Fizz
Rookie
- Nov 20, 2009
- 4,391
- 350
- 0
- Banned
- #21
☭proletarian☭;1884198 said:Could you source that, please?
http://www.civil.northwestern.edu/p...TC Collapse - What Did & Did Not Cause It.pdf
I believe he was looking for an official explanation not one of your pdf files that go nowhere
i'm not sure how much success someone that thinks a pdf file "goes nowhere" will have understanding it... but here is part of the pdf file that goes nowhere but others can also see it. mauybe its a huge government conspiracy.

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 = 4 ahc, 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 = vent ratio = ratio of unobstructed (open) area of the perimeter walls to their total area ( 1). The initial mass of air within one story is ma = aa2hc, where a = 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 a. 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 Aw(vat) = Va. Solving this equation gives the average velocity of escaping air just outside the tower perimeter: va = Va Awt = az 4 hc (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 4 × 3.69m = (205m/s (459 mph or 0.60 Mach) for = 1
340m/s (761 mph or 1.00 Mach) for = 0.604 (8) The vent ratio (which is < 1) is hard to estimate. It surely varies from story to story, and also during the crushing of one story. Its effective, or average, value could be much less than
1 (because some of the perimeter area is doubtless still obstructed early in the crushing of one story, and because much of the air escapes only after the story height has been reduced greatly). In spite of these uncertainties, it is clear that the exit air speed is of the order of 500 mph and that its fluctuations must reach the speed of sound. This must, of course, create sonic booms, which are easily mistaken for explosions (supersonic speeds are virtually impossible since the venting would require an orifice shaped similarly to a convergent-divergent nozzle).
There are other phenomena that can cause va to differ from the estimate in Eq. (8). The air pressure surely exhausts the load capacity of the floor slab for a few microseconds before it is impacted by the layer of compacted debris. So, the floor slab must crack before the story height is reduced to h, and the air must begin to leak through the cracked floor slab into the underlying story, thus increasing the air mass in that story. Obviously some air must also leak
into the ceiling which behaves as a porous layer of compacted gravel (it is impossible for the ceiling and the floor to remain flat and leak no air since otherwise the air pressure would tend to infinity as the ceiling impacts the floor).
All these complex inter-story interactions must cause rapid and large random fluctuations of internal air pressure and exiting air velocity. On the average, however, what matters is the simple fact that the air must, in one way or another, get expelled from each story of the tower within a very short time interval, which is only 0.07 s near the end of crush-down of North 6 Tower. This fact inevitably leads to the average exit velocity estimate in Eq. (7). The high
velocity of air jetting out also explains why a large amount of pulverized concrete, drywalls and glass was ejected to a distance of several hundred meters from the tower (Fig. 3a). Resisting Forces Due to Ejecting Air and Solids
The air mass within the confines of one story, which is aa2hc, gets accelerated from 0 to velocity va as it exits the tower perimeter. The kinetic energy acquired by the escaping air of one story just outside the tower perimeter is Ka = 1 2va 2(a2hc) where a2hc = initial volume of air within the story. The energy dissipated by viscosity of flowing air and by boundary friction is estimated to be negligible. Therefore, virtually all of the kinetic energy of escaping air must be supplied by gravity, and since the spatial derivative of energy is a force (called the material force or configurational force), the vertical resisting force caused by air ejection is Fa(z, z˙
= K h = aa4 32 2hch z˙2 (9)
Solution of Eq. (2) shows that, at the end of North Tower crush-down, Fa 103 MN, which represents about 8.01% of the total resisting force Fc at the end of crush-down. When the first story under the aircraft impact zone gets smashed, Fa contributes only 2.75% of Fc. During the crush-down, the ratio Fa/Fc starts increasing and reaches the maximum of 12.79% when the 83rd story gets impacted, and then decreases due to the increase of Fb at lower stories. Fa < 5% of Fc up to the 3rd story crushed, and < 10% up to the 7th story crushed (by contrast, in building demolitions, which are conducted in the crush-up mode, the crush-up motion begins with zero velocity, and Fa < 5% of Fc for buildings up to about 20 stories tall, in which case the air resistance can be neglected). The maximum of Fa/Fc is 30.27%, which occurs during the crush-up phase in which Fb is small and the velocity high.
The average over-pressure of air within the tower is pa = Fa/a2 above the atmospheric pressure, which gives, for the North Tower, 7.55 kPa (0.075 atm), 14.31 kPa (0.141 atm) and 25.15 kPa (0.248 atm), respectively, during the crushing of the 80th story, 50th story, and at the end of crush-down, respectively. The last pressure value is enough to break up the floor slab. The pressure peaks near the end of squeezing of a story are doubtless much higher, as
already mentioned, and thus must contribute to the break up of many floor slabs (theoretically, the pressure in a thin layer of viscous gas between two colliding parallel flat slabs approaches infinity at the end). The mass that is shed from the tower, characterized by out, exits at various velocities ranging from nearly 0 to almost either the air ejection velocity, for fine dust, or to roughly z˙, for large steel pieces. Instead of complicating our model by some distribution of these velocities, we will simply assume that a certain fraction, eout, gets ejected in any direction (horizontal, inclined downward or upward, or almost vertical) at velocity z˙, while the remaining mass (1 − e)out is shed at nearly vanishing velocity. For a certain empirical value of e, this must be energetically equivalent to considering the actual distribution of velocities of ejected solids. As the crushing front advances dz, the mass of solids (dust plus large fragments) that is ejected at velocity z˙ is eoutμ(z)dz and has kinetic energy eoutμ(z) dz(z˙2/2). This must
be equal to Fedz, i.e., to the work of the resisting force Fe over distance dz. It follows that Fe = 1 2 eoutμ(z) z˙2 (10) The computation results shown in figures have been run for e = 0.2; however, a broad range of e has been considered in computations, as discussed later. For the crush-up, e must be ignored because the compacted layer is stationary.
An upper bound on area through which the air initially contained within every story gets expelled (Fig. 3a) is Aw = 4 ahc, 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 = vent ratio = ratio of unobstructed (open) area of the perimeter walls to their total area ( 1). The initial mass of air within one story is ma = aa2hc, where a = 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 a. 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 Aw(vat) = Va. Solving this equation gives the average velocity of escaping air just outside the tower perimeter: va = Va Awt = az 4 hc (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 4 × 3.69m = (205m/s (459 mph or 0.60 Mach) for = 1
340m/s (761 mph or 1.00 Mach) for = 0.604 (8) The vent ratio (which is < 1) is hard to estimate. It surely varies from story to story, and also during the crushing of one story. Its effective, or average, value could be much less than
1 (because some of the perimeter area is doubtless still obstructed early in the crushing of one story, and because much of the air escapes only after the story height has been reduced greatly). In spite of these uncertainties, it is clear that the exit air speed is of the order of 500 mph and that its fluctuations must reach the speed of sound. This must, of course, create sonic booms, which are easily mistaken for explosions (supersonic speeds are virtually impossible since the venting would require an orifice shaped similarly to a convergent-divergent nozzle).
There are other phenomena that can cause va to differ from the estimate in Eq. (8). The air pressure surely exhausts the load capacity of the floor slab for a few microseconds before it is impacted by the layer of compacted debris. So, the floor slab must crack before the story height is reduced to h, and the air must begin to leak through the cracked floor slab into the underlying story, thus increasing the air mass in that story. Obviously some air must also leak
into the ceiling which behaves as a porous layer of compacted gravel (it is impossible for the ceiling and the floor to remain flat and leak no air since otherwise the air pressure would tend to infinity as the ceiling impacts the floor).
All these complex inter-story interactions must cause rapid and large random fluctuations of internal air pressure and exiting air velocity. On the average, however, what matters is the simple fact that the air must, in one way or another, get expelled from each story of the tower within a very short time interval, which is only 0.07 s near the end of crush-down of North 6 Tower. This fact inevitably leads to the average exit velocity estimate in Eq. (7). The high
velocity of air jetting out also explains why a large amount of pulverized concrete, drywalls and glass was ejected to a distance of several hundred meters from the tower (Fig. 3a). Resisting Forces Due to Ejecting Air and Solids
The air mass within the confines of one story, which is aa2hc, gets accelerated from 0 to velocity va as it exits the tower perimeter. The kinetic energy acquired by the escaping air of one story just outside the tower perimeter is Ka = 1 2va 2(a2hc) where a2hc = initial volume of air within the story. The energy dissipated by viscosity of flowing air and by boundary friction is estimated to be negligible. Therefore, virtually all of the kinetic energy of escaping air must be supplied by gravity, and since the spatial derivative of energy is a force (called the material force or configurational force), the vertical resisting force caused by air ejection is Fa(z, z˙

Solution of Eq. (2) shows that, at the end of North Tower crush-down, Fa 103 MN, which represents about 8.01% of the total resisting force Fc at the end of crush-down. When the first story under the aircraft impact zone gets smashed, Fa contributes only 2.75% of Fc. During the crush-down, the ratio Fa/Fc starts increasing and reaches the maximum of 12.79% when the 83rd story gets impacted, and then decreases due to the increase of Fb at lower stories. Fa < 5% of Fc up to the 3rd story crushed, and < 10% up to the 7th story crushed (by contrast, in building demolitions, which are conducted in the crush-up mode, the crush-up motion begins with zero velocity, and Fa < 5% of Fc for buildings up to about 20 stories tall, in which case the air resistance can be neglected). The maximum of Fa/Fc is 30.27%, which occurs during the crush-up phase in which Fb is small and the velocity high.
The average over-pressure of air within the tower is pa = Fa/a2 above the atmospheric pressure, which gives, for the North Tower, 7.55 kPa (0.075 atm), 14.31 kPa (0.141 atm) and 25.15 kPa (0.248 atm), respectively, during the crushing of the 80th story, 50th story, and at the end of crush-down, respectively. The last pressure value is enough to break up the floor slab. The pressure peaks near the end of squeezing of a story are doubtless much higher, as
already mentioned, and thus must contribute to the break up of many floor slabs (theoretically, the pressure in a thin layer of viscous gas between two colliding parallel flat slabs approaches infinity at the end). The mass that is shed from the tower, characterized by out, exits at various velocities ranging from nearly 0 to almost either the air ejection velocity, for fine dust, or to roughly z˙, for large steel pieces. Instead of complicating our model by some distribution of these velocities, we will simply assume that a certain fraction, eout, gets ejected in any direction (horizontal, inclined downward or upward, or almost vertical) at velocity z˙, while the remaining mass (1 − e)out is shed at nearly vanishing velocity. For a certain empirical value of e, this must be energetically equivalent to considering the actual distribution of velocities of ejected solids. As the crushing front advances dz, the mass of solids (dust plus large fragments) that is ejected at velocity z˙ is eoutμ(z)dz and has kinetic energy eoutμ(z) dz(z˙2/2). This must
be equal to Fedz, i.e., to the work of the resisting force Fe over distance dz. It follows that Fe = 1 2 eoutμ(z) z˙2 (10) The computation results shown in figures have been run for e = 0.2; however, a broad range of e has been considered in computations, as discussed later. For the crush-up, e must be ignored because the compacted layer is stationary.
Last edited: