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 = 9.8 ms^(-2))`.

Explanation

Given, height of water h = 300 m
Rate of flow of water `V = 100 m^(3)//s`
efficiency `eta = 60%`
`g = 9.8 m//s^(2)`.
As we know that input power is required to raised the water up to height
h = 300 m
Power `= (mgh)/(t)`
`= ("volume"xx"Density"xx g xx h)/(t)`
`P_("in")=100xx100xx9.8xx300`
`= 2.94xx10^(8)W`
Suppose, the power output is `P_("out")` which is equal to the power available from the plant. The efficiency of generator
`eta = (P_("out"))/(P_("in"))`
`(60)/(100)=(P_("out"))/(2.94xx10^(8))`
`P_("out")=176.4MW=1764xx10^(5)W`