Section 4: Soil and Water Conservation Engineering
Hydrology
Subtopic: Hydrological cycle and measurement of its components
Question Statement:
A 10 ha watershed received 100 mm uniformly distributed rainfall. Land use pattern consists of 25% residential area with soil group C and curve number 82, good meadow condition in 50% of the area with soil group D and curve number 78, There is also good open space condition in 25% of the area with soil group D and curve number 80. Assuming AMC-II condition, the volume of runoff in m³ from the watershed will be
Options:
(A) 4955
(B) 5705
(C) 5755
(D) 6555
Official Answer Key:
A
Step-by-Step Solution & Explanation
SCS-CN method: S = (25400/CN) − 254 (mm); Q = (P−0.2S)²/(P+0.8S) for P>0.2S, with P = 100 mm rainfall. Runoff volume is computed area-wise and summed: • CN=82 (25% of 10 ha = 2.5 ha): S=55.71, Q=54.59 mm → Vol = 0.05459×25000 m² = 1364.8 m³ • CN=78 (50% of 10 ha = 5 ha): S=71.69, Q=46.66 mm → Vol = 0.04666×50000 m² = 2332.8 m³ • CN=80 (25% of 10 ha = 2.5 ha): S=63.5, Q=50.54 mm → Vol = 0.05054×25000 m² = 1263.5 m³ Total runoff volume = 1364.8+2332.8+1263.5 = 4961.1 m³ ≈ 4955 m³, matching option (A) (small residual difference is rounding in intermediate S/Q values).
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