Question

Each second, 1200 m3 of water passes over a waterfall 100 m high. Assuming that three-fourths of the kinetic energy gained by the water in falling is converted to electrical energy by a hydroelectric generator, what is the power output of the generator? (Note that 1 m3 of water has a mass of 1000 kg).

Answer 1

`P =90* 10^7\ W`

Step-by-step explanation:

Given that

Volume flow rate ,Q= 1200 m³ each seconds

So we can say that

Q= 1200 m³ /s

From energy conservation

(Potential energy +kinetic energy ) at top = (Potential energy +kinetic energy ) at bottom

U₁+ KE₁ = U₂+ KE₂

ρ Q g h + 0 = 0 + KE₂

KE₂= ρ Q g h

For water ,ρ = 1000 kg/m³

KE₂=1000 x 1200 x 10 x 100 W ( take g= 10 m/s²)

KE₂=120 x 10⁷ W

Only three forth of the energy converted into electrical energy

Therefore power P

`P=(3)/(4)* KE_2`

`P=(3)/(4)* 120* 10^7\ W`

`P =90* 10^7\ W`