Ground Zero fires could have heated the steel by up to 4 °F

Introduction

An abundance of forensic evidence proves the official 9/11 conspiracy theory is scientifically and physically impossible. If the official theory were true, the laws of thermodynamics and thermochemistry would need to be rewritten and people would survive months of burial in sand. Air flow through the Ground Zero pile is found to be more than three orders of magnitude short of that required to sustain smoldering combustion except for within voids, and sufficient to raise the temperature of the steel by 2.2 °C in five months. The maximum sustainable heat release rate is equivalent to one burning cigarette every 32 square feet. Steel that was heated to cherry-red would have cooled within hours.

The official conspiracy theory of nineteen Arabs with box cutters is easily refuted with a few short steps that evaluate the physics and the physical evidence. The death blow has already been dealt by combining 1) and 2) with 4) or 5) below, and several further points are provided for diehard reality deniers. A standalone set of evidence described in 8) is more than capable of rebutting the "Arabs did 9/11" thesis, and the same could be said of 6). Part 3) is necessarily lengthy, in order to dispel an aspect of the legend that is much misunderstood and has taken on a quasi-religious tone. Those who prefer to sidestep the (elementary) math can go on to part 4). Or go to the index to review a topic of interest.

What should not have been found at the WTC, if the official conspiracy theory were true

1) The official 9/11 narrative asserts that three steel-framed high-rises at the World Trade Center (WTC) collapsed following office fires in which no accelerants of any kind, e.g., thermite, were employed by the 9/11 conspirators.

2) The temperatures generated by office fires, such as those of 9/11/01 at the WTC buildings, are "most definitely not capable of melting steel", or iron, which has a melting point of 1,535 °C, or iron(III) oxide, which has a melting point of 1,565 °C. The various structural steels allowable for use in the WTC construction had a carbon content of up to 0.3% (e.g., the Japanese JIS G3101-73 grade SS50; see for example NIST NCSTAR 1-3A or its Interim Report Appendix E). The iron-carbon phase diagram shows that 0.3% carbon barely lowers the melting point to below that of iron. Given the steel's maximum sulfur content of 0.05%, its melting point was at least 1,426 °C or 2,600 °F.

Note that steel temperatures in office fires are always below the maximum (upper layer) gas temperatures. It takes time for heat to be transferred to the cooler, absorbing object, and the rate of heat transfer decreases as the temperature of the absorbing object approaches that of the heat source. NIST admitted that the "maximum upper layer air temperatures" would have been "about 1,000 degrees Celsius (1,800 degrees Fahrenheit) in the WTC towers", and also confirmed that "at any given location the combustibles needed about 20 minutes to be consumed". NIST said (NCSTAR1CollapseOfTowers, 2005 p.179) in reference to gas, not steel temperatures: "At any given location, the duration of temperatures near 1,000 °C was about 15 min to 20 min. The rest of the time, the calculated temperatures were near 500 °C or below." That would make the mean gas temperature about 600 °C for a given point in the fire zone. A method outlined here for estimating the mean gas temperature from the heat flux predicted an average temperature of 627 °C. And according to a study commissioned by Silverstein Properties, the temperatures ranged from 750 °F to 1300 °F (about 400 °C to 700 °C).

3) Similarly, oxygen-depleted underground fires that have been smothered with tens of thousands of tons of concrete dust, hosed with three million gallons of water and exposed to another million gallons of rain all within the first ten days, and sprayed with thousands of gallons of the fire retardant foaming agent Pyrocool FEF - of which 500 gallons sufficed to extinguish the Nassia oil tanker fire within a mere twelve-and-a-half minutes - should not generate temperatures at or above the melting point of iron or steel and mysteriously continue to burn at very high temperatures for more than three months, unless accelerants such as thermite have been employed.

A report on tritium levels at the WTC states: "It was determined that 3 million gallons of water were hosed on site in the fire-fighting efforts between 9/11 and 9/21 (the day of the tritium measurement, sequences 6,7 in Table 1) (Headley, 2002). In addition, there were 2 episodes of rain during the same 10 day period after the attack: on 9/14 and 9/20,21 (NOAA, 2001), totaling 0.9 million gallons of water in the Bathtub area. Considering the neighboring area, one can take 1 million gallons from the rain. Therefore, a total of 4 million gallons of water percolated through the debris in the first 10 days and collected at the bottom of the Bathtub."

In all, thirty million gallons collected at the bottom of the Bathtub over the first ten days and flowed through the north PATH train tunnel to Jersey City, whereupon it was having to be pumped out at a rate of thousands of gallons per minute over a 12-hour day. The north PATH tunnel was "completely filled" at the low midriver point. The source of the extra 26 million gallons was assumed by the tritium report authors to have been leaks from the Hudson River and broken mains, whereas the engineer who oversaw the construction of the Bathtub believed that "most" of the total came from the "vast amounts of water that were poured onto the debris to extinguish continuing fires". Thus, the figure of 4 million gallons percolating through the entire height of the pile over ten days may be too low, but can be taken as a conservative estimate. Throughout that time, there is no doubt that the bottom of the Bathtub was a giant lake.

Two more points of interest from the tritium report are (i) the authors' honest assessment of the underground fires, which they described as: "smoldering fires of much lower temperature than the explosive and high-temperature fires up in the Towers (with the exception of possibly the WTC 7 fire)", and (ii) their estimate of 0.1 for the air porosity in the debris pile: "Considering that the Bathtub was at least 50% destroyed and filled with the debris from the buildings (Post, 2001c), one could assume its air porosity of 0.1 (note: for instance, a porosity of close packed spheres is 0.26)." They go on to assume an air porosity of 0.3 for the B6 level with the PATH tunnel, "since the damage was less".

(A word of warning: ignore any claims that you may see about "nuclear" demolitions, "directed energy" beam weapons, "holographic" planes, "TV fakery", or "lizards". Intelligence agency assets have been promoting those red herrings for years, in a desperate attempt to mislead truth seekers and to portray critics of the government-approved conspiracy theory as illiterate imbeciles and raving lunatics who are taken in by absurd nonsense. Not surprisingly, the disinformation agents never support their wild flights of fancy with any form of quantitative analysis. The 2011 Toronto Hearings on 9/11 examined what they termed "the best available evidence", which was a polite way of saying that they would not countenance junk science and hare-brained speculation about "nukes" and "missiles" that were "faked" to look like planes.)

So, the WTC office fires were "most definitely not capable of melting steel", and anyone who wasn't trying to delude themselves or others would accept that the underground fires were, at most, merely "smoldering fires of much lower temperature" than the WTC office fires.

Proponents of the official 9/11 legend have tried to create a mystique about fires that "raged for a hundred days" in the WTC debris pile. They employed the same tactic regarding the office fires, with wild, excited babble about "intense heat" from "the burning jet fuel" - ignoring the fact that it all burnt up within a couple of minutes. The mass burning rate for free burning pool fires is on the order of 50 grams per square meter per second, so the burn time in seconds of a 4,000 gallon, 12,400 kg pool fire is 12,400 / (0.05 × Area in m2), or in this case is equal to 248,000 s.m2 divided by the area in square meters. If spread over all of a single floor, then the area, deducting 10% to allow for vertical shafts and structures, would be 3,617 m2 and the burn up time 68.6 seconds. If spread over four floors, each covering a quarter of the floor area , the time would be the same; if spread more thinly, it would have burnt up in even less time.

And there is no great mystery about underground fires in the ruins of collapsed steel-framed skyscrapers. If such fires generate extraordinarily high temperatures, e.g. "more than 2,800 °F" or 1,538 °C according to daily thermal measurements taken by helicopter, and persist at very high temperatures for months on end despite being smothered with concrete dust and hosed with millions of gallons of water, it is not because of some malign influence of the evil spirits of "Muslim" terrorists. Any method of staging deceptive controlled demolitions that are designed to resemble fire-induced collapses will leave some telltale signs. The best, stealthiest method would employ some variant(s) of thermite and leave evidence of very high temperatures, which the perpetrators would hope to blame on the "intense heat" from "the burning jet fuel". In the thermite reaction the aluminum grabs pure oxygen from the iron oxide. Thus, thermite not only has its own oxygen source and cannot be extinguished by water, it can generate temperatures well in excess of those associated with hydrocarbons burning in air of up to 21% oxygen.

Controlled demolitions that relied upon conventional "bombs" or high explosives would be all too obvious, and it would not take long before the perpetrators were brought to justice. Deceptive demolitions based on thermitic materials offered the conspirators the best prospect of getting away with the crime for the remainder of their lives, notwithstanding the fact that in an age when information could be freely exchanged via the internet, it would take no more than ten years before structural engineers, architects, physicists,﻿ mechanical engineers, metallurgical engineering graduates, PhD scientists, fire protection engineers, fire-fighters, electrical design engineers, explosives technicians, psychologists, etc., were explaining why the official conspiracy theory was believable only to the ignorant, the indifferent and the ill-informed. And various experts would tear apart the wild claims and speculation of the government account at an international hearing in which the panelists were a top judge and various professors.

Proponents of the official conspiracy theory claim that thermite would have all burned up in minutes. However, the official theory not only fails to account for the observed very high temperatures, it also flounders in its attempts to explain the observed long-lasting fires. As will be shown, even smoldering fuel would burn up within an hour to a day, and so the official theory requires an extraordinary series of local burning of pulverized office materials under soaking, smothered conditions followed by transmission of heat that improbably ignites soaked, smothered office materials elsewhere, with the cycle necessarily repeating hundreds or even thousands of times over the following months, and with dozens of these series running in parallel. Under smothered, soaking conditions, no more than a tiny proportion of the office combustibles could continue to smolder. In contrast, accelerants used for the controlled demolitions could continue to burn in water and without an independent supply of oxygen, going on to generate extraordinarily high temperatures, with this heat subsequently igniting accelerants elsewhere by various means.

Paint chips on rusty steel from the WTC were found to be contaminated with an accelerant that included elemental aluminum as the fuel. Iron spheres were generated when the chips were heated to 700 °C, indicating flame temperatures above the melting point of iron. Such a material would be ignited at irregular intervals whenever the clean-up brought it into contact with cooling zones that had been superheated from thermite that had already burned elsewhere, or when its molten iron product flowed down into the pile in the vicinity of unreacted nanocomposites, or after sufficient heat had been conducted through the concrete or steel. Some active accelerant would be ignited soon after the collapses by smoldering office combustibles. The vast majority of the time, the thermite is not burning; it has been consumed in one part of the pile and there is a delay before heat has been conducted over the required distance through the concrete or steel, or the molten iron flows down through the pile to ignite active thermitic material elsewhere, or mechanical diggers bring unreacted thermite into contact with debris that is still at a few hundred degrees or more. Then the next zone is ignited to form another superheated area, and the cycle repeats.

An Al-Fe 2 O 3 -based energetic nanocomposite using nanowires embedded in a 50 nm aluminum film was found to ignite at 410 °C and to produce a flame temperature of ~4,000 °C, which is even higher than the theoretical adiabatic flame temperature of 3,250 °C for a stoichiometric Al-Fe2O3 mixture. The flame temperature was measured by recording the blackbody spectrum from the light flash. It is believed that the reaction is so fast in this material that the heat is unequally distributed amongst the products.

As will be demonstrated, office combustibles relying upon the available oxidizer flux within the pile would have been capable of raising the temperature of the steel by a mere 2.2 °C. And that's assuming all of the heat remained trapped in the pile and continued to build up for a full five months, and all of this heat was transferred to the steel rather than the concrete, aluminum or glass, etc. If we include the requirement to heat the concrete, the potential temperature increase is only 0.7 °C - barely more than 1 °F. Energetic nanocomposites provide the only plausible explanation for the phenomenon of mysteriously persistent fires that generated extraordinarily high temperatures along with a molten iron product that was frequently seen "dripping", "flowing", and "running", e.g., "down the channel rails". To go straight to one of the main conclusions for this section, click here; to see how it was arrived at, read on.

According to NIST, a typical WTC floor had 4 psf of combustible materials. If this is applied to 208 by 208 feet, which is 43,264 square feet, each floor contained 173,056 pounds or 86.5 tons of combustibles such as paper, carpets and workstations. (It also had ~56 tons of fireproofing.) Thus, 110 floors above grade and six below contained 10,037 tons of combustibles - say, 10,040 after allowing for those from the plane, e.g., seats. The oft-quoted 500,000 tons per tower is a high estimate that assumes the live load is almost 100% of the design maximum, and so the combustibles constituted up to 3% of the WTC debris pile by mass. That pile mostly consisted of concrete dust and steel members, along with aluminum window panels, glass and sprayed fire-resistive material - and plenty of water percolating through it. Moreover, at the time of collapse most of the floors were not affected by fire, leaving the vast majority of combustibles located on floors much lower (or higher in the case of WTC7) than the burning floors as the buildings collapsed. This potential fuel was therefore cold and isolated from hot spots in the pile. Other combustibles had already been consumed, and the few remaining pockets of burning paper, carpets or workstations - around 0.1 to 0.2% of the total mass of the pile - would have been isolated from other fuel sources. The pile was subjected to "a nearly constant jet of water" that effectively created a "giant lake".

FEMA's report on WTC1 and 2 quoted Culver's findings that typical office-type occupancies had fuel loads, expressed in terms of the equivalent weight of wood, ranging from 4 to 12 psf with the mean slightly less than 8 psf. NIST's 4 psf is lower than the Culver figure because the calorific value of wood, at up to 16 - 17 MJ/kg, is less than that of other combustibles such as plastics. It is also necessary to allow for the fact that the effective heat of combustion in ventilation-limited fuel-rich compartment fires is around 35% lower than could be obtained under an optimum (stoichiometric) fuel-air ratio that allowed for complete combustion. Kawagoe and others such as Magnusson and Thelandersson used a value of 10.78 MJ/kg as the calorific value of wood in fully developed ventilation-limited fires.

Thus, a fire load of 8 psf or 3.63 kg per square foot of wood in a compartment fire yields some 3.63 × 10.78 = 39.1 MJ per square foot or 421 MJ/m2 or 1.69 TJ over the 208 x 208 feet of a WTC tower floor. If this energy was all released over the typical 20 minutes (as quoted by NIST) for local combustibles to be consumed in a compartment fire before the flames move along to a fresh fuel source, the power output would be an impressive 1.69×1012 / (20 × 60) = 1.41 GW per floor, or 32.6 kW/ft2 or 351 kW/m2. (About 40% of that is vented out and drives the smoke plume - FEMA quoted "one third to one half".)

However, if the energy is released over 100 days instead of 20 minutes, the mean power output decreases by a factor of 7,200 to 48.7 watts per square meter, which is less than half of the 120 W/m2 required to qualify for the World Meteorological Organization's definition of sunshine, and more than an order of magnitude below the 1 kW/m2 received at a point in the tropics when the sun is directly overhead on a clear day. Even if we allow for each floor being compressed into about eight inches, the equivalent of hazy sunlight shining on eight inches of mostly concrete dust is not going to ignite wood, plastics, etc at a depth of a few inches.

And those very high temperatures in the pile were in fact sustained for five months, according to Bronx fire-fighter Joe "Toolie" O'Toole. He reported seeing a crane lift a deeply buried steel beam that was "dripping from the molten steel" as late as February, 2002.

An oxygen-starved, smoldering underground fire would be even less efficient than a fully-developed ventilation-limited compartment fire. The assumption of 65% efficiency and 10.78 MJ/kg of wood in a compartment fire is consistent with an averaging of a range of equivalence ratios from 1 (stoichiometric) to 2 (fuel-rich) at various points within the compartment - from relatively complete combustion near the vents to incomplete elsewhere. (Integrating the range gives a mean equivalence ratio of 1.386 - the natural log of 4.) The smoldering underground fire would be expected to involve more incomplete combustion, with a greater proportion of CO product in place of CO 2 . The ratio of CO 2 to CO in smoldering is typically around unity, compared to about ten times as much CO 2 in a fully developed compartment fire. However, for the Ground Zero fires, the most important figure is the heat released per unit of oxygen consumed.

Flaming combustion of typical hydrocarbons is associated with a heat release of around 13.1 kJ/g of oxygen consumed, whereas the corresponding value for smoldering combustion is about 5 kJ/g of oxygen (5,000 kJ/kg). Bar-Ilan et al (2004) obtained 5,880 kJ/kg-O 2 for opposed (reverse) smolder and 4,550 kJ/kg-O 2 for forward smolder, compared to the 3,900 kJ/kg-O 2 of Torero et al (1993) and 4,550 kJ/kg-O 2 of Walther et al (1998).

Smoldering combustion is typically self-sustained at heat fluxes around 6.5 kW/m2 or more. Bar-Ilan et al (2004) found that, given their sample size and test conditions of forced air mass fluxes, "there exists a minimum critical oxidizer mass flux to achieve a self-sustaining smolder propagation". They placed the value of this minimum mass flux at around 0.6 grams of oxygen per square meter per second under normal gravity. (The minimum flux was only half as much, 0.3 g/m2.s-O 2 under microgravity conditions, apparently because "the presence of buoyant heat losses in normal gravity hinders the smolder reaction".) At 5 kJ/g of oxygen consumed, an oxidizer mass flux of 0.6 g/m2.s-O 2 would correspond to a power density of 3 kW/m2.

Dr. T. J. Ohlemiller (of NIST, ironically) says that a reverse smolder front of a few kW/m2 from a source 0.1 to 0.15 m. in diameter would typically generate gas temperatures of 670 to 970 K (400 to 700 °C), and would be so "weak" that the buoyant plume might not even reach the ceiling of a room.

According to Darcy's law, the total discharge, Q (units of volume per time such as m3/s), of fluid through a medium is equal to the product of the permeability of the medium, k (m2), the cross-sectional area to flow, A (e.g. m2), and the pressure drop, P1 - P2 (Pa), all divided by the viscosity, µ (Pa.s) of the fluid and the length, L (m) that the pressure drop is taking place over.

Q = -k × A × (P1 - P2) / (µ × L)

The permeability k of beds of close packed spheres can be predicted from the porosity and the sphere diameter. Dr. Frank Greening estimates the distribution of particle diameters of the pulverized WTC concrete as 30% within the range 10 cm to 1 cm, 20% within 1 cm to 1 mm, 15% within 1 mm to 100 µm, 10% within 100 µm to 10 µm, and 25% at less than 10 µm. Taking the mean of the logs of each size range in microns (and ignoring the nm range), we have 30 × 4.5 + 20 × 3.5 + 15 × 2.5 + 10 × 1.5 + 25 × 0.5 totalling 270, which is divided by 100 to give the mean 2.7, which places the typical particle diameter at around 500 µm.

This is consistent with sand, of which 0.25 to 0.5 mm grains is classed as "medium" sand, and 0.5 to 1 mm is "coarse" sand. The WTC concrete dust's range of sizes includes all of the silt range (0.0625 mm down to 0.004 mm diameter) and all of gravel (2 mm up to 64 mm). Given the large range of sizes, 0.1 is a fair estimate for the WTC porosity. The smaller particles fill the voids between larger particles, whereas with spheres of equal size the maximum packing density is about 0.74.

Both the Carman-Kozeny model and Rumpf and Gupte's equation predict the permeability k (m2) by squaring the sphere diameter d (m) and then multiplying by some terms that incorporate the porosity (dimensionless, denoted by epsilon ε or phi φ). Carman-Kozeny have d2 multiplied by ε3 / [180×(1-ε)2] which in the case of ε=0.1 is d2 times 6.859×10-6. In Rumpf and Gupte's equation d2 is multiplied by (ε5.5) / 5.6 which is d2 times 5.647×10-7. It is said that the Rumpf-Gupte equation "replicates better experimental data". Nevertheless, let us use the Carman-Kozeny model, which predicts a higher permeability and thus is more favorable to the official 9/11 conspiracy theory. So for a mean particle diameter of 0.5 mm, as derived from Greening's distribution of particle sizes, the permeability of the WTC concrete dust in the pile is 0.00052 × 6.859×10-6 = 1.71×10-12 m2 or 1.71×10-8 cm2. This is well within the range of sand, rather than gravel or silt.

Source: Associated Press

And it is consistent with photographic evidence showing the pile to have the consistency of sand, with a few larger pieces mixed in.

According to Table 1 in the Lioy et al report on the WTC Dust/Smoke Aerosol, the aerodynamically separated samples ranged from 52.21% to 63.6% by mass of particles greater than 53 µm diameter, and the sieved samples had 16% to 23% by mass of particles greater than 300 µm diameter. This suggests that the typical dust particle was smaller than 500 µm diameter, which would make the permeability lower than 1.71×10-12 m2. It's still consistent with poorly-sorted sand of low porosity.

Given the permeability, we can now calculate the Darcy velocity and discharge rate for various fluids such as water and air.

Q = -k × A × (P1 - P2) / (µ × L)

The total discharge rate through a given cross-sectional area is directly proportional to the permeability, the area and the pressure drop along the length of the porous medium under consideration, and inversely proportional to the dynamic viscosity of the fluid and the length of the medium. Dividing the pressure drop (P1 - P2) by the length L gives the pressure gradient or the change in pressure per unit length ΔP, and dividing both sides by the area gives the discharge or flux per unit area. So Darcy's law can also be written as:

q = -k × ΔP / µ

...where q is the discharge per unit area (m/s), k is the permeability of the medium (m2), ΔP is the pressure gradient (Pa/m), and µ is the dynamic viscosity of the fluid (Pa.s). This discharge per unit area is the Darcy velocity, and is equal to the total discharge rate in cubic meters per second for a 1 m2 cross-sectional area. For a porosity of 0.1, the real velocity of the fluid through the pores is ten times greater than the Darcy velocity.

In the case of a gravity-driven downward flow, the pressure gradient derives from the fluid density (and from the gravitational acceleration constant). For a hydraulic head of water, each foot of height gives about 0.43 psi of pressure, irrespective of the area of the pipe or dam. A pipe of 1-meter cross-section filled to a height of 1 meter contains 1 m3 of water, which has a mass of 1,000 kg. The gravitational force acting on the 1,000 kg that would cause it to accelerate in freefall at 9.802 m/s2 (in New York City) is 9,802 N, and so the pressure acting at the bottom is 9,802 N/m2 = 9,802 Pa. (Dividing the 1,000 kgf by 3.2808 to convert to a height of 1 foot, dividing by 39.372 to convert to a 1 in2 area, and multiplying by 2.2046 to convert to lbf gives the 0.43 psi for a 1 ft head of water.) Thus, the pressure gradient ΔP is 9,802 Pa/m.

The dynamic viscosity of water at 10 °C is 0.001308 Pa.s. Thus, we now have the necessary variables to calculate the Darcy velocity q, which is equal to the permeability 1.71×10-12 m2 times the pressure gradient 9,802 Pa/m divided by the dynamic viscosity 0.001308 Pa.s, giving the Darcy velocity as 1.281×10-5 m/s. The total discharge rate for a 1 m2 area is 1.281×10-5 m3/s, and the actual flow velocity in the 10% spaces between the spheres is 1.281×10-4 m/s. This is not far off the approximation shown on this page of 2 hours for water to travel 1 meter down through sand, which is a rate of 1.389×10-4 m/s.

The known flow of four million gallons through the Bathtub in ten days can be used to set a lower limit for the permeability at Ground Zero. The Bathtub area was 510 ft by 980 ft, which is 46,434 m2. Four million gallons is 15,100 m3, and that volume over ten days is equal to 0.0175 m3/s. Rearranging:

Q = -k × A × ΔP / µ

...gives

-k = Q × µ / (A × ΔP)

...and so the permeability k is a minimum of 0.0175 × 0.001308 / (46,434 × 9,802) = 5.03×10-14 m2 (or 34 times lower than the 1.71×10-12 m2 obtained above). This would correspond to a typical particle diameter of 0.5 mm / √(34) = 86 µm.

At a density (and thereby pressure gradient) some three orders of magnitude lower than water and a dynamic viscosity around two orders of magnitude greater than water, the Darcy flow and velocity for air is about one-tenth that of water. Moreover, any air that might flow down into the WTC "sand" not only meets resistance from the solid spheres, but from water and foam being hosed onto the pile. The water flows along the surface of the solids, thus coating the combustibles as it is held by surface tension, whereas the air flows between the outer surfaces of the water, isolated from the combustibles.

For air, the pressure difference per meter of height is the gravitational acceleration constant 9.802 m/s2 times the density of air, which is 1.247 kg/m3 at around 10 °C for 1 atm. So each meter height of air has a mass of 1.247 kg for every square meter, and gravity acting on its mass gives it a weight of 1.247 kg/m2. The 1.247 kgf per m2 translates to 2.749 lbf or 12.22 N, and a pressure of 12.22 N/m2 is 12.22 Pa. Thus, the pressure gradient ΔP for air is 12.22 Pa/m. (At 20 °C air is less dense, so the pressure gradient is only 11.76 Pa/m.) And the dynamic viscosity for air at 10 °C is 1.787×10-5 Pa.s.

From:

q = -k × ΔP / µ

...we have a value of 1.71×10-12 m2 × 12.22 Pa/m / 1.787×10-5 Pa.s to give a Darcy velocity of 1.169×10-6 m/s, a discharge rate of 1.169×10-6 m3 for each square meter cross-section, and an actual velocity in the pores of 1.169×10-5 m/s. To obtain the air mass flux we multiply the discharge rate by the density of air of 1.247 kg/m3, which gives 1.458×10-6 kg/m2.s-air or 0.001458 g/m2.s-air. Given that the oxygen proportion is 23.2% by mass, the mass flux of oxygen available is 0.000338 g/m2.s-O 2 , which at the oxygen heat release of 5 kJ/g-O 2 for smoldering combustion could support a power density of 1.69 W/m2.

This is more than three orders of magnitude short of the minimum 0.6 g/m2.s-O 2 for self-sustained smoldering, which is associated with a power density of 3 kW/m2.

The pressure gradients in air tend to be greater in the horizontal plane, resulting in the winds and breezes that are normal meteorological phenomena. By rearranging:

q = -k × (P1 - P2) / (µ × L)

...to get:

P1 - P2 = q × µ × L / -k

...we can find what pressure difference would be required to force air through the pile formation at a discharge rate that might sustain smoldering combustion, if for example there was a higher atmospheric pressure over Jersey City that might have forced air through the south PATH tunnel - which apparently wasn't completely flooded - and up through the pile and out into Manhattan. (We don't enter in the area, as we are interested in the Darcy flow per m2, but we put in the length of the formation over which the pressure drop occurs. This will take account of the required pressure gradient, which needs to be about 1,775 times greater than 12.22 Pa/m, i.e., 21,690 Pa/m, which is ominously high!)

The required oxidizer mass flux is 0.6 g/m2.s-O 2 , which requires 0.6 / 0.232 = 2.59 g/m2.s-air, which requires an air discharge rate of 0.00259 kg / 1.247 kg/m3 per second per square meter = 0.00207 cubic meters of air per second per square meter, or a Darcy velocity of 2.07×10-3 m/s. The length (height) of the formation is taken as 21.3 m (70 ft).

So the required pressure difference P1 - P2 is 0.00207 m/s × 1.787×10-5 Pa.s × 21.3 m / 1.71×10-12 m2 = 460,764 Pa. This is 4,608 mbar, 4.547 atm, or 66.83 psi. Such a pressure difference would lead to some strong winds above grade, to say the least!

In fact, the WTC towers would probably have been the only buildings in Manhattan and Jersey City that could have survived such winds, since they were designed to withstand hurricanes, earthquakes, and a 600 mph impact with a jetliner that dumped all of its fuel into the building creating a "horrendous fire" that "killed" a "lot of people" but left "the building structure" in place. The exterior columns had "tremendous reserve strength", in that "live loads on these columns" could "be increased more than 2,000% before failure" occurred. Even the infamous Popular Mechanics publicized the fact that the WTC would "withstand winds up to 150 mph with no ill effects", although their moments of candor were in 1986 and 1993, long before the need for a massive cover-up.

Darcy's law may be applied to fluids including water or air, provided the Reynolds number is not too high, which would indicate turbulent rather than laminar flow. In the case of flow of fluid through a bed of approximately spherical particles of diameter d in contact, taking the voidage as ε and the (superficial) Darcy velocity as q, then a Reynolds number can be defined as:

Re = ρ × q × d / (µ × (1 - ε))

For the above case of water with a density ρ of 1,000 kg/m3, a predicted Darcy velocity q of 1.281×10-5 m/s, a mean particle size d of 0.0005 m, a dynamic viscosity µ of 0.001308 Pa.s, and a voidage or porosity ε of 0.1, plugging in the variables gives Re = 0.00544. For air of density 1.247 kg/m3, predicted Darcy velocity of 1.169×10-6 m/s and dynamic viscosity of 1.787×10-5 Pa.s, then Re = 0.0000453. Laminar flow applies up to Re = 10, and fully turbulent flow applies for Re = 2,000 and up. Thus, laminar flow holds and Darcy's law is applicable in both cases.

The predicted oxidizer mass flux through the WTC pile, at 0.000338 g/m2.s-O 2 , is short of 0.6 g/m2.s-O 2 by a factor of 1,775. The 0.6 g/m2.s-O 2 is required to obtain a bare minimum 3 kW/m2 smoldering power density, which would still be too low to sustain smoldering for a range of potential fuel. And the Rumpf and Gupte model's prediction falls short by an additional factor of 12!

Since the permeability is proportional to the particle diameter squared, a 42-fold increase in d would be required to raise the permeability to around the required (minimum) value. This would place the mean particle diameter at 21 mm or 0.83 inches and squarely into the category of pebbles, even coarse gravel, which it clearly was not. The dust was reported to be "like a powder similar to baking flour". Since there was a wide range of grain sizes with finer particles inevitably filling the gaps between coarse particles, it was the very antithesis of well-sorted. Thus, it had low porosity, and was likely to have a lower permeability than well sorted sand.

Let's suppose for a moment that the fuel in the pile was somehow capable of sufficient air intake and exhaust output to continue burning for months.

As previously shown, the fire load of 8 psf of wood equivalent fuel per floor of a WTC tower translates to 39.1 MJ/ft2, which is 421 MJ within a particular square meter zone of a compressed tower floor. A typical heat release rate for smoldering is 6.5 kW/m2, so if we take that as a typical heat release rate for pieces of smoldering fuel within an oxygen-poor debris pile fire, then the 421 MJ/m2 WTC fuel load would typically be consumed in 421,000 / 6.5 = 64,769 seconds which is ~18 hours, for any given location. The 20 minutes typical burn up time for local combustibles in an office fire has increased by a factor of 54, which is the ratio of the 351 kW/m2 flaming combustion of the office fire to the 6.5 kW/m2 smoldering fire. (An analysis below based on sizes of fuel shows that burn times could range from more than a day down to seconds, since the combustible materials were pulverized along with the concrete. Thus, although the largest pieces could smolder for over 18 hours, the 18 hours is associated with relatively large pieces of fuel in a particular zone.)

Since the heat release rate is on the order of 6.5 kW/m2 for all of these hypothetical smoldering pockets of fuel in the months following 9/11, the time for fuel to be consumed at any given point remains the same irrespective of our arbitrary area of the block under consideration. Rather like the office fires that have any given point burning out after 20 minutes, the time typically taken in the smoldering case is ~18 hours, if all fuel in the block is assumed to have started burning at the same time. And in fact, the first pockets of smoldering fuel would have already been part consumed in the office fires before collapse, so this fuel would have burned out in even less time.

Thus, the official theory requires a bizarre scenario of localised pockets of soaking, smoldering office combustibles that burn at around 6.5 kW/m2 whilst enclosed within concrete dust, and then these smothered, drenched, Pyrocool foam-treated pieces of burning paper, plastic or carpet somehow manage to transfer sufficient heat to ignite adjacent pockets of buried plastic or carpet in a wet, oxygen-depleted environment, with this most improbable ignition process occurring not just once, but hundreds or thousands of times over the course of five months throughout the debris pile, continuing to generate temperatures well in excess of the typical 600 °C associated with smoldering combustion. (And these bits of smothered wet paper, etc., have enough energy left over to partly evaporate steel and heat it to cherry-red, and leave such an abundance of "flowing", "dripping" molten steel or iron and possible other metals that the melt forms "little rivers" and looks "like a foundry"!!!)

A typical series of one smoldering pocket igniting an adjacent pocket of combustibles would need to involve ~200 instances of heat transfer and ignitions over five months. Given the persistence of dozens of hot spots at various points in the pile for months on end, there would have to be dozens of parallel sequences that included a similar series of ignitions. If fresh combustibles were enclosed within the same void as already smoldering fuel, heat could be transferred via the hot gases within the void. However, these would be expected to ignite and smolder at roughly the same time as the original burning fuel. Thus, subsequent series of heat transfer and ignition would be as a result of conduction through the concrete dust or the steel, which leads to some interesting calculations on how far sufficient heat could be conducted to ignite adjacent pockets of fuel before fuel in one zone burned out and the source began cooling. The answer turns out to be: not very far!

Of course, there was not sufficient air intake and exhaust to oxidize the fires; at least, nothing beyond a paltry 1.69 W/m2, equivalent to a reasonably sized mono AM radio or a third of a burning cigarette per square meter. The fact that fires are extinguished by smothering in non-combustible material such as sand or pulverized concrete (whether or not they're soaked with water and treated with foam), and the fact that this is confirmed by calculation of the Darcy flux rate for air flowing through the medium, is corroborated by the fact that humans do not survive weeks or months of burial in sand.

For smaller 0.01 m2 blocks of 0.1 x 0.1 meters within the collapsed towers' footprints, the heat release rate is typically 0.01 × 6.5 kW = 65 W per block. At the same power density, but with one-hundredth the fuel of a 1 m2 block, the fuel in the smaller block is consumed in the same time, at 4.21 MJ per 0.01 m2 divided by 65 W = 64,769 seconds which is ~18 hours. That's if it has oxygen!

Each 0.1 x 0.1 x 0.2 m pocket that is smoldering at 65 W has a volume of 2 liters, and given the porosity of 0.1 in the debris pile, had up to 0.2 liters of air (after allowing for water and Pyrocool foam). At 20 °C (sea level) the density of air is ~1.2 grams per liter, so 0.2 liters of air weighs 0.24 grams. Since air is 23.2% oxygen by mass, the 0.24 g of air contains 0.056 g of oxygen. Assuming a heat release in smoldering of 5 kJ/g of oxygen consumed, and with 65 J released each second, then the oxygen required each second is 65 / 5,000 g = 0.013 grams. Thus, the 0.056 g of oxygen would be consumed in ~4.3 seconds. Thereafter the Darcy discharge rate of 1.169×10-6 m3 for each square meter cross-section provides 0.000338 g/s of oxygen per m2, which is 3.38×10-6 g/s of oxygen for the 0.01 m2 block. (And even that's supposing that the fresh air could rapidly diffuse horizontally through the dust, which it couldn't. Even at the vertical Darcy velocity of 1.169×10-6 m/s, it takes 5.94 hours to travel 0.025 m.) The fire is extinguished.

In comparison, the power output of a human being is on the order of a 100 watt light bulb. The human averages about 15 breaths a minute, and exhaled air has about 4 to 5% less oxygen than inhaled air. Thus, if each breath averages 500 mL of air, the volume of oxygen consumed is around 0.5 × 15 × 0.045 = 0.337 liters per minute. The density of oxygen gas at 20 °C and 1 atm is 1.331 grams per liter, so the 0.337 liters / minute is ~0.45 grams / minute or 0.0075 grams / second of O 2 . The human uses oxygen more efficiently than the smoldering fire, since inhaled air is rich in oxygen and the combustion products are CO 2 and H 2 O rather than CO and H 2 . So if the human achieves around the 13.1 kJ/g of oxygen associated with complete combustion, the energy released each second is ~98 J; a power output of 98 W.

The average human has a mass of about 70 kg and a density around that of water, placing the volume of the average human at 70 liters. And the average body surface area of an adult human is 1.73 m2. Fortunately, the "light bulb" human is of course oxidising much more slowly than the typical smoldering rate of 6.5 kW/m2, especially when considering the additional area of internal organs.

Being buried alive has been used as a form of execution, murder, torture, and genocide of those deemed to be "sub-human". Without a fresh air supply, the individual would be breathing air with insufficient oxygen and too much CO 2 within 1 to 2 hours (if buried in a coffin thereby providing an initial volume of air), and death would follow within minutes. Harry Houdini was almost killed in one stunt involving burial under six feet of earth without a casket, and in another attempt managed to remain in a sealed casket submerged in a swimming pool for an hour and a half. A consumption of 0.337 liters of O 2 per minute would total around 40 liters in two hours. The person would be dead before the oxygen had run out, due to high CO 2 levels. Similarly, a smothered fire will extinguish itself before the oxygen gets down to zero. A match would not stay alight below about 15% oxygen. The products of combustion - CO 2 and H 2 O - are used in fire extinguishers, and the bucket of sand is another useful extinguishing method.

For each person who was buried alive, there would need to be sufficient air permeability to allow at least 20 liters of oxygen per hour to reach them, whereas each 0.01 m2 smoldering pocket requires 0.013 grams which is 0.0098 liters per second or 35.2 liters of oxygen per hour.

Let's pretend again for the moment, for the sake of argument, that the office combustibles did receive enough oxygen.

If we consider an "average" cubic meter within the debris pile, it turns out to have around 97.65 kg of actual combustibles, 965 kg of steel and 1.18 tonnes of concrete. Each tower's footprint was 208 by 208 feet or 63.4 by 63.4 meters, which is 4,019 m2. Each compressed floor is assumed to be 0.2 meters, giving a total depth for 116 floors of 23.2 meters; thus the total volume is 93,240 m3. (In the previous calculation on the pressure difference required to force air from Jersey City and the PATH tunnel up through the pile, the height of "the formation" is taken as 21.3 m (70 ft). The 0.2 m per floor here is an approximation, but two or three meters can be allowed for the depth of the flooding at the bottom of the Bathtub at the B6 level, as tens of millions of gallons leaked from the river and from water mains. The tritium report assumes about ten feet of flooding, since it says that "the hose and rain water flowed" through "6/7 of the void volume of the Bathtub".)

Each tower had about 110,000 tonnes of concrete and 90,000 tonnes of steel, leaving 110,000 / 93,240 tonnes = 1.18 tonnes of concrete and 90,000 / 93,240 tonnes = 965 kg of steel per cubic meter in the pile. For the actual combustibles, we take the NIST figure of 4 psf, and allow double the 10.78 MJ/kg associated with FEMA's or Culver's 8 psf of fuel expressed as the equivalent weight of wood. So 4 psf is 19.53 kg/m2 (for one floor). The 1 meter depth is five compressed floors, making the total combustibles 97.65 kg within the meter cube, and at 21.56 MJ/kg, the total potential heat release from the fuel is an impressive 2.105 GJ.

Taking specific heats of 800 J/kg for the concrete and 550 J/kg for the steel (which allows for higher values at higher temperatures), the energy required for each degree Celsius increase in the steel plus the concrete in the 1-meter cube is:

800 × 1,180 + 550 × 965 = 1.475 MJ.

(For simplicity, and to be fair to the government's conspiracy theory, we ignore the aluminum, glass, fireproofing, air conditioning and other loading. This places the combustibles at over 4% of the total mass, when in fact they were 3% or lower.) Thus, the potential temperature increase for the concrete and steel within the one-meter cube is 2,105 / 1.475 = 1,427 °C.

Of course, such temperatures would not be reached, partly because office fires do not get up to such a temperature, and partly because smoldering combustion temperatures are only around 600 °C, even when we ignore the fact that the fuel in the debris pile was too starved of oxygen to support even smoldering. A good supply of pure oxygen could generate steel-melting temperatures - but then the situation is getting close to a thermite scenario. Fuel is necessary, but not sufficient, for combustion. This is fortunate for life on earth, since there is a tremendous fuel load on the earth's surface, most of which isn't burning at any given time. (Although such a benign state of affairs may not endure, if the murderous, unscrupulous psychopaths behind 9/11 are allowed to carry on pulling the strings of puppet presidents and puppet prime ministers for much longer.)

As the fire triangle tells us, "Without sufficient heat, a fire cannot begin, and it cannot continue", and "Without sufficient oxygen, a fire cannot begin, and it cannot continue".

If 1% (a high estimate) of the debris pile was initially at 1,000 °C and the remainder at 10 °C, then the mean temperature within the pile was initially at (1,000+99×10)/100 = 19.9 °C.

Given the typical time to burn up when smoldering of 18 hours, the proportion of the fuel that was smoldering at any particular time could be no more than ~0.5% in order to sustain fires that persisted for five months. If all of the fuel in a 1 m3 volume of the pile was burning at the same time (which it wasn't because of lack of oxidizer flux), then no more than a few hundred out of these 93,240 one cubic meter volumes could be smoldering. Or alternatively, 1/200 of the fuel could have been burning at any particular time within each 1 m3 volume, moving on to adjacent fuel hours later. (As will be seen, even this could not be sustained.) Thus, if the hot spots are at 600 °C and the cold spots are at 10 °C, then the mean temperature across the debris pile cannot be any higher than (600+199×10)/200 which is about 13 °C.

Or that would be 3 °C, when the cold spots were at freezing point. That is inconsistent with the experiences of those who were actually at the site and found that "on the cold days, even in January, there was a noticeable difference between the temperature in the middle of the site than there was when you walked two blocks over on Broadway", and "You could actually feel the heat". Three degrees, a "noticeable difference" that you could "actually feel"?

(The government loyalist may cry, "But I don't care about the average temperature - some of those burning zones could still have been very hot, even if they were only 0.5% or whatever of the pile. Think of all that fuel, and how the heat was trapped in." Never mind, the temperatures go down even lower when we take account of the lack of oxygen. And this time, they are all low!)

If we pretend the impossible occurred, then occasionally there might be a group of adjacent 1-meter cubes that were smoldering. However, after eighteen hours, these would have burned up and the whole zone would cool towards the mean pile temperature. The great majority of the time, a few little bits of burning paper or carpet, etc., would smolder away for hours in some isolated 1-meter block, with the adjacent blocks barely above ambient temperature. When all fuel in the smoldering block had burned up, its heat would be conducted away in both directions along each of the x, y and z axes. A temperature increase of less than one-seventh of 1,427 degrees (204 °C) in adjacent blocks would exceed the auto-ignition temperature of hardly any potential fuel, making even a much smaller than 200 series of conductions and ignitions highly improbable.

Thus, even when we pretend the impossible was possible, the government account remains highly improbable. And when we take into account the lack of oxygen and the abundance of water, in reality these bits of burning paper or carpet would not be smoldering, making such a five-month sequence physically impossible.

So how much air or oxygen would be required to support temperatures of around 600 °C in the steel and the concrete? For a 600 degree temperature increase in the steel and the concrete in the average 1-meter cube, the energy required is 600 times 1.475 MJ, about 885 MJ. Each cubic meter of air has a mass of 1.2 kg and an oxygen content of 0.232 × 1.2 = 278.4 grams. Each gram of O 2 releases 5 kJ in smoldering combustion, so each cubic meter of air can release 278.4 × 5 kJ = 1.392 MJ. Thus, 1 m3 of air would be sufficient to raise the temperature of the concrete and steel by 1.392 / 1.475 = 0.944 °C, and the volume of air required to support a 600 degree increase in the steel and the concrete is 600 × 1.475 / 1.392 = 885 / 1.392 = 636 m3 or 636,000 liters of air.

Since the volume of air required is 636 times the volume of the pile that is to be heated by 600 degrees Celsius, the volume of air required to raise one tower's entire pile by 600 degrees is 93,240 × 636 = about 59,300,000 m3, which is a cube of 1,279 feet (almost the height of the Towers). Assuming a smoldering time of 18 hours as calculated above, each one-meter cube within the pile requires a total of 636 m3 of air in 18 hours, which is 1 m3 every 1.7 minutes. (Or more, of course, when allowing for the aluminum, glass, fireproofing, air conditioning plant, etc. It's also worth noting that, just to bring a single one-meter cube in the pile up to 600 °C, the entire air supply in 6,360 m3 or 6.82% of the pile would be required after allowing for the 0.1 porosity, and this isn't even allowing for the water and foam in the pores. Thus, all of the air contained in a 93,240 m3 pile could only bring 14 one-meter cubes up to 600 °C - even if the air was somehow capable of rapidly flowing through the pile, which it wasn't, and even if the water and foam wasn't there, which was not the case.)

A person who was buried alive in concrete dust or sand would clearly soon run out of oxygen. However, if the individual was being supplied with a cubic meter of fresh air every 1.7 minutes, they wouldn't even get a headache!

The official 9/11 conspiracy theory requires that people could quite happily be buried alive yards deep in concrete dust or sand (e.g. with some water bottles and rations, in a casket to relieve the pressure - but with holes drilled in top and bottom to let water trickle through) for weeks on end as millions of gallons of water percolated through the dust. Thus, this provides inspiration for an alternative to waterboarding that could be utilized on 9/11 suspects who were brought in for questioning. To find out what really happened on 9/11 within next to no time, they could be provided with a communications device in order to signal when they were ready to talk, and warned that if they subsequently lied or kept quiet, there would be no second chance...

Let's consider the possibility of natural 'chimneys' that were left in the pile and extended continuously up to the surface immediately after collapse, along with the more remote possibility of underground voids connecting the bottom of these vertical shafts with the PATH tunnels (although the north tunnel was completely filled to the west at the low midriver point). Air is drawn along from the subway and driven to the surface by buoyant convection. There is no evidence that any such continuous, open voids existed that could support the flow of air at any significant rate, but let's suppose for the moment that they did exist and there were some burning combustibles that just happened to have landed so that they were sticking out into this void. In reality, very little was left intact; office materials such as computers, telephones, chairs, tables and books were mostly pulverized into dust or powdered debris, along with the concrete. And only a tiny proportion of combustibles - less than 5% - was burning at the time of collapse.

A formula approximating the flow rate from stack effect in a chimney is:

Q = Cd×A×√(2×g×H×(Ti-Te)/Ti)

...where: Cd is a discharge coefficient generally taken as 0.62 to 0.7, A is the cross-sectional flow area of the chimney, g is the gravitational acceleration, H is the chimney height, and Ti and Te are interior and exterior temperatures respectively in Kelvin. In SI units of length, Q is in m3/s and g equals 9.802 m/s2 (at New York City). Assuming Ti is 600 °C or 873 K and Te is 283 K, the flow rate for a chimney of cross-section 1 m2 and height 21.3 meters works out at:

Q = 0.65 × 1 × √(2 × 9.802 × 21.3 × (873 - 283) / 873) = 10.9 cubic meters of air per second. At 1,200 g/m3 this is 13,100 g/s of air and 13,100 × 0.232 = 3,040 g/s of oxygen. At the 13.1 kJ/g-O 2 of complete combustion in a flaming fire, the potential heat release rate is 39.8 MW. For such a massive chimney, bits of fuel sticking out from the walls would burn up very quickly within minutes of the collapse - in the improbable event that any of the tiny fraction of fuel that was already burning at collapse time wasn't extinguished by the crushing, smothering, shredding, pulverization, scattering, etc. And any remaining smoldering as the fuel burned away into the walls would soon be extinguished by the hundreds of thousands of gallons of water that was hosed on the pile each day over the first ten days, with the dust washing over the surface and completely smothering the smoldering combustible.

The air flow rate is likely to be less, given that the tortuous path for incoming air at the bottom hardly resembles an open window or a vent in a building, and that the flow rate is related to the hydraulic diameter. The effective area determined by the hydraulic diameter can be much lower than the actual cross-sectional area. For example, a long narrow chimney might have been formed with its long sides bounded by steel columns. If there was a chimney of 1 inch by 10 feet by 70 feet deep, the hydraulic diameter D H is a mere 2 inches, since D H amounts to twice the width in the limiting case for a vent of width much smaller than its length. Although the basic formula for stack effect is not too far out with its prediction of a flow velocity of 10.9 m/s, its predicted flow rate is about 80 times too high. Using Cd = 0.65, g equals 9.802 m/s2, H = 21.3 m, Ti = 873 K and Te = 283 K, the stack effect formula predicts a flow rate of 10.9 m/s times the cross-sectional flow area A. For a rectangular duct or tube of width 0.0254 m and length 1 m and up, the hydraulic diameter is approximately 0.05 m as per this online calculator. Say 0.0508 m.

Inputting the values outside temperature = 10 °C, inside temperature = 600 °C, height = 21.3 m, duct hydraulic diameter = 0.0508 m, duct length = 21.3 m and minor loss coefficient = 0, this online calculator predicts a duct velocity of 10.5 m/s and an air flow of 76 m3/h, which is 0.021 m3/s, i.e., 10.5 m/s × π × (½ × 0.0508)2. Various alternative values of up to 1 for the minor loss coefficient bring about a minor reduction in the flow rate. So the 0.021 m3/s, at 1,200 g/m3, is 25.2 g/s-air and 5.85 g/s-O 2 . Thus, at the 13.1 kJ/g-O 2 of flaming combustion, the potential heat release rate is 76.6 kW. Or if smoldering at 5 kJ/g-O 2 , the maximum heat release rate is 29.2 kW. This compares with examples provided by FEMA, such as 100 kW for a medium wastebasket with milk cartons, or 17 kW for a crumpled double sheet newspaper, 22 g.

And the flow rate already assumes a temperature of 600 °C at the bottom of the chimney, which is unlikely given that only a few percent of the fuel was burning at collapse time. If the burning fuel was, say, halfway up the chimney, the height would be halved - the neutral plane is almost 3/4 of the way up instead of almost halfway up - and the air flow rate reduced by around 29.3%.

A "chimney fire" equivalent to a newspaper or a few milk cartons burning is not going to produce "molten streams" or a "little river" of any metal, and it isn't going to keep burning for months on end. It will burn out within minutes. NIST should have carried out some experiments to test their theory. For example, dig a channel in concrete dust or sand, throw in some aluminum sheets, steel beams that were certified to ASTM E119 by Universal Laboratories, pulverized carpet, paper, computer casings and milk cartons, try to ignite it, pour in plenty of water and fire retardant foam, and see if any metal melts over the following months. And see if people's rubber boots start melting when they walk over the top of the pile. Or NIST could have sponsored some tests on WTC-style floor assemblies, subjected them to greater heat exposure and protected them with less fireproofing compared to the WTC on 9/11, and observed whether or not the test specimens "were able to sustain the maximum design load" without collapsing for as long as the tests were run: 2 to 3 1/2 hours. Wait a minute! NIST did sponsor those floor assembly tests, and the results were inconsistent with the government's theory.

But the fact that the floor assemblies didn't collapse in NIST's tests is at least consistent with the fact that no collapses were observed in the Cardington fire tests of 1995 to 1997 that examined "the behaviour of a multi-storey steel framed building subjected to a fire attack", and found that no collapses occurred despite the columns being only "lightly protected up to a height of 20 mm below the connections", with the beams and the connections "totally exposed", and the "final test arrangement" being "over four times longer than the normal Standard fire resistance furnace" and "believed to be the largest gas fired test furnace ever constructed within a steel framed building". The main report, dated June 2000, for those Cardington tests appears to have provided NIST with the inspiration for its preposterous, unworkable 'theory' of the WTC7 collapse about "unrestrained thermal expansion" that "causes ends to move apart". And NIST had to ignore the report's conclusion, which stated that "The results from both, the Cardington experiments and the computer modelling of those experiments described in the previous chapters, demonstrate that the composite steel framed building tested exhibited inherently stable behaviour under the applied fire scenarios due to the highly redundant nature of the structural system." Neither did collapse occur in the 2003 Cardington test, in which unprotected steel was heated to 1,000 °C.

If there were any smoldering pieces of fuel burning away into the dust down in natural chimneys, then the rain and hose water would have flowed down into those chimneys, washing the dust along with it and smoothing over any holes in the walls leading to the fuel, smothering it. When someone digs a hole in the sand at the beach, the hole rapidly vanishes as soon as the first few waves flow over it when the tide comes in. The fire-fighters weren't fools; whenever they saw smoking holes in the pile they'd have directed the hoses at those holes. Thus, the smoldering fuel would be soaked and smothered, and would be extinguished.

Speculative scenarios of office combustibles burning in a void at the basement level and receiving air from the subway fail to work, because the fire zone in the Towers was towards the top of the buildings: around floor 93 and up for WTC1 and floor 79 and up for WTC2. Thus, any pieces of burning plastic or wood, etc., not only needed to avoid being extinguished by ten or more seconds of smothering / pile-driving / crushing / shredding / pulverization and scattering, but also needed to overtake 80 or 90 undamaged, non-burning floors beneath them on the way down (which would be especially ridiculous given that plastic, wood, paper, etc., are lighter than steel) to end up almost at the bottom of the pile, but not so low that they ended up in the bottom of the Bathtub that was flooded with 30 million gallons of water over ten days.

The old IRT Cortlandt Street/WTC station and subway, which was 40 feet below street level, sustained significant damage when the WTC collapsed. It was reportedly "located directly beneath the Twin Towers", and "the Towers essentially fell on the existing station". The following photos were released the week of September 24, 2001, and confirm that the subway was severely damaged by the collapses. Sections of the tunnels were blocked. None of the photos of these underground voids at the nycsubway.org link below shows the slightest evidence of any fire - no flames, and no smoke. Fires in that area would have rapidly depleted the oxygen and produced a lot of smoke and toxic fumes, making it too hazardous for photographers. Those who descended underground and reported seeing molten metal or molten steel - such as Leslie Robertson, William Langewiesche, Philip Ruvolo, and Richard Garlock - would not have been able to do so had raging fires of hydrocarbons burning in air been responsible for the molten metal. But their testimony is entirely consistent with molten iron produced by thermitic accelerants that contain their own supply of oxygen, with the ensuing extremely high temperatures possibly leading to a little smoldering and smoking of local hydrocarbons that are within voids containing sufficient air to sustain minutes of smoldering or seconds of flaming combustion.

IRT Cortlandt St/WTC station south end of northbound platform. Photo by MTA New York City Transit.

IRT 1/9 tunnel north of Cortlandt St. station. Photo by MTA New York City Transit.

IRT 1/9 tunnel looking north on southbound track at Liberty Street. Photo by MTA New York City Transit.

IRT 1/9 tunnel looking north on northbound track at Liberty Street. Photo by MTA New York City Transit.

Source: nycsubway.org

Any raging underground fires would have rapidly become starved of oxygen, and even if a few bits of wood or plastic were burning and sticking out into voids such as subway tunnels or chimneys that extended for twenty feet or so, they would not produce a "little river" of molten metal, neither would they partly evaporate steel. The interesting thing about these so-called "fires" is how the heat was consistently concentrated in the steel. For example, Larry Keating said, "Sometimes the steel could explode when the buried ends were exposed to the air. You saw some of the thickest steel I’ve ever seen bent like a pretzel, and you just couldn’t imagine the force that that took. The grapplers were pulling stuff out, big sections of iron that were literally on fire on the other end. They would hit the air and burst into flames, which was pretty spooky to see."

With conventional smoldering fires in a pile of debris, the burning debris would be the hottest materials, and any secondary heating of metal would necessarily be to a lower temperature. And in an oxygen starved fire, the smoke should have been dark. Instead, it was white, which is just how aluminum oxide (Al 2 O 3 ) appears when dispersed into the atmosphere after fine aluminum particles are burned in the thermite reaction.

The driving force for stack effect is the pressure difference that results between lighter, warmer air and air of ambient temperature. A formula providing an approximation of this pressure difference is:

(P1 - P2) = C×A×H×(1/Te-1/Ti)

Wikipedia shows the pressure difference associated with a stack of given height as ΔP; it's shown here as (P1 - P2) to avoid confusion with the pressure gradient ΔP, the change per meter, used above for Darcy's law. C is a coefficient (0.0342 for SI units), A is the atmospheric pressure (Pa), H is the stack height (m), and Ti and Te are interior and exterior temperatures respectively in Kelvin.

Let's assume there is a hot, smoldering zone at 873 K, the exterior is at 283 K, the atmospheric pressure is 101,325 Pa, and the stack height is 21.3 m. So the pressure difference is:

0.0342 × 101,325 × 21.3 × (1/283 - 1/873) = 176 Pa. This is 8.27 Pa/m, compared to the 12 Pa/m normal difference with altitude for air of about 15 °C.

So not surprisingly, from q = -k × (P1 - P2) / (µ × L)

...where the fluid is air of dynamic viscosity 1.787×10-5 Pa.s, a pressure difference of 176 Pa across a medium of permeability 1.71×10-12 m2 and length 21.3 m would induce a Darcy velocity of 7.91×10-7 m/s, an air flux of 7.91×10-7 m3/s per m2 or 9.49×10-4 g/m2.s, and an oxidizer flux of 2.2×10-4 g/m2.s, more than three orders of magnitude below the 0.6 g/s required to sustain smoldering combustion.

Let's consider how much combustion could be supported by the air in typical voids within the pile just after collapse - the sort where fire-fighters could "get down below" and they'd "see molten steel, molten steel, running down the channel rails." Each void of 10 m3 would contain some 12 kg of air, with 23.2% of this initially being oxygen; hence, 2.784 kg of O 2 . At first there would be sufficient oxygen to maintain flaming combustion (at 13.1 kJ/g of O 2 consumed). So 2.784 kg of oxygen could support the release of up to 36.5 MJ, although the fire would extinguish itself long before the oxygen had run out. This compares with an average fuel load in each 1 m x 1 m x 0.2 m compressed 'floor' of 8 psf or 3.63 kg per square foot or 39.06 kg/m2 of wood equivalent fuel that could release some 10.78 MJ/kg in this compartment fire that would rapidly become fuel-rich and oxygen-depleted. There is a potential 421 MJ worth of fuel in an area of 1 x 1 m and a depth of 0.2 meters, or a potential 2.105 GJ within a one-meter cube, or a potential 21.05 GJ within 10 m3. So the potential 36.5 MJ from the oxygen within this 10 m3 void is a mere 1 / 577 of the 21.05 GJ of potential fuel. Or for smoldering combustion, the potential heat release at 5 kJ/g-O 2 is 13.92 MJ, which is 1 / 1,512 of the potential fuel. Various little bits of fuel - paper, plastic, etc., - that were sticking out into the void might burn under flaming combustion for minutes or smolder for hours, but they certainly wouldn't be capable of melting metal for weeks or months on end.

So, any burning material that was totally smothered by the concrete dust would simply be extinguished immediately. Where chimneys reached the surface, there might have been some flaming combustion for a matter of minutes and smoldering combustion for hours. However, as the smoldering combustibles at the chimney boundary burned out, the situation would revert to that of cold or cooling combustibles that were smothered by tens of thousands of tons of concrete dust, even if the fuel had improbably escaped being soaked by some of the millions of gallons of water that percolated down through the dust and along any chimneys, or being coated with fire retardant foam that had been diluted to increase its range, or being completely smothered by dust that the water washed over it. Much of the heat would be vented out to drive the smoke plume, and remaining energy would be conducted away through the concrete and massive steel members, with the mean debris pile never being more than 10 °C above ambient temperatures (very briefly, immediately after collapse) and no more than 3 °C above ambient for most of the time, if the government account were true. (Those figures of course refer to the 208' x 208' footprint of a Tower. The area of the Bathtub, at 980' x 510', was nearly six times greater than the combined WTC1 and 2 footprints.)

It is known that underground coal fires can reach high temperatures, e.g. Australia's Burning Mountain reportedly reaches 1700 °C deep underground. The operative word here is coal. And these fires self-extinguish when the overburden becomes too thick and fractures fail to reach the surface to draw in more air. The 1700 °C quoted in popular reports is rather high compared to underground coal fire temperatures stated in the literature, e.g. "up to 1200 °C". Glenn B. Stracher's Geology of Coal Fires states temperatures of 550-1450 °C for coal fires, 550-1650 °C for bitumen fires, and 550-2400 °C for oil fires.

However, underground fires of coal, bitumen or oil are all most unlike the WTC debris pile, which was by mass about 97% non-combustibles such as tightly compacted concrete dust, steel columns, floor decking, air conditioning and refrigeration plant, aluminum, glass, and fireproofing. Note how the potential fire temperature, e.g. up to 2400° C for oil, is strongly correlated with the heat of combustion. As you move up from lignite through to bituminous coal to oil, the heat of combustion - and the adiabatic flame temperature - increases. Coal yields up to 35 MJ/kg, whereas the WTC debris pile would struggle to average 1 MJ/kg.

Crucially, even underground coal fires require an oxygen source, and there was no such source within the WTC debris pile. Stracher's Geology of Coal Fires has this to say about how the fires are aerated:

The two necessary components for combustion are fuel and oxygen. Rugged relief and steeply dipping fuel-bearing strata provide ideal conditions for natural coal bed fires. Coal beds usually are ignited in the steep walls of river valleys and on steep hillsides that are not protected by Quaternary sedimentation. Fuel-bearing layers that are shielded by alluvial sediments commonly are not ignited (Burg et al., 1991, 1999). When fuel-bearing strata burn, a flame front develops along steeply dipping beds, and the fire descends quickly to depth. Fires occur mostly in beds that dip from 30° up to subvertical (see Table 1). A unique example of a fire spreading within near-horizontal strata is in the Powder River Basin, Wyoming, United States (Heffern and Coates, 2004). When a fire starts, it first spreads along an outcrop. With time, it burns deeper into the hillside and causes the overlying rocks to subside progressively into the burned-out void. The resulting fracturing allows air to enter and gas to escape, so that the fire continues (Fig. 1). These fires extinguish themselves naturally at the point when the overburden becomes so thick that fractures from the collapse fail to reach the surface to draw in more air, or when the fire burns down to the water table in the coal.

(In the above, the third requirement for combustion - heat - could have been included, along with fuel and oxygen.)

And another hint is provided by this source:

The rubble zone and the fault closest to the burning coal front are the main path ways to a deep coal fire, as these highly permeable zones act respectively as an air intake and exhaust under practical heterogeneous permeability conditions.

So, there are the questions: Was the WTC debris pile on a mountain or a steep hillside? Was most of the WTC pile comprised of a fuel, e.g. coal or oil, as opposed to concrete dust, fireproofing, steel beams, glass, etc? Did the WTC pile have fuel bearing strata? Was potential fuel in the pile not shielded by the equivalent of alluvial sediment, e.g., tens of thousands of tons of fine-grained concrete dust that had been deposited on top of it? Was there subsidence of overlying rock resulting in fracturing and exchange of gases so that the fires could continue? Was it unlikely that the overburden was so thick that any fractures or chimneys would fail to reach the surface to draw in more air and the fires would consequently extinguish themselves naturally? Did the pile have suitable channels that could have acted efficiently as air intake and exhaust?

If your answers are mostly no, then the answer to the following will also be "no".

Were conditions in the heavily watered, foamed and smothered WTC pile suitable for an underground fire that continued to burn for months on end and was like a "foundry" and like "lava" with "molten steel running down the channel rails"?

Finally, we have enough information to describe the reality of the 'fires' at Ground Zero. Early into the collapses, most or all of the tiny fraction of fuel burning at collapse time is extinguished by the supposed "pile-driver" effect of the collapse as the office combustibles are shredded, pulverized and smothered whilst crushed under the weight of tens of thousands of tons of upper floors for ten seconds or more. (This compares with a 960 kW fire in a 1 m3 void - not totally smothered! - that takes 3.79 seconds to consume all of the oxygen.) The maximum possible temperature increase from friction during the collapse - if all the energy is converted to heat and remains within the material - is given by gh/c where g = 9.802 m/s2 (the gravitational acceleration constant for New York City), h = 1,368 / 3.2808 m (the maximum height over which the collapse occurs), c (specific heat) = 450 J/kg.K for steel (at around ambient temperature) and c = 800 J/kg.K for concrete, which leads to 9.08 °C for the steel and 5.11 °C for the concrete.

Immediately after collapse, the system potentially starts with combustion at several remaining points along the x, y and z axes, albeit these points comprise no more than ~0.5% of the total volume - even if none of the fire was extinguished by the collapse. At points that are completely smothered, any remaining fires are unequivocally extinguished. Even smoldering cannot be sustained, since the oxidizer flux is at least 1,775 times below the minimum requirement of 0.6 g/m2.s-O 2 .

In the case of natural chimneys that happen to reach the surface of the pile, any intact combustibles that improbably happen to be burning and half sticking out into the chimney burn very well at first, but quickly burn away into the concrete dust. Over the first ten days, the bottom ten feet or so of the Bathtub (the actual height extends at least to the level at which it is being pumped out at Jersey City, as the north PATH tunnel is "completely filled" at the low midriver point) is flooded by 26 million gallons that leaks from the Hudson River and from broken mains, together with 1 million gallons of rain and 3 million gallons hosed onto the pile by fire-fighters. Any burning material that falls down to the bottom is immediately extinguished. The Bathtub is 980 ft x 510 ft and 3 million gallons of water has a volume of 401,042 ft3, so 3 million gallons per day per day makes for a height of 401,042 / (980 × 510) = 0.802 feet per day, which after allowing for 0.1 porosity corresponds to 100% saturation over a height of ten feet in barely more than a day.

Fire-fighters direct their hoses at any smoking holes in the ground, and so these chimneys certainly get their fair share of the four million gallons of water that flows right down through the pile from top to bottom within the first ten days, along with thousands of gallons of Pyrocool FEF fire retardant foam. Above the water level, the running water from the hoses washes the dust along until it covers any holes in the chimney walls leading to any smoldering plastic, paper, computers, wood, etc., totally smothering the sodden materials and extinguishing any remaining vestiges of smoldering.

Within voids that are formed in the pile, e.g., underneath massive columns, the fuel can initially burn in flaming combustion (but only for a few seconds), and thereafter a minuscule proportion of the fuel might continue to smolder. Since only ~5% of the fuel is burning at the time of collapse, and much, if not all, of that would have been extinguished by collapse, and the fuel is less than 10% of the volume, the most likely scenario is that none of the fuel that happens to settle at the bottom of a void is burning at collapse time, any fuel that is burning is completely smothered elsewhere in the dust and thereby extinguished, and thermitic material that should not have been present is responsible for all subsequent generation of heat, including possible temporary ignition of office combustibles.

Let's cut to the chase. The mass flux of oxidizer through the debris is calculated above at 0.000338 g/m2.s-O 2 , which at 5 kJ/g-O 2 for smoldering combustion can support a heat release rate of 1.69 W/m2. It doesn't matter whether the fire load is 20 MJ or 200 MJ of fuel per m2. The consumption of oxygen within the void cannot exceed the supply of oxygen for any longer than it takes for oxygen levels to drop too low to sustain smoldering. The oxygen originally within the void provides some reserve capacity, but only for a matter of minutes. 1.69 W/m2 is barely more than 1/1,000th the level of overhead sunshine, and wouldn't be enough to warm your little finger. It's equivalent to a third of a burning cigarette per square meter. It obviously cannot translate into "a noticeable difference between the temperature in the middle of the site" above thousands of tons of concrete dust and steel compared to "when you walked two blocks over on Broadway"; heat that "you could actually feel". In January, that is, four months after the collapses!! And it isn't going to melt any metal - whether steel, aluminum, tin or lead - or heat steel to cherry-red (about 750 to 900 °C) in six or seven weeks. (We'll confirm this in a moment!)

In theory, smoldering could be maintained within a void for some time, but the long-term power output or heat release rate of the fire could never exceed 1.69 W/m2. For example, if pulverized fuel is lying on the floor of a 2 m x 1 m x 0.5 m high void, mixed with concrete dust, taking up one-tenth of the area and all smoldering at a typical rate of 6.5 kW/m2, the total heat release rate is 2 × 0.1 × 6,500 = 1.3 kW. Given the void volume of 1 m3, the air within initially has a density ~1.2 kg/m3, so multiplying 1.2 kg by 23.2% oxygen by mass we have 278 g of O 2 , which at the 5 kJ/g-O 2 of smoldering combustion can release 1.39 MJ. A consumption rate of 1.3 kW would use up 1.39 MJ worth of oxygen in 1,069 seconds = 17.8 minutes. The 1.3 kW over 2 m2 is 650 W/m2. Thus, smoldering can only be sustained if on average no more than 1.69 / 650 = 1/385th of the fuel front (by surface area directly facing the oxidizer) is smoldering.

In this example of a void with a 2 m2 floor area and fuel taking up 10% of this area, smoldering can only be sustained in 0.2 m2 / 385 = 5.2 cm2 of fuel. If a typical piece of fuel is a 1 mm cube and they're all neatly laid out one deep, then there would be 200,000 cubes of fuel, but the oxygen supply could only support the smoldering of 520 of these pieces (at any one time). Over 1 m2, half the floor area, the 260 one mm by one mm fuel fronts totals 0.00026 m2 of fuel per m2, so if this fuel has a heat release rate at its smolder front of 6.5 kW/m2, the square meter area of the void has a total heat release rate of 0.00026 × 6,500 = 1.69 W. Or with each little cube having an output of 6.5 mW, with 260 of them smoldering per m2 of void floor, the total output is 1.69 W/m2.

In the general case of fuel pieces in various shapes and sizes scattered randomly, the maximum proportion of fuel front that can burn (dimensionless) equals the heat release rate that can be supported by the oxidizer flux (e.g. W/m2 which is derived from g/m2.s of O 2 times heat released in J per g of O 2 ) times the floor area (m2) divided by the mean burning rate of the fuel (W/m2) and the area of the fuel (m2). But we don't need to know the details of the fuel in order to disprove the government's conspiracy theory. It's the low oxygen supply rate of 0.000338 g/m2.s that dictates the low maximum possible heat release rate of 1.69 W/m2 from fuel that requires an oxidizer, as opposed to accelerants - e.g. thermite - that contain their own oxidizer.

The point about fuel is made because in order to get the government's 'fires' theory to work just to the point of producing 1.69 W/m2, it is necessary for exactly the right proportion of the fuel to be smoldering. If it's too low, the heat release rate doesn't even get up to 1.69 W/m2. If it's too high, the oxygen in the void is exhausted, the smoldering fuel is extinguished, and the air refresh rate is so low that the fuel cools down and has no hope of re-igniting by the time oxygen levels have been replenished.

Under flaming combustion the air in the void is consumed in seconds. In the example above, the oxygen is fully consumed in 17.8 minutes when all the fuel is smoldering. But combustion will cease before the oxygen gets to zero. The "reserve capacity" is less than ten minutes, although it does mean that within this air pocket (as opposed to a region that's completely smothered by concrete dust) there is time for the air to diffuse to the fuel so that smoldering can be sustained - and that's only provided the oxygen consumption is sufficiently low, and until the fuel burns out. If the oxygen gets used up, the refresh rate is 0.000338 g/m2.s, and so in this example where we have a floor area of 2 m2 and a height of only 0.5 m, it takes 278 / (2 × 0.000338) = 411,243 seconds = 4.76 days to replenish the 278 g of oxygen from the original 1 m3 of air. Or if combustion stops when the oxygen drops by half, it takes 11.4 hours just to return to 60%, by which time the fuel has also cooled down.

Moreover, the oxygen is only sufficient to allow smoldering of the tiny fraction of fuel, e.g. 1/385th, at one particular height (z co-ordinate) - the first (highest) level at which it encounters smoldering fuel as it permeates down into the pile. The air cannot "rejuvenate" itself to go on and oxidize any of the fuel at lower levels. Or alternatively, if less than the calculated proportion, e.g. 1 in 385, of the fuel at some particular level (for some particular area in the horizontal plane) is smoldering, there would be some remaining oxygen for additional fuel above or below.

For example, there could be a 1.69 W/m2 void with smoldering fuel at the (x, y, z) co-ordinates (10, 20, 10) that extended possibly for several meters horizontally, and another higher up at (15, 25, 15). But a void at (15, 25, 10) could not support smoldering, since the oxygen would have already been used up by smoldering in (15, 25, 15) above.

Horizontal diffusion of the incoming air should have been easily capable of sustaining such low-level combustion within a void, in contrast to the fact that no smoldering could be sustained in the dust. The 1.69 W/m2 is 1,775 times short of the minimum smolder power density of 3 kW/m2, or the oxidizer flux at 0.000338 g/m2.s-O 2 is 1,775 times short of the minimum 0.6 g/m2.s-O 2 . And horizontal air permeation through the dust would similarly be too low. In a void, the oxygen has plenty of time to diffuse to the smolder fronts. If the fire's power output is only 1.69 W/m2, then oxygen consumption equals the oxygen supply. The Darcy velocity of the incoming air is predicted to be 1.169×10-6 m/s, and the actual velocity through the 0.1 porosity is 1.169×10-5 m/s. If the void is a one meter cube, the air initially had a reserve capacity of 1,200 g × 0.232 × 5,000 J/g / 1.69 W = 823,669 seconds = 9.53 days at this consumption rate.

As for venting away the exhaust gases, that is more likely to pose a problem for the 'fires'. The exhaust would probably have to diffuse sideways, which would inhibit smoldering in adjacent voids, but these could possibly be at some other height (z co-ordinate). With a Darcy velocity of 1.169×10-6 m/s and ten times that for the actual velocity allowing for the 0.1 porosity, if the exhaust gases permeate through at a comparable velocity, they would take 23.8 hours to ascend one meter, so would surely cool and stop ascending well before clearing the pile. After all, NIST's Dr. Ohlemiller describes a reverse smolder front of a few kW/m2 from a source 0.1 to 0.15 m in diameter as so "weak" that the buoyant plume might not even reach the ceiling of a room. The plume from a 1.69 W/m2 source wouldn't manage to ascend through yards and tons of concrete dust!

The probability that voids of smoldering fuel could encompass the entire 4,019 m2 of footprint per Tower (and all of Building 7's), and all of them would have exactly the right proportion of fuel smoldering, and the fuel in each void could somehow maintain smoldering for five months, or there could be a bizarre series of heat conductions from cool 1.69 W/m2 voids through concrete dust to ignite fuel in even colder voids elsewhere such that the output was maintained across the entire 4,019 m2 for five months, may be safely stated as zero.

Even under this zero probability scenario, the maximum sustainable heat release rate of the 'fires' within a collapsed Tower footprint of 4,019 m2 could be 1.69 W/m2, which is a total of 6.792 kW. Let us suppose that the impossible occurs: that this is in fact sustained for a full five months, and is maintained over the entire 4,019 m2 of the footprint, and that all of the heat remains in the pile (despite photographic evidence to the contrary, e.g. white smoke). So over five months (September - February), the total heat released is 153 × 24 × 3,600 × 6,792 = 89.8 GJ. There was 90,000 tonnes of steel per Tower. Firstly, let's suppose that the entire 89.8 GJ is used to heat the steel, and forget about the concrete. Pretend it's self-heating or something. Or pretend it's just a bizarre coincidence that the heat is somehow mysteriously concentrated in the steel - just as if it had been targeted with thermite.

In order to raise the temperature of 90,000 tonnes of steel by 1°C, the energy required is:

90,000,000 kg × 450 J/kg.K = 40.5 GJ.

Thus, the entire heat output from the 'fires' in the pile, if locked in for five months, could increase the temperature of the steel by 89.8 / 40.5 = 2.2 °C. This is hardly going to turn it cherry-red!

Now we include the heat required to raise 110,000 tonnes of concrete by 1°C.

110,000,000 kg × 800 J/kg.K = 88 GJ.

Thus, the entire heat output from the 'fires' in the pile, if trapped in for five months, could increase the temperature of both the steel and the concrete by 89.8 / (40.5 + 88) = 0.7 °C.

This is barely more than 1 Fahrenheit degree.

And the (im)possibility of the (non-existent) 89.8 GJ over five months and 4,019 m2 of footprint scales down to a (non-existent) 22.3 MJ per m2 over 153 days, with the mass of steel, concrete, aluminum, etc., on average scaling down in a similar proportion. In reality, a few smoldering pieces of burning plastic or wood in an oxygen-starved underground void are going to extinguish within minutes or hours. A 1.69 W/m2 'fire' is equivalent to one burning cigarette every three square meters, or a burning candle every 47.3 square meters - e.g., a single candle in a 22.6 feet by 22.6 feet room.

Even if we include the unlikely "chimney fires" as described earlier, a chimney of hydraulic diameter two inches, with a potential heat release rate under flaming combustion of 76.6 kW equivalent to a few milk cartons, is going to burn out within minutes. According to Babrauskas, a 6.6 liter waste basket of empty milk cartons can produce a peak heat release rate of 70 kW, and the time of intense burning (at over 50 kW) is 200 seconds. So the total heat released will be ~14 MJ. Thus, it would take 89,800 / 14 = 6,414 of these highly improbable chimney fires per Tower footprint - in which any burning fuel prior to collapse wasn't extinguished by more than ten seconds of crushing and smothering - just to match the 0.7 degree increase of the 1.69 W/m2 'fires'.

We could include the maximum possible temperature increases of 9.08 °C for the steel and 5.11 °C for the concrete from friction during the collapse, although in this case we have to pretend that the entire gravitational energy was used to heat the steel and the concrete, and pretend that none of it was converted to sound, or used to eject massive steel beams laterally, pulverize the concrete, and smash the intact, cold, non-burning lower floors, including the breaking of hundreds of steel connections per second. And even then, no metal is going to melt.

A 1.69 W/m2 'fire' is most definitely not going to melt any metal. Period.

The official theory has to clear an impossible series of hurdles even before it can achieve the above maximum possible temperature increase in the pile of 0.7 °C, which is before we allow for the heat vented out with the smoke and the requirement to heat the 43,600 aluminum curtain wall panels of around 100 pounds, the glass, 56 tons of fireproofing per floor, 200,000 pounds of the nonflammable, noncorrosive, nontoxic refrigerant Freon 22, etc. For example, early into the collapses of the Towers, the upper floors (where the fires were located) were being crushed as if by a pile-driver, according to Ronald Hamburger, a contributor to FEMA and NIST. Hamburger had previously said (before learning that a cover-up was required), "It appeared to me like charges had been placed in the building."

Fuel burns much faster under flaming combustion compared to the typical 6.5 kW/m2 of smoldering. According to FEMA, for example, "the associated energy release rate per unit surface area [in typical office building fires] ranges from 320 to 640 kW/m2".

Let's take 480 kW/m2, and assume 13.1 kJ/g of O 2 consumed. A 1 m3 void contains 1.2 kg of air including 278 g of oxygen. Thus, if we have a void of 2 m x 1 m x 0.5 m deep as in the example above and fuel is burning in flaming combustion at a heat release rate of 480 kW/m2, the 960 kW over 2 m2 consumes oxygen at a rate of 73.3 g/s and the oxygen is fully consumed in 3.79 seconds. Even taking a minimum estimate for the collapse time of ten seconds, it is not very credible that materials burning at 480 kW/m2 could have been crushed, shredded, pulverized, scattered, and could remain burning for ten seconds or more in the absence of oxygen whilst crushed under the weight of tens of thousands of tons of upper floors, going on to end up protected in a void in which they could burn for another few seconds. After all, when some drunken lout ignites newspapers and places them on several seats on a bus, the fires are all extinguished when the driver stamps on them a few times. And people with burning clothes - even after being sprayed with oil and gasoline - can roll over to extinguish the fire.

Given that just about everything apart from huge, twisted pieces of steel - computers, telephones, chairs, tables, filing cabinets, books and gypsum along with the concrete - was pulverized into dust or powdered debris, it makes most sense to consider the various combustibles as being of a similar size to the concrete dust. If the distribution of sizes is similar for the fuel and the concrete dust, then there are 30% within the range 10 cm to 1 cm, 20% within 1 cm to 1 mm, 15% within 1 mm to 100 µm, 10% within 100 µm to 10 µm, and 25% at less than 10 µm.

The speed at which the smoldering fire front advances is given by the power density divided by the heat of combustion and the density of the fuel. So kW/m2 / [(kJ/kg) × (kg/m3)] ===> m/s. In the smoldering case of 6.5 kW/m2, assuming 21.56 MJ/kg and 200 kg/m3 we have:

6.5 / (21,560 × 200) = 1.51×10-6 m/s = 5.43 mm/hour = 97.7 mm in the 18 hours typical time for the pocket of fuel to be consumed. So a fuel density of 200 kg/m3 and maximum size of 10 cm fits in with the ~18 hours smoldering time assumed above. Some fuel would have a lower density and so would burn up in less time (for the same maximum size). 6 pcf or 96 kg/m3 is regarded as a "high" density for polyurethane foam, for example. Combustible material compressed within the concrete dust would have a higher density, but combustion there is precluded by total smothering.

Or in the flaming combustion case of 351 kW/m2, the velocity of the fire front is 8.14×10-5 m/s, 54 times faster, again at 97.7 mm, in this case over the 20 minutes typical time for the fuel to be consumed.

Let's consider a 1 mm cube as a typical size. The volume is 10-9 m3, and assuming 200 kg/m3, the mass is 2×10-7 kg. The 1 mm cube has a smolder front area of 10-6 m2 and a heat release rate of 0.0065 W. The average heat of combustion of the fuel is taken as 21.56 MJ/kg, and so the combustion of one of these cubes will release 4.312 J. Thus, the 4.312 J would be released over 4.312 / 0.0065 = 663 seconds which is ~11 minutes. Dividing 5 months into 11 minutes, there would have to be a series of 20,029 smolderings and transfer of heat to ignite the next piece of fuel.

At a smoldering rate of 5.43 mm/hr, a 3 cm cube of fuel burns up in 5.52 hours, a 3 mm cube in 33.1 minutes, 300 µm in 3.31 minutes, 30 µm in 19.9 seconds, and 3 µm in 1.99 seconds. Pieces of 15 µm and smaller would have burned out before the Towers' fire floors even reached the ground. Even a 1,000 kg/m3 fuel with a length of 10 cm and smolder velocity of 1 mm/hr, say, would burn up within 100 hours or ~4 days.

If too much fuel is burning, oxygen levels would be expected to drop too rapidly to allow a few pieces of fuel to remain smoldering. Even if a sustainable proportion of fuel was smoldering, the 1.69 W/m2 within the void is orders of magnitude below the kilowatts/m2 required to ignite smoldering combustion, and even further short of the 20 kW/m2 required for piloted ignition of a a range of common combustibles. At 1.69 W/m2 and 21.56 kJ/g, the mass burning rate is 1.69 / 21,560 = 0.0000784 g/m2.s. Under equilibrium conditions in which the smoke doesn't crowd out the oxygen, the concentration of smoke would be very low and the upper gas layer, marginally less cold than the incoming air, would be very thin. Gas temperatures would be well below the autoignition temperature of any available fuel. If the fire burns more rapidly and smoke generation exceeds the supply of fresh oxygen, the fire extinguishes.

And in order to sustain the less than lukewarm 1.69 W/m2 for months, the government conspiracy theory requires a vastly improbable scenario of adjacent voids containing fuel, with an extraordinary series of conduction of heat through the intervening concrete dust and subsequent ignition of fuel in the next void. Moreover, since there were a number of hot spots in the pile at any given time, the official theory requires that there are many of these most improbable series running in parallel, for months on end. The problem is that the smoldering voids generate very little heat in the first place, but even if they were hot enough, insufficient heat would be conducted across the concrete to the adjacent void before the original void had burned out and started cooling.

If there is a hot spot at a particular point in the pile, the speed at which heat is conducted through the pile is related to the thermal diffusivity of the conducting material. The time t required to heat up a point at a distance x to a temperature midway between the temperatures of the hot spot and ambient temperature, is related to the square of the distance. Conversely, the distance x is related to the square root of the time t.

The concrete or steel is treated as a semi-infinite solid or bar. If a bar of semi-infinite length with insulated sides is "instantly" heated at one end, the distance x at which half the temperature rise occurs is given by:

x = 2 × z × √(α × t)

where z is the value required to yield a Gauss error function of 0.5, t is the time in seconds, and α is the thermal diffusivity of the material in m2/s. Since erf(0.477) is close to 0.5, z is taken as 0.477. Thermal diffusivity (α) is the thermal conductivity (k) divided by the volumetric specific heat capacity (ρ × C), where ρ is the density and C is the specific heat by mass. It is also useful to find the time t, given the distance x.

So rearranging x = 0.954 × √(α × t), we have near enough to:

t = 1.1 × x2 / α

(See this page for example, where their Eq. 4.3 can be used to derive those equations for solving x and t.)

This underestimates the time required for a given distance, because the sides of the "bar" are not insulated. Heat is being conducted in other directions, except in the case of a massive wall of fire, in which case it would be relatively accurate at the center. At Ground Zero, there should have been no great walls of raging fire.

The thermal conductivity k for lightweight concrete and normal weight concrete is 0.1 to 0.3 W/m.K and 0.4 to 0.7 W/m.K respectively. Assuming the concrete dust is packed so as to be closer to the original density of lightweight than normal weight, we'll take the density as 1,750 kg/m3, and the thermal conductivity as 0.3 W/m.K.

So for the concrete, we have:

α = 0.3 W/m.K / [1,750 kg/m3 × 800 J/kg.K] = 2.14×10-7 m2/s

And for steel:

α = 40 W/m.K / [7,860 kg/m3 × 550 J/kg.K] = 9.25×10-6 m2/s

So for concrete that is 610 °C at the hot end and an ambient temperature of 10 °C, the time taken for a point 1 meter distant to reach 310 °C is 1.1 × 12 / 2.14×10-7 = 5,140,187 seconds = 59.5 days. Or for a point 40 mm distant, we have 1.1 × 0.042 / 2.14×10-7 = 8224 seconds = 2.284 hours. Or the time for a point 10 mm distant to reach 310 °C is 1.1 × 0.012 / 2.14×10-7 = 514 seconds = 8.57 minutes. The time associated with one meter is way beyond the time that any smoldering combustibles within a void would take to burn up. A wall of concrete dust that was only 40 mm thick would have collapsed.

For steel, the corresponding times are shorter, or the distances are greater, as one would expect from its higher thermal diffusivity. The time taken for a point 1 meter distant to reach 310 °C is 1.1 × 12 / 9.25×10-6 = 118,919 seconds = 33 hours. Or for a point 0.5 m distant, the time is 1.1 × 0.52 / 9.25×10-6 = 29,730 seconds = 8.26 hours.

From x = 2 × z × √(α × t)

...we can find the distances associated with typical smoldering times, say, 18 hours. So for the concrete, x = 0.954 × √(2.14×10-7 × 64,800) = 0.112 m. However, the 1 mm fuel cube burns up in 663 seconds (11.05 minutes). This is associated with a distance of 0.954 × √(2.14×10-7 × 663) = 11.4 mm = 0.45 inch.

Or for the steel at t = 11.05 minutes, x = 0.954 × √(9.25×10-6 × 663) = 74.7 mm = 2.94 inches. At 18 hours, x = 0.738 m.

(For those who are curious as to how the above formulae compare with a rough approach that estimates the heat conduction rate and the energy required to heat the material, and divides the latter by the former to obtain the time, the standard one-dimensional equation for the quantity of heat conducted per unit time is:



q = -k × A × (T1 - T2) / x



...where q is in J/s or W, k is the thermal conductivity in W/m.K, A is the area of the conducting material in m2, T1 and T2 are the temperatures at the ends of the conductor in °C or K and x is the thickness of the conductor. So (T1 - T2) / x gives the thermal gradient in K/m. The minus sign in k just denotes the direction of heat flow.

Then there is the energy required to raise the mass of material by the required temperature:

E = A × L × ρ × C × (T1 - T2)

...where A is the area of the "bar", L is its length, ρ is its density, and C is its specific heat. If the cold end goes up from ambient to the mean of ambient and the hot side, the average temperature difference could be taken as three-quarters of the difference between hot and ambient for purposes of calculating q. Recalculating the difference every second is more accurate, and adds about 4% to the predicted time for a particular distance. And on average the temperature of the material is increased by three-quarters of the difference between hot and ambient. This also underestimates the time taken, since it does not allow for the fact that the material must also be heated at distance