3 replies to this topic

[craag]

Here is the problem statement.

n-Hexane is burned with excess air. An analysis of the product gas yields the following dry-basis molar composition: 6.9% CO2, 2.1% CO, 0.265% C6H14 (+O2 and N2). The stack gas emerges at 760 mm Hg. Calculate the percentage conversion of hexane, the percentage excess air fed to the burner, and the dew-point of the stack gas, taking water to be the only condensable species.

I don't even know how to begin.

I tried to draw the schematic, but it is a mess. I have 2 streams coming in (hexane and air) and 3 streams coming out (the dry-basis stuff listed; water and oxygen; and I put nitrogen by itself because I thought it would make it easier to balance)

It doesn't have any flow rates, so I know that I can assume a basis, but that's about as far as I get.

It says that the gas emerges at 760 mm Hg, so I was thinking that I can start by calculating the partial pressures. But I can't tell if that is for the 3 species listed (carbon dioxide, carbon monoxide, hexane) or for everything (including oxygen, water, nitrogen).

Furthermore, I don't understand where the carbon monoxide even comes from, because I balance the chemical reaction (combustion of hexane) the only products are water and carbon dioxide.

Please help!

[craag]

Okay, I made awesome progress.

I assumed the product gas to be 100 mol. Therefore, I now have:

6.9 mol CO2
2.1 mol CO
0.265 mol C6H14
90.735 mol Air (which is composed of N2 and O2)


Then, since all of the carbon in the system comes from the hexane, I was able to solve for the Hexane going into the system.

I was then able to solve for the amount of H2O leaving the system, because I know how much hexane was reacted.

I was also able to solve for the amount of Theoretical Air, because I know how much hexane needed to be burned.



But I have hit another road block... How can I solve for the amount of air fed?

If I do an atomic balance for oxygen I have 2 unknown variables (moles of air fed; mole fraction of O2 out)

Perhaps I can use the given pressure of 760 mm Hg somewhere?


If anyone is reading this and can help me out it would be appreciated

Thanks!

Edited by craag, 04 March 2011 - 12:56 AM.

[breizh]

Air = O2+N2

O2/N2 = 21/79
will it help?
Breizh

[Art Montemayor]

Craag:

This is nothing more than a “down-and-dirty”, simple stoichiometric combustion problem in 1^st year Chem Engineering. You are doing good progress in buckling down and applying your common sense and intuition. These type of problems are typical of engineering know-how requirements because they do not require any differential or integral calculus or any chemistry and physics beyond simple burning – or “combustion” as we call it. Most all engineering problems involve simple and basic principles. You’ve done well in setting a basis for your flue gas and finding the basis for establishing a mass balance. You have used your common sense to spot the basic simple fact that the only way that CO₂ and CO can appear in the outlet flue is because they came from the combustion of the hexane fuel. It was that simple. The rest of the problem is also THAT simple also – but with a slight twist.

Note that the oxygen appearing in both oxides also has to come from one source: guess what? Breizh has given you the basic clue. Use it to find the amount of air used in that combustion.

HOWEVER, also make note that you are not only consuming oxygen, but you also are reporting the presence of excess AIR (composed of oxygen & nitrogen) in the flue as well. Use your common sense to contemplate what happens when you take air (a mixture of O₂ & N₂) and use only the O₂ to produce some oxides. Correcto-mundo! You will have an inert left over and joining the other components in the flue. Therefore, the “excess air” that you report in the flue gas exiting the combustion chamber IS NOT REALLY EXCESS AIR. It simply can’t be, using common sense. Regardless of what you may be told, the inert nitrogen portion of the air that contributed to the combustion of hexane MUST COME OUT IN THE FLUE GAS. You may also be feeding what is called “excess air” to the combustion chamber – but that mixture will remain as such and exit in the same quantity (& composition) as it entered --- since it was an “excess” and didn’t participate in the combustion.

Therefore, using God-given common sense, it is obvious that the products of combustion leaving the chamber will have the oxides, un-burnt fuel, PLUS the excess air fed to the chamber (that didn’t react) AND the inert (nitrogen?) that was left over from the portion of air that DID participate in the combustion. You must fully understand and know how to apply this reasoning by knowing and understanding what happens during the combustion of a fuel. Knowing that, then you apply the engineering logic and algorithm necessary to identify and resolve the unknown quantities I’ve described above. By doing that, you can complete the mass balance around the combustion chamber and go on to resolve the rest of the problem – which is a lot easier.

Good luck.