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In the Niagara Falls hydroelectric generating plant, the energy of falling water is converted into electricity. The height of the falls is about 52 meters. Assuming that the energy conversion is highly efficient, approximately how much energy is obtained from one kilogram of falling water? Therefore, approximately how many kilograms of water must go through the generators every second to produce a megawatt of power (106 watts)?

1 Answer

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Answer:

1960.32306 kg/s

Step-by-step explanation:

m = Mass of water = 1 kg

g = Acceleration due to gravity = 9.81 m/s²

h = Height from which the water will fall

Potential Energy


PE=mgh\\\Rightarrow PE=1* 9.81* 52\\\Rightarrow PE=510.12\ J

One megawatts of power is required

So, flow rate


\dotm=(P)/(PE)\\\Rightarrow \dotm=(1* 10^6)/(510.12)\\\Rightarrow \dotm=1960.32306\ kg/s

1960.32306 kg/s is required to produce a megawatt of power

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