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For a 265 ha watershed with a time of concentration of 120 minutes and an excess rain of 42 mm, the peak discharge using a typical SCS synthetic hydrograph is:

A. [tex]Q_p = 17.30 \, \text{m}^3/\text{s}[/tex]
B. [tex]Q_p = 173.04 \, \text{m}^3/\text{s}[/tex]
C. [tex]Q_p = 23.07 \, \text{m}^3/\text{s}[/tex]
D. [tex]Q_p = 1.73 \, \text{m}^3/\text{s}[/tex]
E. [tex]Q_p = 11.59 \, \text{m}^3/\text{s}[/tex]

Answer :

The peak discharge determined using a typical SCS synthetic hydrograph is not able to be determined.

Given data:

Area of the watershed (A) is 265ha.

Time of concentration [tex](T_c)[/tex] is 120min.

Excess rain is 42mm.

The excess rain is termed to be the runoff of the rain water which happens when the absorption of water by the soil reaches the maximum level.

The peak discharge is calculated from [tex]Q_p = (C_i *A)/T_c[/tex].

On conversion of units,

[tex]1ha =10000m^2[/tex]

[tex]1mm = 10^{-3} m[/tex]

Calculation of peak discharge:

[tex]Q_p =(C*10^{-3} *265 *10^{4})/120*60[/tex]

To find the value of C, additional information such as land use, soil type, and slope. The runoff coefficient varies depending on these factors.

Therefore, the exact value of the peak discharge is not able to be determined.

To know more about run-off, visit:

https://brainly.com/question/18800262

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