Answer:
The right answer is "5.105×10⁸ m³/sec".
Explanation:
The given values are:
Catchment area,
A = 3 km²
Length to watershed,
L = 1100 m
Average watershed slope,
S = 0.9% i.e., 0.009
Curve number,
CN = 60
Rainfall duration,
D = 20 min
Let,
Now,
⇒ [tex]t_1=\frac{L^{0.8}\times (\frac{1000}{CN} -9)^{0.7}}{19000S^{0.5}}[/tex]
On substituting the values, we get
⇒ [tex]t_1=\frac{1100^{0.8}\times (\frac{1000}{60} -9)^{0.7}}{19000\times 0.009^{0.5}}[/tex]
⇒ [tex]=0.625 \ hours[/tex]
then,
⇒ [tex]T_p=\frac{D}{2}+t_1[/tex]
⇒ [tex]=\frac{0.33}{2}+0.625[/tex]
⇒ [tex]=\frac{0.33+1.25}{2}[/tex]
⇒ [tex]=\frac{1.58}{2}[/tex]
⇒ [tex]=0.79 \ hr[/tex]
hence,
The peak flow will be:
⇒ [tex]Q_p=\frac{484A}{t_p}[/tex]
⇒ [tex]=\frac{484\times 3}{0.79}[/tex]
⇒ [tex]=\frac{1452}{0.79}[/tex]
⇒ [tex]=1837.97 \ km^3/hr[/tex]
or,
⇒ [tex]=5.105\times 10^8 \ m^3/sec[/tex]
