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FEMA's report quotes Culver (1977) stating that typical office-type

occupancies contain fuel loads (described in terms of the equivalent weight

of wood) of 4 to 12 pounds per square foot, with the mean slightly less than

8. However, NIST's reports of June and October 2004 by S. Shyam Sunder

state that the typical (WTC) floor contained an average of 4 psf of

combustible materials. We shall also allow for remaining aircraft

combustibles.

The fuel load in the core area was relatively light, with much of this zone

taken up by service shafts. The remaining space between shafts would have

contained a limited fuel load, with the great majority of combustible

material such as desks and workstations being outside the core. Let's

suppose there was on average 7 psf or 34.18 kg/m^2 (close to Culver's

geometric mean) outside the core, and 2 psf or 9.765 kg/m^2 within it. The

core area was 137' x 87' out of 208' x 208', or 1107 m^2 out of 4019 m^2,

leaving 2912 m^2 as the non-core area. However, above the floor 78 sky

lobby, some 40% of the core space is taken up by elevator shafts,

stairwells, etc, as detailed below in the section on calculating the amount

of steel. So the core space on the upper floors of the fire zone was

effectively 0.6 * 1107, i.e. about 664 m^2. Hence, the combustibles per

floor, excluding aircraft debris, are 34.18 * 2912 + 9.765 * 664 = 106,016

kg = 116.9 tons, expressed in terms of the equivalent weight of wood.

Averaging over the whole floor including core and shafts space, this works

out around 5.4 psf which is still above NIST's 4 psf average for a typical

WTC floor. The NIST figure may well represent the actual weight of

combustibles, and so would be less than the equivalent weight of wood, given

that some combustibles have a higher calorific value. Other substances such

as plastics would exhibit little difference in the heat released per unit of

oxygen consumed. Since the combustion inefficiency factor associated with

fuel-rich fires derives from the calorific value of the oxygen rather than

the fuel, our inefficiency factor calculated above at around 0.68 is

applicable to a mix of various fuels.

A typical aircraft contains 3,300 to 8,400 kg of combustibles, according to

the FAA (the statement is pre-9/11 and hence more credible). Aluminium is

not counted as a combustible. Unless finely divided such as shavings,

flakes or pure powder, its ignition temperature is around 2000 C. As was

calculated above, flame temperatures would have been well below this. Some

of the aluminium may have melted, but it did not burn. If we allow an

additional 7,000 kg of combustibles distributed over two floors, then these

two worst cases have an extra 3.3% combustibles to give a total of 109,516

kg or 120.7 tons per floor.

For an upper bound on the mass burning rate, we can consider the fire

compartment as a single two layer zone, six storeys tall. If the floors and

ceilings were non-existent enabling unimpeded gas flow throughout the zone,

Hv of the ventilation factor would extend over six floors. The lower bound

would treat each floor as an individual fire compartment, with the height of

its vent being the 10 feet from floor to ceiling, each floor having its own

hot and cold layers. Since the ventilation factor is proportional to

SQR(Hv) as well as the vent area, the 6-floor compartment, assuming 6 times

12 feet between storeys for Hv, would have at least SQR(7.2) times the mass

burning rate (per floor) of the single-floor compartment model. An extra

7.2 / 6 accrues with the 6-floor vent area being more than 6 times higher

than the 1-floor model. With some damage to floors allowing some throughput

of gases over a limited proportion of floorspace, the effective Hv would fit

between these two limits. We shall calculate the mass burning rate for each

limit.

The aircraft entry opening on the north facade of 1WTC can be seen to extend

approximately 1/4 the width of the building for each of the worst affected

floors, with the centre staggering to the right for ascending floors.

Openings did exist on other sides but were much smaller, perhaps some 1/16

of total area per floor. We shall take 30% of the building width as opening

area for the north facade and 7% of the width for each remaining side.

WTC 6-storey fire compartment:

72' x 208' per wall

Hv = 72 / 3.2808 = 21.95 m

Aw = 72 * 208 / 3.2808 ^ 2 = 1391 m^2

Av = (0.3 + 0.07 * 3) * 1391 = 709.4 m^2

At = 2 * 63.4 ^ 2 + 4 * 1391 = 13603 m^2

The total enclosing area At and wall area Aw are used in some methods, so

listed above. Kawagoe's formula:

mass pyrolysis rate, wood (kg/s) = 0.092 * Av * SQR(Hv)

predicts a mass burning rate of 0.092 * 709.4 * SQR(21.95) = 306 kg/s, or

51 kg/s per floor.

Excluding the kerosene which we know burned up within a few minutes, the

worst floors had 109,516 kg equivalent weight of wood. So the burning time

assuming the entire floor's combustibles were involved, is 109,516 / 51 =

2147 seconds = 35.8 minutes! Unfortunately, 1WTC took 102 minutes to

collapse. The mass burning rate was clearly well under 51 kg/s per floor.

WTC 1-storey fire compartment:

10' x 208 ' per wall

Hv = 10 / 3.2808 = 3.048 m (excludes 24" ceiling-floor area)

Aw = 10 * 208 / 3.2808 ^ 2 = 193.24 m^2

Av = (0.3 + 0.07 * 3) * 193.24 = 98.55 m^2

At = 2 * 63.4 ^ 2 + 4 * 193.24 = 8812 m^2

From:

mass pyrolysis rate (kg/s) = 0.092 * Av * SQR(Hv)

we obtain 0.092 * 98.55 * SQR(3.048) = 15.83 kg/s per floor.

(This is down by a factor of SQR(7.2) * 7.2 / 6 compared to the 6-storey

version.)

The kerosene burnt for a few minutes, then on the worst floors assuming the

entire fuel load was consumed, the fire duration is 109,516 / 15.83 = 6918

seconds = 115.3 minutes.

Even this mass burning rate is stretching it. Partitions provided

substantial resistance to fire spread, according to NIST. We could suppose

that one floor burnt for 51 minutes until stopped by the partitions, then

the adjacent floor above burnt for 51 minutes until stopped by the collapse.

But this is less favourable to fire collapse theories. The core columns on

the first floor would be cooling at the time of collapse, and on the next

floor would have absorbed a fraction of the potential energy. Ideally, the

entire fire load of two adjacent floors would burn throughout the 102

minutes, and we will be concentrating on such a worst case scenario.

Alternative theories on mass burning rate can differ in their predicted

burning rate per floor by a factor of more than two. For example, the Law

correlation takes into account an "omega" factor, which is related to the

inverse of the ventilation factor. This method predicts a relatively high

(too high) mass burning rate for the WTC single storey fire compartment

model, and actually gives a lower rate per floor for the 6-storey version.

The 6-storey compartment is not consistent with video evidence, which shows

the worst affected floors to have been near the top of the fire zone around

floor 97. Thus, the neutral point would also have been towards the top of

the zone.

Generally, air throughput in enclosed fires is assumed to derive from stack

effect and the buoyancy forces acting on the hot gases. Wind induced

ventilation would not increase the total air flow rate unless the wind

induced rate exceeded that from stack effect. Consider a building with four

vents - on two opposite sides and at two levels. With stack effect only,

air flows into the lower openings and out through the top. For a small wind

induced rate, the air is still flowing into the bottom and out of the top;

what the wind adds to the lower windward vent and the upper leeward vent

flow is annulled by a decreased flow through the other openings. If the

wind induced rate is very large in relation to stack effect, the air will

then flow in at the windward side and out at the leeward side, with the

boost from stack effect into lower windward and upper leeward vents balanced

by resistance to flow through the remaining vents. So the air throughput is

approximately equal to the rate from either stack effect or wind induced,

whichever is the greater.

Rockett's formula (slightly adjusted for 1200 feet altitude):

air mass inflow rate (kg/s) = 0.4825 * Av * SQR(Hv)

taking Av and Hv from the WTC 1-story compartment gives

0.4825 * 98.55 * SQR(3.048) = 83 kg/s of air inflow. To find the wind

induced rate, the formula for calculating the area of the effective

equivalent vent for the case of two vent areas in series at opposite sides

of the building is:

1 / Av ^ 2 = 1 / (A1 + A2) ^ 2 + 1 / (A3 + A4) ^ 2

where Av is the effective equivalent vent area, (A1 + A2) is the total vent

area on one side and (A3 + A4) is the total vent area on the opposite side.

When the vent area on one side is large in relation to the other side, the

effective equivalent vent area is marginally less than the area of the

smallest side, as intuition would suggest. When the area on opposite sides

is equal, the effective area is 1 / SQR(2) of the area per side.

Taking the opening on one side as 30% x 2080 ft^2 and the opposite side as

7% x 2080 ft^2, the effective equivalent opening is 6.817% x 2080 ft^2 =

141.8 ft^2. The velocity of air flow into the vent is restricted by the

orifice discharge coefficient Cd, which is typically assumed to be about

0.65, so a 15 mph or 22 fps wind is effectively reduced to 14.3 fps.

The pressure gradient developed across each opening is approximately that

value which would induce air flow through the opening at the velocity of the

wind that gave rise to those pressure gradients multiplied by Cd. The

pressure coefficients slightly deviate from unity, but if we assume a flow

velocity of 14.3 fps through a 141.8 ft^2 vent we obtain from the product

2028 cfs = 57.4 m^3/s. Given 1.16 kg/m^3 as the density of air at 1WTC fire

altitude and ambient temp, the flow rate is 66.6 kg/s of air which is

somewhat less than the Rockett derived figure of 83 kg/s.

And this assumes wind direction most favourable to airflow. The actual wind

direction was diagonal to the building. With some component of the flow

being from a 7% opening side to the opposite 7% opening side (east to west,

equivalent to a single 4.95% vent area), wind induced ventilation rate would

be less than 66.6 kg/s. A reasonable formula for the wind direction being

60 degrees off the north facade would be:

cos^2 (60) * 6.817% * 2080 ft^2 * 14.3

+ sin^2 (60) * 4.95% * 2080 ft^2 * 14.3

= 1611 cfs = 52.9 kg/s of air inflow.

The stack effect 83 kg/s assumes the 1-storey compartment; for a compartment

and vent of greater height the rate would increase. So with stack effect

clearly exceeding wind induction, the former is the determinant of

ventilation rate. Given the uncertainty in vent area and the variation in

predicted mass pyrolysis rate amongst alternate theories, it is entirely

possible if not probable that the mass burning rate per floor was close to

109,516 kg / 102 minutes = 17.89 kg/s.

For the maximum possible energy density, we will suppose that the mass

burning rate coincidentally just happened to be such that the entire floor's

combustibles and the jet fuel were consumed over 102 minutes. One would

hardly expect a match, to within a minute, of the time required for global

collapse and the time for total combustion on the storey which initiated

collapse. But at 1 in 102 for the timing it is not too unreasonable. The

much likelier possibility would have about half of the combustibles

consumed. But let us suppose 109,516 kg equivalent weight of wood plus 1484

gallons of jet fuel (767 / 2067 of 4000 which we assumed for the worst

floor) all burnt up on a single storey.

Allowing for combustion inefficiencies, the effective calorific value of the

kerosene was 29.7 MJ/kg, with 11.4 MJ/kg for the wood.

1484 * 3.1 kg/gallon * 29.7 MJ/kg = 1.37 * 10^11 J

+

109516 * 11.4 = 1.25 * 10^12 J

This gives a total of 1.39 * 10^12 J for the worst floor, including heat

which flowed out of the building. The kerosene adds some 11% to the 109,516

kg, all expressed as the equivalent weight of wood. The worst floor has

106,016 kg (wood, plastics, paper, etc; regular office contents) + 3,500 kg

(aircraft combustibles) + 12,018 kg (kerosene) = 121,534 kg. So

121534 / (102 * 60) = 19.9 kg/s mass burning rate. If we assume the vent

area is 30% for the north side and 7% for the remaining sides giving 98.55

m^2 per floor as in the 1-storey compartment above, then from:

mass pyrolysis rate, wood (kg/s) / (0.092 * Av) = SQR(Hv)

...the vent height is 19.9 / (0.092 * 98.55) = 4.817m = 15.8 feet.

This would assume the impedance to gas flow from the ceiling-floor resulted

in the compartment being effectively between the above 6-storey and single

storey examples, with the average vent height working out at a little over

one storey.

Of the 1.39 * 10^12 J, some 60% - i.e. 8.34 * 10^11 J - is retained in the

building, with the building absorbing the greatest proportion when cold.

For this worst case storey, the kerosene and aircraft combustibles are added

to the total floor area excluding shafts. The total floor area comprised

the non-core of 31,345 ft^2 and the 60% of 11,919 ft^2 core space - i.e.

7,151 ft^2, to give a total area excluding shafts of 38,496 ft^2 or 3,576

m^2. With the kerosene yielding the equivalent of 12,018 kg of wood,

12018 kg / 3576 m^2 is an extra 3.361 kg/m^2 or 0.688 psf. 3,500 kg of

aircraft combustibles adds 0.979 kg/m^2 or 0.2 psf. So we have:

Non-core combustibles = 7 + 0.688 + 0.2 = 7.888 psf = 40.8 MJ/ft^2

Core combustibles = 2 + 0.688 + 0.2 = 2.888 psf = 14.93 MJ/ft^2

These figures will be the basis of some computations we shall carry out,

involving summing the radiant heat contributions from each square foot/metre

of floor space, as viewed at some specified point on the floor. Final

temperatures at 102 minutes are predicted for each core column, along with a

selection of other members.