Question

One kg of octane (C₈H₁₈) is burned with 200% theoretical air. Assuming complete combustion determine : 

(i) Air-fuel ratio 

(ii) Dew point of the products at a total pressure 100 kPa.

Answer 1

. The equation for the combustion of C₈H₁₈ with theoretical air is

For 200% theoretical air the combustion equation would be

C₈H₁₈ + (2) (12.5) O₂ + (2) (12.5) \(\left(\cfrac{79}{21}\right)\) N₂

→ 8CO₂ + 9H₂O + (1) (12.5) O₂ + (2) (12.5)   \(\left(\cfrac{79}{21}\right)\) N₂

Mass of fuel = (1) (8 × 12 + 1 × 18) = 114 kg/mole

Mass of air = (2) (12.5) \(\left(1+\cfrac{79}{21}\right)\) 28.97 = 3448.8 kg/mole of fuel

(i) Air-fuel ratio :

Air-fuel ratio,

(ii) Dew point of the products, t_dp : 

Total number of moles of products

= 8 + 9 + 12.5 + (2) (12.5) \(\left(\cfrac{79}{21}\right)\) = 123.5 moles/mole fuel

Mole fraction of H₂O = \( \cfrac{9}{123.5}\) = 0.0728

Partial pressure of H₂O = 100 × 0.0728 = 7.28 kPa 

The saturation temperature corresponding to this pressure is 39.7°C which is also the dewpoint temperature. 

Hence t_dp = 39.7°C