8.6 Advection-Diffusion Method
315
C(x, t) =
V
2 A
√
π Dt
e
−(x−um ·t) 2
4Dt
C(50, 180) =
⎛
⎝
0.3861 m
3
2 · 9 m 2 ·
π · 17
m 2
s
· 180 s
⎞
⎠ e
−(50 m−0.6 m
s ·180 s) 2
4·17 m 2
s ·180 s
= 1.66 × 10
−4
This is equal to 166 ppm, which is greater than the acceptable safe limit for this
gas of 10 ppm.
(c) The point (x) where the highest concentration of gas is found corresponds to
the maximum value of Eq. 8.25. This value is obtained by taking the derivative
of Eq. 8.25 and setting it to zero:
x − u m t = 0; x = u m t
So, for t = 180 s (3 min), x = 108 m.
Then, the maximum concentration for t = 180 s is:
C(108, 180) =
⎛
⎝
0.3861 m
3
2 · 9 m 2 ·
π · 17
m 2
s
· 180 s
⎞
⎠ e
−(108 m−0.6 m
s ·180 s) 2
4·17 m 2
s ·180 s
= 2.188 × 10
−4
= 218.8 ppm
(e) Distribution of the contamination of NO 2 in the gallery after 180 s:
x (m)
C(x, t = 180) (ppm)
0.0
84.4
5.0
92.0
6.3
93.9
8.3
97.2
12.5
103.8
25.0
124.6
33.3
138.7
50.0
166.2
75.0
200.1
100.0
217.6
150.0
189.4
200.0
109.6
250.0
42.1
(continued)
315
C(x, t) =
V
2 A
√
π Dt
e
−(x−um ·t) 2
4Dt
C(50, 180) =
⎛
⎝
0.3861 m
3
2 · 9 m 2 ·
π · 17
m 2
s
· 180 s
⎞
⎠ e
−(50 m−0.6 m
s ·180 s) 2
4·17 m 2
s ·180 s
= 1.66 × 10
−4
This is equal to 166 ppm, which is greater than the acceptable safe limit for this
gas of 10 ppm.
(c) The point (x) where the highest concentration of gas is found corresponds to
the maximum value of Eq. 8.25. This value is obtained by taking the derivative
of Eq. 8.25 and setting it to zero:
x − u m t = 0; x = u m t
So, for t = 180 s (3 min), x = 108 m.
Then, the maximum concentration for t = 180 s is:
C(108, 180) =
⎛
⎝
0.3861 m
3
2 · 9 m 2 ·
π · 17
m 2
s
· 180 s
⎞
⎠ e
−(108 m−0.6 m
s ·180 s) 2
4·17 m 2
s ·180 s
= 2.188 × 10
−4
= 218.8 ppm
(e) Distribution of the contamination of NO 2 in the gallery after 180 s:
x (m)
C(x, t = 180) (ppm)
0.0
84.4
5.0
92.0
6.3
93.9
8.3
97.2
12.5
103.8
25.0
124.6
33.3
138.7
50.0
166.2
75.0
200.1
100.0
217.6
150.0
189.4
200.0
109.6
250.0
42.1
(continued)