Question

Consider a 3-km2 urban catchment. The main channel has a slope of 0.9% and a Manning n of 0.10. The catchment is 50% impervious. The distance along the main channel from the catchment boundary to the outlet is 1100 m. The urban catchment has an average curve number of 60. Determine the peak flow (m3 /s) of the NRCS unit hydrograph for a 20-min rainfall excess.

Answer 1

The right answer is "5.105×10⁸ m³/sec".

Step-by-step 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,

• Time form beginning of the rainfall will be "
  `t_p`".
• Lag time will be "
  `t_1`".

Now,

⇒
`t_1=(L^(0.8)* ((1000)/(CN) -9)^(0.7))/(19000S^(0.5))`

On substituting the values, we get

⇒
`t_1=(1100^(0.8)* ((1000)/(60) -9)^(0.7))/(19000* 0.009^(0.5))`

⇒
`=0.625 \ hours`

then,

⇒
`T_p=(D)/(2)+t_1`

⇒
`=(0.33)/(2)+0.625`

⇒
`=(0.33+1.25)/(2)`

⇒
`=(1.58)/(2)`

⇒
`=0.79 \ hr`

hence,

The peak flow will be:

⇒
`Q_p=(484A)/(t_p)`

⇒
`=(484* 3)/(0.79)`

⇒
`=(1452)/(0.79)`

⇒
`=1837.97 \ km^3/hr`

or,

⇒
`=5.105* 10^8 \ m^3/sec`