Water drops over $49$ $49$ m high Niagara Falls at the rate of $6.0 \cdot 10^{6}$ $6.0 * 10^6$ kg/s. If all the energy of the falling water could be harnessed by a hydroelectric power plant, what would be the plant's power output?

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Answer 1 · 2015-11-13T01:49:29

$2881.2 MW$ $2881.2"MW"$

The potential energy is given by:

$P E = m g h$ $PE=mgh$

In 1 second the potential energy of $6 \times 10^{6} kg$ $6xx10^(6)"kg"$ of water is:

$6 \times 10^{6} \times 9.8 \times 49 = 2881.2 \times 10^{6} J$ $6xx10^(6)xx9.8xx 49=2881.2xx10^(6)"J"$

This means that this amount of energy is being produced every second.

This is what power output refers to. 1 Watt = 1 Joule per second.

So the plant's power output is :

$2881.2 \times 10^{6} W$ $2881.2xx10^(6)"W"$

or :

$2881.2 MW$ $2881.2"MW"$

Water drops over 4949 m high Niagara Falls at the rate of 6.0⋅1066.0 * 10^6 kg/s. If all the energy of the falling water could be harnessed by a hydroelectric power plant, what would be the plant's power output?