Question

An area (catchment) in the city of Ekurhuleni has been developed The catchment is divnded into two sub-catchment areas, ie the lower lying sub-catchment is sem permeable grassland with an average slope of 4 % and the second sub-catchment is an urban area, which has a distribution of land use as presented in Table 4 1 The size of the total catchment is 20 km2 while the urban area is (11 km2) Table 41 Land use of urban area Sıze 21 Sub-urban shopping centers Residential areas (houses) 11A | 2 6 Parks with sandy soils (3 % slopes) Streets, pavements 41 Calculate the runoff coefficient for the whole catchment 42 If the size of the total catchment is 20 km2, the rainfail is 800 mm and the rainfall intensity is 60 mm per hour, calculate the peak runoff 43 What will be the reduction in the peak flow if the rainfall intensity is 20 mm (2) (26)

Answer 1

Ans) Given,

Total area of catchment = 20 km²

Area of urban catchment = 11 km²

Area of lower lying catchment = 20 - 11 = 9 km²

a) Determination of Runoff coefficient (C) :

C = C₁A₁ + C₂A₂.......C_nA_n/ (A₁ + A₂....A_n)

where, C₁ , C₂........C_n are runoff coefficients of different regions respectively

A₁ , A₂.......A_n are Area of different regions respectively

For lower lying catchment area ,with grassland and 4% slope , C₁ = 0.20

For Urban area,

C₂ for sub urban shopping center = 0.50

C₃ for residential area = 0.40

C₄ for parks with sandy soil ans 3% slope = 0.12

C₅ for street,pavements = 0.75

=> C = [9(0.20) + 0.50(0.21 x 11) + 0.40(0.11 x 11) + 0.12(0.26 x 11) + 0.75(4.62)] / 20

=> C = 0.216

b) We know,

Q = C i A / 360

where, Q = Discharge ( in m³/s)

C = coefficient of runoff

i = rainfall intensity (mm/hr)

A = catchment area (in hectares)

=> Q = 0.216 x 60 x 2000/360 ( 1 km² = 100 hectares)

=> Q = 72 m³/s

c) We know,

Q = C i A / 360

=> Q = 0.216 x 20 x 2000/360

=> Q= 24 m³/s

Reduction = 72 - 24 =48 m³/s

% reduction = (48/72) x 100

= 66.67 %