Section 4: Soil and Water Conservation Engineering
Hydrology
Subtopic: Infiltration – indices and equations
Question Statement:
The ordinates of a 6-hour S-hydrograph of a catchment are given in Table below. The catchment has phi-index of 0.25 cm h⁻¹ and baseflow of 10.5 m³s⁻¹. The peak of the flood hydrograph generated from this catchment due to a storm of 45 mm received during the first 6 h in m³s⁻¹ is [Time(h): 0,6,12,18,24,30,36,42,48; Ordinate(m³/s): 0,30,90,180,252,306,342,360,360]
Options:
(A) 259.5
(B) 270.0
(C) 280.5
(D) 349.5
Official Answer Key:
C
Step-by-Step Solution & Explanation
Step 1 — derive the 6-hr Unit Hydrograph (UH) from the S-curve by lagging the S-curve by 6 h and subtracting (UH ordinate at t = S(t)−S(t−6)):
t(h): 0, 6, 12, 18, 24, 30, 36, 42, 48
UH (m³/s): 0, 30, 60, 90, 72, 54, 36, 18, 0
Peak UH ordinate = 90 m³/s (at t=18 h), corresponding to a 6-h unit (1 cm) rainfall excess.
Step 2 — rainfall excess: total rainfall = 45 mm = 4.5 cm over 6 h; φ-index loss = 0.25 cm/h × 6 h = 1.5 cm; rainfall excess = 4.5−1.5 = 3.0 cm.
Step 3 — scale the UH linearly (UH ordinates scale directly with excess rainfall depth in cm since UH is defined per 1 cm): Direct Runoff peak = 90 × 3.0 = 270 m³/s.
Step 4 — add baseflow: Total flood peak = 270 + 10.5 = 280.5 m³/s. Answer: (C).
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