Energy may be stored by pumping water to a high reservoir when demand is low and then releasing it to drive turbines (Fig. 20–15) during peak demand. Suppose water is pumped to a lake 105 m above the turbines at a rate of 1.00 x 105 kg/s for 10.0 h at night. If all this energy is released during a 14-h day, at 75% efficiency, what is the average power output?
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9. Work & Energy
Power
Problem 66a
Textbook Question
(II) Water is stored in an artificial lake created by a dam (Fig. 20–22). The water depth is 48 m at the dam, and a steady flow rate of 35 m³/s is maintained through hydroelectric turbines installed near the base of the dam. How much electrical power can be produced?
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Determine the potential energy of the water at the top of the dam. The potential energy per unit mass is given by the formula: , where is the mass of the water, is the acceleration due to gravity (9.8 m/s²), and is the height of the water (48 m).
Relate the mass flow rate to the volume flow rate. The mass flow rate is given by , where is the density of water (approximately 1000 kg/m³) and is the volume flow rate (35 m³/s).
Calculate the power available from the potential energy of the water. Power is the rate of energy transfer, so it can be expressed as , where is the mass flow rate, is the acceleration due to gravity, and is the height of the water.
Account for the efficiency of the hydroelectric turbines. If the turbines are not 100% efficient, the actual electrical power output will be less than the calculated power. Let the efficiency of the turbines be (a decimal value between 0 and 1). The electrical power output is then given by .
Substitute the known values into the equations. Use kg/m³, m³/s, m/s², and m. If the efficiency is provided, include it in the calculation; otherwise, assume ideal conditions ().

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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Hydrostatic Pressure
Hydrostatic pressure is the pressure exerted by a fluid at equilibrium due to the force of gravity. It increases with depth and is calculated using the formula P = ρgh, where P is the pressure, ρ is the fluid density, g is the acceleration due to gravity, and h is the depth of the fluid. In this scenario, the depth of the water at the dam contributes to the pressure that drives the turbines.
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Flow Rate
Flow rate is the volume of fluid that passes through a given surface per unit time, typically measured in cubic meters per second (m³/s). In this case, the steady flow rate of 35 m³/s indicates how much water is available to generate electricity. The flow rate is crucial for determining the potential energy converted into electrical power by the turbines.
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Power Generation
Power generation in hydroelectric systems involves converting the potential energy of water into electrical energy. The power (P) produced can be calculated using the formula P = ηρgQh, where η is the efficiency of the turbines, ρ is the water density, g is the acceleration due to gravity, Q is the flow rate, and h is the height of the water column. Understanding this relationship is essential for calculating the electrical power output from the dam.
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