S curve hydrograph

Described S-Curve Hydrograph

1. S- CurveHydrograph Prepared by Prof. S. G. Taji Dept. of Civil Engineering S.R.E.S’s Sanjivani College of Engineering, Kopargaon

2. Type-III (B): Derivation of req. T-hr UH from Given D-hr UH (where T is not multiple integral of D) In this type, S-curve approach should be used for calculation. S-curve or S-Hydrograph A ‘S’ hydrograph is nothing but a hydrograph generated by a continuous effective rainfall occurring at an uniform rate for an indefinite period. It is called ‘S’ hydrograph because the shape of the hydrograph comes out like alphabet ‘S’ though slightly deformed. Figure shows a typical ‘S’ hydrograph. It can be derived by summation of the ordinates of an infinite series of unit hydrographs of same .unit duration spaced at the same unit duration apart and hence the name summation hydrograph. It is a curve which rises continuously in the form or shape of the letter S, till a constant discharge value i.e. equilibrium is reached.

3. The discharge of the S-curve at the time of equilibrium, A = Area of catchment in hectares I = Constant rate of effective rainfall in cm/hour If ‘A’ is taken in Sq. Km., then above eq. becomes The ordinates of S-curve can be plotted by using eq. S(t ) = U (t) + S (t-D) Where , S(t)= req. ordinate of S-curve hydrograph at time duration t. U(t ) = Ordinate of Given UH at time duration t. t = required duration (hr) D = Given duration of UH. Therefore , S(t-D) = s-curve ordinate of preceding interval

4. Q. Derive S-curve ordinates from given UH

5. Ordinates of S-curve S(t) = U (t) + S (t-D) For S (0): At t = 0, U (t) = 0, S(t-D) = 0, Thus, S (0) = 0 (D=4hr i.e. given duration of UH) For S (4): At t = 4, U (t) = 10 m3/s, S(t-D) = S(4-4) = S (0) = 0, Thus, S (4) = 10 + 0 = 10 m3/s For S (8): At t = 8, U (t) = 30, S(t-D) = S(8-4)= S (4) = 10, Thus, S (4) = 30 + 10 = 40 m3/s For S (12): At t = 12, U (t) = 25, S(t-D) = S(12-4)= S (8) = 40, Thus, S (4) = 25 + 40 = 65 m3/s ……… continue till the end. These calculation can also be perform in table as shown below .

6. Q. Determine 12 hr UH from given 4 hr UH by using S-curve method. Soln : In this case, D=given duration= 4 hr and T=req. duration = 12hr Step-1: Draw two columns next to given UH (i.e. col. 3 and col. 4). These two columns are filled consequently as follows: Initially, mark dash in first rows up to preceding interval of D hr of 3 rd col (i.e. in this case D=4hr, thus mark dash up to t=0 only) Step-2.1: Now, make addition col.2 and col. 3 and write it in col. 4. Step-2.2: Then, write that result at next row in col. 3 (as shown by arrow). Step-2.3: Again make addition col.2 and col. 3 and write it in col. 4. Step-2.4: Then again, write that result at next row in col. 3 (as shown by arrow ).

7. For ex. At time 8 hr, drag last ordinate from col.4 in col. 3 (i.e. 20 as shown by arrow), now do addition of 80 + 20 = 100 and write in col. 4. Continue this operation till the end of col. 3 and col. 4. Step-3: In col. 4, we get S-curve ordinates which are nothing but the running summation of flow. Now, lag these s-curve ordinates by T-hr as shown in col. 5 (in this case, T=12hr) Note: Always be lag your S-curve ordinate by that duration for which you are going to derive UH (for ex. If you going to plot 8 hr UH from given 24 hr UH, then in step-3 lag your S-curve ordinates by 8 hr and vice versa)

8. Step-4: Now deduct that lagged column from S-curve ordinates (i.e. col. 4 – col. 5 = col. 6) This is necessary because we have a summed valued which should be converted into single flow value. This will give us DRH of T/D cm ER (i.e. in this case, T/D = 12/4 = 3cm ER) Thus, col. 6 is DRH of T/D = 12/4 = 3 cm ER. Step-5: This is the last step, in which divide last column by ratio of T/D. (i.e. in this case, T/D = 12/4 = 3 cm) for getting req. 12 hr Uh .

9. Thank You……….