/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 42 Water falls over a dam of height... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Water falls over a dam of height \(h\) with a mass flow rate of \(R,\) in units of \(\mathrm{kg} / \mathrm{s} .\) (a) Show that the power available from the water is $$\mathscr{P}=R g h$$ where \(g\) is the frec-fall acceleration. (b) Each hydroclectric unit at the Grand Coulce Dam takes in watcr at a rate of \(8.50 \times 10^{5} \mathrm{kg} / \mathrm{s}\) from a height of \(87.0 \mathrm{m} .\) The power developed by the falling water is converted to clectric power with an efficiency of \(85.0 \% .\) How much electric power is produced by each hydroelectric unit?

Short Answer

Expert verified
Part (a): The power available from the falling water is given by \( \mathscr{P} = R g h \). Part (b): Each hydroelectric unit at the Grand Coulee Dam can produce around 6.25 x 10^8 Watts of electric power.

Step by step solution

01

Understand the formula for kinetic energy

The kinetic energy of the falling water can be described by the equation \( KE = \frac{1}{2} m v^2 \), where \( m \) is the mass of the water and \( v \) is its velocity. Since the water starts from rest at the top of the dam, its velocity at the bottom will have been gained through falling under gravity. Therefore, \( v = \sqrt{2gh} \). Substituting \( v \) in the kinetic energy equation gives \( KE = mgh \).
02

Relate kinetic energy to power

Power is the rate of doing work or the amount of energy transferred per unit time, which can be expressed as \( \mathscr{P} = \frac{KE}{t} \). Since the mass flow rate \( R = \frac{m}{t} \), we can rewrite the power equation in terms of \( R \), giving \( \mathscr{P} = Rgh \). This is the power available from the water.
03

Find the actual electric power

In the second part of the exercise, we're given that each hydroelectric unit has a mass flow rate of \( 8.50 × 10^5 \mathrm{kg/s} \), dam height of \( 87.0 \mathrm{m} \), and the efficiency is \( 85.0\% \). We can calculate the power available from the water using the derived formula in step 2, and then use the given efficiency to get the electric power produced, which is \( \mathscr{P}_{\mathrm{electric}} = \eta \mathscr{P}_{\mathrm{water}} \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Mass Flow Rate
The concept of mass flow rate is essential in understanding how much mass passes through a given area in a specific amount of time. In the context of hydroelectric power, it refers to the mass of water flowing over the dam each second. Imagine each drop of water carrying some energy with it, and the more water flows, the more energy is carried.
  • Mass flow rate is measured in kilograms per second \( \mathrm{kg/s} \).
  • It provides the basis for calculating how much potential energy is available to be converted to electricity.
In terms of mathematical representation, if you have a total mass \( m \) of water moving over time \( t \), the mass flow rate \( R \) is given as:\[ R = \frac{m}{t} \]This lets us determine the power output available, provided we know other variables like gravitational acceleration \( g \) and dam height \( h \).
Mass flow rate is a crucial input in the calculation of potential energy and hence power in systems like hydroelectric plants.
Gravitational Potential Energy
Gravitational potential energy is the energy stored due to an object's position in a gravitational field. When water is held at a certain height, such as in a dam, it possesses potential energy because of gravity.
  • This energy can be quantified using the formula \( E_p = mgh \), where \( m \) represents mass, \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \mathrm{m/s^2} \) on Earth), and \( h \) is the height above the reference point.
  • The greater the height \( h \) the water falls from, the more potential energy it initially holds.
We can transform this potential energy into other forms. For hydroelectric power, the focus is on converting it to mechanical energy and then electrical energy. The formula \( \mathscr{P} = Rgh \) explains the power available, where \( R \) is the mass flow rate. This concept is vital for calculating how much energy from falling water can be harnessed effectively.
Energy Efficiency
In real-world applications, energy efficiency tells us how much of the total energy can be successfully converted into useful work. When discussing a hydroelectric plant, we are interested in how well the potential energy of the falling water is turned into electric energy.
  • Efficiency is typically expressed as a percentage, showing the ratio of useful output energy to input energy.
  • An efficiency of 100% would mean that all the potential energy is transformed into electricity, but such ideal systems rarely exist.
For instance, if a hydroelectric unit operates at an efficiency of 85%, it means 85% of the water's potential energy is converted to electricity. The rest could be lost due to factors like friction in machinery or energy dissipation as sound or heat.
The efficiency formula for calculating electric power from available water power is given by:\[ \mathscr{P}_{\mathrm{electric}} = \eta \mathscr{P}_{\mathrm{water}} \]where \( \eta \) is the efficiency of conversion. Understanding efficiency helps in maximizing energy use and minimizing losses in any energy conversion process.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The small piston of a hydraulic lift has a cross-scrional area of \(3.00 \mathrm{cm}^{2},\) and its large piston has a cross-sectional area of \(\left.200 \mathrm{cm}^{2} \text { (Figure } 14.4\right) .\) What force must be applied to the small piston for the lift to raise a load of \(15.0 \mathrm{kN} ?\) (In service stations, this force is usually exerted by compressed air.)

\- A \(10.0-\mathrm{kg}\) block of metal measuring \(12.0 \mathrm{cm} \times 10.0 \mathrm{cm} \times\) \(10.0 \mathrm{cm}\) is suspended from a scale and immersed in water as in Figure P14.25b. The 12.0-cm dimension is vertical, and the top of the block is \(5.00 \mathrm{cm}\) below the surface of the water. (a) What are the forces acting on the top and on the bottom of the block? (Take \(P_{0}=1.0130 \times 10^{5} \mathrm{N} / \mathrm{m}^{2}\) ) (b) What is the reading of the spring scale? (c) Show that the buoyant force equals the difference between the forces at the top and bottom of the block.

A hypodermic syringe contains a medicine with the density of water (Figure \(\mathbf{P} 1 \mathbf{1} . \mathbf{5 3}\) ). The barrel of the syringe has a cross sectional area \(A=2.50 \times 10^{-5} \mathrm{m}^{2},\) and the needle has a cross-sectional area \(a=1.00 \times 10^{-8} \mathrm{m}^{2} .\) In the absence of a force on the plunger, the pressure everywhere is 1 atm. \(A\) force \(\mathbf{F}\) of magnitude \(2.00 \mathrm{N}\) acts on the plunger, making medicine squirt horizontally from the needle. Determine the speed of the medicine as it leaves the needle's tip.

The weight of a rectangular block of low-density material is 15.0 N. With a thin string, the center of the horizontal bottom face of the block is tied to the bottom of a beaker partly filled with water. When \(25.0 \%\) of the block's volume is submerged, the tension in the string is \(10.0 \mathrm{N}\) (a) Sketch a free-body diagram for the block, showing all forces acting on it. (b) Find the buoyant force on the block. (c) Oil of density \(800 \mathrm{kg} / \mathrm{m}^{3}\) is now steadily added to the beaker, forming a layer above the water and surrounding the block. The oil exerts forces on each of the four side walls of the block that the oil touches. What are the directions of these forces? (d) What happens to the string tension as the oil is added? Explain how the oil has this Effect on the string tension. (c) The string breaks when its tension reaches \(60.0 \mathrm{N}\). At this moment, \(25.0 \%\) of the block's volume is still below the water line; what additional fraction of the block's volume is below the top surface of the oil? (f) After the string breaks, the block comes to a new equilibrium position in the beaker. It is now in contact only with the oil. What fraction of the block's volume is submerged?

The Bernoulli effect can have important consequences for the design of buildings. For example, wind can blow around a skyscraper at remarkably high speed, creating low pressure. The higher atmospheric pressure in the still air inside the buildings can cause windows to pop out. As originally constructed, the John Hancock building in Boston popped window panes, which fell many stories to the sidewalk below. (a) Suppose that a horizontal wind blows in streamline flow with a speed of \(11.2 \mathrm{m} / \mathrm{s}\) outside a large pane of plate glass with dimensions \(4.00 \mathrm{m} \times 1.50 \mathrm{m}\) Assume the density of the air to be uniform at \(1.30 \mathrm{kg} / \mathrm{m}^{3} .\) The air inside the building is at atmospheric pressure. What is the total force exerted by air on the window pane? (b) What If? If a second skyscraper is built nearby, the air speed can be especially high where wind passes through the narrow separation between the buildings. Solve part (a) again if the wind speed is \(22.4 \mathrm{m} / \mathrm{s}\), twice as high.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.