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Phi-index and rainfall excess duration from mass curve

Short answer

By trial, φ-index is 0.889 cm/h for 0.5-h intervals; rainfall excess lasts 4.5 hours, producing 3.8 cm runoff.

The mass curve of an isolated storm over a watershed is given below.

Time from start (hr)00.511.522.533.544.55
Cumulative rainfall (cm)00.61.41.92.83.75.46.277.88.2

If the storm produced a direct run off of 3.8 cm at the outlet of the watershed, estimate the ϕ\phi-index of the storm and duration of rainfall excess.

Step 1: Rainfall in each 0.5-h interval (cm)

Interval (h)0-0.50.5-11-1.51.5-22-2.52.5-33-3.53.5-44-4.54.5-5
Rain (cm)0.60.80.50.90.91.70.80.80.80.4

Total rainfall P=8.2P = 8.2 cm; direct runoff R=3.8R = 3.8 cm; total loss =4.4= 4.4 cm over 5 h.

Step 2: Trial for ϕ\phi (let ϕ′\phi' be the loss per 0.5 h) Assume only the 0.4 cm interval is below ϕ′\phi' (the other 9 intervals exceed it):

(8.2−0.4)−9ϕ′=3.8⇒ϕ′=4.09=0.444 cm per 0.5 h(8.2 - 0.4) - 9\phi' = 3.8 \Rightarrow \phi' = \frac{4.0}{9} = 0.444\ \text{cm per 0.5 h} ϕ=0.4440.5=0.889 cm/h\phi = \frac{0.444}{0.5} = 0.889\ \text{cm/h}

Step 3: Check The first nine intervals (0 to 4.5 h) each have rain of at least 0.5 cm, which is more than 0.444 cm, and the last interval (0.4 cm) is below it. Sum of excess =7.8−9×0.444=3.8= 7.8 - 9\times0.444 = 3.8 cm. Valid.

Step 4: Duration of rainfall excess Excess occurs in 9 intervals of 0.5 h, from 0 to 4.5 h:

te=9×0.5=4.5 ht_e = 9\times0.5 = 4.5\ \text{h}

Answer: ϕ\phi-index = 0.889 cm/h (about 0.89 cm/h); duration of rainfall excess = 4.5 h.

✓ Answer checked by Claude Sonnet 5.5

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