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

176.4

Explanation

Here, height of water, h = 300m
Rate of volume flow of water, `(dV)/(dt) = 100m^2//s`
Turbine-generator efficiency, `eta = 60% = 0.6`
PE of water of mass m at height h, i.e., W = mgh
Thus, hydroelectric power,
`P_h=(dW)/(dt)=d/(dt)(mgh)`
`=d/(dt) (Vrhogh)" "(as m = Vrho)`
`= rho gh xx(dV)/(dt)`
`(1000 kg//m^3) (9.8m//s^2) (300m) (100m^3//s)`
`= 29.4 xx10^7 W`
Electric power `= eta xx P_h = 0.6xx(29.4 xx10^7W)`
` = 176.4 xx10^6W = 176.4 MW.`