Jet Fuel

[M. J. Powell]

Does anyone know the products of combustion of, say, a ton, of jet fuel?

Mike
--
M.J.Powell

[Anonymous]

M. J. Powell wrote in message >...
>
>Does anyone know the products of combustion of, say, a ton, of jet fuel?

Yes :-

Heat energy
Light energy
Kinetic energy

A ton of it, too ! ;o)

[Keith Willshaw]

"M. J. Powell" > wrote in message
...
>
> Does anyone know the products of combustion of, say, a ton, of jet fuel?
>
> Mike
> --
> M.J.Powell

Its a complex subject and the results depend on many factors
including grade of jet fuel , engine design, altitude etc

see the aircraft emissions data bank at
http://www.qinetiq.com/aircraft.html

and

http://www.metrokc.gov/health/seatac/jetfuel.pdf
http://www.grida.no/climate/ipcc/aviation/130.htm

Keith

[M. J. Powell]

In message >, Anonymous >
writes
>
>M. J. Powell wrote in message >...
>>
>>Does anyone know the products of combustion of, say, a ton, of jet fuel?
>
>Yes :-
>
>Heat energy
>Light energy
>Kinetic energy
>
>A ton of it, too ! ;o)

No wonder you stay anonymous...

Mike
--
M.J.Powell

[Mark]

But it WAS a good laugh!!!

Mark


"M. J. Powell" > wrote in message
...
> In message >, Anonymous >
> writes
> >
> >M. J. Powell wrote in message >...
> >>
> >>Does anyone know the products of combustion of, say, a ton, of jet fuel?
> >
> >Yes :-
> >
> >Heat energy
> >Light energy
> >Kinetic energy
> >
> >A ton of it, too ! ;o)
>
> No wonder you stay anonymous...
>
> Mike
> --
> M.J.Powell

[Ron Parsons]

In article >,
"M. J. Powell" > wrote:

>Does anyone know the products of combustion of, say, a ton, of jet fuel?
>

Hmmm.... hydrocarbons plus oxygen.... about two tons of carbon dioxide
and water vapor?

--
Ron

[M. J. Powell]

In message >, Ron Parsons
> writes
>In article >,
> "M. J. Powell" > wrote:
>
>>Does anyone know the products of combustion of, say, a ton, of jet fuel?
>>
>
>Hmmm.... hydrocarbons plus oxygen.... about two tons of carbon dioxide
>and water vapor?

Oxygen?

How much water vapour for one ton of fuel?

Mike
--
M.J.Powell

[Tarver Engineering]

"M. J. Powell" > wrote in message
...
> In message >, Ron Parsons
> > writes
> >In article >,
> > "M. J. Powell" > wrote:
> >
> >>Does anyone know the products of combustion of, say, a ton, of jet fuel?
> >>
> >
> >Hmmm.... hydrocarbons plus oxygen.... about two tons of carbon dioxide
> >and water vapor?
>
> Oxygen?

Or Chlorine, or sulfur.

> How much water vapour for one ton of fuel?

Which kind? Combustion only burns the end of the hydrocarbon into water
(for oxygen) and jet fuel comes in molecules of different lengths. From
coal oil to kerosene and mixes in between.

[M. J. Powell]

In message >, Tarver Engineering
> writes
>
>"M. J. Powell" > wrote in message
...
>> In message >, Ron Parsons
>> > writes
>> >In article >,
>> > "M. J. Powell" > wrote:
>> >
>> >>Does anyone know the products of combustion of, say, a ton, of jet fuel?
>> >>
>> >
>> >Hmmm.... hydrocarbons plus oxygen.... about two tons of carbon dioxide
>> >and water vapor?
>>
>> Oxygen?

Oxygen is a product of combustion?
>
>Or Chlorine, or sulfur.
>
>> How much water vapour for one ton of fuel?
>
>Which kind? Combustion only burns the end of the hydrocarbon into water
>(for oxygen) and jet fuel comes in molecules of different lengths. From
>coal oil to kerosene and mixes in between.

What is the typical output of water vapour for typical fuels, then?

Mike
--
M.J.Powell

[M. J. Powell]

In message >, Clark
> writes
>"M. J. Powell" > wrote in news:pKqxbJmQDe2
:
>
>> In message >, Ron Parsons
>> > writes
>>>In article >,
>>> "M. J. Powell" > wrote:
>>>
>>>>Does anyone know the products of combustion of, say, a ton, of jet fuel?
>>>>
>>>
>>>Hmmm.... hydrocarbons plus oxygen.... about two tons of carbon dioxide
>>>and water vapor?
>>
>> Oxygen?
>
>Yup, that's what "burning" requires. Generally, the atmosphere supplies it in
>the form of O2.

I was asking about the products of combustion.
>>
>> How much water vapour for one ton of fuel?
>
>Hydrocarbons have about 2 hydrogens (CnH2n+2 for alkanes, CnH2n for Alkenes,
>etc.) for every carbon. Using decane (C10) or undecane (C11) to represent jet
>fuel may be reasonable - the density looks about right. You can figure an
>average molecular weight somewhere in the large neighborhood of 150 #/#-mole
>(that'll get you a rough number for molecules in a ton). The chemical balance
>and actual computation are left as an exercise for the student...

Do I understand that you don't know?

Mike
--
M.J.Powell

[Tarver Engineering]

"Clark" <stillnospam@me> wrote in message > >

> > Do I understand that you don't know?
> >
> No, you do not understand.

Yes he does, Clark is an idiot.

[mg]

Two mains products is used correctly:

About 12 minutes of flying time and whole lot a fun!

MG


"M. J. Powell" > wrote in message
...
>
> Does anyone know the products of combustion of, say, a ton, of jet fuel?
>
> Mike
> --
> M.J.Powell

[Keith Willshaw]

"M. J. Powell" > wrote in message
...

> >Hydrocarbons have about 2 hydrogens (CnH2n+2 for alkanes, CnH2n for
Alkenes,
> >etc.) for every carbon. Using decane (C10) or undecane (C11) to represent
jet
> >fuel may be reasonable - the density looks about right. You can figure an
> >average molecular weight somewhere in the large neighborhood of 150
#/#-mole
> >(that'll get you a rough number for molecules in a ton). The chemical
balance
> >and actual computation are left as an exercise for the student...
>
> Do I understand that you don't know?
>

Actually Mike he just gave you the information you
need to work put the answer

Knowing the hydrogen/carbon ratio lets you work out the
water/co2 balance in the products of combustion
and with the atomic weights you can figure out the
rough masses , assuming complete conbustion.

Keith

[M. J. Powell]

In message >, Clark
> writes
>"M. J. Powell" > wrote in
:
>
>> In message >, Clark
>> > writes
>>>"M. J. Powell" > wrote in news:pKqxbJmQDe2
:
>>>
>>>> In message >, Ron Parsons
>>>> > writes
>>>>>In article >,
>>>>> "M. J. Powell" > wrote:
>>>>>
>>>>>>Does anyone know the products of combustion of, say, a ton, of jet
>>>>>>fuel?
>>>>>>
>>>>>
>>>>>Hmmm.... hydrocarbons plus oxygen.... about two tons of carbon
>>>>>dioxide and water vapor?
>>>>
>>>> Oxygen?
>>>
>>>Yup, that's what "burning" requires. Generally, the atmosphere supplies
>>>it in the form of O2.
>>
>> I was asking about the products of combustion.
>
>Yup. If you know what goes into the reaction then you know what comes out.
>
>>>>
>>>> How much water vapour for one ton of fuel?
>>>
>>>Hydrocarbons have about 2 hydrogens (CnH2n+2 for alkanes, CnH2n for
>>>Alkenes, etc.) for every carbon. Using decane (C10) or undecane (C11) to
>>>represent jet fuel may be reasonable - the density looks about right.
>>>You can figure an average molecular weight somewhere in the large
>>>neighborhood of 150 #/#-mole (that'll get you a rough number for
>>>molecules in a ton). The chemical balance and actual computation are
>>>left as an exercise for the student...
>>
>> Do I understand that you don't know?
>>
>No, you do not understand.
>
>Do your homework Mike and you will be able to answer your question.

I asked because I don't want to do more homework!

I looked up the previously given references and they told me everything
except what I want to know.

Mike
>
>

--
M.J.Powell

[M. J. Powell]

In message >, Keith Willshaw
> writes
>
>"M. J. Powell" > wrote in message
...
>
>> >Hydrocarbons have about 2 hydrogens (CnH2n+2 for alkanes, CnH2n for
>Alkenes,
>> >etc.) for every carbon. Using decane (C10) or undecane (C11) to represent
>jet
>> >fuel may be reasonable - the density looks about right. You can figure an
>> >average molecular weight somewhere in the large neighborhood of 150
>#/#-mole
>> >(that'll get you a rough number for molecules in a ton). The chemical
>balance
>> >and actual computation are left as an exercise for the student...
>>
>> Do I understand that you don't know?
>>
>
>Actually Mike he just gave you the information you
>need to work put the answer
>
>Knowing the hydrogen/carbon ratio lets you work out the
>water/co2 balance in the products of combustion
>and with the atomic weights you can figure out the
>rough masses , assuming complete conbustion.

I can't. I was looking for some kind soul to help me.

Mike
-
M.J.Powell

[John Mullen]

Clark wrote:
> "M. J. Powell" > wrote in news:Uj2uRaHhqv2$EwF0
> @pickmere.demon.co.uk:
>
>
>>In message >, Clark
> writes
>
>
>>>Do your homework Mike and you will be able to answer your question.
>>
>>I asked because I don't want to do more homework!
>>
>
> So, how's that working out for ya?
>

Assume jet fuel is pure decane, to simplify the calculation.

C10H22 + 15.5O2 -> 11H20 + 10CO2

1 mole C10H22 = 142g
1 mole H2O = 18g
1 mole CO2 = 44g

So for every tonne (1000kg) of decane burned, assuming 100% combustion,
you will have

1000000/142 * 11 * 18g = 1394360g = 1394kg water
1000000/142 * 10 * 44 = 3098590g = 3098kg carbon dioxide

Which comes to about 4492kg of products. Where does the extra weight
come from?

Obviously from the oxygen used to burn the fuel, which comes to

1000000/142 * 15.5 * 32g = 3492957g = 3492kg

The assumptions made won't have much effect on the calculation, and
neither will rounding error. The biggest error probably comes from the
100% combustion; from the smell they make, most jets give out quite a
bit of unburnt fuel. Right ball park though.

HTH

John

[M. J. Powell]

In message >, John
Mullen > writes
>Clark wrote:
>> "M. J. Powell" > wrote in news:Uj2uRaHhqv2$EwF0
>> @pickmere.demon.co.uk:
>>
>>>In message >, Clark
> writes
>>
>>>>Do your homework Mike and you will be able to answer your question.
>>>
>>>I asked because I don't want to do more homework!
>>>
>> So, how's that working out for ya?
>>
>
>Assume jet fuel is pure decane, to simplify the calculation.
>
>C10H22 + 15.5O2 -> 11H20 + 10CO2
>
>1 mole C10H22 = 142g
>1 mole H2O = 18g
>1 mole CO2 = 44g
>
>So for every tonne (1000kg) of decane burned, assuming 100% combustion,
>you will have
>
>1000000/142 * 11 * 18g = 1394360g = 1394kg water
>1000000/142 * 10 * 44 = 3098590g = 3098kg carbon dioxide
>
>Which comes to about 4492kg of products. Where does the extra weight
>come from?
>
>Obviously from the oxygen used to burn the fuel, which comes to
>
>1000000/142 * 15.5 * 32g = 3492957g = 3492kg
>
>The assumptions made won't have much effect on the calculation, and
>neither will rounding error. The biggest error probably comes from the
>100% combustion; from the smell they make, most jets give out quite a
>bit of unburnt fuel. Right ball park though.
>
>HTH

It does indeed. Thank you very much.

Mike
--
M.J.Powell

[Tarver Engineering]

"M. J. Powell" > wrote in message
...
> In message >, John
> Mullen > writes
> >Clark wrote:
> >> "M. J. Powell" > wrote in
news:Uj2uRaHhqv2$EwF0
> >> @pickmere.demon.co.uk:
> >>
> >>>In message >, Clark
> > writes
> >>
> >>>>Do your homework Mike and you will be able to answer your question.
> >>>
> >>>I asked because I don't want to do more homework!
> >>>
> >> So, how's that working out for ya?
> >>
> >
> >Assume jet fuel is pure decane, to simplify the calculation.
> >
> >C10H22 + 15.5O2 -> 11H20 + 10CO2
> >
> >1 mole C10H22 = 142g
> >1 mole H2O = 18g
> >1 mole CO2 = 44g
> >
> >So for every tonne (1000kg) of decane burned, assuming 100% combustion,
> >you will have
> >
> >1000000/142 * 11 * 18g = 1394360g = 1394kg water
> >1000000/142 * 10 * 44 = 3098590g = 3098kg carbon dioxide
> >
> >Which comes to about 4492kg of products. Where does the extra weight
> >come from?
> >
> >Obviously from the oxygen used to burn the fuel, which comes to
> >
> >1000000/142 * 15.5 * 32g = 3492957g = 3492kg
> >
> >The assumptions made won't have much effect on the calculation, and
> >neither will rounding error. The biggest error probably comes from the
> >100% combustion; from the smell they make, most jets give out quite a
> >bit of unburnt fuel. Right ball park though.
> >
> >HTH
>
> It does indeed. Thank you very much.

Just remember not to confuse a metric ton with 2000 lbs.

[M. J. Powell]

In message >, Tarver Engineering
> writes
>
>"M. J. Powell" > wrote in message
...
>> In message >, John
>> Mullen > writes
>> >Clark wrote:
>> >> "M. J. Powell" > wrote in
>news:Uj2uRaHhqv2$EwF0
>> >> @pickmere.demon.co.uk:
>> >>
>> >>>In message >, Clark
>> > writes
>> >>
>> >>>>Do your homework Mike and you will be able to answer your question.
>> >>>
>> >>>I asked because I don't want to do more homework!
>> >>>
>> >> So, how's that working out for ya?
>> >>
>> >
>> >Assume jet fuel is pure decane, to simplify the calculation.
>> >
>> >C10H22 + 15.5O2 -> 11H20 + 10CO2
>> >
>> >1 mole C10H22 = 142g
>> >1 mole H2O = 18g
>> >1 mole CO2 = 44g
>> >
>> >So for every tonne (1000kg) of decane burned, assuming 100% combustion,
>> >you will have
>> >
>> >1000000/142 * 11 * 18g = 1394360g = 1394kg water
>> >1000000/142 * 10 * 44 = 3098590g = 3098kg carbon dioxide
>> >
>> >Which comes to about 4492kg of products. Where does the extra weight
>> >come from?
>> >
>> >Obviously from the oxygen used to burn the fuel, which comes to
>> >
>> >1000000/142 * 15.5 * 32g = 3492957g = 3492kg
>> >
>> >The assumptions made won't have much effect on the calculation, and
>> >neither will rounding error. The biggest error probably comes from the
>> >100% combustion; from the smell they make, most jets give out quite a
>> >bit of unburnt fuel. Right ball park though.
>> >
>> >HTH
>>
>> It does indeed. Thank you very much.
>
>Just remember not to confuse a metric ton with 2000 lbs.

I'm used to 2240 lbs/ton.

Mike
--
M.J.Powell

[B2431]

>From: John Mullen

>> "M. J. Powell" > wrote in

>>>In message >, Clark
> writes
>>
>>
>>>>Do your homework Mike and you will be able to answer your question.
>>>
>>>I asked because I don't want to do more homework!
>>>
>>
>> So, how's that working out for ya?
>>
>
>Assume jet fuel is pure decane, to simplify the calculation.
>
>C10H22 + 15.5O2 -> 11H20 + 10CO2
>
>1 mole C10H22 = 142g
>1 mole H2O = 18g
>1 mole CO2 = 44g
>
>So for every tonne (1000kg) of decane burned, assuming 100% combustion,
>you will have
>
>1000000/142 * 11 * 18g = 1394360g = 1394kg water
>1000000/142 * 10 * 44 = 3098590g = 3098kg carbon dioxide
>
>Which comes to about 4492kg of products. Where does the extra weight
>come from?
>
>Obviously from the oxygen used to burn the fuel, which comes to
>
>1000000/142 * 15.5 * 32g = 3492957g = 3492kg
>
>The assumptions made won't have much effect on the calculation, and
>neither will rounding error. The biggest error probably comes from the
>100% combustion; from the smell they make, most jets give out quite a
>bit of unburnt fuel. Right ball park though.
>
>HTH
>
>John
>
Does PETA know you use moles in chemistry?

Dan, U. S. Air Force, retired

[John Mullen]

M. J. Powell wrote:
> In message >, Tarver Engineering
> > writes
>
>>
>> "M. J. Powell" > wrote in message
>> ...
>>
>>> In message >, John
>>> Mullen > writes
>>> >Clark wrote:
>>> >> "M. J. Powell" > wrote in
>>
>> news:Uj2uRaHhqv2$EwF0
>>
>>> >> @pickmere.demon.co.uk:
>>> >>
>>> >>>In message >, Clark
>>> > writes
>>> >>
>>> >>>>Do your homework Mike and you will be able to answer your question.
>>> >>>
>>> >>>I asked because I don't want to do more homework!
>>> >>>
>>> >> So, how's that working out for ya?
>>> >>
>>> >
>>> >Assume jet fuel is pure decane, to simplify the calculation.
>>> >
>>> >C10H22 + 15.5O2 -> 11H20 + 10CO2
>>> >
>>> >1 mole C10H22 = 142g
>>> >1 mole H2O = 18g
>>> >1 mole CO2 = 44g
>>> >
>>> >So for every tonne (1000kg) of decane burned, assuming 100% combustion,
>>> >you will have
>>> >
>>> >1000000/142 * 11 * 18g = 1394360g = 1394kg water
>>> >1000000/142 * 10 * 44 = 3098590g = 3098kg carbon dioxide
>>> >
>>> >Which comes to about 4492kg of products. Where does the extra weight
>>> >come from?
>>> >
>>> >Obviously from the oxygen used to burn the fuel, which comes to
>>> >
>>> >1000000/142 * 15.5 * 32g = 3492957g = 3492kg
>>> >
>>> >The assumptions made won't have much effect on the calculation, and
>>> >neither will rounding error. The biggest error probably comes from the
>>> >100% combustion; from the smell they make, most jets give out quite a
>>> >bit of unburnt fuel. Right ball park though.
>>> >
>>> >HTH
>>>
>>> It does indeed. Thank you very much.
>>
>>
>> Just remember not to confuse a metric ton with 2000 lbs.
>
>
> I'm used to 2240 lbs/ton.
>
> Mike

Well you can just scale it up, it's a ratio. 1 ton of fuel + 3.5 tons
oxygen gives 1.4 tons water and 3.1 tons carbon dioxide, whichever
version of the ton/tonne you use, and there are at least three that I
know of!

John

[M. J. Powell]

In message >, John
Mullen > writes
>>>
>>>
>>> Just remember not to confuse a metric ton with 2000 lbs.
>> I'm used to 2240 lbs/ton.
>> Mike
>
>Well you can just scale it up, it's a ratio. 1 ton of fuel + 3.5 tons
>oxygen gives 1.4 tons water and 3.1 tons carbon dioxide, whichever
>version of the ton/tonne you use, and there are at least three that I
>know of!

Yes, thank you. It was just the round figures that I wanted to bolster
up my general theory about the weather.

Mike
--
M.J.Powell

[Harry Andreas]

In article >, John
Mullen > wrote:

> The assumptions made won't have much effect on the calculation, and
> neither will rounding error. The biggest error probably comes from the
> 100% combustion; from the smell they make, most jets give out quite a
> bit of unburnt fuel. Right ball park though.

You only smell them when they're on the ground (for the most part)
and operating at partial throttle and low Tit.
When they're operating at higher throttle settings and higher
temps the combustion is more complete.
The higher combustion temp also forms NOx's.

--
Harry Andreas
Engineering raconteur