- #1
PCal
- 31
- 2
- Homework Statement
- A fuel gas consists of 75% butane (C4H10), 10% propane (C3H8) and 15% butene (C4H8) by volume. It is to be fed to the combustion chamber in 10% excess air at 25ºC, where 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. Data: Net calorific value (MJ m–3) at 25ºC of: Butane (C4H10) = 111.7 MJ m–3 Butene (C4H8) = 105.2 MJ m–3 Propane (C3H8) = 85.8 MJ m–3 Air is 21% oxygen, 79% nitrogen by volume and 23.3% oxygen and 76.7% nitrogen by mass. Atomic mass of C = 12, O = 16, N=14 and H = 1.
Calculate actual fuel:air ratio by volume and by mass
I've done this a few times and got different answers every time, I think I might be correct with this one but I'm not sure so any help would be appreciated!
- Relevant Equations
- Balanced reactions
Butane C4H10 + 6.5O2 = 4CO2 + 5H20
Propane C3H8 + 5O2 = 3C02 + 4H2O
Butene C4H8 + 6O2 = 4CO2 + 4H20
By Volume:
Butane 0.75x6.5= 4.875 moles of O2
Propane 0.1x5= 0.5 moles of O2
Butene 0.15x6= 0.9 moles of O2
Total O2 4.875+0.5+9= 6.275 moles
Air required 6.275/0.21=29.88 moles/m^3
With 10% excess air 29.88x1.1= 32.868moles
1 mole fuel:32.87 moles of air
By Mass:
CO2 Moles (0.75x4)+(0.1x3)+(0.15x4)= 3.9
H2 Moles (0.75x10)+(0.1x8)+(0.15x8)= 9.5
O2 Moles (0.75x13)+(0.1x10)+(0.15x12)= 12.55
Nitrogen (12.55/21)x79= 47.21
Moles x Molecular mass
CO2 3.9x12= 46.8
H2 9.5x2= 19
O2 12.55x32= 401.6
Nitrogen= (401.6/23.3)x76.7=1322
Total Fuel= 46.8+19=65.8
Total Air= 401.6+1322=1723.6 +10% excess= 1895.96
1895.96/65.8= 1 mole fuel:28.81 mole of air
Butane 0.75x6.5= 4.875 moles of O2
Propane 0.1x5= 0.5 moles of O2
Butene 0.15x6= 0.9 moles of O2
Total O2 4.875+0.5+9= 6.275 moles
Air required 6.275/0.21=29.88 moles/m^3
With 10% excess air 29.88x1.1= 32.868moles
1 mole fuel:32.87 moles of air
By Mass:
CO2 Moles (0.75x4)+(0.1x3)+(0.15x4)= 3.9
H2 Moles (0.75x10)+(0.1x8)+(0.15x8)= 9.5
O2 Moles (0.75x13)+(0.1x10)+(0.15x12)= 12.55
Nitrogen (12.55/21)x79= 47.21
Moles x Molecular mass
CO2 3.9x12= 46.8
H2 9.5x2= 19
O2 12.55x32= 401.6
Nitrogen= (401.6/23.3)x76.7=1322
Total Fuel= 46.8+19=65.8
Total Air= 401.6+1322=1723.6 +10% excess= 1895.96
1895.96/65.8= 1 mole fuel:28.81 mole of air