how many kilograms of water must go through the generators every second to produce a megawatt of power (10^6)?

Question

The height of the falls is about 51 meters. Assuming that the energy conversion is highly efficient, approximately how much energy is obtained from one kilogram of falling water?

Answer 1

In this problem we assume that gravitational Potential Energy (PE) possessed by the water at the top of the waterfall is completely converted into ‘output’ energy.

Gravitational Potential Energy can be found as,
PE = mgh
Where m is the mass of the water, g is the gravitational acceleration of the water, and h is height of the water above the ‘ground’.

When the water falls from a height, h, to the ground, it looses its Potential Energy. This change in Potential Energy is,
ΔPE = mgΔh
For a constant mass and gravitational acceleration.

Since we can assume the gravitational acceleration to be constant (g = 9.81 m/s^2) and we know the distance the water falls down,
We can solve for the mass of water needed to fall per second in order to get a desired power output.

Remember that power = energy per unit time
Power = Energy / time

In this case,
Power = ΔPE / Δt

We want a power output of 1 Megawatt = 1 E6 Watts = 1 E6 Joules per second

(1 E6 Joules per second) = m(9.81 m/s^2) * (51 meters) / (1 second)
Solving for m,
m = (1 E6 Joules per second) * (1 second) / (9.81 m/s^2) * (51 meters)
m = 1998.8 kg

About 2000 kg of water is required to fall a distance of 51 meters in order to generate 1 Megawatt of power (assuming 100% conversion efficiency).


Similarly, to find the energy derived from 1 kg of water falling 51 meters,
ΔPE = (1 kg) * (9.81 m/s^2) * (51 m)
ΔPE = 500.31 Joules