Irrigation Engineering and Hydraulic Structures 3031335511, 9783031335518

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S. K. Ukarande

Irrigation Engineering and Hydraulic Structures

Irrigation Engineering and Hydraulic Structures

S. K. Ukarande

Irrigation Engineering and Hydraulic Structures

S. K. Ukarande K J Somaiya Institute of Technology Affiliated to University of Mumbai Mumbai, India

ISBN 978-3-031-33551-8 ISBN 978-3-031-33552-5 (eBook) https://doi.org/10.1007/978-3-031-33552-5 Jointly published with ANE Books India The print edition is not for sale in South Asia [India, Pakistan, Sri Lanka, Bangladesh, Nepal and Bhutan] and Africa [all countries in the African subcontinent]. Customers from South Asia [India, Pakistan, Sri Lanka, Bangladesh, Nepal and Bhutan] and Africa [all countries in the African subcontinent] please order the print book from: Ane Books Pvt. Ltd. ISBN of the Co-Publisher’s country edition: 978-93-8365-689-9. 1st edition: © Author 2018 © The Author(s) 2023 This work is subject to copyright. All rights are solely and exclusively licensed by the Publisher, whether the whole or part of the material is concerned, specifically the rights of translation, reprinting, reuse of illustrations, recitation, broadcasting, reproduction on microfilms or in any other physical way, and transmission or information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed. The use of general descriptive names, registered names, trademarks, service marks, etc. in this publication does not imply, even in the absence of a specific statement, that such names are exempt from the relevant protective laws and regulations and therefore free for general use. The publishers, the authors, and the editors are safe to assume that the advice and information in this book are believed to be true and accurate at the date of publication. Neither the publishers nor the authors or the editors give a warranty, expressed or implied, with respect to the material contained herein or for any errors or omissions that may have been made. The publishers remain neutral with regard to jurisdictional claims in published maps and institutional affiliations. This Springer imprint is published by the registered company Springer Nature Switzerland AG The registered company address is: Gewerbestrasse 11, 6330 Cham, Switzerland

Dedicated to My Parents, Wife Asmita and Daughter Prajakta

Preface This book, written in a very concise form as a textbook of Irrigation Engineering and Hydraulic Structures will be useful to undergraduate students. It will also be useful for elective subjects like: 1. Applied Hydrology and Flood Control and 2. Design of Hydraulic Structures. It is written in a simple language, so that average student will understand the subject with self-study. Unlike a research document, a textbook covers all basic useful information and gives in a palatable form to the students in a self-explanatory manner. In doing so, I have utilized existing state of knowledge available in the form of textbook, treatises and research publication by earlier contributors in these fields which I gratefully acknowledge. A list of such contributors is given in the references at the end. I do not claim any originality of basic concepts in this book. The manuscript is checked thoroughly to make this book, near to perfect. I will be grateful to the users/readers of this book to point out mistakes, if any, so that they can be rectified in the next edition of this book. It would be difficult to mention names of all the persons who have given their valuable help in making this project successful. However, I would like to place on record my gratitude towards Dr. P. R. Mehta, Retired Principal of SVRCE, Surat and Ane Books Pvt. Ltd. for publishing this book. Dr S. K. Ukarande

vii

Contents Prefaceiv 1.  Introduction

1–4

● 

Practice of Irrigation in India  ● Scope ●  Status of Irrigation in India   ●  Impact of Irrigation on Human Environment  ●  Irrigation Systems  ●  Command Area Development  ● Development of an Irrigation Project

2.  Water Requirement of Crops 5–25 ● Crops

and Crop Seasons ● Definitions of Terms Used in Calculation of Irrigation Demand  ●  Quality of Irrigation Water  ●  Soil Water Relationship  ●  Soil Characteristics From Irrigation Considerations  ●  Consumptive Use  ●  Field Irrigation Requirement (FIR)  ● Frequency of Irrigation  ●  Irrigation Methods  ●  Exercises

3.  Hydrology

26–68

● 

Scope of Hydrology  ● Rainfall ●  Surface Run–Off  ●  Flood Flow Calculation  ●  IUH, S and Synthetic Unit Hydrograph  ●  To Get Unit Hydrograph of Different Duration from that of Given Duration  ●  Rainfall/Flood Frequency Analysis  ●  Exercises

4.  Ground Water

69–94

● 

Ground Water Occurrence and Resources  ●  Well Irrigation  ●  Well Hydraulics ● Recuperation Test  ●  Well Shrouding  ●  Well Development  ●  Collector or Radial Well  ●  Well Losses  ● Unsteady Groundwater Flow Toward Wells  ●  Exercises

5.  Reservoir Planning

95–119

● Introduction  ●  Types

of Reservoirs  ●  Investigation for Reservoir Sites  ●  Zones of Storage in a Reservoir  ●  Sedimentation in Reservoir  ●  Useful Life of Reservoir  ●  Factors Affecting Selection of Reservoir Site  ●  Determination of Reservoir Capacity from Mass Curves  ● Reservoir Losses  ● Reservoir Flood Routing ● Fixing Height of Dam from Reservoir Working Table  ●  Exercises

6.  Design of Gravity Dams

120–154

● Introduction ● 

Forces Acting on Gravity Dams  ●  Theoretical Profile of Gravity Dam  ● High and Low Gravity Dam  ●  Factor of Safety for Design of Gravity Dam  ●  Practical Profile of Gravity Dam  ●  Drainage Gallery  ●  Joints in Gravity Dam  ●  Foundation Treatment  ●  Types and Function of Galleries  ●  Exercises

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x  Irrigation Engineering and Hydraulic Structures

7.  Spillways and Gates

155–177

● 

General   ●  Ogee Spillway  ●  Design of Stilling Basins  ●  Design of Ski-Jump Bucket and Roller Buckets  ●  Siphon Spillways  ●  Limiting Head of Siphon Spillway  ●  Spillway Crest Gates  ● Design of Air Vent for Vertical Lift Gates Provided in Supply Sluices of the Dam  ●  Cavitation and Vibration Problem in Hydraulic Structures  ●  Exercises

8.  Arch and Buttress Dams

178–189

● Introduction ● 

Types of Arch Dams  ●  Design of Arch Dam by Thin Cylinder Theory  ● Other Varieties of Arch Dam  ●  Buttress Dams  ●  Exercises

9.  Earth Dams

190–217

● 

Earth Dam and its Component Parts  ●  Estimation of Seepage Through Homogeneous Earth Dam  ● Design Considerations ● Causes of Failure of Earth Dam ● Filters: Criteria and Design  ●  Exercises

10.  Canal Head Works

218–240

● 

Barrage and Weirs  ●  Different Units of Canal Head Works  ●  Design of Weir  ●  Bligh’s and Khosla’s Theories  ●  Khosla’s Method of Independent Variables  ●  Salient Features of Design of Weir on Pervious Foundation  ●  Numerical Problems on Weir Design:  ●  Sediment Control at Canal-Head Works  ●  River Training Works for Canal Head Works  ●  Exercises

11.  Distribution System

241–262

●  General

  ●  Canal System  ●  Alignment of Canals  ●  Losses and Types of Cross-Section for Canal  ● Design of Alluvial Channels ● Waterlogging ● Canal Lining ● Design of Lined Canals  ●  Exercises

12.  Canal Structures

263–279

● 

Distribution and Measurement Structures for Canal Flows  ●  Canal Outlets  ● Cross-Drainage Works (C.D. Works)  ●  Exercises

13.  Sediment Transport in Alluvial Canals

280–295

●  Transport

of Sediment in Alluvial Canals  ●  Threshold Movement  ●  Shield’s Entrainment Function  ●  Bed Load Formula by Einstein  ●  Bed Load Formula by Meyer–Peter  ● Du-Bois Bed Load Formula  ●  Suspended Load Concentration  ●  Tractive Force Method of Design of Stable Channel  ●  Exercises

14.  River Training Works and Flood Control ● General ● 

296–303

Classification of Rivers  ●  River Training Works  ●  Flood Control  ●  Exercises

15.  River Flow Measurement

304–316

● 

Need for Discharge Measurement in River Section  ●  Current Meters  ●  Cross-Sectional AreaVelocity Method to Measure Discharge  ●  Slope Area Method  ●  Stage-Discharge Relationship

References 317–319 .  Standard Books and Treatises A B.  Technical Papers and Reports

Subject Index

317 318

321-322

1 Introduction 1.1  PRACTICE OF IRRIGATION IN INDIA Water is the most essential commodity required on earth without which both human and animal life cannot survive and that is why nature has covered three-fourth of earth’s surface with water in the form of oceans. It is evaporation from oceans and appropriate wind forces that bring rain on land and this becomes chief source of potable water for human as well as animal race. In addition to this, water in form of ice on high peaks of mountains, dew on open ground and ground water below earth’s surface is also available but human race prefers rain, as it is available freely as a gift from nature. But this gift of nature may not prove to be a boon each year, for it's uneven distribution and does not occur as and when required. There will be large regions of scanty and untimely rainfall every year. This makes cultivators helpless and sometimes they prefer to abandon farming. For an excellent yield from farms, what is required is occurrence of rain as and when needed by crops. This is seldom possible. Hence, it is from prehistoric time, man has attempted to store rain water and supply it to crops as per crop-requirement, resulting in good yield of crop. This practice of supplying stored water on to farms is known as ­irrigation and is being practiced since dawn of civilization.

1.2 SCOPE Modern times have made irrigation practice more systematic and reliable by constructing huge reservoirs and canal network with the help of hydraulic structures such as dams and spillways. Thus, irrigation engineering deals with storage and supply works to be constructed for the purpose of making water available to farms as and when needed and in the quantity required for optimum growth of the crops. A well-irrigated farm can produce hundred per cent yield from land whereas farms left to availability of rains will give yield as per success of monsoon in that region. Hence, farming as industry can develop only if the farms are lying in region covered with irrigation water supplies. Proper management of supply of © The Author(s) 2023 S. K. Ukarande, Irrigation Engineering and Hydraulic Structures, https://doi.org/10.1007/978-3-031-33552-5_1

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irrigation water will certainly give returns that will recover expenditure incurred on irrigation works as well as farming system. Thus, scope of irrigation engineering is as high as any other industry in this country. Recent emphasis is therefore more on managerial aspects rather than merely on storage and distribution of irrigation water. A well-managed system will utilize minimum water and will give maximum crop yield from same area of culturable land.

1.3  STATUS OF IRRIGATION IN INDIA India is the richest agricultural country as regards its ratio of agricultural land to geographical land is concerned. This ratio is 45% as against 10% of USSR and 25% of USA. Its total culturable land is 148 million hectares which is third largest in the world, 1st being USSR with 224 million hectares and 2nd being USA with 193 million hectares. The amount of annual surface water available in our country is 1800 × 103 MCM (Million cubic meter) of which utilizable amount is 660 × 103 MCM. Out of this utilizable amount, we have harnessed only 50% so far; remaining 50% still needs to be harnessed. One can imagine our progress if we can harness all of our water resources fully. One of the western economists has made an observation in 1950, that India is a rich country inhabited by poor people. We have yet to prove that this observation is wrong. Pre-British period, India used to practice irrigation on a small scale i.e., with the help of wells, tanks and bandharas. During British period some major projects like Upper Ganga Canal, Upper Bari Doab Canal and Krishna and Godavari Delta systems were taken up and at the time of pre-independence era more works like Sarda Canal, Krishnarajsagar dam etc. were taken up. After Independence, irrigation engineering got highest preference at government level and first five “Five Year” plans were irrigation projects oriented. Bhakhra Nangal, Hirakud, Chambal, Nagarjunsagar dam, Damodar Valley project, Ukai, Koyna and Idiki Dam projects were some of the major works, which got completed during first five “five year” plans i.e., from 1951 to 1975. With the completion of major irrigation projects the irrigated area which was 22.6 mha (million hectares) in 1951 has increased to 68 mha upto 1986, which is yet 45% of cultivable area of the country. While on irrigation front we have done noteworthy progress, yet on hydropower project works our rate of progress is very slow. At the beginning of first five year plan per capita power consumption in India was 15 kWh per head, the lowest in the world, which has subsequently increased to 175 kWh per capita during 1982-83, whereas it is 5300 kWh per capita in USA and 3040 kWh per capita in UK. We have yet to achieve a lot to come to world figures as regards hydropower development is concerned.

Introduction  3

1.4  IMPACT OF IRRIGATION ON HUMAN ENVIRONMENT Unscientific and uncontrolled use of irrigation water may result in ill-effects of irrigation. This creates a tremendous impact on human environment. The excess use of irrigation water may lead to creation of waterlogged areas and may breed mosquitoes. This in turn will pollute human environment. Climatic conditions may also change. Damp climate may prevail over areas where excessive and uncontrolled irrigation is being practiced. Excess water results in return of water to either ground water or to surface water i.e., rivers. The return water may be containing toxic chemicals left on account of use of pesticides on farming. This return water may therefore create pollution of river and ground water supplies. In order that irrigation impact on human environment does not create imbalance in natural environment, irrigation water should be used scientifically and in a controlled manner with the provision of proper escape for return water. Suitable land drainage should also be provided so that water logging does not take place.

1.5  IRRIGATION SYSTEMS There are two types of irrigation systems in use: 1. Flow irrigation system; 2. Lift irrigation system. 1. Flow irrigation system covers large areas under irrigation and hence is also known as major irrigation system. This system consists of storage works such as dams, spillways, weirs and head works and also distribution works such as canals, cross drainage works and outlets. It caters need of large command areas and overall development of irrigated land is very fast. 2. Lift irrigation system consists of well irrigation and area covered is small compared to flow irrigation and hence is also known as minor irrigation system. Large irrigation works create problem of land submergence and displacement of people whereas minor irrigation system is free of this nuisance.

1.6  COMMAND AREA DEVELOPMENT Irrigation potential on account of major and minor schemes implemented after independence has been increased from 22 Mha in 1950 to 80 Mha in 1995 and is likely to reach a target of 110 Mha in ­coming decades. However, available irrigation potential remains under utilized due to lack of co-ordination between various agencies existing in a command area. The second irrigation commission and national commission on agriculture recommended an integrated command area development programs for ­optimizing benefits from available irrigation potential.

4  Irrigation Engineering and Hydraulic Structures

Command Area Development Authorities (CADA) have been set up in 1973 to fulfill this need. Main objectives of the program are: 1. Optimization of yield by best crop pattern. 2. Avoiding wastage and misuse of water 3. Increasing area of irrigated land by land development and water managed program. 4. Providing of field channels for equitable distribution of water to farmer’s land. Integrated command area development is the major field of work of CADA. Different agencies, like, State Department of Irrigation, Agriculture, Animal Husbandry, Agriculture Finance Corporation and Commercial Banks etc. should work in co-ordination with CADA.

1.7  DEVELOPMENT OF AN IRRIGATION PROJECT Irrigation Projects aim at supply of controlled amount of waters to cultivable land in order to raise its yield and betterment of the command area. These projects consist of hydraulic structures, which collect and store water from catchment areas and supply it by means of gravity flow to command area to increase agriculturable outputs. These projects are considered feasible only when the total amount of estimated benefits of the projects exceed its total estimated cost. The feasibility of irrigation project is determined on the basis of preliminary estimates of area of land to be irrigated, its water requirements, available water supplies, productivity of irrigated land, and required engineering works. A large irrigation engineering project may take 10-20 years for its completion and the results of its estimated benefits will be felt there after.  

 

2 Water Requirement of Crops 2.1  CROPS AND CROP SEASONS There are three crop seasons in India : Kharif, Rabi and Hot Weather. Besides these three seasons, there are other two overlapping seasons like eight months season which overlaps hot weather and kharif and perennial season which overlaps all three seasons i.e., Kharif, Rabi, and Hot weather. (i) Kharif season starts around middle of June and ends around middle of October, having a base period of 120 days and has major crops like rice, maize, jawar, bajra (millets), groundnuts etc. (ii) Rabi season starts from middle of October and ends around middle of February having a base period of 120 days. It has major crops like wheat, gram, potatoes, tobacco, pulses etc. Rabi crops require less water than Kharif crops. (iii) Hot weather season starts from middle of February and ends around middle of June. It has a base period of 120 days and its major crops are vegetables, fodder crops, fruits etc. (iv) Eight months season starts from March and lasts upto October with a base period of around 200 – 220 days, and its chief crop is cotton. (v) Perennial season spreads over whole year and has a base period of around 300 – 330 days, its chief crop is sugarcane.

© The Author(s) 2023 S. K. Ukarande, Irrigation Engineering and Hydraulic Structures, https://doi.org/10.1007/978-3-031-33552-5_2

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6  Irrigation Engineering and Hydraulic Structures

Table 2.1:  Salient Features of Crop Seasons. Crop Season with base period *

Crops

Seed Requirement (kg/hect.)

Yield (kg/hect.)

Delta (cm)

Kharif (15 June to 15 Oct.) B = 120 days

Rice Maize Jowar Bajra Groundnut

40 40 30 10 100

4000 3000 3000 3000 2500

200 20 20 20 25

Rabi (15 Oct. to 15 Feb.) B = 120 days

Wheat Gram Potato

100 30 1500

3000 3000 30000

30 30 50

Hot weather (15 Feb. to 15 June) B = 120 days

Fodder Crop

_

_

20

Eight Month (March – October) B = 220 days

Cotton

15 – 20

200 – 500

50

Perennial (Nov. – Oct.) B = 300–330 days

Sugarcane

500

25000

100

* Dates given for beginning and end of irrigation season are on average basis of large variation at country level, however, base period in days, season-wise, is almost same throughout the country. Similarly Delta, Yield etc. are on average basis.

2.2 DEFINITIONS OF TERMS USED IN CALCULATION OF IRRIGATION DEMAND 1. Crop Pattern and Crop Ratio: Crop pattern indicates percentage distribution of different crops area-wise in a crop season. It depends on soil characteristics, climatic conditions and local requirement of crop. Command Area Development Authority (CADA) may fix percentage of Kharif and Rabi crops known as crop ratio. Crop ratio =

Area allocated to kharif crop Area allocated to rabi crop = 1 : 2 (usually), i.e., kharif area is half the rabi area

2. Catchment and Command Areas: Catchment area is the area over which whatever rain occurs, flows by gravity towards a river or a reservoir for storage and then distribution is carried out for irrigation purposes.

Water Requirement of Crops  7

Command area is the area to which stored water is made available by canal network for irrigation purposes. Thus, catchment area indicates source whereas command area indicates utilization. 3. Kor Period: First watering given to crop after it has attained height of few cm is called kor watering and specific period over which kor watering must be applied to crop is kor period. It is 2 to 4 weeks for rice and 3 to 6 weeks for wheat. 4. Base Period (B in days): Base period of a season is counted in number of days commencing from pre-ploughing watering to last watering before harvesting. 5. Crop Rotation: When same crop is repeated season to season, soil may suffer of fertility and numerous insects will get developed. To check this, crop rotation is adopted i.e., same crop is not raised again and again on same land but different crops by rotation are raised such that land does not loose its fertility and pest control is also achieved. Crop rotation, thus, helps in increasing crop yield and keeps the land fertile. Following are some suggested crop rotation: (i) Wheat – Jowar – Gram

(ii)  Rice – Gram

(iii) Cotton – Wheat – Gram – Sugarcane. 6. Paleo Irrigation: At the time of seeding for Rabi crops, weather may be dry and hot and as such soil may be dry. Hence, watering is required prior to ploughing which is known as paleo irrigation or simply paleo. 7. Duty (D): Duty is defined as irrigating capacity of one cumec of water supplied throughout the base period of the given crop. If one cumec of water is supplied through out the base period to irrigate 700 hectares of land for the crop rice, then duty of water for rice is 700 hectares / cumec. Thus, duty helps in arriving at total requirement of water for maturity of given crop over the given area and this helps in fixing canal carrying capacity and also irrigation demand for fixing reservoir capacity.It is generally expressed in hectare/cumec. 8. Delta (Δ): Total depth of water required from seeding to harvesting of a given crop is defined as Delta (Δ) and is expressed in centimeter or meter. 9. Relationship of Duty, Delta and Base Period: If a crop, grown on D hectares of land is supplied with water continuously over its base period of

8  Irrigation Engineering and Hydraulic Structures

B days at the rate of 1 cumec then total volume of water supplied to the crop is given by V = 1 × 60 × 60 × 24 × B m3 = 86400 B m3



and area = D hectares



= D × 104 m2

If total depths of water over this area is Δ m then ∆=

∴∆ =

86400 B D × 104 8.64 B m D



Where, B = base period in days



and   D = duty in hectare / cumec.

If Δ is expressed as total depths in cm, then ∴ ∆=

864 B cm D

10. Outlet Factor: Field channels or water courses supply water to the land to be irrigated. Field channels receive their discharge through canal outlets. The duty of water at the outlet is known as outlet factor, and is expressed in ha/cumec. 11. Capacity Factor: A canal is designed to carry certain maximum discharge, but it may not carry the discharge at all the time. Ratio of supply discharge of a canal to its maximum discharge capacity is defined as capacity factor. It is around 0.8. 12. Time Factor: During watering period, canal must supply water for all days of watering, but it may so happen that due to certain unavoidable circumstances it may not work for all these days but may work for less number of days. Thus, the ratio of actual number of days the canal work during watering period to the total number of days of watering period is defined as time factor. For example, if total number of days for watering period is 20, but the canal has actually run for 12 days then time factor is (12/20). 13. Overlap Allowance: Crop of one season sometimes overlaps into next season and in that case some extra water in the overlapped period may be required, which is known as overlap allowance it is expressed as 5 to 10%.

Water Requirement of Crops  9

14. Water Conveyance Efficiency: It is the ratio of water supplied on field to the quantity of water diverted to canal from the reservoir. ηc =

wf × 100 wr

Here ηc = Water conveyance efficiency Wf = Water supplied on field Wr = Water diverted to canal from the reservoir

q  SOLVED EXAMPLES Example 2.1:  The base period of rice is 120 days, and duty is 900 Ha/cumec. Find delta. Solution: D=

864 B 864 ´120 = =115 cm. D 900

Example 2.2:  A head regulator releases water at the rate of 5 cumecs. If duty at the field is 100 Ha/cumec and transit losses are 30%, find area of land irrigated. Solution:  With transit loss of 30%, One cumec will irrigate 0.7 (100) hectares of land; hence, 5 cumec will irrigate 5 × 70 = 350 hectares of land. Example 2.3:  A reservoir with a live storage of 300 MCM is able to irrigate a land of 40,000 Ha, with 2 fillings each year. The crop season is 120 days. What is duty? Solution: D=

40000 6

 (2 × 300 ×10 )     (120 × 60 × 60 × 24) 

=

40000 × 120 × 86400 2 × 300 × 106

= 691.2 Ha / cumec

2.3  QUALITY OF IRRIGATION WATER Generally, potable water is considered suitable for irrigation purposes. Rain water which is stored in ­reservoir and made available for irrigation purposes is suitable and safe if it is free of following impurities: (i) Sediments in suspension (ii) Soluble salts (iii) Higher proportions of sodium ions. (iv) Presence of toxic elements present in water.

10  Irrigation Engineering and Hydraulic Structures

Table 2.2:  Standards for Irrigation Water Class of Water and Suitability Class I Suitable for Irrigation Class II Suitable for Permeable soils Class III Not suitable for irrigation

Electrical Conductivity in Micro MHO / cm

Total Dissolved Salts in ppm

Sodium % of Cations

Baron (ppm)

0 – 1000

0 – 700

0 – 60

0 – 0.5

1000 – 3000

700 – 2000

60 – 75

0.5 – 2.0

3000 and more

2000 and higher 75 and higher

Over 2.0

Well water having high contents of salinity may not be useful for irrigation purposes. River waters are generally free of salinity problems. The following classification given in Table 2.2 of irrigation water as published by USBR may be useful to judge the quality of irrigation water.

2.4  SOIL WATER RELATIONSHIP Irrigation water supplied to land is held in soil mass in three different forms: (i) Gravitational Water

(ii)  Capillary Water

(iii) Hygroscopic Water (i) Gravitational water drains out freely under the influence of gravity and is not useful for plant growth. (ii) Water content retained in the soil after the gravitational water has drained off is known as capillary water, which is held in the soil by surface tension. Plant roots absorb this water gradually and thus it is capillary water in root zone of plant that is responsible for plant growth. Upper limit of capillary water is known as field capacity. Limit upto which capillary water is available for plant growth is known as permanent wilting point (PWP). After wilting point has reached water is held by adhesive forces of soil particles and hence not available to plant. This water in soil mass is known as hygroscopic water. Field Capacity =

Weight of moisture content of soilsample Weight of dry sample

Water Requirement of Crops  11 .......................................... .......................................... .......................................... .......................................... .......................................... .......................................... .......................................... .......................................... .......................................... .......................................... .......................................... .......................................... .......................................... .......................................... .......................................... .......................................... .......................................... .......................................... .......................................... .......................................... .......................................... .......................................... .......................................... .......................................... .......................................... .......................................... .......................................... .......................................... .......................................... .......................................... .......................................... ..........................................

Water readily available to plant

GL Gravitational water Field capacity Capillary water

Wilting point Water not available to plant

Hygroscopic water

Fig. 2.1:  Soil water relationship Let d = depth of root zone ws = weight density of soil w = weight density of water S = specific gravity of soil = ws / w for unit area of soil, weight of soil = ws × 1 × d weight of water held by soil = w × 1 × depth of moisture FC = Field Capacity =

w × 1× depth of water held bysoil w s ×1× d

Depth of water held by soil of unit area of field capacity, w = s d (FC) = S d (FC) w Similarly, depth of water held by soil at permanent wilting point (PWP) = S d (PWP) Depth of available water for plant growth = S × d × (FC – PWP)

q  SOLVED EXAMPLE Example 2.4:  Root zone of a soil has FC of 30 % and wilting point 12 %. What is the depth of moisture in root zone? How much water is available if root zone depth is 1.0 m, specific gravity of soil is 1.4? Solution:  Depth of moisture = S d (FC) = 1.4 × 1 × (0.30 / 100) = 0.42 m.  

Depth of available water = S d [FC – PWP] = 1.4 × 1 × [0.30 – 0 .12] = 0.252 m  

12  Irrigation Engineering and Hydraulic Structures

2.5 SOIL CHARACTERISTICS FROM IRRIGATION CONSIDERATIONS Soil consists of finely divided organic and mineral matters and they are classified as: 1. Alluvial Soil: This is the largest and most important type of soil of India. It consists of alluvials derived from deposition caused by Indus, Ganga and Brahmaputra river systems. Such soils are very fertile and if provided with irrigation facilities, their production is remarkably very high. For example, after construction of Bhakra – Nangal Project, Punjab and Haryana and North Rajasthan have attained highest agricultural yield and these areas are known for green revolution that they have undergone due to irrigation facilities, over post independence era. 2. Black Soils: This is a typical soil derived form Deccan Trap and is known as “black cotton soil”. Maharashtra, M.P., Andhra Pradesh, and south Gujarat region possess this type of soil. These soils are suitable for cultivation of rice, sugarcane and cotton. It swells when wet and cracks when dry and has poor drainage property. 3. Red Soils: They are formed due to weathering of crystalline rocks. Such soils are found in Tamil Nadu, Karnataka, Goa, M.P., Odisha, and Bihar. It is also found in West Bengal and UP. Its natural fertility is quite high, as can be seen from large forest areas that have developed in such soil regions. 4. Laterite Soils: Large regions of southern states like Tamil Nadu, Kerala, Karnataka possess these soils. It is good for rice and coconut cultivation. 5. Soil Characteristics: Irrigation practices are generally influenced by soil characteristics, such as physical properties, chemical properties and soil water relationship. 6. Physical Properties: Permeability of soil is as important as adequate supply of nutrients and water. Permeability of soils refers to movement of water through soil and hence it is dependent on porosity. Porosity is dependent on soil texture and soil structure. Soil texture is decided by its particle size, for example clayey soils have clay particles, (smaller in size than 0.002 mm), in abundance and sandy soils have sand (0.05 to 1.0 mm) particles predominating over other particles. Loam soils possess equal amount of sand, silt and clay particles. Texture of soil affects flow of water, aeration of soil and rate of chemical transformation. 7. Soil Structure: Soil structure is decided by pore space between soil particles. Fine textured soils offer favorable soil structure permitting retention of water, proper movement of air and penetration of roots, which is essential for plant growth. Porosity of soil is a good measure of its structure. Porosity of soils is

Water Requirement of Crops  13

the ratio of volume of voids present in soil sample divided by total volume of soil sample i.e. Porosity, n =

V void G =1 − b , V sample Gs

Where, Gb = Bulk density of soil and



Gs = Relative density of soil

q  SOLVED EXAMPLE Example 2.5:  A moist soil sample has volume of 484 cm3 in natural state and weight of 7.94 N. Dry weight is 7.36 N, specific gravity 2.65. Find porosity, soil moisture content, degree of saturation, S. and volumetric moisture content. Solution: G b = Bulk density of soil sample = Porosity, n = 1 −



7.36 484 × 10−6 × 9810

=1.55

Gb 1.55 =1− = 41.5% Gs 2.65

Weight of moist soil sample − Weight of dry soil sample Weight of dry soil sample 7.94 -7.36 = Soil moisture = 0.0788 7.36 Volumertic moisture = Soil moisture × Bulk density of soil sample Volumetric moisture = 0.788 x1.55 Soil moisture =

= 12.21% Volumetric moisture Degree of saturation = S = Porosity 12.2 = = = 29.43% 41.5

2.6  CONSUMPTIVE USE Water evaporated from leaves and stems of a plant is called transpiration and that evaporated from adjacent soil around the plant is known as evaporation. Water requirement of crop thus consists of evapotranspiration or consumptive use. Consumptive use is thus defined as the amount of water needed to meet the water loss through evapotranspiration and is measured as depth of water on irrigated area. Knowledge of consumptive use helps to determine irrigation requirement of crop, which is the difference between consumptive use and effective

14  Irrigation Engineering and Hydraulic Structures

precipitation. Evapotranspiration, ET, can be correlated to pan evaporation PE by ET = K (PE) Where K = Crop factor for given period. PE is available from Pan evaporation data published by meteorological department. Values of K are 0.5 for sugarcane and 0.2 for rice, wheat, maize etc Blaney – Criddle Formula gives consumptive use of crop and is used extensively. Let u = consumptive use of crop in mm then, u = 25.4 kf k = empirical consumptive use coefficient. f = consumptive use factor, given by p (1.8t + 32) f= 100 here, p = percentage of day light hours of the year occurring during the crop period and t = mean temperature in °C for the period.

2.7  FIELD IRRIGATION REQUIREMENT (FIR) Plant growth can be divided into three stages, (i) Vegetative, (ii) Flowering and (iii) Fruiting. Consumptive use of plant continuously increases during vegetative stage, attains peak during flowering stage and ­decreases during fruiting stage. Vegetables are harvested during vegetative stage and crops like rice, beans, potatoes, bananas etc. are harvested during fruiting stage. Not all the rainfall can be stored in root zone of the soil. Part of it goes away as surface runoff, and a part goes back to atmosphere as evaporation. Thus, effective precipitation is only that part of ­precipitation, which is available to plants as soil moistures; and this is obtained by subtracting run off, evaporation and percolation losses from the total precipitation. If consumptive use exceeds effective precipitation, and that is the normal case, the difference has to be met by irrigation water. Thus, FIR, field irrigation requirement, is given by FIR (mm) =

Det − (D p − Dlosses ) Ea

Where, Det = Depth of evapotranspiration, mm Dp = Depth of precipitation, mm Dlosses = Run off, evaporation and percolation into grounds. Ea = Efficiency of irrigation application. Average depth added to root zone storage = Average depth applied to field

= 60 to 80% depending on type of soil.

Water Requirement of Crops  15

 (Dp – Dlosses) = effective rainfall = 0.8 Dp – 25, if Dp > 75 mm per month = 0.6 Dp – 10, if Dp < 75 mm per month.

q  SOLVED EXAMPLES Solved Examples for calculation of evapotranspiration loss, and field irrigation requirement,: Example 2.6:  Determine evapotranspiration loss and field irrigation requirement for wheat of irrigation efficiency 0.8 and average percentage P of monthly sunshine hours is 0.7 from following data:

Months

Pe (effective Rainfall in cm)

t˚C p(4.6t + 81.3) (cm) (monthly temp) f =

100

November

2.8

20

12.11

December

3.0

16

10.84

January

3.5

14

10.20

February

2.7

15

10.52

∑ f = 43.67

∑ Pe = 12.00

Solution:

V = evaporation loss = k∑f, (k = 0.8, given)



= 0.8 (43.67) = 35 cm FIR =



V − ΣPe 35 − 12 = = 28.75 cm 0.8 η

Example 2.7:  Nine litres of water were required to be added to evaporation pan, 1.22 m in diameter to make up water level. If recorded rainfall during this period was 4.4 mm, find e­ vaporation from pan. Solution:

9 lit =

π 9 cum = (1.22) 2 × h 4 1000 ∴ h = Depth of water in pan = 7.7 × 10–3 m = 0.77 cm

4.4 10 = 1.21 cm

∴ Evaporation from pan = 0.77 +

16  Irrigation Engineering and Hydraulic Structures

Example 2.8:  Using the data given in the following table, find FIR month-wise if irrigation efficiency is 60% (i.e., Ea = 0.6) Given Month col. (1)

To Find Out

K col. (2)

PE (mm) (3)

(Dp – DLosses) mm (4)

Det (mm) = K (PE) (5)

FIR (mm) = [(5) – (4)] / Eff.

Nov.

0.2

118

6.0

23.6

29.33

Dec.

0.36

96

16.0

34.5

30.93

Jan

0.75

90

20.0

67.5

79.17

Feb

0.9

105

15.0

94.5

132.50

Mar.

0.8

140

2.0

112.0

183.33

Total FIR = 455.26 mm Δ = 4.5 m

Det = K (PE) = 0.2(118) = 23.6

Solution: FIR = =

Det − (D p − Dloss ) Ea

23.6 − (6) 17.6 = = 29.33 mm (for Nov.) 0.6 0.6

Similarly for remaining months, calculate FIR and enter in the last column of the table. Total FIR = 455.26 mm.

2.8  FREQUENCY OF IRRIGATION Frequency of irrigation is defined as interval of days between two successive waterings. Irrigation water is to be applied before the soil moisture has depleted below permissible limit, which is generaly 75% of available moisture. Frequency of irrigation is then equal to depth of water stored in root zone during each watering (D) divided by consumptive use of water per day, i.e.  

D = depth of water stored in root zone during each watering. = Sd [0.75 (FC – PWP )]

where,  

S = specific gravity of soil

 

d = effective depth of root zone.

Water Requirement of Crops  17

PWP = permanent wilting point   FC = field capacity Frequency =

D Consumptive use per day

q  SOLVED EXAMPLES Example 2.9:  After how many days, watering is required in order to ensure healthy growth of a crop, given the following data: (i) Field capacity, FC = 29% (ii) Wilting point, PWP = 11% (iii) Specific gravity of soil = 1.3 (iv) Effective depth of root zone, d = 700 mm (v) Daily consumptive use =12 mm Solution:

D = Depth of water stored in root zone during each waterings



= Sd [ 0.75 (FC – PWP)]



= 1.3 (0.7) [0.75(0.29 – 0.11)]



= 1.3 × 0.7 × 0.75 (0.18)



= 0.122 m



= 122 mm

Daily consumptive use = 12 mm Watering interval or frequency of irrigation =



=

D in mm. Daily consumptive use in mm. 122 = 10 days (approx) 12

Example 2.10:  A crop has consumptive use of water as 2.8 mm/day. Determine irrigation and depth of water to be applied when the amount of water available in soil is 25% of maximum depth of available water in root zone, which is 80 mm. Irrigation efficiency, is 65%. Solution: 80 (1 − 0.25) = 21days 2.8 80 (1 − 0.25) (ii ) Depth of water to be applied = = 92.3mm. Say 93mm. 0.65 (i ) Frequecy of irrigation =

18  Irrigation Engineering and Hydraulic Structures

2.9  IRRIGATION METHODS Technique of applying water on to the farms from source is known as method of irrigation. Source may be field channel in case of flow irrigation system or a well in case of lift irrigation system. Methods of applying water may be surface methods or subsurface methods or sprinkler method. These are given in detail below: q Surface Methods of Irrigation In this type, water is allowed to spread on farms either in a controlled manner or uncontrolled manner. Uncontrolled methods are employed only when irrigation system is of inundating type that is canal system gets its supply directly from rivers when they have rise in water level due to floods. If irrigation water is available from perennial irrigation system i.e., canals get water form reservoir

Border strip Field channel

Border strip

From distributary

Fig. 2.2:  Border strip method of irrigation or head works, the controlled methods of surface irrigation are adopted. Chief among this type are: 1.

Border Strip Method

3. Check Basin Method

2.

Furrow Method

4. Contour Method

5.

Sprinkler Method

6. Drip Irrigation Method

1. Border Strip Method: In this method farm is divided into number of strips of size (3 × 100) to (20 × 400) sqm. Slope of the strip should be between 0.002 to 0.004. This method is generally suitable to all types of crops but requires preparation of land for making strips of suitable size. This makes the method costly. 2. Check Basin Method: In this method entire field is divided into number of plots by constructing surrounding levees. Water is admitted from farmer’s water course to these plots turn by turn. Loss of water can be minimised and irrigation of entire farm can be carried out satisfactorily. Irrigation efficiency is higher in this method but constant supervision and field work is required.

Water Requirement of Crops  19

3. Furrow Method: In surface irrigation methods such as border strip and check basin methods, flooding of entire land surface is carried out. As an alternative to this, number of small field channels be provided in the farms known as “furrows” or “corrugation”. Water flowing through these furrows will infiltrate into the land to saturate the soils. Furrow depths vary from 8 cm to 20 cm depending upon whether soil is pervious or less pervious respectively. Their lengths may also be kept around 100 to 200 m and slope be kept between 3 to 6 per cent. In this method only 50 % of land surface is wetted directly and so evaporation losses are minimum but labour requirement is higher. Spacing of furrows is kept between 60 cm to 90 cm. The method is suitable for row crops such as maize, sugarcane, potato, ground nut, tobacco etc. sectional view of furrows is given in following figure.

Fig. 2.3:  Furrow method of irrigation 4. Contour Method: In hilly areas having steep land slope, this method is practiced. Here land is divided into series of terraces which are aligned to follow different contours at a vertical interval of 40 to 60 cm. At the outer end of each strip an earthen bund is provided so that irrigation or rain water does not leave the strip. Each strip can then have its irrigation by furrow or check basin method. 5. Sprinkler Method: This method started with irrigation of nurseries and orchards, and became popular in fully developed and developing countries. PVC flexible pipes are provided with overhead sprinkles which rotate on account of pressure of water leaving the sprinkler. Water gets sprayed over surrounding areas and gives the benefit of rainy water to orchards. Amount of water required is small but at moderately higher pressure. It does not interfere cultivation. Evaporation losses are high but percolation losses are minimum. Initial cost is high but labour requirement is minimum. 6. Trickle or Drip Irrigation Method: Trickle or drip irrigation consists of network of small diameter pipes which deliver water at root zone of plants, through emitters connected to pipes. Lateral PVC pipe lines, 30 mm in diameter are provided with emitters at a spacing suitable to crop type and soil condition. Due to its ability of maintaining constant soil moisture in the root zone, trickle irrigation can have highest efficiency, as high as 100% and results in greater crop yield. Fruits and vegetables respond very well to trickle irrigation. This method is not suitable for closely planted crops

20  Irrigation Engineering and Hydraulic Structures

like wheat and cereal grains. Clogging of emitters is a major problem of the system.

q  SOLVED EXAMPLES Example 2.11:  Following data pertain to wheat growing area : Field capacity, FC

= 25%

Permanent Wilting Point, PWP = 9% Root zone depth

= 2 m

Soil density

= 1400 kg/m3

Effective Precipitation

= 36 mm

Daily consumptive use

= 16 mm

Find frequency of irrigation and FIR, if irrigation efficiency is = 90% Solution: Available Moisture = (FC – PWP) = 25 – 9.0 = 16% Readity available moisture = 0.75 (16) = 12% D = Depth of water stored in root zone Sd = 0.75 ( FC – PWP )  = 1.4 ( 2 )( 0.12 ) = 0.336 m = 336 mm 336 = 21 days 16 Det − (effective rain fall) 336 − 36 300 = = = 333mm FIR = Irrigation efficiency 0.90 0.9

Frequency of irrigation =

Example 2.12:  Find out capacity of a reservoir for culturable command area of 50,000 hectare for following data: Crop

B (days)

D = Duty at field (Hect / cumec)

Intensity of Irrigation

Wheat

120

1800

20%

Rice

120

800

40%

Sugarcane

360

1700

20%

Cotton

180

1400

20%

Assume canal loss as 20% and reservoir loss as 10%

Water Requirement of Crops  21

Solution: Volume of water required crop wise: (i) Wheat, Volume of Water required = A ×∆ Volume of Water required =A×D  8.64B  = 0.2(50, 000)    D   8.64 × 120  =10, 000   = 10, 000 × 0.576 = 5760 Ha-m  1800  (ii) Rice : Volume of water required = A × ∆  8.64 × 120  = 0.4(50, 000)   = 25920 Ha -m  800  (iii) Sugarcane : Volume of water required = A ×∆  8.64 × 360  = 0.2(50, 000)   =18296 Ha-m  1700  (iv) Cotton : Volume of water required = A × ∆  8.64 × 180  = 0.2(50, 000)    1400  = 1,1108 Ha-m

 

Total volume of water required by corps = 5760 + 25920 + 18296 + 11108 = 61084 Ha-m 61084 0.8 × 0.9 = 84838 Ha-m

Required capacity of reservior =

Example 2.13:  A canal is required to cater the need of irrigating following crop pattern: Crop

B (Days)

Area (H)

Duty (Ha/cumec)

Sugarcane

320

900

600

Wheat

120

800

1600

Bajra

120

600

2000

Vegetables

120

320

600

Cotton

180

700

1400

22  Irrigation Engineering and Hydraulic Structures

Taking time factor of canal as 12/20 and capacity factor of canal as 0.8, find design discharge of the canal. Solution:

Q for Sugarcane =

   900 = 1.5 cumec (Perennial) 600

Q for Wheat =

800 = 0.5 cumec   (Rabi) 1600

Q for Bajara =

600 = 0.3 cumec   (Kharif) 2000

Q for Vegetables = Q for Cotton =

320 = 0.533 cumec   (Hot Weather) 600 700 = 0.5 cumec   (Eight Month) 1400

Since sugarcane is perennial crop, it will require water during Rabi, Kharif and Hot weather and cotton is eight month crop and so its water is required in hotweather and Kharif Thus

Q for Rabi

= 0.5 + 1.5

= 2.0 cumec



Q for Kharif

= 0.3 + 1.5 + 0.5

= 2.3 cumec



Q for HW

= 0.533 + 1.5 + 0.5 = 2.533 cumec

Thus, Q for canal is 2.533 that of hot weather, highest of any three season. Time factor of a canal is the ratio of number of days canal has actually run to number of days it was required to be run. Hence, Q canal = 2.533 × 20 / 12 = 4.22 cumec

Design Q canal =

4.22 4.22 = = 5.27 cumec Capacity Factor 0.8

Example 2.14:  A canal has Gross Command Area (GCA) of 25,000 Ha, of which 80% is culturable. Intensity of irrigation for Rabi is 40%, and 20% for Rice. If Kor period for Rabi is 4 weeks and 2.5 weeks for rice, find outlet discharge. Outlet duty for Rabi is 1600 Ha/cumec and for Rice 800 Ha/cumec. Calculate delta for each case.

Water Requirement of Crops  23

Solution:

GCA = 25,000 Ha CCA = 0.8 × 25000 = 20,000 Ha 40 (20,000) = 8000 Ha 100 8000 = 5.0 cumec ∴ Outlet discharge = 1600 20 Area under Rice = (20,000) = 4000 Ha 100 4000 = 5cumec ∴ Outlet discharge = 800 8.64 B 8.64(2.5 × 7) = 0.189 m. ∆ for rice = = 800 D 8.64 (4 × 7) ∆ for Rabi = = 0.15 m. 1600 Area, A under Rabi =

Example 2.15:  Determine the reservoir capacity and main canal discharging capacity for following data: (i) Culturable Command Area, CCA = 50000 Ha. (ii) Canal losses 10%, Reservoir losses 10%. (iii) Time Factor – 0.7, capacity factor – 0.7 (iv) Details of crop : Crop

Base period B in days

Duty D in Ha/ cumec

Intensity of Irrigation –(I)

Wheat

120

1800

20%

Cotton

180

1500

20%

Rice

120

 700

20%

Sugarcane

320

1700

10%

Vegetables

120

 700

10%

Solution: Sl. No. Crop Season (i) (ii)

Wheat (Rabi) Cotton (E.M.) (i.e., HW + Kharif)

Area under B Delta(m) ∆ = × 8.64 Irrigation(Ha) D 50,000 × 0.2 = 10,000 10,000

=

120 1800

∆=

× (8.64) = 0.576 m

180 1500

(8.64) = 1.036 m

Volume of water (ha.m) = A(∆) 5,760 10,360

24  Irrigation Engineering and Hydraulic Structures

Sl. No. Crop Season

Area under B Delta(m) ∆ = × 8.64 Irrigation(Ha) D

(iii) Rice (Kharif)

10,000

(iv) Sugarcane (Perennial) (Rabi + HW + Kh) (v)

Vegetable (HW)

5,000

5,000

∆= ∆=

∆=

120 700 320 1700 120 700

(8.64) = 1.48 m (8.64) =  1.62  m

(8.64) = 1.48 m

Volume of water (ha.m) = A(∆) 14,800 8,131

7,405

Total water Requirement = 46,456 Ha. m

Reservoir capacity =

46456 (as canal and reservoir losses each = 10%) 0.8

∴ Reservoir capacity = 58070 Ha-m Q required in canal = Rabi : Q =

Area + due to perennial drop D

5000 10, 000 + = 8.49 cumec (Due to wheat and sugarcane) 1700 1800

Kharif : Q =

5000 10000 10000 =+ + = 23.88 cumec (Due to cotton, rice 1700 700 1500

and sugarcane)

5000 HW : Q = + 2.94 + 6.66 = 16.74 cumec (Due to cotton, sugarcane and 700 vegetables) ∴ Q is highest in kharif = 14.28 + 2.94 + 6.66 = 23.88 cumec 23.88 ∴ Canal Capacity = = 48.73 cumec 0.7 × 0.7

EXERCISES 1. Define: Duty, Delta, Base Period and obtain relationship between them. 2. What is the effect of transit losses on duty? Will duty of rice be same for field and canal head work? Why? 3. Define: Paleo irrigation, crop ratio, kor period, crop – rotation, Field capacity, PWP and Consumptive use.

Water Requirement of Crops  25

4. Find frequency of irrigation requirement for a crop having following data: Root zone depth 100 cm. FC = 22%, PWP = 12%, specific gravity of soil = 1.5, Consumptive use = 12.5 mm / day.

(Ans: 4 days)

5. A field has FC = 27% and PWP = 13%. If wheat is required to be grown on this field, find the storage capacity in 80 cm. depth of soil if dry density of soil is 1.5 gm/cc. If irrigation water is to be supplied when soil moisture falls to 18%, find FIR for wheat if irrigation efficiency is 80%.

(Ans: 16.8 cm, FIR =13.5 cm)  

 

3 Hydrology 3.1  SCOPE OF HYDROLOGY Hydrology is a science which deals with rainfall, run-off and floods occurring in nature as a result of hydrologic cycle. One-fourth of earth’s surface is land and rest is ocean. Evaporation from oceans, lakes and ponds generate cloud formations, which are transported landward by wind and its condensation results in rains. Rains produce rivers and tributaries which again meet sea and cycle gets repeated. Each river and tributary has its own catchment area over which rain occurs and its resulting surface run-off forms river and tributary. Study of catchment characteristics helps in knowing quantum of surface run-off and its concentrated effects like flood. Thus, scope of hydrology can be defined as: (i) Water yield from a catchment which enables engineers to design storage dams, water supply schemes and hydropower projects. (ii) Floods, their frequency and intensity, which enable civil engineers to design spillways, bridges, culverts etc.

3.2 RAINFALL Rainfall occurring in moderate quantities over catchment areas forms surface-runoff which feed rivers and nalas, but rainfall occurring with great intensity, which is defined as storm, results in formation of floods in rivers. Study of surface run-off helps in getting reliable quantities of water available in a year whereas study of floods helps in finding out suitable methods for its safe disposal, so that hydraulic ­structures such as dams, spillways, bridges etc. do not suffer any damage and loss of life and property can be prevented. Thus, most valuable contribution of science of hydrology is to know the reliable quantum of water supply for irrigation and power projects and intensity of floods and their frequencies for safe and suitable design of storage works.

© The Author(s) 2023 S. K. Ukarande, Irrigation Engineering and Hydraulic Structures, https://doi.org/10.1007/978-3-031-33552-5_3

26

Hydrology  27

3.2.1  Types of Rainfall Rainfall or Precipitation can be of four different types as described below: 1. Convectional Rainfall: Air in contact with warm earth surfaces in summer rises higher up into the sky and gets cooled. If it carries vapour with it in sufficient quantities, then its cooling process results into rainfall. This type of rainfall is convectional or thermal precipitation. 2. Frontal Precipitation: When two air masses of different temperatures and densities collide, resulting rainfall is known as frontal precipitation. 3. Orographic Precipitation: If moist air masses are lifted up due to barriers like mountain ranges then resulting rainfall is known as orographic precipitation. 4. Cyclonic Precipitation: On account of pressure difference in atmospheric air created by unequal heating of earth surface, moist air gets lifted up into a low pressure belt resulting in cyclonic precipitation. There are two types of cyclonic precipitation (i) tropical cyclone, 300 to 1500 km in diameter and causing heavy precipitation along with high wind and (ii) extra tropical cyclonic precipitation of diameter upto 3000 km with wide spread frontal type precipitation.

3.2.2  Measurement of Rainfall Rainfall measurement is carried out with the help of network of rain gage station, spread out over given catchment area. In plain area, one rain gage station is provided to command 500 km² area. For hilly and elevated area one rain gage station is provided per 130 km² to 260 km². There are two types of rain gages: 1. non-recording type and 2. recording type. 1. Non-recording Type: Symon’s rain gage (see Fig. 3.1) is a non-recording type and consists of a funnel of 12.7 cm diameter put into a glass bottle of 10 cm diameter. This bottle is mounted on a concrete pedestal of 60 cm × 60 cm × 60 cm. The funnel top should be 30 cm higher than the surrounding ground. If the receiving bottle is full, it measures 1.25 cm of rainfall. Rain water collected in the bottle is measured by a measuring glass cylinder. Symon’s rain gage can give measurement of previous 24 hours rains that has occurred during morning 8.30 am to next morning 8.30 am.

28  Irrigation Engineering and Hydraulic Structures

Fig. 3.1:  Symon’s rain gage 2. Recording type Rain gage: Automatic or self-recording type of rain gage consists of a funnel, catch bucket, pointer and graph drum and is installed such that funnel top is 75 cm above ground surface (see Fig. 3.2).

Fig. 3.2:  Self-recording type rain gage

Average Annual Rainfall (AAR) Average Annual Rainfall is the mean of rainfall observed over past 35 years. If average annual rainfall is less than 40 cm, the region has arid climate, if AAR is greater than 40 cm, but less than 75 cm, it is semi-arid climate and if AAR is greater than 75 cm, it is humid climate.

Hydrology  29

Estimate of missing Data of Rainfall record: It is generally worked out by simple proportion method, illustrated by following example.

q  SOLVED EXAMPLES Example 3.1  Rain gage station O was in operation for past few weeks, during which a storm occurred. The storm rain recorded at surrounding stations P, Q, R, was 9, 6, 10 cm respectively. If AAR values for the station are 81, 90, 70 and 90 cm respectively, find storm rainfall at O. Solution:  By Simple Proportion Let P0 is storm rainfall at station P.

P0

P 9 6 10 = = = o 81 90 70 90 19 6 10  =  × 90 + × 90 + × 90  3  81 90 70  1 29 = [10 + 6 + 13] = = 9.66 cm. 3 3

For large area (greater than 50 Km²), mean rainfall is worked out by any of following three methods: 1. Arithmetic Mean Method 2. Thiessen Polygon Method 3. Isohyetal Method They are given in detail with illustrative examples below: 1. Arithmetic Mean Method: Pav =

∑ Pi n

where, ΣPi = sum of rainfall at individual station n = number of rain gage station. This method is good if rain gage stations are uniformly distributed over the area. 2. Thiessen Polygon Method: If rain gage stations are not distributed uniformly over the given area then the method used is Thiessen Polygon Method. It is also known as weighted area method. ∑ Ai Pi Pav = ∑ Ai

30  Irrigation Engineering and Hydraulic Structures



ΣAi = total area of basin with i no. of rain gage stations. ΣAiPi = sum of product of area covered by rain gage station and rainfall P measured at that station. The area is obtained by constructing polygon as indicated in the illustrative example (1) below:

3. Isohyetal Method: In this method point rainfalls are plotted on a graph paper and lines are drawn through points representing equal rainfalls. These lines are called “Isohyetals” such as P0, P1, P2, P3 over a given area of A km2

Fig. 3.3:  Isohytes Pav =

 P0 + P1   P1 + P2  (Area bet. iso. 0 and1) +  (Area bet. iso. 1, 2 )    2   2    P +P  +  2 3  (Area bet. iso. 2,3)  / (Total Area A)  2  

Example 3.2  Figure 3.4 shows a catchment area of 400 sq km with following six number of rain gage stations with rainfall records: P1 = 6 cm P2 = 7 cm P3 = 8 cm

P4 = 9 cm P5 = 10 cm P6 = 11 cm

Note – 1. P1, P2---P6 – Rain gage stations 2. ID, HE, GA etc perpendicular bisectors 3. ABCDEA is polygon for station P6 , A B F G is polygon (III) fordstation P2 etc

Fig. 3.4:  Theissen polygon method

Hydrology  31

Calculate Average Rainfall by (i) Arithmetic Mean Method (ii) Thiessen Polygon Method Solution: (i) By Arithmetic Mean Method Pav =

P1 + P2 + P3 + P4 + P5 + P6 51 = = 8.5cm. 6 6

(ii) By Thiessen Polygon Method: Join rainfall stations P1, P2, P3 etc. Draw perpendicular bisector for lines P1P2, P2P3, P3P4, P4P5, P5P1, P1P6, P2P6, P3P6, P4P6, P5P6. These perpendicular bisectors shall intersect each other forming of polygon, such as ABCDE and other surrounding polygons. The central polygon will give area for rain gage station P6. Area of polygon ABFG is for station P2 etc. The areas are given below for each station. The areas can be calculated by planimeter or graph paper. Rainfall

Station

Polygon

11 cm

P6

ABCDE

90 sq km

A6

7 cm

P2

ABFG

120 sq km

A2

8 cm

P3

BCJF

100 sq km

A3

9 cm

P4

CDIJ

90 sq km

A4

10 cm

P5

DEHI

90 sq km

A5

P1

EAGH

130 sq km 640 sq km

A1

11 cm

Pav =

Area

P1A1 + P2 A 2 + P3 A3 + P4 A 4 + P5 A5 + P6 A 6 A1 + A 2 + A3 + A 4 + A5 + A 6

(130 × 11) + (120 × 7) + (100 × 8) + (90 × 9) + (90 × 10) + (90 × 11) 640 1430 + 840 + 800 + 810 + 900 + 990 5770 = = = 9.0 cm. 640 640 =

32  Irrigation Engineering and Hydraulic Structures

Example 3.3  Figure 3.5 shows a basin of 80 sq km area and Isohyetals indicating rainfall of 17,18, 19 and 21 cm are drawn there of. Areas between Isohyetals are given below: 17 cm.

18 cm.

19 cm.

21 cm.

* 20.5 cm. 10 sqkm. B

* 18.7 cm. * 16 cm.

15 sqkm.

* 22 cm.

20 sqkm. 20 sqkm. A

15 sqkm. * 17 cm.

P2

P2

P3

P4

Fig. 3.5:  Isohyetal method Between cover point A and Isohyetal 17 cm, Area = 20 sq km. Between P1 and P2, 15 sq km. Between P2 and P3, 20 sq km. Between P3 and P4, 15 sq km. Between P4 and cover point B, 10 sq km. Find average rainfall by Isohyetal Method. 20 + 15 = 17.5 sq km. 2 15 + 20 Area Represented by P2 = = 17.5 sq km. 2 20 + 15 Area Represented by P3 = = 17.5 sq km. 2 15 + 10 Area Represented by P4 = = 12.5 sq km. 2 17.5 × (17) + 17.5(18) + 17.5(19) + 12.5(21) Hence Pav = 80 297.5 + 315 + 332.5 + 262.5 1107.5 = = = 13.8cm. 80 80 Area Represented by P1 =

Adequacy of rainguage stations: Let N = number of rain gage station required, then

2

 Cv  N=  (3.8)  e 

Hydrology  33

Where e = percentage error in adequacy of number of stations required Cv =

σ x

x = mean value of rainfall recorded at given number of stations say n,



σ=

(x − x)

2

(3.9)

n −1

Example 3.4  Rainfall recorded at seven stations are given below in cm. Take permissible error as 10% in estimation of average rainfall, check adequacy of existing number of stations. x = 154, 109, 139, 144, 124, 112, 172 cm Solution: Station

x–x

x

(x – x)2

1

154

154–136=18

(18)2 = 324

2

109

–27

729

3

139

03

09

4

144

08

64

5

124

–12

144

6

112

–24

576

7

172

36

1296

Σx = 954

∑ (x – x)

∴x =

954 = 136 7 1

∴σ =

Σ (x − x)2  3038  = = (506) 2 = 22.5  7 − 1  n −1

∴ Cv =

σ 22.5 = = 0.1682 x 136

 Cv  N = no. of station required =    e 

 0.1682 × 100  =  10   N = 2.72 = 3.

2

2

2

= 3038

34  Irrigation Engineering and Hydraulic Structures

Hence, existing station are in excess by 7 – 3 = 4 Nos. Thus, 4 number of stations are not required. Double Mass Curve Technique or Test for Consistency of Rainfall Record: If conditions relevant to the recording of rain gage stations have undergone a significant change, during this period of record, then inconsistency would result in rainfall data of that station. This can be checked by double mass curve technique.

Cumulative Annual Rainfall at Station x

Method A minimum of 5 number of stations having reliable data are chosen as base stations. The average annual rainfall for these stations is procured for at least 5 to 10 years and its cumulative record is prepared. A similar cumulative record of rainfall at stations X for which consistency check is required is also prepared. Then a graph between cumulative rainfall of base station on X axis and cumulative rainfall of station X on Y axis is drawn such that latest year record comes first and subsequent years record comes last. If station X has consistent data, this graph will be a straight line or else it will deviate from the year from which inconsistency in data has arises. This can be set right by considering the slope of the straight line if it has deviated and slope of undeviated straight line. This will be clear from the following Fig. 3.6.

45 50 55

Year of Deviation

60 65

a c

70 75 80 90 19

85

c Correction ratio = a c Corrected value for year 1955 = x55 × a

Cumulative Annual Rainfall of 10 station average 3 Σpav of station (in 10 cm units)

Fig. 3.6:  Double mass curve Example 3.5  Test the consistency of rainfall data from following Table for 12 years for station X and 12 surrounding stations using double mass curve technique.

Hydrology  35

S. No.

P at station X (cm)

1

32

32

24.3

24.3

2

40.4

72.4

36.5

60.8

3

45.2

117.6

31.7

92.5

4

33.7

151.3

28.8

121.3

5

28.9

180.2

26.7

148.0

6

33.5

213.7

29.7

177.7

7

50.8

264.5

37.6

215.3

8

29.9

294.4

29.9

245.2

9

26.1

320.5

26.6

271.8

10

23.3

343.8

25.1

296.9

11

29.7

373.5

34.8

331.7

12

18.8

392.3

24.9

356.6

Σ Px (cm)



Pav at Surrounding station (cm)

Σ Pav of surrounding station (cm)

Plot Σ Pav on X-axis Σ Px on Y-axis

Change in data from Sr. No. 7 will be noted as under : (i) Slope of original line = 1.25 (ii) Slope of deviated line = 0.90

∴ Correction =

0.9 1.25

Cumulative rainfall at Sr. No. 6 = 213.7 ∴ its corrected value

= 213.7 ×

0.9 1.25

= 153.87 2. Weighted average of four station method:

1 1 1 1   2 P1 + 2 P2 + 2 P3 + 2 P4  r2 r3 r4  r Px =  1 1 1 1 1 + + + r12 r22 r32 r42

36  Irrigation Engineering and Hydraulic Structures

Where Px is missing rainfall at station X surrounded by four different station at radial distances r1, r2, r3, r4 etc. Example 3.6  Station X failed to report rainfall during a storm. The co-ordinates of four surrounding station are (10,10), (–9,6), (–12, –10) and (5, –12) km respectively. Find missing rainfall at X if rainfall at the four surrounding station was 70, 90, 68, 60 mm respectively. Solution: P1 = 7 cm, P2 = 9 cm, P3 = 6.8 cm, P4 = 6.0 cm

r12 = (10)2 + (10)2 = 200 r22 = (−9)2 + (6)2 = 117 r32 = (−12)2 + (−10)2 = 224 r42 = (5) 2 + (−12)2 = 169  P1 P2 P3 P4   2 + 2 + 2 + 2  r r2 r3 r4  ∴ Px =  1 1 1 1 1  2 + 2 + 2 + 2   r1 r2 r3 r4  7 9 6.8 6 + + + 200 117 224 169 Px = (5 × 10 −3 ) + (8.5 × 10−3 ) + (4.46 × 10−3 ) + (5.9 × 10−3 )

= =

0.035 + 0.076 + 0.03 + 0.035 23.86 × 10−3 0.176 23.86 × 10−3

= 7.37 cm

3.3  SURFACE RUN–OFF Rain occurring over a given basin or catchment area will flow over the surface as per general slope of the catchment area and part there of will penetrate the ground. Some part of flowing water may get detained because of unevenness of area and some part may get evaporated and also transpirated. Thus, resulting overflow is always less than the total rain water that has fallen on the catchment area. This resulting surface water flows by gravity and forms a stream. Part of the ground water sometimes comes back to stream. The stream flow is thus result of surface run-off from the catchment area and hence its measurement gives the magnitude of surface

Hydrology  37

run-off. There are some empirical formulae developed by various researchers to give measure of surface run-off (R) in cm. They are given below: 1. Inglis formula (a) R = 0.85 P – 30.5 for ghat areas. Here R indicates run-off in cm and P is average rainfall in cm for all formulae mentioned below. (P −17.8) P for plain area (b) R = 254 2. Lacey Formula for Indo–Gangetic Plain Area: R=

P 304.8 F 1+ PS

Here F = Monsoon duration factor = 0.5 to 1.5 S = Catchment factor = 0.25 to 3.5 3. Khosla’s Formula for North India: R =P−

T 3.74

Here, T = mean annual temperature in °C for the given catchment area. 4. Rihand Formula: R = P – 1.17 P0.85

3.3.1  Strange Curves Based on empirical observations, strange developed curves giving values of surface run-off R in cm for given average annual rainfall P. They are given in the Fig. 3.7. R = Run-off in cm. P = Rainfall in cm.

Wet catchment

R in cm

Average catchment

Dry catchment

P in cm

Fig. 3.7:  Strange runoff curves

38  Irrigation Engineering and Hydraulic Structures

3.3.2  Stanford – Watershed Model Lansley and Crawford developed a computer model at Stanford University. This model is known as Stanford Water–shed model and gives hourly stream flow if hourly precipitation data and evapotranspiration losses are fed as inputs.

3.3.3  Rational Approach (a) A rational method to obtain yield from given catchment area is based on finding appropriate catchment coefficient C and then using the rational relation. Yield = C × A × P. Here yield is total water available from catchment for the purpose of storage reservoir, A is catchment area and P is average annual rainfall of the catchment area. Value of C depends on catchment characteristics such as type of soil, general slope of the ground, vegetable and temperature prevailing in the area. C = 0.3 is general value adopted. (b) Surface runoff calculations can also be carried out by ascertaining the losses that may take place before water from catchment appears as river flow. These losses are due to infiltration into the ground, interception due to uneven surface of the ground and temporary storages due to depression in the ground. Losses are also due to evaporation from water masses and evapotranspiration from ground as well as vegetation in the catchment area. In order to calculate losses, catchment characteristic should be thoroughly investigated. These characteristics are :

(i) Shape and size of catchment area



(ii) Type of soil and density of vegetation



(iii) Temperature prevailing in the catchment area.



(iv) Wind velocity prevailing in the catchment area.



(v) Month-wise percentage of sunshine hours available into catchment area. Thus, if losses can be calculated, hydrological equilibrium can be written as Rexccss = Run off = P – I – S – E – ET,

where P = mean precipitation as available from rain gage station network in catchment area. I = Infiltration loss into the ground S = Loss due to interception and storage in depressions in catchment area. E = Loss due to evaporation from water bodies ET =Loss due to evapotranspiration

Hydrology  39

(c) Losses: Determination of all these losses are discussed in detail as under: 1. Infiltration Loss: Rain water entering into soil is infiltration loss for calculation of surface run-off where as it is a gain for ground water storage. Rate of infiltration will depend on type of soil and slope of ground. A ground with no slope or gentle slope will have more infiltration loss if the soil is sandy and loose; but if a ground has appreciable slope and also the soil is either rocky or clayey, then infiltration loss will be less. Initially when the rain starts ground surface is dry and it itself will absorb water, but once this is done, water will start entering voids, cracks and cuts and then initially infiltration rate will be high, but as time passes and as the ground soil gets saturated, there will be fall in infiltration rate. This has been studied by Horton who has given following equation: Horton Equation f = fc + (fo – fc)e–kt(3.1)



where, f = rate of infiltration at any time t. fo = initial rate of infiltration fc = constant rate of infiltration Equation 3.1 can be rewritten as under: f – fc = (fo – fc) e–kt ∴ ∫ (f − f c )dt = ∫ (f o − f c )e − kt dt Now fc = area of shaded portion in the Fig. 3.8 = ∫ (f − f c )dt

= no. of square units if graph is down

Infiltration rate

Shaded Area = fc Horton’s curve

fO ft

fc = constant Rate of infiltration. t=0

time (hrs) t

Fig. 3.8:  Horton's curve (infiltration capacity curve)

40  Irrigation Engineering and Hydraulic Structures

Now, ∫ (f o − f c )e− kt dt = Thus, k =

(f o − f c ) k

fo - fc Fc

As an alternative, k can be decided by taking natural logarithms on both sides of equation (3.1), ln (f – fc) = ln (fo – fc) – kt Equation (3.2) on plotting will give a straight line of the form y = ax + c Here y = ln (f – fc) And ax = –kt and c = ln (fo – fc). Slope of the line will give value of K and intercept on y-axis will give fc.

q  SOLVED EXAMPLE Example 3.7  A 24 hour storm occurred over a catchment area of 2 sq km and total rainfall observed was 12 cm. Infiltration curve has fo = 1 cm/hr and fc = 0.3 cm/hr after 15 hours, with Horton’s k value = 5 per hr. An evaporation pan installed indicates a fall in level by 0.6 cm during 24 hours. If pan coefficient is 0.7 find run-off and volume of run-off from the catchment. Solution:

Fp =

24

∫ [fc + (fo − fc )e

− kt

]dt

0

24



= ò [0.3 + (1.0 - 0.3)e-5t ]dt 0



24 0.7   0.7  é  0.7 ù ú = 0.3 × 24 − 5×24  − 0 − 0  = ê 0.3t + 5t êë 5e 5e     -5e úû 0



 0.7  = 7.2 − + 0.14 = (7.2 – 0.107) + 0.14 = 7.233 cm. 5(1.3)  



Run-off = P – Fp – E = 12.0 – 7.233 – 0.7(0.6) = 4.347 cm

Volume of run-off =

4.347 (2 × 106 ) = 8.694 × 104 = 86940 m3 100

2. Infiltration Indices:  If rainfall hyetograph i.e., rain in cm on y axis and time of duration on x-axis is drawn as shown in Fig. 3.9, then infiltration curve as obtained by Horton’s equation can be superimposed on it and infiltration loss and run-off can be obtained as shown in Fig. 3.9.

Hydrology  41

P (cm)

Run-off Horton’s Curve φ index

fo fc 1

2

3

4

5

6

Fig. 3.9:  Infiltration indices Here the difficulty will arise if the curve line is above part of hyetograph and has a variation. To avoid this difficulty, average rate of infiltration is worked out and is expressed as a straight line on hyetograph (dotted horizontal line). Thus, ϕ index can be defined as average rainfall intensity above which rainfall volume is equal to run-off volume. P −Q −S Windex = (3.2) t Here, P = total precipitation corresponding to time interval t, Q = run-off S = interception and storage losses in catchment Thus Windex = ϕindex – S

q  SOLVED EXAMPLE Example 3.8  The rate of rainfall for successive interval of 1 hr, of 6 hr storm are 1.5, 3,5, 2.5, 2, 1.0 cm/hr. The corresponding surface run-off is 6.0 cm. Find ϕ and W indices. Solution: Σ(p − φ)t = R e xcess = 6 cm.

∴[(1.5 − φ) + (3.0 − φ) + (5 − φ) + (2.5 − φ) + (2 − φ) + (1 − φ)] × 1 = 6 cm.

∴15 − 6φ = 6



∴φ = 1.5 cm/hr (1st trial) Ptotal = 1.5 + 3 + 5 + 2.5 + 2 + 1 = 15 cm ∴ Windex =

P − Q 15 − 6 = = 1.5 cm/hr t storm 6

If ϕ = 1.5 as obtained in 1st trial is correct.

42  Irrigation Engineering and Hydraulic Structures

7 P cm/hr

5.0 5 φ index

3.0 3 1.5

2.5

2

1

2

3

4

5

6

7

t hrs

Fig. 3.10:  ϕ-Index Then Rexcess as per shaded area in Fig 3.10 should be equal to 6.0 cm. but it is (1.5 + 3.5 + 1 + 0.5) = 6.5 cm ∴ As a second trial, take Q = 1.6 cm then (1.4 + 3.4 + 0.9 + 0.4) = 6.1 cm and hence it can be accepted . ∴ ϕindex = 1.6 cm/hr Note : thus ϕ index can be obtained by trial and error. 3. Loss on Account of Evaporation and Evapotranspiration: Loss on account of evaporation can be measured either by using evaporation pans on the concerned site of the catchment area or by making use of empirical equations. (a) Evaporation Pans: It is a cylindrical vessel of about 1.25 m in diameter and 0.3 m depth and is open at top and is made up of GI sheet or copper sheet. It is placed about 15 to 20 cm above ground level and water level inside the pan will be 5 to 10 cm below vessel’s rim. The change in water level over specified duration of time is noted and loss of volume of water due to evaporation is worked out, which when multiplied by pan coefficient will give rate of evaporation at the given side.

Pan coefficient =

Actual evaporation from site Measured evaporation from Pan = 0.7, generally

Hydrology  43

There are four varieties of the pans available for measurement of evaporation loss: 1. U.S. class A evaporation Pan: It is a land Pan and is shown in Fig. 3.11 given below: 121 cm dia.

5 cm 30 cm

25 cm

GI sheet or copper sheet depth of water

15 cm

G.L

Fig. 3.11:  US class a evaporation pan 2. IMD Pan: Pan developed by Indian metrological department which is similar to above. 3. Colorado Sunken Pan:  It is 920 mm × 920 mm square in plan and 460 mm deep, and is buried into the ground such that 50 mm is above ground level and water level in the pan is kept at ground level. 4. Floating Pan developed by U.S. Geological survey department:  It is a 900 mm square and 450 mm deep and kept on drum floats which are kept floating in reservoir. (b) Impirical Equations : 1. Dalton’s law of evaporation : E = k(ew – ea) According to Dalton, evaporation is proportional to difference in vapour pressures of water mass and surrounding air mass about 2 m above water surface. They are denoted by ew and ea in the above equation. Higher the temperature and wind velocity, greater is the evaporation, however if greater is humidity, smaller is evaporation, this can be inferred from Meyer’s equations given below: 2. Meyer’s Equations:

 V  E = k m (e w − ea ) 1 + 9  mm / day (3.3)  16 

Here Km – is coefficient and is generally 0.36 ew = saturated vapour pressure at water surface in mm of mercury. ea = saturated vapour pressure of air mass about 2 m above the water surface   = (% of relative humidity of air mass) V9 = monthly mean wind velocity in km/hr at about 9m above the ground and can be worked out by : V9 = V2(9)1/7 where V2 = Wind velocity in km/hr at 2m above the ground or Vh = v2(h)1/7 where h is any height in meter above the ground

44  Irrigation Engineering and Hydraulic Structures

q  SOLVED EXAMPLE Example 3.9  A reservoir with a surface area of 300 Ha had following data for a week : Water temperature = 22.5°C Relative humidity = 40% Ew = 20.44 mm of Mercury Coefficient km = 0.36 in Meyer’s equation  V  E = k m (e w − ea ) 1 + 9  mm / day  16  Estimate daily evaporation from the reservoir and volume of evaporation loss from the reservoir during the week. V2 = 16 kmph, wind velocity at 2m from ground. Solution: ew = 20.44 mm of mercury ea = (relative humidity %) (ew) = 0.4 (20.44) V9 = V2 (9)1/2 = 16(9)1/7 V9 = 22 kmph. km = 0.36  V  ∴ E = 0.36(e w − ea ) 1 + 9  mm / day  16   22  = 0.36(20.44)(0.6) 1 +  = 10.5 mm/day  16  Evaporation loss for reservoir during the week

= 7 × (300 × 104 ) ×

10.5 = 0.22 MCM 1000

3. Methods to Reduce Reservoir Evaporation Loss: (i) Reservoir site should be chosen such that it gives least area of surface exposed to atmosphere at given height. This is highly impossible. (ii) By growing tall trees all around reservoir. (iii) By spraying chemicals on reservoir surface. This is effective. Cetyl alcohol (C16H33OH) (hexadecanol) be applied on the reservoir surface and it develops a thin film of 0.015 micros and will not allow water molecules to pass to atmosphere but wind v­ elocity should be within tolerable limit. 2.2 kg of the cetyle alcohol is required to spread on 1.0 Ha of reservoir surface area.

Hydrology  45

4. Estimation of Evapotranspiration Loss or Determination of Consumptive Use of Water: (a) Blaney Craddle Formula: U = Σkf , (in cm )

where



p(4.6t + 81.3) cm 100 k = monthly consumptive use coefficient. f = monthly consumptive use factor t = mean monthly temperature, ˚C p = monthly percentage of sunshine hours



then, FIR = field Irrigation requirement =

f=



U − ΣPe η η = water application (irrigation) efficiency Pe = effective rainfall for given month. (b) Penman’s Equation: Evapotranspiration (ET) loss can also be worked out by making use of Penmen’s equation, which gives potential evapotranspiration (PET) as under: PET =

AH n + E L γ mm / day. A+γ

Here, A = slope of a graph of vapour pressure v/s temperature. Hn = net radiation in mm of evaporable water in a day. ew = saturated vapour pressure at given temperature ea = (% of relative humidity)   γ = Psychromatic constant = 0.49 EL = Evaporation loss  V      = k m (e w − ea ) 1 + 2  mm / day  16  V2 = wind velocity in kmph at 2 m from G.L.

q  SOLVED EXAMPLE Example 3.10:  Calculate PET for following data: Sunshine hours – 9 Wind velocity at 2m from G.L. = 16 kmph A = Slope of

ew curve = 1 mm/ºC temp

46  Irrigation Engineering and Hydraulic Structures

Hn = 1.99 mm of water per day Relative humidity = 75% ew = 16.5 mm of mercury r = 0.49 Solution:  ea = (75%) (ew)      = 0.75 (16.5) mm of mercury    EL = as per Meyer’s equation with Km = 0.36  v      = Km (ew – ea) 1 + 2   16   16  ∴ EL = 0.36 (ew – 0.75 ew) 1 +  = 0.36 (16.5)(0.25) (2) = 2.97 mm/day  16 

∴ PET =

=

AH n + E L γ (1)(1.99) + 2.97(0.49) = A+γ 1 + 0.49

3.45 = 2.3 mm / day 1.49

3.4  FLOOD FLOW CALCULATION Flood estimation is carried out by (i) Using flood Formula developed empirically (ii) Using unit hydrographic method.

3.4.1  Empirical Formulae for Flood Calculation 1. Dicken’s Formula: Q = C A3/4 m3/sec where, C = catchment constant varying from 11 to 25 A = catchment area in sq km Q is in cumec and gives flood discharge in cumec This formula is used in North and Central India. 2. Inglis Formula: Q=



124 A A + 10.4

cumec

A = catchment area in sq km.

This formula is used in Madhya Pradesh, Maharashtra and Gujarat.

Hydrology  47

3. Ryve’s Formula for South India: Q = CA 2/3 cumec where, C = 6.8 for areas within a distance of 80 km from Coast. = 8.3 for areas between 80 to 24 km from coast. = 10 for areas near hills. and A is in sq km. 4. Fuller’s Formula: This is developed in USA, however it has its importance as it considers concepts of probability of flood and its return period. Qav = CA0.8 , here C = 0.18 to 1.3 QT = Qav ( 1 + 0.8 loge T ) QT = flood occurring once in T years. And a is in sq km and Q is in cumec.

3.4.2  Unit Hydrograph Method In 1932, L.K. Sherman developed unit hydrograph method and is found extremely useful for flood estimation. It not only gives flood peak value but also gives entire flood hydrograph corresponding to anticipated storm. Unit hydrograph method is much more rational than empirical flood formulae. Unit Hydrograph, here after referred as UH, can be defined as the direct run-off hydrograph corresponding to unit effective precipitation, i.e., 1 cm rainfall excess, in a specific time duration, occurring uniformly over the given catchment. Three important conditions are implied in above definition of UH: (i) Unit Effective Precipitation, i.e., one cm of rainfall excess, i.e., net rainfall of 1 cm available as run-off. (ii) Time duration is to be specified for each storm. i.e., UH for 2 hr means hydrograph resulting from 1 cm rainfall excess for a storm of 2 hr duration. It could be 3 hr duration or 4 hr duration or any duration. The unity is for rainfall excess only. (iii) UH does not include base flow, thus while preparing UH, base flow must be deducted from direct run-off hydrograph.

Assumptions: (i) Base period of a direct run-off hydrograph corresponding to storms of same duration but different intensities, for a given catchment, is always constant. (ii) UH can be looked upon as the catchment response to an input of 1 cm rainfall for some specified time.

48  Irrigation Engineering and Hydraulic Structures

(iii) Principle of linearity and principle of superposition are applied. This is explained with the help of following illustration: In the above Fig. 3.12, two storm hydrographs A and B, of same duration but of different intensities are shown. If the intensity of A is just double that of B, then resulting hydrograph is also doubled, this is principle of linearity and is shown in following Fig. 3.13.

Storm A

Q

Storm B

Base period T

Fig. 3.12:  Storm hydrograph Figure 3.14 shows two storms of same intensity and same duration but with a time lag of T hr, generating two identical hydrographs. The composite hydrograph ordinates will be equal to sum of ordinates of the two hydrographs. This is the principle of superposition.

Construction of UH: (i) For a given storm of given intensity and given duration of time, plot the flood hydrograph. (ii) Separate out base flow from the flood hydrograph. T Hyetograph (Indicates rain fall excess)

Storm A Ordinate of hydrograph of storm A = 2 (a1)

Q

Storm B a1 Base period T

Fig. 3.13:  Storm hydrograph (iii) Obtain ordinate of direct run-off (DRO) by deducting base flow ordinate from flood hydrograph i.e., from Q1 subtract Q2.

Hydrology  49

Q

3 hrs. 3 hrs.

3 hrs. Shift in time T 3 hrs.

Fig. 3.14:  Storm hydrograph with time lag Step I:

Find “direct Run-off Depth” by following formula. Σ o× t cm Direct run-off depth = 0.36 A



Σo = sum of direct run-off ordinates t = time internal in hours between successive ordinates. A = catchment area in sq km.

Step II: Then, ordinate of UH =

Ordinate of direct run-off Direct run-off depth

Base Flow Separation: Locate a point E (Fig. 3.15) on recession limb of the flood hydrograph such that run-off after point E is only due to base flow. Similarly locate point A on rising limb. Join AE. Line AE represents base flow separation. To locate point E, a rough guides line is to find out N, given by N = 0.84 A0.2 Here N is in days. Just directly below point N is point E on the hydrograph as shown in Fig. 3.15. tp

N B

N C

Q

Flood hydrograph Q1 3

3 E

Q2

A

E F

2

1

T

Base flow 2

Fig. 3.15:  Base flow separation

50  Irrigation Engineering and Hydraulic Structures

DRH Q

Ordinate of DRH UH Ordinate of UH T

Fig. 3.16:  DRH and UH Alternatively, draw a tangent at A to meet the vertical from B at F. Join FE, and then AFE represents base flow separation line. In this case separation line falls from A to F and rises from F to E, as shown in Fig. 3.15.

Limitation of UH: (i) Applicable only for catchment areas less than 5000 sq km. (ii) Catchment characteristics have to be uniform and storm should occur with uniform intensity over entire catchment area, which rarely happens and so only small catchment areas are suitable for UH. (iii) Selection of design storm and its time duration has to be done carefully or else UH will not give correct results.

q  SOLVED EXAMPLES Example 3.11:  For a catchment area of 250 sq km, first two columns of the following Table indicate time in hour and flood flow in cumec. Assuming base flow of 10 cumec, derive and plot 6 hr. unit hydrograph. Also calculate rainfall excess for the storm. Time (hrs.) Col. No. (1)

00 06 12 18 24 30 36 42 48

Direct Run-off Ordinate (o) (cumec) (4)

UH Ordinate (cumec)

Q (cumec) (2)

Base Flow (cumec) (3)

10 190 305 227 148 94 81 35 10

10 10 10 10 10 10 10 10 10

0 180 295 217 138 84 71 25 0

0 20.6 33.9 24.9 15.8 9.6 8.1 2.8 0

Σ o = 1010

Col.(5) =

Col (4) 8.7

Hydrology  51



Direct Run-off Depth =



=

0.36 ∑ o × t A 0.36 ×1010 × 6 = 8.7 cm. = Rexcess 250

Column (5) indicate UH ordinate which can be plotted against time to give UH for 6 hr. Example 3.12  A storm gave rainfall excess of 4 cm., 5 cm and 6 cm at successive 4 hr. interval. Work out storm hydrograph for 4 hr. storm from the UH ordinates given below. Assume constant base flow of 10 cumec. Time

04

08

40

44

48

52

4 hr. UH ordinate

0

150 350 500 470 390 300 210 130 80

40

20

0

12

16

20

24

28

32

36

Solution: 4 hr. Ordinate UH Rainfall Base for storm Time ordinate excess Surface run off from Rexcess flow Hydrogph hrs. (Cumec) (cm) (cumec) (cumec) (cumec) Col. Col. (2) Col. (3) (4) = Col. (5) = Col. (6) = Col. Col. (7) Col. (8)* (1) (2) × 4 (2) × 5 (2) × 6 – – – 4 cm 5 cm 6 cm – – 04 0 4 0 – – 10 10 08 150 5 150 × 4 0 – 10 610 = 600 12 380 6 1400 150 × 5 0 10 2160 = 750 16 500 2000 1750 150 × 6 = 900 10 4660 20 470 1880 2500 2100 10 6490 24 390 1560 2350 3000 10 6920 28 300 1200 1950 2820 10 5980 32 210 810 1500 2340 10 4660 36 130 520 1050 1800 10 3380 40 80 320 650 1260 10 2240 44 40 160 400 780 10 1350 48 20 80 200 480 10 770 52 0 0 100 240 10 350 56 0 120 10 130 60 0 10 10

*Note: Col. ( 8 ) = Col. ( 4 ) + Col. ( 5 ) + Col. ( 6 ) + Col. ( 7 ) Plot Col. (8) against time to give storm hydrograph.

52  Irrigation Engineering and Hydraulic Structures

Example 3.13:  Two successive storms of rainfall excess 2.5 cm and 3 cm occurring at 6 hr interval give a hydro graph of which the ordinates against time are given below. Derive and plot 6 hr UH, for the catchment of 116.8 sq. km area Time (hr)

0

15

18

21

24

27

30

33

Hydrograph (cumec)

0 25 70 110 126 119

66

38

23

12

6

0

3

6

9

12

Solution: Total run off hydrograph is obtained by multiplying the ordinates of shifted UH by corresponding rainfall excess i.e., 2.5 cm and 3 cm, and adding. Thus, if U0, U1, U2, U3 etc. are ordinates of UH and Q0, Q1, Q2, Q3 are ordinates of total run off hydrograph, then Q3 Run-off

Q2

Q

U3

U3 U2

Q1

U0 0

U1

U1 3

UH of 6 hr each with a shift of 6 hr.

U2

6

9

12

15

18

21

24

27

30

33

Time (hrs.)

Fig. 3.17:  Run-off hydrograph U0

= 0

Q0

= 0

2.5 U1

= Q1

= 25

∴

U1

= 10

2.5 U2

+ 3 U0 = Q2

= 70

∴

U2

= 28

2.5 U3

+ 3 U1 = Q3

= 110

∴

U3

= 32

2.5 U4

+ 3 U2 = Q4

= 126

∴

U4

= 17

2.5 U5

+ 3 U3 = Q5

= 119

∴

U5

= 9.7

2.5 U6

+ 3 U4 = Q6

= 66

∴

U6

= 6.0

2.5 U7

+ 3 U5 = Q7

= 38

∴

U7

= 4.0

2.5 U8

+ 3 U6 = Q8

= 23

∴

U8

= 2.0

2.5 U9

+ 3 U7 = Q9

= 12

∴

U9

= 0

ΣU

= 108.7

Hydrology  53

Here Q3 = 2.5U3 + 3U1 Where U1 and U3 are UH ordinates. This is represented in the following table. Time

DRO

0

0

Q1

03

25

Q2

06

70

UH1 × 2.5

UH2 × 3.0

U1

–

U2

0

*

Q3

09

110

U3

U1

Q4

12

126

U4

U2

Q5

15

119

U5

U3

Q6

18

66

U6

U4

Q7

21

38

U7

U5

Q8

24

23

U8

U6

Q9

27

12

0

U7

Q10

30

06

0

U8

Q11

33

0

0

0

* DRO = 110 = 2.5 U3 + 3 U1 = 2.5 (32) + 3 (10) = 80 + 30 = 110 Volume of U1, U2 etc. can be plotted with a shift of 6 hr, to give two UH of 6 hr. duration each. To check whether UH hydrograph ordinates U1, U2, U3 etc. are correct,

A = 116.8 sq. Km. t = 3 hrs. since there are two 6 hr UH and so t = 6/2 = 3

Rexcess =

0.36 ΣU × t 0.36 ×108.7 × 3 = A 116.8

= 1.0 cm., which is correct, since UH represents 1 cm Rexcess. Example 3.14: Stream flows due to 3 successive storms of 3.5, 4.5 and 2.5 cm of duration 6 hr. each, and given below for a catchment area = 45.4 sq km. Assume base flow of 10 cumec and average storm loss of 0.25 cm/hr. Find 6 Hr UH ordinate. Time

0

Hydrograph ordinate

10 14 18 32 46 54 58 49 36 25 17 12 11 10

3

6

9

12 15 18 21 24 27 30 33 36 39

Solution: Steps: (1) Here storm loss of 0.25 cm/hr will give P1 = 3.5 – 0.25(6) = 2 cm  P2 = 4.5 – 0.25 (6) = 3 cm  P3 = 2.5 – 0.25 (6) = 1 cm

54  Irrigation Engineering and Hydraulic Structures

(2) Subtract BF = 10 cumec from given value of hydrograph-ordinates then calculate ordinate of unit hydrograph (UH) by formula,   Qn = P1Un + P2Un–2 + P3 Un–4 i.e., Q1 = P1U1 ∴ 14 – 10 = 4 = 2U1

∴ U1 = 2

  Q2 = P1U2 + P2U0 ∴ 18 – 10 = 8 = 2U2 + 3(0)  

∴ U2 = 4

Q3 = P1U3 + P2U3–2

32 – 10 = P1U3 + P2U1 = 2U3+3(2) ∴

22 = 2U3 + 6

∴ U3 = 8

Q4 = P1U4 + P2U4–2 + P3U4–4 46 – 10 = 2U4 + 3U2 + (1) U0 ∴ 36 = 2U4 + 3(4) ∴ U4 = 12 54 – 10 = 44 = 2U5 + 3U3 + 1U1 ∴ 44 = 2U5 + 3(8) + 1(2) ∴ U5 = 9 58 – 10 = 48 = 2U6 + 3U4 + 1U2 ∴ 48 = 2U6 + 3(12) + 1(4) U6 = 4 49 – 10 = 39 = 2U7 + 3(U5) + 1(U3) = 2U7 + 3(9) + 1(8) 39 = 2U7 + 27 + 8 ∴U7 = 2 36 – 10 = 26 = 2U8 + 3U6 + 1U4 26 = 2U8 + 3(4) + 1(12) ∴ U8 = 1 27 – 10 = 17 = 2U9 + 3U7 + 1U5 17 = 2U9 + 3(2) + 1(9)

∴ U9 = 1

12 – 10 = 2 = 2U10 +3 (U9) + 1(U8) i.e., U10 will be negative and hence U9 can be taken as zero. Thus; UH-ordinates are U1 = 2 cm; U2 = 4 cm; U3 = 8 cm; U4 = 12 cm; U5 = 9 cm; U6 = 4 cm; U7 = 2cm; U8 = 1 cm; U9 = 0. Σ U = 42 cm

Hydrology  55

Check, direct run-off = 0.36(ΣU)t A A = catchment area = 45.4 sq km 0.36(42)3 t = given time interval = 3 hrs = = 0.99 cm = 1 cm = Rexcess for UH 45.4

3.5  IUH, S AND SYNTHETIC UNIT HYDROGRAPH IUH, Instantaneous Unit Hydrograph, IUH was first developed by Clark in 1945. The development of IUH has removed the difficulty of using a unit hydrograph of specified time duration. IUH is a unit hydrograph as the result of an instantaneous application of 1 cm rainfall excess to a given catchment.

S Hydrograph: S, or Summation Hydrograph is used when unit hydrograph of any other duration from an available unit hydrograph of given duration is required to be obtained. S hydrograph corresponds to rainfall excess falling for an infinitely long duration. The S hydrographs have a typical S type rising limb and then they become asymptotic to a constant value Q. If an S hydrograph is displaced through T1 hours the difference in the ordinates between original S hydrograph and displaced S hydrograph would give ordinate of unit hydrograph of T1 hour duration as indicated in Fig. 3.18.

S hydrograph Q

S hydrograph with a lag of T1 hr

Ordinate of UH of T1 hr duration

T1 hours

Time

Fig. 3.18:  S-Hydrograph

Synthetic Unit Hydrograph: This is developed by Snyder in 1938 for region where no flood data is available. By making use of empirical relationship for similar watershed flood data values are worked out and unit hydrograph is prepared, which is known as Synthetic unit Hydrograph.

56  Irrigation Engineering and Hydraulic Structures

3.6 TO GET UNIT HYDROGRAPH OF DIFFERENT DURATION FROM THAT OF GIVEN DURATION If from a given unit hydrograph of N hour duration, another unit hydrograph of N’ hour duration is ­required to be obtained (here N’ =N) then S-curve technique is most suitable. Let NhUH represents N hour Unit hydrograph and N’hUH represents N’ hour Unit hydrograph. To get N′hUH, let S1 curve be drawn from NhUH and S2 curve be drawn with same ordinates as S1 but with a lag of N′ hour, as shown in Fig. 3.19. 1 Let two hyetographs as shown in Fig. 3.19 be drawn, each representing N cm/hr rain intensity. The difference between two hyetographs represents a storm N′ 1 of duration of N’hour with intensity of cm/ hr; and therefore a rainfall cm. N N Let S1 and S2 be two S graphs drawn with a lag of N′ hour. Both S1 and S2 represent NhUH but with a lag of N′ hour. The difference between ordinate of S1 and S2 has N′ to represent direct run-off due to rainfall of cm. Hence, if this difference of N ordinate is multiplied 1/N cm/hr

1 lag of N′ hr

2

1/N cm/hr

Q (cumec) S2 S1

S(t)

Ordinate for N′ hUH N′ = [(S(t)) – S(t – N′)] = N lag of N′ hr.

Time T. hrs

Fig. 3.19:  UH of different duration  N′  by   then, it represents 1 cm run-off and so ordinate of NhUH can be calculated N N′ from formula = [ (s(t)) − s(t − N′) ]. This will be clear from following numerical N problem.

Hydrology  57

q  SOLVED EXAMPLES Example 3.15: 4 hrUH ordinates against time are given in the following Table. Find out ordinates of 8 hour UH. Solution:  4hr UH ordinates Time (cumecs) (2) (1) U(t) 0

0

4

20

8

80

+

12

130

+

16

150

20

+

130

24

S-curve additions (3)

S-curve ordinates (4)

s(t – 4)

s(t)

S-curves Difference 8 hr UH lagged by of colm (4) ordinates colm 8 hr = (5) & (5) = (6) (6) × 4/8 =(7) s(t – 8)

–

0

–

0

0

0

20

–

20

10

→ 100 = → 230 = = →

100

0

100

50

230

20

210

105

380

100

280

140

380

510

230

280

140

90

510

600

380

220

110

28

52

600

652

510

142

71

32

27

652

679

600

79

40

36

15

679

694

652

42

21

40

5

694

699

679

20

10

44

0

699

699

694

5

2.5

48

–

699

699

699

0

0

20

Here N = 4 hr and N’ = 8 hrs i.e., N’ > N. 4 Ans: is given in col (7) i.e., ordinates of 8 hr UH = Column (6) × 8 Note: Col(3) S curve additions are obtained from col(4) as shown by arrows. Col(4) = col(3) + Col(2) = ordinates of S curve' col(5) is a s curve lagged by 8 hrs and so it starts with zero against 8 hr time and all subsequent ordinates of S curve col (4) are then obtained in col(5) as shown by arrows.

58  Irrigation Engineering and Hydraulic Structures

Example 3.16:  Derivation of 2 hr UH from 4 hr UH i.e., N' < N. Ordinates of 4 hr UH are given in the table below, find out ordinates of 2 hr UH by S curve technique. Solution: Time (t) hrs Ordinates S-curve S-curve (1) of 4hr UH additions ordinates (2) U(t)

(3) s(t–U)

(4) s(t)

S-cuve lagged by 2 hrs

Difference col (4)–Col. (5)

Ordinates of 2hr UH

(6)

(7)

(5) s(t–2)

Col.(6) +

0

0

0

–

0

2

36

36

0

36

72

4

91

0

91

36

55

110

6

93

36

129

91

38

76

8

68

91

159

129

30

60

10

49

129

178

159

19

38

12

34

159

193

178

15

30

14

23

178

201

193

8

16

16

13

193

206

201

5

10

18

6

201

207

206

1

2

20

1.5

206

207.5

207

0.5

1

22

0

207

207

207

0

0

24

0

207

207

207

0

0

+

4 2

4 = 2 hrUH ordinates 2 Column showing S-curve additions can be explained by this argument: Let S curve be derived from N hr UH, then if we take difference of two S-curves lagged by N hr, the difference graph is unit hydrograph of N hr UH. Hence, ordinate of N hr UH at any time t is given by Col(7) = col(6) ×

U(t) = s(t) – s(t – N) ∴ s(t) = U(t) + s(t – N) i.e., ordinates of S graph = ordinates of given UH + ordinates of S graph with a lag of N hrs. For example: Ordinates of S graph at 6hr col(4) = Valule of UH at 6 hr in col(2) + value of s ordinate at (6 – 4) hr in col(4) i.e., 129 = 93 + 36. The second term of the RHS of the equation given above i.e., s(t – N) is called S curve additions entered in col(3).

Hydrology  59

Example 3.17:  The 3hr UH of a catchment area = 20 km2 at one hour interval are given below. Obtain 2hr UH and find value of Qc for the S graph. 0, 0.40, 1.4, 4, 7.70, 10.0, 9.2, 6.6, 4.5, 3.8, 2.7, 2, 1.30, 0.9, 0.4, 0. Solution: Ordinate of 2 hr UH Col  (6) × 3/2

3hr UH m3/s

S-curve additions

S-curve ordinate

S-curve lagged 2 hrs

Diff. of Col.(4) – Col. (5)

(1)

(2)

(3)

(4)

(5)

(6)

0

0

0

0

1

0.4

0.4

0.4

0.6

2

1.4

3

4

4

7.7

5

10

6

Time

(7) 0

1.4

0

1.4

2.1

0

4.0

0.4

3.6

5.4

0.4

8.1

1.4

6.7

10.0

1.4

11.4

4.0

7.4

11.1

9.2

4.0

13.2

8.1

5.1

7.65

7

6.6

8.1

14.7

11.4

3.3

4.95

8

4.5

11.4

15.9

13.2

2.7

4.05

9

3.8

13.2

17.0

14.7

2.3

3.45

10

2.7

14.7

17.4

15.9

1.5

2.25

11

2

15.9

17.9

17.0

0.9

1.35

12

1.30

17.0

18.30

17.4

0.9

1.35

13

0.90

17.4

18.30

17.9

0.4

0.6

14

0.40

17.9

18.30

18.30

0

0

15

0

18.38

18.30

18.30

0

0

+ + +

Qc =

2.78A 2.78(20) = = 18.53m3 / sec N 3

Plot time v/s col(7), it gives 2 hr Unit hydrograph.

3.7  RAINFALL/FLOOD FREQUENCY ANALYSIS Hydrologic events like rainfall, floods can be analysed by applying statical methods and maximum probable rainfall or flood can be worked out and its probability can also be calculated.

60  Irrigation Engineering and Hydraulic Structures

Suppose a given sample size is N (number) i.e., N no. of values are given for either rainfall or flood year wise, and it is required to calculate how many times a particular flood value will occur or exceeded in an interval of say 100 years, then arrange the given number of events in descending order, i.e., start with highest value first, lower value next and so on such that lowest value is the last. Let m = descending order number N = sample size i.e., total number of events, then Probability, P =

or, Recurrence Interval = T =

m N +1

1 N +1 = P m

i.e., it will repeat after an interval of T yrs. The probability of occurrence of event r times in n successive years is given by Pr,n =

n! p r q (n − r) (n − r)!r!

Here q = 1 – p Also, probability of event not occurring at all in n successive years is given by Po, n = qn = (1 – p)n Probability of events occurring once in n successive years is P1, n =



n! (p)1(q)(n −1) (n − 1)!

P1 = 1 – qn = 1 – (1 – p)n = risk factor Here P1 = probability of event occurring at least once in n successive years. Van T Chow (1951) has expressed that probability distribution function applicable to hydrology is

x T = x + k T σ N −1



Here: xT = value of variate x of random hydrologic series with a return period of T.

Σx N    N = total number of events x = mean variate =

σN–1 = standard deviation of variate = KT = frequency factor =

YT − Y n sn

Σ(x − x)2 N −1

Hydrology  61

 T   YT = reduced vairate for a given T = − log e  log e  − 1   T   Y n = reduced mean = 0.533 for N between 20 to 30 = 0.577 for N > 100 Sn = reduced standard deviation = 1.2825.

For calculation of recurrence interval T following formulae are available  (i)

California method : T =

 (ii)

Allen-Hazen formula : T =

N m

2N 2m − 1 N +1 Weibull formula : T = m N +1 Gumbel formula : T = m + c −1

 (iii)  (iv)

but with c = 1, (Gumbel’s correction factor), It will be same as Hazen’s formula i.e., T = If not mentioned, T =

N +1 m

N +1 is generally acceptable value. m

q  SOLVED EXAMPLES Following solved examples will help in understanding above statistical approaches. Example 3.18:  Estimate flood peak with return period of 75, 125 and 900 years by Gumbel’s method from following 13 years data of flood. Values: 3300, 4200, 1300, 3250, 2700, 1900, 1950, 4510, 3175, 2640, 2840, 3504, 1775 (cumec) and N = 13 Solution:

Values in descending order (m)

T=

N + 1 13 + 1 = = 14 m 1

4510

14

4200

7

3504

4.65

3300

3.5

3250

2.8

62  Irrigation Engineering and Hydraulic Structures

Values in descending order (m)

T=

N + 1 13 + 1 = = 14 m 1

3175

2.3

2840

2.0

2700

1.75

2640

1.55

1950

1.40

1900

1.25

1775

1.16

1300

1.07

ΣQ = 37044 ∴Q =

ΣQ 37044 = = 2850 N 13

 Σ(Q − Q) 2  σ N −1 =    N − 1 

1

2

= 950 Now

KT =

For T = 75

YT − Y n , T = 75, 125 & 900 Sn

 T   But, YT = − log e  log e  T − 1     y75 = 4.31

∴ K 75 =

4.31 − 0.533 = 2.91 1.2825

∴ Q75 = Q + K 75σ N −1 = 2850 + 2.91(950)   

 = 5616 cumec

Similarly Q125 = 2850 + (3.5)(950) = 6175 and Q900 = 2850 + (4.896) (950) = 7501

Hydrology  63

Example 3.19:  Recorded annual rainfall year-wise starting from 1951 are given. Find the probability of rainfall equal to or exceeding 11 cm. Year

1951 1952 1953 1954 1955 1956 1957 1958 1959 1960 1961 1962

Rainfall(cm)

12

7.6

14.5

16

9.5

8

12.5 11.2

8.9

9

7.5

6

Solution:  Arrange given rainfall values in descending order and work out return period T = N + 1 where m is order number of rainfall m Return period T =

N + 1 12 + 1 = m m

– (R – R)2

S. No.

Rainfall (cm)

1

16

13

33.4

2

14.5

6.5

18.3

3

12.5

4.33

5.2

4

12

3.25

3.16

5

11.2

2.60

1.0

6

9.5

2.16

0.49

7

9

1.85

1.48

8

8.9

1.625

1.74

9

8

1.44

4.32

10

7.6

1.33

6.86

11

7.6

1.18

7.39

12

6

1.08

Σ (R – R) = 101.25

ΣR = 122.7

∴R =

17.8 2

101.15 122.7 = 10.22 cm & σ N −1 = = 3.03 12 − 1 12 R T = R + k T σ N −1

T  12    T = 12   ∴ YT = − ln ln = − ln ln  = 2.44  T − 1 11   

KT =

YT − 0.533 2.44 − 0.533 = = 1.489 1.2825 1.2825

∴ R12 = 10.22 + 1.489 (3.03) = 14.33 cm 1 For R = 11 cm what is P? i.e., = ? T

64  Irrigation Engineering and Hydraulic Structures

R T = R + K T σ N −1



∴ 11 = 10.22 + KT(3.03)



∴ KT =

11 − 10.22 = 0.257 3.03

YT − 0.533 1.2825 ∴ YT – 0.533 = 0.257 (1.2825) = 0.329 ∴ YT = 0.862 ∴ = 0.257 =



T   ∴ 0.862 = −  ln ln  T −1 

∴

T = 1.525 T −1

∴ T = 1.525T – 1.525 0.525T = 1.525

1.525 0.525 T = 2.9 1 1 ∴P = = = 0.344 T 2.9 ∴T =

P = 34.4% Example 3.20:  Flood values for a river are (i) Q = 600 m3/sec with return period of 50 years. (ii) Q = 700 m3/sec with return period of 100 years. Using Gumbel’s method estimate flood value for a return period of 150 years and – return period for a flood value of 900 m3/sec. Given : Yn = 0.54, Sn = 1.1284. Solution: (i) For return period of 50 years



 50  YT = Y50 = − ln  ln  = 3.9  49 

∴ K T = K 50 =

K50 =

YT − Y n Yn

3.9 − 0.54 = 2.97 1.1284

Hydrology  65

(ii) For return period of 100 years æ100 ÷ö = 4.6 YT = Y100 = - ln ln çç çè 99 ÷÷ø



K100 =

4.6 − 0.54 = 3.59 1.1284

∴ Q50 = Q + K 50 σ



   600 = Q + 2.97σ …(1)    700 = Q + 3.59σ  Subtracting (2) from (1) (3.59 – 2.97) σ = 700 – 600 = 100 100 ∴σ = = 161.2 0.62 Q = 600-2.97 × 161.2



∴ Q = 600 − 479 = 121 cumec



∴ Q150 = Q + K150 σ = 121 + K150 (161.2)

But, K150 =

  150   Y150 − Y 0 and, Y150 = − ln ln    = +5.0 Sn   140  

∴ K150 =



5.054 = 3.95 1.1284

∴ Q150 = 121 + 3.95 (161.2) = 757.74 m3/sec Now T = ? for Q = 900 m3/sec ∴ Q = 121 + KT (161.2) ∴ 900 = 121 + KT (161.2) 900 − 121 ∴ KT = = 4.83 161.2 4.83 =

YT − YN Sn

  ∴ 4.83 = YT − 0.54 1.1284   ∴ YT = 0.54 = 5.453 é æ T öù \ YT = 5.99 = - ln ê ln çç ÷÷ú êë çè T - 1÷øúû

…(2)

66  Irrigation Engineering and Hydraulic Structures

T = 1.002 T −1 ∴ T = (1.002)T – 1.002 ∴ 6.002T = 1.002 ∴

T=

1.002 = 501 years 0.002

Example 3.21:  Analysis of a 50 year flood data for a river gave Q = 1200 m3 /s and Standard deviation = 700 m3/sec.For what discharge would you design a structure on this river to provide 90% assurance that it will not fail in next 60 years. Given Yn = 0.54 and Sn = 1.1284. Solution:

Risk factor = 1 = (1 – p)n ∴ 100 – 90 = 10% i.e., 0.1 = 1 – (1 – p)60



(1 – p)60 = 0.9



∴ (1 – p) = (0.9)1/60 = 0.998



∴ p = 0.002



∴T =

1 = 500 yrs p

Now find Q500 by Gumbel’s method

∴ Q500 = Q + K 500  500  YT = Y500 = − ln ln   = 6.213  499 

K T = K 500 =    K

500

=

Y500 - Yn Sn

6.213 − 0.54 = 5.027 1.1284

Now, QT = Q + KTσ ∴ Q500 = 1200 + 5.027(700) = 4719.23 m3/sec

Example 3.22:  The 50 year-24 hour record of maximum rainfall at a station is 16 cm. Determine the probability of 24 hour rainfall of magnitude exceeding 16 cm, occurring at the station.

Hydrology  67

(i) once in 10 successive years (ii) twice in 10 successive years (iii) at least once in 10 successive years. Solution:  We have Pr,n =

n! (p) r (q) n − r (n − r)!r!

Here q = 1 – p and p =

1 1 = = 0.02 T 50

(i) P1,10 for case (i) P1,10 =

10! 1 (p )(1 − p)9 9!(1!)

= 10(0.02)(0.98)9 = 0.2(0.83)



= 0.166 i.e., 16.6% Probability

(ii) p 2,10 =

10! (p 2 )(2 − p)8 (8!)(2!)



= 45 (0.02)2 (0.98)8 = 0.0153



= 1.53%

(iii)

P1 = 1 – (1 – p)n



= 1 – (0.98)10



= 1 – 0.817



= 18.3%

EXERCISES 1. (a) Define: Orographic, frontal and convectional precipitation. (b) Explain how you would calculate missing precipitation data? (c) State permissible area per rain gage station 2. (a) Define: Hyetograph and Isohyet. (b) A square area of sides AB, BC, CD, and DE of 10 km length each, has rain gage stations at A, B, C, D and at O, which is ht center of the diagonal. The rainfall recorded by these station is 10, 12, 16, 22 and 25 cm respectively. Find mean precipitation by: (i) Arithmetic mean method and (ii) Thiessen polygon method

68  Irrigation Engineering and Hydraulic Structures

Draw the Thiessen polygons showing areas incharge of each station on a graph paper. (Ans: (i) 15 cm , (ii)14.7 cm) 3. A catchment of 2300 sq km. gave following hydrograph for a 6 hr storm. Derive and plot 6 hr. UH. What will be the rainfall excess due to the storm? Time (hrs)

00

06

12

18

24

30

36

42

48

Flow (Cumecs)

15

190

305

227

148

94

61

35

15

Base Flow (Cumecs)

15

10

5

7

8

9

11

13

15

Ans.: (1) UH Ordinates

0

19.22

32

23

15

9

Ans.: (2) Rexcess =

= 0.36

( ΣU)t A

=

0.36 × 997 × 6 230

5.35 2.35

0

= 9.36 cm.

4. (a) What is UH? What are its limitations? (b) In a catchment area, three successive storms of 2.5 cm, 4 cm. and 3.5 cm occurred at six hour interval. Draw the storm hydrograph when 6 hr. UH ordinates are given below. Take a constant base flow of 12 cumecs. Time (hrs.)

06

12

18

24

30

36

42

48

54

60

66

72

6 hr. UH ordinates (Cumecs)

0

255

570

750

705

585

450

315

195

105

30

0

Ans.:

12

650 2457 5059 6769 6919 5944 4646 3334 2156 1189 899

 

 

4 Ground Water 4.1  GROUND WATER OCCURRENCE AND RESOURCES Rain water that infiltrates into the ground get stored either in confined or unconfined aquifers. Pervious soil layer above an impervious soil stratum forms unconfined aquifer if it has water with free surface known as water table. If ground water occurs in a pervious layer, which is sandwitched between two impervious, layers then it is confined aquifer and water level in piezometric tubes represent imaginary ground water table. Thus, ground water is found below earth surface either in confined aquifer or in ­unconfined aquifer. Estimate of present ground water resources in India is of order of 650 cubic kilometers as against 1880 cubic kilometers of surface water. GL.

Capillary zone Water table Ground water in un-confined aquifer

Sand, Gravel etc.

Impervious layer Confined aquifer Impervious layer

Fig. 4.1:  Aquifers

4.2  WELL IRRIGATION Open dug wells or tube wells penetrating confined or unconfined aquifers form source of ground water and power is required to lift this water from the ground to surface so as to be useful for irrigation purposes and in that case it is lift irrigation or well irrigation. © The Author(s) 2023 S. K. Ukarande, Irrigation Engineering and Hydraulic Structures, https://doi.org/10.1007/978-3-031-33552-5_4

69

70  Irrigation Engineering and Hydraulic Structures

4.3  WELL HYDRAULICS A well is a hydraulic structure, which gives water from aquifers. When water is pumped from well, water table in case of unconfined aquifer or piezometric surface in case of confined aquifer gets lowered down and form a cone of depression around the well. Horizontal distance from the center of well to the point where cone of depression ends on the water table is known as radius of influence of well (g0). Difference between initial water table and its lowered level due to pumping from the well, is known as draw down. Well yield per unit draw down in the well is known as specific capacity of well. Ground water moment was first studied by Darcy in 1856, and according to him velocity of ground water is given by, V = Ki, (Darcy’s law) ∴   Q = AV = KAi Here: A = cross sectional area of aquifer through which ground water movement occurs, at a velocity of V m/sec, under a hydraulic gradient i, h − h1 ∴ i = 2 ∴    L where h2 and h1 refers to depth of water table at two points, L meter apart, If c/s area of aquifer is unity and hydraulic gradient i is also unity then, Q=k×1×1=K i.e., permeability of aquifer or hydraulic conductivity K, can be defined as rate of flow through the porous media under unit hydraulic gradient through unit crosssectional area and has dimensions of velocity i.e., L/T. It is expressed either in m/ day or cm/sec. Darcy’s Law is valid for laminar how through porous media having value of reynold number, R ≤ 1. However upto R = 10, Darcy’s law is assumed as applicable without appreciable deviation in the results. Here Reynold number R is given by R=

ρVD μ

where D = mean diameter of soil particle V = Velocity of flow through soil particles. μ = dynamic viscosity of water and ρ = mass density of water.

Ground Water  71

4.3.1  Steady State Discharge from Unconfined Aquifer Let p be a point on the cone of depression at radius r and depression head h then applying Darcy’s Law, Q = K A i, here K is permeability of aquifer in, m/s = K (2πrh) ∴Q

dh dr

dr = 2πkhdh r o = discharge from well GL r0

r

Cone of depression

p s h0 h

hw Impervious layer 2 rw = d = diameter of well

Fig. 4.2:  Case of unconfined aquifer r0

∴Q=

∫

rw

∴Q =

ho

 dr    = 2π k ∫ hdh  r  h w

π k ( h o2

− h 2w

r  ln  o   rw 

)

=

1.36 k (h o2 − h 2w ) r  log  o    rw 

(4.1)

If s = draw down at the well = ho – hw, then h o = s + h w Note :here h w = strainer length ho + h w = s + 2 h w h o2 − h 2w = (h o + h w ) ( h o − h w ) = (s + 2 h w ) (s) ∴Q =

π ks (s + 2h w ) 1.36 ks (s + 2h w ) =  ro  r  ln   log  o   rw   rw 

(4.2)

72  Irrigation Engineering and Hydraulic Structures

Since s = h o − h w and if s is small, then h o ≅ h w Now h o2 − h 2w = (h o + h w ) (h o − h w ) = 2h o (h o − h w ) = 2h o s, Putting this valuein equation (4.1), we get 2πKh o s Q = r  ln  o   rw 





(4.3)

For an unconfined, the coefficient of transmissivity is given by of transmissivity is given For an unconfined aquifer, the coefficient T = kh o (4.4) ∴Q =

 T 2πk   s  k

=

r  ln  o   rw 

2πTs r  ln  o   rw 

Co-efficient of transmissivity or transmissibility T is defined as the rate of flow through vertical strip of aquifer of unit width and extending for full saturated height under unit hydraulic gradient. If aquifer thickness is b, then, dh = unity dr ∴T = k b for confind aquifer of thickness b = k h o for unconfind aquifer of thickness h o T = kAi = k ( b ×1) (i), but, i =

Note: 1. Value of ro falls in range of 150 m to 300 m 2. ro can be calculated from Sicherdt expression, ro = 3000 s

4.3.2  Steady State Discharge from Confined Aquifer From Darcy’s law,

Q = kAi = k (2πrb) Q

dr = 2πkb dh r

dh dr

ro

ho

w

w

 dr  Q ∫   = 2πkb ∫ dh r  r  h

k

Ground Water  73



Q=

2πkb(h o − h w ) r  ln  o   rw 

..... (Thiem's Equation) (4.5)

If s = draw down = ho – hw, and k b = T, then Q=

2π Ts 2.72 T s =  ro  r  ln   log  o  r   rw   w  (4.6)

If there are two observation wells at radius r1 and r2 giving piezometric water level at h1 and h2 then Q=



2πkb (h 2 − h1 ) (4.7)  r2  ln    r1  Q = discharge from well GL

r0 Water table (Imaginary)

r

Cone of depression s

h0

Impervious layer

h hw

Confined aquifer of thickness b Impervious layer 2 rw = d = diameter of well

Fig. 4.3:  Confined aquifer Again s 2 = h o − h 2 and s1 = h o − h1

∴Q =

2πkb (s1 − s 2 ) 2πT (s1 − s 2 )  = r  r  ln  2  ln  2   r1   r1 

(4.8)

Here s1 and s2 are draw down in observation well no. 1 and 2 respectively, and T transmissibility of the aquifer.

74  Irrigation Engineering and Hydraulic Structures Q = discharge from well GL r0

r2 s2 s1

r1 s

h2

h0

h1 hw

Impervious layer

b Impervious layer

2 rw = d = diameter of well

Fig. 4.4:  Case of confined aquifer

q  SOLVED EXAMPLES Example 4.1:  A tube well of 30 cm diameter penetrates fully an unconfined aquifer, calculate its yield from following data: Draw down = 3 m Strainer length = 10m k = 0.05 cm / sec Radius of influence = 300 m Solution:  As per equation (4.2), Q=

πks (s + 2h w ) r  ln  o   rw 

=

1.36 ks(s + 2h w ) r  ln  o   rw 

here, h w =10 m , s = 3m, k = ro = 300 m and rw =

Q=

.05 m/s, 100

15 m 100

1 × 3(3 + 2 ×10) 1.36 ×15 × 23×10−4 100 = 3.3  300 ×100  log   15  

1.36 × .05 ×

=142 ×10−4 = 0.0142 cumec =14.2 lit/s

Ground Water  75

Example 4.2:  A tube well penetrates fully a confined aquifer of thickness 30 m and k = 4.6 × 10-4 m/s. Find well diameter if its yield has to be .04 m3/s, under a draw down of 4.1 m. Take radius of influence = 250 m. Solution:  Using equation (4.6), Q=

2.72 kbs 2.72 × 4.6 ×10−4 × 30 × 4.1 =  ro   250  log  log     rw   rw   

 250  ∴log   = 3.8474  rw  250 ∴ = 7037.849 rw 250 = 0.0355 m 7037.849 diameter of well = 0.07 m = 7 cm

∴rw =

Example 4.3:  A pumping test on a tube well of 20 cm dia. penetrating unconfined aquifer gave following data: Q = 200 m3/hr. RL of water level in well before pumping = 138 m RL of water level in well at constant pumping = 134 m RL of impervious strata at well bottom = 108 m. RL of water in observation well = 137 m Radial distance of observation well = 40 m Find (i) Permeability of aquifer (ii) Radius of influence of the well Solution: Q=

(

2 2 1.36 k h 222 − h112

 r2  log  2   r11 

)

20 r11 = =10 cm = 0.1m = rw w 2 r22 = 40 m h0 = 138 – 108 = 30 m h1 = 134 – 108 = 26 m h2 = 137 – 108 = 29 m Q = 200 m3/hr

76  Irrigation Engineering and Hydraulic Structures

1.36 k (292 − 262 )  40  log    0.1  200 × 2.6 ∴k = = 2.3m/hr = 6.4 ×10−4 m = 55 m/day 1.36 (3× 55) ∴200 =

Also,

Q=

200 =

1.36k(h o2 − h 2w ) r  log  o   rw  1.36 (2.3) (302 − 262 )  r  log  o   0.1 

 r  1.36 × 2.3× 56 × 4 = 3.5 log  o  = 200  0.1  ∴

ro = 3186 ∴ ro = 318.6 m. 0.1

Example 4.4:  Constant pumping rate from a well of 30 cm dia. penetrating unconfined aquifer is 1800 LPM. A test well 30 m away gives draw down of 2 m and another test well 60 m away gives draw down of 1 m. Depth of water in well before pumping = 50 m. Find radius of influence and transmissibility. Q=

Solution:

∴Q =

1.36 k(h 22 − h12 ) r  log  0   γ1 

1.36 k (502 − 482 ) 1.36 k (502 − 492 ) =  r0   r  log  log  0   30    60 

 r  log  0   30  (98 × 2) = (99 × 1)  r0  log   60  1.97 =

log (r0 ) − log (30) log (r0 ) − log (60)

∴1.97 =

x − 1.47 ; here x = log ( r0 ) x − 1.77

1.97 (x − 1.77) = x − 1.47 1.97 x − 3.5 = x − 1.47 0.97 x = 2.03

Ground Water  77

1.97 (x – 1.77) = x – 1047 1.97 (x – 3.5 = x – 1.47) 0.97 x = 2.03 2.03 = 2.09 = log (r0 ) 0.97 ∴ r0 =123m x=

Now, Q = 1800 lpm = 0.03m3 /s. ∴0.03 =

∴k =

1.36 k (502 − 492 )  123  log    60 

0.03× 0.31 = 6.94 ×10−5 m/s 1.36 × 99 ×1

Transmissivity = T = kb = 6.94 ×10−5 × 50 ,since b = 50 ∴ T = 3.47 ×10−3 m 2 /s Example 4.5:  A tubewell penetrates confined aquifer fully the thickness of aquifer is 25 m and K = 40 m/day. If yield from well is 30 LPS under a draw down of 3.5 m, find radius of well. Radius of influence can be taken as recommended by Sicherdt. Solution: As per equation (4.6) Q=

2.72kbs r  log  0   rw 

Here S = 3.5 m, r0 = 3000s K = 226m where K = 40 m/day =

40 = 4.62 × 10–4 cm/sec, b = 25 24 × 3600

Q=

Now Q = 30 LPS =

2.72 × 4.62 × 10−4 × 25 × 3.5  226  log    rw 

30 = 0.03 m3/sec. 1800

∴ 0.03 =

2.72 × 4.62 × 10−4 × 25 × 3.5  226  log    rw 

78  Irrigation Engineering and Hydraulic Structures

 226  ∴ log   = 3.685  rw  ∴

226 = anti log (3.665) = 4623 rw

Radius of tube well = rw =

226 = 0.048 m = 48 mm . 4623

Example 4.6:  A 20 cm diameter fully penetrating well in an unconfined aquifer 30 m deep was pumped at 2800 lit/sec, and draw downs in two test wells at 50 m and 110 m from the pumped well are 4 m and 2m respectively. Find R and T. Solution:  Here draw down S1 = 4 m at r1 = 50 m and S2 = 2m at r2 = 110 m and T = KH = Transmissibility Now,





Q=

2.72KH(S1 - S2 ) (as per equation 4.9) log10 (r2 / r1 )

2800 2.72(T)(4 − 2) = 1000  110  log10    50 

 110  2.8log    50  = 0.176 m 2 /sec ∴T = 2.72 × 2 = KH = K(30) K=

0.176 30

= 5.86 × 10–3 m/sec Draw down at R (radius of influence) will be zero. ∴ S2 at r2 = R will be zero in above equation

∴Q =

∴ log(R / r1 ) =

2.72T(S1 ) log(R / r1 )

2.72 × 0.176 × 4 = 0.683 2.8

Ground Water  79



∴ R/r1 = 4.828

R = 4.828 × 50



R = 241m

Example 4.7:  A 30 cm diameter well peretrates 25 m below ground water table. If steady state discharge from the well, is 90 LPS giving a draw down of 0.53m at 90 m and 1.11m at 30 m from the main well, find T, R and draw down in the main well. Solution:  As well penetrates below ground water table, it is a case of unconfined aquifer ∴Q =

2πKH(S1 − S2 ) r  ln  2   r1 

Here H = 25m, S1 = 1.11m at r1 = 30 m and S2 = 0.53 m at r2 = 90 m and Q = 90 LPS.

∴



90 2πK(25)(1.11 − 0.53) = 100 ln(90 / 30)

0.09 =

157K(0.58) 1.098

∴K =

0.09 × 1.098 = 1.08 × 10−3 m / sec. 157 × 0.58

∴ Transmissibility, T = KH = 1.08 × 10–3 × 25 = 0.027 m2/sec To find draw down in the well, Here h1 = 25 – S1= 25 – 1.11 = 23.89 and r1 = 30 m, rw = 0.15 Q=

0.09 =

πk(h12 − h 2w ) r  ln  1   rw 

π(1.08 × 10−3 )(23.892 − h w 2 )  30  ln    0.15 

∴ 23.892 − h 2 w =

0.09 × 5.298 π × 1.08 × 10−3

= 0.14 × 103

∴hw2 = 570 – 140 = 430 ∴ hw = 20.73 m

80  Irrigation Engineering and Hydraulic Structures



∴ draw down in well = 25 – 20.73 = 4.27 m

To find R: Q=



πK(H 2 − h w 2 ) ln(R / rw )

 π(1.08 × 10−3 )(625 − 430) = 0.09 

R ∴ ln   rw



=

0.661 = 7.34 0.09



\

R = 1547 hw



∴ R = 1547 × 0.15 = 232 m

4.4  RECUPERATION TEST In recuperation test water is pumped from the well so that sufficient depression head is developed. Then water level is allowed to rise to the level which was in the well prior to pumping and time taken is noted. Then yield of the well can be worked out as under:Referring to Fig. 4.5 Let h1 = depression head when pumping was stopped h2 = depression head at time t after pumping was stopped. Thus water level in well recuperated = h1 – h2 in time interval t. dh = small recuperated head, generating volume dv = Adh, in interval of time = dt If Q is recharge rate into the well at time t, under depression head h, then Q GL Original water level h2 h h1

dh Water level when pumping stopped hw

b Cross sectional area at bottom =

Fig. 4.5:  Recuperation test

Ground Water  81

dv = Q dt, here Q = C h , (C is constant ) But,

dv = –Adh,

(–ve since upward movement of water ) ∴ dt = −

A dh C h

Let h = h 2 after time t, then

h

t=−

∴

A 2  dh  A  h1     = + ln   ∫ Ch  h  C  h2  1

(4.9)

h  C 2.3 = log  1  A t  h2 

C = Specific yield or specific capacity of open well which is defined as A volume of water that percolates into the well per unit time under unit depression head, which is expressed as cubic meter per hour per sq. m. of area of well under unit depression head.

4.4.1  Safe Yield Under a Constant Depression Head C Knowing the value   as above, the rate of yield from a well under a constant A depression head ‘h’ may be found out: C Q = Ch = (A.h) A π C Now put value of   as per eqn (4.9) and given A = d 2 4 A   where d = diameter of well and h = given depression head

 2.3  h   π  Q= log  1    d 2  (h)   t  h 2    4 

(4.10)

4.4.2  Interference of Wells If two or more wells are located in such a may that their draw down curves intersect each other they are known as interfering among themselves. Due to this interference

82  Irrigation Engineering and Hydraulic Structures

discharge of each individual well is ­reduced. Discharge from interfering wells can be calculated by making use of Muskat formulae given below: 1.  For confined aquifers of thickness b: Q1 = Q 2 =



2πKb(H − h w )  R2  ln    rw D 

where, R = Radius of influence

b = thickness of confined aquiter

               rw = well radius           D = distance between two wells. 2.  For unconfined aquifers: πK(H 2 − h w 2 ) Q1 = Q 2 =  R2  ln    rw D  Note that R > D, as shown in Fig. 4.6. Q2

Q1

H

H D

hω

rω

R

Fig. 4.6:  Interference of wells

q  SOLVED EXAMPLES Example 4.8:  Two tube wells of 25 cm diameter each are spaced 80 m apart and penetrate fully a confined aquifer of depth 12 m. Find discharge if only one well is discharging under a draw down of 2.5 m. what will be percentage decrease in discharge if both well discharge under draw down of 2.5 m. Take R = 230 m and k = 60 m/day.

Ground Water  83

Solution:  Here S = 2.5 m, b = 12m 25 = 12.5cm 2 2.72kbs Q from one well = log(R / rw ) rw =



   =



=

2.72 × 60 × 12 × 2.5 4896 = = 1500 m3/day 3.264  230  log    0.125  1500 = 0.01736m3 /sec = 17.36 LPS. 24 × 3600

When both wells are discharging under a draw down of 2.5 m, Q1 = Q 2 =

2πKb(H − h w )  R2  ln    rw D 

But, H - hw = S = 2.5 m ∴ Q1 = Q 2 =

2π × 60 × 12 × 2.5 11304 = = 1318 m3/day 2 8.573  (230)  ln   0.125 × 80    = 15.2 LPS

∴ % Increase in Q =

17.36 − 15.2 = 0.124 = 12.4% 17.36

Example 4.9:  A well penetrates an unconfined aquifer of thickness 100 m. If 12 m draw down gives a discharge of 250 LPM, find discharge at draw down of 18 cm. Solution:  Case I: for draw down of 12 m, hw = 100 – 12 = 88 m.

Q= ∴ 250 =

∴

πK(H 2 − h w 2 ) 2.3log10 (R / rw ) πK 2.3log10 (R / rw )

πK 2.3log10 (R / rw )

=

(1002 − 882 )

250 (188 × 12)

84  Irrigation Engineering and Hydraulic Structures

Case II: draw down = 18 m, hw : 100 – 18 = 82m

∴Q =



=

(

πK 1002 − 822

)

2.3log10 (R / rw ) 250 × 182 × 18 188 × 12

= 363 LPM       

4.5  WELL SHROUDING Shrouding is a layer of course material such as gravels and rubbles packed around strainer so that fine material of aquifer does not clog the strainer. (See Fig. 4.7) Shrouded well has greater specific capacity than the one without shrouding. Shrouding material is introduced between casing pipe and well pipe and then casing pipe is removed. GL. Soil Well pipe Impervious layer

Aquifer

Impervious layer Shrouding Strainer Concrete plug at the bottom

Fig. 4.7:  Well shrouding

4.6  WELL DEVELOPMENT Well development is the process of removing fine material from the aquifer surrounding the strainers. It increases specific capacity of well, prevents aquifer sand flowing into the well, and life of well can be increased. It is carried out by any of the following methods. (i) By pumping and surging. (ii) By back washing

Ground Water  85

(iii) By compressed air (iv) By making use of chemicals.

4.7  COLLECTOR OR RADIAL WELL Collector or radial well is used when high rate of discharge from a relatively thinner aquifer is desired. They are normally provided near river bank or in the river bed by providing horizontal strainers along circumference of well casing and these strainers penetrate horizontally into the aquifers, causing high rate of yield. Baroda city water supply system is dependent on such wells driven in bed of River Mahi.

q  SOLVED EXAMPLES Example 4.10:  A 30 cm diameter well fully penetrates an unconfined aquifer. For an effective length of strainer of 12 m, draw down of 3.7 m, radius of influence of 280 m, and permeability coefficient of 42 m/day, find yield of the well. Solution: Q=

1.36 ks (s + 2h w ) r  log  o   rw 

here, k = 42 m per day = 4.8 × 10−4 m / sec, s = 3.7 m, h w = 12 m, r0 = 280 m and rw = 15 m Q=

=

1.36 × 4.8 ×10−4 × 3.7 × (3.7 + 2 (12))  280  log    0.15  24.15 ×10−4 (27.7) = 204.5 × 10−4 m3 /s = .02 m3 /s or 20.4 lit/sec 3.27

Example 4.11:  A 10 cm dia tube well is working under a draw down of 4 m, and has 10m long strainer. The thickness of aquifer is 15 m, K = 30 m/day, find yield. Also Calculate yield for 100% increase in diameter and 50% increase in draw down. Solution:  From Sichardt’s expression, ro = 3000 s k = 3000 (4) 30 / (24 × 3600) = 223 m. k = 30 m / day = 3.47 × 10−4 m/s.

86  Irrigation Engineering and Hydraulic Structures

Q=

1.36 ks (s + 2h w ) 1.36 × 3.47 ×10−4 (4) × (4 + 2 × 10) = r   223  log  log  o    .05  r  w

             

=

1.36 × 3.47 × 10−4 × 4 × 24 = 124 × 10−4 m3 /s 3.65

= 0.0124 m3 / s =12.4 lit/sec

2nd case : New dia = 2 × 10 = 20cm s =1.5 × 4 = 6m ro(new) = 3000 (6) (0.018) = 324 Q

new

=

1.36 × 3.47 ×10−4 (6) × (6 + 20) 736 × 10−4 = 3.51  324  log    0.1 



= 209 × 10 −4 = 0.0209 m3 /s or 20.9 LPS

4.8  WELL LOSSES Actual draw down is slightly higher than the draw down due to logarithmic equation for yield from the well. The difference is known as well loss. This is due to the fact that logarithmic equation represents loss due to movement in aquifer only, whereas the loss that is occurring on account of flow through strainer and turbulent condition near the well is not accounted in that equation. Hence, actual draw down is higher and this difference between actual draw down and theoretical draw down is the well loss (see Fig. 4.8).

Fig. 4.8:  Well losses

Ground Water  87

4.9  UNSTEADY GROUNDWATER FLOW TOWARDS WELLS If cone of depression does not vary with respect to time, it is steady ground water flow towards wells but, if cone of depression changes with respect to time it is a case of unsteady ground water flow towards wells (see Fig. 4.9). Mr. Thies in 1935 developed partial differential equation governing unsteady flow toward's wells, (4)s = draw down at radial distance r at time t after pumping started    Jacob obtained value of s as  (4.10) s=

2.3Q  2.25Tt  log  2   r S  4πT Q GL

s

Water table (Piezometric surface)

r P

Pais a point at radius r at time t after pumping started H

h

Impervious strata Confined aquifer Impervious strata

Fig. 4.9:  Unsteady flow towards well Storage coefficient, S : It represents the volume of water released by a column of confined aquifer of unit cross - sectional area under a unit decrease in piezometric head & S and T are known as formation constants of an aquifer and play very important role in unsteady flow through porous media. Here S is storage coefficient or storativity, T = transmissivity, r is the radius at which point P on the depression curve is chosen and h is piezometric head. Solution of the equation as obtained by Thies is H−h =s=

∞

Q e− u Q × W(u) du = ∫ 4πT u u 4πT

 r 2S , 4Tt (2) h = H for t = 0 i.e. before pumping started. (1) here, u =

∞ −u

(3) W(u) = ∫ u

e

u

du

(4.11)

88  Irrigation Engineering and Hydraulic Structures

(4)s = draw down at radial distance r at time t after pumping started Jacob obtained value of s as

   s =

2.3Q  2.25Tt  log  2   r S  4πT



(4.12)

s=

 Q   4Tt 0  ln  2  − 0.5772   4πT   r S  

(4.13)

If s1 and s2 are draw down at time

(s 2 − s1 ) =

Q   t 2  ln     4πT   t1  

(4.14)

If the draw down S is plotted against t on a semilog paper (S on Y-axis, ordinary scale and t on x-axis log scale) the plot will be a straight line. The slope of this line enables to determine storage coefficient – S i.e., when S = 0, t = to in the eqn (4.13)  Q   4Tt o  ln  2  − 0.5772   4πT   r S  

i.e.,

0=



 4Tt  ∴ ln  2 0  = 0.5772  r S 



∴S =

4rt o 2 0.5772

r e

=

2.25Tt o r2



(4.15)

here r is distance between observation well and pumped well. In using equation (4.15) value of T is required, which can be obtained as under: From the graph as plotted above, take any two values s2 and s1 at time t2 and t1 respectively and then using equation (4.14), for given Q, T can be calculated. After this, use of equation (4.15) gives storage coefficient S. This is illustrated in following problem.

q  SOLVED EXAMPLES Example 4.12:  A 20 cm well penetrates a confined aquifer fully and is pumped at constant rate of 1000 LPM. The draw downs at an observation well at 30 m from pumping well are observed as under: Time (t) min

1.0

2.5

5

10

20

50

100

500

1000

Draw down S (m)

0.2

0.5

0.8

1.2

1.8

2.5

3.1

4.4

5.0

Calculate aquifer parameters, S and T.

Ground Water  89

Solution:  Figure 4.10 shows the plot of T v/s S which is a straight line for t ≥ 10 min and when this straight line cuts X-axis time t = t0 at s = 0 is noted. At s = 0, t = to = 2.5 min from graph (see Fig. 4.9) s1 = 3.1 m at t1 = 100 min t (min) 10

100

1000

0

log scale

1 2

Extend this line to cut X axis so that t = t0 at S /=0

3 S

4 5

T

Fig. 4.10:  Plot of S vs T s2 = 5.0 m at t2 = 1000 min Q = 1000 LPM = 0.016 m3/sec. From Eq. (4.14) s 2 − s1 = ∴ 5 − 3.1 =



∴T =

t  Q ln  2  4πT  t1 

0.016  1000  0.016(4.6) ln  = 4πT  10  4πT

0.016(4.6) = 3.08 × 10−3 m 2 / sec 4π(1.9)

2.25T t 0

=



= 1.155 × 10–3

r2

=

2.25(3.08 × 10−3 )(2.5 × 60)



(30) 2

∴ Storage coefficient = 0.001155 here r = 30 m = radial distance of observation well from pumped well.

90  Irrigation Engineering and Hydraulic Structures

Example 4.13:  The time draw down data from an observation well at 64 m from pumping well are given below. Find S and T if Q = 1.9 m3/min. Time (min)

1.5

3

4.5

6

10

20

40

100

S(m)

0.2

0.65

1.05

1.45

2.45

3.75

5.15

6.95

Solution:  We know s1 = 2.45 m at t = 10 min s2 = 6.95 m at t = 100 min Q = 1.9 m3/min = 0.0316 m3/sec t  Q ln  2  4πT  t1 

s 2 − s1 = 6.95 − 2.45 =

∴T =

0.0316  100  ln   4πT  10 

0.0316 × 2.3 = 1.28 × 10−3 4π × 4.5 S=

2.25T t o r2

Here to = 2.5 min = 150 sec at s = 0 (from graph see Fig. 4.11), r = 64 m ∴S =

2.25 × 1.28 × 10−3 × 150 432 = 64 × 64 64 × 64 S = 1.05 x 10–4

Y Ordinate Value 8 6 S cm

4 2 x log scale t0

2.5 10 min at

100

1000

time (min)

S=0

Fig. 4.11:  Plot of T vs S

Ground Water  91

Example 4.14:  During a recuperation test water level was depressed by 3 m and it recuperated to 2 m in 90 min. Find discharge for a depression head of 3.3 m in a 6 m diameter well and size of well to yield 15 LPS for a draw down of 2.8 m. Solution:  Here s2 – s1 = 3 – 2 = 1 m in 90 m. C=

2.3 s  3 2.3 log   = 7.5 × 10–5 log  2  = T 2  s1  90 × 60

π 2 π 2 d = (6) = 9π. 4 4 Q = CAH = (7.5 × 10–5) (9π) (3.3) = 0.007 m3/sec

A = cross-sectional area of well = Case (2):

Q = 15 LPS = 0.015 & H = 2.8m. ∴ 0.015 = CAH = (7.5 × 10–5) (A) (2.8) π ∴ A = 71.4 = d 2 4 ∴ d = 9.5 m.

Example 4.15:  Estimate Q from a well in a confined aquifer of 30 m depth. Distance of observation well from main well is 100 m and draw down values are 1.5 m after 4 hrs and 2.0 m after 16 hrs. Take S = 0.0003 Solution:  Plot graph of S v/s t and get to = 3.6 min, which is the value of t at S = 0 2.25T t 0 ∴S = r2 0.0003 =

2.25T(3.6 × 60) (100) 2

∴ T = 6.13 × 10−3 m 2 / sec

s 2 − s1 =

t  Q ln  2  4πT  t1 

Here s1 = 1.5 m at t1 = 4 hrs and s2 = 2.0 m at t2 = 16 hrs. ∴ 2 − 1.5 =

Q

 16  ln   4π(6.13 × 10 )  4  −3

∴ Q = 0.0278 m3/sec

92  Irrigation Engineering and Hydraulic Structures

Note:  Just as draw down values in observation well against time gives aquifer constant S and T, similarly draw down values in observation wells located at given distances from the main well will also give S and T will be clear from following numerical problem. Example 4.16:  A well in a confined aquifer was pumped for 2 hrs at a constant rate of 1600 LPM and the draw down in the seven nearly observation wells are given below. Determine aquifer constants, S and T. Observation wells

A

B

C

D

E

F

G

Distance (m) from main wall

5

10

20

40

80

120

200

5.35

4.35

3.35

2.35

1.4

0.8

0.3

Draw down (m)

Solution:  Plot distances (m) on log scale on X axis and S(m) on Y-axis in ordinary scale. For S = 0, ro = 210 m from the graph as shown in Fig. 4.12 Also from the graph ∆s = 3.25 m per log cycle of r (distance) i.e., s value at 10 = 4.35 m and s value at 100 = 1.10 m ∴ Δs = 3.25 m 7 6 5 4 draw down 3 S (m) 2 1

r = 210 m for S = 0 log scale 10 Distance (m)

100

1000

Fig. 4.12:  Plot of distance vs draw down Now for data giving different values of s in wells situated at different values of r, use the formula:

∆s =

2.3Q 2πT

Ground Water  93

 1.6  2.3    60  ∴ 3.25 = 2πT



∴ T = 3 ×10−3 m 2 /sec 



S=



2.25T t r02

(Here, ro = 210 m as available from graph and t = 2 hrs given). =



2.25(3 × 10−3 )(2 × 60 × 60) (210) 2

= 1.1 × 10–3

EXERCISES 1. A 30 cm diameter well penetrates fully an unconfined aquifer of 25 m depth. After steady state pumping at the rate of 90.0 LPS, from a test well at 90 m from the main well draw down observed was 0.53 m and in another test well at 30 m distance from main well gave draw down of 1.11 m. Find transmissibility of the aquifer and draw down in the main well. [Ans: (i) 1.68 m2/min and (ii) 13.0 m]

2. A bore well of diameter 15 cm. Penetrates fully a confined aquifer of 20 m. depth. If well gives a maximum discharge of 20 LPS, what will be the strainer length which has 15% area of opening and safe entrance velocity for the given aquifer is 0.02 m/s? Hint : Q = 0.02 m3/s Ao = Area of opening per m length of strainer = π D (0.15) = 0.070 m² / m length of strainer. Vo = Safe entrance velocity = 0.02 m/s.   Q 0.02 = = 14.2 m.   Ans: strainer length = Vo A o 0.02 × 0.070  

94  Irrigation Engineering and Hydraulic Structures

3. In an artesian aquifer (confined aquifer) of 8 m. thickness, a 10 cm, diameter well gives a steady state discharge of 6000 LPH. Two observation well located at 10 m. and 50 m from the main well gave steady state draw down of 3 m and 0.05 m respectively. Find transmissivity T and hydraulic conductivity K (i.e., permeability) of the aquifer. [Ans: T = 12.5 m² / day K = 1.563 m / day] 4. A well of 50 cm diameter penetrates a confined aquifer of 20 m thickness having hydraulic conductivity equal to 8.2 × 10-4 m/s. What is the maximum yield that can be expected from this well if draw down in the well is not to exceed 3m? Take radius of influence as 250 m. [Ans: 45 LPS.] 5. (a) Define: Transmissibility, permeability and storativity aquifers. (b) A recuperation test was conducted on an open well of diameter 6.5 m, giving following data: (i) RL of water table = 238 m. (ii) RL of water level when pumping stopped = 230 m. (iii) RL of water level after 2.5 hours after pumping was stopped = 234 m.

Find safe yield of the well if working head is 3 m.

[Ans: 7.5 LPS.]  

 

5 Reservoir Planning 5.1 INTRODUCTION Rivers have very little flow during post-monsoon and high flow during monsoon months. In order that high flow water is retained to take care of period of scarcity, usual procedure is to construct a dam and create a reservoir just upstream of it so that it can supply water during low flow or during drought. Thus, reservoirs are useful to regulate flow in rivers, prevent flood havoc and are useful during drought.

5.2  TYPES OF RESERVOIRS Reservoirs are classified as follows: 1. Multipurpose Reservoirs : Reservoirs catering to needs like irrigation, water supply, industrial requirement of water, hydropower production and navigation are called multipurpose reservoirs. 2. Flood Control Reservoirs : Reservoirs constructed for single purpose of controlling water during flood and releasing when the flood intensity becomes tolerable, then such reservoirs are known as flood control reservoirs. These are of two types: (a) Retarding and (b) Detention reservoirs. (a) Retarding Reservoirs : These are meant for holding flood water such that discharge passing over ungated spillway is not causing damage to downstream side. (b) Detention Reservoirs : In such reservoirs outlets and spillways are controlled by means of gates to allow quick disposal of flood water and can also be closed to detain requisite amount of flood water, which may be made useful during other period of demand. 3. Distribution Reservoirs : These are small storage reservoirs used for water supply to city and industrial areas. During peak demands, these need to be supplied by pumping water from external sources. During low demand period these can be used for storage. © The Author(s) 2023 S. K. Ukarande, Irrigation Engineering and Hydraulic Structures, https://doi.org/10.1007/978-3-031-33552-5_5

95

96  Irrigation Engineering and Hydraulic Structures

5.3  INVESTIGATION FOR RESERVOIR SITES Three types of surveys are carried out for investigating suitable site for reservoirs. viz., 1. Engineering Surveys, 2. Hydrological Surveys and 3. Geological Surveys. 1. Engineering Surveys : The area of dam site and reservoir site i.e., valley region just upstream of the dam is surveyed and contour plan is prepared. From this contour plan, area-elevation and volume–elevation curve (Fig. 5.1) of the reservoir sites are prepared by plotting. (i) Reservoir site areas against various elevations for dam site, known as area-elevation curve. (ii) By knowing contour areas of reservoir sites at various elevations, one can work out volume by making use of either of the following formulae: (a) Trapezoidal Formula h ( A1 + A 2 ) 2        where, h is contour inerval between contour areas A1 and A 2 . V=∑

(b) Cone Formula V=∑

h A1 + A 2 + A1A 2 3

(

)

(c) Prismoidal Formula: h V = ∑ ( A1 + 4A 2 + A3 ) 6 By plotting cumulative volume against elevation, volume-elevation curve is obtained.

2 1

Elevation in m.

1.  Volume (ha-m) 2.  Area (ha)

Fig. 5.1:  Area elevation and volume-elevation curves

Reservoir Planning  97

With the help of area-elevation curve and volume-elevation curve, height of the dam for requisite volume of water to be stored in reservoirs can be fixed. These curves help in determining evaporation losses at various elevations by knowing areas at these elevations and respective evaporation loss from experimental data. These curves also help in deciding area of submergence of land for storage at various elevations. Thus, these two curves are starting point for design of a dam and reservoir. 2. Hydrological Surveys : Reservoir planning depends on hydrological surveys, and hence they are required to be carried out to know the following: (i) Study of run-off pattern of river at the proposed dam site to determine storage capacity and height of dam. (ii) Study of flood hydrographs at the dam site to determine spillway capacity and flood control measures. Area of reservoir at the height of the dam is known as water spread area, whereas the area up stream of water spread area, which feed rainwater to reservoir site, is known as Catchment area. Hydrological surveys are therefore required to include the surveys for Catchment areas also. This will provide reliable run-off data and flood flow data at dam site. The hydrological data if available for at least past thirty-five years, then future pattern of floods and run-off can also be predicted by flood frequency analysis. This is a must for any multipurpose reservoir planning. 3. Geological Surveys : Geological investigations of dam and reservoir sites are carried out to determine, (a) Suitability of foundation for dam site. (b) Water tightness of reservoir basin. (c) Location of quarry site for materials of construction required for dam and spillways. (d) To ascertain whether dam site is in seismic zone or not. (e) Whether sound rock foundation is available or not is the prime information required to determine the type of the dam. Hence, detailed foundation explorations are carried out and faults, cavities, fractures etc. if present in rock strata are located and ascertained whether they are permissible or not.

5.4  ZONES OF STORAGE IN A RESERVOIR The zones of water storage in reservoir are as follows: 1. Dead Storage: Volume of water held below Dead Storage Level (DSL) is known as dead storage. This volume may get filled up by sedimentation in reservoir that may take place year by year during life of reservoir. This storage is not useful and hence is known as dead storage.

98  Irrigation Engineering and Hydraulic Structures

2. Live Storage: Water stored between maximum water level (i.e., FRL, full reservoir level) and minimum water level (LWL, lowest water level) is known as live storage or useful storage. If upper part of reservoir i.e., between spillway crest level and FRL is supposed to hold flood water for some time to moderate peak flood value then volume of water between FRL and spillway crest level is known as surcharge storage or flood absorption capacity of reservoir, and in that case live storage is only between spillway crest level and LWL. Live storage also includes evaporation and seepage losses from reservoir. FRL

Free board

Spillway crest

Surcharge storage

Spillway

Level Live storage Supply sluice

LWL Carry over storage DSL Dead storage

Scouring sluice

Fig. 5.2:  Zones of storage in reservoir In absence of suitable data, dead storage is taken as 10% of live storage and losses are taken as 5% of live storage. 3. Carry Over Storage: Volume of water stored between DSL and LWL is known as carry over storage, and is useful for bad year or delayed monsoon period. This provision is not made always, but it depends on the type of reservoir. 4. Gross storage: It is the sum of all types of storage i.e., dead, carryover and live. It gives tentative height of the dam by using volume-elevation curve.

5.5  SEDIMENTATION IN RESERVOIR Catchment characteristics such as soil type, land slope, vegetal cover, climatic condition like temperature, intensity of rainfall have great significance in the production of sediment load in the form of suspended particles and in the form of movable bed load (Fig. 5.3). Sediment load, ppm =

weight of sediment in sample ×106 weight of sample

Reservoir Planning  99

If Qs = Suspended sediment load transported by the flowing water, and Q = Discharge of flowing water, then   Q s = k Qn ∴ log Qs = log k + n log Q Qs = k, when Q = 1; where, n is between 2 to 3. When the sediment laden water reaches reservoir, coarse sediment starts settling down at the mouth or entry to reservoir and fine sediment along with a part of coarse sediment gets settled down in dead storage zone and if sediment load is very high, by opening out scouring sluice gates, the heavily laden sediment water can be removed from reservoir, see Fig. 5.3. Mean annual sediment production rate vary from 250 to 2000 tons/km² of catchment, and reservoir loses its storage capacity around one percent annually. Floating debris sluice

FRL

Density currents sluice LWL

C se oar di se m en t

Supply sluice

se Fine dim en t DSL

Scouring sluice

Fig. 5.3:  Sedimentation in reservoir

5.6  USEFUL LIFE OF RESERVOIR The sedimentation in a reservoir is measured in terms of TRAP efficiency, h, which is defined as percentage of inflowing sediment that is retained in the reservoir, thus

 Reservoir Capacity  C η = Trap Efficiency = f   =f  I  Inflow to Reservoir  n



 1  = 100 1 −   (1 + ax)  Capacity here x = ratio, and a and n are constants Inflow

100  Irrigation Engineering and Hydraulic Structures

The useful capacity of reservoir lost each year by sediment deposition is Vs = Qs ηav Vs = volume of useful capacity of reservoir lost each year. Qs = annual sediment inflow into reservoir While allocating space for dead storage i.e., space for sediment deposition during life of reservoir, trap efficiency is taken as 95% to 90%.

q  SOLVED EXAMPLES Example 5.1: Find probable life of a reservoir with an initial capacity of 4000 ha-m and average annual inflow is 8000 ha-m, and average annual sediment inflow is 2 × 106 KN. Specific weight of sediment is 11.2 KN/m3. Useful life will terminate when 80% of initial capacity is filled with sediment. Values of trap efficiency and capacity-inflow ratio are given below: Capacity Ratio Inflow

Trap efficiency , η

0.1

87

0.2

93

0.3

95

0.4

95.5

0.5

96

0.6

96.5

0.7

97

0.8

97.3

0.9

97.4

1.0

97.5

Solution:   Annual sediment inflow = 2 × 106 KN Vs = Volume of sediment inflow =

2 ´106 = 1.786 ´105 m3 = 17.86 ha- m 11.2

Initial capacity of reservoir = 4000 ha-m Average annual inflow = 8000 ha-m C 4000 = 0.5 Initial Capacity / Inflow = = I 8000

Reservoir Planning  101

Capacity Trap eff Inflow η – from ratio given (2) table (3)

Av value of trap eff. ηav (4)

Vs × Vol. of ηav vol. capacity Years of sed. interval to fill trapped (h-m) col.6 (5) (6) col.5 (7)

Capacity %

Cap. ha-m col.(1)

100

4000

0.5

96

96 + 95.5 2

(17.86 × 0.956)

4000– 3200

800 17

80

3200

0.4

95.5

= 95.7

= 17.0

= 800

= 47

60

2400

0.3

95

95.2

17.0

800

47

40

1600

0.2

93

94.0

16.79

800

47.64

20

800

0.1

87

90.0

16.07

800

50.00

Life of Reservoir = 191.64 years.

Example 5.2:  A reservoir has capacity of 400 ha-m. Catchment area is 130 Km2 and 12 cm runoff. If sediment production per year is 0.03 ha-m/km2 what is probable life of reservoir before its capacity is reduced to 20% of initial? Relation between trap efficiency and (capacity/inflow) ratio is given below: C/I ratio η trap eff.

0.05

0.1

0.2

0.3

0.4

0.5

0.6

0.7

1.0

77

87

93

95

95.5

96

96.5

97

97.5

Solution:

Average annual inflow =

12 130 × 106 × = 1560 ha-m 100 104

C 400 = = 0.256 → η = 93.5,say I 1560 Qs = .03 × 130 = 3.9 ha-m

Vs = Qs η = 3.9 × 0.935 = 3.64 ha-m Similarly calculations are done and entered in following table.

C

Loss of R Vs = Qs × ηav capacity = ΔC

C/I

h

h av

400

0.256

94

 94 + 93     2 

0.256 × 0.935

400-320

320

0.2

93

93.5

= 3.64

= 80

No. of years to fill the loss ΔC = Vs years

800 17 80 = 22 3.64

102  Irrigation Engineering and Hydraulic Structures

Loss of R Vs = Qs × ηav capacity = ΔC

No. of years to fill the loss ΔC = Vs years

C

C/I

h

h av

240

0.15

90

91.5

3.57

80

80 = 22.4 3.57

160

0.1

87

88.5

3.45

80

80 = 23.2 3.45

80

0.05

77

82.0

3.20

80

80 = 25.0 3.2

Total

92.6 years

Life of Reservoirs = 93 Years

5.7 FACTORS AFFECTING SELECTION OF RESERVOIR SITE 1. Reservoir site should be chosen such that run off from catchment area is maximum, with least sediment load. 2. Reservoir site should be such that it gives minimum seepage loss i.e., site should have sound rock formation with least percolation. 3. Reservoir site must be such that it has a suitable dam site i.e., dam axis, such that dam construction cost is minimum for maximum storage of water. 4. Valley opening should be narrow. 5. Site should be accessible with least cost for its approach and maintenance. 6. Land submergence due to flood should be minimum. 7. Evaporation loss should be minimum. This is possible when reservoir site is deep. i.e., area of waterspread exposed is minimum with respect to elevation. 8. Minerals and salts contained in rock formation of reservoir site should not be objectionable, spoiling the quality of water stored.

5.8 DETERMINATION OF RESERVOIR CAPACITY FROM MASS CURVES Procedure: 1. From inflow hydrograph prepare mass inflow curve at least for 35 years. 2. Prepare demand curve by calculating demands such as irrigation, water supply etc per year. 3. From mass curve apex such as A1, A2 etc. draw tangents parallel to demand curve.

Reservoir Planning  103

4. Measure maximum vertical intercepts such as E1D1, E2D2 etc. The vertical interval indicates volume by which inflow falls short of demand for example, C1E1 indicate demand, whereas C1D1 indicate inflow, Hence, E1D1 has to be provided from reservoir storage i.e., it is shortage. Hence, highest of the ordinates E1D1, E2D2 etc. indicate reservoir storage capacity. Elevation against this capacity is read from capacity–elevation curve and that is the height of the dam.

E2 D2

Demand curve

C2

E1 Mass inflow curve

A1

D1

Demand Q

Inflow Q

A2

C1

Demand curve Time (Years)

Time (years)

Fig. 5.4:  Mass curve

5.9  RESERVOIR LOSSES The loss of water from reservoir is due to evaporation, absorption and percolation. Evaporation loss mainly depends on surface area exposed to atmosphere, temperature humidity and wind velocity. It can be measured by making use of evaporation pan and appropriate pan co-efficient. Month-wise evaporation loss values in mm are given in Table 5.1: Table 5.1  Monthly Evaporation Loss Month Evaporation loss in mm in North India In south and central India in mm Jan

70

100

Feb

90

100

March

130

180

April

160

230

May

270

250

June

240

180

104  Irrigation Engineering and Hydraulic Structures

Month Evaporation loss in mm in North India In south and central India in mm July

180

150

Aug

140

150

Sept

140

150

Oct

130

130

Nov

90

100

Dec

80

100

Loss of water due to absorption depends on type of soil forming reservoir basin. This is large in beginning but gradually reduces as pores get saturated. Percolation or seepage loss is very small in most of the cases, however it may be significant if severe leakage takes place through hills or base of dam due to presence of fissured rock. Grouting should be carried out to prevent such loss.

5.10  RESERVOIR FLOOD ROUTING Consequent upon heavy rains in catchment area, a river may start swelling i.e., its rate of flow may increase and if continues to do so, it will start overflowing, due to this people living on banks may suffer because of floods in river. Certain reservoirs do make a provision of holding flood water for some interval of time thereby giving chance to down stream people to get shifted to safe places, but all reservoirs can not do so. However, regulated release of water from reservoir can be planned and executed. This is known as flood routing. Thus, flood routing means control of water levels in reservoir and release of water from reservoir i.e., it’s a matter of controlling “inflow and outflow” of reservoir during flood time. Flood routing is thus regulation of reservoir storage and its outflow and also its means regulation of flood flow in rivers or channels. Thus, flood routing is divided into two separate studies: 1. Reservoir routing 2. Channel routing Reservoir routing makes use of hydrologic methods developed by LG Puls , Goodrich and others , where as channel routing method is known as Muskinghm method which was first tried on river Muskingm in USA. Reduction in Peak values of flood due to routing technique is known as “attenuation”. This is shown in Fig. 5.5. It is the difference in peak value of inflow (flood) hydrograph and outflow hydrograph. Further peak of outflow occurs after peak of inflow, thus

Reservoir Planning  105

giving difference in time between occurrence of two peak values. This is known as “time lag”. T ime lag = Difference in time between peak value of graph (1) and (2)

Q m 3/sec

Attenuation = difference between peak of graph (1) and (2) 1

2

Outflow hydrograph

Inflow hydrograph

T ime (hrs)

Fig. 5.5:  Attenuation and peak lag The attenuation and time lag of flood hydrograph at a reservoir are two important aspects of reservoir operation under flood control criteria. The storage capacity of reservoir and characteristics of spillway and other outlets from reservoir control the flood for some time so that downstream population can be warned. Total flood control is not possible but its detention for certain interval of time is possible. Reservoir routing is a process of calculating changes in reservoir levels, its volume and outflows for a given inflow hydrograph representing a flood approaching from catchment area. This is based on hydrologic method, which can be expressed mathematically as

I − O = ± ∆s .....(5.1)

Where : I = Inflow O = Outflow ΔS = Changes in reservoir storage. By choosing a proper time interval (Δt) between different values of outflow and inflow, equation (5.1) can be written as :

 I1 + I2   O1 + O 2   2  ∆t −   ∆t = S2 − S1 (5.2) 2    

where, I1 and I2 are inflow at a Δt interval, O1 and O2 are outflow at Δt interval and S1, S2 are corresponding storage value in reservoir.

106  Irrigation Engineering and Hydraulic Structures

Equation (5.2) can also be expressed as :  2S   2S  (I1 + I2 ) +  1 − O1  =  2 + O 2  (5.3)  ∆t   ∆t 



Inflow – Storage – Discharge (ISD) method developed by L.G. Puls, makes use of equation (5.2) and Goodrich method or modified Puls method makes use of equation (5.3). By both the methods results obtained do not vary significantly and hence any method can be adopted. In following numerical problem Goodrich method is used to illustrate how reservoir routing can be carried out .

q  SOLVED EXAMPLES Example 5.3:  Following flood hydrograph is to be routed using modified puls method. Find out attenuation and time lag by plotting inflow and outflow hydrographs. Time (hrs) 3

Q (m /s)

0 60

6

12

18

24

30

36

42

48

54

60

66

72

170 280 350 420 360 310 220 165 80

70

25

20

Outflow from reservoir just before arrival of flood = 200 m3/s. Also given following values of outflows from reservoir against storage: Q (outflow) (m3/sec)

0

40

70

85

100

110

250

490

800

S (cumec-day)

10

52

105

163

225

290

365

450

530

Solution:

1 ∆t = 6hr = day 4 \

2S = S´ 8 m3 /day Dt

 2S  First step is to plot Q v/s  + Q  Hence from given data of Q and S, prepare  ∆t   2S  following table and plot Q on Y axis and  + Q  on X axis.  ∆t  Q (m3/sec)

S (cumec-day)

2S = s × 8 (m3/sec) ∆t

 2s  3  + Q m /sec ∆t

  0

 10

   80

   80

 40

 52

 416

 456

  70

105

  840

 910

Reservoir Planning  107

Q (m3/sec)

S (cumec-day)

2S = s × 8 (m3/sec) ∆t

 2s  3  + Q m /sec ∆t

  85

163

1304

1389

100

225

1800

1900

110

290

2320

2610

250

365

2920

3170

490

450

3600

4090

800

530

4240

5040

The above values are plotted as shown in Fig 5.6.

800 700 600 Q

500 400 300 200 100 1000

2000 2S t

3000

4000

Q

 2S  + Q & Q Fig. 5.6:  Graph of   ∆t  Now given value of Q = 200 m3/sec at entry of flood. 2S Hence, read  + Q  from graph of Fig. 5.6, it is 3000 m3/sec against  ∆t  Q = 200 m3/sec.  2S   2S  ∴  − Q  =  + Q  − 2Q  ∆t   ∆t  = 3000 – 2 (200) = 2600 m3/sec.

108  Irrigation Engineering and Hydraulic Structures

Now I1 = 60 and I2 = 170 (given in data) ∴ (I1 + I2) = 60 + 170 = 230  2S  ∴ (I1 + I2 ) +  − Q  = 230 + 2600 = 2830  ∆t 



 2S  =  + Q ∆ t  

 

 2S  Read Q for  + Q  = 2830 from graph of Fig 5.6  ∆t  ∴ Q = 160 m3 / sec From this value of Q = 160,  2S   + Q  = 2830 − 2 × 160 = 2510  ∆t  The steps are continued further and the results are given in Table below: Time (hr)

I (m3/ sec)

(I1 + I2) m3/sec

 2S  – Q (m3/sec)   ∆t

 2S  + Q (m3/sec)   ∆t

Q(m3/sec)

(1)

(2)

(3)

(4)

(4)

(6)

0

60

–

–

3000

200

6

170

230

2600

2830

160

12

280

450

2510

2960

200

18

350

630

2560

3190

230

24

420

770

2730

3500

300

30

360

780

2900

3680

350

36

310

670

2980

3650

340

42

220

530

2970

3500

300

48

165

385

2900

3285

250

54

80

245

2785

3030

210

60

70

150

2610

2760

150

66

25

95

2460

2555

120

72

20

45

2315

2360

110

Plot Time v/s I i.e., column(1) v/s column(2) and get in flow graph. Plot time v/s Q i.e., column(1) v/s column(6) and get outflow graph. These graphs are given in Fig. 5.7. From Fig 5.7, attenuation = 70 m3/sec and time lag = 9 hours.

Reservoir Planning  109 time lag = 9 hours

400

3

70 m /sec = Attenuation

300

200 Outflow as per column (6)

Q 100

Inflow as per column (2) 6

12 18 24 30 36 42 48 54 T (hrs) as per column (1)

Fig. 5.7:  Plot of T vs. Q Example 5.4:  Route the following flood by ISD method and find attenuation and reservoir lag. Data : 1. In flow Hydrograph: Time (hrs)

0

Q (m3/sec.)

60

6

12

18

24

30

36

42

48

54

60

66

72

170 280 350 420 360 310 220 165

80

70

25

20

2. Outflow at time of arrival of fluid = 200 m3/sec 3. Values of outflow against storage. Q(m3/sec)

0

40

70

85

100

110

250

490

800

S cumec-day

0

52

105

163

225

290

365

450

530

Q  Q    Solution:  Calculate  S + ∆t  and  S − ∆t  for Q values given and plot the 2  2   

graph (Fig. 5.8) calculations are given in the table below: Δt = 6 hours = 0.25 day ∴

∆t = 0.125 day 2

110  Irrigation Engineering and Hydraulic Structures

S

Q

Q ∆t = 0.125 Q 2

Cumec-day

m3/sec

cumec-day

S+

Q ∆t 2

S–

Q ∆t 2

cumec-day

cumec-day

0

0

0

0

0

52

40

5.00

57.00

47.00

105

70

8.75

113.75

96.25

163

85

10.62

173.62

152.38

225

100

12.50

237.50

212.50

290

110

13.75

303.75

276.25

365

250

31.25

396.25

333.75

450

490

61.25

511.25

388.75

530

800

100.00

630.00

430.00

For routing the flood hydrograph (i) I1 = 60 m3/sec, I2 = 170 m3/sec. ∴

I1 + I2 = 115 m3 / sec. 2

 I1 + I2  Q1 = 200 m3/sec, Δt = 0.25 day,   ∆t = 115 × 0.25 = 28.75  2  S

Q∆ t 2

800 700

S+

600

Q∆ t 2

500 400 300 200 100 100

200 300 400 500 600 700 S

Q∆ t 2

 

Fig. 5.8:  Polt of  S −

S+

Q∆ t 2

Q∆t  Q∆t    vs. Q and  S +  vs. Q 2 2 

Reservoir Planning  111

For Q1 = 200 m3/sec from graph of Fig. 5.8, Q ⋅ ∆t    S1 −  = 325 cumec − day 2   Q ∆t   I + I  Q ∆t    ∴  S2 − 2  =  1 2  ∆t +  S1 − 1  = 28.75 + 325 = 353.75 cumec-day 2   2  2    3 This gives Q2 = 160 m /sec   From Graph (2) of Fig. 5.8 Now calculations for routing the flood can be entered into following tabular form: Time 0 6 12 18 24 30 36 42 48 54 60 66 72

I(m3/sec) 60 170 280 350

cumec-day – (28.75) 56.25 78.75

420 360 310 220 165 80 70 25 20

cumec-day – 325 300 300

96.25 97.50 83.75 66.25 48.13 30.63 18.75 12.00 7.00

cumec-day – 353.75 356.25 378.75

Q m3/sec 200 160 161 200

421.25 447.50 453.75 441.25 408.00 370.63 338.75 302 282.00

275 330 340 300 250 195 150 110 105

325 350 370 375 360 340 320 290 275

400

3

70 m /s 9 hrs.

300 3

Q m /s Outflow

200

100 Intflow

6

12

18

24

30

36

42

48

54

60

Time (hrs.)

Fig. 5.9:  Graph of T vs. Q and T vs I

66

72

112  Irrigation Engineering and Hydraulic Structures

Plot : Time v/s Q and Time v/s I as shown in Fig 5.9. From the graph read attenuation = 70 m3/sec (420 – 350) One time lag = 9 hours. Which is the same as obtained by modified Pul’s method of previous problem.

5.11 FIXING HEIGHT OF DAM FROM RESERVOIR WORKING TABLE As explained in article 5.8, height of dam can be fixed with the help of mass inflow and mass outflow curves, But in absence of such data it can be tentatively fixed on the basis of reservoir working table as explained below: 1. First workout irrigation, water supply and industrial demand for a year, 5% of this demand can be assumed as losses due to evaporation and seepage. Total of this (demand + losses) forms live storage. 2. To this live storage add dead storage as 10% of live storage as a space for accumulation of sediment. Total of live storage and dead storage will give gross storage. 3. Corresponding to this gross storage, read elevation from volume – elevation curve (Fig. 5.1) this is first approximate height of dam. 4. On the basis of reservoir full against gross storage in the month of October, prepare reservoir working table as given in Table 5.2. 5. Generally the reservoir is full in the month of October. Deduct the demand for irrigation, water supply and losses for October month from the inflow due to rain in catchment area for month of October. Thus difference between month wise outflow and inflow is to be either added or subtracted as the case may be from gross storage and resulting storage will give starting elevation for next month. 6. Like this proceed month wise; around May or June there will be minimum elevation of water in reservoir. If this level is equal to level on account of dead storage, height of dam against gross storage is all right, but if this level is higher than DSL, then volume of water between minimum level in reservoir and DSL is surplus water. This surplus volume of water is to be removed from gross storage and entire procedure is to be repeated till there is no surplus water in reservoir. This is the economic height of dam. The entire procedure is explained by way of a working example and preparation of reservoir working table given below. Data Catchment area = 1200 sqkm Average annual rainfall = 55 cm Loss (infiltration etc) = 5 cm

Reservoir Planning  113

Rexcess = 50 cm = (50/100) m Runoff = C × Rexcess × Catchment area   (Where C = Coefficient of catchment = 0.3)   Runoff = 0.3 × (50/100) × 1200 × 106      = 180 × 106 m3      = 180 MCM The above yield is called monsoon yield. Assume Riparian rights as 2 MCM per month Note: Riparian rights- Water required to be allowed to flow in river through out the year for the benefit of down stream people. It is around 10% of annul runoff. Here it is taken as 2 MCM/month. 90. % of total yield (0.9 × 180 = 162 MCM) is assumed as total yield in monsoon months of (June, July, August and September) which is distributed month-wise in the following percentages: Table 5.2:  Yield in Monsoon Months Month

Percentage (%)

June

15

July

25

August

30

September

30

The remaining 10% of 180 MCM (i.e., 0.1 × 180 = 18) is taken as yield available in the post monsoon months i.e., October to May, which is distributed month-wise in the following percentages: Table 5.3:  Yield in Post-Monsoon Months Month

Percentage (%)

October

20

November

20

December

15

January

15

February

10

March

10

April

5

May

5

114  Irrigation Engineering and Hydraulic Structures

Calculation of Irrigation Demand Table 5.4:  Irrigation Demand D = Duty B = Base Period 8.64B D = Delta = m (Ha/cumec) (Given) (Days)  D

Season

Crop

Rabi

Wheat

1800

120

0.567

Kharif

Rice

800

120

1.296

Hot Weather

Fruit/Vegetable

700

120

1.480

Eight Months Cotton

1400

180

1.110

Perennial

1700

300

1.524

Sugar cane

Assuming Cultivable Command Area = 3000 hect. and 80% is intensity of irrigation; Area under irrigation = 0.8 × 3000 = 2400 hect. Table 5.5:  Irrigation Demand Season

Crop percentage Duty Ha/cumec Base period Delta Area Volume (%) (Given) (Given) (days) (m) (1) Ha (2) (1) × (2)Ha-m

Rabi

25

1800

120

0.567

600*

345.6

Kharif

25

800

120

1.296

600

777.6

Hot Weather

10

700

120

1.480

240

355.2

Eight Months

15

1400

180

1.110

360

399.6

Perennial

25

1700

300

1.524

600

914.4

Total

=2792.4

where,

25 x2400 * × 2400==600 600* 100

Total irrigation demand = 2792.4 Ha-m = 2792.4 × 104 m3 = 27.92 MCM  = 30 MCM/year  (rounded figure) Calculation of month-wise irrigation demand: For perennial season: Total demand for perennial season = 914.4 Ha-m (from Table 5.5)

= 914.4 × 104 m3



= 9.414 MCM (million cubic meter)

Similarly, for other season month-wise irrigation demand can be calculated, which is given in Table 5.6.

Reservoir Planning  115

Table 5.6:  Month-wise Irrigation Demand Month

Eight month (MCM)

Rabi (MCM)

Oct Nov Dec Jan Feb March

Perennial (MCM) 0.762 0.762 0.762 0.762 0.762 0.762

0.4995 0.4995 0.4995 0.4995 0.4995

0.864 0.864 0.864 0.864

April May June July Aug Sept

0.762 0.762 0.762 0.762 0.762 0.762

0.4995 0.4995 0.4995

Kharif (MCM) 1.944

Hot weather (MCM)

0.888 0.888 0.888 0.888 1.944 1.944 1.944

Total (MCM) 2.706 2.1255 2.1255 2.1255 2.1255 2.1495 2.1495 2.1495 2.1495 2.706 2.706 2.706

Calculation of Water Supply Demand Storage for water supply demand can be considered on per capita requirement. Assume Population = 2.5 lakh Water consumption = 200 litre / capita / day Water supply demand = 200 × 365 × 2,50,000 = 18.25 × 109 litre (yearly) = 18.25 × 106 m3 = 18.25 MCM = 20 MCM (rounded figure on higher side) Water supply demand per month = 20/12 = 1.667 MCM Live storage = Irrigation demand + Water supply demand = 30 + 20 = 50 MCM Dead storage = 10% of Live storage = 5 MCM Losses = 10% of Live storage = 5 MCM Carry Over Storage (requirement for weak monsoon) = 50% of Live storage = 25 MCM Gross storage = L.S + D.S + Losses + C.O.S = 50 + 5 + 5 + 25 = 85 MCM Seepage loss = (5% of Live storage) / 12 = 2.5/12 = 0.2083 MCM/month Lowest Water Level (L.W.L) = Dead Storage (D.S) + Carry Over (C.O) = 5 + 25 = 30 MCM Elevation of LWL against storage of 30 MCM = 243.75 m (From volumeelevation graph; see Fig. 5.1) Now to get the required height of the dam, trial reservoir working tables are worked out till difference of water level at minimum value over 12 month and low water level fixed for reservoir are coinciding with each other, in case they do not coincide, then volume of water equivalent to difference in these values is required to be removed from gross storage and one more trial working table should be prepared. This is to be repeated till the difference is negligible as can be seen from Tables 5.6, 5.7 and 5.8.

85

81.8

79.25

75.82

72.41

68.07

63.65

58.29

52.79

70.75

85

85

NOV

DEC

JAN

FEB

MAR

APRIL

MAY

JUNE

JULY

AUG

SEP

(3) 13 9 8 7

161 × 104

158 × 104

4

9 13

148 × 104 16 27 24 18 14 14

4

4

165 × 104

4

265.5

286.25

286.25

277.25

165 × 10

150 × 10

133 × 10

143 × 10

138 × 104

272.5

269.0

152 × 10

4

155 × 10

(5)

(4)

165 × 104

4

275.5

278.25

280.5

282.5

284.25

286.25

(cm)

0.231

0.231

0.2709

0.3192

0.3740

0.2296

0.1924

0.1368

0.1085

0.1268

0.1449

0.2145

(6)

(MCM)

0.2083

0.2083

0.2083

0.2083

0.2083

0.2083

0.2083

0.2083

0.2083

0.2083

0.2083

0.2083

(7)

(MCM)

2.07060

2.7060

2.7060

2.1495

2.1495

2.1495

2.1495

2.1255

2.1255

2.1255

2.1255

2.7060

(8)

(MCM)

1.667

1.667

1.667

1.667

1.667

1.667

1.667

1.667

1.667

1.667

1.667

1.667

(9)

(MCM)

2

2

2

2

2

2

2

2

2

2

2

2

(10)

(MCM)

6.8123

6.8123

6.8522

6.344

6.3988

6.2544

6.2112

6.1376

6.1093

6.1276

6.1457

6.7958

(11)

48.6

48.6

40.5

24.3

0.9

0.9

1.8

1.8

2.7

2.7

3.6

3.6

(12)

(14)

MCM

41.7877

41.7877

* Overflow will take place only when (I-O) added to reservoir at the beginning of month become more than 286.25 m, the starting FRL in the month of October.

41.7877

41.7877

33.6478 24.6478*

17.956

–5.4988

–5.3544

–4.4172

–4.3376

–3.4093

–3.4276

–2.5457

–3.1958

(13)

(MCM) (MCM) (MCM)

Volume of Irrigation Water Evaporation Evaporation Seepage on Supply Riparian Total Inflow – Loss Loss Loss Demand Demand Demand Outflow Inflow Outflow Overflow

Since minimum level = 265.5 m; Corresponding volume = 52.79 MCM and LWL = 243.75 m Corresponding volume = 30.00 MCM Surplus water in reservoir = 52.79 – 30.00 = 22.79; hence Gross storage (new) = 85 – 22.79 = 62.21 MCM FRL is to be obtained from graph of volume –elevation for new value of Gross storage. i.e., 62.21 MCM, corresponding level is 271.5

(2)

OCT

(Ha-m)

(MCM)

(m)

Area

(1)

Month

Flood Reservoir Gross Level Storage (FRL)

Table 5.7  Reservoir Working Table – I (first trial)

116  Irrigation Engineering and Hydraulic Structures

62.21

59.04

56.51

53.10

49.70

45.38

40.99

35.67

30.24

48.27

62.21

62.21

OCT

NOV

DEC

JAN

FEB

MAR

APRIL

MAY

JUNE

JULY

AUG

SEP

271.5

271.5

262.25

244

250.75

255.5

259.75

263.5

265.5

267.5

269.5

271.5

(3)

(m) 13 9 8 7 9 13 16 27 24 18 14 14

4

4

130 × 104

124 × 104

4

4

4

4

4

142 × 104

142 × 10

128 × 10

101 × 10

111 × 10

118 × 10

133 × 10

136 × 10

139 × 10

142 × 10

4

(5)

(cm)

4

(4)

(Ha-m)

Area

0.1988

0.1988

0.2313

0.2424

0.2997

0.1888

0.1619

0.117

0.0931

0.1088

0.1251

0.1846

(6)

(MCM)

0.2083

0.2083

0.2083

0.2083

0.2083

0.2083

0.2083

0.2083

0.2083

0.2083

0.2083

0.2083

(7)

(MCM)

2.7060

2.7060

2.7060

2.1495

2.1495

2.1495

2.1495

2.1255

2.1255

2.1255

2.1255

2.7060

(8)

(MCM)

1.667

1.667

1.667

1.667

1.667

1.667

1.667

1.667

1.667

1.667

1.667

1.667

(9)

(MCM)

2

2

2

2

2

2

2

2

2

2

2

2

(10)

(MCM)

6.7801

6.7801

6.8126

6.2672

6.3245

6.2136

6.1867

6.1178

6.0939

6.1096

6.1259

6.7959

(11)

48.6

48.6

40.5

24.3

0.9

0.9

1.8

1.8

2.7

2.7

3.6

3.6

(12)

(MCM) (MCM)

Volume of Irrigation Water Evaporation Evaporation Seepage on Supply Riparian Total Loss Loss Loss Demand Demand Demand Outflow Inflow

Since minimum level = 244 m => volume = 30.24 MCM LWL = 243.75m => volume = 30.00 MCM Surplus = 0.24 (Not required) Hence, Gross storage (new) = 62.21 – 0.24 = 61.97 MCM, corresponding level is 271.45.

(2)

(1)

(MCM)

Flood Reservoir Gross Level Month Storage (FRL)

Table 5.8  Reservoir Working Table – II (second trial)

41.82

41.82

33.6874

18.0328

–5.4245

–5.3136

–4.3867

–4.3178

–3.3939

–3.4096

–2.5259

–3.1959

(13)

(MCM)

41.82

41.82

24.4374

(14)

MCM

Inflow – Outflow Overflow

Reservoir Planning  117

61.97

58.8

56.27

52.86

49.46

45.14

40.75

35.43

30.00

48.03

61.97

61.97

OCT

NOV

DEC

JAN

FEB

MAR

APRIL

MAY

JUNE

JULY

AUG

SEP

271.5

271.5

262.0

243.75

250.75

255.5

259.75

263.5

265.5

267.75

269.5

271.45

(3)

27 24 18 14 14

111 × 104

100 × 104

4

4

4

142 × 10

142 × 10

128 × 10

118 × 10

16

13

4

124.5 × 10

9

130 × 104

4

7

133 × 104

136.5 × 10

8

9

4

4

13

142 × 104

139 × 10

(5)

(4)

Area (Ha-m)

0.1988

0.1988

0.2304

0.2412

0.2997

0.1888

0.1619

0.117

0.0931

0.1092

0.1251

0.1846

(6)

0.2083

0.2083

0.2083

0.2083

0.2083

0.2083

0.2083

0.2083

0.2083

0.2083

0.2083

0.2083

(7)

2.7060

2.7060

2.7060

2.1495

2.1495

2.1495

2.1495

2.1255

2.1255

2.1255

2.1255

2.7060

(8)

1.667

1.667

1.667

1.667

1.667

1.667

1.667

1.667

1.667

1.667

1.667

1.667

(9)

2

2

2

2

2

2

2

2

2

2

2

2

(10)

6.7801

6.7801

6.8117

6.266

6.3245

6.2136

6.1867

6.1178

6.0939

6.1100

6.1259

6.7959

(11)

48.6

48.6

40.5

24.3

0.9

0.9

1.8

1.8

2.7

2.7

3.6

3.6

(12)

41.8199

41.8199

33.6883

18.034

–5.43

–5.3136

–4.3867

–4.3178

–3.3939

–3.41

–2.5259

–3.1959

(13)

41.8199

41.8199

24.2686

(14)

Volume of Irrigation Water Evaporation Evaporation Seepage on Supply Riparian Total Inflow – Loss Loss Loss Demand Demand Demand Outflow Inflow Outflow Overflow (cm) (MCM) (MCM) (MCM) (MCM) (MCM) (MCM) (MCM) (MCM) MCM

Surplus = 0.0; hence height of the dam = F.R.L – River Bed Level = 271.45 – 200 = 71.45 m

(2)

(1)

Flood Reservoir Gross Level Month Storage (FRL) (MCM) (m)

Table 5.9  Reservoir Working Table – III (third trial)

118  Irrigation Engineering and Hydraulic Structures

Reservoir Planning  119

EXERCISES 1. (a)  What are different types of reservoirs and what are their functions? (b) State sectors those are required to be considered for selection of (i) Reservoir site and (ii) Dam site. 2. (a)  Explain how height of dam is fixed? (b)  What are mass curves? (c)  What is the importance of (i) area – elevation and (ii) volume – elevation curves? 3. What is trap efficiency? What is its role in deciding life of reservoirs? 4. A reservoir has a capacity of 600 ha-m, the catchment area is 150 sq km and average annual run-off is 12 cm. If annual sediment production is 0.03 ha-m/sq km how many years will be required to have 10% reduction in reservoir capacity by sedimentation? Take ηav = 0.95 for C/I ratio between 0.3 and 0.35. (Ans.: 14 years) 5. A reservoir has storage capacity of 740 ha-m. Catchment area is 80 sqkm, from which annual sediment inflow to reservoir is 0.12 ha-m per sq km. Of C.A. Taking ηav = 0.80, find annual capacity loss due to sediment in reservoir. (Ans.: 1.03 % per year) 6. Find probable life of a reservoir having capacity equal to 30 MCM. And annual inflow of 60 MCM. The annual sediment inflow is 0.2 × 106 tonnes. Specific gravity of the sediment is 1.2; the useful life of reservoir will be over when 80% of its capacity is filled with sediment. C/I ratio and η values are given below : C/I Ratio

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

η%

87

93

95

95.5

96

96.5

97

97

97

97.5

(Ans.: 153 years)  

 

6 Design of Gravity Dams 6.1 INTRODUCTION Dam is an obstruction constructed across a river valley with a specific purpose of creating storage reservoir on its upstream side. Its length is divided into two zones, non-overflow and overflow. The overflow part of the length of the dam is defined as spillway and rest is known as non-overflow section of the dam. When a dam resists the force of water pressure solely by its own weight, it is known as Gravity Dam. It can be constructed with masonry or concrete and as such it can be classified as rigid dam. In general, gravity dam is a huge massive masonry or concrete structure facing water pressure as its main force. In addition to this, there are other forces that play their role in the design of gravity dams. They are described in detail below.

6.2  FORCES ACTING ON GRAVITY DAMS Determination of force, its magnitude and line of action, its effect on shape of its profile and its importance whether it is basic or secondary are the essential features the designer should know thoroughly for design of a dam, as it is a very responsible task that a civil engineer is required to perform. Essentially, entire irrigation engineering is nothing but design of civil engineering structures, which deal with storage and distribution of water for irrigation and allied purposes. The forces can be enlisted as under: A. Basic Forces: (i) Weight of body of dam, W (ii) Water pressure, P (iii) Uplift force, U B. Secondary Forces: (i) Wind pressure (ii) Wave pressure (iii) Silt pressure (iv) Ice pressure © The Author(s) 2023 S. K. Ukarande, Irrigation Engineering and Hydraulic Structures, https://doi.org/10.1007/978-3-031-33552-5_6

120

Design of Gravity Dams  121

C. Periodic Type: (i) Earthquake force (ii) Cyclonic conditions Top width FRL

F.B U/s and d/s face vertical (zone l) Pv1 U/s vertical d/s sloping (zone ll)

w2

U/s vertical d/s both sloping R

Pv2

Cuts middle 3rd (zone lll)

PH w3

Heel w1

(h/3)

Re

h

Rf

Toe

High gravity condition R cuts inside middle 3rd (zone lV)

A/3

A = (ch) U

Fig. 6.1:  Forces acting on a gravity dam

Detailed Description of Forces: A. Basic Forces: 1. Self-weight, W: Self-weight W consists of W1, W2 and W3 as shown in Fig. 6.1 W1 = u/s triangular portion due to u/s slope W2 = middle rectangular portion due to top width W3 = d/s triangular portion due to d/s slope W1, W2, W3 act vertically downward through respective center of gravity of each portion of the section of dam, and their magnitude is equal to: W1 = (s γ) V1 W2 = (s γ) V2 W3 = (s γ) V3

122  Irrigation Engineering and Hydraulic Structures



Where V1, V2, V3 are volumes of respective portion and (S γ) is weight density of concrete. Here s is specific gravity of concrete, generally 2.4 and γ is weight density of water, generally 1000 kg for 1.0 tonne.

2. Water Pressure, P: Calculation of water pressure is carried out in two parts: for PH and PV. PH = horizontal pressure acting on vertical projection of upstream side face of the dam and its magnitude = ½ γ h². where, h = height of FRL from river bed level, and is acting at 1/3 h, giving displacing moment about toe of the dam. Pv = (PV1+ PV2)V due to weight of water on u/s sloping face of dam Pv1 = Wt. of water in rectangular portion

Pv2 = Wt. of water in triangular portion



Pv1 and Pv2 act vertically downward through respective centre of gravity of rectangular and triangular portion.



Hence, P = Total pressure =



If u/s face is vertical from top to bottom, then Pv does not exist, and P = PH



PH2 + PV2

Calculation of PH, PV, W, U etc. are carried out for unit length of dam and hence volume.

3. Uplift Force, U: Uplift force is due to a possible presence of water below dam. Ordinarily it is not present as foundation of gravity dam is a sound rock formation, which is free of cracks, crevices and faults. However, these may develop at a later stage and hence as a precautionary measure, it is assumed that certain percentage of hydrostatic force may be acting upward on the base of the dam. This is known as uplift force, U and is given by

U = ½ (B) (C γ h), where



C = Coefficient of uplift

   = Fraction of hydrostatic force    = 0.5, generally, as it can vary from zero to 1.0

U is acting vertically upward through centre of gravity. of the triangular uplift pressure diagram shown below base of dam in Fig. 6.1, i.e., at B/3 from heel of the dam. Here down stream (tail) water depth is assumed as zero since even if it is present, it will be at the toe of spillway portion and not on non-overflow section of the dam.



If drainage gallery provides means of release of uplift pressure then, uplift force gets reduced to one third of its value at the heel to the point where pressure release arrangement is provided. This will be described later on

Design of Gravity Dams  123





when drainage gallery is described under the portion of practical profile of dam. Hence, this effect is shown by means of dotted line in triangular portion of the uplift diagram below the base. Effect of uplift force is to reduce the weight of the dam and also to create displacing moment at the toe of the dam. In the calculation of forces for elementary or theoretical profile of gravity dam, only these three forces i.e., W, P and U are considered and hence they are known as basic forces. Forces under portion B of the list are known as secondary since they need to be considered at a later stage as check on the section of the dam designed on basis of three basic forces. Secondary forces are now described below.

B. Secondary Forces: 4. & 5. Wind and Wave Pressures: Wind with a velocity of 100 to 150 kmph may cause exposed face of the dam to have additional horizontal force either acting u/s to d/s or vice versa depending on direction of wind velocity and its effect is to be resisted by self-weight of the dam, which it is capable to do so for it is designed to resist water pressure which is much more higher than wind pressure. But effect of wind is also to generate waves on u/s side as it blows over reservoir surface exposed to wind effect. This may cause an impact force at FRL level to the body of the dam and also may create conditions of overflow of water if sufficient free board (FB) is not provided. Hence, FB is decided on the basis of wave height, hw, which is given by Moliters formula: h w = 0.032 VF, for (F > 32 km) here, h w is in meter, V is wind velocity in kmph, F is fetch in km Crest FB

FRL hw Pw F (3/8 hw

Trough

DSL

Fig. 6.2:  Wind and Wave pressure acting on a dam.

124  Irrigation Engineering and Hydraulic Structures

Fetch (F) is defined as maximum length of water surface at FRL level from dam axis exposed to wind effect i.e., it is the length of the farthest point on FRL contour from dam axis in km. If value of F is greater than 32 km, above formula is applied but if it is less than 32 km, then, h w = 0.032 VF + 0.763 − 0.271 4 F,

for (F < 32 km)

h w = wave height,

FB = Free Board = 1.5 h or 3 m , whichever is higher.



FB is also decided as per IS 6512 of 1972 as,



FB = [1.33h + wind set-up] or 3 m,



where, wind set-up in m is given by, V2F 62000 D

where, D = average depth of reservoir in m.

Now a days, top width of dam provided is always kept either equal to one or two traffic lane i.e., 3 to 6m, which is sufficient to take care of any impact load at FRL level Wave Pr essure can be calculated by, 5 Pwave = ½(23.5h w ) h w 3 =19.62 h 2w KN / m length of dam, h w in m. and acts at 3 / 8 h w from trough or sill level of wave, see Fig. 6.2

6. Silt Pressure: Over the reservoir portion known as dead storage, silt may get accumulated in course of life span of the reservoir and hence its pressure on u/s face of dam be considered as a check force in the design of dam. This is taken as φ  1 − sin φ  Psilt = ½ (γs) (h1 )2    1 + sin φ  where, γs = Submerged weight of silt

h1 = Depth of silt deposited

    f = Angel of internal friction of silt and is acting horizontally at 1   h1  from base. 3 

7. Ice Pressure: This is generally not required to be considered in India, however as per provision of IS 6512 – 1972 its magnitude Pice = 245 KN/m² is taken as acting over contact area of dam face at FRL level.

Design of Gravity Dams  125

C. Periodic Forces: Periodic forces e.g., earthquake force and forces due to cyclonic conditions may come in effect for a small interval of time in a life span of a reservoir and hence they are considered as check forces for the adopted design. However, sometimes earthquake force is required to be considered if it is suspected that dam site is in seismic zone. As such, full description of earthquake force is required which is given below. As regards cyclonic conditions are concerned, wind velocity of 200 kmph if considered in the design, it will be sufficient. 8. Earthquake Force: It can be defined as that force which structure situated on earth’s crust experiences when some rupture or collision of rock plates below earth’s crust has taken place suddenly. The point where such a rupture or collision has occurred is defined as focus of earthquake and its depth below earth’s crust is focal depth. Corresponding location of focus on earth’s surface is epicentre as shown in Fig 6.3. The energy released by rupture of rock plate travels towards earth’s surface in the form of:

1.  Primary or P waves, which are compression or longitudinal waves



2.  Shear or transverse or S waves



3.  Raleigh or love waves.

Dam Epicenter Earth surface or crust Focal depth Focus Waves of acceleration

Plane of rupture or collision of rock plates

Fig. 6.3:  Earthquake force.

The energy E released at the focal area is quantified as magnitude M on a logarithmic open ended scale, first introduced by Richter and hence known as Richter scale representing earthquake intensity M and is defined as M = log10 (Amplitude of P waves recorded in mm at 100 km from epicenter) Hence, log10 (E) = 11.4 + 1.5 M Here E = Energy released in ergs. M = Richter’s Magnitude of earthquake force.

126  Irrigation Engineering and Hydraulic Structures

However, Engineer’s scale for calculation of earthquake force is not Richter scale, but is based on adoption of a coefficient, denotes as a, and given by α =

acceleration due to earthquake acceleration due to gravity

Hence, horizontal acceleration that structures will have to face on account of earthquake is given by (ag), where a is a fraction varying from 0 to 0.4. As per IS 1893 of 1984, are given in Table 6.1. Table 6.1:  Effect of Earthquake on Structure α Values

Effect of Earthquake on structure

0.01 to 0.05

Negligible

0.05 to 0.1

Tolerable

0.1 to 0.15

Damage to structure on small scale may take place

0.15 to 0.2

Structure may get damaged on large scale

0.3 and above

Full destruction may occur

Hence, a adopted for checking of an adopted design of dam is generally taken as 0.15 A. Effect of Earthquake Force on Self-weight W (a) Horizontal Force due to earthquake =   (α g)  g  = aW, where, W = Wt. of dam. This force (aW) is an additional horizontal force due to earthquake and is acting horizontally through centre of gravity of the dam section causing on additional displacing moment about toe of the dam. (b) Vertical Effect of (ag) on W: (i) Force = (W) (1 + a)↓ acts vertically downward if (ag) is upward. (ii) Force = W (1 – a)↓ acts vertically downward if (ag) is downward For an upward acceleration (ag), the inertia force (aW) would be acting downwards and hence it would result in an increase in the weight of the dam from W to W (1 + a). On the other hand, if acceleration (ag) is acting downwards, the inertia force (aW) will act upward and would result in reduction of the weight of the dam for W to W (1–a). Hence, for ag upward, wt.(new) = W (1 + a) and ag downward, wt.(new) = w(1 – a) both new weights will act vertically downward through their centre of gravity’s since they are weights. This effect of increase or decrease in weight is apparent and is for interval of time over which earthquake waves influence lasts at dam site.

Design of Gravity Dams  127

B. Effect on Water Pressure: Effect of earthquake on water pressure is illustrated in Fig. 6.4. FRL

y Pe h Pe

0 Parabolic pressure diagram due to effect of earthquake on water pressure

Fig. 6.4:  Effect of earthquake force on water pressure

Additional intensity of pressure at depth y from FRL is given by Zanger in 1952 as, pe = C α ( γ y)  θ here, C = 0.735    90 

y  y y y  2−   2 −  +   h h h    h

If u / s face of dam is vertical

θ = 90°, and if y = h, then h  h h h  C = 0.735(1)   2 −  + 2 −   h h h    h  = 0.735 × 2

i.e. , C = 1.47 for y = h and θ = 90° if we assume α = 0.1, p h = 1.47 (0.1) ( γ h) Pe = Total pressure due to earthquake force upto depth y is 0.726 py (y) and  4 acts at   y from base of pressure diagram.  3π 

128  Irrigation Engineering and Hydraulic Structures

6.3  THEORETICAL PROFILE OF GRAVITY DAM Theoretical profile of a gravity dam can be assumed as right angled triangle as it has to face horizontal pressure of water, which is also a right angle triangle in shape as shown in Fig. 6.5. The uplift pressure diagram is also triangular one and U and W are coplanar forces. FRL d/s face of dam

h P

River bed

Vertical from top

W

H

B/3

D

h

R

Rv

R

h/3

E

PH

B (ch )

B/3 U

Fig. 6.5:  Theoretical profile of a gravity dam

Hence, Rv = W – U, here Rv is resultant vertical force Rv = W – U, but W = ½ BH (s γ) and U = ½ B (c g h) Hence, Rv = ½ B γ h (s–c) acts vertically downward at (B/3) from vertical face PH = ½ γ h² , acts at (h/3) from base. Taking moment about 2nd middle 3rd point i.e., E on base, (Fig. 6.5)



h  B PH   = R v   3 3 h  B ∴ ½ γ h²   = ½B γ h (s − c)   3  3  (6.1) ∴ h² = B²(s − c) h ∴B = s−c

Design of Gravity Dams  129

The equation (6.1), i.e., B =

h

, satisfies requirement of no over turning s−c moment about second middle third point. For stability of theoretical section of the dam, safe design criteria can be listed as under: (i) No Overturning (ii) No Sliding (iii) No Crushing and (iv) No Tension From equation (6.1) it can be said that criteria (i) and (iv) both get satisfied, since resultant cuts the base such that it does not go out of first and second middle third point. This is evident since if reservoir is empty, resultant W cuts first middle third point i.e., D and if reservoir is full, resultant cuts second middle third point i.e., E For satisfaction of condition number (ii), i.e., no sliding, let us consider equilibrium of horizontal forces: ΣH = µΣR v , here µ = coefficient of friction at the base ∴ PH = µ(R v )

       ∴ ½ γ h² = µ [½B γ h(s − c) ]

 (6.2)

∴ h = µB(s − c) h ∴B = µ(s − c) If µ = 0.7, s = 2.4 and c = 0.4, h h = we se e that B = from equation (6.2) 0.7(2) 1.4 h h h = = , from equation (6.1),       and also B = 1.4 2.4 − 0.4 2 h hence B = also s−c satisfies condition of no sliding.

For satisfaction of condition (iii) i.e., ‘no crushing’ let us consider stresses that may develop at foundation plan i.e., base. By considering both direct and bending effect, we get normal stress as,

pn =

Rv B

where, e = eccentricity of R from midpoint

 6e  1 ± B  (6.3)  

130  Irrigation Engineering and Hydraulic Structures

B 2  B B =  B −  for R at E =  −  for R at D 2 3 3 2 B i.e., e = ± 6 Substituting this in equation (6.3), we get 2R v and  B = 0



p n max =

(6.4)



p n min

(6.5)

FRL d/s face of dam R

RV h W

H

River bed

B/3

R

h/3

PH

Ø

P

Vertical from top i.e., B

h

E

Pr

Ø

C D B (ch )

90 Ø

B/3 Pn max

B

U

Fig. 6.6:  Principal stress in a gravity dam

By substituting value of Rv = ½ B g h (s–c) in equation (6.4), we get

2(½B γ h(s − c)) (6.6) B = γ h(s − c)

p n max = p n max

Now referring to Fig. 6.6, we consider equilibrium of small triangular prism ABC at the toe of the base and get, pi (AC × 1) cos φ = p n max (AB × 1) where, pi = inclined stress acting normally to plane (AC × 1) = principal stress and p n max = maximum normal stress at toe acting on plane (AB × 1)

Design of Gravity Dams  131

 AB  1 ∴ pi = p n max    AC  cos φ = p n max sec ²φ,



since

∴ pi = p n max (1 + tan ²φ ) But tan φ = =



AB 1 = (6.7) AC cos φ

pH ½ γh² = PV ½B γ h(s − c) h h but B = (6.8) B(s − c), s − c)

∴ tan φ =

1 s−c

Substituting (6.8) in (6.7), we get, 1   pi = p n max 1 + , but p n max = γh(s − c)  s − c 



 s − c + 1 ∴ pi = γ h(s − c)   = γ h(s − c + 1)  (s − c)  pi (6.9) ∴h = , but for no crushing σ ≥ pi γ (s − c + 1) where, σ = permissible compressive stress of conerete ∴h ≤

σ γ (s − c + 1)

If equation (6.9) is satisfied, there will not be crushing of the dam material. Hence, all four criteria of stability of theoretical section of gravity dam get satisfied if h B= and s−c σ h≤ γ (s − c + 1)

6.4  HIGH AND LOW GRAVITY DAM σ is also known as limiting condition for height of low γ (s − c + 1) gravity dam i.e., if actual height, H of tan is greater than limiting height h, then it Condition h ≤

is known as High Gravity Dam and if it is equal to or less than h ≤ it is known as Low Gravity Dam.

σ then γ (s − c + 1)

132  Irrigation Engineering and Hydraulic Structures

If a given dam is falling in the category of low gravity dam, then resultant can be allowed to cut the base exactly on first and second middle third point and stress conditions are satisfied in an optimised manner, however if given dam falls in category of high gravity dam, resultant is required to be kept inside the middle third such that maximum stress (pi) developed is lower than permissible stress of material, s. The portion of dam (H – h1) as shown in Fig 6.7 falls in category of high gravity dam and should be designed on the basis of satisfaction of maximum stress criteria i.e., pi ≤ γ H(s – c+1), and even if R cuts Low gravity dam height R exactly on E

h1 R

H

E

D

Zone of high gravity condition

R

D

E

Fig. 6.7:  High and low gravity dam

the base little inside of 2nd middle third point, it should be tolerated. However for low gravity dam design, R must cut 2nd point exactly and no gap should be tolerated for as it will lead to wastage of concrete on a large scale. Therefore the portion of the dam (H – h1) is designed by strip method in which stresses are calculated for Rfull and Rempty conditions and are compared with Σ. In actual design of dam top width is always provide and as such there may not be any requirement of providing up stream slope. If for reservoir empty condition the resultant cuts outside of 1st middle third point i.e., towards u/s face of dam, then there is a need for providing u/s slope such that R cuts the first middle third point when reservoir is empty. Thus, fixing u/s and d/s slope by strip method is a matter of trial and error and is based on the requirement that R must cut inside the middle third points as well as the maximum stresses developed are within permissible limit. This will be clear from the following illustrative example.

q  SOLVED EXAMPLE Example 6.1: 

Design a gravity dam for following data: RL of river bed = 100.0 m FRL = 197.0 m Fetch = 40 km,

Design of Gravity Dams  133



wind velocity = 100 kmph.



Top width = 6m.



s = 2.4, c = 0.5, µ = 0.7, λ = 300 t/m2



Consider P, W and U only. State whether it is high or low gravity dam and work out stresses.

Solution:  First, let free board be calculated using Moliter's formula, h w = 0.032 VF , since F > 32 km here V = wind velocity in kmph =100 F = fetch in km = 40 6m FB=3m FRL Zone l, no slop on u/s and d/s

y1

Zone ll, u/s vertical and d/s slope assumed as 0.7: 1 upto y2

y2

Fig. 6.8:  Design of gravity dam

∴h w = 0.032 100 x 40 = 2.00 m ∴ FB = 3m. Total height = 97 + 3 =100 m.

Limiting height for low gravity down =



=

σ , neglecting c. γ (s + 1) 300 = 88m 1.0(2.4 + 1)

Given height 100 m > 88 m Hence, high gravity dam. Design upto height = 88 m as low gravity dam and 100 – 88 = 12 m part be designed as high gravity dam.

134  Irrigation Engineering and Hydraulic Structures

Let dam height be divided into following zones: Zone I, u/s and d/s both faces vertical Zone II, u/s vertical, d/s sloping  one III, u/s sloping, d/s sloping upto Z low gravity height 88m  one IV (100–88) = 12m part as high Z gravity type Zone I: Both u/s and d/s face vertical upto y1 W = 6 (3 +y1) 2.4 = 14.4 (3+ y1) 1 P = (1) (y1 ) 2 = 0.5 y12 , here 2

Fig. 6.9:  Design of gravity dam, zone I.

γ = 1 tonne/m3 1 U = (0.5) (1) (6) (y1 ) =1.5 y1 2 Take moment about reference line 0-0: ΣM =14.4 (3 + y1 ) x 3 + 0.5 y12 x

y1 −1.5y1 x 2 3

=130 + 43 y1 + 0.16 y13 − 3 y1 ΣV = W − U =14.4 (3 + y1 ) −1.5 y1

= 43 + 14.4y1 −1.5 y1 = 43 + 13y1

X f = 4m =

ΣM , ΣV

where, X f indicates position of resultant from reference line 0-0 when reservoir is full ∴ 4(43 + 13y1 ) = 130 + 40y1 + 0.16y13 ∴ 42 + 12y1 = 0.16y13 262.5 + 75y1 = y13 Try y1 = 10.0 1012 ≅ 1000 Hence, y = 10 m may be considered as depth upto which no d/s slope is required.

Design of Gravity Dams  135

Zone II: 6m o FB=3m FRL

y1 = 10 m

W2 07 1

3m

y2 = 10 m

o

6m

4.9 m

Fig. 6.10:  Design of gravity dam, zone -II

Let u/s face be vertical upto 20 m from top and d/s face be assumed as sloping at 0.7: 1 from y1 W1 = 6(3 + 10 + 7) 2.4 = 288 t

M1 = W1 × 3 = 864 t m

W2 = ½(4.9) (7) (2.4) = 41 t

M 2 = W2 × (6 + 1.6) = 312 t m

1 U = (.5)(1)(10.9)(17) = 46.32 t 2 1 P = (1)(17)2 = 144.5 t 2 Reservoir full condition : ΣV = W1 + W2 − U = 283

8 M v = 46x = 167 t m 3 17 M p = 144.5x = 818 t m 3

∴ Xf =

ΣM 1827 = = 6.45 ΣV 283

ΣM = 1176 + 818 − 167 = 1827

 2B  ≤ = 2.76  3  

Reservoir empty condition: ΣV = W1 + W2 = 329M ∴ Xe

ΣM = M1 + M 2 = 1176 t m

ΣM 1176 B  = = 3.60 =  = 3.60  ΣV 329 3 

and  S V = W1+W2 – U = 283 t

136  Irrigation Engineering and Hydraulic Structures

Zone III: o

FRL

6m

3m 10 m

PV1

7m 0.7 1

W1

PV2 3m

68 m

W2

1

75 m

0.1 W3

6.8 m

6m

52.5 m

B = 65.3 m

Fig. 6.11:  Design of gravity dam, zone III

From the section onward, provide u/s slope of 0.1 : 1 and d/s slope of 0.7 : 1 upto 88 m W1 = 6 × 88 × 2.4 = 1267

M1 = 1267 × 3 = 3801

1 W2 = (52.5)(75) × 2.4 = 4725 2

52.5 ) 3 = 4725 × 23.5 = 111037 6.8 = −1257 M 3 = 554 × −3 6.7 M PV1 = 115 × = −393 2 6.8 × 2 M PV2 = 231 × = −1048 3

1 W3 = (6.8)(68) × 2.4 = 554 2 PV1 = 17 × 6.8 × 1 = 115 1 PV2 = (6.8)(68) × 1 = 231 2 1 U = (.5)(1)(65.3)(85) 2

1 P = (1)(85) 2 = 3612 2

M 2 = 4725 × (6 +

 13  M u = 1387.6 ×  − 6.8  3  = 1387.6 × (21.76 − 6.8) = 1387 ×15 = −20758 85 M P = 3612 × = 102354. 3

Design of Gravity Dams  137

  SV = 1267 + 4725 + 554 + 115 + 231 + 1387 = 5505 SM = 102354 – 20758 – 393 – 1048 – 1257 + 111037 + 3801 = 193736 ∴ Xf =

ΣM 193736 = = 35.2 ΣV 5505

 2B  ≤ − 6.8  ≤ 37 m  3 

Reservoir empty condition:

ΣV = W1 + W2 +W3



= 1267 + 4725 + 554 = 6546 t ΣM = M1 + M2 + M3



= 3810 + 111037 – 1257 = 113581 t m ∴ Xe =

113581 B  = 17.35 m ≥  − 6.8  ≥ 15 m 6546 3 

Zone IV: o

FRL

3m 13 m

10 m Low gravity zone 88 m

6m

PV1

7m

h=97 m

W1

PV2

0.7 1

3m

68 m

W2

1 0.1

87 m

W3

High gravity zone 8m

6m

61 m

B = 75 m

Fig. 6.12:  Design of gravity dam, zone IV.

Consider full height of dam 100 m, and continue same u/s and d/s slope. Check for stresses and also Xf, and Xe, the positions of the resultant for the reservoir full and empty conditions respectively:

138  Irrigation Engineering and Hydraulic Structures

W1 = 6 × 100 × 2.4 = 1440 t

M1 = 1440 × 3 = 4320 tm

1 W2 = (61) (87) × 2.4 = 6368 t 2

M 2 = 6368 × (6 +

1 W3 = (8) (80) × 2.4 = 768 t 2 PV1 = 17 × 8 × 1 = 136 t 1 PV2 = (8)(80) × 1 = 320 t 2 U=

61 ) 3 = 6368 × 26.3 = 167690 tm 8 = − 2048 tm −3 = 136 × 4 = −544 tm

M 3 = 768 × M PV1 M

PV2 = 320 ×

8× 2 = −1706 tm 3

1 B  (0.5) (1) (75) (97) = 1818 t M u = 1818 ×  − 8 3  2 = −30918 tm

1 97 (1) (97) 2 = 4704 t, M P = 4704 × = 152112 tm 2 3 ΣV = 1440 + 6368 + 768 + 136 + 320 − 1818 = 7214 t ΣM = 4320 + 167690 − 2048 − 544 − 1706 − 30918 + 152112 P=

= 288906 tm

Reservoir Full Condition : ΣM 288906  2B  = = 40 ≤  − 8  ≤ 42 m ΣV 7214  3  Reservoir empty condition: Σ V = W1 + W2 +W3 = 1440 + 6368 + 768 = 8576 t Σ M = M 1 + M2 + M3 = 4320 + 167690 – 2048 = 169962 t-m 169962 B  Xe = = 19.8 m ≥  − 8 = 17 m 3  8576 ∴ Xf =



Check for Stresses: Reservoir full condition: R  6e  p n = v 1 ±  B  B B  e = Xf −  − 8  2   75  = 40 −  − 8   2  = 40 − 29.5 = +10.5

R V = ΣV = 7214 m where, B = 75

Design of Gravity Dams  139

7214  6 × 10.5  1+ 75  75  7214 = × 1.84 = 176.9 t/m² 75 pi = p n max sec ²φ = 176.9 (1 + tan ²φ)

∴ p n max =

= 176.9(1 + 0.49), here, tan φ = 0.7 = 263.5 t / m² < 300 t/m² hence safe.

Reservoir empty condition : pn =

8576  6e  1+  75  B

B  e =  − 8  − X e = 29.5 − 19.8 = 10.7 2   8576  6 × 10.7  p n max = 1+ 75  75  8576 = × 1.857 = 212 t/m² 75 tan φ = 0.1, pi = p n max (1 + tan ²φ) = 212 (1 + 0.01) = 212.2 t / m² < 300 t/m², hence safe.

Fig. 6.13:  Gravity dam (section)

Hence, zone falling in high gravity category has got stresses within limit ∴ Design of dam is safe.

140  Irrigation Engineering and Hydraulic Structures

6.5  FACTOR OF SAFETY FOR DESIGN OF GRAVITY DAM Factor of safety can be defined as ratio of stabilizing moment or force to displacing moment or force, i.e., S.M. S.F. F.S. = or D.M. D.F. There are in general three factors of safety to be considered in the design of gravity dam. 1. Factor of safety against overturning is defined as the ratio of stabilizing moment to overturning moment. stabilizing moment FS against overturning = overturning moment 2  (W − U)  B   3  , (taking moment about toe) F.S. = 1  PH  H  3  1  2   2 BH s γ − ½ cγ HB  3 B B(s − c) 2B 2B² (s − c) = = = H² H² 1  1  H² H γ    2  3  H But, B = s−c H² (s − c) 2 ∴ FS = =2 s − c H²

Hence, theoretical value of F.S. against no overturning is 2 Therefore, for practical profile, it can be prescribed greater than 2. 2. Factor of safety against sliding is defined as the ratio of stabilizing force to overturning force. SF FS against no sliding = DF µ(W − U) = , take µ = 0.7 PH = =

0.7 [½BHsγ − ½cγHB] ½ γH² 0.7(s − c)B 0.7(s − c)H H = , sin ce B = H H s−c s−c

= 0.7 s − c = 0.7(1.4) , taking s = 2.4, c = 0.4, s − c = 1.4 = 0.98 ≅ 1.0

Design of Gravity Dams  141

Hence, theoretically F.S. against sliding should be equal to 1.0 or greater than 1.0. For practical profile, shear–friction factor of safety is considered, which is greater than 4. µ (w − u) + Bq Shear-friction F.S. = PH Here q = shear stress at base of dam = p n max tan φ =

2R V B

 PH  W − U  

2(W − U)PH 2PH = B(W − U) B ∴ Bq = 2PH =

∴ Shear Friction FS =

µ (W − U) 2PH + PH PH

= 1.0 + 2.0 = 3.0

Hence, shear friction F.S. for practical profile is always greater than 3.

6.6  PRACTICAL PROFILE OF GRAVITY DAM Theoretical profile is triangular one, but practical profile cannot be so since top width is always provided which serves useful purpose of transportation of machinery, maintenance gang and also public transport facility. Also free board has to be provided to prevent over topping of water. Thus, theoretical profile gets modified as shown in the Fig. 6.14. The shaded portion is addition to theoretical section, and hence to balance its effect, small triangular portion on u/s side in the form of u/s slope is added. Also drainage and inspection galleries and other openings are provided in the dam section, hence, following Fig. 6.15 illustrates the practical profile fully:

Fig. 6.14  Practical profile of gravity dam.

142  Irrigation Engineering and Hydraulic Structures

6.7  DRAINAGE GALLERY A drainage gallery or inspection gallery is always provided in the body of dam section and is normally below Low Water Level and above Dead Storage Level, also its u/s vertical face may be along a vertical passing through mid point of top width. A porous vertical drainage line checks the seeping water in body of dam and makes it seep into the gallery. The gallery in turn throws it out of dam body via cross-gallery or pumping arrangement. Air vent in the form of a 20 to 30 cm diameter pipe connecting gallery to d/s face of dam above FRL is provided. This provides ventilation in gallery as well as just behind sluice gates. A cut off and dented bed at foundation level is provided to cut down seepage and to increase bed friction. Grouting holes from gallery to foundation can also be provided if curtain grouting is required to be carried out to decrease or cut down seepage. Top width 4 m to 6m

F.B.

F.B. = 1.5 hw or 3m; whicever is higher

F.R.L

Live storage

d/s slope (0.7 ro 0.8 H to 1.0V) Air vent for ventilation in gallery and sluices

L.W.L

Porous drain line, meeting gallery Supply sluice

Carry over store

Drainage or inspection gallery

D.S.L.

Cross gallery Dead storage Secondary sluice Cut-off

Dented bed to increase friction

Fig. 6.15:  Practical profile of a gravity dam

6.8  JOINTS IN GRAVITY DAM Joints can be defined as ‘designed crack’ for if joints are not provided, cracks will develop in the body of the dam in an uncontrolled manner. Also, gravity dam is a massive concrete work and hence lot of heat of hydration on account of setting of cement will get released and hence there is a very high temperature gradient between

Design of Gravity Dams  143

the surface and in the interior of the dam section. This calls for proper release of heat of hydration as well as providing suitable joints. The massive concrete work cannot be taken up all at a time and hence joints are required to provide constructional facilities also. Hence, joint is a must and serves following useful purposes. 1. Constructional facilities just as suitable blocks of concrete length-wise and suitable vertical lift height-wise, here type of joints are key ways and benching. 2. Temperature control by providing contraction joints between block to block of concrete work length-wise, elaborate arrangement to stop leakage water is required, which is illustrated in Fig. 6.16. Joint gap to take care of expansion or contraction in concrete block due to heat of hydration. Copper plate 10 mm thick, to prevent leakage. It is further protected by asphalt poured in space on u/s and d/s of plate. joint gap to take care of expansion or contraction in concrete block due to heat of hydration u/s

Asphalt

Dam back width in plan

Unhealed Copper plate, 10 mm thick to prevent leakage It is further protected by Asphalt poured in space on u/s and d/s of plate

Block A

Block B

d/s 2 to 3 mm

Fig. 6.16:  Contraction joint

Depending upon the magnitude of concreting work, suitable number and type of joints are adopted.

6.9  FOUNDATION TREATMENT Most of gravity dam failures are due to poor foundation treatment. Following cases of failures indicate that foundation treatment is very important and must be carried out with utmost care. (i) Francis dam in California, 62 m high and 210 m long. failed soon after its construction. This was due to presence of conglomerate in one of the abutments which was weakened after the reservoir got filled.

144  Irrigation Engineering and Hydraulic Structures

(ii) Malpasset Dam in France, 60 m high arch dam failed due to presence of clay seam in the abutment rock. (iii) Austin Dam on Colorado River in Texas failed in 1900 due to large cavities got dissolved in its limestone foundation. Hence, foundation treatment in the form of pressure grouting is extensively carried out prior to dam construction. Pressure grouting may be: 1. Consolidation Grouting,

2.  Curtain Grouting,

3.  Contact Grouting

1. Consolidation Grouting: Entire Foundation bed of dam is consolidated by pressure grouting. Shallow holes are drilled into foundation bed upto a depth of 10 to 15 m and at 5 m to 20 m centre to centre. Cement slurry is forced at a pressure of 3 to 4 kg/cm2 through these holes. This is known as low pressure grouting technique. 2. Curtain Grouting: It forms a barrier or curtain against seepage through foundation. Holes are drilled at 1.2 to 1.5 m c/c upto a depth of 30 to 40% of u/s water head. Grouting is carried out under high pressure of 0.250 kg/cm2 where D = depth of grouting hole in m. 3. Contact Grouting: To develop a strong bond between abutment areas of concrete dam with the adjacent rock, contact grouting is carried out. It is also carried out to prevent seepage along the contact plane of concrete and adjacent rock produced on account of shrinkage of concrete on setting.

6.10  TYPES AND FUNCTION OF GALLERIES Galleries are provided either parallel or normal to dam axis at various elevation, and are interconnected by vertical shafts filled with lifts. Functions of galleries are given below : 1. Drainage: Seepage from u/s water face of dam is interconnected by means of vertical drain lines or hollow vertical shafts as shown in Fig 6.15. They will join the drainage gallery and seeping water will be removed by pumping. Drilling into foundation to cut down seepage through foundation is also carried out from drainage galleries.

Design of Gravity Dams  145

2. Inspection: Study of structural behaviour of dam in post construction period is takes up through galleries and actual stresses developed in the body of the dam are measured by providing strain gauages and electronic measuring devices. 3. Operation of Gates: Galleries provide access to mechanical equipments for operation of gates. 4. Foundation Gallery: This is very near to foundation rock surface and its usual size is 2.0 × 2.5 m. It provides space for drilling and grouting of foundation holes from the floor of the gallery. Thus, there are three main types of galleries (i) Drainage, (ii) Inspection and (iii) Foundation.

q  SOLVED EXAMPLES Example 6.2:  Find out maximum height of theoretical profile of a dam if permissible value of compressible stress σ = 350 tonnes/m2. Take specific gravity of concrete s = 2.4. Find height H and base width B if uplift coefficient c = 0.6. Solution:  Limiting height of dam is given by σ 350 H= = = 125 m γ (s − c + 1) (1)(2.4 − 0.6 + 1)

and

B=

H s-c

=

125 2.4 - 0.6

=

125 = 93 m. 1.34

Example 6.3:  Draw uplift pressure diagram for a dam holding 60 m depth of water on u/s and top and base widths are 10m and 40 m respectively. Consider uplift as effective on 60% of base area and tail water depth = 6m. What will be the change in uplift pressure diagram if drainage gallery is provided at 6m from its u/s face? Solution: Uplift at heel = 0.6 (60) = 36 m Uplift at toe = 0.6 (6) = 3.6 m. 1 Uplift at gallery = 3.6 + (36 − 3.6) = 3.6 + 10.8 = 14.4 m  3

146  Irrigation Engineering and Hydraulic Structures 10 m

FRL

60 m

Tail water Gallery 6m 6m

40 m

3.6 m

14.4 m 36 m

Fig. 6.17:  Uplift pressure diagram.

Example 6.4:  Determine pressure, line of action and wave height for a dam 120 m high having a fetch of 30 km and wind velocity 100 kmph. Find required free board. Solution:  As fetch = 30 km < 32 km use formula 1/4    h w = 0.032 FV + 0.763 - 0.27(F)

Design of Gravity Dams  147

\ h w = 0.032 32 ´100 + 0.763 - 0.27(30)1/4   

= 1.75 + 0.763 – 0.63

 

= 1.883 m (wave height)

Free Board (FB) = 1.5 hw

= 1.5 (1.883)



= 2.82 m ≃ 3 m.

Wave pressure acting on dam at FRL per meter length of dam

= 19.62 h2w KN/m



= 19.62 (1.883)2 = 69.56 kN/m = 6956 kg/m

and its line of action is at 0.375 hw above FRL

= 0.375 (1.883)



= 0.7 m above FRL.

Example 6.5:  Find out base width of a theoretical profile of a gravity dam holding 60 m, water for given following data: s = 2.4, c = 0.5, µ = 0.7, γ = 1.0 tonne/m3. Also find principal and shear stresses at the toe. Solution:

B=



W=

H 60 = = 45.11 m. µ(s − c) 0.7(2.4 − 0.5) 1 1 B H S γ = (45.11)(60)(2.4)(1) = 3248 tonnes. 2 2

U = Uplift =

Cg hB 1 = (0.5)(1)(60)(45.11) = 676.65 tonnes. 2 2

∴Rν = W − U = 3248 − 676.65 = 2571.35 tonnes. P=

1 1 g h 2 = (1)(60) 2 =1800 tonnes. 2 2

148  Irrigation Engineering and Hydraulic Structures

ΣM = Moment at toe 2  h = Rv  B − P  3  3 2   60  = 2571 × 45.11 − 1800   = 77140 − 36000 = 41140 tonnes-meter 3    3  ΣM 41140 = = 16.0 m. Rv 2571



XF =



Eccentricity e =

B 45.11 - Xf = - 16 = 6.55m < B / 6 hence safe. 2 2

p n = normal stress at toe

=

RV B

é 6e ù 2571 é 6(6.55) ù 2571(1.871) ê1 + ú = ê1 + ú= êë B úû 45.11 êë 45.11 úû 45.11

p n = 106.6 tonnes/m 2 ∴ principal stress at toe = Pn sec 2 φ = p n (1 + tan 2 φ), here, tan φ = = p n (1.5625) = 106.6(1.5625)

4511 = 0.75 60

= 166.56 t m 2 < 300 t m 2 hence safe. τ = shear stress at toe



= p n tan φ = 106(0.75) = 80t/m 2 .

Example 6.6:  A concrete gravity dam stores water to a depth of 70 m and has top width 8m, fetch 40 km, wind velocity 150 kmph, upstream face vertical and d/s face starts sloping at 10 m from top at 0.7 Horizontal to 1.0 vertical. If coefficient of uplift c = 0.5. Specific gravity of concrete (s) = 2.4, friction factor (µ) = 0.7, compressive stress (σ) = 300 t/m2, shear stress (q) = 140 t/m2. Find: (i) Whether dam is high or low gravity dam (ii) Principal stresses at toe and heel for Rfull and Rempty. (iii) Factor of safety

(a) against overturning



(b) against shear-friction.

Design of Gravity Dams  149

Assume location of drainage gallery at 8 m from vertical face and at 10 m above the base of the dam, and tail water nil. 8m FB = 4 m

10 m

h = 70 m 4m

H = 74 m

W1 64 m

P

1 rh2 2

W2

30 m

23.3 m 10 m Toe

45 m

8m 4m U1

U3 30 m

35 m

2.7 m

U2 1 30 (35) 3

12 m

Fig. 6.18  Forces acting on concrete gravity dam.

Solution:  Ordinates of uplift at heel = cγh = 35 t/m2 ordinate at cross-section of 1 drainage gallery = (35) = 12 m (approx) 3 B = 8 + 0.7(64) = 8 + 44.8 = 8 + 45 = 53m. Fetch, F = 40 km > 32 km v = 150 kmph.

150  Irrigation Engineering and Hydraulic Structures

∴ h w = 0.032 ∨ F = 0.032 150(140) = 2.48 m ∴ FB = Free board

 

= 1.5h w = 1.5(2.48) = 3.71  4m.

(i) Limiting height of dam =

σ 300 = = 103m > 74m r(s − c + 1) 2.9

∴ Given cross-section of dam is low gravity dam. (ii) Forces acting per meter length of dam, 1. W1 = weight of rectangular portion (8 m × 74 m)     = 2.4 × 8 × 74 = 1421 tonnes(t)

Acts at 4 m from u/s face and at 49 m form toe.

2. W2 = weight of triangular portion of base 49 and height 64 m. 1 = (45)(64)(2.4) 2     2 = 3456 t acting at (45) i.e. = 30 m from toe. 3 3. P =

1 1 g h 2 = (1)(70) 2 2 2

   = 2450 tonnes acting horizontally at 4. Uplift:

1 (70) = 23.3 m from base. 3

Ordiantes at heel = cγh = 0.5(1) (70) = 35 tonnes/m2 Ordinate at cross-section of drainage gallery =

1 (35) = 12 t/m2 (approximately) 3

∴ u1 = 12 × 8 = 96 t at 4 m from vertical face of dam. 1 1    u 2 = (8)(35 − 12) = (8)(23) = 92 t at 2.7 m from vertical face of dam. 2 2  

1 u 3 = (12)(45) = 270 t 2

æ 2 ö ççat ´ 45 i.e., 30 m from toe÷÷. çè 3 ø÷

(iii) Moment about toe: Moment due to weight is stabilizing and hence treated as +ve (anticlockwise) and that due to P is –ve (clockwise) about toe.

Design of Gravity Dams  151

Sr. No. Force (t)

Lever arm from toe (m)

Moment (t – m)

1.

W1 = 1421

49.0

1421 × 49 = +69629

2.

W2 = 3456

30.0

3456 × 30 = +103680

3.

P = 2450

23.3 (Fresh base)

2450 × 23.3 = –57085

4.

  Uplift U1 = 96

49.0

96 × 49 = –4704

U2 = 92

50.3

92 × 50.3 = –4628

U3 = 270

30.0

270 × 30 = –8100

Σ M = +98792 t.m

W – U = Σ Rv = 1421 + 3456 – 96 – 92 – 270 = 4419 t ∴ Xf =

ΣM 98792 = Rv 4419

= 22.36 m from toe > B/6 i.e., 17.66 hence safe. B 53 = = 26.5 From toe. 2 2 B − X f = 26.5 – 22.36 2 B = 4.14 m < (i.e., 8.33) 6

∴e =

Hence safe.

R

Heel

B 3

B 3

B 3

B 2

B 2 X

First middle third point

toe Second middle third point

Fig. 6.19:  Position of resultant in full condition

(R) Position of resultant when R full is at X f from toe and is within middle third part of the base, hence safe R  6σ  Maximum normal stress = p n(max) = v 1 + B  B  4419  6(4.14)  1+ 53  53 



=



=122.46 t/m2 (Compressive at toe hence safe)

152  Irrigation Engineering and Hydraulic Structures

∴ p = Maximum principal stress at toe = p n(max) (1 + tan 2 φ)

= (122.46) (1 + (0.7)2); as tan φ = 0.7



= 183 t/m2 > 30 t/m2 hence safe

Factor of safety against overturning (RFull) =

SM Moment due to weight = DM Moment due to P and U



69629 + 103680 (57085 + 8100 + 5633 + 4704)

=

173309 75522 = 2.92 >2, hence safe. =



µ(W − U) + Bq or BT P 0.7(4399) + 53(140) 3079 + 7420 = = 2450 2450

Factor of safety against shear friction =



= 4.25 > 4, hence safe.

(iv) Calculation of principal stress at heel when reservoir empty R empty mean P = 0, V = 0 only weight present

∴ Xe =

=

ΣM ΣW

173309 173309 = 1421 + 3456 4877

= 35.55 m from toe > 35.32 ∴ not safe but difference is only marginal and hence OK. (R) Position of resultant when Rempty e = Xe − B / 2 = 35.53 – 26.5 = 9.03 m p n(heel) =



=

Rv B

 6σ  1 + B   

4877  6(9)  4877 × 2 1+ = 53  53  53

 = 184 t/m2 < 300 t/m hence safe.

Design of Gravity Dams  153

pi = p n (1 + tan 2 f), as u/s face vertical tan φ = tan(0) = 0   = pn   = 184 t/m2 hence OK. Overall result : Dam section is safe. R 17.66

17.66

17.66

Heel

Toe 35.32 X e = 35.53

Fig. 6.20:  Position of resultant in empty condition

EXERCISES 1. Distinguish between “low and high” gravity dam. Derive an expression for base width of elementary profile of a low gravity dam satisfying all the criteria of the dam. 2. Describe forces acting on a gravity dam. 3. Distinguish between practical profile and theoretical profile of a gravity dam. 4. Workout the stress at toe and heel of a gravity dam for reservoir full condition for following data: •

Top width - 6m.

•

U/s slope 0.1 H to 1 V and starts at 20 m from FRL.

•

D/s slope 0.8 H to 1 V and starts at 10 m from top of the dam.

•

Free Board – 3 m.

•

Height of the dam – 53 m.

•

Density of concrete – 24 KN/m3.

•

Uplift Coefficient, C – 0.5



(Ans: At toe pn = 56.6 t/ m2, at heel pn = –1.0 t/ m2, pi = 93 t/m2)

5. Define factor of safety for gravity dam and work out it for dam data of Question – 4, if m = 0.7. (Ans: FS against overturning = 2.0 and FS against sliding = 1.38) 6. Explain various purposes of drainage gallery. Give sketch of a gallery and air vent.

154  Irrigation Engineering and Hydraulic Structures

7. What are joints in concrete gravity dam? Describe them with sketches. 8. In a storage reservoir for irrigation, maximum fetch is 40 km. The wind velocity over water spread area is 100 km/hr and wind set up of 0.2 m is expected at FRL. Find minimum free board that should be provided above FRL. (Ans: FB – 3.23 m) 9. For an elementary gravity dam profile, the limiting height of the low dam without considering uplift is ho, and with consideration of full uplift it is h1. h Find ratio of 1 . Take specific gravity of concrete as 2.5. h0 h (Ans: 1 = 1.4) h0 10. Find the maximum theoretical height of concrete gravity dam, if permissible compressive stress of concrete is 30 kg/cm2 and coefficient of uplift 0.6 and specific gravity of concrete 2.4. 

(Ans: 107.14 m)  

 

7 Spillways and Gates 7.1 GENERAL Spillway is an overflowing part of the length of the dam. Generally, in accordance with site condition, central part of the length of dam is chosen as overflowing section of a dam and is known as spillway. Spillway is a vital part of dam and serves as safety valve to reservoirs in the time of flood. Spillway should be designed for adequate capacity such that design flood passes over it safely without causing any damage on u/s and d/s sides. This is possible by providing sufficient net length of spillway and head over spillway crest. Design flood is chosen as per importance of the project from maximum flood value decided by hydrological analysis. Spillway discharge is given by following formula: Qdesign = C Le He 3/2 2 where, C = Cd 2g = 2 to 3 and is known as spillway coefficient and its usual 3 value is 2.2. Le = Effective length of spillway = Net length – 2 (N Kp + Ka) He N = Number of piers Kp = Pier contraction coefficient = 0.01 Ka = Abutment contraction coefficient = 0.1 V2 He = Effective head = H+ a 2g H = Head over crest = FRL – Crest Level q Va = Velocity of Approach = (H + h) q = Discharge per unit net length of spillway Q = Le h = Height of spillway crest from river bed. © The Author(s) 2023 S. K. Ukarande, Irrigation Engineering and Hydraulic Structures, https://doi.org/10.1007/978-3-031-33552-5_7

155

156  Irrigation Engineering and Hydraulic Structures

Types of Spillways: (i) Ogee or Standard crested spillway (ii) Side channel spillway (iii) Shaft or Morning glory spillway. (iv) Siphon Spillway (v) Chute or Trough Spillway. (vi) Tunnel or Conduit Spillway.

7.2  OGEE SPILLWAY This is the most suitable types of spillway and is provided at all the major project sites in India. It has two different types of curves at crest level, d/s curve and u/s curve. F. R. L a

Crest point, origin curve

Crest level

U/s crest with radii R1 and R2

R2

R1 b

d/s curve upto (dy/dx) = V/H

y x

Curve meets d/s slope tangentially H

h u/s Face vertical

V R y2

River bed 90o Heel

R

R=Bucket radius

End sill

Toe

Fig. 7.1:  Ogee spillway with bucket

Upstream curves has radius R1 from crest point upto a point at horizontal distance (a) from crest point, and then after it has radius R2 upto a point where it meets u/s vertical face of the Spillway. Values of a, b, R1 and R2 are given below. (Note : The crest point is at a horizontal distance of b from u/s face and this crest point is the origin of the d/s curves as well as u/s curves.) a = 0.175H, H = Head over crest b = 0.298H, R1 = 0.5H, R2 = 0.2H D/S Curves follows WES equation, developed by US army corps of engineer, which is given below: x1.85 = 2 y (H)0.85 , H = Head over crest.

Spillways and Gates  157

In its general forms,

x n = KyH n −1



 1  n ∴y =  x  KH n −1 



∴

dy  1  n −1 V = = = D/S slope of spillway body.  nx dx  KH n −1  H

d/s curve starts from origin at crest point and meets tangentially d/s face of the spillway such that dy V = dx H Then after d/s slope (V:H) continues upto Spillway toe. At the toe suitable dissipation works such as 1. Roller bucket 2. Ski-jump bucket 3. USBR basins 4. IS (Indian Standard) Basins, etc. may be provided to dissipate maximum part of Kinetic Energy of flowing sheet of water, Energy dissipation can be related to Froude Number at toe, F1. F1 =

V1 gy1

, here V1 =

q Q , and q = y1 Le

Energy dissipation due to hydraulic jump can be worked out by knowing y1, y2 and F1. ½ 1   y 2 = y1  8F12 + 1 − 1 2  

(

)

3

( y − y1 ) m ; where y1 = pre-jump depth and and energy head lost = 2 4y1y 2 y 2 = post-jump depth Table 7.1:  Relationship of F1 and Hydraulic Jump F1

Type of Jump and % energy dissipation

F1 < 1

Jump will not occur

F1 > 1, 9

Strong jump head loss 85%

158  Irrigation Engineering and Hydraulic Structures

7.3  DESIGN OF STILLING BASINS Energy dissipating works designed on the basis of formation of hydraulic jump are: 1. USBR basins and 2. Indian Standard Basins

Baffle block

y1

End sill

y2

Chute block l.5 y1 Toe of spillway

l.2 y1

0.8 y1

L= apron lenght

Fig. 7.2:  Hydraulic jump type stilling basin

Figure 7.2 shows a typical hydraulic jump type stilling basin developed by USBR. Water from d/s face of spillway enters the apron with depth equal to y1 and F1 greater than one. Tail water depth, y2, should be sufficient for formation of hydraulic jump and it should be equal to ½y1 ( 8F1 ² + 1 − 1)  and then only energy dissipation as per values shown in Table 7.1 will take place. If tail water depth is greater than y2, jump gets submerged and if it is less than y2, jump gets swept off to d/s side. This creates problems of erosion of river bed. Hence, success of stilling basin is dependent on formation of hydraulic jump within basin. USBR type basin are classified into four categories i.e., type I, type II, type III and type IV for different ranges of F1 values and arrangement of length of apron, chute block, baffle block and end sill is made type wise such that hydraulic jump occurs in the basin, giving expected energy dissipation. IS basin type I type II are also designed for various ranges of F1 on similar lines. Types of basin and ranges of F1 values are given in Table 7.2. Table 7.2:  Stilling Basin Types and Values of F & L/Y2 Types

USBR Basin

IS Basin

Value of L/Y2

Type I

8.5 < F1 > 4.5

F1 ≤ 4.5

4 to 6

Type II

4.5 < F1 > 9.5

F1 > 4.5

2.3 to 2.8

Type III

9.0 < F1 > 17

3.85 to 4.3

Type IV

F1 > 17

3.85 to 4.3

Spillways and Gates  159

Relationship of Jump Height Curve (JHC) and Tail Water Rating Curve (TWRC): JHC can be plotted by calculating various values of Y2 for different values of q, discharge per unit net length of spillway and TWRC can be plotted by obtaining field observation for various values of Tail water depth for different values of q, their possible combination are given in Fig. 7.3. (1) (1)

JHC (1)

(2) JHC = TWRC

TWRC (2)

JHC (1) TWRC (2)

(2) JHC > TWRC

q

q

(ii)

(I)

JHC lower thanTWRC for high ranges of q

(1) (2) (1) JHC above TWRC for low ranges of q

(2) JHC (1)

(1) JHC < TWRC

TWRC (2)

q

JHC (1) TWRC (2)

q (iii)

(iv)

Fig. 7.3:  (i) to (iv) showing possible combinations of JHC and TWRC against q

Case I where JHC and TWRC coincide for all ranges of q rarely occurs and if so, it is ideal and a simple apron as energy dissipation work is sufficient, but usually JHC may be found either above or below TWRC as shown in Fig. 7.3 (ii), (iii) and (iv) and in that case provision of stilling basin as per different ranges of Froude number F1 should be made. Last case where JHC and TWRC cross each other is found generally occurring in most of spillway energy dissipation works.

7.4  DESIGN OF SKI-JUMP BUCKET AND ROLLER BUCKETS If TWRC is always below JHC and river bed consists of sound rock formation, then ski-jump bucket type energy dissipation work is economical and more suitable, which is shown in Fig. 7.4 As shown in Fig 7.4 the bucket is turned upward such that its invert is always lower then Tail Water Depth, and its lip makes on ∠θ with the horizontal at a height from invert level equal to R (1 – cos θ).

160  Irrigation Engineering and Hydraulic Structures F.R.L Crest level H Trajectory curve R = Radius of bucket

h

R

River bed level

Solid rock bed of river y

y1

TWD

Invert of bucket

Fig. 7.4:  Ski-jump bucket

The lip angel is kept between 30° to 45° and water shoots from the lip in the form of trajectory curve, with edge of lip as origin, is the equation of trajectory y = x tan θ −

x² 4KE1 cos 2 θ

where,   K = air entertainment coefficient = 0.9 to 0.95 E1 = y1 +

V12 = specific energy head at invert of the bucket 2g

V1 = 2g(H + h) q=

Q = discharge per unit length of spillway net length of spilling

y1 =

V q , F1 = 1 V1 gy1

y max = maximum height of trajectory curve =K

V12sin 2 θ 2g

x = maximum horizontal range = K

V12sin 2 θ g

The energy dissipation is due to air-entertainment and bubble formation and impact of trajectory with rocky river bed having some tail water. If ski-jump bucket is not suitable on account of non-availability of sound rock at river bed, then solid bucket with one roller falling on incoming water and one roller going out into river bed will be a suitable type of energy dissipating device.

Spillways and Gates  161

Here radius is same as that of ski-jump bucket, but invert of bucket is kept at river bed level and lip angle is greater than 45°, preferably 60°, as shown in Fig. 7.5: F.R.L. Crest level

H

Roller falling on the incoming jet H dv/dx1

V

Outgoing roller

R River bed level y1

R (1-cos )

TWD > JWC

Invert of bucket R = Roller bucket radius

Fig. 7.5:  Roller bucket



R = Roller Bucket Radius

  = 0.6 Hh or R= 0.305 (10)p, where p = 1.9, Ven Te Chow’s coefficient Roller bucket is preferred when TWD > JHC. Both these types of buckets remain submerged in tail water and hence are also known as submerged energy dissipaters. Radius of bucket R can also be worked out by using Varshney’s formula.

æRö F1 = 0.09 ççç ÷÷÷ + 1.98, è y1 ø÷ where F1 = Froude number at toe. or Varshney and Bajaj formula, F1 =13(R)1/4 - 19.5 (Note : Formulae for radius of bucket for Ski-jump bucket and for Roller bucket are same) Roller buckets were first used for Grand Coulee dam on river Colarodo (USA) and have become quite popular since then. In India, a number of dams have used roller buckets. Table 7.3 gives. Table 7.3:  Data used for Roller Bucket in India Sr. No. Name of Dam

q (m3/s/m)

Radius of bucket in m

Lip Angle

92

21.3

34°

1

Nagarjun Sagar dam

2

Shri Shailam

121.0

21.3

20°

3

Rana Pratap Sagar

61.2

16.76

40°

4

Hirakud

104.1

15.24

40°

162  Irrigation Engineering and Hydraulic Structures

7.5  SIPHON SPILLWAYS 7.5.1  Saddle Siphon A spillway designed and constructed on the basis of siphonic principle is defined as siphon spillway. It is automatic in action and has advantage of high rate of discharge for same head H over crest level compared to ogee type spillway, but this has a limit that difference between HFL and tail water level should not exceed 7.3  m otherwise its dome may suffer of cavitation and vibration problems. This is proved in next section (7.6). High head dam projects can not satisfy this condition and hence this type of siphon spillways are suitable to low head medium sized projects. It consists of two legs: Inlet leg kept on u/s side and its entrance should be below FRL and discharging leg kept on d/s and its outlet should be always under tail water level and should never be exposed to atmospheric air. On the summit at crest level there hood or crown attached with deprimer pipe having its leg on u/s side above FRL so that when water level rises in reservoir, deprimer gets locked, no air can enter or leave from it. Water rushes through the inlet leg and starts over flowing the crest and getting the air wiped out from discharging leg when Deprimer Unit Hood or Crown air vent Spill way

HFL H

FRL

Crest level X axis

Inlet

Exit end always between tail water Tail water level

Y axis 0.3 H

Fig. 7.6:  Siphon spillway (saddle type)

all the air is removed, siphonic action starts. To speed up the removal of air from discharging leg several arrangement have been tried, such as provision of baby siphon below entrance leg or a plate with spring action on d/s of crest so that water is thrown to strike the upper surface of the discharging pipe and forcing the air to go out. With all such attempts, this difficulty is not overcome totally, hence other type of spillway known as volute siphon has evolved. When water level goes down on u/s sides deprimer leg on u/s gets exposed to atmospheric air first and so air enters the summit and siphonic act gets stopped.

7.5.2.  Volute Siphon A volute siphon spillway consists of a dome with a funnel of D1 diameter and throat of D2 diameter such that volutes are provided over (D1 – D2)h space which is shown in section and plan given in Fig. 7.7.

Spillways and Gates  163

Dome is resting on supports leaving a clear cut entry of water through the Funnel. Deprimer ends above FRL and dome end below FRL, so that when water in reservoir rises above FRL, water starts overflowing into the throat via volutes and thereby a forced vortex flow is generated which will force out air through the exist, Deprimer Above FRL FRL Dome Funnel (inlet) Below FRL

Volute

air entry or exit D1

D2

h D2 D1 Plan of Funnel

Support to Dome Throat

Exit D2

90° Bend

Fig. 7.7:  Siphon spillway (volute type)

the air which gets locked in dome excerts extra pressure on forced vortex motion and so removal of air from exist is achieved in a very short interval of time, thus this type is having a clear advantage over the saddle type of quick siphonic action. However, it suffers off if any clogging in the annular space around funnel. Discharge Formula: Q = CA 2g(H − h f ) Here, H = maximum operating head, which is difference between water level above FRL on u/s and level of centre line through the exist pipe on d/s, and its limit is 7.3 m. A = crest section area of exit pipe, A =

π (D 2 ) 2 4

hf = friction loss in the pipe and bend. With this fall in water level in reservoir below FRL, air enters through deprimer into the dome and siphonic action stops. Maintenance problems are heavy and so this type is not preferred for high head and medium head projects.

164  Irrigation Engineering and Hydraulic Structures

7.5.3  Chute Spillway Normally on earthen dam site, if construction of saddle weir on either side of the dam is possible then a chute i.e., a channel on sloping side of weir is constructed to pass the discharging water to join the river on d/s. This type of site is rarely available and so in practice these types of spillway are very few at global level, and is as shown Figure 7.8. Saddle Weir

Main Dam

Section Weir Dam Side wall Chute Spillway (Channel on sloping river → side of weirs)

Divide wall Plan

Fig. 7.8:  Chute spillway

7.5.4  Side Channel Spillways This type differs marginally from chute type in having the side weir not in alignment of the main dam but at either right angle or at some acute angle with the main dam axis so that overflow takes place from reservoir rim and is guided by means of side channel to meet the river on d/s, as shown in Fig. 7.9. (1) Ogee weir on rim of reservoir

Guide wall

U/S

Side Channel Spilway (2) may be 90° or less

DAM d/s river

Fig. 7.9:  Side channel spillway

The weir provided along the rim of reservoir at ∠θ may be designed as ogee or straight drop type and side walls are provided to guide the water flow to join the river on d/s. This is also not a common type due to non-availability of proper site for it.

Spillways and Gates  165

7.5.5  Shaft Spillway It is similar to volute siphon type with a difference that it has a circular weir accommodating on u/s protection of the main dam and water falling over the rim of the weir joins a tunnel which might have been used as diversion tunnel to divert the river flow at the time of construction of the main dam. This type is chosen only for this purpose that diversion tunnel may be used even after completion of the construction of main dam. Except the ogee spillway and siphon spillway all other types are rarely chosen as sites for these types are rare and also their working needs continuous attention and hence maintenance.

q  SOLVED EXAMPLES Example 7.1:  Design a ski-jump bucket type energy dissipater for following data: (i) Discharge from spillway 9000 m3/s = Q (ii) Height of spillway above bed h = 60 m (iii) Design head H = 10 m (iv) Net length of spillway = 90 m (v) Rock is found at 2 m below river bed (vi) Air resistance coefficient = K = 0.95 (vii) Value of p in Ven Te chow’s expression = 1.9 Assume TWD < JHC Find (i) Radius of bucket 1. Length and height of trajectory 2. Depth of invert of bucket. Solution: R = 0.305(10) p = 0.305(10)1.9 = 24.22m R by 0.6 Hh = 0.6 10 × 60 = 14.7 m Adopt lip angle as 30°∴θ = 30°

Length of trajectory = K

V12 sin 2θ =114.8 m g

V1 = 2g(H + h ) = 2 × 9.81 (60 + 10) = 37.0 m/s. y1 = F1 =

Q 9000 = = 2.7m Length of spillway × V1 90 × 37 V1 gy1

=

37 9.81 × 2.7

=

37 = 7.18 5.15

166  Irrigation Engineering and Hydraulic Structures

Hence, adopt R = 21m é F + 19.5 ù 4 æ 7.18 + 19.5 ÷ö4 ú = çç R by Varshney and Bajaj = ê 1 ÷ = 17.74 m çè êë 13 úû ø÷ 13 or R by Varshney, æRö F1 = 0.09 çç ÷÷÷ + 1.98 çè y1 ÷ø æRö 7.18 = 0.09 çç ÷÷÷ + 1.98 çè 2.7 ø R=

0.699 ´ 2.7 = 20.9 m 0.09

y max = Height of Trajectory = K

V12 sin 2 q 2g

= 0.95 x max = K

(37) 2 (sin 30) 2 = 16.5 m. 2´ 9.81

V12 sin (2q ) (37) 2 (sin 60) = 0.95 = 114.8 m. g 9.81

Depth of invert of bucket from lip = R(1 − cos θ) = 21(1 − cos 30) = 2.79m.

Example 7.2:  The siphon spillway of a dam has 20 vents of size 2m × 1m. The maximum water level is 160.0 m from river bed, and tail water Level is 154.0 m. from river bed. Spillway crest is at 158.0 m. Cd = 0.62. Find discharge of the spillway. If same discharge is to pass over ogee spillway of length = 40 m, find design head, take C = 2.2. Which type would you recommend? And why? Solution: Refer to Fig. 7.10, H = 160 – 154 = 6 m. Q = Discharge through siphon spillway = Cd A = 0.62 (20 × 2 × 1)

2g × 6.0 = 0.62 × 40 × 10.84 = 268.8 m3/s

Q over ogee = C Lnet H3/2 268.8 = 2.2 (40) H3/2

2gH

Spillways and Gates  167

Deprimer hood M.W.L. Lip of deprimer F.R.L. Entrance lip

Lower leg

H=160-154=6 m

Exit lip below TWL

h

T.W.L.

Fig. 7.10:  Siphon spillway

M.W.L.

1

M.W.L.

O

F.R.L. Entrance lip h

Lower leg

H

Exit lip 3

O

T.W.L.

Fig. 7.11:  Siphon spillway

H = (3.05)2/3 = 2.11 m. = head required over crest is higher than that of siphon spillway. Hence, siphon spillway is recommended.

168  Irrigation Engineering and Hydraulic Structures

7.6  LIMITING HEAD OF SIPHON SPILLWAY Let point (1) be at MWL, (2) at throat of spillway and (3) at TWL. [Fig. 7.6 (i)] Apply Bernoulli’s theorem between points 2 and 3, p V2 p 2 V22 + + H = a + 3 + h f 2 −3 γ γ 2g 2g p 2 = pressure at throat of siphon spillway. pa = atmospheric pressure at TWL ∴

pa = 10.35m of water, absolute γ

V2 = velocity at throat, Vapour Pressure =

pv = 2.43 m of water, absolute γ

V3 = velo city at exit lip If pv = absolute vapour pressure at which vaporization occurs, then

æp p ö æ V 2 - V22 ö÷÷ H = çç a - v ÷÷÷ + ççç 3 ÷ + h f 2-3 çè γ γ ÷ø èç 2g ø÷÷ = (10.35 - 2.43) + h f 2-3 ,sin ce (V3 = V2 ) = 7.92 + h f ,

h f = (1 - C2v ) H , C v = 0.96,

\ h f = 0.08H, neglected \ H £ 7.92m, limiting value, for no cavitation at throat Apply Bernoulli 's Theorem between po int s (1) and (2): pa p 2 V22 = + γ γ 2g æp p ö V2 = 2g çç a - 2 ÷÷÷ = V3. çè γ γ ø÷ Limiting value of Velocity at throat = 2g ´ 7.92 = 12.5 m/s. = V3 = value of velocity at exit

Spillways and Gates  169

q  SOLVED EXAMPLES Example 7.3:  Design an ogee spillway to pass a designed flood of 6000 m3/s with a head over crest equal to 10 m. Height of spillway crest from rocky river bed = 60 m. D/s slope is 1.0 V to 0.75 H. Crest coefficient = 2.2. What is the probable area of catchment from which flood is coming? Inglis formula may be used. Solution:

Fig. 7.12:  Ogee spillway with roller bucket

Step I: Refer Fig. 7.12 Q = CLe H3/2 − − − − > neglecting Velocity of Approach 6000 = 2.2 (Le )(10)3/2 . ∴ Le =

6000 = 86.2m. 2.2 × 31.62

L net = Le + 2(NK p + K a )H Assume 12 gates, hence N = no. of piers =11 Assume k p = 0.01 and K a = 0.1 ∴ L net = 86.2 + 2(11 × 0.01 + 0.1)10

= 86.2 + 4.2 = 90.4 m Assume pier width 3m Hence, total length of spillway = 90.4 + 3×11 = 123.4 m.

170  Irrigation Engineering and Hydraulic Structures

Step II: For d/s crest curve, WES equation is x1.85 = 2 y (H)0.85 (1) Assume x and calculate y., draw curve from crest point as origin and upto x 0.85 0.85 dy dy = 11 = 1.85(x) 1.85(x)0.85 = 1.85(x) 1.85(x)0.85 = = H 0.85 = (10)0.85 dx dx 0.75 0.75 H 0.85 (10)0.85 0.85 11 (10) 0.85 (10)0.85 5.10 \ \ xx 0.85 = = 0.75 ´ ´ 1.85 = = 5.10 0.75 1.85 1 1 0.85 xx = = (5.1) (5.1) 0.85

\ = \ = 6.72m. 6.72m. = i.e. , from origin, upto x i.e., from origin, upto x = 6.72 6.72 m, m, crest curve on d/s be drawn using crest curve on d/s be drawn using equation equation (1). (1). Step III: Draw u/s curve by taking R1 = 0 .5H = 0.5 × 10 = 5m R2 = 0.2H = 0 .2 × 10 = 2m a = 0.175 H = 1.75m b = 0.29 H = 2.9m H × h = 0.6 10 × 60 = 14.69 = 15m.

Step IV: Bucket Radius = 0.6 tan ϕ =

dy 1 = dx 0.75

∴ϕ = 53o Adopt lip angle θ = 45o , and height of lip of bucket = R(1 − cos θ) = 15(1 − 0.7) = 4.5m. Step V: Catchment Area = A sq km. By Inglis formula, 124A

Q= ∴ 6000 =

A + 10.4

124A A + 10.4 2

≈ 124 A

 6000  A=  = 2341 sq km.  124 

Spillways and Gates  171

Example 7.4:  Vent size of a siphon spillway is 3m × 1m, and there are 10 vents. The head over crest is 2 m and crest height from bed is 10 m. Tail water depth is 2 m above riverbed. Find discharge through siphon spillway such that vapour pressure at throat is not less than 2.43 m absolute. Solution: R 2=Radius of crown L=3 m R2

Entrance lip

1m R1

R 1=Radius of crest

r

H=MWL-TWL =10 m

h

TWD=2 m

Fig. 7.13:  Siphon spillway

Refer Fig. 7.13. Since crest and crown are curved, vortex motion through vent of spillway gets established and hence: V1R1 = V2R2 = v r = Constant, V1 = velocity at crest level, v = velocity at strip level. 1 V2 velocity at crown level ∴ v = V1R1 But V1 = 2gH r Here, H permissible head =10.35 – 2.43 =7.92 ∴ V1 = 2g × 7.92 = 12.5 m / sec. dq = L × dr × v , here dq = discharge passing through strip of area (dr × L)

dq = L × dr ×

1 V1R1 r

  Q = Cv 12.5 (LR1)

R2

∫

R1

dr r

172  Irrigation Engineering and Hydraulic Structures

R  = C v 12.5 (LR1 ) log e  2   R1  R  = C v 12.5 (LR1 ) 2.3log10  2   R1  2 = C v 12.5 (3 × 1) 2.3log10   1 = 23.36 m3 /s per vent, (C v = 0.9) There are 10 vents ∴ Q total = 23.36 × 10 = 233.6 m 2 /s If Q = Cd A 2gH relationship is used. then Q = 0.62 (30) (12.5) = 232.5 m3 / sec. Example 7.5:  A reservoir catchment of 4500 sqkm. is in hilly area. Using Inglis formula find the flood discharge. If an overflow ogee section with crest gates of 12m × 6.5 m is to be provided, find the number of gates. Neglect velocity of approach and take coefficient C for discharge as 2.1. Solution: Q by Inglis formula =

124 A

=

124(4500)

A + 10.4 4500 + 10.4 = 8309 Cumec.

Assuming H = 6m, Q = CLH3/2 (since gate size given is 12 m ´ 6.5 m , H can be assumed as 6 m.) 8309 = 2.1 L(6)3/2 8309 \L= = 269m. let , n = no. of gates, n ´12 = 269 2.1´14.7 269 \n= = 22.4 say 23 12 = Number of gates = 23

7.7  SPILLWAY CREST GATES Gates on the crest of spillways provide additional storage capacity of reservoirs known as “flood absorption capacity” of reservoirs or surcharge storage. In the event of flood, the crest gates can be opened out as per requirement of reservoir flood routing scheme, and flood water is allowed to pass over the spillway crest.

Spillways and Gates  173

Economics of gated and ungated spillway is based on comparison of 1. Cost of gates and cost of additional land that may get submerged. 2. Benefits that can be made available due to additional storage capacity of reservoirs due to gates. Type of Gates: (i) Vertical lift type (ii) Radial or Taintor gate (iii) Drum gates (iv) Stop log or needle gates Hoisting arrangment Hoisting wire

Skin plate

M.W.L.

L-section to hold rubber Hard rubber seal

Arm of taintor gate

(Head on crest)

I-Beam embedded in block A at crest of spillway

Trunnion of gate Bracing

Crest lavel A

Block A

Fig. 7.14:  Section of taintor gate

Taintor Gates: (Refer Fig. 7.14) Among all four different types, taintor gate or radial gate is preferred for medium and large scale projects. Taintor gates have advantage over vertical lift type as resultant water pressure line will pass through trunion of the gate about which gate rotates, hence water pressure will not exert any moment on the gate and opening and closing of gate will be smooth compared to vertical lift type. Also hard rubber seals at bottom of the skin plate and sides of skin plate will not allow leakage of water. The gate consists of curved skin plate supported by I beams which are connected to arms of the gate. The arm rotates about a trunion embedded in pier. The opening and closing of gate is done mechanically by winch and hoist arrangement provided at the top of the dam.

174  Irrigation Engineering and Hydraulic Structures

7.8 DESIGN OF AIR VENT FOR VERTICAL LIFT GATES PROVIDED IN SUPPLY SLUICES OF THE DAM Vertical lift gates are usually provided in the supply and scouring sluices of the dam. Just at the back of these gates, air vent connection should be provided to keep pressure at the back of gate atmospheric or at least not to allow negative pressure to develop. If gates are partially opened or closed, the cavity formed at the back of gate will develop negative pressure, which will lead to development of cavitation and vibrations. Air vent properly designed will prevent this. Function of air vent is to nullify vacuum effect just d/s of leaf gates of the sluices. Capacity of air vent should be adequate and is fixed on the basis of volume of air needed to nullify development of negative pressure. For example, at Mica project in Canada, airflow of 200 cumec was required in air vent at a maximum air velocity of 53 m/s to allow a discharge of 1000 cumec of water through sluice opening in the dam. Empirical relationship between Qair and Qwater are given below: (i) Qa = 400 C A P here C = 0.7, A = air vent c/s area P = differential pressure between atmospheric condition and that just at d/s of gate leaf in kg / cm² (ii) (a)  Air vent area = 10% of sluice gate area (b)  Air vent velocity = 40 m/s to 90 m/s (iii) As per USBR recommendations, Air ventilation

Vertical lift gate H

Fig. 7.15:  Air vent

Air vent discharge = 25 % of Qwater i.e., Qair = 0.25 Qwater

Spillways and Gates  175

iv) (iv)

Qair = α Q water here α = 0.03(F1 − 1)1.06 for openings in which jump occurs. and α = 0.0066(F1 − 1)1.4 for opening in which jump does not occur F1 = Froude Number of Flow through opening of size (y × b) ∴ F1 =

V gy

,

Qw = v × y × b

q  SOLVED EXAMPLE Example 7.6:  Design an air vent for gate opening of size 2 × 3.3m. The gate in front of the opening is vertical leaf type and is under a head of 42 m. The gate bottom lip angel is 45° and maximum air demand occurs at 80% of vent opening. Cd for gate lip = 0.81 Solution:

Air ventilation 47 cm diameter

Vertical lift gate H = 42 m

y

1.6 m 2 m x 3.3 m

Fig. 7.16:  Air vent

Refer Fig. 7.16, Y = 80% (2 m), gate size 2 m × 3.3 m.   = 1.6 m

176  Irrigation Engineering and Hydraulic Structures

V = Cd 2gH = 0.81 2g ´ 42 = 24m / s F1 =

V gy

=

24 9.81´1.6

= 6.4

a = 0.0066 (F1 - 1)1.4 = 0.07 Qa = aQ w = 0.07[24 ´1.6 ´ 3.3] = 8.8 Q = AV p = d 2 (50), here 50 m / sec is air velocity through vent. 4 ∴ Diameter of air vent d = 4.7 m. Hence, provide air vent of diameter = 47 cm.

7.9 CAVITATIONAND VIBRATION PROBLEM IN HYDRAULIC STRUCTURES 1. Ungated Spillway Crest: Spillway crests are designed as per WES equation, xn = KH dn −1 (y) , When effective head on spillway crest, He, is less than or higher than Hd (here Hd = design head) then the lower nappe of water will compress against or will get lifted from crest profile respectively and hence will exert positive pressures or negative pressures on crest profile .The negative pressure if falls below vapour pressure (2.43 m absolute) then, cavitation may take place and crest profile may get damaged. This can be controlled by providing adequate ventilation. Tests have shown that sub atmospheric pressure is of the order of 0.2 Hd where Hd = 0.75 He. Ungated spillway crest is shown in Fig. 7.17.

Sub-atmospheric pressure on spillway Hd

Fig. 7.17:  Ungated spillway crest

Spillways and Gates  177

2. Gated Spillway Crests: Tests have shown that crests of gated spillway may also develop negative pressure just downstream of gate and its magnitude may be Hd if H d = 0.75 H e 10 3. Cavitations problems may develop in throat section of saddle siphon spillways if pressure in throat falls below vapour pressure. 4. Cavitations and vibration may also occur just downstream of vertical leaf gates provided in supply sluices or scouring sluices in the dam section. This can be controlled by providing atmospheric air just d/s of gate by means of air vents. Gates may also suffer of vibrations due to instability of flow passing through them. Instability of flow may occur due to crest profile may not be truly congruent to lower nappy of water sheet flowing over it. Velocity and pressure fluctuation may get amplified due to instability of flow and hence may lead to induced vibrations in the structure. This can be controlled by adopting proper shape of crest profile of opening of gate and lining it with material, which can withstand cavitation effect.

EXERCISES 1. State different types of spillways and describe in detail any one type. 2. State different types of gates and distinguish between vertical lift type and radial type. 3. A spillway discharges 5400 cumecs through 6 spans of 10 m each and is a part of concrete gravity dam of height 66 m and FB = 3m. Find crest level of spillway and Catchment area Pier thickness is 0.75 m, slope of d/s glacis beyond tangent point is 0.8 H to 1.0 V. Spillway coefficient = 2.5, Kp = 0.01, Ka = 0.1. Find the coordinates of lower nappe and also u/s curvature of spillway surface. (Ans:. 50 m above river bed) 4. Sketch a stilling basin if Froude number at toe is 4.5 5. Sketch and write down specification for radial gate. 6. Design an air vent for a sluice of size 3.3 m × 2.2 m height, closed by vertical lift gate operating under a head of 42 m. Maximum air demand occurs when the gate is lifted to 78% of its height, take lip angle = 45o, Cd = 0.78, Cv = 0.95(Ans:. Diameter of air vent = 0.6 m.) 7. Explain the terms: Cavitations and vibration in hydraulic structure. 8. Siphon spillway of a gravity dam has 25 vents of size 3 m × 1 m height. The MWL for spillway is 160 m for design head, crest of siphon at RL of158 m, and RL of tail water = 154 m, Cd = 0.62, find discharging capacity of siphon spillway.(Ans:. 25 × q = 25 × 20.1)  

 

8 Arch and Buttress Dams 8.1 INTRODUCTION Gravity dams resist water pressure by self-weight and hence are known as gravity dams. Similarly, Arch dams resist water pressure by arch action and pass the load to valley sides, i.e., abutments. Basic requirement for arch dam is therefore sound rock formation along valley line of the dam site. In our country we have constructed only one arch dam so far because non-availability of suitable site for this type of dam. Resistance to water pressure by arch action and transferring the load to valley line at different levels reduces its thickness considerably. Gravity dams need as high as 85% of its height as base width, whereas arch dams need only 5% to 10% of height as its thickness, and hence if arch dam is possible to construct, first preference should be given to it compared to Gravity type. The only arch dam in India, so far, is Iddiki dam situated near Cochin in Kerala State. It is constructed in narrow V-shape gorge and is having double curvature. It has a base width of 19.8 m and top width of 7.6 m over a height of 169 m. Its length is 366m. It is one of the arch dams of world, which has thinnest section.

8.2  TYPES OF ARCH DAMS There are four main types of arch dams depending upon mode of its design. They are: (i) Constant radius arch dam (ii) Constant angle arch dam (iii) Variable radius and angel type (iv) Double curvature type The constant radius arch dam has upstream face in the form of a vertical cylinder and hence radius of extrados remains constant from top arch to bottom arch, however intrados of arches will have variable radius as the thickness of arch from top to bottom increases (Fig. 8.1). © The Author(s) 2023 S. K. Ukarande, Irrigation Engineering and Hydraulic Structures, https://doi.org/10.1007/978-3-031-33552-5_8

178

Arch and Buttress Dams  179

180 160 80 m Ph

V shape (elevation)

140 120 100 A

Extrados for all arches B Intrados at level 140 m Intrados at level 160 m Intrados at level 180 m

Plan

Intrados at level 120 m Intrados of bottom most arch, level 100 m

Fig. 8.1:  Constant radius arch dam

8.3  DESIGN OF ARCH DAM BY THIN CYLINDER THEORY The arches are designed by thin cylinder theory. The constant radius arch dam has relatively large thickness of arch and it increases uniformly with depth, as can be seen from the theoretical analysis based on thin cylinder theory: (Fig. 8.2) U/s water pressure

Abutment RV r sin ( /2)

r sin ( /2)

R sin ( /2) = Rv

RH /2

r /2

Fig. 8.2:  Constant angle arch dam

180  Irrigation Engineering and Hydraulic Structures

PH = pressure force on projected area of arch at depth h from water line. θ = γ h × 2r sin × 1.0, Here, radius of intrados of arch = r 2 ht. of arch = unity, thickness of arch = t. density of water = γ Now PH = 2R v here R v = Component of reaction R in direction opposite to that of Water Pressure = R sin

θ 2

θ θ  ∴ γ h  2r sin  = 2R sin 2 2  ∴ R = ( γ h ) r here, γ h = intensity of pressure at depth h, = σ × ( t × 1)

∴σ =

( t × 1 = A = cross-section of arch )

( γh ) r t

here, σ = permissible compressive stress of concrete.  γ h ∴ t =  r  σ ∴ t α r and also t α h Cross -sectional area of arch = A = t × 1 ∴A α r ∴ A = k r, K = constant For unit height of arch, For unit height of arch, volume of of concrete concreteininarch, arch, volume V = rθ A = Kθ r 2  θ  span of Arch, L L = Sin   = 2r 2r 2 L ∴r = θ 2 sin   2    L  ∴V = Kθ    2sin θ   2

2

Arch and Buttress Dams  181

=

∴

KL2 4

θ θ sin 2   2

dV θ θ θ  = sin 2   − θ  sin cos  = 0 dθ 2 2 2 

θ ∴ tan   = θ 2 ∴θ = 133°34' Hence, volume of concrete for a given arch of height unity and thickness t at a depth h from FRL is least if its subtended angle θ = 133° 34´. In practice this angle may vary from 120° to 140°.

8.3.1  Limitations of Thin Cylinder Theory This cylinder theory of Arch Dam has following limitations: 1. It does not take into account “Rib Shortening.” An arch ring of arch dam is a part of circular ring supported at its ends and on account of uniform radial loads of water pressure, curved length of arch tends to be shortened. This is known as “rib shortening”. It ends are rigid, span of arch does not change and consequently arch ring is deformed, which produces additional stresses in the arch. These are known as rib shortening stresses and should be considered in the design. 2. Stress distribution along the thickness of arch cross sections are assumed to be subject to similar stress distribution, whereas the formulae gives higher stress at the intrados. This is true only near the abutments, while at the crown, stresses are reversed. Thus, top arches and bottom arches, conditions are quite opposite. 3. Thin cylinder theory is based on no yielding of abutments, but actually abutments are not rigid they spread apart slightly under arch action producing additional stresses in the arch. 4. Stresses produces due to temperature changes and shrinkages of concrete are not considered in this cylinder theory. 5. This cylinder theory assumes that the valley is symenetrical, whereas actually it is not so. 6. Earthquake force and silt pressure may not act radially. On account of limitations as stated above, thin cylinder theory is adopted only as preliminary design and final design is carried out by 1. Elastic arch theory or 2. Trial load analysis.

182  Irrigation Engineering and Hydraulic Structures

8.3.2.  Elastic Arch Theory This theory assumes transfer of entire load to abutments through arch actions only and the entire dam consists of horizontal arch rings placed one over the other, with their ends fixed into the abutments and dS

B H

H MA V

B f Arch A Span o

x

MB

Fig. 8.3:  Elastic arch theory each ring is independent of the other. Rib shortening effect, temperature changes, shrinkage and yielding of abutments are considered is this theory. Thin cylinder theory takes into account effect of uniform radial pressure, but elastic theory considers variable loads along the length of the arch. Refer Fig 8.3 for cross-section of Elastic Arch. At a cross section x, moment M is given by M = µ + M A + (M B − M A )

x + Hy L

here, µ = Bending moment (BM) on a straight horizontal freely supported beam. M = BM at any point x on arch MA and MB = fixed end moments at A and B. H = horizontal thrust at abutments, V = vertical thrust at abutments, such that R = resultant = H 2 + V 2 By bending stress formula,

f M f M =  and strain = = y I E EI

Hence, E = modulus elasticity and I = moment of inertia of the crest section under consideration x

x

x

x

x

µy ds (M A − M B ) xy ds y2 M ∴ ∫   y ds = ∫ ds + µ ∫ y + H ++ + ∫ EI ds = 0 ∫ EI EI  EI EI 2 0  0 0 0 0

Arch and Buttress Dams  183

For symmetrical arch , MA = MB

∴∫

M µy yds y 2 ds + H∫ = 0 (8.1) yds = ∫ ds + M A ∫ EI EI EI EI



M = MA + Hy MA ∫



ds y + H ∫ ds = 0  EI EI

(8.2) (8.3)

By solving equations (8.1), (8.2) and (8.3) Arch can be designed as per elastic theory.

8.3.3  Trial Load Analysis The assumption of thin cylinder arch theory and elastic theory that any horizontal arch ring is independent of arch ring above and below it is not valid. Trial load analysis developed by USBR overcomes this difficulty by assuming that the dam is made up of two systems of elements: (i) horizontal arches transmitting thrust to abutments and (ii) vertical cantilever, fixed at foundation as shown in Fig 8.4. Loading is assumed to be divided between them such that deflection at the common element is same. Dam dimensions fixed by following either thin cylinder theory or elastic arch theory are analysed by this method and modified to arrive at final design. The trial load analysis method is widely used in USA for arch dam designs and finite element method is applied to solve the equations involved on high speed electronic computers. In Fig 8.4, AB is horizontal arch, CD is vertical cantilever and is common element on AB and CD. Deflection of this common element should be same by assuming load distribution between arch and cantilever action. If it is not same, then new load distribution is tried. C A

D

Arch Dam B

Fig. 8.4:  Cross-section of arch dam

184  Irrigation Engineering and Hydraulic Structures

8.4  OTHER VARIETIES OF ARCH DAM 8.4.1  Constant Angle Arch Dam In constant angle type of arch dam, subtended angle of arch remains constant but radii of arch go on decreasing as depth of water from FRL goes on increasing towards bottom of the valley. This results in double curvature of up stream face of the dam. Hence, this type of dam does not have vertical cylindrical face at its up stream. Constant angle arch dam has got benefit of least consumption of concrete if the subtended angle adopted is 133°.

8.4.2  Variable Radius (or Angle) Arch Dam The third variety of arch dam is a mix of the above two types and is known as variable radius or variable angel type of arch dam. The arch thickness also increases towards abutment, i.e., splayed arch, which is more suitable rather than arch of constant thickness.

8.4.3  Doubly Curved Type Arch Dam If design procedure adopted considers curvature in both horizontal as well as vertical direction, the dam is known as Double curvature type Arch Dam. Here part load of water pressure is resisted by arch action and remaining part by cantilever from foundation. Hence, it can also be said to be Arch cum Gravity type of dam. USBR has given guidelines for design of such dams. Besides thin cylinder theory, arch dam can also be designed by Trial load method and method based on Theory of shells. The design procedure adopts Finite Element Technique to arrive at solution of differential equation involved. As this is beyond scope of undergraduate studies, it is not included here.

8.5  BUTTRESS DAMS Just as counter forted retaining walls retain earth mass on its up side, similarly, if water is retained on upstream by water tight membrane supported by series of counterforts or buttresses normal to dam axis, then the type of dam is known as Buttress Dam (Fig. 8.5). In India, so far no dam of this type has been constructed. Instead of flat slab as face slab, arches can be provided supported by buttresses, and in that case it is known as ‘multiple arch dam’. (Fig. 8.6) The design of buttress is similar to that of gravity dam, but as buttresses are thinner in sections, they have to be checked against lateral buckling and hence struts should be provided horizontally as well as vertically in between buttresses, (Fig. 8.4)

Arch and Buttress Dams  185 Road way F. R. L.

u/s

Face slab Struts

Buttress spacing

Fig. 8.5:  Buttress dam plan

Fig. 8.6:  Multiple arch dam in plan

q  SOLVED EXAMPLES Example 8.1:  From the profile of a valley, following data is available: Height from top of dam (m)

Width of Valley (m)

Remark Top of dam

0

168

10

142

20

130

30

110

40

96

50

82

Bottom of dam

For a central angle of 133° 34´, find (i) ri = radius to inner surface of arch (ii) t = thickness of arch dam. Solution:  Assume net compressive strength = λ − p = σ λ = 3600 KN / m 2 P = wh w = 9.8 KN / m3 = weight of water per m3 h = height in m from top.

Draw c/s of the arch dam, Refer to Fig. 8.7.

186  Irrigation Engineering and Hydraulic Structures

168 m 0

Top

t L

10

142 m

20

130 m

30

110 m

ri

ri 0/2

96 m

40

0

50

Bottom 82 m

Fig. 8.7:  Cross-section of arch dam

( ri )top =

L 2Sin

θ 2

=

168 168 = = 91.6 m. 2 (Sin 66.5) 2 × 0.917

σ = λ − p = λ − wh , (ri )10 =

142 = 77.4 m. 1.834

σ10 = 3600 − 9.8 × 10 = 3502 KN / m² , 9.8 × 10 × 77.4  whr  t10 =  = 2.16 m.  = σ 3502  10 130 = 70.8, ( σ20 = 3600 − 9.8 × 20 ) = 3404 ( ri )20 = 1.834 9.8 × 20 × 70.8 = 4.07 m. t 20 = 3404 110 = 60m. ( ri )30 = 1.834 9.8 × 30 × 60 9.8 × 30 × 60 = = 5.33m t 30 = 3600 − (9.8 × 30) 3306 96 = 52.3m. ( ri )40 = 1.834 9.8 × 40 × 52.3 20501.6 = = 6.4m t 40 = 3600 − (9.8 × 40) 3208 82 = 44.7m. ( ri )50 = 1.834 9.8 × 50 × 44.7 21903 = = 7.02m 3600 − (9.8 × 50) 3120 Pr ovide top width as 3 m. t 50 =

Arch and Buttress Dams  187 0

10

2.16m

20

4m

30

5.33m

40

6.4 m

50

ri = 77.4m

re = 79. 56m

ri = 70.8m

re = 74. 8m

ri = 60. 0m

re = 65. 33m

ri = 52.3m

re = 58. 7m

ri = 44.7m

re = 51. 72m

7. 02m

Fig. 8.8:  Cross-section of arch dam Example 8.2:  Design and draw section of constant radius arch dam for the data given below: Depth from top (m)

0

10

20

30

40

50

60

Width of valley (m)

180

155

135

122

108

103

86

Consider, Top thickness of arch = 5.1 m , σ = 3600 kN/m2, angle subtended at top = 150˚. Solution:  For constant radius arch dam, extrados radius remains constant, but intrados radius goes on changing due to changes in arch thickness at various levels, and subtended angle also changes with increases in depth from top of the dam. At top, to = 5.1 m, h = 0 Span of arch Lo = 180 m ri = re as t = 0 at top theoretically θ  150  L = 2re sin   = 2re sin   2  2  = 2 re (0.965)



  \ re =

L0 180 = = 93.26m 2(0.965) 2(0.965)

3 fc = net compressive strength = s - g h, g = 10kN / m = σ − 10h = 3600 – 10h



t=

g hg fc

188  Irrigation Engineering and Hydraulic Structures

at top t = 5.1

5.1 =



∴h =

10(h)(93.26) 3600 − 10(10) 3500 × 5.1 = 19.1 m 10 × 93.26

i.e., upto h = 19.1 m say 20 m, top width t = 5.1 m can be considered ri = 93.26 – 5.1 = 88.16 m from top to h = 20 m. r (h)(10) at h = 20, t 20 = i 3400 93.26(20)(10) 3400 = 5.5 m

=

ri at 20 = 93.26 – 5.5 = 87.76 m and subtended ∠ θ will be given by θ L 20 = 135 = 2(87.7)sin   2 135 θ ∴ sin   = = 0.769 2 2(87.7)   θ ∴ = 50.3° 2 ∴ θ = 100.6° Similarly make calculations for L = 122m and h = 30 m and so on. Results are tabulated below: Given

Calculated

Depth from top (m)

width of valley cms

t(m)

r1 = re – t intrades radius r(m)

subtended angle

(1)

(2)

(3)

(4)

(5)

0

180

t0 = 5.1

88.1

150

10

155

t1 = 5.1

88.1

120

20

135

t2 = 5.5

87.76

100.6

30

122

t3 = 8.5

84.76

92

40

108

t4 = 11.6

81.66

82.79

50

103

t5 = 15

78.26

82.30

60

86

t6 = 18.6

74.66

70.33

Arch and Buttress Dams  189

Columns (1), (2) are given in data, Columns (3), (4), (5) are answers. Cross-section is drawn below in Fig 8.9. 5.1 m t t1 t2

re 60 m

t3 t4

ri

0/2

0/2 = 75

o

re

U/S

t5 t6 18.6

Fig. 8.9:  Cross-section of arch dam

EXERCISES 1. Distinguish between (i) Masonry dam and arch dam (ii) Arch dam and buttress dam (iii) Constant angle and constant radius arch dam. 2. Derive an expression for thickness of an arch dam of constant radius based on thin cylinder theory. Also find condition for least volume of concrete for arch dam. 3. Draw the cross-section of a constant radius arch dam for following data: Height of dam 80 m, σ = Allowable stress in concrete = 300 T/m², Top width = 3 m, Radius of dam = 75 m. 4. Explain briefly the trial load analysis of the design of arch dam.  

 

9 Earth Dams 9.1  EARTH DAM AND ITS COMPONENT PARTS Earth dams are constructed of non-rigid materials i.e., earth, gravel, pebbles, rock pieces etc. If reservoir site does not have suitable foundation for gravity dam, then earth dams are preferred on such site. Earth dams are of two types: 1. Rolled Fill Earth Dam and 2. Hydraulic Fill Earth Dams Only rolled filled type is more common in our country. Earth dams are also known as per type of their section i.e., homogeneous earth dam, zoned earth dam and diaphragm type earth dam. For same height of dams, earth dam’s initial cost of construction is far less than that of gravity dam but maintenance is heavier. Earth dam projects provide more employment potential and hence serve as socio-economic solutions of country’s unemployment problems. Component Parts and their functions (Fig. 9.1). In general, following are the components parts of zoned earth dam: 1. Core 2. Cut-Off 3. Curtain Grouting 4. u/s and d/s Shells 5. Filters 6. Internal Drain 7. Rock Toe 8. U/s Blanket 9. Cross Drain 10. Pitching or Rip Rap on u/s Slope 11. Downstream Slope with Sod 12. Top Width 13. Free Board

(12) Top width F.R.L

F B (13) (10)

D.S.L.

(5) (4) (1)

(5) (11) (4) (6) (7)

(8)

(2)

(9)

(3) Semi impervious layer Impervious (rock) stratum

Fig. 9.1:  Earth dam section

© The Author(s) 2023 S. K. Ukarande, Irrigation Engineering and Hydraulic Structures, https://doi.org/10.1007/978-3-031-33552-5_9

190

Earth Dams  191

Functions of Component Parts 1. Core: This is central impervious component part of earth dam and is responsible for cutting down seepage from u/s to d/s to a minimum. It consists of 85% to 90% clay and about 15 % to 10% silt. 2. Cut-off: Core extended upto foundation plane is known as cut-off. It consists of same material as core and has function of preventing seepage from u/s to d/s in foundation plane. Cut off rests on impervious stratum but if it is at very deep level then it is provided upto semi pervious stratum and from that level to imperious stratum. 3. Curtain Grouting: Curtain grouting is provided such that there is no seepage from u/s to d/s in the foundation soil. 4. U/s and d/s Shells: It consists of coarse material such as coarse sand, pebbles, gravels etc. and it provides structural support to the core. The shell material must have excellent drainage properties so that water drains out of it immediately. 5. Filters: Filters are provided between core and shells both on u/s and d/s side and also on u/s side of rock toe and all around internal and cross drains. Filters have important function of not allowing migration of coarse material to finer and finer to coarser, thus it is a buffer zone between two different types of material and maintains their stability. Filters need special design criteria so that migration of filter material is prevented. 6. Internal Drain: Internal drain is provided at toe of core such that whatever seepage takes place through core is tapped here and is thrown out via cross drains. They consist of rock pieces surrounded by filter. 7. Rock Toe: Rock toe is also a drain consisting of rock pieces with a filter on u/s side. This is provided at toe of dam and hence is known as rock toe. Its function is to throw out rain water that has seeped through d/s slope into d/s shell and also it makes dam toe stronger against any failure due to slunghening of d/s dam material. 8. U/s Blanket: U/s blanket is provided at u/s side horizontally between u/s shell and foundation soil such that water from u/s shell does not seep into foundation but in the event of sudden drawdown of water level in reservoir, water in u/s shell is forced to rush out through open joints of the rip-rap provided along u/s slope. 9. Rip-Rap: Rip–Rap also protects the u/s slope from getting washed out due to wave impact on u/s slope at FRL. This is therefore provided upto level higher than FRL on upper side and lower than DSL on lower side of u/s slope. 10. Downstream Slope with Sod: Sod is provided along d/s slope to protect it from getting washed away due to heavy down pour of raining water on d/s side of dam. Sod is a jungle growth of grass along the d/s slope. D/s slope has also got berms, so that maintenance party can work on d/s slope side and maintains it nicely.

192  Irrigation Engineering and Hydraulic Structures

11. Top Width: Top width is provided to allow for at least two traffic lanes on top of dam for free moment of heavy earth moving machinery. It is therefore kept around 6 to 8 m. 12. Free Board: FB is provided such that waves due to exposed water surface at FRL does not over top the dam and washing out of u/s slope at FRL is prevented by proper maintenance at that level. FB is either 1.5 (hw) or 10 ft. (3 m), whichever is higher, and hw is given by hw = 0.032

VF m… (for F > 32 km)

here V = wind velocity in kmph and F = fetch (in km), it is horizontal distance of farthest point on contour at FRL level. Thus, all above 13 component parts have specific function, but in a given earth dam all of them need not be provided. Site conditions will decide which one is required and which is to be dropped. 0.3 b

Arc with radius AF and centre at A to intersect CJ at C

F.R.L A

C

I

Directrix H

P(X.Y) y

H o

G 90 J F

b x 0.7b

D

S L

Horizontal filter of L at toe of dam

D = Floor distance between (A) and (F)

Fig 9.2:  Seepage through homogeneous earthen dam section with horizontal filter at the toe

9.2 ESTIMATION OF SEEPAGE THROUGH HOMOGENEOUS EARTH DAM Seepage calculations make use of flow net analysis. For this drawing of top most pheratic line and potential line is essential. This is explained below with refrence to different boundary conditions at the base of the dam.

9.2.1 Homogeneous Earth Dam Section with Horizontal Filter at Toe Let P be any point having coordinate (x, y) on the base parabola APG as shown in Fig. 9.2 and F is a point on front edge of horizontal. filter and is regarded as origin for coordinate (x, y) of P.

Earth Dams  193

As per property of base parabola, any point on the parabola is equidistant from directrix and focus of the parabola. ∴ PG = PH ∴ x 2 + y2 = x + s ∴ x 2 + y 2 = (x + s)2 = x 2 + 2xs + s 2 



∴ y = 2xs + s 2 dy s ∴ = dx 2xs + s 2

(9.1) (9.2)

Assuming various values for x, y can be worked out and parabola can be plotted. Starting point of parabola A is required to be shifted to B, a point at FRL and u/s slope intersection. Base parabola being Ψ line and u/s slope of earth dam when water upto FRL is present, is Φ line hence both must intersect at 900 at B. A line from B be drawn such that it meets base parabola APG tangentially at I and hence B to I be regarded as ingress correction to the parabola. Then BIPG is the phreatic or top seepage line above which dam section is dry and below it dam section is wet.

Let, q

= Seepage per m length of dam at P. = KAi , (as per Darcy 's Law ) = K (y´1)



= K(y ´1) =K

(

dy dx

s

2xs + s 2

2xs + s 2

)

(

s 2xs + s 2

)

\ q = Ks

(9.3)

= permeability of dam material ´ focal length of filter at base. Hence, seepage q per m length of homogeneous dam section can be worked out. When x = D and y = H, focal length s can be worked out from equation (9.1) as,



s =  D2 + H 2 − D    Here, D = (length of base of dam) − 0.7b − L ,

(see Fig.9.2)

∴ q = K  D2 + H 2 − D  (9.4)  

194  Irrigation Engineering and Hydraulic Structures

9.2.2 Homogeneous Earth Dam Section without Horizontal Filter at Toe

Base parabola AJP cuts the d/s slope at J as shown in Fig. 9.3. However the phreatic line must emerge out at K, a point on d/s face, such that it meets the d/s face tangentially. KF is then known as discharge face making an angle α with horizontal at the toe. The correction Δa by which the base parabola is to be shifted downwards, ∆a is found by the value given by Casagrande for various values of a as given (a + ∆a) in Table 9.1 below: 0.3 b A

B

a J

H

a K  F

b

G

D

Fig. 9.3:  Egress correction base parabola Table 9.1:  Egress Correction α

Δa (a + Δa)

30° 60° 90° 120° 150° 180°

0.36 0.32 0.26 0.18 0.10 0.00

Ifα < 30°, then a =

d d2 H2 − − cos α cos 2 α sin 2 α

(9.7)

If α > 30° but < 60°, then a = H 2 + d 2 − d 2 − H 2 Cot 2 (α) 180° − α ∆a = a + ∆a 400 180° − α ∆a but if α > 90° and < 180° then = a + ∆a 300 Alternatively if α < 90°,

(9.5)

(9.5A)  (9.6)

Earth Dams  195

9.2.3  Phreatic Line for Zoned Earth Dam A zoned earth dam consists of core and u/s and d/s shells as shown in Fig. 9.3(a). Phreatic line will be a base parabola ABFG, which is horizontal from A to B and then BFG is parabolic. It cuts the d/s core F.R.L

A

B a

F H

Core

Shell

a

R

N

P

G

L

d

M

Fig. 9.3(a):  Zoned earth dam at F and FR = Δa, egress correction. As per correction Δa, parabola must meet d/s of core tangentially at R. Seepage loss q can be worked by q = KS, where S = H 2 + d 2 − d and K is permeability of core material. Here d is PM as shown in Fig. 9.3(a).

9.3  DESIGN CONSIDERATIONS Stability analysis of earth dam section is carried out for following cases: 1. d/s slope stability for steady state of seepage 2. u/s slope stability for sudden withdrawal of water from reservoir, a case known as “Sudden Drawdown” 3. Stability of d/s and u/s slope against shear 4. Stability of foundation against shear. Method of stability analysis is Swedish slip circle Method, which is explained as under: Centre of slip circle

A o

b Radius of slip circle Wedges of equal width b end wedges width =b1 at bottom =b2 at top

b2

b R

cg

R

h 0 b1

N T

W

Fig. 9.4:  Slip circle method

T= W sin (0) N= W cos (0) W= (bh x 1)y5

196  Irrigation Engineering and Hydraulic Structures

Chose a center as at A and draw an arc of radius R as shown in Fig. 9.4. Divide the arc in wedges of equal width ‘b’, but width of bottom triangular wedge is b1 and that of upper triangular wedge is b2 Take central wedge for analysis, let vertical representing weight of wedge W be passed through c.g. of the wedge and let it be resolved along normal and tangential components as N and T, then Displacing or Driving Moment = M d = RΣT Stabilizing or Resisting Moment = M s = R [ cL + ΣN tan φ]

Here L = length of arc of slip circle 2πRδ 360 φ = angle of repose of earth dam material. = cohesive strength of soil. δ = angle subtended by arc length at center A. θ = angle between vertical through c.g. of the wedge and N. Hence, Factor of Safety (FS) against sliding

=

=

MS cL + ΣN(tan φ) = ≥ 1.5 Md ΣT

Here, ΣT = ΣW sin θ = Σ (bhγ )sin θ This can be obtained from T diagram, see Fig. 9.4(a) h1 sin 01

h2 sin 01

etc.

b

Fig. 9.4(a):  T diagram T1 = ( γb)h1 sin θ1 T2 = ( γb)h 2 sin θ2 etc γ = weight deinsity of dam material and ΣN = ΣW cosθ , this can be obtained from N diagram see Fig. 9.4 (b) = Σ(bhγ ) cos θ = ( γb)Σ(h1 cos θ1 + h 2 cos θ2 etc)

T1 = ( γb)h1 sin θ1 T2 = ( γb)h 2 sin θ2 etc γ = weight deinsity of dam material

Earth Dams  197

and ΣN = ΣW cosθ , this can be obtained from N diagram see Fig. 9.4 (b) = Σ(bhγ ) cos θ = ( γb)Σ(h1 cos θ1 + h 2 cos θ2 etc) h1 cos 01

h2 cos 01 etc.

b

Fig. 9.4(b):  N diagram Line AE as shown in Fig. 9.5 can be obtained as under: At top of dam, draw a line making ∠a2 = 35° with horizontal and along BC, u/s slope of dam, draw a line at ∠a1 = 25°, for values of a1 and a2 see Table 9.2

Fig. 9.5:  Locus of critical slip circle These lines will intersect at A. Draw line DE at H (m) below base of dam upto 4.5 H from D giving a point E. Then join AE, center of critical slip circle shall lie on this line AE. Chose 4 to 5 centers and calculate FS for each slip circle for

198  Irrigation Engineering and Hydraulic Structures

these centres; and then find least value of FS, from the plot of FS Vs. location of center of slip circle. This should be > 1.5, adopt it for design. This is the critical slip circle. Table 9.2:  Values of slope and angles Slope of u/s or d/s face

Directional

Angle

α1

α2

1:1

28

27

1.5:1

26

35

2:1

25

35

3:1

25

35

4:1

25

35

5:1

25

35

9.3.1  Case of d/s Slope for Steady State Seepage Let the arc of slip circle cut phreatic line at A and D as shown in Fig. 9.6. Take points B, C along this line and measure the vertical height a1, a2 etc. Then at B, draw perpendicular FB = γa1 and at C draw perpendicular CE = γa 2 , where γ = weight density of water. Then join A, F, E, D which gives pore-pressure diagram, find its area, which gives U = pore pressure of zone where phreatic line and slip circle cut each other then FS =

cL + tan φ Σ(N − U) ≥ 1.5 ΣT

F.R.L A Phreatic line

a1 B

a1 a2

F

Are of slip circle a2 C

D

E Filter

Fig. 9.6:  Steady state see page

Earth Dams  199

Here, N and T should be worked out by taking value of γdry for dam material and considering wedge width b as explained earlier, i.e.,

N = W cos θ = (bhγ dry ) cos θ T = W sin θ = (bhγ dry )sin θ

Effect of uplift pressure present in zone below phreatic line is deducted from N, i.e., tan ϕ Σ(N − U) is considered in above formula, here U = uplift pressure, which is the area of uplift pressure diagram ABCDEF as explained with the help of Fig. 9.6. Alternately, for dam material below phreatic line if submerged weight is taken then no need of calculating U separately.

9.3.2  Stability of Foundation against Horizontal Shear (Fig. 9.7)

Fig. 9.7:  Stability of foundation  h 2 − h 22  φ  Total horizontal shear, P = γ m  1 tan 2  45 −   2  2    where, γ m =

γ d (h1 − h 2 ) + γ f h 2 h1

γ d = density of dam material γ d = density of foundation material. γ f = density of water Average unit shear = Sa =

P B× 1

200  Irrigation Engineering and Hydraulic Structures

Smax = maximum unit shear = 1.4 Sa and it occurs at L, which is at 0.4B from J as shown in the Fig. 9.7. Shear strength below toe, X = =S S11 = =C C+ + ((g gff hh 22 )) tan tan f f Shear h ) tan f Shear strength strength below below J, J, S S22 = = C+ C+ ((g gm m h11 ) tan f S S1 +S +S2 S=Average S=Average shear shear strength strength = = 12 2 2

S S3 \ \ F.S.= F.S.= S3 1.5 1.5 for for stability stability of of foundation foundation shear shear Saa Sr Sr3 FS FS at at maximum maximum shear shear point point = = S 3 1.0 1.0 Smax max Here gav hh )) tan f Here Sr Sr = =C C+ + ((g tan f av

(ã h + ã h ) ã av = ( ã dd h + ã ff h 22 ) ã av =

(h (h + + hh 22 )) hh = = 0.6 0.6 (h (h1 - hh 2 )) 1

2

9.3.3  Stability of Earth Dam against Horizontal Shear at the Base of the Dam W = overall weight of dam W tan φ = Shear resistance Fig. 9.8:  Section of earth dam

∴ FS =

W tan φ ≥ 2. P 0.4.B

γh 2 2

d

Phreatic line d / s slope

h=0.6 H



P = Total Horizontal water pressure =

H

Point of maximum shear

h1

Bd

Fig. 9.9:  Section of earth dam (d/s slope)

Earth Dams  201

9.3.4 Stability of d/s Slope against Horizontal Shear at Base (refer Fig. 9.9)  γ av H 2  2  2 ϕ  γ w h12  H − 2 + ϕ γ w h12  = = Earth essure H tan γ av 45 Earth Pr Pressure =  =H =  d Earth Pr essure tan 45 −  2   2 2 + 2   2 d   2   h ) γ 1h1 + γ 2 (H − 1 γ h + γ 2 (H − h1 ) γ av = γ av H = 1 1 H Sd = average shear strength onstrength d / s side at /base = S average shearstrength s side at base Sd d= average shear onon d/sdside at base =

Hd Bd × 1

Smax d

= max shear on d/s at base = 1.4Sd

Rd

= resisting shear force at d/s = (Wed ) tan φ + (Bd × 1)

\ FSd/s =

Rd ³2 Hd

FS at point of maximum shear =

g h tan f + C Smax d

9.3.5  Stability of u/s Slope against Horizontal Shear at Base

Fig. 9.10:  Section of earth dam (u/s slope)  γ H2 φ γ h2   Total horizontal shear H u =  sat tan 2  45 −  + w 1   2 2   2 Hu Su = Bu × 1 Smu = Smax u /s = 1.4Su R u = Weu tan φ + cb u Weu = total effective wt of dam below u / s slope FSu =

Ru ≥2 Hu

FS at pt of

max shear

=

h tan φ1 + C Smu

u

Smu = Smax u /s = 1.4Su R u = Weu tan φ + cb u

202  Irrigation Engineering and Hydraulic Structures

Weu = total effective wt of dam below u / s slope

FSu =

Ru ≥2 Hu

FS at pt of

max shear

here tan φ1 =

=

h tan φ1 + C Smu

γ sat h1 tan φ + C γ sat h1

q  SOLVED EXAMPLES Example 9.1:  A homogeneous earth dam has following data: H = height of dam = 30m Top width = 6m u/s slope 4 : 1, d/s slope 3 : 1 Foundation is pervious to a depth of 20m γ sat = 2160 Kgf / m3 γ sub = 1160 Kgf / m3 C = 1200 Kgf / m 2 φ = 24° Check the stability of foundation against shear for u/s portion of dam. 0.4 bu

H = 30m

4:1

3:1

18m = 0.6 (30)

bu = 120 m

120 m

90 m

Point of maximum shear

Fig. 9.11:  Earth dam section Solution:

 h 2 − h 22  ϕ  P = total horizontal. shear = γ sat  1 tan 2  45 −   2  2    γ h tan ϕ + C = 0.4563 tan φ1 = sat 1 γ sat h1 h1 = 30 + 20 = 50 m and h 2 = 20 m

Earth Dams  203

\ j1 = 24.53° \ P = 9.373´105 kgf Sav =

P 9.373´105 = = 7.8 ´103 kgf / m 2 bu 120

s max = 1.4Sav 10.9 × 103 kgf / m 2 s1 = C + γ f h 2 tan φ = 1200 + 1160 (20) tan 24o = 11.5 × 103 kgf / m 2 s 2 =1200 + 1160 (50) tan 24o = 27 × 103 kgf / m 2

s = FS =

s1 + s 2 = 19.2 × 103 kgf / m 2 2 s sav

=

19.2 × 103 7.8 × 103

FSat pt.of max .shear

= 2.4 〉 2

hence OK

C + 1160(18 + 20) tan 24o = 1.9 〉1.5 s max

hence OK

Note : h at point of maximum shear = 0.6 (30) = 38 m Example 9.2:  A 24m high earth dam has slopes at u/s 4:1 and d/s 3:1, top width 6.4 m, check the stability of u/s portion against horizontal shear, seepage line is at 4.0 m below top of u/s shoulder. γ sat = 21KN / m3

γ sub = 11.2 KN / m3 C = 35 KN / m 2 φ = 25° F. R. L.

4m

4:1

3:1 20m

H = 24m

bu = 96m

Fig. 9.12:  Section of earth dam (stability of u/s past)

204  Irrigation Engineering and Hydraulic Structures

Solution: Hu =

1 ϕ 1  γ sat H 2 tan 2  45° −  + γ w h12 , here H = 24 m, and h1 = 24 − 4 = 20 m 2 2 2 

1 24°  1  2 = (21) (24)2 tan 2  45° −  + (10) (20) 2 2  2 

= 2540 + 2000 = 4540 KN

Hu 4540 = = 47.3 KN / m 2 b u × 1 96 × 1

Su =

Smu = 1.4Su = 66.2 KN / m 2 R u = Weu tan φ + cb u , here, Weu =

1 γ sub (96) (24) × 1 2

1 (11.2) (96) (24) tan (25) + 35 × 96 2 = 6016 + 3360 = 9376 KN =

R u 9376 = = 2.06 > 2 , H u 4540

FSu =

FS at point of maximum shear =

hence OK

γ sub h tan φ + C Smu

(11.2) (0.6 × 24) tan 25° + 35 66.2 75.2 + 35 = = 1.66 >1.0 , hence 66.2 =

OK

Example 9.3:  A homogeneous earth dam section has top width 6m, u/s slope 3.5 : 1, d/s slope 3 : 1, height = 75 m, FB = 3 m. Properties of soil of earth dam: Specific gravity, G = 2.6, void ratio = 0.7 Dam section consists of 55% saturated zone

35% wet zone

and

10% dry zone

For wet zone assume average degree of saturation = 0.5 At base of dam, C = 40 KN / m², φ = 28°. Check the overall stability of dam section.

Earth Dams  205

Solution: F B 3m

3.5:1

bu = 262.5m

75m

6m

3:1

bd = 225.5m

B = 493.5m

Fig. 9.13:  Section of earth dam (overall stability) Note: (This means finding FS against horizontal shear at the base of dam). A = Total area of dam section 1 = ½ (262.5) 75 + 6 × 75 + (225) (75) 2 = 9844 + 450 + 8437 = 18731 sq.m. G = 2.6, η = 0.7

γ (G + η) 1000(2.6 + 0.7) ∴ γ sat = w = = 1.94 1000 = 1940 kgf/m3 γ w (G + η) ×1000(2.6 + 0.7) η) G = 2.6, η(1=+0.7 ∴ γ sat(1.7) = = = 1.94 × 1000 = 1940 kgf/m3 (1 + η) (1.7) 1940 + 940 γ sub = , [ hereit is divided 2 as degree of saturation given is 50 %] 1940 + by 940 2 γ sub = , [ hereit is divided by 2 as degree of saturation give 2 3 =1440 kgf/m =1440 kgf/m3 = 0.55(18731) (1940) = 199.8 × 105 kgf = 200 × 105 kgf , say Wsat = 0.55(18731) (1940) = 199.8 × 105 kgf = 200 × 105 kgf , say = 0.35(18731) (1440 = 94 × 105 5   Wwet = 0.35(18731) (1440 = 94 × 10 Wdry = 0.1(18731) (2.6 × 1000) = 2600

6, η = 0.7

= 48.7 × 105 kgf = 48 × 105 kgf , say Wtotal

= (200 + 94 + 48) × 105 = 340 × 105 kgf , (say)

∴ W tan ϕ = 340 × 105 (tan 28°) = 180 × 105 kgf ∴ Shear resistance at base = W tan φ + C(B × 1) R = 180 × 105 + 4000 × 493.5 = 180 × 105 + 20 × 105 = 200 × 105 Horizontal Force due to water pressure: 1 1 P = γ w h 2 = (1000) (72) 2 = 26 × 105 kgf 2 2

206  Irrigation Engineering and Hydraulic Structures

FS against shear at base =

R 200 × 105 = = 7.69 > 1.3 hence P 26 × 105

F B 4m

40m

3.1

2.5.1 40m

6m

120m

100m 226m

Fig. 9.14:  Section of earth dam (FS) Example 9.4:  Determine seepage loss through earth dam having height = 40 m, FB = 4 m, u/s slope 3:1, d/s slope 2.5:1, horizontal filter length at base = 40 m., Top width 6m., K = 25 m / year. Solution: S = Focal Length =

H 2 + D2 − D ,

here D = Base width – 0.7b – Horizontal filter length, = 226 – 75.6 – 40, [where, b = 36 × 3 = 108m = horizontal projection of u/s slope at FRL, and ∴ 0.7b = 0.7 × 108 = 75.6m] ∴S =

( 40 − 4)2 + (110.4)2 − 110.4 = 116.4 − 110.4 = 6 m.

25   q = KS =  (6) = 48 × 10 −7 m3 / sec/ m length of dam  365 × 24 × 3600 

Fig. 9.15:  Earth dam (seepage loss through dam)

Earth Dams  207

Example 9.5:  Check the stability of d/s portion of earth dam against horizontal shear at the base of the dam. Height of dam 24m., top width 6m, u/s slope 4:1, d/s slope 3:1, seepage line at 6 m below d/s top shoulder, area of d/s portion below seepage line 550 m². Solution: γ sat = 19.62 KN/m3 or 2000 Kgf/m3 γ sub = 1160 Kgf/m3 , φ = 26° 6m 6m

24m

3.1 18m

bd = 72 m

Fig. 9.16:  Section of earth dam (d/s part stability) C = 4000 Kgf/m²

But

 γ H2 φ γ h2   H d =  av tan 2  45 −  + w 1   2 2   2 γ h + γ sat (H − h1 ) γ av = sub 1 H 1160(18) + 2000(6) = = 1370 kgf / m3 24 ∴ Hd =

1 26  1  (1370) (18) 2 tan 2  45 −  + (1000) (18) 2  2 2 2

= 86556 + 162000 = 0.86 × 105 + 1.6 × 105 = 2.46 × 105 kgf

R d = Resisting force = Wed tan φ + cbd Area of d/s portion of dam =

1 (72) (24) = 864 m 2 2

∴ Area above seepage line = 864 – 550 = 314 m2

208  Irrigation Engineering and Hydraulic Structures

∴ Wed = 550 × 1160 + 314 × 2000 = 6.38 × 105 + 6.28 × 105 = 12.66 × 105 Kg(f ) ∴ R d = 12.66 × 105 tan 26 + 4000(72) = 6.17 × 105 + 2.88 × 105 = 9.05 × 105 ∴ ( F.S.)d =

R d 9.05 × 105 = = 3.67 > 2, H d 2.46 × 105

d/s average shear stress = Sd =

H d 2.46 × 105 = bd 72

= 3.4 × 103 kgf /m 2 d/s m ax . shear = Smd = 1.4Sdav = 1.4 × 3.4 × 103 = 4.76 × 103 kgf/m²

SR = Resisting shear strength = γ sub h tan ϕ + C where h = 0.6 H = 0.6 × 24 = 14.4 m ∴ SR = 1160(14.4) tan 26° + 4000 = 8147 + 4000 = 12147 = 12.147 × 103 kgf/m² FS(d)max shear =

12.147 × 103 4.76 × 103

= 2.55 > 1.5

Example 9.6:  In order to find factor of safety of d/s slope during steady state seepage, a scale of 1 cm = 10 m was adopted for drawing homogeneous dam section and following results were obtained on a slip circle: N diagram = 16.4sq.cm. T diagram = 7.45 sq.cm. U diagram = 3.15 sq.cm ϕ = 30°, C = 4000 kgf / m 2 γ soil = 2100 kgf / m3 , γ w = 1000 kgf / m3 Solution: L = Length of trial arc = 12 cm. Find F.S. of the d / s slope

F.S. =

CL + Σ(N − U) tan φ ΣT

Earth Dams  209

Σ(N − U) = 16.4 × (10) 2 × 2100 − 3.15(10) 2 × 1000 = 31.35 × 105 kgf . ∴ FS = =

4000(12 × 10) + 31.35 × 105 tan 30° 7.45 × 102 × 2100 4.8 × 105 + 18.1 × 105 15.6 × 105

22.9 = 1.468 15.6 = 1.5 say (just safe) =

Example 9.7:  Calculate seepage through earth dam having anisotropic permeability, for the data given below:

Fig. 9.17:  Section of earth dam (anisotropic permeability)

Height of dam = 60m u/s slope 2.75 : 1 d/s slope 2.5 : 1

FB = 2.5m. Topwidth = 8m. Filter length = 120m.

K x = coeff . of permeability in x direction = 4 × 107 m / s. K y = coeff . of permeability in y direction = 1 × 107 m / s. Solution:  As it is a case of anisotropic permeability, first step is to transform the original section by transformation parameter, Xt

= Xo K y / K x = X o 1 × 10−7 / 4 × 10−7 =

with respect to transformed section,

x0 2

210  Irrigation Engineering and Hydraulic Structures

D = 46.0 and h = 60 – 2.5 = 57.5 K = K x K y = 4 × 10 −7 × 1 × 10 −7 = 2 × 10 −7 m/s S = (46) 2 + (57.5) 2 − 46 = 73.6 − 46 = 27.6 q = KS = 2 × 10 −7 × 27.6 = 55.26 × 10−7 m3 /s/m Example 9.8:  When the foundation consists of pervious stratum and extends over large depth yf, a horizontal u/s ­impervious blanket is provided to reduce the seepage through foundation. The thickness (yb) and length (x) of such blanket depend on permeability of the blanket material, which is usually between 0.6 to 3.0 m per day. F.R.L.

H Shell yb

Core

Shell

U/S X

Xd Pervious stratum of thickness

Yf

Fig. 9.18:  Design of u/s blanket Solution: Let Kf = permeability of foundation soil   Kb = permeability of blanket material. Then value of constant for u/s blanket is given by a=

Kb 2 and hence length of the blanket x = a K f Yb Yf

X r = equivalent resistance =

e2ax − 1

a(e2ax + 1)

 Xr  h o = head dissipated through blanket =  (H)  X r + x d   Xr  = percentage reduction in seepage due to blanket here,   X r + x d 

Earth Dams  211

Example 9.9: Design u/s blanket for earth dam section shown in the figure below. Blanket thickness = 1.5 m, Xd = 51 m. 6m

2:1

H= 12m

15m

2:1

1.5m X

51m Pervious stratum of thickness

Yf = 10m

Fig. 9.19:  u/s blanket Solution: Kb = 0.08 m/day  Kf = 67 m/day. a=

Kb 0.08 1 = = K f Yb Yf 67 × 1.5 × 10 112

Length of blanket = x = Xr =

e2ax − 1 a(e2ax + 1)

2 = 112 2 = 159 m. a

= 99

h o = head dissipated through blanket =

Xr (H) X r + Xd

99 (12) = 0.66 × 12 = 7.9m and 99 + 51 % reduction in seepage due to blanket = 0.66 × 100 = 66% =

Example 9.10:  Find out factor of safety for the slip circle arc shown in the following figure for sudden draw-down condition of u/s slope, given the following data:

γ sat = 2.10 KN / m3 γ sub = 1.10 KN / m3 C = 2.45 KN / m 2 ϕ = 35°

212  Irrigation Engineering and Hydraulic Structures

∠θ made between N and W, arc length, and area of different slices 1 to 7 are given in first four columns of the table given below R= Radius of slip circle

35o

R

1 3 b 7

6

5

4 O o N w 90 T

2 h Slip surface for u/s slope of the dam, 2H: 1V

Fig. 9.20:  Slip circle

Slice No.

θ°

Arc Area Length of slice (m) (3 ) (4)

T component W=bhγsat KN/m (5)

Wsin q KN/m (6)

W=bhγsat (7)

W cos θ KN/m (8)

(1)

(2)

1

54.5

6.7

12.26

25.74

20.9

13.48

7.8

2

41.0

3.8

19.51

40.97

26.8

21.46

16.2

3

31.0

3.5

21.37

44.87

23.1

23.50

20.1

4

22.0

3.35

20.90

43.89

16.4

23.00

21.3

5

13.0

3.05

19.97

41.93

9.4

22.00

21.4

6

5.00

3.05

16.72

35.11

3.0

18.40

18.3

7

-3.5

3.05

12.08

25.36

–1.5

13.20

13.2

L = total arc length = 26.50



N component

ΣT = 98.1

Factor of Safety =

ΣN = 118.3

CL + tan φΣN ΣT

(2.45) (26.5) + tan 35°(118.3) 98.1 64.9 + 82.8 = = 1.506 > 1.5, hence safe. 98.1 =

Example 9.11:  The following table gives various slices and the angle of the tangents drawn at the base of these slices. Mid ordinate and width of each slice are also given. Find factor of safety given γ = average unit weight of soil = 18KN / m3, C = 20 KN / m², ϕ =30°, Arc Length = 130 m.

Earth Dams  213

Data No. of Slice

1

2

3

4

5

6

7

8

Width of each slice (b) m.

15

15

15

15

15

15

15

15

Mid ordinate (h) m.

6

14

20

24

26

25

20

10

–10

–3

5

14

25

35

45

58

θ in degree Solution: W = bh γ = wt. of each slice

T Component = W sin θ (KN)

1620 3780 5400 6480 7020 6750 5400 2700 –281 –198

470

1567 2966 3871 3818 2289

N Component = W cos θ (KN) 1595 3774 5379 6287 6362 5529 3818 1430

Adding T and N components we get, ΣT = 14500 KN ΣN = 34194 KN CL + tan φ ΣN ΣT (20 × 130) + tan 30(34194) = 14500 2600 + 19741 = = 1.54 > 1.5 14500

FS =

Example 9.12:  In a zoned earth dam of height 50m, water is stored upto 47m. Top width 6m, u/s slope 3.5 H : 1 V and d/s slope 3 H to 1 V. Core section has top width of 3m and height of 48m with side slope as 0.1:1 on both sides. If K of shell is 3 × 10–3 cm/sec and K of core is 1.0 × 10–4 cm/s. Determine seepage flow through core section. Solution: See Fig. See − 9.3(a) Fig. − 9.3(a) L = Length horizontal projection of u/s slope of slope core of core L =of Length of horizontal projection of u/s L = 47 ×L0.1 = 4.7m. = 47 × 0.1 = 4.7m. CoreBase = (48 × 0.1) × 2×+0.1) 3 =×12.6m. CoreBase = (48 2 + 3 = 12.6m. d = 12.6 − 4.7 = 7.9 m

= 12.6 − 3.29 = 9.31 S = d2 + H2 − d S = (7.9) 2 + (47)2 − 7.9 = 47.6 − 7.9 = 39.76 q = KS  1 × 10−4  =  (39.76)  100  = 39.76 × 10 −6 m3 / S / m

S = d2 + H2 − d S = (7.9) 2 + (47)2 − 7.9 = 47.6 − 7.9 = 39.76

214  Irrigation Engineering and Hydraulic Structures

q = KS

 1 × 10−4  =  (39.76)  100  = 39.76 × 10 −6 m3 / S / m

9.4  CAUSES OF FAILURE OF EARTH DAM Many things can be learned from failures. Failures reveal weaknesses entered during design and construction. Proper study of failures will lead to better design and better achievements of construction and maintenance. Earth dam failures can be grouped as under: (i) Failures due to hydraulic reasons. Remove underline such as overtopping, erosion of u/s slope due to wave action and wind set up, erosion of d/s slope by wind and rain, sloughening of material at toe. (ii) Seepage or piping through dam and foundation may lead to failures. (iii) Failures due to structured defects such as sliding of u/s and d/s slopes, liquefaction slides, earthquake etc. Remedial Measures: 1. Overtopping of water should never be allowed during lifetime of earth dam. For this adequate free board and adequate spillway capacity should be provided. Free board kept should be higher than 1.5 (hw + wind set up S), or 3 m whichever is higher. Wind set up S, is the result of piling up of water on one end of reservoir on account of horizontal driving force of the blowing wind. The magnitude of the rise of water surface is called the wind set up or wind tide. The wind set-up S is calculated from Zuider Zee formula, S=

V2F m 62000 D

here V = Velocity of wind in kmph F = Fetch in km D = Average depth of Reservoir in m along Fetch line 2. Rock toe should be provided at toe of dam to prevent sloughening of dam material. Rock toe is a drain provided at toe, it must be provided with a filter on its u/s side. 3. Piping: Seepage through body of dam and its foundation, if uncontrolled, results in piping in d/s portion of the dam. Piping is the progressive erosion of embankment material due to leaks, which develop in body of dam and in foundation. This is the result of poor construction, inferior compaction adjacent to concrete outlets or pipes which have weak bond with surrounding. Weak bond between embankment and abutment is also responsible for leaks. When forces resisting erosion, (mainly cohesion, interlocking stresses, weight of soil

Earth Dams  215

particles) are less than the erosive forces of seepage of water, soil particles get washed away along leaking water and piping begins. Earth slides or sloughening are related to piping.

Fig. 9.21:  Gravel packed relief well Sloughening begins when a small amount of material at the d/s toe has eroded leaving behind steeper slope, which eventually gets saturated due to seepage of water and slumps again. This may continue to cause complete failure of the dam. To check the effect of seepage, seepage line, should be kept well within the d/s face and seepage water is collected and disposed off without causing dam material loss. Core, cut-off, rock toe, relief wells, grouting etc are remedial measures to check ill effect of piping. 4. Relief Wells: Relief wells (see Fig. 9.21) are provided when pervious strata of embankment foundation are too deep to be penetrated by cut – off or toe drains. Relief wells can penetrate most pervious strata and relieve uplift pressures effectively. Spacing between relief wells should be small enough to lower water pressure to desired value. It has interior perforated pipe of minimum 15 cm diameter or larger. The annular space is packed with gravel, however near the surface impervious compacted back fill is provided so that upward flow of water outside relief well may not occur.

216  Irrigation Engineering and Hydraulic Structures

5. Chimney Drains (see Fig. 9.22):They combine the advantage of horizontal blanket drain as well as toe drain. These drains keep seepage line well within the d/s face. These drain material must have high value of permeability so that it can discharge seepage water without excessive head and hence no piping may take place.

Fig. 9.22:  Chimney drain

9.5  FILTERS: CRITERIA AND DESIGN Soil particles lying in embankment surrounded by other soil particles cannot move due to seepage forces but finer soil particles near the boundaries may get washed away into void space of coarser particles. To prevent such migration of finer to coarser particles, fillers are provided, for example between core and shell, shell and rock toe, drains and adjacent zones etc. Filters have to satisfy following criteria: (i) Filter layer must be more pervious than the protected soil so that filter acts as a drain, and (ii) Particle size of filter layer must be small enough to prevent migration of particles from adjacent layers into the voids of the filter layer. (iii) Filter layer, should be sufficiently thick to provide good distribution of all particle sizes throughout the filter. Following filter criteria are generally adopted while designing filter layer: (i)

D15

D15 of filter ≥5 of protected soil

(ii)

D15 of filter ≤5 D85 of protected soil

(iii)

D50 of filter ≤ 25 D50 of protected soil

(iv) Gradation Curve of filter should have approximately same shape as that of protected soil. (v) Where protected soil contains large percentage of gravels, filter should be designed on the basis of gradation curve of the portion of material which is finer than one inch sieve.

Earth Dams  217

(vi) Filter should not contain 5% of finer materials passing through no. 200 sieve. Here D15, D50, D85 represent particle sizes, which are respectively coarser than finest 15, 50, and 85% of the soil by, weight. These filter criteria are based on studies with non-cohesive soils and take into consideration only the grain size of the protected soil. Theoretically thickness of horizontal filter required is 15 cm for sand and 30 cm for gravel. However, a minimum of 1.0 m is desirable. For vertical or inclined filters minimum width should be 2 to 3 m for convenience of construction.

EXERCISES 1. (a) What are rigid and non-rigid dams? (b) Distinguish between:

(i)  Earth Dam and Gravity Dams.



(ii)  Rock filled and Roll-filled Dams.

2. (a) What are causes of failures of earth dam? (b) What are remedial measures for same/ 3. (a) What precautions should be taken during and after construction of earth dam? (b) Explain Swedish slip circle method and location of critical slip circle. 4. Draw neat sketch of an earth dam and show there on all component parts. State the function of these components parts. 5. What design criteria are adopted for safe design of earth dam? 6. (a) Define and draw sketch of ‘phreatic line’ in homogeneous earth dam section with and without horizontal filter at its base. (b) What are egress and ingress corrections to be applied to base parabola? 7. Prove that for a homogenous earth dam section with a horizontal filter at its base of focal length equal to S, seepage loss per meter length of dam is given by q = KS, where K is permeability of dam material. What correction factor is required to be applied to this expression if dam material is non-homogeneous or anisotropic? 8. Explain with sketches: (a) Rock Toe

(b)  Chimney Drain

(c) Relief wells

(d)  Openings through earth dam.

9. What is ‘pore-pressure’ and its role in design and construction of earth dam? 

 

 

10 Canal Head Works 10.1  BARRAGE AND WEIRS A weir is an obstruction to river flow provided with a specific purpose of raising water level in river such that canal can receive water in it by gravity flow. To increase water level, weir crest is raised above river bed and if need be, shutters are also provided on the top of the weir. These shutters may get dropped down during flood so that afflux is minimum. Afflux is defined as difference in water level between upstream and downstream of weir under free flow condition. Controlling of levels by means of shutters becomes difficult when the crest level is higher than 2.0 m. In such cases, a fully gated weir, known, as barrage is preferred. Barrage and weir are similar structures with a difference that barrage has low level crest and gates all along its length. Barrages are considered better than weirs as they keep upstream region free of sediment deposition so that sediment free water enters the canal (See Fig. 10.1).

Fig. 10.1:  Sectional view of barrage

10.2  DIFFERENT UNITS OF CANAL HEAD WORKS Canal head works are basically diversion works consisting of a weir or a barrage and a canal head regulator, as shown in Fig. 10.2. © The Author(s) 2023 S. K. Ukarande, Irrigation Engineering and Hydraulic Structures, https://doi.org/10.1007/978-3-031-33552-5_10

218

Canal Head Works  219

A deep stilling pocket of water upstream of weir wall is created and through canal head regulator water enters the canal. This pocket is separated from other part of river by means of divide wall. A fish ladder is

Fig. 10.2:  Canal head works in plan provided near the divide wall. Sediment excluder near the canal head regulator is also sometimes provided. This arrangement may be either on left or right bank or both banks of the river, as shown in Fig. 10.2.

10.3  DESIGN OF WEIR A weir consists of following component parts: (i) Weir wall (ii) Crest gates (iii) u/s and d/s aprons (iv) u/s, d/s and intermediate sheet piles These are considered in detail as under: Weir crest is normally flat and it is then known as broad crested weir. Crest width should be 2.5 times head over weir or a minimum of 2 m. U/s slope of weir wall may be kept as 2 H: 1V and d/s slope 3H: 1V The d/s horizontal pucca apron (horizontal floor) should be such that maximum dissipation of energy takes place through stable hydraulic jump on it. Total length of impervious floor includes d/s basin length, glacis, weir crest, and u/s floor. Glacis means u/s and d/s sloping part of weir wall. The impervious floor in conjunction with d/s cut off (sheet pile) should result in safe exit gradient. Length of d/s horizontal floor is kept 5 to 6 times height of jump i.e., difference of depth before jump and after jump.

220  Irrigation Engineering and Hydraulic Structures

The u/s and d/s sheet piles are always provided to guard against scouring and piping effects. The depth of sheet piles should be such that its bottom is lower than the level of possible flood scour at that section. The scour depth is given by Lacey’s equation,  q2  R = 1.35    f  here

1/3

q = flood discharge per unit width of river f = silt -factor R = socur depth measured below HFL

The length of horizontal floor should be such that exit gradient GE, is less than safe exit gradient for the soil under consideration. The recommended values of safe exit gradient are: 1 1 (i ) to for gravel 4 5 1 1 (ii ) to for coarse sand 5 6 1 1 (iii ) to for fine sand 6 7 The thickness of impervious floor is decided by uplift pressure acting on the floor which can be decided either by Bligh’s creep theory or Khosla’s method, given hereafter.

10.4  BLIGH’S AND KHOSLA’S THEORIES 10.4.1  Bligh’s Creep Theory According to Bligh (1910), water percolating through pervious foundation below weir wall, creeps along the base profile of the structure as shown by dotted line in the Fig. 10.3 giving a total creep length L as, L = b + 2(d1 + d 2 + d3 ) Weir wall H

d1 Creep length (dotted line)

Hydraulic gradient = H/L

d2 b (Full length of apron)

d3 (Depth of sheet pile)

Fig. 10.3:  Weir section showing creep length

Canal Head Works  221

Here d1, d2, d3 are u/s, intermediate and d/s sheet pile depths, and b = total length of the floor. Hydraulic gradient will then be equal to H H = L (b + 2d1 + 2d 2 + 2d3 ) As hydraulic gradient is constant, and if L1 = creep length upto point 1 on the floor, then residual uplift pressure at that point will be H1 as given below.  L  H  H1 = H −  L1  = H 1 − 1  L L   The reciprocal of hydraulic gradient L H

C, Hence, C =

L , is known as Bligh’s Creep Coefficient, H

Bligh has recommended safe values for C for different soils in order to get safe creep length L L = CH Table 10.1:  Values of C and GH for Different Soils. Sr. No.

Soil

Value of C

GH = safe hydraulic Gradient = 1/C

1.

Mud

18

1/18

2.

Fine sand

15

1/5

3.

Coarse Sand

4.

Gravel

12

1/12

5 to 9

1/9 to1/5

The floor thickness, t, is given by 4 H1 t= 3 (G − 1) Here, G = specific gravity of floor material  L  H  H1 = H −  L1  = H 1 − 1  L L    L1 = creep length upto a point on floor where residual pressure head is H1 Limitation of Bligh’s Creep Theory: 1. Bligh made no distinction between horizontal and vertical creep length. 2. Bligh did not consider safe exit gradient. 3. Bligh did not make any distinction between outer and inner faces of sheet piles.

222  Irrigation Engineering and Hydraulic Structures

4. According to Bligh, d/s sheet pile is not essential. 5. Exit gradient may not be safe even if hydraulic gradient of Bligh (1/C) is safe.

10.4.2  Lane’s Weighted Theory On account of number of limitation to Bligh’s creep theory, Lane put forward his weighted theory in 1932. According to Lane, horizontal creep length, b, is less effective compared to vertical creep length due to piles. Hence, 1  L=  b  +2(d1 +d 2 +d3 ), i.e., weightage of horizontal creep length is reduced to 3  one third compared to its original value whereas value of vertical creep length is retained same.

10.4.3  Khosla’s Theory Dr. A. N. Khosla and his associate published their findings in 1954 for structures, which were designed as per Bligh’s theory but have not given satisfactory performance According to Khosla, (i) Outer faces of end sheet piles were more effective than inner ones. (ii) Intermediate sheet piles if smaller in length than outer ones, are ineffective. (iii) Undermining of floors starts from tail end. If hydraulic gradient at exit is more than permissible (critical) gradient for that soil, the soil particles will start getting washed away with seeping water causing progressive degradation of subsoil. This is known as piping and it leads to failure of structure. (iv) d/s sheet pile is more effective to control piping. Khosla also gave formula for safe exit gradient, GE =

H 1 d π λ

here, G E = safe exit gradient for soil H = head difference between water level on u/s and d/s side of weir wall, d = depth of sheet pile b = total length of pucca floor λ = ½ 1 + 1 + α 2    b α= d

Canal Head Works  223

If P = uplift pressure head at a point on pucca floor, and H= total seepage head, then f=

æ x ö÷ P 1 ÷÷ = cos-1 çç H p èç b 2 ø÷

Pressure variation along apron length given by above formula for f is shown in Fig. 10.4. From this figure, it is clear that d/s half of the floor has more uplift pressure than that given by Bligh.

Fig. 10.4:  Variation of pressure head along apron

H

H

C1

E

b

b

d

d

D1

D

Fig. 10.5 (i):  Sheet pile at u/s end

Fig. 10.5 (ii):  Sheet pile at d/s end

H H

E C b2

b D b

Fig. 10.5 (iii):  Case of intermediate   sheet pile

d D’

b

D’

Fig. 10.5 (iv):  Case of depressed      floor

224  Irrigation Engineering and Hydraulic Structures

10.5  KHOSLA’S METHOD OF INDEPENDENT VARIABLES Seeping water does not creep along the bottom contour of pucca apron as envisaged by Bligh, but it moves along a set of elliptic streamlines. The case of steady seepage in a vertical plane for homogeneous soil can be mathematically expressed by Laplace equation, ∂ 2φ ∂ 2φ + =0 ∂x 2 ∂y 2 here φ = Kh, Where K is permeability of soil and h is residual pressure head at point whose coordinates are (x, y). Dr. Khosla used method of independent variable and obtained solution for Laplace Equation for simplified cases as illustrated in Fig. 10.5 (i) to (iv). Case (i) and (ii): For sheet pile either at u/s or d/s end, Khosla’s solution gives

φE =

1 λ−2 cos −1   , Fig 10.5 (ii) π  λ 

φD =

1  λ −1 cos −1   , for Fig 10.5 (ii) π  λ 

= 100 10.5 φCφ1 C=1 100 − φ−EφEforfor FigFig 10.5 (i)(i) = 100 10.5 φDφ1 D=1 100 − φ−DφDforfor FigFig 10.5 (i)(i) 1 1 (1 + 1 +2α 2 ) λ =λ = (1 + 1+ α ) 22 bb αα == dd Case (iii): For sheet pile at intermediate point on the pucca floor (see Fig. 10.5 (iii)),  λ −1 1 φE = cos −1  1  π  λ2 

φD =

λ  1 cos −1  1  π  λ2 

φC =

 λ +1 1 cos −1  1  π  λ2 

Here, 1 λ1 =  1 + α12 − 1 + α 22   2  1 λ 2 =  1 + α12 + 1 + α 22   2  b b α1 = 1 and α12 = 2 d d

1 1 + α12 − 1 + α 22    2 1 λ 2 =  1 + α12 + 1 + α 22   2  b b α1 = 1 and α12 = 2 d d λ1 =

Canal Head Works  225

Case (iv) : For case of depressed floor, i.e., pucca floor having depth d from bed level, 2 3 φ D ' = ϕD − ( φ E − φ D ) + 2 3 α

φ

D1'

= 100 − φ

D1

Here, fE and fD are same as given in case (ii) Note: C, D, E and C1, D1, E1 are key points in all of above cases for which f values are obtained as per above formulate. In these solutions for f, floor thickness t is not considered. If actual values with consideration of floor thickness are expected, then following corrections should be applied to values as obtained above. A. Correction due to Floor Thickness: (i) Corrections for pile at u/s end [see Fig. 10.6 (i)]

 φ − φC  Correction for pressure at C1 =  D  t1 , + ve  d1 

 φ − φC  Corrected value of pressure at C1 = φc1 = φc +  D  t1  d1  E

t1 = floor thickness

C C1

+ve

d1

d1 = Depth of pile at u/s end D

Fig. 10.6 (i):  Pile of u/s end (ii) Correction for pile at intermediate point E E1 d2

C

t2

C1

-ve

+ve

D

Fig. 10.6 (ii):  Pile at intermediate point

226  Irrigation Engineering and Hydraulic Structures

∴ Corrected value at pt E1,

 φ − φD  φE1 = φE −  E  t2  d2  and corrected value at C1 is given by  φι − φC  = φc1 = φc +  D  t2 d2  

 φ − φD  (iii) Correction for pile at d/s end =  E  t 3 , ( − ve), ( Fig. 10.6 ,iii)  d3  E

d3

C

t1

E1 -ve

D

Fig. 10.6 (iii):  Pile at d/s end

 φ − φD Corrected value at po int E1 = φE1 = φE −  E  d3

  t3 

B. Correction for Mutual Interference of Piles: [see Fig. 10.6(iv)]

Fig. 10.6:  (iv) Mutual interference of pile.

Canal Head Works  227

Formula for correction due to mutual interference of pile is C = 19

D d+D   b'  b 

b' = distance between piles under consideration b = pucca Floor length D = Depth of pile whose influence is to be considered on the neighboring piles. The correction C is additive to points in rear and negative for points on forward side of the pile whose influence is to be considered. For example, if influence of pile no. (2) is to be considered on pile no. (1), then correction C is to be added to φC1, value and if influence of pile 2 is to be considered on pile 3, then correction C is to be subtracted from φ­E3 value but in this case b’ will be distance between pile 2 and 3 and D will be same as d3. C.  Corrections in φ Values for Sloping Apron: The correction for the sloping apron is applied to f values of key points of that pile which is fixed at either the beginning or at the end of the slope. [See Fig.10.6 (iv)]. The correction is positive for slope decreasing in direction of flow and negative for slope increasing in direction of flow. For example, correction due to slope is applicable to pile no. 2 in Fig. 10.6 (iv) and is additive for f value at E2. If bs is horizontal length due to sloping apron, and b1 is horizontal distance between two piles between b  which the sloping apron is located, then slope correction C = CS  1s  . The value b  of Cs depends on slope of the apron and its values are given in Table 10.2 below: Table 10.2:  Cs Values for Different Slopes. Slope V : H

1:1

1:2

1:3

1:4

1:5

1:6

Cs, % of φ value

11.2

6.5

4.5

3.3

2.8

2.5

q  SOLVED EXAMPLES Examples 10.1:  A weir on pervious foundation is shown in Fig. 10.7. Find hydraulic gradient according to Bligh and Lane. Also find uplift pressure at points A, B, C. Take G = 2.65. Find floor thickness also.

228  Irrigation Engineering and Hydraulic Structures

H = 6m B

A 6m

C

d2 = 4m

d1 = 6m d3 = 6m

6m 12m

6m

24m

Fig. 10.7:  Weir on pervious foundation Solution: According to Bligh, Creep Length, L = 2 ( d1 + d 2 + d3 ) + b = 2 (6 + 4 + 12) + 24 = 68m H 6 = L 68 1 1 = < 11.3 6

∴ Hydraulic gradient =

hence safe if soilis coarse grain

According to Lane, Creep Length

b 3 24 = 2(6 + 4 + 12) + = 52m 3 H = L 6 1 1 = = < 52 8.66 6  L   18  = H 1 − A  = 6 1 −  = 4.41m   68  L  = 2(d1 + d 2 + d3 ) +

∴ Hydraulic gradient

h A = Uplift pressure at A L A = Creep length upto A

= 2 × 6 + 6 = 18

Floor thickness at A = tA

4  h  4  4.41  =  A =   = 3.65m 3  G − 1 3  2.65 − 1

Canal Head Works  229

hB = Uplift pressure at B:

L B = 2 × 6 + 2 × 4 + 12 = 32m  32  ∴ h B = 6 1 −  = 3.17m  68  4  3.17  tB =   = 2.56m 3  1.65 

hC = Uplift pressure athC: C = uplift pressure at C : − LC = 2 × 6 + 2 × 4 + 18 = 38 m  38  h C = 6 1 −  = 2.64m  68  4  2.64  tC =   = 2.13m 3  1.65  Example 10.2:  Using Khosla’s method of independent variables, obtain residual pressures at key points for the weir given in Fig. 10.8 and plot Hydraulic Gradient line (HGL). Pond level RL 170m (a) RL 268.32m (b) RL 266.92m (b1) RL 266.51m V:H 1:2

RL 266m RL 265m

H

1m

(c) 266.50m

V

RL 265m d

E1 C 1

u/s pile 2m

5m

2m

bs=7m

1m E C

E C 5m

D1 RL 260m b1=16m

Intermediate pile D

d/s pile 258m RL

b=21m

Fig. 10.8:  Cross-section of weir Solution: (i) Upstream pile, depth d = 266 – 260 = 6 m, b = 21 m b 21 1 α= = = 3.5, λ = 1 + 1 + α 2  = 2.32  d 6 2 

1 λ−2 cos −1   π  λ  1 1 π 82 = (82°) = (82) = = 0.45 = 45% π π 180 180

φE =

258m RL

230  Irrigation Engineering and Hydraulic Structures

φC1 = 100 − ϕE = 100 − 45 = 55% 1  λ − 1  55 cos −1  = 0.3 = 30% = π  λ  180 = 100 − ϕD = 70

φD = φD1

Thickness correction for φD − φC1 70 − 55 φC1 = 1 t= (1) = 2.5% , + ve d 6 Correction due to interference of intermediate pile on f value at C1 = +19

D d+D 266 − 258  6 + 8    = +19   b1  b  16  21 

 14  = +19 ( 0.7 )   = 9 % + ve  21  ∴φC1 corrected = 55 + 2.5 + 9 = 66.5% H = 270 – 265 = 5 66.5 (5) = 3.32 ∴φC1 = 100 ∴ RL of residual pressure at C1 = 265 + 3.32 = 268.32 shown at (a) in Fig 10.8 (ii) Intermediate pile, depth d = 265 – 258 = 7m b1 = 16m, b 2 = 5m b1 16 b 5 = = 2.3, α 2 = 2 = = 0.7 d 7 d 7 1 2 2 λ1 = 1 + α1 − 1 + α 2  2  1 = [ 2.5 − 1.2] = 0.65 2 1 1 λ 2 =  1 + α12 + 1 + α 22  = [ 2.5 + 1.2] = 1.85    2 2 α1 =

φE =

 λ −1 1 1  0.65 − 1  cos −1  1  = cos −1   = 40.4 % π  1.85   λ2  π

φD =

λ  1 1  0.65  cos −1  1  = cos −1   = 38.5% π λ π  1.85   2

φC =

 λ +1 1 1  1.65  cos −1  1  = cos −1   = 15% π λ π  1.85   2 

Canal Head Works  231

 φ − φD  Thickness correction for φE =  E  (1) 6   40 − 38.5 = (1) = 0.25 % , − ve 6 Slope correction for slope of 1V : 2H, Cs = 6.5 from Table 10.2 b  7 ∴ Slope correction = CS  s  = 6.5    16   b1  = 2.84, + ve Correction for interference of u/s pile on φE 264 − 260  4 + 6    = 4.5 − ve 16  21  φE corrected = 70 − 4 + 2.84 − 4.5 = 19

= 64.34% 64.34 (5) 100 = 265 + 3.2

Corresponding RL for φE = 265 +

= 268.2 m shown at (b) in Fig.10.8

φ D − φC t d 38.5 − 15 (1) = 7 = 3.35, + ve

Similarly thickness correction for φc =



Correction for interference of d/s pie on C = 19



6 ( 6 + 6) = 11.86, + ve 5 21

∴φc corrected = 15 + 3.35 + 11.8 = 30.21 %

Corresponding RL of φc = 265 + 0.3021 (5) = 266.51, shown as b1 in Fig. 10.8 (iii) D/s pile: d = 6m

b = 21m b 1 = 3.5, λ = 1 + 1 + α 2  = 2.32  d 2  1  λ − 2  82 φE = cos −1  = 45% = π  λ  180

α=

232  Irrigation Engineering and Hydraulic Structures

1 55  λ −1 1 −1 cos −1  = 30%  = cos (0.56) = π 180  λ  π 45 − 30 thickness correction for φE = × 1 = 2.5% , − ve 6

φD =

Interference of intermediate pile on φE = 19

6 6+6   = 11.86 % − ve 5  21 

∴φE corrected = 45 − 2.5 − 11.86 = 30.64 % ∴ RL for φE at d / s pile = 265 + .3(5) = 265 + 1.5 = 266.5, shown at (c) in Fig. 10.8 HGL is shown as (a b b1 c) in Fig. 10.8 Example 10.3:  A weir is 4 m high and has 16 m long, 1 m deep apron on u/s side and 36m long, 2 m deep apron on d/s side. The depths of u/s and d/s piles are 8m and 8.5 m respectively. Find uplift pressure u/s and d/s of piles at top and bottom by Khosla’s Theory. Present in tabular form data required to plot HGL. Refer to Fig. 10.9,

H = 4m 36m

16m El d1 = 8m

C1

t1 = 1m

t2 = 2m

E

C

C1 E d2 = 8.5m

D1 b = 52m D

Fig. 10.9:  Section of weir (pressure at u/s & d/s piles) Solution:

b = 16 + 36 = 52 b 52 α1 = = = 6.5 d1 8 λ1 =

1 1 + 1 + α12  = 3.7    2

Canal Head Works  233

φE =

λ −2 1 1 −1  3.7 − 2  cos −1  1  = cos   = 34% π  3.7   λ1  π

 λ −1 1 1  3.7 − 1  cos −1  1  = cos −1   = 24% π  3.7   λ1  π = 100 − ϕE = 100 − 34 = 66%

φD = φC1

φD1 = 100 − ϕD = 100 − 24 = 76%  φD − φC1  Correction for thickness in value of φC1 =  1  t1 d1   76 − 66 = (1) = 1.25, (+ ve) 8 Correction due to interference of pile at d/s end = 19

∴φ

C11

8.5  8 + 8.5  d+D   = 19   , here b1 = b = 52 52  52   b  = 19 × 0.40 × .32 = 2.41 = corrected φC1 = 66 + 1.25 + 2.4 = 69.66%

D b1

d/s pile : − d 2 = 8.5 ∴α 2 = λ2 = φD =

b 52 = = 6.11 d 2 8.5

1 1 + 1 + α 22  = 3.6    2

 λ −1 1 1 −1  3.6 − 1  43.76 cos −1  2 = 0.243 = 24.3%  = cos  = π  3.6  180  λ2  π

φE =

λ −2 1 cos −1  2  π  λ2 

1  3.6 − 2  63.6 cos −1  = 0.353 = 35.3% = π  3.6  180 Correction in φE due to thicknes t 2 = 2m =

=

( φE − φD ) d2

t2

35.3 − 24.3 ×2 8.5 = 2.58 % , (− ve)

=

(− ve)

234  Irrigation Engineering and Hydraulic Structures

Correction due to interference of u/s pile

D d+D   b1  b 

= 19

8  8 + 8.5    52  52  = 19 × .39 × .32 = 19

= 2.38 (− ve) φE corected = 35.3 − 2.58 − 2.38 = 30.34% Table for Plot of HGL: U/S pile H

φE1

4m Residual Pressure

φD1

D/s pile

φC1

φE

φD

φC

100 %

76 %

69.66 %

30.34 %

24.3 %

00

4m

3.04m

2.78m

1.212m

0.97m

00

Example 10.4:  An impervious floor of a weir on permeable soil is 16 m long and has sheet piles at both the ends. The u/s pile is 4 m deep and d/s pile is 5 m deep. The weir creates a net head of 2.5 m; neglecting thickness of apron, calculate the uplift pressures at the junction of the interfaces of the pile with the weir floor, using Khosla’s theory. Refer to Fig. 10.10,

H = 2.5m

E C1

E C 16m 4m

5m

D1

D

Fig. 10.10:  Weir on impervious f loor. Solution: u/s pile d = 4m, b = 16m b 16 α= = =4 d 4 1 λ = 1 + 1 + α 2 = 2.56 2 φC1 = 100 − ϕE

(

φE =

)

1 λ−2 cos −1   π  λ 

u/s pile d = 4m, b = 16m b 16 = =4 d 4 1 λ = 1 + 1 + α 2 = 2.56 2 φC1 = 100 − ϕE

Canal Head Works  235

α=

(

φE =

)

1 λ−2 cos −1   π  λ 

1  2.56 − 2  77 = 0.429 = 43% cos −1  = π  2.56  180 = 100 − 43 = 57%

= φC1

Correction due to d/s pile interference:  D  d+D = +19  1    b   b  5  4+5   = 19 × .559 × .56 = 5.98 = 6% + ve 16  16  ∴φC1 Corrected = 57 + 6 = 63% = +19

∴ uplift pressure at c1 = 0.63(2.5) = 1.57m (2) d/s pile: d = 5 m, b = 16 m, α = b d =

16 = 3.2 5

1 1 + 1 + α 2  = 2.17  2  1 λ−2 1 −1  0.17  85.5 φE = cos −1  = 0.475 = 47.5%  = cos  = π λ π    2.17  180 λ=

Correction due to u/s pile interference 4 5+4   = −5.3% 16  16  = 47.5 − 5.3 = 42.2 %

= −19 φE corrected

uplift pressure at E = 0.422(2.5) = 1.05 m

10.6 SALIENT FEATURES OF DESIGN OF WEIR ON PERVIOUS FOUNDATION (Refer to Fig. 10.11)

1/3

 q2  R = 1.35    f    L = Creep length = CH = b + 2d1 + 2d 2 b = horizontal creep length = L 2 + B + L1 4 h  L  here h = H 1 − 2  3 (G − 1) L   L 2 = u/s pucca apron length = L − L1 − (B + 2d1 + 2d 2 ) t = apron thickness =

1/3

236 

 q2  R = 1.35    f   and  Hydraulic Structures Irrigation Engineering L = Creep length = CH = b + 2d1 + 2d 2 b = horizontal creep length = L 2 + B + L1 4 h  L  here h = H 1 − 2  3 (G − 1) L   L 2 = u/s pucca apron length = L − L1 − (B + 2d1 + 2d 2 ) t = apron thickness =

L1 = d/s pucca apron length = 2.21 C

H 10

2

V /2g

TEL

HFL

(HFL-Afflux) K

Crest width

H=12m 1.5R

Weir wall

2R

h 1.5dl Pucca apron

L3

Base width L2

d1

L1

B

(2)

(1)

d2

Fig. 10.11:  Weir on pervious foundation L3 = d / s apron length (pucca + loose) = 18C

H q × 13 75

with crest shutter

= 18C

H q × 10 75

without crest shutter

V2 2g K = shutter depth from TEL

TEL = HFL +

 q  =   1.7 

2/3

Crest Level = TEL – K Level of d/s TEL = HFL before construction +

V2 2g

Loose apron

Canal Head Works  237

Level of u/s TEL = d/s TEL + Afflux V2 Level of u/s HFL = u/s TEL – 2g Level of bottom of (u/s) pile = (u/s) HFL – 1.5 R = HFL (u/s), Level of bottom of (d/s) pile = (d/s) HFL – 2R

10.7  NUMERICAL PROBLEMS ON WEIR DESIGN: A weir on sandy bed has top width 2 m, height 7 m, and base width 9 m. Depth of water on u/s side is 12 m above river bed. Discharge per unit length of weir is 25.7 cumec u/s pile length is 5 m and d/s pile length is 8 m. Design u/s and d/s solid and loose apron. Take C = 12 (Bligh’s creep coefficient). HFL 2m H=12m Weir wall 7m

L3 9m d1=5m

L2

L1

B

d2=8m

Fig. 10.12:  Weir on sandy bed As per Bligh: L = CH = 12 × 12 = 144m = b + 2d1 + 2d 2 = b + 2(5) + 2(8) = b + 26 ∴ b = 144 − 26 = 118m = Length of horizontal pucca appron La = u / s pucca apron length = L − L1 − (b + 2d1 + 2d 2 ) L1 = d / s pucca apron length = 2.21 C

H 10

= 2.21 × 12 × = 29m

12 10

238  Irrigation Engineering and Hydraulic Structures

∴ L 2 = 144 − 29 − (9 + 10 + 16) = 80 m L3 = Length of pucca + loose apron on d/s = 18C

H q × 13 75 12 25.7 × = 18 ×12 × (0.56) =121 m. 13 75

= 18 × 12 × ∴

Length of loose apron on d/s = 121 – 29 = 92 m …



Length of loose apron on u/s = 1.5 d1 = 1.5 × 5 = 7.5m …



t = Thickness of apron below weir wall 4 h  =  = 3  G −1

  L2   H 1− 4   L   4 (.45 × 12) = = 4.36 m 3  2.65 − 1  3 1.65    

10.8  SEDIMENT CONTROL AT CANAL-HEAD WORKS (Note: Fig. 10.13, which shows sectional view of head regulator, for plan please refer Fig. 10.2) In canal head works, head regulator for canal (Fig. 10.13) is provided just upstream of the weir wall and canal draws water from the pond created between weir, divide wall and the bank. The purpose of creating the pond is to exclude entry of excessive silt into the canal. Undersluices provided in the weir wall remove the silt deposited in the pond. However, even with this arrangement, if silt entering into the canal is excessive, then silt excluder is provided just in front of canal head regulator, and if after entry of water into the canal, silt load is found to be excessive than silt ejector is provided just little d/s of canal head regulator in the canal reach so that water flowing into the canal is carrying least silt load. This is known as sediment control at canal head works. It is achieved by silt excluder and silt ejector provided on u/s and d/s side of head regulator respectively. H.F.L.

Approach road Pier of head regulator

Gate

Pond level

Canal

Divide wall

Silt excluder tunnels

Silt ejector

Fig. 10.13:  Canal head regulator

Canal Head Works  239

10.9  RIVER TRAINING WORKS FOR CANAL HEAD WORKS River training structures for canal head works are provided to satisfy following purposes: 1. To prevent overflow from structures 2. To prevent flooding of land u/s of barrage or weir 3. To prevent entry of silt into canal and to pass silt load to d/s of weir. For this, following structures are required to be constructed at the canal head works. (Fig. 10.2) (i) Guide banks (ii) Divide wall (iii) Weir wall with under slucies (iv) Pond (v) Head regulator with silt excluder tunnels (vi) Groynes or spurs to protect the river banks (vii) Fish ladder to pass fish d/s of weir

EXERCISES 1. (a)  D  istinguish between Khosla’s theory and Bligh’s creep theory for seepage below weirs on permeable foundations. (b)  How does Lane’s theory differ from Bligh’s theory? 2. Briefly out line Khosla’s theory on design of weirs on permeable foundations. Enumerate the various corrections that are needed in the application of this theory. 3. (a)  Explain briefly Khosla’s exit gradient concept (b)  What is piping? What are the precautionary methods to avoid the ill effects of piping? (c)  Distinguish between a weir and a barrage. 4. Design a weir on coarse sand strata having following data making use of Bligh’s creep theory: Flood discharge = 300 cumecs. Value of Bligh’s C = 12 for coarse sand Length of weir = 30 m.

240  Irrigation Engineering and Hydraulic Structures

Top of weir is at 2m alone LWL And height of falling shutter = 0.6 m. Top width of weir 2 m., and bottom width 3.5 m. 5. (a)  Draw a sectional view of a weir and show there of various parts.

What is exit gradient? How does it affect the design of a weir?

(b)  Find maximum flood discharge that weir having following data can pass safely: Number of vertical gates = 51 Span of each gate = 10 m. FRL

= 110 m

Crest Level

= 106 m

Kp

= 0.02,

Ka

= 0.1

C

= 1.7 in Bligh’s Formula.

Neglect velocity of approach. 

[Ans:. 6818 cumecs]  

 

11 Distribution System 11.1 GENERAL Irrigation engineering projects consist of storage works and distribution works. Storage works are in the form of reservoirs, dams, spillways and distribution works are head works, canals, regulation and cross drainage works. Success or failure of project depends on efficient management of distribution system.

11.2  CANAL SYSTEM A canal is defined as a channel constructed on ground to carry water from river or reservoir by gravity flow. It has an open surface and is normally unlined and hence suffers of losses due to evaporation, transpiration and seepage of water. Depending upon functional role, canals can be classified as: (i) Irrigation, power or feeder canals (ii) Main, branch, distributary and field channel or watercourse. An irrigation canal carries water from its source to agricultural fields situated in its command area. Command area is demarcation of agricultural area in which canal water can be provided by gravity flow. Power canals carry water from source to electrical power generating hydro plants. After utilization for power generation, water is again fed back to canal system for irrigation purposes. Feeder canal feeds two or more canals. Irrigation canal system is subdivided into main, branch, distributary and field channels as per its function. Main canal feeds branch canals through cross regulators and they in turn supply water to distributaries. Field channels receive its share of water through regulation works such as outlets provided in distributary canals. Canals may have to cross natural drainages such as rivers, tributaries etc. and works provided at such crossing are known as cross drainage works.

© The Author(s) 2023 S. K. Ukarande, Irrigation Engineering and Hydraulic Structures, https://doi.org/10.1007/978-3-031-33552-5_11

241

242  Irrigation Engineering and Hydraulic Structures

11.3  ALIGNMENT OF CANALS Based on alignment, (see Fig. 11.1) canals are known as: (a) Contour Canal (b) Ridge Canal (c) Side slope or watershed canal If a canal has its alignment parallel to a contour of the topography of the land then it is known as contour canal and if the alignment is along a ridge line it is known as ridge canal. Canal running along a sloping ground is known as side slope or watershed canal. A canal has to be terminated into a canal escape, feeding extra water back to tributary or to river. Ridge canal has advantage over contour canal that it can irrigate land on both sides. Tributary Contours

Contour canal

Head Regulator Weir

RL 100

R C.D. Work

Cross regulator RL 99

E R

side slope canal RL 98

RL 97

I V

Ridge canal

RL 96

RL 95

Fig. 11.1:  Alignment of canal Alignment of canal is generally carried out such that it has to cross minimum number of tributaries or natural drainages and also such that topography of land has gentle ground slope. If ground slope is more or changing suddenly along the canal alignment, a fall is required to be provided such that the canal bed slope upstream of fall and downstream of fall remains same. Structure known as canal fall takes care of change in ground level by providing a weir wall and cistern. Falls are discussed separately. Alignment of canal should be such that filling and cutting of ground is balanced or is kept minimum. A canal may be fully above ground level and is known as canal in high embankment. It may

Distribution System  243

be partly above ground and partly in cutting and sometimes fully in cutting. [See Fig. 11.3(i) to (iii)].

11.3.1  Curves on Canal Alignment As far as possible curves should be avoided in the alignment of canals because they become source of siltation or scouring in the canal bed and change its cross sectional area and create disturbance in the working of the canal system. However, if they are unavoidable, they should be gentle and according to IS 5968 of 1970 the radii of the curves should be 3 to 7 times the width of the canal at FSL (full supply level) subject to the minimum values given ahead: (i) Unlined Canals Table 11.1:  Radius of Curves(Lined Canals) Capacity in Cumecs

Minimum Radius of curve in meter

80 and above

1500

30 to 80

1000

15 to 30

600

3 to 15

300

0.3 to 3

150

(ii) Lined Canals Table 11.2:  Radius of curve (Unlined canal) Capacity in cumecs

Minimum Radius of curve in meter

280 and above

900

280 to 200

750

200 to 140

600

140 to 70

450

70 to 40

300

40 to 10

200

10 to 3

150

3 to 0.3

100

It may be noted that idle length of canal should be kept minimum while considering the alignment of canal and rock strata should be avoided.

244  Irrigation Engineering and Hydraulic Structures

11.4  LOSSES AND TYPES OF CROSS-SECTION FOR CANAL 1. Losses in Canals:  Right from head works where water enters the canal to its destination upto fields where it is utilized for irrigation purposes, canal water undergoes loss, which is known as transit, or transmission loss and is around 25 to 45 per cent of water that enters the canal at the head works. These losses are due to (i) Evaporation (ii) Transpiration through vegetation on banks (iii) Seepage through (a) percolation and (b) absorption Loss on account of evaporation and transpiration is minor compared to seepage loss, it is of the order of 1% minimum and 10% maximum of the total transit losses. Depending upon permeability of soil and position of water table, seepage loss may occur in two ways: (a) Percolation: When Ground water table is close to ground level, top flow line meets the water table forming a zone of fully saturated soil below canal water way section. This is percolation loss and it is dependent on difference between FSL and GWT. (See Fig. 11.2).

GL

HG Top flow line

F.S.L HG D

Fully saturated soil Ground water table

Fig. 11.2:  Percolation loss in canal (b) Absorption: When the water table is at a considerable depth, top flow line does not join it, but a small zone of saturation is formed around the water way section of the canal, and seepage loss will depend only on FSD (full supply depth) of the canal. This is loss on account of absorption and is small compared to percolation loss. Percolation loss may be around three times higher than absorption loss. Canal losses can be measured by a method known as inflow and outflow method. In this method a long reach of canal is chosen and except main flow all other outlets are closed. Discharge measurements are taken at inlet and outlet of the reach and difference between these values of discharge is the loss occurring in the reach. In the alluvial soil zone of North India, values of total transmission losses are 17% in main canal and branch canal, 8% in distributaries and 20% in field channels making a total of 45% of water supplied at the canal head works.

Distribution System  245

2. Types of Canal Sections:  Canal sections are usually of following three varieties depending upon site conditions of the track: (a) Canal in full cutting (b) Partly in cutting and partly in filling (c) Fully in filling i.e., either in low embankment or in high embankment These are illustrated in the Fig. 11.3(i) to (iii). A canal is said to be in cutting when ground level is above FSL of the canal. A canal is said to be in filling when canal bed level (CBL) is higher then ground level (GL). A canal is partly in cutting and partly in filling when GL is in between FSL and CBL. Top width of bank as recommended in IS 7112 of 1973 are as shown in Table 11.3.

Fig. 11.3 (i):  Canal section in full cutting Road width

0.5m Dowel F.S.L Free board

m

0.5:1 D 2m

2D

1.5:1

1:1

1:1

GL 1.5:1

2D

C.B.L. B

Fig. 11.3 (ii): Canal section partly in cutting and partly in filling Road width

0.5m Dowel F.S.L. 1.5:1

1.5:1

Free board

1.5:1 D 3D

GL

C.B.L.

B

Hydraulic gradient line

1.5:1

1.5:1

3D

C.B.L. is higher than GL

Fig. 11.3 (iii):  Canal in high embankment Table 11.1:  Top Width of Bank Q cumec

Inspection Bank (Road)

Non-inspection bank

Upto 7.5

5.0 m

1.5 m

7.5 to 15

6.0 m

2.5 m

15 to 30

7.0 m

3.5 m

246  Irrigation Engineering and Hydraulic Structures

3. Balancing Depth: (see Fig. 11.4) A canal section partly in filling and partly in cutting is preferred so that material excavated may be used for filling the banks. If for a canal section, depth of cutting is such that quantity of excavation is equal to earth filling required in the banks for a given reach then this depth of cutting is known as balancing depth. This is quite economical and saves time of construction. t

n:1

t Filling

n:1

n:1

F. S. L. n:1

h m:1

y

D

Cutting m:1

B

Fig. 11.4:  Balanced depth of cutting Let y be the depth of cutting and (h–y) be depth of filling for a given reach of unit length.[see Fig. 11.4]. Area of cutting = ( B + my ) y Area of filling = 2 [ t + n(h − y)] (h − y) Equating these two areas we get

( B + my ) y = 2 [ t + n(h − y)] (h − y) ∴ y 2 (2n − m) − (B + 4nh + 2t)y + h(2t + 2nh) = 0 A canal has usually side slope of 1:1 in cutting and hence m = 1, whereas a side slope of 1.5:1 in filling, i.e., n = 1.5 Putting these values, we get 2y 2 − (B + 6h + 2t)y + h(2t + 3h) = 0 By putting given values of B, h, t etc, y = balancing depth can be obtained from above equation.

11.5  DESIGN OF ALLUVIAL CHANNELS A lined canal is a rigid boundary canal and can be designed as per laws of hydraulics but an unlined canal has movable bed and sides and as such has to be designed as per principles of loose boundary hydraulics. A change in discharge will cause only a change in depth of flow for rigid boundary canals whereas in case of non-rigid

Distribution System  247

boundary canals it will cause changes in cross-section, bed slope, bed forms and roughness coefficient. The application of the theory of rigid boundary canals to lose boundary canals therefore is not possible. Empirical approach based on experimental observations is therefore adopted. Number of talented PWD engineers have worked extensively on this problem during British regime, Chief among them are: (i) R. G. Kennedy (1895) (ii) Lindley (1919) and (iii) Gerald Lacey (1945) A channel or river flowing through non-cohesive sediment material generally consisting of silt, sand and gravel is called an alluvial channel or river. Upper Bari Doab canal system situated in west Punjab, is a­ lluvial type and RG Kennedy made number of observations on this canal system and arrived at f­ ollowing conclusions: (i) Sediment in alluvial channel is kept in suspension solely by vertical component of the eddies which are formed along channel bed. (ii) Eddies formed due to sides of the channels do not have sediment supporting power. (iii) Sediment supporting power is thus proportional only to depth of channel and not to its wetted perimeter.

11.5.1  Kennedy’s Method According to Kennedy, Vo = 0.55 D0.64 Where Vo = critical velocity in m/s = non-silting and non scouring velocity and D = depth of flow in m. According to him bed with of canal does not influence Vo, but silt grade has some effect. To take care of effect of silt grade, he introduced in his above equation, a factor m called critical velocity ratio, CVR; ∴ Vo = 0.55 m D0.64 Value of m = 0.8 to 1.2 as per silt grade For calculation of mean velocity V, Kennedy suggested use of Kutter’s equation or Manning’s equation. As per Kutter’s equation, 1  0.00155  +  23 +  N  S  V= 0.00155  N  1 +  23 +  S   R

RS

248  Irrigation Engineering and Hydraulic Structures

A = hydraulic radius P P = Wetted perimeter

Here R =

S = bed slope N = rugosity coefficient = 0.025 to 0.03 1 2 3 12 R S n Here n = Manning's coefficient which is usually 0.014 to 0.02 As per Manning 's , V =

Limitation of Kennedy’s theory: B (i) Kennedy did not consider   ratio in the design of alluvial channels. D (ii) No equation for bed slope is given which may lead to varied designs. (iii) Silt charge or silt grade was not given due weightage. (iv) Value of m decided arbitrarily. (v) Design of channel by his method is based on trial and error i.e., one has to assume D and find V by V = 0.55 mD0.64 and then using mean velocity equation of Kutter or Mainning he has to prove that assumed value of D is alright.

11.5.2  Lindley’s Equation In view of limitation of Kennedy’s theory, Lindley in 1919 gave following equation for stable design of alluvial channels:   V = 0.567 D0.57 B = 7.86 D1.61 He gave equation for bed width B, but did not give equation for bed slope. However, he was first to put forward the concept of regime for alluvial channels. As per concept of regime, the dimensions: bed width B, depth of flow D, and bed slope S of alluvial channel carrying given discharge are all function of time. British engineers working in Punjab around 1910 to 1920 introduced the term “regime”. They studied behaviour of some such structures of existing canals where the bed was in a state of stable equilibrium. These stable reaches had not required any sediment clearance for several years of canal operation. They called such canals as “regime channels”. This means that the term regime was used to express no change in initial condition of bed of channel even after working for several years. In other words when a channel is constructed in alluvial soil, carrying silt laden water, its bed and sides will silt and scour until the width, depth and bed slope attain a state of

Distribution System  249

balance and the channel is then said to be in “regime”. Lacey followed this concept in development of his theory.

11.5.3  Lacey’s Regime Theory Based on his observation, Lacey (1945) found that eddies generated from bed as well as from sides of the channel, in the direction normal to surface of generation, keep the silt particles in suspension. So Lacey considered hydraulic mean depth (R =A/P) as more pertinent variable rather than depth of flow as considered by Kennedy. He also introduced a silt factor, f. Hence, according to Lacey, 2 f R 5 V = mean velocity of flow in m / s V=

(11.1)

f = silt factor A R = = hydraulic mean depth P A = c / s area of flow P = wetted perimeter He also gave second equation, Af 2 = 140V5 Multiplying both sides by V,

(11.2)

∴ AVf 2 = 140V 6 = Qf 2  Qf 2  ∴V =    140 

1

6

(11.3)

P = 4.75 Q  q2  R = 1.35    f 

(11.4) 1

3

(11.5)

In equation (V), if q = flood discharge per unit width of canal or river then R stands for scour depth. f

5

3



S=



S = bed slope

 1  3340  Q 6   



(11.6)

With a set of above six equations, design of stable alluvial channel can be carried out accurately.

250  Irrigation Engineering and Hydraulic Structures

q  SOLVED EXAMPLES Example 11.1:  Design an irrigation canal to carry 36 cumecs with B/D ratio of 2.5. Take CVR = 1, and Kutter’s regosity coefficient =0.025. Solution: Assuming bed slope as l in 4000, and using Kennedy’s equation, V0 = 0.55 m D0.64 Here, m = CVR = 1. = 0.55 (1) D0.64



1/2:1 n:1

1/2:1 D n:1

B

Fig. 11.5:  Unlined alluvial canal (trapezoidal section) For trapezoidal section, as shown in Fig. 11.5, A = (B + nD)D D 1  =  B +  D, for sideslope of H :1V 2 2  Now Q = AVO D  36 = D  B +  Vo 2  B = 2.5 and Vο = 0.55 D0.64 D ∴ B = 2.5D

(

D  ∴ 36 = D  2.5D +  0.55 D.64 2 

)

= D 2.64 (3) (0.55) 1

 36  2.64 ∴D =    3 × .55  = (21.8)0.378 = 3.2m ∴ B = 8.00m 3.20   2 ∴ A =  8.00 +  3.20 = 30.72m  2 

Distribution System  251

 5  P =  B + 2 D  = 15.15m 2   A 30.72 = 2.02 = 2 R= = P 15.15 1 0.00155   +  23 +  .023  S  RS , V= 0.00155  N  1 +  23 +  S   R 1 with S = 4000 43.5 + 23  1  V= × 1.41 ×   0.023  63.24  1 + (23) 1.41 66.5 1.41 68.4 = × = = 1.08 m / s 1.37 63.24 63.24

V0 = .55(D)0.64 = .55(3.2)0.64



= 0.55 × 2.1 = 1.15 > 1.08 Hence, try S as 1 in 3600, in equation (1) above. 68.4 V = = 1.14, which is nearer to value of V0 =1.15 60 Hence, S as 1 in 3600 is suggested. Example 11.2:  Design a regime channel to carry a discharge of 40 cumecs and silt factor 1.0. Solution: Use Lacey’s theory, 1

1

 Q f 2  6  40 × 1  6 = V=  = 0.81 m / s  140   140    Q 40 A= = = 49.38 m 2 V 0.81 5  V 2  5  0.812  R=  =   = 1.64 m 2  f  2  1  P = 4.75 Q = 4.75 ( 40 )

1

2

= 30m

For trapezoidal channel having side slope as P = B + 5D and A = (B + .5D)D

1 :1, 2

252  Irrigation Engineering and Hydraulic Structures

∴ 30 = B + 2.23D ∴ B = 30 − 2.23D D  49.38 =  B +  D 2  D  Subtituting for B, 49.38 =  30 − 2.23D +  D 2  49.38 = 30 D − 1.7D 2 ∴ D 2 − 17.64 D + 29 = 0

(

)

∴ D = 17.64 ± 311 − 116 / 2 17.64 − 13.96 , positive sign will be unfeasible. 2 ∴ D = 1.84 m B = 30 − 2.23(1.84) = 25.9 m. =

S=

f

5

3

3340 Q

1

6

=

1 3340 (40)

1

6

=

1 1 = 3340 × 1.849 6176

Example 11.3:  Design an irrigation channel by Kennedy’s theory to carry 1 20 cumecs. Take n = 0.02550 in Manning’s equation, m = 1, B/D = 5, side slope :1 . 2 Solution: D  A =  B +  D, 2 

Vo = 0.55 D.64

here B = 5D,

= 0.55(2)0.64

∴ A = (5D + 0.5D) D

= 0.85 m / s.

∴A = 5.5 D 2

Q = AV ∴ 20 = (5.5D 2 ) (0.55 D0.64 ) = (3)D 2.64 1

hence,

 20  2.64 D=  = 2m  3 

B = 5D = 10m A = 5.5D 2 = 22, and P = B + 5D = 14.47 A 22 R= = = 1.52 P 14.47 1 2 1 V = R 3S 2 n

Distribution System  253

0.85 =

2 1 1  0.021  (1.52) 3 S 2 ∴ S =   0.025  1.32 

2

= l in 3951 B =10 m, D = 2 m Example 11.4:  Design a channel based on Lacey’s theory for a design discharge of 12 cumecs. Take silt factor = 1.0 Solution:

∴ D = 1.28 m. ∴ B = 16.45 − 2.23(1.28) = 13.6 m. S=

f

5

3

3340 Q

1

6

=

1 1 3340(12) 16

= 1 in 5053

11.6 WATERLOGGING Waterlogging is due to rise of ground water table beyond permissible limits. With increase in irrigation potential, this problem gets developed. Land subjected to water-logging does not give expected yield. About 5 million hectares (3.3 %) of culturable area in India suffer of this problem. A land is said to be waterlogged when soil strata within the root zone of plants gets saturated, this causes insufficient air circulation and plant’s growth get affected adversely. Impermeable soil stratum has higher capillary rise and so gets waterlogged more often than permeable soil stratum. If ground water table is within 1.5 m from ground surface, waterlogging may result. Water table depth of 0.9 m to 1.2 m is considered very dangerous for plant growth and needs immediate remedial measures for control of waterlogging. Presence of alkali salts further worsens the situation.

11.6.1  Causes of Waterlogging Natural balance between inflow and outflow get greatly disturbed due to excessive irrigation facilities introduced in that area. Percolation of surface water on fields for irrigation and also seepage from canal system add to ground water reservoir resulting in rise of water table. Effective methods of cultivation and irrigation practices and blockage of natural drainage of surrounding ground add to intensify the problem of waterlogging.

11.6.2  Effects of Waterlogging (i) Waterlogging generally creates conditions of poor aeration in root zone of soil and is responsible for accumulation of salts near ground surface. Uncontrolled weed growth and difficulties in cultivation process are

254  Irrigation Engineering and Hydraulic Structures

additional ill effects of water logging. Surrounding area may suffer of mosquito growth resulting in poor health condition of persons staying in such areas. Water logging prevents free circulation of air in the root zone and thereby affects chemical process and bacterial activities, which are essential for proper growth of plant, and hence crop yield is reduced considerably. Also preparation of land i.e., tillage etc in wet condition is difficult and hence expensive. Weed growth at the cost of plant growth may further decrease crop yield. (ii) Salt efflorescence: High water table may cause upward rise of capillary water and this water will get evaporated early as it is nearer to the surface. As a result, chemicals, particularly alkaline salts brought by water from deeper soil depths get accumulated in root zone of soil masses, making the soil infertile. Thus waterlogging first creates marshy land and then after if unattended results in baren land. Hence, waterlogging should be avoided at any cost.

11.6.3  Remedial Measures to Control Waterlogging Waterlogging can be prevented if: 1. Methods adopted for irrigation should supply only that much water, which is required for healthy plant growth. This is possible by adopting drip irrigation system rather than irrigation by furrow methods. 2. Intercepting drains must be provided to cause effective drainage of irrigated lands. 3. FSD of canals be kept as low as possible. 4. Obstruction in natural drainage of areas is removed. 5. Running canals by rotation and adopting varabandhi will minimize wastage of water and hence reduce unwanted percolation in irrigable areas. 6. Rotation of crops will also generate fertility of soils. 7. Drains should be provided along canal lengths with high embankments so that they collect seeping water and prevent its entry into the ground so that undue rise in ground water table may not take place. 8. For every crop there is optimum water requirement for maximum yield, and farmers should be made aware of this fact and advised to practice irrigation accordingly. 9. Above all good management of irrigation water is surest means of controlling waterlogging, for, prevention is always better than cure. 10. If canal lining is carried out for entire canal system of the project, seepage losses will be minimised and water logging can be avoided.

Distribution System  255

11. Closed drain system provided below ground water table such as to see that water table does not rise above permissible limits is effective remedial measures for waterlogging but it is very costly.

11.6.4 Design of Closed Drain System for Effective Drainage of Waterlogged Areas Q If Q D = total discharge per unit length of drain then D is the discharge that enters 2 the drain from either side. (see Fig. 11.6). L  −x QD  2 ∴ Qy =   2  L   2  QD = (L − 2x) 2L dy Q y = Ky according to Darcy 's Law dx

Fig. 11.6:  Drainage system to reduce waterlogging Above equation for QD shows that: 1. Spacing (L) of closed drains is independent of drain size. 2. Increase in drain diameter will make only a little increase in drain discharge. 3. Spacing of drains (L) is usually 1.5 m to 4.5 m c/c and drain diameter is between 20 cm to 30 cm.

11.7  CANAL LINING Canals for irrigation purposes have been kept unlined since the inception of irrigation projects in India. This is mainly due to exorbitant cost towards lining, for in any irrigative project total length of canal system is quite large and proportion of project cost consumed by canal system is also as large as 60% to 70% of the total cost of project.

256  Irrigation Engineering and Hydraulic Structures

Considering advantage of unlined canals, low initial cost is the only factor in its favour, but disadvantages are many: 1. In unlined canals, non-scouring and non-silting velocity is required to be maintained very low, around 0.6 m/s to 0.8 m/s, and hence cross-section area of such canals are quite large. 2. B/D ratio is kept high so that bed width B is around ten times depth of flow, D. This helps in keeping FSD low, but because of large bed width land acquisition required is large and is responsible for heavy seepage losses. 3. Weed growth on unlined canal is very heavy and retards its flow and calls for repeated maintenance works. 4. Bank may get breached due to erosion or due to burrowing animals or due to both. As against these disadvantages of unlined canals, lined canals have got many advantages: 1. It reduces seepage losses and hence reduces chances of waterlogging. 2. Since losses are minimized, more water is available for irrigation purposes. 3. It is quite safe against breaching and weed growth is prevented to a minimum. 4. Hence, maintenance cost is minimum. 5. For well irrigation, lined canals are provided to reduce cost of pumping of water. 6. Flow velocity permissible is around 1.5 m/s to 2.5 m/s and hence cross-sectional area required is low.

11.7.1  Economics of Canal Lining 1. Annual Benefits: Let C1 be water charge for irrigation purposes per cumec of water used. If m cumces of water is saved due to lining of canals, then benefit due to lining is mC1 Rs. Lining will also reduce maintenance cost. From previous records, if C2 = annual cost of maintenance of unlined channels, then saving achieved due to lining is nC2, where n = percentage reduction in maintenance cost on account of lining. Hence, total annual benefits = (mC1 + nC2) Rs. 2. Annual Costs: If capital expenditure required on lining is C Rs. and lining life is Y years, annual depreciation will be C Y Rs. If i is the rate of interest i on capital investment, then annual interest charges = C . 100 This interest will have impact of decreasing value of capital due to depreciation and hence considering average rate of interest,

( )

Distribution System  257

Annual interest charges =

C i    2  100 

Hence total annual investment for lining =

C C i  +   y 2  100 

∴ Benifit cost Ratio = =

Annual Benefits Annual Costs

mC1 + nC2 1 i  C +  y 200  

≥ 1. For project feasibility, benefit cost ratio should be greater than one, or at least one. In addition to above direct benefits of savings of water and reduction in maintenance cost, there will be intangible benefits such as prevention of water logging, reduced costs of drainage of lands, reduced costs of breaching work and prevention of public health hazards such as malaria. Hence, canal lining is economical.

11.7.2  Types of Lining Lining work can be of following different types: 1. Concrete Lining: This is the best type of canal lining. It can last around 30 years, and hence though initial cost is high, it is cheaper in long run. Canal banks slopes are kept self supporting i.e., 1.5 H: 1 V so that lining work is not required to bear earth pressure. Thickness may vary from 7.5 cm to 12 cm, but in India minimum thickness of 10 cm is generally adopted. Concrete linings are laid without formwork and so workability of concrete should be quite good. Reinforcement provided is upto 0.4% of area in longitudinal direction and upto 0.2% of area in transverse direction. Reinforcement in lining prevents cracking in lining and provides a tie between adjacent section of lining. Lining may suffer of shrinkage cracks but this can be overcome by proper construction joint. Cracks can be sealed with asphaltic compounds. Present practice is not to provide reinforcement in lining but well constructed concrete only will do the job. However, for large canals lining needs reinforcement. 2. Shotcrete Lining: Shotcrete lining is constructed by applying cement mortar pneumatically to the canal surfaces. Thickness is around 5 cm and this type of lining is good for small canals. No subgrade is required and reinforcement is also not provided. Hence, it is the cheapest type. It is preferred on curved alignment of canals.

258  Irrigation Engineering and Hydraulic Structures

3. Brick Lining: Bricks are laid in layers of two with about 1.25 cm of 1:4 cement mortar sandwitched in between. This type has been used extensively in North India and is quite cheap compared to concrete lining. The Sarda Power canal has been lined with bricks. 4. Asphaltic Lining: Most commonly used asphaltic linings are: (i) Asphaltic concrete and (ii) Buried asphaltic membrane (i) Asphaltic concrete is a mix of asphalt cement, sand and gravel mixed at a temperature of 110° C and is placed with laying equipments. The lining is compacted with heavy iron plates while hot. Asphaltic concrete lining is smooth, flexible and erosion resistant. It is not suitable for hot climatic conditions. (ii) Buried asphaltic membrane can be of two types: (a) Hot sprayed asphaltic membrane, (b) Prefabricated asphaltic membrane (a) A hot sprayed asphaltic membrane is constructed by spraying hot asphalt on the subgrade to make a layer of 6 mm thick on which when it is cool a 30 cm layer of earth is provided and due to this, canal sections are over excavated. The lining is flexible and gets adapted to the subgrade surface. (b) Prefabricated asphaltic membrane is prepared by coating rolls of heavy paper with 5 mm layer of asphalts or 3 mm of glass fibrereinforced asphalt. These rolls are laid on subgrade and then covered with earth material. Maintenance cost of such type of lining is high. 5. Earth Linings: Different types of earth linings are being used for irrigation canals. Their initial cost is low but requires high maintenance expenditure. These earth linings can be of following types. (a) Stabilized earth lining (b) Compacted earth lining (c) Buried bentonite membrane (d) Soil cements linings. Use of earth linings is restricted to small irrigation canals of capacity lower than 10 cumecs and flow velocity permissible is only 1 m/s.

Distribution System  259

11.8  DESIGN OF LINED CANALS There are two types of sections, which are adopted for lined canals: (i) A triangular section with circular bottom (ii) A trapezoidal section with rounded corners. For canals having discharge less then 85 cumecs and velocity of flow limited to 1.8 m/s, triangular section with rounded bottom is adopted, whereas for canals having discharge greater then 85 cumecs trapezoidal section is adopted. IS 10430 of 1982 recommends only trapezoidal section with rounded corners for all discharges. For concrete lined canals limiting velocity of flow is recommended as 2.7 m/s and for bricks lined canals it is allowed upto 1.8 m/s. Value of Manning’s constant n, for lined canals may vary from 0.015 to 0.03. (i) Trapezoidal section with rounded corners: (see Fig. 11.7)

Q = AV

D = FSD

1 2 1 V by Manning = R 3 S 2 , n = 0.015 to 0.03. n  Vn  R= 1  S 2

3

2

A Q and A = P V A = central rectangular potion + triangular portion on each end R=

1 1  A = BD +  θD.D 2 + ( D cot θ D ) 2 2  2 ∴ A = BD + D 2θ + D 2 cot θ = BD + D 2 (θ + cot θ) and P = B + 2Dθ + 2D cot θ = B + 2D(θ + cot θ) F. S. L

D cot θ

90O

Dθ

θ

D

B

θ

90O

Dθ

Fig. 11.7:  Lined canal (trapezoidal section) Since A and P are known, B and D can be determined.

D cot θ

260  Irrigation Engineering and Hydraulic Structures

(ii) Triangular section with rounded bottom: (see Fig. 11.8) F. S. L. 0

D

D 90O

D

90O D cot 0

D0

Fig. 11.8:  Triangular section with rounded bottom For ∆lar section, A = D 2 (θ + cot θ) P = 2D(θ + cot θ) D = FSD R=

A D = P 2

q  SOLVED EXAMPLES Example 11.5:  Design trapezoidal shaped concrete lined canal to carry 100 cumecs of flow at a bed slope of 3 × 10-4. The side slopes are 1.5 : 1 and N = 0.015 with limiting value of flow velocity = 2 m/s. Solution:  Vn R = 1  2 S A=

   

3

3

2

=

Q 100 = = 50m 2 , A 2

  2  2 × 0.015  = 2.28m  1   3 × 10−4 2    A 50 = 22m. P= = R 2.28

(

)

For trapezoidal shaped section, θ = cot −1 (1.5) = 0.56 radians A = BD + D 2 θ + D 2 cot θ 50 = BD + D 2 (0.56) + D 2 (1.5) = BD + D 2 (0.56 + 1.5) ∴ 50 = BD + 2D 2 P = B + 2D (θ + cot θ) 22 = B + 2D (0.56 + 1.5) = B + 4D ∴ B = 22 − 4D Substituting value of B in eqn (i), 50 = (22 − 4D)D + 2D 2 = 22D − 4D 2 + 2D 2 = 22D − 2D 2

A = BD + D θ + D cot θ 50 = BD + D 2 (0.56) + D 2 (1.5) = BD + D 2 (0.56 + 1.5) ∴ 50 = BD + 2D 2 P = B + 2D (θ + cot θ) 22 = B + 2D (0.56 + 1.5) = B + 4D ∴ B = 22 − 4D Substituting value of B in eqn (i),

Distribution System  261

50 = (22 − 4D)D + 2D 2 = 22D − 4D 2 + 2D 2 = 22D − 2D 2

…(1)

…(2)

D 2 − 11D + 25 = 0 11 ± 121 − 100 2 11 − 4.58 = , neglect positive sign = 3.2m 2 ∴ B = 22 − 4D = 22 − 4(3.2) = 9.2m ∴D =



Example 11.6:  Design concrete lined canal having triangular shape section with rounded bottom to carry a discharge of 20 cumecs. Take side slope as 1.25 : 1 and Manning’s N = .016 and S = I in 8100 Solution: A D R= = , P 2 cot θ = 1.25 , (since side slope given is 1.25:1) A = D 2 (θ + cot θ) = D 2 (.675 + 1.25 ) A = 1.925D 2 V= =

…(1)

1 2 3 12 R S n 1 D   0.016  2 

2

3

 1     8100 

1

2

D 3 2.29 Q 20 Also V = = A 1.925D 2 ∴V =

2

D 3 20 ∴ = 2.29 1.925 D 2 ∴D ∴D

8

3

= 23.8 = ( 23.8 )

3

8

= 3.2m.

2

…(2) …(3)

…(4)

262  Irrigation Engineering and Hydraulic Structures

EXERCISES 1. What are different types of canal sections? Give sketches and explain their importance. 2. State factors that influence determination of canal alignment. What is importance of fall in canal alignment? 3. State types and purposes of canal linings. 4. Compare and contrast Kennedy’s and Lacey’s silt theories. Which is more practicable? Why? 5. Design an irrigation canal to carry 18 cumecs. Use Kennedy’s theory. Take Manning’s n = 0.0225, CVR = 1, B / D = 5, side slopes ½ H to IV. [Ans: B = 9.8 m, D = 1.96 m, So in 2570.] 6. What is balancing depth of cutting? 7. How curves are provided in canal alignment ? 8. Design a channel using Lacey’s theory to carry a discharge of 60 cumecs. Take silt factor = 1.0 and side slope ½ : 1. [Ans: B = 32 m, y = 2.16 m, S is l in 6608]  

 

12 Canal Structures 12.1 DISTRIBUTION AND MEASUREMENT STRUCTURES FOR CANAL FLOWS The flow of a main canal bifurcating into a branch canal with the rest flowing downstream is controlled with the help of a cross regulator across the parent canal and a head regulator across the branch canal. At times, the flow of a canal divides into two or three smaller branch canals without any regulating structure by designing the entrance of the canals in such a way that the flow enters each branch canal proportionate to its size. Again, from a canal, outlet structures may take out water for delivery to the field channel or water courses belonging to cultivators. These outlet works, of course, are generally not provided on the main canal and branches, but are installed in the smaller distributaries. Apart from these, there could be a need to measure the flow in a canal section and different structures have been tried, mostly based on the formation of a hydraulic jump and calibrating the discharge with the depths of flow. Typical structures of these kinds are graphically represented in Fig. 12.1 and this lesson deals with each type in detail.

12.2  CANAL OUTLETS Cultivators get water for irrigation purposes from distributary channel to their watercourse. An outlet is a structure through which water is admitted from distributary into a water course or a field channel. It also acts as a water measuring device. Cultivators prefer an outlet which supplies water at constant discharge, whereas, the canal management would prefer outlets that can supply water at variable discharge so that tail end of distributary is neither flooded nor dried. Usual types of canal outlets are: 1. Non-modular Outlets or simple pipe outlet 2. Semi-modular Outlets and 3. Modular Outlets

© The Author(s) 2023 S. K. Ukarande, Irrigation Engineering and Hydraulic Structures, https://doi.org/10.1007/978-3-031-33552-5_12

263

264  Irrigation Engineering and Hydraulic Structures

1. Non-modular Outlets: These types of outlets supply water on the basis of difference of water levels in distributary and the field channel. Thus, discharge through non-modular outlets varies over a wide range with variation in the water levels of distributary and watercourse. A shutter at its upstream end controls such an outlet. It cannot ensure equitable distribution of water to cultivators. Loss of head in non-modular outlets is less than that in modular outlets. 2. Semi-modular Outlets: Discharge through these outlets depend only on water level of distributary and is independent of water level in field channel subject to a minimum working head required for its working. A semi-module is thus more suitable for equitable distribution of water to cultivators. It has however a greater loss of head than in other types. 3. Modular Outlets: These outlets work independently of water levels in distributary and that in water course within a reasonable working limit. They may have or may not have moving parts. They are known as rigid modules if they do not have moving parts. These outlets are preferred on a branch canal for supplying water to distributary.

12.2.1  Parameters Controlling Behaviour of Outlets 1. Flexibility, F: It is a ratio of rate of change of discharge at outlet to rate of change of discharge at distributary. dQ 0 / Q π Q0 = discharge through outlet ∴F = Q = discharge through distributary dQ / Q Q = C1D n

and Q0 = C2 H m

Here D = depth of flow in distributary And H = head on the outlet. C1 , C2 are constants and m and n are indices ∴

dQ dD =n Q D

and

dQ0 dH =m Q0 H

m D dH × × n H dD For semi-modular outlet, dH = dD

∴F =

 m  D ∴F =      n   H

When a certain change in the distributary discharge causes a proportionate change in outlet discharge, the outlet is said to be proportional and is semimodular type. ∴ For Proportinal semi-module, F = 1 H m ∴ = D n

Canal Structures  265

H is a measure of location of outlet and is termed setting. D If F = 1, outlets are classified as proportional

The ratio

F > 1, hyper proportional and F < 1, sub-proportional 2. Sensivity, S.: The Sensivity, S, is defined as ratio of rate of change of discharge at outlet to rate of charge of depth of flow in the distributary. dQ0 / Q0 dD / D dQ0 / Q0 F= = dQ / Q

∴S =

dQ0 / Q0 n (dD / D)

1 S n ∴ S = nF =

Flexibility of a modular outlet is zero and hence its sensivity is also zero.

12.2.2  Non-Modular Pipe Outlets 1.5:1

1.5:1

F. S. L.

F. S. L. of Water course

(H)

Distributing Q

D

Pipe of dia. d

Q0

L

Fig. 12.1:  Non-modular pipe outlet

Let, H = difference in water level of distributary and watercourse then

H = 0.5 H=

V2 2g

V 2 f LV 2 V 2 + + 2g 2gd 2g fL   0.5 + d + 1  

d   ∴ V = 2gH   1.5d + fL 

1

2

Q0 = discharge of pipe outlet d π    = AV =  d 2  2gH   4   1.5d + fL  ∴Q0 = CA 2gH so that

1

2

H=

0.5 + + 1 2g  d  1

d   2 ∴ Vand = Hydraulic 2gH  Structures 266  Irrigation Engineering  1.5d + fL  Q0 = discharge of pipe outlet d π    = AV =  d 2  2gH   4   1.5d + fL 

1

2

∴Q0 = CA 2gH so that d   C=  1.5d + fL 

1

2

q  SOLVED EXAMPLE Example 12.1:  A non-modular pipe outlet is to carry a discharge of 25 lit / sec, with a minimum head of 0.3 m from the distributary. Pipe length is 12 m, and f = 0.01. Find diameter of the pipe. Solution: d   Qo = CA 2gH =   1.5d + fL   25 = cumec. 1000 ∴

1

2

π 2  d  2g (0.3) 4 

  d =  1.5d + 0.01(12) π 2g(0.3)   0.025 × 4

7.2 × 10−3 × (1.5d + 0.12 )

1

2

1

2

(d 2 )

= d 2.5

By trial and error method, If d = 0.12, 3.9 × 10−3 ≠ 4.9 × 10−3 Try, d = 0.10, 3.7 × 10−3 ≈ 3.2 × 10−3 Try d = 0.11, 3.8 × 10−3 ≈ 4 × 10−3 , which can be accepted Hence d = 0.11m =11cm

12.2.3  Kennedy’s Gauge Outlet This is a semi-modular type outlet, developed by Kennedy in 1906. It consists of a bell mouth orifice discharging into a long expanding delivery pipe 3 m long with a vertical air column above the throat. This arrangement makes the discharge through the outlet independent of water level in field channel. The cultivators can easily temper this outlet by blocking the air vent pipe to increase discharge through the outlet. It is costly and hence not much in use.

Canal Structures  267 Air vent pipe h1 Supports

H0 Distributary Orifice Throat

Cast iron pipe Field channel

Fig. 12.2:  Kennedy’s gauge outlet (semi-module)

12.2.4  Modular Outlet There are following two types of rigid module outlets: 1. Gibb’s Rigid Module: Inlet pipe receives water from distributary and through a rising spiral pipe it is supplied to eddy chamber. This arrangement provides free vortex motion, so that V × R = Constant Hence in eddy chamber, outer curve of spiral pipe will have less velocity and inner one has larger velocity, keeping, V × R = Constant. Baffles are provided in eddy chambers to cut down excessive energy of flow and thus ensure constant discharge into field channel. Thus, discharge through outlet is kept independent of levels of distributory and field channel. Hence, it is a modular type of outlet and since there are no moving parts it is known as Gibb’s rigid module. But it is very costly and not much in use. (See Fig. 12.3). Baffle

Eddy chamber

Distributary

o

Rising spiral pipe (180 )

Field channel

Inlet pipe

Baffles Distributary

Rising spiral o pipe (180 ) Field channel

Fig. 12.3:  Gibb’s module

268  Irrigation Engineering and Hydraulic Structures

2. Khanna’s Rigid Module: It is similar to orifice semi-module with an additional provision of sloping shoots fixed in arched roof covering. These shoots cause back flow and thereby keep the outlet discharge constant. If shoots are blocked, outlet will continue to function as a semi module. (See Fig. 12.4). 1.5:1

1.5:1

(H)

F. S. L.

F. S. L. of

water course

Distributing QD

pipe of dia. d

Q0

L Fig. 12.4:  Khanna’s rigid module

q  SOLVED EXAMPLE Example 12.2:  A semi-modular pipe outlet of 20 cm diameter, receives water from distributary having bed level 200.3 and FSL 201.5 m. Maximum water level in field channel is 201.15. Set outlet for maximum discharge and find its value. Take C = 0.62 Indicate type of setting. Solution: d H = ( 201.5 − 201.15 ) −   2 H is measured from centre line of the pipe outlet 0.20 2 = 0.25m

H

= 0.35 −

Q

= CA 2gH π  = 0.62  d 2  2g ( 0.25) ) 4  π  = 0.62  (0.20) 2  2g (0.25) 4 

= 0.043m3 / s m D F = Flexibility = n H 1 5 m = and n = ; D = 201.5 − 200.3 = 1.2 2 3 m ∴ = 0.3 and H = 0.25 n 1.2 ∴ F = 0.3 × = 1.44 0.250 Since F is greater than one, it is hyper proportional module

m D n H 1 5 Canal Structures  269 m = and n = ; D = 201.5 − 200.3 = 1.2 2 3 m ∴ = 0.3 and H = 0.25 n 1.2 ∴ F = 0.3 × = 1.44 0.250 Since F is greater than one, it is hyper proportional module F = Flexibility =

setting =

H 0.25 = = 0.208m D 1.2

Falls: Whenever natural ground slope is steeper than canal bed slope, the difference between these two slopes is adjusted by a structure known as fall (Fig. 12.5). The necessity of a fall is due to the fact that ground slope usually exceeds designed bed slope of a canal. In its head reach, irrigation canal is usually in cutting and soon after it meets with a condition when it has to be in filling. Irrigation canal in filling is very costly both in initial cost as well as its maintenances. Hence, structures known as falls are provided to keep the balance between cutting and filling.

Fig. 12.5:  Fall

Types of Falls (i) Cylindrical fall or well type fall (ii) Sarda type fall or vertical drop type fall and (iii) Glacis fall 1. Cylindrical or Well Type Fall:  It consists of an inlet well with a pipe at its bottom, carrying water to downstream well to cistern. The d/s well is required if fall is greater then 1.8 m or discharge is higher than 0.29 cumecs (See Fig. 12.6).

270  Irrigation Engineering and Hydraulic Structures

Fig. 12.6:  Cylindrical or well type fall

2. Sarda Type Falls:  A raised crest fall with a vertical impact was first introduced on Sarda Canal system in UP. In that area a thin layer of sandy clay was found above a stratum of pure sand. If canal bed is to be cut deep and touches sandy stratum seepage losses would be high. Hence to keep depth of cutting low, large number of falls were required. Vertical drop or Sarda type being economical is adopted. Here water falls into a cistern and then enters d/s canal (See Fig. 12.7). 3. Glacis Fall:  This differs from above as in the case, that after raised crest, weir wall has a gentle slope of 2:1, known as glacis, such that hydraulic jump occurs and energy dissipation due to hydraulic jump takes place.

Fig. 12.7:  Vertical drop fall or sarda type fall

12.3  CROSS-DRAINAGE WORKS (C.D. WORKS) Canals in head reach are normally canals in cutting and are aligned as contour canals. Here they are required to cross a number of natural drainages or tributaries, and structures are required to be provided at such crossing are known as cross-drainage works. After initial reach of cutting, canal soon takes up as ridge canal and is in filling. Here it is not required to cross many natural drainages but distributaries from this canal may again be parallel to contour and hence distributaries may have

Canal Structures  271

to cross tributaries. Thus, number of cross-drainage works are provided on branch and distributary canals. These cross drainage works are usually of three types: (i) Super passage (ii) Aqueducts (iii) Level crossing If a cross-drainage work is such that natural drainage is taken over the canal then it is known as superpassage and if the natural drainage has to pass below the canal then it is known as aqueduct. If natural drainage and canal are crossing each other at same level, it is known as level crossing. Superpassages are normally found in initial reach of canal head works whereas aqueducts are found after the main canal bifurcates into branch and distributary canal system. Level crossing are rarely provided. Among all the three types, the most commonly found crossdrainage work is aqueduct.

12.3.1 Aqueducts Aqueducts are of two different types: 1. Free Aqueducts: Free aqueducts have no uplift pressure on canal bed from below due to HFL of natural drainage as HFL is considerably below CBL, but if HFL is slightly above CBL, water of natural drainage will exert an uplift pressure on canal bed and in that case aqueduct is known as siphon aqueducts. Natural drainage floor is depressed below canal crossing so as to minimize uplift pressure on canal bed and water way does not get obstructed due to siltation of natural floor. (See Fig. 12.8)

Fig. 12.8:  Siphon aqueduct

Aqueducts and siphon aqueducts are further classified into following three types: Type I: Here no fluming in canal bed width is carried out and canal banks are carried forward as they are. In this type width of CD work is large but expenditure on canal wings and bank connections is saved. It is suitable for small streams of low width and low flood discharge.

272  Irrigation Engineering and Hydraulic Structures

Type II: This type of aqueduct is similar to type I with a provision of retaining walls to terminate outer slopes of canal banks there by reducing the length of the CD work. This type is preferred for streams of medium magnitude. Type III: In this type, canal banks are discontinued over its crossing length and canal water is carried in rectangular RCC trough. The canal banks are connected to respective trough walls on u/s and d/s side of the natural drainage by means of wing walls. The canal bed width is also reduced and canal section from trapezoidal type is transformed to rectangular trough. Transition lengths with expansion and contraction are provided before and after the crossing as shown in Fig. 12.9.

12.3.2  Super Passage CD work: where canal water passes below a natural drainage or stream the CD work is known as super passage. Here FSL of canal is much lower than stream bed level and if canal water is carried through a pipe it is known as irrigation siphon. Super passage with canal FSL free of the bottom of stream are just aqueducts with the only difference that here stream is above and canal is below. Irrigation siphons are just pipe culverts carrying canal water below natural streams. Thus, irrigation siphon is just opposite of siphon aqueduct.

Fig. 12.9:  Aqueduct type III (Note Le=1.5 Lc)

Canal Structures  273

q  SOLVED EXAMPLES Example 12.3:  Design a suitable C-D works for following data: Canal

Natural Drainage (ND)

Discharge = 18 cumecs

Q = 200 cumecs

Full supply level = 213.5

HFL = 210

CBL = 212.0 and CB width = 10 m

ND bed level = 207.5

Side slope = 1.5 H : 1.0 V

G.L. = 212.5

Solution: Since CBL is above HFL by 2m and ND is large, type III aqueduct is suitable: (See Fig. 12.10) Design: Step I: Design of ND waterway P = 4.75 Q = 4.75 200 = 67.2 m Using 8 bays of 7m width, Lnet = 56 m. Taking number of piers as 7, provide pier width = 1.5 m ND bed width = 56 + (7 × 1.5) = 66.5 m Provide end bays with 1.75 m width, so that ND bed width = 66.5 + (2 × 1.75)

= 70 m , which is greater than P = 67.2 m , hence OK.

Step II: Design of Canal water way : Canal bed width = 10 m = Bo Let it be flumed to 5 m = Bf Taking splay of 2 : 1, contraction length Lc will be,  10 − 5  Lc = 2   = 5m and expansion length Le will be,  2  Le = 1.5Lc = 5 × 1.5 = 7.5m

Lc and Le represent length of transition curves.

274  Irrigation Engineering and Hydraulic Structures

Step III: Design of Transition curves: (Fig. 12.10) (ii) Contraction length, Lc = 5m Using Mitra’s Hyperbolic Transition formula, Bx

=

Bo Bf Lc Lc Bo − x ( Bo − Bf )

=

(10) (5) (5) 250 = 5(10) − x(10 − 5) 50 − 5x

Assume x values between 0 to 5 and worked out Bx values are given in the table below: x Bx

0 5

1 5.5

2 6.25

3 7.14

4 8.33

5 10

Fig. 12.10:  Channel transitions

(iii) Expansion length, Le = 1.5 Lc = 7.5m. Bx = =

Bo Bf Le Le Bo − x (Bo − Bf )

(10) (5) (7.5) 375 = 7.5(10) − x (10 − 5) 75 − x(5)

Assume × values between 0 to 7.5 and worked out Bx values are given in the table below: x Bx

0 5

1 5.36

2 5.77

3 6.25

4 6.81

5 7.5

6 8.33

Step IV: Design of trough: Assuming constant depth of flow u/s to d/s, d = depth in trough = FSD of canal   = 213.5 – 212 = 1.5 m V=

Q 18 18 = = = 2.4 m / s. < 2.5 m/s A 1.5 × 5 7.5

7 9.375

7.5 10

Canal Structures  275

Example 12.4:  Design a suitable type of CD work for following data: Canal Q = 20 cumecs B = 20 m FSD = 1.5 m CBL = 206 Side slope 1½ : 1

ND HFQ = 250 cumecs HFL = 207 m Bed level = 204 m GL = 206.5 m

Solution: Since HFL is higher than CBL, siphon aqueduct is to be provided (see Fig. 12.8, 12.9 and 12.11) Design Step I: Waterway P = 4.75 Q = 4.75 250 = 75m. Provide nine clear span of 7m each and pier width = 1.5 m ND bed width = 8 × 1.5 + 9 × 7 = 75 m, which is equal to P, hence Let V = 2 m/s of ND then waterway area =

250 = 125m 2 2

125 = 1.98m ≅ 2m 9×7 HFL = Bed Level + depth of flow = 204 + 2 = 206 which is less than given HFL of 207 and hence

depth of flow =



Step II: Canal Waterway Bed width, B of canal = 20m Let it be flumed to 10m, following a fluming ratio of 1:2 Provide a splay of 2:1 for contraction length 20 − 10 Lc = × 2 = 10 unit m. 2 and a splay of 3:1 for expansion, 20 − 10 Le × 3 = 15m. = 1.5Lc , OK 2 Bo Bf Lc Bx c = Lc Bo − x ( Bo − Bf ) =

20 (10) (10) 10(20) − x(20 − 10)

=

2000 200 − 10x

=

200 20 − x

Bx e =

20 (10) (15) 15(20) − x(20 − 10)

=

20 (10) (10) 10(20) − x(20 − 10) 2000 200 − 10x

276  Irrigation Engineering and Hydraulic Structures =

= Bx e =

200 20 − x 20 (10) (15) 15(20) − x(20 − 10)

=

3000 300 − 10x

=

300 20 − x

Assume × and work out Bxc, Bxe and enter in the following table: x

Bx c

Bx e

0

10

10

2

11.1

10.7

4

12.5

11.5

6

14.2

12.5

8

16.6

13.6

10

20

15

12

–

16.6

14

–

18.75

15

–

20

Step III: Design of CBL at Various sections (4 – 4, 3 – 3, 2 – 2 and 1 – 1 as shown in Fig. 12.11)

Fig. 12.11:  Channel transitions

Canal Structures  277

At section 4 – 4, When canal returns to normal section, i.e., beyond 4 – 4, A = (B + 1.5D)D

  = (20 + 1.5 × 1.5)1.5



  = 33.375 m² V4 =

Q 20 = = 0.6 m / s A 33.375

V42 (0.6)2 = = 0.0183 m. 2g 2g RL of canal bed at 4 – 4, given as 206 RL of water surface at 4 – 4 = 206 + 1.5 = 207.5 RL of TEL at 4 – 4 = 207.5 + 0.018 = 207.518 At section 3 – 3 Assume constant depth of 1.5 m throughout Area of trough = 1.5 × 10 = 15 m² V3 =

20 = 1.33m / s 15

V32 = 0.09 m. 2g  V 2 − V42  Loss of head in expansion = 0.3  3  2g   

 1.76 − 0.36  = 0.3   = 0.07 × 0.3 = 0.021 2g  

RL of TEL at 3 – 3 = RL of TEL at 4 – 4 + Loss in expansion

= 207.518 + 0.021 = 207.521

RL of water surface at 3 – 3 = 207.521 – 0.09 = 207.431 RL of bed at 3 – 3 = 207.431 – 1.5 = 205.931 At section 2 – 2, Loss between 2 – 2 and 3 – 3 as per Manning’s HL =

n 2 V32 L R4

=

3

(0.016) 2 (1.33) 2 (75) (1.15) 4

3

2.56 × 10−4 × 1.76 × 75 1.2 = 0.0281 =

2

2

n VEngineering 278  Irrigation and Hydraulic Structures 3L HL =

R4

      =

3

(0.016) 2 (1.33) 2 (75)

   {Where, Manning’s n = 0.016 A 15 R= = −4 P 10 + 2(1.5) 2.56 × 10 × 1.76 × 75 } = 15 1.2 = = 1.15 = 0.0281 13 (1.15) 4

3

RL of TEL at 2 – 2, = RL of TEL at 3 – 3 + Loss between 2 – 2 and 3 – 3.

= 207.521 + 0.028 = 207.549

RL of water surface at 2 – 2 = 207.549 – 0.09

= 207.459

RL of bed at 2 – 2 = 207.459 – 1.500

= 205.959

At section 1 – 1 : Loss due to contraction  V 2 − V12  = 0.2  2  2g    2  1.33 − 0.62  = 0.2    2g    1.4  = 0.2    2g 

:( Where, V1 = V4 = 0.6 and V2 = V3 = 1.33)

= 0.0143 RL of TEL at 1 – 1 = 207.549 + 0.0143 = 207.563 RL of water surface at 1 – 1 = 207.563 – 0.018 = 207.545 RL of bed at 1 – 1 = 207.545 – 1.5 = 206.045 Step IV: Uplift Pressure on slab of trough : Assume trough slab thickness = 0.3 m RL of bottom of trough slab = CBL – slab thickness

= 2.6 – 0.3 = 205.7

Assume afflux = 0.33 m, and loss of head on account of entry of flood water below trough V2 (2) 2 = 0.5 = 0.5 = 0.1 2g 2g ∴ Uplift on trough slab = HFL + afflux – entry loss – RL of bottom of slab trough

= 207 + 0.33 – 0.1 – 205.7



= 1.53 m of water



= 1.53 t/m²

Canal Structures  279

Downward load of concrete slab = 0.3 × 2.4 = 0.72 t / m3 (sp. gr. of concrete of slab = 2.4) The balance of uplift pressure = 1.53 – 0.72 = 0.81 t / m², has to be resisted by reinforcement to be provided at top of the slab thickness. If there is no flood water in ND, then the slab of trough has to be designed for water load in canal + trough dead weight, and this design is to be checked for HFL condition.

EXERCISES 1. Distinguish between (a) Aqueduct and Superpassage (b) Irrigation siphon and syphoned aqueduct. (c) Modular outlets and non-modular outlet (d) Fall and Weir (e) Setting and Flexibility. 2. Design a suitable C-D work for following data :



(Ans: Syphon Aqueduct)

Canal

Drainage

Q = 350 cumecs

HFQ = 4500 cumecs

B = 28 m

Bed Level 195 m.

FSD 6 m

HFL 198.5 m

FSL 204 m

Natural GL 198 m

CBL 198

 

 

13 Sediment Transport in Alluvial Canals 13.1  TRANSPORT OF SEDIMENT IN ALLUVIAL CANALS Sediment transport is a natural process coupled with flow of water either in rivers or in canals. Transporting power of flowing water is always beyond prediction: it can transport anything, right from small sediment particles to boulders, trees, dead animal bodies and debris. The ill effect of this transportation is scouring of bed and banks of canals and rivers and erosion of surfaces of hydraulic structures such as spillways and bridge piers. A systematic study of transport of sediment in canal is therefore of utmost importance. Many researchers have contributed in this field such as Kennedy, Lacey, Ven Te Chow, Einstein, Shield, Peter – Meyer, Du Bois etc. Early works are due to Kennedy and Lacey as has been pointed out in Chapter 11 on Distribution System. In this section, summary of further work in this field is detailed What makes sediment move? The basic mechanism responsible for sediment motion is drag force exerted by fluid flow on individual grains. The cumulative effect of all such drag forces is retarding shear stress exerted by contact surfaces on the flow. Determination of this bed shear is of prime importance in the problem of sediment transport in alluvial channels.

13.2  THRESHOLD MOVEMENT Threshold condition of movement of sediment particles along the bed cannot be defined with absolute precision. The approach is based on dimensional analysis and experimented studies. This has been carried out by (i) Shield, (ii) Einstein, (iii) Meyer – Peter and (iv) Du – Bois which are given below: Basic parameters involved in sediment transport are: (i) τ0 = Shear Stress at Bed (ii) ρs = Mass Density of Sediment Grain (iii) ρf = Mass Density of Fluid Particle (iv) d = Grain Diameter © The Author(s) 2023 S. K. Ukarande, Irrigation Engineering and Hydraulic Structures, https://doi.org/10.1007/978-3-031-33552-5_13

280

Sediment Transport in Alluvial Canals  281

(v)   g = Acceleration due to Gravity (vi) µ = Fluid Viscosity Six variables as listed above contain three fundamental dimensions in them i.e., all L, M and T are present among these six variables. Hence, according to Buckingham’s π theorem, (6-3) = 3 pie terms can be formed out of these six variables. They are: τ0 , ρf gd



d τ0 ρf

ρs , ρf

µ

τ0 = shear velocity, ρf

Let v* =

= Velocity which would give rise to shear stress, τ0 From Darcy – Weishbach equation for pipe flow, it can be shown that v* f = v 8

where v is flow velocity and f is friction factor.

The first π1 term is ∴π1 = π2 =

τ0 τ but 0 = (v* ) 2 ρf gd ρf

(v* ) 2 is a kind of Froude Number, F* for sediment flow gd

ρs = specific gravity of sediment = Ss ρf

For channel flow , τ0 = γ RS0 , where R=

A = hydraulic mean depth P

S0 is the bed slope and γ is weight density of fluid. The third pie term, π3 =

d τ0 ρf µ

 1  Multiply and divide the 3rd pie term by   so that  ρf  d π3 =

τ0 ρf

=

dv* = R* ν

 µ     ρf  here ν (nu) = Kinematic vis cos ity µ = ρf

282  Irrigation Engineering and Hydraulic Structures

∴ 3rd pie term, π3 represents a kind of Reynold number for sediment flow in channels. Thus the phenomenon of sediment transport in alluvial channel is governed by F*, R* and Ss.

13.3  SHIELD’S ENTRAINMENT FUNCTION Shield studied experimentally the relationship between these π terms and developed a function known as Shield’s entrainment function, given by Fs =

τ0 γ d (Ss − 1)

Shield has obtained the relation of his function Fs with R*, and is given in following graph of Fs v/s R* on log-log scale given in Fig. 13.1. 1.0

Threshold movement

FS (Iog)

Laminar condition

Turbulent condition 0.1

FS = 0.056 at R+= 400

0.01 1.0

10

100

400 1000

R+ (Iog scale)

Fig. 13.1:  Plot of Fs v/s R* Shield obtained value of Fs = 0.056 at R* = 400 from his above curve given in Fig. 13.1 and accordingly Fs =

τ0 = 0.056 γ d(Ss − 1)

Ss = 2.56 which is usual value of specific gravity for sediment of size not greater than 6 mm. Now τ0 is also equal to γ RS0

γ RS0 = 0.056 γ d(2.65 − 1) ∴ d = 11RSo ∴ Fs =

(13.1)

Above equation gives minimum size of bed material d, which will remain at rest in channel of hydraulic radius R and bed slope S0. Here d is D-75 size of sediment grain.

Sediment Transport in Alluvial Canals  283

q  SOLVED EXAMPLE Example 13.1:  A wide rectangular canal is to carry 2.7 cumecs through a track of coarse alluvial of size 6 mm, and canal bed slope is 0.001. Assume that banks are protected by grass, find minimum width of the canal (Take n = 0.015). Solution:

d = 11RS0 ∴R =

d 0.006 = = 0.54 m = y = depth of flow (11)So 11(0.001)

since given canal is wide rectangular section and so R = Y, depth of flow. By Manning’s Formula, 1 2 3 12 R So n 2 1 1 (0.54) 3 (0.001) 2 = 0.015 = 1.36 m / s

v=

q = discharge per unit width of Canal = vy = 1.36 (0.54) = 0.73 m3 /s Q 2.7 ∴ canal bed width B = = = 3.7 m q 0.73

13.4  BED LOAD FORMULA BY EINSTEIN Einstein assumed that every particle after it is River dislocated on the bed, travels a certain minimum distance which is proportional to its grain size width before coming to rest. Refer Fig. 13.2 below. Unity [Where L = River length as shown in the above L figure (between two given section)] Fig. 13.2  Grains dislocation L ×1 Number of grains dislocated = K1d 2 and volume of each grain K2d3 here, K1, K2 = coefficients giving c/s area and volume of grain size d, p = probability of dislocation of grain qs = volume of sediment transported per unit width per sec. K2 L p d3 = K1

284  Irrigation Engineering and Hydraulic Structures

Einstein assumed value of p as proportional to shield’s Function, Fs divided by K2,   F  K  τ0 p = K s  =    K 2  K 2  γ d(Ss − 1)  Here K = proportionality constant. Now, qs =

K2 L p d, K1

K   K  K Fs d  = LFs d ∴ qs =  2 L    K1   K 2  K1 τ0 K L d, here τ0 = γ RSo = K1 γ d(Ss − 1) ∴ qs = =

γ RSo K L d K1 γ d(Ss − 1) RSo K L K1 ( Ss − 1)

But velocity of flow in channel as per Chezy is given by v = C RSo , where C = Chezy 's constant ∴ RSo =



∴ qs =

v2

(13.2)

C2

K L K1

v2

C2 ( Ss − 1)

From equation (13.2) it can be deduced that (i) Sediment carrying capacity of river or canal increases with increase in flow velocity. (ii) Floods carry more sediment since velocity of flow is high during floods. (iii) Branching of canal from main canal reduces discharge of main canal more rapidly than designed value. Einstein also gave alternative equation for bed load calculation:

 

é ù3 to ê ú qs = 40W0 d ê g d(S - 1) ú s ë û 2 36u 2 36 u 2 Here Wo = G gd(Ss - 1) and G = + 3 3 gd (Ss - 1) gd3 (Ss - 1)

(13.2a)

Sediment Transport in Alluvial Canals  285

here υ is Kinematic viscosity of water = 1 × 10−6 m 2 / s  n'  and τo = γ R So Here R = R   n   '

'

3

2

13.5  BED LOAD FORMULA BY MEYER–PETER According to experimental findings of Meyer and Peter, τo (bed shear) is reduced due to ripple formation at bed. 3    n'  2  ∴τeffective = τo   − τc   n      1

d 6 24 n = 0.02 if Q > 11cumecs

here, n ' =

= 0.0225 if Q < 11 cumecs τc

= 0.07 d

d = grain size in mm Let τ0 = RS0 ,     qs = volume of sediment transported in kg per unit weight per hour, (13.3) then G s = qs γ Ss 3    n'  2  = 4700 τo   − τc   n     

3

2

kg / unit weight / hr

13.6  DU-BOIS BED LOAD FORMULA According to Du-Bois, volume of bed load transported is proportional to the excess of tractive force over critical shear and is given by q s = K [ τo ( τo − τc ) ] (13.4) = Volume of sediment transported per unit width of canal per sec. 0.173 K = Proportionality constant = 3 (d) 4 d = Grain size in mm τo = γ RSo = tractive force τc = Critical bed shear = 0.07d, here d = d50 or d60 size in mm.

286  Irrigation Engineering and Hydraulic Structures

q  SOLVED EXAMPLES Example 13.2:  Design a wide rectangle channel carrying 30 cumecs with a bed load concentration of 50 ppm by weight. The bed material consists of grains of average size 2 mm. Use Lacey’s regime equations and Meyer–Peter formula. Take n = 0.02. Solution:

 50  Bed load transported =  6 × 1000  × 30 = 1.5 kg/sec.  10  P by Lacey 's equation = 4.75 Q = 4.75 30 = 4.75 × 5.47 = 26 m.

Given Channel is wide rectangular section ∴ R = y = depth of flow Let depth of flow be assumed as 1m A ∴R = = 1 ∴ A = P = 26 m 2 = By P ∴ B = 26 m, since y = 1 ∴ G s = wt.of bed load transport per unit width per sec 2.5 = 0.096 Kg/m/se c = 0.096 × 3600 26 = 346 kg/m/hr =

By Meyer–Peter formula, 3

3  2   n'  2  G s = 4700  τo   − τc  kg/m/hr    n    1

 2 6   0.35 1000  n' = = = = 0.0147 24 24 24 1 (d) 6

n ' 0.0147 = = 0.739 n 0.02 τc = 0.07d = 0.07 × 2 = 0.14

∴

3

3  2 1 ∴ G s = 4700  γRSo ( 0.739 ) 2 − 0.14  × kg / m / sec 3600  

Sediment Transport in Alluvial Canals  287 3

3

(0.63) − 0.14 ) 2 = 1.3 (1000RS = 1.3 (1000RS o (0.63) −o 0.14 ) 2 3

3

= 1.3 ( 630RS − 0.14 ) 2 = 1.3 ( 630RS o − 0.14 )o2

3 3 0.096 =×1.3(630 1 × So2− 0.14) 2 ∴ 0.096 ∴ = 1.3(630 1 × So −×0.14)

∴ 630So = ( 0.0738 )2 3 + 0.14 = 0.314 ∴ So =

0.314 1 = = 1 in 2006 630  630     0.314 

Now, A = 26 y, and according to Manning’s formula, 2 1

2

1

Q 1 1  1 2 (y) 3  = V = R 3 So2 =  A n 0.02  2006  2

30 y3 , = 26y 0.895 here, by putting y =1, we get LHS = 1.15 and RHS = 1.12 ∴ y = 1 m, assumed as above is correct ∴ B = 26m, y = 1 m and S0 is 1in 2001 Example 13.3:  Find rate of bed load transport in wide rectangular channel for the following data: Depth of flow, y = 3 m Velocity of flow, v = 1 m/sec. Bed slope So = 1 in 5000 Average size of grain, d = 1.14 mm. Kinematic Viscosity of water ν = 1 × 10−4 Stokes Make use of Einstein formula. Solution:  Using Manning’s equation, 2

1

1 (R) 3 (S) 2 n Now, for wide rectangular channel, R = y = 3m and v = 1m / sec.

V=

288  Irrigation Engineering and Hydraulic Structures 1

2 1  1 2 ∴1 = ( 3) 3   n  5000  2.08 ∴n = = 0.029 70.7 1

1

1 1  1.14  6 0.32 n′ = (d) 6 = = 0.13   = 24 24  1000  24 3

 n′  2 ∴ R′ = R    n 3

 0.013  2 =3  = 3 × 0.316 = 0.95m.  0.029 

Wo = G=

fall velocity = G gd(Ss − 1) 2 36υ2 36υ2 + 3 − 3 gd (Ss − 1) gd3 (Ss − 1)

36υ2 gd3 (Ss − 1)

=

(

36 × 1 × 10−6 3

)

2

 1.14  9.81  ( 2.65 − 1)  1000 

, (here υ in m 2 /s =1 × 10−6 )

= 1.5 × 10−3 2 + 1.5 × 10−3 − 1.5 × 10−3 3 = 0.817 − 0.038 = 0.78

∴G =

(

)

∴ Wo = 0.78 9.81 × 1.14 × 10−3 (1.65 ) = 0.78 × 0.135 = 0.105 m/s According to Einstein,  γ R 'So  qs = 40 × w o d    γ d(Ss − 1) 

3

1   0.95 ×   −3 5000 = 40 × 0.105(1.14 × 10 )   −3  1.14 × 10 (1.65) )   

(

3

)

∴ qs = 4.23 × 0.00114 × 1 × 10 −3 = 4.82 × 10 −6 m3 / s / m

Sediment Transport in Alluvial Canals  289

Bed load in terms of N/s/m = qs γ Ss = 4.82 × 10−6 × 1000 × 2.65 kg / s / m = 12.77 × 10−3 kg / s / m = 12.77 × 10−2 N / s / m

13.7  SUSPENDED LOAD CONCENTRATION Suspended load concentration, C, can be calculated from Hunter Rouse’s equation, given below. Limitation for using of this equation is that suspended load concentration at a point y (measured from channel bed) can be calculated provided suspended load concentration, Ca, at a point “a” above channel bed is known.

 W0   

C  a(D − y)   KV*  (13.5) = Ca  y(D − a)  here D = Full supply depth of flow in channel W0 = Fall velocity of sediment grain in still water, which is usually 0.02 m/s K = Karman Constant = 0.4 V* = Shear velocity =

τo / ρ

q  SOLVED EXAMPLES Example 13.4:  In a wide rectangular channel, suspended load sample taken at 0.5 m height from bed had a value of 1000 ppm of sediment by weight. The full supply depth of channel is 5 m and its bed slope is 1 in 4000. Fall velocity of sediment grain is 2 cm/sec. Find suspended load concentration at: (i) Mid depth and (ii) At surface. Solution:  Using Rouse equation,  W0   

C  a(D − y)   KV*  = Ca  y(D − a)  here, a = 0.5m., D = 5 m., y = 2.5 m, Ca =1000 ppm, Wo = 0.02 m/s, K = 0.4  0.02 

 0.5(5 − 2.5)   0.4V8  C ∴ = 1000  2.5(5 − 0.5)  V* =

τo = ρ

γ RSo ρ g RSo = ρ ρ

= g RSo = 9.81 × 5 ×

R = D, Since wide rect. channel = 5m, 1 = 0.11 m/s 4000

 0.02 

 0.5(5 − 2.5)   0.4V8  C ∴ = 1000  2.5(5 − 0.5)  τ o Hydraulic γ RSoStructures ρ g RSo 290  Irrigation Engineering and * V =

ρ

=

ρ

=

ρ

= g RSo

R = D, Since wide rect. channel = 5m,

= 9.81 × 5 ×

∴

Wo KV

*

=

1 = 0.11 m/s 4000

0.02 = 0.45 0.4(0.11)

 0.5 × 2.5  C = (1000 ppm)    2.5 × 4.5  = (1000 ppm) [ 0.37 ]

0.45

= 370 ppm at mid depth C at surface = 0 since D − y = 0 as y = D at surface. Example 13.5:  Determine the bed load transport in a wide alluvial stream for following condition: Depth of flow = 2.5 m, velocity of flow = 1.5 m/s Average size of sediment = 5.0 mm Specific gravity of sediment = 2.65 Solution: Using Meyer - Peter equation,

3

3  2   n'  2  1 qs = 4700  τ o   − τ c  × 3600   n   

kg / m / sec

1

 5 6   0.4135 1000  n' = = = = 0.0172 24 24 24 1 d6

n = Manning 's Constant = 0.0225 n ' 0.0172 = = 0.764 n 0.0225 τo = γ RSo

wide rectangular channel, R = D = 2.5m.

= 1000(2.5) (8 × 10−4 ),

Where So = free surface slope = bed slope for uniform flow = 8 ×10−4

Sediment Transport in Alluvial Canals  291

= 2 Kgf / m 2 τc = 0.07d = 0.07 × 5 = 0.35 3

3 2 4700   2(0.764) 2 − 0.35 ∴ qs = 3600     3

= 1.3[ 2 × 0.667 − 0.35] 2 =1.3(0.985)1.5 =1.3 × 0.9775 =1.27 Kgf / m / sec

13.8 TRACTIVE FORCE METHOD OF DESIGN OF STABLE CHANNEL Critical tractive force method of design does not allow the shear stress anywhere in the channel to a value less than the critical shear stress of bed material.

A M

P T

Consider a particle P on side of a channel making an angle θ with the horizontal, see Fig. 13.3 (A) and (B). M

N

C/S of channel B

Fig. 13.3 (A):

Top of side bank of channel N P FD drag force on particle

90° WS cos T

Side Slope of channel

WS sin T WS T

Flow Direction

L

T, Sloping side makes

Flow Channel Bed R

Fig. 13.3 (B):  Isometric view of side MNLR Let d = size of particle so that its c/s area, A = K1 d2 Where K1 = coefficient. Ws = submerged weight of the particle = K2 ( γ s − γ ) d3 , where K2 = coefficient.

292  Irrigation Engineering and Hydraulic Structures

Ws can be resolved into two components, (Ws) Sin θ along side slope and (Ws) Cos θ normal to side slope. Due to flow a shear stress τw acts on the particle P. The drag force FD due to flow, acting on the particle P is given by FD = τw (A) , here A = K1d2 This drag force is parallel to side slope and is in direction of flow. Thus particle P is influenced by drag force FD, and displacing force (Ws) Sin θ. The stabilizing force is, ( Ws ) Cosθ tan ϕ, where ϕ is angle of response of material, thus ( Ws ) Cosθ tan ϕ = FD2 + Ws2sin 2θ Drag Force FD = τ w A and

(13.6) (13.7)

Ws tan ϕ = τ b A

(13.8)

Substituting values of FD and Ws tan ϕ in equation (13.6), we get τb A Cosθ = τ2w A 2 + Ws2 Sin 2 θ 2

τ A ( τw A Cosθ )2 = τ2w A 2 +  b  sin 2 θ  tan ϕ  2

τ  sin 2 θ ∴  w  = Cos 2 θ − tan 2 ϕ  τb 





1

 tan 2 θ  2 τ ∴ w = Cos θ 1 − 2  τb  tan ϕ 

(13.9)

Equation 13.9 can be re-arranged as under 2

 τw  sin 2 θ 2   = cos θ − tan 2 ϕ  τb  = cos 2 θ + sin 2 θ − sin 2 θ −

sin 2 θ 2

tan ϕ

by adding and subtracting sin 2 θ

 1  =1 − sin 2 θ 1 + , since sin 2 θ + cos 2 θ = 1 2    tan ϕ   1 + tan 2 ϕ  = 1 − sin 2 θ   tan 2 ϕ    = 1 − sin 2 θ 2

sec2 ϕ tan 2 ϕ

τ  sin 2 θ ∴ w  = 1 − 2 sin ϕ  τb 

 1  2 2 =1 − sin 2 θ 1 +  , since sin θ + cos θ = 1 2   tan ϕ   1 + tan 2 ϕ  Sediment Transport in Alluvial Canals  293 = 1 − sin 2 θ   tan 2 ϕ    = 1 − sin 2 θ

sec2 ϕ tan 2 ϕ

2

τ  sin 2 θ ∴ w  = 1 − 2 (13.10) sin ϕ  τb 



∴

τw sin 2 θ = 1− 2 τb sin ϕ = K3

Thus, shear stress due to flow of water on particle P on side slope is always less then shear stress at bed τb if θ is less than ϕ . Now bed shear τb = γ RSo and hence τw Can be worked out τw = K 3 τb here K 3 = 1 −

τ sin 2 θ or w = 1 − 2 τb sin ϕ

sin 2 θ sin 2 ϕ



(for θ < ϕ)

(13.11)

Example 13.6:  Design a stable non-erodible wide rectangular channel to carry a discharge of 12 m3/s. Channel bed consists of gravel of size 10 mm and its bed slope is 1 in 1250, and side slope 2 H to 1 V, Take angle of repose of bed material as 32°. Solution:

1 Side slope 2 H to 1 V means tan θ = = 0.5 2 ∴θ = 26.56°

,

∴ sin θ = 0.447 and cos θ = 0.894 ϕ = 32o ∴ sin ϕ = 0.529 τ c = 0.9(d mm )

= 0.9 × 10 = 9

N / m2

τ b = 0.9(τ c ) = 8.1 N / m 2 τ b = γ RSo , here R = yo = depth of flow as channel is wide rectangular = γ yo so

294  Irrigation Engineering and Hydraulic Structures

      ∴ yo =

τB 8.1 = = 1.04 m. (13.12) γ so 10000(0.0008)

K3 = 1 −

Sin 2θ Sin 2 ϕ

= 1−

(0.447)2 (0.529) 2

= 1 − 0.714 = 0.533       ∴ τ w = K 3 τ b = 0.533 (8.1) = 4.36 N / m 2 = 0.75 γ yo so ∴ yo =



4.36 = 0.743 0.75 × 10000 × 0.0008

(13.13)

From (13.12) and (13.13), adopt lower value of yo = 0.743 m. A=By0 = 0.743B,

(B is bed width and A is c/s area) 2 1

Q 12 1 V= = = yo3 So2 A 0.743B n 2

1

1  1 2 (0.743) 3  =  0.022  1250  = 1.05 12 ∴B = = 15.3 m. 0.743 × 1.05 Bed width B = 15.3 m., depth of flow = 0.743m.

EXERCISES 1. In a wide rectangular stream, suspended load sample taken at 0.3m height from bed had a concentration of 900 ppm. The full supply depth is 5 m and So is 1 in 4000. If fall velocity of bed material is 2 cm/s, find suspended load concentration at mid depth. (Ans: 270 ppm.) 2. A canal is carrying a discharge of 600 cumecs. Bed slope is 1 in 1000. Bed material size is 5 cm, and its angle of repose 37o. Design canal section if it is unlined and trapezoidal with side slope 3 :1. Manning’s n = 0.025. (Ans: Y = 2.8m, B = 7m.)

Sediment Transport in Alluvial Canals  295

3. A wide irrigation channel has hydraulic mean depth of 3m. and bed slope of 1 in 6250. The bed material is 0.3 mm size, and has specific gravity of 2.65. Take Manning’s n = 0.02. Find rate of bed load transported by the channel in N/s/m width of channel. (Ans: 0.7 N/s/m) 4. Prove that for stable non-erodible channel, τw Sin 2 θ , = 1− τb Sin 2 ϕ

if θ < ϕ.

where τw is shear stress due to influence of flow on sloping side and τb is shear stress at bed, θ is the angle the sloping side makes with the horizontal and ϕ is angle of repose.  

 

14 River Training Works and Flood Control 14.1 GENERAL Every monsoon we find certain rivers swell due to heavy flood flow and cause damage on either side of banks and sometimes we find that river changes its course also. Such situations demand for river training works and arrangement for flood control. River training works in the form of guide banks, embankments, levees etc. keep the river confined to its usual track and flood control measures provide warning time so that loss of life and material can be prevented or minimized.

14.2  CLASSIFICATION OF RIVERS Right from its mouth to tail, a river passes through four different types of stages: (1) Rocky, (2) Boulder, (3) Trough and (4) Deltaic. Along its flow path it may undergo meandering. Rivers flowing through alluvial plains can therefore be classified as: 1. Meandering type rivers 2. Aggrading type rivers 3. Degrading type rivers

ss

in

g

Apex

Cr o

Meandering width Mw Meandering length, ML

Fig. 14.1:  Meandering type river

© The Author(s) 2023 S. K. Ukarande, Irrigation Engineering and Hydraulic Structures, https://doi.org/10.1007/978-3-031-33552-5_14

296

River Training Works and Flood Control  297

1. Meandering Type River: A part of meandering type river is shown in Fig. 14.1. Depth of cross – section over apex part is greater than that over crossing part. Meandering width MW, meandering Length ML and river width vary as per square root of river discharge Q according to Inglis. 2. Aggrading Type Rivers: Aggrading type rivers have sediment load in excess of their carrying capacity and hence this extra sediment load settles along its bed and consequently changes the bed slope of river. 3. Degrading Type Rivers: Degrading type rivers have less sediment load than its carrying capacity and hence they suffer from erosion of their bed due to high velocity of flow. Thus, meandering, aggrading and degrading type of rivers undergo changes in their bed slope and this may force them to change their course. Rivers Brahmaputra, Koshi and Sone in north-east region of our country exhibit this type of behavior and are well known for their flood havocs.

14.3  RIVER TRAINING WORKS Main objective of river training works is to keep the river within its channel section and make it flow along its established alignment, however few special types of objectives are listed below: 1. Safe disposal of flood flow without over flooding the banks. 2. Out flanking of bridge or aqueduct, which may result as a heavy flood in river, is to be prevented by river training works. 3. Protection of banks from erosion and efficient transport of sediment load are also considered important objectives of river training works. To meet with above requirement of river training, any of the following works can be adopted: (i) Marginal bunds or levees (ii) Guide banks or guide bunds (iii) Groynes or spurs (iv) Cut-offs (v) Pitched islands (vi) Bandalling

14.3.1  Marginal Bunds or Levees In order to keep flood waters within the river section, embankments parallel to river bank and of nominal height should be provided. These are known as marginal bunds or levees. They are usually designed as per design of embankments for canals in filling. If the sloping side of the embankment is likely to come in contact with floodwater, it may be provided with pitching. The effect of these bunds is to raise the water level during flood and to dispose off flood water quickly.

298  Irrigation Engineering and Hydraulic Structures

14.3.2  Guide Banks or Guide Bunds Guide banks are constructed such that they provide a straight approach towards a work such as bridges, weirs, culvert or C–D works across the river. They also prevent river from changing its course. Bell was first to construct these types of bunds in India and hence they are also known as Bell’s bunds. Fig. 14.2 (A) shows half sectional plan and bund along one of the banks of the river and Fig. 14.2 (B) shows half section of the bund. River flow L/2

0.125 L

d/s nose of guide bank

0.25 L

1.25 L A 45o

A 120o

u/s nose of guide bank

Fig. 14.2 (a):  Half plan of guide bank

Fig. 14.2 (b):  Half section of guide bank along AA

Length of waterway, L, is worked out from Lacey’s formula Q = 4.75 Q ∴ L = Centre to center distance between bunds along the river banks = 4.75

Q.

here Q = flood discharge in the river. Length of bunds can be kept as 1.25L and length of u/s nose can be kept as 0.25 L whereas that of d/s nose is kept as 0.125L. Side slope of the bund can be kept as 2:1 and u/s side be pitched.

River Training Works and Flood Control  299

14.3.3  Groynes or Spurs The main job of river training is done by groynes or spurs. They are also a type of bund but are provided normal or inclined to the path of river flow and extend from river bank into the river flow path. They protect the river bank from erosion by keeping the flow away from it. According to the angle they make with the river bank, they are classified as 1. Attracting Groyne or Spur,q > 90° 2. Repelling Groyne or Spur,q < 90° 3. Deflecting Groyne or Spur,q = 90° They are shown in Fig. 14.3 (a), (b) and (c).

>90°

Fig. 14.3 (a):  Attracting Groyne

90°

Fig. 14.3 (c):  Deflecting Groyne

1. Attracting Groyne: A groyne making θ > 90° i.e., pointing d/s tends to attract the river flow towards the bank at its back. They are not useful for bank protecting but can safe guard a part length of the bank on its d/s side. 2. Repelling Groyne:  A groyne that makes θ < 90° , i.e., pointing u/s tends to repel the river flow away from the bank on which it is provided and so are known as repelling groyne. Head of repelling groyne needs strong protection since it is subject to direct attack of whirling motion. 3. Deflecting Groyne: A groyne making θ = 90° with the bank tends only to deflect the flow without repelling it and gives local protection to the adjoining bank length on either side.

14.3.4 Cut-Off A river flowing along a meandering path may sometimes abandons particular bend and follows a straight path between successive bends, creating a path known as cut-off. Following this natural process artificial cut-offs can also be provided for river training works in which diversion of flow from curved path is required. This is shown in Fig. 14.4.

300  Irrigation Engineering and Hydraulic Structures Cut-off

Fig. 14.4:  Cut-off

14.3.5  Pitched Islands A pitched island is created artificially in river bed to cause redistribution of velocity and hence tractive force. This arrangement will attract the river currents towards itself to reduce erosion of river banks. A deep scour occurs around the pitched island causing heavy concentration of flow around it. Stone pitching along the sloping side of island and launching apron along river bed from its toe are provided to protect against maximum scour condition. (Fig. 14.5)

Fig. 14.5:  Pitched island

River Training Works and Flood Control  301

14.3.6 Bundalling It is also one of the methods of river training used for navigation purposes. A low level flow in a river can be confined in a single track for maintaining required depth for navigation purpose. After recession of floods and when water level starts falling, bandalling work is carried out. This work consists of framework of bamboos driven into river bed at 0.6 m center to center. On these bamboos horizontal ties and strutting consisting of bamboos are provided. This arrangement is carried out at an angle of 30° to 40° with downstream for creating a narrow channel of required depth, see Fig. 14.6. Bandals check the flow and cause deposition of sand behind themselves. Thus, a channel confined between bandals is formed with sand banks on either side and entire river flow is diverted through this channel. In case a pitched island is formed in river bed, it will be an additional help to bandalling work as shown in Fig. 14.6. Vertical bamboos provided @ 6m c/c and of 3m to 6m length

3m to 6m

Horizontal supports

Flow

Strutting

Bundalling in plan

Flow

Pitched island

Bundalling in plan

Fig. 14.6:  Plan and sectional view of bundalling

14.4  FLOOD CONTROL If heavy storms occur in catchment area either consecutively or at very close interval the magnitude of run-off will be large and its concentration in river flow will result in a flood. Floods are result of natural processes and cannot neither be avoided nor controlled, but their effect can be suitably modified by following flood routing techniques. A flood wave entering a reservoir can be routed by ISD Pul’s method as explained in Chapter 5. Flood peak can be reduced by working out attenuation and time lag of occurrence of peak values by using reservoir routing techniques. By

302  Irrigation Engineering and Hydraulic Structures

adjusting out, flow from reservoir, flood wave monitoring is possible. Similarly river flow capacity can also be utilized to allow safe disposal of flood water by following channel routing techniques. Thus, flood routing process for reservoirs and channels can help to modify the effect of floods i.e. to reduce peak values and to provide time lag for occurrence of peak values. This will enable the authorities to give a warning time to d/s occupants and loss of life and property can be prevented or minimized.

14.4.1  Flood Absorption Capacity of Reservoirs Reservoirs have large capacity to withhold flood water over spillway crest level and spillway gate height and this capacity can be utilized to hold flood water in reservoir by following reservoir routing technique. This enables the authority to have long interval of time to give warning and to carry out temporary evacuation works on d/s side of reservoir. Similarly channel improvements also help in increasing flood carrying capacity of rivers. Construction of levees along river banks also helps in increasing flood carrying capacity of rivers.

14.4.2  Flood Forecasting Methods Flood forecasting can be carried out by making use of any of following methods : (i) Application of unit hydrograph technique for quick estimation of maximum flood that may result due to storms of high intensities in the catchment areas. (ii) Flood frequency analysis by statistical methods. This method will help to forecast highest value of flood for a given recurrence interval. This method however requires at least 35 years past record of rain storms and their resulting floods. (iii) Empirical flood formulae such as Dicken’s, River’s, Inglis etc. can be used to predict maximum flood for a given catchment area.

14.4.3  Channel Improvement A river can carry higher discharge at lower levels or stages by improving its hydraulic conditions: (i) Increase in water way area by increasing its cross-sectional area. (ii) Increase in velocity of flow by deepening, straightening and shortening the channel length by cut-offs, removing barriers in channel section and lining the channel sides to improve its coefficient of rugosity. It may be noted that channel improvement program if not carried out all along its length will merely shift the point of location of flooding of waters in channel section from one reach to another reach of the channel. Hence, implementation of channel improvement should be carried out for the entire channel reach.

River Training Works and Flood Control  303

EXERCISES 1. (a) Define (i) Meandering type of rivers. (ii) Aggrading type of rivers. (iii) Degrading type of rivers. (b) Explain how guide banks are useful in controlling the river track. 2. Explain the following: (i) Groynes of various types (ii) Pitched Island (iii) Bundalling 3. What is flood control? What are the methods of flood forecasting? 4. Explain how flood routing is useful for flood control purposes.  

 

15 River Flow Measurement 15.1 NEED FOR DISCHARGE MEASUREMENT IN RIVER SECTION Flood hydrographs are essential for flood flow calculations. These graphs can be prepared by collecting values of flood flow in river at a chosen section against specified time intervals. A river may have deep or shallow sections and they can be wide or narrow. Discharge measurement is done by dividing the river section into small segments and measuring the velocity into small segments by current meters or float or salt dilution technique. Use of current meter is more reliable and can be easily handled.

15.2  CURRENT METERS It works on the principle of rotation of its runner being proportional to the rate of flow. The runner may have vertical axis or horizontal axis and is that case known as: (i) Vertical Axis Meter (ii) Horizontal Axis Meter.

15.2.1  Vertical Axis Meter A series of conical shape cups are mounted around a vertical axis, see Fig 15.1. The cups rotate in horizontal plane and a cam shaft attached to the vertical axis – spindle records generated signals proportional to the revolutions of the cup assembly. The measurement of velocity is possible in the range of 0.15 to 4.0 m/ sec with 1% accuracy. If river flow has appreciable vertical components of velocity, then this type of meter is not suitable.

© The Author(s) 2023 S. K. Ukarande, Irrigation Engineering and Hydraulic Structures, https://doi.org/10.1007/978-3-031-33552-5_15

304

River Flow Measurement  305

Hoist & Electrical Connectors

Vertical Axis Conical Cup

Stabilising Unit

Sounding Weight

Fig. 15.1:  Vertical axis meter

15.2.2  Horizontal Axis Meter A propeller mounted on horizontal axis gets rotated due to flow and its rotational number per sec are converted into velocity of flow by electrical instruments. This type is stable against vertical shocks of flow and measures velocity over range of 0.15 to 4.0 m/sec with 1% accuracy. This type is widely used.

Hoisting Propeller

Sounding Weight Stabilizing Unit

Fig. 15.2:  Horizontal axis meter

A current meter records rotation of the cup unit or propeller and these rotations are converted into corresponding velocity of flow by the equation V = aN + b Where N = revolution per second and a and b are meter constants usually a = 0.65 and b = 0.03, V is velocity of flow in m/sec.

306  Irrigation Engineering and Hydraulic Structures

15.3 CROSS-SECTIONAL AREA-VELOCITY METHOD TO MEASURE DISCHARGE At a chosen section on a river length, its cross-section is drawn by taking readings of depths at regular interval of width along the section. The depths are measured by sounding rods or sounding weights. Its river section is having more depth, then eco-depths recording is used. The verticals representing river depths are drawn on the section and making use of current meter, velocity at either 0.6d, 0.8d, or 0.2d is measured. (Refer Fig. 15.3) If a plot of values of velocities along a vertical representing depths d of the section is drawn it will be observed that velocity is maximum not at free surface of flow but at 0.2d from free surface and average velocity for shallow depths occurs at 0.6d, if depths are lesser than average velocity VAV =

V0.2d + V0.8d 2

(i.e., velocity at 0.2b + velocity of 0.8d)/2 Gives average velocity for d more than shallow depth, shallow depth is usually less than 2.0 m. Vertical Free surface of river 0.2 d 0.6 d 0.8 d Depth (d)

River Bed.

Fig. 15.3:  Velocity at different depths­­

From a station A on left bank, river depths at regular width of ‘b’ are measured upto station B on right bank and cross-section is drawn as in Fig 15.4. If d ≤ 2m, reading taken at 0.6d for velocity measurement are if d > 2 m, readings are taken at 0.2d and at 0.8d, and average value of V is obtained. 1 1  b on left and area upto  b  on right is takes as 2 2  representative area for the depth at y and ∆Q is worked out as At a vertical y, area upto

+ V0.8d  V ∆Q = bd  0.2d  2  

River Flow Measurement  307 l tica Ver b

b

Y

b

V at 0.2 d

Left Bank A

B Right Bank

V at 0.6 d

V at 0.8 d

Fig. 15.4:  Area velocity method

All these ∆Q added will give Q at this section, and this method is known as midsection method. In mean sections method, consecutive depths of sections are chosen and velocity measurement at these section is carried out, then  d + d 2   V1 + V2  ∆Q =  1  b1 etc  2   2 

Where d1 = depth as section 1 and   v1 = velocity at section 1 d2 and v2 are the depths and velocity at section 2   b = width between section 1 and 2. This is illustrated in the following numerical problem:                               

q  SOLVED EXAMPLES Example 15.1:  A current meter was used to measure velocity at 0.6d of flow. Compute discharge if given calibration equation is

V = 0.32 N + 0.32



Where N = revolutions per second and



V = velocity in m/sec

Distance from bank

0

Depths

0 0.375 0.75 1.24 2.0 2.3

N rev/sec

0

1.5

3

4.5

6

9

12

18

15

22

23

24

1.9

1.7

1.8 1.55

1.3

0.8

0

0.33 0.54 0.79 1.1 1.16 1.01 0.91 0.95 0.76 0.71 0.58

0

Solution: V1 = 0.32 N + 0.32   = 0.32 m/sec at 0 from bank V2 = 0.32 (0.33) + 0.32   = 0.425 m/sec at 1.5 m from bank V3 = 0.32 (0.54) + 0.32   = 0.492 m/sec at 3 m from bank

20

308  Irrigation Engineering and Hydraulic Structures

Sample Calculations: ΔQ = by mean section method  d + d3   V2 + V3  = 2  (b)   (where b = 3 − 1.5 = 1.5m 2  2   



 0.375 + 0.75   0.425 + 0.492  =  (1.5)   2 2     = (0.562)(1.5)(0.4585)



∆Q = 0.386 m3/sec. Similarly all values of ΔQ are worked out and they are entered in tabular form given below (Table 15.1): Table 15.1  Discharge Calculation Distance from bank (m) depth d (m) N riv/sec 0 0 0 1.5 0.375 0.33 3.0 0.75 0.54 4.5 1.24 0.79 6.0 2.0 1.1 9.0 2.3 1.16 12.0 1.9 1.01 15.0 1.8 0.95 18.0 1.7 0.91 20.0 1.55 0.76 22.0 1.30 0.71 23.0 0.80 0.58 24.0 0.00 0.00

V = 0.32 N to 32 m/sec 0.32 0.425 0.492 0.57 0.67 0.69 0.64 0.62 0.61 0.56 0.54 0.50 0.32

bm

∆Q m3/sec

1.5 1.5 1.5 1.5 3.0 3.0 3.0 3.0 2.0 2.0 1.0 1.0

0.09 0.386 0.795 1.5 4.38 4.06 3.49 3.22 2.89 1.54 1.00 0.04 Q = ΣΔQ = 23.34 m3/sec

Example 15.2:  Velocity in m/sec at one meter interval starting from river bed are given below. Calculate Q per unit width of the stream, depth of flow at measuring section was 5.0 m. V (m/sec) d (m) from bottom (bed level)

Solution:

0 1

0.6 2

V0.2d + V0.8d 2 0.9 + 0.6 = = 0.75 m/ sec 2

Vmean =

0.8 3

0.9 4

0.9 5

River Flow Measurement  309

Q per meter width = (bd) Vmean = (1 × 5) (0.75) = 3.75 m3/sec/m Note: In this table depths are from bottom to top whereas in formula for Vmean, d is from top ∴ V0.8d = 0.9 and V0.2d = 0.6 Example 15.3:  Calculate stream flow using mid-section method from the following data : Current meter rating equation is V = 0.04 + 0.75 N, where N is revolution/sec Distance from bank

0.8

1.4

2.2

3.2

4.0

4.7

5.2

5.8

Depth

1.3

4.3

5.8

6.8

4.8

2.8

1.3

0

N

0.3

0.55 0.91

0.87 1.11

0.94 1.3

0.61 1.02

0.54 0.8

0.43

–

Note: Single reading of N is at 0.6 d double readings are at 0.8d, and 0.2d respectively. Solution: Table 15.2  Discharge Calculation ΔQ = b×d ×Vmean or

Distance from bank

b

d depth

N

V0.6d

V0.8 d

V0.2d

Vmean

1   bd Vmean  2

0

–

–

–

–

–

–

–

–

0.8

0.8

1.3

0.3

0.265

–

–

0.265

0.1378

1.4

0.6

4.3

0.55 .9

–

0.452 0.722 0.587

1.515

2.2

0.8

5.8

0.81 1.11

–

0.692 0.872 0.782

3.629

3.2

1.0

6.8

0.94 1.3

–

0.745 1.015 0.880

5.984

4.0

0.8

4.8

0.61 1.02

–

0.4975 0.805 0.650

2.500

4.7

0.7

2.8

0.54 0.8

–

5.2

0.5

1.3

0.43

0.362

5.8

0.6

0

–

–

0.445 – –

0.64

0.542

1.06

–

0.362

0.117

–

– ΣΔQ = 14.942 m3/sec

310  Irrigation Engineering and Hydraulic Structures

Sample Calculation: For 1st section, depth = 1.3, b = 0.8, it is Δlar ∴∆Q =

1 bdV 2

1 ∴∆Q = (0.8)(1.3)(0.265) = 0.1378. 2 where, V0.6d = 0.4 + 0.75(0.3) = 0.265 m/sec For 2nd section, d = 4.3, b = 0.6 0.8d

= 0.04 + 0.75(.55) = 0.452 m/sec

V0.2d = 0.04 + 0.75(0.91) = 0.722 m/sec ∴ Vmean = (0.452 + 0.722)/2 = 0.587 m/sec Similarly, calculate ∆Q and enter values in Table 15.2.

15.4  SLOPE AREA METHOD This is a hydraulic method in which manning’s formula for velocity is applied at ends of given reach of a river. The discharge is calculated by trial and error method as under:

(i) Assume V1 = V2, apply Bernoull’s theorem between sections (1) and (2): Z1 +

V12 V2 + d1 = Z2 + 2 + d 2 + h f 2g 2g hf = (Z1 + d1) – (Z2 + d2), as V1 = V2 (assume) = fall in water surface levels between section 1, 2. ∴

hf = slope of energy line = Sf L

Now K = conveyence factor =

∴

Q sf

=

1 AR 2/3 by definition n

h f Q2 = L k2

∴ Q can be calculated using above equation Now calculate V1 =

Q Q and V2 = A1 A2

River Flow Measurement  311

hf = Sf 2 1

V /2g V22/2g

d1

1 2 d2

So Z1

Z2

L

Fig. 15.5:  Slope area method

with these values of V1 and V2 find hf Repeat the procedure till value of hf is constant. This method is illustrated in following problem: Example 15.4:  A 10 m wide rectangular channel has depth of flow as 4.0 m and 3.8 m at two section 100 m apart. If drop in water surface elevation is 0.15m, and n = 0.025, find discharge passing through the channel by slope - area method. Solution: Section 1 Section 2 d1 = 4.0 m d2 = 3.8 m A1 = 10 × 4 = 40 m2      A2 = 3.8 × 10 = 38 m2 P1 = 10 + 2(4)                  P2 = 10 + 2(3.8) = 18 m      = 17.6 m A A 38 R1 = 1 R2 = 2 = P1 P2 126 =

40 18

= 2.22 K1 =

1 A1R12/3 n

= 2.15m 1 (38)(2.15) 2/3 0.25 = 2538

∴ K2 =

1 (40)(2.22) 2/3 0.025 = 2730

=

Average K = K1K 2 = 2632, ∴ Q = 2632 sf− hf = given value of fall in water surface between (1) and (2) = 0.15 m

312  Irrigation Engineering and Hydraulic Structures

h f 0.15 = = 0.0015 L 100



sf− =



∴ sf 0.038







∴Q = 2632 × 0.038 = 101.9  Q   2632 sf    V12  A1   A1 = = 2g 2g 2g  Q    2 V2  A 2  = 2g 2g

2

   2  = (2.548) = 0.33 2g

2

=

(2.682) 2 = 0.366 2 × 9.81

 V2 V2  ∴ h f = fall +  1 − 2   2g 2g   



= 0.15 + (0.33 − 0.366) = 0.15 − 0.036

hf = 0.11 2nd step: sf− =

(1)

h f 0.114 = = 0.00114, 2 100

∴ sf = 0.033 ∴ Q = 88.86  2632  × 0.033   40  = 2g 19.62

2

V12

=

4.935 = 0.251 19.62 2

 2622  × 0.033  v 2 2  38  = 0.266 = 2g 2g ∴ hf = fall + (0.251 – 0.266) = 0.15 – 0.015 = 0.135

(2)

River Flow Measurement  313

3rd step: sf− =

h f 0.135 = = 0.00135 L 100

sf− = 0.036 ∴ Q = 2632 × 0.036 = 96.7 m3/sec  96.7  v12  40  = 2g 19.62

2

= 0.297 Q

O

log scale (h – a)

Fig. 15.6:  Stage-discharge curve

 96.7  2   v2 38  = 2g 19.62

2

= 0.33 ∴ hf = 0.15 + (0.297 – 0.33)= 0.117 ∴ Q = 2632 sf− 1/2

 0.117  = 2632    100 

= 2632 × 0.0342 = 90.0 m3/sec ∴ Q = 90.0 m3/sec can be adopted as average fall value remains constant around 0.12

314  Irrigation Engineering and Hydraulic Structures

15.5  STAGE-DISCHARGE RELATIONSHIP The relationship between stream discharging ‘Q’ are gauge height ‘h’ can be expressed as Q = k(h – a)n Here, h = gauge reading when discharge = Q A = gauge reading when discharge is zero and k, n are constants to be determined for given stream. Taking log on both sides of the equation log Q = log K + n log (h – a) This is similar to y = mx + c, equation of straight line. Hence assume values of ‘a’ and plot log Q v/s log (h – a) if it is a straight line assumed value is OK or else change value of ‘a’ till you get a straight line. The slope of the line gives value for ‘n’ and value of Q for (h – a) = 1, gives K.

q  SOLVED EXAMPLES Example 15.5:  For various stages, discharges in a river section are given below, obtain an equation for stage discharge relationship and find value of Q for a stage of 5.0 m and 10 m. Stage (m)

1.8

2.0

2.3

2.9

3.7

4.5

5.5

6

7

8

Q

1.0

1.5

2.5

5.5

11.7

20.2

33

44

70

90

Solution: Q = k (h – a)n Note: ‘a’ is stage value for Q = 0; is stage value for gives Value of Q, K and n and 'a' are required to be determined. Assume ‘a’ and plot (h – a) on x-axis (log scale) and Q on Y axis (log scale). If this graph is a straight line, assumed value of ‘a’ is correct, if not change it till you get a straight line. To get ‘a’ value following method may be followed a=

h1h 3 − h 22 (h1 + h 3 ) − 2h 2

Choose h1, h2, h3 such that corresponding values of Q1, Q2, Q3 follow: Q 2 = Q1 Q3 Hence, from given data

Let h1 = 2 giving Q1 = 1.5 h2 = 2.9 giving Q2 = 5.5 h3 = 4.5 giving Q3 = 20.2

River Flow Measurement  315

\ Q 2 = Q1Q3 = (1.5)(20.2) = 5. 5 ∴a =

( )

h1h 3 − h 22

h1 + h 3 − 2h 2

=

2(4.5) − (2.9) 2 2 + 4.5 − 2(2.9)

=

9 − 8.41 0.59 = = 0.84 6.5 − 5.8 0.70

Assume ‘a’ = 0.9, and plot Q v/s(h – a) It gives a straight line, see fig. 15.7. In absence of a log-log paper, log values be taken and plotted: see Fig 15.7 For Fig. 15.7, n = 2.2 and k = 1.2, Q = 1.2(h – a)2.2 = 1.2(h – 0.9)2.2 is the equation for stage – discharge relationship. For h = 5m, Q = 26.7 m3/sec and for h = 10m, Q = 154. m3/sec. Example 15.6:  Develop an equation of the form Q = k(y – a)b from following stage discharge data : Stage y(m)

216.46

217.54

219.74

3

55

260

1260

Q (m /sec)

Solution: Here Q1 = 55, Q2 = 260, Q3 = 1260 ∴

Q1 55 = = 0.21 and Q 2 260 Q 2 260 = = 0.81 Q3 1260

i.e., Q 2 = Q1Q3 ∴a =

Now,

y1y3 − (y 2 )2 (y1 + y3 ) − 2y 2

=

(216.46)(219.74) − (217.54) 2 (216.46 + 219.74) − 2(217.54)

=

47565 − 47323 242 = = 201.6 436.2 − 435 1.2

Q 2 k(y 2 − a)b  y 2 − a  = =  Q3 k(y3 − a)b  y3 − a 

b

316  Irrigation Engineering and Hydraulic Structures

∴

260  217.54 − 201.6  =  1260  219.74 − 201.6 

b

b

 15.94  b ∴ 0.21 =   = (0.878) 18.14   ∴ log (0.210) = b log (0.878) ∴b = −

0.677 = 11.98 0.565

∴ b  12 Also,

K=

Q3 (y3 − a)

b

=

1260 12

(214.74 − 201.6)

=

1260 1.26 × 105

= 9.92 × 10–13 ∴ Q = 9.92 × 10–3 (y – 201.6)12  

 

References A.  STANDARD BOOKS AND TREATISES S. No. Title 1. Water Resources Engineering

2.

Ground Water Hydrology

3.

Irrigation Enginnering Vol I to Vol V

4.

Open Channel Hydraulics

5. 6. 7. 8. 9.

10. 11. 12. 13. 14. 15.

Author and Publication R. K. Linsley and J B Franzini Mc Graw Hill Book Co., New York, 1974 D. K. Todd John Wiley and Sons, New York S. Leliavsky Chapman and Hall, London

Ven Te Chow Mc Graw Hill Book Co., International Students Edition. Water Wealth of India Dr. K. L. Rao Longman Publication Engineering for Dams Coreager, Justin and Hinds Vol I, II and III John Wiley and Sons, New York,1945 Earth and Rock Dams Sherard, James and Others, John Wiley and Sons,USA,1963 Dams and Control Works USBR Publication 1954 Irrigation Vol. I to VII Khushlani K. B. and Manohar Khushlani Oxford and IBH Publishing Co., New York Hydrology H. M. Raghunath Wiley Eastern Ltd., New Delhi Engineering Hydrology K Subramanaya Tata – Mc Graw Hill Co., New Delhi. Irrigation Engineering G. L. Asawa Wiley Eastern Ltd., New Delhi, 1993 Irrigation Engineering and S. K. Garg Hydraulic Structures Khanna Publication, New Delhi Irrigation, water Resources and P. N. Modi Water Power Engineering Standard Book House, New Delhi Design of Small Dams USBR Publication 1960

© The Author(s) 2023 S. K. Ukarande, Irrigation Engineering and Hydraulic Structures, https://doi.org/10.1007/978-3-031-33552-5

317

318  Irrigation Engineering and Hydraulic Structures

S. No. Title 16. Text Book of Hydrology 17.

Water Power Engineering

18.

Embankment Dams

19.

Flow in Open Channels

20.

Hand book of Applied Hydrology Theory and Design of Irrigation Structures

21.

22. 23.

24.

Fundamentals of Irrigation Engineering INDIA 2020 A Vision for the New Millennium River Engineering

Author and Publication Jayarami Reddy Laxmi Publication Ltd., New Delhi Dandeker and Sharma Vikas Publication, Delhi Sharma H. D., Oxford and IBH Publishing Co., New Delhi, 1991 Subramanya K. Tata – Mc Graw Hill, New Delhi Ven Te chow, Mc Graw Hill, New York, 1964 Varshney, Gupta and Gupta Nem Chand and Bros., Roorkee (U.P., India) Bharat Singh, Nem Chand and Bros., Roorkie,1983. Abdul Kalam and Rajan, Penguin Books Pvt. Ltd., New Delhi, 1998. Peterson M.S., Prentice-Hall, 1986

B.  TECHNICAL PAPERS AND REPORTS S. No. Title 1. Symposium on Optimum Requirements and Utilization of water for irrigated crops. 2. Regime Flow in Incoherent Alluviums 3.

Canal Lining in India

4.

Symposium on “Water Logging – Causes and Remedial Measures” Design of Weirs on Permeable Foundations Tubewell and Ground Water Resources

5. 6.

Author and Publication Publication No. 93 Central Board of Irrigation and Power, Delhi, 1963 (CBIP) Publication No. 20 by Gerald Lacey, Central Board of Irrigation and Power, Delhi 1939. Kanwar Sain, Proceedings of International Commission on Irrigation and Drainage, 3rd congress, San Francisco 1957 Publication No. 118, CBIP, Delhi, 1972 A. N. Khosla, et al, Publication No. 12, CBIP, Delhi, 1965 Publication No. 54, CBIP, Delhi, 1965

References  319

S. No. Title 7. IS 1893 of 1962 Recommendations for earth quake resistant design of structures 8. Ranga Raju and Asawa, “Viscosity and Surface Tension Effects on Weir Flow” 10. Garde and Ranga Raju, “Regime Criteria for Alluvial Streams” 11. Meyer – Peter and Muller, “ Formulae for Bed Load Transport” 12. Einstein, “Bed Load Functions for Sediment Transportation in open Channels” 13. Keenedy R. G., “Prevention of silting in Irrigation Canals” 14. Casagrande, “Seepage Through Dams”. 15. Vittal and Porey, “Design of Cascade stilling Basins” 16. Report of Irrigation Commission 17. Proceedings of Workshop on Sprinklers and Drip Irrigation 18. Varshney and Bajaj, “Ski-jump Buckets on Indian Dams” 19. Jogalekar D.V., “Mannual of River Behavior, Control and Training”

Author and Publication Bureau of Indian Standards, New Delhi.

Journal of Hyd. Div. Pros. ASCE, Oct. 1977 Journal of Hyd. Div. Proc. ASCE, Nov. 1963 Proceedings, 2nd IAHR Congress, Stockholms, 1984 USDA Tech. Bulletin No. 1026, Sept. 1950 Paper No. 2826, Proc. ICE, London, Vol. 119, 1895 Journal of New England Water works Association, June 1937 Journal of Hyd. Div. Proc ASCE, 1987

Publication of Ministry of Irrigation and Agriculture, 1972, New Delhi Publication of CBIP, New Delhi. Journal of Irrigation and Power, Oct. 1970. CBIP Publication No. 60,1971.

 

 

Subject Index A Alluvial channels  246, 248, 280 Aqueduct  271-273, 275, 279 Arch Dam  144, 178, 179 Constant angle  178, 179, 184 Constant radius  178, 179, 187 Thin cylinder theory  179, 181, 182 Trial load analysis  181, 183, 189 Asphaltic concrete lining  258 B Balancing depth  246, 262 Bundalling  301, 303 Barrage  218, 239 Base period  5-9, 23, 24 Bed load  98, 283-287, 289 Berms 191 Bligh’s creep theory  220-222, 239 Border strip method  18 Branch canals  241, 263 Ski-jump  157, 159, 160 C Canals  3, 18, 241 Alignment  164, 242, 243 Contour  18, 19, 96 Distributary  241, 263, 264 Lining  177, 254, 255 Losses  9, 14, 19 Ridge  242, 270, 280 Watershed  38, 55, 242 Casagrande 194 Catchment Area  4, 6, 7 Cavitation  162, 168, 174 Chute spillway  164 Command area  3, 4, 6 Command area development authority (CADA) 6

Consumptive use of water  16, 17, 45 Critical slip circle  197, 198, 217 Curtain grouting  142, 144, 190 D Divide wall  219, 238, 239 Draw down  70, 71, 73 Duty of water  7, 8 E Earth dams  190 Causes of failure  214, 217 Design criteria  129, 191, 217 Seepage line  193, 203, 207 Stability analysis  195 Earthquake force  121, 125-127, 181 Escape  3, 242 Evaporation losses  19, 97 Exit gradient  219-222, 239, 240 F Falls  25, 50, 72 Various types  303 Fetch  123, 124, 132 Field irrigation requirement (FIR)  14 Filters  190, 191, 216 Flexibility  264, 265, 268 Flood control  97, 105, 296 Flood frequency  59, 97, 302 Foundation treatment  143, 144 Free board  123, 124, 133 Froude number  157, 159, 177 Furrow method  18, 19, 254 G Gates  99, 142, 145 Vertical lift type  173, 177 Gibb’s module  267 Gravity dams  120, 178, 217 Groynes  239, 297, 299

© The Author(s) 2023 S. K. Ukarande, Irrigation Engineering and Hydraulic Structures, https://doi.org/10.1007/978-3-031-33552-5

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322  Irrigation Engineering and Hydraulic Structures

H Head regulator  9, 218, 219 Head works  3, 18, 218 Hot weather  5, 6, 22 Hydraulic jump  157, 158, 219 Hydrograph  46-56, 58, 59 Flood  18, 26, 46 Unit  11, 46, 47

Synthetic 55

Hyetograph  40, 41, 48 Hygroscopic water  10 I Ice pressure  120, 124 Impervious blanket  210 Irrigation methods  18, 19 Irrigation siphon  272, 279 K Kennedy’s theory  248, 252, 262 Key ways  143 Kharif crops  5 Khosla’s theory  222, 232, 234 L Lacey’s theory  253, 262 Lane’s weighted theory  222 Launching apron  300 Levees  18, 296, 297 M Main canal  23, 241, 244 Marginal bunds  297 Mass curve  34, 102, 119 Modular outlets  263, 264, 279 O Ogee spillway  156, 165, 166

P Paleo  7, 24 Permeability  12, 70, 71 Perennial season  5, 114 Permanent wilting point  10, 11, 17 Pipe outlets  265 Piping  214-216, 220, 222 Pitched islands  297, 300 Pore pressure  198, 217 Q Quality of irrigation water  9, 10 R Rabi season  5 Regime  247-249, 251,286 Regulator  9, 218, 219 Cross  3, 70, 87 Head  2, 3, 9 Relief wells  215, 217 Reservoirs  1, 96, 97 S Ski-jump bucket  157, 159-161, 165 Silt factor  220, 249, 251 Siphon aqueduct  271, 272, 275 Super passage  271, 272 T Transpiration  13, 241, 244 U Uplift force  120, 122, 123 W Waterlogging 253-256 Weirs  3, 218, 239