Question

at certain times the demand for electric energy is low and electric energy is used to pump water to a reservoir 45 m above the ground and then used to drive turbines during high demand periods. if this process is 70.0% efficient, how much water must be raised to produce 150 MW of power for 3 hours?

Answer 1

The mass of water that must be raised is $5.25\cdot 10^7 kg$

Explanation:

Since the process is 70% efficiency, the power in output to the turbine can be written as

$P_{out} = 0.70 P_{in}$

where $P_{in}$ is the power in input.

The power in input can be written as

$P_{in} = \frac{W}{t}$

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

$W=mgh$

where

m is the mass of water

$g=9.8 m/s^2$ is the acceleration of gravity

h = 45 m is the height at which the water is lifted

Combining the three equations together, we get:

$P_{out} = 0.70 \frac{mgh}{t}$

Where

$P_{out} = 150 MW = 150\cdot 10^6 W$

And solving for m, we find:

$m=\frac{Pt}{0.70gh}=\frac{(1.50\cdot 10^6)(10800)}{(0.70)(9.8)(45)}=5.25\cdot 10^7 kg$

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