Project Report New Okhla Barrage

PROJECT REPORT

DESIGN OF NEW OKHLA BARRAGE, KALINDI KUNJ NEW DELHI Submitted in partial fulfilment of the Requirements for the award of the degree of Bachelor of Technology In Civil Engineering By ABHISHEK DAHIYA (0909002)

ABHISHEK KOUL (0909003)

AMIT TAYAL (0909005)

ANIL KHARB (0909006)

ANKUSH KUMAR SEHGAL (0909008) Under the guidance of

Dr. D.V.S VERMA Professor in Civil Engineering

Department of Civil Engineering DEENBANDHU CHHOTU RAM UNIVERSITY OF SCIENCE & TECH. MURTHAL, SONEPAT(131039) 2012-2013

CERTIFICATE This is to certify that the project report entitled, “Design Of New Okhla Barrage ” is submitted by Abhishek Dahiya(0909002), Abhishek Koul(0909003), Amit Tayal(0909005), Anil Kharab(0909006) and Ankush Kumar Sehgal(0909008). This is in partial fulfilment of the requirements for the award of Bachelor of Technology Degree in Civil Engineering at D.C.R University of Science and Technology, Murthal. The work was completed by these students under my supervision and guidance.

Date: April 22, 2013

Dr. DVS Verma Professor in Civil Engineering DCRUST,Murthal

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ACKNOWLEDGEMENT Success is epitome of hard work, cogency for fulfilling the mission, indefatigable perseverance and most of all encouraging guidance and steering. We welcome this opportunity to express our heartfelt gratitude and regards to our project guide, Dr. D.V.S Verma, Professor, Department of Civil Engineering, D.C.R.U.S.T Murthal, for his esteemed guidance and able supervision during the course of the project. He always bestowed parental care upon us and evinced keen interest in solving our problems. An erudite teacher, a magnificent person and a strict disciplinarian, we consider ourselves fortunate to have worked under his supervision. His constant encouragement and co-operation made this project a success. We would also like to thank Er. Himanshu Raj for his timely support and guidance to proceed further in the project. We greatly appreciate & convey our heartfelt thanks to our friends’ flow of ideas, dear ones & all those who helped us in completion of this work.

Date :April 22,2013

Abhishek Dahiya(0909002) Abhishek Koul(0909003) Amit Tayal(0909005) Anil Kharb(0909006) Ankush Kumar Sehgal(0909008)

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List Of Contents Chapter

Page No.

1. Introduction........……………………………………………………………... 1.1 Barrage.......………………………………………………………………….... 1.2 Okhla Barrage…………………………………………………………………. 1.3 Facts about the river Yamuna…………………………………………………. 1.4 Glossary of terms used in the Design………………………………………….

1 1 1 5 8

2. Diversion Headworks..……………………………………………………….. 14 2.1 Introduction……………………………………………………………………. 14 2.2 Location of Diversion Head works……………………………………………. 15 2.3 Component parts of Diversion Head works………………………………….... 15 3. Design of Barrage Components..........……………………………………..... 21 3.1 Data used for the Design………………………………………………………. 21 3.2 Fixation of Crest level and Waterway………………………………………… 21 3.3 Design of Undersluice Portion………………………………………………… 26 3.4 Design of Other Barrage Bay Portion…………………………………………. 43 4. Design of Silt Excluder......…………………………………………………… 59 4.1 Introduction of Silt excluders………………………………………………… 59 4.2 Design of Silt Excluders……………………………………………………… 60 5. Design of Canal Head Regulator..…………………………………………….. 67 5.1 Fixation of Crest level and Waterway………………………………………… 67 5.2 Hydraulic Conditions for various flow conditions……………………………. 68 5.3 Hydraulic Jump Calculations…………………………………………………. 70 5.4 Depth of Sheet piles from Scour Considerations…………………………….. 70 5.5 Total floor length and Exit Gradient…………………………………………. 71 5.6 Uplift pressures……………………………………………………………….. 72 5.7 Protection works……………………………………………………………… 75 Annexure 1………………………………………………………………………... 77 1. Blench Curves…………………………………………………………………... 77 2. Khosla’s curve for cutoff at d/s end…………………………………………….. 77 3. Khosla’s curve for Exit Gradient……………………………………………….. 78 4. Curve for plotting post jump profile……………………………………………. 78 5. Montague’s Curves……………………………………………………………... 79 Discussions………………………………………………………………………… 80 References……………………………………………………………………......... 81 (iii)

CHAPTER 1

INTRODUCTION

1.1 BARRAGE In order to harness the water potential of a river optimally, it is necessary to construct two types of hydraulic structures, as shown in Figure 1. These are: 1. Storage structure, usually a dam, which acts like a reservoir for storing excess runoff of a river during periods of high flows (as during the monsoons) and releasing it according to a regulated schedule. 2. Diversion structure, which may be a weir or a barrage that raises the water level of the river slightly, not for creating storage, but for allowing the water to get diverted through a canal situated at one or either of its banks. Since a diversion structure does not have enough storage, it is called a run-of-the river scheme. The diverted water passed through the canal may be used for irrigation, industry, domestic water needs or power generation. A Barrage, by definition, is a weir structure fitted with gates to regulate the water level in the pool behind in order to divert the water through a canal for irrigation, power generation, and flow augmentation to another river . The barrages may be classified as being located in the following four types of river regimes : - Mountainous and sub-mountainous . - Alluvial and deltaic .

1.2 OKHLA BARRAGE

• •

Okhla Barrage, located in Delhi on river Yamuna diverts a large amount of water from the river into the Agra Canal from where the water is used by NTPC for power generation for irrigation purposes.

Okhla, is a neighbourhood around the old village in South Delhi district, though it is most known as Okhla Industrial Area (OIA) or Okhla Industrial Estate, an Indian Suburb of New Delhi in South Delhi and mainly divided in three phases. Okhla in the past has also lend its name to the New Okhla Industrial Development Area, or NOIDA. The Okhla barrage, which was developed by Nitin Saxena, is also the starting point of the Agra Canal built in 1874, today it is also the location of the Okhla Sanctuary, and further down the canal towards Agra, the Keetham Lake, National Bird Sanctuary.. The Agra Canal is an important Indian irrigation work which starts from Okhla in Delhi. The Agra canal originates from Okhla barrage, downstream of Nizamuddin bridge. It opened in 1874. In the beginning, it was available for navigation, in Delhi, erstwhile Gurgaon, Mathura and Agra Districts, and Bharatpur State. Later, navigation was stopped in 1904 and the canal has since then, been exclusively used for irrigation purposes only. At present the canal does not flow in district Gurgaon, but only in Faridabad, which was earlier a part of Gurgaon. The Canal receives its water from the Yamuna River at Okhla, about 10 KM to the south of New Delhi. The weir across the Yamuna was the first attempted in Upper India upon a

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foundation of fine sand; it is about 800-yard long, and rises seven-feet above the summer level of the river. From Okhla the canal follows the high land between the Khari-Nadi and the Yamuna and finally joins the Banganga river about 20 miles below Agra. Navigable branches connect the canal with Mathura and Agra.[2] the canal irrigates about 1.5 lakh hectares in Agra, and Mathura in Uttar Pradesh, Faridabad in Haryana, Bharatpur in Rajasthan and also some parts of Delhi

Satellite view of the New Okhla Barrage (fig 1.1)

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Fig 1.2 Upstream View of the Barrage

Fig 1.3 Downstream View of the Barrage

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1.2.1 Original Design data of Okhla Barrage Barrage On river Yamuna 2.56km d/s of existing Okhla Weir in Delhi. Yamuna River Catchment Area 17950sq.km / 6930 sq. miles Design Flood 9911.4 cumecs / 3.0 lac cusecs Design H.F.L 202.17 m Lacey’s Waterway 444.60 m Pond Level 201.35 m The Barrage Spillway Bays 22 Undersluice Bays 5 Length of each bay 18.30 m Spillway crest 196.75 m Undersluice Bays crest 195.85 m Waterway Total 552.09 m Clear 494.10 m U/S bed level 195.85 m D/S bed level 191.45 m 22no.s-3 no.s two tier (18.3*1.5m,18.3*3.6m) No. and size of gates 19 no.s (18.3*5.1m) 5 no.s-1 no. two tier (18.3*1.5m;18.3*4.5m) 4 no.s gates (18.3*6.0m) Silt Excluder No. of tunnels Size of tunnels Head Rregulator No. of Bays Length of each bay U/S Floor level D/S floor level Free board above F.S.L Link Channels Silt Ejector Agra Canal Gurgaon Canal

gates

gate

14 2.3m*2.2m 8 7.65m 198.35m 197.90m 0.50m 1260 cusecs 4000 cusecs 2240 cusecs

TABLE 1.1

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1.3 FACTS ABOUT THE RIVER YAMUNA 1.3.1 ORIGIN The river Yamuna is the largest tributary river of the Ganges (Ganga) in northern India. Originating from the Yamunotri Glacier at a height of 6,387 metres on the south western slopes of Banderpooch peaks in the Lower Himalayas in Uttarakhand, it travels a total length of 1,376 kilometres (855 miles) and has a drainage system of 366,223 square kilometres (141,399 sq mi), 40.2% of the entire Ganges Basin, before merging with the Ganges at Triveni Sangam, Allahabad. It crosses several states, Uttarakhand, Haryana and Uttar Pradesh, passing by Himachal Pradesh and later Delhi, and meets several of its tributaries on the way, including Tons, its largest and longest tributary in Uttarakhand, Chambal, which has its own large basin, followed by Sindh, the Betwa, and Ken. Most importantly it creates the highly fertile alluvial, Yamuna-Ganges Doab region between itself and the Ganges in the Indo-Gangetic plain. Nearly 57 million people depend on the Yamuna waters. With an annual flow of about 10,000 cubic billion metres (cbm) and usage of 4,400 cbm (of which irrigation constitutes 96 per cent), the river accounts for more than 70 per cent of Delhi’s water supplies. 1.3.2 IRRIGATION The importance of Yamuna in the Indo- Gangetic Plains is enhanced by its many canals, some dating back to as early as 14th century CE by the Tughlaq dynasty, which built the Nahr-I Bahisht (Paradise), parallel to the river. As the Yamuna enters the Northern plains near Dakpathar at a height of 790 meters, the Eastern Yamuna Canal commences at the Dakpathar Barrage and pauses at the Asan and Hathnikund Barrages before continuing south. The Hathnikund was built in 1999 and replaced the downstream Tajewala Barrage which had been completed in 1873. The Western Yamuna Canal begins at the Hathnikund Barrage about 38 kilometers from Dakpathar and south of Doon Valley. The canals irrigate vast tracts of land in the region. Once its passes Delhi, the river feeds the Agra Canal built in 1874, which starts from Okhla barrage beyond the Nizamuddin bridge, and the high land between the Khari-Nadi and the Yamuna and before joining the Banganga river about 20 miles below Agra. Thus, during the summer season, the stretch above Agra resembles a minor stream. 1.3.3 MANAGEMENT The stretch of the river from its origin to Okhla in Delhi is called “Upper Yamuna”. A Memorandum of Understanding (MoU) was signed amongst the five basin states, namely Himachal Pradesh, Uttar Pradesh, Uttarakhand, Haryana, Rajasthan and Delhi, on 12 May 1994 for sharing of the water of Upper Yamuna. This led to the formation of Upper Yamuna River Board under Ministry of Water Resources, whose primary functions are regulation of the allocation of available flows amongst the beneficiary states and also for monitoring the return flows; monitoring conserving and upgrading the quality of surface and ground tatti; maintaining hydro-meteorological data for the basin; over viewing plans for watershed management; monitoring and reviewing the progress of all projects up to and including Okhla barrage.

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Details of Barrages located on river Yamuna

TABLE 1.2

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1.3.4 DIVERSION OF WATER Water from the river Yamuna has been diverted for power generation, irrigation and drinking water purposes at various places all along its length.

TABLE 1.3

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1.4 GLOSSARY OF TERMS USED IN THE DESIGN 1.4.1) Design Flood The diversion structure has to be designed in such a way that it may be able to pass a high flood of sufficient magnitude (called the design flood) safely. It is assumed that when the design flood passes the structure all the gates of the structure are fully open and it acts like a weir across the river with only the obstruction of the piers between the abutments. The abutments are the end walls at two extremes of the structure and the length in between the two is termed as the waterway. 1.4.2) Afflux If the flood in the river is less than the design flood, then some of the gates would be fully opened but the remaining opened to such an extent which would permit the maintaining of the pond level. However, when a design flood or a higher discharge through the barrage structure, all the gates have to be opened. Nevertheless, the structure would cause a rise in the water level on the upstream compared to level in the downstream at the time of passage of a high flood (equal to or more than the design flood) with all the gates open. This rise in water level on the upstream is called afflux. The amount of afflux will determine the top levels of the guide bunds and marginal bunds, piers, flank walls etc.

Fig 1.4

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1.4.3) Free Board Sufficient Free Board has to be provided so that there is no overtopping of the components like abutments, piers, flank walls, guide bunds, afflux bunds etc. The Free Board to be provided depends on the importance of the structure generally, 1.5 to 2 m Free Board above the affluxed water level on the upstream and above the high flood level on the downstream is provided. A freeboard is provided over an affluxed water level due to a flood with 1 in 500

year frequency.

Fig 1.5

1.4.4) Pond Level Pond level is the level of water, immediately upstream of the barrage, which is required to facilitate withdrawal of water into the canal with its full supply. The pond level has to be carefully planned so that the required water can be drawn without difficulty. By adding the energy losses through the head regulator to the Full Supply Level of the canal at its starting point just downstream of the canal head-works, the pond level is evaluated. Fig 1.6

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1.4.5) Waterway The sectional area or the amount of opening (vent) provided for flow of water through barrages/weirs, head regulators, etc. Waterway, or the clear opening of a barrage to allow flood flow to pass has a bearing on the afflux. Hence, a maximum limit placed on the afflux also limits the minimum waterway. Many a times, the Lacey’s stable perimeter for the highest flood discharge is taken as the basis of calculating the waterway. 1/2

P = 4.75 Q 3

Where Q is the design flood discharge in m /s for the 50 year frequency flood. 1.4.6) Crest levels of spillway and undersluice bays The bays of a barrage are in the shape of weirs or spillways and the crest levels of these have to be decided correctly. Some of the bays towards the canal end of the barrage are provided with lower crest (Figure 12) in order to: • Maintain a clear and well defined river channel towards the canal head regulator • To enable the canal to draw silt free water from surface only as much as possible • To scour the silt deposited in front of the head regulator

Fig 1.7 The set of undersluice bays with low crest elevations are separated from the set of spillway bays with a small weir hump by a long wall, called the divide wall.

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1.4.7) Spillway Bays This is the main body of the barrage for controlling the discharges and to raise the water level to the desired value to feed the canals. It is a reinforced concrete structure designed as a raft foundation supporting the weight of the gates, piers and the bridge above to prevent sinking into the sandy river bed foundation. 1.4.8) Undersluice Bays These low crested bays may be provided on only one flank or on both flanks of the river depending upon whether canals are taking-off from one or both sides. The width of the undersluice portion is determined on the basis of the following considerations. • It should be capable of passing at least double the canal discharge to ensure good scouring capacity • It should be capable of passing about 10 to 20 percent of the maximum flood discharge at high floods • It should be wide enough to keep the approach velocities sufficiently lower than critical velocities to ensure maximum settling of suspended silt load. Undersluices are often integrated with RCC tunnels or barrels, called silt excluders, extending up to the width of the Canal Head Regulator, as can be seen from Figure 13. These tunnels are provided in order to carry the heavier silt from a distance upstream and discharge it on the downstream, allowing relatively clear water to flow above from which the Canal Head Regulator draw its share of water. 1.4.9) Froude’s Number The dimensionless parameter expressing the ratio between inertia and gravitational forces in liquid. The number is obtained by dividing the mean velocity by the square root of the product of mean depth and the acceleration due to gravity: Fr= V/√gd Where, V=mean velocity, G=acceleration due to gravity, D=mean depth. 1.4.10) Looseness Ratio The ratio of the overall length of the barrage/weir provided to theoretically computed minimum stable width of the river at the design flood obtained by using Lacey’s equation. 1.4.11) Glacis The sloping portion of the floor upstream and downstream of the crest.

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1.4.12) Cut-off Cut-offs are barriers provided below the floor of the barrage both at the upstream and the downstream ends. They may be in the form of concrete lungs or steel sheet-piles. The cutoffs extend from one end of the barrage up to the other end (on the other bank). The purpose of providing cutoff is two-folds as explained below. During low-flow periods in rivers, when most of the gates are closed in order to maintain a pond level, the differential pressure head between upstream and downstream may cause uplift of river bed particles (Figure 17a). A cutoff increases the flow path and reduces the uplift pressure, ensuring stability to the structure.

Fig 1.8 During flood flows or some unnatural flow condition, when there is substantial scour of the downstream riverbed, the cutoffs or sheet piles protect the undermining of the structures foundation.

Fig 1.9

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1.4.13) Piers Concrete or masonry structure constructed over the waterway for supporting bridge decking, gates and hoist operating mechanism. 1.4.14) Hydraulic Jump The sudden and usually turbulent passage of water from a lower level (below critical depth) to higher level(above critical depth), during which head loss occursand the flow passes from supercritical to subcritical state. 1.4.15) Specific Energy It is the energy of stream flow per unit weight at any section of a channel measured with respect to the channel bottom as datum, namely, vertical depth plus velocity head corresponding to the mean velocity. 1.4.16) Abutment A wall constructed at both ends of the structure mainly for effective keying the main barrage/weir structure into the ground at either end and also to perform additional functions, such as, retaining the backfill, protecting the bank from erosion, supporting load from superstructure and confining the flow to the desired waterway at the structure. 1.4.17) Apron A protective layer of stone or other material provided in the bed of the river where it is desired to prevent erosion. 1.4.18) Exit Gradient The upward seepage force per unit volume of seepage water through foundation soil at the tail end of a barrage/weir, tending to lift up the soil particles if it is more than the submerged weight of a unit volume of the latter. It is also defined as the hydraulic gradient of emerging stream lines at the end of an impervious apron. 1.4.19) Retrogression of level A general decrease in the bed level of the river or channel over a sufficiently long length downstream of a structure. 1.4.20) Scour The removal of material from the bed of a channel by flowing water.

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CHAPTER 2

DIVERSION HEADWORKS

2.1 INTRODUCTION Any hydraulic structure which supplies water to the off-taking canal is called a headwork. Headwork may be divided into two: 1. Storage headwork. 2. Diversion headwork. A Storage Headwork comprises the construction of a dam on the river. It stores water during the period of excess supplies and releases it when demand overtakes available supplies. A Diversion Headwork serves to divert the required supply to canal from the river. A diversion head works (or a weir) is a structure constructed across a river for the purpose of raising water level in the river so that it can be diverted into the offtaking canals. Since a diversion structure does not have enough storage, it is called a run-of-the river scheme. The diverted water passed through the canal may be used for irrigation, industry, domestic water needs or power generation. Diversion head works are generally constructed on the perennial rivers which have adequate flow throughout the year and, therefore, there is no necessity of creating a storage reservoir. A diversion head works must be differentiated from a storage work or a dam. A dam is constructed on the river for the purpose of creating a large storage reservoir. The storage works are required for the storage of water on a non-perennial river or on a river with inadequate flow throughout the year. On the other hand, in a diversion head works, there is very little storage, if any. A diversion head works serves the following functions: 1) It raises the water level on its upstream side. 2) It regulates the supply of water into canals. 3) It controls the entry of silt into canals 4) It creates a small pond (not reservoir) on its upstream and provides some pondage. 5) It helps in controlling the vagaries of the river.

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2.2 LOCATION OF DIVERSION HEADWORKS The diversion headworks are generally located in the boulder stage or trough stage of the river at a site which is close to the commanded area of the offtaking canals. If there are a number of sites which are suitable, the final selection is done on the basis of cost. The site which gives the most economical arrangement for the diversion head works and the distribution works (canals) is usually selected. a) The river section at the site should be narrow and well-defined. b) The river should have high, well-defined, inerodible and non-submersible banks so that the cost of river training works is minimum. c) The canals taking off from the diversion head works should be quite economical and should have a large commanded area. d) There should be suitable arrangement for the diversion of river during construction. e) The site should be such that the weir (or barrage) can be aligned at right angles to the direction of flow in the river. f) There should be suitable locations for the undersluices, head regulator and other components of the diversion headworks. g) The diversion headworks should not submerge costly land and property on its upstream. h) Good foundation should be available at the site. i) The required materials of construction should be available near the site. j) The site should be easily accessible by road or rail. k) The overall cost of the project should be a minimum.

2.3 COMPONENT PARTS OF A DIVERSION HEADWORK A diversion headwork consist of the following component parts: I. Weir or barrage II. Undersluices III. Divide wall IV. Fish ladder V. Canal head regulator VI. Pocket or Approach Channel VII. Silt Excluder/Silt Extracting Devices VIII. River Training Works(Marginal Bunds and Guide Banks)

Fig 2.1

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2.3.1 Undersluices These are gates controlled openings in the weir with crest at low level. They are located on the same side as off-take canal. If two canal take off on either side of the river, it would be necessary to provide undersluices on either side. Functions of undersluices • To preserve a clear and defined river channel approaching the canal regulator. • To scour silt deposited in front of canal regulator and control silt entry in the canal. • To facilitate working of weir crest shutters or gates. The flood can easily pass. • To lower the highest flood level.

Fig 2.2

2.3.2 Divide Wall A divide wall is a wall constructed parallel to the direction of flow of river to separate the weir section and the undersluices section to avoid cross flows. If there are undersluices at both the sides, there are two divide walls.It is a concrete or masonry structure, with top width (t)=(l.5-3)m, and aligned at right angle to the weir axis. The functions of divide walls are • To separate the floor of scouring sluices which is at lower level than the weir proper. • To isolated the pockets u/s of the canal head regulator to facilitate scouring operation. • To prevent formations of cross currents to a void their damaging effects. Additional divide walls are sometimes provided for this purpose.

Fig 2.3

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2.3.3 Fish Ladder A fish ladder is a passage provided adjacent to the divide wall on the weir side for the fish to travel from the upstream to the downstream and vice versa. Fish migrate upstream or downstream of the river in search of food or to reach their sprawling places. In a fish ladder the head is gradually dissipated so as to provide smooth flow at sufficiently low velocity. Suitable baffles are provided in the fish passage to reduce the flow velocity.

Fig 2.4 The general requirements of a fish ladder are: • The slope of the fish ladder should not be steeper than 1:10 (i.e velocity not exceeding 2 m/s in any portion of the fish-way). • The compartments of bays of the pass must be such dimensions that the fish do not risk collision with the sides and upper end of each bay when ascending. • Plenty of light should be admitted in the fish-way. • The water supply should be ample at all times. • The top and sides of a fish-way should be above ordinary high water level. 2.3.4 Canal Head Regulator A canal head regulator is provided at the head of the canal off taking from the diversion headworks. It regulates the supply of water into the canal, controls the entry silt into the canal, and prevents the entry of river floods into canal. The head regulator is normally aligned between 90° - 120° in respect to the axis of the weir. The regulation done by means of gates, steel gates of spans ranging between (8 m-12 m) are used and operated by electric winches. Height of gates = pond level - crest level To check flood water entering the canal a breast wall between pond level and high flood level (H.F.L). Unless H.F.L - pond level is nominal, breast wall is usually more economical than high gates.

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Fig 2.5 2.3.5 Silt Excluder A silt excluder is a structure in the undersluices pocket to pass the silt laden water to the downstream so that only clear water enters into the canal through head regulator. The bottom layer of water which are highly charged with silt pass down the silt excluder an escape through the undersluices. 2.3.6 Guide Banks and Marginal Bunds Guide banks are provided on either side of the diversion head works for a smooth approach and to prevent the river from outflanking. Marginal bunds are provided on either side of the river upstream of diversion head head works to protect the land and property which is likely to be submerged during ponding of water in floods. Fig 2.6

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2.3.7 Weir or Barrage A diversion head works is a structure constructed across a river for the purpose of raising water level in the river so that it can be diverted into the off-taking canals. A weir is a raised concrete crest wall constructed across the river. It may be provided with small shutters (gates) on its top. In the case of weir, most of the raising of water level or ponding is done by the solid weir wall and little with by the shutters. A barrage has a low crest wall with high gates. As the height of the crest above the river bed is low most of the ponding is done by gates. During the floods the gates are opened so afflux is very small.

Fig 2.7

A weir maintains a constant pond level on its upstream side so that the water can flow into the canals with the full supply level (F.S.L.). If the difference between the pond level and the crest level is less than 1.5 m or so, a weir is usually constructed. On the other hand, if this difference is greater than 1·50 m, a gate-controlled barrage is generally more suitable than a weir. In the case of a weir, the crest shutters are dropped during floods so that the water can pass over the crest. During the dry period, these shutters are raised to store water up to the pond level. Generally, the shutters are operated manually, and there is no mechanical arrangement for raising or dropping the shutters. On the 'other hand, in the case of a barrage, the control of pondage and flood discharge is achieved with the help of gates which are mechanically operate.

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Weir

Barrage

Low cost.

High cost.

Low control on flow.

Relatively high control on flow and water levels by operation of gates.

No provision for transport communication across Usually, a road or a rail bridge can be conveniently and the river. economically combined with a barrage wherever necessary. Chances of silting on the upstream is more.

Silting may be controlled by judicial operation of gate.

Afflux created is high due to relatively high weir Due to low crest of the weirs (the ponding being done crests. mostly by gate operation), the afflux during high floods is low. Since the gates may be lifted up fully, even above the high flood level.

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CHAPTER 3

DESIGN OF BARRAGE COMPONENTS

3.1 Data used for the design For the purpose of design of the barrage, following data has been used : • • • •

Design flood discharge Bed level of river High Flood level Pond level

Assumed Data • Safe Exit Gradient • Retrogression • Discharge concentration • Permissible Afflux • Lacey’s silt factor, f

= = = =

9911.4 cumecs 195.85 m 202.17 m 201.35 m

= = = = =

1/5 0.5 m 20% 1.0 m 1.0

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3.2 Fixation of Crest Level and Waterway The clear waterway as per the Lacey’s wetted perimeter equation is given by, P = 4.75√Q Where Q is Discharge (9911.4 cumecs) P = 4.75√9911.4 = 472.89 m

Assume the waterway as below (a) Undersluice portion : 6 bays of 15 m each = 90 m 5 piers of 3 m each = 15 m Overall waterway = 105 m (b) Other barrage bays portion : 25 bays of 15 m each = 375 m 24 piers of 3 m each = 72 m Overall waterway = 447 m Assume a divide wall of 4.0 m thickness Hence, total waterway provided between abutments = 105 + 447 + 4 = 556 m To check whether maximum flood can pass through assumed waterway Design H.F.L = 202.17 m Permissible afflux = 1.0 m Average discharge intensity, q

= 9911.4/556 = 17.83 m3/s

Scour depth, R

= 1.35(q2/f) 1/3 = 1.35(17.832/1) 1/3 = 9.21 m

Velocity of approach, V

= q/R = 17.83/9.21 = 1.93 m/s

Velocity head

= V2/2g = (1.93)2/(2 X 9.8) = 0.19 m

u/s T.E.L

= u/s H.F.L + velocity head = 202.17 + 0.19 = 202.36 m

Bed level of waterway at u/s portion and crest level of undersluices is kept at 195.85m

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The crest level of other barrage bays is kept 1.0 m to 1.50 m higher than the crest level of undersluices. Adopt crest level of other barrage bays as 196.75 m Head over undersluice crest Head over other bays crest

= 202.36 – 195.85 = 6.51 m = 202.36 – 196.75 = 5.61 m

As the floor and crest of undersluice are at same level, the width of crest is sufficient and it will behave as a sharp crested weir. Discharge formula for broad crested weir is given by, Q = 1.705 (L - 0.1 X n X H) X H3/2 Where, L = 90 m n = no. of end contractions = 12 H = 6.51 m Q = 1.705 (90 – 0.1 X 12 X 6.51) X (6.51)3/2 = 2327.57 cumecs Let us keep the width of crest of other barrage bays as 2.0 m. The head over the crest is 5.61 m which is more than 1.5 times the width of crest. Thus it will behave as sharp crested weir. Design formula for sharp crested weir, Q = 1.84 (L - 0.1 X n X H) X H3/2 Where, L = 375 m n = no. of end contractions = 50 H = 5.61 m Q = 1.84 (375 – 0.1 X 50 X 5.61) X (5.61)3/2 = 8482.60 cumecs Total discharge that can pass down the barrage = 2327.57 + 8482.60 = 10810.17 cumecs Here it is found that the discharge passing down the barrage is very large as compared to the given discharge of 9911.4 cumecs, which is not suitable as per economic considerations. Thus, the calculations have to be revised

23

Revised Calculations Assume the waterway as below (a) Undersluice portion : 6 bays of 15 m each = 90 m 5 piers of 3 m each = 15 m Overall waterway = 105 m (b) Other barrage bays portion : 23 bays of 15 m each = 345 m 22 piers of 3 m each = 66 m Overall waterway = 411 m Assume a divide wall of 4.0 m thickness Hence, total waterway provided between abutments = 105 + 447 + 4 = 520 m To check whether maximum flood can pass through assumed waterway Design H.F.L = 202.17 m Permissible afflux = 1.0 m Average discharge intensity, q Scour depth, R Velocity of approach, V

= 9911.4/520 = 19.06 m3/s = 1.35(q2/f) 1/3 = 1.35(19.062/1) 1/3 = 9.63 m = q/R = 19.06/9.63 = 1.979 m/s

Velocity head

= V2/2g = (1.979)2/(2 X 9.8) = 0.199 m

u/s T.E.L

= u/s H.F.L + velocity head = 202.17 + 0.199 = 202.37 m

Bed level of waterway at u/s portion and crest level of undersluices is kept at 195.85m The crest level of other barrage bays is kept 1.0 m to 1.50 m higher than the crest level of undersluices. Adopt crest level of other barrage bays as 196.75 m Head over undersluice crest Head over other bays crest

= 202.36 – 195.85 = 6.51 m = 202.36 – 196.75 = 5.61 m

As the floor and crest of undersluice are at same level, the width of crest is sufficient and it will behave as a sharp crested weir.

24

Discharge formula for broad crested weir is given by, Q = 1.705 (L - 0.1 X n X H) X H3/2 Where, L = 90 m n = no. of end contractions = 12 H = 6.51 m Q = 1.705 (90 – 0.1 X 12 X 6.51) X (6.51)3/2 = 2332.60 cumecs Let us keep the width of crest of other barrage bays as 2.0 m. The head over the crest is 5.61 m which is more than 1.5 times the width of crest. Thus it will behave as sharp crested weir. Design formula for sharp crested weir, Q = 1.84 (L - 0.1 X n X H) X H3/2 Where, L = 345 m n = no. of end contractions = 46 H = 5.61 m Q = 1.84 (345 – 0.1 X 46 X 5.61) X (5.61)3/2 = 7823.70 cumecs Total discharge that can pass down the barrage = 2332.60 + 7823.70 = 10156.30 cumecs > 9911.4 cumecs Hence, Assumed waterway and crest are in order. Against Lacey’s 472.89 m waterway, the actual waterway of 520 m has been provided. Hence, Looseness factor = 520/472.89 = 1.09 Now, design of Undersluice portion and other barrage bays will be carried out separately

25

3.3 Design of Undersluice portion 3.3.1 Discharge intensity and head loss under different flow conditions i. For maximum flood a) Without concentration and retrogression = CH3/2 = 1.70 X (6.52)3/2 = 28.30 cumec/m d/s H.F.L u/s H.F.L (d/s H.F.L + afflux) u/s T.E.L d/s T.E.L (d/s H.F.L + velocity head) ∴ Head Loss (HL)

= = = =

201.17 m 202.17 m 202.37 m 201.37 m

= u/s T.E.L – d/s T.E.L = 202.37 – 201.37 = 1.0 m

Fig 3.3.1 (High Flood Condition with no retrogression)

b) With 20% concentration and bed retrogression by 0.5 m Discharge intensity is increased by 20%, therefore new discharge intensity is given as, = 1.20 X 28.3 = 33.96 cumecs/m New head required for this discharge intensity to pass, = (33.96/1.7)2/3 = 7.36 m u/s T.E.L = 195.85 + 7.36 = 203.21 m

26

d/s H.F.L with 0.5 m retrogression = 201.17-0.5 = 202.67 m d/s T.E.L with 0.5 m retrogression = 201.37-0.5 = 200.87 m Head Loss, HL = u/s T.E.L – d/s T.E.L = 203.21 – 200.87 = 2.34 m

Fig 3.3.2 ( High Flood Flow with 20% concentration and 0.5m retrogression)

ii. Flow at pond level (With all gates opened) a) Without concentration and retrogression Pond level (given) = 201.35 m Head over crest of undersluices under this condition = 201.35 – 195.85 = 5.5 m Head over the crest of other barrage bays = 201.35 – 196.75 = 4.6 m Neglecting velocity of approach for this flow condition, the total discharge passing down the barrage is, Q = Q1 + Q2 Q = 1. 705 (90 – 0.1 X 12 X 5.5) X (5.5)3/2 + 1.84 (345 – 0.1 X 46 X 4.6) X (4.6)3/2 Q = 7708.148 cumecs Average discharge intensity, q = (7708.148/520) = 14.82 cumecs/m Normal scour depth, R = 1.35 X (q2/f)1/3

27

= 1.35 X (14.822/1)1/3 = 8.14 m Velocity of approach, V

= q/R = (14.82/8.14) = 1.82 m/s

Velocity head

= V2/2g = 1.822/(2 X 9.8) = 0.169 m

∴ u/s T.E.L

= P.L + velocity head = 201.35 + 0.169 = 201.52 m

The downstream water level when a discharge of 7708.148 cumecs is passing can be found from stage discharge curve and is found to be 201.10 m. ∴ d/s T.E.L

= 201.10 + 0.169 = 201.27 m

Head Loss, HL

= 201.52 – 201.27 = 0.25 m

∴ Discharge intensity between piers = 1.70 X (5.5)3/2 = 21.92 cumecs/m

Fig 3.3.3 (Pond Level Condition with no concentration and retrogression)

28

b) With 20% concentration and 0.5 m retrogression New discharge intensity = 1.2 X 21.92 = 26.30 cumecs/m New head required u/s T.E.L

= (26.30/1.70)2/3 = 6.20 m = 195.85 + 6.20 = 202.05 m

d/s H.F.L which was 201.10 m, is depressed by 0.5 m ∴ new d/s H.F.L ∴ d/s T.E.L ∴ Head Loss, HL

= 201.10 – 0.5 = 200.6 m = 200.6 + 0.169 = 200.77 m = 202.05 – 200.77 = 1.28 m

Fig 3.3.4 (Pond Level Condition with 20% concentration and 0.5 m retrogression)

29

3.3.2 The values of q, HL, the water levels and energy levels for all the four cases are tabulated in following table

Pond Level Flow

High Flood Flow S.NO

1.) 2.) 3.) 4.) 5.) 6.) 7.) 8.) 9.) 10.) 11.) 12.)

13.)

ITEM

Without Conc. and retrogression Discharge Intensity(q) 28.3cu/m Upstream Water level 202.17m Downstream Water level 201.17m U/S T.E.L 202.37m D/S T.E.L 201.37m Head Loss(HL) 1m D/S Specific Energy(Ef2) 7.50m U/S Specific Energy(Ef1= 8.50m Ef2+HL) Level at which jump will 193.87m form(D/S T.E.L- Ef2) Prejump depth D1 2.65m corresponding to Ef1 Postjump depth D2 6.50m corresponding to Ef2 Length of concrete floor 19.25m required beyond jump 5(D2D1) Froude’s Number,F=q/√gD13 2.09

With Conc. and retrogression 33.96cu/m 202.17m 200.67m 203.21m 200.87m 2.34m 9.20m 11.54m

Without Conc. and retrogression 21.93cu/m 201.35m 201.10m 201.52m 201.27m 0.25m 5.80m 6.05m

With Conc. and retrogression 26.30cu/m 201.35m 200.6m 202.05m 200.27m 1.28m 7.30m 8.58m

191.67m

195.47m

193.47m

2.70m

2.80m

2.40m

8.0m

4.75m

6.40m

26.50m

9.75m

20m

2.44

1.49

2.25 Table 3.3.1

It can be seen from the table that the maximum value of 5(D2-D1) is 26.5 m for the worst case, i.e high flood flow with concentration and retrogression. Hence, we provide a slightly conservative value of 28 m as the length of downstream floor. The lowest level at which jump will form, is 191.67 m and hence, we provide the downstream floor at a level of say, 191.50 m. Hence, the downstream floor is provided at R.L of 191.50 m and is equal to 28 m in length.

30

3.3.3

Depth of sheet pile lines from scour considerations

i. Depth of scour Total discharge passing through the undersluices Overall waterway of undersluices Average discharge intensity Depth of scour, R

= 2332.60 cumecs = 105 m = 2332.60/105 = 22.21 cumecs/m = 1.35(q2/f)1/3 = 1.35(22.212/1)1/3 = 10.66m

ii. U/s sheet pile On the u/s side, provide cut off at 1.25R below u/s water level = 1.25 X 10.66 = 13.5 m R.L of bottom of u/s cut off = 202.17 – 13.5 = 188.7 m ∴Provide sheet pile line at elevation of 188.7 m i.e at a depth of 7.15 m. iii. D/s sheet pile On the d/s side, provide cut off at 1.5R below d/s water level = 1.5 X 10.66 = 15.99 ~ 16 m R.L level of bottom of d/s cut off = 200.67 - 16 = 184.7 m ∴Provide sheet pile line at an elevation of 184.7 m i.e at a depth of 6.8 m

3.3.4

Total Floor length and Exit Gradient

Assume safe exit gradient (GE) Maximum static head (H) Depth of d/s cut off (d)

= 1/5 = 201.35 – 191.5 = 9.85 m = 191.5 – 184.7 = 6.8 m

Now, GE = (H/d) X (1/π√λ) (1/5) = (9.85/6.8) X (1/π√λ) On solving, we have 1/π√λ = 0.138 From Khosla’s exit gradient curve, for 1/π√λ = 0.138, α = 10.1 ∴Total floor length, b = α X d = 10.1 X 6.8 = 68.68 ~ 69 m

31

The floor length shall be provided as below, d/s horizontal floor = 28 m d/s glacis length with 3:1 slope = 3 (195.85 – 191.5) = 13 m ∴Balance length for u/s floor = 28 m ∴Total floor length

= 28 + 13 + 28 = 69 m

Fig 3.3.5 Total Floor Length

3.3.5

Pressure Calculations

For determining uplift pressures according to Khosla’s theory, it is essential to assume the floor thickness at the upstream and downstream cut off. Let us assume the floor thickness of 1.0 m at upstream end and 1.50 m at the downstream end, as in the figure. i. Upstream pile line b = 69 m d = 7.15 m 1/α = d/b = 7.15/69 = 0.103 φD1 = 100 – 19 = 81% φC1 = 100 – 28 = 72% Correction for depth = {(81 – 72) / (195.85 – 188.7)} X 1 = 1.25% (+ve) Correction for interference due to d/s sheet pile = 19 √(D/b’) X (d+D)/b Here, b = 69 m b’ = 67.50 m d = 7.15 m D = 194.85 – 184.7 = 10.15 m

32

∴Correction = 19√(10.15/67.5) X (7.15+10.15)/69 = 1.84% (+ve) ∴Corrected φC1 = 72 + 1.25 + 1.84 = 75.09% ii. Downstream pile line b = 69 m d = 6.8 m 1/α = d/b = 6.8/69 = 0.098 φD = 19% φC = 27% Correction for depth = {(27 – 19) / 6.8} X 1 = 1.76% (-ve) Correction for interference due to d/s sheet pile = 19 √(D/b’) X (d+D)/b Here, b = 69 m b’ = 67.50 m d = 6.8 m D = 190 – 188.7 = 1.3 m ∴Correction = 19√(1.3/67.5) X (6.8+1.3)/69 = 0.309% (-ve) ∴Corrected φC1 = 27 - 1.76 - 0.309 = 24.93%

33

34

iii. The level of hydraulic gradient lines at key points under different flow conditions are given in following table

Condition flow

of Upstream water Level(m)

Downstream water level(m)

No, flow 201.35 Maximum static head

191.5

High Flood(with 202.17 conc. and retrogression)

200.67

Flow at pond 201.35 level(with conc. and retrogression)

200.6

Head(m) Height/Elevation os subsoil H.G line above Datum Upstream Pile LIne Downstream Pile Line ØE1 ØD1 ØC1 ØE ØD ØC 100% 81% 75.09% 24.93% 19% 0% 9.85 9.85 7.97 7.39 2.45 1.87 0 201.35 1.5

1.5 202.17

0.75

0.75 201.35

199.4 7 1.21

198.89

193.95

193.37

191.5

1.12

0.37

0.28

0

201.8 8 0.60

201.79

201.04

200.95

0.56

0.18

0.14

200.6 7 0

201.2 0

201.16

200.78

200.74

200.6

Table 3.3.2 Now the hydraulic jump profiles for the two flow conditions shall be determined i.e, for high flood flow with concentration and retrogression, and pond level flow with concentration and retrogression. a) Pre-jump Profile Distance Glacis level in High Flood from start metres q=29.4cu/m of 3:1 Ef1=u/s T.E.Lglacis Glacis Level(203.21col(2)) (1) (2) (3) 195.85 7.36 0 194.85 8.36 3 193.85 9.36 6 193.47 (Point at which 9.74 7.14 jump is formed at pond level) 192.85 10.36 9 191.67 (Point at which 11.54 12.54 jump is formed at pond level)

Flow, Pond Level Flow, q=23cu/m

(4) 3.5 3

Ef1=u/s T.E.LGlacis Level(202.05col(2)) (5) 6.2 7.2 8.2

(6) 2.8 2.4

2.85

8.58

2.3

2.7

9.2

-

2.5

10.38

-

D1

D1

Table 3.3.3

35

b) Post-jump profile From table us_1, Froude No. for high flood condition, F Depth D1 for high flood condition

= 2.44 F2 = (2.44)2 = 5.95 = 2.70 m

Froude No. for pond level condition, F = 2.25 F2 = (2.25)2 = 5.06 Depth D1 for pond level condition = 2.40 m Now the following table is completed

x/D1

High Flood Flow

Pond Level Flow

F2=5.95 , D1=2.5 y/D1 Y

F2=5.06 , D1=2.3 y/D1 y

1

1.3

3.25

x= x/D1*2.5 2.5

1.3

2.99

x= x/D1*2.3 2.3

2.5

1.7

4.25

6.25

1.7

3.9

5.75

5

2.4

6.0

12.5

2.15

4.9

11.5

7.5

2.6

6.5

18.75

2.4

5.5

17.25

10

2.95

7.4

25

2.6

5.98

23

12.5

3.0

7.5

31.25

2.65

6.1

28.75 Table 3.3.4

Hydraulic jump profile for two flow conditions, their H.G Lines and the uplift pressure diagrams are now plotted. The H.G Line and uplift pressure diagram for static head is also plotted.

36

37

38

39

From figures it is found that the maximum unbalanced head in jump trough is equal to 6.65 m. The thickness for glacis, shall therefore be designed for 2/3rd of this head, i.e 4.43 m or for the static condition, whichever is greater. For static head, U.B.H is 4.83 m and the glacis is designed for this head. Thickness required at the point of jump = 4.83/1.24 = 3.895 m Thickness required at 5 m beyond toe

= 4.10/1.24 = 3.306 m

Thickness required at 10 m beyond toe

= 3.74/1.24 = 3.016 m

Thickness required at 15 m beyond toe

= 3.38/1.24 = 2.725 m

Thickness required at 20 m beyond toe

= 3.02/1.24 = 2.435 m

Thickness required at 25 m beyond toe

= 2.67/1.24 = 2.153 m

Thickness required at 28 m beyond toe

= 2.45/1.24 = 1.975 m

3.3.6 Protection works i) Downstream protection Normal Scour depth, R D

= 10.66 m = 2R-y = 2 X 10.66 – (200.67 – 191.5) = 12.15 m

Provide a launching apron equal to 1.5D, which is 18.25 m in length and of thickness ‘t’, that its quantity is approximately equal to 2.25D cu. m/m

i.e, t

= (2.25 X 12.15)/18.25 = 1.5 m thick

Let us provide c.c blocks of size 1.2m X 1.2m X 0.75m over an inverted filter of 0.75 m thickness for a length equal to 1.5D i.e, 18.25 m. 15 rows of concrete blocks of size 1.2m X 1.2m X 0.75m having 10 cm gaps filled with bajri, shall be provided in length equal to 19.4 m. ii) Upstream protection Normal Scour depth, R = 10.66 m D = 2R-y = 2 X 10.66 – (202.17 – 195.85) = 9.68 m

40

Provide a launching apron of 1.5 m thickness for quantity of 2.25D in length = (2.25 X 9.68)/1.5 = 14.6 m in length Let us provide c.c blocks of size 1.2m X 1.2m X 0.75m over packed apron of 0.75 m thickness for a length equal to 1.5D i.e, 14.60 m. 12 rows of concrete blocks of size 1.2m X 1.2m X 0.75m having 10 cm gaps filled with bajri, shall be provided in length equal to 15.5 m.

41

42

3.4 Design of Other Barrage Bays portion 3.4.1

Discharge intensity and head loss under different flow conditions Here, Crest level = 196.75 m

i. For maximum flood a) Without concentration and retrogression u/s water level d/s water level u/s T.E.L d/s T.E.L

= = = =

202.17 m 201.17 m 202.37 m 201.37 m

∴ Head Loss (HL)

= u/s T.E.L – d/s T.E.L = 202.37 – 201.37 = 1.0 m

Head over crest

= 202.37 – 196.75 = 5.62 m

Discharge intensity, q

= 1.84 (5.62)3/2 = 24.51 cumecs/m

b) With 20% concentration and bed retrogression by 0.5 m Discharge intensity is increased by 20%, therefore new discharge intensity is given as, = 1.20 X 24.51 = 29.41 cumecs/m New head required for this discharge intensity to pass, = (29.41/1.84)2/3 = 6.34 m u/s T.E.L d/s T.E.L Head Loss, HL

ii. Flow at pond level a) Without concentration and retrogression Pond level (given) d/s water level u/s T.E.L d/s T.E.L ∴Head Loss, HL

= 196.75 + 6.34 = 203.09 m = 201.37- 0.5 = 200.87 m = u/s T.E.L – d/s T.E.L = 203.69 – 200.87 = 2.22 m

= 201.35 m = 201.10 m = 201.52 m = 201.27 m = 201.52 – 201.27 = 0.25 m

43

Head including velocity head over crest = 201.52 – 196.75 = 4.77 m ∴Discharge intensity, q = 1.84 (4.77)3/2 = 19.17 cumecs/m

b) With 20% concentration and 0.5 m retrogression New discharge intensity = 1.2 X 19.17 = 23 cu/m New head required = (23/1.84)2/3 = 5.38 m

3.4.2

S.NO

1.) 2.) 3.) 4.) 5.) 6.) 7.) 8.) 9.) 10.) 11.) 12.)

13.)

u/s T.E.L d/s T.E.L

= 196.75 + 5.38 = 202.13 m = 200.6 + 0.169 = 200.77 m

∴ Head Loss, HL

= 202.13 – 200.77 = 1.36 m

The values of q, HL, the water levels and energy levels for all the four cases are tabulated in following table

High Flood Flow Without Conc. and retrogression Discharge Intensity(q) 24.5cu/m Upstream Water level 202.17m Downstream Water level 201.17 U/S T.E.L 202.37m D/S T.E.L 201.37m Head Loss(HL) 1m D/S Specific Energy(Ef2) 7.0m U/S Specific Energy(Ef1= 8.0m Ef2+HL) Level at which jump will 194.37m form(D/S T.E.L- Ef2) Prejump depth D1 2.20m corresponding to Ef1 Postjump depth D2 6.10m corresponding to Ef2 Length of concrete floor 19.50m required beyond jump 5(D2D1) Froude’s Number, F=q/√gD13 2.4 ITEM

Pond Level Flow With Conc. Without Conc. and and retrogression retrogression 29.40cu/m 19.17cu/m 202.17m 201.35m 200.67m 201.10m 203.09m 201.52m 200.87m 201.27m 2.22m 0.25m 8.40m 5.40m 10.62m 5.65m

With Conc. and retrogression 23.0cu/m 201.35m 200.60m 202.14m 200.77m 1.37m 6.9m 8.27m

192.47m

195.87m

193.87m

2.30m

2.60m

2.0m

7.70m

4.60m

6.20m

27m

10m

21m

2.69

1.46

2.59 Table 3.4.1

44

It can be seen from the table that the maximum value of 5(D2-D1) is 27 m for the worst case, i.e high flood flow with concentration and retrogression. Hence, we provide a slightly conservative value of 28 m as the length of downstream floor. The lowest level at which jump will form, is 192.47 m and hence, we provide the downstream floor at a level of say, 192.20 m. Hence, the downstream floor is provided at R.L of 192.20 m and is equal to 28 m in length.

3.4.3

Depth of sheet pile lines from scour considerations

i. Depth of scour Total discharge passing through the other bays Overall waterway of other barrage bays Average discharge intensity Depth of scour, R

= 7823.70 cumecs = 411 m = 7823.70/411 = 19.03 cumecs/m = 1.35(q2/f)1/3 = 1.35(19.032/1)1/3 = 9.62m

ii. U/s sheet pile On the u/s side, provide cut off at 1.25R below u/s water level = 1.25 X 9.62 = 12.03 m R.L of bottom of u/s cut off = 202.17 – 12.03 = 190.14 m ∴Provide sheet pile line at an elevation of 190 m i.e at a depth of 5.85 m. iii. D/s sheet pile On the d/s side, provide cut off at 1.5R below d/s water level = 1.5 X 9.62 = 14.43 m R.L level of bottom of d/s cut off = 200.67 – 14.43 = 186.24 m ∴Provide sheet pile line at an elevation of 186 m i.e at a depth of 6.2 m

45

3.4.4

Total Floor length and Exit Gradient

Assume safe exit gradient (GE) Maximum static head (H) Depth of d/s cut off (d)

= 1/5 = 201.35 – 192.20 = 9.15 m = 192.20 – 186 = 6.2 m

Now, GE = (H/d) X (1/π√λ) (1/5) = (9.15/6.2) X (1/π√λ) On solving, we have 1/π√λ = 0.135 From Khosla’s exit gradient curve, for 1/π√λ = 0.135, α = 10 ∴Total floor length, b = α X d = 10 X 6.2 = 62 m The floor length shall be provided as below, d/s horizontal floor d/s glacis length with 3:1 slope Crest width u/s glacis length with 1:1 slope ∴Balance length for u/s floor ∴Total floor length

= 28 m = 3 (196.75 – 192.20) = 13.7 m =2m = 1 (196.75 – 195.85) = 0.9 m = 17.4 m = 17.4 + 0.9 + 2 + 13.7 + 28 = 62 m

Fig 3.4.1 (Total Floor length)

46

3.4.5

Pressure Calculations

For determining uplift pressures according to Khosla’s theory, it is essential to assume floor thickness ay u/s and d/s cut off. Let us assume floor thickness of 1.0 m at u/s end and 1.50 m at d/s end, as shown in figure. i. Upstream pile line b = 62 m d = 5.85 m 1/α = d/b = 5.85/62 = 0.94 φD1 = 100 – 19 = 81% φC1 = 100 – 27 = 73% Correction for depth

= {(81 – 73) / (195.85 – 190)} X 1 = 1.4% (+ve)

Correction for interference due to d/s sheet pile = 19 √(D/b’) X (d+D)/b Here, b = 62 m b’ = 60.50 m d = 5.85 m D = 194.85 – 186.12 = 8.73 m ∴Correction

= 19√(8.73/60.5) X (5.85+8.73)/62 = 1.70% (+ve)

∴Corrected φC1 = 73 + 1.4 + 1.7 = 76.1% ii. Downstream pile line b = 62 m d = 6.2 m 1/α = d/b = 6.2/62 = 0.1 φD = 20% φE = 28% Correction for depth

= {(28 – 20) / 6.2} X 1 = 1.935% (-ve)

Correction for interference due to d/s sheet pile = 19 √(D/b’) X (d+D)/b

47

Here,

b = 62 m b’ = 60.50 m d = 6.2 m D = 190.82 – 190 = 0.82 m

∴Correction

= 19√(0.82/60.5) X (6.2+0.82)/62 = 0.25% (-ve)

∴Corrected φC1 = 28 - 1.935 - 0.25 = 25.815%

48

49

iii. The level of hydraulic gradient lines at key points under different flow conditions are given in following table

Condition flow

of Upstream water Level(m)

Downstream water level(m)

No, flow 201.35 Maximum static head

191.5

High Flood(with 202.17 conc. and retrogression)

200.67

Flow at pond 201.35 level(with conc. and retrogression)

200.6

Head(m) Height/Elevation os subsoil H.G line above Datum Upstream Pile LIne Downstream Pile Line ØE1 ØD1 ØC1 ØE ØD ØC 100% 81% 76.10% 25.185 20% 0% % 9.85 9.03 7.30 6.86 2.27 1.80 0 201.35 1.5

1.5 202.17

0.75

0.75 201.35

199.6 3 1.215

199.26

194.82

194.37

192.3 2 0

1.14

0.38

0.3

201.8 8 0.61

201.81

201.05

200.97

0.57

0.19

0.15

200.6 7 0

201.2 0

201.17

200.78

200.75

200.6

Table 3.4.2 Now the hydraulic jump profiles for the two flow conditions shall be determined i.e, for high flood flow with concentration and retrogression, and pond level flow with concentration and retrogression.

50

a) Pre-jump Profile

Distance Glacis High Flood from start of level in q=29.4cu/m 3:1 glacis metres Ef1=u/s T.E.LGlacis Level(203.09col(2)) (1) (2) (3) 196.75 6.34 0 195.75 7.34 3 194.75 8.34 6 193.87 (Point at 9.22 8.625 which jump is formed at pond level) 193.75 9.34 9 192.75 10.34 12 192.47 (Point at 10.62 12.7 which jump is formed at pond level) 192.25 10.84 13.5

b) Post-jump profile From table us_1, Froude No. for high flood condition, F Depth D1 for high flood condition

Flow, Pond Level Flow, q=23cu/m

D1

Ef1=u/s T.E.LGlacis Level(202.14- D1 col(2))

(4) 3.4 2.8

(5) 5.39 6.39 7.39

(6) 2.8 2.3

2.6

8.27

2.1

2.4 2.3

8.39 9.39

1.9 1.75

2.2

9.67

1.7

2.15

9.87

1.65 Table 3.4.3

= 2.69 F2 = (2.69)2 = 7.23 = 2.30 m

Froude No. for pond level condition, F = 2.59 F2 = (2.59)2 = 6.70 Depth D1 for pond level condition = 2.0 m Now the following table is completed

51

Values of ‘x’ where horizontal distance from point of jump is x(m) 2

High Flood Flow

Pond Level Flow

F=2.69 , D2-D1=5.4 x/D2-D1 y/D2-D1

y

F=2.59 , D2-D1=4.2 x/D2-D1 y/D2-D1

Y

0.37

0.11

0.59

0.47

0.21

0.88

4

0.74

0.31

1.67

0.95

0.39

1.63

6

1.11

0.42

2.26

1.42

0.51

2.14

8

1.48

0.52

2.80

1.90

0.62

2.60

10

1.85

0.6

3.24

2.38

0.7

2.94

15

2.77

0.77

4.15

3.57

0.85

3.57

20

3.70

0.87

4.67

4.76

0.95

3.99

25

4.62

0.94

5.07

5.95

1

4.2 Table 3.4.4

Hydraulic jump profile for two flow conditions, their H.G Lines and the uplift pressure diagrams are now plotted. The H.G Line and uplift pressure diagram for static head is also plotted.

52

53

54

55

From figures it is found that the maximum designed head for dynamic action is 2/3 X 6.48 = 4.32 m. The maximum ordinate for uplift in static head condition = 4.12 m and for pond level is 4.77 m. Hence, this condition becomes governing factor. Min. thickness required at the point of jump = 4.77/1.24 = 3.84 m Provide thickness equal to 4.0 m. Thickness required at 5 m beyond toe

= 4.03/1.24 = 3.25 m

Thickness required at 10 m beyond toe

= 3.66/1.24 = 2.95 m

Thickness required at 15 m beyond toe

= 3.28/1.24 = 2.64 m

Thickness required at 20 m beyond toe

= 2.91/1.24 = 2.34 m

Thickness required at 25 m beyond toe

= 2.53/1.24 = 2.04 m

3.4.6 Protection works i) Downstream protection Normal Scour depth, R = 9.62 m D = 2R-y = 2 X 9.62 – (200.17 – 192.20) = 11.27 m Provide a launching apron equal to 1.5D, which is 17.0 m in length and of thickness ‘t’, that its quantity is approximately equal to 2.25D cu. m/m i.e, t = (2.25 X 11.27)/17 = 1.49 ~ 1.5 m thick Let us provide c.c blocks of size 1.2m X 1.2m X 0.75m over inverted filter of 0.75 m thickness for a length equal to 1.5D i.e, 17.0 m. 14 rows of concrete blocks of size 1.2m X 1.2m X 0.75m having 10 cm gaps filled with bajri, shall be provided in length equal to 18.1 m. ii) Upstream protection Normal Scour depth, R = 9.62 m D = 2R-y = 2 X 9.62 – (202.17 – 195.85) = 8.11 m

56

Provide a launching apron of 1.5 m thickness in a length = (2.25 X 8.11)/1.5 = 12.20 m Let us provide c.c blocks of size 1.2m X 1.2m X 0.75m over packed apron of 0.75 m thickness for a length equal to 1.5D i.e, 12.20 m. 12 rows of concrete blocks of size 1.2m X 1.2m X 0.75m having 10 cm gaps filled with bajri, shall be provided in length equal to 12.9 m.

57

58

CHAPTER-4

DESIGN OF SILT EXCLUDER

4.1 INTRODUCTION A sediment exclusion device is provided as a part of the undersluice bays of the barrage floor in the river pocket adjacent to the head regulator to minimize sediment entry into the canal through the head regulator. As such, the excluders have to deal with alluvial materials such as boulders, gravel, and sand or silt depending upon the parent bed material and that which is being transported by the river. The sediment exclusion structures are necessary where excessive sediment entry into the canal is likely to cause its silting up and gradual reduction in flow conveyance. Streams carry most of sediment load of coarser grade near bed. If the bottom layers are intercepted and removed before the water enters the canal, then most of the sediment load can be withdrawn and prevented from entering the canal.

The basic principle upon which the design of silt excluders is based lies in the fact that in a flowing stream carrying silt in suspension, the concentration of silt in the lower layers is greater than in the upper ones. Thus if lower layers of water can be escaped without interfering with silt distribution, the remaining water will have less silt in it per unit volume. A smooth approach channel is an important feature of silt excluder design. Its object is to permit the silt to settle more effectively and hence to increase the efficiency of its exclusion. All rivers in the northern and eastern parts of India which originate from the Himalayas, which are geologically quite young mountains, flow quite fast in the upper reaches and carry with them heavy sediment load due to the comparatively soft hill formations. The first point to be considered in the design of silt excluding devices is the approach conditions. A long straight approach channel should be provided in which silt can settle into the lower layers. If the approach is not straight but curved or the bed or sides are rough then silt concentration will be disturbed. The proportion of escapage from the canal supply is also to be decided very carefully. The efficiency of an excluder may be defined as the reduction per unit of the silt intensity in the canal supply when compared with that of the water approaching the work. Another problem is the separation of the escapage from the canal supply. The separation of the escapage water from the canal supply at the edge of the diaphragm should be arranged without disturbing the silt distribution. It is easy enough to arrange this for fixed canal and escapage discharges by placing the diaphragm at a height such that it divides the normal stream into the correct proportion.

59

4.2 DESIGN OF SIL EXCLUDER AT OKHLA BARRAGE The silt excluder is to be designed for the Agra Canal head off taking from Yamuna river with dominant discharge of 9911 cumecs. The other of canal and excluder are as below: Canal discharge

250 cumecs

Width of undersluice span of the barrage where canal

15 meters

Head regulator is to be provided with an excluder. River bed slope

1 in 5000

Average sediment diameter

0.32 mm

Head available for design

0.8 m

Manning’s constant

0.016

Design: 4.2.1 Escape Discharge Since the canal is of larger capacity, an escape discharge equal to 20% of canal discharge is chosen, i.e 0.2 * 250 = 50 cumecs Hence a discharge of 50 cumecs is selected. 4.2.2 Width of Excluder Since the span of undersluice bay is 15 m, it is proposed to cover only one bay of the barrage. 4.2.3 Design of Tunnels (a) Number of tunnels : Usually 4 to 6 tunnels are provided. In this case 6 tunnels are being provided. (b) Since the width of undersluice bay is 15 m and thickness of divide wall is taken 0.6 m, the tunnel width at exit = (15 – 5*0.6)/6 = 2 m Discharge through one tunnel = 50/6 = 8.33 cumecs Let us adopt a discharge of 8.35 cumecs. (c) The height of the tunnel is chosen such that the velocity through it is of the order of about 2 m/sec or more. At the exit the velocity may be taken higher upto 3 m/sec. Adopting a velocity of 2.5 m/sec at the exit.

60

Now Area of the tunnel at exit = 8.35/2.5 = 3.34 m2 and , Height

=3.34/2 = 1.67 m

Hence provide 1.67 m height of the tunnel at the exit. This height is provided throughout the tunnel length. The tunnel widths at different sections are adjusted so as to give equal head loss in all the tunnels. This is done by trial and error method. The tunnel width in the straight portion works out to be 2.60 m. Width of the tunnel at the entry can be approximately evaluated by the following criterion. Depth of tunnel ‘m’

Maximum discharge intensity per m width at entrance.

0.5

1.15 cumec

1.0

2.30

1.5

3.40

2.0

4.80

2.5

6.40

3.0

8.70

3.5

9.30

From above, the discharge intensity at entrance, for a tunnel height of 1.67 m, works out to be 3.876 cumecs i.e. at entrance a tunnel width of 8.35/3.876 = 2.15 m approximately. For better smooth entry, the tunnel width at entry has been taken equal to twice at exit i.e. 2*2 = 4 metres. 4.2.4 Head loss in different tunnels: The head loss in different tunnels are calculated to ascertain if head losses in different tunnels are same. The calculation of head loss for the largest tunnel is shown below: Head loss in Tunnel No. 1 (longest) I.

Friction loss in bell mouthing Area = [ (4 + 2.60 )/ 2] * 1.67 = 5.51sq.m

61

Wetted Perimeter = (4 + 2.60) + 2 * 1.67 = 9.94 m R = A / P = 5.51 / 9.94 = 0.554 m Average Velocity, Q / A = 8.35 / 5.51 = 1.52 m/sec Friction loss by Manning’s formula: = (V2Ln2) / R4/3 = [(1.52)2*3.8*2.56*10-4 ] / 0.445 =0.00494 m II.

Friction loss in straight reach Area = 2.6 * 1.67 = 4.342 sq.m Wetted Perimeter = 2( 2.6 + 1.67 ) = 8.54 m R = A / P = 4.342 / 8.54 = 0.51 m Velocity, V = 8.35 / 4.342 =1.923 m/sec. Hf = [(1.923)2 * 80 * (0.016)2 / (0.51)4/3 = 0.186 m

III.

Friction loss in bend Average area = [(2..6 + 2) / 2]*1.67 = 3.84 sq.m Wetted perimeter = 2.6 + 2 + 2 * 1.67 = 7.94 m R = A / P = 3.84 / 7.94 = 0.48 m Velocity, V = 8.35 / 3.84 =2.17 m/sec. Hf = (V2Ln2) / R4/3 = [(2.17)2 * 14 * (0.016)2 / (0.48)4/3 = 0.045 m

IV.

Friction loss in remaining length of tunnel Area = 1.67 * 2 = 3.34 sq.m Wetted perimeter = 2( 1.67 + 2 ) = 7.34 m R = A / P = 3.34 / 7.34 = 0.455 m Velocity, V = 8.35 / 3.34 =2.5 m/sec. Hf = (V2Ln2) / R4/3 = [(2.5)2 * 1 * (0.016)2 / (0.455)4/3 = 0.0046 m Total friction loss = I + II + III + IV =0.00494 + 0.186 + 0.045 + 0.0046 = 0.24054 M

V.

Loss at entry He = 0.2[(V12 – V22) / 2g]

62

Velocity at the entry = 8.35 / ( 4 * 1.67 ) = 1.25 m/sec He = 0.2[(1.9232 – 1.252) / 2*9.8] = 0.0218 m

VI.

Loss due to bend Hb = F * (V2 / 2g) * (Ø/180) Where, F = a coefficient which varies with radius and width of tunnel F = 0.124 + 3.104 *(S /2R)1/2 Ø = angle of deviation = 180 R = radius = 45 m S = width of tunnel = (2.6 + 2) / 2 = 2.3m F = 0.124 + 3.104 *(2.3 /2* 45)1/2 = 0.62

Hence head loss at the bend = 0.774 * (2.17)2 /19.6 *18 / 180 =0.0186 m VII.

Head loss due to change in velocity Hbe = 0.3 (V12 – V22) / 2g = 0.3 (2.52 – 1.9232) / 2 * 9.8 = 0.026 m

Hence total loss through tunnel 1 =0.24 + 0.0218 + 0.0182 + 0.026 = 0.306 m

The various head losses for all the tunnels are similarly calculated and are given in the next table.

63

Tunnel No.

Cross Head loss due to Section (m*m) Friction Entry (m) (m)

Bend (m)

*1.67

Total head loss Velocity (m) change in bend (m)

Average velocity inside tunnel (m/sec)

Max. velocity inside tunnel (m/sec)

1

2.6

0.24

0.0218

0.0186

0.026

0.306

1.92

2.5

2

2.45

0.247

0.0265

0.016

0.032

0.323

2.04

2.5

3

2.3

0.2475

0.032

0.0175

0.0135

0.3105 2.17

2.5

4

2.15

0.235

0.039

0.0188

0.0885

0.3015 2.32

2.5

5

2.0

0.216

0.0478

0.0203

0.00

0.284

2.5

2.5

6

1.85

0.185

0.0584

0.022

0.0159

0.28

2.7

2.5

Average Total Head loss = 0.32 m 4.2.5 Tunnel Layout To trap a major portion of coarse material, the tunnel bed level is kept such that the top slab is flush with the sill at the head regulator. If the thickness of top slab is 0.2 m then the bottom of the tunnel is 1.73 m from the regulator sill. (ii) Approach To increase the zone of suction at the upstream mouth, bell mouthing of the tunnels has been done according to x2 / (0.75)2 + y2 / (0.25)2 =1 The radius of bell mouthing in plan varies from 2 to 8 times the tunnel width, the radii increasing for unnels away from canal head regulator. (iii) Exit The tunnels have been throttled at the exit to increase the velocity to prevent sediment deposits.

64

(iv) Bend Radius It is kept 8 to 18 times the tunnel width. In this design it is kept varying from 10 to 17 times the tunnel width. (v) The top slab has been protruded into the river by about 1.07 metre at the entry, to increase suction effect of the tunnels to draw in more sediments. The protrusion has been extended and elliptically shaped at the entry. 4.2.6 Escape Channel NO special outfall channel is required as the sediments and escape discharge will pass down the barrage.

65

66

CHAPTER 5

DESIGN OF CANAL HEAD REGULATOR

5.1 Fixation of Crest level and water way Full Supply discharge Anticipated maximum full supply level of canal Bed level of canal Safe exit gradient for canal bed material

= 250 cumecs = 201.10 m = 196.0 m = 1/5

The crest level of canal head regulator is kept 1.2 - 1.5 m higher than crest level of undersluices. The crest level of undersluices = 195.85 m Pond level = 201.35 m u/s H.F.L = 202.17 m The crest level of regulator is kept 1.5m high than undersluices. As silt excluder is used, raise crest level by 1 m and further by 1.05 m. Crest level of regulator = 195.85 + 1.5 + 1 + 1.05 = 199.4 m

Fig 5.1 (Fixation of Crest Level and Waterway) Now fix the waterway for regulator, such that the full supply discharge of 250 cumecs can pass through it.

Discharge ‘Q’ through regulator is given as,

67

Q = 2/3 C1 X L X √2g {(h+h1)3/2 – ha3/2} + C2 X L X h1X √(2g X (h + ha)) Here, C1 = 0.577, C2 = 0.80 Neglecting head due to velocity of approach, ‘ha’ Here, Q = 250, h = 0.25, h1 = 1.7 Now, 250 = 2/3 X 0.577 X L X √(2 X 9.8)2 X (0.25)3/2 + 0.80 X L X 1.7 X √(2X9.8X0.25) 250 = 0.212 L + 0.3010 L ∴L = 77.59 ~ 77.6 m Provide 10 bays of 7.8 m each, giving a clear water way of 78 m. Provide 9 piers of 1.5 m each ∴Overall water way of regulator = 78 + (9 X 1.5) = 91.5 m

5.2 Hydraulic conditions for various flow conditions (i) Full supply discharge passing down regulator during high flood When u/s water level is 202.17 m, water shall pass over the regulator and the gated opening provided between the sill level and pond level shall have to be adjusted by partially opening this gate.

Fig 5.2 (Hydraulic Conditions) Let the gate opening be ‘x’ meters. The discharge can then be calculated by submerged orifice formula i.e, Q = Cd X A X √(2gh) Here, Q = 250 cumecs A = L X x = 78.x m2 Cd = 0.62 h = head causing flow = 202.17 – 201.10 = 1.07 m

68

∴250 = 0.62 x 78.x X √(2X9.8X1.07) x = 1.13 m Velocity of flow through opening, v = 250/(78 X 1.13) = 2.83 m/s Loss of head at entry

= 0.5 X v2/2g = 0.5 X (2.832/(2X9.8))= 0.204m

T.E.L just u/s of gate

= 202.17 + 0.199 = 202.37 m

T.E.L just d/s of gate

= 202.37 – 0.204 = 202.16 m

d/s water level

= 201.10 m

Head Loss, HL

= 202.16 – 201.10 = 1.06 m

Discharge intensity, q

= 250/78 = 3.20 cumecs/m

(ii) Full supply discharge passing down regulator at pond level Head Loss, HL = 201.35 – 201.10 = 0.25 m Discharge intensity, q = 3.20 cumecs/m

69

5.3 Hydraulic jump calculations for two conditions are tabulated as follows :

ITEM

High Flood Pond Level Flow Condition Flow Condition

1.)

Discharge Intensity(q)

3.20cu/m

3.20cu/m

2.)

Upstream Water level

202.17m

201.35m

3.)

Downstream Water level

201.10m

201.10m

4.)

U/S T.E.L

202.17m

201.35m

5.)

D/S T.E.L

201.10m

201.10m

6.)

Head Loss(HL)

1.06m

0.25m

7.)

D/S Specific Energy(Ef2)

2.05m

1.9m

8.)

U/S Specific Energy(Ef1= Ef2+HL)

3.11m

2.15m

9.)

Level at which jump will form(D/S T.E.L- 199.05m Ef2)

199.2m

10.)

Prejump depth D1 corresponding to Ef1

0.5m

0.6m

11.)

Postjump depth D2 corresponding to Ef2

1.80m

1.7m

12.)

Length of concrete floor required 6.5m beyond jump 5(D2- D1)

5.5m

13.)

Froude’s Number,F=q/√gD13

2.19

S.NO

2.89

Table 5.1

5.4 Depth of sheet piles from scour considerations Discharge intensity, q

= 3.20 cumecs/m

Depth of scour, R

= 1.35 X (q2/f)1/3 = 1.35 X (3.202/1)1/3 = 2.93 m

(i) d/s sheet pile Provide d/s cutoff upto 1.5R below d/s water level = 1.5 X 2.93 = 4.39 m ∴R.L of bottom of d/s cutoff = 201.10 – 4.39 = 196.71 m

70

This value gives only 199 – 196.71 = 2.29 m deep cut off, which is small. So, let us provide d/s cutoff to a bottom level of 194.7 m i.e,4.3 m deep below the d/s floor level. (ii) u/s sheet pile Provide u/s sheet pile line to the elevation of 188.7 m (i.e, same as that of undersluices) which is 7.15 m deep below level. 5.5 Total floor length and exit gradient The worst condition of flow occurs when the maximum flood is passing in river (u/s water level = 202.17 m) and there is no water in the canal (i.e, canal is completely closed). The bed level of d/s floor = 199 m. Maximum static head under this condition, H = 202.17 - 199 = 3.17 m Depth of d/s cutoff (d) = 4.3 m GE = 1/5 Now, = (H/d) X (1/(π√λ)) GE 1/5 = (3.17/4.3) X (1/(π√λ)) = 0.180 ∴1/(π√λ) From Khosla’s exit gradient curve, for 1/(π√λ) = 0.180, α = 4.5 ∴ b = α X d = 4.5 X 4.3 = 19.55 Adopt total floor length = 20 m The floor length shall be provided as below : d/s horizontal length =7m d/s glacis length with 3:1 slope = 3(199.4 – 199) = 1.2 m crest width =2m balance length provided at u/s floor = 9.8 m

Fig 5.3 (Total Floor Length)

71

5.6 Uplift Pressures Let us assume 1.0 m floor thickness on u/s and 1.5 m on d/s as shown in above figure. (i) u/s pile b = 20 m d = 7.15 m 1/α = d/b = 7.15/20 = 0.357 φD1 = 100 – 33 = 67% φC1 = 100 – 49 = 51% Correction for depth

= {(67 – 51) / 7.15} X 1 = 2.23% (+ve)

Correction for interference due to d/s sheet pile = 19 √(D/b’) X (d+D)/b Here, b = 20 m b’ = 18.5 m d = 6.15 m D = 194.85 – 194.7 = 0.15 m ∴Correction = 19√(0.15/18.5) X (6.15+0.15)/20 = 0.405% (+ve) ∴Corrected φC1 = 51 + 2.23 + 0.405 = 53.63% (ii) d/s pile b = 20 m d = 2.3 m 1/α = d/b = 2.3/20 = 0.115 φD = 20% φE = 30% Correction for depth

= {(30 – 20) / (199 – 194.7)} X 1.5 = 1.948% (-ve)

Correction for interference due to d/s sheet pile = 19 √(D/b’) X (d+D)/b Here, b = 20 m b’ = 18.5 m d = 2.8 m D = 197.5 – 188.7 = 8.8 m

72

∴Correction

= 19√(8.8/18.5) X (2.8+8.8)/20 = 4.60% (-ve)

∴Corrected φC1 = 30 - 1.48 – 4.60 = 23.92% The levels of the H.G lines at key points for different flow conditions are as tabulated in following table :

Condition flow

of Upstream water Level(m)

Downstream water level(m)

Head(m) Height/Elevation os subsoil H.G line above Datum

No, flow 202.17 Maximum static head

199.0 (No Water)

3.17

High Flood on 202.17 barrage and full supply discharge through regulator Flow at pond 201.35 level

201.10 (Canal F.S.L)

1.07

201.10 (Canal F.S.L)

0.25

Upstream Pile LIne

Downstream Pile Line

ØE1

ØD1

ØC1

ØE

ØD

ØC

100%

67%

53.63%

23.92%

20%

0%

3.17

2.12

1.70

0.758

0.634

0

202.17

201.1 2 0.716

200.70

199.75

199.63

199

0.573

0.255

0.214

0

202.17

201.8 1

201.67

201.35

201.31

201.1 0

0.25

0.167

0.134

0.059

0.05

0

201.35

201.2 6

201.23

201.16

201.15

201.1 0

1.07

Table 5.2

Floor thickness The maximum static head on the floor occurs for the worst condition, i.e, when high flood is passing down the barrage but the canal is empty and regulator gates are closed. ∴maximum static head = 202.17 – 199 = 3.17 m The subsoil H.G line shall be drawn for the maximum static head only, since the floor thicknesses are governed by this critical condition.

73

74

Thickness required at d/s end of floor = 0.758/1.24 = 0.62 m Thickness required at 2 m from end As it is very less, provide thickness

= 0.92/1.24 = 0.75 m = 1.75 m

Thickness required at 4 m from end As it is very less, provide thickness

= 1.08/1.24 = 0.9 m = 3.25 m

Thickness required at 7 m from end As it is very less, provide thickness

= 1.33/1.24 = 1.1 m = 5.25 m

5.7 Protection works (i) u/s protection It shall be equal to what was provided in the u/s of undersluices. (ii) d/s protection Normal scour depth, R Anticipated scour The level of d/s scour hole

= 2.93 m = 2R = 2 X 2.93 = 5.86 m = d/s water level – 2R = 201.10 – 5.86 = 195.24 m Scour depth (D) below d/s floor = 199 – 195.24 = 3.76 m Let us provide a launching apron in a length equal to 1.5 D (say 5.65 m) and of thickness = = (2.25 X 3.76)/5.65 = 1.49 ~ 1.5 m Use 6 m length of 1.5 m thick launching apron. Let us use 1.3m X 1.3m X 0.75 m c.c blocks in a length equal to approx. 1.5D i.e, 5.65 m. Hence, use 4 rows of 1.3m X 1.3m X 0.75m c.c blocks with 10 cm gaps filled with bajri, in a total length of 5.5 m.

75

76

ANNEXURE 1

CURVES USED IN DESIGN

1. Blench Curves

2. Khosla’s curve for cutoff at d/s end

77

3. Khosla’s curve for exit gradient

4. Curve for plotting post jump profile

78

5. Montague’s curve

79

DISCUSSIONS 1. The barrage has been designed independently by only considering the required data. 2. There has been some changes made in the original specifications of the barrage as per the requirements. 3. The dimensions and number of gates of the undersluice and other barrage bays have been changed corresponding to the economic conditions. 4. The crest level of head regulator has been increased more as a silt excluder has to be provided. 5. In the design of silt excluder, the tunnels have been provided in one bay of undersluice only, in contrast to the two bays in the original design. As, the silt content in the Yamuna river decreases to some extent upon reaching that portion of Delhi.

80

REFERENCES

1. Irrigation Engineering and hydraulic Structures by S.K Garg. 2.Theory and design of Irrigation Structures by Varshney and Gupta. 3. Irrigation and Water Power Engineering by B.C Punmia and Jain & Jain. 4. Indian Standard codes 4410, 6966 and11130. 5.References from Internet.

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