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Reservoir drawn down
                                                                               Depletion
                                                                                                          of
                                                                                  of
                                                                                storage
                                                                                                        storage
                                                   600
                                                                                                                    full
                                                   500
                                                  Cumulative runoff and draft (MG)  700  period at point Reservoir full  1. Draw cumulative draft Σ D  Reservoir empty  Replenishment  Reservoir  2.4 Design Storage  25
                                                                                                             Must intersect
                                                                       parallel to rate of draft A
                                                                                                             runoff curve,
                                                                       and tangent to curve B
                                                   400
                                                                                                              if reservoir
                                                                                                              is to refill
                                                                                  3. Measure maximum
                                                                                                        Curve B.
                                                         Start of dry
                                                                                  deficiency cumulative
                                                                                                       Cumulative
                                                                                   ΣD–ΣQ = 124 MG
                                                   300
                                                                                                        runoff, ΣQ
                                                         of tangency
                                                                        Storage, S
                                                                                                      End of dry
                                                   200                ΣQ–ΣD = 124 MG                 period at point
                                                                                                      of tangency
                                                                            2. Draw parallel to line A
                                                                             and tangent to curve B
                                                   100                                  Line A, slope = rate of draft
                                                             Must intersect runoff curve,
                                                               if reservoir is to be full
                                                                at start of dry period
                                                     0
                                                      0   1    2   3    4   5    6   7   8    9   10   11  12  13   14  15
                                                                                 Order of months
                                                Figure 2.7 Mass diagram method for the determination of storage required in impounding reservoirs.
                                                                                    2
                                                                          2
                                                (A constant draft of 750,000 gpd/mi = 23 MG/mi for a month of 30.4 days is assumed.) Conversion
                                                                                                  2
                                                                                                              2
                                                                                                                      2
                                                factors: 1 MG = 1,000,000 gal = 3.785 ML = 3,785,000 L; 1 MG/mi = 1.461 ML/km ; 1 gpd/mi =
                                                          2
                                                1.461 L/d/km .
                                    intervals of time—from monthly to daily values, for exam-  generally suffices in the well-watered regions of the United
                                    ple. The additional storage identified by such a shift may be  States. Water is plentiful, stream flows do not vary greatly
                                    as much as 10 days of draft.                      from year to year, reservoirs generally refill within the annual
                                       Assuming that inflow and drafts are repeated cyclically,  hydrologic cycle, and it does not pay to go in for advanced
                                    in successive sets of T years, a procedure called sequent  or complete development of catchment areas. In semiarid
                                    peak was developed for determining minimum storage for no  regions, on the other hand, water is scarce, stream flows fluc-
                                    shortage in draft based on two needed cycles. Example 2.1  tuate widely from year to year, runoff of wet years must be
                                    illustrates the procedure.                        conserved for use during dry years, and it pays to store a large
                                       For variable drafts and inclusion of varying allowances  proportion of the mean annual flow. In these circumstances,
                                    for evaporation from the water surface created by the  operational records of adequate length become important,
                                    impoundage, the analytical method possesses distinct  along with computational aids.
                                    advantages over the graphical method. The principal value  Given a series of storage values for the flows observed or
                                    of the mass diagram method, indeed, is not for the estimation  generated statistically, the engineer must decide which value
                                    of storage requirements, but for determining the yield of  he will use. Will it be the highest on record, or the second,
                                    catchment areas on which storage reservoirs are already  third, or fourth highest? Obviously, the choice depends on the
                                    established.                                      degree of protection to be afforded against water shortage.
                                                                                      This must also be considered in terms of drought experience,
                                                                                      which is a function of the length of record examined. To
                                    2.4 DESIGN STORAGE
                                                                                      arrive at a reasonable answer and an economically justifiable
                                    Except for occasional series of dry years and very high den-  storage design, the engineer may resort to (a) a statistical
                                    sity housing and industrial developments, seasonal storage  analysis of the arrayed storage values and (b) estimates of
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