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Ch 09: Work and Kinetic Energy
Chapter 9, Problem 10

In a hydroelectric dam, water falls 25 m and then spins a turbine to generate electricity. (b) Suppose the dam is 80% efficient at converting the water's potential energy to electrical energy. How many kilograms of water must pass through the turbines each second to generate 50 MW of electricity? This is a typical value for a small hydroelectric dam.

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1
Calculate the potential energy (PE) of the water using the formula PE = mgh, where m is the mass of the water in kilograms, g is the acceleration due to gravity (approximately 9.8 m/s^2), and h is the height the water falls (25 m).
Determine the total potential energy converted to electrical energy per second by using the power output and the efficiency. Since the dam is 80% efficient, the electrical energy output per second (power) is 80% of the potential energy per second. Use the formula Power = 0.8 \times mgh.
Rearrange the formula to solve for m (mass of water per second). This can be done by isolating m on one side of the equation: m = \frac{Power}{0.8 \times g \times h}.
Substitute the values for Power (50 MW or 50,000,000 watts), g (9.8 m/s^2), and h (25 m) into the equation to find the mass of water needed per second.
Convert the mass from kilograms per second to a practical unit like tons per second if necessary, to better understand the scale of water flow through the turbines.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Potential Energy

Potential energy is the energy stored in an object due to its position in a gravitational field. In the context of a hydroelectric dam, the potential energy of water is determined by its height above the turbine. The formula for gravitational potential energy is PE = mgh, where m is mass, g is the acceleration due to gravity, and h is the height. This energy is converted into kinetic energy as the water falls, which is then used to spin the turbine.
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Efficiency

Efficiency in a system refers to the ratio of useful output energy to the input energy, often expressed as a percentage. In this case, the dam's efficiency of 80% means that 80% of the potential energy of the falling water is converted into electrical energy, while the remaining 20% is lost, typically as heat or sound. Understanding efficiency is crucial for calculating how much energy is actually available for conversion into electricity.
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Power and Energy Conversion

Power is the rate at which energy is transferred or converted, measured in watts (W), where 1 watt equals 1 joule per second. In this scenario, the dam needs to generate 50 MW (megawatts) of electrical power, which is equivalent to 50 million joules per second. To find out how much water is needed, one must relate the power output to the potential energy of the falling water, taking into account the efficiency of the system.
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