Question

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))`.

Answer 1

Here, `h = 300 m, volume//sec, V = 100 m^(3) s^(-1), eta = 60%`, Electric power = ?, `g = 9.8 m//s^(2)`
Hydroelectric power = `("work")/("time") = ("force" xx "dist.")/("time") = Force xx velocity = ("Pressure" xx "area") xx "velocity"`
But `"area" xx "veloctiy" = "Volume/sec" = V` `:.` Hydroelectric power `= P xx C`
As efficincy is 60% , therefore Power available` = (60)/(100) PV = (3)/(5) (h rho g) V`, where `rho = 10^(3) kg//m^(3)`, for wate
`:. Power = (3)/(5) xx 300 xx 10^(3) xx 9.8 xx 100 = 1.764 xx 10^(8) W = 176.4 MW`