Question

1. Given a strip commercial development near Columbia, SC, with the following characteristics: overland flow length = 180 ft on 2.1% grade over asphalt cover (n=0.011); shallow concentrated flow length of 1000 ft on 1.8% grade (paved gutter); drainage area = 5 acres; runoff coefficient = 0.60

1. Sheet flow travel time (minutes) =                           
2. Shallow concentrated flow travel time (minutes) =                              
3. Watershed time of concentration (minutes) =                         
4. 10-year design rainfall intensity (inches/hour) =                                 
5. Runoff peak (cfs) for 10-year design rainfall =                                   

2. Given a 28-acre watershed with the following land use mix: 8 acres of SFR (38% impervious) on HSG-B soils, 5 acres of SFR (30% impervious) on HSG-C soils, and 15 acres of open space (park and playground) in fair condition on HSG-C soils. Watershed time of concentration is 21 minutes. The design return period 24-hour rainfall depth is 5.25 inches.

1. The runoff-weighted average curve number (CN) =                             
2. The area-weighted unit hydrograph peak rate factor (PRF) =                                                                 

c)    The runoff volume from a rainfall of 3.25 inches (watershed inches) =                             

d)   The runoff volume from a rainfall of 0.47 inches (watershed inches) =                             

Answer 1

Ans a) We know, time of concentration for sheet flow can be given as,

Tc = 0.007 (NL)^0.8 / [(P)^0.5 (S)^0.4)]

where, N =Manning roughness coefficient = 0.011

L = Length = 180 ft

P = 2 yr 24 hr rainfall = 4.2 in/hr as per IDF curve of Columbia,SC

S = Slope = 2.1 % or 0.021

Putting values,

=> Tc = 0.007(0.011x180)^0.8 / [(4.2)^0.5 x (0.021)^0.4)]

=> Tc = 0.027 hr or 1.66 min

Ans b) According to TR-55 ,time of concentration for shallow concentrated flow can be given as,

Tc' = L / V

where, V = Velocity in ft/s

For paved gutter flow, V = 20.328 S^0.5

=> V = 20.328 (0.018)^0.5 = 2.72 ft/s

=> Tc' = 1000 ft / 2.72 ft/s = 368 sec or 6.12 min

Ans c) Total time of concentration of watershed ,T

T= Tc + Tc'

=> T = 1.66 + 6.12 = 7.78 min

Ans d) According to IDF curve of Columbia,SC rainfall intensity for T = 7.78 min and 10 yr design rainfall , i = 6.6 in/hr

Ans e) According to Rational method, peak discharge (Qp) ,

Qp = C i A

where, C = Runoff coefficient = 0.60

i = Rainfall intensity = 6.6 in/hr

A = Drainage area = 5 acres

Putting values,

=> Qp = 0.60 x 6.6 x 5

    => Qp = 19.8 cfs