AFR Calculator (Air-Fuel Ratio)
Calculate air-fuel ratio for combustion engines and determine optimal fuel mixture for complete combustion
Calculate Air-Fuel Ratio
Formula: CH₄
AFR: 17.19:1
Molar Mass: 16.04 g/mol
Calculate required air mass
AFR Results
Calculation Details
Fuel: Methane (CH₄)
Stoichiometric AFR: 17.19:1
Formula: CH₄
Mixture analysis: Ideal mixture for complete combustion
Combustion Analysis
Example Calculation
Gasoline Engine Example
Fuel: Gasoline (C₈H₁₈)
Stoichiometric AFR: 14.7:1
Fuel consumed: 5 kg
Question: How much air is needed for complete combustion?
Step-by-Step Solution
Formula: Required air = fuel mass × AFR
Calculation: Required air = 5 kg × 14.7 = 73.5 kg
Answer: 73.5 kg of air needed for complete combustion
Volume: ≈ 60.4 m³ of air at STP (1.217 kg/m³)
Common AFR Values
Lambda (λ) Values
Rich (λ < 1.0)
Excess fuel
Higher power, more emissions
Stoichiometric (λ = 1.0)
Perfect balance
Complete combustion
Lean (λ > 1.0)
Excess air
Better efficiency, lower power
Key Concepts
AFR Formula
mass air / mass fuel
Air-fuel ratio by mass
Stoichiometric
λ = 1.0
Perfect air-fuel balance
Lambda
λ = actual/stoichiometric
Equivalence ratio
Understanding Air-Fuel Ratio
What is Air-Fuel Ratio?
The air-fuel ratio (AFR) indicates the amount of air needed to achieve complete combustion of a given quantity of fuel. It's expressed as the mass ratio of air to fuel required for stoichiometric (complete) combustion.
Why is AFR Important?
- •Determines combustion efficiency and power output
- •Affects exhaust emissions and fuel consumption
- •Critical for engine performance optimization
- •Influences catalytic converter operation
Combustion Chemistry
General Hydrocarbon Combustion
CₐHᵦ + (a + b/4)O₂ → aCO₂ + (b/2)H₂O
Complete combustion with oxygen
Air Composition
21% O₂ + 79% N₂ (by volume)
Air contains ~23% oxygen by mass
Lambda Calculation
λ = (actual AFR) / (stoichiometric AFR)
Equivalence ratio for mixture analysis