Some electric power companies use water to store energy. Water is pumped from a low reservoir to a high reservoir. To store the energy produced in 1.0 hour by a 180-MW electric power plant, how many cubic meters of water will have to be pumped from the lower to the upper reservoir? Assume the upper reservoir is an average of 380 m above the lower one. Water has a mass of \(1.00 \times {10^3}\;{\rm{kg}}\) for every \(1.0\;{{\rm{m}}^3}\).

The rate at which work is done by a device is termed the power of that device. Work done by the device is generally equal to the change in mechanical energy of the system.Thus,

\(P = \frac{{{\rm{Work}}}}{{{\rm{Time}}}} = \frac{{{\rm{Change in energy}}}}{{{\rm{Time}}}}\) .

Power of electric power plant is \(P = 180\;{\rm{MW}} = 180 \times {10^6}\;{\rm{W}}\).

The average height to which water is raised from lower reservoir to upper reservoir is h = 380 m.

The density of water is calculated as follows:

\(\begin{aligned}{c}\rho = \frac{{1.00 \times {{10}^3}\;{\rm{kg}}}}{{1.0\;{{\rm{m}}^3}}}\\ = 1.0 \times {10^3}\;{\rm{kg/}}{{\rm{m}}^{\rm{3}}}\end{aligned}\)

Time taken:

\(\begin{aligned}{c}t = 1\;{\rm{h}}\\ = \;\left( {1\;{\rm{h}}} \right) \times \left( {\frac{{60\;\min }}{{1\;{\rm{h}}}}} \right) \times \left( {\frac{{60\;{\rm{s}}}}{{1\;\min }}} \right)\\ = 3600\;{\rm{s}}\end{aligned}\)

Let m be the mass and V be the volume of water raised by a height hin timet. So, the density of water can be written as follows:

\(\begin{aligned}{l}\rho = \frac{m}{V}\\m = \rho V\end{aligned}\)

When water is pumped from the lower reservoir to the upper reservoir, the gravitational potential energy of the water gets increased.

Thus, the increase in gravitational potential energy of the water is as follows:

\(\begin{aligned}{c}\Delta PE = mgh\\ = \rho ghV\end{aligned}\)

The power of an electric power plant is equal to the rate of increase in gravitational potential energy of the water, i.e.,

\(\begin{aligned}{l}P = \frac{{\Delta PE}}{t}\\P = \frac{{\rho ghV}}{t}\end{aligned}\)

Thus, the volume of water pumped from the lower reservoir to the upper reservoir in one hour is as follows:

\(\begin{aligned}{c}V = \frac{{Pt}}{{\rho gh}}\\ = \frac{{\left( {180 \times {{10}^6}\;{\rm{W}}} \right)\left( {3600\;{\rm{s}}} \right)}}{{\left( {1.0 \times {{10}^3}\;{\rm{kg/}}{{\rm{m}}^{\rm{3}}}} \right)\left( {9.8\;{\rm{m/}}{{\rm{s}}^2}} \right)\left( {380\;{\rm{m}}} \right)}}\\ = 174.01 \times {10^3}\;{{\rm{m}}^{\rm{3}}}\\ = 1.7 \times {10^5}\;{{\rm{m}}^3}\end{aligned}\)

Thus, \(1.7 \times {10^5}\;{{\rm{m}}^3}\)of water has to be pumped from the lower reservoir to the upper reservoir to store the energy produced in 1.0 hour.