April 4, 2009 update: With the publication of a paper in a peer-reviewed journal proving that active thermitic materials were discovered in the WTC dust, there is no longer any doubt that a thermite variant was used in the controlled demolitions. The material found was highly energetic and included particles on the nanometer scale. Fleshing out details of where it was placed, and whether conventional thermite or thermate was also used, remains to be seen.

See here for links to the paper and associated discussion.

WTC MOLTEN STEEL - THE 9/11 SMOKING GUN

Introduction

A novel approach to determining gas temperatures in compartment fires is employed, in order to establish whether the observed molten steel or iron in the ruins of the World Trade Center could have been formed as a natural consequence of aircraft impact, combustion of jet fuel and office furnishings, and fire-induced building collapse. The information provided below will enable any Windows-based PC user to readily calculate temperatures after inputting several variables. We find that not only is there zero probability of the office fires melting any steel, but formation of molten steel or iron would have been impossible in the post-collapse fires within the debris pile ... unless accelerants had been installed as a means of ensuring global collapse.

The Evidence

There are many corroborating sources that reported molten steel in the ruins of the World Trade Center. Another excellent summary of reports, ranging from "cherry red steel" (about 745 degrees C) at a depth of six stories underground that was found six weeks after the attacks, to a steel beam "dripping with the molten steel" to "streams" and "rivers" of molten steel, may be found here at the George Washington blog. The latter includes a link to a New York Times archived article (pay-per-view) stating that Dr. Jonathan Barnett, a professor of fire protection engineering, said that an uncontrollable fire could not explain steel members in the WTC 7 debris pile that appear to have been partly evaporated in extraordinarily high temperatures. The History Bytes blog has yet another first-class summary with a number of named first-hand eyewitness accounts featuring quotes such as "massive amounts of molten steel".

Mark Loizeaux of Controlled Demolition, Inc. of Phoenix, MD., who was involved in the removal of the rubble, was reported as saying that molten steel was found at the basement levels in the ruins of World Trade Center buildings 1, 2 and 7. This was confirmed by journalist William Langewiesche who personally witnessed areas where "steel flowed in molten streams" after "descending deep below street level" (subscribers only link here, but see above George Washington blog for the quotes). As late as February 2002, Bronx firefighter Joe "Toolie" O'Toole saw a crane lift a steel beam vertically from deep within the catacombs of Ground Zero. "It was dripping from the molten steel," he said. Some molten material could have been aluminum, but analysis confirmed significant quantities of iron. Since the existence of molten steel or iron cannot be denied, NIST has even made a (failed) attempt to address the matter in its recently released "FAQ"- scroll down to #13 at this link. Molten steel or iron was also observed pouring from the side of WTC 2 in the minutes prior to its collapse [see pp. 10-16 in the paper at this link].

The Science

The temperatures generated by a hydrocarbon-fuelled office compartment fire are not capable of melting steel. A stoichiometric combustion of kerosene, generally regarded as dodecane, for example...

C12H26 + 18.5O2 +69.595N2 ===> 12CO2 + 13H2O + 69.595N2 + 7518 kJ

...would have at best the above 7,518 kJ locked into 2.712 kg of combustion products to yield an adiabatic flame temperature (AFT) of some 2,398 K or 2,125 C (it is about 130 K less after including the inevitable endothermic dissociation reactions at these temperatures in order to reach an equilibrium state). At standard temperature and pressure, one mole of an ideal gas occupies 22.4 liters; at 2,398 K the volume would have expanded by a factor of 2398/273, i.e. nearly 9 times, to 196.8 liters/mol. The 7.518 MJ is spread amongst 94.595 moles or 18.62 cubic meters of hot gaseous products. At some 0.146 kg/m^3, these are much lighter than air at STP. In order to melt a mere 1 kg of steel that would take up a volume of only 0.127 liters, at least 1 MJ would be required. If 1/7.518 of the enthalpy (heat) in the products was directed at the kilogram of steel, the temperature rise would be reduced from 2,105 K to a little above 1,825 K which would result in a temperature of a little over 1,845 C and still potentially hot enough to melt some steel. But in real-world conditions, flame and upper layer hot gas temperatures are well below the AFT, typically barely reaching 1,000 C and certainly well below the melting point of steel. Heat radiates out in all directions, and would not be concentrated just at some particular 0.127 liter lump of steel. And the jet fuel fires in the WTC would have burnt out within a few minutes. The remaining combustible office material would have released lower quantities of heat per unit mass of fuel, and the associated AFTs would have been lower. Consequently, flame temperatures would have been lower for a given proportion of heat losses to the surroundings.

Moreover, the oxygen-starved fuel-rich environment of the compartment fire would have reduced the temperature increase and the heat yielded per unit of fuel by a factor of about one-third. And a report, authored by a team of engineers commissioned by Silverstein Properties, concluded that fire temperatures in the WTC were even lower than typical fully-developed office fires. The fires were inhibited by dust and debris distributed by the impacts, and the study placed the temperatures at between 750 F and 1300 F (about 400 C to 700 C). Following collapse, low air exchange rates in the rubble pile would lead to low mass burning rates, poor combustion efficiency and low power output. A few smoldering pieces of wood and plastic could not create temperatures capable of melting steel.

If it were possible for hydrocarbon fires to melt significant quantities of steel, then burnt-out cars would regularly be observed consisting primarily of solidified pools of molten steel on the ground, and kerosene heaters would be about as useful as a chocolate teapot!

Nevertheless, it should be possible to input values for several variables and ascertain some probable figures for gas temperatures. It is clear that these will lie between ambient temperature and adiabatic flame temperature. The greater the proportion of energy that remains locked in the gas products, the closer the gas temperature approaches the AFT. But from the Stefan-Boltzmann equation for radiative heat transfer:

P / A = e * sigma * (Te^4 - Ta^4)

where P is the rate of radiative heat transfer, A is the exposed area of the absorbing body, e is the emissivity, sigma is the Stefan-Boltzmann constant of 5.6703 * 10^-8 W/m^2.K^4, and Te and Ta are the absolute temperatures of emitter and absorber respectively, it is well established that the heat radiated out to the surroundings is proportional to the difference between the fourth powers of these absolute temperatures. If the gas temperatures are high, then a lot of heat will be radiated out from the products, and hence the temperatures would not be high. On the other hand, if the gas products are not much warmer than ambient, then very little heat would have been emitted to the surroundings at those temperatures, and consequently the gas temperatures would not have got so low. There will be an equilibrium temperature, close to the point where the Stefan-Boltzmann radiation curve intersects the nearly linear, negative gradient function that shows the proportion of heat radiated from the products versus temperature.

The mass burning rate times the heat of combustion of the fuel determines the heat release rate. The quantity of combustibles per unit area of floor space - often expressed in terms of the equivalent weight of wood (psf) - determines the heat release rate per unit area. If the gas temperature is midway between ambient and adiabatic, then the flux radiated from the products is approximately half of the total power density. In this case it is actually a little more than half, since specific heat is greater at higher temperatures. A reasonable correction would be to add 10%. When the gas temperature is close to ambient, the correction factor of 1.1 reverts to 1.0; at close to AFT the correction factor is more like 1.2. For our purposes, the following equation will suffice:

P / A = 1.1 * Ftot * (Taft - Teq) / (Taft - Tamb)

where: Ftot is the total heat release rate per unit area of floor space in kW/m^2, Taft is the mean expected adiabatic flame temperature in Kelvin over a range of stoichiometries from phi = 1.0 to phi = 2.0, Tamb is the ambient temperature in K, Teq is the equilibrium mean upper layer gas temperature in K, and P/A is the heat flux transmitted from the gas products in kW/m^2. Teq is assumed to be the emitter temperature Te, and Tamb is taken as the absorber temperature Ta in the Stefan - Boltzmann equation. The two functions intersect at:

1.1 * Ftot * (Taft - Teq) / (Taft - Tamb) = e * sigma * (Teq^4 - Tamb^4)  [equation #1]

We assume emissivity e = 1, and take 5.67 * 10^-8 W/m^2.K^4 for the constant sigma. Dividing both sides by 1,000, this becomes 5.67 * 10^-11, as it is more convenient to input heat flux (Ftot) as kW/m^2 rather than W/m^2. There are three known variables: Ftot, Tamb and Taft, and one unknown variable: Teq. (See below for how the "known" values can be derived for the WTC fires; the reader will be able to input various figures and observe the effect on Teq.)  After rearranging into orders of descending Teq, we have:

[5.67*10^-11(Tamb-Taft)]Teq^4 - 1.1FtotTeq + 1.1FtotTaft + 5.67*10^-11(TaftTamb^4-Tamb^5) = 0 [equation #2]

Now the unknown variable Teq is 'x' in a quartic equation of the form:

ax^4+bx^3+cx^2+dx+e=0

There are no x^3 or x^2 terms, so b and c are equal to zero. The next step is to simplify by dividing throughout by a, so that a equals one. We now have a=1, b=0, c=0, and:

d = -1.1*Ftot / [5.67*10^-11*(Tamb-Taft)]

e = 1.1*Ftot*Taft / [5.67*10^-11*(Tamb-Taft)] + (Taft*Tamb^4-Tamb^5) / (Tamb-Taft)

Teq may be directly solved by inputting the values a to e into an online quartic equation calculator. This yields four roots X1 to X4. Ignore the complex or negative real X2 to X4; Teq (in degrees K) is X1, the positive real root. With the values for a, b, and c remaining constant at 1, 0, and 0 respectively; it is only necessary to calculate and input d and e for various values of Ftot, Tamb and Taft. (The value for e ranges from around -10^10 to -10^17; d is a positive number smaller than e.)

The simplest way to input lots of values for the three variables and quickly find an associated likely Teq value, avoiding manual calculations, is to program a loop that steps through from, say, 273 K to 4,000 K in one degree steps, and records the temperature at which equation #1 comes closest to balancing. The following code does that, and also calculates the d and e values ready for inputting into the quartic equation calculator. As a final check, it can be seen that when the latter's obtained value for X1 or Teq is inputted as x in the quartic equation, the result is close to zero and rapidly diverges for other x values. 

*****Qbasic code, copy and paste*****

REM****CALCULATE MEAN EQUILIBRIUM GAS TEMP IN COMPARTMENT FIRES
PRINT : minresult# = 1D+20
INPUT "Enter Tamb (C), Taft (C), Ftot (kW/m^2)"; tamb#, taft#, ftot#
REM***convert temps from celsius to kelvin
tamb# = tamb# + 273.15: taft# = taft# + 273.15
REM***Find equilibrium temperature
FOR teq# = 273 TO 4000
out1# = .0000000000567# * (teq# ^ 4 - tamb# ^ 4)
out2# = ftot# * 1.1 * (taft# - teq#) / (taft# - tamb#)
result# = ABS(out1# - out2#)
IF result# < minresult# THEN minresult# = result#: eq# = teq#
NEXT teq#
PRINT "Equilibrium temperature ="; eq# - 273; "C ="; eq#; "K": PRINT
REM***Check by converting to quartic equation
a# = 1: d# = -1.1 * ftot# / (.0000000000567# * (tamb# - taft#))
e1# = d# * -taft#
e2# = (taft# * tamb# ^ 4 - tamb# ^ 5) / (tamb# - taft#)
e# = e1# + e2#
PRINT "To calculate the exact temperature (K), we have a quartic equation"
PRINT "of the form: ax^4 + bx^3 + cx^2 + dx + e = 0, where:"
PRINT "a ="; a#; ", b = 0 , c = 0 , d ="; d#; ", e = "; e#: PRINT
PRINT "Input these variables into a quartic equation calculator to solve x."
PRINT "Then enter the obtained value for x, to check that this value gives"
INPUT "a result closest to zero (Enter '-1' to exit loop)"; x#
IF x# = -1 THEN RUN
GOSUB DISPLAYRESULT: PRINT
REM***Partial Quartic Checker
EXITCHECK: INPUT "Enter x ('-1' to exit loop)"; x#: IF x# = -1 THEN RUN
GOSUB DISPLAYRESULT: GOTO EXITCHECK
DISPLAYRESULT: PRINT a# * x# ^ 4 + d# * x# + e#: RETURN

For those without QBasic, downloads are available here or here. It is best run in a DOS ("Command Prompt") window. Create a directory such as C:\qbasic and unzip after downloading. Using an editor, create a text file and paste the above code into it. Save as eq.bas or whatever you choose to call it; it must have a .bas extension. It should then be possible to open the file after launching Qbasic.exe. Shift-F5 to run, and Ctrl-C to exit to the integrated development environment.

For the WTC fires, the values initially assumed were Tamb = 25 C, Taft = 1,488 C (1,761 K), and Ptot = 56.8 kW/m^2. The derivation of the Taft value is detailed at this page. To summarise, adiabatic flame temperatures for dodecane and wood at various stoichiometries were obtained from graphs in figures 3 to 5, Chapter 1 of Combustion (Third Edition), Irvin Glassman. These values (degrees Kelvin) were as follows:

phi 0.5  0.75 1.0  1.1  1.5  2.0  fuel

AFT 1490 1960 2265 2290 1960 1610 kerosene

AFT 1380 1810 2120 2075 1660 1210 wood

The jet fuel pool fires burnt up within a few minutes of the impact, so the values for wood were taken. To allow for the fuel-rich conditions of the compartment fire, it was supposed that the fuel : oxygen equivalence ratio (phi) ranged from 1.0 nearest to the vents to 2.0 at the most oxygen-starved zones. Averaging across this AFT versus phi function, the mean value was found to be 1,761 K. The above link also shows how combustion efficiency, averaging across this range of stoichiometries was calculated at around 68%, with 11.4 kJ/g (as opposed to 16.75 kJ/g  at phi = 1.0) of wood equivalent fuel. 11.4 kJ/g is in line with the 10.75 kJ/g assumed by Kawagoe and others such as Magnusson and Thelandersson as the net effective heat of combustion for wood equivalent fuel in fully developed compartment fires (see p. 4 at this link).

A worst-case scenario for WTC 1 supposed that on the floor with the severest fires, the entire fuel load of office materials was consumed over the 102 minutes from impact to collapse, and the energy released from an above average share of the jet fuel and some aircraft combustibles was added to the total. The total heat released for this worst-case floor, including energy that was vented out to drive the smoke plume, was calculated at 1.39 * 10^12 J (see this page for details of the calculations). This total corresponds to an average of some 5.4 psf of wood equivalent fire load excluding jet fuel and aircraft combustibles (cf. NIST's estimate of "about 4 psf of combustible materials" on a "typical floor", p. 18 at this link; given that some fuel has a higher calorific value than wood, the equivalent weight of wood could be a little higher than 4 psf).

The 1.39 * 10^12 J total was then divided by 102 * 60 seconds, obtaining 227 MW for the entire floor, and this was divided by the floor area of 4,000 m^2 to arrive at 56.8 kW/m^2. This value for Ftot, the total flux or the total heat release rate per unit area, is used together with Taft = 1,488 C and Tamb = 25 C to estimate a mean temperature for a typical point on the floor, averaged over the full 102 minutes. At any given point, the local combustibles would typically take some 20 minutes to be consumed, after which flames would move along to another zone (see p. 19 at the above link to NIST material). NIST estimated maximum temperatures of about 1,000 C for 20 minutes and 500 C or less for the remaining time (82 minutes for WTC 1). This would suggest mean temperatures for any given point to be around 600 C, which is the sort of result we should expect from inputting 56.8 kW/m^2, 25 C and 1,488 C into our formula.

Alternatively, a peak value for temperatures as the local combustibles are consumed over the 20 minutes may be derived, which could be nearer 1,000 C than 600 C. We increase the Ftot value to 7 psf of wood equivalent combustibles outside the core consumed over 20 minutes. 7 psf is 34.18 kg/m^2 and at 11.4 MJ/kg would release 389.65 MJ/m^2. This amounts to 324.7 kW/m^2 over 20 minutes.

We could have taken a higher value for Tamb as the environment is heated, although the Teq value relates to the upper layer gas temperatures, and a cooler layer of fresh air is continually being drawn in to feed the fires. The Ftot value is a rather high estimate, and it was found that the predicted value for Teq was more dependent on the inputted value for Ftot. Increases in Tamb and Taft led to relatively small increases in Teq.

Results and Discussion

For the worst-case WTC floor, taking Ftot as 56.8 kW/m^2 averaged over 102 minutes, Tamb as 25 C, and Taft as 1,488 C, the predicted value for the mean Teq is 627 C or 900 K. The Silverstein Properties study team quoted temperatures of 400 to 700 C. The mean of their range, 550 C, would correspond to 36.25 kW/m^2 which is 145 MW for a whole floor, and 8.874 * 10^11 J over 102 minutes. As local combustibles are consumed over 20 minutes, Ftot is raised to 324.7 kW/m^2 and the temporary Teq increases to 958 C. There could well have been short-lived peaks in excess of 1,000 C.  With NIST's estimate for mean temperatures at around 600 C, the predictions of the formula are in line with generally accepted models.

Table 1 has Tamb set to 25 C throughout. The first column shows various values assigned to Ftot, and the remaining columns show the calculated values for Teq in Celsius for particular combinations of Ftot and Taft. As Ftot approaches zero, Teq approaches Tamb; at very high values for Ftot, Teq approaches Taft.

Table 1

Table 2 has eight columns with Tamb set to 225 C, and the final two have Tamb at 525 C. The values for Taft have been increased in line with the Tamb increase. In fact, for a given reaction, flame temperatures would not rise as much as Tamb. At temperatures over 1000 C, endothermic dissociation reactions such as CO2 back to CO + 1/2O2 occur. As the temperature increases further, a higher proportion of endothermic reactions lower the equilibrium heat of reaction. Consequently, if the Taft of 2400 C example below represents the same reaction as 2200 C Taft when Tamb is 25 C, the Teq temperatures shown are too high, particularly in the temperature range of molten steel.

Table 2

In general, in compartment fires, the flux will peak in the hundreds of kilowatts per square meter range, and decrease to tens of kilowatts per square meter or lower when combustion does not include fuel in the immediate vicinity. From Table 1, it can be seen that temperatures capable of melting steel (1500 C and higher), from a typical hydrocarbon reaction that could yield an adiabatic flame temperature of 2,000 C under fairly optimum conditions but lacking pre-heated air, would require a flux in excess of 2,000 kW/m^2. Even with air pre-heated to 225 or 525 C, the flux requirement would be 1,000 kW/m^2 or more depending on the fuel.

Fires may be fuel-controlled or ventilation-controlled. Even the former, when the fuel involves regular hydrocarbons, would be very unlikely to produce massive fluxes in the megawatts per square meter range. And if they did, such peaks could not be sustained for more than a matter of a few seconds, which would not suffice to melt steel. A study comparing the heat release rates of wood-plastic composites found, as expected, that HRRs were higher with increasing concentrations of plastic. The highest peak HRR was obtained with a mix of 80% high density polyethylene and wood flour (mainly pine); this was still under 900 kW/m^2. The study mentions that Elliot and others tested medium density polyethylene and obtained a peak of 1,550 kW/m^2 at 450 seconds, up from 500 kW/m^2 at 350 seconds. In the case of the WTC debris pile, most if not all of the combustion was ventilation-controlled, and we can determine the maximum possible flux densities by imagining tiny chimneys within the debris as miniature 'compartments' and calculating the air throughput.

Suppose a regular compartment such as a floor of the World Trade Center has horizontal dimensions of x and z, and a vertical dimension of y. We imagine air flowing through the compartment with a velocity v in the direction of the x dimension. Here, it is convenient to work in units of feet, since each cubic foot of air supplies sufficient oxygen to release some 100 kJ in typical hydrocarbon combustion. This derives from the fact that, for a range of hydrocarbon reactions, each gram of oxygen releases about 13.1 kJ under ideal conditions. At ambient temperatures a little above the 0 C of STP, we may take the density of air at around 1.2 kg/m^3. Each cubic meter contains 21% oxygen by volume, but 23.2% by weight, which is 0.232 * 1.2 kg = 278 grams per cubic meter = 7.87 grams per cubic foot.

7.87 * 13.1 kJ ~ 100 kJ at marginally less than ideal conditions.

In the case of the WTC, x and z were 207 feet, and the product was assumed to be the total area of the heat flux, with the y dimension relatively low at 12 feet. 1.39 * 10^12 J in 102 minutes for a complete floor is a heat release rate of 227 MW. With each cubic foot of air providing 100 kJ, the air flow rate required to sustain a 227 MW fire on a single floor would be 2,270 cfs. If a 12 ft by 207 ft side was fully vented and the air flowed directly along the x-axis, the necessary velocity v would be only 2,270/(12*207) = 0.914 fps. (Of course, the vents did not extend over the total wall area for any floor, so the inlet velocity would have been higher than that. And the internal air velocity would have been greater towards the center of a fire zone floor compared to, say, the north-west and north-east corners of WTC 1. Also, the floor/ceiling barriers were not intact, so the principal fire zone was effectively intermediate between six or more compartments of height 12 feet and a single 72-feet high compartment.) The maximum power output would be related to the velocity times the vertical cross-section:

P = vyz*100 kW

...and dividing by the flux area xz we obtain Maximum Ftot = 100 * 3.2808^2 * vy/x or:

Ftot(max)  = 1076 * vy/x [equation 3]

where v, y, and x are inputted in fps or feet to obtain the maximum sustainable flux total in kW/m^2. In the above example, we have 57 kW/m^2 after rounding errors, in line with the original 56.8 kW/m^2.

We now consider the situation within the debris pile. Suppose there is a vertical 'chimney' extending continuously from underground voids up to the surface. In effect, we have another potential compartment fire; the 'compartment' has been rotated by 90 degrees. Air is supplied from the old Port Authority Trans-Hudson (PATH) railroad tubes, and driven to the surface by buoyant convection. A formula approximating the flow rate from stack effect in a chimney is:

Q = Cd*A*SQR(2*g*H*(Ti-Te)/Ti)

...where: Cd is a discharge coefficient generally taken as about 0.65, A is the cross-sectional area of the chimney, g is the gravitational acceleration, H is the chimney height, Ti and Te are interior and exterior temperatures respectively in Kelvin. Sticking to feet as the units of length, g equals 32.17 f/s^2 and Q is in ft^3/s (cfs). The velocity of the exhaust gases through the chimney is equal to Q / A, so:

v = Cd*SQR(2*g*H*(Ti-Te)/Ti)

This corresponds to velocity v in equation 3 above, and the height H corresponds to the x length of the compartment. To simplify, we shall assume a square chimney, with SQR(A) corresponding to the height y of the original compartment, although A has already been eliminated from the equation. Substituting for v in equation 3, we obtain:

Ftot(max) = 1076*y*0.65*SQR(2*32.17*H*(Ti-Te)/Ti)/H

...which simplifies to:

Ftot(max) = 5610*y*SQR[(Ti-Te)/(Ti*H)] [equation 4]

...with y the length of a side of the square chimney in feet, H is the length or height of the chimney in feet, Ti and Te are the interior and exterior temperatures respectively in Kelvin, and the maximum potential flux Ftot is given in kW/m^2. Assuming Ti is 1,000 C or 1,273 K and Te is 298 K, the maximum Ftot that could be supported by the air throughput in the case of a one foot square chimney of height 50 feet works out at 694.3 kW/m^2. A one inch by one inch chimney could only sustain one-twelfth of this flux, at 57.86 kW/m^2. This would only correspond to Ti = 630 C, assuming Taft = 1,488 C. Changes to the temperatures have a relatively small effect on the flow rate; lowering Ti by 370 degrees will lower the gas velocity from 32.26 fps to 30.11 fps. This would correspond to 54.12 kW/m^2 and Ti = 619 C. The debris pile was reported as "smoldering" for weeks, which would imply an even lower flux and mean gas temperature.

Unlike a blast furnace, the incoming air had not been pre-heated to 900 or 1,000 C. And over a range of stoichiometries from phi = 1.0 to 2.0 in the variously oxygen-depleted conditions of a compartment fire, the mean adiabatic flame temperature would not exceed some 1,600 C for a range of woods and plastics. In Table 1 above, a reaction with Taft = 1,600 C at Tamb = 25 C was calculated as requiring a total heat release rate per unit area of compartment space of some 10,000 kW/m^2 in order to sustain temperatures in excess of 1,500 C that could potentially melt steel. Within the debris pile, the 'chimney' required to provide sufficient air throughput in order to support this would need to be in the order of 14.4 feet square! And these estimates of flux and temperature are on the high side, since 50 feet is a rather low estimate for H. The debris pile below and above ground was generally regarded as being rather more than 4 stories deep.

Let's just suppose for the moment, for the sake of argument, that a chimney existed in the debris pile that supplied just as much oxygen as the fire required, with plenty of combustible materials in the sides and conveniently positioned steel members with their ends projecting into the chimney at various points above the fuel, and the fire improbably managed to sustain gas temperatures of 1,500 C for one minute, after a steel member had been uniformly heated to 1,200 C (of course, it never would have). Initially, radiative transfer would be equal to:

5.67 * 10^-8 * (1773^4 - 1473^4) = 293 kW/m^2, assuming emissivity = 1. (Convective transfer is relatively low, and in any case is more than balanced out by the fact that emissivity is less than one.) In fact, if the steel heated up to 1,500 C over that time, the mean temperature difference would be half as much, and the mean heat transferred would be a little over half at some 167 kW/m^2. So over 60 seconds, the total energy transferred to the steel is about 10 MJ/m^2.

Thermal diffusivity is directly related to thermal conductivity and inversely related to volumetric specific heat capacity, i.e. alpha equals conductivity divided by the product of density and specific heat. For a semi-infinite steel bar, if one end is instantly exposed to a particular temperature, the point at which half the temperature increase occurs will vary, naturally, with the exposure time. After one minute, the distance x from the exposed end to the point where half the temperature rise has occurred would be about 0.9 inches. (After 102 minutes - the duration of the WTC 1 fire - it would be nearly 9 inches. The distance is related to the square of the time.) The actual area of the steel bar is largely irrelevant, since the absorbed energy and the energy required to heat a given depth of steel are both linearly related to the area. Heat losses to the sides would act to lower the temperature increase in the steel.

Suppose we wanted to heat a 1 m^2 area of 0.025 m or about 1 inch thick steel from 1,200 to 1,500 C. Its mass is 0.025 m^3 times 7,860 kg/m^3 = 196.5 kg. At 1,200 C the specific heat of steel is some 725 J/kg.K, well up on the 450 J/kg.K at 25 C. So the energy required would be:

196.5 * 725 * 300 = 42.7 MJ. This is considerably over the 10 MJ supplied per square meter. Another way of looking at this is to predict the heat conducted through the steel if the 300 degrees temperature difference is applied across a length of only 0.025 meters. Taking thermal conductivity k for steel at 40 W/m.K, from:

q = dQ / dt = -k * A * dT / dx

...heat conduction through the steel would be at a rate of 40 * 300 / 0.025 = 480 kW/m^2, which is considerably over the 167 kW/m^2 available from radiative transfer as calculated above. The steel would never have reached anywhere even close to a uniform 1,200 C to start with, and it should be clear that after a mere 60 seconds' exposure to 1,500 C, large temperature gradients would turn the steel into a massive heat sink, with heat flowing away before the fire can even begin to melt the end of the member.

Experimental tests could be conducted to determine whether a chimney of 70 feet long and 16 feet diameter can melt the ends of steel beams protruding from its walls, following the ignition of wood and other combustibles beneath the steel. However, there is little doubt as to the outcome. Fireclay, invented in the 15th century, is routinely employed to line flues, stacks, and fireplaces. The melting temperature of fireclay at about 1,540 C is very similar to that of steel. Fireclay does not turn into molten pools at the bottom of a chimney after a few hours or days of exposure to a regular hydrocarbon fire.

As for the kinetic energy available from the massive collapse, this is given by:

KE = 0.5mv^2 where v = SQR(2gh) which leads to KE = mgh

...and if all of this was converted to heat and remained within the material, the temperature increase is found by dividing by the mass and heat capacity:

T2 - T1 = mgh/(mc) leading to gh/c

So for g = 9.807 m/s^2, h = 1368 / 3.2808 m, and c = 450 J/kg.K for steel at around ambient temperature, the mean temperature increase is 9.09 degrees Kelvin. The concrete would be cooler; its specific heat is nearly twice that of steel.

This is already a very high estimate for the mean, since that supposes the entire building mass was dropped from 1368 feet. There should be a reducing factor of more than 2, considering the steel was much heavier grade at the bottom, and a considerable amount of the building's mass was below ground.

400,000 tonnes distributed over an area of more than 4,000 square meters averages less than 100 tonnes per square meter. There would not be spots where a few pieces of steel would experience temperature increases that were hundreds of times greater than the mean increase. If a couple of trucks collided head-on, tens of tonnes of mass would be distributed over a cross-section of a few square meters and the impact would be very sharply concentrated over tens of milliseconds as opposed to ten seconds or more. Such collisions do not result in puddles of molten metal on the road. And unfortunately for the "gravitational energy melts steel" theory, even if the steel was already hot, it still requires some 250 KJ/kg for the latent heat of fusion to melt it, which is over 60 times the energy needed to raise from, say, 25 C to 34 C.

As for the idea that molten aluminum from a plane might have reacted with water from fire hoses and produced an exothermic reaction and just might have transferred all this heat to steel in the rubble, this theory is not only inapplicable to WTC 7, but fails to work for the towers. If ten tonnes of aluminum was in the form of a cube of volume 3.7 cubic meters and sides of 1.547 meters, the problem for the official theory is that little of it would melt in the fires, due to the relatively low surface area to mass ratio and the relatively low emissivity of aluminum at around 0.1. Alternatively, suppose this ten tonnes had been smashed into 3,700,000 cubes of length 0.01 meters and volume 1 cc. If they happened to be dispersed without lying on top of each other so as to present the maximum surface area for absorption, they might indeed all melt if the fires were sufficiently severe. But even if surrounded by concrete debris, they would not remain molten for more than a matter of minutes. Another contradiction is that the steel must always be near enough to be melted by the aluminum / water reaction, but never near enough so as to conduct heat away from the molten aluminum.

Molten iron or steel was not only observed flowing in the debris, but also filmed pouring out of the side of WTC 2 in the minutes leading up to its collapse. "You were only supposed to cut through the bloody columns, not shoot the molten iron out the windows in front of the cameras!" (Irate arch-villain.) A comparison of the video evidence with this video, of thermite being used to melt through a car's engine block and then through the fuel tank with rather irreversible results, shows a strong similarity. Molten aluminum would have appeared silvery because of its low emissivity, and the fires could not have produced sufficient mass of molten metal to account for the observed streams. The aluminum did not burn and generate heat, since that would have required it to have been in the form of flakes or a fine powder.

Suppose we take an (on the high side) estimate of 1000 C for flame and hot gas upper layer temperatures in very close proximity to a plate of aluminum alloy just as it is supposed to have melted. If the aluminum was at the low end of the aircraft fuselage skin thickness range at some 0.03 inches or 0.000762 m and its area was 0.485 m^2, this would place the volume of the piece around 0.00037 m^3 and hence its mass would be 0.00037 * 2700 kg/m^3 ~ one kilogram. The heat required to melt this is given by the temperature increase of (660 - 25) which is 635 degrees K times the heat capacity of 900 J/kg.K (it's actually more than this over the range up to 660 C) which comes to 571,500 J, plus 397 kJ for the latent heat of fusion, to give a total requirement of 968.5 kJ. After allowing for elements such as zinc in the alloy, the melting point, specific heat and enthalpy of fusion would be slightly lower.

So we have this piece of aircraft debris which just happens to have settled such that its 0.485 m^2 area side is squarely facing the radiant heat from the flames of burning office stationery, carpets, workstations, etc, and in a vertical position against the wall at the top of a window. Suppose it has already reached 660 C (or slightly less for the alloy), after absorbing the initial 571.5 kJ. The rate of radiant heat transfer is related to the difference in the fourth powers of the absolute temperatures of emitter and absorber:

P = e * lowercasesigma * A * (Te^4 - Ta^4)

A major problem with the "melting aluminum" theory is that aluminum is a poor absorber of radiant heat. Although it is not unreasonable to take 1 as the emissivity of flames, aluminum has an emissivity of only about 0.1. This is why it is used as a reflector in infrared heaters. Also, for a thin plate of aluminum, some radiation would be transmitted through it. Moreover, the location by the window would result in a further 40% or so reduction. At 1000 C (flame) and 660 C (aluminum) the potential aluminum absorption is proportional to 1273^4 - 933^4, i.e. 1.868*10^12, and it would be likely to transmit an amount to the exterior proportional to 933^4 - 298^4, i.e. 7.499*10^11. If in close contact with the wall, it would conduct heat to it.

Most of the radiant heat striking the aluminum would be reflected back, and would probably end up being vented out to drive the smoke plume.

Taking the Stefan-Boltzmann constant (lowercasesigma) as 5.6703 * 10^-8 W/m^2.K^4, we have:

P = 0.1 * 5.6703 * 10^-8 * 0.485 * (1273^4 - 933^4) = 5.138 kW.

(If the emitter temperature is lowered from 1000 to 900 C, P drops to 3.122 kW.)

At 397 kJ required to melt the 1 kg after it has reached the melting point, the time required is 397 / 5.138 = 77.26 seconds, or slightly less for the alloy, or almost 127 seconds assuming 900 C for the emitter temperature. Yet there was enough melting liquid to be clearly visible, and to sustain a four-second flow before subsiding. Even if the whole 1 kg had somehow resisted melting for 77 seconds whilst receiving heat at 660 C, and then the entire piece suddenly melted within a mere 4 seconds, aluminum only expands by about 12% when molten to a density of 2400 kg/m^3. So the 1 kg would have a volume of 417 ml which is barely more than the contents of a can of Coke. The video evidence shows that the flow was much more than 100 ml per second, and at times more like a "waterfall". And as NIST admitted...

Many such liquid flows were observed from near this location in the seven minutes leading up to the collapse of this tower.

...which would be many, many kilograms of aluminum, and would require a greater energy source than a few burning papers and carpets in an office compartment fire.

There was aluminum cladding on the exterior columns, but this was on the three exterior sides and on the wrong side of fire-resistant plaster. The flows of molten liquid support the theory that thermite or thermate played a part in weakening columns over the space of several minutes leading up to each collapse. It is hardly surprising that guards were immediately placed at the crime scene, Ground Zero, to prevent independent investigators from getting hold of samples of steel, and a couple of weeks later Mayor Giuliani even banned photography at the site.

Conclusion

We conclude that some WTC structural steel melted, and molten iron was formed, by a type of combustion such as a thermite reaction, as opposed to hydrocarbon fires of office combustibles ignited by jet fuel. WTC 1, 2 and 7 were subjected to controlled demolitions involving pre-installed accelerants. As corroborating evidence, pre-2001 studies found that the Towers would survive a 600 mph impact from a jetliner, and predicted that the buildings would remain standing even if all of the jet fuel had been dumped inside and ignited a horrendous fire that killed many people.

(This article) Revised: December 14, 2006