Direct runoff from a flood over a basin of area 50 hectare was observed as $25000 m^{3}$ during a storm. The equivalent depth of rainfall over the basin was 97 mm. The time distribution of the rainfall during the storm is given below. Determine the $ϕ - i n d e x$ for the storm.
$\begin{matrix} Time(hr.) & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 Rainfall(mm) & 0.0 & 6.0 & 11.0 & 34.0 & 28.0 & 12.0 & 6.0 & 0.0 \end{matrix}$
8.75

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Solution

The correct option is A 8.75

$Direct runoff volume = 25000 m^{3}$

$Area of basin = 50 h a$

$Direct runoff depth = \frac{V o l u m e}{A r e a}$

$= \frac{25000}{50 \times 10^{4}} = 50 m m$

Total rainfall depth = 97 mm

Infiltration = 97- 50
= 47 mm

Time of rainfall = 6 hr

Considering the whole period of rainfall to be the rainfall excess

$ϕ - i n d e x = \frac{47}{6} = 7.83 m m / h r$

Since the time period (0 to 1 hr & 5 to 6 hrs) are not the period of rainfall excess, these periods are not considered in calculation of $ϕ - i n d e x$

Again,
$ϕ - i n d e x = \frac{47 - 6 \times 1 - 6 \times 1}{6 - 1 - 1}$

$ϕ - i n d e x = 8.75 m m / h r$

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ϕ - i n d e x

\begin{matrix} Time from & Incremental rainfall start (hour) & in each hour (cm) 1 & 0.4 2 & 0.8 3 & 1.2 4 & 0.9 5 & 1.8 6 & 1.0 7 & 0.3 8 & 0.8 \end{matrix}

\begin{matrix} Time from & Incremental start(h) & Rainfall in cm 1 & 1.0 2 & 2.0 3 & 2.8 4 & 3.3 5 & 2.5 6 & 1.8 7 & 0.6 \end{matrix}

ϕ - i n d e x

k m^{2}

Time from start of storm (h)	0	2	4	6	8
Accumulated rainfall (cm)	0	0.5	3.0	4.1	6.0

ϕ

600 k m^{2}

ϕ - i n d e x

333 k m^{2},

20 \times 10^{6} m^{3}

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