Objective: To estimate reservoir capacity by using a mass curve (Rippl diagram)

Reservoir sizing (Calculation of storage volume)

Consider a reservoir on the stream whose mass curve is plotted in the figure below. If it is assumed that the reservoir is full at the beginning of a dry period, i.e. when the inflow rate is less than the withdrawal (demand) rate, the maximum amount of water drawn from the storage is the cumulative difference between supply and demand volumes from the beginning of dry season. Thus the storage required S is

S = maximum of (VD - VS)

where VD = demand volume, VS = supply volume. The storage, S which is the maximum cumulative deficiency in any dry season is obtained as the maximum difference in the ordinate between mass curves of supply and demand. The minimum storage volume required by a reservoir is the largest of such S values over different periods.

Figure 1. Flow mass curve

Consider the line CD of slope Qd drawn tangential to the mass curve at a high point on the ridge. This represents a constant rate of withdrawal Qd from a reservoir and called demand line. If the reservoir is full at C (at time tc_, then from point C to E the demand is larger than the supply rate as the slope of the flow-mass curve is smaller than the demand line CD. Thus the reservoir will be depleting and the lowest capacity is reached at E. The difference in the ordinates between the demand line CD and a line EF drawn parallel to it and tangential to the mass curve at E (S1 in the figure) represents the volume of water needed as storage to meet the demand from the time the reservoir was full. If the flow data for a large time period is available, the demand lines are drawn tangentially at various other ridges (e.g. C’D’ in the figure) and the largest of the storages obtained is selected as the minimum storage required by a reservoir. The example given below explains this use of the mass curve.

Example 1. The following table gives the mean monthly flows in a river during 1981. Calculate the minimum storage required to maintain a demand rate of 40 m3/s.

Month Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov DecMean 60 45 35 25 15 22 50 80 105 90 80 70Flow(m3/s)

Solution : From the given data the monthly flow volume and accumulated volumes are calculated as in the table given below. The actual number of days in the month are used in calculating the monthly flow volume. Volumes are calculated in units of cumec.day.

Table 1. Calculation of mass curveMonth Mean flow (m3/s) Monthly flow volume

(cumec-day)Accumulated volume(cumec-day)

JanFebMarAprilMayJuneJulyAugSepOctNovDec

6045352515225080105908070

186012601085750465660155024803150279024002170

18603120420549555420608076301011013260160501845020620

A mass curve of accumulated flow volume against time is plotted. In this figure all the months are assumed to be of average duration of 30.4 days. A demand line with slope of 40 m3/s is drawn tangential to the hump at beginning of the curve; line AB in the figure. A line parallel to this line is drawn tangent to the mass curve at the valley portion; line A’B’. The vertical distance

S1 between these two parallel lines is the storage required to maintain the demand. The value of S1 is found to be 2100 m3/s.days = 181.4 million m3.

Figure 2. The flow mass curve from the example

Example 2 Work out the example 1 through arithmetic calculation without the use of the mass curve.

The inflow and demand volumes in each month are calculated in Table 2. Column 6 indicates the departure of the inflow volume from the demand. The negative values indicates the excess of demand over the inflow and thus these have to be met by storage. Column 7 indicates the cumulative excess demand (i.e. the cumulative excess negative departure). The maximum value in this column represents the minimum storage necessary to meet the demand requirements. The storage requirement in the present case is 1920 m3/s day.

Table 2. Calculation of storage

Month Mean Flow rate(m3/s)

Volume Flow (cumec.day)

DemandRate(m3/s)

Demand volume (cumec.day)

DepartureCol 3-5

CumulativeExcessDemandVolume(cumec.day)

CumulativeExcessFlowVolume(m3/s)

1 2 3 4 5 6 7 8JanFebMarAprMayJunJulyAugSept OctNovDec

6045352515225080105908070

186012601085750465660155024803150279024002170

404040404040404040404040

124011201240120012401200124012401200124012001240

+620+140-155-450-775-540+310+1240+1950+1550+1200+930

-155-605-1380-1920

620760

31015503500605072508180

Column 8 indicates the cumulative excess inflow volume starting from each withdrawal from storage. This indicates the filling up of reservoir and the volumes in excess of storage (in this case 1920 m3/s.day) represent spillover. The calculation of this column is necessary to know whether the reservoir fills up after a demand, and if so, when. In this example, the cumulative excess inflow volume will reach 1920 m3/s in the beginning of September. The reservoir will be full again and will start filling after that time.