Jet Fuel

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

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

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