How much power is developed when a hydropower plant operates under an effective head of 20 m and a discharge of 30 m3/sec?

Question

RRB Paper · Accepted Answer

The power developed in a hydro-power plant is given by the formula

P = ρ × Q × g × H

where

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:

P = ρ × Q × g × H

= 1000 × 30 × 9.81 × 20
= 5,880,600 W
= 5.88 MW

Therefore, the correct answer is 5.88 MW.

3.86 MW