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Phi-index, excess rainfall hyetograph, runoff volume

Short answer

With φ = 0.40 cm/h, excess depths in successive 2-h periods are 0, 1.4, 1.6, 0.6, 0.1, 0.9, 0 cm. Total excess = 4.60 cm, direct runoff volume = 2.3×10⁵ m³.

A catchment area of 5 km2^2 had the following rainfall pattern.

Time (h)02468101214
Cumulative rainfall (cm)00.602.805.206.607.509.209.60

If ϕ\phi-index is 0.40 cm/hr, construct the excess rainfall hyetograph and also find the volume of direct runoff.

Rainfall in each 2-hour period is the difference of cumulative rainfall; the intensity is depth/2 h. Loss rate ϕ=0.40\phi = 0.40 cm/h.

Period (h)Rain depth (cm)Intensity (cm/h)Excess rate = i - 0.40 (cm/h)Excess depth (cm)
0-20.60.30 (below ϕ\phi)0
2-42.21.10.71.4
4-62.41.20.81.6
6-81.40.70.30.6
8-100.90.450.050.1
10-121.70.850.450.9
12-140.40.20 (below ϕ\phi)0
Total9.64.6

Excess rainfall hyetograph: bars of excess depth for 2-h periods (bar heights are labelled, not drawn to scale):

 excess (cm)
 1.6 |               ####
 1.4 |        ####   ####
 0.9 |        ####   ####                 ####
 0.6 |        ####   ####   ####          ####
 0.1 |        ####   ####   ####   ####   ####
     +-------------------------------------------------
       0-2    2-4    4-6    6-8    8-10   10-12  12-14  h

(Bar heights, 2-4 h: 1.40; 4-6 h: 1.60; 6-8 h: 0.60; 8-10 h: 0.10; 10-12 h: 0.90; there is no excess in 0-2 h and 12-14 h.)

Volume of direct runoff:

V=4.60 cm×5 km2=0.046 m×5×106 m2=2.30×105 m3V = 4.60\ \text{cm}\times 5\ \text{km}^2 = 0.046\ \text{m}\times 5\times10^6\ \text{m}^2 = 2.30\times10^5\ \text{m}^3

Answer: Total excess rainfall = 4.60 cm; volume of direct runoff = 2.3 × 10⁵ m³ (0.23 million m³).

✓ Answer checked by Claude Sonnet 5.5

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