orweathering]

gneiss 1.6 1.7granite 1.5 1.9Mica schist 3.0 1.3Sandstone 1.0 1.7

Inferior conditions Gneiss unaltered 0.6 1.0Granite weathered 0.3 1.3Grey wacke < 0.1 0.6Lime stone open jointed 0.3 0.3Mica schist 0.4 0.7Sand stone 0.1 0.6

Critical foundation features Faults/crush zone materials < 0.2 < 0.3Clay seams / joints infill < 0.1 < 0.2

Shale dry 0.2 0.4saturated 0.0 < 0.2

5.6.3 Stress Analysis

Concrete dams are designed to ensure:

• No tension in any part of the concrete,

• Compression stresses are within the maximum permissible limit (in the elastic range)

• Shear stresses are within the maximum permissible limit (in the elastic range)

Stress evaluation is made for every plane in the section for: (Fig. Navak p-103)

i. vertical normal stress, σz

ii. horizontal normal stress, σ on horizontal plane

y

iii. horizontal and vertical shear stress τ on vertical planes

zy and τ

iv. Principal stresses σ

yz

1 and σ3

Vertical normal stress, σ

(for direction and magnitude)

This is given as (Novak et al):

z

I y M

AW

z

′Σ±Σ=*

σ

where ΣW = sum of vertical loads excluding uplift, ΣM*=sum of momentswith respect tocentroid of plane, y’ = distance from the neutral axis to the point where the stress is beingdetermined, I = second moment of the area A of the plain w.r.t. its centroid (for a rectanglesection area A = T x d and I = d T 3/12, where T = thickness from u/s to d/s side, d = section

width and for d of unit width (d = 1), A = T, I = T3

/12,. Thus:




3

12T

yeW T W

z

′Σ±Σ=σ

where e is eccentricity of the load R. At u/s or d/s face, y’=T/2 and

±

Σ= T

eT W

z6

1σ

The stresses at two faces are as below.

U/s face, i.e. heel:

−Σ=

T e

T W

u z

61:σ ,

D/s face, i.e. toe:

+Σ=

T e

T W

d z

61:σ

The eccentricity e = ΣM*/ΣV and ΣM*= (ΣMR – ΣMO

Horizontal and vertical shear stress τ

). For e > T/6, the u/s face becomes

under tension. Therefore the resultant R must intersect the plain d/s of its centroid forreservoir full conditions. The vertical normal stress varies linearly from u/s to d/s faces.

zy and τ

The horizontal and vertical shear stresses are numerically equal and complimentary and aregenerated due to variation in vertical normal stress over a horizontal plain. The boundaryvalues at u/s and d/s faces are as:

yz

uu zwu p φ σ τ tan)( :−= and d d zd φ σ τ tan:= where uφ and

d φ are upstream and downstream slope angles from vertical, u z:σ and d z:σ are vertical normal

stresses at u/s and d/s faces and p w

uu zu uplift Max φ σ τ tan)( : −=

is external hydrostatic pressure. The maximum shear

stress is at u/s face as:.

The shear stress variations from τu and τd

Horizontal normal stress, σ

depends on rate of change of vertical normal stressand usually have a parabolic distribution.

These operate on vertical plains. Boundary values are as:

y

U/s face: uwu zwu y p p φ σ σ 2:: tan)( −+=

and d/s face: d d zd y φ σ σ 2:: tan= [p w = γw

Principal stresses σ

h]

1 and σ

These are determined from σ3

z and σy

Major:

and the Mohr circle diagram. Major stresses are:

max1 2τ

σ σ σ +

+= y z and Minor: max2

3 τ σ σ

σ −+

= y z where 2max 2/)( τ σ σ τ +−=

y z

The u/s and d/s faces are each plains of zero shear and therefore plains of principal stress.Boundary values are as:

U/s face: uwuu zu p φ φ σ σ 22

::1 tan)tan1( −+= and wu p=:3σ




D/s face: )tan1( 2::1 d d zd φ σ σ += and 0:3

=d σ

Punmia defined minor principal stress as: ( ) uw zuu p φ σ σ tan3 −−= and d zd d φ σ σ tan3

= .

Horizontal cracks:

Horizontal cracks may appear on u/s face if vertical stress σzu

t

t wd zu F

zK

'''

Min σ γ

σ −

=

is small. The minimum stress

required to stop cracks is as: where K’ d = drainage factor (~ 0.4 if

relief drains are effective and 1.0 if no drains), σ’t = tensile strength of concrete across ahorizontal joint surface, F’ t

Permissible stresses:

= factor of safety.

Compressive stresses are generally low as 2-3 MN/m 2, but may exceed for largestructures. Factor of safety F c

Load combination F

≥ 3.0 is usually prescribed. In some cases both factor of safety

and maximum stress are defined as:c (concrete) F r

Normal 3.0 (σ (rock)

max ≤ 10 MN/m2

Unusual 2.0 (σ) 4.0

max ≤ 15 MN/m2

Extreme 1.0 1.3

) 2.7

5.7 GRAVITY PROFILE DESIGN

The shape of gravity dam consists of a triangular profile and a vertical u/s face is invariablyassociated with a mean d/s slope of the order of 0.75H:1V. The primary load regime for agravity dam of given height is fixed. Little scope exists to modify the standard triangular

profile. Design of small dams is based on adopting such a geometry and checking itsadequacy for stability against sliding, overturning and material stress and making anynecessary minor modifications. Large dams require a unique profile to match the site specificconditions applicable. Multi-stage and single-stage approach is used.

Multi-stage: A profile is defined whereas u/s and/or d/s slope are altered at suitable intervals.Design commences from crest level and descends down. Each stage is proportioned so as tomaintain stress levels within acceptable limits.

Single stage:

η γ γ φ −= wcd //1tan

A suitable profile of uniform d/s slope is defined. The apex of the theoreticaltriangular profile is set at or above the maximum retention level or DFL and an initialrequired base width T is determined for each loading condition for safety against overturning.The critical value of T is then checked for sliding stability and modified, if necessary.Subsequently heel and toe stresses at base level are compared withmaximum permissible. Approximate definition of d/s slope (angle tovertical) giving no tension in u/s vertical face is as:

For no tension the resultant force vector R should act at the ‘inner third point’ under reservoir empty condition and through ‘outer third point’ i.e.

Base width by Stress criteria

M1 M 2

R




point M 2 under reservoir full condition. Taking moment of all forces about M 2

032

132

132

1 2 =×−×+× T hT

T hc

hh cww γ λ γ

:

, from wherec

hT

c −

=γ

, where c = uplift

pressure coefficient (c~1 in the absence of any grouting, cutoff, u/s blanket etc). For γc

hT 83.0==2.45

and c =1, .

For no sliding to occur, ΣH forces causing sliding should be balanced by the frictional force.Thus γ

Base width by sliding stability criterion

w h2 / 2 = f’(W – U) = f’ (γc T h/2 – c T γw )(' c f hT c −= γ h) and thus where f’ =

friction factor concrete to base rock.

The maximum stress is given as:

Criteria for maximum stress

±Σ=

T

e

T

W z

61σ .For x

T e −=

2

and

V

M M x O R

Σ

Σ−Σ=

and ΣV = (W – U). For full reservoir conditions: )( ch cw z −= γ γ σ and stress at heel will be

zero. The principal stress at toe is as: f ch cw ≤+−= )1(1 γ γ σ from where

)1( +−≤

c f

hcw γ λ

where f is maximum allowable stress. The shear stress at toe is as:

d d zd φ σ τ tan:= ch cw

−= γ γ

Allowable unit shear stress for concrete ~ 14 kg/m

Allowable unit tensile stress for concrete ~ 4.2 kg/m

2

Unit weight of concrete ~ 2400 kg/m

2

Permissible stress ~ 2 – 3 MN/m

3

Adopted safety factor ~ ≥ 3.0

2

The dam height is given as:

Limiting height

)1( +−≤

c f

hcw γ λ

. For c = 0, γw = 1000 kg/m 3, γc = 2.4, the

dam height h ≤ 0.295 f. For f = 300 ton/m 2, h = 300/[1(2.4 – 0 + 1)] = 88 m. For h > 88 m, the

dam is called as high dam, and stresses for high dam could increase from the allowablestresses. To accommodate this, the dam section is given extra slope to u/s or d/s faces or usehigher strength concrete.

Adding top width for roadway etc induces tension in the u/s face due to shift of centroid of

the dam mass. A batter is provided below depth y’ (below triangle apex) and

Top width

ca y c −= γ 2'

where a = top width. Crest width is determined by practical considerations as need of travelacross the dam, access to gate operating mechanism, climatic conditions, highway crossing,etc.




SHAFTS/GALLERIES (USACE 1995, EM 1110-2-2200)A system of galleries, adits, chambers, and shafts is usually provided within the body of thedam to furnish means of access and space for drilling and grouting and for installation,operation, and maintenance of the accessories and the utilities in the dam. The primaryconsiderations in the arrangement of the required openings within the dam are their

functional usefulness and efficiency and their location with respect to maintaining thestructural integrity.a. Grouting and drainage gallery . A gallery for grouting the foundation cutoff will extendthe full length of the dam. It will also serve as a collection main for seepage from foundationdrainage holes and the interior drainage holes. The location of the gallery should be near theupstream face and as near the rock surface as feasible to provide the maximum reduction inoverall uplift. A minimum distance of 5 feet should be maintained between the foundationsurface and gallery floor and between the upstream face and the gallery upstream wall. It has

been standard practice to provide grouting galleries 5 feet wide by 7 feet high. Experienceindicates that these dimensions should be increased to facilitate drilling and groutingoperations. Where practicable, the width should be increased to 6 or 8 feet and the height to 8feet. A gutter may be located along the upstream wall of the gallery where the line of groutholes is situated to carry away drill water and cuttings. A gutter should be located along thedownstream gallery wall to carry away flows from the drain pipes. The gallery is usuallyarranged as a series of horizontal runs and stair flights. The stairs should be provided withsafety treads or a non slip aggregate finish. Metal treads are preferable where it is probablethat equipment will be skidded up or down the steps since they provide protection againstchipping of concrete. Where practicable, the width of tread and height of riser should beuniform throughout all flights of stairs and should never change in any one flight. Furtherdetails on the grouting and drainage gallery are covered in EM 1110-2-3506.b. Gate chambers and access galleries . Gate chambers are located directly over the serviceand emergency sluice gates. These chambers should be sized to accommodate the gate hoistsalong with related mechanical and electrical equipment and should provide adequateclearances for maintenance. Access galleries should be sufficient size to permit passage of thelargest component of the gates and hoists and equipment required for maintenance. Drainagegutters should be provided and the floor of the gallery sloped to the gutter with about 1/4inch/foot slope.

FILTERS AND DRAINS

FREE BOARD

WEEP HOLES

SPILLWAY AND TUNNEL LAYOUTFOUNDATION REQUIREMENT

Solid rock foundations

Gravel foundations

Silt or fine sand foundation

Clay foundation

Foundation treatments

ARRANGEMENTS FOR CONTROL OF SEEPAGE




EXAMPLE

For the gravity dam shown in thefigure, given γC = 2400 Kg/m 3,γw = 1000 Kg/m 3, friction factorf ′ = 0.75, maximum shear stressat concrete-rock interface = 14Kg/m 2

Solution

, determine the (i) Factor ofsafety against overturning, (ii)Factor of safety against sliding,(iii) Shear friction factor, (iv)Compressive, shear and principalstresses at toe and heel at base,(v) Compressive, shear and

principal stresses at toe and heelat 4 m above the dam base.

(1) Loads:

(a) Self weight = W d

(b) Weight of water column over u/s sloping face = W

= 2400*[10*2/2 + 10*2 + 6.25*10/2] = 24000 + 48000 + 75000 =147,000 Kg

w

(c) Uplift pressure = U = 1000 * [10*10.25/2] = 51,250 kg

= 1000*[10*2/2] = 10,000 kg

(d) Vertical net weight = ΣV = Wd + W w

(e) Horizontal water pressure load W - U = 147000 + 10000 - 51250 = 105,750 kg

h = 1000*[10 2

(f) Total horizontal load =ΣH = W/2] = 50,000 kg

h

(2) Moment due to various forces about d/s toe

= 50,000 kg

(a) M d

(b) M

= 24000*(2*1/3+2+6.25) + 48000*(1+6.25) + 75000*(2/3*6.25) = 214,000 +348,000 + 312,500 = 874,500 kg-m

w

(c) M

= 10000*(2*1/3+2+6.25) = 95,833 kg-m

U

(d) M

= 51250 * (2/3*[2+2+6.25]) = 350,208 kg-m

h

(e) Overturning moments M

= 50000*[1/3*10] = 166,667 kg-m

O = M U + M h

(f) Resisting moment = M

= 350,208 + 166,677 = 516,875 kg-m

R = M d + M w

(3) Factor of safety against overturning = FS = M

= 874,500 + 95,833 = 970,333 kg-m

R /M O

(4) Factor of safety against sliding: F

= 970,333 / 516,875 = 1.88

S

(5) Shear friction factor at base: F

= f’* ΣV / ΣH = 0.75 * 105,750 / 50,000 = 1.58

SF = [f’ ΣV + c b] / ΣH ( b = base width = 10.25 m ) =[0.75*105,750+14*10.25]/50,000 = 1.59

6.25 m2 m2 m

10 m

NCL

10 m Uplift

B C

A D TWL




(6) Stresses at bottom:

(a) Location of resulting loads (weight, u/s water column weight, water pressure) = x =[M d + M w + M h] / [W d + W w

(b) Eccentricity = e = b/2 -

] = 874,500 + 95,833 + 166,667 ] / [147,000 + 10,000] =5.46 m (from toe)

x = 10.25/2 – 5.46 = - 0.31 m (load is acting 0.31 m left ofsection centre i.e. towards heel)

(c) Compression stress at toe = σzd = ΣV/T (1+6e/T) [Note: ΣV = Wd+W w = 147,000 +10,000 = 157,000 kg ], and σzd = 157000/10.25*(1+6*-0.31/10.25) = 12,537 kg/m 2

(d) Compression stress at heel = σ

=12537*9.81/1000 = 123 KPa

zu = ΣV/T (1-6e/T) = 157000/10.25*(1-6*-0.31/10.25)= 18,096 kg/m 2

(e) U/s and d/s angles: Tan φ = 177.5 KPa

u = 2/10 = 0.2, Tan φd

(f) p

= 6.25/10 = 0.625

w = 10 m * 1000 kg/m 3 = 10,000 kg/m

(g) Horizontal shear stress

2

u/s face: τu = (p w - σzu) Tan φu = (10,000 – 18,096) * 0.2 = - 1,620 kg/m 2

d/s face: τ

= - 15.9KPa

d = σzd * Tan φd = 12,537 * 0.625 = 7,835 kg/m 2

(h) Principal stresses:

= 76.9 KPa

σ1u = σzu(1+Tan 2φu) – p w Tan 2φu = 18,096*(1+0.2 2) – 10,000 * 0.2 2 = 18,419 kg/m

σ

2

3u = p w

σ = 10,000 kg/m2

1d = σzd * (1 + Tan 2φd) = 12,537 (1 + 0.625 2) = 17,434 kg/m 2

Σ.

3d

= 0




5.11 CONCRETE ARCH DAMSArch dams with u/s curvature are suitable for sites where the geologic strata of the

abutments are strong enough to take up the forces of dam coming on the abutments (Figs.5.21 and 5.22). The arch dams are suitable for tall rocky gorges and represent a stablestructural form given that the integrity of the supporting abutments is assured. As the damthrust is taken by the abutments, the concepts of overturning and sliding stability have littlerelevance to the arch or cupola dam. Arch dams can fail for overstress only. Arch dam designis therefore centered largely upon the definition of the arch geometry and stress analysiswhich avoids local tensile stresses and/or excessive compressive stresses.

Arch dams were designed earlier as cylindrical shell element. These assumed thatwater load is entirely carried by individual arches and only the weight of the structure wasconsidered to be transmitted to the foundation. This practice did not consider the cantileveraction arising from resistance from foundation and is satisfactory for small dams. For large

dams complex 3-d analysis is required. For Cupola dams the compression due to dead loadcompensate for the cantilever tensions caused by water load.

Figure 5.21: Arch dam: El-Atazar dam, Madrid, Spain (Lozoya River). (Source: Canal de Isabel @

http://www.pbs.org/wgbh/buildingbig/dam/el_atazar.html ). Height = 134 m, reservoir = 424 Mm 3. basewidth at foundation = 52.3 m . Construction 1972. Repaired for 150 ft deep crack in 1979.Repaired cracking in foundation rock in 1983.

http://www.pbs.org/wgbh/buildingbig/dam/el_atazar.html






Figure 5.22: Pacoima dam California USA. (Source: Top: http://www-socal.wr.usgs.gov/scign/group/pacoima_dam, bottom: Google-earth)

Vaiont Dam, Italy. Height = 162 m, crest length = 190 m. Completed in 1961. During firstfilling it was damaged by a massive land slide in the reservoir when 314 Mm 3

Highest arch dam is Inguri dam, Georgia, Soviet Union, with 272 m high and 680 m long; theconcrete volume is 3.9 Mm

material slidinto reservoir causing a surge of 260 m on opposite side of valley forcing a 100 m high floodwave overtopped the dam, resulting in severe damages in the d/s valley including 2500 lossof life. The dam with top width of 3.4 m stood strong with superficial damage to dam.

3.

http://rds.yahoo.com/_ylt=A9G_bHNr_sNHJP8AhYajzbkF/SIG=1280o52me/EXP=1204113387/**http%3A/www-socal.wr.usgs.gov/scign/group/pacoima_dam







Figure Morrow Point arch dam. [http://simscience.org/cracks/advanced/arch_hist1.html]

5.12 DEFINITIONS

Terminology used in the layout and analysis of arch and cupola dams are as under.

Plan: A plan is an orthographic projection on a horizontal plan, showing principal features of

a dam and its appurtenant works with respect to topography and geologic data.Profile: A profile is a developed elevation of the intersection of the dam with the ground

surface, rock surface or the excavated surface along the axis of the dam, or u/s face, ord/s face, or other designated location.

Section: A section is representation of a dam as it would appear if cut by a plane.

Dam axis: This a vertical reference plane coincident with the u/s edge of the top of dam; in anarch dam axis is curved horizontally.

Length of dam: This is distance measured along the axis at the top of the main dam bodyfrom abutment to abutment, including any overflow spillway section but excludingany spillway located adjacent to the dam on its abutment.

Arch or Arch Unit: The arch unit refers to a portion of the dam bounded by two horizontal planes unit height apart. Arches may have uniform or variable thickness. Arches atdifferent levels may have same or varied location (in plan) of centers.

Cantilever or Cantilever unit: This refers to portion of dam contained between two verticalradial planes unit length apart. The cantilever may have straight or curves surface onu/s and d/s side.

Extrados and Intrados: This refers to the u/s and d/s curved surface or face of the arch / arch

unit of the dam. These terms are used for horizontal (arch) units; the faces ofcantilever are referred as upstream and downstream, as appropriate.




Central angle: This is the angle bounded by lines radiating from the arch extrados center to point of intersection of the arch centerline with the arch abutments.

Crown cantilever: This is a cantilever element located at the point of maximum depth in thecanyon.

Reference Plane: This is vertical plan which pass through the crown cantilever and the centerfor the axis radius.

Line of Centers: This is the loci of centers for circular arcs used to describe the face or portion of dam.

Figure 5.23 : Typical arch unit and cantilever unit. (USCoE 1994) p:1-3

5.13 TYPE OF ARCH DAMSThe types of arch dams may be based on mode of curvature, on symmetry, and

thickness, as under.

Types based on curvatur e

Single curvature arch dam: These dams are curved in plan only. Cantilevers have u/s verticaland d/s straight slope faces, or may also be curved with the limitation that no concreteoverhangs the concrete below.

Double curvature cupola dams: Cupola dams have curved surface both in plan and crosssection. This type of dam utilizes the concrete weight to greater advantage than singlecurvature dam, and consequently less concrete is needed resulting in thinner moreefficient dam.

Types based on symmetry

Symmetrical arch dams: The dams are termed as symmetrical if the arch length on each sideof reference plane differ by less than 5% between 0.15H and 0.85H (H = maximumdam height).

Non-symmetrical arch dams: Non-symmetrical sites result in dams with longer arch lengthson one side of the crown cantilever than the other. Such dams will often have two

reference planes, one for each side of the crown cantilever with same or different axisradius. Symmetry of dam may be improved by excavating deeper in appropriate




places, by construction an artificial abutment, or by reorienting and relocationg thedam.

Figure 5.24: Single and double curvature arch dams. USCOE 1994, p:1-10.

Types based on thickness

Dams are classified as Thin, medium thick, Thick and Very thick dams based on basethickness (T) to height (H) ratio; Very thick dams are also termed as Arch-gravity

dams, as under.Thin-arch dams. These dams have maximum cantilever thickness less than 0.2H. (T < 0.2H)

Medium-Thickness arch dams. These dams have maximum cantilever thickness between0.2H and 0.3H. (T ~ 0.2 to 0.3H)

Thick arch dams. These dams have maximum cantilever thickness of more than 0.3H. (T >0.3H)

Arch-gravity dams: These dams have maximum cantilever thickness of more than 0.5H. (T >0.5H)

Uniform thickness dams:

Variable thickness dams:

Types based on ar ch geometry and prof ile (radius and central angle)

Constant radius profile: This has the simplest geometry, combining a vertical upstream faceof constant radius with a uniform radial downstream slope (Fig. Novak p-117,118,Punmia p-338 fig 9.1). The downstream face radius therefore varies with elevation.The extrados and intrados radius differ by the arch thickness only and centers of archelements at various elevations lie on a single vertical line. The central angle reaches a

maximum at crest level. In a symmetric valley the minimum dam volumes result fromcentral angle of 133.34 degrees. However the considerations of the abutment entry




angle limit the central angle to less than 110 degrees. This profile is most suited torelatively symmetrical narrow-U shaped valleys.

Variable radius profile: The profile radius for extrados and intrados vary at various elevations being maximum at the crest and certain minimum at its bottom. The central angle also

varies to obtain maximum arch efficiency. The dam has vertical or overhanging faceson u/s side near the abutments and on d/s side near the crown The centers of archelements at various elevations do not lie on a single vertical line, hence it is alsoknown as variable center arch dams.

Constant angle profile: The central angle for all arch elements is constant with varied radiusresulting in a complex geometry. This results in considerable upstream overhang nearthe abutments. This profile is most suited for symmetric narrow-V shaped valleyswith steep side slopes.

5.14 ARCH SHAPES

Several configurations are available to describe the horizontal arch shapes. These are as:

Single centered dam: For symmetric valleys a single centered profile with vertical or curvedline of centers is used. The loci of the centers are the same for both sides of the damand lie on the reference plane.

Two-centered dams: For nonsymmetrical valley, single centered profile cannot besatisfactorily fitted to the site. A two centered scheme can be used; such profile hastwo separate set of lines of centers, one for each side of the dam. Each line of centersmust lie along the reference plan to maintain continuity, but axis radius may bedifferent on the two sides.

Three centered profile: For wide V or U valleys the arch-abutment joint may not be efficientto achieve required thrust angle. The three centered or elliptical arches are used. Thethree centered dam has a shorter radii in the central part and longer radii in the outersegments. The loci of centers for central portion lie on the reference plan but those ofouter portion do not. These have inherent characteristics of conforming more nearly tothe line of thrust.

Fillet: This is enlargement of arch cross section towards the abutment. For sites whereabutment thickening is desirable, short radius fillet may be added to uniform thicknessarches on the d/s sides. The fillet radii at each side of arch need not be same.

5.15 VALLEY SHAPES

The geometry of the arch evolved gradually over the years due to better understandingof its behavior under loads. As the arch dams transmit the impose loads by arch action intothe valley walls, the arch geometry and shape of the dam is dependent upon the cross sectionand geometry of the valley itself (Fig. 5.25). The canyon shape is described by canyon width

between the abutments (L C ) to height H ratio (L C

Narrow- V site have canyon L

:H) and the canyon wall slope. These parameters affect the central angle, shape of the profile and type of the dam layout. Thevalley may be classified as:

C :H of 2:1 or less. Canyon walls are generally straight with fewundulations and converge to a narrow streambed. Arch dams are such sites will




transmit the applied load to the abutments by the arch action. The lower arches arerelatively short and greater portion of the load is carried by arch action. A thin singlecentered arch dam of uniform thickness is suitable for this.

Wide-V site have canyon L C

Narrow-U site have near vertical canyon walls in the upper half of the canyon. The lowerarches have approximately the same chord length as those near the top. The streambedis fairly wide perhaps one-half the canyon width at the crest. The water load over thelower quarter height will be transferred to foundation and supported by cantileveraction towards the lowest point and balance upper load will be transferred to abutmentrock by arch action. A single centered arch dam with uniform thickness upper archesand variable thickness lower arches is suitable.

:H of 5:1 or more to as much as 10:1. Canyon walls have more

pronounced undulations but will become generally straight after foundationexcavation converging to a less pronounced V-section streambed. Most of live loadwill be transferred to abutments by arch action. A three-centered medium thicknessarch dam with increased thickness near the abutments is suitable option.

Wide-U sites have very wide streambed with near vertical walls above it. Much of the liveload is carried by cantilever action and arches are relatively long. These sites ate mostdifficult for an arch dam because long flexible arches carry relatively load, and thickcantilevers are needed support the increased water pressure. Arch thickness isgenerally uniform for arches near c rest and variable near streambed with transitionnear the upper one-third level. Three centered dams can be used advantageously insuch case.

An arch dam must be given first consideration for a site with length-height ratio ofless than 3 or less. For sites with length-height ratio of 3 to 6, an arch dam may still providemost feasible structure depending upon the extent of foundation excavation required to reachsuitable materials. The effects of factors other than length-height ratio become much more

predominant in the selection process for dam sites with length-height ratio of more than 6.

Narrow-V

Narrow-U

Wide-V

Wide-U

Figure 5.25: Canyon valley shapes.

LC

H




Figure 5.26: Typical single centered variable thickness arch dam in a symmetrical site.(USCOE 1994) p:1-5

Figure 5.27 : Typical two centered variable thickness arch dam in a non-symmetrical site.(USCOE 1994) p:1-6




Figure 5.28: A three centered variable thickness arch dam. (USCOE 1994, p:1-7)




Figure 5.29a: Monar Dam. (http://www.corestore.org/DeanieMonar.htm)

Figure 5.29b: Plan and elevation of Monar dam(Source: htto://www.corestore.org/MonarDraw1.jpg)




Figure 5.29c: Section of Monar dam (Source: htto://www.corestore.org/MonarDraw2.jpg)

5.16 ARCH-ABUTMENT CONTACT

The horizontal angle of arch thrust must be transferred into the abutment at a safeangle, i.e. one which will not promote abutment yielding or instability (Fig. 5.30). Thehorizontal thrust is considered to distribute into the rock with an included angle of 60° (30°on either side from tangent to extrados and intrados). The thrust must not be aligned tooclosely with the valley sound rock contours or with any major discontinuity. This suggests

abutment entry angle β (tangent of extrados and approx trend of sound rock contours alongvalley sides) between 40 and 70°.




Figure 5.30a: Angle between arch thrust and the rock contour. (USCOE 1994, p:2-2)

Figure 5.30b: Arch abutment types. (USCOE 1994, p:2-3)

5.17 CHARACTERISTICS OF ARCH DAMS

1. Arch dams utilize strength of the dam’s material to counteract the imposed loads asopposed to gravity dam which rely on the weight of the materials.

2. Arch dams are more economical because less material is needed for construction.

3. V-shaped valleys are preferred arch dam construction sites.

4. Reinforcement is generally not required in arch-gravity and thick-arch dams. Its use isfavored in thin arch dams, but for high dams the reinforcement cost become

prohibitively high to negate the adoption of such dams.

5. Uplift is usually not important in thin arch dams, but suitable internal drainage is provided in thick-arch dams.






Figure 5.32: Thrust block at right abutment. (USCOE 1004, p:3-27)

Figure 5.33: Straight and curved thrust block, plan view. (USCOE 1994, p:3-27)




Figure 5.34: Developed profile of arch dam.

Figure 5.35: Empirical values of L 1 and L 2

i. Select a suitable location where geologic foundation is available of requisite standard.

(USCOE 1994, p:5-3)

ii. Mark abutment location where arch can be placed. Measure straight line distance L 1

iii. Determine R

between the crest level topographic contours (or to abutment excavated to assumedrock quality at crest).

axis = 0.6 L 1

iv. Draw an arc of radius = 0.6 L1 on a transparent sheet on same scale as topographicmap.

.

v. Overlay the arc on the topographic map. Adjust the arc position to produceacceptable/ optimum position. The acceptable position provides (i) shortest dam




length along the arc, (ii) the central angle between the abutments do not exceed 110°,and (iii) the incident angle of the arc β be between 40 and 70 degrees. If angle > 120degrees, then increase radius.

vi. Two or three centered profile may be used for unsymmetrical canyon shapes.

vii. Trial and error procedure is used to obtain suitable location and orientation of thecrest arch.

viii. Locate crown cantilever, the reference plane, etc.

ix. Define the suitable single or double curvature cross section.

x. Select the line(s) of centers and complete plan location of lower arches ensuring thegeometric constraints.

xi. The geometry of crown cantilever controls the shape of entire dam. Trial dam sectionthickness of crown cantilever at crest (T c), at base (T B) and at 0.45 height (T 0.45H

( )12.101.0 L H T c +=

) as:

( )3 400/21 400/0021.0 H

B H L L H T = B H T T 95.045.0 =

xii. Dam axis is set at u/s end of crest. Crest width is set d/s of axis. Width at base is taken 0.67 and 0.33 on u/s and d/s of axis. Width at 0.45H is takenu/s of axis (Fig. 5.37).

xiii. Carryout requisite stress analysis.

xiv. Accept or modify the design depending on computed stress levels.




Figure 5.36: Layout of dam axis. (USCOE 1994, p:5-4)

Figure 5.37: Projections of crown cantilever and definition of u/s and d/s faces. (USCOE1994, p:5-6)

Figure 5.38 : Contact points between dam and foundation at crest and crowncantilever.(USCOE 1994, p 5-7)




Figure 5.39 : Development of the line of centers. (USCOE 1994 p 5-11)

Figure 5.40: Upstream contact line at dam-foundation interface (USCOE 1994, p:5-8)




Figure 5.41: Arch sections. (P:5-18)

Figure 5.42: Cantilever sections. (p:5-17)




Figure 5.43: Spillway and outlet works for arch dam.

5.21 DAM STRESS ANALYSIS

Dam stress analysis can be done under different conditions as: (i) static conditionswhen all loads except earthquake loads are considered, (ii) pseudo-static conditions when allloads including earthquake loads are considered, and (iii) dynamic conditions, when suddentemporal earthquake loads of finite duration are superimposed over the static loads. The trialsmay be based on static or pseudo-static analysis, and final selected design must then be re-evaluated under dynamic loads.




The input data to stress analysis includes concrete properties (modulus of elasticity,Poisson ratio, unit weight, permissible maximum compressive and tensile strength,coefficient of thermal expansion), Dam geometric data (curvatures, radius, height, arch lengthand thickness, head and tail water levels, dam spillway and other openings, thrust blocks,expected temperature changes at site, sediment, wind, ice, seismic and other loadings), and

Foundation and abutment properties (modulus of elasticity, Poisson ratio) etc.5.22 STATIC ANALYSIS

This section describes analysis and evaluation procedures required for assessing thestructural stability of arch dams and their abutment foundation under static loads. Theacceptable methods of analysis for computing deflections and stresses developed in the daminclude Thin Cylinder Theory, Trial Load Method and three dimensional finite element (FE)and in certain cases continuum solution procedures, as applicable.

5.22.1 Thin Cylinder Theory

The thin shell/cylinder subjected to uniform water loads will develop compressive

stresses (considered to be uniform over the section). The maximum stress will develop nearthe abutment as: T Rhw /γ σ = where h is dam height, T is arch thickness, and R is archradius. If the maximum permissible compressive stress is f n

nw f RhT /γ =, then required arch thickness is

given as: . The thin cylinder theory give very simplistic picture as: (i) it do notconsider bending and shear in concrete, (ii) it consider water load only, (iii) it do not accountfor stresses induced due to rib shortening and abutment yielding.

5.22.2 Finite Element Analysis

The finite element procedure is the numerical method most often used for thestructural analysis of arch dams. This guideline assumes that the reader is already familiar

with the general theory of finite element analysis of elastic solids (Zienkiewics, 1971; Batheand Wilson, 1976). The FE stress analysis should be conducted by developing an accuratethree-dimensional model of the dam-foundation system. The manner by which various staticloads are applied should be described. The results of analyses should be presentedappropriately in order to facilitate examination, interpretation, and evaluation of the findings.FE models as SAP2000 can be used for 3-D structural analysis of arch dams (Sarwar 2005).The following remarks are intended only to point out some special considerations in theapplication of this technique to arch dam analysis.

Structur al M odeli ng Assumpti ons

The finite element analysis of arch dams is based on the same assumptions that underlie all

finite element analyses. This being the case, the basic principles that govern elementformulation, mesh construction, and load application are as valid in the analysis of arch damsas they are anywhere in structural mechanics. There are, however, certain specialconsiderations in the use of the finite element in arch dam analysis:

1. The body of the dam is typically assumed to be bonded to the foundation rock throughoutits contact with the canyon. However, the validity of this modeling assumption is oftenwhat the analysis is seeking to determine. If this assumption results in excessive shear ortensile stresses on the foundation contact, this modeling assumption may requiremodification.

2. The dam is typically assumed to be a monolithic structure with linear elastic and isotropic

material properties. In reality, the typical arch dam is divided by construction joints, con-




traction joints, and pre-existing cracks. In addition,concrete by its nature is not isotropic because itscompressive strength is typically 10 times itstensile strength.

3. The foundation rock is assumed to be monolithicwith linear elastic and isotropic material

properties, when in reality it is jointed with non -linear characteristics. The use of a "deformationmodulus" instead of the actual Young's modulus isan attempt to deal with the complex character ofthe typical foundation.

Dam M odel

The basic geometry data for developing a 3Dfinite element mesh for the dam can be obtained fromthe construction drawings (Fig. 5.44). In somesituations, however, it may be necessary to confirm the accuracy of such data by visualinspection, and possibly by field surveys, to ensure that the existing conditions of the dammatches the as-built drawings. For example, a severely deteriorated layer of concrete near thedam surface may have lost its strength, suggesting that a reduced dam thickness or a reducedeffective modulus of elasticity might better represent the actual conditions.

In other situations, structural modifications may have increased both stiffness andmass of the dam. Critical gravity abutment thrust blocks that may exist at one or both ends ofan arch dam should be included in the dam model. Smaller and less important thrust blocksmay be considered as part of the foundation rock, and not modeled separately. The FE modeldeveloped for the dam should closely match the dam geometry and be suitable for applicationof the various loads.

The basic results of a finite element analysis include nodal displacements and elementstresses. As a minimum, nodal displacements and surface stresses should be presented for thestatic loading combinations in clear graphical form. Surface stresses should be presented inthe local arch and cantilever directions. Additionally, since nodal loads can be obtained fromfinite element analyses on an element by element basis, dam thrust needed for the rock wedgestability analysis can be determined from loads acting on elements having a common surfaceor a common edge with the dam/foundation contact surface.

Maximum tensile and compressive stresses in an arch dam usually occur at the facesof the dam, therefore evaluation of stresses on the faces of the dam is required. The surfacestresses resolved into arch and cantilever stresses are usually presented in the form of stresscontours on each face of the dam, while surface principal stresses are displayed in the form ofvector plots, as illustrated in Fig. 5.45.

In addition to the arch and cantilever stresses, the magnitudes of the shear stressescaused by the bending and twisting moments should be examined, especially for very thinarch dams and those with cracked sections. These include radial cantilever shear stressesacting radially on a horizontal plane and radial arch shear stresses acting radially on a vertical

plane. Also, excessive tangential shear stress acting on the foundation can be a cause forconcern.

Figure 5.44: Model of dam andfoundation.




Fig. 5.45 Arch and cantilever stress contour and principle stress vector plots.

A concrete arch dam under static loading conditions is considered to be safe fromover stressing failure if the allowable stresses are not exceeded in any extensive area.Allowable stresses of concrete are obtained by dividing the strength capacities by theappropriate safety factors. This requirement is easily satisfied for a well designed arch dam

which resists the loads by developing essentially compressive stresses with very little tension(Fig. 5.45). In other cases compressive stresses usually meet the criteria but tensile stressescaused by temperature loads, or other unfavorable situations may be significant. Whensignificant tensile stresses are indicated, sections of the arches and cantilevers subjected toexcessive tension are assumed to be cracked. This cracking will result in the re-distribution ofstresses and loads.

5.22.3 Trial Load Method

The trial load method is based on the assumption that an arch dam is made of twosystems of structural members: horizontal arch units and vertical beams or cantilever units(Fig. 5.46); that the water load is divided between the arch and cantilever units in such a way

that the resulting arch and cantilever deflections and rotations at any point in the dam areequal (Fig. 5.47). The preceding agreement is accomplished by subjecting arch and cantileverunits to a succession of self-balancing trial-load patterns and solving the simultaneousequations involved. The solution is normally obtained by computers using a trial load

program such as ADSAS developed by the US Bureau of Reclamation. The resulting loaddistributions required to achieve geometric continuity are then used to compute stresses in thedam.




Fig. 5.46 Translations and rotations of archand cantilever units.

Progressing from the simplest to the most comprehensive, a trial load analysis mayconsist of crown-cantilever adjustment, radial deflection adjustment, or the completeadjustment, which includes adjustments for the radial and tangential translations as well asrotations. The crown-cantilever and radial deflection analyses are usually used for the

preliminary and feasibility studies of new dams. For safety evaluation of existing arch damsonly the complete trial load analysis should be attempted.

Many comparisons with measurements from actual dams and scale models as well aswith 3D fintite element analyses have shown that ADSAS gives reliable results for one-, two-or three-centered dam layouts, subjected to standard static loads. It has been usedsuccessfully in the design of new dams over many decades, but its use in the evaluation ofexisting dams is limited to the geometry configurations just described and to static loadingonly. Complex geometry and material property variation, the effects of openings within the

body of the dam, and nonradial abutments cannot be analyzed by ADSAS.

Unlike FEM, ADSAS does not permit analysis of the effects of rapid changes in thedam geometry where detailed stress information may be required. Analysis of the effects ofunusual loads, special boundary conditions, and seismic loading is not possible. The use oftrial load method and ADSAS should be limited to the geometry configurations describedabove so far as the computed static stresses are not excessive. Existing dams located inseismic regions requiring dynamic analysis should be evaluated by the finite element method.

5.23 Sliding on the Abutment Contact

Usually, sliding stability along the dam-foundation contact of a concrete arch dam isunlikely because of the wedging produced by arch action. However, arch dams withrelatively flat abutment slopes, or in cases where the concrete is not thoroughly bonded to thefoundation rock and adequate drainage is not provided, the benefit of wedging will bereduced. In these situations the possibility that a portion of the dam might slide along thedam-foundation contact should be evaluated. The potential for sliding can be evaluated bycomparing computed shear forces with the shear resistance along the dam-foundation contactsurface.

5.24 Buckling Failure Modes

Over and above the determination of stresses and displacements in arch dams, under

some extreme dam geometries such as thin, single curvature dams with large radii, thequestion of buckling stability of an arch dam structure may arise. Figure 5.48 and the

Figure 5.47: Arch and cantilever elements




corresponding equation describe the buckling mechanism of circular arches subject touniform compressive load, qcr

Figure 5.48. Buckling of a simply supported circular arch under uniform load.

5.25 DYNAMIC ANALYSIS

All dams in seismic zone 3 and higher should be evaluated using dynamic analysistechniques. Dams in zone 2 may also require dynamic analysis on a case by case basis.Currently, three-dimensional linear-elastic finite-element analysis is the most commontechnique used for dynamic analysis. A linear-elastic dynamic analysis of arch dams typicallyconsists of the following four basic steps:

1. Determination of design or evaluation earthquakes and the associated ground motions;

2. Development of appropriate three-dimensional finite-element models including dam-foundation and dam-water interaction effects;

3. Specification of dynamic material properties, damping, and reservoir-bottom absorption, ifapplicable; and

4. Computation of the earthquake response and presentation, interpretation, and evaluation ofthe results.

The requirements for development of design orevaluation earthquakes and earthquake ground motionsare obvious. The design or evaluation earthquake for archdams is the maximum credible earthquake (MCE). Theearthquake ground motions include the horizontal andvertical response spectra, or three components ofacceleration time histories. They are applied uniformly atthe fixed boundaries of the foundation model.




Figure 5.49: Gallery system in right side of arch dam. (p:3-21)




Figure 5.50: Gallery system in left side of arch dam.




REFERENCES / BIBLIOGRAPHY

Golze. 1977. Handbook of Dam Engineering.

Pumnia, B.C. and Pande B.B. Lal. 1979. Irrigation and Water Power Engineering, 5 th

Sarwar, Murtaza. 2005. Study of Concrete Arch Dam for Zarwam Reservoir on KurramRiver. M.S. WRE Thesis. Centre of Excellence in Water Resources Engineering,University of Engineering & Technology, Lahore,

edition.Standard Publisheres Distributors. Delhi.

USBR. 1976. Design of Gravity Dams. Denver.

Zee, C. and R. Zee. 2006. Earthquake hydrodynamic pressure on dams. J. HydraulicEngineering, ASCE. 132(11):1128-33. Nov 2006.

www.dur.ac.uk/~des0www4/cal/dams/conc/acone.htm .

www.ferc.gov/industries/hydropower/safety/guidelines/eng-guide/chap11.pdf

www.usace.army/mil/usace-docs/eng-manuals/em-1110-2-2200

http://www.dur.ac.uk/~des0www4/cal/dams/conc/acone.htm


http://www.ferc.gov/industries/hydropower/safety/guidelines/eng-guide/chap11.pdf


http://www.usace.army/mil/usace-docs/eng-manuals/em-1110-2-2200











Figure 5.61: 106 m high Buttress dam, Iran.




Figure 5.62: 106 m high Buttress dam, Iran.




Figure 5.63 : Elmali Dam (Buttress) Turkey (http://www.ce.metu.edu.tr/~ce471/links.htm )

http://www.ce.metu.edu.tr/~ce471/links.htm







.Figure: 5.64: Spaulding Arch Dam, USA.

Figure 5.65 : Porsul Dam (Concrete gravity) Turkey.(http://www.ce.metu.edu.tr/~ce471/links.htm )








Figure 5.66 : Sariyar Dam (concrete gravity) Turkey(http://www.ce.metu.edu.tr/~ce471/links.htm )

Figure 5.67: Gokcekaya Arch dam Turkey (http://www.ce.metu.edu.tr/~ce471/links.htm )












Figure 5.68a: Loch Laggan Dam, Scotland (masonary dam) u/s view L=700’/213 m, H =170’/ 52m (Constructed in 1934) (http://www.flickr.com/photos/graeme_smith/1147054638/ )

Figure 5.68b: Loch Laggan Dam, Scotland (masonary dam) d/s view spillway and outletsworking (www.dcs.st-and.ac.uk/~rd/remote/dam.jpg )

http://www.flickr.com/photos/graeme_smith/1147054638/



http://www.dcs.st-and.ac.uk/~rd/remote/dam.jpg








Figure 5.69: Cahora Bassa concrete arch dam, Mozambique (L = 303 m, H = 171 m,Reservoir is 240 km long with volume capacity as 51MAF~ 63BCM) (http://www.epoch-suite.com/images/casestudies/cahora_01.JPG )

http://www.epoch-suite.com/images/casestudies/cahora_01.JPG







Figure 5.70 : Gordon arch dam. http://en.wikipedia.org/wiki/File:Gordon_Dam.jpg




Fig 5.71 : A steel dam. http://en.wikipedia.org/wiki/File:088808pv.jpg




Figure 5.72: Spillway from Llyn Brianne Dam, Waleshttp://en.wikipedia.org/wiki/File:Llyn_Brianne_spillway.jpg



Tariq. 2008. Dam and Reservoir Engineering 6.1Chapter 6 - Dam Spillways

6-1

Chapter - 6

DAM SPILLWAYS6.1 INTRODUCTION

Spillways are required for storage dams to pass surplus or floodwater flows, whichcannot be contained in the allotted storage space. The excess water is drawn from top of damand conveyed through an artificial waterway back to the river.

• Failure of dam can result due to improper design of spillway (due to overtopping)especially for earthfill and rockfill dams.

• Larger capacity spillway required for earthfill and rockfill dams to avoidovertopping

• Overtopping of concrete dams (gravity, arch) is less dangerous to the damstructure, but yet remain an undesired condition.

• Spillway is to be located such that spillway discharge will not erode or underminetoe of dam

• Spillway surface be able to withstand high scouring velocities as 50+ m/s

• Due to very high exit velocity a device is required to dissipate excess energy at theend of spillway.

• At large storage dams spillway are used infrequently; For small storage dam,spillway use more often or even constantly

• The flood water also carries with it varying sizes of debris (logs and trees

uprooted by the flood water in the catchment areas). Arrangements are required toensure spillway is not blocked by the floating debris and continued performanceof spillway; especially if the flow path include a closed flow section as tunnel.Debris control boom is usually provided.

6.2 LAYOUT/LOCATION

Every dam is provided with one or more spillways to evacuate flood waters. Thelocation of a spillway depends on type of the dam (EF, RF, Concrete gravity, Arch, Buttress).For earthfill and rockfill dams the spillway is usually located separate from main damembankment at an adjacent saddle. For a concrete gravity dam the spillway can be made

integral part of the dam embankment and weir is located at the dam crest and d/s face of theembankment forms the chute. Spillway can also be carried as tunnel through abutments. Archdams may have a spillway separate or part of the dam depending upon dam height andthickness. Spillway for buttress dam is placed across one or more buttress. In fact the choiceof the type of dam and type of the spillway are dependent on each other. Figs. 6.1 to 6.6 showlayout of spillway for some dams.




6-2

Figure 6.1: Hoover dam and spillway layout and debris control boom (top) and side spillwayand tunnel (bottom).

Debris control boom




6-3

Figure 6.2: Tarbela Dam: Embankment and spillway layout (top) and service and emergencyspillway showing control structure, chute, and plunge pool (bottom).




6-4

Figure 6.3: Tarbela dam service and emergency spillway (Source: Pakistan picture gallery).

Figure 6.4: Spillway for Warsak Dam, Pakistan. Source:http://flickr.com/photos/toufeeque/330744180/ , (Dec 23, 2006)

http://flickr.com/photos/toufeeque/330744180/






6-5

Figure 6.5: Main and auxiliary spillway of Simly dam. Note double stilling basin for mainspillway.

Figure 6.6: Horse-shoe shaped mass gravity concrete s pillway with nape aeration.Source: www.evn.co.za/photo_005.htm

http://www.evn.co.za/photo_004.htm







6-6

6.3 SPILLWAY DISCHARGE CAPACITY

Spillways are provided to safeguard dam safety against small or large floods; thusspillway discharge capacity is selected to ensure safety of the structure against damages thatcould occur due to dam failure/breach on overtopping for exceptionally high floods of veryrare occurrence. Floods of any extra ordinary magnitude are real possibility and could causedevastation by its own. The system is designed in consideration of the additional andincremental damages caused by breach of the structure. ICOLD 1992 noted that the spillwaydesign discharge has direct bearing on structure safety on one side and project costs on theother side; and ideally this should be based on engineering and economic considerationsrelevant to the site and its environment.

6.3.1 Risk factors

ICOLD 1992 discussed the various risk factors affecting the choice of spillwaydischarge capacity are:

i) Dam height: Smaller height dams reflect smaller risk. The structure is classified as ofsmall, intermediate and large

ii) Storage volume : Small storage volume reflects lesser hazard.

size for dam heights of < 12.25, 12.25-30.5 and morethan 30.5 m.

iii) D/s loss of life due to dam breach: Loss of any human life is unacceptable in generalsense, but occasional some loss of life may be tolerated.

iv) D/s economic loss due to dam breach: The risk of economic loss is classified as high,significant or low

v) Type of dam : Earthfill and rockfill dams are likely to be subjected to severe damages andfailure risk in the event of overtopping of dam. The concrete gravity dam can sustainovertopping to some extent (due to having factor of safety greter than 2 againstoverturning, sliding and material stresses), and arch dam can withstand overtopping toeven larger extent owing to wedge effects (stresses due to water loads are smaller thanstresses for temperature loads). The bridge deck and machinery placed at dam crestare liable to some damage on structure overtopping.

depending on the extent of damage to housing, industrial,

commercial or agricultural facilities, public utilities, communication network, the damitself, repair or replacement of dam or its some components, etc.

vi)) Consequences of dam failure: Loss of some structure structures can not be accepted asthis may be so vital (e.g. water supply) for the sustenance of the community. In othercases the dam failure may cause economic losses only for temporary curtail of theoutput (hydropower, irrigation water).

vii)) Replacement/repair costs: This considers the extent of damages caused by theovertopping; e.g. loss of whole or part of dam embankment, damage to spillway only,failure of one or more panels of concrete dams, etc.

viii) Opportunity for dam repair/replacement activities: This refers to the extent and possibility of shutting down the river system and making necessary repairs,

replacements.




6-7

6.3.2 Design Flood and Safety check flood.

In most cases two distinct discharges are set: (1) the design flood that must bedischarged through the spillway structures under normal conditions with a safety of margin

provided by the free board; this is usually taken flood of selected recurrence

probability/return period. (2) the safety check f lood , which is the discharge which can be passed by the crest structure, the waterway and the energy dissipater on the verge of failure but to exhibit marginally safe performance. The design flood is selected to ensure adequatesafety of structure under most conditions. The safety check flood is taken a higher magnitudeflood or even the PMF depending on the importance of the dam. For Mirani dam (CFRD)floods of 200 and 10,000 years return period was taken as design flood and safety checkflood, respectively. Flood of 200 years return period and PMF was taken as design flood andsafety check flood, respectively, for Patrind Hydropower Project Weir (concrete gravitydam); since historic flood at the site exceeded the 200 year design flood, the design flood wasupgraded to flood of 1000 years return period

6.3.3 Design Inflow Flood

Selection of design inflow flood is a policy decision. This depends on evaluation oflosses/ damages that can occur in the event of dam failure.

• If damage/loss of human life expected, select design flood equal to probablemaximum flood (PMF).

• If rare-overtopping tolerable: 0.5-1 PMF or T > 10,000 years, whichever is higher.

• No loss of human life, but heavy property damage, some risk accepted: 0.3-0.5PMF, or T = 1000-10000 years, whichever is higher.

• Negligible risk to life and property: 0.2-0.3 PMF, or T = 150-1000 years,whichever is higher.

• Very limited flood damages: 0.2 PMF, or T = 100-150 years, whichever is higher.

• The safety check flood may be taken as next higher design discharge or the PMF.

6.3.4 Spillway Design Discharge

The design inflow flood is partially stored temporarily in the reservoir flood surchargespace above the normal conservation level. This results in reduction of the peak flooddischarge. The peak flow of the outflow flood for which spillway is designed is determined

by routing the design inflow flood through the reservoir (Fig. 6.7) for a pre-selected spillwayconfiguration and the discharge rating curve (i.e. discharge capacity vs. reservoir water levelsor depth). Suitable routing procedure e.g. a level pool routing procedure explained in chapter2 – “Hydrology and Sedimentation” may be used. The spillway configuration and/ordischarge rating curve is modified until the maximum flood surcharge equals the allowableflood surcharge. The resulting peak outflow is then taken as spillway design flood discharge.




6-8

Figure 6.7 : Design inflow, outflow flood hydrographs and reservoir water levels.

6.4 CLASSIFICATION

Spillways may be classified according to:

1. Utility

• Main or service spillway-to pass more often flows

• Auxiliary/subsidiary or emergency spillway to rare/PMF floods

• Combined service and auxiliary spillway

2. Control of flow

• Gated spillway: gates used to control / regulate outflow

• Free/ungated spillway: outflow is direct function of reservoir water levels

• Free/ungated spillway with a breaching dyke or collapsible rubber dam: Theoutflow is delayed until the water level reaches the top of breaching dyke/rubberdam and subsequently outflow is direct function of reservoir water levels

3. Type of flow

• Orifice flow: Outflow is through an orifice, e.g. Mangla dam

• Weir flow: outflow is over a weir, Tarbella dam, Simly dam. For full gateopenings the flow is free flow and the structure works as weir flow. For partialgate opening flow is under pressure and flow is as orifice flow through the opensection of the structure.

4. Structural arrangements for auxiliary spillway (which are mostly ungated)

• Regular constructed free flow section. The flow through the section is initiated as

soon the water level rises above the sill of the flow section.

KTD PMF Routing for Spillway Design

0

20000

40000

60000

80000

100000

120000

140000

160000

180000

200000

0 5 10 15 20 25 30 35 40 45 50Time (hrs)

F l o w

( c f s )

2100

2101

2102

2103

2104

2105

2106

2107

2108

2109

2110

R e s e r v o

i r w a

t e r e

l e v a

t i o n

( f t )

Inflows

Outflow

Res Water Elev




6-9

• Regular constructed free flow section with a breaching section e.g. an earthen fuse plug placed across the regular sect ion. The fuse plug will be eroded quickly oncethe water level rises above the top of the fuse plug, and subsequently wholesection conducts the flood flow.

• Regular constructed free flow section with a collapsing section e.g. a rubber dam placed across the regular section. A rubber dam is anchored onto the section bottom and is inflated to form a raised dam like profile/shape. The rubber damwill detain water as long as the water level remains below the top of the rubberdam. On small increase of water level above the crest of the rubber dam, the flowglides over the rubber dam surface as a broad crested weir. On further increase ofwater level, the rubber dam is deflated and lies at the bottom of the flow sectionmaking all section available for flood flows.

• Regular constructed free flow section with stop logs or gates used to close theflow section.

6.5 SPILLWAY COMPONENTS

Most spillways have four components as: (a). Approach channel, (b). Outflowstructure, (c). Discharge channel / chute (d). Terminal energy dissipation structure and (e)Exit channel.

a. Approach channel

• This is channel leading from main reservoir body to the spillway crest ordischarge/control point.

• Large flow reaching spillway crest through approach channel

• Undesirable conditions in approach channel can reduce spillway capacity, producetroublesome disturbances, contribute to possibility of cavitations, prevent passageof floating debris, produce erosion and undermining of u/s portion of spillwaystructure.

• Maximum velocity at critical elevation and discharge not to exceed scouringvelocity of channel materials.

• Curvature if any be gradual to avoid differential water depths.

• End walls to guide flow to control section to extend upstream from crest to avoiddevelopment of vortex (which would be carried over the crest). Walls to be flaredor curved to form a streamlined entrance design of control section (cost basis).

• Approach channel lengthens the seepage path under the spillway structure; thushelps the reduce seepage losses and more importantly the uplift pressure under thecrest and chute structures.

• A floating debris boom may be used to stop flow of large debris (trees and logs)and block the spillway especially if spillway takes water from the top and a closedconduit (e.g. tunnel) forms the part of discharge channel/chute.

b. Outflow Structure

• This forms the discharge control point.• Permit outflow from the reservoir and regulation/ control of outflow




6-10

• Consists of sill, weir, orifice, tube or pipe

• Discharge head relation fixed for ungated structure and variable for gatedstructure (due to variable gate opening)

• Takes various positioning and shapes as orifice and weir.

• Overflow crests can be straight, curved, circular, semi-circular, U-shaped, Wshaped

• Orifice control in horizontal, inclined or vertical position

• Tubes or pipes may be horizontal, vertical or inclined and flow section can becircular, square, rectangular, horse shoe or other shapes in cross section.

• Overflow can be over sharp crested, ogee shape, broad crested or varied cross-section or bell mouth shape.

• Fully contracted or suppressed jet over the crest

• Tubes can flow full or part, and flow control may be located at u/s or d/s end oftube/pipe.

c. Discharge Channel

• Flow from outflow structure carried to river bed in a discharge channel orwaterway

• This may be d/s face of concrete dam, an open excavated channel, a closed cutand cover conduit cut through or under a dam or a tunnel excavated throughabutment

• Profile variably flat or steep• Cross-section-rectangular, trapezoidal, circular or other shape.

• Channel dimensions are set according to hydraulic requirements

• Channel profile, sections, shapes depend on geologic and topographicconsideration

• Open channels follows closely the ground surface profile

• Channels straight or curved in profile and plan

• Section divergent or convergent or combination

• Discharge channels be lined to resist scouring, cavitations and be structurallystrong to withstand forces from backfill, uplift and water loads. May be unlined if

pass through sound rock

d. Terminal Structure

• Outflow posses considerable energy due to large fall from pond level to river bed.(At Tarbela dam spillway flows carries flow energy in excess of ------ MW.)

• Static head converted to kinetic head and high velocity.

• High velocities if impeded result in large pressures (chute blocks of Mangla damservice spillway are each subjected to thrust of 7.2 million pounds.)




6-11

• Scour and erosion protection required at terminal end of discharge channel; thusstilling basin, roller bucket, flip bucket, plunge pool provided for dissipatingexcess energy.

• Direct exit of high velocity jet over strong rocks of bed/abutment permitted

• Also if erosion location farther from dam and no damage likely to dam• Flip buckets, cantilever extensions/deflectors to throw jet away from the structure

• Flaring jet by deflectors reduce impact and bed erosion

• For likely severe scour, plunge pool provided with bed and sides lined with riprap.

• Flow may be allowed to erode a natural pool (for small installations), protectedriprap may be provided later, if needed.

• Adequate cutoff to ward off undermining and uplift

• For serious erosion hazard provide stilling basin/hydraulic jump basin, a roller bucket, sill block apron, impact baffles etc.

e. Exit Channel

• An exit channel may be required from d/s of stilling basin to the river channel.

6.6 DESIGN APPROACH

Spillways are designed using established flow equations and associated flowcoefficients. As these coefficients vary with operating head vis-à-vis the design head, theapproach and exit conditions, the accuracy of the theoretical design may be lowered. Thedesigned spillway is tested for flow conditions using a scale model, and final design valuesand flow characteristics are therefore established. This is very necessary for large structureswhere significant cost savings and/or design safety improvements can be achieved due toimprovement in design with smallest possible structural sizes.

For small structures it may be more economical to provide generous low velocityapproach conditions (use lower and safer coefficients’ values) rather than invest in a modelstudy.

6.7 SPILLWAY TYPES

Main types of spillways are as under:

1. Free overfall or straight drop spillway

2. Overflow or ogee spillway (e.g. at Tarbela, Simly dam)3. Side channel spillway (Hoover Dam, Glen Canyon Dam)

4. Open channel or trough-chute spillway (Tainpura Dam)

5. Conduit or tunnel spillway

6. Drop inlet + shaft spillway (morning glory, multiple rose petals)

7. Siphon spillway (Baran dam)

8. Culvert spillway with inlet or outlet control

9. Labyrinth spillway with U, V or W shaped crest

10. Baffle apron spillway (in combination with open channel or culvert spillway)




6-12

11. Stepped spillway (in combination with open channel or culvert spillway)

6.8 OVERFALL –STRAIGHT DROP SPILLWAY

• Low height weir crest as control section (Fig. 6.8).

• D/s face vertical or nearly vertical, thus usable with concrete thin arch or buttressdams only.

• Water drops freely from crest

• Underside of nape ventilated

• No pulsating jet

• Crest may be extended to form an overhanging lip

• Water drops into a stilling basin, plunge pool, onto a sound rock, or a concreteapron

• Not recommended for high head as impact can generate vibrations causing cracksin structure

• Danger of toe scour

Figure 6.8: Overfall spillway and plunge-pool Picture.

• FERC 1999 defined the scour depth D as (Fig. 6.8):

θ Sinq H D 54.0225.032.1= (6.1)

Where D = max scour depth below tail water level (ft)

q = unit discharge (cfs)

θ = angle of inclination of the jet at water surface.

• Mason and Arumugam 1985 compared results from more than 30 formulae forestimating scour depth with data from 47 models and 26 prototypes and definedthe scour depth as:

10.030.0

015.02

05.060.0

27.3d g

h H q D = (6.2)

Where d = characteristic size of bed material




6-13

D = max scour depth below tail water level (m)

q = unit discharge (m 3

H = Head difference between reservoir level and tail water level (m)

/s/m)

h2

6.9 OGEE OVERFLOW SPILLWAY

= tail water depth

6.9.1 Spillway Profile

• Control weir is ogee or S shaped in profile

• Ogee profile conforms and corresponds to lower nape of a sharp crested weir

• Flow is made to adhere to the surface of ogee and prevent access of air to theundesired of the water sheet.

• Flow at design discharge glides over the crest and ogee profile with nointerference from boundary surface. This results in maximum dischargeefficiency.

• Small negative pressure created along the profile at low discharges. Largenegative pressures can lead to cavitation damage. Air is provided in the chutefloor to circumvent negative pressure and decrease cavitation potential along thechute.

• The spillway chute as tangent to the ogee surface to support the sheet on the faceof the overflow.

• A reverse curve at bottom of slope turns the flow on to a stilling basin.

• A broader crest width (more than that required for ogee shape) causes positivehydrostatic pressure along the contact surface. This supported sheet causes a backwater affect on the crest discharge and reduces efficiency of discharge.

• For a sharp crest, water sheet tends to pull away from the crest and produces asub-atmospheric pressure along the contact surface. Negative pressure results inincrease of effective head and thereby increases the discharge. Negative pressureincrease cavitation hazard, but this may be eliminated by providing air at placesalong the chute.

• Ogee crest could be used as outflow structure of other spillway type (e.g. sidechannel spillway, drop inlet spillway). Ogee crest + apron may work as fullspillway.

• The ogee spillway may be equipped with gates or free flowing (no gates). Thegated spillway works as orifice for partial gate opening and as free flowing weirfor full gate openings.

• The crest of free flowing spillway usually coincides with the normal conservationlevel. The crest of gated spillway is usually set below the normal conservationlevel to achieve higher operating head.

• Because of high discharge efficiency, ogee profile is used for most spillways.

• The u/s face of the spillway may be vertical, or have 1H:3V, 1H:2V, 1H:1Vslopes




6-14

• The ogee profile d/s of the ogee crest is defined by equation as:n

H x

K H y

−=

00

(6.3)

where K and n are constants defined as function of h a/H 0 and u/s inclination; h a =velocity of approach head. Generally K = 0.5, n = 1.85 (for u/s face vertical); thusy = -0.5 x 1.85 H -0.85

• Upstream shape of crest profile is defined by compound circular curves of radiusR

(Wei, ). The factors K and n are given in Fig. 6.9 (takenfrom Fig. 247 P-347of USBR, 2001). K value increases with slope of u/s face (Khigher for vertical or 1:3, less for 3:3) and n value decreases (n higher for verticalor 1:3, lower for 3:3). Morris and Wiggert described the coefficients K and n as:(0.516 and 1.836), (0.515 and 1.810), (0.534 and 1.776) for other u/s face slopesof 1v:3h, 2v:3h and 3v:3h respectively.

1 and R 2 with lip of the crest inlet offset by amount X c, Y c . The R 1 , R 2 , X c, Y c depend on h a/H o

• The approximate profile shape for a crest with vertical u/s face, negligible velocityof approach and approach height P equal or greater than H

and u/s face inclination. (Fig. 6.10)

0

• The lower part of the profile is formed as chute of selected constant slope of2H:1V to 0.7H:1V. The origin of the chute is determined by equating the dy/dx ofEq. (6.3) equal to the selected slope. For y = - 0.5 x

/2 may be given in theform of a compound curve (Fig. 6.11)

1.85 H -0.85 , dy/dx = - 0.5 * 1.85x0.85 H -0.85 . Let desired chute slope = 1H:1V (dy/dx = slope = 1.0) and H = 36 ft,then X T = 39.46 ft and Y T

6.9.2 Flow Over Ogee Spillway For Ungated Condition

= 81.4 ft. Thus chute starts at a horizontal distance of

39.46 ft and vertical distance of 81.4 ft from crest top.

The discharge for ungated (or gated with fully open gates) ogee spillway at the designhead H 0

Q is as

0 = C 0 L H 03/2

where

(6.4)

Q 0 = spillway discharge (Q 0

C

= design discharge)

0 = coefficient of discharge (variable) at the design head H 0 (dimensionL./T). The discharge coefficient is written as C for head different fromdesign head. The C 0

L = crest effective length

is corrected for inclination of the u/s face of crest.

H 0 = Total design head on crest including static head and velocity ofapproach head (H 0 = h 0 + h a

h

) but excluding entry and friction losses inthe approach channel.

0 = Static head upstream from weir (Note h 0

h

do not include entrance orfriction losses in the approach channel.)

a = velocity of approach head




6-15

Let P = depth of approach channel below spillway crest level, L C = spillway gross crestlength (including piers). The flow depth in the approach channel = h 0 + P. Then approachvelocity V a = Q / [L C (h 0 + P)] and velocity of approach head h a = V a

2

/2g.

Figure 6.9 Spillway ogee crest profile and factors K, n for defining nape shape d/s of crest

(Source: USBR, 2001 p-374).




6-16

Figure. Factors R 1 , R 2 , X c, Y c

for defining nape shape profiles u/s of crest.

Figure 6.10: Factors R 1 , R 2 , X c, Y c for definition of nape-shaped crest profile u/s of crest(Source: USBR 2001, p-375)




6-17

Figure 6.11 : Ogee crest profile defined by compound curves.

6.9.3 Effective Length of spillway

Usually a number of piers are placed to support structural components as gates, road bridge etc. across the spillway length. Presence of piers and contraction along piers and theabutment decrease the spillway effective length.

L = L′ – 2 (N K p + K a) H e

where

(6.5)

L = effective length of spillway crest excluding piers, and contraction along abutments and piers.

L′ = net length of spillway crest excluding pier width = L C – N W

L

p

C = Total crest length including piers,




6-18

N = number of piers

W p

K

= Width of each pier

P

K

= pier contraction coefficient

a

H

= abutment contraction coefficient

e = total effective head on crest (Note H e may be equal to, smaller than or greater thanH 0

The coefficient K

for varied flow depth h and discharges Q other than design conditions)

p is affected by: i. Shape and location of pier nose, ii. Thickness of pier,iii. Head in relation to design head, and iv. Approach velocity. For design head, K p

- Square nose (with corner rounded to r = 0.1 of pier thickness W

forvarious u/s pier nose conditions is as under:

p

- Round nose pier K

) = 0.02

p

- Pointed nose pier K

= 0.01

p

The shape of the d/s pier nose has no effect on the discharge, but it effect on d/s jetdisturbance, cross shock waves and roaster tail formation.

= 0.00

The coefficient K a is affected by: i. shape of abutment, ii. angle between abutment wall

and u/s approach / axis of flow, iii. head in relation to design head, and iv. velocity ofapproach. For design, head, average K a

- Square abutment, head well at 90

as (vertical wall):o to direction of flow: K a

- Rounded abutment, head well at 90

= 0.20o to direction of flow (abutment radius r = 0.15-

0.5 H 0): K a

- Rounded abutment, head well at < 45

= 0.10o (r > 0.5 H o): K a

The abutment wall continues as straight vertical wall.

= 0.00

6.9.4 Coefficient of Discharge for free flow conditions

The discharge coefficient C0 or C is influenced by depth of approach channel P below

crest level, relation of actual crest shape to ideal shape, u/s face slope, d/s apron interference,d/s submergence. The coefficient C o (i.e. coefficient at design head H 0) is related to P/H o

P/H

asunder (Fig. 6.12):

0.00 0.1 0.2 0.3 0.4 0.5 1.0 1.5 2.0 3.0

C 3.087o 3.4 3.57 3.68 3.76 3.79 3.88 3.92 3.935 3.95

Ef fect of u/s face slope

The discharge coefficient also varies with the inclination of the u/s face as given inFig. 6.13. For small P/H o, C increases with decreasing slope (vertical 0:1 to inclined 1:1). For

large P/H o C decreases with decreasing slope (for relatively flat slopes only).

Pier Nose Shapes Abutment Shapes




6-19

Ef fect of vari ed fl ow depth

The coefficient C varies for heads H e other than design head H o

H

as: (Fig. 6.14)

e/H 0.10 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.1 1.2

C/C 0.821 o 0.853 0.878 0.90 0.92 0.939 0.957 0.972 0.987 1.0 1.013 1.026

Generally C/C 0 = 0.802 + 0.257 (H e/H 0) – 0.057 (H e/H 0 )2. Note both H e and H 0 includevelocity of approach head h a

.

Figure 6.12 : Discharge coefficient for vertical face ogee crest.

Figure 6.13 : Discharge coefficient for inclined u/s crest face. (Source: USBR 2001 p-379)




6-20

Figure 6.14 : Coefficient of discharge for different ratios of effective head to design head.

Economy of D esign

Use design head (H 0) less than the maximum expected head (H M) and determinecoefficient C 0 corresponding to H 0 . Generally H 0 maximum <= 1.33 H M . Higher heads(H F/H 0 > 1) increases discharge coefficient and thus discharge for same spillway effectivelength. Determine C/C 0 for H M /H 0 and C corresponding to H M . Determine spillway widthrequired to pass the design flood Q 0 corresponding to H M . e.g. let H M = 35 ft. P = 15 ft, letH 0 = 30 ft, P/H 0 = 0.5, C 0 = 3.80, H M /H 0 = 35/30 = 1.167, C/C 0 = 1.02, and C = 1.02 * 3.80= 3.876. But for H 0 = H M = 35 ft, P/H 0 = 15/35 = 0.429, C 0 = 3.773, C/C 0 = 1.0, and C =3.773. Thus selecting H 0 = 30 ft lead to 2.73% higher C and higher Q. But the shape of the

profile for H 0 = 30 ft will be different than for H 0

At higher head sub-atmospheric conditions develop over crest (Fig. 6.15). This sub-atmospheric head is less than 0.5 of design head if design head > 75% of maximum head.Such small negative pressures do not induce cavitations. Unevenness of surface (abruptoffsets, depressions, projections) will amplify negative pressure to cavitation level thussmooth surface must be ensured.

= 35 ft, and may induce small backwatereffects.

Sub-atmospheric pressure zone H

M =

1 . 3 3 H

0

H0

ha

Xc H0

H0/2

Figure 6.15 : Sub-atmospheric pressure over crest for H 0/HM = 0.75.




6-21

Exampl e 6.1:

Determine the maximum flood discharge for an ogee gated spillway for followingconditions: Normal conservation level = 640.24 m, Spillway crest level = 631.09 m,maximum flood surcharge = 2.0 m, Gates fully open, Number of spans = 6, Span width =11.18 m, Pier width = 2.0 m, Approach channel floor level = 624.993 m, length = 100 m, U/sface inclination = 2H:3V. Pier have rounded nose. Abutment radius = 5 m, and approach wallat 30 o

Solution:

from flow axis. Draw discharge-rating curve. Compute spillway discharge if crest levelis at normal conservation level NCL and P = flood surcharge.

A: Gated spil lway:

Step I

1. P = 631.09 – 624.993 = 6.097 m. Approx H

: Determine approximate discharge over the spillway,

M

2. L

= (640.24 – 631.09) + 2.0 = 11.15 m

C

3. Select K

= 6 x 11.18 + 5 x 2.0 = 67.08 + 10.0 = 77.08 m.; L = 6 x 11.18 = 67.08 m.

p = 0.10, K a

4. Let C

= 0.00, then L = 67.08-2(5x0.10+0.0)x11.15 = 55.93

0 = 4.04 (assumed), then approximate spillway Q = C L H 3/2 = 4.04 x 55.93 x11.15 3/2 = 8412 m 3/s.

Step-II

5. Now channel flow depth d = H+P = 11.15+6.097 = 17.247 m, channel flow area A =d x L

: Determine head loss in approach channel, and net head over spillway crest.

C = 17.247 x 77.08 = 1329.4 m 2, perimeter = 77.08 + 2 x 17.247 = 111.574 m,

hydraulic radius = 1329.4/111.574 = 11.915 m. The approach velocity in channel V a

6. Velocity of approach head in channel h

= Q/A = 8412/1329.4 = 6.3277 m/s,

a = 6.3277 2

7. For approach channel let n = 0.0225, then hydraulic slope is computed by Manning’sformula as: S = (V n / R

/2x9.81 = 2.041 m.

2/3)2 = (6.3277 x 0.0225 / 11.915 2/3)2 = 0.000744. Theapproach channel friction losses ∆h f

8. Head loss on entrance into approach channel ∆h

= hydraulic slope x channel length = 0.000744 x100 = 0.0744 m.

e ≈ 0.1 h a

9. Total losses in approach channel ∆h

= 0.1 x 2.041 = 0.204 m

t

10. Net maximum head H

= 0.0744 + 0.2041 = 0.2784 m.

M = 11.15 – 0.2784 = 10.872 m.

Step-III

11. Let H

: Determine corrected flow over spillway

0 = 0.75 x H M

12. P/H

= 8.154 m

0 = 6.097/8.154 = 0.7477. For P/H 0 = 0.75, C 0

13. For P/H

= 3.824

o = 0.75, C inclined / C vertical = 1.011 and C 0

14. Now H

= 3.824 x 1.011 = 3.866

e = H M = 10.872 and H e/H 0

15. For H

= 10.872/8.154 = 1.333 (at max flow)

e/H 0=1.333, C/C 0 = 1.05 and C 0

16. Corrected effective crest length = L = 67.08-2(5x0.10+0.0) x 10.872 = 56.208 m

= 3.866 x 1.05 = 4.0593

17. Q = 4.0593 x 56.208 x 10.872 1.5 = 8,179 m3/s.




6-22

Step-IV: (optional) The discharge may be further corrected by adjusting head loss inapproach channel. [This depends on how close C 0 in step-4 is taken in comparisonto C 0

a) Channel d = 6.097 + 10.872 = 16.969, A = 1307.97, P = 110.018, R = 11.782. V

computed in step-15.]

a

= 8179/1307.97 = 6.2532, h a = 1.993, S = 0.000737, ∆h f = 0.0737, ∆he = 0.199,∆h t = 0.2727. H e

b) The discharge coefficient remain same = 4.0593

= 11.15 – 0.2727 = 10.8773.

c) Effective length = 67.08 – 2 x 5 x 0.1 x 10.8773 = 56.203

d) Q = 4.0593 x 56.203 x 10.8773 1.5 = 8,184 m 3

18. The discharge rating for water levels other than flood level are computed in Table 6.1and shown in Fig. 6.16. It should be noted that spillway will be operated only whenreservoir water level rises above the normal conservation level.

/s. which is close to previous results.

Table 6.1 : Spillway discharge for free flow conditions.Crest El = 631.09 m; Ch floor = 624.99 m; P = 6.097 m; H 0 = 8.175 m; C 0

N = 5, W

=3.866;

p =2.0 m; W s =11.18 m; LC = 77.08 m; L’ = 67.08 m; K a = 0.10, K p = 0.00

B: Spil lway Discharge F or Crest L evel = N CL

1. Approx H M

2. L = 67.08-2(5x0.10+0.0) x 2.0 = 65.08 m

= 2.0 m, and P = 2.0

3. Let C = 3.86, then approximate spillway Q = CLH 3/2 = 3.86 x 65.08 x 2 3/2 = 710 m 3

4. Now d = 2+2 = 4m, A = 77.08 * 4 = 308.32, P = 85.08 m, R = 308.32/85.08 = 3.624

/s.

5. Channel V a = Q/A = 710/308.32 = 2.303 m/s, and velocity of approach head h a =2.303 2

6. For approach channel let n = 0.0225, R = 3.624; then hydraulic slope is computed byManning’s formula as: S = (V n / R

/2 x 9.81 = 0.27 m.

2/3)2 = 0.00048. The approach channel frictionlosses h f

= hydraulic slope x channel length = 0.00048 x 100 = 0.048 m.




6-23

Spillway Discharge Rating Curve

630

632

634

636

638

640

642

644

0 2000 4000 6000 8000 10000

Discharge (m3/s)

R e s e r v o

i r w a

t e r

l e v e

l ( m )

Figure 6.16 : Spillway discharge rating curve for free flow conditions.

7. Head loss on entrance into approach channel h e ≈ 0.1 h a

8. Total losses in approach channel h

= 0.1 x 0.27 = 0.026 m

t

9. Net total head H

= 0.048 + 0.027 = 0.075 m.

M = 2.0 – 0.075 = 1.925 m. Let H 0 = H M

10. P/H

= 1.925 m

0 = 2.0/1.925 = 1.039; For P/H 0 = 1.039, C 0

11. For P/H

= 3.837

o = 1.039, C inclined / C vertical = 1.007 and C 0

12. Corrected effective crest length = L = 67.08-2(5x0.10+0.0) x 1.925 = 65.155 m

= 3.837 x 1.007 = 3.8638

13. Q = 3.8638 x 65.155 x 1.925 1.5

14. Thus 12 times longer spillway is required to pass the flood for ungated spillway.

= 672 m3/s.

Example 6.2:

Given Q = 2000 cfs, head at crest = 5 ft, u/s crest face slope = 1V:1H, approach channellength = 100 ft, pier width = 1.5 ft, pier nose = round, abutment radius = 5 ft and approachchannel at 30 degrees from flow. Determine required spillway width with span not more than20 ft between the piers.

Solution:

1. Let P = 2 ft, P+H = 2+5 = 7 ft, P/H = 2/5 = 0.4,

2. Let C = 3.7, q = Q/L = CH 3/2 = 3.7 * 5 3/2

3. v

= 41 cfs/ft

a = 41/7 = 5.9 fps, h a = 5.9 2

4. For approach channel let n = 0.025, R = d = 7, then S = [(5.9 * 0.025)/(1.486*7

/2*32.2 = 0.5 ft2/3)]1/2

5. Let entrance losses = 10% = 0.1 * 0.5 = 0.05 ft

= 0.0006 and channel friction losses = 0.0006 * 100 = 0.06 ft

Normal conservation level = 640.24 m




6-24

6. Total approach channel losses = 0.06 + 0.05 = 0.11 ft

7. Effective head = 5 – 0.11 = 4.89 ft

8. P/H0 = 2/4.59 = 0.41, and C0 = 3.77

9. For 1:1 u/s face and P/H0 of 0.41 C inc /C v

10. Now 2000 = 3.77 L 4.89

= 1.018 and C = 1.018 * 3.77 = 3.843/2, and thus required clear water way width L =

[2000/(3.77*4.89 3/2

11. For span < 20 ft, No. of spans = 48.2/20 = 2.4, select 3 spans and number of piers = 2

)], = 48.2 ft

12. For round pier nose contraction coefficient = 0.1, and abutment coefficient = 0.0

13. Gross spillway opening L’ = L + 2(2*0.01+0.0)*4.89 = 50.15 ft, and span length =40.15/3 = 16.72 ft. Total spillway width L* = L’ + 2 * 1.5 = 53.15 ft.

6.9. 5 Gated Ogee Spillway

The ogee overflow spillway is usually provided with gates to control the flow. Thus the

gates are opened partially or fully to discharge the requisite flood flows. The opening ofspillway gates is chosen to ensure the evacuation of design inflow flood under selected floodsurcharge. For ungated spillway large crest length is required to pass the design flood due tosmall flood surcharge. Gated spillway allows setting the spillway crest level below thenormal conservation level, and obtaining a large working head. Gates are kept closed as longas reservoir water level remains below the normal conservation level. On rise of water levelsgates are opened partially or fully to obtain requisite outflow capacity.

• For partial gate opening, flow is as low-head orifice flow.

• Free discharging trajectory follow the path of a jet issuing from an orifice

• For a vertical orifice, path of jet expressed as a parabolic equation: y = - 0.25 x 2/H(H = head on the center of opening)(For ungated: y = - 0.5 x 1.85 /H0.85

• For orifice inclined at angle θ from vertical y = x tan θ + x

)2/4H cos 2

• To avoid sub-atmospheric pressure, ogee profile to conform to trajectory profilefor orifice flow (part of ogee d/s from the gate).

θ

• For small gate opening, large head result in negative pressure over spillway profile.

• For ideal ogee shape (for maximum H o

• If ogee profile is taken as orifice trajectory it results in a wider crest and reduceddischarge efficiency for full gate opening.

), negative heads remain less than 1/10 of

design head

• If wider crest may be needed for structural stability, than adopt orifice trajectory profile

• Gate sill is placed 0.2 H 0

• The spillway discharge is given as:

d/s of crest to minimize negative pressure due to idealnape profile. This causes d/s inclination of jet for small openings allowing ogee

profile for orifice flow condition to be closer to ogee profile for full open gatecondition.




6-25

Q = 2/3 √(2g) C g L (H 13/2 – H 2

3/2

• where H

) (6.6)

1 and H 2 are respectively total head including velocity head to the bottomand top of orifice (Fig. 6.17), C g

• Spillway effective length L = L’ – 2 (N K

is the discharge coefficient of gatedspillway, and L is spillway effective crest length.

p + K a) H• Coefficient of discharge for gated spillway C

1

g

• Top contraction for vertical leaf gate differs from curved inclined radial gate andthus will slightly effect the discharge coefficient.

differs from free flow condition.

• U/S profile affect bottom contraction of jet and d/s profile affect back pressureand thus effective head over the spillway.

• Cg as function of gate opening (d = gate opening = H 1 – H 2

d/H

) is given as (Fig.6.18): Coefficient corrected for u/s face inclination.

.051 .1 .2 .3 .4 .5 .6 .7 .8 0.9

Cg .73 .713 .690 .688 .677 .666 .656 .646 .634 .625

In simple form:

C g = 0.7206 - 0.1077 d/H 1

C

d/H > 0.15

g = 0.7535 - 0.5838 d/H 1 + 1.7455 (d/H 1)2 0.05 ≤ d/H 1 ≤ 0.15

H2 H1

d

ha

Figure 6.17 : Gated ogee spillway.

0.2 H 1




6-26

Figure 6.18 : Coefficient of discharge for flow under gates.

Example 6.3

Determine discharge of gated spillway of Example 6.1 with 4.573 m (15 ft) gateopening for maximum flood water level.

Solution

1. H 1 = H M = 11.15 m, H 2 = H 1

2. d/H

– d = 11.15 – 4.573 = 6.577 m. P = 6.097 m

1 = 4.573/11.15 = 0.4102. For d/H 1

3. For 2:3 u/s face inclination, V

of 0.41, C from Fig. 6.14 is 0.6765.

inclined /C vertical

4. Crest effective length L = 67.08 – 2(5*0.1 + 0.0)*11.15 = 55.93 m

= 1.05. Thus corrected C = 1.05*0.6765 =0.7103

5. Approximate Q = 2/3*(2*9.81) 0.5*0.7103*55.93*(11.15 1.5 – 6.577 1.5) = 2389 m 3

6. Channel flow depth = 6.097+11.15 =17.247 m, A = 1329.4 m

/s2

7. Channel V

, P = 111.574 m, R =11.915 m

a = 2389/1329.4 = 1.797, h a = 1.797 2/2*9.81 = 0.1646 m, h e = 0.1 h a

8. S = (1.797 * 0.0225 / 11.915

=0.016 m

2/3)2 = 0.00006, h f = 0.00006*100 = 0.006 m. and h t

9. H

=0.016 + 0.006 = 0.022 m

1 = 11.15 –0.022 = 11.128 m, H 2 = 11.128 – 4.573 = 6.555 m, d/H 1

10. L = 67.08 – 2(5*0.1 + 0.0)*11.128 = 55.952 m

= 0.411, and C= 0.6762, C corrected = 1.05 * 0.6762 = 0.7100

11. Q = 2/3*(2*9.81) 0.5*0.7100*55.952*(11.128 1.5 – 6.555 1.5) = 2387 m 3/s




6-27

12. The spillway discharge for various water level and gate opening (of d = 4.573 m) arecomputed in Table 6.2 and shown in Fig. 6.19. The discharge rating curve for variouswater levels and gate openings are given in Table 6.3 and shown in Fig. 6.20.

Discharge Rating Curve of Partial Gate Opening

632

634

636

638

640

642

644

0 500 1000 1500 2000 2500

Discharge (m3/s)

R e s e r v o

i r W a t e r

L e v e l ( m

)

Figure 6.19 : Spillway rating curve for partial gate opening (d=4.573 m).




6-28

Table 6.3: Discharge rating curve for various gate openings d (m).

WL Full d=1 d=2 d=3 d=4 d=5 d=6 d=7 d=8

632.00 184

633.00 568 234

634.00 1079 311 518635.00 1694 370 653 842

636.00 2397 417 759 1022 1198

637.00 3175 457 846 1167 1414 1579

638.00 4020 491 921 1288 1591 1825 1981

639.00 4921 521 985 1392 1741 2029 2251 2399

640.00 5870 547 1041 1482 1871 2203 2478 2690 2831

641.00 6858 571 1089 1561 1983 2354 2672 2936 3139

642.00 7879 592 1132 1629 2081 2485 2841 3147 3399

642.24 8179 596 1141 1644 2102 2514 2878 3193 3456

Spillway Discharge Rating Curve

0

1000

2000

3000

4000

5000

6000

7000

8000

9000

630 632 634 636 638 640 642 644

Reservoir Water Level (m)

D i s c

h a r g h e

( m 3 / s )

Fulld=1d=2d=3d=4

d=5d=6d=7d=8

Figure 6.20 : Spillway discharge rating for full and partial gate openings d (m).Tur bulent B oundary L ayer

Self aeration of nape starts at a distance L i from the crest as : L i = 14.7 q 0.53 ≈ 15 √q; (Figs.6.21 and 6.22) where q = specific discharge (Q per unit spillway crest length). Growth ofturbulent boundary layer is as: δ/L = 0.0212 (L/H s)0.11 (L/k) -0.10 where H s

= potential flowvelocity head and k = equivalent roughness value. Uniform aerated flow at considerabledistance from crest. Side wall height to be determined for aerated flow conditions. Aerationfacilities are placed in the floor




6-29

Backwater affects

The ogee spillway will operate freely for the condition of (h d +d)/H e > 1.7 (Figs. 6.23-24). Ifthis limiting condition is violated then d/s flow depth over the chute/apron will cause

backwater effects leading to lowering of the discharge coefficient and C should be correctedas per Figs. 6.24/6.25.

Figure 6.21: Growth of turbulent boundary layer for gated ogee spillway.

Non aeratedflow region

Partiallyaerated flow

region

Fully aeratedflow region

Fully turbulentflow region

Li

Radial gate




6-30

Figure 6.22: Boundary layer development in overflow weirs. (http://www.tpub.com )

http://www.tpub.com/







6-31

Figure 6.23 : Effect of downstream influences on flow over weir crest. (USBR 2001, p-380)

Figure 6.24 : Ratio of discharge coefficient due to apron effects. (USBR, 2001 p-381)




6-32

Figure 6.25 : Ratio of discharge coefficient due to tail water effects. (USBR, 2001 p-382)

6.10 SYPHON SPILLWAYS From: Novak p-170.

• Closed conduit in the form of an inverted U.

• Types as: as Saddle or Drop inlet / Shaft / Volute• Parts, inlet, short upper leg, throat (control section), lower leg/barrel and outlet.

Bottom and top of throat section are termed as crest and summit, respectively.

• Crest is placed about 0.2 ft (0.061 m) above the normal conservation level.

• Low flows-siphon operates as weir (free flow) and throat as control. Under increasingdischarge it hydraulically behaves as weir flow (free surface). At further increase indischarge, priming occurs and flow become pressurized; the transition between freesurface and pressure flow depends on mainly the aeration and deaeration of the siphoncrest.

• Throat as rectangular with depth d and width b (area = b*d)

• As u/s water level rises, flow increases, velocity increases, and flow in lower legexhaust air from top of siphons

• On complete priming, flow is pressurized and full pipe flow begins quickly andcontinues until de-primed.

• Transition between free and pressurized flow depends mainly on the aeration anddeaeration of siphon crest.

• As water level drop below the de-priming pipe inlet, air enters into the throat, break

suction and flow ceases suddenly.




6-33

• Full flow as pipe flow and discharge given as:

Q = C d gH 2A C d

where A = throat area, H = difference of reservoir water level and tail water level pastthe outlet, C

≈ 0.65 (6.7)

d

4321

1

K K K K C d +++

=

is discharge coefficient given as:

(6.8)

where K’s are head loss coefficients at entry, bend, exit and friction loss in the barrel(usually 0.2, 0.42 (for radius < 2.5d), 0.1).

Figure 6.26b. Arrangements for lower leg of saddle siphon spillway.

Figure 6.26a : Saddle siphon spillway.

H

Maximum reservoir water level

Normalconservation

levelThroat

De- priming pipe

Deflectors

Tudal point

Siphon breaker ventSummit

CrestWater seal

Throat section

d

b




6-34

Figure 6.26c : Alternate arrangement of siphon spillway.

Figure 6.27: Drop inlet or Shaft or Volute siphon spillway.

• Inlet below water level to exclude floating debris, ice and prevent vortex formationand work as water seal against air drawing in.

• The minimum height of inlet tip from approach channel bed is as h in = 2d. This givesinflow velocity as: v in = q/h in and velocity head h vin = v in

2

• Minimum hydraulic seal is as: s = 1.5 h

/2g.

vin

• Deflectors (tudals or steps) or reverse slope of lower leg to seal the lower leg and provide a more positive priming

+ 0.152 (in m).

• Sub atmospheric pressure, pipe to be rigid to enable withstand collapsing forces

• To total drop of siphon limited to 20 ft to prevent absolute pressure within the conduitapproaching cavitation pressure and water vaporization.

• Automatic operation.

• Erratic make-and-break action of siphon results in sudden surges and stoppages ofoutflow, this may cause radical fluctuations in the d/s river stages.

• Priming pipe attached at summit for priming and later de-priming of barrel. De- priming caused to admit air, break siphon action and cease flow.

• More discharge than ogee for same surcharge and throat width due to extra headdifference of outlet




6-35

• Multiple siphons at increasing crest levels and/or set the siphon breaker vents atgradually increasing reservoir heads. This will more closely balance the outflow andinflow.

• Siphon spillway best suited as a service spillway in conjunction with auxiliary or

emergency spillway• Flow cross section same as throat. The crest and summit curves radius as: R CL = 2 d,

R C = R CL – d/2, R S = R CL + d/2. where R CL = radius of barrel centerline, R C = radiusof crest, R S = radius of summit, (R S = R C

• Minimum throat height d is 0.6 m (2 ft ), select a value which will provide requisite

unit discharge q;

+ d), d = throat depth.

gH d C q d 2= , the throat width is determined as: b = Q/q.

• Flow at throat section restricted to maximum allowable sub-atmospheric pressure as(vortex equation): (USBR 2001, p-438)

c sC rs R R Rh g BQ ln2max η = and Q ≤ Q max

where η = efficiency coefficient, B = Throat Width, R

(6.9)

s and R c = radius of summit andcrest of throat, and h rs = allowable maximum sub-atmospheric pressure. [Atmospheric

pressure at mean sea level h atm = 10.35 m = 34 ft]. The atmospheric pressure at anyelevation EL is corrected as h atm ~ 10.35 – 0.00105 * EL (metric units). Usually h rs ≈ 0.7 h atm (7.92 m or 26 ft corrected for the site elevation) Thus Q max = 12.45 B R c lnR s/R c for metric units and Q max = 40.92 B R c ln R s/R c for imperial/fps units. Theselected Q ≤ Q max

• Main siphon waterway of constant section

.

• Siphon breaker cross section area is taken as 1/24 th

• H ≤ 7.92 + h

of barrel area, determine required breaker diameter.

l , where H = difference of elevation of throat and outlet end and h l

• The transition losses: Diverging flow = 0.2 (V

=head loss due to friction between throat and exit end

12/2g – V 2

2/2g); Converging flow = 0.1(V 1

2/2g – V 22/2g); Entry losses = 0.2 V 2/2g, Bend losses = 0.42 V 2/2g (for radius <

2.5 D), Exit losses = 0.1 V 2/2g, Friction losses = 0.25 V 2

• The outlet area same or increased to convert some of velocity head to potential head.

Then head H is as: (e = exit and T = throat)

/2g.

+++≤

g V

g H

T e

22V

h7.92 22

l (6.10)

and area is increased less than above. At low head outlet is divergent and at high headoutlet area is convergent. See Fig. 4.18 on p-171 of Noval et al.

Example: Punmia p-454

Design siphon spillway for: Reservoir normal level = 435.0 m, Siphon outlet = 429.6 m, HFL= 435.85 m, Max Q = 600 m 3/s




6-36

Solution:

1. Let width b = 4 m, height d = 2 m. then throat area A = 2 x 4 = 8 m

2. H = 435.85 – 429.6 = 6.25 m

2

3. Q = C.A √(2gH) = 0.65 * 8 * √(2 * 9.80 * 6.25) = 57.8 m 3

4. No. of barrels required = 600/57.8 = 10.38 ≈ 10 and flow per barrel = 600/10 = 10m

/s

3

5. The required H to generate selected discharge = 6.785 m.

/s

6. Maximum flood level = 429.6 + 0.061 + 6.785 = 436.446 m.

7. R CL = 2 * 2 = 4 m, R C = 4 – 2/2 = 3 m, R S

8. H

= 4 + 2/2 = 5 m

atm = 10.35 – 0.00105 * 435 = 9.893 m, h rs

9. Q

= 0.7 * 9.893 = 6.925 m

max = 4 * (2*9.81*6.925) 0.5 * 3 * ln (5/3) = 71.45 m 3/s and Q < Q max

10. Let inlet depth = 2 d = 2 * 2 = 4 m

, thus designOK

11. Inlet velocity = 60/(4*4) = 3.75 m/s

12. hvin = 3.75 2

13. Minimum seal height = 1.5 * 0.717 + 0.152 = 1.228 m

/2*9.81 = 0.717 m

14. Bed level of approach channel = 435.0 – 1.228 – 4.0 = 429.772 m.




6-37

6.11 STEPPED / CASCADE SPILLWAY

The stepped spillway includes a flow control section (e.g. ogee or broad crested weiror culvert or channel), a chute incorporating concrete steps for energy dissipation (Figs 6.28to 6.30), and occasionally a stilling basin/roller bucket at the end of chute to dissipate any

surplus energy. This type of spillway is useful for concrete gravity dams as well asembankment dams. Spillway chute is formed into small steps to dissipate energy of theflowing water. The chute is formed on the downstream face of embankment or concrete damor at a separate location. The concrete dam will have steep chute (1V:0.75H) andembankment dams will have flatter chute (1V:3H); flatter chutes results in better energydissipation. The stepped spillways are best suited for small heads and small dischargecapacity (Q ~ 3 m 3

The chute steps are designed to operate in a skimming flow regime. The stepscontribute to a substantial flow resistance and most of energy dissipation as a form of drag

process (Chanson 2001). Both flow acceleration and boundary layer development affect theflow properties on the stepped chute. Flow computations are tedious. The steps affect free-surface aeration of the flow significantly. Additional energy dissipation features (e.g. flip

bucket and plunge pool) may be needed at the downstream end to dissipate any residualenergy. Model studies are recommended for detail design of stepped spillways.

/s/m). The stepped structure can be formed by use of individual blocksinterlocked with the next element or concrete steps may be in-situ formed of concrete. Thestepped chute blocks have flexibility of allowing differential setting of the embankment. Thestepping blocks are laid over a lower drainage layer flanked by suitable geotextile membraneto relieve uplift pressure along the chute. [See Vischer and Hager 1998. p-147-51].

Figure 6.28: Stepped spillway.




6-38

Figure 6.30: Stepped spillway with a culvert as control section. (source:http://www.uq.edu.au/~e2hchans/pictures/russian3.jpg)

Figure 6.29: Stepped spillway for Melton overflow embankment dam, Australia (source:http://www.uq.edu.au/~e2hchans/over_st.html)

http://www.uq.edu.au/~e2hchans/pictures/russian3.jpg

http://www.uq.edu.au/~e2hchans/pictures/melton2a.jpg




6-39

6.12 BAFFLE APRON DROP SPILLWAY

• Include a flow control section (ogee or broad crested weir) followed by a slopingchute

• Multiple rows of baffles provided on the chute to dissipate energy, thus no stilling

basin required, no tail water requirements.• Most useful where fall is small.

• Chute on 2:1 or flatter slope (steeper slopes need to be model tested)

Figure 6.31: Typical baffle apron spillway. (USBR 2001, p- )

• Baffle piers obstruct flow and dissipate energy

• Lower end well below channel floor to prevent damage from degradation or scour.

• Recommended entrance velocity as low as practical. Ideal condition (curve D Fig.

6.32) when V = ( 3 gd – 5) for discharge < 70 cfs/ft.

• Higher velocity – jet air born after striking 1st baffle

• Critical depth for rectangular chute D c3 2 /gq =

• Baffle block length H ∼ 0.9D c – 0.8 D c

• Baffle block width ≈ 1.5 H and spacing < H

.

• Row spacing = H ÷ slope, and inter row spacing as 2H

• U/s face of baffle at right angle to chute floor

• At least 4 rows of baffles for full control of flow




6-40

• Other rows needed to maintain u/s established control

• At least 1 row below channel bed level to protect against scour

• Chute training walls 3 x H this will contain flow and most splash

• Riprap at d/s end of training walls for protection of bank scour

Figure 6.32 : Baffle drop spillway.

6.13 SPECIAL SPILLWAYS

• For an ungated spillway with small flood surcharge large crest length needed to passrequisite discharge. Ignoring huge cost of such long spillway structure, such spacemay not be available at the site.

• Alternative is to increase the totallength of crest by forming the crestin different shapes as: semi-circular,circular (morning glory,) Rose petal(sunflower shape).

• Other possibilities are to shape crestas Labyrinth triangle or trapezia, U,V, W shapes. Thus crest lengthextended for small horizontal space.

6.13.1 MORNING GLORY SPILLWAY

• This type of spillway is basically agiant cement funnel. Rather thanspilling over the dam, high waters spill Figure 6.33. Semi-circular crest.




6-41

into the funnel. Morning glory spillways are also known as bell-mouth spillways ordrop-shaft spillway.

• Anti-vortex fixtures or pier guide vanes may be needed.

• Crest is given an ogee shape

• Channel as closed conduit or an open channel; Closed conduit combination ofvertical, inclined and horizontal segments. Conduit is placed through abutment orembankment. Seepage collars are provided if conduit is placed in the embankmentfill. Tunnel or cut and cover to place the conduit in the abutment.

• D/s energy dissipation may be required.

• Some problems with nape aeration and noise.

• At small discharges flow control is at crest (free flow), at large discharge flow controlsection shift below the crest in the vertical tube (orifice flow) and on further increasein discharge the lower horizontal pipe become full and flow is controlled by the pipefull flow hydraulics (pipe flow control as pipe flowing full).

• Discharge for crest control free flow condition by using ogee crest formula and usingcircular discharge coefficient.

• Discharge for orifice control (crest submerged) free flow condition by using orificeflow formula.

• Discharge for pipe control condition by using pipe flow formula.

Figure 6.34: Layout of morning glory spillway for Montecello dam (source:

http://www.usbr.gov/mp/berryessa/images/morning_glory_spillway_large.jpg)

http://www.usbr.gov/mp/berryessa/images/morning_glory_spillway_large.jpg

http://www.usbr.gov/mp/berryessa/images/morning_glory_spillway_large.jpg




6-42

Figure 6.35: Layout of morning glory spillway for Montecello dam CA USA. (a): Location,(b): working, (c): end portal, (d): close up of end portal with ski-boarder, (e): Nonworking state (note interior fins). The glory hole is located about 200 feet from the dam.The funnel's largest diameter is 72 feet and narrows to about 28 feet. (source:http://www.daviswiki.org/Morning_Glory_Spillway and http://www.usbr.gov/mp/

berryessa/ facts.html

Figure 6.36: Morning glory spillway during construction (source:http://www.rovni.co.yu/en/dam.html)

http://www.daviswiki.org/Morning_Glory_Spillway

http://www.usbr.gov/mp/

http://www.rovni.co.yu/en/dam.html

http://www.rovni.co.yu/en/dam.html

http://www.usbr.gov/mp/

http://www.daviswiki.org/Morning_Glory_Spillway




6-43

Figure 6.37: Anti-vortex fins for morning glory spillway. Left: Apanas Dam, Nicaragua(Source: http://www.therandymon.com/nicaragua/Apanas.htm) and Right: South HolstonDam. (Source: http://cgi.ebay.com)

Figure 6.38. Morning glory spillway with circular crest.

http://www.therandymon.com/nicaragua/Apanas.htm

http://cgi.ebay.com/

http://cgi.ebay.com/

http://www.therandymon.com/nicaragua/Apanas.htm




6-44

Figure 6.39: Circular crest discharge coefficient for morning glory spillway.

Figure 6.40: Morning glory spillway (USCOE 1990)




6-45

The flow control of the spillway may be due to the crest, orifice control or channel control.

Figure 6.41: Nature of flow and discharge characteristics of a morning glory spillway(USBR 2001, p-414)




6-46

6.13.2 LABYRINTH SPILLWAY

The spillway crest length is extended by giving a special shape to the crest (Figs. 6.42 to6.45). The flow passes over the crest as free overfall flow. The flow past the crest is carriedthrough an open channel or chute with suitable energy dissipation on or at the end of the

chute.

Figure 6.42: Labyrinth spillway.

Figure 6.43 : Arrangements for crest and channel layout for labyrinth spillway.

The discharge of long crested weirs for design purposes can be estimated using the followingequation:

Q = C B H (3/2) *(2g)^ (1/2)

where Q = discharge over the weir (m

(6.11)

3/sec), C = discharge coefficient, B = crest length (m),and H = height of water above the weir crest (m). Kraatz and Mahajan (1975) give thefollowing estimates of the discharge coefficients for various types of long crest weirs and fortwo different crest types: Unrounded Crest: c = 0.31, Rounded Crest: c = 0.34.(http://library.wrds.uwyo.edu/wrp/93-13/ch-02.html).




6-47

Figure 6.44: Labyrinth spillway crest.

Figure 6.45 : Labyrinth spillway (http://projects.ch2m.com/standley_public/pres/sld017.htm)

http://projects.ch2m.com/standley_public/pres/sld017.htm

http://projects.ch2m.com/standley_public/pres/sld017.htm




6-48

6.14 SIDE CHANNEL SPILLWAY

In some cases only a short spillway could be provided parallel to the dam axis.Spillway crest length then may be extended by providing crest at a reservoir side at an angleto the dam axis where space is not limited. The spillway discharges into a side channel, turns

at right angle and flows parallel to spillway crest. The side channel transforms into thespillway discharge channel.

Ogee crest of side spillway.

Dam axis A p p r o a c h

c h a n n e l

C r e s t

C h u t e

S i d e c h a n n e l

Figure 6.46 : Side channel spillway.

PLAN SECTION




6-49

6.15 BOX-CULVERT-CHANNEL SPILLWAYA channel is lead from the reservoir. An inlet box (flush or raised) or a culvert may be

placed at the u/s end to work as control point and to regulate the flow. The flow in thechannel may be varying or uniform depending on the hydraulics of the channel (slope), andd/s boundary condition. The channel outfalls over solid rocks thus d/s energy dissipationsystem is not provided. Any occasional damage to d/s location may be repaired at small costin comparison to the cost of elaborate energy dissipation system.

Open channels are often used as the emergency spillway and sometimes as the principal spillway for small dams. For dams with pipe conduit principal spillways, an openchannel emergency spillway is almost always required as a backup in case the pipe becomes

clogged. Open channels are usually located in natural ground adjacent to the dam and can bevegetated, rock-lined, or cut in rock. High velocity flows and/or frequent operation can causesevere erosion and result in a permanently lowered lake level if not repaired. Proper design ofan open channel spillway will include provisions for minimizing any potential erosion. Oneway to minimize erosion is to design a flatter channel slope to reduce the velocity of the flow.Earthen channels can be protected by a good grass cover, an appropriately designed rockcover, concrete or various types of erosion control matting. Rock-lined channels must haveadequately sized riprap to resist displacement and contain an appropriate geotextile fabric orgranular filter beneath the rock. Guide berms are often required to divert flow through openchannels away from the dam to prevent erosion of the embankment fill. If an open channel isused for a principal spillway, it must be rock-lined or cut in rock due to more frequent orconstant flows. Grass-covered channels should be mowed at least twice per year to maintain a

Figure 6.47 : Side channel spillway.




6-50

good grass cover and to prevent trees, brush and weeds from becoming established. Trees and brush must be removed from the channel. Erosion in the channel must be repaired quicklyafter it occurs. Channel way should be kept clear of all obstructions. Deterioration due toweathering should be repaired. (Source: http://www.in.gov/dnr/water/dam_levee/inspection_man/pdf/ part4-FactSheets/03-19OpenChannelSpillway-EarthRock.pdf)

Figure 6.48: Schematic View of Open Channel Spillway(http://www.dnr.state.oh.us/water/pubs/fs_div/fctsht49.htm)

Figure 6.49: Open channel spillway for Tainpura dam. (Left: entrance end, Right: outfallend). The channel is cut in medium to hard sandstone. Gabion mating is used to

protect d/s end and the bottom impact area from erosion damage.

http://www.in.gov/dnr/water/dam_levee/%20inspection_man/pdf/







6-51

Figure 6.50: Open channel spillway with a control weir section.[http://members.optusnet.com.au/~richardw2/projects.html]

6.16 CHUTE SPILLWAY




6-52

Figure 6.52: Burrinjuck Dam (92 metres high, concrete gravity dam) on the MurrumbidgeeRiver near Yass. The chute spillway is cut in abutments. The d/s end of spillway chutewas destroyed in a flood due to erosion of the rocks.http://members.optusnet.com.au/~richardw2/projects.html






6-53

6.17 CREST CONTROLS OF FLOW

• Uncontrolled spillway release water whenever water rises above crest level.

• Most dependable and eliminate need of constant attendance. No regulation by anoperator. (History: dam failed as operators could not reach the spillway to open thegates.)

• No repair/maintenance of gates

• No advance lowering of water level in anticipation of flood

• No blockage to floating debris.

• No lowering of reservoir water level below crest level due to delayed closure of gatesfor ungated spillway.

• Require a much longer spillway as h is usually small. h increased by height of gates

• Controlled spillways more flexible but demanding.

• Require experienced operator

• Require dependable availability of electricity (+ back up), operating mechanisms,operating bridges etc

• Control selected based on factors as economy, adaptability, reliability, efficiency,frequency of operation, urgency of control over short/long time width.

• Types include flash boards, stoplogs, needles, bear trap gate, tilting hinged leaf gates,drum gates, vertical or inclined rectangular lift gates, bulk heads, roller gates andradial gates.

• Backup power electricity arrangements necessary in case of power failure.

6.17.1 Flashboard, Stoplog, Needle

• Individual boards beams/girders supported by vertical pins anchored to the crest.

• Stoplogs flash boards span horizontally between grooves in the piers

• Required adequate mechanism for removing and placing/crew

• May be placed/removed manually or mechanically

• Removed by failing automatically.

• Cannot be restored while flow passing over the crest• Repeated replacement costly

• Most suitable for closely spaced piers

• Most suitable for infrequent use

• Needles supported at crest and by a top beam




6-54

Figure 6.50: Stoplog. Top: rear view with locking arrangements. Bottom: front view.




6-55

Figure 6.51: Stoplog lowering. Note the pier slot in top picture.




6-56

Fig 6.52. Lift gate (top ↑) and roller train (right →)

6.17.2 Vertical Lift Gates• Span horizontally between guide grooves in

supporting piers; Width 5 to 15 m

• Made of wood/concrete but mostly steel.

• Placed vertical or inclined and seal at bottom &sides on u/s side

• Placed in slots on the pier.

• Opens upward.

• Sliding, fixed wheal, stony gate/roller wheal train fixed on piers as support system.(pic)

• Lifting by overhead hoist mechanism (manual or electric)

• Range of heads 1 to 15 m

• Slots recessed/beveled by slope 20:1 of d/s side to reduce cavitation.

6.17.3 Drum Gate [Novak p-204,5,6,7]

• Skin plate profile of ogee

• Hinged at u/s or d/s (usually u/s)

• In open form gate hides in housing

• No lifting gear

• Ease of passing ice, debris over the gate.• Water discharges over the gate for small flows.

• Automatic control with optional manual operation is provided for raising andlowering the gates. When in raised position a gate may be held continuously in that

position by the pressure of water against its bottom, until the water surface of the

reservoir rises above a fixed point, when by action of a float the gate is automaticallylowered. As the flood peak decreases, the gate can be operated manually so as to

Drum gates rotate backward, loweringtheir tops and permitting a measured flow over the top of the gate.




6-57

gradually empty the flood control portion of the reservoir without creation of floodconditions down stream.

• Hoower dam each spillway has four steel drum gates, each 100 feet long and 16 feethigh. These gates can't stop the water going into the spillway, but they do allow an

additional 16 feet of water to be stored in the reservoir. Each gate weighsapproximately 5,000,000 pounds. (http://www.usbr.gov/lc/hooverdam/History/essays/spillways.html)

Figure 6.53 : Roller mounted lift gates assembly and roller and seal details (source: http:

//www. armtec. com/ catalogue/Water_Control/Roller_Gates.pdf)

Figure 6.54 : Drum gate (http://www.cee.mtu.edu/~watkins/ ce4630/ presentations/ gates.pdf)

http://www.cee.mtu.edu/~watkins/







6-58

Figure 6.55: Radial gate and hoisting for Mangladam.

6.17.4 Radial Gates (Tainter Gates)• Made of steel or steel + wood for small size.

• Segment of a cylinder formed by skin plate and horizontal and vertical support frame.

• Gate held by radial struts pivoted at the pier by a pin

• Struts moved by a hoisting mechanism.

• Entire thrust of water taken by trunion pin

• Small moment needed to lift raise/lower gate.

• Gate counter weighted to ease in operation

• Hand operation practical on small installations

• Small hoisting forces → easyautomation

• Does not need slots in the pier or bottom.

• Multiple gates operatedsuccessively

• Gate sealing on edges (bottom,sides, top)

• Gate seat at 0.2H d/s from crest.

6.17.5 Flap/Tilting Hinged Leaf Gates

• Gate hinged at bottom d/s of crest

• Lifting by a hydraulic push lever

• When lowered, match with crest profile and gate and lifting mechanism fit within adesigned recess.

Figure 6.56: Flap gate with d/s jack. Source:(http://www.dur.ac.uk/~des0www4/cal/dams/spill/f4.htm)




6-59

6.17.6 Roller Gate

• Cylinder of large dia (equal to opening)

• Gear teeth at edges and inclined rack with gear teeth on inclined rock at piers

• Gate rolled up by hoist/cable

• A cylindrical segment at lower portion of dam fully closes the spillway opening

Roller gates are large cylinders that move in an angled slot. They are hoisted with a chainand have a cogged design that interfaces with their slot

6.17.7 Fusegate:

Fusegates are mechanical equivalent offuse plug. (Carroll ----). Multiple gates areinstalled on a spillway crest approximating the

shape of a labyrinth weir. For moderate waterheights, the water spills over the fuse gate as overa labyrinth weir. When reservoir level exceeds a

predetermined value, the fusegate overturns byrotating about its d/s edge. Each gate is set tooverturn at progressively higher reservoirelevation. At design discharge, all gates tip andentire crest length is available to pass the flow.

Figure 6.58: Labyrinth crested fusegatesat the Terminus dam USA.

Figure 6.59 : Working of fusegate (source: www.hydroplus.com)

Flashboards - Individual lengths of timber, concrete, or steel anchored to the crest of a spillway toraise the retention water level but which may be quickly removed in the event of a flood either by atripping device or by deliberately designed failure of the flashboard or its supports. To provideadequate spillway capacity, the flashboard must be removed before the floods occur, or they must bedesigned or arranged so that they can be removed while being overtoppedStoplogs - Wooden boards, timber, or steel beams or panels spanning horizontally between slots orgrooves recessed in the sides of supporting piers placed on top of each other with their ends held in

http://www.hydroplus.com/

http://www.hydroplus.com/




6-60

guides on each side of a channel or conduit providing a temporary closure versus a permanentbulkhead gate.Trash Rack - A screen located at an intake to prevent the ingress of debris. A trash rack is typically astructure of metal or reinforced concrete bars located at the intake of a waterway, designed to prevententrance of floating or submerged debris of a certain size and larger.Bulkh ead Gate —A gate used either for temporary closure of a channel or conduit to empty it forinspection or maintenance or for closure against flowing water when the head differential is small,e.g., a diversion tunnel closure. Although a bulkhead gate is usually opened and closed under nearlybalanced pressures, it nevertheless may be capable of withstanding a high pressure differential whenin the closed position.Flap Gate —A gate hinged along one edge usually either the top or bottom edge. Examples of bottom-hinged flap gates are tilting gates and fish belly gates, so-called due to their shape in cross section.




6-61

6.18 DISCHARGE CHANNEL OR CHUTE

6.18.1 Channel features

• Discharge channel or chute receives flow from the crest.

• Discharge channel connect crest to stilling basin.

• Discharge channel/chute as tangent to crest profile or with a concave curve beginningat location where h d +d/H e > 1.7 where d is flow depth, h d is discharge head and H e

• Flows passes from critical flow at crest to supercritical at discharge channel

iseffective head over the crest [else the backwater affect and weir C will decrease)

• No flow jump allowed in discharge channel, thus flow must continue at supercriticalvelocity

• The flow may remain uniform or accelerating or decelerating (depend on chute slope)

• Flow specific energy = d + h v

• Velocities and depths fixed by selectinggrade and flow area

= (total head drop – losses in u/s portion)

• Bernoulli theorem – conservation ofenergy follows.

• ∆Z + d 1 + h v1 = d 2 + h v2 + ∆hL

• ∆h

(Fig 6.60)

L

s = average friction slope by Manning Eq.

= s ∆L = friction head loss

s = (q n/1.486 R 2/3 )

• ∆h

2

L

• For concrete lined channel n – 0.017 for depth calculation, n – 0.008 for specific

= maximized for evaluating depth offlow and minimized for evaluating energycontent of flow.

energy calculations.

6.18.2 Discharge channel Profile

• Profile selected to conform to

topography and foundation conditions• Straight segments joined by convex or

concave vertical curve

• Avoid sharp horizontal or verticalcurves

• For horizontal curve provide super elevation

• Convex curves to maintain minimum +ve pressure

• Concave curves to minimize dynamic force

d 1

V2

∆Z

V1

d 2 ∆L

hV1

hV2

Figure 6.60: Channel energy balance.

Concavecurve

Convexcurve

Section

Figure 6.61: Convex and concave




6-62

• Convex curve : - y = x tan θ + x 2 / K[4(d+h v)Cos 2θ] where θ = slope angle offloor at beginning of the curve, d = flow depth h v = velocity head, constant K ≥ 1.5

(to assure +ve contact), USACE 1990 defined convex curve as: - y = x tan θ + (gx2) / {2(1.25 V) 2Cos 2

• Concave curvature - radius: R ≤ 2qV/P or = 2dV

θ} based on theoretical equation of a free trajectory issuingfrom an inclined orifice.

2/P, where R = minimum radius in ft,q = unit discharge cfs/ft, V = velocity at entrance fps, d = flow depth, P = normal

permissible dynamic pressure exerted on floor (P ≈ 100 lb/ft 2) allowed by material properties. In no case R should not be less than 10 d . Reverse curve of R ≈ 5d(i.e. R not less than 5d) at lower end of ogees is acceptable.

Figure 6.62: Discharge channel energy.

6.18.3 Convergent and divergent chutes.

• Chute may be widened or narrowed to vary unit discharge conforming to energydissipation system requirements.

• Straight walls preferred as convergence and divergence will induce crosscurrents/waves.

• The inclination angle limited by: F 31tan =α where F is Froude number at beginning of transition inclination. [ gd V F = where V and d is average of values

at beginning and end of transition]

6.18.4 Cavitation and air entrainment

• Small chute surface irregularities can cause cavitation in the event of high velocityflow.

• Air is admitted at the chute bottom to decrease cavitation potential.

• Air entrainment will cause bulking of flow depth, thus higher chute walls arerequired.

13F Flow




6-63

6.18.5 Channel Free Board

• Depend on surface roughness, wave action, air bulking, splash and spray (depends ond and flow energy head)

• Empirical relation: Free board (feet) = 2.0 + 0.025 v d 1/3

6.18.6 Forces on Spillway Channels

.

• Water weight against floor and walls

• Walls also take active earth pressure due to backfilling

• Design side walls as retaining walls

• Floor subjected to water hydrostatic load

• Boundary drag forces due to frictional resistance

• Dynamic forces due to flow impingement

• Uplift pressure due to sub atmospheric pressures along the boundary• Uplift pressure caused by leakage through joints/cracks

• Uplift pressure due to leakage and under seepage along bottom surface from reservoirwater.

• Uplift pressure due to high groundwater conditions near tail end

• Floor/walls subject to temperature expansion/contraction, freezing/thawing.

• Open to weathering and chemical deterioration e.g. boulders from adjacent hill fallingonto wall/floor.

• Flowing debris can cause damage to floor and walls• Effects of settlement & buckling

• Provide under-drain, anchors, cutoffs, temperature + other reinforcement to stabilizefloor

• Provide water stop at all joints

• Perform force analysis for worst conditions

• Gallery below crest for drainage and/or grouting

• Gravel layer below floor for impervious foundation

• Joints made with lower slab lip ½” below upper slab lip-to avoid high build up ofdynamic head at the joint. The dynamic head could introduce water at high pressureunder the slab, which would result in uplift or dislodgement of floor panels. Enoughreinforcement to make the two slabs move/settle equally, if any

• Cutoff provided at u/s end of spillway to minimize seepage and uplift pressures

• Cut off at d/s end to safeguard against erosion and undermining.

• Intermediate cutoff may also be provided to lengthen seepage path.

• Provide lining or riprap on floor of inlet supply channel to protect against any scour.




6-64

6.18.7 Channel Loss (USBR p-401-557

Channel head loss h f depend on chute length and flow velocity v 1 (or h v

23/4

3/5

486.18

)2(3

=

nq

h g Lh v

f

)

Where q is unit discharge, L is channel length, h v = V 2/2g, but v is unknown and d isunknown. Generally h v = (1 - α)H o

• Basin directly d/s from crest, chute length < hydraulic head. consider no loss α = 0

for rough determination of basin depth

• Channel length between 1-5 times hydraulic head. Consider 10% loss of head, α = 0.1

• Channel length > 5 times of hydraulic head – consider 20% of loss α = 0.2

6.18.8 Joints, under drainage and anchoring

These are shown in Figs. 6.63 and 6.64.

Figure 6.63: Floor lining and drains on firm foundation.

Figure 6.64: Transverse joints in chute lining with anchors.




6-65

6.19 ENERGY DISSIPATION

• Energy dissipation system required to cater for energy of spillway flow water. Typicalinstallations include:

• Stilling basin; Roller bucket; Flip bucket or ski jump; Stepped spillway; Baffle apronspillway. A suitable plunge pool is required for roller bucket and flip bucket.

• Choice depends upon peak flow rate, differential head, site conditions, etc.

6.19.1 Stilling Basin

• The spillway crest flow taken to a stilling basin by discharge channel or chute

• Supercritical flow in chute enters the stilling basin where a hydraulic jump forms andflow is converted to a subcritical flow

• Excess energy lost by turbulence and intermingling of flow on hydraulic jump

• For unit discharge

Q = C L H c3/2 or q = C H c

3/2 and H c = (q/C) 2/3

• Setting basin invert level is iterative process.

(6.21)

• Select initial stilling basin floor level below the tail water level

• Energy at inlet into stilling basin (Fig. 6.65)

E 1 = crest height above stilling basin floor + H c – friction losses in chute = d 1 + h v1

• Ignoring losses:

=reservoir water level – apron level (for no losses in the chute).

21

21

1

21

11 22 gd q

d g

V d E +=+= (6.22)

• Pre-jump velocity - supercritical

• Find pre-jump depth d 1

E

by trials by using Eq. 6.22. as

1 = d 1 + C/d 12 [; C = q 1

2

• Find pre-jump velocity: v

/2g] (6.22b)

1 = q/d 1 (6.23)

Crestheight Z

HC Loss in channel

d 1

hv1 E1

E2 d 2

d C VC

2/2g

y2

Figure 6.65: Symbols for spillway hydraulic jump.

Apron level




6-66

• Froude No. F = 11 d g/v or 31

2

d g q

F = (6.24)

• Determine post jump depth d 2

as: [also see Fig. 6. ]

[ ] −+=−+= 1812

1812 3

1

2121

12 d g

qd F d d (6.25)

• Compare d 2 with tail water depth y 2

• Check basin invert level = tail water level – d

for q.

• The basin invert may be at, below or above the river bed level. But the post jumpwater level in the basin must be equal or above the river water level.

2

• If computed and initially selected basin invert levels match, design is OK else redothe design with computed invert level in the last trial.

• Redo the design for other discharges from zero to q and determine invert basin for allflow conditions.

• Select the lowest invert level of all flows.

• Determine post jump velocity as: v 2 = q/d 2

• Stilling basin floor sloping or level; a slope of less than 1V:6H is considered as level.

• Efficiency of stilling basin to dissipate energyas 20 to 60% for various conditions.

Eff = (E 1-E 2)/E 1

( )21

312

4 d d d d −

~ (6.26)

For F> 2, Eff ~ ( )2/21 F − (6.27)

• Length of hydraulic jump Lj as:

Lj ~ 8.0 d 1 F 1 for F 1

Lj ~ 3.5 d

> 5 and

1 F 11.5

• Chute blocks, baffle blocks and end sill addedfor additional energy dissipation

for F between 2 and 5

• Jump formation and stilling basin designdepend on Froude Number and tail water depth

• D/S flow condition

(i) JHC ≈TWC, (ii) JHC lower than TWC forall q, (iii) JHC above than TWC for all q, (iv) JHC lower than TWC at small q, aboveat large q, (v) JHC above than TWC at small q, lower at large q [JHC = post jumpheight, TWC = tail water depth]

F = 1.7 to 2.5 Form A pre-jump stage (Fig. 6.66)

2.5 – 4.5 B Transition stage

4.5 – 9 C Well balanced jump

A: F 1.7 to 2.5

B : F 2.5 to 4.5

C : F 4.5 to 9.0

D: F > 9.0

Figure 6.66: Post jump conditions.




6-67

> 9 D Effective but rough jump (rough surface carriedd/s of basin into exit channel)

Figure 6.67: Velocity distribution of post jump flow in stilling basins.(www.tpub.com/contents/ coastalhydrauliclaboratory/100-b/index.htm)

Basin Design

F < 1.7

d 2 ≈ 2 d 1 ≈ 1.40 d

v

c

2 ≈ 0.5 v 1 0.70 v

No special basin or use Type I basin

c

Jump length ∼ 4d

F 1.7 to 2.5 (Fig. 6.66 A)

2

Type-I basin (or No special basin), results in weak jump

No baffles or sills

Basin sufficiently long L = 4d 2 ≤ L ≤ 4.75 d

F 2.5 to 4.5 Transition flow stage (Fig. 6.66 B)

2

• True hydraulic jump does not fully develop (oscillatory jump)

• Stilling basin less effective in dissipation due to wave action not being controllable byusual basin devices.

• Energy loss efficiency = 20 – 48%

• Waves generated persist beyond end of basin• Type IV basin relatively effective

http://www.tpub.com/contents/







6-68

• Auxiliary wave dampeners, suppressers provided for smooth d/s flow

• Water depth in basin to be 10% larger than d 2 (d 2 * ≈1.1 d 2

• Plan to alter inlet Froude Number of this range (widen basin etc).

)

F 4.5 to 9 (Fig. 6.66 C, D)

• Adjust E1 to account for friction losses in chute / discharge channel

• Forms a true hydraulic jump (steady jump)

• Energy loss efficiency 48 to 68% without devices; increases with devices

• For V 1

• For V1 > 50 fps, use Type-II basin.

< 50 fps, provide Type-III basin (Chute slope ~ 0.6-0.8H:1V).

• Type-II basin has basin length greater than for Type-III basin.

• Chute block, impact baffle block + end sill stabilize jump and lead to short basin

length• Chute block height d 1 , spacing and width = d

• Minimum y

1

0 ~ d 2 , L ≤ 2.68 d 2

• Baffle block height ≤ 2.2 d

.

1 located at 0.8 d 2 from chute. End sill height = 1.5 d

• Cavitation possible due to baffle blocks this limited velocity (V < 50 fps)

1

• For V> 50 fps or no baffle blocks, provide Type-II basin

yo ≈ 1.05 d 2 ; L ≤ 4.23 d 2

F > 9 (Fig. 6.66 D)

(dentated sill)

• Use Type-II or III with additional d/s bank protection against d/s waves

• Use double / multiple basins at different levels.

Stilling Basin Free Board

Wall not to be over topped by surges, splashes and spray and wave action

Free board = 0.1 (V 1 + d 2

Cascade stilling basins

)

For certain conditions one stilling basin at the end of discharge chute may not provideadequate energy dissipation. The more than one stilling basins are provided in cascade toenhance the energy dissipation. Mangla dam and Simly dam are equipped with two stilling

basins. Vittal and Porey (1987) describe the design of cascade stilling basin. Bakhtyar et al.2007 describe an approach to optimal design of cascade stilling basins.

Example 6.4:

Crest level = 2070 ft, Tail water level in exit channel at design discharge = 1925 ft, Spillwaydischarge = 1,55,000 cfs, chute width = 200 ft, H C

Solution:

= 36.5 ft

• q = 155000/200 = 775 cfs/ft,




6-69

Fi rst trial

• Let d 2

• Ignoring losses in chute: E

= 60 ft, then stilling basin level = 1925 – 60 = 1865 ft,

1

• E

= (2070-1865)+36.5 = 241.5 ft. (Alternatively somelosses as 10% may be considered in the chute depending upon its length etc)

1 = d 1 + q 2/2gd 12 ; 281.5 = d 1 + 775 2/(2*32.2*d 12) = d 1 + 9326.5/d 1

2

• By trials: (i) For d

1 = 6.0 ft, RHS = 265.06 -- not ok, (ii) For d 1 = 6.1 ft, RHS = 256.7-- not ok, (iii) For d 1 = 6.2 ft, RHS = 248.8 -- not ok, (iv) For d 1 = 6.25 ft, RHS =245.5 -- not ok, (v) For d 1 = 6.3 ft, RHS = 241.3 -- ok ; thus d 1

• v

= 6.30 ft.

1 647.830.62.32

01.123 =×

== gd

V F = 775/6.3 = 123.01 fps and

• d 2 = 6.3/2 [(1 + 8 * 8.647 2)0.5 – 1] = 73.87 ft. and y 0 = 1.05 d 2

• Required tail water level = 1865 + 77.6 = 1942 ft. which is higher than TWL of 1925ft; design to be revised by lowering apron level.

= 1.05 * 73.87 = 77.6ft

Second tr ial

• Let basin level = 1925 – 77 = 1848 ft, E 1

• E

= (2070-1848)+36.5 = 258.5 ft

1 = 258.5 = d 1 + 775 2/(2*32.2*d 12) = d 1 + 9326.5/d 1


2

1 = 6 ft, RHS = 265.06 -- not ok, (ii) For d 1 = 6.1 ft, RHS = 256.7 --not ok, (iii) For d 1 = 6.08 ft, RHS = 258.4 -- ok ; thus d 1

• v

= 6.08 ft.

1 11.908.62.32

46.127 =×

== gd

V F = 775/6.08 = 127.46 fps and

• d 2 = 6.08/2 [(1 + 8 * 9.11 2)0.5 – 1] = 75.35 ft. and y 0 = 1.05 d 2

• Required tail water level = 1848 + 79.12 = 1927.12ft. which is higher than TWL of1925 ft; design to be revised

= 1.05 * 75.35 = 79.1ft

Thir d trial

• Let basin level = 1925 – 79 = 1846 ft, E 1

• E

= (2070-1846)+36.5 = 260.5 ft

1 = 260.5 = d 1 + 775 2/(2*32.2*d 12) = d 1 + 9326.5/d 1


2

1 = 6.07 ft, RHS = 259.2 -- not ok, (ii) For d 1 = 6.06 ft, RHS =260.02 -- ok ; thus d 1

• v

= 6.06 ft.

1 155.906.62.32

89.127 =×

== gd

V F = 775/6.06 = 127.89 fps and

• d 2 = 6.06/2 [(1 + 8 * 9.155 2)0.5 – 1] = 75.49 ft. and y 0 = 1.05 d 2

• Required tail water level = 1846 + 79.26 = 1925.26ft. which is ~ TWL of 1925 ft OK

= 1.05 * 75.49 =79.26 ft

• v2 = q/d 2

• Use basin of Type-II.

= 775/75.49 = 10.27 fps.




6-70

• From Fig. 6.70, jump length L/d 2

• Free board = 0.1 (v

= 4.3, thus basin length L = 4.3 * 77.25 = 332 ft.

1 + d 2

• Basin wall height = d2 +FB = 79.26 + 20 = 99.26 ft ~ 100 ft.

) = 0.1(127.89 + 75.49) = 20.33 ~ 20 ft.

(design to be completed for L, h1, h2 )




6-71

Figure 6.68: Type-IV stilling basin. (USBR, 2001 p-398)




6-72

Figure 6.69: Type-III stilling basin (USBR, 2001 p-399)




6-73

Figure 6.70: Type-II stilling basin (USBR 2001, p-400)




6-74

Figure 6.71: Stilling basin depths versus hydraulic heads for various channel losses. (USBR2001 p-402 Fig. 268)




6-75

Figure 6.72: Tail water reduction due to end-sill.




6-76

Jump Depth vs Tail Water Depth

• Tail water depth TWD (y 0

• 5 conditions

) determined from flow rating curve of channel d/s ofstilling basin

1. y o = d 2

2. y

for all q

o > d 2

3. y

for all q

o < d 2

4. y

for all q

o > d 2 for small q and y o > d 2

5. y

for large q

o > d 2 for small q and y o < d 2

• Basin floor level selected to provide jump depths which most nearly agree with thetail water depth at all discharge rang.

for large q.

• Basin depth selection corresponds to conjugate depth needed for a perfect jump

• If tail water depth less → jump moves ahead onto stream bed (sweep out) of the basin

• TWD more, jump recedes and drowns

Protection for vari ous condi tions

1. Ideal condition - jump forms at toe of spillway chute

Simple horizontal apron of length 5 x (d 2-d 1

2. Jump completely submerged (y

)

0 > d 2

- No visible standing waves, litt le energy dissipation

)

- Use a sloping apron. Jump form on sloping glacis- Or use roller bucket

- Provide end sill/baffles to dissipate energy by impact and friction

3. y o < d

- Basin floor is deepened below river bed by (d

2

2-y o

- Baffles provided

)

- OR provide a low secondary weir and baffles

1. yo < d 2 at small q and y o > d 2

- Sloping apron partly above and partly below river bed

at high q.

- Jump forms over sloping apron for low q

- Jump at level basin at high q

- Provide chute blocks, baffles, end sill

5 y o > d 2 for small q and y o < d 2

for large q.




6-77

Example (USBR 2001, p-405)

Design a stilling basin for an overflow dam for max discharge of 2000 cfs/ft. crest elevation =1000 ft, tail water as 985, 981, and 978 for Q of 2000, 1000, and 500 second-ft. River bedelevation = 970 ft.

Solution

2. Q = 2000 cfs, Let L C = 20 ft, then q = 2000/20 = 100 cfs. Head over crest H C =(q/C) 2/3

3. h

= 8.7 ft [for C=3.9]. Reservoir water elevation RWL = 1000+8.7 = 1008.7 ft.

v1

4. Tail water velocity = (2g x 23.7)

= RWL – TWL = 1008.7 – 985 = 23.7 ft (assuming no losses in chute)1/2 = (2 x 32.2 x 23.7) 1/2

5. Tail water depth = q/v = 100/39.1 = 2.56 ft

= 39.1 fps

6. F at tail water = 39.1 / [(32.2 x 2.56) 0.5

7. Conjugate depth d

] = 4.3

2 [from Fig. 6.71 for q = 100, H T

8. Required apron elevation = 985 – 16.7 = 968.3 ft

= 23.7 ft, alpha = 0] = 16.7 ft

9. Specific energy at u/s end of basin E 1

10. E

= 1008.7 – 968.3 = 40.4 ft

1 = 40.4 = d 1+q 2/2gd 1 2 = d 1+100 2/(2*32.2* d 12) = d1 + 155.28/ d 1

11. By trials: For d

2

1 = 2.01, RHS = 40.44 ≈ 40.4, thus d 1

12. V

= 2.01 ft.1 = 100/2.01 = 49.75 ft/s. F = V 1/[(g d 1)0.5] = 49.75 / [(32.2 x 2.01) 0.5

13. Required is Type-III basin

] = 6.18 ~ 6.2

14. d 2 = d 1 /2[(1+8F2) 0.5 – 1] = 2.01/2[(1+8x6.18 2)0.5

15. Required basin invert = 985 – 16.59 = 968.41 ft. which is same as initial trial.

– 1] = 16.59 ft

16. Basin length = L = 2.48 x 16.6 = 41.17 ft ≈ 42 ft.

17. Basin free board = 0.1(V 1+d 2

18. Basin wall height = 16.6 + 6.64 = 23.25 ft

) = 0.1(49.75 + 16.6) = 6.64 ft

19. Chute block height h 1 = d 1

20. Baffle block height h

= 2.0 ft

3 = 1.82 x d 1

21. End sill height = 1.65 d

= 1.82 x 2.01 = 3.66 ft

1

22. Chute block width = h

= 1.65 x 2.01 =- 3.32 ft

1

23. Baffle blocks: w = 0.75 h

= 2.0 ft; Spacing = width = 2 ft; Numbers: (2+2)*N=20. N=5

3

24. Distance of baffle blocks from chute = 0.8 d

= 0.75 x 3.66 = 2.75 ft, Spacing = width = 2.75 ft; Numbers : (2*2.75)N=20, N = 3.64 ≈ 4, then width = 20/(4*2*w), w = 2.5 ft,and spacing = w = 2.5 ft.

2

25. Post jump velocity V

= 0.8 x 16.6 = 13.28 ft.

2 = q/d 2

26. E

= 100/16.6 = 6.02 fps

2 = d 2 + V 22

27. Energy loss efficiency = (E1 – E2)/E1 = (40.4 – 17.16) / 40.4 x 100 = 56.4%

/2g = 16.6 + 6.022/2*32.2 = 17.16 ft

28. Remaining energy will be lost due to chute blocks and baffle blocks.






6-79

Figure 6.73: Roller bucket. Left: details, Right: Hydraulic action.

Slotted

• Velocity jet leave lip at flatter angle

• Small jet part reach surface

• Less violent roller and smoother flow

• Slotted provide better energy dissipation

- Over a small range of TWD

- To be preferred over solid if TWD allows

- For deficient TWD, surface roller swept out of bucket by incoming jethigh velocity flow d/s

- At more depth, sweep out and submergence of surface roller alternatively prevail.

- Excessively deep TWD-jet become depressed and dives to the river bed/dissipation very loss.

• Radius of curvature, T min , T max

• Calculate v

and TWD from Fig. 6.75 (Fig. 274 USBR P. 407)

1 , h v1 , d t, and E t (E t = d t

• Find R/E

+ hv1) at tangent point where chute coincideswith tangent to bucket radial profile.

t and R from figure for known F(d) or else as: R min = 51.9x E t / F t1.64

• Find T

.

min , T max

• Find lowest bucket invert level = TWL - T

from figure (a,b)

• Find highest bucket invert level: = TWL – T

max

min






6-81

Figure 6.75 : Limiting criteria for roller slotted bucket design. (USBR 2001 p-407)




6-82

Figure 6.76: Chart for roller bucket roller height.




6-83

Figure 6.77 : Chart for Roller bucket surge height.




6-84

Figure 6.78: Roller bucket energy dissipation sample calculations for high overflow dams.[Source:www.tpub.com/contents/coastalhydrauliclaboratory/100-b/index.htm]

Example:

Design a roller bucket for given data. Q=2000 cfs, crest length L c

Solution:

= 20 ft, q = 100 cfs/ft,

Reservoir flood water level = 1008.7 ft, River water level = 985 ft.

1. Let bucket invert level = 968.3 ft, E 1

2. From E

= 1008.7-968.3=40.4 ft

1 = d 1+v 12/2g d 1 = 2.01 ft and v 1 = q/d 1

3. F = 49.75/(32.2*2.01)

= 100/2.01 = 49.75 fps0.5

4. Now R = 5.19 * 40.4 / 6.2

= 6.21.64

5. From graph R/E

= 10.5 ft;

1

6. Let select R = 12 ft. R/E = 12/40.4 = 0.297 ~ 0.30

= 0.275 and R = 0.275*40.4 = 11.11 ft.

7. For F = 6.2 and R/E = 0.30, T min /d 1 = 9.5 and thus T min = 9.5 * 2.01 = 19.1 ft andhighest invert level = 985 – 19.1 = 965.9 ft




6-85

8. For F = 6.2 and R/E = 0.3, T max /d 1 = 13.0 (for case-II – bed 0.05R below apron tip)and thus T max

9. Select invert level between two limiting values as: Invert = 963 ft. This value is

slightly lower than last assumed value. Thus a new trial may be performed to obtain better values.

= 13.0 * 2.01 = 26.13 ft and lowest invert level = 985 – 26.13 = 958.87ft.

10. h1 = 1008.7 – 963 = 45.7 ft; h 2 = 985 – 963 = 22 ft, h 2/h 1

11. P = q*10

= 22/45.7 = 0.48,3 2/3h g / = 100*1000/(32.2 0.5 * 45.7 3/2

12. From figure 6.76 for P=57 and h

) = 57

2/h 1=0.48, h b/h 1 = 0.33 h b

13. From figure 6.77 for P=57 and h

= 0.33 * 45.7 = 15.08 ft

b/h 1=0.33, h s/d 1 = 0.57 h s = 0.57 * 45.7 = 26.05 ft

Figure 6.79: 95 m high Platanovrisi RCC dam, Greece with a ski jump.

(Source: http://www.industcards.com/hydro-greece.htm)

6.19.3 FLIP BUCKET / SKI JUMP+ PLUNGE POOL (Novak P 185-6)

Flip Bucket

• The end of chute is formed into a bucket of radius R and lip exit angle β

• Called as flip bucket for β > 0, it is called as ski jump if β = 0.

• Design if geological and topographical conditions permit

• Tailor made for a project and designs developed with aid of models




6-86

• Jet direction changed without changing flow conditions (d, v) and flow mingling

• Jet leaving the lip fall into a plunge pool; jet splinters at lip tip.

• Air bubbles are drawn into jet and flow is well aerated

• Additional air sucked into at time of the impact into d/s pool

• Energy dissipation due to compression of air bubbles on impact

• Key design parameters: approach flow depth and velocity, bucket radius and lip angleβ, bucket height h (jet fall Y), jet horizontal throw X H

• Minimum bucket radius as:

, jet impact angle θ, scour depthof plunge pool D.

d P d v

RT γ

ρ −

=2

where ρ = mass den sity of water, γ =

specific weight of water, v and d = velocity and depth of flow at entry into bucket, P T

= theoretical unit maximum load on the bucket invert (USACE 1990). Adopt suitable

value.

• Free jet trajectory of inclined orifice as: y = - x tan β - x 2/K[4(d + h v) cos 2 β]with K ~ 0.9, x = horizontal distance from lip, y = vertical distance below lip, d andhv

• USACE 1990 provide jet throw as: X

= depth and velocity head of flow at the bucket exit, [DHAH, p 16.47]

H = h v sin 2 β + 2 cos β[h v(h v sin 2β+Y)] 1/2 whereY is vertical fall below bucket lip to the water surface, h v

• If water level coincides with lip level (Y=0), Jet throw length as L = v

= velocity head at lip2

• A jet angle of β = 45° result in maximum trajectory distance.

/g sin 2 β.

• Jet strike angle = θ = tan -1[sec β (sin 2β + Y/h v)1/2

Figure 6.80: A flip bucket and plunge pool for energy dissipation.

]

β

R

Plunge pool

Flip bucket

D

River water level

Original river bed level

θ

Jet throw X H Chute

Dam d/s face

Jet fall Y




6-87

• Minimum lip height h as: h > h min , and h min = R – R cos (Φ – tan -1

d Rd Rd

−−=Φ

2/1)]2([

S) where S is chute

slope above bucket, and

• The resulting lip angle is as: β = cos -1

• If Φ > tan

[(R – h]/R]-1S, then h min

• No affect of air resistance for V ≤ 20 m/s, but significant affect for higher velocities

0 and lip height h is defined by selected lip angle as: h = R – R cos β.

• Throw reduces by 30% for V > 40 m/s

• Keep impact zone as far as possible from dam toe to protect structure against netregressive erosion.

• Jets can be planner, 3-D, straight or skew

• Excessive pressure on flow at bucket

• All small q, a jump form in bucket

• At large q the jet sweep out and proper jet throw formation starts

• Useful when additional stream bed protection not required

• Flip bucket good for high v and low q conditions. Successful design in the past for qin excess of 1000 cfs/ft.

• Length required for total disintegration L = 5.89 q 0.139 ~ 6 q

• Heller et al. 2005 described that ski jump is the only structure to accomplishsatisfactory energy dissipation for takeoff velocity in excess of 20 m/s. Study wasconducted to establish head and pressure distribution along circular bucket, takeoffcharacteristics and impact characteristics. Typically 40% energy is dissipateddepending upon lip angle, relative bucket curvature, and relative fall height.

0.139

• Novak 2006 mentioned that energy dissipation on a ski jump occurs in 5 stages: (1):on the spillway and bucket, (2): in the free falling jet, (3): at impact into the tail water,(4): in the plunge pool or stilling basin, and (5): at the exit into the d/s river flow.

Ski jump

This is similar to flip bucket with the difference that it has large curvature radius with jetdirected horizontal or slightly downward into the plunge pool. Thus jet throw will besmall.

Plunge Pool

• Pools excavated fully or partially, or scoured in stream bed

• Scour depth D depends on unit discharge q, jet fall height Y, lip angle β, jet strikeangle θ, and bed particle size d.

• In simplified form general form of scour as: D = 1.5 q 0.6 Y 0.1

• OR D = 1.32 q

0.54 Y 0.225 sin θ




6-88

• The scour depth is as: y s = 6 y cr tan β1 ; ys = y s’ – y 0 , where y s = scour depth below

bed level, y cr = critical depth, y s’ = scour depth below tail water level, y 0 = tail waterdepth, β1

• The y

= u/s angle of scour hole (14 < β ≤ 24 degrees).

s’ = c q x H *y βw / d z where c = coefficient, (0.65 – 4.7), x = 0.5 – 0.67, y = 0.1 –

0.5, w = 0 – 0.1, z = 0 – 0.1. Thus y s’ has a wide range.

Figure 6.82: Chute and flip bucket spillway for 656 ft high San Roque dam, Phillipines.Water drops through a fall of 70 ft into the plunge pool.(http://www.gomaco.com/resources/ worldstories/ world31_1/raytheon.html)

Figure 6.83 : Flip bucket spillway for MicaDam, Canada, The spillway comprises a 3gate ogee overflow weir, spillway chuteand flip bucket. (http://www.york-net.com/60s/mica_dam.asp)

Figure 6.84: Flip bucket for Crystal dam.Foundation consists of Precambrianmetamorphic rocks intruded by

pegmatite dikes(http://www.usbr.gov/dataweb/dams/co00387.htm)

http://www.gomaco.com/resources/


http://www.york-net.com/60s/mica_dam.asp


http://www.usbr.gov/dataweb/dams/

http://www.usbr.gov/dataweb/dams/







6-89

Figure 6.85: Lostock Dam on the Paterson River, near Newcastle (38 metres high an earthand rock fill embankment dam) with concrete lined flip bucket spillway. [source:http://members.optusnet.com.au/~richardw2/projects.html].




6-90

Figure 6.86: Chute and flip bucket spillway of 173 m high arch gravity Karakaya dam,Turkey. The 6x300 MW power station achieves an exceptional layout compactness for

being incorporated adjacent to the dam toe under the overflow spillway flip-bucket. Thetwo spillway chute aerators could also be skillfully integrated: the air ducts of the firstone utilizing the spillway piers, while that of the second optimally solves the problem ofthe joint between the dam body and the powerhouse structure. For minimizing riverscouring in the impact area of the spillway jet, the flip-bucket is equipped with splittersoptimized by physical model testing. [http://www.electrowatt-ekono.co.uk/projects/karakaya.pdf]

Example: (USACE 1990, p:F-1)Design a flip bucket for given data. Q=66,200 cfs, chute and flip bucket width = 88 ft, chuteslope = S = 0.05 ft/ft, depth of flow entering bucket = d = 9.5 ft, bucket invert elevation =1375 ft, design flood tail water elevation = 1330 ft, allowable foundation bearing pressure = 2kips/ft.




6-91

Solution:1. v = q/d = Q/(w d) = 66200/(88 * 9.5) = 79.2 ft and hv = 79.2 2

2. Bucket radius R = [1.94 * 79.22 * 9.5] / [2000 – 62.4 * 9.5] = 82 ft; use R = 100 ft/64.4 = 97.4 ft

3. Φ = tan -1{[9.5 * (2 * 100 – 9.5)] 0.5

4. Minimum bucket height h/[100 – 9.5] = 25.2’

min = 100 – 100 * cos(25.2 – tan -1

5. Lip elevation = 1375 + 7.5 = 1382.5 ft

0.05) = 7.48 ft; Use h =7.5 ft

6. Trajectory angle: β = cos -1

7. Vertical drop = Y = 1375 – 1330 = 45 ft[(100 – 7.5)/100] = 22.3°

8. Horizontal jet throw = 97.4 sin (2*22.3°) +2 cos 22.3° [97.4 (97.4 sin 222.3° + 45)] 1/2

9. Impact angle = θ = tan

= 208.6 ft

-1[sec 22.3° (sin 222.3° + 45/97.4) 1/2

10. If the jet throw is less than desired, then the lip height and lip angle may be increasedto a higher value.

] = 33.7°




6-92

Figure 6.87: Flip bucket throw distance. (http://www.tpub.com/contents/coastalhydraulic laboratory/100-b/index.htm)

http://www.tpub.com/contents/coastalhydraulic







6-93

6.20 CAVITATION

• Cavitation occurs when water pressure drops below the value of pressure of thesaturated water vapor P v

• Also due to exclusion of gases from the water due to the low pressure.

at prevailing temperature

• Water bubbles when carried to regions of high pressure vapor quickly condensesand bubble explode.

• Cavities filled suddenly by the surrounding water

• Noisy process, and disruption inflow pattern

• Cavity exploding against a surface

• Violent impact of water particles at high pressure in quick succession

• Substantial damage to surface (steel, concrete) and/or complete failure of thestructure.

• Pitting often accompanied by violent vibration

• Low pressure (below atmospheric) occur at separation of water flowing alongside boundaries/especially for high velocity flow.

• Cavitation number (form of Euler number)

• Cavitation starts of J follow below critical J

• 3 mm offset cause cavitation at U > 11 m/s

• 3 mm recess at U > 32 m/s

• Roughness 1 mm +• Avoid pressure below 7 m vacuum (3 m absolute)

• Model/prototype studies to ascertain cavitation danger

• For possible cavitation

Change design

Change mode of operation

Provide other safeguard

• Introduction of air at the endangered parts (i.e. artificial aeration).

• Prevent extremely low pressures

• Provide smooth surface (epoxy mortars) on concrete surface

6.21 SPILLWAY DRAINAGE AND UPLIFT CONTROL

Drainage below the spillway crest structure is minimized by providing long approach channelfloor, by grouting the rock mass under the crest. Any uncontrolled seepage is intercepted by

proving drainage/pressure relief wells located in a drainage gallery near the structure base.The seepage and resulting uplift pressure is controlled by providing drainage interceptor

pipes below the chute at suitable intervals and disposing it at some lower elevations undergravity (usually into the stilling basin). High uplift pressure are countered by providing thickchute floor, by tying the floor with tie rods embedded into the rock below the chute etc.




6-94

6.23 SPILLWAY DESIGN STEPS

1. Mark alignment of spillway on topography sheet (approach channel + crest + chute +discharge channel + stilling basin + exit channel). A straight alignment is preferred.Horizontal curves of permissible criteria may be allowed for approach channel,

discharge channel, and exit channel, if necessary to reach the reservoir or river. Exitchannel is to connect stilling basin with the river/creek.

2. In marking alignment, leave space for other utilities as tunnels, penstock, powerhouse, energy dissipation basin, etc. The layout become more critical for rivers indeep gorge conditions when there is small space to house all these utilities.

3. The spillway crest may be inline, u/s or d/s of dam axis. The crest can be parallel oroblong to the dam axis.

4. Mark station RD for the selected layout with 0+000 RD for crest.

5. Read out NSL from the topo map for various RD’s. Draw a profile/section (El vs RD)

for NSL at suitable scale.6. Mark contour of NCL to topo map and on profile. (e.g. NCL = 2100 ft)

7. Note peak of inflow hydrograph, e.g. 1,80,000 cfs

8. Estimate (guesstimate!) spillway maximum outflow Q max

9. Select design Q (Q

from consideration of peakof inflow hydrograph and the storage space available above NCL. e.g 1,55,000 cfs.

d ) as 70-80% of Q max

10. Select gate height, e.g. 25 ft.

. , e.g. 1,16,000 cfs

11. Determine crest elevation EL c (EL c = NCL – gate height for gated spillway and EL c

= NCL for ungated spillway). Practical gate height is 20 to 40 ft. e.g. EL c

12. Select target food elevation EL

= 2100 – 25= 2075 ft.

F (EL F = NCL + flood surcharge = 2100 + 6.5 =2106.5 ft) and water surface elevation at design discharge EL d (EL d < EL F), e.g. EL d

13. Determine effective head over crest at flood level = 2106.5 – 2075 – 0.2 = 31.3 ft(considering 0.2 ft losses in approach channel)

= 2104.5 ft.

14. Select depth of approach channel P below crest level. (e.g. P = 10 ft) Mark this on profile and draw the channel bottom to meet the NSL profile. Determine the length of

approach channel L A , e.g. L A

15. Select crest length L = 250 ft.

2 (including pier width, number of piers) e.g. L 2 = 250 ft. Findeffective crest length L (excluding pier width and pier/abutment end contractions) asL’ = L – 2 (N K p + K a) H 0 . Let Pier width = 5 ft, No. of bays = 5, No. of piers = 4.Select K a (e.g. 0.01 for round nose) and K P

16. Determine unit discharge q = Q

(e.g. 0.0 for head wall parallel to flow).Thus L’ = 250 – 4 * 5 – 2(4*0.1 + 0) * 31.3 = 205 ft.

d /L 2 , (= 1,55,000/250 = 620 cfs/ft) flow velocity V =q/D, (D = flow depth in approach channel = P + EL F – EL c = 10 + 2106.5 – 2075 =41.5 ft) V = 620/41.5 = --------. Determine friction losses (from Manning formula) andother losses HL A in approach channel. e.g. HL A = 0.22 ft.




6-95

17. Determine effective head over crest at design discharge H d = EL d – EL c - HL A

18. Determine P/H

=2104.5 – 2075 – 0.22 = 29.28 ft.

d ( e.g. = 10/29.28 = 0.342) and find discharge coefficient C 0

19. Determine required crest length from eq. Q

from Fig249 of USBR (e.g. 3.66).

d = C 0 L H d 1.5

20. For selected crest length L (=205 ft) and coefficient C

[e.g. L = 116000/(3.66 *29.28^1.5) = 200 ft). Compare this with L in step 15. If different, then redo steps 15to 19. Let final selected L is 205 ft.

0 (=3.66) find the spillwaydischarge for NCL <= water elevation <= EL F (e.g for water elevation of 2100.5,2101.0, ….., 2105.0, 2105.5, 2106.0, 2106.5). For each level find H/H d and find C/C 0 form fig 250 of USBR, and discharge coefficient C. Note H = water EL – EL c - HL A .Q = C L H 1.5 . (e.g. for el = 2103. H = 27.78, H/H d

21. Select gate opening d for various water levels, e.g. gate opening of d = 2 ft for WL =2100.5, d = 5 ft for WL = 2101.0, etc.

= 0.95, C/C0 = 0.992, C = 3.66 *0.992 = 3.5712, Q = 3.5712 * 205 * 27.78^1.5 = 1,07,200 cfs.)

22. Determine spillway flow coefficient C for partial open flows for selected d from Fig.257 of USBR and determine Q = 2/3 (2g) 0.5 C L (H 1

3/2 – H 23/2 ) where H 1 = WL –

EL c – HL A and H 2 = H 1

23. Develop a EL vs. spillway Q policy from steps 20 and 22 as given in Box 1 below.

– d. Determine Q for all selected (EL, d) values (e.g. at WL= 2102.0, d = 10 ft, H1 = 2102.0 – 2075 – 0.22 = 26.78, H2 = 26.78 – 10 = 16.78,d/H1 = 10/26.78 = 0.373, C = 0.681, Q = 52,168 cfs.)

Box 1: Selected spillway discharge rating curve

Water surfaceelevation EL

Flood depth D Gateopening

Discharge

2100 0 0 0

2105.5 0.5 2

2101.0 1.0 5

2101.5 1.5

2102.0 2.0 10 52,168

2102.5 2.5

2103.0 3.0 Full open 1,07,200

2103.5 3.5 Full open


2104.5 4.5 Full open 1,16,000







6-96

24. Fit a curve to Q vs flood depth D. A single curve for all data or more than one curveto separate parts of the curve (e.g. one for full open data and other for partial openflows) is ok.

25. For the known flood inflow hydrograph, reservoir elevation. vs. storage volume

relationship, perform the flood routing using the spillway el-vs-Q data curve. Find themaximum spillway outflow discharge and maximum water level EL max reached. Theselected spillway design is considered acceptable if maximum water level does notexceed (preferably match) the target flood surcharge levels i.e. EL max <= EL F

26. If EL

.

max > EL F

27. If EL

, then do following. i. Lower crest level (and increase H over crest), ii.Increase crest length, iii. Provisions which result in decrease of pier/abutmentcontractions and increase of discharge coefficient, e.g. increase P, flatten u/s slope ofcrest approach (Fig 251 USBR) etc. iv. Increase gate opening to affect largerdischarge at a selected elevation e.g. increase d from 2 to 5 ft for EL = 2100.5 ft etc.Redo steps 16 to 25.

max << EL F

28. When design of crest is complete, mark it on profile. Draw ogee profile. At d/s ofogee chute (2-3 H

then do inverse of step 26. Redo steps 16 to 25.

c

29. Draw a blow up map/section of crest and ogee details.

below crest) provide concave curve (in vertical) to reach close to NSL and then convex curve to attain the same chute slope as of average NSL.Determine flow depth from energy equation and provide wall height for dischargechannel accordingly (include free board.). Continue along the NSL with minimumvertical curves.

30. Provide stilling basin at river level. Determine length, depth, and other details.




6-97


Bakhtyar, R., S.J. Mousavi and A. Afshar. 2007. Dynamic programming approach to optimaldesign of cascade stilling basin. J. Hydraulic. Eng. ASCE. 133(8):949-54. Aug-07.

Boes, R.M. and W.H. Hager. 2003. Two phase flow characteristics of stepped spillways. J.Hydr. Engg. ASCE. 129(9):661-70, Sep 03. and Discussions by: Chanson, H.;Amador et al.; Andre et al.; and Closure. J. Hydr. Engg. 131(5):419-29. May 05.

Bollaert, E.F.R. and P.J. Mason. 2006. A physically based model for scour prediction atSrisailam dam. Hydropower and Dams. 13(4):96-103.

Carroll, John. ----. Fusegate spillway for Terminus dam at lake Kaweah, Tulare county,California. (http://www.hec.usace.army.mil/misc/watershed_conference/PDF_files/carrol_John.pdf)

Chanson, H. 2001. THE HYDRAULICS OF STEPPED CHUTES AND SPILLWAYS.Balkema, Lisse, The Netherlands.

FERC 1999. ENGINEERING GUIDELINES FOR THE EVALUATION OFHYDROPOWER, PROJECTS; CHAPTER 11 - ARCH DAMS. Federal EnergyRegulatory Commission, Division of Dam Safety and Inspections, Washington, DC20426, October, 1999.

Heller, V., W.H. Hager and H. Minor. 2005. Ski jump hydraulics. J. Hydr. Engg. ASCE.131(5):347-55. May 2005. and Discussions by Katsuia, RM,; Novak, P.; Heller et al.J. Hydr. Engg. ASCE, 132(10):1115,-1117. Oct. 2006.

Heller et al. 2006. Closure of discussion on: [Heller, V., W.H. Hager and H. Minor. 2005. Ski jump hydraulics. J. Hydr. Engg. ASCE. 131(5):347-55.] J. Hydr. Engg. ASCE,132(10):1115,-1117. Oct. 2006.

ICOLD 1992. Selection of Design Flood: Current methods.International Commission onLarge Dams (ICOLD) Bulletin No. 82., Paris, France.

Katsuia, RM. 2006. Discussions on: [Heller, V., W.H. Hager and H. Minor. 2005. Ski jumphydraulics. J. Hydr. Engg. ASCE. 131(5):347-55.] J. Hydr. Engg. ASCE,132(10):1115,-1117. Oct. 2006.

Mason R.J. and K. Arumugan. 1985. Free jet scour below dams and flip buckets. J. Hydr. Engg.ASCE. 111(2):

Morris and Wiggert. . Applied Hydraulics in Engineering.

Novak, P. 2006. Discussions of:[ Heller, V., W.H. Hager and H. Minor. 2005. Ski jumphydraulics. J. Hydr. Engg. ASCE. 131(5):347-55.] J. Hydr. Engg. ASCE,132(10):1115,-1117. Oct. 2006.

Savage, B.M. and M.C. Johnson. 2001. Flow over ogee spillway: Physical and numericalmodel case study. J. Hydr. Engg. ASCE. 127(8):640-49, Aug. 2001.

USBR 1978. HYDRAULIC DESIGN OF STILLING BASIN AND ENERGYDISSIPATOR. Engineering Monograph No. 25, US Bureau of Reclamation, Denver.

http://www.hec.usace.army.mil/misc/watershed_conference/

http://www.hec.usace.army.mil/misc/watershed_conference/




6-98

USBR 1987. DESIGN OF SMALL DAMS. US Dept of Interior, Bureau of Reclamation. AWater Resources Technical publication, 3 rd

USBR 2001. DESIGN OF SMALL DAMS. US Dept of Interior, Bureau of Reclamation. AWater Resources Technical publication. Oxford and IBH Publishing Co. Pvt. Ltd.

New Delhi.

edition. (PDF edition available at:www. usbr .gov/pmts/hydraulics_lab/pubs/manuals/ SmallDams .pdf)

Vittal, N., and P.D. Porey. 1987. Design of cascade stilling basins for high dam spillways. J.Hydraulic Eng., ASCE. 113(2):225-237.

Wei, C.Y. Spillways and stream bed protection. In: Davis’ Handbook of Applied Hydraulics.P-16.8.




6-99

Figure 6.101: Hoover dam: dam arch, intake towers, side channel spillways and tunnels.

Figure 6.102 : Ski jump Spillway of Menzelet Dam, Turkey(http://www.ce.metu.edu.tr/~ce471/links.htm )








6-100

Figure 6.103: Spillway of Hasan Ugurlu Rockfill dam, Turkey(http://www.ce.metu.edu.tr/~ce471/links.htm )

Figure 6.104: Ski jump spillway and power penstock for Keban Dam (Rockfill), Turkey.(http://www.ce.metu.edu.tr/~ce471/links.htm )












6-101

Figure 6.105: Ataturk Dam (http://www.ce.metu.edu.tr/~ce471/links.htm )

Figure 6.106: Spillway face of Ataturk Dam (http://www.ce.metu.edu.tr/~ce471/links.htm )












6-102

Figure 6.107: Oymapinar Dam (Arch) http://www.ce.metu.edu.tr/~ce471/links.htm

Figure 6.108: Oymapinar Dam (Arch) (http://www.ce.metu.edu.tr/~ce471/links.htm )












6-103

Figure 6.109: Altinkaya Dam (RF) Turkey (http://www.ce.metu.edu.tr/~ce471/links.htm )

Figure 6.110: Simly dam ungated ogee type emergency spillway.








6-104

Figure 6.111: Simly Dam: Plan and profile of 2-stage spillway and diversion tunnel.

Figure 6.112: Lower basin of Simly Dam spillway. Seen also plugged tunnel end.

Diversion tunnel(abandoned and)

1st Stage

2nd Stage

Crest




6-105

Figure 6.113a: The Craig Goch Dam (masonary dam constructed in 1890) in Elav valley,Wales (d/s view) (http://www.riderz.co.uk/pics/north_wales_07/3_elan_dam.jpg ) and

(http://history.powys.org.uk/history/rhayader/craig.html )

Figure 6.113b: http://www.fotolibra.com/gallery/378594/elan-valley/

http://www.riderz.co.uk/pics/north_wales_07/3_elan_dam.jpg



http://history.powys.org.uk/history/rhayader/craig.html



http://www.fotolibra.com/gallery/378594/elan-valley/









6-106

Figure 6.114: Grahamstown dam labyrinth spillway(http://www.ipenz.org.nz/nzsold/NZSOLD-Newsletter-46-Text.pdf )

Figure 6.115: St Marry dam flip bucket spillway

http://www.amec.com/uploadfiles/stmary.pdf











6-107

Figure 6.116: San Roque dam flip bucket(http://www.gomaco.com/Resources/worldstories/world31_1/photos/raytheon/CL-010301-

D10.jpg )

The spillway is located on the right-hand abutment of the dam and guides the flow of excess water supplied tothe reservoir. At 328 feet (100 m) wide and 1722 feet (525 m) long, the San Roque Dam spillway is designed fora flow of 452,028 cubic feet (12,800 m 3

The flip bucket runs the full 328 feet (100 m) width along the bottom edge of the spillway. Its 70 foot (21 m)drop dissipates the force of the water flowing down the chutes. Its reinforced concrete slabs are 33 feet (10 m)thick to withstand the constant pounding it takes from the force of the water flowing down the slope. The railsystem the C-700 traveled on was formed to match the radius of the flip bucket.

) per second. The spillway is separated into three sections by two 26 feet(8 m) high longitudinal walls. Each section is 98 feet (30 m) wide and runs the entire length of the spillway,

from the ogee at the top to the flip bucket at the bottom. The C-700 was set up to finish each slab within eachsection in two 49 feet (15 m) wide passes.

One of the important design aspects of the San Roque spillway included the installation of six air duct galleriesthat run underneath and transversely across the entire width of the spillway. As you look at the pictures of thefinished spillway, the galleries look like six steps going up the spillway chute. The purpose of the galleries is toinject a protective cushion of air into the water. The galleries are designed with natural ventilation and, as thewater travels over them, a vacuum is created and injects the air right into the water. The air reduces the wearingof the concrete as the massive amounts of water flow down the chute, thereby extending the operating life of thespillway and reducing maintenance costs. The concrete mix included a high-quality, low alkali, sulfate resistantPortland cement with a moderate heat of hydration. A Type II cement was specified because it generates lessheat as it cures. More than 1.8 million bags of cement were used to produce the concrete for the spillway.

The mix design had a strength of 6000 psi (41 MPa). Concrete slump averaged between three to 3.5 inches (76to 90 mm).

http://www.gomaco.com/Resources/worldstories/world31_1/photos/raytheon/CL-010301-D10.jpg









6-108

Figure 6.117: Closeup of side channel spillwayhttp://members.optusnet.com.au/~richardw2/img/bjuck9b.jpg

Figure 6.118: Ski-jump spillway at Tocantins dam, Para, Brazil (http://ponce.sdsu.edu/legacy_tales_a_great_mans_glory.html )

http://ponce.sdsu.edu/legacy_tales_a_great_mans_glory.html







6-109

Figure 6.119: Spillway radial gates . (http://www.ce.metu.edu.tr/~ce471/links.htm )

Figure 6.120: Baffled chute (http://www.uq.edu.au/~e2hchans/pictures/irago05.jpg )




http://www.uq.edu.au/~e2hchans/pictures/irago05.jpg








6-110

Figure 6.121: Ski jump spillway at Poechos Dam, Peru(http://ponce.sdsu.edu/poechos_spillway.html )

http://ponce.sdsu.edu/poechos_spillway.html






Tariq 2008 Dam and Reservoir Engineering 7-1Chapter 7 Dam Outlet Works

CHAPTER 7

DAM OUTLET WORKS

7.1 PURPOSES AND FEATURES

• Outlets serve to regulate or release water impounded by a dam. There are one ormore outlets structures to requisite discharge capacity.

• Most outlets are gated but some may be gated or ungated (as for flood retard)

• Release at retarded rate for detention dam (outflow < inflow)

• Diversion dam-release into canals or pipelines (canal outlet), release to ahydropower plant (pressure pipe outlet) or release into river for d/s uses orenvironmental purposes (river outlet)

• Open channel outlets for small heads and closed conduit for large heads (tunnel or

cut & cover)• Conduit flow as under pressure (full or part length) or free gravity flow

• Elevation of intake of outlet works usually just below dead storage level. Low-level outlets have its intake at very low elevation, usually close to original river

bed level.

• Elevation of outlet terminal structure depends on water uses (at river bed level forriver outlets, at/above canal bed level for irrigation, at penstock or turbine inlet for

power outlets).

• Outlets may be used to pass diversion flood flows (large capacity) Or capacity justenough to release present / future / target releases (small capacity)

• Low level outlets may be used to sluice or flush out reservoir sediments.

• Used to release water temporarily stored in flood control storage space.

• Used to lower water below NCL in anticipation of flood.

• Used to empty reservoir to permit inspection, to make needed repairs/maintenanceof structures that are normally inundated.

7.2 OUTLET CAPACITY

• Outlet capacity based on target releases (maximum) for various purpose(s).

• Irrigation/water supply/power production: Demand based Q, with minimumfriction losses required for hydropower.

• Diversion flood: Capacity enough to keep flood water level well below coffer damheight (less free board).

7.3 INTAKE LEVEL VS RESERVOIR STORAGE LEVEL

• This depends on discharge capacity. Outlets are placed sufficiently belowminimum reservoir operating level to provide head for required flow capacity.

• Outlet to be low enough below lowest water levels to avoid swirling / vortexformation above inlet.




• Depend on purpose. Outlets for diversion flood passage have inlet near river bedlevels.

• Sedimentation: Outlet intake high enough to be not affected by futuresedimentation.

• Intake position is important consideration, high enough to prevent interferencefrom sediments and low enough to permit drawdown below top of inactive storagespace (dead storage).

• Intake at multiple levels possible if more than one outlets. Tower intake.

• One low level outlet to river level (e.g. for flushing) and other as pressure outlet/canal outlet.

• Diversion tunnel usually converted to permanent outlet works after completion ofconstruction phase.

• The intake at river bed level for diversion outlet may be closed later and revisedintake provided at a higher level for regular outflows. The lower level intake may

be subsequently closed/ plugged.

7.4 COMPONENTS

Outwork works included following components (Fig. 7.1):

1. Intake section consisting of: (i) Trash rack to stop trash entry into outlet, and (ii)Inlet structure. Occasionally a fish rack will be required for the case power tunnel.

2. A fluid way (tunnel or open canal)

3. Outlet exit with: (i) Energy dissipation structure for canal or river outlet, or (ii)Transition into penstocks for hydropower outlet or pressure pipe for water supply

4. Control devices including (i) One or more flow control/regulating gates or valves(ii) Bulkhead or stoplog at entrance mouth, (iii) Guard gate

7.5 LAYOUT

• Usually a straight alignment

• Fluid way may be curved in plan and profile to suit inlet and outlet locations

• May open into spillway stilling basin/chute if outflow is into a river. Thusadditional energy dissipation structure will not be required.

• Outlet into canal head works for irrigation outlet.

• Topography, geology (rock, type, over burden), type and design of fluid wayaffect location and type (tunnel or cut-and-cover conduit).

Stilling basin

Damembankmentor abutment

Intake

Gate/valve

Trash / fish rack

Fluidway River/Canal

R e s e r v o

i r

Gate/valve

Figure 7.1 : Typical outlet layout.

Turbine




• For earthfill and rockfill dams, outflow is usually carried through the rockabutments as tunnel or as pre-cast pipes across the embankment with suitableseepage retarding collars.

• For gravity dam outlet is usually carried through the dam body using pre-cast

materials (e.g. embedded steel pipes) or cast-in-place.• Outlet fluid way as tunnel for whole length or central part as tunnel plus cut-and-

cover for entrance and exit sections may be used.

• For quality constraints, (silt, algae, debris, fresh) multiple level inlet to draw waterfrom different levels of the reservoir at different seasons of the year.

• Intake at abutment face for embankment dams and at dam face for concrete dams

• Inlet may be located in a tower (away from face) and bridge deck or boat providedto reach.

7.6 OUTLET CONTROL WORKS

• Control gates and valves installed across fluid way/conduit to regulate release.

• Operating gates and valves to control and regulate outflow and operate in any position from closed to fully open (partial open to full open)

• Guard gates near u/s end as full open or full closed condition to effect closure inthe event of failure of operating gates/valves or for unwatering of conduit forinspection/repair of the operating gates.

• Stoplogs or bulkhead at entrance to conduit for closure.

• Control gates may be located at u/s end, at an intermediate point, or at lower d/s

end of structure.• For free discharge conduit tunnel portion u/s of gate is under pressure thus controllocation affect conduit design.

• For discharge into closed pressure pipe, gates serves only to regulate flow, full pipe flow under pressure both u/s & d/s of control. Gate location have little affecton conduit design.

7.6.1 Control on U/S end

• For free flow conduit, part or full flow along whole length.

• Operating head and conduit slope determine flow regime (uniform, varying, sub

critical, critical or super critical, usually supercritical).• Conduit subject to external loading of full water depth near u/s end (where rock

cover is small). Conduit wall near u/s end designed to withstand this large pressure.

• Fish screen, trash rack, guard gates, stoplog slots, regulating valves all combinedin a single intake structure.

• Entire conduit can be dewatered readily (except some leakage into tunnel fromabutment above)

7.6.2 Control at Intermediate Point

• Pressure in conduit u/s of control equal to full reservoir depth/head.




• External and internal hydrostatic pressure on conduit shell same, thin tunnellinings.

• Water tightness of u/s portion less important

• Leakage through joints/cracks to outside of pipe can start piping along outside of

pipe leading to saturation and later sloughing of dam or abutment.• U/S portion of conduit provided with steel liner.

• For likely settlement/movement/cracking of conduit wall, a separate small steel pipe carried inside the larger conduit and control placed at suitable place.

7.6.3 Control at d/s End

• Suitable for conduit discharging freely into canal or river and stilling basin.

• All conduit length under pressure

• Guard gates at u/s end for un watering conduit.

7.7 WATER WAYS7.7.1 Open Channel

A channel or flume is placed through the embankment to carry flow from reservoir toan open canal. A gated culvert or orifice controls outflow. Used for small reservoirs.

7.7.2 Tunnels

• Tunnel dug through abutment

• Geology affects location and design

• Tunnels excavated by tunnel boring machine, drill + blast, controlled blasting ormanual force + donkey machine (Fig. 7.3)

• Inside surface usually provided lining of smooth finish. Lining for surfacesmoothness or for strength purposes.

• Lining must in soft rock zones

• No lining required in competent rock (except for surface smoothing)

• Reinforced lining for pressure tunnels to withstand external/internal pressures ofrock load, hydrostatic, etc.

• Minimum tunnel size limited by as required for construction limitations (D > 5 ft).

• Tunnel shape round, horseshoe, modified horseshoe (Fig. 7.4)

• Pressure tunnels usually circular/round, free flowing of horseshoe section.

7.7.3 Cut-and-Cover conduit

• Small size tunnel cannot be constructed.

• Foundation geology may not support tunnel

• Then cut-and-cover conduit is best option

• Conduit passes in shallow depth in abutments under the dam or through the dam

• Conservative and safe design against danger of settlement, cracking, leakage, piping along outside periphery, lateral or longitudinal displacement

• Carried over hard rock or at places with minimum overburden over rock.




Figure 7.2 : Schematic of typical outlet arrangements.




• Conduit cast in place or prefabricated metal/concrete pipes.

• Must have large strength towithstand full load overthe conduit.

• Conduit construction before embankment.

• Seepage collars to reduce piping hazard.

Circular Horseshoe Modified horseshoe

Figure 7.4 : Tunnel shapes.

Figure 7.3a : Diversion tunnel at Split Rock Damexcavated in sedimentary breccia, greywacke andmudstone. (Source: http://members.optusnet.com.au/~engineeringgeologist/page12.html)

Figure 7.3b Tunnel support form work forlining




7.8 TUNNEL DESIGN

• Tunnel bottom at some slope

• Tunnel flow as open channel flow for small discharge and head

• Tunnel as pipe flow at large Q & u/s head

• Tunnel may discharge freely at atmospheric pressure (H = 0) into canal or under pressure (H > 0) as for power plant or pressure water supply.

7.8.1 Open Channel Flow• For low depth d (d < D), determine Q by Manning formula (uniform flow),

include losses due to surface frict ion and other installation (gates, valves etc).

• Full A = π r 2 = 2 π (D/2)

• For a circular tunnel the Flow area A = (β – sin β)D

2 2

• n = 0.013 for smooth concrete surface

/8 where β is angle subtendedin radians, D is tunnel diameter; Perimeter P = 0.5 β D; and hydraulic radius R =0.25[(1 – sin β)/β] D. Note β≤ π for d ≤ R and β > π for d > R and R = radius =D/2

7.8.2 Full Flow H > Dia (pipe flow)

• For free flow d/s H = 0 (e.g. flow intostilling basin)

• For pressure flow (H > 0) Flow byWilliam Hazen formula

V = 1.318 C R 0.63 S

C = 125; S = (H

0.54

1-H 2)/L where H 1 = reservoir water level, H 2 = water level atexit (usually H 2

R = A/P = πD

< Dia), L = tunnel length.2

Q = A V

/4 πD = D/4

• The tunnel discharge may be determined by using Darcy-Weisbach formula as:

h f = f L/D V 2

where f is friction loss coefficient [generally f = 116.5 n

/2g2/r 1/3 = 185 n 2/D1/3] Use

Moody diagram to select f.

H1 L

R

dα

β

R

d




Figure 7.5: Mangla dam power tunnels during construction/lining.

Figure 7.6: A tunnel boring machine (TBM)




7.8.3 Head Loss for Flow Less than Maximum Flow

• Tunnel designed for large Q (e.g. 30,000 cfs for diversion flood)

• Release Q for e.g. hydropower is small (e.g. 2000 cfs)

• For selected tunnel dia for Q = 30,000 (e.g. D = 20 ft)

V for small flow = 2000/ π x 10 2

h

= 6.37 fps

f = f L / D V 2

L = 1200 ft

/2g

f = 185 x 0.013 2/20 1/3

h

= 0.0115

f = [0.0115 x 1200 / 20] x [6.37 2

Head available for hydropower = H

/ 2 x 32.2] = 0.44 ft only

1

7.9 THE OUTLET SYSTEM

- 0.44 ft – head loss in other fittings

7.9.1 Intake Layout Davis HAH p:22.6-7

1. Vertical intake on dam face or abutment2. Small slope intake on dam face or abutment

3. Large slope intake on abutment

4. Tower intake (rectangular gate or cylinder gate) Fig. 7.7

5. Shaft or submerged intake

• Intake submerged or extended as tower above maximum reservoir level

• Entrance be vertical, inclined or horizontal

• Conduit entrance rounded or bell mouthed to reduce entrance losses

• Rounding on bottom, sides, roof• Trash rack and / fish rack in front of entrance with suitable arrangement for rack

cleaning

Figure 7.7a: Intake tower and access bridge.




Figure 7.7b: Intake tower with access platform for Tainpura-I dam.

Figure 7.7c: Submerged tower inlet for Simly dam. Temporary arrangements to pump outwater from below dead storage level in 2001 from a pontoon are also shown.




7.9.2 Control Arrangements

• Top free radial gate-in open channels

• Top seal radial gate-installed atentrance or within a culvert outlet

followed by open channel or conduit• Slide gate for open channel or culvert

outlet (head wall structure).

• Slide gate for tower structure,worked from an operating deck

• Wet wall shaft - for slide gate or topseal radial gates (extends fromconduit to crest/platform for access)

• Water tight bonnets over the gate

slots, gate operated from dry shaft ora operating chamber located aboveconduit.

• Dry well to reach valves (u/s and d/send of valve encased in concrete.Operating platform reached fromcrest top or from d/s end of conduitthrough a separate chamber (doomedchamber).

• Housing around controls to houseoperating equipment under adverseweather, to enclose top of the access shaft, and to accommodate auxiliary equipmentas ventilating fans, heaters, flow measuring and recording meters, air pumps, powergenerator set, M&R accessories)

Submergence for non-vortex formation

Vortices at intake structures can affect intake efficiency and create a safety hazard. Vorticesare associated with high discharges and shallow intakes, these can form at intakes submerged

as much as 60 to 100 ft (USACE 1980, EM-1110-2-1602 p.4-3). The intensity of circulationaround an intake is function of submergence of the intake, the discharge and the intake andapproach channel geometry. Undesirable vortices can draw air and debris into the structurereducing its max capacity. Anti-vortex devices are installed to reduce vortices. When vortex

prevention devices are used the critical submergence (ratio of water depth above the top ofinlet to the inlet diameter – both dimensions at entrance to inlet bell mouth) should equal orexceed the inlet flow Froude number to provide vortex free formation; otherwise it shouldequal or exceed Froude number plus one. Model studies may be needed in some instances.

Unsymmetrical approach: S ≥ 0.4 V D

Symmetrical approach: S ≥ 0.3 V D

½ ½

Figure 7.7d: Outlet towers for Hoover Dam. Towersize = 82 ft at base, 63 ft at top and 30 ft dia inside fluidway[ @http://www.usbr.gov/lc/hooverdam/images/D009a.jpg]




Figure 7.8a : Typical intake layout and gating arrangements.




Figure 7.8b . Typical inlet layout and gating arrangement.







7.9.4 Terminal Structures

• Plunge basin-for tunnels that end up in flip bucket or d/s valve,

• Stilling well – (Riser well) (energy dissipated by fillets and diffuser blocks, andturbulence), Rising velocity 1-3 ft/s

• Stilling basin (for hydraulic jump)

• Impact type stilling basin

I mpact Basin

A concrete impact basin is an energy dissipating device located at the exit of theoutlet in which flow from the discharge conduit strikes a vertical hanging baffle. Discharge isdirected upstream in vertical eddies by the horizontal portion of the baffle and by the floor

before flowing over the end sill. Energydissipation occurs as the discharge strikes the

baffle, thus, performance is not dependent ontailwater. ODNR (1999)

Figure 7.9a : Impact Basin (ODNR 1999)

USBR Type II and Type III Stilling Basins

Type II and Type III basins reduce the energy of the flow discharging from the exit of anoutlet or spillway and allow the water to exit into the outlet channel at a reduced velocity.

Type II energy dissipators contain chute blocks at the upstream end of the basin and adentated (tooth-like) endsill. Baffle piers are not used in a Type II basin because of the highvelocity water entering the basin.

Type III energy dissipators can be used if the entrancevelocity of the water is not high. They contain baffle

piers which are located on the stilling basin aprondownstream of the chute blocks. Located at the end of

both the Type II and Type III basins is an endsill. Theendsill may be level or sloped, and its purpose is tocreate the tailwater which reduces the outflowvelocity. If any of the severe defects associated with

Figure 7.9b : Baffle blockimpact stilling basin

Figure 7.9c: Type III Basin




concrete structures are observed, a registered professional engineer should be contacted toevaluate the stability of the basin.

Abutment

Stop log grooveTrash rack slot

Fig. 7.12 : Front entrance intake

Groove for stop log and/or bulkhead gate. Extends alongface of dam to hoist on crest

Fish screen,trash rack

abutment

Fig. 7.11 : Intake with sloping entrance.

Fi ure 7.14 : A dr well for ate valve controlFigure 7.13 :Submerged shaft inlet

Figure 7.10 : Stilling wells: L. Impact well, R. riser well energy dissipater.

C o n

t r o l w e l l

Dentatedsill




Figure 7.15 : Inclined intake and wet well control.

Figure 7.16 : Vortex formation above a horizontal inlet. (Visher and Hager p.223)

Abutment

Embankment




Figure 7.17 : Plunge pool energy dissipater for river outlets.7.10 FLOW CONTROL

The outflow from outlets is controlled/regulated according to needs. This is done byinstalling gates and valves in the fluid passage.

7.10.1 Nomenclature

Gate: Gate is a closure device in which a leaf or closure member is moved across the fluidway from an external position to control the flow of water.

Valve: Valve is a closure device in which the closure member remain fixed axially withrespect to the fluid way and is either rotated or moved longitudinally to control flowof water.

Guar d gate/valve: Operates fully open or closed and functions as secondary device forshuttling off the flow in case of primary closure device become inoperable. Guardgate or valve is operated under balance pressure (no flow conditions) except inemergencies

Regul ating gate/valve: Operate under full pressure and flow conditions to throttle and varythe rate of discharge.

Bulk head gate: Installed at entrance to drain fluid way for inspection, repair. Opened orclosed under balanced pressures (no flow conditions)

Stop logs: As smaller bulk head sections to permit easier handling.




7.10.2 GATE TYPES

1 Conduit slide gate

2 Ring follower gate

- Upper bulk head

- Lower ring3 Ring seal gate

Similar to ring follower gate with roller train and wheals to reduce friction

4 Jet flow gate to reduce slot affects

5 Wheal and roller mounted slide gates (on sloping u/s face of dam/abutment).

6 Cylinder gates (a cylindrical shell lowered to control flow through radial openingsinto a circular vertical intake structure). Located inside or outside of circular structure.When lowered, closes all circumferential openings/slots

7. Tainter / radial gates (located in middle or d/s end of tunnel)

8. Bulk head gates and stop logs

- Placed in slots in front of entrance (for u/s end control)

- Seals provided on all sides with slot face (music note or J-seal).

- Stop log is steal structure with rubber seals

- Placed under balanced pressure i.e. no-flow conditions

- Bulkhead closes opening and stoplog closes approach section.

- Bulk head composed of maximum two closure sections and cover only theentrance opening

- More than 2 units, units fill to a height above the existing reservoir surface – stop log.

- Placement by crane and lifting frame.

- Infrequent use – when conduit need emptying

- Lifting frame (with latching device operated by a tag line) permitconnecting and disconnecting units for lowering or raising.

- Bulkhead gate at Hoover dam is of size 50 x 50 ft weighing 3 million pounds 90 Metric tons.

Flow conditions below slide gate

The flow conditions below slide gate opening into closed conduits as tunnels dependsupon the gate opening, d/s conduit hydraulics and air supply. These are as under forincreasing gate openings:

i. Spray flow

ii. Free flow

iii. Foamy flow

iv. Hydraulic jump due to tailwater submergence

v. Hydraulic jump due to pressurized tailwater

vi. Fully pressurized flow.




Figure 7.18: Vertical lift gates.




Figure 7.19: Ring follower gate. (Davis HAH p.22.16)




Figure 7.20: Ring seal gate. (Davis HAH p22.20)




Figure 7.21 : Jet flow gate. (Davis HAH p-22.22)




Figure 7.23: Wheel/Roller mounted slide gate. (Davis HAH p.22.26)




Figure 7.24: Cylinder gate. (Davis HAH p-22.32)




Figure 7 .25: Cylinder gate. (Davis HAH p-22.34)




Figure 7.26: Typical bulkhead gate with lifting arrangements. (Davis HAH p-22.35)

Figure 7.27: Typical stop log installation and lifting arrangement. (Davis HAH p.22.36)




Figure 7.28 : Needle valve. (Davis p-22.40)

Figure 7.29: Tube valve (p.22.43) Figure 7.30 : Jet flow valve (p-22.50)

Figure 7.31: Sphere/cylinder valve. (Davis p-22.64)




Figure 7.32: Howel-Bunger valve (Source: Novak p-208) (Ex Mangla dam)

7.10.3 VALVES

• Valves generally installed in pipe lines or pressure tunnels

1. Needle valves

Operated by water pressure (by filling and emptying of various chambers).

2. Tube valves

(small cavitation)3. Fixed cone (Howell-Bunger) Valves (also called cone dispersion valve)

- Operating system exterior to valve body

- Discharge into atmosphere as jet. (example Mangla dam irrigationvalves)

- Installed at d/s end of pipe e.g. into canal (no stilling basin, plunge poolrequired)

- May have an external hood (cover) to direct flow in forward direction

4. Hollow jet valve

- Free discharge into atmosphere (can operate partially submerged).

Hood




- Placed at d/s of pipe/tunnel

5. Sleeve type valve (similar to cylinder gate)

6. Butterfly valve

7. Sphere valve

10.8 BELL-MOUTH ENTRANCE (Davis p:22-68)• Entrance circular or rectangular in cross section

• Rectangular entrance transitioned into circular section.

• Entrance may be fully flared for circular entrance on all sides and for rectangularentrance on 4-w, (all sides) 3-w (sides and top), 1-way (top)

• Flare tangent at 2 ends (90 o

• Eliptical curve (single) D is conduit dia (circular), conduit height for vertical

apart)

curve, and width for horizontal curve.

Figure 7.33 : Entrance transitions. (Davis p-22.68)




• Circular flare as: X 2/(0.5D) 2 + y 2/(0.15 D) 2

• Rectangular 4-w, 3-way X

= 12/D2 + y 2/(D/3) 2

• Rectangular 1-w X

= 12/(1.5D) 2 + y 2/(0.5 D) 2

• Compound elliptical curves for high head and low back pressure (low back pressure – static head d/s of entrance < 0.1 H

= 1

v

• Required back pressure

)

V fps < 50 60 80 100 120 140 200

BP (ft h) N/A 5 10 15 23 30 63

10.9 FLUIDWAY SURFACE (Davis p:22-70)

• Smoothness, waviness, alignment become increasingly important as velocityincrease and back pressure decrease.

• At 29 fps velocity offset 5/16” (sharp corned) create cavitation.

• At high V, 1/32” (0.8 mm) offset can produce cavitation and damage.

• Smooth surface required d/s regulating gates and valves

• Surface smoothness of 250 µ in (0.006 mm) in required d/s of valves.

• Machined steel surface, smooth troveled concrete provide such smoothness butmass concreting become slow

• Provide metal lining for some distance d/s of gates.

• Stainless steel coating best-no painting required

• Flat surface required to eliminate waviness

• Offset at joints to be eliminated d/s member outward offset by 1/16” eliminatecavitation

• Small raids or bug holes on concrete surface can cause cavitation

• Manholes and pipe openings in regions of pressure gradients above atmospheric

10.11 EXIT PORTAL PRESSURE

The conduits flowing full become partially flowing near the exit end. The intersectionof the hydraulic grade line is function of Froude number of the conduit flow. The valuesof Y p/D are also dependent on the condition of support of issuing jet and tailwatercondition; a good approximation is two-thirds the vertical dimension above the exit portalinvert.




Figure 7.34: Exit portal pressure at end of full flowing tunnels (USCOE 1980).

10.12 CAVITATION

• Formation of vapor filled cavities in a liquid (pressure in localized regionsreduced to vapor pressure). Collapsing of cavities near a fluid way surface resultin putting and damage to surface.

• This is important factor in design of high velocity outlet works.• Fluid way alignment and energy surface discontinuity are potential source of

producing cavitation.

• Follow following design approaches

• Keep alignment and boundary surface as straight and free of irregularities as possible.

• Hold pressure gradient as high as possible at potential cavitation locations

• Introduce air, if possible, at points where sub-atmospheric pressures exist in theflowing water (usually behind gates/valves).




• Provide definite and adequate spring points for flow separation, such as in asudden enlargement and for various type of gates.

• Cavitation index K = (H 2-H vp ) / (H T-H 2

• H

)

2

• H

= static pressure head downstream from gate/valve

T

• H

= total head (static + velocity) in u/s conduit

vp

• Incipient cavitation index: gate valve 1.5, butterfly

= vapour pressure head of water at given temp.

• ( ) 2// 200 V P P ρ σ −= , where P = local pressure and P 0 and V 0

• Irrigation tunnel lining blown out due to cavitation behind the u/s gates in Tarbeladam

reference values

of u/s flow.

• The spillway tunnel in Glen Canyon dam scoured to 31 ft deep pit over 100 long

section due to cavitation problem in 1993. ( http://www.popsci.com/ popsci/science/ 5917359b9fa84010vgnvcm1000004eecbccdrcrd.html

)

Figure 7.34: Gate slot types for high velocity outlets. (Davis p-22.74)

http://www.popsci.com/%20popsci/science/







• The gate slots produce a discontinuity in the side wall which may cause cavitation.Metal plate liners are used down stream of the gate slots to protect the concrete fromthe erosive action of the cavitation.

• Gate slots are streamlined to to a taper d/s of slot to check the potential incipient

cavitation.•

10.13 AIR VENTS

• Control valves and gates located a considerable distance from the exit end (i.e. donot discharge into the atmosphere) require air vents.

• Emergency gates located immediately u/s of air vented service gates do notrequire air vents. However emergency gates should normally not be used forregulation.

• Maximum air demand occurs at 80% gate opening.

• Air vents for 2 purposes

a) Breathers lines – to vent or admit air during filling or emptying/draining offluid way

b) To deliver continuous air to a discharging gate or valve to eliminate cavitation.

• Air vent area as 0.5-1% of area of fluidway (rule of thumb)

• Vents not controlled with valves and extend above maximum water surface.

• High filling rates could produce very large water hammer pressure – larger vents be provided.

• Slow filling recommended.• Air demand for conduit filling below gate: Q a = Q w x 0.03 (F-1) 1.03

• Vent area for air V of ≤ 150 fps

, F = Froud No. at vena contracta.

• Air vent to assure spread of air across the full width of the conduit.

• Air vent designed so that head loss do not exceed 0.5-1.0 ft of water.

• Protect vent intake from public with screens/grill due to rushing air to avoiddamage to human life near air intake.

10.14 Tunnel LiningFor smoothness 15-20 cm thickness

For strength 0.5 to 1.5 m thick. Provide reinforcement

Required smooth workmanship to avoid cavitation at very high velocities.

Cavitation largest hazard below valves, thus provide lining for some length d/s of gates andvalves.

10.15 SYSTEM LOSSESSystem losses are generally as: H T = K L V 2

1. Trash rack losses/2g. Various losses are as:

2. Entrance losses




3. Bend losses (horizontal and vertical bends)

4. Contraction losses (due to reducer and other fittings)

5. Expansion losses

6. Gate/valve losses

7. Skin friction losses

8. Velocity head exit loss at outlet.

The H T

• free discharging outlet = center of outlet valve/gate or opening

is measured from reservoir water surface to

• jet supported onto d/s floor as for stilling basins – to top of emerging jet at the point of greatest contraction

• submerged outlet portal – to tail water level

10.15.1 Trash Rack Losses

H l = K t v2/2g where K t = 1.45 – 0.45 (a n/a g) – (a n/a g)

where a

2

n = net face area a g

Assume 50% area clogged (for maximum loss) and no clogging (for minimum loss)

= gross face area

Losses are usually small as 0.3 and 0.5 ft for velocity of 1.5 and 2.0 fps.

Losses depend on bar shape (round, rectangular or aerofoil) and spacing

Bar spacing/opening not to be larger than minimum openings in valves/gates, turbines etc tostop oversize debris.

Arrangements for cleaning/racking need to be provided.

Rack resonance can be a problem

Fritz (1984) gave trash rack losses as: g V Sinbt ht 2/)/(3.2 2θ = where t = bar thickness, b =

bar spacing, θ=angle of rack inclination from horizontal, and v = aperture velocity. [Fritz, J.J. 1984. Small and Mini Hydropower System. CEWRE Lib Ac# 2048]

h l f o r e n

t r a n c e a n

d

t r a s h r a c k ,

f i s h s c r e e n

h l f o r e x p a n s i o n

h l f o r c o n

t r a c t i o n

h l f o r g u a r d g a t e

h l f o r v a l v e

h l f o r

b e n

d

Totalheadloss

Reservoir water level

Energy line

Figure 7.18 : Pictorial representation of head loss in outlet conduits flowing full under pressure.

Horizontal bend




10.15.2 Entrance Losses

Entrance losses may be considered as equivalent to losses in a short tube. If Q = C a√2gH, then V = C √2gH and

H = 1/C 2 V2/2g H = sum of velocity head h v and entrance loss h

A weighted loss coeff K

e

e = (1/C 2 – 1) and H v + h e = K e v2

Circular bell mouth entrance K /2g

e

Square or rectangular = 0.10 – 0.16

= 0.05 (average)

Inward projecting entrance = 0.80

Gate in thin wall (unsuppressed contraction) = 1.50

Gate in thin wall (bottom sides suppressed) = 1.0

Gate in thin wall (corner rounded) = 0.50

10.15.3 Bend losses

• Bend losses are function of bend radius (R b

• Generally H), pipe dia (D), bend angle.

l = K b V 2/2g K bR

= K × Bend angle factor

b 1/D 2 3 4 8 10 12 16

K 1 0.4 0.18 0.14 0.11 0.10 0.08 0.075

Bend angle (°) 20 40 60 80 90 100 120

Factor 0.36 0.64 0.83 0.95 1.00 1.04 1.13

10.15.4 Transition Losses (contraction / expansion)• Depend on area ratio and angle/length of transition

h c = K c(V 22/2g – V 1

2

h

/2g),

ex = K ex (V 12/2g – V 2

2

K

/2g)

c

Flare angle 2

= 0.1 gradual contraction, = 0.5 abrupt contraction, = 0.2 gradual expansion, = 0.5 abruptexpansion

o 5o 10 o 12 o 15 o 20 o 25 o 30 o 60

K

o

ex

Flare angle = angle of side wall with center line

0.03 0.04 0.08 0.10 0.16 0.31 0.40 0.49 0.72

10.15.5 Gate and Valve Losses

• Gate at entrance-wide open-no losses

• Upstream gate in thin wall-losses included in entrance

• Gate in conduit with u/s & d/s walls continuous-loss due to slot = K g V 2/2g, K g

• Partial open gate – depend on top contraction - K

≈ 0.1

g

• Wide open gate valve - K

approaching 1.0

g

• Partial open gate valve K

0.19

g

• Butterfly wide open Kg = 0.15 (vary 0.1 – 0.5)

1.15 (3/4 open); =5.6 (1/2 open) and 24.0 (1/4 open)

• Spherical valve Kg ≈ 0.0

Flare angle




Gate Discharge Formula: Q = C A √(2gH)

Head loss

H l ∼ (1/C d 2 – 1) H v H v

Slide gate C

= √2gH

d

Ring follower ∼ 1.0

= 0.95 – 0.97

Jet flow gate 0.80 – 0.84

Cylinder gate 0.80 – 0.90

Needle valve 0.45 – 0.60

Tube valve 0.50 – 0.55

Howell-Bunger fixed cone valve 0.85

Hollow jet valve 0.70

Sleeve valve (submerged) 0.85

Butterfly valves 0.60 – 0.80Sphere valve ∼ 1.0

10.15.6 Exit Losses

K v = 1.0 h l = K v V 2

(no recovery of head from free discharging pipe)

/2g

• For diverging tube, partial recovery, particularly if outlet is submerged example –draft tube K T = (a 1

2/a 2)2, a 1to increased area a

2




Figure 7.35: Abrupt transition losses.




Figure 7.36: Conical transition losses.




Figure 7.37: Moody diagram for pipe flow resistance.

Figure 7.38: Tunnel parameters. Figure 7.39: Horseshoe tunnel.




10.16 OUTLET ENERGY DISSIPATION

I mpact Basin http://www.dnr.state.oh.us/water/pubs/fs_div/fctsht51.htm

A concrete impact basin is an energy dissipating device located at the outlet of the spillway in

which flow from the discharge conduit strikes a vertical hanging baffle. Discharge is directedupstream in vertical eddies by the horizontal portion of the baffle and by the floor beforeflowing over the endsill. Energy dissipation occurs as the discharge strikes the baffle, thus,

performance is not dependent on tailwater. ODNR (1999)

Figure 7.40: Impact Basin

Baff led Chute

Baffled chutes require no initial tailwater to be effective and are located downstream of thecontrol section. Multiple rows of baffle piers on the chute prevent excessive acceleration ofthe flow and prevent the damage that occurs from a high discharge velocity. A portion of the

baffled chute usually extends below the streambed elevation to prevent undermining of thechute. If any of the severe problems associated with concrete that are referenced in theopening paragraphs are observed, a registered professional engineer should be contacted toevaluate the stability of the outlet.

Figure 7.41: Baffled Chute Basin




Plu nge Pool

A plunge pool is an energy dissipating device located at the outlet of a spillway. Energy isdissipated as the discharge flows into the plunge pool. Plunge pools are commonly lined withrock riprap or other material to prevent excessive erosion of the pool area. Discharge from the

plunge pool should be at the natural streambed elevation. Typical problems may includemovement of the riprap, loss of fines from the bedding material and scour beyond the riprapand lining. If scour beneath the outlet conduit develops, the conduit will be left unsupportedand separation of the conduit joints and undermining may occur. Separation of the conduit

joints and undermining may lead to failure of the spillway and ultimately the dam. Aregistered professional engineer should be contacted to ensure that the plunge pool isdesigned properly.

Figure 7.42: Plung Pool




10.17 STILLING BASIN ENERGY DISSIPATION FOR TUNNEL OUTLETS

10.17.1 General:

The circular tunnel portal of diameter D with exit velocity v is transitioned into rectangularchannel chute by providing fillet over a transition length L F = 1.5 D; this fillet transition

length continues with the slope of the tunnel. The chute width is same over a distance L T

gD

andthen flared out at a radius R = 5D. Subsequently the diverging and sloping transition chute ofhorizontal distance X and vertical drop Y. The transition flare is taken as 1:ΔL where ΔL ~

2F (F is Froude number of flow at tunnel exit portal = v/ ) with minimum of 1:6. This is

done to increase the basin width W b and to reach the basin invert level. The L T is given as:LT

θ θ 22

2

cos)25.1(2tan

v x g

x y −−=

= R tan φ/2 where φ=tan -1(1/ ΔL). The profile of the sidewall transition is set by a parabolic form given as:

where θ is angle of the slope of the tunnel as θ=tan -1(S) S is

tunnel slope (ft/ft).

Figure 7.45: Hydraulic jump stilling basin (USBR Type-II and III) for tunnel outlets.

The stilling basin is located at the end of parabolic transition with the basin width equal to thetransition width at X. The width W b is given as: W b = D + 2 (X+L F -L T

The flow velocity (v

)/ ΔL and depe nds onselected value of Y and X.

1) and depth (d 1) at the entrance of stilling basin depends on the energyE 1 at this point. But the energy E1 depends on the flow energy at the tunnel exit portal E 0 andthe elevation difference between the tunnel exit portal invert and the stilling basin invert.




Figure 7.46: Stilling basin and exit channel: Plan and profile.

For full flowing tunnels the pressure head at tunnel exit portal is given as: P 0 = y p * D wherey p

E

is the exit portal pressure coefficient (Fig ). The tunnel exit portal total energy is as:

0 = P 0 + v 2/2g = y pD+v 2

E

/2g.

1 = E 0

But E

+(Elev tunnel invert – Elev basin invert) – head losses in the transitions.

1 = d 1 + v 12/2g. However Q = W b * d 1 * v 1 , and thus d 1 = Q/(W b * v 1

E

), therefore:

1 = Q/(W b * v 1) + v 12/2g and few iteration swill provide value of v 1 . Now determine d 1 and

F1

−+= 181

22

11

2 F d

d

of the flow entering the basin. Now determine value of post jump conjugate depth d2 as:

The tail water level should be equal to d 2 (minimum of 0.85 d 2). Thus required tail waterlevel = selected basin invert level + 0.85 d 2

. If this agrees with the actual tail water leveldetermined from the rating curve of the exit channel, the design is accepted. Else completeanother trial with revised basin invert level.

Example: (Source: USCOE 1980 Appendix-F, p:F-1)

Q = 12,320 cfs, D = 14 ft, conduit slope = 0.01 ft/ft, Exit portal invert elevation = 100 ft amsl,Exit channel invert = 90 ft amsl, TW rating curve (Q, El):

Q 0 500 1000 1500 4000 8000 12320

El 90 91.5 92.5 93.2 95.7 98.2 100.2

Solution:

1. Conduit invert slope = tan -1

2. Conduit area A

(0.01/1) = 0.573°

0 = π14 2

3. v

/2 = 154 sft

0 = Q/A = 12320/154 = 80 ft/s




4. Exit portal pressure head = y p

5. F

*D = 0.57 * 14 = 8 ft above invert.

0

6. Flare ratio for parabolic transition ΔL = Max (2F

= 80/(32.2*14)0.5 = 3.77

0 , 6) = 2*3.77 = 7.54, and transitionangle from flow centerline φ = tan -1

7. Length of fillet transition = L

(1/7.54) = 7.56°

f

8. Radius for fillet transition = R = 5*14 = 70 ft= 1.5 * 14 = 21 ft.

9. Tangent length L t

10. Elevation drop in fillet transition = 0.01 * 21 = 0.21 ft

= R tan (φ/2) = 70 * tan 7.56/2 = 4.61 ft

11. Invert elevation at end of fillet transition = 100 – 0.21 = 99.79 ft amsl

12. Equation for parabolic transition: y = - x * tan 0.573 –x2[32.2/2*(1.25*8) 2*cos 20.573] = - 0.01 x – 0.00161 x 2 and is shown in figure belowfor interpolation. The basin invert is placed Y below and X away from the end of fillettransition (both X and Y to be determined as part of design computations). The Y =invert at end of fillet section – selected basin invert and corresponding X isdetermined from the above parabolic equation (or graph below)

Figure 7.47: Transition bed profile.

13. Basin width W b = D + 2 * [(X + L f – L t

14. Flow depth at entrance into the basin = d

)/ΔL] = 14 + 2 *[(X + 21 – 4.61)/7.56} =14+(X+16.39)/3.77

1 = Q/(W b*v 1) = 12320/( W b*v 1

15. E)

0 = 100 + 8 + 80 2

16. E/2*32.2 = 207. 37 ft.

1 = basin selected invert + d 1 + v 12/2g = 12320/( W b *v 1) + v 1

2/2g = E 0

17. 12320/( W

= 207.37(ignoring any losses in fillet and parabolic transitions and thus:

b *v 1) + v 12/2g = 207.37 - basin selected invert, (an iterative procedure is

required to determine v 1

Tri al 1:

)

18. Let basin inert = 80 ft amsl19. Then Y = 99.79 – 80 = 19.79 ft and X = 107.84 ft (from graph)




20. W b

21. E = 207.37 – 80 = 127.37 = 12320/( W

= 14+2*(107.84+16.39)/3.77 = 46.94 ft

b *v 1) + v 12

(i) let v

/2g, Iterative solution is as:

1 = 50, RHS = 12320/(46.94*50)+50 2

(ii) let v

/64.4 = 44 (not ok)

1 = 60, RHS = 12320/(46.94*60)+60 2

(iii) let v

/64.4 = 60.3 (not ok)

1 = 80, RHS = 12320/(46.94*80)+80 2

(iv) let v/64.4 = 102.66 (not ok)

1 = 90, RHS = 12320/(46.94*90)+90 2

(v) let v

/64.4 = 128.7 (not ok)

1 = 89.5, RHS = 12320/(46.94*89.5)+89.5 2

22. Thus v

/64.4 = 127.31 ( ok ),

1 = 89.5 fps and d 1

23. F = 12320/(46.94*89.5) = 2.93 ft

1 = 89.5/(32.2*2.93) 0.5

24. d

= 9.22

2 = 2.93/2*[(1+8*9.22 2)0.5

25. 0.85 d2 = 0.85 * 36.76 = 31.25 ft

– 1]= 36.76 ft

26. Required tail water level = 80+31.25 = 111.25 ft (which is higher than actual TWL of

100.2 ft for Q=12320 cfs), thus trial basin level is not ok.Tri al 2:

27. Let basin inert = 65 ft amsl

28. Then Y = 99.79 – 65 = 34.79 ft and X = 143.98 ft (from graph)

29. W b

30. E = 207.37 – 65 = 142.37 = 12320/( W

= 14+2*(143.98+16.39)/3.77 = 56.54 ft

b *v 1) + v 12

(i) let v


1 = 90, RHS = 12320/(56.54*90)+90 2

(ii) let v

/64.4 = 128.2 (not ok)

1 = 95, RHS = 12320/(56.54*95)+95 2

(iii) let v

/64.4 = 142.43 (not ok)

1 = 94.9, RHS = 12320/(56.54*94.9)+94.9 2

(i) let v

/64.4 = 142.14 (not ok)

1 = 94.95, RHS = 12320/(56.54*94.95)+94.95 2

31. Thus v

/64.4 = 142.29 ( ok )

1 = 94.96 fps and d 1

32. F

= 12320/(56.54*94.96) = 2.295 ft

1 = 94.96/(32.2*2.295) 0.5

33. d

= 11.05

2 = 2.295/2*[(1+8*11.05 2)0.5

34. 0.85 d2 = 0.85 * 34.73 = 29.52 ft

– 1]= 34.73 ft

35. Required tail water level = 65+29.52 = 94.52 ft (which is lower than actual TWL of100.2 ft for Q=12320 cfs), thus trial basin level is not ok.

Tri al 3:

36. Let basin inert = 70 ft amsl

37. Then Y = 99.79 – 70 = 29.79 ft and X = 133.0 ft (from graph)

38. W b

39. E = 207.37 – 70 = 137.37 = 12320/( W

= 14+2*(133.0+16.39)/3.77 = 53.63 ft

b *v 1) + v 12

(i) let v


1 = 90, RHS = 12320/(53.63*90)+90 2

(ii) let v

/64.4 = 128.33 (not ok)

1 = 91, RHS = 12320/(53.63*91)+91 2

(iii) let v

/64.4 = 131.11 (not ok)

1 = 93, RHS = 12320/(53.63*93)+93 2

(i) let v

/64.4 = 136.77 (not ok)

1 = 93.25, RHS = 12320/(53.63*93.25)+93.25 2

40. Thus v

/64.4 = 137.48 ( ok )

1 = 93.25 fps and d 1 = 12320/(53.36*93.25) = 2.464 ft




41. F1 = 93.25/(32.2*2.464) 0.5

42. d

= 10.47

2 = 2.464/2*[(1+8*10.47 2)0.5

43. 0.85 d2 = 0.85 * 35.27 = 29.98 ft

– 1]= 35.27 ft

44. Required tail water level = 70+29.98 = 99.98 ft (which is same as actual TWL of

100.2 ft for Q=12320 cfs), thus trial basin level is ok.Check:

45. Check the selected basin invert level of 70 ft for other discharges.

Basin fi xtur es:

46. As TWL is set for 0.85 of d2, baffle blocks (two rows are added)

47. Baffle block height h = Min (d 1 , d 248. Baffle block width w = h = 2.5 ft

/6) = Min (2.464,35.27/6=5.48) = 2.5 ft

49. Location of 1 st row >= 1.5 d 2

50. Location of 2

= 1.5 * 35.27 = 52.9 ft; select at 60 ftnd

row at d 2

51. End sill height = h/2 = 2.5/2 = 1.25 ft with 1:1 u/s slope./2 = 35.27/2 = 17.6 ft from first row; select 18 ft.

Exi t channel r ipr ap and grades:

52. Velocity at end of sill: v = Q/(actual TWL – top of end sill) = 12320/(100.2 – 71.25) =8 f/s

53. Select riprap size for v = 8 f/s as W 50 = 45 lb and D 50 = 0.8 ft ~ 10 inches; use D 50

54. Designed scour hole length = 0.5 d

= 1ft.

2 = 35.27/2 = 17.7 ft, and invert dip = 0.15 d 2

55. The width of riprap and exit channel is increased by 0.3 d2 = 0.3*35.27 = 10.6 ft andtotal width = 53.6 + 10.6 = 64.2 ft.

=0.15 * 35.27 = 5.3 ft at 70 – 5.3 = 64.7 ft.

56. The exit channel invert is raised at 1:10 grade from 64.7 ft amsl at end of riprap to 90ft at start of outlet channel.

F inal design:

57. The selected design for the stilling basin and exit channel are shown below.




Figure 7.48: Selected transition plan and profile.

Figure 7.49: Basin design.




http://www.popsci.com/popsci/science/5917359b9fa84010vgnvcm1000004eecbccdrcrd.htmlWater Vapor Almost Busts Dam

By Brian Fortner | March 2003

A strange phenomenon was shredding Glen Canyon Dam. Here's how it was saved.

THE SITUATION Late spring, 1983. Heavy snowmelt and steady rainscreate the worst flooding in nearly a century in the Colorado River basin.Lake Powell, a 185-mile-long reservoir on the Utah-Arizona border, is thehardest hit. Both spillways at the reservoir's 710-foot-high Glen CanyonDam must be opened for the first time to prevent the reservoir frombreaching its top.On June 6, rumbling sounds begin emanating from the left spillway. It'sthe calling card of cavitation, a little-understood phenomenon involvingthe formation of vapor cavities in high-velocity water columns. Thesecavities are short-lived, imploding with enough force to scour concretefrom the spillways. If they eat into the dam abutments, the structurecould give way, unleashing 17 billion cubic meters of water on thecanyon below.Bureau of Reclamation engineers temporarily shut down the left spillwayto assess the damage. Philip Burgi, then a hydraulic engineer with thefederal agency, notes "five holes in a Christmas tree pattern," startingsmall at the top and getting larger as they go down. The only positive: They are ripping into themountain below rather than toward the dam abutments on the side.THE RESPONSE Operators install 4-foot-high wooden flatboards on top of the dam to give LakePowell additional capacity. These are replaced by 8-foot-high metal flashboards as the reservoircontinues to rise. By mid-June, however, concrete chunks are blasting 60 to 80 feet in the air out ofthe bottom of the left spillway, along with sandstone-colored water. Operators reduce flow through theleft spillway and crank it up on the right. In July, the water level peaks just inches from the top of theflashboards.LESSONS LEARNED When engineers finally enter the left spillway to begin repairs, they find a crater32 feet deep and 180 feet long at its elbow, and the holes Burgi discovered in June are now cavities10 feet deep and 20 feet long. What's more, nearly 300 cubic yards of concrete, reinforcing steel andsandstone have been deposited in the deflector bucket at the base of the spillway. The right spillwaysuffers similar, if less severe, damage. "The spring runoff would come again in 1984," says Burgi. "Wehad to get this thing up and operational, and we only had one year to do it."Contractors blast away damaged concrete, fix tunnel linings and fill holes with 3,000 cubic yards ofconcrete. Engineers, meanwhile, begin their own race to retrofit the dam with aeration slots, a newtechnology that introduces small amounts of air into rushing water, cushioning the blow of implodingvapor cavities. The plan works. The '84 runoff sets more records, but the spillways show no sign ofcavitation.This success leads the Bureau of Reclamation to retrofit aerators to two other large dams, Hooverand Blue Mesa. "It was a defining moment in dam design," says Burgi. "The world was watching howwe were going to solve this problem." As it turns out, the world did more than watch -- aeration slotsare now standard from the Tarbela Dam in Pakistan to the Infiernillo Dam in Mexico.

HOW AERATION WORKSRushing water skips over an air supply slot, creating a low pressure zone under the water's bottomsurface. This draws air into the water column from the upper portion of the slot, softening the impactof cavitation




Tunnel Boring Machine (TBM)

TMB include cutter, conveyors, shields, rails and jacks.

Figure: A TBM. (http://www.robbinstbm.com/_img/products/cd_mb_906x365.jpg)

The equipment (http://www.soundtransit.org/documents/pdf/projects/link/central/FACT_TBM-4.pdf) The Tunnel Boring Machine at its full size is approximately the length of a football fi eld. Themachine includes “trailing gear” such as supply tanks, electrical support, exhaust fans and aconveyor belt. • The weight of the TBM and trailing gear is approximately 642 tons. • A 21-foot in diameter cutterhead (the blue part) is positioned at the front of the machine. Equipped




with various cutting tools, the cutterhead turns around at the rate of 0.1 to 2.5 revolutions perminute. Excavated material goes through openings in the face of the machine.• The spoils are brought into the machine by a corkscrew-like screw conveyor located behindthe cutter head. The spoils are then taken out the back on a conveyor system to betemporarily stored on site and then loaded into dump trucks.• Foam is usually added to condition the soil cuttings into a paste in order for it to passthrough the conveyor for removal. Water, bentonite or polymers may also be used,depending on the soil type, groundwater and other factors. The conditioning agents are bio-degradable.• The machine is propelled and “steered” with 16 hydraulic jacks that are located around theperimeter of the machine. The operator steers the machine using sophisticated positioningtechnology that is accurate to within an inch.• The machine was manufactured by Mitsubishi Heavy Industries in Kobe, Japan and arrivedin Seattle by ship. It took 25 truck loads to deliver the boring machine in pieces to theconstruction site.Lining the tunnel with concrete• Tunnel liner segments made out of pre-cast concrete are brought into the tunnel. The

machine positions each segment into place using an “arm” or erector, creating a ring. Thencement grout is placed behind the ring, forming the tunnel’s permanent liner. The segmentsare manufactured in Tacoma by Technopref Industries/Concrete Technology Corporation. 10inches thick, the segments include rubber gaskets and are water-tight. • The TBM then usesthe surface of the liner ring to propel itself forward, by pushing against it.




REFERENCES

Davis, . HANDBOOK OF APPLIED HYDRAULICS.Fortner, Brian. 2003. Water Vapor Almost Busts Dam. A strange phenomenon was shredding

Glen Canyon Dam. Here's how it was saved . @ www.popsci.com/popsci/science/

5917359b9fa84010vgnvcm1000004eecbccdrcrd.html Novak et. Al. ----. HYDRAULIC STRUCTURES.

ODNR 1999. Dam Safety: Outlet Erosion Control Structures (Stilling Basins) by Ohio Dept.of Natural Resources Fact sheet # 99-51. @http://www.dnr.state.oh.us/water/pubs/fs_div/fctsht51.htm .

USACE. 1980 Hydraulic Design of Reservoir Outlet Works: Engineering and DesignReport., Engineer Manual. EM 1110-2-1602 (15 Oct 1980) US Army Corps ofEngineers.

http://www.popsci.com/popsci/science/



http://www.dnr.state.oh.us/water/pubs/fs_div/fctsht51.htm







Figure 7.61 : Inside of the Simly dam diversion tunnel (now abandoned and plugged).

Figure 7.62 . Schematic diagram of spillways and controlled-outlet facilities at Trinity Dam.(http://www.usbr.gov/pmts/hydraulics_lab/pubs/PAP/PAP-0830.pdf)




Figure 7.63: Typical examples of short penstocks in a rockfill dam (Atatürk dam, Turkey,http://www.waterpowermagazine.com

Figure 7.64: Shasta dam penstock (http://www.flickr.com/photos/gord99/1026444760/ )

http://rds.yahoo.com/_ylt=A9G_bI8EV1ZInU0AFHWjzbkF/SIG=12hjqg7k2/EXP=1213704324/**http%3A/www.waterpowermagazine.com/storyprint.asp%3Fsc=2031304


http://www.flickr.com/photos/gord99/1026444760/







Tariq. 2008. Dam and Reservoir Engineering 8-1Ch-8: Dam Safety and Instrumentation

Chapter 8

DAM SAFETY AND INSTRUMENTATION

8.1 GENERALDams are usually very large and important structures and utmost care is needed to

ensure safety of the structure during construction and subsequent operation. Safety failureswill result in colossal loss of life and property in the event of dam breach due to flooding andinundation of d/s areas, expenses required to bring the structure back to the operationalcondition (requiring large repairs or complete reconstruction), losses on account of non-accruing of benefits during the period of dam rebuilding but continued loan repayments, etc.

8.2 HAZARDS, RISK, FAILURES

Present national loss statistics from dam failure fully justify the need for dam ownersto better understand the public risks involved with dam ownership, the kinds of hazards that

promote these risks and the reasons why dams fail. Public risk is high because people have been allowed to settle below dams in potential inundation zones and because new dams are being built in less than ideal sites.

Other elements of risk include natural phenomena such as floods, earthquakes andlandslides. These hazards threaten dam structures and their surroundings. Floods that exceedthe capacity of a dam's spillway and then erode the dam or abutments are particularlyhazardous, as is seismic activity that may cause cracking or seepage. Similarly, debris fromlandslides may block a dam's spillway and cause an overflow wave that erodes the abutmentsand ultimately weakens the structure.

The International Commission of Large Dams (ICOLD) has determined that the threemajor categories of dam failure are 1) overtopping by flood, 2) foundation defects, and 3)

piping. For earthen dams, the major reason for failure is piping or seepage. For concretedams, the major reasons for failure are associated with foundations. Overtopping has been asignificant cause of dam failure primarily in cases where there was an inadequate spillway.

8.3 DAM SAFETY PROGRAM

The safety of dams is encompassed in all three stages of planning and design,construction and operation. The damage to dam usually stem from inadequatecharacterization and handling of design flood, foundation failure, structural failure,geotechnical failure etc. Adequate evaluation is required for following during various phases.

1. Plann ing and Design stage:

1. Layout planning of various components

2. Determination of maximum reservoir level during floods and assurance of reservoirrim closure.

3. Physical model studies to establish adequacy of spillway and outlet works.

4. Determination of foundation conditions through elaborate geotechnical field andlaboratory investigations.




5. Seismo-techtonic studies and selection of seismic design parameters

6. Slope stability analysis to ensure safety under all loading conditions.

7. Seepage analysis of dam and foundation to select appropriate seepage controlmeasures.

8. Design of instrumentation system to monitor in-situ conditions during constructionand operation of the project.

2. Constructi on stage:

1. Review of actual foundation conditions with information obtained during constructionexcavation and modifying design as per new site conditions.

2. Ensuring required construction code and material properties.

3. Monitoring of safety instruments and analysis of data.

4. Selection of appropriate design modifications based on results of review of actual

foundation conditions and instrumentation data.3. Operation stage:

1. Close monitoring during first reservoir filling and operation of various dam structures.

2. Continuous monitoring of instrumental data and installation of additional instruments,if needed.

3. Detailed periodic safety inspections.

4. Safety inspections and analysis of instrumental data after events of special and extra-ordinary nature (severe earthquake, exceptionally high floods and water levels,operation of certain dam features e.g. spillway subsequent to its maximum and near-to-design loadings.

8.4 SAFETY INSPECTION

Dam safety is ascertained by conducting site inspections and evaluating availableinformations.

Inspection Timings

Safety inspections are carried out at various intervals as:

1. Casual/routine inspections: The observations are recorded for all unusual observationson every / any time dam structure is visited by the inspectors/dam operators.

2. Periodic / annual inspection. The dam is visited by the dam operators and/or dammonitoring organization once a year to record all features and perform criticalanalysis of instrumentation data gathered during the last year.

3. Periodic 5-yearly inspection. The dam is visited by a panel of experts engagedspecifically for this purpose out of professional/academic community having thoroughunderstanding and experience of their areas of expertise. The inspection is carried at aminute detail level where operational record and instrumentation recode is analyzedfor acceptance of dam operational procedures adopted, dam stresses, distresses, damunusual response to usual as well unusual loading conditions, etc. The surfacecondition of all concrete structures is evaluated for marks of cavitation pitting orerosion; particularly for the spillway chute stilling basin, baffle blocks etc.




4. Special inspections. The dam is inspected and its performance evaluated at its firstfilling and later after every condition of unusual loadings e.g. major earthquake, extra-ordinary flood flows, high water levels, first operation of various structures asspillways, stilling basins, outlets.

Each inspection ends with the preparation of performance and safety report. The report isdiscussed with all stake holders and any discrepancies / shortfall / omissions completed.

Inspection Parameters

Safety inspections carried at any interval include:

I. Physical inspections of A: Embankments for cracks, depressions, holes, movementssettlements, deflection, slides, free boards, (i) condition of crest, (ii) condition of slopes, (iii)condition of d/s toe in terms of boils, excessive seepage, piping, pore pressure, water oozing,etc., B . Concrete structures for (i) cracking, (ii) cavitation marks. C . Reservoir rim for signsof slips, cracks, slope failures, seepage rates D . External and internal deformations, u/s

blanket for sink holes, slope sloughing, slope condition.II. Analysis of monitoring data for piezometers, seepage data for all relief wells, seepagegallery, etc in terms of seepage quantity and condition (water is clear, muddy, etc) andmovement data of horizontal and vertical movements and closure data for tunnels and u/sworks, safe evacuation of floods,

III. Special problems

8.5 DAM INSTRUMENTATION

specific on project i.e. exceptionally high floods, major earthquakes etc.

"Instrumentation of a dam furnishes data to determine if the completed structure isfunctioning as intended and to provide a continuing surveillance of the structure to warn of

any developments which endanger its safety" (ICOLD, 1969).The means and methods available to monitor phenomena that can lead to dam failure

include a wide spectrum of instruments and procedures ranging from very simple to verycomplex. Any program of dam safety instrumentation must be properly designed andconsistent with other project components, must be based on prevailing geotechnicalconditions at the dam, and must include consideration of the hydrologic and hydraulic factors

present both before and after the project is in operation.

Instruments designed for monitoring potential deficiencies at existing dams must takeinto account the threat to life and property that the dam presents. Thus, the extent and nature

of the instrumentation depends not only on the complexity of the dam and the size of thereservoir, but also on the potential for loss of life and property downstream of the dam.

An instrumentation program should involve instruments and evaluation methods thatare as simple and straightforward as the project will allow. Beyond that, the dam ownershould make a definite commitment to an ongoing monitoring program or the installation ofinstruments probably will be wasted . This chapter discusses deficiencies in dams that may bediscovered and the types of instruments that may be used to monitor those deficiencies. Table8.1 describes deficiencies, their causes and generic means for detecting them. Increasedknowledge of these deficiencies acquired through a monitoring program is useful indetermining both the cause of the deficiency, the necessary remedy and adequacy of thr

remeady. Involvement of qualified personnel in the design, installation, monitoring, andevaluation of an instrumentation system is of prime importance to the success of the program.




Table 8.2 Minimum recommended instrumentation for proposed dams Type OfMeasurement

Low hazardPotential -

All DamTypes

Significant or High-Hazard Potential DamsEmbankm

ntConcreteGravity

Arch Buttress SeparateSpillway

Outlet

IntegralPowerHouse

Visual Observation x X X X X X XReservoir Level X X X X X XTail water Level X X X X X XDrain Flow,Seepage, Leakage

X X X X X X

Pore/Uplift Pressure X X X XSurface Settlement XSurface Alignment X X X X X XInternal Movement X X XJoint/CrackDisplacement

X X X X X

FoundationMovement

X X X X X X

Temperature X X XSeismic Loads x X X X X XLoads In PostTensioned Anchors

X X X x X

• Visual observation consists of walking tours of the crest, toes, abutments, etc.• For concrete dams greater than about 100 feet high.• Only on structurally significant joints or cracks that have visible displacement.• Should be considered for dams on compressible or weak foundations.• Should be considered on a case-by-case basis for dams in seismic zones.• Loads should be measured in anchors that are required to meet stability criteria

8.6 PHILOSOPHY OF INSTRUMENTATION AND MONITORING (FERC 2007) The purpose of instrumentation and monitoring is to maintain and improve dam safety

by providing information to 1) evaluate whether a dam is performing as expected and 2) warnof changes that could endanger the safety of a dam.

8.6.1 Dam Failures

The causes of dam failures and incidents have been catalogued (ASCE 1975 and1988, Jansen 1980, National Research Council 1983, ICOLD 1992).

The common causes of concrete dam failures and incidents are:• Overtopping from inadequate spillway capacity or spillway blockage resulting in

erosion of the foundation at the toe of the dam or washout of an abutment oradjacent embankment structure;

• Foundation leakage and piping in pervious strata, soluble lenses, and rockdiscontinuities; and

• Sliding along weak discontinuities in foundations.• Abutment yielding (arch dam)

The principal causes of embankment dam failures and incidents are:• Overtopping from inadequate spillway capacity, spillway blockage, or excessive

settlement resulting in erosion of the embankment;• Erosion of embankments from failure of spillways, failure or deformation of outlet

conduits causing leakage and piping, and failure of riprap;




• Embankment leakage and piping along outlet conduits, abutment interfaces,contacts with concrete structures, or concentrated piping in the embankment itself;

• Foundation leakage and piping in pervious strata, soluble lenses, and rockdiscontinuities;

• Sliding of embankment slopes due to overly steep slopes, seepage forces, rapiddrawdown, or rainfall;

• Sliding along clay seams in foundations;• Cracking due to differential settlements; and• Liquefaction of embankment and/or sediment deposition u/s of the embankment.

8.6.2 Reasons for Instrumentation

Instrumentation and proper monitoring and evaluation are extremely valuable indetermining the performance of a dam. Specific reasons for instrumentation include:

• Warning of a Problem - Often instruments can detect unusual changes, such as waterfluctuations in pressure that are not visible. In other cases, gradual progressive

changes in say seepage flow, which would go unnoticed visually, can be monitoredregularly. This monitoring can warn of the development of a serious seepage problem.

• Analyzing and Defining a Problem - Instrumentation data is frequently used to provide engineering information necessary for analyzing and defining the extent of a problem. For example, downstream movement of a dam because of high reservoirwater pressure must be analyzed to determine if the movement is uniformlydistributed along the dam, whether the movement is in the dam, the foundation, or

both, and whether the movement is continuing at a constant, increasing or decreasingrate. Such information can then be used to design corrective measures.

•

Proving Behavior Is as Expected - Instruments installed at a dam may infrequently (oreven never) show any anomaly or problem. However, even this information isvaluable because it shows that the dam is performing as designed and provides peaceof mind to an owner. Also, although a problem may appear to be happening orimminent, instrument readings might show that the deficiency (say increased seepage)is normal (merely a result of higher than normal reservoir level) and was foreseen inthe dam's design.

• Evaluating Remedial Action Performance - Many dams, particularly older dams, aremodified to allow for increased capacity or to correct a deficiency. Instrumentreadings before and after the change allows analysis and evaluation of the

performance of the modification.

8.6.3 Instrumentation Protocol

Instrumentation and monitoring, combined with vigilant visual observation, can provide early warning of many conditions that could contribute to dam failures and incidents.For example, settlement of an embankment crest may increase the likelihood of overtopping;increased seepage or turbidity could indicate piping; settlement of an embankment crest or

bulging of embankment slopes could indicate sliding or deformation; inelastic movement ofconcrete structures could indicate sliding or alkali-aggregate reaction. Conversely, lack ofnormally expected natural phenomena may also indicate potential problems. For example,

lack of seepage in a drainage system could indicate that seepage is occurring at a locationwhere it was not expected or contemplated by the designer.




Instrumentation and monitoring must be carefully planned and executed to meetdefined objectives. Every instrument in a dam should have a specific purpose. If it does nothave a specific purpose, it should not be installed or it should be abandoned. Instrumentationfor long-term monitoring should be (1) rugged, (2) easy to maintain, (3) should be capable of

being verified or calibrated, (4) easy to read out, (5) capable of automation and remotereadout.

The primary function of monitoring is to ensure the longevity and safety of the dam.Monitoring must ensure the timely detection of any behavior that could deteriorate the dam,

potentially result in its shutdown or failure, in order to implement corrective measures.Monitoring also plays a fundamental role during construction. It enables the verification ofdesign hypotheses and may affect the construction rate of certain works. Monitoring is

particularly crucial during the initial filling of the reservoir, a critical phase in the life of adam.

Instrumentation is used to accurately quantify the certain parameters of structural behavior over t ime and to monitor their rate of change. Thus instruments should be based onthree major selection criteria: (1) Reliability of the measurements obtained (accuracy,resolution, precision and drift). (2) Instrument longevity, supported by numerous references –as instruments, if fail to perform, can not be replaced; at Mangla as much as 50% of theinstruments for pore water pressure are down in 40 years of dam service. (3) Ease of readoutautomation, essential for efficient data collection and interpretation.

The scope of the monitoring methods employed depends on potential risk associatedwith the dam and site characteristics, including: i. Dam height and type, ii. Extent of potentialdamage to people and structures located in flood zone, iii. Reservoir and spillway capacity,

iv. Site seismicity, and v. Foundation weakness zones.8.6.4. Monitoring Parameters

Various parameters to be monitored for safety of any dam include following:

a) Physical condition – slopes, surfaces, reservoir rim, sand boils in the dam toe area,

b) Discharges – inflows, outflows, spillway flow

c) Seepage flow – pressure relief wells, toe drain, drainage gallery under spillway,abutments

d) Water level – Reservoir and tail water levels, wave amplitude and frequency

e) Movements – crest settlement, deflection, lateral movement; inclination of concrete damu/s face, dam interior sett lement/ deflection, abutment movement/ deflection

f) Pore water pressure – at various locations in the dam core, shell, foundations, uplift pressure under structures, etc.

g) Movement of joints, cracks

h) Point loads, stress, strains – in the concrete dam and various structures and

i) Temperature – in the concrete dam body and exteriors, in shafts, passage ways, reservoirwater temperatures

j) Water quality – depth wise quality of reservoir water, seepage water (for suspended anddissolved substances) and chemical contents




k) Meteorological / weather parameters – max-min temperatures, wind velocity/direction,rainfall, ice, relative humidity, pan evaporation rate, solar radiation

l) Siesmic activity – earthquake, ground acceleration

8.7 INSTRUMENT TYPES AND USAGE

A wide variety of devices and procedures are used to monitor dams. The features ofdams and dam sites most often monitored by instruments include: Visual observations;Movements: (horizontal, vertical, rotational and lateral); Pore pressure and uplift pressures;Water level and flow; Seepage flow; Water quality; Temperature; Crack and joint size;Seismic activity; Weather and precipitation; Stress and strain. Instruments for pore water anduplift pressure and movements form the core of the monitoring activity.

Various instrumentation used to measure these pore water and uplift pressurecomprise various types of piezometers (a) Standpipe piezometers (casagrande type) (b)

Electric piezometers (vibrating wire type) (c) Pneumatic piezometers, (d) Hydraulic piezometers (e) Pressure transducer The type of piezometer is selected as per need foraccessibility for readout, ease, accuracy etc. Instruments used for monitoring movement(horizontal, vertical, rotational and lateral); include (a) Inclinometers, (b) extensometers (c)Joint meters, (d) movement/settlement markers. Earthquake is monitored by using strongmotion accelerographs.

8.7.1 Piezometer:

Piezometer measure the water pressure at its screened tip. These are

embedded in a bore hole or driven-in torequisite level. Standpipe piezometersare used where accessibility is easy forits readout. Vibrating wire, pneumatic,electric piezometers can be arrangedfor a remote readout, e.g. in theinstrumentation monitoring room.

8.7.2 Extensometer

Extensometers are used formeasuring differential settlement in thedam body. Tape extensometers areused to measure deformation ofconduits.

8.7.3 Pendulum

Pendulum measure verticalinclination of selected points.

Figure 8.1a: A standpipe piezometer




Figure 8.1b: Schematic of observation well Figure 8.1c: Schematic of open standpipe piezometer installed in a borehole

Figure 8.1d: Electric probe for water level depth measurements in open standpipeobservation well and piezometers.

re contact en




Figure 8.1e: Schematic of vibrating wire pressure sensor.

Figure 8.1f: Vibrating wire electric piezometers.




Figure 8.1g. Schematic of pneumatic device (USACE 1995)

Figure 8.1h : Pressure transducer, cables and monitoring unit for measuring piezometric head.




Figure 8.1j: Readout unit for electrical piezometers.

Figure 8.1k: Readout units for piezometers at Simly dam.




Figure 8.2: Deep water segmental sample collection cylinder.

Figure 8.3: Inverted and hanging pendulums.




“http://www.slopeindicator.com/

instruments/ext-magnet.html”

Rod extensometers.

Magnetic extensometer Figure 8.4a: Extensometer.

http://www.slopeindicator.com/







Figure 8.4b: Extensometer. Cross-arm gage, pipe arrangement, and measurement probe.

Figure 8.5: Tape extensometer.




8.7.4 Inclinometers (Goins, 1995.)

Introduction –

An inclinometer is an instrument, which is used to measures ground movements. This

instrument measures ground movements in directions perpendicular to the axis of a drill holewhere a grooved casing has been installed. An inclinometer monitors horizontal movementsin near vertical drill holes and through the use of data reduction programs can provide acomplete and detailed profile of displacements along the drill hole. This detailed profile willgive localized ground movements wherever they occur. An inclinometer system formeasuring these detailed profiles consists of a probe fitted with guide wheels and containing

biaxial gravity-operated tilt sensor connected by an electrical cable to a power source andreadout unit.

For obtaining horizontal ground movements at only selected locations within a drillhole, a fixed-in-place inclinometer is used. A fixed-in-place inclinometer consists of a seriesof probes fitted with guide wheels and each containing a biaxial gravity-operated tilt sensor.The probes are joined by articulated rods and suspended down a guide casing within the drillhole. Electrical cables connect each sensor to the ground surface where they are attached to a

power supply and readout unit. The in-place inclinometer can provide ground movementinformation perpendicular to the drill hole axis at the depths selected where the probe is setwith the guide casing. The in-place inclinometer will provide continuous information aboutthat selected depth in the guide casing. These instruments may be automated or an alarm can

be added to the unit if required.

How the Inclinometer Works –

The inclinometer is an electronic instrument that measuresthe averaged horizontal movement of a vertical casing usuallyinstalled within the ground. The inclinometer consists of four

pieces of equipment, the grooved casing, the inclinometer probe,the inclinometer cable, and the inclinometer readout. Theinclinometer is read by an operator moving the probe within thecasing and recording the data. Two values of deflection arerecorded at right angles to each other. The inclinometer reads in

both positive and negative directions according to which direction. For maximum accuracy,two sets of readings are taken by rotating the borehole probe 180E so that the spring-loadedwheels travel in opposite grooves of the casing. This process will eliminate or minimizeerrors contributed by casing irregularities, depth measurements, and instrument calibrations.The inclinometer does not measure displacement directly. Instead, it measures the tilt of thecasing. The tilt is converted to a lateral distance using computer software.

8.7.5 Joint meters

Joint meters are used to monitor the relative movement of two sides of a joint. Thetwo ends of the joint meter are cast-in the structure. Any movement is remotely read-out.

8.7.6 Crack meters

Cracks can develop across various surfaces. These need to be monitored by Crackmeters whose two sides are attached to the opposing sides of the crack.




8.7.7 Survey reference markers

Survey reference markers are installed at dam crest in a straight alignment, and tied totwo permanent markers installed on the right and left abutments. Periodic survey foralignment control and elevation of the markers will provide information about crest

deflection and settlement.

Figure 8.6a: Inclinometer casing and probe.




Figure 8.6b: Typical inclinometer plots.(Source: http://www.rizzoassoc.com/NewsData/Listening-to-Dam-Final.pdf)

Figure 8.6c: Inclinometer detail at surface. Figure 8.6d: Inclinometer and casing.

Figure 8.6e: Plot of inclinometer readings.

http://www.rizzoassoc.com/NewsData/Listening-to-Dam-Final.pdf

http://www.rizzoassoc.com/NewsData/Listening-to-Dam-Final.pdf




Figure 8.7: Joint meter.




8.7.8 Borros Anchor Settlement PointThe settlement point consists of an anchor and twoconcentric riser pipes. The inner pipe, which is connected tothe anchor, can move freely within the outer pipe. A changein the distance between the top of the inner pipe and the topof the outer pipe indicates movement.

Figure 8.9: Installation of permanent reference/measurement point for monitoring settlement on surfaceof embankment dams.

Reservoir area

Dam axis

Instrument station(Located where it

will not move whendam moves)

Points marked by rebar

Target station

Line of sightestablished outside traffic area

Dam crest

Figure 8.10: Survey markers for surface settlement and surface alignment.

Figure 8.8: Anchorsettlement oint.




Figure 8.11: Monitoring cracks on embankments.




8.8 MEASURING INSTRUMENTS

8.8.1 Instruments

Various instruments used to make measurements include:

a) Discharges – Depth gage and rating curve,

b) Seepage flow – flow meters, weirs, flumes

c) Water level – Automatic water level monitor, staff gage (inclined or vertical)

d) Crest settlement – surface reference markers/monuments, Extensometers

e) Deflections: - Inclinometers, Pendulums, Survey monuments, Surface inclinometer

f) Pore water pressure – Piezometers

g) Uplift pressures: - Remote readable piezometers

h) Joint movement – Joint meters

i) Tunnel deflections – tape meters, Point studs j) Cracks – Crack meters, crack monitoring bar

k) Point loads, - Load cells, Strain gages

l) Temperature – Thermometers

m) Reservoir water quality – Differential depth water sampler

n) Meteorological / weather parameters – Climatic instruments

o) earthquake–Strong motion accelerograph

8.8.2 Visual observations:

The visual observations by the dam owner or the owner's representative may be themost important and effective means of monitoring the performance of a dam. The visualinspections should be made whenever the inspector visits the dam site and should consist of aminimum of walking along the dam alignment and looking for any signs of distress orunusual conditions at the dam.

8.8.3 Movements:

Movements occur in every dam. They are caused by stresses induced by reservoirwater pressure, unstable slopes (low shearing strength), low foundation shearing strength,settlement due to compressibility of foundation and dam materials, thrust due to arching

action, expansion resulting from temperature change, and heave resulting from hydrostaticuplift pressures. They can be categorized by direction:

a) Horizontal Movement – Horizontal or translational movement commonly happens in anupstream downstream direction in both embankment and concrete dams. It involves themovement of an entire dam mass relative to its abutments or foundation. Dam crest anddeeper layers can move. In an embankment dam, instruments commonly used for monitoringsuch movement include: Extensometers, Multi-point extensometers, Inclinometers,Embankment measuring points, Shear strips, and Structural measuring points. For a concretedam, instruments for monitoring horizontal movements may include: Crack measuringdevices, Extensometers, Multi-point extensometers, Inclinometers, Structural measuring

points, Tape gauges, Strain meters, Plumb lines, Foundation deformation gauges.




b) Vertical Movement : Vertical movement is commonly a result of consolidation ofembankment foundation materials resulting in settlement of the dam. Another cause is heave(particularly at the toe of a dam) caused by hydrostatic uplift pressures. In an embankmentdam, vertical movements may be monitored by Settlement plates/sensors, Extensometers,Vertical internal movement devices, Embankment measuring monuments. In a concrete dam,vertical movement monitoring devices may include Settlement sensors, Extensometers,Foundation deformation gauges

c) Rotational Movement : Rotational movement is commonly a result of high reservoir water pressure in combination with low shearing strength in an embankment or rotation and mayoccur in either component of a dam. This kind of movement may be measured in eitherembankment or concrete dams by instruments such as: Extensometers, Inclinometers,Tiltmeters, Surface measurement points Pendulums (concrete only)

d) Lateral Movement : Lateral movement (parallel with the crest of a dam) is common inconcrete arch and gravity dams. The structure of an arch dam causes reservoir water pressureto be translated into a horizontal thrust against each abutment. Gravity dams also exhibitsome lateral movement because of expansion and contraction due to temperature changes.These movements may be detected by: Structural measurement monuments, Tilt meters,Extensometers, Crack measurement devices, Pendulums, Inclinometers, Joint meters

8.8.4 Pore pressure and uplift pressure:

In spite of various seepage control measures, a certain amount of water seeps through,under, and around the ends of all dams. The water moves through pores in the soil, rock, orconcrete as well as through cracks, joints, etc. The pressure of the water as it moves actsuniformly in all planes and is termed pore pressure. The upward force (called uplift pressure)

has the effect of reducing the effective weight of the downstream portion of a dam and canmaterially reduce dam stability. Pore pressure in an embankment dam, a dam foundation orabutment, reduces that component's shearing strength. In addition, excess water, if noteffectively channeled by drains or filters, can result in progressive internal erosion (piping)and failure. Pore pressures can be monitored with Piezometers (electrical, open stand pipe,casagrande, pneumatic, hydraulic, porous tube, slotted pipe), Pressure meters & gauges andLoad cells

8.8.5 Water Level and Flow:

For most dams, it is important to monitor the water level in the reservoir and the

downstream pool regularly to determine the quantity of water in the reservoir and its levelrelative to the regular outlet works and the emergency spillway. The water level is also usedto compute water pressure and pore pressure; the volume of seepage is usually directlyrelated to the reservoir level. It is also important to establish the normal or typical flowthrough the outlet works for legal purposes.

Water levels may be measured by simple elevation gauges - either staff gauges or numbers painted on permanent, fixed structures in the reservoir - or by complex water level sensingdevices. Flow quantities are often computed from knowledge of the dimensions of the outletworks and the depth of flow in the outlet channel or pipe.




Table 8.2: Advantages and limitations of common water level and pressure instruments.

Type Advantages LimitationsStaff Gage Simple device, inexpensive, reliable. Cannot be automated. Float-TypeWater LevelGage

Simple device, inexpensive, reliable.Easily automated.

Requires readout device. Sensor must be inwater. Must be protected from ice

UltrasonicWater LevelSensor

Simple device, inexpensive, reliable.Sensor does not touch water. Easilyautomated

Requires readout device. Must be corrected forair temperature. Debris, foam, and ice can causefalse readings

Bubbler Simple device, inexpensive, reliable.Easily automated.

Requires readout device. Sensor must besubmerged inwater.

ObservationWell

Simple device, inexpensive. Easilyautomated.

Applicable only in uniform materials, notreliable for stratified materials. Long lag time inimpervious soils.

OpenStandpipePiezometer

Simple device, inexpensive, reliable.Simple to monitor and maintain.Standard against which all other

piezometers are measured. Can besubjected to rising or falling head tests toconfirm function. Easily automated.

Long lag time in impervious soils. Potentialfreezing problems if water near surface. Poroustips can clog due to repeated inflow and outflow.

Not appropriate for artesian conditions where phreatic surface extends significantly above topof pipe. Interferes with material placement andcompaction during construction. Can bedamaged by consolidation of soil aroundstandpipe.

ClosedStandpipePiezometer

Same as for open standpipe piezometers. Same as open standpipe piezometer butappropriate for artesian conditions.

Twin-tubeHydraulicPiezometer

Simple device, moderately expensive,reliable, long experience record. Shortlag time. Minimal interference withconstruction operations.

Cannot be installed in a borehole, therefore,generally not appropriate for retrofitting.Readout location must be protected fromfreezing. Moderately complex monitoring and

maintenance. Periodic de-airing required.Elevation of tubing and of readout must be lessthan 10 to 15 feet above piezometric elevation.Can be automated, but moderately complex

PneumaticPiezometer

Moderately simple transducer,moderately expensive, reliable, fairlylong experience record. Very short lagtime. Elevation of readout independentof elevation of tips and piezometriclevels. No freezing problems.

Moderately complex monitoring andmaintenance. Dry air and readout devicerequired. Can be automated, but not over longdistances. Sensitive to barometric pressure.Automation is complex. Moderately expensivereadout.

VibratingWirePiezometer

Moderately complex transducer. Simpleto monitor. Very short lag time.Elevation of readout independent of

elevation of tips and piezometric levels. No freezing problems. Frequency outputsignal permits transmission over longdistances. Easily automated

Lightning protection required. Expensivetransducer and readout. Sensitive to temperatureand barometric pressure changes. Risk of zero

drift, but some models available with in-situcalibration check

BondedResistanceStrain Gage(Electronic)Piezometer

Moderately complex device, expensive.Simple to monitor. Very short lag time.Elevation of readout independent ofelevation of tips and piezometric levels.

No freezing problems. Easily automated.

Lightning protection required. Subject to zero-drift, therefore, not recommended for long-termmonitoring. Expensive transducer and readout.Voltage or current output signal sensitive tocable length, splices, moisture, etc.




8.8.6 Seepage Flow:

Seepage must be monitored on a regular basis to determine if it is increasing,decreasing, or remaining constant as the reservoir level fluctuates. A flow rate changingrelative to a reservoir water level can be an indication of a clogged drain, piping, or internal

cracking of the embankment. Seepage may be measured using the following devices andmethods: Weirs (any shape such as V-notch, rectangular, trapezoidal etc.), Flumes (such as aParshall flume), trajectory methods (for free flowing pipes and conduits), Timed-bucketmethods (volumetric method) and Flow meters

8.8.7 Water Quality:

Seepage comes into contact with various minerals in the soil and rock in and aroundthe dam. This can cause two problems: the chemical dissolution of a natural rock such aslimestone, or the internal erosion of soil.

Dissolution of minerals can often be detected by comparing chemical analyses of

reservoir water and seepage water. Such tests are site specific; for example, in a limestonearea, one would look for calcium and carbonates, in a gypsum area, calcium and sulfates.Other tests, such as pH can also sometimes provide useful information on chemicaldissolution. Internal erosion can be detected by comparing turbidity of reservoir water withthat of seepage water. A large increase in turbidity indicates erosion.

8.8.9 Temperature:

The internal temperature of concrete dams is commonly measured both during andafter construction. During construction, the heat of hydration of freshly placed concrete cancreate high stresses which could result in later cracking. After construction is completed and

a dam is in operation, it is not uncommon for very significant temperature differentials toexist depending on the season of the year. For example, during the winter, the upstream faceof a dam remains relatively warm because of reservoir water temperature, while thedownstream face of the dam is reduced to a cold ambient air temperature. The reverse is truein the summer temperature measurements are important both to determine causes ofmovement due to expansion or contraction and to compute actual movement. Temperaturemeasurements can be made by using any of several different kinds of embeddedthermometers or by making simultaneous temperature readings on devices such as stress andstrain meters which provide means for indirectly measuring temperature of the mass.

8.8.10 Crack and Joint Size:

A knowledge of the locations and widths of cracks and joints in concrete dams and inconcrete spillways and other concrete appurtenances of embankment dams is important

because of the potential for seepage through those openings. Even more, it is important toknow if the width of such openings is increasing or decreasing. Various crack and jointmeasuring devices are available, and most allow very accurate measurement. Some usesimple tape or dial gauges, while others use complex electronics to gain measurements.

8.8.11 Seismic Activity:

Seismic measuring devices (strong motion accelerograph) record the intensity andduration of large-scale earth movements such as earthquakes. Almost all large dams use these

instruments for seismic recordings. It may or may not be necessary for a private dam to




contain any seismic devices depending upon whether it is in an area of significant seismicrisk. Seismic increments can also be used to monitor any blasting conducted near a dam site.

8.8.12 Weather and Precipitation:

Monitoring the weather at a dam site can provide valuable information about the day-

to-day performance and developing problems. A rain gauge, thermometer, and wind gauge,evaporation pan can be easily purchased, installed, maintained and monitored at a dam site.

8.8.13 Stress and Strain:

Measurements to determine stress and/or strain are common in concrete dams and to alesser extent, in embankment dams. The monitoring devices previously listed for measuringdam movements, crack and joint size and temperature are also appropriate for measuringstress and strain. Monitoring for stress and strain permits very early detection of movement.Various kinds of load cells are used for this purpose.

8.9 FREQUENCY OF MONITORINGThe frequency of instrument readings or making observations at a dam depends on severalfactors including:

• Relative hazard to life and property that the dam represents• Height or size of the dam• Relative quantity of water impounded by the dam• Relative seismic risk at the site• Age of the dam

• Frequency and amount of water level fluctuation in the reservoirIn general, as each of the above factors increases, the frequency of monitoring shouldincrease. For example, very frequent (even daily) readings should be taken during the firstfilling of a reservoir, and -more frequent readings should be taken during high water levelsand after significant storms and earthquakes. As a rule of thumb, simple visual observationsshould be made during each visit to the dam and not less than monthly. Daily or weeklyreadings should be made during the first filling, immediate readings should be takenfollowing a storm or earthquake, and significant seepage, movement, and stress-strainreadings 'should probably be made at least monthly.




Table 8.3: Typical monitoring schedule for significant and high-hazard potential dams Type ofMeasurement

Frequency Of Measurements Constructio First

FillingFirst Year AfterFilling

Second And ThirdYears

Long-TermOperation

Visual Observation Daily Daily Weekly Monthly Monthly Reservoir Level - Daily to

Weekly Semi-monthly

and at same timeas any other

measurements

Monthly and atsame time as any

othermeasurements

Monthly toquarterly and at

same time as anyother

measurements Tailwater Level - Weekly Semi-monthly

and at same timeas any other

measurements

Monthly and atsame time as any

othermeasurements

Monthly toquarterly and at

same time as anyother

measurements Drain Flow - Daily to

Weekly Weekly tomonthly

Monthly Monthly toquarterly

Seepage/ LeakageFlow

Monthly Daily toWeekly

Weekly tomonthly

Monthly Monthly toquarterly

Pore Pressure/Uplift

Daily toWeekly

Daily toweekly

Monthly Monthly Monthly toquarterly

Surface Settlement - Monthly Quarterly Semi-annually toannually

Semi-annually toannually

Surface Alignment - Daily tomonthly

Quarterly Semi-annually toannually

Semi-annually toannually

Internal Movement - Weekly toMonthly

Monthly toquarterly

Monthly to semi-annually

Monthly toannually

Joint/CrackDisplacement

- Weekly toMonthly

Monthly toquarterly

Monthly to semi-annually

Monthly toannually

FoundationMovement

Weekly Weekly toMonthly

Quarterly Semi-annually Semi-annually toannually

Temperature Hourly toweekly

Weekly Semi-monthly Monthly Typically notrequired

Loads InPosttensionedAnchors

Typicallynot

required

Typicallynot

required

Annually Typically notrequired

Quinquennially




Figure 8.12a: Instrumentation layout at Kurram Tangi Dam. (source Sohail, 2003)




Figure 8.12b: Kurram Tangi Dam: Instrumentation at Sections 1-1 (top) 2-2 (middle) and 3-3 (bottom).




Figure 8.13: Instrument layout for a Earth Core Rockfill Dam (ECRD) (USACE 1995, p:9-2).




8.10 INSTRUMENTATION AT LOS VAQUEROS DAM(Lawton et al. 2007)Description of DamLos Vaqueros Dam will be a zoned embankment dam with a crest length of about 1000 feet and amaximum height of 197 feet. The design includes a thick central clay core supported by sandstoneand claystone shells. The dam foundation consists of a dipping sedimentary rock foundationcomposed of sandstone, siltstone, and claystone. A grout curtain is placed beneath the core keytrench.The appurtenant works at the dam include the outlet works to be used when filling the reservoir andfor releases to the CCWD system, and a spillway. The outlet works consists of a 5-port sloping intakestructure located on the right abutment of the dam, a 1300-foot long outlet tunnel through the rightabutment intake/outlet tunnel, and an outlet works control building at the downstream toe of the damwith associated gates to divert water to the transfer pipeline to the CCWD’s system or divert water tothe Kellogg Creek. The spillway will consist of a concrete lined approach channel, concrete-ogeecrest, and a concrete-lined discharge channel, located on the left abutment.InstrumentationGeotechnical parameters to be monitored by the instrumentation system are:• Pore Pressures—Monitor the pore pressures within the dam to check design assumptions for

construction pore pressures, embankment consolidation, and post-construction seepage control.• Embankment Deformation—Monitor embankment deformations within embankment zones and at

slope surfaces to assess design strength assumptions.• Embankment/Foundation Seepage—Monitor for seepage following reservoir filling and after any

earthquakes.• Tunnel Performance—Monitor the ground control for tunnel and portals during construction and

monitor the tunnel during operation to observe for seepage.• Strong Motion Measurements—Measure and record the time history of any seismic motions

imposed on the dam site. Intensive monitoring of selected instruments will be started whenmotions are detected above a predetermined level.

• Structure Monitoring—Monitor survey measurement points on the spillway and intake structure todetect any post-construction movements.

• Meteorological Monitoring—Monitor a standard set of meteorological conditions (temperature,rainfall, wind speed/direction, relative humidity) near the Intake Control Building.

The plan of the instrumentation for the dam and a cross-section of the dam with instrumentation areshown in Figures and , respectively.

The instrumentation monitoring system designed for the Dam and reservoir is summarized inTable 1 . Dam survey monuments, intake and spillway survey markers, inclinometers and settlementsystems, tunnel convergence points, and open standpipe piezometers will be read manually. All of theremaining instrumentation will be monitored by an automatic data acquisition system

Of the instruments listed in the above table, the most important types for assessing the safetyof the dam are the piezometers and the seepage gauge.

(ADAS). Thestrong motion instruments will be self-contained, automatic recording, solar powered units. The storeddata will be read using a portable computer after each earthquake which produces motions above apre-determined threshold.

Table 1 – Instrumen tation Summ aryInstrument Type Number ADASDam Vibrating Wire Piezometers 33 XDam Open Standpipe Piezometers 3Tunnel Piezometers 4 XTunnel Convergence Points 5Tunnel Portal Inclinometers 2Dam Survey Monuments 31Structure Survey Markers 9SLEX Inclinometer/Settlement System 1Reservoir Level Piezometers 2 XDownstream Toe Seepage Gauge 1 XSingle Point Settlement Sensors 6Tube Profile Settlement Gauges 2Seismometers 3 X Seismic trigger connected to the ADAS).Meteorological Station 1 X




Figure 8.14a: Instrumentation design of Los Vaqueros Dam - Plan




Figure 8.14b: Instrumentation design of Los Vaqueros Dam (cross section)




8.12 Roctest Telemac Instruments (www.roctest.com )

http://www.roctest.com/







Types of Measurements




Earth and Rockfill Dams

Arch Dams

Concrete Gravity Dams




Symbols

Figure 8.15a: Symbols for instrumentation design.

Figure 8.15b: Instrumentation for concrete gravity dam.




Figure 8.15c: Instrumentation for Arch or Multiple Arch Dam.




Figure 8.15d: Instrumentation for Earth or Rock Fill Dam (Top – d/s section, bottom – u/s

section).




Figure 8.16a: Instrumentation for a 46 ft high homogeneous embankment dam (Source: FERC 2007). Legend: Δ = Reservoir or tail waterlevel indicator; ▲ = drain flow, seepage and leakage; ○ = Pore pressure / uplift; = Survey monument for surface settlement and surfacealignment; ● = Internal movement; □ = Joint and crack di splacement; = Foundation movement; ■ = Temperature.




Figure 8.16b: Instrumentation for a 46 180 ft high zoned EF/RF embankment dam (bottom) (Source: FERC 2007). Legend: Δ = Reservoir ortail water level indicator; ▲ = drain flow, seepage and leakage; ○ = Pore pressure / uplift; = Survey monument for surface settlement andsurface alignment; ● = Internal movement; □ = Joint and crack displacement; = Foundation movement; ■ = Temperature.




Figure 8.17a : Minimum instrumentation for a 88 ft h igh concrete g ravity d am (FERC 200 7)). Legend: Δ = Reservoir or tail water levelindicator; ▲ = drain flow, seepage and leakage; ○ = Po re p ressu re / u plift; = Survey monument for surface settlement and surfacealignment; ● = Internal movement; □ = Joint and crack displacement; = Foundation movement; ■ = Temperature.




Figure 8.17b : Minimum instrumentation for a 120 ft high concrete gravity dam (FERC 2007) ). Legend: Δ = Reservoir or tail water levelindicator; ▲ = drain flow, seepage and leakage; ○ = Pore pressure / uplift; = Survey monument for surface settlement and surfacealignment; ● = Internal movement; □ = Joint and crack displacement; = Foundation movement;■ = Temperature.




8.13 SISGEO DAM SAFETY INSTRUMENTS Source: www.sisgeo.com Dam monitoring instruments include piezometers, pressure cells, extensometer, tilt meters, straingages, joint meters, pendulums, thermometers, flow meters, inclinometer, settlement gages, and loadcells. Some of these are described below.

MAGNETIC EXTENSOMETERSMagnet extensometer is a system - based onBritish Building Research technique - formeasuring either settlement or heave at variousdepths in soil, embankments, earthfill dams anddikes.The system consists of access tube with externalcorrugate pipe, magnet rings, telescopic bottomsection with datum ring and suspension head.Magnet rings (targets)are fixed, externally to theaccess tube in the ground where movement mayoccur. Magnet rings move together withsurrounding soil along the axis of the access tube.

Readings are obtained with a portable readout,lowering the reed switch probe through the accesstube. Comparison of surveys taken over time provide profiles of ground settlement or heave.

IN-PLACE EXTENSOMETERIn-place extensometers are used in conjunctionwith flush-coupled inclinometer casing for themeasurements of settlement or heave. The in-place system is designed to be left inside thecasing to permit automatic or continuousmonitoring. Strings of in-place extensometersensors are joined together with stainless steelwire. The sensors can be located at differentdepths where the settlement occur. The modelequipped with biaxial tilt sensor is also availablefor 3-D borehole profile monitoring. A dataloggerprovides automatic monitoring and by means ofGSM module it is also possible to have on-linemonitoring of unattended locations.

HYDRAULIC ANCHOR LOAD CELLSHydraulic anchor load cells are used to monitorloads in tiebacks, rock bolts and cables. Theyconsist of two ring-shaped stainless steel plateswelded together around their periphery. Theannular space between the plates is filled undervacuum by de-aired oil.

The load is directly measured by a Bourdonmanometer connected to the cell body. Themanometer is calibrated in laboratory to allowdirect readings in KN. A very stiff distribution plateis supplied, in order to ensure that the load isapplied equally over the loading surface of the cell.

Abutment plate may be not required if adequateprovision has been incorporated into theinstallation design.

ELECTRIC AND VIBRATIG WIREPIEZOMETERS

Vibrating wire and electrical piezometers are usedto measure soil pore pressure or water table level

http://www.sisgeo.com/







in boreholes. Applications include control of over-pressure in silt and clay soils, automaticmeasurement of ground water levels, measurement and control of permeability and monitoring upliftpressure and hydraulic gradients in dams and in natural or cut slopes. Output signals are easily read,easily automated and suitable for transmission over long distances. The sensor is housed in a smallstainless steel sealed body with a porous filter tip. Filters with different porosity are available to suitspecific applications.

CASAGRANDE AND STANDPIPE PIEZOMETERS Casagrande filter unit is used to measure thewater pressure in permeable soil. Filter unit ismade in synthesized high density polyethylene. Itis available in different models to suite all thecustomer applications. Filter units have threadedcap joint with two 0.5" twin tubes or with a 1.5"single tube. Standpipe piezometers are used tomonitor the ground water table. The standpipefilter unit consists of a slotted tube covered bygeotechnical fabric for filtered water entry.Stainless steel push-in filter unit is also availablefor drive-in piezometer installation in soft soils.

AUTOMATIC WATER LEVEL MONITORINGThis is an integrated measuring system designedfor water level and temperature monitoring insidewells, standpipes and Casagrande piezometers. Itconsist of a submergible probe, vented cable andbattery operated miniature datalogger. Thesubmergible probe is equipped with a smallpassive pressure sensor which provides highaccuracy and long-term stability. A vented cablewith reference tubing connects the probe to thedatalogger located at the ground level.

CRACKMETERS AND JOINTMETERS

The measurements of superficial movements areimportant for the assessment of the behaviour ofcivil structures and historical buildings. For thispurpose several models of crackmeters and

jointmeters have been designed. Each instrumentconsists of two parts: the sensor housing and thetarget. The sensor housing and target are mountedonto two anchors. Typically the anchors are fixedon the opposite sides of the joint or crack. Thedisplacement transducer housed in the sensorbody is positioned across the joint/crack enablingto measure the changes in the distance betweenthe anchors. Applications include:

• - Monitoring construction joints in dams, bridges and structures• - Monitoring submerged joints in concrete dams• - Monitoring superficial fissures in concrete, brick and masonry• - Monitoring the surface of shaft linings and underground openings




IN-PLACE INCLINOMETERS In-place inclinometers are designed for automatedor remote long term monitoring of inclinometercasings. The in-place system consists of a stringof linked inclinometer probes installed within agrooved inclinometer casing. SISGEO in-place

inclinometers, being linked together, measuredifferential movements and not simply rotations aswith other types.SISGEO manufacture in-place inclinometers withservo-accelerometer and magneto resistive sensoroptions. A string of in-place inclinometers can bereadily connected to a data acquisition unit for realtime monitoring.

SURFACE CLINOMETER SISGEO S500 surface clinometer measures thechanges of tilt of the surface of rock or civilstructures. Surface clinometer measures the

change of angle of the instrument sensors in theirmeasurement axis with reference to gravityvertical datum. The sensor is either a magneto-resistive or servo-accelerometer type and isavailable in uniaxial or biaxial options. Modelspecification, including measuring range, shouldbe selected to suit particular application. Outputreadings are obtained with a suitable SISGEOreadout unit. Automated reading collectionprovides continuous real time monitoring with tiltalarms on preset thresholds.

PENDULUMS Pendulums are designed to monitor the horizontalmovements in dams, dam foundations, abutmentsand to determine the structural and foundationmovements of bridge piers towers and tall buildings.The direct pendulum consists of a steel wireanchored at the upper end to the structure, with atensioning weight suspended at the lower end whichis free to move in a tank filled by a damping fluid.The inverted pendulum use the identical wireanchored in firm soil beneath the structure, with afloating unit to its upper end. The float is free tomove in a water tank and it allows to tension thewire and keep it vertical.

Sisgeo floating unit allows to install two or moreinverted pendulums on the same vertical havingtheir anchors grouted at different depths in a singleborehole.

VIBRATING WIRE AND RESISTIVE STRAIN-GAUGESVW strain-gauges are used to monitor strain in steelor in reinforced concrete and massive concretestructures. Arc-weldable VW strain-gauge isdesigned for arc welding to steel structure such astunnel linings, piles and bridges. The gauges aresupplied with mounting blocks.Embedment VW strain-gauges are directlyembedded in concrete for strain measurements ofpiles, foundations, dams, tunnel linings, etc. VW




strain-gauges are particularly rugged and thermally aged to minimize long-term drift and changes incalibration.Resistive strain-gauge are designed for dynamic measurement in concrete or steel structure and arecompensate for both temperature and bending effects.Rebar strain-meters are designed to be embedded in concrete for the purpose of measuring concretestrain due to imposed load. Application of rebars strain-meters are in concrete structures such as

piles, bridge, tunnel lining, foundation, retaining walls, etc.EARTH PRESSURE CELLS Earth pressure cells are used to monitor totalpressure in earthfill dams and embankments orplaced at the interface between the structure andthe wall of excavation.The earth pressure cells are constructed from twostainless steel plates, welded together around theirpheriphery. The anular space between these platesis filled under vacuum by deaired oil.The pressure pad is connected via a stainless steeltube to the transducer forming a closed hydraulicsystem.The stress is then converted to an electrical signal

and may be remotely read on a variety of portablereadout units or dataloggers.

HYDRAULIC PRESSURE CELLS

Hydraulic pressure cells are designed to measurestress in mass concrete or placed at the interfacebetween the structure and the wall of excavation.The pressure pad consists of two steel plateswelded together around their periphery and spacedapart by a narrow cavity saturated under vacuumwith de aired oil which guarantee maximum rigidity

Documents

Dam and Reservoir Engineering