How much power is developed when a hydropower plant operates under an effective head of 20 m and a discharge of 30 m3/sec?
The power developed in a hydro-power plant is given by the formula
P = ρ × Q × g × H
H is the falling height or head
Q is the water discharge
g is the acceleration due to gravity
ρ is the density of water.
Using the given data, we have:
H = 20 m
Q = 30 m3/s
g = 9.81 m/s2
ρ = 1000 kg/m3
Substituting these values in the formula, we get:
= 1000 × 30 × 9.81 × 20
= 5,880,600 W
= 5.88 MW
Therefore, the correct answer is 5.88 MW.