Storm hydrograph from 6-h UH, 15 cm storm

The storm lasts 3 h while the UH is for 6 h, so the 3-h UH is derived by the S-curve method and then applied to the rainfall excess.

Step 1: Rainfall excess

Initial loss = 0.5 cm, then infiltration at φ = 1 cm/h during the 3-h storm.

$$ER = 15 - 0.5 - (1\times 3) = 11.5 \text{ cm}$$

Step 2: 3-h UH from the S-curve

The S-curve at 6-h points is the running sum of the UH ordinates (0, 6, 18, 43, 53, 61, 67, 70, 70). Draw a smooth S-curve through them, read it at 3-h intervals, lag it by 3 h, and take the difference:

$$S(t) = U_6(t) + S(t-6), \qquad U_3(t) = \frac{6}{3}\,[S(t)-S(t-3)]$$

Step 3: Storm hydrograph

$$Q(t) = 11.5\,U_3(t) + 4$$

The S-curve becomes constant at 70 m³/s (sum of the 6-h UH ordinates).

Answer: peak of the storm hydrograph ≈ 297.1 m³/s at t = 15 h; it returns to the base flow of 4 m³/s at t = 45 h.