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Each second, \(1200 \mathrm{~m}^{3}\) of water passes over a waterfall \(100 \mathrm{~m}\) high. Three-fourths of the kinetic energy gained by the water in falling is transferred to electrical energy by a hydroelectric generator. At what rate does the generator produce electrical energy? (The mass of \(1 \mathrm{~m}^{3}\) of water is \(1000 \mathrm{~kg}\).)

Short Answer

Expert verified
The generator produces electrical energy at a rate of 882 MW.

Step by step solution

01

Determine the mass flow rate

We know that each second, the volume of water flowing is \( 1200 \, \text{m}^3 \). Since the mass of \( 1 \, \text{m}^3 \) of water is \( 1000 \, \text{kg} \), the mass flow rate \( \dot{m} \) is calculated as follows: \[ \dot{m} = 1200 \, \text{m}^3/s \times 1000 \, \text{kg/m}^3 = 1,200,000 \, \text{kg/s} \]\.
02

Calculate gravitational potential energy change per second

The gravitational potential energy \( U \) gained by the water when it falls is given by the formula \( U = mgh \), where \( m \) is the mass, \( g \) is the acceleration due to gravity \( 9.8 \, \text{m/s}^2 \), and \( h \) is the height. Therefore, the energy per second is: \[ U = 1,200,000 \, \text{kg/s} \times 9.8 \, \text{m/s}^2 \times 100 \, \text{m} = 1.176 \times 10^9 \, \text{J/s} \] which simplifies to \( 1.176 \times 10^9 \, \text{W} \) or \( 1.176 \text{GW} \).
03

Calculate the electrical energy generated

Since three-fourths of the kinetic energy is converted to electrical energy, the rate at which electrical energy is produced is \( \frac{3}{4} \) of the gravitational potential energy change. So, the electrical power \( P_e \) is: \[ P_e = \frac{3}{4} \times 1.176 \times 10^9 \, \text{W} = 0.882 \times 10^9 \, \text{W} = 882 \text{MW} \].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Gravitational Potential Energy
Gravitational potential energy (GPE) is the energy an object has due to its position in a gravitational field. When water is stored at a higher elevation, such as in a reservoir or behind a dam, it possesses potential energy because of its height above ground level. This stored energy can be released when the water is allowed to flow downwards.
In mathematical terms, the gravitational potential energy of an object is calculated using the formula:
  • \[ U = mgh \]
where:
  • \( m \) is the mass of the object (or water in kilograms),
  • \( g \) is the acceleration due to gravity (approximately \( 9.8 \, \text{m/s}^2 \) on Earth), and
  • \( h \) is the height of the object above a reference point (usually the ground or the lowest point in a system).
In our case, the water at the top of a 100-meter waterfall has gravitational potential energy because of its height. As it falls, this energy converts into kinetic energy, which can then be harnessed for doing work, such as generating electricity in hydroelectric power stations.
Energy Conversion
Energy conversion is a critical process in systems like hydroelectric power plants. It involves changing one form of energy into another. In the context of hydroelectric power, we primarily deal with the conversion of gravitational potential energy into kinetic energy, and then further into electrical energy.
When water falls from a height, its potential energy due to gravity is transformed into kinetic energy. This is because energy cannot be created or destroyed, only converted from one form to another (according to the law of conservation of energy). This conversion happens as water accelerates downward through the force of gravity, increasing its speed and kinetic energy.
Most hydroelectric plants use turbines, which are turned by the flowing water. As the turbine blades spin, they drive a generator that converts kinetic energy into electrical energy, which can then be distributed and used to power various applications. In our example, three-fourths of this kinetic energy is successfully converted into usable electrical energy, indicating the efficiency of the system in capturing and transforming energy resources.
Mass Flow Rate
The mass flow rate is a crucial concept in fluid dynamics and relates to the amount of mass moving through a system per unit of time. In hydroelectric plants, it refers to the mass of water passing through the system every second. This measurement is essential for determining how much energy can be generated from a hydroelectric plant, as it directly affects both the potential and kinetic energies involved.
For our waterfall problem, given a volume flow rate of 1200 cubic meters per second and knowing the density of water is 1000 kilograms per cubic meter, the mass flow rate \( \dot{m} \) can be calculated using the formula:
  • \[ \dot{m} = \text{Volume flow rate} \times \text{Density} \]
Plugging in the values, we find:
  • \[ \dot{m} = 1200 \, \text{m}^3/ \text{s} \times 1000 \, \text{kg/m}^3 = 1,200,000 \, \text{kg/s} \]
This mass flow rate tells us how much water, in terms of mass, is flowing past a point every second, allowing us to calculate the potential energy available for conversion into electricity.

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Most popular questions from this chapter

A \(60 \mathrm{~kg}\) skier leaves the end of a ski-jump ramp with a velocity of \(24 \mathrm{~m} / \mathrm{s}\) directed \(25^{\circ}\) above the horizontal. Suppose that as a result of air drag the skier returns to the ground with a speed of \(22 \mathrm{~m} / \mathrm{s},\) landing \(14 \mathrm{~m}\) vertically below the end of the ramp. From the launch to the return to the ground, by how much is the mechanical energy of the skier-Earth system reduced because of air drag?

A locomotive with a power capability of \(1.5 \mathrm{MW}\) can accelerate a train from a speed of \(10 \mathrm{~m} / \mathrm{s}\) to \(25 \mathrm{~m} / \mathrm{s}\) in \(6.0 \mathrm{~min} .\) (a) Calculate the mass of the train. Find (b) the speed of the train and (c) the force accelerating the train as functions of time (in seconds) during the 6.0 min interval. (d) Find the distance moved by the train during the interval.

A \(25 \mathrm{~kg}\) bear slides, from rest, \(12 \mathrm{~m}\) down a lodgepole pine tree, moving with a speed of \(5.6 \mathrm{~m} / \mathrm{s}\) just before hitting the ground. (a) What change occurs in the gravitational potential energy of the bear-Earth system during the slide? (b) What is the kinetic energy of the bear just before hitting the ground? (c) What is the average frictional force that acts on the sliding bear?

A certain spring is found \(n o t\) to conform to Hooke's law. The force (in newtons) it exerts when stretched a distance \(x\) (in meters) is found to have magnitude \(52.8 x+38.4 x^{2}\) in the direction opposing the stretch. (a) Compute the work required to stretch the spring from \(x=0.500 \mathrm{~m}\) to \(x=1.00 \mathrm{~m} .\) (b) With one end of the spring fixed, a particle of mass \(2.17 \mathrm{~kg}\) is attached to the other end of the spring when it is stretched by an amount \(x=1.00 \mathrm{~m} .\) If the particle is then released from rest, what is its speed at the instant the stretch in the spring is \(x=0.500 \mathrm{~m} ?\) (c) Is the force exerted by the spring conservative or nonconservative? Explain.

Two blocks, of masses \(M=2.0 \mathrm{~kg}\) and \(2 M,\) are connected to a spring of spring constant \(k=200 \mathrm{~N} / \mathrm{m}\) that has one end fixed, as shown in Fig. \(8-69 .\) The horizontal surface and the pulley are frictionless, and the pulley has negligible mass. The blocks are released from rest with the spring relaxed. (a) What is the combined kinetic energy of the two blocks when the hanging block has fallen \(0.090 \mathrm{~m} ?\) (b) What is the kinetic energy of the hanging block when it has fallen that \(0.090 \mathrm{~m} ?\) (c) What maximum distance does the hanging block fall before momentarily stopping?

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