Air - Civil and Environmental Engineering | SIU

Report
FE Review for
Environmental
Engineering
Problems, problems, problems
Presented by L.R. Chevalier, Ph.D., P.E.
Department of Civil and Environmental Engineering
Southern Illinois University Carbondale
FE Review for Environmental Engineering
AIR
Problem
Strategy
Solution
Estimate the amount of soda ash required per m3 of exhaust gas to neutralize
20,000 g/m3 of SO2 at 20° C.
Problem
Strategy
Solution
• Review governing reaction
• Determine the mass of soda ash needed from mole ratio
The neutralization reaction is:
SO 2  Na 2 CO 3  Na 2 SO 3  CO
MW: SO2 = 64
Na2CO3 = 106
2
The mass of soda ash is then:
m Na 2 CO 3
 106 g Na 2 CO 3
 20 . 0 mg SO 2 
64 g SO 2

 33 . 125 mg
Therefore, for each m3 of exhaust gas, we require
33.1 mg Na2CO3




Problem
Strategy
Solution
Consider a box model for an air shed over a city 1 x 105 m wide with a mixing
depth of 1200 m. Winds with no SO2 blow at 4 m/s against one side of the box.
SO2 is emitted in the box at a rate of 20 kg/s. If SO2 is considered to be
conservative, estimate the steady state concentration in the air shed.
Report your answer in g/m3.
Problem
Strategy
Solution
• Draw a schematic of your system
• Consider mass balance
Solution
L
H = 1200 m
wind
u = 4 m/s
Co = 0 g/m3
W = 1 x 105 m
wind
u = 4 m/s
Ce =?
Emission
20 kg/s
Problem
Strategy
Solution
L
H = 1200 m
wind
u = 4 m/s
Co = 0 g/m3
W = 1 x 105 m
wind
u = 4 m/s
Ce =?
Emission
20 kg/s
Ce = Co +(qL)/(uH)
L
H = 1200 m
wind
u = 4 m/s
Co = 0 g/m3
W = 1 x 105 m
wind
u = 4 m/s
Ce =?
Ce = Co +(qL)/(uH)
Emission
20 kg/s
Ce 
 20 kg s 

L
 LW 


 4 m s 1200 m 
 10
9 g
kg
 41 . 67
g
m
3
Problem
Strategy
Solution
Consider the emission of SO2 from a coal fired power plant, at a rate of 1,500 g/s.
The wind speed is 4.0 m/s on a sunny afternoon. What is the centerline
concentration of SO2 3 km downwind (Note: centerline implies y=0).
Stack parameters:
Height = 130 m
Diameter = 1.5 m
Exit velocity = 12 m/s
Temperature = 320°C (593° K)
Atmospheric conditions: P=100 kPa T=25° C (298° K)
• Review data
• Estimate effective stack height
vs d 
H 
1 . 5 
u 

 Ts  Ta   
2
 2 . 68  10  P 
 d  
 Ts   

• Determine stability class
• Calculate sy and sz
• Review terms and apply governing equation

C  x , y ,0 , H   
 

Q
  exp
s y s z u  

2


1 y  
   exp
 
 2  s y    


 1  H 2 
 
  

2
s

 z   
12 m s 1 . 50 m  


 593  298 
2
H 
1 . 5   2 . 68  10
100 
1 . 5  

 4 .0 m s  
593




  4 . 5 m  3 . 43   15 . 44 m
H = effective stack height
= h + H
= 130 m + 15.4 m = 145.4 m
Problem
Strategy
Solution
Atmospheric stability class: Class B
sy = ax0.894 = 156(3)0.894 = 416.6 m
sz = cxd + f = 108.2(3)1.098 + 2 = 363.5 m
1

 1  145 . 44
1500
C  x, y, z, H  
exp  0  exp   
  416 . 6  363 . 5  4 . 0 
 2  363 . 5
4

exp   0 .08 
4
0 .923 
 7 . 88  10
 7 . 88  10
 7 . 28  10
4
3
g m of SO 2
 728 . 6  g m
3
2

 
 
Example
Solution
Simplify the Gaussian dispersion model to describe a ground level source with
no thermal or momentum flux, which is the typical release that occurs at a
hazardous waste sites. In this situation, the effective plume rise, H, is
essentially 0.

C  x , y ,0 , H   
 

Q
  exp
s y s z u  

2


1 y  
   exp
 
 2  s y    



C  x , y ,0 , H   
 

Q
  exp
s y s z u  

2


1 y  
  exp 0 
 
 2  s y   



C  x , y ,0 , H   
 
2



Q
1 y 
 
 exp   
s y s z u 
 2  s y  


 1  H 2 
 
  

2
s

 z   
Problem
Strategy
Solution
Consider soil under a single story house that emits 1.0 pCi/m2·s of radon gas.
The house has 250 m2 of floor space, and average ceiling height of 2.6 m, and
an air change rate of 0.9 ach. Estimate the steady state concentration of
radon in the house, assuming that the ambient concentration is negligible.
For radon, k = 7.6 x 10-3 hr-1
• Review data
• Review SS equation
Volume = (250 m2)(2.6 m) = 650 m3
Q/V = 0.9
Ventilation rate = Q = (650 m3)(0.9) = 585 m3/hr
Ci = 0
E = (1.0 pCi/m2·s)(250 m2)(3600 s/hr) = 900000 pCi/hr
Q
Ce  V
Ci 
Q
V
E
V
k
90000
pCi
hr
3
650 m
Ce 
1
0 . 9 hr  0 . 0076 hr
 1525
pCi
m
3
1

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