- #1
Mitch1
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C4H10 + 4.5O2 -> 4Co2 + 5H2O1. Homework Statement
A fuel gas consists of 75% butane (C4 H10 ), 10% propane (C3 H8 ) and butene (C4 H8 ) by volume.
It is to be fed to the combustion chamber in 10% excess air at 25ºwhere it is completely burnt to carbon dioxide and water. The flue gases
produced are to be used to generate 5 bar steam from water at 90ºC.
Determine the maximum flame temperature.
Actual oxygen supplied = theoretical x excess air
Heat content = enthalpy x no of moles
lower valvue for propane 2046 KJ/mol, butane 2660 KJ/mol
(.75*2660)+(.1*2046)+(.15*2660)
1995+204.6+399=2598.6KJ = 2598.6KJ/mol fuel
C4H10 + 4.5O2 -> 4Co2 + 5H2O
1mol 4.5mol 4 mol 5 mol
.75 mol 3.375mol 3mol 3.75mol
C3H8 + 5O2 -> 3Co2 + 4H2O
1mol 5mol 3 mol 4 mol
.1 mol .5mol .3mol .4mol
C4H8 + 6O2 -> 4Co2 + 4H2O
1mol 6mol 4 mol 4 mol
.15 mol .9mol .6mol .6mol
4.775 O2 Req
actual O2 = theo x excess air
4.775 x 1.1 = 5.2525 mol
N2 = 5.2525 x 3.76=19.749 moles in flue gas (3.76 is 79%/21% conversion factor from O2 to air to N2 )
the problem is the next bit once i work out the heat content the enthalpy table goes up to 2200 Deg C however when i work out the heat content it it below what i worked out in the first steps (2598.6KJ/mol fuel) could someone check my calcs above to ensure there are correct as i believe it should be somewhat less to tie into the enthalpy table to give me a max flame temp.
thanks
A fuel gas consists of 75% butane (C4 H10 ), 10% propane (C3 H8 ) and butene (C4 H8 ) by volume.
It is to be fed to the combustion chamber in 10% excess air at 25ºwhere it is completely burnt to carbon dioxide and water. The flue gases
produced are to be used to generate 5 bar steam from water at 90ºC.
Determine the maximum flame temperature.
Homework Equations
Actual oxygen supplied = theoretical x excess air
Heat content = enthalpy x no of moles
The Attempt at a Solution
lower valvue for propane 2046 KJ/mol, butane 2660 KJ/mol
(.75*2660)+(.1*2046)+(.15*2660)
1995+204.6+399=2598.6KJ = 2598.6KJ/mol fuel
C4H10 + 4.5O2 -> 4Co2 + 5H2O
1mol 4.5mol 4 mol 5 mol
.75 mol 3.375mol 3mol 3.75mol
C3H8 + 5O2 -> 3Co2 + 4H2O
1mol 5mol 3 mol 4 mol
.1 mol .5mol .3mol .4mol
C4H8 + 6O2 -> 4Co2 + 4H2O
1mol 6mol 4 mol 4 mol
.15 mol .9mol .6mol .6mol
4.775 O2 Req
actual O2 = theo x excess air
4.775 x 1.1 = 5.2525 mol
N2 = 5.2525 x 3.76=19.749 moles in flue gas (3.76 is 79%/21% conversion factor from O2 to air to N2 )
the problem is the next bit once i work out the heat content the enthalpy table goes up to 2200 Deg C however when i work out the heat content it it below what i worked out in the first steps (2598.6KJ/mol fuel) could someone check my calcs above to ensure there are correct as i believe it should be somewhat less to tie into the enthalpy table to give me a max flame temp.
thanks