How is runoff found from a hyetograph with φ = 2.5 cm/h?

Sum of (i - φ)Δt over intervals where i > φ

Total rainfall minus φ x total time

Which intervals give no runoff (φ = 2.5 cm/h)?

0-30 (2.5), 45-75 (2.0) and 125-150 (1.8) min

Maximum excess rate (2.8 cm/h) x area, converted to m3/s

Maximum rainfall rate (5.3 cm/h) x area

Short answer

Total rainfall = 10.69 cm, runoff = 3.73 cm, peak discharge = 3.89 m³/s.

For a storm of 3 hours on a 50 ha catchment, the rainfall rates are as follows:

Time of rain from beginning (min)	0	30	45	75	100	125	150	180
Rainfall rate (cm/hour)	0	2.5	3.5	2.0	4.8	5.2	1.8	5.3

If the $\phi$ index of this basin is 2.5 cm/hour, calculate total rainfall, runoff in (cm) and peak discharge.

Each rate is taken as the rainfall rate during the interval ending at the stated time. Area $= 50$ ha $= 5\times10^5$ m².

Interval (min)	Duration (h)	Rate (cm/h)	Rain (cm)	Excess rate (rate - 2.5)	Runoff (cm)
0-30	0.5	2.5	1.25	0	0
30-45	0.25	3.5	0.875	1.0	0.25
45-75	0.5	2.0	1.0	0	0
75-100	0.4167	4.8	2.0	2.3	0.958
100-125	0.4167	5.2	2.167	2.7	1.125
125-150	0.4167	1.8	0.75	0	0
150-180	0.5	5.3	2.65	2.8	1.4
Total			10.69		3.733

Total rainfall $= 10.69$ cm.

Runoff $= \sum (i - \phi)\Delta t$ for $i > \phi$ $= 0.25 + 0.958 + 1.125 + 1.40 = 3.73$ cm.

Peak discharge: the maximum excess rate is $5.3 - 2.5 = 2.8$ cm/h (last interval):

Q_{p} = \frac{0.028\ \text{m/h}\times5\times10^5\ \text{m}^2}{3600} = 3.89\ \text{m}^3/\text{s}

Answer: Total rainfall = 10.69 cm; runoff = 3.73 cm; peak discharge = 3.89 m³/s (for the direct runoff rate, ignoring catchment lag).

✓ Answer checked by Claude Sonnet 5.5