Question

S. N- Bu tane Fuel is burned with 200 percent theoretical air during a combustion process Determine the mass fraction of each product .Calculate air- fuel ratio on mole basis and volumetric analysis of the products

Answer 1

Butane

C₄H₁₀

Balenced equation for combustion is

200 % theretical air :

$C_4H10+13(O_2+3.76N_2) \rightarrow 4CO_2+5H_2O+6.5O_2+13*3.76N_2$

Mass of CO₂ = 44*4 = 176

Mass of H₂O = 18*5 = 90

Mass of Oxygen = 6.5*32 = 208

Mass of Nitrigen = 3.76*13*28 = 1368.64

total mass = 1842.64

Mass fraction of CO₂ = 0.0955

Mass fraction of H₂O = 0.0488

Mass fraction of Oxygen = 0.1129

Mass fraction of Nitrigen = 0.7428

Molar airfuel ratio = 13

Volumetric analysis:

Total moles in products = 4+5+6.5+3.75*13 = 64.38

Molar ratio of CO₂ = 4/64.38 = 0.0621

Molar ratio of water= 5/64.38 = 0.07766

Molar ratio of oxygen= 6.5/64.38 = 0.1010

Molar ratio of nitrogen= 4/64.38 = 0.7592