At a hydroelectric power plant, the water pressure head is at a heightof 300 m and the water flow available is 100 m3s–1. If the turbinegenerator efficiency is 60%, estimate the electric power availablefrom the plant (g = 9.8 ms–2 ).
Given: The height of water pressure head is 300 m , the volume of water flowing per second is 100 m 3 / s , the efficiency of the turbine generator is 60 % and the acceleration due to gravity is 9.8 m / s 2 .
The electric power is given as,
P = η × h ρ g V
Where, the height of water pressure head is h , the volume of water flow per second is V , the efficiency of turbine generator is η , the acceleration due to gravity is g and the density of water is ρ .
By substituting the given values in the above equation, we get
P = 0.6 × 300 × 10 3 × 9.8 × 100 = 176.4 × 10 6 = 176.4 MW
Thus, the electric power available from the plant is 176.4 MW .
Given: The height of water pressure head is
The electric power is given as,
Where, the height of water pressure head is
By substituting the given values in the above equation, we get
Thus, the electric power available from the plant is
