In a hydroelectric power plant, 65 m3/s of water flows

Chapter 2, Problem 122P

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The demand for electric power is usually much higher during the day than it is at night, and utility companies often sell power at night at much lower prices to encourage consumers to use the available power generation capacity and to avoid building new expensive power plants that will be used only a short time during peak periods. Utilities are also willing to purchase power produced during the day from private parties at a high price.

Suppose a utility company is selling electric power for \(\$ 0.05 / \mathrm{kWh}\) at night and is willing to pay \(\$ 0.12 / \mathrm{kWh}\) for power produced during the day. To take advantage of this opportunity, an entrepreneur is considering building a large reservoir \(40 \mathrm{~m}\) above the lake level, pumping water from the lake to the reservoir at night using cheap power, and letting the water flow from the reservoir back to the lake during the day, producing power as the pump-motor operates as a turbine-generator during reverse flow. Preliminary analysis shows that a water flow rate of \(2 \mathrm{~m}^{2} / \mathrm{s}\) can be used in either direction. The combined pump-motor and turbine-generator efficiencies are expected to be 75 percent each. Disregarding the frictional losses in piping and assuming the system operates for \(10 \mathrm{~h}\) each in the pump and turbine modes during a typical day, determine the potential revenue this pump-turbine system can generate per year.

FIGURE P2–123

Equation Transcription:

Text Transcription:

$0.05/kWh

$0.12/kWh

2 m^3/s

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