Question

If the normal discharge at a given state under steady uniform flow is 120 $\frac{m^3}{s}$. The rate of change of stage is 1.2 $\frac{m}{s}$ and the velocity of the flood wave 15 $\frac{m}{s}$. The channel slope is 1 in 500, then the actual unsteady flow is:

• A. 180.64$\frac{m^3}{s}$
  
• B. 207.85 $\frac{m^3}{s}$
  
• C. 110.5$\frac{m^3}{s}$
  
• D. 203.6 $\frac{m^3}{s}$
  

Answer 1

The correct option is B 207.85 $\frac{m^3}{s}$


The normal discharge under steady and actual unsteady discharge is related as
$\frac{Q_m}{Q_n}=\sqrt{1+\frac{1}{V_w{S}_0}\frac{dh}{dt}}$
Where $Q_n$= Normal discharge under steady condition.
$Q_m$= Actual discharge under steady condition.
$S_0$=Channel slope
$\frac{dh}{dt}=Rateofchangeofstage.$
$V_w$=Velocity of the flood wave
Hence, $\frac{Q_m}{120}=\sqrt{1+\frac{1}{300\times \frac{1}{500}}\times 1.2}$
$\Rightarrow \frac{Q_m}{120}=\sqrt{3}$
$Q_m=207.84\frac{m^3}{s}$