ABE TUTORIALS

AGRICULTURAL AND BIOSYSTEMS ENGINEERING

PROBLEM:
DETERMINE THE AIR TO FUEL RATIO BY WEIGHT AND BY VOLUME FOR THE COMPLETE COMBUSTION OF PROPANE (C3H8). ASSUME AIR IS 79% NITROGEN AND 21% OXYGEN.

---

1. The air-to-fuel ratio by weight is (in kg air/kg fuel):

a. 3.6
b. 0.275
c. 15.6
d. 0.0064
e. nota

Click to View Solution

Step 1: Write the balanced combustion equation.
C₃H₈ + 5(O₂ + 3.76N₂) → 3CO₂ + 4H₂O + 18.8N₂

Step 2: Calculate Molar Masses.
Fuel (C₃H₈) = (3 × 12) + (8 × 1) = 44 kg/kmol
Air = 5 moles O₂ + 18.8 moles N₂ = (5 × 32) + (18.8 × 28) ≈ 686.4 kg

Step 3: Calculate AF Ratio by Weight.
AF weight = Mass of Air / Mass of Fuel
AF weight = 686.4 / 44 ≈ 15.6

Result: c. 15.6

2. The air-to-fuel ratio by volume is:

a. 23.8
b. 5.0
c. 0.2
d. 0.0042
e. nota

Click to View Solution

Step 1: Understand Volume Ratio (Avogadro's Law).
For gases at the same temperature and pressure, the volume ratio is equal to the mole ratio.

Step 2: Calculate total moles of air per mole of fuel.
Moles of Oxygen = 5
Total Moles of Air = 5 / 0.21 ≈ 23.8

Step 3: Calculate AF Ratio by Volume.
AF volume = 23.8 moles air / 1 mole fuel = 23.8

Result: a. 23.8