Question

An isolated 4 hour storm occurred over a catchment of 12 ha area is as follow

$\begin{array}{cc}\text{Time}& \text{Intensity (cm/hr)}\\ \text{10.00 AM to 10:30 AM}& 0.9\\ \text{10:30 AM to 11:00 AM}& 1.2\\ \text{11:00 AM to 12:00 PM}& 1.6\\ \text{12:00 PM to 12:45 PM}& 2.2\\ \text{12.45 PM to 1.15 PM}& 1.8\\ \text{1.15 PM to 2.00 PM}& 0.7\end{array}$

The $\varphi -index$ for the catchment is 1.1 cm/hr. The estimated runoff volume $\left(in{m}^3\right)$ from the catchment due to above storm is

• A. 1704
• B. 2070
• C. 4640
• D. 4800

Answer 1

The correct option is B 2070

Only the storm between 10:30 AM to 1:15 PM will contribute to the runoff

$\therefore \text{Runoff}=$

$(1.2-1.1)\times \frac{30}{60}+(1.6-1.1)+(2.2-1.1)\times \frac{45}{60}+(1.8-1.1)\times \frac{30}{60}$

$=1.725cm$

Runoff volume

$=(1.725\times {10}^{-2})\times (12\times {10}^4)m^3$

$=2070m^3$