2-hr UH from figure: area, flood, Tc, two storms

(i) Catchment area

Ordinates of the 2-h UH read from the curve at 2-h intervals (m³/s per cm):

A UH represents 1 cm of runoff, so

$$\begin{aligned} \sum Q &= 70.83 \text{ m}^3/\text{s} \\ V &= 70.83 \times 2\times 3600 = 509,976 \text{ m}^3 \\ A &= \frac{V}{0.01\ \text{m}} = 50,997,600 \text{ m}^2 \end{aligned}$$

Answer: A ≈ 51 km² (readings from the figure are approximate).

(ii) Flood hydrograph for 1.5 cm/h for 2 h

The duration (2 h) equals the UH duration, so scale the UH by the net rain depth:

$$ER = 1.5\times 2 = 3.0 \text{ cm}, \qquad Q = 3.0\,U + 2$$

Answer: peak flood = 3.0 × 12 + 2 = 38 cumecs at t = 10 h.

(iii) Time of concentration

Time of concentration is the time from the end of the effective rain to the end of direct runoff, i.e. the time taken by water from the farthest point to reach the outlet: $t_c = t_b - D$.

$$t_c = 27 - 2 = 25 \text{ h}$$

Answer: $t_c \approx$ 25 h.

(iv) Two storms of 1 h with φ = 0.12 cm/h

Excess rain in each hour = intensity − φ:

• Storm 1: 0.47 − 0.12 = 0.35 cm
• Storm 2: 0.64 − 0.12 = 0.52 cm

The two successive 1-h storms together make 2 h (equal to the UH duration), with excess 0.35 + 0.52 = 0.87 cm. The peak occurs at the UH peak (12 cumecs):

$$Q_p = 0.87\times 12 = 10.44 \text{ cumecs}$$

Answer: peak discharge ≈ 10.44 cumecs (about 12.44 cumecs if the 2 cumecs base flow of part (ii) is added).