Reservoir Intro

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Physical Characteristics of Reservoirs The most important physical characteristic of reservoirs is the Storage Capacity. For reservoirs on natural sites, this may be estimated from topographic surveys. First, the area-elevation curve may be constructed by measuring the area within each contour elevation within the reservoir site. Next, the area-elevation curve may be integrated to obtain the elevation-storage capacity curve. The incremental storage between two elevations can be computed by using the prismoidal formula or by multiplying the average of the areas at the two elevations by the difference in elevation between them. The summation of these increments below any elevation is equal to the volume stored upto that elevation. If the topographic maps are not available, then, a survey is to be undertaken and the vertical cross-sections are to be obtained and the storage capacity is to be computed using the prismoidal formula. A typical set of Elevationstorage and elevation-area curves is shown in Fig.1.

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Fig.1 Typical storage area curves .   The surface area at each elevation is important from the evaporation loss computation during different months or periods, especially in semi-arid and arid regions. Computing the storage upto each elevation assumes importance in: reservoir capacity decisions flood routing computations allocation and reallocation of storage pools in multipurpose reservoirs. reservoir operation studies

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Zones of Storage: Figure 2 shows the different zones of storage in a reservoir. These are: dead storage, useful storage, surcharge storage and valley storge. In order to designate the classification of these storages, it is necessary to define a few pool elevations, namely: Normal Pool Level: The maximum elevation (pool level) to which the reservoir water level will rise during normal operating conditions. Mostly, this level will correspond to the spillway crest level or the level of the top of the spillway gates.Minimum Pool Level Minimum Pool Level: Lowest elevation to which the pool may be drawn under normal conditions. Tis elevation will correspond to the level of the lowest outlet in the dam. In case of hydroelectric systems, this may correspond to the lower limit of the operating efficiency of the turbines. Pool Level during floods, will be above the normal pool level and the excess flow will be discharged over the spillway. This may go as unutilised at the reservoir site. For high flows in the river, the water surface cannot be assumed to be horizontal (especially in shallow and narrow reservoirs) and the wedge shaped element of storage above a horizontal will have to be considered. The water surface profile computation is to be done using non-uniform flow, file:///D|/dr.srini old system backup/d drive backup/ce303/CE303_BTech_WRE/Water_Resources/Reservoirs/zones.htm[11/12/2010 8:02:26 PM]

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in such a case.  

Fig.2 Zones of storage in a reservoir.   Dead Storage: Water stored below the minimum pool level is referred as "Dead Storage". Except in extreme drought conditions, this storage will not be drawn upon. The sediment brought by the river/stream is trapped in this zone. Useful Storage: The storage volume between the minimum and the normal pool levels, is termed as "Useful Storage". This can be further subdivided into conservation storage and flood control storage, in case of multi-purpose dams. Surcharge Storage: The wedge storage above the spillway crest (normal pool level ) is file:///D|/dr.srini old system backup/d drive backup/ce303/CE303_BTech_WRE/Water_Resources/Reservoirs/zones.htm[11/12/2010 8:02:26 PM]

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termed as "Surcharge Storage". This is available only during flood times and not after. Bank Storage: During the filling of the reservoir, some water enters the soil in the reservoir banks and drains out as the level is lowered. This storage in the soils of the reservoir banks is known as "Bank Storage". This storage increases the capacity of the reservoir above that given by the storage-elevation curve. Depending on the geologic conditions, this amount may be significant. Valley Storage: This is a variable volume (with respect to time) stored in the channel in the natural state. At a given time instant, if the inflow and outflow volumes are known in a reach of the channel considered, the valley (channel) storage can be computed. Over small intervals of time, average values of valley storage can be found out. The net increase in storage due to the construction of a reservoir will be: (reservoir capacity - valley storage). This has special significance in case of flood control reservoirs.   Computation of water surface profile is an important part of the reservoir design. The variation in water surface elevation along the length of the reservoir has a special significance with regard to the acquisition of land and the flowage rights, relocation of the existing developments, location of docks, residential areas, storm drainage outlets, roads, bridges and other infrastructural developments along the banks of the reservoir. Many times, a great file:///D|/dr.srini old system backup/d drive backup/ce303/CE303_BTech_WRE/Water_Resources/Reservoirs/zones.htm[11/12/2010 8:02:26 PM]

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deal of forest and fertile agricultural land may also get submerged by the construction of the reservoir. Hence, the information on water surface profiles for various inflow rates and water surface elevations at the dam, will be useful in deciding the level of the top of the dam, spillway crest and other important levels.  

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Reservoir Yield: Yield is the amount of water that the reservoir can supply during a specified interval of time when it is put to operation for a particular purpose. The interval of time varies from a day in case of small distribution reservoirs for municipal water supply to a year or more in case of large storage reservoirs. The relation between storage capacity and yield is very vital in the storage reservoir design. Safe Yield (or Firm yield) refers to the maximum quantity of water that can be supplied even during the most critical dry period. However, the critical period that has been experienced in history may be surpassed in the future. That is, a more dry period may be experienced in future that would correspond to a lower safe yield than the one decide based on the historical inflow record. Hence, the safe yield cannot be determined with certainty. It is to be treated as a probabilistic variable. The theoretical maximum yield during a specified period is: (Mean inflow during that period Losses during that period). Since there is an associated risk with regard to a specified target yield, it would be more appropriate to consider the reliability of realizing that yield with the given storage capacity. Hence, it is more meaningful to consider the storage - reliability - yield relations than just the storage - yield relation. It is preferable to fix the target yield to be low for municipal water supply systems, since the risk of not realizing the same is to be minimal. The risk can be only of the order of 1 - 5% in such systems. On the other hand, for irrigation systems, the admissible risk can be of the order file:///D|/dr.srini old system backup/d drive backup/ce303/CE303_BTech_WRE/Water_Resources/Reservoirs/reservoir yield.htm[11/12/2010 8:03:01 PM]

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of 20%. In case of hydropower production, a certain amount of energy is to be supplied to the consumers on a firm basis and hence that quantity will be somewhat low. However, during periods of high flows, it may be possible to produce some surplus energy that can be sold at a lower price, as and when available. This is known as the secondary energy.

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Estimating the capacity of a distribution reservoir for a city water supply system The required yield is based on an estimate of the maximum daily demand by the consumers. The inflow rate is fixed by the decision to pump at a uniform/variable rate. The reservoir capacity must be sufficient to supply the demand at times when the demand exceeds the pumping rate.  

Fig.3 Graphical illustration of the computation of required reservoir capacity. An example is presented below:   Example: The water supply for a city is pumped from wells to a distribution reservoir. The estimated hourly water requirements for a maximum day is given in the 2nd column of the table below. If file:///D|/dr.srini old system backup/d drive backup/ce303/CE303_BTech_WRE/Water_Resources/Reservoirs/distributionreservoir.htm[11/12/2010 8:03:40 PM]

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the pumps are to operate at a uniform rate, what distribution reservoir capacity is required? Hour Ending

Demand, m3/h

Pumping rate,m3/h

0100 0200 0300 0400 0500 0600 0700 0800 0900 1000 1100 1200 1300 1400 1500 1600 1700 1800 1900

200 250 250 300 400 450 500 600 700 700 800 750 600 650 700 800 650 500 450

500 500 500 500 500 500 500 500 500 500 500 500 500 500 500 500 500 500 500

Required from reservoir, m3/h 0 0 0 0 0 0 0 100 200 200 300 250 100 150 200 300 150 0 0

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2000 2100 2200 2300 2400 Total

400 450 400 300 300 12,000

500 500 500 500 500 12,000

0 0 0 0 0 1,950

Average Pumping rate( over the day) = 12,000 / 24 = 500m3/h Total Quantity of storage required =

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Selection of Capacity for a River Reservoir This is done by an operation study that involves simulating the reservoir operation for a period (critical period from historical record) using some rules for the operation (based on practical considerations). This can also be performed using synthetic information on inflows generated. A mass curve or a sequent peak algorithm can be used for this purpose. The sequent peak method is explained below..  

Fig.4 Illustration of sequent-peak algorithm. Reservoir Storage capacity Sequent Peak Algorithm

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If there is any evaporation / seepage loss, the same may be included in the demand A double cycling over the period of record may be necessary in case the end of record shows a possibly continuing critical period ( if further record is taken) Refer example1  

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Example Given that the inflows during the most critical period of record are as follows. t

1

2

3

4

5

6

7

8

9

10

11

12

Dt (106m 3)

50

50

50

50

50

75

75

75

75

80

80

80

Qt (106m 3

30

40

50

70

80

65

70

55

90

75

60

135

Estimate the storage capacity required using sequent peak algorithm. Indicate emptying, filling & spilling periods. If the demands in each period will increase by 30% in future, what will be the reservoir capacity required? Solution: 50 50 50 50 50 75 75 75 75 80 80 80

30 40 50 70 80 65 70 55 90 75 60 135

20 30 30 10 0 10 15 35 20 25 45 0 Next Eg

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Example 2: Estimate the specific weight (dry) of deposited sediment that is always submerged. The sediment is 25% sand, 35% silt and 40% clay by weight. Find out how the specific weight of the deposited sediment varies with time (for 1, 2, 5, 10, 20, 50 years). Also find the volume occupied by 500 5 of first- year deposit. How does the situation change if the reservoir operation corresponds to a moderate reservoir drawdown.   Solution:

For the sediment deposit that is always submerged, W1= 93 ; W2= 65 ; W3= 30 B 1= 0 ; B 2= 5.7 ; B 3= 16 i.e W = 0.25(93 + 0) + 0.35(65 + 5.7 logT) + 0.40(30 + 16logT) Using the above equation, Years For T = 1 year For T = 2 years

W(pcf)

W( N/m 3)

58

9111.8

60.53

9508.8

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For T = 5 years For T = 10 years For T = 20 years For T = 50 years

63.87

10,033.6

66.395

10,430.7

68.922

10,827.7

72.263

11,352.5

Therefore , volume occupied by 500 t of first-year deposit =

If the reservoir operation corresponds to a moderate reservoir drawdown, then W1= 93 ; W2= 74 ; W3= 46 B 1= 0 ; B 2= 2.7 ; B 3= 10.7 Repeat the same exercise for this condition also. Previous Eg Next Eg             file:///D|/dr.srini old system backup/d drive backup/ce303/CE303_BTech_WRE/Water_Resources/Reservoirs/workedex5.htm[11/12/2010 8:05:25 PM]

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Example 3: A proposed reservoir has a capacity of 500ha-m. The catchment area is 125km 2, and the annual streamflow averages 12 cm of runoff. If the annual sediment production is 0.03 ha-m/km 2, what is the probable life of the reservoir before its capacity is reduced by 10% of its initial capacity by sedimentation? The relationship between trap efficiency (%) 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

(%)

43

60

74

80

84

87

93

95

96

97

Solution: Av. annual stream flow = 12cm of runoff Area of catchment = 125km 2                            = 125 * 106m 2 Therefore, Annual Flood inflow = ( 125 * 106 ). 12 / 100 m 3                                                   = 15 * 106  m 3                                                           = 15 M.m3 (Mcum) Annual sediment inflow= 0.03 ha-m / km2 of catchment                                   = 0.03 * 125 ha-m                                   = 0.03 * 125 * 104  m 3                                                                            = 3.75 *  104  m 3                                     = (3.75 / 100 )   *   106 m 3                                 = 0.0375 M-m 3 (Mcum) Reservoir capacity = 500 ha-m 4

3

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                           = 500 * 10 m                                = 5 * 106m 3                                = 5 M. m 3   (Mcum) It means that 0.0375 Mcum of sediment flows every year into the dam/ reservoir site, but tyhe quantum of this, which is trapped in the reservoir , depends on the average trap efficiency ( ) during that year, and this trap efficiency, inturn, depends upon capacity / inflow ratio. In the question, the total capacity to be filled up by sediment is 10% of the initial reservoir capacity. i.e 10% * 5 Mcum = 0.5 Mcum Now, we have to calculate the time during which this 0.5 Mcum of sediment will get deposited in the reservoir, as follows : Capacity of reservoir at the start = 5 Mcum Capacity of reservoir at the end i.e when 0.5 Mcum of sediment is filled up = 4.5 Mcum Therefore, Capacity / inflow at start = 5 Mcum / 15 Mcum = 0.333   at start = 95% Capacity / inflow at end = 4.5 / 15 = 0.30 at the end of interval = 95% Average = 95% Therefore , sediment load trapped / yr = 0.0375 * 95% = 0.035625 Number of years during which 0.5 Mcum of sediment will get trapped =

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  Hence after 14.04 years , the 10% reservoir capacity will get filled up. Previous Eg Next Eg

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Worked Example 4: Following information is available regarding the relationship between trap efficiency and capacity-inflow ratio Capacity inflowratio Trap Efficiency (%)

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

86

93

94.5

95.5

96

96.5

97

97

97

97.5

Given that the mean annual flood inflow is 100 Mm 3 , the mean annual sediment inflow is 3,00,000 tonnes, and the initial reservoir capacity is 50Mm 3 . Find out the useful life of the reservoir assuming the specific weight of the sediment to be 12 KN / m 3 throughout the life. Also find the useful life of the reservoir , if the specific weight of thesediment is assumed to vary with age.Assume that the useful lfe of the reservoir will end when 80% of the initial capacity is filled with sediment. Solution: Volume of the mean annual sediment inflow = For filling the first 20% of the storage capacity (10Mm3) let us find out the number of years. This means , the volume of the sediment deposited annually till that 20% of the capacity is filled, is to be found. From this, the number of years required to fill that 20% of the capacity (10Mm3) can be found out. Initial reservoir capacity = 50Mm 3 Annual flood inflow = 100Mm3 Capacity to inflow ratio at the beginning of the 1st time interval = 50 / 100 = 0.5 Trap efficiency (beginning of the 1st time interval) = 96% Capacity -inflow ratio at the end of 1st time interval = 40 / 100 = 0.4 Trap efficiency (end of the 1st time interval) = 95.5% Average Trap efficiency during the first time interval = (96+95.5) / 2 = 95.75% file:///D|/dr.srini old system backup/d drive backup/ce303/CE303_BTech_WRE/Water_Resources/Reservoirs/workedex7.htm[11/12/2010 8:09:01 PM]

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Volume of sediment deposited annually till the 20% capacity is filled = 0.245 * 0.9575 = 0.2346Mm3 No. of years during which the first 20% capacity(10Mm 3) will be filled up =

Proceeding on the same lines, the number of years required to fill the next 20% of the capacity(2nd interval) can be found. The calculations are presented in the following table:     Previous

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Reliability Concept: Traditionally, water resources engineers involved in reservoir planning were working with the "safe yield" concept. In other words, for a selected reservoir capacity, a particular yield can be satisfied, for the most critical inflow sequence that has occurred in the past. The logic behind this concept of "safe yield" is acceptable, as far as the future flows are not going to be more critical than the most critical flows of the past. But, since the streamflows are uncertain, we cannot accept this concept. This means that for a selected reservoir storage capacity, there is a probability corresponding to each yield for a "no failure" situation. This probability is known as the "Reliability" for the given storage and the yield. Thus, it is useful to construct or develop StorageReliability-Yield relationships than just the Storage-Yield relationship as done traditionally. One can easily sense that low yields can be supplied with a higher reliability by the same reservoir storage capacity, compared to higher yields. Hence, when firm yields are to be supplied as in case of city water supply systems, the yields cannot be high. If higher yields are to be satisfied on a firm basis, then, the storage capacities required will be enormous, which may be prohibitive from economic and other considerations. However, the potential yield is limited by the "mean annual flow" at that site. For city water supply systems, the reliability may be of the order of 0.950.99. On the other hand, in irrigation systems, a lower reliability such as 0.80-0.85 may be acceptable.   file:///D|/dr.srini old system backup/d drive backup/ce303/CE303_BTech_WRE/Water_Resources/Reservoirs/reliabilityconcept.htm[11/12/2010 8:09:58 PM]

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Fig.8 A reservoir reliability curve.        

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Sediment Transport by Streams: Larger solids - are moved along the stream bed as bed load. Finer solids - are carried in suspension. Specific gravity of soil materials (particles) is about 2.65 and hence they tend to settle towards the channel bottom.On the other hand, upward currents in the turbulent flow counteract the gravitational settling. When the sediment - laden water reaches a reservoir, the velocity and the turbulence are very much reduced. Larger suspended particles and most of the bed load are deposited as a "Delta" at the head of the reservoir. Smaller particles remain in suspension longer and are deposited farther down the reservoir near the heel. Smallest particles remain in suspension for a long time and some may pas through the sluiceways or turbines or over the spillways. Suspended sediment Load - measured by sampling the water, filtering to remove the sediment, drying and weighing the filled material. Expressed in : parts per million (ppm). Sediment Rating Curve Qs = KQn n varies between 2 & 3. Bed load may vary from zero to several times the suspended load. Common range : 5% - 25% Depends on : size distribution of the bed material and streamflow file:///D|/dr.srini old system backup/d drive backup/ce303/CE303_BTech_WRE/Water_Resources/Reservoirs/sedimentation.htm[11/12/2010 8:10:22 PM]

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rate. Total sediment transport = suspended sediment transport + bed load transport. Mean annual sediment - production rates(sediment yield) generally range from 70 - 1400 t/km2 Absent of suspended sediment data - comparison with similar watersheds whose sediment transports have been studied. Reservoir Sedimentation: The specific weight of the settled sediments seems to vary with the age of the deposit and the character of the sediment. Specific weights (dry) of sediment samples from reservoirs : 650 1500 kg/m3 Mean value for fresh sediments : 800 kg/m3 Mean value for old sediments : 1300 kg/m3 Specific weight of suspended sediment can be estimated using the equation :

% of sand, silt and clay on weight basis. W1,W2,W3 : Specific weights of sand,silt and clay respectively at the end of the I year. B 1,B2,B3 : Relate to compaction characteristics of these types. file:///D|/dr.srini old system backup/d drive backup/ce303/CE303_BTech_WRE/Water_Resources/Reservoirs/sedimentation.htm[11/12/2010 8:10:22 PM]

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Values of W must be calculated for each year, since deposition occurs during the life of the reservoir. Compaction occurs with time and hence older sediments will occupy less space per unit weight. Reservoir Trap efficiency: The percentage of the inflowing sediment that is retained in a reservoir, is known as "Trap efficiency" This is related to the ratio of reservoir capacity to total inflow. The trap efficiency of a reservoir decreases with time(age), as the reservoir capacity is reduced due to the sediment accumulation. Useful life of the reservoir is terminated when the capacity occupied by sediment is sufficient to prevent the reservoir from serving its intended purpose.useful life can be computed - Total time required to fill the critical storage volume. Sediment simulation - can be added to a continuous hydrologic simulation model - calibration through sediment sample collection on a daily basis for a number of years.

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