To develop this problem we will calculate the volume from the given flow (volume per unit of time). This process will be accompanied by transforming the units given in the International system.
Through the density-volume-mass ratio, we will calculate the mass. Finally we will proceed to calculate the total power generated.
1 feet = 0.3048 m
[tex]\dot{V} = 84760ft^3/s[/tex]
[tex]\dot{V} = 84760ft^3/s (\frac{(0.3058m)^3}{(1ft)^3})[/tex]
[tex]\dot{V} = 2400m^3/s \text{and the Volume each second is } V= 2400m^3[/tex]
Now the mass of water falling down each second would be
[tex]\rho = \frac{m}{V} \rightarrow m = \rho V[/tex]
[tex]m = (1000kg/m^3)(2400m^3)[/tex]
[tex]m = 2.4*10^6kg[/tex]
Now Height in meters is
[tex]h = 167 ft (\frac{(0.3058m)}{(1ft)})[/tex]
[tex]h= 50.9 m[/tex]
The power produced is equivalent to the work done by gravity over a certain time. At this case that can be expressed as,
[tex]P = \frac{W}{t}\\P = \frac{F*h}{t}\\P = \frac{mgh}{t}\\P = \frac{2.4*10^6*9.8*50.9}{1}\\P = 1.197*10^9 Watts[/tex]
The power of Niagara Falls then would be 1.197GW while the power generated by the Three Gorges Dam is 22,500 MW. This indicates that the Niagara Falls, if it were a dam that covered 100% of the fluid it carries, would exceed the production of the dam in China by thousands.
