Question

2. A 10-minute unit hydrograph (UH) of a catchment is given in the table below. What is the catchment area? (5 marks) Time step (10 minutes) 1 2 3 4 Discharge (mº/{s.cm)) 8 18 30 10 For a storm partern given in the table below, determine and plot runoff hydrograph at a 5-minute interval using the above 10-minute UH. Determine the time of concentration of the catchment. Assume the o index value for the catchment to be 12 mm/hr, and the baseflow to be a constant of 2.5 m/s. (Hint: Use the linear interpolation method to convert a 10-minute UH from a 10-minute time interval to a 5-minute time interval.] (15 marks) Time step (5 minutes) 1 2 3 4 5 6 7 Rainfall depth (mm) 0.5 6.0 6.0 1.0 3.5 3.5 3.5

Answer 1

10 mm un 1 2. 3 4 Time step 10 min discharge mileecl cm 8 16 30 Cathshment area=2 unit hydrograph tepth= 1cm. area of an= volume [(8+18) + 4 (18730) + }(20 +10) ] x 60X60 34200 m3/cm Volume area of catchment X depth of un- AX1cm 34200 mlom A = 34200 m2' xjö? А 342x109 m² area of catchment = 34RX109 m2 on.

5min ush 1 linear inter, 8 13 18. was e tub २५ 30 20 z 4 TO 5min ay (5min) m3 reclam 8 1 2. 113 3 18 24 5 30 6 20 7 to 3 4 5 2 Time smin 13.5 3.5 6,0 1.0 3.5 142 42 Rham internity x mm)bert Rohre Kermelhor 42 ܐܬܐܩܚ16

Pindex .. 12 mm/hr for Osist 5 min no runoff (Rainfull < &index) remoff rainfall - inda) ho & also 4th 5min 5 min hydrograph smm = 0,5cm for 2,3,4th 2, 3, 4th effective depth (6- 60 2.5mm 0,25cm for 5,614 effective depth (3.5 - (15) 60

Time / UH smin Ist Rain and Brd 5thIGH 19th Ist total reth 0.5 cm istem (0) Baseflood flood hydra 1 st-ith DRM cm 8 О O 0 es o o 2 13 88.5=4 o 6.5 R.S 0 o 3 13 18 6.5 les 19 10.5 9 0 18 3.5 6.5 15.5 5 30 0 12. 9 0 2 23 asis 19.5 २० 15 12 O 32. PS 34.75 34,75 137.25 2 10 lo IIS 14.5 3.3572 2.50 6 16 0 16 250 $1.25 4.5 3.95 28.75 6 23 15 11.5 4.5 2.5 Rsis lo 5 o 7.5 18.5 25 2.5 2.5 17.5 7.5 15 7,5 P.S 5 lo 5 12 13 14 PS 12.5 flood hydrograph shows 5 min Runoff hydrograph time of concentration maym Runoff hydrograph ordinate 7 stepsie - 785 to ២ an.