/*! 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 31 A Cottrell precipitator consists... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A Cottrell precipitator consists of a column containing electrodes that are connected to a high-voltage direct current (DC) source. The Cottrell precipitator is placed in smokestacks to remove smoke particles from the gas discharged from an industrial plant. Explain how you think this works.

Short Answer

Expert verified
The Cottrell precipitator uses electrostatic precipitation to charge and collect smoke particles, reducing industrial air pollution.

Step by step solution

01

Understanding the Purpose

A Cottrell precipitator is designed to remove smoke particles from gases emitted by industrial plants. The goal is to minimize air pollution and enhance air quality by filtering out harmful particles before they are released into the atmosphere.
02

Explaining the Basic Mechanism

The Cottrell precipitator works using the principle of electrostatic precipitation. It utilizes high-voltage electrodes installed within a column to generate an electrical field.
03

Details of Particle Charging

As smoke and dust particles move through the precipitator, they become charged due to the interaction with the electric field created by the electrodes.
04

Understanding Particle Collection

Once charged, these particles are attracted to and collected on plates or other surfaces within the precipitator that carry an opposite charge. This collection prevents the particles from exiting the stack into the air.
05

Describing Results and Benefits

By capturing these particles, the Cottrell precipitator significantly reduces the emission of pollutants into the atmosphere, thus reducing environmental pollution and improving air quality around industrial sites.

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.

Electrostatic Precipitation
Electrostatic precipitation is a filtration process that uses electrical forces to remove particles from a gas flow. When smoke or dust-laden gas passes through the precipitator, the particles become electrically charged as they interact with the high-voltage electrodes. This charging process is crucial because once particles carry an electrical charge, they begin to behave differently in the presence of an electric field.
Charged particles are then drawn towards and deposited on collector plates with the opposite charge. This mechanism efficiently captures pollutants, preventing them from being released into the atmosphere. The beauty of electrostatic precipitation is its ability to clean large volumes of gas continuously and effectively, making it an ideal choice for industrial emission control.
  • High-voltage electrodes create an electric field.
  • Particles become charged upon entering the field.
  • Opposite charges attract, causing particle collection on plates.
This method is highly efficient in removing fine particulate matter from the gas, which could otherwise contribute to environmental pollution. As such, electrostatic precipitators play a crucial role in air pollution control strategies.
Air Pollution Control
Air pollution control is the practice of reducing harmful emissions released into the atmosphere from industrial processes. The goal is to safeguard both human health and the environment by minimizing the level of contaminants in the air. Cottrell precipitators are a vital component in this effort, specifically designed to manage and reduce smoke emissions from industrial smokestacks.
Utilizing technologies like electrostatic precipitation, these devices capture smoke and dust particles, preventing them from entering the external environment. This is imperative in maintaining air quality, especially around urban and industrial areas where the concentration of factories and plants can significantly impact local air quality.
  • Improved air quality through reduction in pollutants.
  • Essential in urban and industrial locales.
  • Protects environmental health and public well-being.
By implementing such technologies, industries can not only comply with environmental regulations but also play a part in preserving the planetary ecosystem.
Industrial Emission Reduction
Reducing industrial emissions is a crucial step toward mitigating the adverse effects of pollution on the environment and health. Industries are major contributors to environmental pollution, releasing various pollutants into the air. A Cottrell precipitator is an effective tool for reducing these emissions by removing particulates before they can escape into the atmosphere.
Emissions reduction from industrial processes involves capturing pollutants such as smoke and dust that would otherwise contribute to air quality deterioration. This reduction is not only a regulatory requirement but a moral obligation to minimize environmental footprints. Implementing electrostatic precipitation technology allows industries to achieve significant emission reductions.
  • Helps in compliance with stringent environmental regulations.
  • Significantly lowers the emissions of smoke and dust.
  • Contributes to sustainable industrial practices.
Such measures support sustainable development goals and ensure that industrial growth does not come at the cost of ecological health.

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

Equal numbers of moles of two soluble, substances, substance \(\mathrm{A}\) and substance \(\mathrm{B}\), are placed into separate \(1.0-\mathrm{L}\) samples of water. a. The water samples are cooled. Sample A freezes at \(-0.50^{\circ} \mathrm{C}\), and Sample \(\mathrm{B}\) freezes at \(-1.00^{\circ} \mathrm{C}\). Explain how the solutions can have different freezing points. b. You pour \(500 \mathrm{~mL}\) of the solution containing substance \(\mathrm{B}\) into a different beaker. How would the freezing point of this \(500-\mathrm{mL}\) portion of solution \(\mathrm{B}\) compare to the freezing point of the \(1.0-\mathrm{L}\) sample of solution \(\mathrm{A}\) ? c. Calculate the molality of the solutions of \(\mathrm{A}\) and \(\mathrm{B}\). Assume that \(i=1\) for substance \(\mathrm{A}\). d. If you were to add an additional \(1.0 \mathrm{~kg}\) of water to solution B, what would be the new freezing point of the solution? Try to write an answer to this question without using a mathematical formula. e. What concentration (molality) of substances \(\mathrm{A}\) and \(\mathrm{B}\) would result in both solutions having a freezing point of \(-0.25^{\circ} \mathrm{C}\) ? f. Compare the boiling points, vapor pressure, and osmotic pressure of the original solutions of \(\mathrm{A}\) and \(\mathrm{B}\). Don't perform the calculations; just state which is the greater in each case.

A gaseous mixture consists of \(80.0\) mole percent \(\mathrm{N}_{2}\) and \(20.0\) mole percent \(\mathrm{O}_{2}\) (the approximate composition of air). Suppose water is saturated with the gas mixture at \(25^{\circ} \mathrm{C}\) and \(1.00\) atm total pressure, and then the gas is expelled from the water by heating. What is the composition in mole fractions of the gas mixture that is expelled? The solubilities of \(\mathrm{N}_{2}\) and \(\mathrm{O}_{2}\) at \(25^{\circ} \mathrm{C}\) and \(1.00\) atm are \(0.0175 \mathrm{~g} / \mathrm{L} \mathrm{H}_{2} \mathrm{O}\) and \(0.0393 \mathrm{~g} / \mathrm{L} \mathrm{H}_{2} \mathrm{O}\), respectively.

Phosphoric acid is usually obtained as an \(85 \%\) phosphoric acid solution. If it is \(15 M\), what is the density of this solution? What is its molality?

Methanol, \(\mathrm{CH}_{3} \mathrm{OH}\), is a compound that is used in car windshield washer fluid to lower the freezing point of the solution. If an aqueous solution of windshield washer fluid has a mole fraction of methanol of \(0.520\), what is the mass percentage of water in this solution?

Calculate the number of moles of barium chloride in \(427 \mathrm{~g}\) of a \(3.17 \%\) by mass barium chloride solution? a. \(6.50 \times 10^{-02}\) mol b. \(7.83 \times 10^{-02}\) mol c. \(4.31 \times 10^{-02}\) mol d. \(7.81 \times 10^{-02}\) mol e. \(4.27 \times 10^{-02} \mathrm{~mol}\)

See all solutions

Recommended explanations on Chemistry 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.