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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

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

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


    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

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.

Continue to WTC Fires Part Three