Flood hydrograph from 4-h UH, three 2-hr rainfalls

Approach. The rain falls in three 2-h periods, but the given UH is for a 4-h rain. A 4-h UH cannot be used directly for 2-h excess periods, so first derive the 2-h UH by the S-curve method, then apply the three rainfall excesses by superposition.

Step 1: Rainfall excess per 2-h period

Loss per period = φ × 2 h = 1.25 × 2 = 2.5 cm.

Step 2: 2-h UH from the 4-h UH (S-curve)

The S-curve is $S(t)=U_4(t)+S(t-4)$, using the ordinates at 2-h steps. The 2-h UH is

$$U_2(t)=\frac{4}{2}\,[S(t)-S(t-2)]$$

The two interleaved series (t = 0, 4, 8, … and t = 2, 6, 10, …) of the plain sum settle at 480 and 456 m³/s, because the given table is not perfectly consistent. The S-curve therefore zig-zags, and subtracting it directly gives erratic (even negative) 2-h ordinates. So a smooth S-curve is drawn through the points. It must level off at $\sum U_4/2 = 936/2 = 468$ m³/s.

Check: Σ(2-h UH) = 936 m³/s = Σ(4-h UH), so both UHs carry the same runoff volume of 1 cm.

Step 3: Convolution with the 2-h UH

$$Q(t) = 0.70\,U_2(t) + 0.50\,U_2(t-2) + 2.50\,U_2(t-4) + 40$$

Check: Σ(Q − 40) = 3463 m³/s = 3.7 cm (total excess) × 936 m³/s, so the volume is conserved.

Answer: peak flood ≈ 678 m³/s at t = 10 h; the flow returns to the base flow of 40 m³/s at t = 28 h. The last column of the table is the flood hydrograph. The ordinates depend slightly on how the smooth S-curve is drawn, so they are approximate.