What power could be produced from Niagara Falls?

Question

Niagara Falls is about 53 m high. An estimated 6 x 106 kg of water passes over the falls every second. If the energy of the falling water were usefully employed, what power could be produced?

Answer 1

Power = Engergy / Time

Engergy produced here is potential energy = m.h.g
where: m: mass, h: height, g: gravity acceleration
If power = energy / time then Power = m.h.g/t
we know m/t = 6 x 106 kg/sec from your question (I assume this is 6x10^6 six times ten to the power 6, i.e. six millions).
h = 53 m and g = 9.8 m/s^2
Now we substitue in P = m.h.g / t
---> P = 6x10^6 * 53 * 9.8 = 3.1164 x 10^13 Watt.
If you usually round g to 10 m/s^2, the answer will be 3.18x10^9 watt.

Answer 2

if the water filled a tube 53m high the pressure at the bottom would be apx 150lb/in^2
now if a volume of water apx 1m x 1m x 6000m passes per sec which is equal to 6 x10^6 kg per sec then a 1m x 1m paddle on an arm 10meters long displaces
apx 60 cubic meters per revolution so at a speed of 100 rev/sec displaces the apx amount of water from above
now the tourque on that arm is equel to 150lb/in^2 = apx 1400 in ^2 x 150 lbs/ in^2 on a 1m x 1m paddle
or 210,000 lbs x (10m x3) =6,300,000 lbft of torque
and hp= (tor x rpm)/5252
or 6,300,000 x 6000 /5252 =7,000,000
about 7 millon hp or 760 watts= 1hp
or about 5,000,000,000 watts or 5 x 10^9 joules per sec

Answer 3

There is a power plant at the falls that produces hydroelectricity.

we saw it this past summer when we were at the falls.

there are many power producing plants along the Niagara River.

Answer 4

the speed would be

v = (2gh)^0.5 = (2*9.81*53) ^0.5 =32.6m

and E =1/2mv^2 = mgh =6*10^6 *9.81*53 =3.12 10^9J

I have supposed that the mass is 610^6 and not 6*106=636!

Answer 5

BY INSTALLING A WATER WHEEL AT THE BASE OF THE FALLS, HYDROELECTRIC POWER COULD BE PRODUCED......

Answer 6

it is in the form of kinetic energy, convert it to a mechanical energy, and later may be to electrical energy.