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Peak Flow from a Unit Hydrograph

Hardunit-hydrographsuperpositionrunoff
A 1-hour, 1-cm unit hydrograph has discharge ordinates of 0, 2, 5, 3 and 0 m³/s at t = 0, 1, 2, 3 and 4 h (zero afterward). Effective rainfall is 1.0 cm during hour 1 (0–1 h) and 0.5 cm during hour 2 (1–2 h). Ignore baseflow. What is the peak direct-runoff discharge?
  1. A7.5 m³/s
  2. B6.0 m³/s
  3. C6.5 m³/s
  4. D5.0 m³/s
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Answer: B. 6.0 m³/s

Superposition: Q(t) = 1.0·U(t) + 0.5·U(t − 1 h). The first response is [0, 2, 5, 3, 0, 0] at t = 0–5 h. The second is 0.5 × U shifted one hour: [0, 0, 1, 2.5, 1.5, 0]. Sum = [0, 2, 6, 5.5, 1.5, 0], so the peak is 6 m³/s at t = 2 h. 7.5 m³/s adds both responses without shifting (1.5 × 5). 6.5 m³/s adds rainfall depths to the ordinate (5 + 1.0 + 0.5), mixing depth with discharge. 5.0 m³/s ignores the second rainfall block altogether.
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