The mass of water that must be raised is [tex]5.25\cdot 10^7 kg[/tex]
Explanation:
Since the process is 70% efficiency, the power in output to the turbine can be written as
[tex]P_{out} = 0.70 P_{in}[/tex]
where [tex]P_{in}[/tex] is the power in input.
The power in input can be written as
[tex]P_{in} = \frac{W}{t}[/tex]
where
W is the work done in lifting the water
t = 3 h = 10,800 s is the time elapsed
The work done in lifting the water is given by
[tex]W=mgh[/tex]
where
m is the mass of water
[tex]g=9.8 m/s^2[/tex] is the acceleration of gravity
h = 45 m is the height at which the water is lifted
Combining the three equations together, we get:
[tex]P_{out} = 0.70 \frac{mgh}{t}[/tex]
Where
[tex]P_{out} = 150 MW = 150\cdot 10^6 W[/tex]
And solving for m, we find:
[tex]m=\frac{Pt}{0.70gh}=\frac{(1.50\cdot 10^6)(10800)}{(0.70)(9.8)(45)}=5.25\cdot 10^7 kg[/tex]
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