Homework No. 2
Sean Edward Paquette
Rensselaer Polytechnic Institute
Course: Air and Water Pollution Prevention and Control Engineering; MANE-6960H01
Professor Ernesto Gutierrez-Miravete
Fall 2013
Homework (Number 2) {CORRECTED} – Due date September 26, 2013
1.
At the following link http://www.ewp.rpi.edu/~ernesto/F2013/AWPPCE/AdditionalReadings/Silva2013-
ERL-APMortality.pdf
you will find a recent journal article reporting research on mortality associated with air pollution. Read the paper carefully through and create a ~ 250 word critical essay relating the contents of the paper to the notion of NAAQS.
The Clean Air Act, 1990 requires the EPA to set National Ambient Air Quality Standards (NAAQS) for 6 principal pollutants considered harmful to the public health and environment. These 6 pollutants are
“critical” pollutants and their measurable units are parts per million (ppm) and parts per billion (ppb) by volume and micrograms per cubic meter of air (ug/m 3 ). The 6 critical pollutants are Carbon Monoxide,
Lead, Nitrogen Dioxide, Ozone, particle pollution and Sulfur Dioxide. The journal article, Global premature mortality due to anthropogenic outdoor air pollution and the contribution of past climate change provided an assessment of the burden of global anthropogenic air pollution on premature mortality over the past years including climate change. The article focused on two of the “critical” pollutants according to the NAAQS, the pollutants are ozone and particulate matter (PM
2.5
). The data collected was derived using model simulation estimates of both present-day (2000) and preindustrial
(1850) air pollution and conducted under the Atmospheric Chemistry and Climate Model
Intercomparison Project (ACCMIP). The study estimated mortality based on the change in concentration between 2000 and 1850 simulations used epidemiological concentration-response functions. Globally and annually 470,000 premature respiratory deaths occur due to increase in ozone and 2.1 million premature CPD and LC deaths due to PM
2.5
. These estimates are less than those reported by Anenberg
(2010). Furthermore, the most highly polluted areas are India and East Asia while there has been a slight decrease in southeast US and South America.
Sean Paquette Air & Water Pollution Prevention & Control Engineering Semester Fall 2013
MANE-6960H01
Home d

Homework No. 2
2.
The 1-hour NAAQS for carbon monoxide is 40 mg/m^3. Convert this to ppm at standard temperature y pressure.
Carbon Monoxide = CO
The conversion equation is based on 25°C at 1 atmosphere
X ppm = (Y mg/m 3 ) x (24.45) / (Molecular weight)
Or
Y mg/m 3 = (X ppm) x (Molecular weight) / (24.45)
Carbon: 12 (Molecular weight)
Oxygen: 16 (Molecular weight)
Total: 28
Correct answer, professor calculated this problem differently. Therefore,
I recalculated using the professor’s
X ppm = (40 mg/m 3 ) x (24.45) / (28)
CO = 34.9 ppm approach
Concentration = C = 40 mg/m 3 = 0.04 g/m 3
M
CO
= [Carbon (C) = 12.01] + [Oxygen (O) = 16.00] = 28 g/mol c = C/M = (0.04/28.00) = 0.00142 mol /m 3
Ideal Gas Law: V = RT/P
Volume of 1 mol of air = V = RT/P = (8.3144)*(298.15) / 101,325 = 0.0244 m 3
Number of mols in 1m 3 of air is 1/V = 40.87 mol/m 3
Concentration (C ppm
) = (0.0142/40.87)(10e 6 ) = 34.7 ppm
Sean Paquette Air & Water Pollution Prevention & Control Engineering Semester Fall 2013
MANE-6960H01
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Homework No. 2
3.
A solid waste incinerator emits particulate matter at the NSPS rate of 0.18 g/dscm. The incinerator burns 50 metric tons per day and exhausts gases at a ratio of 20 kg per kg of feed at atmospheric pressure and 453 K. Assume that the average molecular weight of the emitted gases is 30 and that they contain 12% CO2 and 10% H2O. What is the daily emission rate of
particulate matter?
PV = nRT
V = (nRT) / P
P = 0.987 atm = atm
T = 453 K
R = Ideal Gas Constant = 8.3145 J/mol K = 0.08205 L atm/mol K
Burning 50 metric tons/day exhausts 20kg per kg of feed
1 metric ton = 1,000 kilograms
(50 metric tons) x (1,000 kilograms) = 50,000 kilograms
Burning 50,000 kilograms/day exhausts = (20kg) x (50,000 kilograms) = 1,000,000kg exhaust/day
(1,000,000kg exhaust/day) x 1,000 = 1,000,000,000 grams exhaust/day
Molar mass = mass / molecular mass
Molar mass = (1,000,000,000) / 30 = 33,333,333 gram moles
Incorrect Calculations:
Corrected Below
V = (nRT) / P
V = ((33,333,333 gram moles) x (0.08205 atm/mol K) x (453K)) / (0.987 atm)
V = 1,255,273,544 standard Liters
V = (1,255,273,544 standard Liters) x (0.001) = 1,255,274 m 3
A solid waste incinerator emits particulate matter at the NSPS rate of 0.18 g/dscm
Dscm = Dry Standard Cubic Meter
Daily emission rate of particulate matter = (1,255,274 m 3 ) x (0.18 g/dscm) = 225,949 grams/day
Solid waste incinerator emits PM at NSPS rate of 0.18 g/dscm = C pm
Dscm = dry standard cubic meter
To convert a gas from dscm to “standard cubic meter”
*There is a given of 10% H
2
O multiply (0.18)(100% - 10%) to get standard cubic meter, also because the temperature is not standard, to get m 3 , need to ratio temperature
(0.18)x(0.90)x(298/453) = 0.1065 g/m 3
PV = nRT :: Ideal Gas Law :: V = RT/P for ONE mol of exhaust gas
V = [(1)(8.3144)(453)] / [101,325] = 0.037 m 3 /mol
Sean Paquette Air & Water Pollution Prevention & Control Engineering Semester Fall 2013
MANE-6960H01
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Homework No. 2
Volume per gram of “emitted gases” = V/MW = 0.037/30 = 0.00123 m 3 /g
Number of m 3 of exhausted gas per day = (0.00123) x (50x10 6 )(20) = 1.230x10
6 m 3 /day
Therefore the amount of PM emitted per day is (0.1065) x (1.230 x10 6 ) = 131,350 g/day
4.
Methanol (CH3OH) is burned in dry air at an equivalence ratio of 0.75.
(a) Determine the fuel-air mass ratio.
(b) Determine the composition of the combustion products.
CH3OH (I) → CO (g) + 2H2 (g)
Atomic Weights
Carbon (C) = 12.01
Oxygen (O) = 16.00
Hydrogen (H) = 1.008
Molecular weight calculation of CH3OH = (C) (12.0107) + (H3) (1.00794 x 3) + (O) (15.9994) + (H)
(1.00794) = 32.04186 g/mol
One oxygen molecule weighs = 15.9994
Correct Answer Below
The fuel-air mass ratio = (15.9994) / (32.04186) = 0.449 kg
So 0.449 kg of oxygen for every 1kg of fuel
Since 23.2 mass-percent of air is actually oxygen, we need: (0.449 kg) x (100/23.2) = 1.94 kg air for every 1 kg of methanol
So the stoichiometric fuel-air ratio of methanol is 1.94
The composition of the combustion products will be H20, CO2 and CO
The simplified combustion reaction of methanol with oxygen is
CH
3
OH + (3/2)O
2
→ CO
2
+ 2H
2
O
Combustion Reaction is = 1.5 / 0.75 = 2
In air, the combustion reaction can be written as
CH
3
OH + 2(O
2
+ 3.78N
2
) → CO
2
+ 2H
2
O + (1/2)O
2
+ 7.56N
2
Air Fuel Mass Ratio = M
CH3OH
/ (2*(M
O2
+ 3.75*M
N2
)) = (12+4+16)/(32+3.78*28)) = 32/272.5 =
0.116
Total number of mols = 11.06
Sean Paquette Air & Water Pollution Prevention & Control Engineering Semester Fall 2013
MANE-6960H01
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Homework No. 2
Y
CO2
= (1/11.06) = 0.0904
Y
H2O
= (2/11.06) = 0.1808
Y
O2
= (0.5/11.06) = 0.0452
Y
N2
= (7.56/11.06) = 0.683
5.
Create a ~250 word essay about your local air quality conditions.
The website I used to determine my local air quality conditions was www.airnow.gov
. I currently live in Rocky Hill, CT although AirNow calculated quality conditions from Middletown,
CT. The Air Quality Index (AQI) for Middletown, CT on September 26, 2013 is a 4 (good).
Today’s high could reach an AQI of 33 (good). AQI pollutant details are the following; Particles
(PM2.5) 33 (good) and Ozone 31 (good). Below is a table, Air Quality Guide for Particle
Pollution, published by the United States Environmental Protection Agency (EPA). Particles in the air can cause or aggravate a number of health problems and have been linked with illnesses and deaths from heart or lung diseases.
6.
Revisit your proposed project idea from last week. Build upon it and the feedback provided to create a refined edited version of your proposal statement. Incorporate additional information gathered from further researching the topic using all resources available.
See project proposal update folder.
Sean Paquette Air & Water Pollution Prevention & Control Engineering Semester Fall 2013
MANE-6960H01
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