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Total rainfall, net runoff, W-index of 140-min storm

Short answer

Convert each 20-minute rate to depth by dividing by 3; total rainfall is 10.0 mm. Rainfall excess over phi = 3 mm/h gives net runoff 4.2 mm, and W-index = (10.0 - 4.2 - 0.8)/(140/60) = 2.14 mm/h.

In a 140-minute storm, the following rates of rainfall were observed in successive 20-minute intervals: 3.0, 3.0, 9.0, 6.6, 1.2, 1.2 and 6.0 mm/hr. Assume the ϕ\phi index value as 3.0 mm/hr and an initial loss of 0.8 mm. Determine the total rainfall, net runoff and W-index for the storm.

Each interval is 20 min = 1/3 h, so depth in an interval = rate ×13\times \frac{1}{3}.

IntervalRate (mm/h)Rain (mm)ϕ\phi loss (mm)Excess (mm)
13.01.0001.0000
23.01.0001.0000
39.03.0001.0002.000
46.62.2001.0001.200
51.20.4000.4000
61.20.4000.4000
76.02.0001.0001.000

The initial loss (0.8 mm) is absorbed in the first interval, where the rain is only 1.0 mm and is already at the ϕ\phi rate, so it does not change the excess rainfall.

Total rainfall

P=3.0+3.0+9.0+6.6+1.2+1.2+6.03=30.03=10.0 mmP = \frac{3.0+3.0+9.0+6.6+1.2+1.2+6.0}{3} = \frac{30.0}{3} = 10.0\ \text{mm}

Net runoff

R=2.0+1.2+1.0=4.2 mmR = 2.0 + 1.2 + 1.0 = 4.2\ \text{mm}

W-index

W=P−R−Iat=10.0−4.2−0.8140/60=5.02.333=2.14 mm/hW = \frac{P - R - I_a}{t} = \frac{10.0 - 4.2 - 0.8}{140/60} = \frac{5.0}{2.333} = 2.14\ \text{mm/h}

Answer: Total rainfall = 10.0 mm; net runoff = 4.2 mm; W-index = 2.14 mm/h.

✓ Answer checked by Claude Sonnet 5.5

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