Jet Fuel

"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.

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

"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

In message , Clark
writes
"M. J. Powell" wrote in
:

In message , Clark
writes
"M. J. Powell" wrote in newsKqxbJmQDe2
:

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

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

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

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

"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.

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

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