At a hydroelectric power plant, the water pressure head is at a height of 300 m and the water flow available is 100 `m^(3)s^(-1)`. If the turbine generator efficiency is 60%, estimate the electric power available from the plant `(g=9.8ms^(-2))`.

Explanation

Input power ( hydroelectric power ) is,
`P_(i) = Fv `
`= Pav` ( Where P = pressure, A = area of cross- section , v = velocity )
`= ( rho g h ) ( V.)` ( Where h = pressure head, V. = Av = volume flux )
`= ( 10^(3) xx 9.8 xx 300 ) ( 100) `
`2.94 xx10^(8)W`
Efficiency (of turbine generator),
`eta=(P_(0))/(P_(i))=("output power obtained (electrical)")/("input power supplied(hydroelectric)")`
`:. P_(0) = eta P_(i)`
`= ((600)/(100)) ( 2.94 xx 10^(8)) `
`=1.76.4 xx 10^(6)W`
`=1.76.4 MW`