The best on-line discussion resource found for these calculations was at Ref. (10). According to this source the inner core consisted of from 44 to 47 box columns (the exact number and layout is not known; the architectural firm had not released the construction drawings). The dimensions of the columns reduced in size with increasing height, changing to I-beams above the 85th floor. The above website article assumes (generously) that each core box column has the following (average) X-section: 12"wide x 36"deep x 2" thick.
If w = box column width, d = depth, and t = thickness, then the X-sectional steel area is given by
A = [d x t + (w-2 x t) x t] x 2. For d = 36", t = 2" and w = 12", then
A = [36" x 2" + (12"-2 x 2") x 2] x 2 = 176 in2 = 1.222 ft2.
Floor height was 12ft, so we choose for discussion sake, a 12′ high box column in these calculations. Note that multiple floors could have had thermite-type compounds placed there. Also, no more than a foot portion, rather than a full 12 ft of column would be necessary to collapse that floor. Also, complete melt of a column portion is not necessary to cause collapse. So, per floor, per column there is a steel volume V = 12′ x 1.222 ft2 = 14.67 ft3. Also, note that the internal X-sectional area of a box column is given by
Aint = [d-(2 x t)] x [w- (2 x t)], and the internal volume by Vint = 12′ x [d - (2 x t)] x [w - (2 x t)].
Here, Vint = 12′ x [36"-2 x 2"] x [12"- 2 x 2"]/(144 in2/ft2) = 12′ x 1.778 ft2 = 21.333 ft3.
The internal volumes will be re-examined later as a possible space to place the thermite.
The website also mentions that the largest box columns used at the core bases had the dimensions of 16" wide x 36" deep x 4" thick. It is not known where exactly the molten steel, that puddled in the WTC basement, originated in the towers. The melt could have occurred some what higher in the columns (where "average" box columns would have been), or at the base where the "largest" box columns were. Molten material would flow down the various WTC shafts to the lowest point possible, 6 stories (some 72′) below ground level. Applying the same formulae as above, we have for these "largest" columns, A= [36" x 4" + (16" - 2 x 4") x 4"] x 2 = 352 in2 = 2.444 ft2. Note that this happens to be twice the area as for the "average" box column assumed above. Again, for a 12′ column, V = 12′ x 2.444 ft2 = 29.328 ft3. Also, here, the internal volume is Vint = 12′ x [36′ - 2 x 4"] x [16′ -2 x 4"]/144" = 18.667 ft3.
In summary, we have for a 12 ft. high core box-column, for a
12" wide x 36"deep x 2" wall thickness (hereafter referred to as an "average" box column), that it has 14.67 ft3 = 0.415 m3 volume of steel, and 21.33 ft3 = 0.604 m3 of internal volume; and
16" wide x 36" deep x 4" wall thickness (hereafter referred to as a "largest" box column), that it has 29.328 ft3 = 0.832 m3 of steel and 18.667 ft3 = 0.529 m3 of internal volume.
Sensible and Latent Heat Energies Needed for Melting a Core Column Section
Knowing the volume of steel involved, we next turn our attention to calculating the energy needed to melt a core column section. We decided to use values for the element iron rather than steel for the following pragmatic reasons:
steel is mostly iron (Fe);
whatever steel is chosen, may be the wrong kind and would be contested: Fe is a given and known quan y, whereas there are many steels;
Fe values found were readily available and reasonably self-consistent;
except for stainless steels, the thermal properties of steel are relatively close to Fe, although the mechanical properties may certainly differ more.
For Fe we will use the following values:
Density = 7874 kg/m3
Melting point = 1811 K = 1538 C
Specific heat = 25.1J/mol K = 449 J/kg K = 0.449 kJ/kg K
Latent heart of fusion = 13,800 J/mol = 2.47 x 10+5 J/kg
Latent heat of evaporation = 347,000 J/mol = 6.21 x 10+3 kJ/kg
mol = gm mole equivalent = 0.0558 kg for Fe
For a 12 ft high core Fe column, we have
for the "average" box column, 0.415 m3 x 7874 kg/m3 = 3267.71 kg Fe; and
for the "largest" box column, 0.832 m3 x 7874 kg/m3 = 6551.17 kg Fe.
Taking 300 K as "ambient" temperature on 9-11, then the temperature difference up to the melting point of Fe is given by
1811 K - 300 K = 1511 K (give or take a few degrees K).
Hence, the energy needed to raise a 12 ft high Fe column to its melting point temperature is given by
for an "average" column, 3267.71 kg x 1511 K x 0.449 kJ/kg K = 2.22 x 10+6 kJ; and
for a "largest" column, 6551.17 kg x 1511 K x 0.449 kJ/kg K = 4.44 x 10+6 kJ.
To actually melt the Fe at 1511 K, we need to provide the latent heat of fusion:
for "average" column, 3267.71 kg x 2.47 x 10+2 kJ/kg = 8.07 x 10+5 kJ; and
for "largest" column, 6551.17 kg x 2.47 x 10+2 kJ/kg = 1.62 x 10+6 kJ.
Thus we see that the sensible heat energies involved are almost a factor of 3 times larger than the latent heats.
Hence, for the total amount of energy needed to melt a 12 ft high Fe column, we need:
for "average" box column, (2.22 + 0.81) x 10+6 kJ = 3.03 x 10+6 kJ; and
for "largest" box column, (4.44 + 1.62) x 10+6 kJ = 6.06 x 10+6kJ
Energies of the Thermite Reaction
An iron oxide/aluminum "thermite" mixture consists of 23.7% Al, 74.7% Fe2O3 by weight, in the reaction
Fe2O3 + 2 Al => Al2O3 + 2 Fe + 849 kJ/mol.
Thus, 849 kJ of energy are released for every g-mole-equivalent (mol) of Fe2O3 that reacts with 2 mol of Al.
For Al, with a density of 2.699 g/cm3, there are 26.98 g/mol.
For Fe2O3, with a density of 5.24 g/cm3, there are 159.70 g/mol.
So then, 159.70 g of Fe2O3 + 53.96 g of Al (213.66 g total) produces 849 kJ of energy, or 3.974 kJ/g = 3.974 x 10+3 kJ/kg (Note that this gives the proper % component mixtures by weight).
For an infinitesimally compacted powder mixture, this would occupy a volume of 159.70g x (cm3/5.24 g) + 53.96 g x (cm3/2.699 g) = (30.48 + 20.0) cm3 = 50.48 cm3.
A separate analysis of a CuO/Al thermite mixture (used to weld copper parts) indicates a powder packing fraction of 0.82 (82%) can be achieved. Let’s assume a powder packing fraction of 0.82. Hence, our Fe2O3/Al thermite mixture would occupy not 50.48 cm3, but 61.5 cm3.
Thus the physical density of our densely-packed Fe2O3/Al thermite mixture is
213.66 g/61.5 cm3 = 3.474 g/cm3 = 3.474 x 10+6 g/m3 = 3.474 x 10+3 kg/m3,
and our energy density (per volume) is given by
849 kJ/61.5 cm3 = 13.805 kJ/cm3 = 1.3805 x 10+7 kJ/m3.
Thus to melt a 12 ft high Fe column, we need
for an "average" column, (3.03 x 10+6 kJ)/(3.974 x 10+3 kJ/kg) = 0.7625 x 10+3 kg = 762.5 kg of thermite. This would occupy a volume of 762.5 kg/(3.474 x 10+3 kg/m3) = 0.219 m3. Note that this volume of thermite is less than the internal volume Vint calculated earlier, 0.604 m3. Actually, the internal volume of the "average" box column could be filled with 0.604 m3/0.219 m3 = 2.76 times more than needed to do the job. Alternatively, the column does not require as high a packing density ( i.e. <0.82) and yet be able to load a sufficient charge of thermite mixture to cause melting
for a "largest" column, (6.06 x10+6 kJ)/(3.974 x 10+3 kJ/kg) = 1524.9 kg thermite. This would occupy a volume of 1524.9 kg/(3.974 x 10+3 kg/m3) = 0.439 m3. Note that this volume of thermFite also is less than the earlier calculated Vint = 0.529, but would require a moderately high packing density, approximately > 0.82 x 0.439/0.529 = 0.68.
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