Question

The drainage area of a watershed is $50k{m}^2$. The ϕ - index is 0.5 cm/hour and the base flow at the outlet is $10m^3/s$. One hour unit hydrograph (unit depth = 1 cm) of the watershed is triangular in shape with a time base of 15 hours. The peak ordinate occurs at 5 hours.
For a storm of depth of 5.5 cm and duration of 1 hour, the peak ordinate (in $m^3/s$ ) of the hydrograph is

• A. 55.00
• B. 82.60
• C. 92.60
• D. 102.60

Answer 1

The correct option is D 102.60

Unit hydrograph is a hydrograph of direct runoff resulting from 1 cm of effective rainfall. So volume of water contained inside the unit hydrograph (i.e., area of unit hydrograph) = 1 cm x catchment area


$\frac{1}{2}\times a\times 15hrs=1cm\times A$

$\frac{1}{2}\times a\times 15\times 3600=0.01\times 50\times {10}^6$

$a=18.52m^3/s/cm\text{of effective rainfall}$

For a storm of 5.5 cm and duration of 1 hour
Effective rainfall = storm depth - $\varphi$ index
$=5.5-0.5=5cm$
Peak ordinate of hydrograph due to 5 cm effective rainfall
$=5\times \text{peak ordinate of UH + base flow}$
$=5\times 18.52+10=102.60m^3/s$