306
8 Secondary Ventilation
t =
3000 m
3
(27 + 0.2)
m 3
s
· ln
0.2
m
3
s
− 0.07 · (27 + 0.2)
m
3
s
0.2
m 3
s
− 0.01 · (27 + 0.2)
m 3
s
= 349.0 s = 5.8 min
(e–c) There is a CH 4 leak of Q g = 0.2 SCMS and the sweeping air has a
concentration B = 0.0025, then:
t =
V
Q + Q g
ln
Q B + Q g − X 0
Q + Q g
Q B + Q g − X
Q + Q g
t =
3000 m 3
(27 + 0.2)
m 3
s
· ln
27 · 0.0025
m 3
s + 0.2
m 3
s − 0.07 · (27 + 0.2)
m 3
s
27
m 3
s · 0.0025 + 0.2
m 3
s − 0.01 · (27 + 0.2)
m 3
s
= 650.3 s = 10.8 min
(f) By substituting values in Eq. 8.11, the following table can be constructed:
Flow rate (SCMS)
t (s)
t (min)
27
650.3
10.8
32
358.7
6.0
38
259.8
4.3
45
200.3
3.3
54
155.9
2.6
64
125.5
2.1
76
101.9
1.7
91
82.5
1.4
Therefore, plotting it we have that:
0
2
4
6
8
10
12
0
20
40
60
80
100
Time (min)
Flow rate (SCMS)