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 head, h = 300 m
rate of water flow, V/t = 100 `m^(3) s^(-1)` (V is the volume and t is the time)
Hydroelectirc power,
p = pressure `xx` Volume second = PV/t
Also, `" " ` P = h `rho`g
` therefore " " P = h rho g xx V = h rho g `V
Given, efficiency of generator = 60% = 0.6
Efficiency= `("output Power ")/("Input") = (P.)/(p)`
Output power
`p. = 0.6 xx `p
` = 0.6 xx 300 xx 10^(3) xx 9.8 xx 100`
= 176.4 MW