i) Draw the hydrograph and find the mean flow

ii) Draw the flow duration curve.

iii) Find the power in MW available at mean flow if the head available is 80m and overall efficiency of generation is 85%. Take each month of 30 Days.

Mean discharge flow =$\frac{ (Total flow for time duration)}{(Total number of months)}$

= $\frac{(580 × 10^6)}{12}$

= 48× 1$0^6$ cu.m/ month

Average flow = $\frac{(48 ×10^6)}{(30 ×24 ×3600)}$ = 18.6471 cu. m /sec

Average power generated at side in kW = $\frac{(ρgHQ Ƞ_g)}{1000}$

= $\frac{(1000 ×9.81 × 80 × 18.6471 ×0.85 )}{1000}$

= 12439.12 kW = 12.43912 MW