Question

A 4 h unit hydrograph of a catchment is triangular in shape with base of 80 h. The area of the catchment is 720 $k{m}^2$ . The base flow and $\varphi$ index are 30 $m^3/s$ and 1 mm/h, respectively. A storm of 4 cm occurs uniformly in 4 h over the catchment.

The peak flood discharge due to the storm is

• A. $210m^3/s$
• B. $230m^3/s$
• C. $260m^3/s$
• D. $720m^3/s$

Answer 1

The correct option is A $210m^3/s$

Catchment area,
$A=720k{m}^2=720\times {10}^6{m}^2$



Excess rainfall = 1 cm
$\text{Peak discharge}=Q_p$
Run off volume = Area under UH
$A\times 0.01=\frac{1}{2}(80\times 3600)\times Q_p$

$\Rightarrow 720\times {10}^6\times 0.01=\frac{1}{2}(80\times 3600)Q_p$

$\Rightarrow Q_p=50m^3/s$

Depth of precipitation = 4 cm
Infiltration in 4 hours = 4 x 1
= 4 mm
= 0.4 cm
Run off depth = 4 - 0.4
= 3.6 cm

peak of DRH of 4 hours
$=3.6\times \text{peak of UH of 4 hrs}$
$=3.6\times 50=180m^3/s$

Peak of flood discharge
= peak of DRH + base flow
$=180+30$
$=210m^3/s$