Question 14.3: Determine the air-fuel ratio for the complete combustion of ......
Determine the air-fuel ratio for the complete combustion of C_8H_{18} with theoretical amount of air.
Given: Reactants—Air and C_8H_{18}
Find out: Air fuel ratio for the combustion of C_8H_{18}.
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_xH_y + \left\lgroup x + \frac{y}{4} \right\rgroup (O_2 + 3.78 N_2) \longrightarrow xCO_2 + \frac{y}{2} H_2O + \left\lgroup x + \frac{y}{4} \right\rgroup (3.78 N_2)
Here for C_8H_{18}, x = 8 and y = 18.
Putting these values in the chemical reaction gives:
C_8H_{18}+ \left\lgroup8 + \frac{18}{4} \right\rgroup (O_2 + 3.78 N_2) \longrightarrow 8 CO_2 + \frac{18}{2} H_2O + \left\lgroup8 + \frac{18}{4} \right\rgroup (3.78 N_2)
Simplifying further:
C_8H_{18} + 12.5 ( O_2 + 3.78 N_2) \longrightarrow 8 CO_2 + 9H_2O + 12.5 (3.78 N_2)
The molar air-fuel ratio:
Mole of air = 12.5 moles of oxygen + 12.5 × 3.78 moles of nitrogen = 59.5
AFR mole = \frac{Mole of air}{Mole of fuel} = \frac{59.5}{1} = 59.5
Mass air-fuel ratio:
AFR mass = \frac{Mass of oxygen + Mass of nitrogen}{Mass of fuel} = \frac{\left[12.5 \times 32 + 3.78 \times 12.5 \times 34\right] }{\left[12 \times 8 + 1\times 8\right] } = 15.1
which means 15.1 kg of air is required for a complete combustion of 1 kg of C_8H_{18}.