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Peak discharge from 6-h UH with minimum infiltration

Short answer

Rainfall excess depth = (18.3/6 − 0.25) × 6 = 16.8 cm. Peak discharge = 16.8 × 96 + 25 = 1637.8 m³/s.

A 6 h unit hydrograph of a basin has a peak ordinate of 96 m3^3/s. When the base flow in the stream is 25 m3^3/s, and when the basin has reached its minimum infiltration capacity of 2.5 mm/h, a 6 h storm with 18.3 cm of total rainfall had occurred on the basin. What is the magnitude of the peak discharge in the flood hydrograph produced by this storm?

Direct runoff = rainfall excess; the peak of the flood hydrograph = (ER depth in cm) × (UH peak) + base flow, because a 6 h storm on a 6 h UH needs no lagging.

Data

  • UH duration = storm duration = 6 h, UH peak = 96 m³/s per cm
  • Base flow = 25 m³/s
  • Infiltration is at its minimum rate, so it is taken as the φ-index = 2.5 mm/h = 0.25 cm/h

Rainfall excess

Rainfall intensity = 18.3 / 6 = 3.05 cm/h

ER=(i−ϕ) t=(3.05−0.25)×6=16.8 cm\begin{aligned} ER &= (i-\phi)\,t = (3.05 - 0.25)\times 6 \\ &= 16.8 \text{ cm} \end{aligned}

Peak discharge

Qp=ER×Up+Qb=16.8×96+25=1637.8 m3/s\begin{aligned} Q_p &= ER \times U_p + Q_b \\ &= 16.8 \times 96 + 25 \\ &= 1637.8 \text{ m}^3/\text{s} \end{aligned}

Answer: peak discharge ≈ 1637.8 m³/s (direct runoff 1612.8 m³/s + base flow 25 m³/s).

✓ Answer checked by Claude Sonnet 5.5

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