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

Hydroelectric power = pressure x rate of flow of water
` = (h rho g) (av) = 300 xx 9.8 xx 10^3 xx (100)`
Electric power ` = 60/100 xx 300 xx 9.8 xx 10^3 xx 100 = 176 MW`