272
7 The Role of Ventilation in Fires and Explosions
Moreover, the y-axis represents the percentage of effective combustible gases,
according to the expression shown in Eq. 7.25:
Y (%) = CH 4 (%) + 0.4 CO(%) + 1.25 H 2 (%)
(7.25)
The expression given above is the classic version, but nowadays it is expanded to
take into account other gases, yielding Eq. 7.26:
Y (%) = CH 4 (%) + 0.4 CO(%) + 1.25 H 2 (%) + 0.6 C 2 H 6 (%) + 0.54 C 2 H 4 (%)
(7.26)
Finally, when a curve is being selected from the diagram, the ratio of CH 4 to total
combustible gases (R) as in Eq. 7.27 must be used:
R =
CH 4 (%)
CH 4 (%) + H 2 (%) + CO(%)
(7.27)
In the case of more recent publications, R is expanded to include further gases
(Eq. 7.28):
R =
CH 4 (%)
CH 4 (%) + H 2 (%) + CO(%) + 0.6 C 2 H 6 (%) + 0.54 C 2 H 4 (%)
(7.28)
The mixtures that fall within each of the areas A 1 , A 2 , A 3 and so on are explosive.
Each R also defines the oblique line that joins the extreme right of each area with the
origin of the coordinates. This line marks the boundary between explosive mixtures
and those that are explosive when mixed with air. An important note is that the
method is not valid if CO (%) > 3.0 and H 2 (%) > 5 (Zabetakis et al. 1959).
Exercise 7.6 Determine the explosiveness of the following gas mixtures using the
USBM graph:
(a) O 2 = 8.7%, CO = 0.94%, CH 4 = 1.8%, CO 2 = 11.2%, H 2 = 3.3%, N 2 =
74.06%.
(b) O 2 = 2.0%, CO = 1.3%, CH 4 = 3.0%, CO 2 = 26.0%, H 2 = 1.2%, N 2 = 66.5%.
Solution
(a)
First, calculate the percentage of N 2 excess:
N 2 excess (%) = N 2 sample (%) − N 2 normal (%) = 74.06-8.7 · (79.04 ÷ 20.93)
= 41.2
Second, calculate the effective inert gases:
X (%) = N 2 excess (%) + 1.5 CO 2 (%) = 41.2 + 1.5 · 11.2 = 58
Third, calculate the effective combustible gases:
Y (%) = CH 4 (%) + 1.25 H 2 (%) + 6.4 CO (%) = 1.8 + 1.25 · 3.3 + 0.4 · 0.94
= 6.3
7 The Role of Ventilation in Fires and Explosions
Moreover, the y-axis represents the percentage of effective combustible gases,
according to the expression shown in Eq. 7.25:
Y (%) = CH 4 (%) + 0.4 CO(%) + 1.25 H 2 (%)
(7.25)
The expression given above is the classic version, but nowadays it is expanded to
take into account other gases, yielding Eq. 7.26:
Y (%) = CH 4 (%) + 0.4 CO(%) + 1.25 H 2 (%) + 0.6 C 2 H 6 (%) + 0.54 C 2 H 4 (%)
(7.26)
Finally, when a curve is being selected from the diagram, the ratio of CH 4 to total
combustible gases (R) as in Eq. 7.27 must be used:
R =
CH 4 (%)
CH 4 (%) + H 2 (%) + CO(%)
(7.27)
In the case of more recent publications, R is expanded to include further gases
(Eq. 7.28):
R =
CH 4 (%)
CH 4 (%) + H 2 (%) + CO(%) + 0.6 C 2 H 6 (%) + 0.54 C 2 H 4 (%)
(7.28)
The mixtures that fall within each of the areas A 1 , A 2 , A 3 and so on are explosive.
Each R also defines the oblique line that joins the extreme right of each area with the
origin of the coordinates. This line marks the boundary between explosive mixtures
and those that are explosive when mixed with air. An important note is that the
method is not valid if CO (%) > 3.0 and H 2 (%) > 5 (Zabetakis et al. 1959).
Exercise 7.6 Determine the explosiveness of the following gas mixtures using the
USBM graph:
(a) O 2 = 8.7%, CO = 0.94%, CH 4 = 1.8%, CO 2 = 11.2%, H 2 = 3.3%, N 2 =
74.06%.
(b) O 2 = 2.0%, CO = 1.3%, CH 4 = 3.0%, CO 2 = 26.0%, H 2 = 1.2%, N 2 = 66.5%.
Solution
(a)
First, calculate the percentage of N 2 excess:
N 2 excess (%) = N 2 sample (%) − N 2 normal (%) = 74.06-8.7 · (79.04 ÷ 20.93)
= 41.2
Second, calculate the effective inert gases:
X (%) = N 2 excess (%) + 1.5 CO 2 (%) = 41.2 + 1.5 · 11.2 = 58
Third, calculate the effective combustible gases:
Y (%) = CH 4 (%) + 1.25 H 2 (%) + 6.4 CO (%) = 1.8 + 1.25 · 3.3 + 0.4 · 0.94
= 6.3