Of all the events on 9/11, the well documented skyscrapers collapse was by far the most important. We can be certain that the collapses did occur, and in a moment we shall analyse whether plane impacts and consequent fires could have been responsible. The Pentagon incident points to Bush Administration complicity; analysis of the Manhattan events will provide a more definitive identification of the principals. Throughout the 100-year history of steel framed high-rise buildings and skyscrapers up to September 10, 2001, there had never been even a single incidence of a collapse due to fire in the US. According to Ronald Hamburger in the Stanford Report, Dec 3, 2001, 7WTC was the "first major structure in the U.S. to collapse because of fire". A report on multiple fatality fires, 1994-1996, has no mention of "collapse". And National Institute of Standards and Technology (NIST), Building and Fire Research Laboratory (BFRL), Investigation of the World Trade Center Disaster reports "The WTC Towers and WTC 7 are the only known cases of total structural collapse in high-rise buildings where fires played a significant role." The worldwide statistics for high-rise buildings since 1970 include only 17 cases of fire-induced collapse, two of these being structural steel buildings, and none in the US. Yet on September 11, 2001, three separate skyscrapers collapsed. We'll be generous to the BLIARS theory - just before a random event occurs the rate of occurrence will tend below the mean; after the event it will tend above the mean. Let's suppose 3 collapses in every 100 years is the average fire-induced collapse rate to be expected for steel framed high-rise buildings in the US (and the fact that the rate remained at zero for so long suggests that this is an overestimate). So the rate is one day every (100 / 3) years = 1 day every 12175 days. Hence, the probability for one such collapse to occur on any specified day is 1 in 12175, and the probability for three collapses is 1 in 12175 ^ 3 = 1 in 1.805 * 10 ^ 12. The BLIARS advocate might suggest that we count the number of fires and collapses rather than the number of days. For the US only, the overall picture for 2001 was 1,734,500 reported fires, as stated in Fire Loss in the United States During 2001 by Michael J Karter, Jr, the National Fire Protection Association (NFPA), September 2002. Many others were unreported, causing further industry and property loss. Of those reported, about 30% were structure fires, 20% were vehicle fires, and 50% occurred outside. Residential fires comprised some 76% of structural fires. With structure fires totalling 521,500 and residential fires at 396,500, the number of non-residential structure fires, which we are concerned with, totalled 125,000. If the 2001 annual statistics are representative of the last 100 years, then the total per century would be about 12,500,000, which is over 300 times greater than 36,525, the number of days. In the medium term, at least, the number of fires shows a downward trend. Over the decade 1993 to 2002, non-residential structure fires averaged 139,200 per annum. And over the decade 1989 to 1998, they averaged 154,300 per annum. Back in 1977, the annual figure for total reported fires was over 3 million, and structure fires numbered 1,098,000, more than double the 2001 figure. So the 2001 statistics understate the medium-term total. Let's stick with the 2001 figures. In order for fire-induced collapse to be less improbable than 1 in 12175, there must have been more than 12,500,000 / 12175 = 1027 fire-induced collapses of non-residential structures over 100 years, or one every 5 weeks (in the US alone). We could then point to three particular collapses, all of which are high-rise steel frames (1, 2 and 7WTC), and conclude that the probability of all three undergoing collapse is greater than 1 in 12175 ^ 3. But such an incidence of collapse, with associated loss of life, would be scandalous and widely reported. Preventative measures would, at least, have been attempted - with any ongoing failure also making news. Although we are now including all non-residential structures as opposed to limiting the set to steel framed high rises, and there may well have been instances of fire-induced collapse amongst this larger set, there is no evidence that the incidence of collapse is anywhere near a rate of ten per annum. The BLIARS advocate may counter that there was something "special" about the WTC fires - they were deliberately and skilfully instigated with collapse in mind, and were particularly severe. The former is surely true; the question as to severity brings us to some of the most fascinating aspects of 9/11: How hot did the steel get? If a floor, or a 15-floor section had collapsed, would this effect global collapse? If global collapse did occur, would the collapse time be close to the free-fall time of around 9 seconds, and would pools of molten steel remain buried deep in the debris for a further five weeks? Would another form of premeditated sabotage provide a better explanation of all the evidence? We shall consider each point in turn. Firstly, the energy output and temperature of the fires. If we assume BLIARS to be true, it is reasonable to accept the official line that Flights AA 11 and UA 175 were carrying about 10,000 gallons of jet fuel at the time of impact with the WTC. For the Boeing 767-223ER (AA 11), maximum fuel capacity is quoted at around 24,000 gallons and maximum range is in the region of 10450 km or 6500 (non-nautical) miles. The shorter range 767-222 (UA 175) is about two-thirds the capacity and range. In order to avoid unnecessary overheads, the planes would not have been carrying significantly more fuel than necessary. Adding a safety factor of 10% to 24,000 / 6500 obtains a requirement of 4.06 gallons per mile. Taking the remaining distance from New York to LA as 2,500 miles implies 10,150 gallons on board. Take off would consume a disproportionate amount of fuel, and both planes had been diverted from their scheduled flight paths. (Flight 11 was in fact not scheduled on 9/11, and a later section deals with this.) The amount of fuel consumed in the initial fireballs is calculated from D = 5.25 * m ^ 0.314 where D is the diameter of the fireball in metres, and m is the mass of fuel vapour in kg. Jet fuel has a density of about 3.1 kg per gallon so 10,000 gallons weighs about 31,000 kg. If all 10,000 gallons had been used up in a single fireball, then the diameter would have been about 135 metres or 443 feet. The diameter is not well-defined, but images from 2WTC and reports suggest three fireballs, each about 200 to 300 feet. The FEMA report quotes (for 2WTC), "...three fireballs...diameters...were greater than 200 feet, exceeding the width of the building". If we assume only 200 feet or 60.96 metres per fireball, from m = D ^ 3.185 / 196.6 we obtain a mass of 2465 kg or 795 gallons per fireball. 10,000 less 3 x 795 leaves 7615 gallons. The FEMA report suggests 3,000 gallons as a reasonable assumption for the fuel initially consumed in fireballs, leaving 7,000 gallons. It assumes about half of this flowed away, either down elevator shafts or out of the building, with approximately 4,000 gallons remaining on the impact floors. It also says, "Barring additional information, it is reasonable to assume that an approximately similar amount of jet fuel was consumed by fireballs as the aircraft struck WTC 1". We have no quibbles with these assumptions, and will take their figure of 4,000 gallons dispersed on the impact floors, which would have burned as a pool fire. There would have been pockets of fire from fuel that flowed down elevator shafts, but there is no evidence of major fires below the impact floors. Depending on the available ventilation and load of ignited combustible fuel, compartment fires are either fuel controlled or ventilation controlled. A typical fire would include a pre-flashover stage with gas temperatures under 500 C, prior to ignition of most of the compartment's combustible contents. The growing fire would obtain sufficient oxygen for combustion, hence would be fuel controlled. Post flashover, the fully developed fire stage would produce the greatest heat release rate, with all combustibles within the compartment involved in the fire. Gas temperatures are typically 800 to 1200 C. In this stage the amount of pyrolized fuel will often exceed the quantity that can be burned given the available oxygen within the compartment, hence the fire is ventilation controlled. In some respects the WTC fires were clearly not typical, one feature being the effective by-passing of the pre-flashover stage, with the jet fuel enabling relatively rapid ignition of much of the combustibles. Commercial jet fuel Jet-A, blended from kerosene with 1 to 2% additives such as anti-static, anti-freeze and corrosion inhibitor, has a flashpoint of 49 C. In contrast to gasoline, jet fuel as a pool would not ignite at normal ambient temperatures. Vapour levels would be too low resulting in a mixture which was too lean for combustion. But the volume of a sphere varies with the cube of the radius, whereas the surface area varies with the square of the radius. Consequently, when fuel is atomised or sloshed into a fine mist following an aircraft impact, the surface area becomes large in relation to the volume of fuel; all of this area can then interact with the surrounding air. The vapour-air mix becomes sufficiently rich that, given an ignition source, a fireball can result. As we shall see, in other respects the WTC fires were typical office fires, albeit rather large office spaces - hence the volume of smoke. The jet fuel played a key role in quickly advancing the fires to the fully developed stage. But the pre-flashover stage has little effect on the structure. As most reports correctly state, the jet fuel would have burnt up within the first few minutes. Most of the total heat released was from typical office contents such as desks, carpets and stationery, rather than the jet fuel. Let's suppose for the moment that there was an unlimited supply of oxygen. The mass burning rate for free burning pool fires is on the order of 50 grams per square metre per second. Each WTC floor was 4019 square metres less space taken up by elevator shafts, escalators, core columns, partitions, etc. If the fuel spread throughout just one floor, and we deduct 10% of the space for vertical structures and shafts, then we have a 4,000 gallons pool fire dispersed over an area of 3617 square metres. With burn time t given by mass of fuel {Kg} / (0.05 * pool fire area {m^2}) = t {secs} and 3.1 x 4,000 = 12,400 kg of fuel we have 12,400 / (0.05 * 3617) = 68.6 seconds to burn up. If the fuel is dispersed over most of the floor space of several floors, the burn up time is still lower. In fact, the pool fire would have covered various patches on several floors. If the fuel had spread across five floors for example, an average area of only 723 m^2 per floor would only provide 68.6 seconds of burn time. The product {pool fire area x burn up time} is equal to 12,400 / 0.05 = 248,000 s.m^2. The NIST report on the WTC Investigation Status as of June 22, 2004 by S. Shyam Sunder, points out that flames in a given location would have typically lasted for twenty minutes before spreading to adjacent areas with unburned combustibles, and partitions - not necessarily the main partitions between tenants - "provided substantial resistance to fire spread in the affected floors". So the fires did not cover entire floors at once, and if we assume a 4,000 gallons two minute pool fire over 2067 m^2, it would have been unevenly distributed over several floors. A distribution over five floors might, for example, have consisted of areas of 100, 300, 400, 500, and 767 square metres respectively, which would make even the largest less than one-fifth of the floor area. It is well established that the quantity of heat released is closely related to the quantity of available oxygen (Huggett, also see FEMA May 2002 report appendix A), with each gram of oxygen releasing in the region of 13.1 kJ for a range of fuels. Air is 21% oxygen by volume, or 23.2% by weight which is the relevant ratio here. Each MJ of heat released requires some 76 grams of oxygen, which is provided by 76 / .232 = 328 grams of air. At 20 C and an altitude of 1200 feet, the density of air is about 1.16 kg/m^3 (20 degrees above freezing accounts for a nearly 7% reduction, and the altitude a 3.5% reduction, compared to sea level at S.T.P.). Thus, 328 grams of air has a volume of 0.328 / 1.16 = 0.283 cubic metres. So each cubic metre of air can release 1 / 0.283 = 3.53 MJ, or each cubic foot can release 100 kJ. Consider the worst affected floor in the penultimate paragraph example. 767 m^2 out of a total area of 2067 m^2, assuming equal pool depth, is 767 / 2067 of the 4,000 gallons, i.e. 1,484 gallons. Given a density of about 3.1 kg/gallon and a calorific value of 44 MJ/kg, the potential energy from kerosene alone on this floor is 2.024 * 10^11 J, which if released over two minutes is 1.687 GW of power - a massive rate for a single floor! Since one cubic foot of air is required to release 100 kJ, 2.024 * 10^11 J would require 2.024 * 10^11 / 100,000 = 2,024,000 cubic feet of air. If one floor of the WTC is 208' x 208' x floor to ceiling height of 10', the volume of air per floor is 432,640 cubic feet. Hence, to achieve a power output of 1.687 GW per floor would require 1.687 * 10^9 / 10^5 = 16,870 ft^3 of air per second (cfs) = one air change every 25.6 seconds. The widely used Kawagoe formula is a useful starting point in mass loss or heat release rate calculations: mass pyrolysis rate (kg/s) = 0.092 * Av * SQR(Hv) where Av is the area of the vent (m) and Hv the height of the vent (m). Kawagoe's formula was based on experiments involving the burning of wood cribs. In the general case, the constant of proportionality will vary with the stoichiometric fuel-air ratio of the particular fuel. E.g., for complete combustion of wood, some 1.26 kg of oxygen or 1.26 / 0.232 = 5.43 kg of air is required for every kg of wood. 0.5 / 5.43 yields Kawagoe's 0.092 constant. Rockett used Kawagoe's ventilation factor [Av * SQR(Hv)] to calculate the mass flow of incoming air. The constant of proportionality was calculated at 0.4 to 0.61, with 0.5 now widely used (hence appearance in derivation of Kawagoe constant above): air mass inflow rate (kg/s) = 0.5 * Av * SQR(Hv) This is an approximation with greatest accuracy in post flashover conditions where the hot gas temperatures exceed 500 C. The derivation includes ambient air density as directly related to air mass inflow, so deducting 3.5% would compensate for the 1200 feet altitude of the 1WTC fire, placing the constant at 0.4825. Other methods, e.g. Law, show that for large compartments such as the WTC, the mass burning rate can be less than predicted by the Rockett / Kawagoe ventilation factor. In determining the heat release rate, a combustion efficiency factor must be allowed for. The fire compartment is assumed to be a well stirred but imperfect reactor. At a given instant, the fire zone is unlikely to cover the entire floor space, thus some of the oxygen throughput will miss the fire. And the effective calorific value of the oxygen and the fuel will vary across the fire zone depending on the deviation of the local fuel-air mix from stoichiometric conditions. The following analysis aims to quantify the actual WTC mean flame temperatures and combustion efficiency of both the kerosene-based and office furniture/fittings stages of the fires. The equivalence ratio (phi) of a given fuel-air mixture is defined as the actual mixture ratio divided by the stoichiometric fuel-air ratio that would provide just the required amount of oxidiser to completely burn a quantity of fuel: phi = (MASSfuel / MASSair) / (MASSfuel / MASSair)st or phi = (VOLfuel / VOLair) / (VOLfuel / VOLair)st Since the stoichiometric ratio derives from the fuel-oxygen ratio, we could be discussing combustion in pure oxygen and have: phi = (MASSfuel / MASSoxygen) / (MASSfuel / MASSoxygen)st or phi = (VOLfuel / VOLoxygen) / (VOLfuel / VOLoxygen)st The fuel and fuel-oxygen ratio determine the heat of reaction. Flame temperature calculations in air, whether normal atmospheric, oxygen-enhanced or depleted, must take account of the quantity of nitrogen. If 5.43 kg of air is required to burn 1 kg of wood volatiles, then a fuel-rich mixture with, say, 1.5 kg of wood volatiles for every 5.43 kg of air would have an equivalence ratio of 1.5. A lean mixture would have phi less than 1. In the case of a lean mixture, energy is wasted heating up surplus oxidiser amongst the products. A rich mixture is less exothermic (releases a lower quantity of heat per mole of fuel) because products such as CO and H2 are included, with heats of formation higher than that of CO2 and H2O. (CO HOF has a smaller negative value compared to the high negative value of CO2 and H2O, and H2 HOF is zero.) The adiabatic flame temperature peaks at between phi = 1 and phi = 1.1 depending on the fuel, because a slightly rich mixture has diatomic rather than triatomic molecular products, the latter tending to have higher molar specific heats. In diffusion flames, small-scale experiments (e.g. with a Wolfhard-Parker fuel jet burner) show that the oxidiser (air) and fuel volatiles will diffuse together at near stoichiometric rates. So an adiabatic situation (no loss of heat to the surroundings) would produce a flame temperature close to the adiabatic stoichiometric flame temperature. However, for a compartment fire in a large confined space such as a floor of the WTC Towers (208 feet square, 10 feet from floor to ceiling, some breaks in the ceiling / floor layer but the rest intact, various partition walls), combustion will proceed at a wide range of equivalence ratios. The side of the fire nearest to the vent will receive fresh air with 21% (by volume) oxygen, and it is towards this air side that the hottest, most efficient burning will occur. As the air flows across, the oxygen level will decrease, and at the fuel-rich oxygen depleted side, oxygen levels might be barely above zero. At room temperatures, flaming combustion would probably self-extinguish at around 14% oxygen levels; at the much higher post-flashover temperatures it could be maintained at very low oxygen levels. Proceeding towards the fuel side, as oxygen levels drop the proportion of products will tip from CO2 and H2O towards CO, H2, C and others. The difference between the reactants' and products' heats of formation will decrease, and with it the heat released per unit mass of fuel and oxygen. Burning under fuel-rich conditions rather than stoichiometric will reduce the effective calorific value of the fuel by a greater amount than that of the oxygen. For example, in the reaction C + O2 ===} CO2, the heat released is around 32.76 kJ per gram of carbon and 12.3 kJ per gram of oxygen. As combustion becomes fuel-rich the proportion of CO in the products increases. The reaction C + (1/2)O2 ===} CO is less exothermic, with some 9.202 kJ released per gram of carbon and 6.908 kJ per gram of oxygen. Breaking a hydrocarbon down to its elements is endothermic and in the case of kerosene or dodecane, for example, costs slightly over 2 kJ/g. Hydrogen comprises less than 20% of the mass, but its capacity to release over 120 kJ/g provides dodecane with a calorific value above that of carbon. It also boosts the heat release per unit of oxygen above the 12.3 kJ/g rate from carbon alone. The reaction occurs locally, step by step, rather than as shown in the global reaction which has the fuel and oxidiser as reactants and CO2 and/or H2O "instantly" formed amongst the products. However, calculations for each local step reaction in turn, summing the energy gained or lost, e.g.: C + (1/2)O2 ===} CO + 110.53 kJ CO + (1/2)O2 ===} CO2 + 282.99 kJ will predict the same total heat release as in the global reaction: C + O2 ===} CO2 + 393.52 kJ In both cases one mole (12.011 g) of carbon was introduced at the start, so 393.52 / 12.011 is 32.763 kJ/g of heat released per unit of carbon. One mole of oxygen (31.999 g) was used overall, so 393.52 / 31.999 is 12.298 kJ/g provided by the oxygen. (The hydrogen content of the fuel will also boost the heat release per unit of oxygen.) The WTC fires consisted of a relatively short period which included jet fuel combustion, and the remaining cellulosic burning of typical office contents. Kerosene is a mixture of alkanes - hydrocarbons of the form CnH2n+2. The n-number can range from 9 to 17, and kerosene should contain five consecutively numbered alkanes. Kerosene is often considered to be dodecane (C12H26), which we shall assume in the following reaction: C12H26 + 18.5O2 + 69.595N2 ===} 12CO2 + 13H2O + 69.595N2 + 7518.26 kJ This assumes complete combustion. A very rough calculation of adiabatic flame temperature (AFT) based on commonly used values of specific heats would greatly overestimate the temperature. One error, of some 20 to 30%, would arise from the fact that specific heat increases with temperature, and can be around 50% higher at 2000 K. Calculations using the JANAF thermochemical data tables eliminate this error. However, relatively simple calculations based on the JANAF tables will still typically overestimate adiabatic temperatures by some 100 K for burning in air. And this remaining error rises substantially for cases involving higher temperatures where the oxidiser is oxygen-enriched air or even pure oxygen. This is due to equilibrium considerations. At temperatures greater than 1250 K, endothermic dissociation reactions occur. Some of the CO2 will break down into CO and (1/2)O2, some H2O will form H2 and (1/2)O2, and even some O2 will break down into 2O. These reactions are two-way: CO2 {===} CO + (1/2)O2 H2O {===} H2 + (1/2)O2 O2 {===} 2O At very high temperatures the reaction will proceed from left to right and is then endothermic because the R.H. products have higher heats of formation compared to the L.H. reactants. At lower temperatures, say, 1000K, the reverse would be the case. Given an oxidiser, the CO and H2 would burn, and oxygen would be stable in diatomic rather than monatomic form. The system will always tend towards an equilibrium state, where a particular proportion of dissociation products remain for a particular reaction. This equilibrium state has a lower temperature than the JANAF-based but equilibrium disregarded prediction. In order to obtain a fairly accurate value for the equilibrium AFT at various equivalence ratios, the most straightforward method is to find the enthalpy content of the reaction (i.e. the heat of formation of the fuel), the H/C ratio, and to consult figures 3 to 5, Chapter 1, Combustion, Third Edition, (Irvin Glassman). These graphs allow one to readily solve flame temperature problems for a wide range of hydrocarbons. For kerosene burning in air, assuming 21% oxygen by volume and a stoichiometric mix, JANAF-based calculations disregarding equilibrium would predict an AFT of 2398 K. The actual equilibrium value is around 2265 K. We are also interested in the reaction for wood combustion which comprised the vast majority of the burning time and total heat release of the WTC fires. For a range of equivalence ratios, and taking the H/C ratios and enthalpies as outlined below, the Glassman graphs predict the following equilibrium AFT (Kelvin) for kerosene and wood: 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 Dodecane (kerosene) has a heat of formation of -2.042 kJ/g or -0.488 kcal/g and a H/C ratio of 26/12 = 2.167 (Note: the Table 1 Heats of Formation at 298 K in the Glassman Combustion book has an error with the incorrect chemical formula for dodecane, and a slight error in the kJ/g heat of formation due to division by an incorrect molecular weight. The molar heat of formation appears to be correct at -347.77 kJ/mol for the liquid. In the above reaction for complete combustion of dodecane, this value predicts a calorific value of 44.14 kJ/g and 12.7 kJ/g of oxygen consumed, which is in line with generally accepted values.) The office contents include carpets, plastics, paper etc as well as wood. However, the fire load is regarded as equivalent to some given density of wood, by mass per unit of floor space. So we calculate the likely combustion efficiency and flame temperatures for wood. The heat released from aircraft combustibles will also be taken into consideration. The flame temperatures for wood are based on a heat of formation of -6.062 kJ/g or -1.448 kcal/g and a H/C ratio of 1.524. These values were obtained through consideration of the reaction for complete combustion of wood in air, which is taken to be: 0.8C6H10O5 + 0.2C18H24O11 + 8.5O2 + 31.976N2 ===} 8.4CO2 + 6.4H2O + 31.976N2 + 3563.2 kJ According to most definitions, wood is not strictly a hydrocarbon. Its oxygen content lowers its heat of formation, and as a result breaking wood down into its elements is more endothermic than for kerosene, costing some 6.06 kJ/g as opposed to 2.04 kJ/g. However, the Glassman graphs take into account the enthalpy and still predict the equilibrium AFT for wood quite accurately, at 2120 K (equivalence ratio phi = 1). A simple calculation using thermochemical data tables predicts 2267 K which is 147 K high. For kerosene the corresponding figures were 2265 K and 2398 K, a 133 K difference. All else being equal, the degree of dissociation and consequent reduction in AFT due to equilibrium considerations would decrease with decreasing AFT. But with wood, the fact that it costs more to break the bonds in the first place means that any subsequent incomplete combustion has a proportionately greater effect on net heat released and flame temperature. Hence, we see that wood AFT decreases more than kerosene AFT under fuel rich conditions, and a smaller degree of dissociation has a slighly greater effect on wood AFT. Cellulose is generally regarded as (C6H10O5)x, a polymer of molecular weight 5,000 - 10,000 which repeats the monomer glucose. Lignin is a very complex high molecular weight polymer which can form random variations, but approximates as (C18H24O11)x. These approximations are close to the actual elemental composition of the substances, so the above reaction predicts the calorific value and stoichiometric fuel-air ratio fairly accurately, even though "one mole" of cellulose or lignin is taken to have a weight of hundreds rather than the actual thousands of grams. Cellulose comprises about 50% of the mass of wood, whereas lignin is some 1/4 to 1/3 of the mass. Lignin's lower oxygen content leads to the calorific value of the wood increasing with its lignin content, along with the quantity of oxygen required for complete combustion. The molar proportions 80% cellulose and 20% lignin assumed in the above reaction equate to a lignin mass nearly 2/3 the mass of cellulose or nearly 1/3 the total. Assuming a per unit heat release from oxygen of 13.1 kJ/g (probably slightly high), the maximum calorific value of the wood comes out at 16.74 kJ/g, with 1.28 kg of oxygen or 5.52 kg of air required to burn each kg of wood volatiles. With 70:30 molar proportions the lignin content would be too high with a mass proportion slightly above that of cellulose. The calorific value would then be 17.16 kJ/g, with 1.31 kg of oxygen required to burn each kg of wood. (For complete combustion, cellulose would release some 15.53 kJ/g, compared with 18.64 kJ/g for lignin.) Kawagoe and others such as Magnusson and Thelandersson assumed 10.75 MJ/kg as the net effective heat of combustion of wood in fully developed compartment fires. Under ideal conditions the heat of combustion of wood is typically regarded as 16.5 MJ/kg. Assuming an efficiency factor of around 0.65 would allow for unburnt oxygen in the imperfect mix, and for the inefficiency of fuel-rich combustion in part of the fire zone. Taking Rockett's 0.5 proportionality constant, with each kilogram of air having the potential to release some 3 MJ of heat, from the air inflow the maximum possible heat release rate is 1.5 MW times the ventilation factor [Av times SQR(Hv)]. After allowing for 65% efficiency, the actual power output in a real compartment fire would be approximately Av times SQR(Hv) in MW. Although variations in stoichiometry have a greater effect on the combustion efficiency of the fuel than that of the oxidiser, the Rockett and Kawagoe proportionality constants can be reconciled. If fuel-rich burning takes, say, 35% off the calorific value of the fuel but only 17% off the heat release per gram of oxygen, this is balanced out by the fact that a greater proportion of oxygen is wasted due to the unmixedness of the compartment compared to the proportion of unburnt fuel. The formula: air mass inflow rate (kg/s) = 0.5 * Av * SQR(Hv) counts the total air inflow, including air that flows out with little or no reduction in its oxygen. Some of this will flow out of vents on the same facade before reaching the fire, and some will flow across from one wall to another but miss the fire zone. In contrast, fuel that is not pyrolised does not count, and most of the pyrolised fuel would burn within the compartment - albeit at a range of equivalence ratios. The proportion of fuel that does not burn until exiting the compartment is likely to be small in relation to the proportion of oxygen that flows through without being consumed by the fire. Although this unmixedness could be considered as a source of fuel-rich inefficiency, we can treat the fuel inefficiency as wholly accounted for by rich burning, whereas for the oxygen the inefficiency partly derives from the unmixedness as well as the stoichiometry. Starting with the reaction for complete combustion of wood (we may ignore atmospheric nitrogen in calculations of calorific value; since its heat of formation is zero it is only relevant in flame temperature calculations): 0.8C6H10O5 + 0.2C18H24O11 + 8.5O2 ===} 8.4CO2 + 6.4H2O + 3563.2 kJ As the mixture becomes richer, some CO2 product is replaced with CO and some H2O is replaced with H2. The proportion of CO2 that is substituted by a less oxidised product tends to be higher than the proportion of lost H2O. After considering some actual examples of equilibrium product compositions at various equivalence ratios, (e.g. Table 3, Chapter 1, Combustion, Third Edition, Glassman) the ratio of replaced CO2 to replaced H2O was found to be about 59 to 41. Soot particulate increases as the system gets richer, but the proportion of C product is relatively low. We shall assume that for every gram-atomic weight or 0.5 mol of oxygen removed from the oxidiser in the above reaction, 0.59 mol of CO2 is replaced with 0.59 mol of CO, and 0.41 mol of H2O is replaced with 0.41 mol of H2. For every mole of CO2 product replaced by CO, the total heat of formation of the products increases by -393.52 + 110.53 or about 283 kJ. For every mole of H2O product replaced by H2, the total heat of formation of the products increases by 241.83 kJ. So for every 0.5 mol of oxygen removed from the original 8.5 mol in the above reaction, the heat of reaction decreases by 0.59 * 282.99 + 0.41 * 241.83 = 266.11 kJ. A 0.5 / 8.5 or 5.8823% decrease in oxygen leads to a 266.11 / 3563.2 or 7.4683% reduction in heat released. So the heat released decreases by 7.4683 / 5.8823 = 1.2696% for each 1% decrease in oxygen. Each gram of oxygen lost takes 16.63 kJ off the heat of reaction. With complete combustion yielding some 13.1 kJ/g of oxygen, the calorific value of the oxygen decreases as the system gets richer, although not as quickly as the decrease in fuel calorific value. After the oxygen/fuel ratio has halved, with 4.25 mol or 136 grams of oxygen removed from the oxidiser, the equation is then: 0.8C6H10O5 + 0.2C18H24O11 + 4.25O2 ===} 3.385CO2 + 5.015CO + 2.915H2O + 3.485H2 + 1301.23 kJ The heat of reaction is 0.36518 times the rate for complete combustion. With the mass of wood in the above reaction taken as 212.8 grams, the original 16.74 kJ/g calorific value has dropped to 6.11 kJ/g. Proceeding from the air side to the fuel side of the fire zone, decreasing temperatures and heat flux would lower fuel mass pyrolysis rates to some extent. So the oxygen-fuel ratio would decrease less rapidly than the oxygen-air ratio. If we suppose that the rates of pyrolysis and air flow provide close to stoichiometric conditions on the air side of the fire with 21% O2, then 4.25 moles of oxygen per "mole" of wood corresponds to somewhat less than 10.5% O2. Ignoring the very low oxygen range and assuming a linear distribution over the reactants range 8.5O2 to 4.25O2, the average value of a little under 15.75% O2 by volume of air or 4.25 oxygen gram-atomic weights lost suggests an average calorific value of wood at (3563.2 + 1301.23) / (2 * 212.8) = 11.43 kJ/g. At 68.26% efficiency this is close to Kawagoe's 10.75 kJ/g net effective heat of combustion of wood. If 8.5O2 in the reactants corresponds to phi = 1 and 4.25O2 corresponds to phi = 2, then calculating the integral of the function shows that a linear oxygen-fuel distribution across this range equates to a mean equivalence ratio of phi = 1.386 (the natural log of 4). It seems reasonable to retain Kawagoe's 0.092 constant but taking 11.4 kJ/g as the effective heat of combustion of wood in a structure fire, postulating a factor of around 0.68 as an overall combustion efficiency factor. This allows for burning under non-stoichiometric conditions and for the unmixedness of the compartment. Rockett's 0.5 constant is retained, and we assume that the net effective heat released is 2 MJ for each kg of air flowing into the compartment. So with combustion efficiency taken into account, the actual power output in a real compartment fire would be approximately Av times SQR(Hv) in MW. This is applicable to the secondary office contents-based stage. As for the primary kerosene-based stage, we shall review the dodecane combustion equation to find the efficiency factor. Ignoring the nitrogen (although if really burning in pure oxygen, the higher flame temperatures would increase dissociation products and reduce the heat of reaction): C12H26 + 18.5O2 ===} 12CO2 + 13H2O + 7518.26 kJ With the molecular weight of dodecane at 170.34 grams, the above case of complete combustion predicts a maximum calorific value of 7518.26 / 170.34 = 44.14 kJ/g. Suppose that for every gram-atomic weight or 0.5 moles of oxygen removed from the oxidiser, 0.59 mol of CO2 is replaced with 0.59 mol of CO, and 0.41 mol of H2O is replaced with 0.41 mol of H2. So for every 0.5 mol of oxygen removed from the original 18.5 mol in the above reaction, the heat of reaction decreases by 0.59 * 282.99 + 0.41 * 241.83 = 266.11 kJ. A 0.5 / 18.5 or 2.7027% reduction in oxygen leads to a 266.11 / 7518.26 or 3.5395% decrease in heat released. So the heat released decreases by 3.5395 / 2.7027 = 1.3096% for each 1% decrease in oxygen. After the oxygen/fuel ratio has halved, with 9.25 mol or 296 grams of oxygen removed from the oxidiser, the equation is then: C12H26 + 9.25O2 ===} 1.085CO2 + 10.915CO + 5.415H2O + 7.585H2 + 2595.14 kJ The heat of reaction is 0.34518 times the rate for complete combustion. With the mass of dodecane in the above reaction taken as 170.34 grams, the maximum 44.14 kJ/g calorific value has dropped to 15.24 kJ/g. So averaging over the oxygen-fuel range 0.5 to 1.0, the mean calorific value of jet fuel is (44.14 + 15.24) / 2 = 29.69 MJ/kg, an overall combustion efficiency of 1.34518 / 2 = 0.67259. Generalizing from the above and given our assumptions on maximum equivalence ratio and change in product composition, it can be seen that the overall combustion efficiency and the ratio of heat of reaction decrease to oxygen decrease derive from the heat release per unit of oxygen. 12.7 kJ/g of oxygen was calculated for dodecane, given its heat of formation. 13.1 kJ/g was assumed for wood; most fuels are within a few percent of this. When the change in product composition results in a 16.63 kJ decrease in heat of reaction for each gram of oxygen lost from the oxidiser, 16.63 divided by the heat released (kJ/g) per unit of oxygen yields the heat of reaction decrease divided by the oxygen decrease. Since 16.63 / Hox = (1 - Eff) / 0.25, it follows that the overall combustion efficiency is given by: Eff = 1 - 4.1575 / Hox where Hox is the maximum heat released per unit of oxygen (kJ/g) consumed for the particular fuel, in complete combustion. A more general formula which allows for adjustments to our postulates of maximum phi = 2 and a 59 : 41 split between CO2 and H2O lost from the products is: Eff = (32 * Hox - 41.17 * c + 41.17 * c / f - 241.83 + 241.83 / f) / (32 * Hox) where c moles of CO2 product are replaced by CO and 1 - c moles of H2O product are replaced by H2 for each gram-atomic weight of O2 removed from the oxidiser, and f is the maximum equivalence ratio at which combustion occurs. We are assuming c = 0.59 and f = 2. Lowering c to 0.5 (with Hox = 13.1) merely raises Eff from 0.6826 to 0.687. At f = 2 we ignore the sub-0.5 oxygen-fuel region, erring in favour of fire collapse theories. To bring about an increase in predicted Eff by 10% to 0.7509 (with c = 0.59 and Hox = 13.1) would require f to be taken as only 1.646, which is too low given the expected temperatures. However, this formula does not hold when f is much greater than 2, since the CO2 product has all been converted to CO and further O2 reductions result in CO converting to C which has less effect on the heat of reaction. We shall assume 29.7 MJ/kg as the mean calorific value for jet fuel, ignoring the one or two percent losses due to additives, and 11.4 MJ/kg as the mean value for wood or a fire load expressed as the equivalent of some particular area density of wood, in a compartment fire. The actual gas temperature in the compartment will depend on the proportion of the heat of reaction that escapes from the combustion products to heat the surrounding building environment. For example, suppose the fuel releases an average of 11.4 kJ/g, after allowing for combustion inefficiencies. If half of this is retained for heating the products of combustion and the other half leaks to the surroundings, then the internal gas temperature would be approximately midway between the adiabatic flame and ambient temperatures, with 5.7 kJ/g available to heat the building. The remaining 5.7 kJ/g locked into the gas temperature eventually gets to heat the exterior air, after being vented out with the plume. In fact, in this case the internal gas temperature would be a little higher than the mean of AFT and ambient temperature, because specific heat increases with temperature. For some dissociation products the reaction will reverse, with another slight boost to temperature by way of increased heat release. If we suppose that more than half of the heat released leaks to the surroundings, then the gas temperature will be lower. Conversely, if less than half of the total heat released is used to heat the building structure, the internal gas temperature is higher, but there is less power available to heat the building. Some heat is transmitted by the flames and some by the upper smoke layer, so the actual flame temperature will be between the AFT and the temperature of the hot gas as it exits the building. Typical upper layer temperatures will be between the flame temperature and the exit temperature. A point to bear in mind regarding the AFTs taken from the Glassman graphs (figures repeated below) is that they assume standard air with around 21% oxygen. We are assuming that the air becomes increasingly depleted of oxygen as one proceeds from the vent or air side of the fire to the fuel side. For the case of oxygen-depleted air in a compartment fire, even after allowing that actual conditions would not be adiabatic, the 21% oxygen temperatures are still too high. When the amount of air simply decreases in relation to the pyrolysis products, the reduced heat of reaction is part compensated by the fact that the fire has less atmospheric nitrogen to heat up. If the quantity of oxygen reduces without a corresponding decrease in nitrogen, the fire is abated to the full extent of the decline in power output. We will refer to the Glassman graph temperatures as the Gordon-McBride predictions; they were calculated from that program. The values shown over the phi range 1.0 to 2.0 are for phi = 1.0, 1.1, 1.5 and 2.0; intermediate values are obtained by linear interpolation. The other method of calculating flame temperature is based on manually looking up JANAF tables; and makes our assumption of 0.59 mol of CO2 converting to CO and 0.41 of H2O converting to H2 for each 0.5 mol of O2 lost from the air. The Gordon- McBride predictions are accurate for phi = 1.0, whereas the manual predictions are some 100 to 150 degrees high, failing to allow for some 3% dissociation. However, the Gordon-McBride predictions are too high in the fuel-rich area, since they have air-fuel ratio as the variable rather than oxygen-air ratio. A comparison of both predictions illustrates these trends. For wood combustion, when phi = 1.0, the manually calculated AFT is 147 degrees higher than the Gordon-McBride prediction. The latter is correct, since dissociation products reduce combustion efficiency. However, at phi = 1.1 the manual prediction is only 11 degrees higher than Gordon-McBride; at phi = 1.333 to 2.0 the manual prediction averages some 75 degrees below Gordon-McBride. Over this range the manual prediction is not only the more appropriate of the two (due to the non-reduction in nitrogen level), it probably remains a little high itself due to disregard of equilibrium and dissociation. However, considering the case of propane burning in normal air at phi = 1.5 for example, manual calculations predict an AFT only 4 K higher than Gordon-McBride's predicted 1974 K. With the assumed 59:41 ratio of CO2 to H2O lost from the products, the predicted product composition is close to that shown in the Glassman table. With dissociation at under 1%, the equilibrium / dissociation error is much less than 147 K. Gordon-McBride equilibrium AFT (Kelvin) for kerosene and wood, repeated from above: 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 For wood, disregarding the phi less than 1.0 region, the simplest single straight line function connecting the phi = 1 and phi = 2 data points: AFT = -910 * (phi - 1) + 2120 yields an average value of (2120 + 1210) / 2 = 1665 K. This is very close to the value for phi = 1.5, but 46 degrees low at phi = 1.1. A sawtooth function of two straight lines based on three data points, predicting a peak at (1.026, 2145) assuming equal absolute gradients either side: AFT = -961 * ABS(phi - 1.026) + 2145 is 30 degrees high at phi = 1.5. Calculation of the area under the function yields an average AFT of 1689 K. An even better fit, four straight lines of similar absolute slope based on all four data points and the same predicted peak, shows an average AFT of 1676 K. We are assuming a linear distribution of oxygen-fuel ratios, from a stoichiometric mixture down to half the oxygen content, i.e., from phi = 1.0 to phi = 2.0. From integration of the reciprocal function, this corresponds to an average phi of 1.386, so the above flame temperature values understate the mean. The best fit function above, with four data points, has an area above the baseline which exceeds that of the single straight line function by 2.4%. Calculating the AFT for phi = 1.386 from the single line function above and adding 2.4% to the excess over 1210 K obtains 1782 K as the mean AFT. This is still marginally low, as most of the undershoot of the single straight line function is in the range towards phi = 1.026, which gets a higher weighting from the linear oxygen level distribution. A computation which steps through oxygen-fuel ratios, calculates the equivalence ratio and, depending on the latter, selects one of four functions to calculate AFT, predicted 1786.2 K as the mean AFT. In a similar exercise for kerosene, the mean AFT was computed at 2057 K, with a peak of 2319 K at phi = 1.065. The kerosene temperatures are less important, since this stage of the fire would have burnt out well before the steel could have absorbed sufficient heat to bring about any appreciable temperature increase. As pointed out above, these functions are based on Glassman graphs calculated from the Gordon-McBride program. These become inappropriate over the fuel-rich range, since we are assuming oxygen-depleted air rather than an excess of fuel pyrolysis over air flow rate as the source of fuel-rich combustion. From phi = 1.333 to 2.0 the Glassman-based functions average about 75 degrees high. The range 1.0 to 2.0 is more heavily weighted toward the 1.0 end, and so it would seem reasonable to take 25 degrees as the average error of these functions. So deducting 25 K from the mean AFT for wood of 1786 K as computed above, obtains 1761 K. Another approach is to imagine a large quantity of points in space each with some random oxygen-fuel ratio between stoichiometric and 0.5 of stoichiometry. We assume the heat of reaction and product molar quantities vary linearly with oxygen level. The mean heat of reaction and the mean molar quantity of each product equates to the 0.75 of stoichiometric oxygen level; these mean values multiplied by the number of sample points in space would represent a total heat release and product composition over a large volume. The AFT for the 0.75 case, from JANAF look-ups, works out at 1747 K. We'll be fair and stick with the above 1761 K. 1761 K is an increase of 1463 K from the ambient temperatures of 298 K usually assumed in AFT calculations. If the increase for the exiting gases is half of the 1463 K due to slightly more than half of the released heat escaping to the surroundings, then this 732 K increase places the exiting gas temperature at 757 C and we have, say, 6 kJ/g available to heat the building floors, spandrels, columns, concrete, etc. The remaining 5.4 kJ/g is then vented out with the plume. If we take the case of combustion products for a stoichiometry that would produce the average adiabatic flame temperature for the fire, calculations based on JANAF tables should provide quite an accurate guide as to the proportion of the adiabatic temperature rise that would remain given some specified proportion of heat outflow. The change in specific heat with temperature is accurately allowed for, and the error from dissociation becomes small as the temperature approaches 1250 K. As we shall see, using the Gordon-McBride / Glassman plots in reverse to find the stoichiometry given the AFT and then using JANAF look-ups to calculate the AFT leads to a discrepancy of some 93 K. This is due to the differing assumptions regarding nitrogen levels as outlined above. However, we shall continue to use the higher figure of 1761 K for the mean AFT, as calculated above. Because specific heat is higher at higher temperatures, a heat outflow of n% would lead to a temperature excess over ambient of more than (100 - n)% of the adiabatic temperature increase. The following is concerned with determining the additional proportionate temperature increase, rather than determining the absolute temperatures. Given the average AFT calculated above at 1761 K, the phi value clearly lies within the range 1.1 to 1.5, where the best fit function is: AFT = -1037.5 * (phi - 1.1) + 2075 To use this function we add back the 25 degree correction previously deducted, retrieving the previously computed mean AFT of 1786.2 K. So for an AFT of 1786.2 K, phi is 1.378 and the oxygen-fuel ratio is 0.7255 of the stoichiometric ratio. The phi value is slightly off the 1.386 mean due to the slight non-linearity of the phi-AFT function which occurs mostly close to a stoichiometric ratio. Compared to the formula for complete combustion: 0.8C6H10O5 + 0.2C18H24O11 + 8.5O2 + 31.976N2 ===} 8.4CO2 + 6.4H2O + 31.976N2 + 3563.2 kJ ...the oxygen level decreases by 2.333 mol or 4.666 gram-atomic weights. So 0.59 * 4.666 = 2.753 mol of CO2 product are replaced with 2.753 mol of CO, and 0.41 * 4.666 = 1.913 mol of H2O are replaced with 1.913 mol of H2. 0.8C6H10O5 + 0.2C18H24O11 + 6.167O2 + 31.976N2 ===} 5.647CO2 + 2.753CO + 4.487H2O + 1.913H2 + 31.976N2 + 2321.51 kJ Combustion efficiency at this stoichiometry is slightly below the mean for the 0.5 to 1.0 (oxygen/fuel) / (oxygen/fuel)st range, at 10.91 kJ/g of wood and 11.76 kJ/g of oxygen. FEMA's report on WTC 1 and 2 (chapter 2) quotes "one third to one half" of the total fire output as the energy that is vented out to drive the smoke plumes. Initially the proportion would be relatively low, increasing as the steel decking, concrete slabs, etc approached the temperature of the hot gases. The degree of this increase would depend on the final temperature difference. Let's take 40% as the typical proportion of the total heat release that remains locked in the gas products until after exiting the building, which allows 60% (mostly radiated) to heat the building structure. The above thermochemical equation for phi = 1.378, assuming just 928.6 kJ or 40% of 2321.51 kJ remains to heat the products, would result in an exiting gas temperature of 903 K or 630 C, a 605 degree increase on ambient temperature. Under adiabatic conditions the same stoichiometry would lead to a flame temperature of around 1693 K, a 1395 K increase from ambient of 298 K. So if 40% of the heat remains locked in the gases, the fractional increase in temperature is 605 / 1395 = 43.37% of the increase for 40% of the energy. The 1693 K figure has not allowed for any dissociation, so is probably a little high. Rounding up to 44% of the increase, then, we can add 10% to the exiting gas excess temperature over ambient. So making an allowance for specific heat increasing with temperature, when 40% of the energy is retained in the products, we should add some 10% to the calculated temperature increase. For 100% of the energy there is clearly no adjustment; by interpolation, when 70% of the energy is retained the adjustment would be about 5%. We shall incorporate these allowances below: Let the adiabatic flame temperature equal a degrees above ambient temperature; let the typical actual flame temperature equal b degrees above ambient; let the typical upper layer gas temperature equal c degrees above ambient; let the typical exiting gas temperature equal d degrees above ambient. Then the heat available for absorption by the building is 1 - d / (1.1 * a) multiplied by the effective calorific value of the fuel after allowing for combustion inefficiency. The proportion of the calorific value transmitted by the flames is 1 - b / (1.05 * a), and the proportion transmitted by the upper gas smoke layer is b / (1.05 * a) - d / (1.1 * a). The c increase on ambient temperature is an intermediate increase between b and d. A 7.5% adjustment would place c a few degrees above the mean (b + d) / 2. Heat losses from gas above the mean would exceed losses from gas below the mean; therefore the average gas temperature c would be a little below the mean of flame temperature and exiting gas temperature. If the average effective heat of combustion of the wood or equivalent fuel is 11.4 MJ/kg and 60% is emitted and absorbed before flowing out with the smoke plume, then 6.84 MJ per kg of wood or equivalent is available for heating the building structure. The jet fuel averaged 29.7 MJ/kg after allowing for combustion inefficiencies, so 60% internally transferred is 17.82 MJ/kg or 55.24 MJ/gallon, leaving 11.88 MJ/kg to drive the external plume. The actual flame temperature is between the AFT and exiting gas temperature, depending on the proportions of heat transmitted by the flames and the smoke layer. Typical flame and hot gas layer temperatures are known to be between 800 and 1200 C. The mean adiabatic flame temperature was 1761 K or 1488 C which is 1463 above ambient. The exiting gas temperature is 40% of this increase plus an extra 10%, which is 0.44 * 1463 = 644 + ambient of 25 C = 669 C. If the mean actual flame temperature was 1000 C then 4.17 MJ/kg was transmitted from the flames and 2.67 MJ/kg was transmitted from the smoke. The mean upper layer gas temperature was 835 C. Setting the flame temperature at 1100 C would raise the mean upper layer temperature by 50 degrees and result in almost equal proportions, i.e. 30% of the 11.4 MJ/kg transmitted from the flames and 30% transmitted from the smoke layer. It is likely that the flames emitted more heat than the smoke, and so 1000 C is a more reasonable figure for actual flame temperature. So we have 835 C to 1000 C as the fire temperatures transmitted to the steel, with 17.82 MJ/kg being available from the kerosene and 6.84 MJ/kg available from the wood. The actual temperatures reached by the steel core column members would depend on their mass and the heat absorbed. The amount of heat absorbed by the core columns rather than the perimeter columns, steel decking, or concrete slabs, depended on their volume. An object extending from floor to ceiling and taking up a great proportion of the floor area would absorb a high proportion of the heat which did not flow out with the plume; a tiny object would absorb a small proportion. Before determining the mass and volume of steel and concrete in the WTC, we shall consider the power output of the fires, and total heat released.