Question

10 5 The total rainfall hyetograph of a storm, which took place over a basin is given below. Ordinates of Total Storm Hydrograph (TSH) obtained for the basin is given in the table below. The baseflow is constant and is 3 m/s. If the depth of excess rainfall is 6 cm, determine: a) depth of direct runoff, i (mm/hr) b) depth of infiltrated water, c) -index, d) equation of surface runoff hydrograph in terms of UH3 and lag times. e) UH3 ordinates of this basin, f) area of the basin. t(hr) 0 3 9 12 TSH (hr) (m /s) 3 3 6 6 13 9 21 12

Answer 1

ilmm/hr) Given, 25 Base flow = 3m3/s Depth of excess Rainfall = 6 cm 10 Ilho) and WikiT, 0 3 9 12 Area of thyetograph - Total precipitation = (1013) + 2516) +513) = 195mm = ln-index = 195-60 OO = 11-25 mmmm 12 . Now, 190 - index = Pe -R 195-60- 110+5)(3) to 6 - 15 mm lho a(0-index > w-index) [: 8-index = TO - Total inflitration during excess Rainfall Time period of rainfall excess / period Givera, R=6um

(d) t TSH URH DRH = ISH- Base flow IDRHY (m/s) • 0.5 ๆ - õ Į Į õ ö wo 167 3 3:17 0.33 ส o - า 92 - 1. 67 . 0 a L * 0.33 * M UHz coordinate of basin Septh of Direct rumoff = 125-15)16) = 60 mm ar Depth of Infiltred water = Total Precijatation - Aepth Direct Runoff s 195-60 = 125mmib, Total volume of Direct Remoff = [0+0+2 (3+10+18+19+14+10+6+2)7 (3)13600) - 22110,800) m3 - 88.56 haam

a Log time = Delay time between maximum rainfall amount & peak discharge = 12-6 =6hr n Direct Runoff hydrograph diego des surface Runnioff hydrograph - Storm Hly clorograph - Base flow e UN are calculated in * UHs ordinates of this basin - the table as ! DRH Unit tey dowograph = Rainfall excess slow, - of basin = I Area Total volume of direct runoff Rainfall excess - 188.562(100) 60 ha =147-6