Air Pollution worksheet MD
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Feb 20, 2024
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Air Pollution work Sheet
Part A – Short Answer
1.
Which is a greater health concern, outdoor or indoor air pollution? Provide evidence that backs up your answer. (2 pts) Indoor because people spend 70-98% of their time indoors, pollutants are higher concentrated indoors compared to outdoors, and you are more likely to find common pollutants inside
homes and buildings instead of outdoors. 2.
Assess Miami’s average air quality in comparison with other United States cities: do you think it is better or worse? Don’t just look up what it is at this moment – think about it in longer terms. What are some reasons that this might be the case? (2 pts) Miami seems to fare better than most US Cities, probably due to the amount of people that go outdoors to the beach, or park. Miami has fairly nice weather year round, and allows for more outdoor activities. 3.
Explain two things that you or your household could do to reduce your contribution to outdoor air pollution. (2 pts)
Get car exhaust inspections twice a year. Use mass transit, or bicycles for transportation.
4.
Explain two things that you or your household could do to reduce your contribution to indoor air pollution. (2 pts)
Do not buy products containing formaldehyde.
Don’t store gas, solvents, and other hazardous chemicals inside the home or garage.
Part B – Criteria Pollutants Use the EPA factsheets in this week’s module, or the information here
, to fill out the below table on the sources, impacts on the body, and impacts on the environment of the six criteria air pollutants. (9 pts)
Name of the
Pollutant
Sources of Pollutant
Impacts on the body
Impacts on the environment
Lead
(Pb)
Ore and metals processing and piston-engine Can affect the nervous system, kidney function, immune system, reproductive and Decreased growth and reproduction of plants and animals, and neurological effects in
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of 6
Name of the
Pollutant
Sources of Pollutant
Impacts on the body
Impacts on the environment
aircraft operating on
leaded aviation fuel.
Waste incinerators, lead-acid battery manufacturers. developmental systems and cardiovascular system. Affects oxygen carrying capacity of blood. Learning deficits, lower IQ.
vertebrates.
Sulfur
Dioxide
(SO₂)
Burning of fossil fuels by power plants and industrial
buildings. Extracting metal from ore. Volcanoes.
Vehicles and equipment that burn fuel with high sulfur content. Negatively affects respiratory system. Can worsen asthma.
Damages foliage and decreases growth in tress and plants. Contributes to acid rain. Carbon
Monoxide
(CO)
Cars, trucks, and vehicles or machinery that burn
fossil fuels. Reduces oxygen in bloodstream.
Causes dizziness, confusion, unconsciousness, and death.
Chest pain.
Contributes to climate change and global warming by adding to the amount of greenhouse gases. Particulate
Matter
(PM)
Construction sites, unpaved roads, fields, smokestacks, fires.
Pollutants from power plants, industries and vehicles.
Premature death
Heart attacks
Asthma
Decreased lung function
Coughing
Acidic bodies of water
Contributes to acid rain
Damages forests and farms
Depletes nutrients in soil
Affects diversity in ecosystems
Nitrogen
Dioxide
(NO₂)
Burning of fuel
Cars, trucks, power plants
Off road equipment
Respiratory disease
Asthma
Coughing
Wheezing
Difficulty breathing
Contributes to formation of particulate matter and ozone.
Acid rain
Air haziness
Nutrient pollution
Ground
Ozone
Created via chemical
reactions between NOx and VOC. Irritates airways
Coughing
Sore, scratchy throat
Difficulty breathing and pain
Lung infection
Asthma
Death
Reduces photosynthesis
Slows plant growth
Increases plant disease and damage Changes to habitat quality, water cycle, and plant assortment
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Part C – Concentrations of Air Pollutants Estimate the concentration of pollutants in a (1) confined valley and an (2) unconfined valley under the two different scenarios provided for each. It is helpful if you first follow along with the example cases provided for both valleys.
Case I - Confined Valley
1 km = 1000 m
1 metric ton = 1000 kg
Volume= Length* Width * Depth m
3
Mass of the air= Volume* Density
Concentration= Mass of the Pollutant
Massof the Air
Part per million (ppm) = mg of the pollutant
kgof the Air
Example Case I
What is the concentration of the pollutant in a valley, if the total mass of pollutants discharged from the factory chimney (located on the bottom of the valley) is 40 metric tons. The dimensions of the valley are length is 1 km, width is 1 km, and depth is 1 km. The density of the air is 1.2 kg/m³.
Express the answer in kg/kg and part per million by mass (ppm).
Solution
Volume (V) = Length* Width * Depth, so
V = 1000 * 1000 * 1000= 1 x 10
9 m
3 Mass of the air = Volume * Density, so
Mass of the air = 1 x 10
9 m
3 * 1.2 kg/m
3
= 1.2 x10
9 kg or 1,200,000,000 kg
Concentration= Mass of the Pollutant
Mass of the Air
Concentration= 40,000
kg
1,200,000,000
kg
= 0.0000333 = 3.3 x 10
-5 kg/kg = 33.3 ppm.
Case I Exercises (4 pts)
1.
What is the concentration of the pollutants in the valley, if the mass of pollutants that are discharged from the factory chimney, which is located on the bottom of the valley, is 20 tons? The dimensions of the valley are as follows: length is 2 km, width is 0.5 km, and depth is 1 km. The density of the air is 1.2 kg/m³. Express the answer in kg/kg and part per million by mass (ppm).
2.
What is the concentration of the pollutants in the valley, if the mass of pollutants that are discharged from the factory chimney, which is located on the bottom of the valley, is 25 tons? The dimensions of the valley are as follows: length is 2 km, width is 1 km, and depth is 0.1 km. The density of the air is 1.2 kg/m³. Express the answer in kg/kg and part per million by mass (ppm).
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Question
Show the concentration calculations and circle the answer
1
2 x 10^4kg / 1.2 x 10^9 kg = 1.67 / 10^5
16.7 / 10^6
16.7 2
2.5 x 10^4 / 2.4 x 10^8 = 2.5/2.4 = 1.04
1.04 / 10^4 multiply top and bottom by 10^2 to get 10^6
104 / 10^6
104
Page 4
of 6
Case II – Unconfined Mixing
Volume of a hemisphere = 2
3
r
3
Mass of the air = Volume * Density
Concentration= Mass of the Pollutant
Massof the Air
Example Case II
What is the concentration of the pollutants in the air on the distance 500 m from the pipe (the radius is 500 m)
if the mass of discharged pollutants is 40 metric tons. The density of the air is 1.2 kg/m
3
.
Solution
V = 2
3
*
3.14* (500)
3
so,
V = 261,799,387.8 m
3
Mass of the air = V *Density, so Mass of the air=
261,799,387.8 m
3 * 1.2 kg/m
3
= 314,159,265.4 kg
Concentration= Mass of the Pollutant
Mass of the Air
so,
Concentration= 40,000
kg
314,159,265.4
kg
= 0.000127324 = 1.27 x 10
-4 kg/kg = 127 ppm
Case II Exercise (4 pts)
1.
What is the concentration of the pollutants in the air at a distance 100 m from the pipe if the mass of discharged pollutants is 1 ton? The density of the air is 1.2 kg/m
3
. Express the answer in kg/kg and part
per million by mass (ppm).
2.
What is the concentration of the pollutants in the air at a distance 50 m from the pipe if the mass of discharged pollutants is 10 tons? The density of the air is 1.2 kg/m
3
. Express the answer in kg/kg and part per million by mass (ppm).
Question
Show the concentration calculations and circle the answer
1
100 / 2,513,274.12 = 0.000397887358184 = 3.97 x 10^-4 kg
Turn the equation into division
3.97 / 10^4 multiply top and bottom by 10^2 to get 10^6
397 / 10^6
Page 5
of 6
397 ppm rounded
. 2
10,000 / 314,159.253589793 = 0.03183099 = 3.183099 x 10^-2
Turn the equation into division
3.183099 / 10^2 multiply top and bottom by 10^4
31,831 / 10^6
31,831 rounded.
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