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

Explanation

`h=300m,(V)/(t)=100m^(3)s^(-1),eta=0.6,g=9.8ms^(-2)`
Power `=(mgh)/(t)=(Vrhogh)/(t) " "rho=m//V,m=V_(rho)`
At an efficiency of 0.6, power available will be = `(0.6Vrhogh)/(t)=0.6xx100xx10^(3)xx9.8xx300`
`=17.64xx10^(7)W`
`therefore` Power available is 176.4 MW