82%ViewDraw10:26 AM Thu Dec 2InsertHomeQ Patrick @ nevada.unr.eduT16.12 For the storm drain system shown in Fig. P16.12, determine the time of concentration atpoint Cusing Figure 16.7.Figure P16.12A = 0.45 sq. miPavedA = 0.23 sq. miarea1200 ft500 ftS = 2.5%S= 7%Baresoil1000 ftV = 5.2 fps3000 ftAV = 8 fpsCD Figure 16.7 Average velocity of overland flow (from U.S. SCS 1975b).2030.5200.o o o 000Velocity (ft/s)Slope (%)2.Forest with heavy ground litter and hay meadow (overland flow)0.20.3Trash fallow or minimum tillage cultivation: contour or strip cropped and woodland (overland flow0.40.5Short grass pasture (overland flow)9'00.70.8Cultivated, straight row (overland flow)-Nearly bare and untilled (overland flow); and alluvial fans western mountain regions,9.81Grassed waterwayPaved area (sheet flow); and smali upland gullies35.8.10.15

Given:- The storm drain system is given The cross-sectional area of the bare soil area = 0.23 sq mi The cross-sectional area of the paved area = 0.45 sq mi To determine:- The time of concentration at point CFor the bare soil area, The length of the farthest point = 500 ft The given slope = S = 7% From the graph given, For slope = 7% and nearly bare area (overland flow), the average velocity will be V1 = 2.6 ft/sec So, The time of concentration at point A will be tA=LengthVelocitytA=5002.6tA=192 sec Similarly, For the paved area, The length of the farthest point = 1200 ft The given slope = S = 2.5% From the graph given, For slope = 2.5% and paved area (sheet flow), the average velocity will be V2 = 3.1 ft/sec So, The time of concentration at point B will be tB=LengthVelocitytB=12003.1tB=387 secThe time taken from A to B will be Length = 1000 ft Velocity = 5.2 ft/sec tAB=LengthVelocitytAB=10005.2tAB=192 sec Similarly, The time taken from B to C will be Length = 3000 ft Velocity = 8 ft/sec tBC=LengthVelocitytBC=30008tBC=375 sec Now, The time of concentration at point C from the bare soil area will be tC = tA + tAB + tBC Putting all the values, tC = 192+192+375 tC = 759 sec Also, The time of concentration at point C from the paved area will be tC = tB + tBC Putting all the values, tC = 387+375 tC = 762 sec Taking the larger value Hence, The time of concentration at point C = 762 sec