Chapter No 4

Chapter No. 4

RESERVIOR PLANNING INTRODUCTION

A reservoir is created with the impounding of runoff form the catchment U/S

by the construction of dam across a river or stream. Storage is done during the period

when the flow is in excess of demand and it is released during the lean supply period

so as to maintain a continuous hydel power generation.

SELECTION OF SITE FOR A RESERVOIR

There are various considerations in the selection of a site for a reservoir:-

1. good run-off from catchment area with minimum percolation losses.

2. run-off free from excessive silt.

3. water tight reservoir perimeter with no leakage from reservoir basin.

4. high and steep hills to ensure deep reservoir with maximum storage but

minimum surface area and hence minimum evaporation losses and lesser

possibility of weed growth.

DEFINITIONS

Effective Stroage or active capacity

It is storage between lowest level of release and highest controlled water surface.

Area Capacity Curve

It is graph between area of water spread and storage volume of reservoir.

Carry-Over Storage

It is storage collected during surplus years to meet demand in lean/dry years.

Coefficient of storage

It is a coefficient to express relation between storage capacity of reservoir and mean

annual inflow in the reservoir.

CLASSIFICATION OF RESERVOIRS

Storage reservoirs

They store surplus water during the period of excess flow and maintain a

continuous supply for irrigation, hydel power generation, municipal water supply and

industrial purposes during period of lean supply in rivers.

Flood control Reservoirs

These reservoirs store water during flood and release it gradually at a safe rate

when the flood reduces.

Detention Reservoir

A reservoir with gates and valves installation at the spillway and at the outlets

is known as detention reservoir.

Retarding Reservoir

A reservoir with fixed ungated outlets is known as retarding reservoir.

Distribution Reservoir

It is a small storage reservoir used for water supply in a city. It caters for the

fluctuations in water supply demand.

Multipurpose Reservoir

In this reservoir the storage and release cater for a combination of two or more

purposes such as irrigation, hydel power generation, flood control, water supply,

navigation, recreation, fisheries etc.

STORAGE ZONES OF A RESERVOIR

Dead Storage

The volume of water before the minimum reservoir level is known as dead

storage. It is used to cater for sediment deposition and is equal to the volume of

sediment expected to be deposited during the designed life of reservoir usually 100Y.

LIVE STORAGE

It is storage capacity of reservoir above dead storage level which constitutes

the usable portion of the total storage.

FLOOD STORAGE

Flood storage is the storage contained between maximum reservoir level and

the full reservoir level. The maximum level to which water rises during the worst

design flood is known as maximum reservoir level.

Bank Storage

It is the volume of water storage in the pervious formation of river banks and

becomes available in whole or in part when the water level drops down in the

reservoir

VALLEY STORAGE

Before the construction of a dam, certain amount of water is stored in the

stream called valley storage.

DESIGN CAPACITY OF RERVOIRS

The design capacity of a reservoir depends:-

1. Long-term precipitation record for the catchment

2. Long term run-off data at or near the site

3. Sediment yield into the reservoir from catchment

4. Area and capacity curves

5. Losses in the reservoir

6. Max requirement of water for different uses from the Reservoir

7. U/S use

8. Density current aspects & location of OUTLETS

MASS CURVE

It is a curve of the accumulated total flow or rainfall (inflow) to

reservoir against time. A mass inflow curve is a plot between accumulated inflow in

the reservoir with time.

DEMAND CURVE

A demand curve is a plot between accumulated demand with time.

The demand curve representing a uniform rate of demand is a straight line having

slope equal to the demand rate. A demand curve may also be indicating variable rate

of demand.

STORAGE CAPACITY DETERMINATION

The storage capacity of a reservoir to meet demand is determined from

discharge data of a river on which dam is to be built. Then for the driest years in as

long a period as is available, say 25 to 30 years, minimum 10 years is considered. The

storage capacity is determined by the following methods.

1. Analytical method

2. Mass Curve method

3. Bar Graph method

These methods are explained in the following example.

EXAMPLE

The yield of water from a catchment area during each successive month is

given below. Determine the minimum capacity of a reservoir required to allow the

above volume of water to be drawn off at a uniform rate assuming that there is no loss

of water over the spillway:-

1.4 2.1 2.8 8.4 11.9 11.9

7.7 2.8 2.52 2.24 1.96 1.68 (×106 m³ )

SOLUTION

1.By analytical method Total inflow =

(1.4+2.1+2.8+8.4+11.9+11.9+7.7+2.8+2.52+2.24+1.96+1.68 )×106 m³

= 57.4×106 m³

Monthly Demand = 57.4×106/12 = 4.78 × 106m³

Storage required = [(8.4+11.9+11.9+7.7)×106]

-[(4.78×106)×4 ] = 20.78 ×106 m³

2.BY MASS CURVE METHOD (i). The cumulative inflows are

(1.4,3.5,6.3,14.7,26.6,38.5,46.2,49.0,51.52,53.76,55.72 and 57.40)×106m³

Plot mass inflow curve as shown by dotted lines.

(ii).The monthly demand is

= 57.54 × 106/12 = 4.78×106m³

The demand curve is shown as solid line. It is a straight line as monthly

demand is uniform.

(iii).From the curve it is evident

(a) S2 is the lowest point on mass inflow curve (b) S3 is the highest point on the curve

(c) From 0 to S2 the slope of mass inflow is less than that of demand curve indicating inflow is less than outflow so reservoir is emptying

(d) From S2 to A slope of mass curve is more indicating that reservoir level is rising (e) At A reservoir level is same as that at 0, i.e. , beginning.

(f) ( S1 - S2) indicates the minimum storage required to cater for deficit during 0 – S2 (g) (S3 - S4) represent storage between initial water level , i. e. , at 0 and full reservoir

level.

( h ) (S1 - S2) plus (S3 - S4) represents the minimum storage capacity of the reservoir to meet demand

(i) From S3 upwards the slope of mass inflow curve is less than demand curve

indicating that inflow is less than the outflow and reservoir is again emptying.

The minimum storage capacity required to meet the demand

= ( S1 – S2 ) + ( S3 – S4 ) = ( 8.04+12.74 ) × 106m³

= 20.78 × 106m³

BY BAR GRAPH METHOD Bar graph (as shown below) indicates storage of 20.78 × 106m³ which is the area

between the bar graph and the average demand graph so that there is no loss of water

over spillway. Minimum storage to meet the demand is 20.70 × 106 as shown hatched.

CLASS WORK:

The flows from a certain stream in each successive month are:

MONTHS

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

FLOW (M m3)

3.0 3.6 6.0 19.6 25.2 25.2 21.6 9.9 7.8 7.2 6.6 6.3

Determine the maximum capacity of the reservoir if the above water is drained off at

a uniform rate and none is lost by flow over spillway. ANS: 44.27 M m3

ASSIGNMENT # 1 ( EXAMPLE 4.2 ) ( a ) The run-off data for a river along with the probable demands is given below. Can

the flow in the river cater to the demand? If so, how ?

( b ) what is the maximum uniform demand that can be met and what is the storage

capacity required to meet the demand?

Month J F M A M J J A S O N D

River Flow ( M.cum ) [ 106m³/sec ]

135+R, 23+R, 27+R, 21+R, 15+R, 40+R, 120+R, 185+R, 112+R, 87+R, 63+R, 42+R

Demand ( M.cum ) 60/+R 55/+R 80/+R 102/+R 100/+R 121/+R 38/+R 30/+R 25/+R

59/+R 85/+R 75/+R

SAFE YIELD

The maximum quantity of water that can be guaranteed during a critical dry

period is known as safe yield or firm yield

YIELD

It is the amount of water that can be supplied from the reservoir in a specified

interval of time.

SECONDARY YIELD

It is quantity of water available in excess of safe yield during periods of high

flood.

AVERAGE YIELD

It is the arithmetic average of the safe and the secondary yield over a long

period of time is called average yield.

DENSITY CURRENTS

The water stored in the reservoir is generally free from silt but inflow during

floods is muddy. There are thus two fluids having different densities resulting in the

formation of density currents: which may be defined as the gravitational flow of one

fluid under another fluid of slightly different density (flow stratification). The density

currents thus separates the turbid water from clearer water and make the turbid flow (a

little bit heavier) along the river bottom in the vicinity of dam. The rate of silting can be reduced by venting the density currents by

properly locating and operating the outlets and sluiceways.

TRAP EFFICIENCY Reservoir sedimentation is measured in terms of Trap Efficiency. Trap

efficiency is defined as the ratio of sediment deposited in the reservoir to the sediment

brought by the water into reservoir, i.e.,

Trap Efficiency (η) = Total Sediment deposited in Reservoir

Total Sediment inflow in Reservoir

In most of the reservoirs, η is 95 to 100% of the sediment load flowing into them.

Even if various silt control measures are adopted it has not been possible to reduce

this trap efficiency below 90% or so.

RESERVOIR SEDIMENTATION

All the rivers carry certain amount of silt eroded from the catchment area

during heavy rain. The extent of erosion and hence the silt load in the stream depends

upon:-

(i) Nature of soil of catchment area

(ii) Topography of catchment area

(iii) Vegetation cover

(iv) Intensity of rainfall

The sediment transported by rivers can be divided into two heads:-

(a) Bed load

(b) Suspended load

The bed load is dragged along the bed of rivers. The suspended load is kept in

suspension because of vertical component of eddies formed due to friction of flowing

water against bed. Bed load is normally 10–15% of suspended load. When stream

approaches the reservoir, the velocity is very much reduced. Thus coarser particles

settle in the head reaches of the reservoir while the finer particles are kept in

suspension for sufficient time till they settle just to the U/S side of the dam. Some fine

particles may pass through sluiceways, turbines or spillways.

Capacity-Inflow Ratio

It is the ratio of reservoir capacity to the total inflow of water annually. Trap

efficiency is the function of capacity-inflow ratio

η= f (Capacity / Inflow)

Graphically, it is shown as follows.

EXAMPLE The following information is available regarding relationship between

trap efficiency and capacity-inflow ratio.

Capacity

Inflow

Ratio

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

Trap %

Efficiency

87 93 95 95.5 96 96.5 97 97 97 97.5

Find the available life of reservoir with an inatial reservoir capacity of 30

Million m³, if the average annual flood inflow is 60 Million m³ and the average annual

sediment inflow is 2,00,000 tons. Assume the specific weight of sediment is equal to

1.2gm per c.c. The usual life of the reservoir will terminate when 80% of the initial

capacity is filled with sediment.

SOLUTION

Average annual sediment inflow = 2,00,000 tons

= 2 × 105 tons = 2 × 108 kg = 2 × 1011gm

Volume of average annual sediment inflow = Weight / Specific weight

Volume of average annual sediment inflow = [2 ×1011 / 1.2] (cm)3

= 2 × 1011 m³ = 1/6×106 = 1/6 Million m³

1.2×106

Initial Reservoir Capacity = 30 Million m³

Annual Flood inflow = 60 Million m³

Let us assume 20% of the capacity, i.e., 6 Million m³ is filled in the first interval.

Capacity-inflow ratio at the start of interval = 30 / 60 = 0.5

Trap Efficiency at start of interval (From table)= 0.96

Capacity- inflow ratio at the end of interval =24/60 = 0.4

Trap Efficiency at the end of interval (From table)= 0.955

Average Trap efficiency during this interval

= 0.96+0.955 = 0.9575

2

Volume of sediment deposited annually till the 20% capacity is filled

=1/6 × 0.9575 Million m³/year

No of years during which 20% of capacity i.e. 6 Million m³ shall be filled

= 6 years = 37.6 years

1/6×0.9575

Calculations dividing the entire reservoir capacity into 20% intervals are

tabulated as follows:

Interval Cap-

Inflow

ratio(at

start of

interval)

Cap-

Inflow

ratio(at

end of

interval)

Trap Ƞ(at start of

interval)

Trap Ƞ(at end of

interval)

No of Years

during 20%

capacity is

filled

1st 0.5 0.4 0.96 0.955 37.6

2nd 0.4 0.3 0.955 0.95 38.8

3rd 0.3 0.2 0.95 0.93 38.3

4th 0.2 0.1 0.93 0.87 40

Total 153.7 years

ANSWER: Total probable life of reservoir till 80% capacity gets filled up is

equal to 153.7 years

Class Work A proposed reservoir has a capacity of 500 ha-m (1 ha =

10,000m²). The catchment area is 125km² and the annual streamflow averages 12cm

of run-off. If the annual sediment production is 0.13 ha-m/km², what is the probable

life of the reservoir before its capacity is reduced by sedimentation to 20% only ? The

relationship between trap efficiency η(%) and capacity inflow ratio (C/I) is as under:-

C/I 0.01 0.02 0.04 0.06 0.08 0.1 0.2 0.3 0.5 0.7

η(%) 48 59 72 78 83 86 92 94 95 96

ASSIGNMENT # 2 Ex. -1 Work out the life of a reservoir before its capacity is reduced to 20% of the

initial capacity from the following sedimentation data. Catchment area is 1,000 sq.

km, reservoir capacity is 10,000 ha m, average annual flow is 15 cm of runoff,

average annual sediment inflow is 0.26 million tons.

Ex. -2 Calculate the life of a reservoir from the following date:

Dead storage capacity = 1.0 × 109 m³

Average Annual suspended load in dead storage = 0.015 × 109m³

Assume coarse silt 1%, medium silt 17% fine silt 82%

RESERVOIR SEDIMENT CONTROL

Following are some of methods used for the control of silting of reservoirs.

(1) Proper selection of reservoir site

A stream collecting water from a catchment area having loose or soft soil and having

steep slopes may carry more silt load. If a certain tributary of the main stream carries

more silt, the dam should be constructed to the U/S of that tributary.

(2) Control of sediment Inflow

Small check dams may be constructed across those tributaries which

carry more silt. Increase of vegetal cover over the catchment area also decreases the

soil erosion and hence sediment inflow is reduced.

(3) Proper designing and reservoir planning

A small reservoir on a big river has lesser trap efficiency. Hence if a dam is

constructed lower in the first instance and is being raised in stages the life of the

reservoir will be very much increased.

During the floods, the sediment carried by the stream is the maximum.

Hence sufficient outlets should be provided in the dam at various elevations so that

the floods can be discharged to the d/s without much silt deposit.

(4) Construction of Undersluices in Dam

The dam is provided with openings in its base so as to remove the more silted

water on the d/s side. Sluices are located at level of higher sediment concentration.

Sometimes water dig channels behind sluices leaving most of sediment undisturbed

so mechanical loosening and scouring of sediment is required simultaneously. But

providing large sluices near bottom of dam is again a structural problem so use of this

method is limited.

(5) Removal of Post Flood Water

The sediment content increases just after floods therefore attempts are

generally made not to collect this water. Hence the provision should be made to

remove the water entering the reservoir at this time.

(6) Mechanical Stirring of Sediment

The deposited sediment is scoured and disturbed by mechanical means so as to

keep it in a moving state and thus help in pushing it towards the sluices.

(8) Erosion control and Soil conservation

This includes all those methods to reduce soil erosion to make it more and

more stable. This is because when soil erosion is reduced, the sedimentation is

reduced automatically. But the methods of treating the catchment in order to minimize

erosion are very costly. The methods of soil conservation are provision of control

bunds, checking gully formation by providing small embankments, afforestation,

regrassing and control of grazing etc. Provision of vegetation screen helps in reducing

the sheet erosion.

RESERVOIR LOSSES

The important reservoir losses are:-

EVAPORATION LOSSES

These losses mainly depend upon the reservoir surface area. The other factors

influencing these losses are temperature, wind velocity, relative humidity. Standard

pan evaporation can be measured and when multiplied by pan coefficient gives the

reservoir evaporation losses.

Pan evaporimeter is an instrument for measuring amount of water lost by evaporation

per unit area in a given interval from a shallow container. The pan evaporation values

are used to interpret the evaporation losses from the water surface of reservoirs.

These losses are expressed in cm of water depth and may very from month to

month.

ABSORPTION LOSSES

These losses do not play any significant role in planning since their amount,

though sometimes large in the beginning, falls considerable as the pores get saturated.

They certainly depend upon the type of soil forming the reservoir.

PERCOLATION LOSSES OR RESERVOIR LEAKAGE

For most of the reservoirs, the banks are permeable but permeability is so low

that the leakage is of no importance. But if the walls of the reservoir are made of

badly fractured rocks or continuous seams of porous strata, serious leakage may

occur. Sometimes pressure grouting may have to be used to seal the fractured rocks.

The cost of grouting to be included in the economic studies of the project if leakage is

large.

Ex.:4.3 Evaporation loss has been measured by a pan evaporimate (Pan coefficient

= 0.7) placed near a big reservoir area 5.2 sq-km. After a year the surface area of

reservoir is reduced to 2.2 sq-km. Determine the annual quantity of water lost by

evaporation when the pan evaporation losses (cm) over the year are known.

Months

]=[\Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec

11.5 11.0 13.2 12.8 13.2 16.2 14.1 11.2 10.2 12.0 12.4 11.0

Solution

Average surface area of

reservoir = 1/3 (A1+A2+√A1A2)

= 1/3(5.2+2.2+√5.2×2.2)

= 3.59 km² = 3.59 × 106m²

Annual Evaporation loss =

(11.5+11.0+13.2+12.8+13.2+16.2+14.1+11.2+10.2+12.0+12.4+11.6) =

149.4 cm = 1.494 m

Annual water loss by evaporation from

reservoir = 3.59 × 106 × 1.494 × 0.7

= 3.75 × 106m³

ASSIGNMENT # 3

An evaporation pan 1.5m in diameter was used to find out evaporation from a

reservoir. The pan was initially filled up with water to a depth of 10 cm During the

period of observation a rainfall of 5cm was recorded. To keep the water level same in

the pan 2.5 cm water depth had to be removed. At the end of the period of

observation, the depth of water in the pan was found to be 8.5 cm. If the pan

coefficient is 0.75, estimate the evaporation from the surface of reservoir.

FLOOD ROUTING (RESERVOIR ROUTING) The hydrograph of a flood entering a reservoir, will change in shape as it

emerges out of the reservoir, because certain volume of water is stored in the reservoir

temporarily and is let off as the flood subsides. The base of the hydrograph, therefore,

gets broadened, its peak gets reduced and, of course, the time is delayed. The extent

by which the inflow hydrograph gets changed due to reservoir storage can be

computed by a process know as flood routing.

Since the flood protection reservoir are generally located many kilometers

upstream of the cities which are to be saved against floods, it is sometimes necessary

to route the outflow hydrograph of the reservoir up to these downstream localities.

The reservoir outflow graph may then be routed through this much length of river

channel, so as to obtain final shape of the hydrograph at the affected cities. This

routing in which the stream itself acts like an elongated reservoir is known as flood

routing through river channels.

The relationship governing this computation is very simple. Over any interval

of time, the volume of inflow must be equal to the volume of outflow plus change in

storage during this interval, i.e., I = O ± ΔS

(increase in storage is +ve and decrease is -ve)

The relations between time and inflow rate, elevation and storage,

elevation and outflow rate cannot be expressed by simple algebraic equations. They

are respectively represented by inflow flood hydrograph, elevation-storage curve and

outflow-elevation curve.

If order to solve this problem, a step by step procedure by choosing a

sufficiently small interval of time may be adopted such that the conditions at the ends

of each interval are successively determined. Step by step computation can be done

either graphically or by trial and method.

Trial and error method is normally adopted as follows:-

TRIAL & ERROR METHOD FOR RESERVOIR ROUTING

Result Desired To obtain the outflow hydrograph and to estimate the maximum water level to

be attained in the reservoir.

Data Given (i) The inflow hydrograph

(ii) Elevation-storage curve(or elevation-area curve)

(iii) Elevation-outflow curve

Procedure (i) Divide the inflow hydrograph into a number of small intervals. The time

interval should be so chosen as not to miss the peak values.

(ii) It is the assumed that the worst design flood enters the reservoir only after

the reservoir is filled upto crest level, i.e., normal reservoir level.

(iii) Work out spillway and discharge rating curve.

(iv) Work out elevation-storage curve for reservoir from elevation-area

curve using cone formula

V = Σh/3[A1+A2+√A1A2]

where h = contour interval

(v) Find out total inflow

I =1/2 (I1+I2 ) × t → (1)

where I1 = Inflow at the start of interval

I2 = Inflow at end of interval

t = Duration of interval

(vi) The reservoir level at the start of the flood (i.e. start of first interval) is known.

Assume a trial value for the reservoir level at the end of interval.

(vii) Compute the total outflow

O = Q1+Q2 ×t → (2)

2

(viii) Using elevation-storage curve, determine the storage S1 at the start and

S2 end of interval corresponding to known and assumed reservoir levels respectively.

Now the amount of flood stored is

ΔS = S2 – S1 → (3)

(ix) Now check if I = O + ΔS then assumed reservoir level is correct

otherwise change it and repeat the procedure till this coincidence is obtained.

(x) All the above steps should be repeated for other time intervals till the

entire flood is routed or till the reservoir level returns to pre-flood pool level.

(xi) Outflow ordinates are plotted so as to obtain the outflow hydrograph.

The point at which it crosses the inflow hydrograph gives the peak outflow rate. From

this time, the rate of outflow begins to fall due to decrease in the inflow rate.

(xii) The time lag between the two peaks is evaluated

EXAMPLE

The inflow flood discharge for a possible worst flood are tabulated as under,

at suitable intervals starting from 0.00 hours on August 20 , 1975.

TABLE 1 Time

in Hrs

0 6 12 18 24 30 36 42 48 51 60 66 78 90 102 114

Inflow

In M

m3

0 50 280 610 1290 1900 2130 1900 1600 1440 1060 780 50 370 220 130

This flood approaches a reservoir with an uncontrolled spillway, the crest of which is

kept at RL 140m. Determine the maximum reservoir level and the hydrograph of the

routed flood. Values of storage and outflow at various elevation above crest are

tabutated in tables 2 and 3 respectively.

TABLE 2 Elevation

in m (y)

140.0 141.0 142.0 143.0 144.0 145.0 146.0

Storage

(above

crest) in

Million

m³ (x)

0.0 15.0 35.0 60.0 95.0 140.0 240.0

TABLE 3 ELEVATION (y) OUTFLOW DISCHARGE

IN CUMECS ( χ )

140.0

141.0

142.0

143.0

144.0

145.0

146.0

0

170

482

883

1360

1905

2500

The elevation-storage curve and elevation-outflow curve are plotted from tables 2 and

3 as shown in Fig (A) and Fig (B)

Fig (A) ELEVATION STORAGE CURVE

Fig (B) ELEVATION OUTFLOW CURVE

The hydrograph of the given flood is plotted in Fig

(C).

Flood routing is carried out by hit and trial method as shown in Table (4).

The outflow hydrograph is plotted from col (7) of Table (4) as shown in Fig (C)

by dotted curve the point at which intersect the inflow hydrograph represent the peak

of routed flood.

RESULTS Peak rate of outflow = 1458 Cumecs

Maximum Reservoir level = 144.18m

Time lag = 15 HRS.

ASSIGNMENT # 04

EXAMPLE NO. 1d The hydrograph of inflow to a reservoir is given as under

Time

(days)

0 2 4 6 8 10 12 14 16 18 20

Flow

(m³/s)

68 115 425 550 440 320 260 206 156 116 68

The reservoir is full at the start of the flood inflow. The storage S is reservoir above

the spillway crest is given in million m³ by: S = 8.64 h Where h is the head in meters

above the crest. The discharge over the spillway Q = 60h. find head over spill having

crest at the end of 8th day of Flood.

EXAMPLE NO .2 A small reservoir has an area of 4000 hectarcs [(40×10)m²] at spillway crest

level. Banks are essentially vertical above the spillway is 50m long and has a

coefficient C of 2.2 in equation q = CH. The inflow to the reservoir is given below:-

Time

from

start

(hrs)

0 6 12 24 30 36 42 48 54

Inflow

(m³/s)

46 346 722 326 192 118 80 56 46

Compute the maximum outflow discharge over the spillway and reservoir level to be

expected if the reservoir level was at the spillway crest at the start.

EXAMPLE NO. 3 The following data (Table 1) pertains to an inflow flood hydrograph whose

flows in 100 cumecs have been recorded at 6 hour interval starting from 0.00 hrs on

June 1, 1959 on a certain stream.

Table # 1 0.47 0.50 0.62 0.93 1.52 2.60

2.72 3.40 3.50 3.38 3.14 2.88

2.63 2.40 1.98 1.70 1.43 1.20

This flood approaches a reservoir with uncontrolled spillway, with elevation

area and elevation-outflow data as shown below:-

Elevation

(m)

100 100.3 100.6 100.9 101.2 101.5 101.8 102.1 102.4 102.7

Area 403 410 418 424 428 436 445 453 460 469

(ha)

Outflow

(m³/s)

0 15.9 42.5 77.5 119 168 216 271 335 403

The water level just reaches the crest level (Elevation 100m) of spillway at 4

hrs on june 1, 1959. Determine the maximum reservoir level and maximum discharge

over the spillway. Draw inflow and routed hydrograph indicating the reduction in

peak flow and peak lag introduced due to routing.

ENVIRONMENTAL EFFECTS OF RESERVOIR

The creation of reservoirs results in far-reaching changes in ecosystem (living

organisms and their surroundings). Sedimentation, soil erosion, stratification, adverse

effect on fish are some major disruptions in the ecosystem which may involve

economic loss. It also affects forests, wild life, ground water, climate and agriculture.

Human environmet is affected in respect of alteration of human settlement,

occupational patterns and water born diseases. An increasing constraining factor in

hydropower development has been the conflict between the expanding electric energy

demands and the concern for environmental effects of reservoirs. Compared to high

head dams, low head hydropower projects are less destructive in equatic life.

However long term use of reservoir has clearly proved the stable ecosystem should

match to the natural environment. The environmental problems arising from water

reservoirs and remedial measures are given below:-

FISH Free passage of migratory fish to and from is disturbed due to dam. However, creation

of reservoirs provide favorable environment to several fish species. The river

discharge during dry season is higher fovouring the development of fisheries in d/s

river reach.

FLORA & FAUNA (plants, birds etc.) Deforestation of the reservoir submerged area and consequent displacement of wild

life population is inevitable. Area submerged in a reservoir constitutes an insignificant

percentage of catchment area upstream of the dam.

However to enhance flora & fauna it is essential to develop more

parks, forests etc.

DISEASES Reservoirs contribute to the spread of water born human diseases.

Planning for reservoir should include consideration of the water quality standards.

RESERVOIR INDUCED SEISMICITY

Reservoirs are considered to trigger earth tremors. While there is no complete

agreement on the exact cause of this seismic activity, a general belief is that

earthquake are caused by the filling of reservoirs at site where natural stresses in the

underlying rock mass have developed to a state very close to rupture. An important

measure is early installation of seismogrophs to provide information on the seismic

potential of the region before construction behaviour.

AQUATIC NUISANCE PLANTS

The growth of nuisance aquatic vegetation is associated with reservoirs.

Concern, however, is limited to those equatic plants which are larger than microscopic

algae. Early planning and action can avoid some of the hazards posed by equatic

plants to public health, fisheries and navigation.

SILT The river water deposits silt into the reservoir which in turn triggers a number

of environmental problems. Silting may be reduced by placing outlets in the dam at

such points that allow some of the silt to escape.

Thus, in future, the multi-purpose irrigation and flood control projects

should take into account, right at the planning stage, appropriate steps for the

protection, preservation and development.

With suitable environmental protection measures, hydropower plants

are considered harmless form environmental view point as compared to other energy

sources.