The capacity of a storage reservoir is determined on the basis of the inflow to the reservoir and the demand of the consumers (or the yield of the reservoir).
The following two methods are generally used for determining the capacity of a storage reservoir:
1. Analytical Method:
In this method an analysis of demand and inflow of water per month of the year is made.
The following data are required:
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(i) Total inflow of the stream during each month of a critical low flow year (or dry year) at the reservoir site.
(ii) Total loss of water due to evaporation, percolation, etc., during each month of the year.
(iii) Total precipitation (if any) during each month of the year.
(iv) Total amount of water required to be released from the reservoir during each month of the year to satisfy the prior water right requirements of the residents on the downstream of the reservoir.
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(v) Total demand of water during each month of the year.
The following procedure is adopted to determine the capacity of the storage reservoir:
(i) From the total inflow of the stream during each month, the total loss of water due to evaporation, percolation, etc., and the total amount of water to be released to meet the downstream requirements during that month are subtracted and the total amount of precipitation (if any) during the same month is added. This gives the adjusted or net inflow of the stream for different months of the year.
(ii) By subtracting the adjusted or net inflow from the demand the deficiency or the amount of water required from the storage to meet the demand for different months is obtained. However, if the demand is less than the adjusted or net inflow it indicates a surplus.
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(iii) The total deficiency during the successive months gives the required capacity of the storage reservoir.
(iv) If provision is to be made for two or three successive dry years, the capacity obtained in step (iii) is increased accordingly.
This method of determining the capacity of a storage reservoir is indicated in Illustrative Example 3.1.
2. Mass Curve Method:
The mass curve method is more commonly used for determining the capacity of a storage reservoir.
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Mass Curve and Determination of Capacity of a Storage Reservoir Required For a Specified Yield or Demand Using Mass Curve:
A mass curve of inflow (or mass curve) is a plot of accumulated flow in a stream against time. As indicated below a mass curve of inflow can be prepared from the flow hydrograph of a stream for a large number of consecutive previous years.
Figure 3.13 (a) shows a typical flow hydrograph of a stream for six consecutive years. The area under the hydrograph from the starting year (i.e., 1953) upto any time t1 [shown by hatching in Fig. 3.13 (a)] represents the total quantity of water that has flown through the stream from 1953 upto t1 time and hence it is equal to the ordinate of the mass curve at time t1.
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The ordinates of the mass curve corresponding to different times are thus determined and plotted at the respective times to obtain the mass curve as shown in Fig. 3.13 (b). A mass curve continuously rises as it shows accumulated flows.
The slope of the curve at any point indicates the rate of flow at that particular time. If there is no flow during certain period the curve will be horizontal during that period. If there is high rate of flow the curve rises steeply. Thus relatively dry periods are indicated as concave depressions on the mass curve.
A mass curve of demand (or demand curve) is a plot between accumulated demand and time (Fig. 3.14). If the demand is at a constant rate then the demand curve is a straight line [Fig. 3.14 (a)] having its slope equal to the demand rate. However, if the demand is not constant then the demand curve will be curved [Fig. 3.14 (b)] indicating a variable rate of demand.
The capacity of a storage reservoir required for a specified yield or demand may be determined by using mass curve of inflow and mass curve of demand or demand curve as indicated below:
(1) A mass curve of inflow is prepared from the flow hydrograph for a number of consecutive years selected from the available stream flow record such that it includes the most critical or the driest period. Figure 3.15 shows a mass curve of inflow for a typical stream for a 6 year period.
(2) Corresponding to the given rate of demand, a demand curve is prepared. If the rate of demand is constant then the corresponding demand curve is a straight line as shown in Fig. 3.15.
(3) Lines such as GH, FJ, etc., are drawn parallel to the demand curve and tangential to the high points G, F, etc., of the mass curve of inflow (or the points at the beginning of the dry periods).
(4) The maximum vertical intercepts X1, Y1 X2Y2 etc., between the tangential lines drawn in step (3) and mass curve are measured. The vertical intercepts indicate the volume by which the total flow in the stream falls short of the demand and hence required to be provided from the reservoir storage. For example assuming the reservoir to be full at G, for a period corresponding to points G and Z1, there is a total flow in the stream represented by Y1Z1 and there is a total demand represented by X1Z1 leaving a gap of volume represented by X1Y1 which must be met with from the reservoir storage.
(5) The largest of the maximum vertical intercepts X1Y1, X2Y2, etc., determined in step (4) represents the reservoir capacity required to satisfy the given demand. However, the requirement of storage so obtained would be the net storage which must be available for utilization and it must be increased by the amount of water lost by evaporation and percolation.
As shown in Fig. 3.15 the vertical distance between the successive tangential lines such as GH and FJ represents the quantity of water which could spill over from the reservoir through the spillway and go as a waste to the downstream side. This is so because between H and F the reservoir would remain full and all inflow in excess of demand would flow through the spill-way to the downstream side.
The tangential lines drawn parallel to the demand curve when extended forward must intersect the mass curve, such as at H, J, etc., so that the reservoir which was full at G and F will be filled again at H and J. However, if the line does not intersect the mass curve, the reservoir will not be filled again. Moreover, if the reservoir is very large the time interval between the points G and H, F and J, etc., may be several years.
For the numerical example indicated in Fig. 3.15, corresponding to a demand of 1.76 × 105 ha-m per year (or 55.8 cumec) the following results are thus obtained.
(i) The required capacity of the reservoir is given by X1Y1 = 1.0 × 105 ha-m
(ii) Assuming the reservoir to be full at G, it would be empty at Y1 and would be full again at H.
(iii) Between H and F the reservoir would remain full and all inflow in excess of the demand would be discharged through the spillway to the downstream side. The spill over from the reservoir would be 1.2 × 105 ha-m.
(iv) Assuming the reservoir to be full at F, it would be depleted to (1.0 × 105-0.64 × 105) = 0.36 × 105ha-m of storage at Y2 and would be full again at J.
The rate of demand has been assumed to be constant. However, the rate of demand may not be always constant, in which case the demand curve will be a curve with its slope varying from point to point in accordance with the variable rate of demand at different times.
In this case also the required capacity of the reservoir can be determined in the same way by superimposing the demand curve on the mass curve from the high points (or beginning of the dry period) till the two meet again. The largest vertical intercept between the two curves gives the required reservoir capacity.
It is, however, essential that the demand curve for the variable demand coincide chronologically with the mass curve of stream-flow, i.e., June demand must coincide with June inflow and so on. Furthermore, in this case also the storage obtained must be increased to account for the water lost by evaporation and percolation.
As shown in Fig. 3.16 if the end points of the mass curve are joined by a straight line AB, then its slope represents the average discharge of the stream over the total period for which the mass curve has been plotted. If a reservoir is to be constructed to permit continuous release of water at this average value of discharge for the entire period then the capacity required for the reservoir is represented by the vertical intercept between the two straight lines A’ B’ and A”B” drawn parallel to AB and tangent to the mass curve at the lowest tangent point C and the highest tangent point D respectively.
If the reservoir having this capacity is assumed to contain a volume of water equal to AA’ at the beginning of the period, then the reservoir would be full at D and it would be empty at C. However, if the reservoir was empty in the very beginning, then it would be empty again at point E and also during the period from F to K. On the other hand if the reservoir was full in the very beginning it would be full again at points F and K, and between points A and E there will be spill of water from the reservoir.
Determination of Yield from a Reservoir of Given Capacity:
Mass curve may also be used to determine the yield which may be obtained from a reservoir of given capacity for which the following procedure may be adopted:
1. A mass curve as shown in Fig. 3.17 is prepared in the same manner as in the above case.
2. Tangents are drawn at the high points G, F, etc., of the mass curve in such a manner that their maximum departure from the mass curve does not exceed the given capacity of the reservoir.
3. The slopes of each of these tangents are measured which indicate the yield which can be obtained in each year from the reservoir of given capacity. The slope of the flattest demand curve is the safe or the firm yield.
For the Numerical Example shown in Fig. 3.17 from a reservoir of capacity 1.2 × 105 ha-m the safe yield of 1.68 × 105 ha-m per year may be obtained. It is given by the slope of the tangent GH which is flatter than that of the tangent FJ.
Illustrative Examples:
Example 3.1:
The following table gives the monthly inflows during a critical low water period at a site of a proposed dam, the corresponding monthly evaporation and precipitation at a nearby station, and the estimated monthly demand of water. Prior water rights require a release of full natural flow or 6.5 hectare-metre per month, whichever is less. Assume that 30 percent of the rainfall on the land area to be flooded by the reservoir has reached the stream in the past. Using a net increased pool area of 520 hectares, find the required useful storage. Take pan evaporation coefficient equal to 0.7.
Solution:
The first five columns are same as given in the problem. Column (6) gives the downstream requirement on account of prior water right and the entries in this column are equal to the river flow or 6.5 hectare-metre whichever is less.
Column (7) gives the quantity of water E which evaporates from whole of the water surface area of reservoir during each month which is computed by the following expression
Column (3) hectare-metre Column (8) gives the precipitation P in hectare-metre, falling over the reservoir area. Since 30% of the precipitation is assumed to have reached the stream it is included in the now given in Column (2) and hence only 70% of the precipitation falling on the reservoir area is considered which is given by the following expression-
Column (9) gives the adjusted inflow I which is computed as follows:
I = Column (2) — Column (6) — Column (7) + Column (8).
Column (10) gives the quantity of water S required from storage to meet the required demand and it is computed as follows.
S = Demand — Adjusted inflow = Column (5)—Column (9)
Only the positive values of S are to be considered and entered in column (10) and where negative values are obtained nil should be entered in column (10). This is so because a positive value of S indicates that water is required from the storage since demand is more than the adjusted inflow, whereas a negative value of S indicates that no water is required from the storage since demand is less than the adjusted inflow.
The sum of the column (10) gives the required useful storage, the value of which in this case is found to be 216.36 hectare-metre.