Construct 3-h UH and flood hydrograph, 524.88 ha

Step 1: Shape of the 3-h UH

Given: $t_b$ = 30 h, peak at 9 h, half-peak at 3 h and 15 h. So the UH is a straight-line polygon through (0, 0), (3, ½$Q_p$), (9, $Q_p$), (15, ½$Q_p$) and (30, 0).

 Q
 Qp   |        /\
      |       /  \
 Qp/2 |    /       \_
      |   /           \__
    0 +--+-----+-----+------+---- t (h)
      0  3     9    15      30

Step 2: Peak ordinate from the area

A UH has 1 cm of runoff over the catchment:

$$V = 524.88\text{ ha}\times 10^4 \times 0.01 \text{ m} = 52\,488 \text{ m}^3$$

Area under the polygon (in units of $Q_p\cdot$h):

$$\tfrac12(3)(0.5) + \tfrac12(0.5+1)(6) + \tfrac12(1+0.5)(6) + \tfrac12(15)(0.5) = 0.75+4.5+4.5+3.75 = 13.5$$ $$Q_p \times 13.5 \times 3600 = 52\,488 \ \Rightarrow\ Q_p = 1.08 \text{ m}^3/\text{s}$$

Step 3: 3-h UH ordinates (3-h interval)

Step 4: Rainfall excess (3-h blocks)

Cumulative rain: 12 cm at 6 h, still 12 cm at 9 h (no rain 6 – 9 h), 23 cm at 12 h. Rain is taken uniform inside each interval.

Step 5: Flood hydrograph

$$Q(t) = 5\,U(t) + 5\,U(t-3) + 0\cdot U(t-6) + 10\,U(t-9) + 20$$

Answer: peak flood ≈ 35.66 m³/s at t = 18 h (the flood hydrograph returns to the base flow of 20 m³/s at t = 39 h).