&
☭proletarian☭
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- #41
The direct answer the to the OP question
Velocity of Air Ejected from the Tower
An upper bound on area through which the air initially contained within every story gets
expelled (Fig. 3a) is Aw = 4ψahc , where 4ahc = area of one perimeter wall, a = 64 m = width
of the side of square cross section of tower, hc = 3.69 m = clear height of one story = distance
from the bottom of a story slab to the top of the underlying slab, and ψ = vent ratio = ratio
of unobstructed (open) area of the perimeter walls to their total area (ψ
≤ 1). The initial
mass of air within one story is ma = ρa a2 hc , where ρa = 1.225 kg/m3 = mass density of air at
atmospheric pressure and room temperature. Just outside the tower perimeter, the air jetting
out (Fig. 3a) must regain the atmospheric pressure as soon as it exits (White 1999, p.149), and
its temperature must be roughly equal to the initial temperature (this is a well-known general
feature of exhausts, e.g., from jet engines (White 1999, p.149) or pipes (Munson et al. 2006)).
So, the mass density of exiting air ρ
≈ ρa .
The time during which the top slab collapses onto the lower slab
≈ ∆t = hc / ̇z = time
during which the air is expelled out (which is only about 0.07 s for stories near the ground).
Conservation of the mass of air during the collapse of one story requires that ρAw (va ∆t) = ρVa .
Solving this equation gives the average velocity of escaping air just outside the tower perimeter:
va = Va
ψAw ∆t =
a ̇z
4ψhc (7)
Since the velocity of the crushing front near the end of North Tower crush-down is, according
to the solution of Eq. (2), ̇z = 47.34 m/s (106 mph), the velocity of escaping air near the end
of crush-down is
va = 64m × 47.34m/s
4ψ
× 3.69m
=
Thanks
I'll someone better at math than I am confirm the math is valid (I have no reason to suspect otherwise, but it never hurts to check). Barring mathematical error, your two posts have answered the question I had.
Thank you.,
Thread Over.