Unit - 4

Reservoir Planning

Q1) What is Yield reservoir?

A1)

• When any type of storage reservoir is to be designed. It is of almost importance to analyse the relationship between the capacity of the reservoir and the yield of the reservoir.
• Yield can be defined as, "It is the actual amount of water that can be supplied in a given period of time from the storage reservoir".
• Depending upon the size of the reservoir the given period of time i.e., the time interval varies from, may be a day to a couple of months or may be to a couple of years. i.e., small sized reservoirs have lesser time interval than the larger sized reservoirs. The yields also vary from a year to another. It depends upon the inflow of a water to the reservoir.

Q2) What is reservoir storage?

A2)

• It is constructed to store water which flows during the high flow, during rainy season and the summer is used, during the dry
• It can serve the purpose of irrigation, hydel power generation, domestic and industrial water supply. In some cases, - these reservoirs also help to control the floods, to safe guard the fields and the settlements along the banks of the river; on the down-stream side of the river.
• So, on the basis of the requirements, there reservoirs can be single purposed or multi purposed.

Q3) What is Reservoir clearance?

A3)

The site selection of reservoir is based on the following factors

1. The proposed location of the reservoir must have suitable dam site.
2. The width of the valley of the stream must be narrow, to the length of the dam wall would he minimums to reduce the cost of construction. At the same the valley must have a large opening towards the upstream to allow to store a large volume of water in the reservoir.
3. The hills which surround the dam to form the rim, must be water-tight, so the loss of stored water through leakage would be least.
4. The basin of the reservoir also should be water tight which will not allow the stored water to escape under the surrounding wills. Non porous strata of the basin is ideal for the reservoirs.
5. The site of the reservoir must be such to have minimum loss of cultivated areas, human settlements, reserves of minerals: through the spread of back waters
6. The water through the tributaries having maximum salt and other sediments should be avoided for the construction of the reservoirs.
7. The site of the reservoir must have the potential capacity to store the required amount of water (This requirement of water will vary as the purpose of the reservoir, i.e., Requirement of domestic water supply will be different than the irrigation. Water and so the site depends up on the purpose of the reservoir also).
8. The site should be selected where the reservoirs will be deep, to accommodate more water in less extensive area, to reduce the land cost per unit of storing capacity will be less the loss of water through evaporation will be less and the development of weeds would be minimum.
9. The bed rocks and the side rocks of the reservoir must not have the soluble minerals which reduce the quality of the stored water mainly for the reservoirs to make supply of domestic water for consumption.
10. The site of the reservoir must be such where the cost of construction of associated works like. Transport linkages and housing colonies would be minimum.

Q4) What Investigation is done for Reservoir Planning?

A4)

When any type of reservoir is to be constructed, various investigations are made through different field surveys. Which are as follow:

1. A: Engineering surveys
2. B: Geological surveys
3. C: Hydrological survey

A: Engineering surveys

• Under these surveys, the site of the reservoir, and other associated work is surveyed and a contoured plan (showing the relief of the area) is prepared.
• From this plan only, the storage capacity of the reservoir and the water spread area, at various levels (elevations) is calculated as shown below.
• The water spread area of the reservoir at any elevation is determined, directly by measuring the area covered by a contour of the elevation, by the help of the planimeter.

Fig.: Elevation area and elevation capacity curve

• This change in the water spread areas at different height can be plotted on a graph capacity on 'x' axis and elevation on 'y' axis. As shown in Fig.
• In the line graph, Maximum pool level, Normal pool level, and minimum pool level is shown.
• By taking the contour areas at equal interval and by summing up various formulae, the storage capacity of any reservoir can be calculated. Following the formulae used for such calculations.
• If 'V' is storage volume and A₁, A₂, A3 ….An are the areas covered by the successive contours, with an interval of 'h'. Three different formulae can be sued for this purpose.

Trapezoidal formula

V=

Cone formula

V=

Prismoidal formula

V =

• By using any above-mentioned formulae, the storage volumes at various elevation are calculated and are plotted in the above-mentioned elevation capacity curves.
• These maps also indicate, the area and the property, which would go under water when, the reservoir is filled upto a certain elevation. This helps to calculate the compensation to be paid to the farmers and other owner of the land and the time. Before they should vacate the area to be submersed under the water of the reservoir.

B: Geological surveys

• These surveys are necessary to confirm the followings:

1. Suitability of foundation for the darn.
2. Water tightness of the reservoir basin.
3. Location and estimation of the store quarry, to be used for the construction work (the cost of transport is added to the total co.t of the reservoir)

• On the basis the type and height of the proposed dam, the exploration of the sub surface is carried out, to calculate, the depth of the over burden to be excavated and removed for the laying in foundation, the natural of rock its formation and the extent of fault zone (if the rock has such fault)
• In case of earthen dams, or low dams such sub-surface exploration, gives information about the soil properties at various depths on the basis of the information collected helps in devising a suitable programme B of foundation treatment (say by grouting) before the actual construction is taken up.
• These geological surveys also help to generate the information about proportion of cavities or pores in the rocks. Which is the basic cause of escape of water through. Such rocks in the adjacent valley, if the proportion of such pores is less it is treated but if it is wide spread, the site of the reservoir is changed.
• For the construction of dam and other works like embankment work, diversion hand work, canal regulation work etc., a huge quantity of building material is required. The availability of material and the distance of such location from the dam site, also is an important criteria to select a site.
• So, the information collected by through such geological field surveys help to calculate the stability and usability of the reservoir.

C: Hydrological surveys

• The whole of the planning of any reservoir is based on the estimates of the total quantity of water to be available in the stream, from season to another season i.e., the total volume of water during summer and rain season and the yearly average availability of water. Through the hydrological surveys investigations are made, two types of

1. Study of runoff pattern of the stream, at the proposed dam site: This helps to determine the storage capacity of the expected reservoir.
2. To study the hydrograph of worst flood: This helps to determine the methods to take care of the excess water from the reservoir like spill ways gates etc.

• So, in short, engineering surveys help to locate the site, the geological surveys help to determine the strength of the foundation and expected losses through the rock pores and the hydrological surveys help to calculate to total amount of water, available in the stream and the danger level of worst floods, etc.
• On the basis of the information collected by their surveys the selection of the dam site is made; for the actual select of the dam site following factors are considered.

Q5) Explain Fixation of Reservoir Capacity from annual inflow and outflow?

A5)

• As we have already seen that the storage capacity of any reservoir depends upon the nature of the inflow the amount of water received from the inflow, depth of the reservoir the rock strata around the reservoir climatic conditions (to determine the rate of evaporation) etc.
• It is necessary to calculate the capacity of the reservoir to meet the demand of water for various purpose. This capacity of the reservoir can be determined by the Mass curve and the demand curve as explain below:

Use of Mass Curve to Determine the Capacity of the Reservoir

• By using the available stream flow record, for a number of consecutive years, the mass curve is prepared, on the base of the flow hydrograph. The span of the period to be considered, must have a critical or driest period of the region (which would) indicate the lower most figures of the stream inflows look at the Fig. 5.4.1 which has expressed a mass curve for period of six years. i.e., from the year 2000 to 2006. The time period in years in given an x axis and the accumulated flow is given on y axis.
• In this Fig. In the downward side, the demand curve has been drawn (for the comparison). It indicates the variation in the rate of demand. If the rate of demand is constant, the demand curve will be a straight line (with no curve) the demand curve is always on rise as the population rises, the consumption of domestic purpose, irrigation purpose. Industrial purpose, hydel power generation purpose always increases.
• On the A-B curve, G-H curve, and F-J doted lines are drawn, parallel to the demand curve and are tangent to the high points like, G and H of the mass curve. These are the point placed at the beginning of the dry period.
• In the mass curve, maximum vertical intercepts i.e., X₁ Y₁, X₂ Y₂ etc. between the tangential lines are drawn to measure the mass curve. These vertical intercepts i.e., x1, y1, or x2, y2, indicate, the volume by which the total inflow of the stream falls short of the demand and so it is necessary to make supply form the reservoir storages. (Refer Fig.).
• Let us assume that the reservoir to be fall at point G (in the Fig.) for a period from G to Z1, (in the year between 2001 and 2002) there is a total inflow in the stream which is represented by Y1, Z1, and at the same time the total demand is represented by X1, Z1, (leaving a gap of volume of water, shown by X1, Y₁). This is needed to be met by using reservoir's storage.
• The largest of the maximum vertical intercepts i.e., X1, Y1, X2, Y2, etc.; represent the capacity of the reservoir, required to satisfy the given demand. Whatever has been calculate in the Net storage' required to meet the demand, so if there is any type of evaporation or percolation loss in the storage of the reservoir, it must be compensated by using storage capacity.
• The vertical distance between two tangential lines, such as GH and FJ, represents the extra amount of water which spills over from the reservoir, through the spillway. It goes as a waste in the dam stream direction. This happens because between height of H and F, the reservoir remains full and so all the inflow in excess of demand will be sent out through the spill way to the downstream side.
• If these tangential lines which are drawn parallel to the demand curve, are extended further, they should intersect' the mass curve e.g., at H or J, so the reservoir which was full at G and F will be filled soon, but these extended lines do not intersect the mass curve it will indicate that the reservoir will not be filled again. If the reservoir is very large, the time interval between the points G and H; F and J etc. may be of several years.

Fig.: A mass curve to determine the required capacity of reservoir

Q6) Explain Fixation of Reservoir Capacity using elevation capacity curve & dependable yield?

A6)

The yield of any reservoir can be classified as-

• A: Safe yield or firm yield
• B: Secondary yield.
• C: Average yield.

A: Safe yield or firm yield

• It can be defined as "It is the maximum quantity of water that can be supplied, during the critical dry period i.e., worst dry period".
• The critical or worst period is the time (in days or months), when the natural inflow of the stream is lowest, on record. In same period of time, this dry period may occur. With the lower yield than the safe or firm yield. Which is calculated on the basis of the past record of the inflow of the natural stream.

B: Secondary yield

• It can be defined as, "It is the excess amount of water available than the safe yield during the period of high flows".

C: Average yield

• It can be defined as "It the arithmetic average of the yield i.e., average of the safe and secondary yield; over a long period of time".

The Yield and Storage Capacity of Reservoir:

• Both, the yield and storage capacity are interdependent factors. The storage capacity of a reservoir depends upon, the inflow to the reservoir. The relationship among the yield. Inflow and storage can be expressed as below.
• Inflow-yield change in the storage so, it is very easy to understand that when the inflow is greater than the yield (supplies made through the reservoir), the storage in the reservoir will 0increase and on the other hand, if the inflow is lesser than the yield, the storage reduced.
• The value of yield which can be adopted for the design of a reservoir is called as 'designed yield'.
• The designed yield is kept as level that, it should meet the consumers demand still it will not affect the storage capacity.
• The value of designed yield depends upon the urgent need of the water to be supplied or on the amount of risk involved in it; when the actual yield is lesser than the designed yield. i.e., in case of the reservoir designed for the domestic water supply must be planned on the basis of safe yield in case of a reservoir for irrigation purpose, should be planned with the designed yield value atleast 20 % greater than the safe yield.
• On the same basis, the power (electricity) commitments to the actual domestic users must be based on the firm basis and it should not exceed the power which can be produced with the safe yield. Unless there is an alternate arrangement of thermal or any other type of power is available to support the Hydel power.
• In case of the demands from the large-scale industries for the hydel power, it should be developed form secondary yield, as and when it is available.

Q7) What are the reservoir losses and its control measures?

A7)

The losses of water from the reservoir are the result of three factors namely:

      A: Evaporation

      B: Absorption

      C: Percolation

These are all natural losses of water, which must be taken into consideration while designing and in the operation stage of any reservoir. Let us check this loss with their proportion to the total losses.

A: The evaporation losses

• The loss of water from the reservoir is proportional to the area of the reservoir. It is expressed in millimetres of depth of water. i.e., It always shows the volume of water lost, per unit area of the exposed surface of water of the reservoir.
• Together with the surface area, other factors, which affect the loss of water through evaporation, are:

1. Temperature
2. Wind Velocity (over the reservoir)
3. Relative Humidity (RH)

i.e., Higher the temperature faster the rate of evaporation.

• Higher the wind velocity higher the rate of evaporation.
• Higher the relative humidity lesser the rate of evaporation.
• A simple technique is used to measure these losses by using "Pan-Evaporation Method".

B: The absorption losses

• These losses depend upon the type of soil the bed of the reservoir.
• It has been observed dapia te begg the absorption is high, as the soil is dry and the pores are open but when they are filled and soil gets saturated, the rate of losses through becomes very low and finally almost insignificant.

C: The percolation losses

• These losses depend upon the soil strata.
• If the soil is porous or there are cavities of fissures in the rocks, the rate of percolation is very high and cause, a lot of economic.

Control measures to reduce the losses of water

• A: To reduce the losses, the reservoir should be designed to have higher depth and lesser surface area.
• B: To reduce the wind velocity, wind breakers i.e., the plants must be developed.
• C: To reduce the rate of evaporation the surface must be covered by a thin, chemical film. (Acetyl alcohol)
• D: To reduce the percolation losses the grouting techniques should be used.
• E: To reduce the losses of water it is better to construct the underground reservoirs. (They need high construction cost and maintenance cost).

Q8) What is Reservoir sedimentation?

A8)

• We already have discussed about the causes of the difference between the capacity of the reservoir and the actual yield of the reservoir. The basic reasons are variation in the 'inflow of the (1. Stream, the variation in the annual total precipitation which indicate the variation in the receipt of the water in the storage of the reservoir.
• Let us say that the receipts of water may be same but due after reasons, like loss of storage water due to evaporation, or (and) the loss of water due to percolation and seepage the capacity to hold water and the actual discharge (yield) have variation.
• There is one more factor which reduces the capacity of the reservoirs to hold the given amount of water. It is the process of sedimentation. Let us study this process of sedimentation, to find out the measures to control it.

Process of Sedimentation:

• Due to physical weathering of the surface in the catchment area, the process of erosion starts. This loose material is transported from the source region to the down-stream areas of the catchment.
• The carrying capacity of the stream depends upon the velocity of the stream (which depends upon the slopes i.e., if they are steep, more load will be carried by the stream) and the amount of surface runoff is if it is more, more load would be carried in the dam stream areas of the catchment
• The relationship between the load and velocity of flow is funny i.e., if the velocity is more than the load carried by the stream would be more, but when the load in the stream increases, the velocity of the stream is reduced the carrying capacity and the load gets deposited (which is called as sedimentation).
• Once it gets deposited, the flow becomes free from the load, so the velocity increases to increase it carrying capacity. The process goes on and on till the cycle of erosion is completed.
• The amount of sediments i.e., load, in the stream depends upon the following factors:
• Nature of the soil in the catchment: If the soil is sandy, loose, and easily erodible, the river will have a large amount of sediments.
• The topography of the catchment: If it has a rough topography, with steep slopes, the stream will have higher velocity of flow. So, the sediments in the river flowing with higher velocity, would have more sediments.
• Vegetation cover in the catchment: If it has thick vegetation cover, the top soil would be held by the root of the plants and the process of weathering will be reduced. So, the catchment has thick vegetation cover, will have less sediments.
• Intensity of rainfall in the catchment: If the intensity of rainfall is higher the total runoff will be more so the sediment carrying capacity will be more and so the sediments in the stream will be more. (But, if the intensity of rainfall is more, the growth of natural vegetation will be more to protect the surface soils and so the sediments will be less).
• So, while considering these factors to affect the proportion of sediment, we must consider the interaction among these factors. It is a complex process of weathering erosion transportation and deposition i.e., sedimentation.

Q9) What are the Effects of Sedimentation of Reservoirs?

A9)

• As already mentioned, the carrying capacity of a stream, depends upon its velocity. When a dam is constructed across a river, the velocity is almost capacity of a stream, depends upon its velocity.
• When a dam is constructed across a river, the velocity is almost 0 till the reservoir is full and starts over flowing, or through the gates or valves.
• So, whatever load i.e., the sediments has been brought from the upstream, gets deposited on the bed of the reservoir and make it shallow.
• Due to over sedimentation, the storage capacity of the reservoir is reduced and so its usefulness to meet the demand for domestic consumption, or irrigation, of hydel power generation etc. is also lost.
• Over sedimentation in the reservoir makes the back water to be spread over a larger region to increase the loss of properties and the bio-mass. Over sedimentation increases the economic and ecological losses of the catchment.
• During floods as the capacity of the reservoir to hold water is reduced, so the excess water is discharged in the downstream, command areas, to create man-made floods. So, it is necessary to control sedimentation.

Q10) What are the Measures to control reservoir sedimentation?

A10)

• To increase the useful life of a reservoir, or atleast to maintain its water storage capacity, as its design; it is necessary to control the deposition of sediments in the reservoir. Following are some of the important methods to be used to control the process of sedimentation.

Site selection of the reservoir:

• The site of the reservoir must be such where the proportion of sediments in the inflow is less. If a river has many tributaries coming from different directions out of which one tributary has maximum sediments it its surface flow.
• So, the dam of the reservoir should be constructed before 'this' tributary meets the main river.

Design of the reservoir:

• A reservoir of a small capacity on a river which has larger inflow rate will have lesser sedimentation than a larger size reservoir on the same river. So, if the capacity of the reservoir is increased in stages, the sedimentation proportion will be reduced.
• If the out lets, at different elevations are provided, the sediments will not be allowed to settle down, mainly during floods having maximum sediments, which would be carried forward and the reservoir would be saved from the danger of over sedimentation.

Control of sediment inflow:

• By providing vegetation screens or check dans the inflow of sediments will be put under control.

Removal of the sediments:

• The sediments which have been already deposited can be removed by excavation or by scouring through sluices. Both of these methods are costly and need have some structural changes.

Control of erosion in the catchment:

• If the measures are taken to control the process of erosion and weathering in the catchment, the development of sediment will be reduced and so the problem of over sedimentation will be put under control.
• This can be done by using the soil conservation methods by afforestation controlling the forestation, controlling over grazing, contour bunding, during gully erosion by contracting small embarkments etc.
• So, by using the methods mentioned above, we can save soil, water and also the ecological and economical loss of the catchment areas.

Q11) What is density currents?

A11)

• Density current can be defined as "It is a gravity flow of one fluid under another fluid having slightly different density".
• The water already stored in the reservoir is naturally settled and so it is clear if compared with the flood water having a lot of sand, clay and muddy material.
• These sediments carrying flood water has higher density than the original water in the reservoir, so it goes down towards the bed.
• This condition remains for a considerable period of time. If this sediment carrying water is taken out by suitable outlets and sluice ways, it will help to reduce the reservoir trap efficiency i.e., the rate of sedimentation (which can be reduced upto 5% to 6%).

Q12) What is the Significance of Trap efficiency?

A12)

• "Trap efficiency is the percentage c inflowing sediment, (through the flood water), which is retained in a reservoir. It is a function of the capacity of the reservoir to the total inflow."
• The inflow of the sediments entering in the reservoir has a seasonal variation i.e., in summer, it is almost nil which in rainy season, it is maximum (due to monsoon floods).
• At the end of the life of any reservoir, it gets completely filled with these sediments. The life of the reservoir, so depends upon the inflow the of the sediments and the capacity of the reservoir.
• If the rate of inflow is greater than the absorbing capacity the useful life of the reservoir would be less.
• The annual inflow of the sediments can be calculated by using the simple equation given below,
• Average annual Transport sediments = (Weight of the sediments/per unit volume of water) × (Average annual inflow of the channel)
• So, in simple terms, the reservoir efficiency can be calculated by the equation given below
• R.E = Capacity (in m³) /Volume of annual in flow (in m³)

Fig.: Trap Efficiency

Fig.: The section of dam the level of sedimentation

Q13) What are the factors affecting Trap Efficiency Value?

A13)

Mainly there are three factors which affect the trap efficiency value of any storage i.e., Pond, Lake, reservoir etc.

1. The detention period
2. The size of the particles
3. The dead storage

To control the trap efficiency following steps should be taken.

1.Reduce the detention period

• If the detention period of storm runoff is reduced, its neaps to control Trap efficiency e.g., if the detention period of the surface runoff is reduced from 30 days to 2 days. The trap efficiency would be reduced up to 80% to 90%.

2.Reduce size of the particles

• The large size particles are the first to get deposited while the fine clay always remains suspended for a longer period of time. So small particles would be easily carried forward with the surface runoff, to reduce the trap efficiency.

3.Elimination of Dead storage

• If the dead storage of the reservoir is reduced, it would help not only to reduce the Trap-efficiency but also to reduce the detention period. So, it does help to reduce the Trap-efficiency of the reservoir.

Q14) What is Useful Life of Reservoir?

A14)

• The useful life of a reservoir can be calculated by following the steps given below,
• Calculate the capacity efficiency of the reservoir (based on the capacity and the average annual inflow.)
• Calculate the trap efficiency of the reservoir by the equation given above
• Divide the total capacity, for some specific period (in e.g., 10% and 90% assume this 10 % capacity has been reduced due to the deposition the sediments. Now, calculate the trap efficiency for the remaining 90% of the original capacity of that reservoir and the inflow.
• Get the mean of these two trap efficiencies; which can be assumed as the average value to the trap efficiency for the 10% of the capacity of the reservoir (which is filled totally with the sediments.)
• Calculate the average annual sediments, transported by the stream, or a channel and multiply this figure by the mean trap efficiency.
• Convert this quantity of the sediments, into weights (cubic contents) i.e., volume in hectare meters (ha.m)
• The 10% of the capacity of this reservoir is to be divided by the volume of sediments deposited. This will help to calculate the total number of years required, to till this 10% capacity of the reservoir by the inflow of sediments.
• The above procedure is to be repeated, to calculate the number of years required to fill next 10% capacity of the reservoir i.e., 80%, 70%, 60%...... Till the capacity of the reservoir becomes 20%.
• The addition of the years, required to till, each of the 10% capacities of the reservoir, will indicate, the useful life of the reservoir.

Q15) What is Apportionment of Total cost?

A15)

• The apportionment of a multipurpose reservoir scheme means. "The sub-division of the various uses of the reservoir, for which it has been planned."
• The total cost of the scheme can be classified as,

      A: The separable cost of the single purpose of the scheme.

      B: The joint

• A: The separable cost is those which are chargeable to a single function, e.g. The cost of the power-house (in case of hydro-power generation) of the cost of locks for a scheme inland navigation. These costs are calculated as the total cost minus the estimated cost of a particular function.
• B: The joint costs: These costs are calculated by subtracting the sum of the separable cost from the total cost of the scheme.

Q16) What is Use of facilities method?

A16)

• The joint cost of the scheme is divided in proposition to the use of storage volume used for the function. This method is easy to apply but it lacks in the required accuracy because the cost of the project is not directly proportional to the reservoir storage.
• In fact, it is true that as the storage increases, the cost per unit of volume of storage, declines. So, the unit of storage cost goes on changing.

Q17) What is Alternative justifiable expenditure method?

A17)

• In this, the joint costs are distributed with reference to the difference in the separable costs and the estimated cost of a single purpose scheme which would provide same services and would be justifiable, economically.
• The drawback of this method is that, it is difficult to estimate. The benefits or (and) to estimate the cost of each function.
• One must note that none of the above methods, is not equally applicable to all the projects, depending upon data availability. The proper method must be used to calculate joint cost of the multi-purpose project.