THE JET FUEL; HOW HOT DID IT HEAT
THE WORLD TRADE CENTER?
The Federal Emergency Management Agency (FEMA) report into collapse of the WTC towers, estimates that about 3,500 gallons of jet fuel burnt within each of the towers. Imagine that this entire quantity of jet fuel was injected into just one floor of the World Trade Center, that the jet fuel burnt with perfect efficency, that no hot gases left this floor, that no heat escaped this floor by conduction and that the steel and concrete had an unlimited amount of time to absorb all the heat. With these ideal assumptions we calculate the maximum temperature that this one floor could have reached.
“The Boeing 767 is capable of carrying up to 23,980 gallons of fuel and it is estimated that, at the time of impact, each aircraft had approximately 10,000 gallons of unused fuel on board (compiled from Government sources).”
Quote from the FEMA report into the collapse of WTC’s One and Two (Chapter Two).
Since the aircraft were only flying from Boston to Los Angeles, they would have been nowhere near fully fueled on takeoff (the aircraft have a maximum range of 7,600 miles). They would have carried just enough fuel for the trip together with some safety factor. Remember, that carrying excess fuel means higher fuel bills and less paying passengers. The aircraft would have also burnt some fuel between Boston and New York.
“If one assumes that approximately 3,000 gallons of fuel were consumed in the initial fireballs, then the remainder either escaped the impact floors in the manners described above or was consumed by the fire on the impact floors. If half flowed away, then 3,500 gallons remained on the impact floors to be consumed in the fires that followed.”
Quote from the FEMA report into the collapse of WTC’s One and Two (Chapter Two).
What we propose to do, is pretend that the entire 3,500 gallons of jet fuel was confined to just one floor of the World Trade Center, that the jet fuel burnt with the perfect quantity of oxygen, that no hot gases left this floor and that no heat escaped this floor by conduction. With these ideal assumptions (none of which were meet in reality) we will calculate the maximum temperature that this one floor could have reached. Of course, on that day, the real temperature rise of any floor due to the burning jet fuel, would have been considerably lower than the rise that we calculate, but this estimate will enable us to demonstrate that the “official” explanation is a lie.
Note that a gallon of jet fuel weighs about 3.1 kilograms, hence 3,500 gallons weighs 3,500 x 3.1 = 10,850 kgs.
Jet fuel is a colorless, combustible, straight run petroleum distillate liquid. Its principal uses are as an ingredient in lamp oils, charcoal starter fluids, jet engine fuels and insecticides.
It is also know as, fuel oil #1, kerosene, range oil, coal oil and aviation fuel.
It is comprised of hydrocarbons with a carbon range of C9 – C17. The hydrocarbons are mainly alkanes CnH2n+2, with n ranging from 9 to 17.
It has a flash point within the range 42° C – 72° C (110° F – 162° F).
And an ignition temperature of 210° C (410° F).
Depending on the supply of oxygen, jet fuel burns by one of three chemical reactions:
(1) CnH2n+2 + (3n+1)/2 O2 => n CO2 + (n + 1) H2O
(2) CnH2n+2 + (2n+1)/2 O2 => n CO + (n + 1) H2O
(3) CnH2n+2 + (n+1)/2 O2 => n C + (n + 1) H2O
Reaction (1) occurs when jet fuel is well mixed with air before being burnt, as for example, in jet engines.
Reactions (2) and (3) occur when a pool of jet fuel burns. When reaction (3) occurs the carbon formed shows up as soot in the flame. This makes the smoke very dark.
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In the aircraft crashes at the World Trade Center, the impact (with the aircraft going from 500 or 600 mph to zero) would have throughly mixed the fuel that entered the building with the limited amount of air available within. In fact, it is likely that all the fuel was turned into a flammable mist. However, for sake of argument we will assume that 3,500 gallons of the jet fuel did in fact form a pool fire. This means that it burnt according to reactions (2) and (3). Also note that the flammable mist would have burnt according to reactions (2) and (3), as the quantity of oxygen within the building was quite limited.
Since we do not know the exact quantities of oxygen available to the fire, we will assume that the combustion was perfectly efficient, that is, that the entire quantity of jet fuel burnt via reaction (1), even though we know that this was not so. This generous assumption will give a temperature that we know will be higher than the actual temperature of the fire attributable to the jet fuel.
We need to know that the (net) calorific value of jet fuel when burnt via reaction (1) is 42-44 MJ/kg. The calorific value of a fuel is the amount of energy released when the fuel is burnt. We will use the higher value of 44 MJ/kg as this will lead to a higher maximum temperature than the lower value of 42 (and we wish to continue being outrageously generous in our assumptions).
For a cleaner presentation and simpler calculations we will also assume that our hydrocarbons are of the form CnH2n. The dropping of the 2 hydrogen atoms does not make much difference to the final result and the interested reader can easily recalculate the figures for a slightly more accurate result. So we are now assuming the equation:
(4) CnH2n + 3n/2 O2 => n CO2 + n H2O
However, this model, does not take into account that the reaction is proceeding in air, which is only partly oxygen.
Dry air is 79% nitrogen and 21% oxygen (by volume). Normal air has a moisture content from 0 to 4%. We will include the water vapor and the other minor atmospheric gases with the nitrogen.
So the ratio of the main atmospheric gases, oxygen and nitrogen, is 1 : 3.76. In molar terms:
Air = O2 + 3.76 N2.
Because oxygen comes mixed with nitrogen, we have to include it in the equations. Even though it does not react, it is “along for the ride” and will absorb heat, affecting the overall heat balance. Thus we need to use the equation:
(5) CnH2n + 3n/2(O2 + 3.76 N2) => n CO2 + n H2O + 5.64n N2
From this equation we see that the molar ratio of CnH2n to that of the products is:
| CnH2n : CO2 : H2O : N2 | = 1 : n : n : 5.64n moles |
| = 14n : 44n : 18n : 28 x 5.64n kgs | |
| = 1 : 3.14286 : 1.28571 : 11.28 kgs | |
| = 31,000 : 97,429 : 39,857 : 349,680 kgs |
In the conversion of moles to kilograms we have assumed the atomic weights of hydrogen, carbon, nitrogen and oxygen are 1, 12, 14 and 16 respectively.
Now each of the towers contained 96,000 (short) tons of steel. That is an average of 96,000/117 = 820 tons per floor. Lets suppose that the bottom floors contained roughly twice the amount of steel of the upper floors (since the lower floors had to carry more weight). So we estimate that the lower floors contained about 1,100 tons of steel and the upper floors about 550 tons = 550 x 907.2 ≈ 500,000 kgs. We will assume that the floors hit by the aircraft contained the lower estimate of 500,000 kgs of steel. This generously underestimates the quantity of steel in these floors, and once again leads to a higher estimate of the maximum temperature.
Each story had a floor slab and a ceiling slab. These slabs were 207 feet wide, 207 feet deep and 4 (in parts 5) inches thick and were constructed from lightweight concrete. So each slab contained 207 x 207 x 1/3 = 14,283 cubic feet of concrete. Now a cubic foot of lightweight concrete weighs about 50kg, hence each slab weighed 714,150 ≈ 700,000 kgs. Together, the floor and ceiling slabs weighed some 1,400,000 kgs.
So, now we take all the ingredients and estimate a maximum temperature to which they could have been heated by 3,500 gallons of jet fuel. We will call this maximum temperature T. Since the calorific value of jet fuel is 44 MJ/kg. We know that 3,500 gallons = 31,000 kgs of jet fuel
will release 10,850 x 44,000,000 = 477,400,000,000 Joules of energy.
This is the total quantity of energy available to heat the ingredients to the temperature T. But what is the temperature T? To find out, we first have to calculate the amount of energy absorbed by each of the ingredients.
That is, we need to calculate the energy needed to raise:
| 39,857 | kilograms of water vapor to the temperature T° C, | |
| 97,429 | kilograms of carbon dioxide to the temperature T° C, | |
| 349,680 | kilograms of nitrogen to the temperature T° C, | |
| 500,000 | kilograms of steel to the temperature T° C, | |
| 1,400,000 | kilograms of concrete to the temperature T° C. | |
To calculate the energy needed to heat the above quantities, we need their specific heats. The specific heat of a substance is the amount of energy needed to raise one kilogram of the substance by one degree centigrade.
| Substance | Specific Heat [J/kg*C] |
|---|---|
| Nitrogen | 1,038 |
| Water Vapor | 1,690 |
| Carbon Dioxide | 845 |
| Lightweight Concrete | 800 |
| Steel | 450 |
Substituting these values into the above, we obtain:
| 39,857 x | 1,690 x (T – 25) | Joules are needed to heat the water vapor from 25° to T° C, |
| 97,429 x | 845 x (T – 25) | Joules are needed to heat the carbon dioxide from 25° to T° C, |
| 349,680 x | 1,038 x (T – 25) | Joules are needed to heat the nitrogen from 25° to T° C, |
| 500,000 x | 450 x (T – 25) | Joules are needed to heat the steel from 25° to T° C, |
| 1,400,000 x | 800 x (T – 25) | Joules are needed to heat the concrete from 25° to T° C. |
The assumption that the specific heats are constant over the temperature range 25° – T° C, is a good approximation if T turns out to be relatively small (as it does). For larger values of T this assumption once again leads to a higher maximum temperature (as the specific heat for these substances increases with temperature). We have assumed the initial temperature of the surroundings to be 25° C. The quantity, (T – 25)° C, is the temperature rise.
So the amount of energy needed to raise one floor to the temperature T° C is
= (39,857 x 1,690 + 97,429 x 845 + 349,680 x 1,038 + 500,000 x 450 + 1,400,000 x 800) x (T – 25)
= (67,358,330 + 82,327,505 + 362,967,840 + 225,000,000 + 1,120,000,000) x (T – 25) Joules
= 1,857,653,675 x (T – 25) Joules.
Since the amount of energy available to heat this floor is 477,400,000,000 Joules, we have that
1,857,653,675 x (T – 25) = 477,400,000,000
1,857,653,675 x T – 46,441,341,875 = 477,400,000,000
Therefore T = (477,400,000,000 + 46,441,341,875)/1,857,653,675 = 282° C (540° F).
So, the jet fuel could (at the very most) have only added T – 25 = 282 – 25 = 257° C (495° F) to the temperature of the typical office fire that developed.
Remember, this figure is a huge over-estimate, as (among other things) it assumes that the steel and concrete had an unlimited amount of time to absorb the heat, whereas in reality, the jet fuel fire was all over in one or two minutes, and the energy not absorbed by the concrete and steel within this brief period (that is, almost all of it) would have been vented to the outside world.
“The time to consume the jet fuel can be reasonably computed. At the upper bound, if one assumes that all 10,000 gallons of fuel were evenly spread across a single building floor, it would form a pool that would be consumed by fire in less than 5 minutes”
Quote from the FEMA report into the collapse of WTC’s One and Two (Chapter Two).
Here are statements from three eye-witnesses that provide evidence that the heating due to the jet fuel was indeed minimal.
Donovan Cowan was in an open elevator at the 78th floor sky-lobby (one of the impact floors of the South Tower) when the aircraft hit. He has been quoted as saying: “We went into the elevator. As soon as I hit the button, that’s when there was a big boom. We both got knocked down. I remember feeling this intense heat. The doors were still open. The heat lasted for maybe 15 to 20 seconds I guess. Then it stopped.”
Stanley Praimnath was on the 81st floor of the South Tower: “The plane impacts. I try to get up and then I realize that I’m covered up to my shoulder in debris. And when I’m digging through under all this rubble, I can see the bottom wing starting to burn, and that wing is wedged 20 feet in my office doorway.”
Ling Young was in her 78th floor office: “Only in my area were people alive, and the people alive were from my office. I figured that out later because I sat around in there for 10 or 15 minutes. That’s how I got so burned.”
Neither Stanley Praimnath nor Donovan Cowan nor Ling Young were cooked by the jet fuel fire. All three survived.
Summarizing:
We have assumed that the entire 3,500 gallons of jet fuel was confined to just one floor of the World Trade Center, that the jet fuel burnt with perfect efficency, that no hot gases left this floor, that no heat escaped this floor by conduction and that the steel and concrete had an unlimited amount of time to absorb all the heat.
Then it is impossible that the jet fuel, by itself, raised the temperature of this floor more than 257° C (495° F).
Now this temperature is nowhere near high enough to even begin explaining the World Trade Center Tower collapse.
It is not even close to the first critical temperature of 600° C (1,100° F) where steel loses about half its strength and it is nowhere near the quotes of 1500° C that we constantly read about in our lying media.
“In the mid-1990s British Steel and the Building Research Establishment performed a series of six experiments at Cardington to investigate the behavior of steel frame buildings. These experiments were conducted in a simulated, eight-story building. Secondary steel beams were not protected. Despite the temperature of the steel beams reaching 800-900° C (1,500-1,700° F) in three of the tests (well above the traditionally assumed critical temperature of 600° C (1,100° F), no collapse was observed in any of the six experiments.”
Quote from the FEMA report (Appendix A).
Recalling that the North Tower suffered no major structural damage from the intense office fire of February 23, 1975, we can conclude that the ensuing office fires of September 11, 2001, also did little extra damage to the towers.
Conclusion:
The jet fuel fires played almost no role in the collapse of the World Trade Center.
So, once again, you have been lied to by the media, are you surprised?
Todays Stock Market Activity Is EXACTLY What Happened Days Before 911, – Video
Today Stock Exchange trading went off the charts with 2000% increases instantly. This is EXACTLY what happened 2 days before 911. It’s happening again!
The stock market was already on edge because of the financial crisis in Greece. Images of mobs demonstrating in Athens were fueling an underlying panic. There’s a growing fear that a financial collapse in Greece could trigger a wave of financial trouble across Europe (and possibly even the world).
Early in the morning, some stocks swung wildly on unusually high trading volume, after what appeared to be technical miscues. The New York Stock Exchange said it was investigating unusual trades in 140 stocks.
Here are some of the stocks that had astronomical activity at 10AM this morning for no reason. I don’t know what the abbreviations stand for, but I do know that at the top of the list is the US Government: This is the same group that increased 800% 1 day before 911 and made BILLIONS OF 911.
Kass also provides a list of stocks that are seeing WAY abnormally high volume compared to normal.
1973% GOV US Equity Who’s increase was 800% 2 days before 911.
1063% PL US Equity
747% MTZ US Equity
672% EXG US Equity
578% N US Equity
518% BG US Equity
420% HOG US Equity
407% TRN US Equity
401% PIR US Equity
388% PPO US Equity
316% KRO US Equity
314% DDD US Equity
305% DOLE US Equity
298% JWN US Equity
262% HLF US Equity
248% MCP US Equity
248% NOK US Equity
247% FRX US Equity
238% FSL US Equity
237% LH US Equity
Unusual Option Activity
Listed below are options with unusually high volume for the most recent trading session compared with average daily volume. The relative level of activity is expressed in the daily volume ratio, which represents the current day’s volume divided by the average daily volume for the past month.
As Of 12:00 P.M. EDT Wednesday, August 01, 2012
Unusual Call Option Activity (intraday) Unusual Put Option Activity (intraday)
Data provided by SchaeffersResearch.com
Ticker
symbol
Put Volume
Average Put Volume
Daily Volume Ratio
DD 12,179 1,450 8.4
HLF 11,726 1,819 6.4
VZ 25,132 3,909 6.4
HOG 7,076 1,192 5.9
XOP 27,471 4,878 5.6
DE 10,258 2,200 4.7
JNPR 4,995 1,310 3.8
FSLR 8,883 2,383 3.7
GDX 29,379 7,902 3.7
GM 7,088 1,920 3.7
NOK 26,227 7,706 3.4
MA 6,343 2,072 3.1
F 21,175 7,672 2.8
ZNGA 5,001 1,935 2.6
FXI 29,145 11,438 2.5
ABX 4,611 1,878 2.5
WFC 16,778 6,949 2.4
SBUX 10,965 4,657 2.4
EFA 12,407 5,370 2.3
GG 2,947 1,291 2.3
The SEC’s Investigation
Shortly after the 911 attacks the SEC circulated a list of stocks to securities firms around the world seeking information. 16 A widely circulated article states that the stocks flagged by the SEC included those of the following corporations: American Airlines, United Airlines, Continental Airlines, Northwest Airlines, Southwest Airlines, US Airways airlines, Martin, Boeing, Lockheed Martin Corp., AIG, American Express Corp, American International Group, AMR Corporation, AXA SA, Bank of America Corp, Bank of New York Corp, Bank One Corp, Cigna Group, CNA Financial, Carnival Corp, Chubb Group, John Hancock Financial Services, Hercules Inc., L-3 Communications Holdings, Inc., LTV Corporation, Marsh & McLennan Cos. Inc., MetLife, Progressive Corp., General Motors, Raytheon, W.R. Grace, Royal Caribbean Cruises, Ltd., Lone Star Technologies, American Express, the Citigroup Inc., Royal & Sun Alliance, Lehman Brothers Holdings, Inc., Vornado Reality Trust, Morgan Stanley, Dean Witter & Co., XL Capital Ltd., and Bear Stearns.
An October 19 article in the San Francisco Chronicle reported that the SEC, after a period of silence, had undertaken the unprecedented action of deputizing hundreds of private officials in its investigation:
The proposed system, which would go into effect immediately, effectively deputizes hundreds, if not thousands, of key players in the private sector.
…
In a two-page statement issued to “all securities-related entities” nationwide, the SEC asked companies to designate senior personnel who appreciate “the sensitive nature” of the case and can be relied upon to “exercise appropriate discretion” as “point” people linking government investigators and the industry. 17
Michael Ruppert, a former LAPD officer, explains the consequences of this action:
What happens when you deputize someone in a national security or criminal investigation is that you make it illegal for them to disclose publicly what they know. Smart move. In effect, they become government agents and are controlled by government regulations rather than their own conscience. In fact, they can be thrown in jail without a hearing if they talk publicly. I have seen this implied threat time and again with federal investigations, intelligence agents, and even members of the United States Congress who are bound so tightly by secrecy oaths and agreements that they are not even able to disclose criminal activities inside the government for fear of incarceration. 18
The New York Stock Exchange said it was investigating unusual trades in 140 stocks. But don’t expect a different result than the SEC!
