Engineering 115: Introduction to
Environmental Resources Engineering
Air Resources Module
•
•
•
•
Introduction
Units/conversions w/ Ideal Gas Law
Indoor Air
Outdoor Air
• Carbon Monoxide, Particulate Matter
• Visibility, Sox, Photochemical Smog
• Air pollution transport
• Guassian Plume Model
*not all topics are covered in any given
semester.
Engineering 115: Introduction to
Environmental Resources Engineering
Air Resources Module
 These slides have been developed by Eileen M. Cashman and
Elizabeth A. Eschenbach for use in an introductory
engineering and environmental science course.
 Supported by NSF CCLI DUE program
 Figures and photos from text and other copyright sources have been
removed.
Agenda
• Air Resources
• How is air a resource?
• How do we manage air as a
resource?
• Clean Air Act
• Units
How do we USE air as
a resource?
What is in the Air We Breathe?
Other Things in the Air…
http://www.epa.gov/air/data
Some Air Pollution Concerns
• Sulfur dioxide (SO2)
• Carbon monoxide (CO)
• Nitrogen oxides (NOx)
• Volatile organic compounds (VOCs)
• Carbon dioxide (CO2) and greenhouse gases
How do we MANAGE air as
a resource?
Classifications
•
•
•
•
•
Gaseous v. particulate
Primary v. secondary
Mobile v. stationary
Point v. area
Local v. regional v. global
Clean Air Act
• Enacted in 1970,
• Amended in 1977, 1990, 1997
• Sets National Ambient Air Quality
Standards (NAAQS)
• Primary and secondary standards
Criteria Pollutants
Ozone Nonattainment
Pollution Standard Index
• PSI is method used to develop general air
quality index for an area
• Calculated using an index number
corresponding to criteria pollutant
concentrations (see Example 7.2 and
homework problem 7-6)
How much air do you breath in a 1-hour
period?
216,000 in3
36 – 38,400,000 grams
54-22,000 liters – 600-2000 most common
432,000 mL
113,000 – 2,000,000 cm3
.45-2400
m3
95-120 ft3
28-1440 gallons
106 milk cartons
Quiz Answers
I estimated that I take a breath about every three seconds.
I am not sure how I would estimate lung capacity maybe in
liters? So I would say between my two lungs I can hold
about two liters of air.
But with each breath I do not fill my lungs to their capacity
so I would say each breath is about a liter of air.
So, with a breath every three seconds I would take 20 breath
a minute and about 1200 breaths and hour so that would
be about 1200 liters of air I breathe in an hour.
O.K. Well, I'm going to say that in one average breath I could inflate a
regular balloon with regular elasticity, red in color, out to a radius of 3.4
inches.
By the equation for the volume of a sphere: V=(4/3)(pi)(radius cubed), the
volume of air inside of this particular red balloon would be
(4/3)(3.1415...)(39.304) = 164.6 cubic in.
So, after timing my breathing for one minute I find that I take 13 breaths
per minute. There are still 60 minutes in an hour yeah? So, (13
breaths)X(60 minutes) = 780 breaths in one hour on average.
With each breath I draw an average of 164.6 cu. in. and this equals 13.72
cu.ft. So, (780 breaths)X(13.72 cu.ft.) = 10,701.4 cu.ft. of air in one
hour...great googly moogly!
Quiz Answers
Theoretical Lung Capacity Formula
v=.041h-.018a-2.69
v= lung capacity in liters, h=height in centimeters, a=age in
years source:
www.regentsprep.org/Regents/math/fsolid/TSolids.htm
[(5.75 ft.)/(h)]=[(3.28 ft.)/(1 m) h=1.75 m=175 cm
v=.041(175)-.018(22)-2.69 v=4.089 liters
Assumption: 1 breath for every 2 seconds. 1800 breaths per
hour v=(4.089)(1800)=7360.2 liters per hour
•
I thought that this paragraph was very interesting. I got it from:
http://www.breathing.com/articles/how-much-air.htm
How much air we breathe in is, as a statistic just by itself, risking being extremely misleading. The answer would
move around somewhere between a per breath 1/2 liter in quiet breathing to 6 liters for a tall singing base baritone
depending on height, weight, posture, parents genes (big lungs, small lungs, small bones vs big bones, activity quality
& intensity) cellular condition, chemistry, emotions (joy, shallow breathing, hyperventilation), breathing skill level,
thinking processes and more. Not a great way to get any real clear conclusions.
Considering this, and the fact that peoples breaths per minute range from 4 to 15 or more, I decided that averages of
these statistics would not give me a very accurate estimation. Therefore, I decided to take my own measurements.
I found a bag, filled it with air, and breathed in that air. I repeated this process until I could determine how full the bag
had to be for me to comfortably breathe all the air in, as naturally as possible. I then filled the bag to that level with
water, and then I measured the water.
I completed this experiment from start to finish three times in a row. The first time I got 1000ml, the second time I got
890ml and the third time I got 930ml. I used the mean of these numbers to assume that I breathe in about 940ml per
breath in a relaxed state.
I then calculated my breaths per minute, by simply counting. I did this several time and came to the assumption that I
breathe in about 10 times per minute in a relaxed state.
Based on these measurements, I concluded that, in a relaxed state, I breathe in about 9.4 liters per minute. Hence, 564
liters per hour.
Lung Capacity Estimates
• Blowing into a plastic bag
• Compare to gallon milk jug
• Blowing up balloons
-estimating volume
-weighing the balloon
• Internet research
• Experiments with roommates
• Rulers
• Guessing
57,600 Liters/Day
Assuming:
20 breaths minute
2 liter lung capacity
Units
• PPM means what for
– Aqueous Solutions
– Gaseous Solutions
Units
• PPM
– Aqueous Solutions (mass concentration)
• Mass per 1,000,000 units of mass
• E.g.,
– 20 ppm = 20 mg/kg = 20 mg/ 1,000,000 mg
– Gaseous Solutions (volume concentration)
• Volume per 1,000,000 units of Volume
• E.g.,
– 20 ppm = 20 ml/m3 = 20 ml/1,000,000 ml
Atmospheric Concentration Units
• percent (x)
• parts per million (y)
• parts per billion (z)
x
y
z
 6 9
100 10 10
You try it...
• The smoke inhaled from a cigarette contains
about 400 ppm of CO. Express this
concentration as a percentage of the air
inhaled.
ANSWER: 0.04%
Examples
• Convert oxygen concentration from 21% to ppm
• Convert argon concentration from 0.9% to ppb
• Convert CO2 concentration from 350 ppm to %
Micrograms per Cubic Meter
3
(μg/m )
• Mass of pollutant / volume of air
1 μg/m3 = 1 ng/L
1 m3 = 1000 Liters
1 μg = 1 x 10-6 grams
1 ng = 1 x 10-9 grams
(See Tables A-1.3 and A-1.4 in your Text)
Regulations
How do we get from one to the
other?
Ideal Gas Law (IGL)
PV  nRT
P = Pressure (absolute)
V = Volume
n = Number of Moles
T = Temperature
R = Ideal Gas Law Constant
Ideal Gas Law
PV  nRT
• Make sure units are consistent!
• T is in Kelvin (K) not 0C
• To convert from 0C to K:
K = 0C + 273
298 K = 250C
Ideal Gas Law - Units
If R = 0.0821 L·atm/K·mol
Then P = 1 atm, V = Liters
If R = 8.31 J/K·mol
Then P = 101,325 Pa, V = m3
Why would ppm or ppb be the
preferred units for standards?
They are independent of
Pressure and Temperature
How much Chloroform is this?
Reference to scientific american
article
www.atsdr.cdc.gov
How do we get from one to the
other?
• Convert 50 ppm Chloroform (CHCl3 ) to
μg/m3.
• Convert 1 ppm Benzene (C6H6) to μg/m3.
How do we get from one to the
other?
• Chloroform (CHCl3 )
– 50 ppm = 243,000 μg/m3
• Benzene (C6H6)
– 1 ppm = 3188 μg/m3
Your Answers
We need to convert 975 micrograms CHCl3/m3 to parts per million.
We do this using the Ideal Gas Law PV = nRT
First we solve for n since we know our amount of CHCl3 in grams.
975mg/m^3 = 9.75E-4g/1000L
Multiply this # by the atomic weight of CHCl3 = 119.37g/mol
= 9.75E-7 * 119.37 = 1.164E-4 mol
This is our n value, now solve for V, V= (nRT)/P
Assume that the temp in a steamy shower is 25 C = 298K
V = ((1.164E-4 mol(0.0831)(298)/1)
V = 2.023E-7 L CHCl3 per 1 L of O2
Now to find out how much CHCl3 this is in ppm we multiply our V value by
10^6,
giving: .202 ppm
Considering that OSHA limits chloroform exposure to 50 ppm I think that the
amount of chloroform we are exposed to in the shower is of little concern.
Your Answers
•
50ppm for 8 hours a day 40 hours per work week...
PV=nRT
Pressure= 1 atm
Volume= 50 liter/1,000,000 liters
Moles= unknown
Constant= .0821
Temperature= 293 k
(1 atm)(50 liter chlorofrom/1,000,000 liters air) = n(.0821)(293)
number of moles = 2.079 moles
I don't know where else to go from here because I don't know the molecular formula
of chloroform, but if I had to take a guess I would say that it is CHCL3, makeing its
molecular wieght, 119.368 grams per mole
2.078 moles x 119.368 grams per mole = 248.05 grams
1000 liters = 1 cubic meter
(248.05 grams/liter) x 1,000,000 grams =
(248,050,000 micrograms/liter)/1000 liters=
248,050 micrograms/cubic meters allowed in 1 week
I really have no idea idea if this is correct... but if it is, there is no need to worry about
excessive exposure to chloroform in your shower.
Your Answers
•
50ppm for 8 hours a day 40 hours per work week...
I don't know if i should be concerned, but because of the bioacumulation concept-probably.
PV=nRT is the equation that I tried to figure this one out with.
I used:
P=pressure at 1atm
V=don't know--something in liters or micrograms or cubic meters.
n=think I'm trying to find this
R=0.0821 constant
T=293 Kelvins
Chloroform molecular weight=48g/mol
I tried to set the problem up a bunch of different ways and since i couldn't figure the
Volume out nothing worked.
•
Next i tried using the "conversion equation" at the bottom of page 505 under table 133 and that didn't work either.
•
Lastly I talked to a chemistry major friend who told me to do this equation (below)
on the basis that
ppm=micrograms/mL.
(900ug/cubic meter)(cubic meter/100cm cubed)(1cm cubed/1 mL)= 28.6ug/mL =
28.6 ppm chloroform
Then I checked one of the problems in Table 13-3 using this conversion equation
(above) and it did not work out, telling me that 28.6 ppm does not equal 900
micrograms.
Convert 50 ppm Chloroform to μg/m3.
• 50 ppm (by volume)
means there are 50
Liters of Chloroform
in 1 million liters of
air.
50 L CHCl3
6
10 L air
• 1000 L = 1 m3 so 106 L = 1000 m3
• So we want to know how many micrograms are in 50 Liters of
Chloroform.
Convert 50 ppm Chloroform to μg/m3.
• Use ideal gas law: PV = nRT
• Assume P = 1 atm, T = 25C=298 K,
R = 0.0821 L·atm/K·mol
PV
(1atm)(50 L)
n

RT (0.0821 atm  L )( 298 K )
K  mol
n  2.04 moles
Convert 50 ppm Chloroform to μg/m3.
• Chloroform has a molecular weight of 119 grams/mole
119 grams
2.04 moles 
 243 grams
mole
10 g
243 g 
 243,000,000 g
1g
6
Convert 50 ppm Chloroform to μg/m3.
• So there are 243,000,000 micrograms of Chloroform in 50
Liters of Chloroform
• There are 243,000,000 micrograms of Chloroform in 1
million Liters of air
Emissions vs Exposure
• Discussion from Scientific American Article
Indoor Air Pollution
Chloroform
• Picture of
person in
shower
• Picture of
person
smoking
Benzene
Indoor Air Pollution
Solutions
• Low tech solutions:
• Door mats, remove
shoes
• Dust sensor on
vacuum cleaner
• Ventilation – use CO2
What are the implications for
Regulation and Management?
CO2 Sources (Indoor)
• CO2 is produced from outdoor sources, indoor
combustion and metabolic generation
• Average CO2 concentration of exhaled human
breath is 4% (40,000 ppm)
• American Society of Heating Refrigeration and
Air-Conditioning Engineers (ASHRAE)
recommend a value of 1000 ppm
• Major health threat at concentrations of greater
than 6% (60,000 ppm)
Impacts
• CO2 is at least an indicator of poor
ventilation
• Minor impacts of “stuffiness”, discomfort,
shortness of breath, fatigue, lack of
attentiveness
• Serious impacts of death
Controls
• Better Ventilation
CO2 data
CO2 level (ppm)
Air Quality Lab - Annex Chapel
1800
1600
1400
1200
1000
800
600
400
200
0
9:50 AM 10:04
AM
10:19
AM
10:33
AM
10:48
AM
Time
11:02
AM
11:16
AM
11:31
AM
CO2 data
CO2 Emissions vs. Time in SCID Room 13
1000
900
800
700
CO2 (ppm)
600
500
400
300
200
100
0
10:00 10:02 10:04 10:06 10:08 10:10 10:12 10:14 10:16 10:18 10:20 10:22 10:24 10:26 10:28 10:30 10:32 10:34 10:36 10:38 10:40 10:42 10:44 10:46
AM
AM
AM
AM
AM
AM
AM
AM
AM
AM
AM
AM
AM
AM
AM
AM
AM
AM
AM
AM
AM
AM
AM
AM
Time
Ventilation Rate < 1 scfm/person
CO2 level (ppm)
Air Quality Lab - Annex Chapel
1800
1600
1400
1200
1000
800
600
400
200
0
9:50 AM 10:04
AM
10:19
AM
10:33
AM
10:48
AM
Time
11:02
AM
11:16
AM
11:31
AM
Ventilation Rate 18-19 scfm/person
CO2 Emissions vs. Time in SCID Room 13
1000
900
800
700
CO2 (ppm)
600
500
400
300
200
100
0
10:00 10:02 10:04 10:06 10:08 10:10 10:12 10:14 10:16 10:18 10:20 10:22 10:24 10:26 10:28 10:30 10:32 10:34 10:36 10:38 10:40 10:42 10:44 10:46
AM
AM
AM
AM
AM
AM
AM
AM
AM
AM
AM
AM
AM
AM
AM
AM
AM
AM
AM
AM
AM
AM
AM
AM
Time
Other Ventilation Rates
Space
Library Study
Ventilation Rate
(scfm/p)
6
Camper Van
1
Sci D 5
22-37 nr door,
15-19 nr windows
16-19
Dean’s Conference
Room
Theater Arts 114
2-3
Ventilation rate: Theater Arts 114
Time
CO2
0
5
8
11
15
18
20
23
25
Decay
541
519
514
495
489
492
490
496
490
0
0.169636
0.212561
0.394883
0.460124
0.426971
0.44895
0.384412
0.44895
Ventilation Rate
Design Capacity
Volume (ft2)
Slope
20
2880
0.0218
Ventilation Rate (scfm/person)
3.1392
Ventilation Rate Analysis
y = 0.0218x
2
R = 0.6613
0.6
Deay Rate
0.5
0.4
0.3
0.2
0.1
0
0
5
10
15
Time (min)
20
25
30
Carbon Monoxide
Sources
• CO is produced from incomplete
combustion of hydrocarbonds
• 77% from transportation
• Standards are 9 ppm over 8 hour period, or
35 ppm over 1 hour period
• CO is a primary pollutant and a precursor to
low-level ozone
Impacts
• CO is an asphyxiant
• Blood has a higher affinity for CO than O2,
so when inhaled, CO forms COHb
• Serious impacts from mental impairment to
death
Controls
• Must increase parameters that allow for
complete combustion of hydrocarbons
• For mobile sources, we can also: drive less
and use alternative fuels
Particulate Matter
Definition
• small particles of dust, soot, fumes, mist,
smoke, etc. found in the atmosphere
• Categorized by diameter in microns
• NAAQS standards for PM-10 and PM-2.5
are 150 and 65 micrograms/m3 respectively
over a 24 hour period
Sources
• Largest EPA tracked sources are
– fuel combustion (45%)
– industrial processes (33%)
– transportation (22%)
• Other sources include forest fires, windblown soil, construction sites
• Settling velocity is defined by a modified Stokes’
equation:
gd 
2
v
18
where,
v = terminal or settling velocity (m/s)
g = gravitational force constant = 9.80 m/s2
d = particle diameter (m)
 = density of particle (g/m3)
 = viscosity of air = 0.017 g/m-s
Impacts
• Serious impacts on human respiratory
functions.
• Smaller the particle, the farther it can get
lodged in the respiratory system
• Larger particles can also attract
hydrocarbons and transport these into our
bodies
Insert figure of human respiratory system
Controls
insert cyclone
• Electrostatic precipitator
• Fabric
filter
baghouse
Visibility
Factors in Visual Air Quality
•
•
•
•
•
•
Viewing direction
Terrain
Time of day
Season
Meteorology
Air pollution
Visibility Metrics
• Extinction Coefficient:
– reduction of image forming light per unit
distance due to scattering
• Visual Range:
– How far away a large black target could be seen
• Deciview scale:
– Perceptual scale based on just-noticabledifferences (analogous to decibels in sound)
SOx
Transport
Utilities
Industrial
Slice 4
Sources of SOx
• SO2 is emitted primarily from coal burning
power plants
S(s) + O2(g) ==> SO2(g)
SO2(g) + OH* ==> HOSO2*
HOSO2*(g) + O2(g) ==> SO3(g) + HO2*
SO3(g) + H2O(l) ==> H2SO4(aq)
H2SO4(aq) ==> 2H+(aq) + SO42-(aq)
Definition of Acid
• Acids are substances that release
hydrogen ions, H+, usually in aqueous
solutions
pH of Natural Rain Water
• Rain water typically has a pH less than 7.0
due to dissolved CO2:
CO2(g) + H2O(l) <==>H2CO3(aq)
H2CO3(aq) <==> H+(aq) + HCO3-(aq)
• This can bring rain pH down to values as
low as 5.6.
• So where does the other acid come from?
Measuring the pH of Rain
insert map of pH distribution is
US
Sources of Acid Rain
• NO2 is emitted from burning and undergoes
the following:
N2(g) + O2(g) ==> 2NO(g)
2NO(g) + O2(g) ==> 2NO2(g)
NO2(g) + OH*(g) ==> HNO3(l)
HNO3(l) ==> H+(aq) + NO3-(aq)
Impacts of Acid Deposition
• Damage to materials (especially limestone):
CaCO3(s) + 2H+(aq) ==> Ca2+(aq) + CO2(g) +
H2O(l)
• Reduced visibility
• Respiratory problems in humans
Impacts on Lakes and Streams
• Direct aquatic impacts
• Indirect aquatic impacts
• Acid Neutralizing Capacity (ANC) which is
a function of geology..
SO2 Controls
• Fuel switching
• Coal cleaning
• Fluidized-bed combustion
2CaCO3(s) + 2SO2(g) + O2(g) ==>2CaSO4(s)
+ 2CO2(g)
• Flue Gas Desulfurization (“Scrubbers”)
Control Policy
• Political implications of SO2 control based
on regionality of pollution sources and
impacts
• Clean Air Act Amendments of 1990 and the
Title IV SO2 Allowance Trading System
Photochemical Smog
Origins
• primarily low-level O3
• Standards in U.S. is 80 ppb over 8 hours
• Produced by NOx and volatile organic
compound (VOC) reactions
• O3 is a secondary pollutant
Sources
• Sources of NO and VOCs are:
– any kind of burning in air with hot flame
– unburned hydrocarbons industrial solvents and
other organic compounds
– non-anthropogenic sources
Formation Reactions
• The general reactions are:
N2 + O2 ==> 2NO (mostly)
NO + [VOCs or CO] ==> NO2
NO2 + hv ==> NO + O
O + O2 ==> O3
• Note in evenings:
NO + O3 ==> NO2 + O2
Impacts
• Mostly local
• Respiratory problems
• May stunt vegetative growth; may reduce
immune systems of vegetation.
Controls
• Must reduce either NOx or VOCs
• Evaporative controls
• Burning at lower temperatures
Controls (con’d)
• Mobile sources:
– driving less (carpools, mass transit)
– cleaner fuels
– evaporative controls
– catalytic converters (three-way converters) can reduce
NO and can oxidize unburned fuel
2H2 + 2NO ==> N2 + 2H2O
Fuel + O2 ==> CO2 + H2O
Stationary
Sources
Precombustion Controls
• Fuel switching
• Coal cleaning
Combustion Controls
• Fluidized-bed combustion
2CaCO3(s) + 2SO2(g) + O2(g)
==>2CaSO4(s) + 2CO2(g)
• Integrated Gasification Combined
Cycle
Post Combustion Controls
• Flue Gas Desulfurization
(“Scrubbers”)
CaCO3 + SO2 + 2H2O ==>CaSO3• 2H2O +
CO2(g)
Air Pollution Transport and
Dispersion
What information (input) might
be required for an air pollution
model?
What information (output)
might an air pollution model
provide?
Air Pollution Meteorology
• Predictions of ambient concentrations use
models that consider: source, chemical
composition, meteorology, and atmospheric
chemical reactions
Air Pollution Transport
• Pollution is transported at the speed and in
the direction of the wind
• Pollutants also disperse under turbulence,
changes in wind direction/speed, and
molecular diffusion
• warm air rises
• cold air sinks
Atmospheric Conditions
• Stable vs. Unstable
– Stable = NOT MOVING
– Unstable = MOVING
Which condition would have the
higher likelihood of higher
concentrations?
Adiabatic Lapse Rate
dT
g

dz
cp
where g is 9.8 m/s2 and cp is 1005 J/kg-K.
Adiabatic Lapse Rate
• Thus, for every increase in altitude of 100
meters, we expect a parcel of air to drop in
temperature 1 degree C.
• This is true for dry air with no heat transfer
across its boundary (adiabatic)
Inversions
• Inversions occur when ambient temperature
actually increases with vertical distance
• In this case, the rising parcel is always
pushed down
• Radiation and subsidence inversions are
possible
Insert plume figure
Gaussian Plume Model
Dispersion Modeling
C x , y , z  
Q
2 y z u
e
1  y

2   y




2
e
1 z
 
2  z



2
where x=0 at source, and y=z=0 at centerline of plume
Dispersion Modeling
C x , y , z  
Q
2 y z u
e
1  y

2   y




2
e
1  Z H
 
2  z



2
Dispersion Modeling
C x , y , z  
Q
2 y z u
e
1  y

2   y




2


e


1  Z H
 
2  z



2
e
1 Z H
 
2  z



2





Assumptions
• These models assume
–
–
–
–
–
–
wind is uniform and average speed used
Gaussian distributions laterally and vertically
No loss by decay, reaction, deposition
Relatively flat, homogeneous surface
Steady state conditions
Pollutants have same density as air around them
Proposed Paper Processing Mill
• Emits 500 kg of H2S per day from a single
stack
• Small town 1700 m NorthEast
• SW winds 15% of the time
• H2S concentration can not exceed 20 ppb or
30.3 μg/m3
Emissions Data
•
•
•
•
•
•
•
Gas exit velocity 20 m/s
Gas exit temperature 122ºC
Stack diameter at top 2.5 m
Ambient air temperature 17 ºC
Wind velocity 2 m/s
Temperature lapse rate 6 ºC/km
Slightly stable atmosphere – category E
At z=0, y=0 this reduces to
C x , 0, 0  
Q
 y zu
e
 H2

 2 2 z





Plume Diffusion Equation
C x , y , z  
Q
2 y zu
e
1  y

2   y
Equation 13-28 in text




2
1 Z H 
  1  Z  H 

 
 2  z 
2  z 
e
e

2
2




Source Strength
500kg / day
3
Q
 5.79 x10 kg / s
86,400s / day
What you should be able to do...
– Understand the role of each variable in
downwind concentration predictions
– Simplify the Gaussian model based on x,y,z
information
– Given variable values, use the Gaussian model
to predict concentrations
– Understand sensitivity of model to changes in
variables
“Improvement in domestic emissions
from space heating is possible by
using district heating systems and
electric heating that transfer
emissions to a different site, where
tall stacks and emission control
systems can be installed.”
(page 510 of your text) – use as a
transition to energy module