Question

The peak discharge, by using Snyder's method of unit hydrograph of standard duration (hour)having catchment area, regional constant and basin lag as 200 $k{m}^2$, 0.6 and 6 hour respectively, will be

• A. $51.28m^3/sec$
• B. $55.6m^3/sec$
• C. $60.75m^3/sec$
  
• D. $64.55m^3/sec$

Answer 1

The correct option is B $55.6m^3/sec$

Peak discharge,
$Q_{PS}=\frac{2.78C_{p}A}{t_p}$
where,
$C_p=0.6$
$A=200k{m}^2$
$t_p=6hours$
$Q_p=\frac{2.78\times 0.6\times 200}{6}=55.60m^3/s$