S-CURVE METHOD

5.6 S-CURVE METHOD

S -curve or the summation curve is the hydrograph of direct surface discharge that would result from a continuous succession of unit storms producing 1 cm in t r –hr (Fig. 5.18). If the time base of the unit hydrograph is T hr, it reaches constant outflow (Q e ) at T hr, since 1 cm of net rain on the catchment is being supplied and removed every t r hour and only T/t r unit graphs are necessary to produce an S-curve and develop constant outflow given by,

. 2 78 A

where Q e = constant outflow (cumec) t r = duration of the unit graph (hr)

A = area of the basin (km 2 )

Given a t r -hour unit graph, to derive a t r ′ -hour unit graph (t r ′ ≥ t r )—Shift the S-curve by the required duration t r ′ along the time axis. The graphical difference between the ordinates of the two S-curves, i.e., the shaded area in Fig. 5.18 represents the runoff due to t r ′ hours rain at

HYDROGRAPHS

an intensity of 1/t r cm/hr, i.e., runoff of t r ′ /t r cm in t r ′ hours. To obtain a runoff of 1 cm in t r ′ hours (i.e., t r ′ -hour UG), multiply the ordinates of the S-curve difference by t r /t r ′ . This tech-

nique may be used to alter the duration of the given unit hydrograph to a shorter or longer duration. The longer duration need not necessarily be a multiple of short.

(cm/hr) 1t r

Successive unit storms of P net = 1 cm

Intensity i =

To obtain t – hr UG

r 480 r multiply the S-curve difference by t /t rr ¢

S-curve 280 (lagged by

s-curve Area s-curve t – hr) Area

Constant outflow Constant outflow 2 (cumec) 240

2.78 A(km ) Q

difference difference

1 1 Unit hydrographs in Unit hydrographs in

e e t (hr) 200

r r r r cm

succession produce succession produce

Q= Q=

constant outflow, Q cumec constant outflow, Q cumec e e Cumec 160

Time t (hr) Fig. 5.18 Changing the duration of UG by S-curve technique (Example 5.5)

Example 5.4 The ordinates of a 4-hour unit hydrograph for a particular basin are given be- low. Derive the ordinates of (i) the S-curve hydrograph, and (ii) the 2-hour unit hydrograph,

and plot them, area of the basin is 630 km 2 .

Time

Discharge (hr)

Solution See Table 5.4

C-9\N-HYDRO\HYD5-2.PM5

Table 5.4 Derivation of the S-curve and 2-hour unit hydrographs. (Example 5.4)

Time

S-curve 2-hr UGO (hr)

4-hr UGO

S-curve additions

(unit storms after every 4 hr = t r )

(cumec)

(cumec)

(cumec) (cumec)

– 10.5 – 21* *Slight adjustment is required to the tail of the 2-hour unit hydrograph.

Col (5): lagged S-curve is the same as col (4) but lagged by t r ′ = 2 hr.

Col (7): col (6) × r ,t = 4 hr. t ′ = 2 hr.

Col (3): No. of unit storms in succession = T/t r = 24/4 = 6, to produce a constant outflow.

. Q e = . 2 78 A 2 78 × = 630 = 437 cumec, which agrees very well with the tabulated S-curve terminal value of 436. The S-curve HYDROLOGY

additions can be written in one column, without having to write in 5 columns successively lagged by 4 hours (= t r ), as is illustrated in the example 5.5.

Plot col (4) versus col (1) to get the S-curve hydrograph, and col (7) versus col (1) to get the 2-hour unit hydrograph, as shown in Fig. 5.19.

HYDROGRAPHS

Example 5.5 The ordinates of a 4-hour unit hydrograph for a particular basin are given be- low. Determine the ordinates of the S-curve hydrograph and therefrom the ordinates of the 6- hour unit hydrograph.

Time

4-hr UGO (hr)

4-hr UGO

Solution See Table 5.5.

Table 5.5 Derivation of the 6–hour UG for the basin (Example 5.5)

Time 4-hour

lagged 2 S-curve 6-hr (hr)

S-curve 1 S-curve

difference UGO (cumec)

(cumec) (cumec)

–9 – 6.0 1—Start the operation shown with 0 cumec after t r = 4 hr.

2—Lag the S–curve ordinates by t r ′ = 6 hr. Plot col (4) vs. col (1) to get S-curve hydrograph and col (7) vs. col (1) to get 6-hr unit hydrograph

as shown in Fig. 5.19.

HYDROLOGY