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Reservoir planning
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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.