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Water is pumped from a lower reservoir to a higher reservoir by a pump that provides \(20 \mathrm{kW}\) of shaft power. The free surface of the upper reservoir is \(45 \mathrm{m}\) higher than that of the lower reservoir. If the flow rate of water is measured to be \(0.03 \mathrm{m}^{3} / \mathrm{s}\), determine mechanical power that is converted to thermal energy during this process due to frictional effects.

Short Answer

Expert verified
Answer: Approximately 6.7605 kW.

Step by step solution

01

Calculate the ideal power required to pump the water

We first need to find the ideal power required to pump the water from the lower reservoir to the upper reservoir. We can use the formula: Ideal power = flow rate * g * height difference Where flow rate is given as \(0.03 m^3/s\), g is the acceleration due to gravity (\(9.81 m/s^2\)), and the height difference is \(45 m\). Ideal Power = \(0.03 m^3/s * 9.81 m/s^2 * 45 m\) Ideal Power = \(13.2395 kW\)
02

Calculate the power lost due to friction

We can now calculate the power lost due to friction using the following equation: Power lost due to friction = Total pump power - Ideal power Total pump power is given as \(20 kW\). We have calculated ideal power as \(13.2395 kW\) in step 1. Power lost due to friction = \(20 kW - 13.2395 kW\) Power lost due to friction = \(6.7605 kW\) The mechanical power that is converted to thermal energy due to frictional effects is approximately \(6.7605 kW\).

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Most popular questions from this chapter

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 \(2 \mathrm{m}^{3} / \mathrm{s}\) 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.

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