SELECTION OF RESERVOIR CAPACITY

16.7 SELECTION OF RESERVOIR CAPACITY

The determination of the required capacity of a storage reservoir is usually called an ‘opera- tion study’ using a long-synthetic record. An operation study may be performed with annual, monthly, or daily time intervals; monthly data are most commonly used.

When the analysis involves lengthy synthetic data, a computer is used and a sequent- peak algorithm is commonly used. Values of the cumulative sum of inflow minus withdrawals taking into account the precipitation, evaporation, seepage, water rights of the downstream users, etc., are calculated, (Fig. 16.15). The first peak and the next following peak, which is greater than the first peak, i.e., the sequent, peak, are identified.

e +v

Sequent peaks

ha-m)

Initial peak

Max Max

Storage Storage

storage storage

Demand), –

Time (months) w

(Inflo S

e –v

Fig. 16.15 Sequent-peak algorithm

HYDROLOGY

The maximum difference between this sequent peak and the lowest trough during the period under study is taken as the required storage capacity of the reservoir.

Example 16.2 The mean monthly flow data for a proposed reservoir site are given be- low:

Month

May June Mean monthly flow (cumec)

Nov. Dec. Mean monthly flow (cumec)

27 29 30 27 31 15 Determine the average discharge that can be expected throughout the year. Draw the

residual mass curve and obtain an expression for the range as developed by Hurst on the basis of the monthly flow data.

Solution

Cumulative Residual monthly

mean flow mass curve Month

flow volume

mothly

throughout (ha-m) (cumec)

flow, x

(ha-m)

inflow (ha-m)

the year (ha-m) 1 2 3 4 5 6 Jan

Mean flow (per month) 47174

= 15 cumec

12 throughout the year

= 3931.2 ha-m

The average discharge that can be expected throughout the year

47174 × 10 4 m 3

= 15 cumec = x

365 86400 × S

The residual mass curve is plotted in Fig. 16.16 and the range, R = 18333 ha-m, which is the storage capacity of the reservoir to maintain the mean flow of 15 cumec throughout the year.

x : 6 3 1 2 7 1 27 29 30 27 31 15 x – x :

MATHEMATICAL MODELS IN HYDROLOGY

(x – x ) 2 :

64 196 144 196 225 144 256 Σ (x – x ) 2 = 1815

R =σ F I HG 2 KJ

Let

F 12 I

= 12.84 (30.4 × 24 × 60 × 60) HG

2 KJ

D Months ha-m) 2

Range, R 8 = 18333 ha-m =

Storage for Q = 15 cumec 10 throughout year

Residual 12

Residual 14 mass curve

Fig. 16.16 Residual mass curve (Example 16.2)

Thus, the expression for range (on the basis of 12 months data) is

n 0.945

R =σ F I ...(16.32)

HG 2 KJ

Usually k varies from 0.5 to 1.0, the average value being 0.73. Usually, a number of years of observation are required. Example 16.3 Given in Table 16.2 (Col. 1, 2, 3, 5 and 6) are the monthly inflows during low- water period at the site of a proposed dam, the corresponding monthly precipitation and pan

HYDROLOGY

evaporation at a nearby station, and the estimated monthly demand for water. Prior water rights downstream require a special release of 6 cumec or the natural inflow, whichever is less. Assuming that only 24% of the rainfall on the land area to be flooded by the proposed reservoir has reached the stream in the past, reservoir area as 6000 ha on an average, and a pan coefficient of 0.7, construct the sequent peak alogrithm and determine the required storage capacity of the reservoir .

Solution Since 24% of the rainfall (P) is runoff, which is already included in the monthly inflows into the reservoir, only 100 – 24 = 76% of the rainfall on the reservoir area is to be included. Reservoir evaporation = 0.7 × pan evaporation (E P ). (0.76P – 0.7 E p ) values have to

be multiplied by the average reservoir area at the beginning and end of each month. The monthly change in storage and cumulative storage (at the end of each month) are

worked out in Table 16.2 and the sequent peak algorithm is drawn as shown in Fig. 16.17 and the required storage capacity of the Reservoir (difference between the initial peak and the lowest trough in the interval) is 13045 ha-m, which is also indicated in the col. (10) of Table

16.2. Actually this process has to be done for 4-5 consesecutive years and the difference be- tween the highest peak and the suceeding lowest trough gives the required storage capacity to meet the specified demand. The required storage capacity is also equal to the sum of the negative quantities (ΣDeficit) in Col (8) of Table 16.2, which is less than the sum of the positive quantities (ΣSurplus) col (8), thus ensuring the filler of the reservoir during mon- soons.

Peak

24 e 20

+v 16 13045 = Reservoir ha-m)

July A

A M June

4 Time months

(Inflo S 8 e 12

–v 16

Fig. 16.17 Sequent-peak Algorithm (Example 16.3)