DESIGN OF HYDRO POWER PLANT WATER CONVEYANS SYSTEM

A PROJECT REPORT

Submitted by

JEYALALITHA.K RAMESH KUMAR.N.S SANTHOSH KUMAR.R

SATHISH KUMAR.N

in partial fulfillment for the award of the degree

of

BACHELOR OF ENGINEERING

IN

CIVIL ENGINEERING

RATNAVEL SUBRAMANIAM COLLEGE OF ENGINEERING AND TECHNOLOGY,

DINDIGUL - 5

ANNA UNIVERSITY :: CHENNAI 600 025

NOVEMBER - 2008

DESIGN OF HYDRO POWER PLANT WATER CONVEYANS SYSTEM

A PROJECT REPORT

Submitted by

JEYALALITHA.K 91305103003 RAMESH KUMAR.N.S 91305103333 SANTHOSH KUMAR.R 91305103335 SATHISH KUMAR.N 91305103338

in partial fulfillment for the award of the degree

of

BACHELOR OF ENGINEERING

IN

CIVIL ENGINEERING

RATNAVEL SUBRAMANIAM COLLEGE OF ENGINEERING AND TECHNOLOGY,

DINDIGUL - 5

ANNA UNIVERSITY :: CHENNAI 600 025

NOVEMBER - 2008

ANNA UNIVERSITY : CHENNAI - 600 025

BONAFIDE CERTIFICATE

Certified that this project report “DESIGN OF HYTRO POWER PLANT

(WATER CONVEYANCE SYSTEM)”. is the bonafide work of

“JEYALALITHA.K, RAMESH KUMAR.N.S, SANTHOSH

KUMAR.R, SATHISH KUMAR.N” who carried out the project work

under my supervision.

SIGNATURE SIGNATURE

Mrs.B. KAMESHWARI,ME.ME.Ph.D., Mr. A.MANI, M Tech.Phd., HEAD OF THE DEPARTMENT ASSISTANT PROFESSOR

Department of Civil Engineering, Department of Civil Engineering Ratnavel Subramaniam College of Ratnavel Subramaniam College of Engineering And Technology, Engineering And Technology, Dindigul. Dindigul. Submitted for viva-voce examination held on INTERNAL EXAMINER EXTERNALEXAMINER

CHAPTER 1

INTRODUCTION

The growing energy demand worldwide on the one hand and the

emerging ecological awareness on the other are leading to an increased demand

for regenerative energy. As a continuously available base-load energy supply

option, hydropower is significant regenerative energy source.

Studies to determine new locations for hydropower plants have

explored innovative avenues, with due consideration of ecological and

economic aspects. There is, however, a strong need for updating the methods for

determining establishing and as far as predictable, future developments. The

hydropower plants that are currently being realized or about to be realized are

predominantly based on old studies, with economic data (investment costs and

revenue) having been updated, but without addressing the general actual issue in

view of energy demand, ecology and globalization.

Water is the most abundant resource in the world, it is important to

utilize the power of flowing water. The most efficient way to harness the power

of water is to collect the potential energy. This is done by damming up a body

of flowing water. A dam is an object that restricts the flow of water. In today’s

hydroelectric dams, the restricted water is diverted to a turbine using a penstock

and exits the turbine through the tailrace.

The turbine is made up of a shaft with blades attached. As a fluid

flows through the blades a rotational force is created. This force causes a torque

on the shaft. The turbine shaft is coupled to a generator, where electricity is

produced. The backbone of most power generation system is the generator. An

electric generator is “any machine that converts mechanical energy into

electricity for transmission and distribution.” The generator works by spinning

a rotor that is turned by a turbine.

The rotor is a shaft that has field windings. These windings are

supplied with an excitation current or voltage. As the rotor turns, the excitation

current creates a magnetically induced current onto a stator. The stator is a

cylindrical ring made of iron that is incased by another set of field windings and

is separated from the rotor by a small air gap.

Hydroelectric generations can vary from 1 watt to 100’s mega-

watts. With today’s technology it is possible to generate power with small scale

parameters. With low flow and low head parameters a micro generator can be

used to produce electric power. From the source of the flowing water, a weir,

small scale dam, can be used to restrict the flow of water. From this the water

can be piped to a turbine. Since the turbine is coupled to the generator, a micro

generator can generate about 1 watt to 100 kilowatts. This generator can be used

to power residential loads.

One of the first steps in planning is to measure the power potential

of the stream. The amount of power that can be obtained from a stream depends

on:

- The amount of water flow

- The height which the water falls (head)

- The efficiency of the plant to convert mechanical energy to

electrical energy.

CHAPTER 2

BACKGROUND

The implementation of the requirements is effected through a

framework plan. When elaborating the framework plan the following relevant

data are incorporated:

Basic data-energy aspects:

� Hydrology: gauge levels from hydrological yearbooks.

� Topography: catchments area sizes and potential head

conditions.

� Existing facilities: surveying using questionnaires and water

registers.

Basic data-ecological aspects:

� General information (e.g. protected areas).

� Linear information such as morphological structure,

distribution of sensitive species along a river stretch.

� Selective in formation such as biological quality of rivers

and streams.

CHAPTER 3

NOMENCLATURE

Ao = Area of orifice or ports

AP = Cross-sectional area of penstocks

At = Area of riser of differential surge tank

A, = Net cross-sectional area of surge tank

A, = Cross-sectional area of head race tunnel

J&h = Thoma area of surge tank

c = Velocity of propagation of pressure wave

D = Diameter of head race tunnel

F = Friction factor governing head loss

[to be taken from IS : 4880 ( Part 3 ) - 1976” ]

F, = Factor of safety over Ath

g = Acceleration due to gravity

H = Gross head on turbines

Ho = Net head on turbines

hr = Total head loss in head race tunnel system

hrp = Total head loss in penstock system

L = Length of head race tunnel

Ls, = Length of riser spill in crest

m = Reciprocal of Poisson’s ratio for rock

P = Power generated

Ph = Pressure due to water hammer in the conduit upstream of

surge tank

Qd = Maximum discharge supplied by the surge tank in case of

specified load acceptance

R1 = Internal radius of the pressure conduit

R2, = Outer radius of the pressure conduit

V’ = Volume of water in surge tank corresponding to Z

Y’ t = Volume of water in the conduit in a given time interval

∆t = V1,At. ∆t

vo* = Velocity of flow in tunnel corresponding to maximum steady

flow, upstream of surge tank

V1* = Velocity of flow in tunnel at any instant, upstream of surge

Tank

V2* = Velocity of flow in conduit at any instant, downstream of

surge tank

Z = Water level in surge tank measured positive above reservoir

level

Zm = Maximum surge level above maximum reservoir level

fi, f2, = longitudinal stresses, N/mm2

C = moment coefficient

E = modulus of elasticity

fi = total circumferential stress, N/mm2

fy = total longitudinal stress, N/mm2

l = height of stiffener ring or ring girder, mm

L = span length of pipe, mm

M = moments

P = internal pressure including water hammer, N/mm2

P1 = total reaction at support, N

q1 = shear stress, N/mm2

Y = radius of pipe shell, mm

rl = mean radius of shell, mm

S = hoop stress in pipe, N/mm2

St = equivalent stress, N/mm2

Sa, S, = principal stresses, N/mm2

I = temperature rise or drop, “C

t = thickness of pipe shell, mm

tl = thickness of stiffener ring or ring girder, mm

W = total distributed weight, that is, self-weight of shell + weight

of water, N/m2

WI = total weight, that is, weight of shell + weight of water, N

z = section modulus of pipe shell, ma

α = coefficient of linear expansion or contraction of pipe shell

material, per “C

µ = coefficient of friction.

CHAPTER 4

OBJECTIVE

The objective is to design a hydroelectric plant utilizing optimal

energy in the water, with minimum submergence and economic costs,

considering seasonal variation in power generation to meet the region's demand

during all seasons.

DEFINITIONS:

Head:

Water level is the highest possible water level at the station intake

in full operation and with zero bypass flow.

Tail:

Water level is the energy head of the water flowing out of the

turbines.

Total (gross) head:

= Vertical distance between head- and tail water.

Gross capacity:

Maximum capacity if all head losses, hydraulic and otherwise, are

considered zero.

Effective head:

It is losses subtracted from the gross head at installed capacity

output.

CHAPTER 5

DESIGN ASSUMPTIONS AND PARAMETERS:

Assumptions:

Detailed geological and topographical investigations carried out by

the Department of Mines and Geology to determine the best site for the dam,

pressure shaft alignment, power house location, etc., can be used for

implementing this design.

Parameters:

Input data consist of site specific data (discharge, water yield,

generation head, evaporation rate, seepage rate etc.), technical data (efficiency

of turbine and generator, and dependability norm for storage capacity, load

factor, etc.) and economic data (civil construction costs for various types and

heights of dam, cost of electrical machinery of various capacities,

environmental costs, rehabilitation costs, etc.).

Decision variables:

The decision variables determine the optimum storage capacity,

installed generation capacity and seasonal power drafts; net energy availability

in the region (objective function) needs to be maximized subject to seasonal

hydrological constraints, and costs and submergence area are to be minimized.

CHAPTER 6

HYDRO POWER

Head is defined as the difference in elevation between two

particular cross sections of the river. Making a head useful for hydropower use

needs a concentration by means of hydropower impoundment, diversion or tail

water lowering. At the point of concentration the powerhouse is situated.

The conversion of the energy potential of the river into electricity

requires a turbine (potential and kinetic energy into mechanical energy)

[rotation] and a generator [rotation into electrical energy]. The output of a

hydropower plant is given in terms of power [kW] and electricity production

[kWh].

Following an equation which computes monthly hydropower

production as a function of volume of water discharged (Q), gross head of this

water (H) and efficiency of the couple turbine generator (r), between 0.7 and

0.85).

Hydropower (kW) is given by:

P (kW) = Q (m3/s) x H (m) x ηtot x 9,81

and approximately = Q x H x 7.8

ηtot = total efficiency

(ηturbine x ηgenerator x ηspeed increaser x ηtrafo)

P = electrical power output

Q = rated discharge

H = net head

Electricity production - the thing we pay for - is power during a

certain time period.

The annual electricity production of a hydropower (HP) station is

approximately calculated as

E (kWh) = P (kW) x 4500 (h)

The head of a HP station is mainly determined by geographical and

topographical parameters. The discharge varies due to the natural flow regime.

Usually a Hydropower station runs at full load for roughly three months. The

rest of the year according to the lesser discharge the station is operated at part

load.

CLASSIFICATION OF HYDROPOWER PLANT:

Hydroelectric facilities range in size from large-scale power plants

to small- and micro-scale plants. Though the definitions may vary, hydropower

plants can be classified as follows:

� Large hydropower plants have a generation capacity of more than

30 000 kW (or 30 MW);

� Small hydropower plants have a generation capacity of between

1000 kW and 30 000 kW;

� Mini hydropower plants have a generation capacity of between

100kW and 1000 kW;

� Micro hydropower plants have a generation capacity of less than

100 kW.

THE BENEFITS OF HYDROPOWER:

Hydropower offers several environmental advantages:

� Hydropower uses the renewable power of naturally flowing water,

With out wasting or depleting it;

� Hydropower produces very few greenhouse gases and no other air

contaminants, and it does not generate any polluting or toxic waste

by-products;

� Hydropower projects with reservoirs can store water to generate

electricity for future use. This storage capacity means that

hydropower can support intermittent renewable sources of

electricity, such as wind and solar power, as well as run-of-river

hydropower;

� The storage capacity combined with unique operational flexibility

allows for optimized use of fossil fuel power plants, and therefore

leads to reduced greenhouse gas emissions and air pollutants from

the electricity sector. In addition to having many environmental

benefits, hydropower facilities can provide the following social and

economic benefits:

� Hydropower is a local resource and is not subject to the drastic

fluctuation of international oil markets;

� Low operating and maintenance costs, and long lifetime make

small hydropower projects virtually inflation proof;

� Hydropower projects have the flexibility to complement other

renewable energy technologies and help support a larger

deployment of the technologies;

� When built in conjunction with existing infrastructure, such as

dams and water level controls, the cost of developing small

hydropower projects are competitive with those of other energy

projects.

� Hydropower is a clean, domestic and renewable source of energy.

Hydropower plants provide inexpensive electricity and produce no

pollution. And, unlike other energy sources such as fossil fuels,

water is not destroyed during the production of electricity. It can be

reused for other purposes.

HOW HYDROPOWER WORKS?

Hydropower converts the energy in flowing water into electricity.

The quantity of electricity generated is determined by the volume of water flow

and the amount of "head" (the height from turbines in the power plant to the

water surface) created by the dam. The greater the flow and head, the more

electricity produced. A typical hydropower plant includes a dam, reservoir,

penstocks (pipes), a powerhouse and an electrical power substation.

The dam stores water and creates the head; penstocks carry water

from the reservoir to turbines inside the powerhouse; the water rotates the

turbines, which drive generators that produce electricity. The electricity is then

transmitted to a substation where transformers increase voltage to allow

transmission to homes, businesses and factories.

TYPES OF HYDRPOWER PLANTS

1. Conventional

Most hydropower plants are conventional in design, meaning they use one-way

water flow to generate electricity. There are two categories of conventional

plants:1 run-ofriver and 2: storage plants.

1.1. Run-of-river plants:

These plants use little, if any, stored water to provide water flow

through the turbines. Although some plants store a day or week's worth of

water, weather changes - especially seasonal changes - cause run-of-river plants

to experience significant fluctuations in power output.

(Figure No 1). The Tazimina project in Alaska is an example of a

diversion hydropower plant. No dam was required.

1.2. Storage plants:

These plants have enough storage capacity to off-set seasonal

fluctuations in water flow and provide a constant supply of electricity

throughout the year. Large dams can store several years’ worth of water.

2. Pumped Storage

In contrast to conventional hydropower plants, pumped storage

plants reuse water. After water initially produces electricity, it flows from the

turbines into a lower reservoir located below the dam. During off-peak hours

(periods of low energy demand), some of the water is pumped into an upper

reservoir and reused during periods of peak-demand.

3. Sizes of Hydroelectric Power Plants

Facilities range in size from large power plants that supply many

consumers with electricity to mall and micro plants that individuals operate for

their own energy needs or to sell power to utilities.

3.1 Large Hydropower

Although definitions vary, DOE defines large hydropower as

facilities that have a capacity of more than 30 megawatts.

3.2 Small Hydropower

Although definitions vary, DOE defines small hydropower as

facilities that have a capacity of 100 kilowatts to 30 megawatts.

3.3 Micro Hydropower

A micro hydropower plant has a capacity of up to 100 kilowatts. A

small or micro hydroelectric power system can produce enough electricity for a

home, farm, ranch, or village.

WATER CONVEYANCE SYSTEM

Power Canals:

Introduction, design-uniform flow, location and construction, unlined

canals in soft ground, lined canal in soft ground, canals in rocks.

Tunnels:

Introduction, geometric design, hydraulic design, cross-sectional forms

and size, rock tunneling procedure, tunnel support, tunnel lining , grouting.

Concrete Pipes:

Applications, pre cast concrete pipes, reinforced concrete pipes.

Penstocks:

Types of Penstocks, stresses in penstocks, economical diameter of

penstocks, design of penstock, number of penstocks and equivalent penstock

diameter, joints in pipeline, anchors and supports, valves.

Surge Tanks:

General function of a surge tank, types of surge tanks, design

consideration of surge tanks, stability of surge tanks, Lined canals, Layout of

lined canals.

CHAPTER 7

TUNNELS

1. INTRIDUCTION:

Tunnels can be designed as underground passages made without

removing the overlying rock or soil.

The headrace tunnel extending from the head pond to the surge

tank has a length of 11860 m. It has a D shaped cross section (20.42 m2) with a

concrete lining. The equivalent hydraulic diameter of the horseshoe section is

approximately D = 4.75 m. The original tunnel layout has a horizontal in the

downstream direction. In order to shorten the construction time of the tunnel it

was thought of working both from the upstream and the downstream end. For

the drilling from the upstream end of the tunnel this brings the question of

evacuation of the water which may come from eventual karstic springs.

Therefore it was proposed to investigate a change in the vertical profile of the

tunnel, where a high point created somewhere in the tunnel will allow to carry

out the drilling works upslope going from both ends. Of course such a solution

can only be feasible if the high point can be placed such that L1≈L2.

Hydraulic calculations were performed for several headrace tunnel

alternatives. It was seen that, it is not possible to create a high point at almost

equal distances from the two adits even by adding a penstock, which was not

envisaged in feasibility study. After all it is decided to keep the headrace layout

in the feasibility study.

CLASSIFICATION OF TUNNELS :

Tunnels may be classified according to their purpose, shape and

supporting arrangements.

Depending on their purpose the following two main groups of

tunnels may be distinguished:

� Traffic tunnels:

� Railway tunnels

� Highway tunnels

� Navigation tunnels

A. Conveyance tunnels:

� Hydroelectric power station tunnels-these shall be referred as

“hydraulic tunnels” in all further discussions.

� Water supply tunnels

� Sewer tunnels transportation tunnels in industrial plants.

B. Hydraulic tunnels can be further sub-divided into the following

categories:

� Pressure tunnels

� Free flowing tunnels

� Free flowing cum-pressure tunnels

C. Depending on their shape, tunnels may be classified as:

� D-shaped

� Horse-shoe shaped

� Circular shaped

� Elliptical shaped

� Square or rectangular shaped

THE SELECTION OF THE TUNNEL CROSS-SECTION IS

INFLUENCED BY:

The clearances specified in view of the vehicles and materials

transported in the tunnel,

� Geological conditions,

� The method of driving the tunnel, and

� The material and strength of tunnel lining.

Tunnels may also be classified as “lined” or “unlined” tunnels. If

the rock conditions are favorable and the tunnel is required to be used for a short

period of time, e.g., a diversion tunnel constructed for the construction of a dam,

the tunnel may be left unlined. However, in most cases, hydraulic tunnels are

invariably lined with cement concrete (Plain or reinforced) or short Crete,

hydraulic tunnels discharging silt - ladden water under high velocities. (E.g. silt

flushing tunnels) are required to be steel-lined.

Depending upon the type of supports, tunnels may be classified as:

i. Tunnels supported by R.S.J sections.

ii. Tunnels supported by rock bolts.

iii. Tunnels supported by short Crete.

iv. Tunnels supported by a combination of (i), (ii), (iii).

TUNNEL CROSS SECTION

Cross – section of a tunnel depends on the following factors:

� Geological conditions prevailing along the alignment,

� Hydraulic requirements,

� Structural considerations, and

� Functional requirements.

D-SHAPED SECTION:

D-shaped section is found to be suitable in tunnels located in good

quality, intact sedimentary rocks and massive external igneous, hard,

compacted, metamorphic rocks where the external or internal pressures.

SURGES IN TUNNELS

Water hammer is created in long closed tunnels by the sudden

closure of the turbine gates. The water hammer pressure provides the necessary

force to retard the flow in tunnel when load is rejected by the turbine. For very

long tunnels, the water hammer corresponding to normal operation of the

turbine may be very great and may require extra ordinary strength of the tunnel

to withstand it and the violent fluctuations of pressure in the tunnel may

seriously interfere with proper turbine regulation. Similarly, for sudden opening

of the gates, the resulting negative water hammer, or reduction of pressure,

provides the necessary force to accelerate the water and is correspondingly

objectionable for very long tunnels.

The simplest means of eliminating the positive and negative water

hammer pressures is to provide a surge tank at the lower end of the tunnel. The

steady state water level in the surge tank fluctuates up and down as the turbine

rejects or accepts the load.

The tunnel should be designed to withstand the maximum excess

pressure that is likely to occur. Similarly, it is very essential to determine the

sub-normal pressures in the surge tank for sudden acceptance of the load. Care

should be taken that the pressure in the tunnel never becomes negative as, under

such conditions, the tunnel is likely to collapse. For head race tunnels full,

maximum pressures occur at the time of load acceptance. However, for tail race

acceptance. Hence, to meet the safety requirements of a tail race tunnel, a surge

tank may be provided downstream of the power house also.

Methods of tunneling:

� Full face attack

� Top heading and benching

� Bottom heading and stopping

� Drift method.

TUNNEL SUPPORT:

When an underground opening is made, it generally becomes

necessary to install supports to hold the rock which has a tendency to drop out

of the roof of the opening .in the earlier day’s timber sections were used as

temporary supported till permanent lining could be placed. With the gradual

availability of steel sections, timber supports have now become almost obsolete.

More recently on the basics of work done by beiniiiawaski , Barton , rabcewicz

and others, even steel supports are being dispensed with and the present trend is

to reinforce the rock by means of rock bolting and shortcreting.

THE NEED FOR TUNNEL SUPPORT:

The necessity of tunnel support arises from the fact that the

excavated rock has a tendency to drop out of the roof of the tunnel. The time

which the loosened rock takes to drop out and also the amount of rock expected

to fall depend upon the “bridge action period” of the rock. The bridge action

period “to” be defined as the time which elapses between blasting and the

beginning of collapse of the unsupported roof. It may range from a few hors to a

few weeks. The bridge action period for cohesion less and or completely

crushed rock is almost zero.

TYPES OF STEEL SUPPORT SYSTEMS:

Tunnel support system by way of steel ribs may be classified into

the following:

(a) Continuous rib,

(b) Rib and post,

(c) Rib and wall plate,

(d) Rib, wall plate and post,

(e) full circle rib; and,

(f) Invert strut in addition to those shown in types (a) to (d).

CONSTITUENTS OF TUNNEL SUPPORTS:

Every type of tunnel support system consists of two or more

different elements, each of which serves a different function, the basic elements

are:

(a) Ribs

(b) Posts

(c) Invert strut

(d) Wall plates

(e) Crown bars

(f) Truss panels

(g) Bracings and spreaders

(h) Blocking

DESIGN PROCEDURE:

� Construction of load diagram

� Construction of force polygon

� Determination of thrusts and

� Computation of stresses in the arch rib

Lining in tunnels is a very important component and makes up for

30 to 40 percent of the total cost of tunnel. Tunnels forming part of water

conductor system have to be invariably lined with cement concrete-plain or

reinforced: or steel lined. However, in cases where it has served the purpose

e.g., diversion tunnels the lining could be avoided.

CONCRETE LINING:

The function of concrete lining is one or more of the following:

� To reduce head losses in the system;

� To protect steel ribs from deteriorating;

� To prevent leakage of water;

� To protect the turbines by preventing loose rock particles falling

into the water and being carried to the turbines;

� To take that part of the internal pressure. Which is not taken by

the rocking.

STEEL LINING

General:

As briefly described earlier, steel lining is provided where the

tunnel has to withstand high pressures. Steel lining consists of a steel plate of

adequate thickness provided to the inner surface of the tunnel and serves the

following purposes:

To prevent water loss fro the tunnel;

� To resist the bursting pressure of water carried by the tunnel;

� To provide protection from seepage of water from the

surrounding mass like rock, concrete etc. and

� To provide a smooth surface for flow of water.

CHAPTER 8

HYDRO PROJECT IN UTTARAKHENT

DESIGN DATAS:

FRL = 1017.00

MWL = 1020.00

MDDL = 1009.50

Length of HRT = 11.86Km

Proposed shape of HRT = D Shape

NO of units = 3

Power = 3 x 33 MW

TWL = 810

Net head = 204.5 – 16.4

= 188.1m

Gross head = 204.5m

Design discharge = 59.60 Cumec

1. HYDRAULIC DESIGN:

Discharge Q = V x A

.’. Q/V = A

= 0.905 D2

= 59.6 / ( 3 x 0.905 )

.’. dia D = 4.68m

D = 4.75m

.’. Vetted perimeter P = 3.58 D

P = 17M

Hydraulic Radius = A / P

= 0.2528 D

= 1.2m

Net head = H = ( MDDL – TWL ) + 2/3 ( FRL – MDDL )

= ( 1009.5 – 810 ) + 2/3 ( 1017 – 1009.5 )

H = 204.5m

Hydro power P = γQHη

= 9.81 x 103 x 59.6 x 204.5 x 0.92

Power = 110 MW

TFL at the end of the tunnel

= total head – friction loss

Mannings Formula:

hf = V2 x N2 x L / R4/3

where

N = Rugosity Co – efficient ( for inner tunnel )

Take N = 0.014

.’. hf = 32 x 0.0142 x 11.86 x 103 / 1.24/3

hf = 16.4m

.’. inner end of the tunnel

TFL = 204.5 – 16.4

= 188.1m = 188m

TYPE OF FLOW

Reynold No = ρVD / µ

R = 1000 x 3 x 4.75 / 1x10-2

= 1.425x106 > 4000

.’. it is a turbulent flow

2. STRUCTURAL DESIGN:

Internal diameter of tunnel = 4.75m

Thick of lining (t) = 0.2m

Internal radius of tunnel (r) = 4.75/2

= 2.375m

Mean radius of tunnel lining (R) = 2.375+0.1

= 2.475m

Young’s modulus of lining material (E) = 2.1x106/13 kg/cm2

= 1.6154x106 t/m2

Moment of inertia (I) = bd3/12

(considering 1m strip of lining) = (1 x 0.23)/12

= 0.0007 m4

Unit weight of water (W) = 1.0 t/m3

Unit weight of concrete (Wc) = 2.5 t/m3

Total rock load on mean dia (P) = 5.15x1.287x2.5x1

Pressure p = (4.75+0.4)x(0.25x

4.75+0.2+0.2)

= 16.57 t

From table 4.2(S.l no:2) = 0.25 x (B)

= 0.25x(4.75+0.22)

φφφφ

Due to uniform vertical load

Due to conduit weight

Due to contained water

Total

00

= 0.125 PR = 0.125 x 17.06 x 2.475 = 5.278

= 0.4406 Wct R2

= 0.4406 x 2.4 x 0.2 x 2.4752 = 1.295

= 0.2203 Wr2R = 0.2203 x 1 x 2.3752 x 2.475 = 3.0755

9.945

450

Zero

= - 0.0334 Wct R2

= - 0.0334 x 2.4 x 0.2 x2.4752

= - 0.0970

= - 0.0167 Wr2R = - 0.33

- 0.427

900

= - 0.125 PR = - 5.278

= - 0.3927 Wct R2

= - 1.1546 = - 0.1963 Wr2R = - 2.7403

- 9.172

1350

Zero

= 0.0334 Wct R2

= 0.097 = 0.0167 Wr2R = 0.33

0.427

1800

= 0.125 PR = 5.218

= 0.3448 Wct R2

= 1.0138 = 0.1724 Wr2R = 2.407

8.698

Now m = Qbd2 Q = 13 b = 100 cm .’. d = 9945/(13 x 100) = 7.68 cm Actual thickness of lining provided = 20 cm Including over break in rock, Total thickness of concrete lining up to the payline = 30 cm

Hence the depth is considered to be adequated.

BENDING MOMENTS

φφφφ

Due to uniform vertical

load

Due to conduit weight

Due to contained water

Total

00

Zero

Zero

Zero

Zero

450

= - 0.25 P = - 4.142

= - 0.8976 WctR = - 1.066

= - 0.4488 Wr2

= - 2.749 - 8.551

900

Zero

= 0.1667 WctR = 0.1980

= 0.0833 Wr2

= 0.510 0.708

1350

= 0.25 P = 4.142

= 0.6732 WctR = 0.739

= 0.3366 Wr2

= 2.062 7.003

1800

Zero

Zero

Zero

Zero

Maximum radial shear = 8.551 t

.’. Shear stress = 8.551/(100 x 20) = 4.28 kg/cm2 < 5 kg/cm2

Hence the design is safe

RADIAL SHEAR

φφφφ

Due to uniform vertical

load

Due to conduit weight

Due to contained water

Total

00

Zero

= 0.1667 WctR = 0.198

= - 1.4166 Wr2 = - 8.678

- 8.48

450

= 0.25 P

= 1.1332 WctR = 1.346

= - 0.7869 Wr2

= 4.82 0.668

900

= 0.5 P

= 1.5708 WctR = 1.866

= - 0.2146 Wr2 = - 1.314

8.837

1350

= 0.25 P = 0.4376 WctR = 0.519

= - 0.4277 Wr2 = - 2.620

2.041

1800

Zero

= - 0.1667 WctR = - 0.198

= - 0.5834 Wr2

= - 3.574 - 3.772

Maximum negative thrust = - 8.48 t ( which indicates tension at the crown)

Tensile stress in concrete at crown

= 8.48 x 1000/(100 x 20) = 4.24 kg/cm2 < 5 kg/cm2

Hence the design is safe

NORMAL THRUST

φφφφ

Due to uniform vertical

load

Due to conduit weight

Due to contained water

Total

00 Zero Zero Zero Zero

450 = 0.01473 x PR3/EI

= 3.17 mm

= O.504 x WctR4/EI

= 0.81 mm

= 0.0252 x Wr2R3/EI

= 1.9 mm

5.88

900 = 0.04167 x PR3/EI

= 8.95 mm

= O.13090 x WctR4/EI

= 2.10 mm

= 0.06545 x Wr2R3/EI

= 5.29 mm

16.34

1350 = 0.01473 x PR3/EI

= 3.17 mnm

= 0.1309 x WctR4/EI

= 0.67 mm

= 0.02108 x Wr2R3/EI

= 1.59 mm

5.43

1800 Zero Zero Zero Zero

Deflections are with in permissible limits.

HORIZONTAL DEFLECTION

φφφφ

Due to uniform vertical

load

Due to conduit weight

Due to contained water

Total

00 Zero Zero Zero Zero

450 = 0.02694 x PR3/EI

= 5.9 mm

= 0.09279 x WctR4/EI

= 1.54 mm

= 0.0464 x Wr2R3/EI

= 3.5 mm

10.94 mm

900 = 0.04167 x PR3/EI

= 9.2 mm

= 0.13917 x WctR4/EI

= 2.31 mm

= 0.06958 x Wr2R3/EI

= 5.26 mm

16.77 mm

1350 = 0.0564 x PR3/EI

= 12.52 mm

= 0.18535 x WctR4/EI

= 3.08 mm

= 0.09268 x Wr2R3/EI

= 6.96 mm

22.56 mm

1800 = 0.0833 x PR3/EI

= 18.50 mm

= 0.2618 x WctR4/EI

= 4.34 mm

= 0.1309 x Wr2R3/EI

= 9.89 mm

32.73 mm

Deflections are with in permissible limits.

VERTICAL DEFLECTION

CHAPTER 9

FOR ECONOMICAL DIMENTIONS FOR TUNNELS

Using the following symbols and assumptions:

D = Tunnel Diameter in meters.

d = Mean he of Lining.

E = Mean unit price of tunnel excavation Rs.290/- per cum

L = Unit price of concrete lining = Rs. 485-per m3

G = Mean unit price of grouting = Rs. 250/- m3

A = Total cost of tunnel per meter length.

C = Contingencies [percentage of total cost = 5%]

S = Supervision charges [percentage of total cost = 15%]

O = Operating and maintenance cost = 10%

Y = Life project in years = 5 years

P = Depreciation factor ( for straight line method = 1/50)

N = Rate of interest ( 8%)

B = Manning’s co-efficient (0.014 for concrete lined tunnel )

R = Hydraulic mean radius ( D/4 for circular tunnel)

Q = Equivalent discharge = 120 m3/sec

U = Value of one unit of power = 0.10

e = Overall efficiency of plant = 0.85

Economical diameter of tunnel as per hand book of hydroelectric engineering by

Dr. P.S.Nigam is given by

D7.33 = (19.35Q3n2e x 4x105)/[(E+0.36)L x O]

D = Diameter of tunnel

Q = Discharge through tunnel = 59.6 m3/sec

n = Rigidity co-efficient = 0.014

e = overall efficiency = 85%

u = cost of power = 16 paisa/unit

CONSTRUCTION COST AND FIXED CHARGES:

(a) Cost of excavation:

= [E(D+2d)2]/4 Rs/m

(b) Cost of lining

= L [ (D+2d)ss-D2]/4 Rs/m

(c) Cost of grouting:

= g(D+2d)

That fore total cost per linear meter.

A = E[D+2d]/4 + L[(D+2d)2-D2]/4 + g(D+2d)

A = [ED2/4] + EDd + Ed2 + Ld2 + LDd + gD + 2gd

A = [ED2/4] + D[Ed + Ld + g] + Ed2+ Ld2 + 2gd

This must be increased by ‘C’ for contingencies and by S for

supervision charges.

Overall cost = A (1+c) x (1+s)/m

Annual charges of tunnel due to supervision & interest

= A (1+c) x (1+s) x (P+N)/m

OPERATION & MAINTENANCE COST:

Gross grass annual cost and expressed operation & maintenance cost

= A (1+c) x (1+s) x O

Value of annual cost:

Loss of head due to friction per M length =

h = n2 v2/R4/3

= n2 Q2/ [(D2/4)2 x (D/4)4/3]

h = 101.12 D-16/3 n2Q2

Power Loss = 9.8 Qh x e

Total number of hours of separation per year = 7925 hours

= 9.8 eTQ(101.12D-16/3 n2 Q2)

= 1000 en2 TQ3D-16/3U

That fore total annual cost.

T1 = A(1+c) x (1+s) x (P+N) + A(1+c) x (1+s) x

O+1000en2TQD-16/3 x w

Where

A = ED2/4 + D[Ed + Ld + G] + Ed2 + Ld2 + 2Gd

For the economy

dT1/dD = 0

Difference the above

[ED/2 + d(E+L) + g] x (1+c) x (1+s) x (p+N) - [(1000 x 16)/3] x

n2Q3TUeD-16/3 = 0

(OR)

[D+2d(E+L)+G]/E = [322000/3] x {[(n2Q3TUeD-19/3)]/

[E(1+c)x(1+s)x(P+N)}

(OR)

(32000/3) en2Q3 x [T.U/E(1+c).(1+s).(p+N)] x D-19/3

= D + [2d(E+L)+G]/E

(OR)

mD-19/3 = D + K

(OR)

loge m D-19/3 = D + K

(OR)

loge m D-19/3 = loge (D + K)

(OR)

loge m (–19/3) x loge D = loge (D + K)

By substituting

loge D = [2(D+1)/(D+1)]

loge (D + K) = loge K + 2D/(2K+D)

then,

loge m-19/3 x 2(D-1)/(D+1) = loge K + 2D/(2K+D)

(OR)

2D/2(K+D)+(19/3) x 2(D-1)/(D+1) = loge m - loge K

By multiplying (2K+D)(D+1)

We obtain

2D(D+1)+(38/3) x (2K+D)(D-1) –

(2K+D)(D+1) x (loge m- loge k) = 0

By substituting the volume assumed we hence,

m = (32000/3) x en2Q3 x [(T x U) / E(1+C)(P+N)(1+S)]

m = (32000/3) x [(0.85x0.0142x56.93x8400x0.16) /

(350x(1+0.05)(0.02+0.07)(1+0.10)]

m = 1.9240 x 108

loge = 19.24

K = [2 x 0.4(350+750) + 250] / 350

loge = 1.172

2D(D+1) + (38/3)(6.526xD)(D-1)-(6.556+D)(D+1)

(19.24-1.425) = 0

Solving this equation we will get.

D = 2.58 m

DESIGN OF TUNNEL C.S.I.R METHOD:

BIENIAWSKI (OR) C.S.I.R METHOD:

(a) Access the various parameters

(b) Access the rating of each the above parameters using table ten

percentages from tunnel design manual.

(c) Once the basic, rock mass value has been arrived at adjust for the joint

orientation.

(d) Arrived at the total rock mass rating and the classify the rock mass

accordingly.

(e) Having accessed the rock mass classification determined the expected

standard time figure 10.8 from tunnel design manual.

(f) After determined the rock mass classification and the stand up time

arrive at the support requirements in the table 10.2 from tunnel design

manual.

BARTON’S N.G.I METHOD:

Design of rock bolts and facing of rock bolts.

(i) Access various rock parameters and there corresponding values

using table 10.3 from tunnel design manual.

Q = [RQD/Jn] x [Jr/Ja] x [Jw/SRF]

(ii) Work out the value of De called the equivalent dimension using

the relationship .

(iii) Roof support.

Proof = (2/Jr)(Q)-1/3

(iv) Depending upon the value of rock mass quality.

DESIGN:

(1) Q = [RQD/Jn] x [Jr/Ja] x [Jw/SRF]

RQD = Rock quality of designation

Jn = Jt, structural number

Jr = Jt, roughness number

Ja = Jt, Alteration number

Jw = Jt, Water reduction factor

SRF = Stress reduction factor

Dc = Equivalent dimension

(ii) Dc = excavation span, Dia & Height/Excavation support ratio

Table 10.4

(iii) Roof pressure

Proof = [2/Jr] x Q-1/3

= (2/2) x (12)-1/3

= 0.44 kg/cm2 �(1)

(v) substituting (1) value for (i)

Q = [RQD/Jn] x [Jr/Ja] x [Jw/SRF]

= (72/6) x (2/2) x (1/1)

= 12 Very good rock.

Using table 10.4 [from tunnel manual]

Value of ESR = 1.6

Excavated diameter of tunnel = 5m

Dc = 3.125

Provide rock bolt 3m long @ 1.5 m c/c.

Determination of thrusts

T = 63000 kg

Computations of stresses in the area Rib

C = Chords length between following point in cm

C = 150 cm In our case

R = Radius of neutral axis of the rib in cm

R = 1750 cm

h = Rise of arc between blocking points in cm

h = R- sq root [R2-(c/2)]2

h = 1750 - sq root (17502 – (150/2)2

h = 1.6 cm

T = Thrust in kg = 6300 kg

Mt = Bending moment in kg.cm

Mt = hT = 1.6 x 63000

= 1,00,,800 kg.cm

M max = Maximum 3m in kg.cm in the rib

= 0.86 x 100800

= 86,608 kg.cm

Z = section modulus of the rib in cm3

Z = 8603.6 cm3 x RST x 300 x 140

A = 56.76 cm2

fr = (T/A) + (M max/Z) kg/cm2

fr = (63000/56.76) + (100800/8603.6)

fr = 1109.94 + 11.72

fr = 1121.76 kg/cm2Since allowable stress in the rib is 1150 kg/cm2

The Design considers being safe.

Hence ok.

ROCK MASS CALCULATION:

i) Q = ( RQD x Jr x Jw ) / ( Jn x Ja x SRF )

Where

RQD = Rock quality designation

Jr = Joint roughness number

Ja = Joint alteration number

Jw = Joint water reduction factor

SRF = Stress reduction factor

Dc = Equivalent dimension

ii) Dc = Excavation span, Dia ( or ) Height / Excavation support ratio

[ESR]

iii) Root Pr Proof = 2 x Q-1/3/ J6

= 2 x 12-1/3/ J6

= 0.44 Kg / Cm2

iv) Q = ( RQD x Jr x Jw ) / ( Jn x Ja x SRF )

= 72 x 2 x 1 / 6 x 2 x 1

= 12, Very good rock

Using table 10.4 value of ESR = 1.6

Excavated dia of tunnel = 5m

Dia = 5 / 1.6 = 3.125

Provide rock bolt 3m long @ 1.5m C/C

CHAPTER 10

SURGE TANK

A surge tank is provided at the end of the headrace tunnel in order

to protect the hydro mechanical equipment and the tunnel lining against a

possible damage from water hammer.

The operation of storage power plants requires a completely free

operation without any restrictions on changes in loading or flow of neither the

pumps nor the turbines. Examinations of traditional, simple Shafter chamber -

type surge tanks show their ineffectiveness due to the required chamber volume

and the resulting costs. This demand led to the development of a more effective

throttling device in connection with dual chamber surge tanks.

SELECTION OF SURGE TANK TYPE:

The principle demand on a surge tank is to compensate the mass

oscillation of the water flow in the pressure tunnel of load changes of turbines

and/or pumps, whereas the construction type in connection with a suitable

throttling device should effect in a most powerful damping of the amplitude

already in the very first period of oscillation. Partial or full-load rejection leads

to on upsurge oscillation, whereby the maximum pressure is limited by the

bearable stress of the concrete lining of the power tunnel. Load demand,

however is followed by a down surge oscillation and the damping effect of the

throttling device should avoid reaction on the turbine or pump. In this case the

minimum pressure must not come below the elevation of the power tunnel.

For the design of the pumped-storage power plant and later for

rebuilding of a new waterway of high-head power plant an investigation for the

most economic type of surge tank fulfilling the operational requirements has

been carried out. Four types of surge tanks with different throttling devices were

investigated with specific computer software developed by Verbundplan and the

results compared.

- (type 1) Shaft surge tank with orifice

- (type 2) Chamber surge tank with symmetric orifice

- (type 3) Differential surge tank with asymmetric orifice

- (type 4) Differential surge tank with reverse flow throttle

The obvious different characteristics and damping effects of

chamber surge tanks and differential types are compared as for example in

figure 1 for a single load case full load rejection. The graphs simply show the

benefit of differential surge tanks due to the much more effectiveness in

damping of the oscillation.

Loading cases for comparison of different surge tanks

Fig: 1 Different Characteristics and Damping Effect

SELECTION OF A SUITABLE THROTTLING DEVICE

Shaft surge tanks (type 1) and simple chamber surge tanks (type 2)

usually are equipped with simple throttle blends or asymmetric orifices. For the

latter the ratio of upsurge to down surge losses varies from 1:2 about to 1:3

depending on the geometric construction.

New methods were required to get this ratio up higher for

economical surge tank design. Such asymmetric reacting throttling devices can

be used in principle only with differential surge tanks. (type 3).

Fig. 2: Surge Tank with Reverse Flow Throttle

These consist of two separate hydraulic systems:

The lower chamber narrows at the end to a ventilation pipe with

much smaller diameter, leading upward into the upper chamber. The second

system consists of the upper chamber and shaft. The throttling device is located

at the bottom of the shaft and dramatically retards emptying of the shaft and the

upper chamber.

The pressure is controlled by the level in the ventilation pipe which

drops very fast, because as it empties suddenly and unhindered into the lower

chamber. The so-called reverse flow throttle was developed based on an idea of

Thoma. It consists of a steel torus similar to a spiral casing of a Francis turbine

(fig. 2). The downsurge oscillation produces a vortex flow which stabilizes

within a few seconds. The water is forced to exit the torus through a small

connection pipe rectangular to the plane of the vortex flow and is discharged

into the lower chamber. This change of flow direction results in very high

pressure losses, these are 20 – 50 times higher than in reverse direction (type 4).

Steady Flow Calculations

For normal operation conditions, neglecting small (automatic)

discharge adjustments by the turbine governor (closing or opening the wicket

gates), the flow in the hydraulic conveyance system can be assumed to be in

steady state. To be able to obtain the available head (which is defined as the

lake level minus the total head loss) at the turbine entrance for a given

discharge, the head losses in the conveyance system must be calculated. By

subtracting the tail water elevation from the computed available head, the net

head for the operation of turbines is obtained. In order to be able to compute the

net head for all possible "incomingdischarge/ turbine-discharge" combinations,

a computation sheet for the steady state calculations was established for all

discharges appearing in the flow duration curve. The results presented in the

computation sheet are then used in obtaining the duration curves for ; turbine

discharges, available head, net head, total power generation, overall turbine

efficiency.

LOAD CASES CONSIDERED FOR SIMULATIONS:

The transient calculations have been carried out, with the program

SIMPIP, for 4 principal load cases:

Load case 1: Full load rejection:

At the beginning of the simulation, all three Francis turbines are

functioning at their nominal discharge, namely 19.87m3/s. The total discharge

passing through the headrace tunnel is 59.6m3/s. After 8 seconds of normal

operation, all three turbines are rapidly closed.

Load case 2: Load demand during power plant start-up:

At the beginning of the simulation all three turbines are stopped.

Everywhere in the system the discharge is null and hydrostatic conditions

prevail. The three Francis turbines are started as follows: The wicket gates of

the all turbines are opened 15% in 3s (This assumes that the wicket gates can be

opened from 0 to 100% in 20s). For the next 60s the turbines operate at 15%

opening while the machines are synchronized with the network. It is admitted

that at the end of this period the synchronization is obtained and the units are

connected to the network. The wicket gates of the units are opened to 100% in

the next 17s. From then on, turbines continue to operate at constant nominal

discharge which is 59.6m3/s.

Load case 3: Combined turbine emergency closure and start-up:

At the beginning of the simulation, all three Francis turbines are

functioning at their nominal discharge, namely 19.87m3/s. The total discharge

passing through the headrace tunnel is 59.6m3/s. After 8 seconds of normal

operation, all three turbines are rapidly closed. After closure, all three turbines

are started following the start-up procedure described in load case 2, when the

water level in the surge tank reaches its minimum level.

Load case 4: Load demand during power plant start-up:

At the beginning of the simulation all three turbines are stopped.

Everywhere in the system the discharge is null and hydrostatic conditions

prevail. The three Francis turbines are started as follows: The wicket gates of

the first turbine are opened 15% in 3s (This assumes that the wicket gates can be

opened from 0 to 100% in 20s). The wicket gates are left at 15% opening for 2s

in order to let the turbine to take up speed. The wicket gates are closed partially

to 10% opening in 1s. For the next 60s the turbines operate at 10% opening

while the machines are synchronized with the network. It is admitted that at the

end of this period the synchronization is obtained and the unit is connected to

the network. The wicket gates of the first unit are opened to 100% in the next

18s. From then on, the first turbine continues to operate at constant nominal

discharge which is 19.87m3/s. The above start-up procedure for the next unit

starts when the preceding machine is synchronized. Therefore during the first

three seconds of the 18s period, during which the first turbine discharges is

increased from 10% to 100%, the wicket gates of the second unit is opened to

15%; and so on.

CHAPTER 11

HYDRAULIC DESIGN OFSURGE TANK:

Ath = Lat/bV12 H0 . V1

2 /2g

` At = 20.41m

L = 11.86km

V1 = 3m/sec

H0 = 204.5m

Ath = (11860x20.41/1.822x32x204.5 ) x (32 / 2x9.81)

= 33.112m

β = hf / V12

hf = 16.40m

V1 = 3m/sec

.’. β = 1.822m

AREA OF SURGE TANK:

As = Ath [ 1-1.5(1-K)]

= 133.94m

SURGE HEIGHT:

Z* = V0√ L/g .At /A s

= 3 √11860/9.81 x 20.41/138.94

= 40.72m

Z* = 40.72m Po = 0.2048

Maximum upsurge = 1- 2/3 Po + 1/9 Po2

= 1-0.137+0.0047

= 0.863 neglecting the 3rd term,

Zm = 0.863 x 40.72

= 35.14m

The following down surge

Zm” = - 1/(1+0.478)

= - 0.676

Zm” = - 0.676 x 40.72

= - 27.53m

Instantaneous complete loading

Zm ’’ = - 1 - 0.0256

= - 1.0256

Zm = - 1.0256 x 40.72

= - 41.76m

Assume 50% load,

LOAD REJECTION

L/2gφb2 VO2 - Zm / b VO

2 - L/2gφb2 VO2 [ e – 2gφ/L(2m+ b VO

2)] = 0

e = 0.92

[turbine constant x 11.86/2x9.81x6.56x1.8222x32 - 35.14/1.822x32

- 11.86x103/(2x9.81x6.56x1.8222x32

[0.92/[(2x9.81x6.56)/(11.86x103)] x [35.14+(1.822x32)] = 0

= 3.08 – 2.143 –3.08 [0.92-0.01085 x (51.54)] = 0

= - 0.174

STRUCTURAL DESIGN OF CIRCULAR SURGE TANK:

Circular tank with fixed base for capacity 4,00,000 litres. The

depth of water is to be 4m.

Area of surge tank = 133.94

Dia = 13.06m.

Use M20 grade concrete and grade – I mild steel.

Permissible direct tensile stress in concrete = 1.2N/mm2

Permissible stress in direct section = 100 N/mm2

Capacity of tank = 400000litres.

Dia = 13.06m.

Free board = 200mm.

Density of water w, = 10KN/m3

Tensile stress in steel = 100N/mm2

σcbc = 7N/mm2

j = 0.841

Q = 14.01

m = 13

overall height of tank = 4+0.2 = 4.2m.

The thickness of walls and base slab are assumed to be 160mm thick.

Bending moment, ring tension and shear

[H2/Dt] = [4.22/(13.06 x 0.16)]

= 8.4

= 8.5

Is : 3370 part (iv) -1967 table 14.2

Max bending moment = -0.0122 wH3 at base

= (-0.0122x8.5x4.2 3)

= -7.683KNm

Max shear = 0.158wH2

= (0.158x8.5x4.22)

= 23.7 KN at base

Max ring tension = 0.608[(WHO/2)] at 0.64 from top

= 0.608 x 8.5 x 4.2 x (13.06/2)

= 141.74 KN

Acting at 2.52 m from top.

Steel for hoop tension

Ast = 141.74x103 /100 = 1417mm2

Using 20mm φ bars spacing

= [(100x314)/1417]

= 221.6m

Use 20mm φ bars at 200mm c/c (As = 1571mm2)

Wall thickness required from hoop stress consider is given by,

{ (141.74x103)/(100t+(13-1)x1571)} = 1.2

T = 99.3mm <150mm adopted

Steel for bending moment:

The thick required from bending moment consideration is usually

very small. Hence the area of steel required will be calculated.

Ast = [(7.683x106)/(100x0.84x1.30)]

= 702.7mm2

Use 12mm φ bars at 130mm centers.

Vertical reinforcements:

Max vertical reinforcements at top

= [(0.3/100)x150x1000]

= 450mm2

Distributing for each face

As = 225 mm2

Provide 12mmφ bars at 400mm c/c on both faces

Base slab reinforcements:

At junction of wall and slab provide 12mmφ at 130mm centers.

At center, top and bottom

As = [(0.3/100)x150x1000]

= 450mm2

Providing top and bottom

As = 450mm2

Use 8mm φbars at 200c/c both at top and bottom and both ways.

Check for shear stress:

Max shear at base v = 23.7 KN/m

Max shear stress τv = (v/bjd)

= [(23.7x103)/(1000x0.841x130)]

= 0.22 N/mm2 < 1.7 N/mm2 permissible.

CHAPTER 12

PEN STOCK

INTRODUCTION:

A penstock is a piping system normally used in hydraulic power

plant for conveying water from the reservoir to the powerhouse where the

turbines are located to generate power. The height maintained between the

powerhouse and the reservoir is the driving potential for the water to flow

through the penstock and reaching to the turbine. When water flows through the

penstock from the reservoir, it induces pressure in the inside surface of the

penstock pipe. This pressure is the essential cause vibration in the pipe.

The cause of vibration may be due to bending of pipes, vortex

shedding, turbulence, water hammering. Vortex shedding occurs when the flow

past an obstacle such a sphere or any other disturbing object, resulting in

vortices behind the object which may induce vibration.

Water hammer normally occurs during the opening or closing of

valves. Severe vibrations may cause failure of the piping system that can cause

the economic loss for the plant and in extreme cases it leads to loss of human

lives. In order to minimize the vibration level to the best possible extent,

“Anchor Blocks (AB)” which is concrete structures provided at the pipe joint

with the ground and the intermediate portions of the pipe are supported by

“Rocker Arms (RA)”.

The aim of this work is to study the effect of flow-induced

vibration in a penstock by coupling the fluid flow and the solid surface through

the forces exerted on the wall by fluid flow.

The fluid flow causes the structure to deform it, and then produces

changes in the flow; as a result feedback between the structure and flow occurs.

This phenomenon is called as fluid structure interaction (FSI). FSI involves

mainly three discipline; namely, computational fluid dynamics, structural

mechanics and to some extent control systems theory.

The simulation is carried out by using a software package ANSYS

where sequentially coupled physics analysis method is used for solving FSI

problems.

PENSTOCKS

Penstocks can be installed over or under the ground, depending on

factors such as the nature of the ground itself, the penstock mate rial, the

ambient temperatures and the environmental requirements. Interred penstocks

should be generally preferred to exposed ones, because of the smaller visual

impact and possible movement barri ers for animals. Nevertheless the burying

of penstocks could have major geological risks connected with the stability of

steep slopes traversed by pipes, both during construction and operation. In fact

during operation water leakage from an interred penstock could trig ger

landslides much more easily than an exposed one.

The following measures help to reduce the environmental impact

of penstocks:

PENSTOCK INTERMENT:

Penstock interment should take place whenever possible. Pipe and

coating technologies have reached a very good reliability level, so that an

interred penstock requires practically no maintenance for decades and on the

other hand the result for the environment and especially for the landscape is

excellent. However to avoid problems connected with steel pipe corrosion, with

eddy currents in the ground and to reduce maintenance, the use of plastic pipes

(glass reinforced plastic or HDPE) is advisable.

UNCOVERED ANCHORING BLOCKS:

The impact of an exposed outdoor penstock can be further reduced

if the uncovered solution for anchoring blocks is adopted. That means that the

penstock is not covered with concrete at the anchoring blocks but is connected

to them by steel beams. This solution reduces the visual impact and allows for

the inspection of the whole pipe resulting in higher construction and operation

reliability.

Schematic diagram of water flow through the penstock.

PENSTOCKS WITHOUT EXPANSION JOINTS:

Where a penstock cannot be interred for some reason, construction

without expansion joints is preferable because it doesn’t require any

maintenance or any associated access tracks or roads to the penstock with the

consequent reduction of environmental impact.

The factors include but are not limited to the following:

a. Type of shutoff at the penstock intake:

A quick closing shutoff at the penstock intake, operable under

emergency conditions, may be an alternative to a shutoff at the powerhouse.

Where maintenance and emergency shutdown can be satisfied with the intake

shutoff, the requirement for powerhouse valves can seldom be justified.

b. Length of penstock:

A long section of penstock downstream of the shutoff will increase

the time required to shut the unit down during an emergency closure, increase

the time required to unwater the unit, and increase leakage losses. Maintenance

and emergency shutdown requirements will usually justify a powerhouse

shutoff when the penstock is several hundred feet long.

c. Head:

A shutoff valve near the unit will reduce the effective head on the

unit, which in turn will reduce the leakage.

d. Multiple units per penstock:

Operational and maintenance flexibility will normally require a

separate shutoff valve for each unit. Generally, maintenance requirements alone

will justify powerhouse shutoff valves for multiple unit penstocks.

e. Type of wicket gate seal:

A tight seal reduces leakage losses. However, deterioration of the

seal with time should be considered when determining the effects of leakage.

Evaluation of the factors should consider their effects on maintenance,

emergency operations, and costs. The factors considered and basis of

determination should be included in the mechanical design memorandum.

PENSTOCK SUPPORTS:

Ring Girders

Ring girders, which are used to support long span elevated

penstocks, are constructed by welding steel plate rings to penstocks. All loads

are transferred from the penstock to the ring girder and support legs. The

support legs are welded to the ring girder, then attached to bearing plates. The

bearing plates are attached to a concrete foundation.

Ring girders should be visually inspected for signs of deterioration

and distortion. Inspection of ring girders should also include the condition of the

.coatings. The potential for premature coating failure is greater at ring girders

than at adjacent smooth penstock surfaces because ring girder surfaces are

irregular.

Often, ring girder supports must allow penstock movement caused

by changes in temperature. This movement is usually accommodated in

bearings located under the support legs. -Rocker, roller, and low friction slide

bearings are commonly used for ring girder support. The bearings should be

inspected to verify theirintegrity. They should be clean and well maintained to

allow full penstock movement throughout the full range of design temperatures.

Clean, well maintained bearings will help minimize forces in the penstock and

anchorages.

Saddle Supports

Saddle supported penstocks typically span shorter distances

between supports than ring girder supported penstocks discussed above. Stress

concentrations occur at the tip of the saddle where "horn stresses" result in the

penstock shell becoming unsupported. Saddles are usually constructed from

reinforced concrete and support the lower 120-degree arc at the penstock invert.

However, saddle supports may also be fabricated from rolled steel plate. Sheet

packing that may be lubricated with graphite can be used as a cushion between

the saddle support and the penstock. The sheet packing also permits limited

movement of the penstock relative to the support as a result of temperature

changes.

If required, the penstock shell at saddle supports is stiffened by

welding steel rings to the shell at each side of the saddle support.

Saddle support inspection should include a coating inspection and

inspection for signs of deterioration and high stress areas similar to ring girders.

Localized buckling or distortion can occur at the penstock's upper contact points

with a saddle support.

In addition, the condition of the concrete saddles should be noted

and investigated for any signs of settlement or concrete deterioration. Inspection

of the surfaces between the saddle and the shell is difficult, but important,

because significant corrosion may be occurring in the contact area.

Anchor/Thrust Blocks

Anchor/thrust blocks are designed to provide restraint to exposed

penstocks at changes in alignment. They should be assessed to verify their

support function has not been compromised. Thrust blocks should be examined

for signs of settlement and movement and for any cracking or spalling of

concrete.

TYPES OF JOINTS:

Unrestrained Joints (Expansion

Joints and Bolted Sleeve-Type Couplings)

Unrestrained joints include

expansion joints and bolted sleeve-type

couplings. Typically, unrestrained joints are

not working if any water is leaking past the

seal or if the joint is seized. Look for leakage,

cracked welds, base metal flaws, loose or

missing bolts, and heavily corroded areas. In

unrestrained joints, scrape marks or unpainted

surfaces may be visible where the pipe has

moved in relation to the follower ring.

Restrained or Fixed Joints

Some basic types of restrained joints include lap welds, butt welds,

flanges, butt straps, and various rubber-gasketed joints. Several methods used to

attach these types of connections include rivets, forged welds, and arc-welds.

Corrosion, erosion, and flaws in the original construction can affect the

condition of structural welds, bolts, and rivets in the penstock.

i. Riveted Joints

In riveted joints, examine the

rivet head, butt strap, plate, and caulked edge

conditions (fig. 5). Look for leakage past the

rivets or the edges of the bands. Rivets may

be missing, broken, or may have corroded or

abraded heads. The base metal may also be

corroded to the extent that rivets can pull

through and be ineffective.

ii. Forge-Welded Joints

Experience with forge-welded joints has not been good. Flaws and

other fabrication defects, such as lack of fusion and slag, may be prevalent. The

welding process used in forge-welded penstocks, in which the steel is heated to

about 2000 EF, produces a loss of carbon, which makes the steel more

susceptible to corrosion. As a result, if forge-welded joints are not well

protected, corrosion may occur faster in the joints than in the base metal.

iii. Welded Joints

A representative portion of all

structural welding performed on the inside and

outside of the penstock is visually examined for

signs of rusting, pitting, or other structural

defects. For welded joints, look for cracked

base metal or welds, surface flaws, etc. Flaws

in welds during construction can occur from

high carbon content of the base material,

embrittlement of the heat affected zone,

improper preheat, and improper rate of cooling

after welding. Typically, these problems are

more likely to occur as the plate becomes

thicker or when the joint is made under adverse

construction conditions.

HYDRAULIC DESIGN OF PENSTOCK

Design discharge for penstock pipe ( 10% over design discharge )

= 59.60 cumec

.’. for single = 19.87 cumec

Full supply level at fore bay = 1020.00

Lowest elevation of penstock = 810.00

Length of penstock (approximately) = 300m

Dia of penstock D1 = 1.12 Q0.45 H-0.12

= 3.72m

Velocity through penstock = 5.5 m/Sec

Area of penstock = 10.87 m2

Head loss through penstock

HL = 0.34 x ( V1.9 / D1.1 )

x ( L / 1000)

Substitute in feet

= 0.34 ( 18.041.9 / 12.211.1 )

x ( 984 / 1000 )

= 5.2 feet

= 1.6m

WATER HAMMER PRESSURE

Ph = ( C V0 / g ) x { 1 / ( 1+ 0.5 φ) x [ 1 + As2 φ /( 4 Ao2 )

– Vo / C x ( 1/ ( 1 + 0.5 φ)) ]}

Where,

Vo = velocity of flow in tunnel corresponding to maximum study

flow, upstream of surge tank.

φ = Ratio of surge tank to that of conduit = As/At = 6.56

As = Net cross section or of surge tank

Ao = Area of the orifice ( or ) port = 20.42m2

Ph = ( 930.16 x 3 / 9.81) x [ 1/( 1 + 0.5 x 6.56 ) x { 1 + 133.942 x

6.56 / ( 4 x 20.422) – 3 / ( 930.16 ) x ( 1 / ( 1 + 0.5 x 6.56 ))}]

= 4.75MN/m2 in upstream side.

Consider C is a thin pipe the thick of which is small as compared to its

diameter.

C = [ ( g / w) / ( 1/ Ew ) + D / E.e] ½

Where

g = acceleration due to gravity ( 9.81 m2 / Sec).

w = Specific weight of water.

Ew = Young’s modulus of elasticity of water.

D = Dia of Head race tunnel.

E = Young’s modulus of elasticity of the conduit wall ( steel )

e = 0.92 ( constitutional joint efficiency )

Propagation of pressure wave:

From clause 5.5.3.6 in IS 7396 ( part – 1 ) 1985

Yo = 204.5

Interpolation:

C = 883 + ( 1000 – 883 ) / ( 300 – 140 )

x ( 204.5 – 140 )

C = 930.16m/Sec

TO FIND THE PRESSURE P:

The internal pressure P is due to Static head + Dynamic head

P = Static Head + Dynamic Head

= Ps + Ph

= γh + Ph

= 9.81 x 204.5 + 4.755x103

= 6.761 N/mm2

LONGITUTINAL STRESS DUE TO BEAM ACTION:

f = M/Z

Weight of water = 9.81 x 22/7 x 3.722/4

= 106.62 kN/m

.’. Maximum moment at center = Wl2/8

= 2.998x106 Nm

Add 15% Extra moment due to self weight = 0.45x106 Nm

.’. M = 3.5x106 Nm

Moment of inertia I = 22 x (3.7564 - 3.724 )/( 7 x 64 )

= 0.369 m4

Y = 1.86 m

Z = I/Y

= 0.1985 m3

.’. f = M/Z

= 17.63 N/mm2

.’. Assume allowable stress f1 = 17.63 N/mm2

CIRCUMFERENTIAL STRESS:

Circumferential Stress f1 = Pd/2t

.’. t = f1 x 2 / P x d

= 0.0014 mm

But we provide 18 mm thick steel pipe

Hence The Design Is Safe

LONGITUDINAL STRESS:

Longitudinal Stress = half of the hoop stress

= 8.815 N/mm2

CHECK:

Longitudinal Stress = 8.815 N/mm2

Circumferential Stress = f1 = 17.63 N/mm2

Max Shear Stress = q = 800 N

Sx = (f1 + f2)/2 + (f1 + f2)2/2 +q2

Sx = 813.24 N/mm2

Sy = - 786.8 N/mm2

Se = Sx2 + Sy2 + 2 Sx Sy

Se = 26.44 N/mm2

For exceptional condition

Se ≥ Maximum Longitudinal stress x 0.8

≥ 8.81 x 0.8

26.44 N/mm2 ≥ 7.048 N/mm2

Hence The Design In Safe In Stress

DIAMETER OF BRANCH PIPE

Y junction is required to be provided at the end of the Penstock for

more then one unit.

The present case having three units,

The area required for each unit = 3.62 m2

Dia of Pensock D2 = D1/20.4

= 2.82 m

Provide 2.82 m internal Dia Penstock of 18 mm thick having Outer Dia of

2.856 m.

Area provided = 6.24 m2 > 3.62 m2

Hence the Design Is Safe

LINER THICKNESS:

.’. Thickness of steel liner as per Clause 8.2 in IS11639 (part 1) – 1986

t = (D + 50)/ 800

= (372+50)/800

= 0.53cm

SUMMARY AND OUTLOOK:

Essential results of the study are:

• Evaluation and assessment of the existing small-scale hydropower plants.

• Evaluation of ecologically justifiable development potential in existing plants that is of interest with regard to energy production

Evaluation of ecologically justifiable development potential in undeveloped river stretches that is of interest with regard to energy production.

On account of the worldwide growing energy demand renewable

energy sources play an increasingly important role. Energy generated from

hydropower which, in contrast to wind energy, ensures base-load supply,

comprises an important component of the diversifications of energy resources.

The interdisciplinary study of existing energy resources permits ecologically

sustainable resources to be systematically developed in future in order to join

the worldwide efforts for environmentally friendly and sustainable construction.

New studies on the erection of new hydropower projects are

needed. Most of the power plant projects currently being designed and built are

based on more than 20-year-old studies. The technical, economic and ecological

boundary conditions have changed significantly when one takes into account the

globalization of the energy market. That is why it is necessary to elaborate

power plant studies which address these changed conditions in order to achieve

the best possible efficiency and effectiveness when implementing new

hydropower projects.