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Air Pollution Tutorial: Ground-Level Concentration Calculations, Assignments of Refrigeration and Air Conditioning

sample of air pollution and useful

Typology: Assignments

2020/2021

Uploaded on 06/12/2022

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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.
SO2 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:
HS=90m, U10=4m/s, Q=400 g/s,
Δh=60 m
a) Ground level, Z=0
He= Hs+
Δh
= 90 + 60 = 150m
Uz=U10 (Z
10 )0 . 2=4( 90/10 )0 . 2=6 . 21 m/S
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Download Air Pollution Tutorial: Ground-Level Concentration Calculations and more Assignments Refrigeration and Air Conditioning in PDF only on Docsity!

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

  1. 2
  1. 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.