Question

The average flow rate in the Niagara River is 6.0 × 106 10. kg/s, and the water drops 50 m over Niagara Falls. If all this energy could be harnessed to generate hydroelec- tric power at 90% efficiency, what would be the electric power output?

Answer 1

`2.65\cdot 10^9 W`

Step-by-step explanation:

The power produced by the water falling down the Falls is

`P=(E)/(t)=(mgh)/(t)`

where

E = mgh is the potential energy of the water, with m being the mass, g the gravitational acceleration, h the height

t is the time

In this problem we have

`(m)/(t)=6.0\cdot 10^6 kg/s` us the mass flow rate

h = 50 m is the height

g = 9.8 m/s^2 is the acceleration of gravity

Substituting,

`P=(6.0\cdot 10^6 kg/s)(9.8 m/s^2)(50 m)=2.94\cdot 10^9 W`

And since the efficiency is only 90%, the power output is

`P_(out) = (0.90) (2.94\cdot 10^9 W)=2.65\cdot 10^9 W`