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The University of Western Ontario
Faculty of Engineering
Department of Chemical and Biochemical Engineering
Air Pollution (CBE 4405/ CBE 9312)
Tutorial # 03
Problem - 1
At noon on a sunny summer day (stability class C) with a surface speed (10 m height) of 4 m/s.
SO
2
is released over rough terrain from a 90 m stack at a rate of 400 g/s in an urban regime.
Assume that the plume rise is 60 m, calculate the ground-level centerline concentration:
a) At 3000 m downwind
b) At 3000 m downwind and 100 m crosswind.
Solution:
H S
=90m, U 10
=4m/s, Q=400 g/s,
Δh = 60 m
a) Ground level, Z=
H e
= H s
Δh
= 90 + 60 = 150m
U
z
= U
10
Z
- 2
- 2
= 6. 21 m / S
Class C and x=3 km,
σ
y
= 290 m
,
σ
z
= 180 m
C =
Q
πU σ
y
σ
z
exp−
(
H
σ
z
)
2
400 g / s
π ∗6.
m
s
∗ 290 m ∗ 180 m
exp−
(
150 m
180 m
)
2
= 5.56 * 10
g/m
3
.
b) x = 3 km, y = 100 m
C =
Q
πU σ
y
σ
z
exp−
(
y
σ
y
)
2
∗exp−
(
H
σ
z
)
2
400 g / s
π ∗6.
m
s
∗ 290 m ∗ 180 m
exp−
(
100 m
290 m
)
2
exp−
(
150 m
180 m
)
2
g/m
3
Problem - 2
Nitric oxide (NO) is emitted at 110 g/s from a stack with physical height of 80 m. The wind
speed at 80 m is 5 m/s on an overcast morning. Plume rise is 20 m. Class D should be assumed
for overcast condition, day or night, calculate:
a) The ground level concentration 2.0 km downwind from the stack
b) The concentration at 100 m off the centerline at the same x distance.
Solution:
Q=110 g/s, H s
=80 m, u=5 m/s,
Δh
=20 m, Class D
a) H e
= H s
Δh
= 80 + 20 = 100 m
Ground level, Z=0, y=0, x=2 km
σ
y
= 150 m
,
σ
z
= 50 m
C =
Q
πU σ
y
σ
z
exp−
(
H
σ
z
)
2
110 g / s
π ∗ 5
m
s
∗ 150 m ∗ 50 m
exp−
(
100 m
50 m
)
2
= 1.26 * 10
g/m
3
.
Problem – 4
A factory emits 20 g/s of SO 2
at height H and the wind speed is 3 m/s. At a distance of 1 km
downwind, the values of σ y
and σ z
are 30 and 20 m, respectively. What are the SO 2
concentrations at the centerline of the plume, and at a point 60 m to the side of and 20 m below
the centerline?
Solution:
Q = 20 g/s, u = 3m/s, x =1 km,
σ
y
= 30 m ,
σ
z
= 20 m
a) At centre line, Z - H e
= 0, y = 0
C =
Q
2 πuδ
y
δ
z
=
20
2 π ∗ 3 ∗ 30 ∗ 20
= 1. 77 ∗ 10
− 3
g / m
3
= 1770 μg / m
3
b) y = 60, (Z-H e
)= -20 m
C =
20
2 ∗ π ∗ 3 ∗ 30 ∗ 20
exp−
1
2
(
60
30
)
exp−
1
2
(
− 20
20
)
¿ 1. 45 ∗ 10
g / m
= 145 μg / m
Problem 5
Determine the atmospheric stability for each of the following temperature profiles:
T (
o
Case Z (m) C)
(i )
(ii )
(iii )
Solution:
i)
ΔT
ΔZ
T
2
− T
1
Z
2
− Z
1
= 0. 015 C / m
temperature inversion
ii)
ΔT
ΔZ
=− 0. 0314 C / m =− 3. 14 C / 100 m
It is more negative than ambient LR, the atmosphere is Unstable.
iii)
ΔT
ΔZ
− 3
C / m =− 0. 187 C /100 m
it is more positive than ambient LR, the atmosphere is stable.
