1

CHAPTER 1

INTRODUCTION

1.1 BACKGROUND

1.1.1 Reasons for focusing on Hydropower Projects

The demand of energy is increasing day by day and the supply is limited, so it has

pronounced negative impacts such as depletion of natural resources, environmental

degradation, etc. This causes depletion in non-renewable and exhaustive sources of

energy, which may invite energy crisis in future. Also, it is essential to meet the concept

of sustainable development .So, energy consumers should keen on using of non

exhaustive and renewable sources of energy. Hydropower, one of the most reliable and

common renewable sources of energy is abundantly available in the hilly regions of

Nepal. Again, hydraulic conveyance circuit can be beneficial for multipurpose use

(irrigation, water supply etc.). Hydropower production does not consume water, so it is

considered as renewable source of energy. Consumption of this energy is environmental

friendly because it uses water as fuel and no harmful byproducts are produced. It does not

emit green house gases that cause ozone layer depletion and global warming.

Because of abundant water resources and potential hydropower sites available, there is

huge possibility of hydropower production. Large projects involve huge amount of funds

and the development period is large hence activities regarding development of Small

Hydropower Projects are accelerating in these days which is technically, financially and

environmentally sustainable at the present scenario.

1.1.2 History of Hydropower Development

Use of energy generated from water has been started since the very beginning of human

civilization. There are evidences of it in Greek and Roman civilization. Though, Michael

Faraday demonstrated that mechanical energy could be converted into electrical energy

and vice versa, in 1831, development and use of electrical energy began gaining

momentum after 1890. By 1900, hydropower plants had become a common source of

obtaining electricity. In the early 19th

century, progress in the hydropower development

was slow because of less efficiency in power transmission over the long distance. The

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pace of hydropower development increased dramatically after 1930. United States made a

policy to invest in water based projects to create jobs for unemployed and to stimulate

economic recovery in the country when it faced severe economic hardship in 1929. In the

former Soviet Union, hydropower was considered synonymous with industrialization and

economic prosperity after 1920.

After 2nd

world war, leaders of African and Asian nations has replicated the western US

model to meet energy and water needs of their own countries and many large scale

hydropower projects were built in India, Pakistan and Egypt between 1950 and 1980.

None of the projects in US, former Soviet Union and India had the objective of exporting

energy to its neighboring to earn revenue for the country. In recent decades, the concept

of production of electrical energy has been changed. Now, it has been traded between two

or more nations after agreement upon certain terms of trade. Exporting electricity to a

neighboring country to earn revenue for the government is one of the stated objectives of

developing large scale hydropower projects in Nepal.

1.1.3 Hydropower development in Nepal

Nepal entered into hydropower development field almost a century ago. The first in this

region was the construction of 500KW Pharping hydropower project, commissioned in

1911 AD followed by Morang Hydro 1918 and Sundarijal Hydro in late thirties. The

planned development approach was initiated in late fifties with the First Five Year Plan.

In about a century only 450MW has been developed through government agency. On the

other hand after liberation (1990), IPPs have developed about 125MW in less than a

decade.

First approach in hydropower development in Nepal was the power generation from the

construction of Pharping Hydropower station (500 KW) in 1911. But the progressive

development was gradual only after the Sundarijal (600 KW) and Panauti (2400 KW)

Hydropower Stations came into operation after long interval of 23 and 29 years.

The completion of Dhankuta Hydropower station (240 KW) in 1971 was regarded as the

bench mark of small hydel development of Nepal. The establishment of small hydel

development board in 1975 was another milestone under which several small hydro

schemes such as Jhupra (345 KW), Doti (200 KW), Jumla (200 KW) etc. were made

during 1975 to 1985. Nepal Electricity Authority (NEA), established 1985, responsible

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for generation, transmission and distribution of electric power brought the revolution in

hydropower development. Many potential sites for hydropower generation had identified

by private consultancies and companies in collaboration with NEA.

Prior to 1960, all the hydropower stations were constructed through grant aid from

friendly countries like the USSR (Panauti), India (Trishuli, Devighat, Gandak, Surajpura-

Koshi) and China (Sunkoshi). Since 1970, hydropower development took a new turn with

the availability of bilateral and multilateral funding sources. The major donor countries in

the period were Japan, Germany, Norway, South Korea, Canada, Finland, Denmark,

Sweden and USA. The financial lending agencies were the World Bank, ADB, JBIC,

Saudi Fund for Development, Kuwait Fund and others.

From 1990s, subsequent to the adoption of the policy of economic liberalization,

hydropower development took yet another turn with the private sector entering the arena.

After formulating Hydropower Development Policy – 1992 by government of Nepal,

many private sectors are involving towards power development. In order to encompass

projects of various scales intended for domestic consumption as well as to export

hydropower, the former policy was replaced by the Hydropower Development Policy

2001 to provide further movement to active participation of private sectors.

Development of hydropower in Nepal is a very complex task as it faces numerous

challenges and obstacles. Some of the factors attributed to the low level of hydropower

development are lack of capital, high cost of technology, political instability, and lower

load factors due to lower level of productive end-use of electricity and high technical and

non technical losses.

Legends for the Power Development in Nepal

Table 1.1 Major Hydropower Plants

Name Capacity(MW) Name Capacity (MW)

Trishuli 24.00 Gandak 15.00

Sunkoshi 10.05 Devighat 14.10

Kulekhani 1 60.00 Khulekhani 2 32.00

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Marsyandi 75.00 Upper Modi 14.00

Khimiti Khola (HPL) 60.00 Jhimruk (BPC) 12.30

Bhotekoshi (BPCL) 36.00 Kaligandaki A 144.00

Chilime (CPC) 20.00

Table 1.2 Small Hydropower Plants

Name Capacity(MW) Name Capacity (MW)

Tatopani, Myagdi 2.00 Panauti 2.40

Seti, Pokhara 1.50 Phewa, Pokhara 1.088

Tinau, Butwal 1.024 Chatara 3.20

Andhikhola(BPC) 5.10 Indrawati (NHPC) 7.50

Piluwa Khola 3.00 Sunkoshi (Sanima) 2.60

Table 1.3 Planned & Proposed Hydropower Plants

Name Capacity(MW) Name Capacity (MW)

Rawa Khola 2.30 Molung Khola 1.20

Naugargad (Darchula) 1.80 Gandigad (Doti) 1.80

Daram Khola (GHP) 5.00 Upper Khimti 4.00

Chaku Khola (A. Power) 1.50 Lower Indrawati 4.60

Lower Nayagdi (BHN) 4.50 Mardi Khola 1.40

(Source: http://www.fncci.org/text/pp-eup.pdf)

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1.1.4 Hydropower potential of Nepal

Due to the steep topography, abundant precipitation and perennial nature of most of the

rivers which originate from the Himalayas of Tibetan plateaus; there exist a tremendous

hydropower potentiality in Nepal. The theoretical potential is estimated to be about 83

GW out of which 42 GW has been considered as financially viable and 44 GW as

technically viable. Approximately 6000 big and small rivers have been identified in

Nepal's territory carrying about 174×109m³ of surface run-off annually (0.5% of total

surface run off of the world).

Table 1.4 Hydropower Potential of Nepal (in million KW)

(Source: Nexant SARI/Energy, September 2002.In topic Hydropower Potential.5-2pp.

Regional Hydro-power Resources: Status of Development and Barriers Nepal)

1.1.5 Justification for Small Hydropower Project

NEA has classified the hydropower projects according to the power output into the

followings groups:

I. Micro Hydro Power Plant : Less than 100 KW

II. Mini Hydro Power Plant : 100 KW – 1MW

III. Small Hydro Power Plant : 1MW – 10 MW

IV. Medium Hydro Power Plant : 10 MW – 300 MW

V. Large Hydro Power Plant : More than 300 MW

(Source: ITDG, 2002)

A small hydropower plant is found to be most feasible than both the micro hydro and

large hydropower in context of Nepal. A small hydro ranges from 1- 10 MW for which

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corresponding head and discharge is easily available than the other hydro electric project.

Investment required for small hydro is affordable to the countries like Nepal.

In remote mountainous area large hydro power seems to be not easy to serve for local

community because power produced from the plant could not get sufficient load to meet

the produced power. Again large hydropower needs heavy capital investment which can't

be solely invested by government. Also cost of transmission line and preliminary works

in large hydropower project is high and not economically viable in Nepal in many cases.

In case of micro hydro, the loss is heavy and cost per unit production is high. Micro hydro

projects are launched as a support to the villagers and are not treated as a revenue

generating projects to the nation. They have been constructed under government's

subsidy. Also, being community based projects, micro hydro projects cannot be launched

in all community.

Small hydro power project is independent to the local community in regards to its

investment and incurs little problems regarding social and community matters. Work

quality in small hydro power project is of intermediate type, thus, it is easy to plan,

construct, monitor and maintain than large projects. The capacity of Nepalese

entrepreneurs to invest is not comparable to those of foreign investors. This motivates

national investors to focus on small hydropower sector. Hence, small hydropower plant is

better suited and justified to generate electricity in Nepal.

1.2 OBJECTIVE

The following are the main objective of the study:

To fulfill the partial requirement of the final year project for the completion of

Bachelor Degree in Civil Engineering.

To study & analyze the geo-hydrological situation of the site.

To design the hydraulic components of the hydropower plant at the site.

To design the structural components of the hydropower plant at the site.

1.3 SCOPE AND LIMITATIONS

The following are the scope of the study undertaken:

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To work throughout the project under the supervision of BPC.

To collect the hydrological and meteorological data from BPC.

To assess the hydrological condition of the project site.

To design all the relevant civil structures for the hydropower plant.

To prepare a volume of comprehensive project report.

The following were the limitations of the project:

Due to time constraint (90 working days), detailed survey was not possible

and geo-hydrological data were also taken from BPC.

The detailed estimation and rate analysis is beyond the scope.

The designs were carried out referring several materials due to lack of unified

guideline. Hence, the design output may not fulfill all the standards.

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CHAPTER 2

LITERATURE REVIEW

2.1 POWER SITUATION IN NEPAL

The present total installed capacity of the INPS, including capacity added through IPPs

amounts to nearly 614MW; of which ,the capacity of ten major NEA hydro plants is

389.150MW while that of the grid connected ten small hydro plants (SHP) is 12.792MW.

Isolated thirty SHP have capacity of 6.176MW and ten IPPs contribute 148.683MW of

power to the system. Similarly, six NEA thermal plants have capacity equal to

56.756MW while two solar plants provide 100 kW of electric power in total.

Besides, during the time of deficit, power up to 50MW is imported from India as per the

Indo-Nepal Power Exchange Agreement. Nepal and India have agreed in principle to

increase this level of exchange from the existing 50 MW to 150MW. Nepal is also

entitled to receive 70 million units of energy annually from Tanakpur in the far west

under the Mahakali Treaty and 10MW power according to Koshi Contract. Power is also

exported to India through some sections of the INPS according to the exchange

agreement. Although the present integrated grid has a total of about 550MW installed

capacity without considering capacity of thermal plants and that available through

Mahakali treaty and Koshi Contract, only about 425MW can be generated from hydro

power stations during the winter season when the power demand is at its peak.

In the area of transmission and sub-transmission of electricity, the INPS has grown to a

network of more than 1800 km of 132kV, more than500 km of 66kV and around 2500 km

of 33 power lines. In order to accelerate the pace of expansion and conduct management

of rural distribution systems in a sustainable manner, NEA has adopted a concept of

community participation in rural electrification schemes.

2.2 DOMESTIC POWER DEMAND AND SUPPLY

The electricity demand forecast to the year 2019/20 is prepared by NEA during the

preparation of its Corporate Development Plan for the year 2004/05-2008/09 in 2005.

This forecast is based on the power consumption data of FY 2004/05. Total energy

requirement in Nepal is projected to go upon by an average of about 8% per annum over

the forecast period, from 2,299.9 GWh in FY 2003/04 to 7894 GWh in FY 2019/20 while

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peak demand is projected to grow from 512.2MW in FY 2003/04 to 1733 MW in FY

2019/20.After the year 2008, the peak power deficit will continue to increase unless

projects of reasonable sizes are not constructed without any further delay.

2.3 LOAD FORECAST

The load forecast for INPS made by NEA according to the power system master plan

studies is presented here under table. The load has been forecasted considering the

country's macro- economic indicators and rural electrification expansion programs. The

forecast revealed that the energy and peak demand is expected to grow more than three

times between 2005 and 2020.

Table 2.1 Load forecast for INPS

Load Forecast

Fiscal

Year

Total

Generation

Requirement

GWh

System Peak

Load

Peak Load

Growth %

2005-06 2774 603.28

2006-07 2897.1 642.2 6.5

2007-08 3136.6 695.3 8.3

2008-09 3428.1 759.9 9.3

2009-10 3698.4 819.8 7.9

2010-11 4057.1 890.6 8.6

2011-12 4423.3 971 9

2012-13 4815 1057 8.9

2013-14 5231.2 1148.4 8.6

2014-15 5673.8 1245.6 8.5

2015-16 6144.7 1336.1 7.3

2016-17 6645.9 1445.1 8.2

2017-18 7179.6 1561.1 8

2018-19 7719.4 1678.5 7.5

2019-20 8296.7 1804 7.5

Average Growth 8.14

(Source: NEA, 2003/4)

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Figure 2.1 Load Forecast

2.4 ENERGY CONSUMPTION PATTERN OF NEPAL

In Nepal traditional energy sources are the biggest contributors having share of 87.7% in

the total energy. These sources comprise of fuel wood (78.1 %), agricultural residues and

animal wastes (9.6 %). Commercial energy sources share 10.5 % having rest to other non

conventional sources. Electricity contributes about 1.8 % of the total energy needs.

(Source: WECS,2006)

2.5 DEVELOPMENT OF THE GRID SYSTEM AND POWER TRANSMISSION

PLAN

At present, the INPS consists of 1,132 Km of 132 KV single circuit, 412.1 Km of 132 KV

double circuit, 231.46 Km of 66 KV single circuit, 161.3 Km of 66 KV double circuit, 22

Km of 66 KV and 132 KV double circuit, 3.37 Km of 66 KV four circuit and 2,362 Km

of 33 KV single circuit transmission line. Total substation capacity of the system is

902.45 MVA. In the field of transmission, NEA is operating at system voltage levels of

132 KV and 66KV.

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Rising load demands have created load saturation in some sectors of these transmission

lines leading to poor regulation and reliability at supply terminals resulting in increase in

technical losses. The completion of projects like the Middle Marshyangdi would require

augmentation in the current carrying capacity of the major 132 KV network and

construction of some 220 KV lines. The urgently needed 220 KV sections are Hetauda -

Bardghat and Khimti – Dhalkebar. The Khimti – Dhalkebar 220 KV transmission line is

going to be constructed with loan assistance of the World Bank, while efforts are

underway with donors for implementation of the 220 KV Hetauda – Bardghat section.

NEA is also constructing three power exchange links namely Butwal – Sunauli,

Parwanipur – Birgunj and Dhalkebar – Bhirttamod at the 132 KV level to enhance the

transfer capability of Nepal – India crossborder interconnections. This will enhance the

quality of grid connected supply to 34,000 consumers including 17,000 new connections.

In the area of distribution system expansion in Mid and Far Western regions of the

country, Swedish Government has conveyed its commitment to provide concessionary

credit of about US $ 20 million. The Rural Electrification, Transmission and Distribution

Project aided by ADB and OPEC has five different components that will in addition to

providing transmission and institutional support, develop the distribution system to

connect about 123,000 rural households of 277 Village Development Committees.

2.6 POWER DISTRIBUTION PLAN

The need to extend distribution over the country is reflected from the fact that 85%

population of the country is not getting electricity as a source of energy. So, the

distribution of electricity should be done strategically. NEA has taken systematic studies

of carrying out rural electrification and distribution system reinforcement feasibility on

district-wise basis. NEA intends to undertake these works with multi-source financing.

Also, Nepal Government contributes to rural electrification scheme on an annual basis

with an increasing magnitude in the year, 1999/2000, outlay being approximately 4.5

million US dollar. NEA and Nepal Government are jointly working for the electrification

of rural areas. To cope with this objective, micro and small hydropower are the better

options in the present scenario. The total capital investment in distribution system

expansion and reinforcement for the fiscal year 1999/2000 to 2007 is estimated at 9,349.2

million NRS.(Source:NEA,2006)

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2.7 LEGAL PROVISION TO INVEST IN HYDROPOWER SECTOR IN NEPAL

Hydropower industry is one of the major industries with wider scope in Nepal. For an

industry to prosper there should be support of government policies and legal provisions.

Only the potential cannot do the development of a nation if the policies cannot be

harnessed. Clearly defined conditions and attractive policy are always essential to harness

the innumerous resources. Realizing this fact, Nepal Government has developed certain

policies.

a. Why to invest in Nepal?

Attractive Investment Features

One-Window Policy

Repatriation of Foreign Exchange

Income Tax Incentives

Fixed Royalty Payments

Import Concessions

Export Opportunities

No Nationalization of Projects

b. Policies, Act and Regulations:

Hydropower Development Policy-1992

Industrial Policy- 1992

Foreign Investment and One Window Policy- 1992

Electricity Act- 1992

Industrial Enterprises Act-1992

Foreign Investment and Technology Transfer Act -1992

Environment Conservation Act – 1996

National Environment Impact Assessment Guidelines – 1993

c. Legal Framework:

Survey License issued within 30 days

Survey License Period up to 5 years

Project License issued within 120 days

Project License period up to 50 years

Exclusive Water Rights

Public Consultation before issuance of Project License

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Government land available on lease

d. Institutional Framework for Electricity Development as "One Window"

Issuance of Survey & Survey licenses

Provision of tax concessions & incentives

Assistance in importing goods, land permits, approvals etc.

Regulation and monitoring of projects

e. Incentive Income Tax

Generation :- 15 years tax holiday

Transmission :- 10 years tax holiday & M Contracts:- 5 year tax holiday

After tax holiday:- 10 percent less than period prevailing

Foreign Lenders:- 50 percent capital cost allowance

Equity Investors:- No tax on interest earned

No tax on dividend

f. Import Facilities :-

Plant and Equipment including Construction Equipment

1% Custom Duty on items not manufactured in Nepal

import License Fee and sales tax exempted

effective from the date of commercial operation

g. Repatriation of Foreign Exchange

Principal and interest on debt

Return on equity

Sale of share equity

Prevailing Market rates

h. Royalty Payments:

For first 15 years

Installed Capacity/annum - NRs. 100/KW

Energy Generated - 2% of energy sales

After first 15 years

Installed Capacity/annum - NRs. 1000/KW

Energy Generated - 10% of energy sales

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i. Market:

Domestic: Nepal Electricity Authority (NEA)

Foreign: India

Under Power Exchange Agreement

Under Power Trade Agreement between two countries

Regional: Government

Probably under the Regional Cooperation especially

Quadrangle concept within SAARC

j. Nepal Government/ NEA Policy on Purchases from Small Project

The private sectors should do the Power Purchase Agreement (PPA) with NEA to sell the

energy produced. To promote the private sectors in national level and to provide the

opportunity to invest in the hydropower sectors for the Nepalese people, NEA has the

provision to purchase the energy of small hydropower plants with first priority.

k. Export Opportunities:

Existing Power Trade Agreement between Nepal and India

Existing Interconnection Facilities with India

Power Deficit in India

Oriented Projects in Nepal

2.8 COMPONENTS OF HYDROPOWER PLANTS

2.8.1 WEIR AND UNDERSLUICE

2.8.1.1 General

Weir is a structure constructed at the head of canal, in order to divert the river water

towards the canal so as to ensure a regulated continuous supply of silt free water with

certain minimum head into the canal. The types of weir and its use depend upon the

topography, geology, discharge, river morphology etc. If the major part or the entire

ponding of water is achieved by a raised crest and smaller part or nil part of it is achieved

by the shutters then it is called weir.

Undersluice is the structure constructed side by the weir for the purpose of flushing the

deposited silt by providing

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Opening provided on the weir portion with crest level positioned at lower level than the

crest of the weir. It creates comparatively less turbulent pocket of water near intake.

Undersluiced length of weir is divided into no. of ways by piers and regulated by gate.

Opening helps in scouring and removing the deposited silt from undersluiced pocket

hence is also called the scouring sluices. Gate-controlled undresluice helps regulating

flow in intake at the dry weather flow and low flow and periodic flushing.

Spillway is a structure constructed at the weir side for effectively disposing of surplus

water from u/s to d/s channel. It doesn’t let the water rise above maximum reservoir level

and prevents the weir from damage. Spillway is essentially a safety value for weir. Types

of spillway according to location, operation, structures etc. are Straight drop, Overflow or

Ogee, Chute, Side Channel, Shaft, Siphon, Orifice, gated etc. Ogee Spillway is an

improvement upon the free overfall spillway and is widely used with concrete, masonry,

arch and buttress dams. Ogee spillway works effectively only on one particular head

called designed head.

2.8.1.2 Design Consideration of Diversion Weir

The design of weir includes computing the elevation of weir crest, length of weir,

computing the forces acting on the weir and checking the safety of the weir from all

aspects like overturning, sliding, crushing etc. They all are explained in the following

articles.

2.8.1.2.1 Elevation of Weir Crest

There are various factors that affect the elevation of the crest, but in our case, diversion of

water is the purpose and the height should be sufficient to pond the water at a level that

can facilitate design flow in the intake. The height of the weir is governed by the height of

intake sill, depth of intake orifice and depth of the river at the intake site.

Four other important considerations to be considered for fixing the crest level of the weir

are as follows:

The height of the crest affects the discharge coefficient and consequently the

water head above the weir as well as the back water curve.

The elevation of the weir crest has to be fixed such that the design flood is

safely discharged to the downstream without severe damage to the

downstream.

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The elevation of the weir determines the head of the power production.

The height of the weir crest affects the shape and location of the jump and

the design of the basin.

The height of the weir crest affects the discharge that can be diverted into

the canal.

The bed level of the river at the headwork is 950.00m. The crest level of weir

provided is 952.00m

2.8.1.2.2 Length of Weir and Undersluice

The length of the weir depends upon the width of the waterway at the intake site. Crest

length should be taken as the average wetted width during the flood. The upstream and

downstream should be properly examined for the protection consideration.

Rise in water level on the upstream of the structures after construction of the weir is

called afflux. Fixation of afflux depends on the topographic and geomorphologic factors.

A high afflux shortens the length of the weir but increases the cost of the river training

and river protection works. For alluvial reaches it is generally restricted to 1m but for

mountainous region it may be high. The water way must be sufficient to pass high floods

with desired afflux. A weir with crest length smaller than the natural river width can

severely interfere the natural regime of flow thus altering the hydraulic as well as the

sediment carrying characteristics of the river.

Generally, the spillway and undersluice lengths are designed so as to safely pass 80 %

and 20 % of the design flood respectively. In our particular design, the spillway and

undersluice is so accommodated that from total water way, 6.2 m is given to undersluice

and remaining 28.8 m is given to spillway. The spillway is so designed that it can

accommodate total flood design. The undersluice portion is designed only for sluicing the

bed load. Hence, the undersluice is design taking for 183.634 m³/sec This will economize

in the construction of energy dissipaters.

2.8.1.2.3 Shape of the Spillway

The spillway has been designed as free over fall Ogee shaped weir. The discharge

capacity of Ogee shaped spillway is maximum as compared to that of other type of weirs.

Ogee shaped weir increases hydraulic efficiency and prevent cavitations. The profile of

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the spillway is made similar to the nappe profile of the free overfall weir to ensure that

there is minimum possibility of negative pressure development along its length.

The provisions of the sheet piles, cutoff walls, impermeable concrete floor and protection

works including launching apron has been designed considering various factors as

presented in the detailed design of the weir structure. The parameters under consideration

are

Hydraulic jump characteristics

Length and the height of formation of jump

Seepage Pressure

Erosion characteristics

2.8.1.2.4 Forces acting on Weir

The main forces which are acting on the weir when it will be operation are: Water

Pressure, Uplift Pressure, Slit Pressure and Weight of the weir.

2.8.1.2.4.1 Water Pressure

It is the major external force acting on the weir. This is called hydrostatic pressure force

and acts perpendicular on the surface of the weir and its magnitude is given by:

2P 0.5 H b Where, γ = Unit weight of water, H = Depth of water, b = Width of

the Weir surface. This pressure force acts on H/3 from the base.

2.8.1.2.4.2 Uplift Pressure

Water seeping through the pores, cracks and fissures of the foundation material, seeping

through the weir body itself and seepage from the bottom joint between the weir and its

foundation exerts an uplift pressure on the base of the weir. The uplift pressure virtually

reduces the downward weight of the weir hence acts against the dam stability. The

analysis of seepage is done using Khosla's Theory. Khosla's Theory is the mathematical

solution of the Laplacian equation and it is easy and accurate method for seepage

analysis.

2.8.1.2.4.3 Silt Pressure

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The silt gets deposited on the upstream of the weir and exerts the horizontal and vertical

pressure as exerted by the water. So, flushing of the silt should be done regularly to

reduce its effect of destabilizing the weir. It is done by the use of undersluice gate. The

silt pressure is given by the relation: 2

silt sub aP 0.5 H K . Where, γsub=Submerged

unit weight of silt; H= Depth of silt deposited and Ka= Coefficient of Active earth

pressure and is given by, a

1 sinK ,

1 sinAngle of internal friction of silt. The silt

pressure force also acts at a height of H/3 from the base.

2.8.1.2.4.3 Weight of Weir

The weight of weir and its foundation is the major stabilizing/ resisting force. While

calculating the weight, the cross section is splited into rectangle and triangle. The weight

of each along with their C.G. is determined. The resultant of all these forces will represent

the total weight of dam acting at the C.G. of dam. Simply, when the sectional area of each

part is multiplied by unit weight of concrete, weight of that part is obtained.

The weir is designed with ogee profile for spilling over its length. Hence weight is

calculated by knowing its section and multiplying by its unit weight.

2.8.1.2.5 Modes of Failure & Criteria for Structural Stability of Weir

2.8.1.2.5.1 Overturning about the toe

If resultant of all the forces acting in the weir passes outside, the weir shall rotate and

overturn about the toe. Practically, this condition will not arise because the weir will fail

much earlier by compression. The ratio of resisting moment to the overturning moment

about the toe is the factor of safety against overturning and it should greater than1.5 for

safety.

2.8.1.2.5.2 Compression or Crushing

While designing the weir section it should be so design that the resultant should pass

through middle 3rd

part of the section to avoid the possible tension on the weir section.

The section should be totally in compression. So, weir should be checked against the

failure by crushing of its material. If the actual compressive stress may exceed the

allowable stress, the dam material may get crushed. The vertical combine stress at the

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base is given by: max/ min

V e1 6

B B, where

B Me x, x

2 V,e= eccentricity of

the resultant force from the centre of the base.

2.8.1.2.5.3 Sliding

Sliding will occur when the net horizontal force above any plane in the weir or at the base

of the weir exceed the frictional resistance developed at that level. Factor of safety against

the sliding is measured as Shear Stability Factor (SSF) and is given by:

V BqSSF

H

Where, μ= Coefficient of friction; q= Average shear strength of the joint.

For safety against sliding, SSF should be greater than 3-5. To increase the value of SSF,

attempts are always made to increase the magnitude of q, which is achieved by providing

the stepped foundation, ensuring the better bond between the dam base and rock

foundation etc.

2.8.1.3 Protection Work for Weir Structure

The weir should be well protected from the flowing river to avoid creep effect. For this,

the wing wall is essential to construct. It should be well anchored into the bed. Similarly,

to protect the channel bed from being eroded, launching apron is used. To protect the weir

body riprap is usually placed. In the site both the banks are rocky hence not any especial

protection shall be introduced. Some sorts of works to protect banks and to confine the

river at up stream may be required. Gabion walls are used as protection works for the

banks which ultimately protect the degradation of the weir. In the downstream side of the

spillway, energy dissipater is designed to the excessive energy of the flood water. Divide

wall is constructed to prevent the cross flow of the weir spillway portion and the

undersluice portion. The undersluiced portion is designed as for the flood flow with

limited opening of maximum 2.5 m and energy dissipators as the submerge flow. To

prevent the seepage effect, sheet piles are inserted at the upstream and downstream.

20

2.8.2 INTAKE STRUCTURE

2.8.2.1 General

An intake can be defined as a structure that diverts water from river or other course to a

conveyance system downstream of the intake. The intake structure is used to trap the

required amount of water for the specific purpose with or without storing. An intake

structure should control the flow of water and prevent the heavy sediment load of the

river from entering the conveyance system. For this purpose, proper selection and sitting

of intakes must be chosen to evacuate necessary amount of water at any regime to the

channel. The peak discharge must be safely evacuated without any damage. To achieve

this, hydrological data must be collected and evaluated and the structures should be

designed accordingly.

Prerequisites of the location of intake structure:

The course of the river should be relatively permanent at the intake site, i.e.

the river should not change its course at the intake location at the time.

The river should not have a large gradient at the intake site.

In case there is a confluence of two rivers in the selected site, the intake

should be located downstream of the confluence to take advantage of the

flow of both rivers.

Advantage should be taken of stable banks such as rock outcrops or armored

boulder banks to protect the intake from erosion.

The intake should be located at the outer bend where flow is deeper and

clearer and towards the downstream end of the bend where the effect of the

secondary currents has fully developed. This limits sediment deposition at

the intake area and also ensures the flow availability during the dry season.

The intake structure is designed for 20% more than design discharge, 10% for flushing in

gravel trap, 10% for settling basin i.e. Qdesign(intake)= 1.2*Qdesign. The three orifice type

intake of width 2.4 m and height of 1.5m is designed which allows the design flow to pass

through it under normal condition but restricts higher flows during floods.

21

2.8.2.2 Design Consideration of Intake Structures

For small hydropower projects it is general practice to use 100 years return period from

probabilistic analysis of flood. A simple and moderately priced construction should be

used to minimize maintenance and repairs. For the small projects with no automation

facilities, hydraulically controlled structures become more feasible than mechanically

controlled units. There must be adequate provision to remove the suspended and bed load

deposited upstream behind the weir. This may be done using intermittent flushing using

sluice gates or allowing some water to flush it continuously. It has been found that entry

of bed load towards diverted canal will be minimum if the intake is located just

downstream of concave bank of the river bend. It not only restricts the bed load, but also

ensures sufficient water depth even at low water condition.

Topography, geology, height of bank, ratio of water diverted to that available, channel

width, routing of diversion canal, ease of diversion of river during construction, stability

of river bank and sides, river protection works governs the selection of the intake location

and type. For steeper gradients with straight reaches of river bottom rack intake is more

suitable. But in rocky banks, winding river, considerable suspended load it is not

desirable. The lateral side intake functions well in such case. Intake sill with 2 m is used

not to allow bed loads to enter the canals. Trash rack is used to prevent the entry of tree

branches, leaves and other coarse materials in the canal. Head is extremely valuable in

hydropower projects so that trash rack should be designed with minimum head loss.

Suitable factor of safety should be employed to design height of intake sill, to ensure

sufficient withdrawal capacity in the future.

2.8.2.3 Protection Work

Protection works are the river protection and river training works to safeguard the intake

structure against flood, debris and sediment. The skimmer wall is constructed to protect

the entry of flood water in the canal at the time of high flood. Trash racks are used to

prevent the entry of trash matters in the canal. To prevent adverse effect of seepage, sheet

pile is used inside the ground below sill.

22

2.8.3 GRAVEL TRAP

2.8.3.1 General

Gravel trap is a basin (pond) close to the intake where gravel and other coarse materials

are trapped and then removed. In the absence of this structure, gravel can settle along the

gentler section of the head race or in the settling basin and reduce the discharge capacity

of canal and ultimately cause the wearing and chocking of the turbine units. The main

design principle for a gravel trap is that the velocity through it should be less than

required to move the smallest size of gravel to be removed. In general gravel trap should

settle coarse particle (>2mm dia). During the high flood season, the river carries

appreciable amount of gravel hence a gravel trap should be provided to trap the design

size of gravel entering through intake.

The gravel trap is designed for maximum discharge of 8.64 m³/sec with continuous

flushing of sediment ≥ 2mm through flushing orifice The flushing channel size is 0.5×0.5

m with flushing velocity of 4.03 m/s and slope 1:25.

2.8.3.2 Design Considerations

Gravel trap should be located either close to intake as possible to minimize blockage of

the headrace and damage due to abrasion in headrace or at a safe place. The dimension of

gravel trap should be sufficient to settle and flush the gravel passing through upstream.

The flow velocity is limited to 0.51m/s so that it does not affect the settling velocity of the

particles. For construction easiness, depth is generally limited to 3m; width is calculated

to satisfy the velocity of flow and length is calculated to ensure desirable efficiency of

settling.

Generally continuous flushing is adopted for gravel trap as the sediment load is high.

Gates are used to control flow at flushing orifice at outlet. The flushing orifice is designed

on the basis of the head to cause flow. Sufficient bed slope and cross slope is required to

make the flushing effective. Standard methods such as Vetter's equation to calculate

efficiency, Camp’s formula to calculate the transit velocity and Newton's formula to

calculate the settling velocity are used.

23

2.8.3.3 Protection Works

Gates are used to control the flow across the gravel trap. Flushing gates are used to flush

the settled matters. The flushing orifices are controlled using the flushing gates. Flushed

water and the excess water are safely diverted to the river using open channel. The side

protection works fencing etc. are carried out.

2.8.4 SETTLING BASIN

2.8.4 General

Suspended particle that is not settled in the gravel trap is trapped in the settling basin. The

basic principle behind settling is that the greater the basin surface area and lower the

through velocity, the smaller the particles that can settle. The geometry of settling basin

must be such as to cause minimum turbulence, which might impair the efficiency. To

ensure uniform flow, transitions are provided at inlet and outlet. Selection of width and

length also depends upon land available. For more reliable operation, more than one

chamber is employed to make whole system running even if one of the chambers has to

be stop for maintenance.

Flushing of deposited matters is essential for smooth operation of settling basin. The

lateral and longitudinal slope may be provided for this purpose. For control of flow in and

from settling tank, gates can be used. Generally periodic flushing system is adopted for

settling basin since e the sediment load is not as high as in gravel trap.

The design discharge for settling basin is taken as 7.92 m³/sec. Settling basin is 14.2m

wide, 4m deep and 86.1m length (30 transition length and 56.1 m settling length).

2.8.4.2 Design Consideration

The settling basin is designed following standard practices. The geometry of inlet, the

width of basin and any curvature must be such as to cause minimum turbulence which

might impair the efficiency. Concentration approach is used to design it. Trap efficiency

is selected as 90% removal of 0.2 mm sized sedimentary particles. Vetter's equation is

used for efficiency calculation. Hazen's equation and various charts are used to compute

the transit velocity and the settling velocity.

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2.8.4.3 Protection Works

Gates are used to control the flow across the settling basin. Flushing gates are used to

flush the settled matters. The flushing orifices are controlled using the flushing gates.

Flushed water and the excess water are safely diverted to the river using open channel.

The side protection works like fencing etc. are carried out.

2.8.5 HEADRACE CANAL

2.8.5.1 General

The water diverted from the intake to the desilting basin of inlet chamber through the

conveyance system is termed as headrace. A high head diversion plant is generally

associated with tunnel to divert water where as a medium head to low head diversion

plants generally employ canal diversion. Geology, topography and hydrology are major

factors to select such options. For small plants with low heads intra basin diversion

having fairly straight reaches of river, canal is the best option. Headrace has to convey

extra discharge for continuously flushing the settling basin.

2.8.5.2 Design Aspect of Conveyance System

The conveyance system used could be non-pressurized (free gravity flow) or pressurized

flow. Depending upon the flow condition inside the tunnel, it could be either pressure

tunnel or the non-pressure tunnel. Flow through canal is non-pressurized. The conveyance

of flow could be made through pipe as pressurized flow where head loss is minimum.

Points to be considered for canal alignment can be summarized as:

The canal alignment should be sufficiently diverted away from the river so

that the risk of flood damage is minimum.

The alignment should be along the level to slightly sloping ground, pass

through stable terrain and follow the shortest reasonable route with a

minimum crossings and a minimum of head loss and minimum seepage loss

because loss in head or discharge is the loss in power production.

From earthwork point of view, the alignment should be selected to balance

cut and fill as far as possible. But this does not mean to balance cut and fill

even by using costly retaining structures. Stability is also a major factor.

25

Manning's equation is the most widely used formula in open channel flow. From

economic point of view, size should be smallest possible and channel section should be

most economical. But, from constructional feasibility the most economical channel

section obtained by minimizing perimeter may not be always economical. Size and bed

slope have inverse relationship. So, choice of bed slope is dependent on the size of canal.

A power canal is generally of rectangular shape and is rigid boundary channels. Lining is

used to reduce roughness and seepage. A larger size not only incurs cost but also

increases the cost of cross drainage works and cost of operation and maintenance and the

risk of failure. The canal should be shortest possible and as straight as possible.

The velocity should be non-scouring and non-silting. To get small sections, a high

velocity which does not scour the lining material is usually desired but to save head, a

mild is preferred. The slope is limited by non-settling criteria. Tractive shear approach is

always used to ensure sufficient shear stress to cause scouring of particle transported by

it. The shear stress induced by water on the particles should be greater than critical

tractive shear stress of the particles.

o c w w s eOr, ;or, RS S 1 F R

dR S

11

For d>6mm, the above relation implies,

Where, Re= Particle Reynold’s Number

2.8.5.3 Protection Measures for Headrace

The alignment of the canal is selected to ensure maximum stability as far as possible. As

the canal follows the contour and the area is flat at most of the sections, not heavy

protection works would be required.For closed flow there is not any risk of failure as

canal section damage, seepage loss, heavy cross drainage work and its protection etc. The

pipe should be anchored at the every change in horizontal and vertical alignment. At cross

drainage, pitching and concreting around the pipe should be done to secure at its position.

26

2.8.6 FOREBAY

2.8.6.1 General

A forebay is a type of hydraulic structure connecting the non-pressure canal with

pressurized flow system, i.e. flow through penstock. It is a storage basin which is

constructed at the end of the headrace canal during peak power demand period. It serves

the following functions:

Regulates the flow into the penstock, particularly through the excess water into a

spillway

Release the surge pressure as the wave travels out of the penstock pipe

Serve as secondary or final settling basin and trap some particles that enter the

head race downstream of settling basin

Forebay is 15m long, 5.6m deep and 14.4m wide. The inlet chamber is followed by

Penstock pipe of 1.9m dia. with inlet velocity of 2.54m/sec.

2.8.6.2 Design Aspect of Forebay

The design of forebay is similar to that of design of settling basin, in general, except that

exit portion is replaced by a trash rack and penstock entrance area. The entrance to the

penstock should fully submerge in its design. The different parts of inlet chamber:

Entrance bay or basin, Spillway, Flushing sluice, Screens, Valve chamber or gate

chamber and Conduit or penstock gate.

It is designed such that 15 secs to 3 mins of the design can be safely stored in the storage

above the minimum pipe submergence level. The limiting velocity in the inlet chamber

should be adopted in between 0.2m/sec to 0.8m/sec and the submergence depth (S)

should be greater than 0.7D; where D= Diameter of penstock. Then total depth of tank =

Free Board + S + D + Settling Zone; where, S = Submergence head and setting zone > 0.3

D. The submergence head should be more than

2v1.5

2g . The width of tank can be found by

using the relation:

QB =

h v ; where, v = Limiting velocity adopted and h = S + D.

27

The volume of inlet chamber is calculated by quantifying the volume of water stored

within the plant startup time such that the depth of the inlet chamber should be enough to

dissipate the overflow during upsurge and drawdown. And effective volume of the inlet

chamber, V = Q t where, t = retention time 15 secs to 3 min

And the length of the tank can be calculated as:

VL =

B h .

The length of the spillway in the inlet chamber is designed by simple weir formula.

Qspillway = Cw *Lspillway *(Hovertop)1.5

Where, Lspillway = Length of the spillway

Hovertop = Head over the spillway

Cw = Coefficient of discharge (depends upon weir profile)

2.8.6.3 Protection measures for Forebay

The forebay is located at a flat area nearby river side. Fencing is done around the inlet

chamber so that no one could enter. The excess water from the inlet chamber is allowed

to spill from the spillway structure constructed on it. This water is safely discharge to the

river using on open channel constructed for the purpose. Gates are used at inlet and outlet

for its safe operation.

2.8.7 PENSTOCK

2.8.7.1 General

The potential energy of the flow at the inlet chamber is converted into the kinetic energy

at the turbine of a hydropower plant via the pipe known as penstock. It has to bear a very

high pressure caused due water hammer effect at the sudden closure of the gate by

governing mechanism of the turbine. Penstock should be smooth enough so as to result

minimum head loss while flowing water and it corrosion resistance from durability

aspect. The thickness should be sufficient to resist hoop stress developed by water

hammer pressure and normal pressure not exceeding the allowable stress. Penstock

alignment must be straight to avoid head loss at bents and the extra cost of anchor block

unless it is mandatory by site condition. The penstock may be either embedded or

28

exposed as per topography, location of inlet chamber/ Surge Tank, Powerhouse and

construction easiness etc. The penstock alignment should start where the ground profile

gets steeper. An ideal ground slope would be between 1:1 and 1:2 (V: H).

2.8.7.2 Design Criteria for Penstock

For a particular head and discharge, there may be several options for the size of penstock

according to continuity equation (Q=AV). Also, head loss increases squarely with

increase in velocity as per Darcy-Weishbach equation:

2

l

flvh

2gd .

So, a smaller size penstock saves cost of construction material but the loss of energy due

to loss of head takes place and vice versa. Due to this fact, we need to deduce optimum

diameter which has minimum cost and minimum loss of energy. Water hammer pressure

in excess of normal water pressure can be expressed in equivalent water column height

as,

om c

Vh = V ×

g , Where Vo= Velocity of water in penstock,

Vc = Velocity of wave

mm

K 1; K

1 D

K tE, Where K = Bulk

Modulus of water, D = Diameter of penstock, t = thickness of penstock, E = Young's

Modulus of elasticity of steel, ρ = density of water.

Also, thickness of pipe

P dt

2 ; Where, P = total pressure in pipe and σ = Permissible

hoop stress of steel in pipe.

For small hydro project, wave velocity is calculated as a = 1440/ √ (1 + (2150*D/ (E*t)))

m/s, and surge head is calculated as, K = (L*V/ (g* h(gross) * T))2

and h surge = (k/2 ±

√ (K+ K2/4))* h (gross)

If the penstock has to feed more than one turbine, various factors govern whether use

independent pipes in number equal to the equal to the no. of turbine or use one pipe and

29

bifurcate it at turbine inlet. Length from inlet chamber to powerhouse, construction

feasibility, reliability, transportation and fabrication feasibility are some important factors

to be considered for this.

2.8.7.3 Optimization

Penstock is one of the costly and important structures in hydropower plant. The larger

size incurs more cost of the structure and a smaller size saves the cost of structure but is

associated with increased head loss (which is ultimately the power loss). So, there is

always an optimum size of penstock for which the total cost of loss and the material is

minimum. To seek this size, optimization technique is used. Increase in size tends to

increase the thickness, as thickness is directly proportional to diameter but this relation is

no more valid as the water hammer pressure decreases with increase in size. The

optimization is carried out considering these aspects.

2.8.7.4 Protection Works for Penstock

Penstock is very sensitive structure and its failure is of fatal nature. Exposed penstock is

susceptible to temperature stress and hence, should be provided with expansion joints.

Anchor blocks are used to resist vertical and horizontal forces in the penstock. They

prevent the yielding of penstock. Expansion joints are provided adjacent to them. To

support at intermediate locations and prevent bending stresses, slide blocks are used. The

inner surface of penstock is galvanized and the outer surface is frequently painted to

prevent from corrosion. Frequent checking of the penstock should be done to ensure its

safe operation and to foresee the faults before failure.

2.8.8 ANCHOR BLOCK AND SLIDE BLOCKS

2.8.8.1 General

An anchor block is and encasement of penstock designed to restrain the pipe movement

and to fix the pipe in place during installation and operation. Anchor blocks tend to

prevent the movement of the penstocks due to steady or transient forces including

expansion and contraction forces and water hammer pressures. They provide necessary

reaction to the dynamic forces at the bends. To provide the necessary degree of stability

to the pipe assembly, anchor blocks find their significance. Anchor blocks are provided at

all horizontal and vertical bends of the pipe.

30

Slide blocks are used to support the pipes at intermediate points so as to prevent excessive

bending stresses in the pipe. They resist the weight of the pipe and water and resist the

lateral movement but allow the longitudinal movement of the pipe. So, these blocks are

lighter in weight than anchor blocks and save the overall cost of the support action.

2.8.8.2 Design Philosophy

Water flowing under pressure when diverted from straight path exerts pressure as the

bends. To resist various forces these blocks are designed. The blocks act as the massive

structures and work as the gravity dams. Sliding, Overturning, tension and crushing are to

be checked for the blocks.

Provision for Slide Blocks (Support Piers)

The support engages less than the full perimeter of the penstock, generally between 90

and 180 degrees of arc, and typically 120°. These are simpler to construct than full

perimeter ring girder supports, but generally are spaced closer together than the ring

girders. It is usually spaced between 6 to 8 m between the anchor blocks. It is constructed

of concrete 1:3:6. Design procedure is same as that of the anchor blocks but only the

combination of load is different.

Provision of Expansion joints

Mechanical joints either expansion joint or bolted sleeve type coupling is used in both

exposed and buried penstocks to accommodate the longitudinal movement caused by the

temperature changes and to facilitate the construction. The joints shall allow for

movement where differential settlement or deflections are anticipated.

Expansion joint permit only the longitudinal movements. The joints are used primarily

with aboveground installations and are located between the supports at the points where

the penstock deflections are of equal magnitude and direction. These joints divide the

barrel shell into separate units, which are watertight, but structurally discontinuous. It

should be provided just below the anchor block. Length of the expansion joints tL

2.8.8.3 Construction

Anchor blocks are the support of the penstock and are constructed to meet this purpose.

As the penstock is circular, the anchor blocks are made to fit the curve surface. Saddle

31

supports are used in it and a sufficient cover is provided above the pipe for adequate

fixity.

2.8.8.4 Mode of failure and safety against

Anchor blocks are designed similar to the gravity dam. The blocks are to be designed to

resist overturning, sliding, crushing and tension failure. A firm foundation is required for

the blocks. The blocks should be prevented from gulley erosion due to rain water.

2.8.9 POWER HOUSE

2.8.9.1 General

Power house accommodates electro-mechanical equipments such as the turbine,

generator, switch gear, control room, engineer's room, reception room and operator's

accommodation. The main function of this building is to protect the electro-mechanical

equipments from the adverse weather as well as possible mishandling by unauthorized

persons. The powerhouse should have adequate space such that all equipments can fit in

and be accessed without difficulty. Basically, there are two types of powerhouse i.e.

surface and underground powerhouse. Surface power house is cost effective and is best

suited when the power house is far away from flood plane. On other hand, the

underground powerhouse is located inside the rock mass which makes it more stable

against flood effects and other external forces. Due to underground construction and high

technological methods, the underground powerhouse is highly costlier than surface ones.

Some powerhouses are located as semi-underground structures being partly on surface

and partly underground.

2.8.9.2 Power House Size

Power house size mainly depends on the discharge, head, type of turbine and generator,

number of units and the general arrangement in the power house. The size of the power

house should be sufficient to house all the components. Sufficient clear space should be

available for installation of various components and for maintenance purpose.

2.8.9.3 Height of Power House

Height of power house is fixed by the dimensions of lower turbine block and its

superstructure. Height of the lower turbine block from the foundation to the floor of the

32

machine hall is to be determined by the dimensions of the turbine. The height of the

power house should be sufficient for the installation of turbine, generator and shaft and

gear mechanism. There should be sufficient space for removal and overhaul of any of the

components without disturbing other components. Sufficient clear space is also provided

for crane operation etc.

2.8.10 TAILRACE

2.8.10.1 General

Tailrace is the channel into which the water is discharged after passing through turbine. If

the power house is close to river, the outflow may be discharged directly into the river.

But when the river is far off from the power house, one may have to construct a channel

or pipes according to topography of the site between the power house and the river. The

tailrace should be designed and maintained properly so that excessive aggravation and

degradation is avoided.

2.8.10.2 Design Criteria

Design of the tailrace is similar to that of headrace channel except that higher velocity can

be allowed in the design without caring for head loss in the channel. High grade of

concrete is required to resist erosion of tailrace channel due to higher velocity. The

downstream end of tailrace must be protected to prevent the river by erosion or by flow

from the tailrace. The discharge should be disposed off over rock or large boulders. If

erodible slopes exist in the vicinity, a stilling basin may be required to dissipate energy.

2.8.11 ELECTRO MECHANICAL UNITS

2.8.11.1General

A hydropower plant requires a great deal of mechanical and electrical equipment. The

major electrical components are: Generator, Exciters and Voltage regulators,

Transformers, Switchgear, Control room equipment including switch boards. Similarly,

mechanical components are: Shaft, bearing coupling etc for generators, oil circuits and

pumps, compressors and air ducts, braking equipments. The arrangements for lighting,

water supply and drainage should also be provided.

2.8.11.2 Turbine

33

2.8.11.2.1 General

Hydraulic turbines are machines which convert hydraulic energy into mechanical energy.

The mechanical energy developed by turbine is used in running an electric generator

which is directly coupled to the shaft of the turbine which in turns converts mechanical

energy into electrical energy. Based on the energy conversion, turbines are classified as

Impulsive or Active and Reactive Turbines.

Impulse Turbine

The turbine, in which pressure head or potential energy of water is converted into the

kinetic energy of water in the form of jet of water issuing from one or more nozzles and

hitting a series of buckets mounted on the periphery of the wheel, at atmospheric pressure

is called impulse turbine. It is used for high head and low discharge. Pelton and Turgo

turbines are the examples of the impulse turbine.

Reactive Turbine

The turbine, in which both kinetic energy and potential energy of water is utilized to

rotate the runner or the turbine is called the reactive turbine. The water flows through the

runner under kinetic and potential energy. The turbine runner is submerged and water

enters all around the periphery of the runner. Water is taken up to the tailrace by means of

a closed draft tube and thus whole passage of water is totally enclosed. Francis, Kaplan,

Propeller, Deriaz turbine etc are the examples of reactive turbines.

2.8.11.2.2 Design Philosophy

Selection of suitable type of turbine for the project depends upon several factors like

head, discharge, power production, load condition and corresponding efficiency, quality

of water, tail water level, size, construction feasibility etc. Selection of turbine is essential

for the layout of the powerhouse, approaching and discharging pipes, conditions of

construction and exploitation and techno economic parameters. The turbine is selected

from the following basic criteria:

Head and discharge:

High head and low discharge -Pelton turbine

Medium head and medium discharge - Francis turbine

34

Low head and high discharge - Kaplan turbine

Specific speed:

10 to 50 - Pelton turbine

80 to 400 - Francis

300 to 500 - Kaplan – Diagonal

450 to 1200 - Kaplan – Axial

For Tadi Small Hydropower Project, Francis Turbine is selected. The selection was made

by using the head vs discharge graph, which is attached in annex. Because of medium

head of 74.12m, we have selected Francis turbine as it has better efficiency than other in

part load operation. Discharge variations in the river would happen as per FDC. For at

least four months, FDC shows that discharge is nearly equal to or less than half of design

discharge. Because of this reason we used two turbines of same capacity so that one will

be operated at part loads not by reducing the efficiency to practically low level. Also, two

sets of turbines increases the degree of reliability. In our project, two Francis turbines are

used with equal capacity. The runner diameter of the turbine is 0.675m.

2.8.11.3 Generator

Generator is 3-phase synchronous machine having the speed range of 70 to 1000 rpm. It

may have either vertical shaft alignment or horizontal shaft alignment. The vertical shaft

alignment is usually preferred for medium and large installation. The stator of generator is

manufactured in a number of segments which are then joined at the site. The entire stator

assembly is embedded firmly in concrete foundation. The generator voltage depends upon

the electrical design (which is bounded by the scope) but the normal range is between 6 to

18 KV.

2.8.11.4 Exciters

The poles of the rotor have to be fed with field current, which is achieved through

excitation system. This is known as a static excitation scheme. The generator bus bars

feed current to a step-down transformer to bring down its voltage 230 V which is

35

converted into d.c. current with the help of thyristors converters. The main requirement of

the exciters is the reliability with a steady and stable excitation current. The whole

excitation system is made automatic to achieve quick and accurate control.

2.8.11.5 Ventilation, Cooling and Lubrication

The generator cooling can be achieved by air circulation through the stator ducts. This

may be a closed circuit air cooling system which feeds air to the blades of fan provides on

the rotor. In such cases about 3 m³/min. of air per KW of the generator loss will be

needed. Cooling by water is the common process. There is an elaborated lubrication

arrangement to provide lubrication to the bearing. It consists of a main and stand by oil

pump which are driven by induction motor. The lubrication circuit is provided with

circulating pumps.

2.8.11.6 Transformer

The generator voltage (6.6 to 11 KV) has to be step up to the transmission voltage level

(33 KV) to minimize loss. This is achieved by the use of step up transformer. The

transformers are oil filled for insulation purpose as well as for cooling purposes.

Transformer problems need a very close scrutiny. It is also a great fire susceptible in view

of the substantial quantity of oil in such a close proximity to power cables. A specially

designed firefighting equipment is always included at the transformer site. The location of

the transformer can be either indoors or outdoors. For surface power stations, outdoor

locations are common. The outdoor location may prove to be more economical and less

hazardous hence the outdoor location is preferred. The numbers of transformer may be as

many as the generating units. Each transformer is sufficiently isolated from other so as to

contain the fire in case of accidents. Besides the main transformer, there should be

auxiliary transformer (step down) for power house lighting and other use.

2.8.11.7 Control Room Equipment

The modern hydroelectric stations have a centralized control for its various components

such as machine starting and stopping, machine loading and frequency control, Generator

and system voltage control, machine running supervision and Hydraulic control. All these

components are controlled entirely from a control room.

36

2.8.11.8 Switchgear

The term switchgear is a general term applied to all the variety of the apparatus in the

power house employed for making and breaking the circuit. It may consist of switches,

isolators, surge arresters and circuit breakers. The cost of switchgear depends upon the

cost of bus bar voltage. The bus bars and switches can be provided at the generating

voltage before the transformers or at the transmitting voltage or after the transformers.

The choice depends up on the current and fault rating which in turn influence the design

and cost of the switchgear. If the switchgear is at the generating voltage, it is normally

preferred to locate it indoors. If it is at transmitting voltage, it is usual to locate it outdoors

in the area known as switchyard. The high voltage circuit breakers usually use oil for

insulation. The size of the switchgear assembly depends up on the individual machine

size as well as the electrical clearance of the air, specified in the standards. Both the

switchgear and transformers, particularly located outside, have to have adequate light

protection.

2.8.12 TRANSMISSION LINE

2.8.12.1 General

Energy generated at the power station has to be carried to the consumer's premises

through a network of transmission and distribution lines. Transmission lines transmit bulk

electrical power from power stations to load centers in the form of either underground

cables or overhead lines. Transmission system of an area is known as grid. The different

grids are interconnected through tie lines to form a national grid. Transmission voltages in

Nepal are 33 KV, 66 KV and 132 KV and planning to transmit at 220 KV. The high

voltage transmission lines transmit electrical power from the sending end sub-station

(Power Station) to the receiving end stations. The transmission facilities affect the cost

and reliability of energy supplied to the consumers to a great extent.

2.8.12.2 Design Aspect of Transmission Line

The choice of the most economical voltage for transmission line requires a detailed study

of many technical and economical factors. The power capacity and distance of

transmission are specified. The detail design includes the line voltage, size of phase

conductor, span, spacing and configuration of the conductors, numbers and size of earth

wires, number of insulators, clearance, sag under operating and erecting condition etc.

37

The transmission line for the purpose of economy is required to be constructed at lowest

cost. This is achieved by optimizing the tower height and the span length. This will

reduce the overall cost of line. While deciding the length various factors such as voltage,

public safety and Government's regulation must be considered.

Once these design features are available, the voltage regulation and efficiency can be

calculated. In case these quantities are not within the prescribed limits, a revision of the

design is necessary. Most of the parameters mentioned above are beyond the scope of this

project work. The cost and performance of the line depend, to a great extent, on the line

voltage. An empirical formula for the optimum voltage is:

0.5L P

V = 5.5 +1.6 100 ; Where,

V = Line voltage in KV; L = Distance in Km; P = Power in KW. A standard voltage

nearest to this value should be adopted. The above formula gives only a preliminary

estimate.

38

CHAPTER 3

DESCRIPTION OF PROJECT

3.1. LOCATION OF PROJECT SITE

The proposed project “Tadi” is located in Thaprek VDC of Nuwakot district in the central

development region of Nepal. The proposed site lies between 27º55’36” N to 27º55’0” N

latitude and 85º21’08”E to 85º19’15” E longitude. The elevation of the proposed site area

is within 796.5 m – 725 m. Tadi khola is one of the snow fed river and there are large

number of rounded boulders deposited in either side of the bank of the proposed site. It

finally merges to Trishuli river.

The catchment area of the river basin at intake site is approximately 247 sq. km. The

hydraulic structures like intake, approach canal and power house is proposed to run

through the right bank of the river. The intake site is located at Chamate-Kavre Dovaan

and the powerhouse site is located near the intake of Simara Irrigation Scheme.

3.2 ACCESSIBILITY

There is direct transportation from Kathmandu to Trishuli Bazar which branches off from

Ganagate to the project site. There is a motorable road(earthen and partially graveled) of

about 24 km from Ganagate to the project site. It takes about 6 hours from Kathmandu

and about 3 hours from Trishuli to reach the project site.

3.3 TOPOGRAPHY AND PHYSIOGRAPHY

The Tadi khola catchment lies within the Gandaki river basin. About, 3% of the

catchment is covered by snow whereas 26% is covered by rock and meadow. Similarly,

about 46% of the area is of agriculture land whereas 25% of the catchment area is covered

by forest. The Tadi khola flows with an average river slope of about 1 in 33. However, it

is about 1 in 20 in the project corridor.

3.4 CLIMATE

The average annual rainfall at the region is estimated about 2755 mm from Theissen

Polygon Method. The project area experiences hot and humid climate during June to

39

September and cold climate during November to January. The mean monthly temperature

varies from 14oC to 25.4

oC. The relative humidity varies from 33% to 93% over the year.

3.5 AVAILABILITY OF CONSTRUCTION MATERIAL

Local construction materials like sand, stones, aggregates etc. are available on the

construction site. Other construction material like cement, steel can be brought from

Kathmandu.

3.6 HYDROLOGY OF SITE

3.6.1 Collection of Hydrological data

The design of any hydropower project primarily depends upon an accurate assessment of

hydrological data. Hydrology is of great importance throughout all the project phases:

from the preliminary planning to the technical design, to the final management. In each

phase of the project the hydrological contribution is never a standard approach, but is a

result of a fine calibration of methodologies according to the requested degree of

approximation and the available data. The longer the hydrological data the more reliable

is the estimation of design parameter for the project.

The collection of hydrological data mainly includes:

Precipitation in the catchment area.

Stream flow measurement.

The collection of hydrological data is required to:

Study precipitation pattern in the catchment area.

Discharging capacity of catchment area.

Prediction of design discharge.

Prediction of high flow & low flow level of the stream.

Tadi Khola is of the main tributaries of Trisuli River. The catchment area of the river lies

mainly in dense mixed forest. The Tadi Khola is gauged river whose gauging station

number is 448 located at Tadi pul, Belkot, Nuwakot. The distance between the gauging

station 448 & the proposed intake area is about 20km.Since,the catchment area

characteristics of the intake site is more or less similar to the gauge station number 448.

40

So the hydrological data for the proposed hydropower station can be taken from the above

station.

3.6.2 Precipitation

Precipitation data acts as a guide for hydrological data. The record of precipitation is

important to understand the nature of catchment and flow of the stream. The precipitation

can be measured by two types of Rain Gauge, viz: Recording Type Rain gauge & Non

Recording Type Rain gauge.

The measurement of precipitation of the catchment of Tadi Khola was done by

Recording Rain Gauge.

The average precipitation of the catchment area was derived by taking rainfall data from

all the marked station spread around the Tadi catchment by using Theissen Polygon

Method.

Accordingly, the annual average precipitation was found to be 2755 mm.

3.6.3 Discharge and Water Level

3.6.3.1 Discharge

The measurement of stream flow i.e. discharge is amenable to fairly accurate assessment

of hydrological data. Moreover, stream flow is the only parts of hydrological cycle that

can be measured accuratey. There are various methods for determining discharge of river,

viz:

Direct method

Further, there are several ways of measuring discharge by direct method. They are:

Area velocity method

Dilution method

Electromagnetic method

Ultrasonic method

Indirect method

Also, there mainly two ways of measuring discharge by indirect method. They are:

41

Hydraulic structures, such as weir, flumes and gated structures.

Slope area method.

The discharge of Tadi khola was measured using Area velocity method.

3.6.3.2 Water level (stage):

The stage of a river is the elevation of top water surface above any arbitrary datum.

Water level of any river is mainly measured by:

Manual gauge (staff gauge, wire gauge).

Automatic stage recorders.

Float gauge recorders.

Bubble gauge.

The water level of Tadi khola was measured by using manual method.

The gauge discharge can be expressed by rating equation as:

Where,

Q =storm discharge.

G=gauge height.

α=gauge reading to zero discharge.

Cr & β are rating curve constant.

3.6.4 Peak discharge calculation:

The design and projection of hydraulic structures of hydropower plant depends on peak

flow. Basically, such flow is considered for the design of diversion weir, settling basin,

gravel trap, drainage work, canal crossing, tailrace canal and power house. The flood flow

value acts as a safety measure for the design and location of hydraulic structure as well as

superstructures.

The annual maximum instantaneous peak flow of the Tadi khola for the period of 1969-

1995 for station 448 tadi khola was obtained from department of hydrology and

42

metrology (DHM) and those data are used to calculate flood frequency analysis for

different return period.

3.6.5 Frequency analysis

Frequency analysis is used to calculate the peak flow of a stream for any number of return

periods. It is a statistical method which is used for prediction of peak flow. For any

predetermined return period, the general equation of hydrologic frequency analysis:

Where,

xt=value of variate (discharge).

= mean of variate.

K= frequency factor depends upon the period T.

σ= standard deviation of variate.

The commonly used frequency distribution functions for the prediction of extreme peak

flow values are:

Gumbel’s Method

Normal

Log-normal II

Log –normal III

Pearson type II

Log Pearson Type III

The annual maximum instantaneous flood discharge of the Tadi Khola for the period of

1969- 1995 for Station 448 . Tadi Khola , a tributary of the Trishuli River , which is

available from DHM . These data are used to calculate flood frequency analysis for

different return period by using the computer programming FREQ. In FREQ , the

following 6 types of statistical distribution area used and the result drawn thereon is

present below :

Table 3.1 Estimation of peak flow at proposed intake by different methods

Method Std.

Error

Return periods(years)

10 20 50 100 500 1000

Normal 114.6 359.34 398.68 443.31 472.82 532.58 555.66

43

LN II 85.6 352.15 413.81 496.27 560.57 716.04 786.77

LN III 76.6 358.58 430.83 531.45 612.77 819.68 918.02

Gumbel 102.1 336.27 425.91 454.66 504.97 620.34 670.27

Pearson II 82.4 357.45 419.48 499.3 559.06 695.61 754.62

Log P III 79.8 353.29 425.16 528.8 614.66 844.26 958.5

From the above table, it can be seen that the lowest standard error is in LN III. Therefore,

the design flood level at proposed headwork is taken as 612.77m3/s for a return period of

100 years.

The catchment area from the available map shows that there is only 9 km2 added on top

of the catchment area with respect to proposed headwork and there are no rivulets that

can contribute significant amount of flash flood . However, the flood flow as determined

using catchment correlation for 100 years return period is 635.1m3/s.

3.6.6 Establishment of design discharge for power calculation

The major approaches for assessing water availability of hydropower projects are mean

monthly flow and flow duration curve.

These curves give realistic estimation of flow and ascertain economic viability of a

project.

Mean Monthly Flow

Availability of adequate water flow for power generation is indicated by estimation of

mean flow. Mean annual flow gives the potential power of stream but stream flow is

usually less than this flow. To calculate mean monthly flow following methods are used.

MIP method

Catchment Correlation Method

Hydest and Modified Hydest Method

MHSP

3.6.6.1 MIP Method

The MIP method is used for estimating the distribution of monthly flow throughout year

for un gauged locations. The MIP methodology uses a database consisting of DHM spot

measurements. MIP presents non dimensional hydrograph of mean monthly flows for 7

44

different physiographic regions. These hydrographs presents monthly flow as ratio of

flows. The measured flow is then used with the regional non- dimensional hydrograph.

3.6.6.2 Catchment Correlation

The mean monthly flow derived at the proposed intake using gauge station no 448 data

and then transferred to station 447.4 at Ruwatar gauging station which is very close to the

proposed power house area. Correlation factor as derived using time series data of station

447.4 and 448 are used to develop mean monthly flow at Ruwatar station 447.4.

3.6.6.3 Hydest and Modified Hydest Method

The Hydest method has been used to estimate mean flow series at the proposed intake.

The method was developed by WECS/DHM in 1990 for evaluating the hydrologic

characteristics of ungaged catchments. For the complete hydrological analysis by this

approach, the catchment area and its distribution in altitude are essenential along with

monsoon wetness index (MWI) of the catchment s. the monsoon wetness index from the

isoheytal map for the project area is taken as 1800 mm. In the present study , this

approach is used to compute long term hydrology and extreme hydrology of the Tadi

khola at the proposed intake. Modified version of HYDEST is also used to analyze the

hydrological parameters of the project.

3.6.6.4 MHSP

The Medium Hydropower study project(MHSP) under NEA in 1997 developed a method

to predict long-term flows, flood flow and flow duration curves at ungauged sites through

regional regression technique. The MHSP method has been used to estimate mean

monthly flow series at the proposed intake site. Daily flows, maximum and minimum

instantaneous flows of 66 hydrometric stations obtained from the DHM are used in the

regression. The input variables are similar to those used in WECS/DHM method.

This approach used both MWI and average precipitation of the area along with the

catchment area of the river.

Table 3.2 Calculation of mean monthly flows by different methods.

Month MIP Catchment

Correlation

HYDEST MODIFIED

HYDEST

MHSP

Jan 9.73 5.82 3.11 4.49 3.66

45

Feb 6.75 5.33 2.65 3.73 3

Mar 4.95 4.88 2.4 3.28 2.77

Apr 3.59 4.41 2.5 3.28 3.45

May 6.75 8 3.42 4.55 3.95

Jun 11.23 16.66 11.92 14.25 12.97

Jul 48.59 41.58 36.58 39.76 38.62

Aug 89.72 75.4 43.95 55.47 45.71

Sep 74.75 54.75 33.79 39.76 35.49

Oct 37.39 24.42 14.73 17.99 16.45

Nov 17.94 13.02 6.55 8.02 7.92

Dec 13.46 6.81 4.24 5.44 5.12

Annual 27.05 21.76 13.82 16.61 14.93

Figure 3.1 Mean Monthly Flows by different methods

Conclusion of graph

The dry monthly flow as obtained using catchment correlation method has given

relatively high value. Thus, this method is preferred.

3.6.7 Flow Duration Curve

The flow duration curve is a probability discharge curve that shows the percentage of

time ; a particular flow is exceeded or equaled. In a runoff river hydropower project, it is

useful to know the variation of flow over the year so as to make ease to select the most

0

10

20

30

40

50

60

70

80

90

100

0 1 2 3 4 5 6 7 8 9 10 11 12

Mea

n M

on

thly

Flo

w(m

3/s

)-->

Months-->

Catchment Correlation

MIP

HYDEST

MODIFIED HYDEST

MHSP

46

appropriate turbine configuration as well as for project optimization. Flow duration Curve

at different probability of exceedence is presented in table 4.2 and in figure 4.2.

Table 3.3 Flow Duration Curve

Month Catchment Correlation

Descending Order

Probability of exceedence(%)

Jan 5.82 75.4 8.333

Feb 5.33 54.75 16.667

Mar 4.88 41.58 25.000

Apr 4.41 24.42 33.333

May 8 16.66 41.667

Jun 16.66 13.02 50.000

Jul 41.58 8 58.333

Aug 75.4 6.81 66.667

Sep 54.75 5.82 75.000

Oct 24.42 5.33 83.333

Nov 13.02 4.88 91.667

Dec 6.81 4.41 100.000

Figure 3.2 Flow Duration Curve

0

10

20

30

40

50

60

70

80

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

Dis

cha

rge(

m3

/s)-

->

Probability of exceedence(%)-->

47

Table 3.4 Annual Energy Calculation

The design discharge is taken as 7.2 m3/s corresponding to 65% exceedence.

3.7 SUSPENDED MATTER AND BED LOAD

3.7.1 General

The sediment transport in the river is a complex phenomenon. A careful study of the

sediment inflow and assessment of deposition is of major importance in planning of any

hydropower project. The mineralogical analysis of the sediment sample is necessary to

determine the presence of hard and soft material contents. The sediment collected from

samples of fine and coarse deposits are generally used for mineralogical analysis.

The size of the sediment greater than 100mm is taken as bed load. The sediment having

size less than 100mm is taken as suspended matter. The major effects of sediment and

minerals for run off river projects are:

Serve abrasion, wear and tear of turbine depending on nature of sediment

overtime.

Reduction of storing capacity of dam due to deposition of sediment.

Cost of construction and maintenance of overall project increases.

Mont

h

Catchment

Correlatio

n

Descendin

g Order

Probability of

exceedence(%

)

Design

Discharge(m3/s

)

Power(MW

)

Energy(GW-

H) Remark

Jan 5.82

75.4 8.333 7.2 4.140 3.020

WET

ENERG

Y

Feb 5.33

54.75 16.667 7.2 4.140 3.020

Mar 4.88

41.58 25.000 7.2 4.140 3.020

Apr 4.41

24.42 33.333 7.2 4.140 3.020

May 8

16.66 41.667 7.2 4.140 3.020

Jun 16.66

13.02 50.000 7.2 4.140 3.020

Jul 41.58

8 58.333 7.2 4.140 3.020 21.142

Aug 75.4

6.81 66.667 6.81 3.915 2.857

DRY

ENERG

Y

Sep 54.75

5.82 75.000 5.82 3.346 2.441

Oct 24.42

5.33 83.333 5.33 3.064 2.236

Nov 13.02

4.88 91.667 4.88 2.806 2.047

Dec 6.81

4.41 100.000 4.41 2.535 1.850 11.431

Total 32.572

48

Degradation of river downstream of the dam resulting in instability on either

side of banks.

The concentration of bed load for any hydropower plant is approximately taken as 30% of

the maximum discharge which is used for sluice gate design. Similarly, the concentration

of suspended matter is taken as 10-20% of the maximum discharge of the river which is

used for the design of settling basin. The realistic and objective assessment of

sedimentation is necessary for both project economic and environment consideration.

3.7.2 Morphological influence upon solid matter transport

The size of the sediment in the river system of our country usually varies from fine sand

to big boulders. Tadi khola is one of the snow fed river and there are large number of

rounded boulders deposited in either side of the bank of the proposed site. At the

meandering section of the Tadi khola, high deposition of sediments takes place on either

side of the river. The bed characteristic of the river is rocky and mobile. The slope of the

river is relatively high as it is in the hilly region and the sediments are transported to a

larger distance.

3.7.3 Measurement of amounts of suspended matter and bed load

The estimation of the sediment is not an exact science, largely due to various processes

which produce the sediment. The best measure of sediment yield from a catchment is:

Observation of sediment volume deposited in reservoir or river meandering

parts

To take samples of suspended sediment to obtain annual yields.

Since there are no reservoir in the river so one has to visualize the sediment type and

nature from outer bend of meandering section of the river.

Tadi khola has a sediment measurement station located some 45km upstream from the

proposed intake. Sediment transport rate are high in Himalayan River and specific

sediment transport rates exceeding 10000t/km2/year have been reported (WEC, 1987).The

average monthly sediment load in the river as recorded in gauging station 447 Betrawats,

Nuwakot which is in close proximity of the project is shown in the table below.

49

Table 3.5 Average Sediment Loads at station 447 (ton/km2)

Area

(km2)

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

4110 19 24 18 41 158 897 2995 4362 74 362 57 28 4

The catchment area of station 447 partly lies in Tibet and thus difficult to estimate

sediment influx. The above table can be used only to make reference for establishment of

sediment load at the proposed project.

A tentative estimation of the sediment transport at proposed site is made using empirical

formula as suggested for Hill Irrigation System.

Where,

(Mm3/100km

2 ) from 100 km

2 of basin area

A = Total basin area.

Considering the density of sediments as 1.5ton/m3, the annual sediment load for Tadi

khola at the proposed headwork is tentatively estimated as 1.8 million ton. The average

sediment concentration for the design of settling basin to make the project is taken as

5000ppm or 5kg/m3.

As per DPR, for mineralogical analysis two samples from the river bed were taken.

The mineralogical analysis showed high content of quartz in suspended (50-60%).Since

there is high percentage of quartz content, there will be higher probability of wearing of

turbine. As such a rigorous settling criterion for the design of the design of settling basin

should be adopted.

3.7.4 Geological Study of the project Area

3.7.4.1 General

Geological and geo-technical study plays an important role in hydropower projct. The

geo-technical issues carries a greatest risk in construction and investigation are necessary

to clarify the magnitude of the risk and how they can be reduced or managed.

50

Geologically and Geomorphlogically, Nepal is divided into five principle morphologic

regions, which reflect the underlying geology, and are continuous as linear belts

throughout the length of the entire country. These five regions are:

Tibetan Plateau

Higher Himalayas

Lesser Himalayas

Sub-Himalayas (Siwalik)

Terai (Indo-Gangetic Plain)

Among these region, Tadi Khola is situated in Nuwakot District which lies on Lesser

Himalayas Region .

3.7.4.2 Lesser Himalayas

The broad belt of Lesser Himalayas lies in South of the Higher Himalayas and separated

from them by the Main Central Thrust (MCT) to the north and Main Boundary Thrust

(MBT) to the south, lies. The MCT and MBT, the two major structures separate younger,

weaker rocks in the Siwalik range from the stronger metamorphosed rocks of the Higher

Himalayas. The average width of the zone is 60 kilometer and is characterized by the

mountains of Mahabharat range in the south.

It consists of meta-sedimentary rocks between which thrust slabs of gneiss and high

grade metamorphic rocks are present as observed in the project area. The zone is folded

and faulted and terminated in the south by the MBT, along which it has been moving

southwards since the Pliocene time.

The varied geologic structures and rock types give rise to irregular high relief terrain

showing short ridges, deep gorges and dissected mountains. River terraces are particularly

conspicuous, especially in the long strike valleys, such as the Arun Valley.

3.7.4.3 Major geological structures

The area under consideration lies in Lesser Himalayan Metamorphic rock with slate and

intercalation of Phyllite and occasional band of Quartzite Lenses. The dip direction of the

51

rock is N or NE and is bounded by Dorkhu some 10-15 km in the west and Likhu thrust

in SE about 20-30 km.

3.7.4.4 Geomorphology of project area

Physiographically, the project area lies in the Midland hills bounded by the Great

Himalayan and has elevation of 200 m to 3000 m. Vertical cliffs can be seen in both

banks. It consist of low hills, river valleys and tectonic basin. The topography is less steep

to gentle. The Tadi khola flows from East to West and minor tributaries joins with Tadi

Khola almost perpendicular flowing from North and South direction. The valley is deeply

incised due to high current flowing rivers. The vegetation cover is high at the left bank as

compared to right bank. The slopes are covered with thin colluvial soil deposit while the

bedrock outcrops are exposed at base of the slopes along banks of river. Slope of the

khola at the proposed headworks and powerhouse area is sloped in the order of 1 in 30 or

so.

3.7.5 Conclusion

The project site consists of bedded schist, phyllites and meta-sandstones with few

quartzite band. The bed rocks are exposed at few locations especially along the river bank

and foot hill.

52

CHAPTER 4

DESIGN OF HYDROPOWER COMPONENTS

4.1 HYDRAULIC DESIGN

4.1.1 Intake

1. Qdesign= 7.2m3/s

2. 10% extra for flushing = 0.72

10% extra for settling basin = 0.72

Qrevised total = 8.64m3/s

3.

From DPR,

Bed slope = 1:20

Bed width of river channel = 7m(approach channel)

Assuming, n = 0.065

Since, the river reach is weedy containing deep pools & sludge.

Also, the river cross section is assumed to be rectangular, as the river is bound by rocky

topography on either side.

From manning’s equation,

Q = A×V

Q = b × h× ×R2/3

×S1/2

8.64 = b × h× × )2/3

×( )1/2

8.64×0.065×(20)1/2

=

2.511 =

53

4.

From iterative calculation,

h=0.57 0.6m

Assuming, vintake = 0.8m/s(0.6 v 0.8m/s)

For width,

A = Q/V

A = 8.64/0.8

= 10.8m2.

Again, Q = A×V

B = = 10.8m.

As the bed width of river channel is 7m.

So, width of intake should not exceed 7m.

So, using b=7m.

Now,

H = = = 1.542m

Then,Actual width (b) = = = 7.2m.

No of bays

A single bay cannot be used as the gate of 7.2m is difficult to operate (design, maintenance

& transport).

Assuming two bays,

Each bay width= = = 3.6m.

54

Again the gate of 3.6m width is also very large to operate.

So, assuming three bays,

Each bay width = = = 2.4m.

Size: 3×(2.4×1.5)m

5. Pier

Using 2 pier each of width 0.5m.

6.

Track Rack

Using 25mm bars @25mm clear spacing at an inclination 80 with horizontal direction.

Velocity of flow in front of track rack = = = 0.8m/s.

From krischmer’s formula,

hLr = kr(t/b)4/3×

sinα.

= 1.67(0.025/0.025)4/3

( sin80 .

= 0.053

Where,

Kr=Loss coefficient depending on shape of track rack (oval shape=1.67)

t=width of track rack bar

b=clear spacing

α=angle of inclination of track rack with horizontal

7. Pond Level

55

Pond Level =RL of bed level + ht. of sill of intake +

ht. of intake + free board

=794+2.5+1.5+0.5

=798.50m.

4.1.2 Diversion Weir

1. Weir height

For 1000 year return period,(hydrological analysis)

Q1000=918.02m3/s. (Source: ICOLD)

R.L. of river bed level U/S = 794.000

From Lacey’s regime flow,

W= 4.75

= 4.75

= 143.91m.

But a diversion weir of 143.91m is not feasible as there is a narrow gauge having width of

35m with rocky base where abutment can be anchored.

So,

L (width) = 35m.

For design discharge,

15% to 20% of flood will pass through sluice gate.

R.L. of weir crest level = Sill level of intake + height of intake opening + free board

= 796.5+1.5+0.5

56

= 798.5m.

Height of the weir = 798.50-794.00

= 4.5m

2. Energy Dissipater (Spillway)

From hydrological calculation,

Q100 = 613m3/s

Assuming 80% of flow through weir,

0.8*Q100 = CL

0.8*613 = 2.1*35*

He = 3.54m.

Now,

h/H e= 4.5/3.54 = 1.3 ~ 1.33.

Since, the ratio is equal to 1.33. So, the velocity head is neglected. Thus,

He = Hd = 3.54m

Initial depth of jump,

Eo = h + Hd

= 4.5+3.54

= 8.04m

Length of weir (L) = 35m

Specific discharge (q) = Q/L

= 613/35

= 17.514 m3/s.

Calculation of Tail water depth:

57

q= S1/2

/n

17.514 = 201/2

/0.065

yo = 2.7m

As 20% of discharge goes through sluice and so energy dissipation for remaining 80% of

discharge is required to be designed.

Q = 0.8*Q100/L = 0.8*613/35 = 14.011m3/s

Applying energy equation at crest & section before jump,

H + Hd = y1 +

4.5+3.54 = y1 +

y1=1.789m

Now,

Velocity (v1) = q/y1 = 14.011/1.789 = 7.83m/s.

Froude number, ( ) =

=

=1.87

As F<2.5, the jump is weak and energy loss is low. No devices are used to dissipate the

energy.

3. Design of weir

i. Weir Profile

Upstream profile

Y = 0.724(x+0.27Hd)1.85

+ 0.126Hd – 0.4315Hd0.375

*(x+0.27Hd)0.625

On Solving,

58

Table 4.1Weir Upstream Profile.

Downstream profile

Xn

= kHd(n-1)

y

K=2.0 , n=1.85

X=2.6Y0.5405

On Solving,

Table 4.2 Weir Downstream Profile.

Y X

0 0

0.5 1.78

1.00 2.60

1.50 3.23

2.0 3.78

2.50 4.266

3.00 4.71

3.50 5.12

4.0 5.50

4.50 5.86

Y X

0 0

0.0985 0.62

0.246 0.998

59

ii. Weir length

Total head loss (HL) = 802.04-793.10

= 8.94m

Therefore, Total creep length (LT) = c*H2

= 9*8.94

= 80.46m

Length of downstream (pucca floor)

L2 = 2.21*c* (for weir having no shutter)

= 2.21*9*

= 18.806

= 18.81m

Downstream talus (Stone pitching)

L2+L3=18c*( * ) (for no shutter)

18.806+L3=18*9*( * ) where, q=Q/B = (Q100*0.8)/35=14.011 m3/s/m

L3 = 47.40m

Upstream stone talus;

L4 = L3/2 = 23.7m

Bottom width of weir,

B = = = 6.35m < 6.858m (OK)

B = 6.9m

60

Lacey’s scour depth (R) = 1.35(1/3

= 1.35*(

1/3

= 7.845 7.85m

Depth of upstream sheet pile, from HFL

= 1.5*R

= 1.5*7.85

= 11.775 11.78m

R.L. of bottom of upstream sheet piles = R.L. of upstream HFL - 11.78

= 802.04-11.8 = 790.26m

Therefore, Depth of upstream sheet pile from upstream bed = R.L. of u/s bed –R.L. of

bottom of sheet pile

= 794.0 – 790.26 = 3.74m

Depth of u/s sheet pile from HFL = 1.5R (1.5R to 2.5R)

= 1.5*7.85 11.78m

R.L. of bottom of d/s sheet pile = R.L. of d/s HFL

= 793.10-11.78

= 781.32m

Depth of d/s sheet pile from d/s bed level

= R.L. of d/s bed level - bottom of R.L. of d/s sheet pile

= 790.4-781.32

= 9.08m

Length of u/s pucca floor(L1) = LT-L2-B-2*sheet pile u/s-2*sheet pile d/s

61

= 80.46-18.81-6.9-2*3.74-2*9.08

= 29.19m

iii. Floor Thickness

=

= *8.94

= 2.51m

D/S floor thickness (td) =

= = 1.57m

Increase thickness by10% for safety

Therefore, Td = 1.1*1.57m = 1.727 m 1.73 m

For U/S,

Provide pucca floor of minimum 1m thickness U/S of the weir body and 1.5m thickness

below the weir.

iv. Check of exit gradient:

Maximum seepage head = HL = 8.94m

Depth of d/s cut off (d2) = 9.08m

Total floor length (tl) = 80.46-2*9.08-2*3.74 = 54.82m

= = = 6.037

Λ =

= 3.56

GE = = 0.166 = 1/6.02 (safe) (1/4<GE<1/7)

62

4. Sluice way:

Q100 = 613m3/s.

Head over the crest = 3.54m

Discharge for sluice gate = 20% of Q100

= 0.2×918.02 = 183.634m3/s.

Assuming, size of sluice gate opening to be 2.5m, to pass boulder of maximum size

greater than 2m.

R.L. of bed level of sluice way = sill level of intake – height of sluice way

= 796.5-2.5 = 794m.

R.L. of water level above weir crest = 798.5+3.54m

= 802.04m.

Tail water depth = 2.7m (Calculation of TWL in Weir Design)

R.L. of d/s water level = R.L. of bed load sluice (d/s bed level) + depth of water at d/s.

= 790.4+2.7

= 793.1m.

Now,

Δh = W.S.L. at u/s – W.S.L at d/s

= 802.04-793.1

= 8.94m.

Then,

Q = c × A (2×g×Δh)1/2

Q100×0.2 = 0.6×A*(2×9.81×8.94)1/2

613×0.2 = 0.6×A*(2×9.81×8.94)1/2

63

A = 15.43m2.

Using h = 2.5m,

So, b = = 6.17m 6.2m

Adopting 2 bays, each of width 3.1m.

Size of sluice: 2×(3.1×2.5)m

4.1.3 Design Gravel Trap

1. Given Parameter:

Q = 1.2×7.2

= 8.64 m3/s

Size of the particle = 2 mm or large

Flushing system:

High flood: continuous

Low flood: Intermediate

2. Settling Velocity:

Iteration 1:

= (For 1000<R<10000;Cd=0.4)

= 0.323m/s

= = 489.393

For R = 0.5 to 1000,

64

Cd = + + 0.34

= + + 0.34 = 0.524

Iteration 2:

ω =

= 0.282m/s

R =

=

= 427.272

For R=0.5 to 1000,

Cd = + + 0.34

= + + 0.34

= 0.541

Iteration 3:

ω=

= 0.28m/s (OK)

3. Transit velocity (Horizontal):

V = a ; (a=36 for 2mm particle)

= 36

= 50.91 51cm/s

4. Vetter’s method:

65

η (efficiency) = 90%

ω = 0.28m/s

Q = 8.64m3/s

0.9 = 1-

0.1 =

ln(0.1)=

As=

As=71.051m2

Let, H=3.5m

Width (b) = (β=correction factor)

= = 7.26m 7.5m

Length (L) =

=

= 9.78m 10.0m

Effective length = = 11.5m

Size: 11.5×7.5×3.5m

5. Transition design

Inlet length: No inlet transition is provided as gravel trap is constructed immediately after

intake.

Outlet length = /

= 9.5m

66

Overall length = 11.5+9.5

= 21m

6. Flushing discharge

Qflushing = 10% of Qdesign

= 10%×7.2

= 0.72m3/s

Size of particle for hydraulic flushing = 10mm.

Let us use 0.5×0.5 sized flushing canal.

Longitudinal slope=1in 25

V = R2/3

S1/2

= ( )2/3

( )1/2

(R= )

= 4.03m/s

Shear stress ( o) = RS

= 9.81×0.166×

= 0.0654KN/m2

= 65.4N/m2

Critical shear stress ( c) = 0.056× ×d×(Gs-1)

= 0.056×9.81×0.01×(2.6-1)

= 8.78N/m2

Since, o> c (OK )

7. Opening for flushing canal

Head over orifice = 3.5+0.2

67

= 3.7m

Q = CdA

0.72=0.6×A×

A=0.140m2

h=

h=0.28m

Size: 0.50 .28m (B × h)

4.1.4 Approach Channel (Gravel trap to Settling Basin)

Assume,

Stone masonry in cement mortar using split stones (1:3),

Height of canal (H) = 1.2m.

Side slope (1:N) = 1:1 (H:V)

Bed slope (s) = 1:500

Discharge (Q) = 1.1×7.2m3/s.

= 7.92m3/s

Manning’s coefficient (n)=0.020

For a trapezoidal section,

Q = 2/32

5/32

])N(12Nn[(B

S)NH(BH

Q = 2/32

5/32

])1(11*20.02[(B

)1.2*1B*(1.2 S

By iterative process,

68

Bottom width (B) = 2.4m.

Take, Free board = 0.3m.

Top width (T) = B + 2HN

= 2.4+2×1.2×1= 4.8m

So, Area (A) = 2.4×1.2+1*1.22

= 4.59m2

Velocity (v) = Q/A

= 7.92/4.59

= 1.725m/s < 3m/s (ok)

Size: Top Width: 4.8m

Bottom Width: 2.4m

Height: 1.5m

Perimeter (P) = 2.4+2*1.2√(1+12)

= 4.59

Hydraulic radius (R) = A/P

= 2.025/3.89

= 0.94.

Check,

Transported material (dmax) = 11RS

= 11×0.94×1/500

= 0.02068m.

= 20.68mm > 0.2mm. (OK)

69

Spillway Design:

For Broad crested weir with rounded edges,

Cw = 1.6

Q = 7.92m3/s

Assume, L = 10m

Now,

Qspillway = CwLspillwayH1.5

overtop

7.92=1.6*10*H1.5

H=0.626m

4.1.5 Design of Settling Basin

Qdesign= Q×1.1(10% for flushing)

= 7.2×1.1

= 7.92m3/s

1. Particle Approach:

Assumption,

Particle size (d) = 0.2mm

Horizontal velocity (v) = a

= 44 (a=44 for 0.1 1mm)

= 19.67cm/s 20cm/s

From Stoke’s law,

Settling velocity (ω) = (ss-1)

70

= (2.6-1)

= 0.0264m/s 0.026m/s

A = where β is the correction factor

=

= 456.923m2

Assuming, = 8 (4 10)

So,

L×B = 456.923

8B2

= 456.923

B=7.55m 7.6m

From deposition criteria,

B = 7.2Q0.5

B = 7.2×(7.92)0.5

= 20.26m ≰ 7.6m

From flushing criteria,

B = 4.83Q0.5

= 4.83×7.920.5

= 13.59 ≰7.6m

So,

L = 8B

=8×7.6

= 60.8m

71

Then,

Q = A × v

7.92 = B×H×0.2 (0.2<v<0.8)m/s

H =

=5.21m 5.3m

2. Vetter’s method:

Sediment Removal Efficiency (η) = 90%

Q = 7.92m3/s

ω = 0.026m/s

We know,

η=1-

0.9=1-

0.1=

Ln0.1=

As=

As=701.4m2

Again,

As = L×B = 701.402

8B2

= 701.402

B = 9.362m 9.4m ≱ 20.26m (deposition criteria)

B = 9.4m ≱ 13.59 (flushing criteria)

So,

L= 8B = 8×9.4 = 75.2m

Then,

72

Q=A × v

7.92 = B×H×0.2

H =

= 4.21m 4.3m

3. Hazen’s method:

Q = 7.92m3/s

ω = 0.026m/s

η = 90%

m = 0.1(0 m 1)

By Hazen,

η=1-[1+mωA/Q]-1/m

0.9 = 1-[-1/0.1

0.1 = [1+ ]

=

= (10)1/10

= 0.259

A = 788.953m2

Again,

A = L×B

8B2

= 788.853

B=9.93m 10.0m ≱ 20.26 (deposition)

≱13.59 (flushing)

And,

L= 8B= 80m

73

Then,

Q = A × v

7.92 = B×H×0.2

H =

= 3.96m 4.0m

Size: (80×10×4)m

Number of chambers:

Area of each chambers (As) = 788.953/2 = 394.476

As = l × b = 394.476

8b2=394.476

b=7.022m and l=8×7.022=56.1m

Again, providing baffle wall of 0.25m,

Size of each chamber: (56.1×7.1×4)m

Therefore, total width = 7.1×2+0.2 = 14.4m

Inlet profile:

Provide inlet expansion ratio of 1 in 5.

Length of inlet transition = ( ×5/2 = 30m

Outlet profile:

Since, the approach length of the canal is only 36.90 m & it is not possible to make

forebay separately in such a short distance. So, forebay is kept adjacent to the settling

basin replacing outlet contraction zone.

4. Height of the sediment deposited

Flushing time (T): 8 Hrs

Design Discharge (Q): 7.2×1.1 =7.92 m3/s

74

Sediment Concentration (ss) = 5000 ppm = 5 Kg/m3

Density of sediment ( ) = 2600 kg/m3

We have,

T =

Where, = Volume of deposition zone

= volume of inflow sediment

So,

= (7.92×8×60×60)×5/2600

= 438.646 m3

5. Orifice:

Discharge (Q) = A × v

= * *R2/3

*S1/2

0.36 = * *(2/3

*(1/2

D = 0.6m.

Orifice is designed for half flow for better flushing and the discharge for each orifice is

half of 10% of Qdesign.

Clearance on either side of pipe = 0.3m

So,

Total width = 0.3+0.6+0.3m

=1.2m

Height = 0.3+06+0.3

=1.2m

Size: (1.2×1.2)m

75

6. Using two Dufour chamber

Longitudinal Slope: 1:50

Lateral Slope: 1:1

Storage Depth (h)

From the geometry of figure (Refer annex III-3)

h = = 3.15m

Check volume,

Volume calculated = ( (0.8+7.1)*3.15*56.1)*2

= 1396.048m3 > Vdeposited. (OK)

4.1.6 Design of Forebay

1. Volume

Diameter of penstock (D) = 1.900m

Length of penstock (L) =2238.73m

Design discharge (Q) = 7.2m3/s

Generation time (T) = 2minutes

Generation volume (V) = Q×T

= 7.2×2×60

= 864m3

Limiting velocity (v) = 0.3m/s (adopted) (0.2<v<0.8)

2. Submergence depth

S ≥1.5D = 1.5×1.9 = 2.85m

≥ = = 6.8×10-3

76

Adopt, S = 2.85m

3. Settling zone = 0.5m (0.3<Sz<1.5m)

4. Adopt free board = 0.3m

5. Total depth = 0.3+2.85+0.5+1.9

=5 .55m

6. Effective depth (h) = S+D

=2.85+1.9 = 4.75m

7. Breadth (B) = 14.4m

(as forebay is adjacent to settling basin)

8. Length (L) =

= =12.63m~13m

9 .Check surge

Area of penstock (Ap) = 1.92

= 2.835m2

Velocity of water in penstock (Vp)= = =2.539m/s (Vp≱5m/s)

Area of cross section (As)=14.4*13=187.2m2

Surge (Zmax)=V

= 2.539

=4.72m 4.8m

hf= = = 1.525 m (Moody’s Chart)

77

p0= = =0.317m

Zup= Zmax(1- +

= 4.8(1- +

= 3.834m.

Zdown=Zmax( )=4.8*

=2.759m < hs=2.85m (OK)

10. Submergence check

Effective depth - max.surge (h-Zmax) = 4.75-4.8=-0.05 0

Since, the calculated maximum surge is greater than effective depth. So, effective height

is revised.

Put,

h=4.8m

So, Total depth= F.B. + 4.8 + settling zone

= 0.3+4.8+0.5

= 5.6m

Also, Lateral check

= =0.88>0.4(OK)

Therefore, Size: 13×14.4×5.6m

11. Spillway Design:

For Broad crested weir with rounded edges,

Cw=1.6

Q=7.2m3/s

78

Assume, L=10m

Now,

Qspillway=CwLspillwayH1.5

overtop

Or,7.2=1.6*10*H1.5

Or,H=0.588m

12. Design of Trashrack(before penstock)

Dimension of Trashrack = 3m×3m

Take, clear spacing of bars = 100mm.

Diameter of bars =20mm

If n is the no. of bars,

20n+100(n-1) =3×103

So, n = 25.83

Take, n = 26nos.

Net width =3-26×20/1000

=2.48m.

Net velocity (v) = 7.2/(2.48×3)

= 0.96m/s.

Now, loss in T.R.(LT) = kv2/2g

Where, k = 1.45-0.45R-R2

Where, R = Anet/Agross

= (2.48×3)/(3×3)

= 0.82

So, k = 0.408

79

So,LT = (0.408×0.962)/(2×9.81)

= 0.019m.

4.1.7 Design of Penstock

For successful operation the size of the pipe for a given discharge may vary between wide

limits, but there is usually one size that will make the greatest economy design. Hence,

the diameter of the penstock is determined from the consideration of economy and is

checked to see that the acceptance velocities are not exceeded.

Figure 4.1 Graph of Cost vs Diameter for Optimization of Penstock

From the graph,

Optimum diameter(d)=2075mm

Thickness(t)=14mm

Velocity (v)= 2.129 m/s(2.5<v<3.5m/s)

Revise

Adopted, diameter(d)=1900mm

Velocity (v)= 2.53m/s(2.5<v<3.5m/s)

(OK)

1,000,000.00

51,000,000.00

101,000,000.00

151,000,000.00

201,000,000.00

251,000,000.00

301,000,000.00

351,000,000.00

800 1000 1200 1400 1600 1800 2000 2200 2400 2600

Co

st(R

s)--

>

Diameter(mm)-->

Penstock Optimization

Total Cost Cost of Energy Loss Cost of Material

80

Table 4.3 Calculation for Optimization of Penstock

Discharge(Q) 7.2 m3/s Cost of Energy(Rs/Kw-h) 5 Bulk modulus of water 1.962E+09 N/m2

Gross head(Hg) 74.1 m Material Cost(Rs/Kg) 180 Design Pressure(P) 7.271 kg/cm2

Length(L) 2239 m Density(ρ) 7850 Kg/m3 Allowable Stress(σ) 1020 kg/cm2

Life(n) 40 years Young's Modulus(E) 2.10E+11 N/m2 Joint Efficiency(η) 0.9

Interest rate(i) 11 % Ultimate tensile strength(S) 3.50E+08 N/m2 Time of closure 10 s

Density of water 1000 kg/m3

Internal Diameter(d) mm 800 1000 1200 1400 1600 1800 2000 2200 2400 2600

Empirical thickness(t) mm 4.68 5.48 6.28 7.07 7.87 8.66 9.46 10.25 11.05 11.85

thickness(t) mm 14.00 14.00 14.00 14.00 14.00 14.00 14.00 14.00 14.00 14.00

Area m2 0.50 0.79 1.13 1.54 2.01 2.54 3.14 3.80 4.52 5.31

Velocity(Vo) m/s 14.324 9.167 6.366 4.677 3.581 2.829 2.292 1.894 1.592 1.356

Frictional Factor 0.012 0.012 0.012 0.012 0.012 0.01 0.01 0.01 0.01 0.01

Head Loss(Hf) m 351.171 115.072 46.245 21.396 10.974 5.075 2.997 1.861 1.204 0.807

Power loss kw 21083.353 6908.593 2776.409 1284.545 658.855 304.681 179.911 111.711 72.302 48.455

Annual Energy Loss kw-h 1.847E+08 6.052E+07 2.432E+07 1.125E+07 5.772E+06 2.669E+06 1.576E+06 9.786E+05 6.334E+05 4.245E+05

Cost of Energy loss Rs 9.235E+08 3.026E+08 1.216E+08 5.626E+07 2.886E+07 1.335E+07 7.880E+06 4.893E+06 3.167E+06 2.122E+06

Wave velocity(Vc) m/s 1130.978 1084.767 1043.794 1007.138 974.092 944.099 916.715 891.585 868.413 846.960

Critical time s 3.959 4.128 4.290 4.446 4.597 4.743 4.884 5.022 5.156 5.287

Surge Head(Hs) m 233.489 149.433 103.773 76.241 58.372 46.121 37.358 30.875 25.943 22.106

Total Head(Ht) m 307.609 223.553 177.893 150.361 132.492 120.241 111.478 104.995 100.063 96.226

Effective thickness(Te) mm 8.606 8.606 8.606 8.606 8.606 8.606 8.606 8.606 8.606 8.606

Safety Factor(S.F.) 2.448 2.695 2.822 2.862 2.842 2.783 2.702 2.608 2.509 2.408

Check reject ok ok ok ok ok ok ok ok reject

Area of Pipe m2 3.580E-02 4.460E-02 5.339E-02 6.219E-02 7.099E-02 7.978E-02 8.858E-02 9.738E-02 1.062E-01 1.150E-01

Weight kg 6.292E+05 7.838E+05 9.384E+05 1.093E+06 1.248E+06 1.402E+06 1.557E+06 1.711E+06 1.866E+06 2.020E+06

Cost of Penstock Rs 1.133E+08 1.411E+08 1.689E+08 1.967E+08 2.246E+08 2.524E+08 2.802E+08 3.080E+08 3.359E+08 3.637E+08

Annual Penstock Cost Rs 1.265E+07 1.576E+07 1.887E+07 2.198E+07 2.509E+07 2.820E+07 3.130E+07 3.441E+07 3.752E+07 4.063E+07

Total Cost Rs 9.361E+08 3.184E+08 1.405E+08 7.824E+07 5.394E+07 4.154E+07 3.918E+07 3.931E+07 4.069E+07 4.275E+07

81

4.1.8 Design of Anchor Block

Head (H) = 74.12 m

Discharge Q = 7.2 m3/sec

Specific wt. of concrete Γconcrete = 25 KN/m3

Specific wt. of masonry Γmasonry = 20 KN/m3

Specific wt. of pipe material Γpipe = 78.5 KN/m3

Specific wt. of soil Γsoil = 20 KN/m3

Soil angle of friction ø = 30o

Young modulus of elasticity E = 210000 N/mm2

Weight of pipe Wp = Π(d+t)tγsteel

= Π(1.9+0.014)*0.014*78.5

= 6.608 KN/m

weight of water Ww = Γw*Πd2/4

= 9.81*π*1.92/4

= 27.78 KN/m

Total weight Tw = 34.4 KN/m

Coefficient of friction

between pipe and support piers (f) = 0.25

Pipe internal diameter (d) = 1.9 m

Wall thickness of penstock (t) = 0.014 m

Maximum temperature change (ΔT) = 5 °C

Length of the pipe (L) = 2238.73 m

Closure time (T) = 10 s

82

(Source:ITDG,2002)

83

There are altogether 27 anchor blocks in Tadi hydropower plant and among them anchor block

no. 19 is one of the critical anchor block having horizontal bend of 47.26o so its stability has

been checked among all other anchor blocks.

19. Anchor block design(1+549.42)

Uphill ground slope (i) = 1o

Upstream penstock angle (α) = 2o

Downstream penstock angle (β) = 2o

Buried depth of block at upstream face (H1) = 2.05 m

Buried depth of block at downstream face (H2) = 1.95 m

Surge head(L*Vo/(9.81*t)) (Hs) = 34.70 m

Total head including surge (Htotal) = 108.82 m

Half distance from anchor centerline d/s of

anchor (L1d) = 63.1 m

Half distance from anchor centerline u/s of

anchor (L1u) = 62.94 m

Distance bet. two consecutive support d/s of

anchor (L2d) = 126.20 m

Distance bet. two consecutive support u/s of

anchor (L2u) = 125.88 m

Distance from anchor centerline to the d/s

expansion jt. (L4d) = 121.20 m

Distance from anchor centerline to the u/s

expansion jt. (L4u) = 120.87 m

upstream exposure depth = 3.07 m

downstream exposure depth = 3.17 m

length upstream of joint = 1.5 m

length downstream of joint = 2.5 m

Width of anchor block (w) = 7.2 m

Volume of Block = 248.93 m³

Coefficient of active earth pressure (ka) = 0.33

Weigth of the Block (Wb) = 4978.51 KN

No of support piers = 0

Table 4.4 Calculation of elevant forces

Forces X-comp. Y-comp.

F1u 2162.959 -75.486 2161.6

F1d 2168.561 -75.682 2167.2

Frictional force 1081.488

F2u(±) 0 0.000 0.000

F3 0 0 0

F4u 27.875 27.858 0.973

84

F4d 0 0 0

F6 (±) 0

0 0

F7u 86.893 86.841 3.033

F7d(-) 90.398 -90.342 -3.155

F8 0 0 0

F9 0 0 0

F10 100.906 100.891 1.761

Wb 4978.512 0 4978.5

ΣH ΣV

At expansion = -25.921 9310.0

At contraction = -25.921 9310.0

Checking stability:

Center of gravity of block from upstream face of block = 3.85 m

sum of horizontal force acting at bend expansion = -126.81 KN

contraction = -126.81 KN

sum of vertical force at bend expansion = 4329.73 KN

contraction = 4329.73 KN

Failure against overturning:

At expansion:

Sum of moments about 'O' with clockwise as

positive (∑M) = 35537.6

Distance of resulting force (X) = 3.817

Eccentric distance (e) = 0.033

Allowable eccentric distance (ea) = 1.283

(Safe)

At contraction:

Sum of moments about 'O' with clockwise as

positive (ΣM) = 35537.6

Distance of resulting force (X) = 3.817

Eccentric distance (e) = 0.033

Allowable eccentric distance (ea) = 1.283

(Safe)

85

Failure on bearing capacity:

At expansion:

Pressure at base =172 KN/mm2

Allowable bearing pressure in soil =200 KN/mm2

(Safe)

At contraction:

Pressure at base =172 KN/mm2

Allowable bearing pressure in soil =200 KN/mm2

(Safe)

Failure on sliding:

Coefficient of friction =0.6

At expansion (F.O.S) =215.5

(>1.5Safe)

At contraction (F.O.S) =215.5 (>1.5Safe)

4.1.9 Design of Bifurcation

1. Design discharge

(Q) = 7.2m3/s

After bifurcation discharge in each pipe = 3.6 m3/s

Velocity (V) = 7.2/(1.92*π/4) = 2.539 m/s

Velocity in pipe before and after the bifurcation is same.

2. Area

(A) = Q/V=3.6/2.539

A = 1.417m2

D = 1.343m.

This is the diameter after bifurcation.

86

3. Calculation of thickness

S = 1020kgf/cm2

Efficiency = 85%

Total pressure in pipe (P) = H*

=9810×68.95

=6.764kg/cm2

Internal radius of pipe(R) = 1.032/2=51.6cm.

ASME formula,

t= 15.0*6.0*

*

PS

RP

= 5.31mm 6mm.

4.1.10 Turbine

1. Selection of turbine

Gross head =74.12m

Design discharge =7.2m3/s.

So, we select Francis turbine (medium head turbine) with horizontal axis.

For one unit,

Design discharge (Qd) = 3.6m3/s

Maximum efficiency = 92%

Assume, Efficiency (η) = 85%

87

2. Calculation of losses

i. Wall loss

hwall loss = [fLV2/2gd]

f = 0.01(Moody’s chart), L = 2238.73m,V = 2.539m/s

hwall loss = 3.87m

ii. Entrance loss = kentrance*(V2/2g)

= 0.2*0.329

= 0.066m

iii. Bend loss

Bend angles

15.45o, 26.28

o, 9.29

o, 10.68

o, 26.51

o, 22.59

o, 8.37

o, 8.59

o, 0

o, 22.07

o,4.87

o, 20.21

o, 10.3

o, 0

o,

6.61o, 0

o, 0.26

o, 0

o, 47.26

o, 1.42

o,40.75

o, 3.16

o, 0

o, 23.66

o, 0

o, 0

o, 0

o.

Due to variation in bend angles ,

So, average value of r/d = 3 is taken.

Table 4.5 Calculation of bend loss

Block Bend Angle r/d k

No. α°

1 15.45 3 0.182

2 26.28 3 0.225

3 9.29 3 0.157

4 10.68 3 0.163

5 26.51 3 0.321

88

6 22.59 3 0.305

7 8.37 3 0.153

8 8.59 3 0.154

9 0 3 0.000

10 22.07 3 0.208

11 4.87 3 0.139

12 20.21 3 0.201

13 10.3 3 0.161

14 0 3 0.000

15 6.61 3 0.146

16 0 3 0.000

17 0.26 3 0.121

18 0 3 0.000

19 47.26 3 0.361

20 1.42 3 0.126

21 40.75 3 0.283

22 3.16 3 0.133

23 0 3 0.000

24 23.66 3 0.215

25 0 3 0.000

26 0 3 0.000

27 0 3 0.000

Kbend 3.754

Bend loss = Kbend×(V^2/2g)

89

(Adam Harvey Table 3.11.3)

Total loss = 5.17m

Net Head (H) = 68.95m

3. Turbine Design

Provide two units,

Discharge (Q)=3.6m3/s

Power (P) = 0.85*9.81*68.95*3.6

= 2069.78kw

= 2810.76HP

Specific speed (Ns) = 2400/√H

= 289.03rpm

Turbine speed(N) = (Ns*H5/4

)/P0.5

= 1083.17rpm

No. of poles = [120f/N]

= 5.53

Take 8,

Corrected N=750rpm

Corrected Specific speed, Ns = [(N√P)/H5/4

]

= 200rpm (60 to 400) (OK)

= 1.234

90

Diameter of runner (D) = N

H6.84 where, Φ=0.0211Ns

2/3=0.721

= 0.675m

4. Cavitation check

σc= 0.0318*

2

100

Ns

= 0.1272

Hs = Hatm – Hvap - σc.H

= 10.3-0.3-(0.1272*68.95)

= 1.229m (OK)

Assuming, Bi/Di = 0.15

Coefficient of velocity(Cv) = 0.98

Therefore, Q=π*Bi*Di*vi

3.6=π*Di*n*Cv*

3.6=π* *0.98

Di=0.46m

=460mm

Therefore, Bi = 0.069m

= 69.055mm

And, Do = Di/2=230mm

Bo = 34.5mm

91

Tangential velocity (ui) = πDiN/60

= π*0.46*750/60

= 18.064m/s

Speed ratio=u/

= 18.064/

= 0.491

5. Design of vane angle

= vfi/( )

=

=

= 2.618

α = 69.099o

The no. of vane varies from 16 to 24.

So, no. of vane=20

And, no. of guide vane=21

4.1.11 Design of Powerhouse

1. Machine hall

92

For horizontal axis Francis turbine,

Unit size:

L= (4D+2.7) = 5.4m

B = (4D+3.7)×2 = 6.4m

Length of erection bay = (4D+2.7)*1

= 5.4m

Length of E.O.T crane = 5m

Total Length = 2x5.4+5.4+5

= 21.2m

Breadth = 12.8m

Height = H1+H2

2. Draft Tube:

Depth of draft tube (H1) = (2.5 to 3)*D

= 2.75*0.675m

= 1.856m

Length of draft tube (L) = (4 to 5)*D

= 4.5*0.675

= 3.037m

Width of draft tube at exit (B) = ( 2.6 to 3.3)*D

= 3*0.675

93

=2.025m

Now,H2 = Lt+h+k

Where,Lt = Lo+(1.5 to 1.6)

= W/(KoDi2N) +1.5=11*1000/(5.7*0.46

2*750) + 1.5=13.6

h=K√Df

=0.65√Do+1.2

=0.65√(0.23+1.2)

=0.777

K = (5.5 to 7.0)m

= 6m.

H2=20.437m

Therefore,

H=1.856m+20.437m

=22.293 22.3m

Size of equipment room: (6.4×5)m

Size of transformer room: (6.4×5)m

4.1.12 Design of Tailrace Canal

1. Before combination of two canal

Discharge=3.6 m3/s

94

Take R.C.C canal (1:1.5:3)(higher velocity can be considered as head loss need not be

minimized, RCC resists erosion due to higher flow velocity)

n=0.020

Permissible velocity (Vc)=3 m/s

Adopting velocity (V)=2.5 m/s

Area(A) = Q/V = 3.6/2.5 = 1.44 m2

Provide Breath (B) = 2.4 m

Height (H) = A/B

= 1.44/2.4

= 0.6 m

Velocity (V) = Q/A

= 3.6/1.44

= 2.52.5<3(OK)

2. After the combination of two canal

Design discharge=7.2 m3/s

Area (A) = Q/A

= 7.2/2.5

= 2.88 m2

For rectangular canal,

Assume, Width = 2.5 m

Height = 1.2 m

95

Provide free board=0.3 m

Total height of canal=0.3+1.2 = 1.5m

Area=1.2x2.5

=3 m2

Velocity (V) = 7.2/3 =2.4<3(OK)

Wetted Perimeter = 2.5+2x1.2

= 4.9m

Hydraulic radius (R) = A/P

= 3/4.9

= 0.612

Using Manning’s Equation

S=

2

3/2

*

R

v= (0.02*2.4/0.612

2/3)2

1 in 226.

4.1.13 Design of Transmission Line

1. Transmission Line

i. Length of transmission (L) = 8.5 km

(from power house to Chaugadha sub-station 33 kv)

ii. Transmission Voltage

V = 5.5*(L/6.6+P/100)

Where,L is in km

P is in kw

96

V is in kv

V = 5.5*(8.5/1.6 + 4200/100)0.5

= 37.831kv (step down to 33kv)

Adopt, transmission voltage = 33kv.

iii. Spacing(s) = 100m (100<S<300)

iv. Number of Tower = L/S = 8500/100=85 no.

2. Generator

Using 3 phase synchronous generator,

Design voltage level (V) = (MVA)0.5

= (Power Generated/P.F)0.5

= (4.2/0.95)0.5

= 2.102 kv (6<V<18)Adopt a generation voltage = 6kv.

4.2 STRUCTURAL DESIGN

4.2.1 Weir

4.2.1.1 Weight of the Weir

Table 4.6 Calculation of weight of weir

SN Area

Lever

Arm Moment at Toe

W1 21.353 23.042 492.016

W2 29.232 40.284 1177.582

W3 10.287 22.238 228.762

W4 21.600 11.61 250.776

W5 7.683 2.205 16.941

W6 1.656 9.21 15.252

91.811 108.589 2181.329

97

Total Area = 91.811 m^2

Moment of Area = 2181.329 m^3

CG from toe = 23.759 m

Unit weight = 25 KN/m^3

Total Weight = 2295.275 KN/m

Coefficient of friction (μ)= 0.6

Total Length L 54.9 m

Area of upstream water A1 238.654 m^2

Area of downstream water A2 26.487 m^2

H1 4.5 m

H2 8.04 m

H3 2.7 m

γ 9.81 KN/m^3

Figure:Weight of Weir

Figure 4.2 Longitudinal Section of Weir

4.2.1.2 Calculation of Forces

Figure 4.3 Pressure Diagram of Weir

98

Table 4.7 Calculation of stability of weir

SN Description Forces Lever Arm Moment about Toe

(KN- m/m)

Vertical Horizontal (m) MR MO

(KN/m) (KN/m) (Counter clock) (Clockwise)

1 Weight of Weir 2295.275 - 23.759 54533.219 -

2 Water Pressure

Ph1=γ*(H2-H1)*H1 - 156.273 2.25 - 351.615

Ph2 = 1/2*γ*H1^2 - 99.326 1.5 - 148.989

Ph3 = 1/2*γ *H3^2 - 35.757 0.9 32.181 -

Pv1 = γ*A1 2341.196 - 40.039 93739.136 -

Pv2 = γ *A2 259.837 - 5.4 1403.122 -

3 Uplift Pressure

U1=γ *H3*L -1454.136 - 27.45 - 39916.041

U2 =1/2* γ*(H2-H3)*L -1437.979 - 36.6 - 52630.040

Total 2004.193 291.357 149707.660 93046.686

∑ Vertical Forces (∑V) 2004.193 KN/m

∑ Horizontal Forces (∑H) 291.357 KN/m

∑ Positive Moment (∑+M) 149707.660 KN/m

∑ Negative Moment (∑-M) 93046.686 KN-m/m

∑ Moment (∑M) 56660.975 KN-m/m

4.2.1.3 Stability Check

1. Check for sliding

FS= µ*∑V/∑H

= (0.6*2004.193)/291.357

= 4.127 >1.5(SAFE)

2.

Check for Overturning

FO =

= 149707.660/93046.686

= 1.61 >1.5(SAFE)

99

3. Tension at Heel

X=

= 56660.975/2004.193

= 28.271 m

Width (B) = 54.9 m

Eccentricity ,e = B/2- X

= (54.9/2-28.488)

= -0.821 m

which is less than B/6(9.15 m)

Therefore, no tension is produced at the heel of the dam.

4. Principle stress check

σmax= ∑V(1+6*e/B)/B

= 33.23 KN/m^2

< 200

KN/m^2 (SAFE)

σmin= ∑V(1-6*e/B)/B

= 39.783 KN/m^2

< 200

KN/m^2 (SAFE)

100

4.2.2 Settling Basin

4.2.2.1 Stability Check

Unit Weight of soil(KN/m3) 18 Allowable bearing Capacity of soil(KN/m2) 150

Unit Weight of concrete(KN/m3) 25 Unit Weight of stone masonry(KN/m3) 20

Unit Weight of water(KN/m3) 9.81 Coefficent of active earth pressure 0.333

Coefficient of lateral friction(μ) 0.6 Unit Weight of water(KN/m3) 9.81

Case 1:Both side backfill with full water in both compartments of the Settling Basin

SN Description Forces Lever Arm

Moment

V (KN/m) H

(KN/m) Mr(KN-m/m) Mo(KN-m/m)

1 W1=8.35*0.5*25 104.375

0.25 26.094

2 W2=8.35*0.5*25 104.375

7.850 819.344

3 W3=8.35*0.5*25 104.375

15.45 1612.594

4 W4=0.7*15.7*25 274.75

7.85 2156.788

5 W5=0.5*2.95*3.15*20 92.925

1.48 137.529

6 W6=0.5*2.95*3.15*20 92.925

6.62 615.164

7 W7=0.5*2.95*3.15*20 92.925

9.08 843.759

8 W8=0.5*2.95*3.15*20 92.925

14.21 1320.464

9 W9=2.95*1.2*20 70.8

1.975 139.830

10 W10=2.95*1.2*20 70.8

6.125 433.650

11 W11=2.95*1.2*20 70.8

9.575 677.910

12 W12=2.95*1.2*21 70.8

13.725 971.730

13 Ps1 =0.5*Ka*18*(8.35+.7)*(8.35+.7)

245.708 3.017 741.218

14 Ps2 =0.5*Ka*18*(8.35+.7)*(8.35+.7)

-245.708 3.017

741.218

15 Pw1= 0.5*y*8.35*8.35

-341.989 3.483

1191.261

16 Pw2= 0.5*y*8.35*8.35

341.989 3.483 1191.261

17 Pw3= 0.5*y*8.35*8.35

-341.989 3.483

1191.261

101

18 Pw4= 0.5*y*8.35*8.35

341.989 3.483 1191.261

19 Wa1= (7.1*4+1.2*1.2+0.5*3.15*(7.1+1.2))*9.81 420.972

4.050 1704.935

20 Wa2= (7.1*4+1.2*1.2+0.5*3.15*(7.1+1.2)*9.81 420.972

11.650 4904.319

Sum 2084.718 0.000

19487.849 3123.740

∑M= 16364.109

B= 15.7

Factor of safety against Overturning(Fo)= ∑Mr/∑Mo= 6.239

>1.5 (Safe)

Factor of safety against Sliding(Fs)= μ∑V/∑H= ∞

>1.5 (Safe)

Check for Tension

X= ∑M/∑V= 7.850

Eccentricity (e)= B/2-X= 0.000 m <B/6(ok)

Safety against Bearing(Fb)

Pmax= ∑V/B*(1+6*e/B)= 132.807 KN/m2 <200 (safe)

Pmin= ∑V/B*(1-6*e/B)= 132.762 KN/m2 <200 (safe)

Case 2:Both side backfill with water inside one compartment(hill side) of Settling Basin

SN Description Forces

Lever Arm Moment

V (KN/m) H (KN/m)

Mr(KN-m/m) Mo(KN-m/m)

1 W1=8.35*0.5*25 104.375

0.25 26.094

2 W2=8.35*0.5*25 104.375

7.850 819.344

3 W3=8.35*0.5*25 104.375

15.45 1612.594

4 W4=0.7*15.7*25 274.75

7.85 2156.788

5 W5=0.5*2.95*3.15*20 92.925

1.48 137.529

6 W6=0.5*2.95*3.15*20 92.925

6.62 615.164

7 W7=0.5*2.95*3.15*20 92.925

9.08 843.759

102

8 W8=0.5*2.95*3.15*20 92.925

14.21 1320.464

9 W9=2.95*1.2*20 70.8

1.975 139.830

10 W10=2.95*1.2*20 70.8

6.125 433.650

11 W11=2.95*1.2*20 70.8

9.575 677.910

12 W12=2.95*1.2*21 70.8

13.725 971.730

13 Ps1 =0.5*Ka*18*(8.35+.7)*(8.35+.7)

245.708 3.017 741.218

14 Ps2 =0.5*Ka*18*(8.35+.7)*(8.35+.7)

-245.708 3.017

741.218

15 Pw1= 0.5*y*8.35*8.35

-341.989 3.483

1191.261

16 Pw2= 0.5*y*8.35*8.35

341.989 3.483 1191.261

17 Wa1= (7.1*4+1.2*1.2+0.5*3.15*(7.1+1.2))*9.81 420.972

4.050 1704.935

Sum 1663.747 0.000

13392.268 1932.479

∑M= 11459.790

B= 15.7

Factor of safety against Overturning(Fo)= ∑Mr/∑Mo= 6.930 >1.5 (Safe)

Factor of safety against Sliding(Fs)= μ∑V/∑H= ∞ >1.5 (Safe)

Check for Tension

X= ∑M/∑V= 6.888

Eccentricity (e)= B/2-X= 0.962 M <B/6(ok)

Safety against Bearing(Fb)

Pmax= ∑V/B*(1+6*e/B)= 144.933 KN/m2 <200 (safe)

Pmin= ∑V/B*(1-6*e/B)= 67.009 KN/m2 <200 (safe)

Case 3:Both side backfill with no water on both compartment of Settling Basin

SN Description Forces

Lever Arm Moment

V (KN/m) H (KN/m)

Mr(KN-m/m) Mo(KN-m/m)

1 W1=8.35*0.5*25 104.375

0.25 26.094

2 W2=8.35*0.5*25 104.375

7.850 819.344

103

3 W3=8.35*0.5*25 104.375

15.45 1612.594

4 W4=0.7*15.7*25 274.75

7.85 2156.788

5 W5=0.5*2.95*3.15*20 92.925

1.48 137.529

6 W6=0.5*2.95*3.15*20 92.925

6.62 615.164

7 W7=0.5*2.95*3.15*20 92.925

9.08 843.759

8 W8=0.5*2.95*3.15*20 92.925

14.21 1320.464

9 W9=2.95*1.2*20 70.8

1.975 139.830

10 W10=2.95*1.2*20 70.8

6.125 433.650

11 W11=2.95*1.2*20 70.8

9.575 677.910

12 W12=2.95*1.2*21 70.8

13.725 971.730

13 Ps1 =0.5*Ka*18*(8.35+.7)*(8.35+.7)

245.708 3.017 741.218

14 Ps2 =0.5*Ka*18*(8.35+.7)*(8.35+.7)

-245.708 3.017

741.218

Sum 1242.775 0.000

10496.072 741.218

∑M= 9754.855

B= 15.7

Factor of safety against Overturning(Fo)= ∑Mr/∑Mo= 14.161 >1.5 (Safe)

Factor of safety against Sliding(Fs)= μ∑V/∑H= ∞ >1.5 (Safe)

Check for Tension

X= ∑M/∑V= 7.849

Eccentricity (e)= B/2-X= 0.001 m <B/6(ok)

Safety against Bearing(Fb)

Pmax= ∑V/B*(1+6*e/B)= 79.180 KN/m2 <200 (safe)

Pmin= ∑V/B*(1-6*e/B)= 79.135 KN/m2 <200 (safe)

Case 4:Both side backfill with water inside one compartments(river side) of the Settling Basin

SN Description Forces Lever Arm Moment

V (KN/m) H (KN/m)

Mr(KN-m/m) Mo(KN-m/m)

1 W1=8.35*0.5*25 104.375

0.25 26.094

104

2 W2=8.35*0.5*25 104.375

7.850 819.344

3 W3=8.35*0.5*25 104.375

15.45 1612.594

4 W4=0.7*15.7*25 274.75

7.85 2156.788

5 W5=0.5*2.95*3.15*20 92.925

1.48 137.529

6 W6=0.5*2.95*3.15*20 92.925

6.62 615.164

7 W7=0.5*2.95*3.15*20 92.925

9.08 843.759

8 W8=0.5*2.95*3.15*20 92.925

14.21 1320.464

9 W9=2.95*1.2*20 70.8

1.975 139.830

10 W10=2.95*1.2*20 70.8

6.125 433.650

11 W11=2.95*1.2*20 70.8

9.575 677.910

12 W12=2.95*1.2*21 70.8

13.725 971.730

13 Ps1 =0.5*Ka*18*(8.35+.7)*(8.35+.7)

245.708 3.017 741.218

14 Ps2 =0.5*Ka*18*(8.35+.7)*(8.35+.7)

-245.708 3.017

741.218

15 Pw3= 0.5*y*8.35*8.36

-341.989 3.483

1191.261

16 Pw4= 0.5*y*8.35*8.37

341.989 3.483 1191.261

17 Wa2= (7.1*4+1.2*1.2+0.5*3.15*(7.1+1.2)*9.81 420.972

11.650 4904.319

Sum 1663.747 0.000

16591.653 1932.479

∑M= 14659.174

B= 15.7

Factor of safety against Overturning(Fo)= ∑Mr/∑Mo= 8.586 >1.5 (Safe)

Factor of safety against Sliding(Fs)= μ∑V/∑H= ∞ >1.5 (Safe)

Check for Tension

X= ∑M/∑V= 8.811

Eccentricity (e)= B/2-X= -0.961 m <B/6(ok)

Safety against Bearing(Fb)

Pmax= ∑V/B*(1+6*e/B)= 67.054 KN/m2 <200 (safe)

Pmin= ∑V/B*(1-6*e/B)= 144.888 KN/m2 <200 (safe)

105

Case 5:Backfill in hill side with water inside both compartments of the Settling Basin

SN Description Forces Lever Arm Moment

V (KN/m) H (KN/m)

Mr(KN-m/m) Mo(KN-m/m)

1 W1=8.35*0.5*25 104.375

0.25 26.094

2 W2=8.35*0.5*25 104.375

7.850 819.344

3 W3=8.35*0.5*25 104.375

15.45 1612.594

4 W4=0.7*15.7*25 274.75

7.85 2156.788

5 W5=0.5*2.95*3.15*20 92.925

1.48 137.529

6 W6=0.5*2.95*3.15*20 92.925

6.62 615.164

7 W7=0.5*2.95*3.15*20 92.925

9.08 843.759

8 W8=0.5*2.95*3.15*20 92.925

14.21 1320.464

9 W9=2.95*1.2*20 70.8

1.975 139.830

10 W10=2.95*1.2*20 70.8

6.125 433.650

11 W11=2.95*1.2*20 70.8

9.575 677.910

12 W12=2.95*1.2*21 70.8

13.725 971.730

13 Ps1 =0.5*Ka*18*(8.35+.7)*(8.35+.7)

245.708 3.017 741.218

14 Pw1= 0.5*y*8.35*8.35

-341.989 3.483

1191.261

15 Pw2= 0.5*y*8.35*8.35

341.989 3.483 1191.261

16 Pw3= 0.5*y*8.35*8.36

-341.989 3.483

1191.261

17 Pw4= 0.5*y*8.35*8.37

341.989 3.483 1191.261

18 Wa1= (7.1*4+1.2*1.2+0.5*3.15*(7.1+1.2))*9.81 420.972

4.050 1704.935

19 Wa2= (7.1*4+1.2*1.2+0.5*3.15*(7.1+1.2)*9.81 420.972

11.650 4904.319

Sum 2084.718 245.708

19487.849 2382.522

∑M= 17105.327

B= 15.7

Factor of safety against Overturning(Fo)= ∑Mr/∑Mo= 8.180 >1.5 (Safe)

106

Factor of safety against Sliding(Fs)= μ∑V/∑H= ∞ >1.5 (Safe)

Check for Tension

X= ∑M/∑V= 8.205

Eccentricity (e)= B/2-X= -0.355 m <B/6(ok)

Safety against Bearing(Fb)

Pmax= ∑V/B*(1+6*e/B)= 114.765 KN/m2 <200 (safe)

Pmin= ∑V/B*(1-6*e/B)= 150.805 KN/m2 <200 (safe)

Case 6:Backfill in hill side with full water in one compartment(hill side) of the Settling Basin

SN Description Forces Lever Arm Moment

V (KN/m) H (KN/m)

Mr(KN-m/m) Mo(KN-m/m)

1 W1=8.35*0.5*25 104.375

0.25 26.094

2 W2=8.35*0.5*25 104.375

7.850 819.344

3 W3=8.35*0.5*25 104.375

15.45 1612.594

4 W4=0.7*15.7*25 274.75

7.85 2156.788

5 W5=0.5*2.95*3.15*20 92.925

1.48 137.529

6 W6=0.5*2.95*3.15*20 92.925

6.62 615.164

7 W7=0.5*2.95*3.15*20 92.925

9.08 843.759

8 W8=0.5*2.95*3.15*20 92.925

14.21 1320.464

9 W9=2.95*1.2*20 70.8

1.975 139.830

10 W10=2.95*1.2*20 70.8

6.125 433.650

11 W11=2.95*1.2*20 70.8

9.575 677.910

12 W12=2.95*1.2*21 70.8

13.725 971.730

13 Ps1 =0.5*Ka*18*(8.35+.7)*(8.35+.7)

245.708 3.017 741.218

14 Pw1= 0.5*y*8.35*8.35

-341.989 3.483

1191.261

15 Pw2= 0.5*y*8.35*8.35

341.989 3.483 1191.261

16 Wa1= (7.1*4+1.2*1.2+0.5*3.15*(7.1+1.2))*9.81 420.972

4.050 1704.935

Sum 1663.747 245.708

13392.268 1191.261

107

∑M= 12201.007

B= 15.7

Factor of safety against Overturning(Fo)= ∑Mr/∑Mo= 11.242 >1.5 (Safe)

Factor of safety against Sliding(Fs)= μ∑V/∑H= ∞ >1.5 (Safe)

Check for Tension

X= ∑M/∑V= 7.333

Eccentricity (e)= B/2-X= 0.517 m <B/6(ok)

Safety against Bearing(Fb)

Pmax= ∑V/B*(1+6*e/B)= 126.891 KN/m2 <200 (safe)

Pmin= ∑V/B*(1-6*e/B)= 85.052 KN/m2 <200 (safe)

Case 7:Backfill in hill side with full water in one compartment(river side)of the Settling Basin

SN Description Forces Lever

Arm Moment

V (KN/m)

H

(KN/m)

Mr(KN-

m/m)

Mo(KN-

m/m)

1 W1=8.35*0.5*25 104.375

0.25 26.094

2 W2=8.35*0.5*25 104.375

7.850 819.344

3 W3=8.35*0.5*25 104.375

15.45 1612.594

4 W4=0.7*15.7*25 274.75

7.85 2156.788

5 W5=0.5*2.95*3.15*20 92.925

1.48 137.529

6 W6=0.5*2.95*3.15*20 92.925

6.62 615.164

7 W7=0.5*2.95*3.15*20 92.925

9.08 843.759

8 W8=0.5*2.95*3.15*20 92.925

14.21 1320.464

9 W9=2.95*1.2*20 70.8

1.975 139.830

10 W10=2.95*1.2*20 70.8

6.125 433.650

11 W11=2.95*1.2*20 70.8

9.575 677.910

12 W12=2.95*1.2*21 70.8

13.725 971.730

108

13 Ps1 =0.5*Ka*18*(8.35+.7)*(8.35+.7)

245.708 3.017 741.218

14 Pw3= 0.5*y*8.35*8.36

-341.989 3.483

1191.261

15 Pw4= 0.5*y*8.35*8.37

341.989 3.483 1191.261

16 Wa2= (7.1*4+1.2*1.2+0.5*3.15*(7.1+1.2)*9.81 420.972

11.650 4904.319

Sum 1663.747 245.708

16591.653 1191.261

∑M= 15400.392

B= 15.7

Factor of safety against Overturning(Fo)= ∑Mr/∑Mo= 13.928

>1.5

(Safe)

Factor of safety against Sliding(Fs)= μ∑V/∑H= ∞

>1.5

(Safe)

Check for Tension

X= ∑M/∑V= 9.256

Eccentricity (e)= B/2-X= -1.406 m <B/6(ok)

Safety against Bearing(Fb)

Pmax= ∑V/B*(1+6*e/B)= 49.012 KN/m2 <200 (safe)

Pmin= ∑V/B*(1-6*e/B)= 162.930 KN/m2 <200 (safe)

Case 8:Backfill in hill side with no water in both compartments of the Settling Basin

SN Description Forces Lever

Arm Moment

V (KN/m)

H

(KN/m)

Mr(KN-

m/m)

Mo(KN-

m/m)

1 W1=8.35*0.5*25 104.375

0.25 26.094

2 W2=8.35*0.5*25 104.375

7.850 819.344

3 W3=8.35*0.5*25 104.375

15.45 1612.594

4 W4=0.7*15.7*25 274.75

7.85 2156.788

5 W5=0.5*2.95*3.15*20 92.925

1.48 137.529

6 W6=0.5*2.95*3.15*20 92.925

6.62 615.164

7 W7=0.5*2.95*3.15*20 92.925

9.08 843.759

109

8 W8=0.5*2.95*3.15*20 92.925

14.21 1320.464

9 W9=2.95*1.2*20 70.8

1.975 139.830

10 W10=2.95*1.2*20 70.8

6.125 433.650

11 W11=2.95*1.2*20 70.8

9.575 677.910

12 W12=2.95*1.2*21 70.8

13.725 971.730

13 Ps1 =0.5*Ka*18*(8.35+.7)*(8.35+.7)

245.708 3.017 741.218

Sum 1242.775 245.708

10496.072 0.000

∑M= 10496.072

B= 15.7

Factor of safety against Overturning(Fo)= ∑Mr/∑Mo= ∞

>1.5

(Safe)

Factor of safety against Sliding(Fs)= μ∑V/∑H= ∞

>1.5

(Safe)

Check for Tension

X= ∑M/∑V= 8.446

Eccentricity (e)= B/2-X= -0.596 m <B/6(ok)

Safety against Bearing(Fb)

Pmax= ∑V/B*(1+6*e/B)= 61.138 KN/m2 <200 (safe)

Pmin= ∑V/B*(1-6*e/B)= 97.178 KN/m2 <200 (safe)

110

4.2.2.2 Structure Design

i. Bending moment and axial force

= 18kN/m3

=150kN/m

2

=25kN/m3

=9.81kN/m3

=20kN/m3

Assuming M20 concrete and Fe 415 steel, Φ=300

Ka = =

Effective height (h)=8.35+.7/2

=8.7m

Width=7.1+0.25*2

=7.6m

AB=FC=ED=8.7

BC=CD=7.6

For purpose of design 1m length of settling basin is considered.

111

CASE 1: Both side backfill with full water in both compartments of the Settling Basin

Active pressure due to soil = Ka*Ysoil*h

= 1/3*18*8.7

= 52.200

Lateral pressure due to water = Yw*h

= 9.81*8.7

= 85.347

Soil pressure at bottom:

∑V = 2084.718

B = 15.2

e = 0.000

Pmax = (∑V/B)*(1+6*e/B)

= 137.153

Pmin = (∑V/B)*(1-6*e/B)

= 137.153

Fig:Loads acting in Frame

Figure 4.4 Loading for case 1

112

Moment of Inertia:

BC(I1) = bd3/12=1*0.5

3/12 = 0.0104 m

4

CD(I2) = bd3/12=1*0.5

3/12 = 0.0104 m

4

A F E

I1 I2

B C D

Now, I1= I2=I

Table 4.8 Calculation of Stiffness Distribution Factor

Joint Member K ∑K D.F.

B BA 0 0.00 EI 0.51 EI 0

BC 4EI1/L1=4E*I/7.85 0.51 EI 1

C CB 4EI1/L1=4E*I/7.85 0.51 EI 1.02 EI 0.5

CF 0 0.00 EI 0

CD 4EI2/L2=4E*I/7.85 0.51 EI 0.5

D DC 4EI2/L2=4E*I/7.85 0.51 EI 0.51 EI 1

DE 0 0.00 EI 0

Fixed End Moments:

MFBA(cantilever moment) = (1/2)*(85.347-52.2)*8.7*8.7*(1/3)

= 418.149 KNm.

113

MFBC = (+137.153*7.6^2)/12

= 660.161 KNm

MFCB = (-137.153*7.6^2)/12

= -660.161 KNm

MFCD = (+137.153*7.6^2)/12

= 660.161 KNm

MFDC = (-137.153*7.6^2)/12

= -660.161 KNm

MFCF(cantilever moment) = (1/2)*(85.347-85.347)*8.7*8.7*(1/3)

= 0 KNm.

MFDE(cantilever moment) = (1/2)*(52.2-85.347)*8.7*8.7*(1/3)

= -418.149 KNm.

Table 4.9 Moment distribution for case 1

Joint B C D

Member BA BC CB CF CD DC DE

Distribution Factor 0.00 1.00 0.50 0.00 0.50 1.00 0.00

F.E.M 418.149 660.161 -660.161 0.000 660.161 -660.161 -418.149

Balance

-1078.310 0.000

0.000 1078.310

C.O.M

-539.155

539.155

Balance

0.000 0.000

0.000 0.000

Final moment 418.149 -418.149 -1199.316 0.000 1199.316 418.149 -418.149

For vertical slab AB,

It is subjected to two uniformly varying loads(U.V.L) having maximum intensity of 52.2KN/m2

and 85.347 KN/m2 at B.

Reactions:

Reaction at B(RB)= ½*(85.347-

52.2)*8.7 = 144.189 KN.

MFBA(cantilever moment) = (1/2)*(85.347-52.2)*8.7*8.7*(1/3)

= 418.149 KNm.

114

At L/2,M(cantilever moment)=wl^2/48 = (85.347-52.2)*8.7^2/48

= 52.269 KNm.

For horizontal slab BC,

It is subjected to a uniformly distributed load(U.D.L) having intensity of 137.153 KN/m2.

Reactions:

Taking moment about C, we have

Rb*7.6 = -418.149-1199.32+137.153*7.6^2/2

=>Rb = 308.355 KN.

Also,∑Fy = 0

=>Rc = 137.153*7.6-308.357

= 734.004 KN.

Simply Supported moment:

At L/2, M = wl^2/8 = 137.153*7.6^2/8

= 990.241

Net Moment:

At L/2,

From property of similar triangle,

x/418.149 = (7.6-x)/1199.32

=>418.149*7.6-418.149x = 1199.32x

=>x = 1.96 from B.

Again,

1199.316/(7.6-1.96) = y/(7.6-1.96-3.8)

=>y = 390.583 KNm.

At L/2, Mnet = 599.658 KNm.

For horizontal slab CD,

It is subjected to a uniformly distributed load(U.D.L) having intensity of 137.153 KN/m2.

Reactions:

Taking moment about D, we have

Rc*7.6 = +418.149+1199.316+137.153*7.6^2/2

=>Rc = 734.004 KN.

Also,∑Fy = 0

=>Rd = 137.153*7.6-734.004

= 308.355 KN.

Simply Supported moment:

At L/2, M = wl^2/8 = 137.153*7.6^2/8

115

= 990.241

Net Moment:

At L/2,

From property of similar triangle,

x/418.149 = (7.6-x)/1199.32

=>418.149*7.6-418.149x = 1199.32x

=>x = 1.96 from D.

Again,

1199.316/(7.6-1.96) = y/(7.6-1.96-3.8)

=>y = 390.583 KNm.

At L/2, Mnet = 599.658 KNm.

For vertical slab CF,

It is subjected to two uniformly varying loads(U.V.L) having maximum intensity of 52.2

and 85.347 KN/m2 at C.

Reactions:

Reaction at C(Rc)= ½*(85.347-

85.347)*8.7 = 0.000 KN.

MFCF(cantilever moment) = (1/2)*(85.347-85.347)*8.7*8.7*(1/3)

= 0 KNm.

At L/2,M(cantilever moment)=wl^2/48 = (85.347-85.347)*8.7^2/48

= 0.000 KNm.

For vertical slab DE,

It is subjected to two uniformly varying loads(U.V.L) having maximum intensity of 52.2

and 85.347 KN/m2 at D.

Reactions:

Reaction at D(Rd)= ½*(52.2-

85.347)*8.7 = -144.189 KN.

= 144.189

KN.

MFDE(cantilever moment) = (1/2)*(52.2-85.347)*8.7*8.7*(1/3)

= -418.149 KNm.

At L/2,M(cantilever moment)=wl^2/48 = (52.2-85.347)*8.7^2/48

= -52.269 KNm.

116

Fig:Bending moment diagram

Figure 4.5 Bending Moment Diagram for case 1

117

Case 2:Both side backfill with water inside one compartment(hill side) of Settling Basin

Active pressure due to soil = Ka*Ysoil*h

= 1/3*18*8.7

= 52.200

Lateral pressure due to water = Yw*h

= 9.81*8.7

= 85.347

Soil pressure at bottom:

∑V = 1663.747

B = 15.2

e = 0.962

Pmax = (∑V/B)*(1+6*e/B)

= 151.024

Pmin = (∑V/B)*(1-6*e/B)

= 67.890

Fig:Loads acting in Frame

Figure 4.6 Loading for case 2

118

The moment of inertia of all the members will be same as in previous case.

Fixed End Moments:

MFBA(cantilever moment) = (1/2)*(85.347-52.2)*8.7*8.7*(1/3)

= 418.149 KNm.

MFBC = (+((109.457*7.6^2)/12+(151.024-109.457)*7.6^2/20))

= 646.900 KNm

MFCB = (-((109.457*7.6^2)/12+(151.024-109.457)*7.6^2/30))

= -606.884 KNm

MFCD = (+((67.89*7.6^2)/12+(109.457-67.89)*7.6^2/20))

= 446.822 KNm

MFDC = (-((67.89*7.6^2)/12+(109.457-67.89)*7.6^2/30))

= -406.806 KNm

MFCF(cantilever moment) = (1/2)*(-85.347)*8.7*8.7*(1/3)

= -1076.652 KNm.

MFDE(cantilever moment) = (1/2)*(52.2)*8.7*8.7*(1/3)

= 658.503 KNm.

Table 4.10 Moment distribution for case 2

Joint B C D

Member BA BC CB CF CD DC DE

Distribution Factor 0.00 1.00 0.50 0.00 0.50 1.00 0.00

F.E.M 418.149 646.900 -606.884 -1076.652 446.822 -406.806 658.503

Balance

-1065.049 618.357

618.357 -251.697

C.O.M

-532.525

-125.848

Balance

0.000 329.186

329.186 0.000

C.O.M

0.000

0.000

Balance

0.000 0.000

0.000 0.000

Final moment 418.149 -418.149 -191.865 -1076.652 1268.517 -658.503 658.503

119

For vertical slab AB,

It is subjected to two uniformly varying loads(U.V.L) having maximum intensity of 52.2 KN/m2

and 85.347 KN/m2 at B.

Reactions:

Reaction at B(RB)= ½*(85.347-52.2)*8.7 = 144.189 KN.

MFBA(cantilever moment) = (1/2)*(85.347-52.2)*8.7*8.7*(1/3)

= 418.149 KNm.

At L/2,M(cantilever moment)=wl^2/48 = (85.347-52.2)*8.7^2/48

= 52.269 KNm.

For horizontal slab BC,

It is subjected to a uniformly distributed load(U.D.L) having intensity of 151.024 KN/m2 at B

and 109.457 KN/m2 at C.

Reactions:

Taking moment about C, we have

Rb*7.6 = -418.149-191.865+109.457*7.6^2/2+0.5*7.6*(151.024-109.457)*7.6*(2/3)

=>Rb = 440.976 KN.

Also,∑Fy = 0

=>Rc = 109.457*7.6+0.5*7.6*(151.024-109.457)-440.976

= 548.854

KN.

Simply Supported moment:

At L/2, M = wl^2/8 +wl^2/16 = 109.457*7.6^2/8+(151.024-109.457)*7.6^2/16

= 940.338 KNm.

Net Moment:

At L/2,

From property of similar triangle,

120

x/418.149 = (7.6-x)/191.865

=>418.149*7.6-418.149x = 191.865x

=>x = 5.21 from B.

Again,

418.149/(5.21) =

y/(5.21-

3.8)

=>y = 113.142 KNm.

At L/2, Mnet = 1053.480 KNm.

Let the maximum sagging B.M. occur at a distance x from C.

Mx + 41.567*x^3/(7.6*2*3)+109.457*x^2/2-548.854x+191.865 = 0

=>Mx = -41.567*x^3/(7.6*2*3)-109.457*x^2/2+548.854x-191.865

For maximum,

dMx/dx= -41.567*x^2/(7.6*2)-109.457*x+548.854

which gives x = 4.5

from C.

Now,

Mmax = -41.567*4.5^3/(7.6*2*3)-109.457*4.5^2/2+548.854*4.5-191.865

= 1086.660 KNm.

For horizontal slab CD,

It is subjected to a uniformly distributed load(U.D.L) having intensity of 109.457 KN/m2 at C

and 67.890 KN/m2 at D.

Reactions:

Taking moment about D, we have

Rc*7.6 = -658.503-1268.517+67.89*7.6^2/2+0.5*7.6*(109.457-67.89)*7.6*(2/3)

=>Rc = 443.550 KN.

Also,∑Fy = 0

121

=>Rd = 67.89*7.6+0.5*7.6*(109.457-67.89)-443.550

= 230.368 KN.

Simply Supported moment:

At L/2, M = wl^2/8 +wl^2/16 = 67.89*7.6^2/8+(109.457-67.89)*7.6^2/16

= 640.221 KNm.

Net Moment:

At L/2,

Average FEM = 963.510 KNm.

Mnet = -323.289 KNm.

For vertical slab CF,

It is subjected to a uniformly varying loads(U.V.L) having maximum intensity of 85.347 KN/m2 at C.

Reactions:

Reaction at C(Rc)= ½*(-85.347)*8.7 = -371.259 KN.

= 371.259

KN.

MFCF(cantilever moment) = (1/2)*(-85.347)*8.7*8.7*(1/3)

= -1076.652 KNm.

At L/2,M(cantilever moment)=wl^2/48 = (-85.347)*8.7^2/48

= -134.582 KNm.

For vertical slab DE,

It is subjected to a uniformly varying loads(U.V.L) having maximum intensity of 52.2KN/m2 at D.

Reactions:

Reaction at D(Rd)= ½*(52.2)*8.7 = 144.189 KN.

122

MFDE(cantilever moment) = (1/2)*(52.2)*8.7*8.7*(1/3)

= 658.503 KNm.

At L/2,M(cantilever moment)=wl^2/48 = (52.2)*8.7^2/48

= 82.313 KNm.

Fig:Bending moment diagram

Figure 4.7 Bending Moment Diagram for case 2

123

Case 3:Both side backfill with no water in both compartments of Settling Basin

Active pressure due to soil = Ka*Ysoil*h

= 1/3*18*8.7

= 52.200

Lateral pressure due to water = Yw*h

= 9.81*8.7

= 85.347

Soil pressure at bottom:

∑V = 1242.775

B = 15.2

e = 0.001

Pmax = (∑V/B)*(1+6*e/B)

= 81.762

Pmin = (∑V/B)*(1-6*e/B)

= 81.762

Fig:Loads acting in Frame

Figure 4.8 Loading for case 3

124

The moment of inertia of all the members will be same as in previous case.

Fixed End Moments:

MFBA(cantilever moment) = (1/2)*(-52.2)*8.7*8.7*(1/3)

= -658.503 KNm.

MFBC = (+81.762*7.6^2)/12

= 393.545 KNm

MFCB = (-81.762*7.6^2)/12

= -393.545 KNm

MFCD = (+81.762*7.6^2)/12

= 393.545 KNm

MFDC = (-81.762*7.6^2)/12

= -393.545 KNm

MFCF(cantilever moment) = (1/2)*0*8.6*8.6*(1/3)

= 0.000 KNm.

MFDE(cantilever moment) = (1/2)*(52.2)*8.6*8.6*(1/3)

= 643.452 KNm.

Table 4.11 Moment distribution for case 3

Joint B C D

Member BA BC CB CF CD DC DE

Distribution Factor 0.00 1.00 0.50 0.00 0.50 1.00 0.00

F.E.M -658.503 393.545 -393.545 0.000 393.545 -393.545 643.452

Balance

264.958 0.000

0.000 -249.907

C.O.M

132.479

-124.953

Balance

0.000 -3.763

-3.763 0.000

C.O.M

0.000

0.000

Balance

0.000 0.000

0.000 0.000

Final moment -658.503 658.503 -264.829 0.000 264.829 -643.452 643.452

125

For vertical slab AB,

It is subjected to a uniformly varying loads(U.V.L) having maximum intensity of 52.2KN/m2 at B.

Reactions:

Reaction at B(RB)= ½*(-52.2)*8.7 = -227.070 KN.

= 227.070

KN.

MFBA(cantilever moment) = (1/2)*(-52.2)*8.7*8.7*(1/3)

= -658.503 KNm.

At L/2,M(cantilever moment)=wl^2/48 = (-52.2)*8.7^2/48

= -82.313 KNm.

For horizontal slab BC,

It is subjected to a uniformly distributed load(U.D.L) having intensity of 81.762 KN/m2.

Reactions:

Taking moment about C, we have

Rb*7.6 = 658.503-264.892+81.762*7.6^2/2

=>Rb = 362.493 KN.

Also,∑Fy = 0

=>Rc = 81.762*7.6-362.493

= 258.895 KN.

Simply Supported moment:

At L/2, M = wl^2/8 = 81.762*7.6^2/8

= 590.318

Net Moment:

At L/2,

Average FEM = 461.666 KNm.

Mnet = 128.652 KNm.

126

For horizontal slab CD,

It is subjected to a uniformly distributed load(U.D.L) having intensity of 81.762 KN/m2.

Reactions:

Taking moment about D, we have

Rc*7.6 = -658.503+264.892+81.762*7.6^2/2

=>Rc = 260.8749836 KN.

Also,∑Fy = 0

=>Rd = 81.762*7.6-260.875

= 360.513 KN.

Simply Supported moment:

At L/2, M = wl^2/8 = 81.762*7.6^2/8

= 590.318

Net Moment:

At L/2,

Average FEM = 454.141 KNm.

Mnet = 136.177 KNm.

For vertical slab CF,

It is subjected to two uniformly varying loads(U.V.L) having maximum intensity of 52.2 and 85.347 KN/m2 at C.

Reactions:

Reaction at C(Rc)= ½*(0)*8.7 = 0.000 KN.

MFCF(cantilever moment) = (1/2)*0*8.6*8.6*(1/3)

= 0 KNm.

At L/2,M(cantilever moment)=wl^2/48 = (0)*8.7^2/48

= 0.000 KNm.

127

For vertical slab DE,

It is subjected to two uniformly varying loads(U.V.L) having maximum intensity of 52.2 and 85.347 KN/m2 at D.

Reactions:

Reaction at D(Rd)= ½*(52.2)*8.7 = 227.070 KN.

MFDE(cantilever moment) = (1/2)*(52.2)*8.6*8.6*(1/3)

= 643.452 KNm.

At L/2,M(cantilever moment)=wl^2/48 = (52.2)*8.7^2/48

= 82.313 KNm.

Fig:Bending moment diagram

Figure 4.9 Bending Moment Diagram for case 3

128

Case 4:Both side backfill with water inside one compartments(river side) of the Settling Basin

Active pressure due to soil = Ka*Ysoil*h

= 1/3*18*8.7

= 52.200

Lateral pressure due to water = Yw*h

= 9.81*8.7

= 85.347

Soil pressure at bottom:

∑V = 1663.747

B = 15.2

e = -0.961

Pmax = (∑V/B)*(1+6*e/B)

= 67.938

Pmin = (∑V/B)*(1-6*e/B)

= 150.976

Fig:Loads acting in Frame

Figure 4.10 Loading for case 4

129

The moment of inertia of all the members will be same as in previous case.

Fixed End Moments:

MFBA(cantilever moment) = (1/2)*(-52.2)*8.7*8.7*(1/3)

= -658.503 KNm.

MFBC = (+((67.938*7.6^2)/12+(109.457-67.938)*7.6^2/30))

= 446.915 KNm

MFCB = (-((67.938*7.6^2)/12+(109.457-67.938)*7.6^2/20))

= -406.946 KNm

MFCD = (+((109.457*7.6^2)/12+(150.976-109.457)*7.6^2/30))

= 646.760 KNm

MFDC = (-((109.457*7.6^2)/12+(150.976-109.457)*7.6^2/20))

= -606.791 KNm

MFCF(cantilever moment) = (1/2)*(+85.347)*8.7*8.7*(1/3)

= 658.503 KNm.

MFDE(cantilever moment) = (1/2)*(52.2-85.347)*8.7*8.7*(1/3)

= -418.149 KNm.

Table 4.12 Moment distribution for case 4

Joint B C D

Member BA BC CB CF CD DC DE

Distribution Factor 0.00 1.00 0.50 0.00 0.50 1.00 0.00

F.E.M -658.503 446.915 -406.946 658.503 646.760 -606.791 -418.149

Balance

211.588 -449.159

-449.159 1024.941

C.O.M

105.794

512.470

Balance

0.000 -309.132

-309.132 0.000

C.O.M

0.000

0.000

Balance

0.000 0.000

0.000 0.000

Final moment -658.503 658.503 -1059.443 658.503 400.940 418.149 -418.149

130

For vertical slab AB,

It is subjected to a uniformly varying loads(U.V.L) having maximum intensity of 52.2 KN/m2 at B.

Reactions:

Reaction at B(RB)= ½*(-52.2)*8.7 = -227.070 KN.

= 227.070 KN.

MFBA(cantilever moment) = (1/2)*(-52.2)*8.7*8.7*(1/3)

= -658.503 KNm.

At L/2,M(cantilever moment)=wl^2/48 = (-52.2)*8.7^2/48

= -82.313 KNm.

For horizontal slab BC,

It is subjected to a uniformly distributed load(U.D.L) having intensity of 109.457 KN/m2 at C and 67.938 KN/m2 at B.

Reactions:

Taking moment about C, we have

Rb*7.6 = 658.503-1059.443+67.938*7.6^2/2+0.5*7.6*(109.457-67.938)*7.6*(1/3)

=>Rb = 258.000 KN.

Also,∑Fy = 0

=>Rc = 67.938*7.6+0.5*7.6*(109.457-67.938)-258.000

= 416.1010006

KN.

Simply Supported moment:

At L/2, M = wl^2/8 +wl^2/16 = 67.938*7.6^2/8+(109.457-67.938)*7.6^2/16

= 640.396 KNm.

Net Moment:

At L/2,

Average FEM = 858.973 KNm.

Mnet = -218.577 KNm.

131

For horizontal slab CD,

It is subjected to a uniformly distributed load(U.D.L) having intensity of 150.976KN/m2 at D and 109.457 KN/m2 at C.

Reactions:

Taking moment about D, we have

Rc*7.6 = 418.149-400.94+109.457*7.6^2/2+0.5*7.6*(150.976-109.457)*7.6*(1/3)

=>Rc = 576.302 KN.

Also,∑Fy = 0

=>Rd = 109.457*7.6+0.5*7.6*(150.976-109.457)-576.302

= 413.344 KN.

Simply Supported moment:

At L/2, M = wl^2/8 +wl^2/16 = 109.457*7.6^2/8+(150.976-109.457)*7.6^2/16

= 940.164 KNm.

Net Moment:

At L/2,

From property of similar triangle,

x/418.149 = (7.6-x)/400.940

=>418.149*7.6-418.149x = 400.940x

=>x = 3.88 from D.

Again,

418.149/(3.88) = y/(3.88-3.8)

=>y = 8.605 KNm.

At L/2, Mnet = 948.768 KNm.

132

Let the maximum sagging B.M. occur at a distance x from C.

Mx + 41.519*x^3/(7.6*2*3)+109.457*x^2/2-576.302*x+400.94 = 0

=>Mx = -41.519*x^3/(7.6*2*3)-109.457*x^2/2+576.302*x-400.94

For maximum,

dMx/dx= -41.519*x^2/(7.6*2)-109.457*x+576.302

which gives x = 4.71

from C

Now,

Mmax =

-41.519*4.71^3/(7.6*2*3)-109.457*4.71^2/2+576.302*4.71-

400.94

= 1004.204 KNm.

For vertical slab CF,

It is subjected to a uniformly varying loads(U.V.L) having maximum intensity of 85.347 KN/m2 at

C.

Reactions:

Reaction at C(Rc)= ½*(85.347)*8.7 = 371.259 KN.

MFCF(cantilever moment) = (1/2)*(+85.347)*8.7*8.7*(1/3)

= 658.503 KNm.

At L/2,M(cantilever

moment)=wl^2/48 = (85.347)*8.7^2/48

= 134.582 KNm.

For vertical slab DE,

It is subjected to a uniformly varying loads(U.V.L) having maximum intensity of 52.2KN/m2 at D.

Reactions:

Reaction at D(Rd)= ½*(52.2-

85.647)*8.7 = -144.189 KN.

= 144.189

KN.

MFDE(cantilever moment) = (1/2)*(52.2-85.347)*8.7*8.7*(1/3)

=

-

418.14940

5 KNm.

At L/2,M(cantilever

moment)=wl^2/48 = (52.2-85.347)*8.7^2/48

= -52.269 KNm.

133

Fig:Bending moment diagram

Figure 4.11 Bending Moment Diagram for case 4

134

Case 5:Backfill in hill side with water inside both compartments of the Settling Basin

Active pressure due to soil = Ka*Ysoil*h

= 1/3*18*8.7

= 52.200

Lateral pressure due to water = Yw*h

= 9.81*8.7

= 85.347

Soil pressure at bottom:

∑V = 2084.718

B = 15.2

e = -0.355

Pmax = (∑V/B)*(1+6*e/B)

= 117.928

Pmin = (∑V/B)*(1-6*e/B)

= 156.377

Fig:Loads acting in Frame

Figure 4.12 Loading for case 5

135

The moment of inertia of all the members will be same as in previous case.

Fixed End Moments:

MFBA(cantilever moment) = (1/2)*(85.347-52.2)*8.7*8.7*(1/3)

= 418.1494 KNm.

MFBC = (+((117.928*7.6^2)/12+(137.153-117.928)*7.6^2/30))

= 623.146 KNm

MFCB = (-((117.928*7.6^2)/12+(137.153-117.928)*7.6^2/20))

= -604.639 KNm

MFCD = (+((137.153*7.6^2)/12+(156.377-137.153)*7.6^2/30))

= 715.682 KNm

MFDC = (-((137.153*7.6^2)/12+(156.377-137.153)*7.6^2/20))

= -697.175 KNm

MFCF(cantilever moment) = (1/2)*(85.347-85.347)*8.7*8.7*(1/3)

= 0 KNm.

MFDE(cantilever moment) = (1/2)*(-85.347)*8.7*8.7*(1/3)

= -1076.65 KNm.

Table 4.13 Moment distribution for case 5

Joint B C D

Member BA BC CB CF CD DC DE

Distribution Factor 0.00 1.00 0.50 0.00 0.50 1.00 0.00

F.E.M 418.149 623.146 -604.639 0.000 715.682 -697.175 -1076.652

Balance

-1041.296 -55.522

-55.522 1773.828

C.O.M

-520.648

886.914

Balance

0.000 -183.133

-183.133 0.000

C.O.M

0.000

0.000

Balance

0.000 0.000

0.000 0.000

Final moment 418.149 -418.149 -1363.942 0.000 1363.942 1076.652 -1076.652

136

For vertical slab AB,

It is subjected to two uniformly varying loads(U.V.L) having maximum intensity of 52.2KN/m2 and 85.347 KN/m2 at B.

Reactions:

Reaction at B(RB)= ½*(85.347-52.2)*8.7 = 144.189 KN.

MFBA(cantilever moment) = (1/2)*(85.347-52.2)*8.7*8.7*(1/3)

= 418.149 KNm.

At L/2,M(cantilever moment)=wl^2/48 = (85.347-52.2)*8.7^2/48

= 52.269 KNm.

For horizontal slab BC,

It is subjected to a uniformly distributed load(U.D.L) having intensity of 137.153 KN/m2 at C and 117.928 KN/m2 at B.

Reactions:

Taking moment about C, we have

Rb*7.6 = -418.149-1363.942+117.928*7.6^2/2+0.5*7.6*(137.153-117.928)*7.6*(1/3)

=>Rb = 237.991 KN.

Also,∑Fy = 0

=>Rc = 117.928*7.6+0.5*7.6*(137.153-117.928)-237.991

= 731.3136326

KN.

Simply Supported moment:

At L/2, M = wl^2/8 +wl^2/16 = 117.928*7.6^2/8+(137.153-117.928)*7.6^2/16

= 920.839 KNm.

Net Moment:

At L/2,

From property of similar triangle,

137

x/418.149 = (7.6-x)/1363.942

=>418.149*7.6-418.149x = 1363.942x

=>x = 1.78 from B.

Again,

1363.942/(7.6-1.78) = y/(7.6-1.78-3.8)

=>y = 472.896 KNm.

At L/2, Mnet = 447.943 KNm.

For horizontal slab CD,

It is subjected to a uniformly distributed load(U.D.L) having intensity of 156.377KN/m2 at D and 137.153 KN/m2 at C.

Reactions:

Taking moment about D, we have

Rc*7.6 = 1363.942+1076.652+137.153*7.6^2/2+0.5*7.6*(156.377-137.153)*7.6*(1/3)

=>Rc = 866.662 KN.

Also,∑Fy = 0

=>Rd = 137.153*7.6+0.5*7.6*(156.377-137.153)-866.662

= 248.752 KN.

Simply Supported moment:

At L/2, M = wl^2/8 +wl^2/16 = 137.153*7.6^2/8+(156.377-137.153)*7.6^2/16

= 1059.643 KNm.

Net Moment:

At L/2,

From property of similar triangle,

x/1076.652 = (7.6-x)/1363.942

=>1076.652*7.6-1076.652x = 1363.942x

=>x = 3.35 from D.

138

Again,1363.942/(7.6-3.35) = y/(7.6-3.35-3.8)

=>y = 143.645 KNm.

At L/2, Mnet = 915.999 KNm.

Let the maximum sagging B.M. occur at a distance x from C.

Mx +19.224*x^3/(7.6*2*3)+137.153*x^2/2-866.662x+1363.942 = 0

=>Mx = -19.224*x^3/(7.6*2*3)-137.153*x^2/2+866.662x-1363.942

For maximum,

dMx/dx= -19.224*x^2/(7.6*2)-137.153*x+866.662

which gives x = 5.99

from C.

Now,

Mmax = -19.224*x^3/(7.6*2*3)-137.153*x^2/2+866.662x-1363.942

= 1276.225 KNm.

For vertical slab CF,

It is subjected to a uniformly varying loads(U.V.L) having maximum intensity of 85.347 KN/m2 at

C.

Reactions:

Reaction at C(Rc)= ½*(85.347-

85.347)*8.7 = 0.000 KN.

MFCF(cantilever moment) = (1/2)*(85.347-85.347)*8.7*8.7*(1/3)

= 0.000 KNm.

At L/2,M(cantilever moment)=wl^2/48 = (85.347-85.347)*8.7^2/48

= 0.000 KNm.

For vertical slab DE,

It is subjected to a uniformly varying loads(U.V.L) having maximum intensity of 52.2KN/m2 at D.

Reactions:

Reaction at D(Rd)= ½*(-85.647)*8.7 = -371.259 KN.

= 371.259

KN.

MFDE(cantilever moment) = (1/2)*(-85.347)*8.7*8.7*(1/3)

=

-

1076.65241 KNm.

At L/2,M(cantilever moment)=wl^2/48 = (-85.347)*8.7^2/48

= -134.582 KNm.

139

Fig:Bending moment diagram

Figure 4.13 Bending Moment Diagram for case 5

140

Case 6:Backfill in hill side with full water in one compartment(hill side) of the Settling Basin

Active pressure due to soil = Ka*Ysoil*h

= 1/3*18*8.7

= 52.200

Lateral pressure due to water = Yw*h

= 9.81*8.7

= 85.347

Soil pressure at bottom:

∑V = 1663.747

B = 15.2

e = 0.517

Pmax = (∑V/B)*(1+6*e/B)

= 131.775

Pmin = (∑V/B)*(1-6*e/B)

= 87.139

Fig:Loads acting in Frame

Figure 4.14 Loading for case 6

141

The moment of inertia of all the members will be same as in previous case.

Fixed End Moments:

MFBA(cantilever moment) = (1/2)*(85.347-52.2)*8.7*8.7*(1/3)

= 418.1494 KNm.

MFBC = (+((109.457*7.6^2)/12+(131.775-109.457)*7.6^2/20))

= 591.308 KNm

MFCB = (-((109.457*7.6^2)/12+(131.775-109.457)*7.6^2/30))

= -569.823 KNm

MFCD = (+((87.139*7.6^2)/12+(109.457-87.139)*7.6^2/20))

= 483.883 KNm

MFDC = (-((87.139*7.6^2)/12+(109.457-87.139)*7.6^2/30))

= -462.398 KNm

MFCF(cantilever moment) = (1/2)*(-85.347)*8.7*8.7*(1/3)

= -1076.65 KNm.

MFDE(cantilever moment) = (1/2)*0*8.7*8.7*(1/3)

= 0 KNm.

Table 4.14 Moment distribution for case 6

Joint B C D

Member BA BC CB CF CD DC DE

Distribution Factor 0.00 1.00 0.50 0.00 0.50 1.00 0.00

F.E.M 418.149 591.308 -569.823 -1076.652 483.883 -462.398 0.000

Balance

-1009.458 581.296

581.296 462.398

C.O.M

-504.729

231.199

Balance

0.000 136.765

136.765 0.000

C.O.M

0.000

0.000

Balance

0.000 0.000

0.000 0.000

Final moment 418.149 -418.149 -356.491 -1076.652 1433.143 0.000 0.000

142

For vertical slab AB,

It is subjected to two uniformly varying loads(U.V.L) having maximum intensity of 52.2 KN/m2 and 85.347 KN/m2 at B.

Reactions:

Reaction at B(RB)= ½*(85.347-52.2)*8.7 = 144.189 KN.

MFBA(cantilever moment) = (1/2)*(85.347-52.2)*8.7*8.7*(1/3)

= 418.149405 KNm.

At L/2,M(cantilever moment)=wl^2/48 = (85.347-52.2)*8.7^2/48

= 52.269 KNm.

For horizontal slab BC,

It is subjected to a uniformly distributed load(U.D.L) having intensity of 131.775 KN/m2 at B and 109.457 KN/m2 at C.

Reactions:

Taking moment about C, we have

Rb*7.6 = -418.149-894.817+109.457*7.6^2/2+0.5*7.6*(131.775 -109.457)*7.6*(2/3)

=>Rb = 370.550 KN.

Also,∑Fy = 0

=>Rc = 109.457*7.6+0.5*7.6*(131.775 109.457)-370.550

= 546.133

KN.

Simply Supported moment:

At L/2, M = wl^2/8 +wl^2/16 = 109.457*7.6^2/8+(131.775 -109.457)*7.6^2/16

= 870.849 KNm.

Net Moment:

At L/2,

From property of similar triangle,

x/418.149 = (7.6-x)/356.491

=>418.149*7.6-418.149x = 356.491x

143

=>x = 4.10 from B.

Again,

418.149/(4.1) = y/(4.1-3.8)

=>y = 30.829 KNm.

At L/2, Mnet = 901.678 KNm.

Let the maximum sagging B.M. occur at a distance x from C.

Mx +22.318*x^3/(7.6*2*3)+109.457*x^2/2-546.133x+356.491= 0

=>Mx = -22.318*x^3/(7.6*2*3)-109.457*x^2/2+546.133x-356.491

For maximum,

dMx/dx= -22.318*x^2/(7.6*2)-109.457*x+546.133

which gives x = 4.69

from C.

Now,

Mmax = -22.318*x^3/(7.6*2*3)-109.457*x^2/2+546.133x-356.491

= 950.569 KNm.

For horizontal slab CD,

It is subjected to a uniformly distributed load(U.D.L) having intensity of 109.457 KN/m2 at C and 87.139 KN/m2 at D.

Reactions:

Taking moment about D, we have

Rc*7.6 = 894.817+0+87.139*7.6^2/2+0.5*7.6*(109.457-87.139)*7.6*(2/3)

=>Rc = 576.238 KN.

Also,∑Fy = 0

=>Rd = 67.89*7.6+0.5*7.6*(109.457-67.89)-576.238

= 170.825 KN.

144

Simply Supported moment:

At L/2, M = wl^2/8 +wl^2/16 = 87.139*7.6^2/8+(109.457-87.139)*7.6^2/16

=

709.710540

6 KNm.

Net Moment:

At L/2,

Average FEM =

716.571600

3 KNm.

Mnet = -6.861 KNm.

For vertical slab CF,

It is subjected to a uniformly varying loads(U.V.L) having maximum intensity of 85.347 KN/m2 at C.

Reactions:

Reaction at C(Rc)= ½*(-85.347)*8.7 = -371.259 KN.

= 371.259

KN.

MFCF(cantilever moment) = (1/2)*(-85.347)*8.7*8.7*(1/3)

= -1076.652 KNm.

At L/2,M(cantilever moment)=wl^2/48 = (-85.347)*8.7^2/48

= -134.582 KNm.

For vertical slab DE,

It is subjected to a uniformly varying loads(U.V.L) having maximum intensity of 52.2KN/m2 at D.

Reactions:

Reaction at D(Rd)= ½*(0)*8.7 = 0.000 KN.

MFDE(cantilever moment) = (1/2)*0*8.7*8.7*(1/3)

145

= 0 KNm.

At L/2,M(cantilever moment)=wl^2/48 = (0)*8.7^2/48

= 0.000 KNm.

Fig:Bending moment diagram

Figure 4.15 Bending Moment Diagram for case 6

146

Case 7:Backfill in hill side with full water in one compartment(river side)of the Settling Basin

Active pressure due to soil = Ka*Ysoil*h

= 1/3*18*8.7

= 52.200

Lateral pressure due to water = Yw*h

= 9.81*8.7

= 85.347

Soil pressure at bottom:

∑V = 1663.747

B = 15.2

e = -1.406

Pmax = (∑V/B)*(1+6*e/B)

= 48.689

Pmin = (∑V/B)*(1-6*e/B)

= 170.225

Fig:Loads acting in Frame

Figure 4.16 Loading for case 7

147

The moment of inertia of all the members will be same as in previous case.

Fixed End Moments:

MFBA(cantilever moment) = (1/2)*(-52.2)*8.7*8.7*(1/3)

= -658.503 KNm.

MFBC = (+((117.928*7.6^2)/12+(137.153-117.928)*7.6^2/30))

= 409.854 KNm

MFCB = (-((117.928*7.6^2)/12+(137.153-117.928)*7.6^2/20))

= -351.355 KNm

MFCD = (+((137.153*7.6^2)/12+(156.377-137.153)*7.6^2/30))

= 702.352 KNm

MFDC = (-((137.153*7.6^2)/12+(156.377-137.153)*7.6^2/20))

= -643.852 KNm

MFCF(cantilever moment) = (1/2)*(85.347)*8.7*8.7*(1/3)

= 1076.652 KNm.

MFDE(cantilever moment) = (1/2)*(-85.347)*8.7*8.7*(1/3)

= -1076.65 KNm.

Table 4.15 Moment distribution for case 7

Joint B C D

Member BA BC CB CF CD DC DE

Distribution Factor 0.00 1.00 0.50 0.00 0.50 1.00 0.00

F.E.M -658.503 409.854 -351.355 1076.652 702.352 -643.852 -1076.652

Balance

248.649 -713.825

-713.825 1720.505

C.O.M

124.324

860.252

Balance

0.000 -492.288

-492.288 0.000

C.O.M

0.000

0.000

Balance

0.000 0.000

0.000 0.000

Final moment -658.503 658.503 -1433.143 1076.652 356.491 1076.652 -1076.652

148

For vertical slab AB,

It is subjected to a uniformly varying loads(U.V.L) having maximum intensity of 52.2 KN/m2 at B.

Reactions:

Reaction at B(RB)= ½*(-52.2)*8.7 = -227.070 KN.

= 227.070 KN.

MFBA(cantilever moment) = (1/2)*(-52.2)*8.7*8.7*(1/3)

= -658.503 KNm.

At L/2,M(cantilever moment)=wl^2/48 = (-52.2)*8.7^2/48

= -82.313 KNm.

For horizontal slab BC,

It is subjected to a uniformly distributed load(U.D.L) having intensity of 109.457 KN/m2 at C and 48.689 KN/m2 at B.

Reactions:

Taking moment about C, we have

Rb*7.6 = 658.503-894.817+48.689*7.6^2/2+0.5*7.6*(109.457-48.689)*7.6*(1/3)

=>Rb = 160.064 KN.

Also,∑Fy = 0

=>Rc = 67.938*7.6+0.5*7.6*(109.457-48.689)-160.064

= 440.889954

KN.

Simply Supported moment:

At L/2, M = wl^2/8 +wl^2/16 = 48.689*7.6^2/8+(109.457-48.689)*7.6^2/16

= 585.531 KNm.

Net Moment:

At L/2,

Average FEM = 1045.823 KNm.

Mnet = -460.292 KNm.

149

For horizontal slab CD,

It is subjected to a uniformly distributed load(U.D.L) having intensity of 170.225KN/m2 at D and 109.457 KN/m2 at C.

Reactions:

Taking moment about D, we have

Rc*7.6 = 894.817+1076.652+109.457*7.6^2/2+0.5*7.6*(170.225-109.457)*7.6*(1/3)

=>Rc = 681.481 KN.

Also,∑Fy = 0

=>Rd = 109.457*7.6+0.5*7.6*(170.225-109.457)-681.481

= 381.311 KN.

Simply Supported moment:

At L/2, M = wl^2/8 +wl^2/16 = 109.457*7.6^2/8+(150.976-109.457)*7.6^2/16

= 1009.653 KNm.

Net Moment:

At L/2,

From property of similar triangle,

x/1076.652 = (7.6-x)/356.491

=>1076.652*7.6-1076.652x = 356.491x

=>x = 5.71 from D.

Again,

1076.652/(5.71) =

y/(5.71-

3.8)

=>y = 360.081 KNm.

150

At L/2, Mnet = 1369.734 KNm.

Let the maximum sagging B.M. occur at a distance x from C.

Mx +60.768*x^3/(7.6*2*3)+109.457*x^2/2-681.481x+356.491= 0

=>Mx = -60.768*x^3/(7.6*2*3)-109.457*x^2/2+681.481x-356.491

For maximum,

dMx/dx= -60.768*x^2/(7.6*2)-109.457*x+681.481

which gives x = 5.22

from C.

Now,

Mmax = -60.768*x^3/(7.6*2*3)-109.457*x^2/2+681.481x-356.491

= 1520.027 KNm.

For vertical slab CF,

It is subjected to a uniformly varying loads(U.V.L) having maximum intensity of 85.347 KN/m2 at

C.

Reactions:

Reaction at C(Rc)= ½*(85.347)*8.7 = 371.259 KN.

MFCF(cantilever moment) = (1/2)*(85.347)*8.7*8.7*(1/3)

= 1076.652 KNm.

At L/2,M(cantilever

moment)=wl^2/48 = (85.347)*8.7^2/48

= 134.582 KNm.

For vertical slab DE,

It is subjected to a uniformly varying loads(U.V.L) having maximum intensity of 52.2KN/m2 at D.

Reactions:

Reaction at D(Rd)= ½*(52.2-

85.647)*8.7 = -144.189 KN.

= 144.189

KN.

MFDE(cantilever moment) = (1/2)*(-85.347)*8.7*8.7*(1/3)

=

-

1076.6524

1 KNm.

At L/2,M(cantilever

moment)=wl^2/48 = (-85.347)*8.7^2/48

= -134.582 KNm.

151

Fig:Bending moment diagram

Figure 4.17 Bending Moment Diagram for case 7

152

Case 8:Backfill in hill side with no water in both compartments of the Settling Basin

Active pressure due to soil = Ka*Ysoil*h

= 1/3*18*8.7

= 52.200

Lateral pressure due to water = Yw*h

= 9.81*8.7

= 85.347

Soil pressure at bottom:

∑V = 1242.775

B = 15.2

e = -0.596

Pmax = (∑V/B)*(1+6*e/B)

= 62.537

Pmin = (∑V/B)*(1-6*e/B)

= 100.986

Fig:Loads acting in Frame

Figure 4.18 Loading for case 8

153

The moment of inertia of all the members will be same as in previous case.

MFBA(cantilever moment) = (1/2)*(-52.2)*8.7*8.7*(1/3)

= -658.503 KNm.

MFBC = (+((117.928*7.6^2)/12+(137.153-117.928)*7.6^2/30))

= 356.531 KNm

MFCB = (-((117.928*7.6^2)/12+(137.153-117.928)*7.6^2/20))

= -338.024 KNm

MFCD = (+((137.153*7.6^2)/12+(156.377-137.153)*7.6^2/30))

= 449.067 KNm

MFDC = (-((137.153*7.6^2)/12+(156.377-137.153)*7.6^2/20))

= -430.560 KNm

MFCF(cantilever moment) = (1/2)*0*8.7*8.7*(1/3)

= 0 KNm.

MFDE(cantilever moment) = (1/2)*0*8.7*8.7*(1/3)

= 0 KNm.

Table 4.16 Moment distribution for case 8

Joint B C D

Member BA BC CB CF CD DC DE

Distribution Factor 0.00 1.00 0.50 0.00 0.50 1.00 0.00

F.E.M -658.503 356.531 -338.024 0.000 449.067 -430.560 0.000

Balance

301.972 -55.522

-55.522 430.560

C.O.M

150.986

215.280

Balance

0.000 -183.133

-183.133 0.000

C.O.M

0.000

0.000

Balance

0.000 0.000

0.000 0.000

Final moment -658.503 658.503 -425.692 0.000 425.692 0.000 0.000

154

For vertical slab AB,

It is subjected to a uniformly varying loads(U.V.L) having maximum intensity of 52.2 KN/m2 at B.

Reactions:

Reaction at B(RB)= ½*(-52.2)*8.7 = -227.070 KN.

= 227.070 KN.

MFBA(cantilever moment) = (1/2)*(-52.2)*8.7*8.7*(1/3)

= -658.503 KNm.

At L/2,M(cantilever moment)=wl^2/48 = (-52.2)*8.7^2/48

= -82.313 KNm.

For horizontal slab BC,

It is subjected to a uniformly distributed load(U.D.L) having intensity of 81.762 KN/m2 at C and 62.537 KN/m2 at B.

Reactions:

Taking moment about C, we have

Rb*7.6 = 658.503-425.692+62.537*7.6^2/2+0.5*7.6*(81.762-62.537)*7.6*(1/3)

=>Rb = 292.624 KN.

Also,∑Fy = 0

=>Rc = 62.537*7.6+0.5*7.6*(81.762-62.537)-292.624

=

255.709181

7

KN.

Simply Supported moment:

At L/2, M = wl^2/8 +wl^2/16 = 62.537*7.6^2/8+(81.762-62.537)*7.6^2/16

= 520.916 KNm.

Net Moment:

At L/2,

155

Average FEM = 542.098 KNm.

Mnet = -21.182 KNm.

For horizontal slab CD,

It is subjected to a uniformly distributed load(U.D.L) having intensity of 100.986KN/m2 at D and 81.762 KN/m2 at C.

Reactions:

Taking moment about D, we have

Rc*7.6 = 425.692+0+81.762*7.6^2/2+0.5*7.6*(100.986-81.762)*7.6*(1/3)

=>Rc = 391.057 KN.

Also,∑Fy = 0

=>Rd = 81.762*7.6+0.5*7.6*(100.986-81.762)-391.057

= 303.385 KN.

Simply Supported moment:

At L/2, M = wl^2/8 +wl^2/16 = 81.762*7.6^2/8+(100.986-81.762)*7.6^2/16

= 659.720 KNm.

Net Moment:

At L/2,

Average FEM = 212.846 KNm.

Mnet = 446.874 KNm.

Let the maximum sagging B.M. occur at a distance x from C.

Mx +19.224*x^3/(7.6*2*3)+81.762*x^2/2-391.057x+425.692= 0

=>Mx = -19.224*x^3/(7.6*2*3)-81.762*x^2/2+391.057x-425.692

For maximum,

dMx/dx= -19.224*x^2/(7.6*2)-81.762*x+391.057

which gives x = 4.47

from C.

156

Now,

Mmax = -19.224*x^3/(7.6*2*3)-81.762*x^2/2+391.057x-425.692

= 467.840 KNm.

For vertical slab CF,

It is subjected to a uniformly varying loads(U.V.L) having maximum intensity of 85.347 KN/m2

at C.

Reactions:

Reaction at C(Rc)= ½*(0)*8.7 = 0.000 KN.

MFCF(cantilever moment) = (1/2)*0*8.7*8.7*(1/3)

= 0.000 KNm.

At L/2,M(cantilever

moment)=wl^2/48 = (0)*8.7^2/48

= 0.000 KNm.

For vertical slab DE,

It is subjected to a uniformly varying loads(U.V.L) having maximum intensity of 52.2KN/m2 at

D.

Reactions:

Reaction at D(Rd)= ½*(0)*8.7 = 0.000 KN.

MFDE(cantilever moment) = (1/2)*0*8.7*8.7*(1/3)

= 0 KNm.

At L/2,M(cantilever

moment)=wl^2/48 = (0)*8.7^2/48

= 0.000 KNm.

Fig:Bending moment diagram

Figure 4.19 Bending Moment Diagram for case 8

157

ii. Design

Design of vertical slab

1. Design constants:

Grade of concrete = M20

For steel in compression zone,

Permissible stress in steel for HYDS (σst)=190 N/mm2

Permissible stress in concrete (σcbc)= 7 N/mm2

Modular ratio(m)= 280/3σcbc

= 13.333

Neutral Axis depth factor(k)= mσcbc/(mσcbc+σst)

= 0.329

Lever arm factor(j)= 1-k/3

= 0.890

Coefficient of moment (R)= (1/2)*σcbc*j*k

= 1.026

2. B.M. and Axial Forces:

The vertical wall is designed for maximum B.M. and direct forces which occurs in case of wall CF.

Table 4.17 Bending Moment and axial forces in wall CF

Case B.M. at centre or B.M. at Direct force

intermediate ends (KN-m) (KN)

point (KN-m)

1 0.000 0.000 1468.008 (compressive)

2 134.582 1076.652 992.403 (compressive)

3 0.000 0.000 519.770 (compressive)

4 134.582 658.503 992.403 (compressive)

5 0.000 0.000 1597.976 (compressive)

158

6 134.582 1076.652 1122.371 (compressive)

7 134.582 1076.652 1122.371 (compressive)

8 0.000 0.000 646.767 (compressive)

2. Section at end of the vertical wall:

The section at end of vertical wall CF is subjected to a maximum B.M. of 1076.652 KNm

and a direct compression of 992.403 KN.

Eccentricity(e)= M/P

1084.894 mm

The thickness(depth) of the vertical wall(D)= 750 mm

∴ e/D= 1.447

(0.25<e/D<1.5)

The eccentricity developed is moderate between 0.25 and 1.5.

Effective depth(d) = 700 mm

Let us reinforce the section with 20 mm υ bars @ 140 mm c/c provided on both

the faces as shown in figure below with nominal cover of 50 mm on either side.

Now,

Asc=Ast= 1000*pi()*20^2/(140*4)

= 2243.995 mm2

Let, n be the depth of N.A. from compression face.

The depth of N.A. is computed from the following expression:(B.C. Punima,Design of R.C.C,Eq.14.17)

((b*n/2)*(D-dt-n/3)+(mc-1)*Asc*1/n*(n-dc)(D-dt-dc)) / ((b*n/2)+(mc-1)*Asc*(n-d)/n-m*Ast*(D-dt-n)/n)

= (e+D/2-dt)

=>((1000*n/2)*(750-50-n/3)+(1.5*13.333-1)*2243.995*1/n*(n-50)(750-50-50))/((1000*n/2)+(1.5*13.333-1)*2243.995*(n-700)/n-m*Ast*(750-50-n)/n)

= (1.205+900/2-50)

=> -166.667 n3 - 354947n

2 - 7.458E+07n + 7.022E+10 = 0

Solving the above cubic equation ,we get

159

n= 332.646 mm

The algebraic sum of internal force must be equal to external force,(B.C. Punima, Design of R.C.C,Eq.14.15)

((b*n*c')/2)+(mc-1)*Ast*c'/n*(n-dc)-m*Ast*c'/n*(D-dt-n) = P

=>((1000*332.646*c')/2)+(1.5*13.333-1)*2243.995*c'/204*(332.646-50)-13.333*2243.995*c'/332.646*(750-50-204) = 992.403*1000

=> c'= 5.855 N/mm2

So, compressive stress in concrete= 5.855 N/mm

2

Stress in compression steel(tc)= 1.5*m*c'*(n-dc)/n

= 99.489 N/mm2

Stress in tension steel(t)= m*c'*(D-dt-n)/n

86.204 N/mm2

Thus, the stresses in both concrete and steel are within permissible limits.

Now,

Area of distribution steel= 0.15*b*D/100

= 1125.000 mm2

Area on each layer = .5*1125

= 562.500 mm2

Provide 10 mm distribution bar of area = 78.540 mm

2

Spacing of 10 mm υ bars = 1000*78.540/562.5

= 139.626 mm

Hence, provide 10 mm υ distribution bars @ 130 mm c/c on each layer.

Now, (B.C. Punima, Design of RCC, pp 95)

For, ϕ = 20 mm

Development length(Ld)= σ* υ /4*τbd = 678.571 mm

The details of reinforcement are shown in the figure in annex III-3.

160

Design of horizontal slab

1. Design constants:

Grade of concrete = M20

For steel in tension zone,

Permissible stress in steel for HYDS bars(σst)= 150 N/mm2

Permissible stress in concrete (σcbc)= 7 N/mm2

Modular ratio(m)= 280/3σcbc

= 13.333

Neutral Axis depth factor(k)= mσcbc/(mσcbc+σst)

= 0.384

Lever arm factor(j)= 1-k/3

= 0.872

Coefficient of moment (R)= (1/2)*σcbc*j*k

1.171

2. B.M. and Axial Forces:

The slab is designed for maximum B.M. and direct forces which occurs in case of slab CD.

Table 4.18 Bending Moment and axial forces in slab CD

Case B.M. at centre or B.M. at Direct force

intermediate ends (KN-m) (KN)

point (KN-m) 0 0

1 599.658 1199.316 144.189 (tensile)

2 323.289 1268.517 144.189 (compressive)

3 136.177 643.452 227.070 (compressive)

4 1004.204 418.149 144.189 (tensile)

5 1276.225 1363.942 371.259 (tensile)

6 950.569 418.149 185.630 (compressive)

161

7 1520.027 1076.652 144.189 (tensile)

8 467.840 425.692 0.000

2. Section at mid-point of the horizontal wall:

The section at mid point of horizontal slab CD is subjected to a maximum B.M. of 1520.027 KNm and a direct tesion of 144.189 KN.

Since,the direct tension is small its effect may be neglected.

Overall depth = 700 mm

Using a nominal cover = 50 mm

∴ Effective depth(d) = 650 mm

The eccentricity developed is very large.

Moment of resistance of the section(Mr)= R*b*d^2

= 494.674 KNm

which is less than 1520.027 KNm

Hence,increase the overall depth(D)= 1500 mm

d= 1450 mm

∴ Mr'= R*b*d^2

= 2461.662 KNm (OK)

Now,

Ast(rq.)= Mr/(σst*j*d)

= 8013.140 mm2

Use 32 mm diameter bar having area= 804.248 mm2

So, spacing = 100.366 mm

Using two layers of 32 mm υ rods 100@ mm c/c.

Actual,Ast Provided = 1000*804.248/100

= 8013.140 mm2 (OK)

Bend half the bars up near supports at distance of L/5 = 7.6/5

= 1.520 m

Area of distribution steel= 0.15*b*D/100

= 2250.000 mm2

162

Provide 20 mm distribution bar of area = 314.159 mm2

Spacing of 20 mm dia bars = 1000*314.159/2250

= 139.626 mm

Hence,provide 20 mm υ distribution bars @130 mm c/c on each layer.

2. Section at support of the horizontal wall:

The section at support of horizontal slab CD is subjected to a maximum B.M. of 1363.942 KNm and a direct tesion of 371.259 KN.

But,the effect of direct tension is neglected as the slab is reinforced both at top and bottom.

Grade of concrete = M20

For steel in compression zone,

Permissible stress in steel for HYDS bars(σst)= 190 N/mm2

Permissible stress in concrete (σcbc)= 7 N/mm2

Modular ratio(m)= 280/3σcbc

= 13.333

Neutral Axis depth factor(k)= m*σcbc/(mσcbc+σst)

= 0.329

Lever arm factor(j)= 1-k/3

= 0.890

Coefficient of moment (R)= (1/2)*σcbc*j*k

1.026

Now,Area of steel required

Ast= Mr/(σst*j*d)

= 5561.456 mm2

Area available from the bent up bars from middle section of top portion

= Ast/2

= 4006.570 mm2

Additional area required = 1554.886 mm2

However,providing 20mm υ bars @ 130 mm c/c throughout the bottom portion.

163

Also,providing 10 mm υ distribution bars @ 130 mm c/c on throughout as in top portion.

Now, (B.C. Punima,Design of RCC,pp 95)

For,ϕ= 32 mm

Development length(Ld)= σ*υ/4*τbd = 1085.714 mm

For,υ= 20 mm

Development length(Ld)= σ*υ/4*τbd = 678.571 mm

The details of reinforcement are shown in the figure in annex III-3.

164

CHAPTER 5

RESULTS AND FINDINGS

Power Output:

Design discharge 7.2 m3/s

Installed capacity 4.14 MW

Natural head available 74.12 m

Net head available 68.95 m

Type of weir Ogee shaped

Size of weir (L×B×H) (126×35×4.5) m

Weir crest level 798.5 m.

River bed level 794 m(U/S).

790.4m(D/S).

Undersluice crest level 796.5 m

Intake:

Intake type Side intake

Intake opening (B×H) 3× (2.4×1.5) m.

Intake sill level 796.5 m.

Gravel Trap:

Size (21×7.5×3.5) m

Flushing type Continuous

Size of flushing canal (0.5×0.28) m

Approach Canal:

Material RCC

165

Length 36.9 m

Top width 4.8 m

Bottom width 2.4 m

Height 1.5 m (with freeboard)

Spillway:

Length 10 m

Height 0.626 m

Settling basin:

Size (L×B×H) (86.1 ×7.1 ×4) m.

Number of bays 2 nos.

Flushing type Periodic

Frequency 8 Hrs.

Size of Orifice (1.2×1.2) m

Forebay:

Size (L×B× H) (13 ×14.4×5.6) m

Spillway:

Height 0.588 m.

Penstock:

Type of material Steel

Length 2238.73 m.

Diameter 1.90 m.

166

Thickness 14 mm.

Anchor Block:

Number 27

Bifurcation:

Diameter 1.343 m.

Thickness 6 mm.

Power house:

Type Sub-Surface

Machine Hall(L×B×H) (21.2×12.8×22.3) m.

Equipment Room(L×B×H) (6.4×5×22.3) m.

Transformer Room(L×B×H) (6.4×5×22.3) m.

Turbine:

Type Francis

Number of turbine 2 nos.

Turbine speed 750 rpm

Diameter of Runner 0.675 m

Tail Race Canal:

Size (2.5×1.5) m

Generator:

Type 3 phase synchronous, 6 KV

167

Transmission Line:

Length 8.5 km

Transmission Voltage 33KV

168

CHAPTER 6

CONCLUSION AND RECOMMENDATION

From the preceding design and analysis, it is found that the project is technically feasible and

structurally safe. However, there is relatively less difference in elevation from forebay to

powerhouse, which is only 68.95m for over 2km length, as a results large no. of anchor blocks

are required due to which the economic cost of the scheme would be higher.

As the plant can be connected to the national grid, it will help to minimize the raising electricity

demand of the country. During installation, it is highly recommended to involve experienced

parties and technicians so that the structure would be technically appropriate.

The study reveled that the project is technically feasible and hence recommended for

construction after performing financial analysis.

169

REFERENCES / BIBLIOGRAPHY

Bansal, R. K., A Text Book Of Fluid Mechanics And Hydraulic Machines, Laxmi Publication

(P) Ltd, New Delhi,2005.

BPCH Design Guidelines.

Civil Works Guidelines for Micro hydropower in Nepal, 2002, BPC Hydroconsult,

ITDG, Kathmandu, Nepal.

Dandekar, M.M. & Sharma, K.N., Water Power Engineering, Vikas Publishing House Pvt.Ltd,

Delhi, 1997.

Detailed Feasibility Study of Shreeram MHP, nec 2008.

Final Year Reports of Senior students.

Garg, S. K, Irrigation Engineering & Hydraulic Structure, Khanna Publishers, Delhi,

2002.

Nexant SARI/Energy, September 2002,Regional Hydro-power Resources: Status of

Development and Barriers Nepal.

Harvey, Adam et. al., 1993 ,Micro-Hydro design manual, A guide to small scales water power

schemes, Intermediate Technology Publications ISBN 1 85339 103 4.

Punima, B.C., Design of R.C.C Structure, Laxmi Publications,Delhi,2000.

Rajput,R.K., Strength of Materials, Khanna Publishers, Delhi, 2002.

Ranjan, Gopal and Rao,ASR, Soil Mechanics And Foundation Engineering, New Age

International Pvt. Ltd. Publishers, New Delhi,2007.

Water Conveyance System Design Guidelines For Hydropower Projects, DED.

170

ANNEXES