Question

The ordinates of a 2 h unit hydrograph at 1 h intervals starting from time = 0 are 0, 3, 8, 6, 3, 2, and 0 $m^3/s$. Use trapezoidal rule for numerical integration, if required.

A storm of 6.6 cm occurs uniformly over the catchment in 3 h. if $\varphi -index$ is equal to 2 mm/ h hand base flow is 5 $m^3/s,$ what is the peak flow due to the storm?

• A. $41.0m^3/s$
• B. $43.4m^3/s$
• C. $53.0m^3/s$
• D. $56.2m^3/s$

Answer 1

The correct option is A $41.0m^3/s$

Duration of storm is diffrent from duration of unit hydrograph (UH)

so, we have to convert 2h UH into 3 h UH (duration of UH equal to storm duration)

Storm duration
= m x duration of UH
$3h=m\times 2h$

As m is not an integer so using s curve method


$\text{Peak of 3h UH}=6m^3/s$

Effective rainfall of storm
$=6.6-3\times 0.2=6cm$

Peak flow due to storm

$=6\times \text{peak of 3h UH}+\text{base flow}$
$=6\times 6+5=41m^3/s$