[Alist_005]

Dear All,

I need a help in combustion calculation. I have the following information:

The following gaseous stream shall be combusted with the following composition:

Total Flow: 368 Nm3/hr, inpt. Temp. 40 °C, Heat content: 150164,7 JK/Nm3, MW of stream: 86 kg/kmol,

Composition in               vol.%    Nm3/hr
C7H16                           10,34     38,16
C6H14                           17,24     63,61
C6H5CH3                       27,58     101,77
C8H10                           18,96      69,97
C12H24                         5,172      19,08
S                                  20,68       76,33

Can you help me in finding the followings: Air required to burn above, outlet gas composition and combustion temperature.

I would be glad if you can help me to solve the problem.

thanks a lot for your help.

Regards,
Alist_005

[Hunter2]

You should show some own work first.

Hint: Develop the combustion equation of each component. Ideal condition convert the volume to mol.
With this you can get the composition and also the required air.

[Alist_005]

You should show some own work first.

Hint: Develop the combustion equation of each component. Ideal condition convert the volume to mol.
With this you can get the composition and also the required air.

The amounts are already in the normal conditions. Here are the following combustion equations.

C7H16 + 11O2 = 7CO2 + 8H2O
C6H14+ 9,5O2 = 6CO2 + 7H2O
C6H5CH3 + 9O2 = 7CO2 + 4H2O
C8H10 + 10,5O2 = 8CO2 + 5H2O
C12H24 + 24O2 = 12CO2 + 12H2O
S + O2 = SO2

I can explain what I can do out of this, to estimate the air requirement. Please, correct me if I am wrong. Based on the equations above I can get fuel to air ratio for each constituent and then by multipliying the amount of each species I can estimate the required air. Would that be correct? here below. Please let me know.

1C7H16+11O2=7CO2+8H2O      15,18
1C6H14+9,50O2=6CO2+7H2O      15,24
1C7H8+9O2=7CO2+4H2O      13,51
1C8H10+10,5O2=8CO2+5H2O      13,68
1C12   H24+24O2=12C1O2   +   12   H   2   O      19,72
1   S   1         +   1   O   2   =   1   S   1   O   2                     4,32

[Alist_005]

You should show some own work first.

Hint: Develop the combustion equation of each component. Ideal condition convert the volume to mol.
With this you can get the composition and also the required air.

The amounts are already in the normal conditions. Here are the following combustion equations.

C7H16 + 11O2 = 7CO2 + 8H2O
C6H14+ 9,5O2 = 6CO2 + 7H2O
C6H5CH3 + 9O2 = 7CO2 + 4H2O
C8H10 + 10,5O2 = 8CO2 + 5H2O
C12H24 + 24O2 = 12CO2 + 12

You should show some own work first.

Hint: Develop the combustion equation of each component. Ideal condition convert the volume to mol.
With this you can get the composition and also the required air.

The amounts are already in the normal conditions. Here are the following combustion equations.

C7H16 + 11O2 = 7CO2 + 8H2O
C6H14+ 9,5O2 = 6CO2 + 7H2O
C6H5CH3 + 9O2 = 7CO2 + 4H2O
C8H10 + 10,5O2 = 8CO2 + 5H2O
C12H24 + 24O2 = 12CO2 + 12H2O
S + O2 = SO2

I can explain what I can do out of this, to estimate the air requirement. Please, correct me if I am wrong. Based on the equations above I can get fuel to air ratio for each constituent and then by multipliying the amount of each species I can estimate the required air. Would that be correct? here below. Please let me know.

1C7H16+11O2=7CO2+8H2O      15,18
1C6H14+9,50O2=6CO2+7H2O      15,24
1C7H8+9O2=7CO2+4H2O      13,51
1C8H10+10,5O2=8CO2+5H2O      13,68
1C12   H24+24O2=12CO2+12H2O      19,72
1S1+1O2=SO2                     4,32

H2O
S + O2 = SO2

I can explain what I can do out of this, to estimate the air requirement. Please, correct me if I am wrong. Based on the equations above I can get fuel to air ratio for each constituent and then by multipliying the amount of each species I can estimate the required air. Would that be correct? here below. Please let me know.

1C7H16+11O2=7CO2+8H2O      15,18
1C6H14+9,50O2=6CO2+7H2O      15,24
1C7H8+9O2=7CO2+4H2O      13,51
1C8H10+10,5O2=8CO2+5H2O      13,68
1C12   H24+24O2=12C1O2   +   12   H   2   O      19,72
1   S   1         +   1   O   2   =   1   S   1   O   2                     4,32

[Hunter2]

For the first compound you need 11 mol O2 . You have 38,16 Nm³ in volume How much moles correspond this. This times 11 gives the actual mole of oxygen. Air containd 21% oxygen so this has also be considered.

[Alist_005]

For the first compound you need 11 mol O2 . You have 38,16 Nm³ in volume How much moles correspond this. This times 11 gives the actual mole of oxygen. Air containd 21% oxygen so this has also be considered.

I convert the Nm3 for C7H16 to mole in the following: (38,16 Nm3/hr)/(22,414 Nm3/kmol)*1000 mol/kmol=1676,64 mol/hr, as per equation C7H16 + 11O2 = 7CO2 + 8H2O this means for heptane the air required will be as following:

C7H16  1676,64 mol/hr * 11 / 0,21 = 87824,22 mol/hr /1000 kmol/mol / 22,414 Nm3/kmol = 1968,49 Nm3/hr

is that correct? please, let me know.

[Hunter2]

I think its correct

[Alist_005]

I think its correct

May I ask a further, how do I calculate the equivalnece ratio. As the air calculated was the stoichiometric, is that right? and the equivalence ratio equals to:

equivalence ratio= air actual / air stoichiometric

Can you tell me how the "air actual" is calclated.

[Hunter2]

You have to caluculate for all compounds and the summery is the actual air.

[Alist_005]

You have to caluculate for all compounds and the summery is the actual air.

In the following as per equations below and respective compositions:
                                                                       
C7H16 + 11O2 = 7CO2 + 8H2O                   
C6H14+ 9,5O2 = 6CO2 + 7H2O                 
C6H5CH3 + 9O2 = 7CO2 + 4H2O
C8H10 + 10,5O2 = 8CO2 + 5H2O
C12H24 + 24O2 = 12CO2 + 12H2O
S + O2 = SO2

             Nm3/hr          mol/hr                                     stoich air in mol/hr         stoich air in Nm3/hr            equivalence ratio
C7H16  38,16         38,16/22,414*1000=1702,5      1702,5*11/0,21=89178,95         1998,85            14735,42/1998,85=7,37
C6H14   63,61        63,61/22,414*1000=2837,95                     128383,83                2877,59                              5,12
C6H5CH3 101,77    101,77/22,414*1000=4540,46                    194591,39               4361,57                              3,37
C8H10     69,97       69,97/22,414*1000=3121,7                       156085,48               3498,5                                4,21
C12H24   19,08       19,08/22,414*1000=851,25                        72964,60                1635,42                               9,0
S             76,33       76,33/22,414*1000=3405,46                     16216,48                 363,47                                40,54
                                                                            TOTAL:       657420,74               14735,42

was that correct? that means I have an actual air of 14735,42 Nm3/hr and stoichiometric air for C7H16 1998,85 and in this case equivalence ratio for C7H16 in the mixture would be 7,37 would that be right?  and if it is ok if the equivalence ratio is much higher such as in Sulphur which is 40,54

[Enthalpy]

The question doesn't make much sense to me - maybe I didn't grasp something.

To me, vol% and Nm³/h suggest gases, and "gaseous stream" even tells it. While C₁₂H₂₄ isn't volatile enough (0.0159 mm Hg at 25 °C for 1-dodecene), I don't imagine gaseous sulphur at 40°C, sorry.

And under conditions where sulphur vapour has a significant pressure, it's not composed of S₁:
"Berkowitz showed that vapor contains all molecules S_n, 2<n<10"
http://web.gps.caltech.edu/~vijay/Papers/Chemistry/Meyer-76.pdf

I don't feel necessary neither to write a balanced reaction equation for each component. Compute how many moles of C, H and (if any sensible) S are in the fuel to evaluate globally the amount of oxygen hence air.

[Alist_005]

The question doesn't make much sense to me - maybe I didn't grasp something.

To me, vol% and Nm³/h suggest gases, and "gaseous stream" even tells it. While C₁₂H₂₄ isn't volatile enough (0.0159 mm Hg at 25 °C for 1-dodecene), I don't imagine gaseous sulphur at 40°C, sorry.

And under conditions where sulphur vapour has a significant pressure, it's not composed of S₁:
"Berkowitz showed that vapor contains all molecules S_n, 2<n<10"
http://web.gps.caltech.edu/~vijay/Papers/Chemistry/Meyer-76.pdf

I don't feel necessary neither to write a balanced reaction equation for each component. Compute how many moles of C, H and (if any sensible) S are in the fuel to evaluate globally the amount of oxygen hence air.

Thanks fo reply, here I undenstand S is actually solvents, I am assuming it to be Sulphur, do you have any suggestion what I can take instead. for the C12H24 I do not know perhaps I made a mistake in taking this compound. But it was actually a gasoline fraction, where I found it to have almost 80 % of C12 molecules, therfore I considered C12H24. here I am free to chosse any of C12 coumpounds, if you have any in you literature, would be glad for your suggestion.

I did not get your last querry. Do you mean to get the air amount out of this below equation? the equation is actually counts also other constitients but here as it was only hydrocarbons I only consider the above.

Oxyden requirement=4,76[*(n+m/4)CnHm]=(7+16/4)*38,16+(6+14/4)*63,61+(7+8/4)*101,77+(8+10/4)*69,97+(12+24/4)*19,08+76,33)=3047,13 Nm3/hr

Air requirement= Oxygen requirement/0,21=3047,13/0,21 = 14510,14 Nm3/hr, is that waht you mean?

can you let me if the equivalence ratio calculations were correct? as I need to also estimate the outlet compositions.

[Enthalpy]

I didn't check your detailed figures, but the complexity of computing the oxygen amount is about what it should be. Or you can total the number of carbon moles contributed by all fuel components, the number of hydrogen moles, and deduce the moles of O₂. This tells you as well the outlet composition.

I still don't understand the "S" in the composition. Nor do I imagine a circumstance where someone voluntarily burns sulphur mixed with hydrocarbons.

The C₁₂ stays a mystery to me. No C₁₂ molecule has a vapour pressure of 0.05atm at 40°C. The vol% might be an analysis result from very dilute vapour in a carrier gas, or maybe a vapour at a very low pressure, but not at 1atm or above, as most combustions are carried.

An important proportion of C₁₂ is unexpected in gasoline. The others C₆, C₇, C₈ fit better, while C₁₂ is an abundant constituent of Diesel oil rather. The composition looks very artificial anyway, as there should be many more formulas, and certainly not a gap between C₈ and C₁₂.