Question 14.4: Determine the mass and mole fraction of CO2 and water-vapour......
Determine the mass and mole fraction of CO_2 and water-vapour if C_3H_8 is burned with theoretical amount of air.
Given: Combustion reactants: Air and C_3H_8
Find out: Mass and mole fraction of CO_2.
Properties: Molar mass of C, H_2 and O_2 and air are: 12 kg/kmol, 2 kg/kmol, 32 kg/kmol and 29 kg/kmol respectively.
Assumption: (1) Combustion is complete and (2) Combustion products are H_2O, CO_2 and N_2 only.
Chemical reaction taking place for the combustion of a hydrocarbon can be written as-
C_3H_8 + a (O_2 + 3.78 N_2)\longrightarrow bCO_2 + cH_2O + d (3.78 N_2)
Applying conservation of mass (atoms) both sides for Hydrogen, Carbon, Oxygen and Nitrogen gives:
H balance: 2 c = 8 or c = 4.
C balance: b = 3.
O balance: 2 a = z b + c or a = 5.
N balance: 2 × 3.78 a = 2d or d = 18.8
Balanced equation for the combustion will be:
C_3H_5 + 5(O_2 + 3.78 N_2) \longrightarrow 3 CO_2 + 4H_2O + 5(3.78 N_2)
C_3H_8 + (5O_2 + 18.8 N_2) \longrightarrow 3CO_2 + 4H_2O + (18.8 N_2)
for CO_2 = \frac{Mass of CO_2}{Mass of products} = \frac{Mass of CO_2}{Mass of CO_2 + Mass of H_2O + Mass of N_2}
= \frac{\left[3 \times 44\right] }{\left[3 \times 44 + 4\times 18 + 18.8 \times 28\right] } = 0.181
Mass fraction
for H_2O = \frac{Mass of H_2O}{Mass of products} = \frac{Mass of H_2O}{Mass of CO_2 + Mass of H_2O + Mass of N_2}
= \frac{\left[4 \times 18\right] }{\left[3\times 44 + 4 \times 18 + 18.8 \times 28\right] } =0.155