OK HERE WE GO.
F.E.M.A estimates that 3,500 gallons of jet fuel were burned in each tower.
Assuming this info lets work on a smaller basis (floor by floor).
1) Assume that the jet fuel was inserted into ONE floor of the WTC.
2) Assume that the jet fuel burned efficiently and oxygen was available in the perfect ratio to contribute to this efficiency.
3) Assume that no gases (hot) left the floor (we're working with ONE floor) by and movement what so ever (motion, conduction, convection, radiation).
4) Assume that the steel and concrete had infinite (available) time to absorb the heat.
5) Assume that the heat distribution is even along the entire floor.
Now assuming all this (1-5) we can calculate how much the jet fuel could've burned.
Assume an efficient chemical reaction:
C(n)H(2n)(hydrocarbon)+3n/2 O2 (Oxygen) ==>
n CO2 (Carbon Dioxide)+n H2O (water vapor)
This has a calorific net value of: 44-42 megajoules(joules*1 million)/kilograms
The
heat energy is used to heat the byproducts resulting from the chemical reaction and the concrete and steel. Nitrogenous particles in the air also absorb heat.
Molar Ratio for Air: O2 (oxygen)+3.76 N2 (nitrogen)
Now using the equation ABOVE the calorific net value.
Cn
H2n+3n/2(O2+3.76 N2) ==> n CO2+ n H2O+5.64n N2
OR CnH2n:CO2:H2O:N2 = (equal) 1:n:n:5.64n moles.
Now we convert the moles to mass. Use the formula n(moles)=m(mass)/M(molar_mass).
M=Molar Mass (below)
M(Hydrogen)=1 : M(Carbon)=12 : M(Nitrogen)=14 : M(Oxygen)=16 OR equal to ==>
14n : 44n : 18n : 28*5.64n Kilograms OR equal to ==> 1 : 3.14286 : 1.28571 : 11.28 Kilograms.
^ Mass when
3500 gallons of kerosene burnt: 10,850 : 34,100 : 13,949 : 122,288 kilograms
BTW
Every one of the twin towers higher floors had 550,000 kilograms of structural steel. PLUS 1,428,300 kilograms of concrete from floor (slabs) and ceiling.
Now we will convert the units from the chemical reaction. The atmospheric temp is (25 C), and we will convert it to a temp known a (T).
1) 13,949 kilograms
(H2O) vapor to T degrees Celsius
2) 34,100 kilograms
(CO2) gas to T degrees Celsius
3) 122,388 kilograms of
Nitrogen(2) gas to T degrees Celsius
4) 550,000 kilograms of
structural steel to T degrees Celsius
5) 1,428,300 kilograms of
concrete to T degrees Celsius
Quote:
And you are already acquainted with heat and heat capacity, so you can use that for your explanation as well.
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Now I WILL. The table below shows the calculated
heat capacity of the substances
bold-ed above.
(H2O)-1,690 J/Kg Celsius
(CO2)-845 J/Kg Celsius
(N2)-1,038 J/Kg Celsius
Structural Steel-450 J/Kg Celsius
Concrete-800 J/Kg Celsius
Now to take the numbers on the right (above) to the aforementioned
T degrees Celsius, we will used the the formula
E=MC(delta)T.
M=Mass
C=Heating Capacity
(delta)T=rise/or difference in temp
We're also going to have to assume that the atmospheric temp on that day was
(delta)T=T-25 degrees Celsius. (Remember atmospheric temp is 25 degrees C)
Now for the conversion.
H2O=13.949*1,690*(
T-25)
CO2=34.1*845*(
T-25)
N2=122.388*1,030*(
T-25)
Steel=550,000*450*(
T-25)
Concrete=1,428,300*800*(
T-25)
The energy in (
E=MC(delta)T) is received from the jet fuel burned (3500 gallons or 10,850 kilograms).
Calorific Value of kerosene:
44 Mj (mega joules) or 44,000,000 joules
So you multiply the kilograms of jet fuel (that day) by the calorific value of kerosene SO: 10,850*44,000,000 = 477,400,000,000 Joules
This is the amount of energy (
477,400,000,000 Joules) that will be used to heat up the FIVE aforementioned substances (water, carbon, nitrogen, steel, and concrete).
Now I'll give you a task. Multiply 477,400,000,000 Joules with the five substances and you'll get (guaranteed)
624 degrees Celsius, MUCH MUCH too low to melt or even weaken structural steel (concrete etc.).