CE 326 Principles of Environmental Engineering
Air Pollution
Shane Rogers
Review of the Ideal Gas Law:
Ideal Gas Law:
[1]
Where:
P
V
n
R
T
= absolute pressure (kPa)
= volume occupied by n moles of gas (L)
= number of moles of gas in question (moles) = mass/molecular wt.
= ideal gas law constant (8.3143 J/(K·mole))
= absolute temperature (K)
Remembering that the density of something is the mass divided by the volume that it occupies,
the volume and the density of n moles of gas at a particular temperature and pressure are related
to one another. This can be expressed:
=
nGMW  M

V
V
Where:
[2]

= density of the gas (gm/L)
GMW = gram molecular weight of 1 mole of gas (gm/mol)
M
= mass of the gas (gm)
And the ideal gas law becomes:
`
[3]
Dalton’s Law of Partial Pressures – The total pressure exerted by a mixture of gases is equal to
the sum of partial pressures exerted by each component of the gas.
In terms of pressure:
Ptotal = P1 + P2 + P3 +…+ Pn
[4]
Pi 
ni RT
Vtotal
Ptotal
 ni RT  n   RT 

  Pi    
   ni 
V
V
i 1
i 1  total 
i 1   total  
n
[5]
n
[6]
In terms of volume:
Vtotal = V1 + V2 + V3 +…+ Vn
[7]
1
Vi 
ni RT
Ptotal
Vtotal
 ni RT  n   RT 

  Vi    
   ni 
P
P
i 1
i 1  total 
i 1   total  
n
[8]
n
[9]
Measuring the concentration of gaseous pollutants:
The concentration of a gas within a mixture of gases can be expressed:
 mass per volume basis (particulate matter or gaseous) = µg/m3
 volume per volume basis (gaseous only) = parts per million by volume (ppmv).
ppmv 
 
Vi
10 6 , where Vi and Vtotal are in the same units
Vtotal
[10]
Concentrations expressed as ppmv are particularly useful for comparing different regions on air
quality because changes in temperature or total system pressure or volume do not affect ppmv
values (assuming the gases behave ideally).
Calculating the µg/m3 of a gas from the partial pressure:
Concentration is mass per volume. By rearranging equation 5, and substituting it in for the total
volume, we can calculate the concentration of gas:
Concentration  C 
Mass
moles(GMW ) ni GMW  Pi GMW 



Volume
Volume
Vtotal
RT
[11]
Example:
The partial pressure of SO2 in the air is 0.000001 atm. Determine the concentration of SO2 in
µg/m3. Assume standard temperature and pressure (STP).
Converting µg/m3 to ppmv:
Substituting equations 8 and 11 into 10, with Ci in µg/m3:
ppmv 
ni GMW 
Vi
n RT
10 6 , Vi  i
, and Ci 
Vtotal
Vtotal
Ptotal
 
2
we get:
ppmv 
 
 
Vi
n RT
10 6  i
10 6 
Vtotal
Vtotal Ptotal
 
Ci RT
10 6

µg 

Ptotal GMW 10 6
g 

[12]
Example 1:
Consider CO in air at 200 µg/m3. Convert this to ppmv. Assume STP.
Example 2:
Recalculate the above at 20ºC and 1.1 atm.
Example 3:
Consider air with 4 ppmv benzene (C6H6) at STP. Calculate Cbenzene in µg/m3.
Modeling the dispersion of pollution from a stack:
Gaussian Dispersion Model:
The Gaussian dispersion model is one way of estimating the ground-level concentration of a
pollutant emitted from a smokestack at any distance down-wind from the stack.
Assuming:
1. Atmospheric stability is uniform throughout the layer into which the gas stream is
discharged.
3
2. Turbulent diffusion is a random activity allowing the dilution of the contaminated gas
stream to be described in the horizontal or vertical direction by a Gaussian (normal)
distribution.
3. Pollutant is emitted into the atmosphere at a height equal to the stack height plus an
effective plume rise based on the temperature, pressure, and velocity of the gases emitted
from the stack
4. The degree of dilution of the effluent plume is inversely proportional to the wind speed
5. Pollutants reaching the ground level are reflected back into the atmosphere
The ground-level concentration of a pollutant emitted from a smokestack at any x,y location
is given by:
 E 
 x, y,0, H   
e
 s y s z u 
 
 1  y
 2 sy
 




2



e
 1 H
 
 2  s z




2



[13]
Where:
(x,y,0,H)
E
sy
sz
u
x, y, z
H
= downwind concentration at the ground level (z = 0), g/m3
= emission rate of pollutant, g/s
= plume standard deviation in the y-direction, m
= plume standard deviation in the z-direction, m
= wind speed, m/s
= distances, m
= effective stack height, m
The effective stack height (H) in equation 13 is the sum of the physical stack height (h) and the
rise in the effluent gas (H):
H
[14]
The rise in the effluent gas from the top of the stack is due to the temperature, pressure, and exit
velocity of the gases emitted:
 sd 

 Ts  Ta  
2

d 
H 
1.5   2.68 10 P 

u 
 Ts  

Where:
vs
d
u
P
Ts
Ta
= stack velocity, m/s
= stack diameter, m
= wind speed, m/s
= atmospheric pressure, kPa
= stack gas temperature, K
= air temperature, K
The plume standard deviations in the y and z directions are described with the algorithms:
4
[15]
s y  ax0.894
[16]
s z  cx d  f
[17]
Where a, c, d, and f are constants that depend on atmospheric conditions and the distance
downwind from the stack (see Table 6-7 in the text), sy and sz are in meters, and x is input in km.
Example:
A coal-fired power plant has a 2 m diameter and 30 m tall smokestack that exhausts a mixture of
gases containing 50 µg/m3 SO2 at a total flowrate of 22 m3/s and a temperature of 315ºC. The
atmosphere is stable (class C) at 101.325 kPa and 20ºC. The wind is blowing at 2 m/s.
Determine the ground level concentration of SO2 at a distance of 0.4 km directly down-wind
(y=0) from the smokestack.
Solution:
from table 6-7 in the text (pg. 506), a = 104, c = 61, d = 0.911, f = 0, and:
s y  ax0.894 
sz  cx d  f 
The velocity of gas coming out of the smokestack is given by the flowrate divided by the crosssectional area of the stack. The emission rate of SO2 is given by the flowrate times the
concentration of SO2 in the flue gas such that:
 m3 
 22 
s 
vs  

 2m 2
4
and
 m3  µg  1g 
E   22  50 3  6  
s  m  10 µg 

Substituting equation 15 into equation 14 and solving for H:
 sd 

 Ts  Ta  
2

d 
H  h  H  h 
1.5   2.68  10 P 

u 
 Ts  

 m
 7 2m 


s
 588K  293K  
2


H  30m  
1
.
5

2
.
68

10
101
.
325
kPa


2m 


m
588
K

 


2
s
H
5
Finally, solving for the downwind concentration at 0.4km using equation 13:
 E 
 400m,0m,0m,59.6m   
e
 s y s z u 
 
 1  y
 2 sy
 




2
2

   1  H  
  2 s  
   z 
e

 
2
2
1
   1
 






  2  45.84m    2  26.47 m  
 

e 

e
  ______ m ______ m  2 m  



 s  
g 
g

  __________ 3  ___ _______   _________ 3
m 
m

µg
 400m,0m,0m,59.6m   __________ 3
m
6