266
7 The Role of Ventilation in Fires and Explosions
where
• L b : Backlayering or distance traveled (m) by the fumes in the opposite direction
to the ventilation airflow. The critical speed (V c ) corresponds to the value of L b
= 0, i.e. no backlayering.
• W: Width of the airway (m).
All other parameters and variables remain the same as before, with the same
definitions, values and units.
Exercise 7.3 Calculate the critical speed for the situation described previously in
Exercise 7.2 using the revised model proposed in NFPA 502:2020. Assume that the
cross section of the gallery resembles an ellipse.
Solution
Where NFPA 502:2020 uses the height (H) and width (W) of the tunnel, the NFPA
502:2017 uses the height (H) and free area of the cross section (A x ). For the purpose
of comparative calculations, the geometric figure of the area is considered to be an
ellipse. The formula applied to calculate the area is, therefore:
A x = π
H
2
W
2
where solving for W we have that:
10 = π
3.5
2
W
2
W = 3.64 m
Parameter A is calculated as:
A =
Q
ρ a C p T a g
1
2 H
5
2
A =
6000
1.204 · 1.005 · (273.15 + 26) · 9.81
1
2 · 3.5
5
2
= 0.23565
Parameter B is obtained from:
B = 0.15
H
W
−
1
4
B = 0.15
3.5
3.64
−
1
4 = 0.15146
7 The Role of Ventilation in Fires and Explosions
where
• L b : Backlayering or distance traveled (m) by the fumes in the opposite direction
to the ventilation airflow. The critical speed (V c ) corresponds to the value of L b
= 0, i.e. no backlayering.
• W: Width of the airway (m).
All other parameters and variables remain the same as before, with the same
definitions, values and units.
Exercise 7.3 Calculate the critical speed for the situation described previously in
Exercise 7.2 using the revised model proposed in NFPA 502:2020. Assume that the
cross section of the gallery resembles an ellipse.
Solution
Where NFPA 502:2020 uses the height (H) and width (W) of the tunnel, the NFPA
502:2017 uses the height (H) and free area of the cross section (A x ). For the purpose
of comparative calculations, the geometric figure of the area is considered to be an
ellipse. The formula applied to calculate the area is, therefore:
A x = π
H
2
W
2
where solving for W we have that:
10 = π
3.5
2
W
2
W = 3.64 m
Parameter A is calculated as:
A =
Q
ρ a C p T a g
1
2 H
5
2
A =
6000
1.204 · 1.005 · (273.15 + 26) · 9.81
1
2 · 3.5
5
2
= 0.23565
Parameter B is obtained from:
B = 0.15
H
W
−
1
4
B = 0.15
3.5
3.64
−
1
4 = 0.15146