In environmental and water treatment, the presence of dissolved gases can pose significant challenges. Oxygen, nitrogen, and carbon dioxide, among others, can lead to corrosion, fouling, and negatively impact water quality. To address this, vacuum deaeration emerges as a powerful technique, effectively removing dissolved gases from liquids, primarily water.
How It Works: The Science Behind the Vacuum
The key principle behind vacuum deaeration is the manipulation of partial pressure. By reducing the pressure above the liquid, the partial pressure of dissolved gases decreases. This creates a gradient where dissolved gases are driven out of the liquid phase and into the vapor phase.
The Mechanism in Detail:
Key Benefits of Vacuum Deaeration
Applications of Vacuum Deaeration in Environmental and Water Treatment:
Conclusion: A Powerful Tool for Clean Water
Vacuum deaeration is a valuable and reliable technique for effectively removing dissolved gases from liquids, especially water. By minimizing corrosion, enhancing water quality, and improving system efficiency, vacuum deaeration plays a vital role in environmental and water treatment processes, contributing to a cleaner and more sustainable future. As we continue to strive for cleaner water sources, understanding and implementing techniques like vacuum deaeration will be crucial in ensuring access to safe and quality water for all.
Instructions: Choose the best answer for each question.
1. What is the primary principle behind vacuum deaeration?
a) Increasing the pressure above the liquid. b) Manipulating the partial pressure of dissolved gases. c) Using a chemical reaction to remove dissolved gases. d) Heating the liquid to release dissolved gases.
b) Manipulating the partial pressure of dissolved gases.
2. Which of the following gases are commonly removed by vacuum deaeration?
a) Oxygen and Nitrogen only. b) Carbon Dioxide and Hydrogen only. c) Oxygen, Nitrogen, and Carbon Dioxide. d) Oxygen, Nitrogen, and Helium.
c) Oxygen, Nitrogen, and Carbon Dioxide.
3. How does vacuum deaeration enhance corrosion protection?
a) By adding chemicals that neutralize corrosive agents. b) By removing dissolved oxygen, a major contributor to corrosion. c) By increasing the pH of the water. d) By preventing the formation of scale on surfaces.
b) By removing dissolved oxygen, a major contributor to corrosion.
4. Which of the following is NOT a benefit of vacuum deaeration?
a) Improved water quality. b) Reduced fouling of equipment. c) Increased system efficiency. d) Increased water temperature.
d) Increased water temperature.
5. Which application is NOT a common use for vacuum deaeration?
a) Industrial water treatment for boiler feed water. b) Municipal water treatment for drinking water. c) Wastewater treatment for odor reduction. d) Agricultural irrigation for crop fertilization.
d) Agricultural irrigation for crop fertilization.
Scenario:
A water treatment plant uses vacuum deaeration to remove dissolved gases from its water supply. The plant has a large storage tank that holds 100,000 gallons of water. The water contains an initial dissolved oxygen concentration of 10 ppm (parts per million). The vacuum deaeration system is designed to reduce the dissolved oxygen concentration to 2 ppm.
Task:
Calculate the total volume of oxygen that needs to be removed from the storage tank to achieve the desired dissolved oxygen concentration.
**1. Calculate the mass of dissolved oxygen in the initial water:** * 10 ppm means 10 mg of dissolved oxygen per liter of water. * Convert gallons to liters: 100,000 gallons * 3.785 liters/gallon = 378,500 liters * Total mass of dissolved oxygen: 10 mg/liter * 378,500 liters = 3,785,000 mg = 3.785 kg **2. Calculate the mass of dissolved oxygen after deaeration:** * 2 ppm means 2 mg of dissolved oxygen per liter of water. * Total mass of dissolved oxygen after deaeration: 2 mg/liter * 378,500 liters = 757,000 mg = 0.757 kg **3. Calculate the total volume of oxygen removed:** * Total volume of oxygen removed: 3.785 kg - 0.757 kg = 3.028 kg **Therefore, approximately 3.028 kg of oxygen needs to be removed from the storage tank to achieve the desired dissolved oxygen concentration.**
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