Learn & Understand

Where 11.5, 34.5, and 4.32 Actually Come From

The constants 11.5, 34.5, and 4.32 in this calculator's formula look like they were pulled from a reference table with no explanation - but each one falls directly out of atomic weights and air's known composition, the same way the combustion balancing formula in this site's advanced combustion category derives from simple atom counting.

Starting From the Balanced Combustion Reactions

Each element in a fuel's ultimate analysis combines with oxygen in a fixed, known atomic ratio: one atom of carbon needs one molecule of O2 to become CO2, four atoms of hydrogen need one molecule of O2 to become two molecules of water, and one atom of sulfur needs one molecule of O2 to become SO2. Converting these atomic ratios into a mass-basis requirement - since the ultimate analysis gives carbon, hydrogen, and sulfur as mass percentages, not mole counts - requires dividing by each element's atomic weight and multiplying by oxygen's molecular weight, which is exactly where each constant in the formula originates.

Deriving the Carbon Constant

Carbon's atomic weight is 12; oxygen's molecular weight (O2) is 32. One kilogram of carbon therefore needs 32/12 = 2.667 kg of oxygen to fully oxidize to CO2. Since air is only about 23.2% oxygen by mass, the oxygen requirement has to be scaled up to a full air requirement: 2.667 ÷ 0.232 ≈ 11.5 - exactly the carbon coefficient in the formula, arrived at purely from atomic weights and air's known oxygen mass fraction.

Deriving the Hydrogen Constant (and Why It Subtracts Oxygen)

Hydrogen's atomic weight is 1; four atoms of hydrogen (4 kg) combine with one O2 molecule (32 kg) to form two water molecules, meaning 1 kg of hydrogen needs 32/4 = 8 kg of oxygen. Scaled up to air the same way as carbon (8 ÷ 0.232 ≈ 34.5), this produces the formula's hydrogen coefficient. The formula subtracts oxygen already present in the fuel itself (O/8) because any oxygen the fuel already contains reduces the additional oxygen that has to be supplied from combustion air - fuel-bound oxygen effectively pre-supplies part of the hydrogen's own oxidation requirement, dividing by 8 for the same atomic-weight reasoning used to derive the hydrogen coefficient itself.

Deriving the Sulfur Constant

Sulfur's atomic weight is 32; one atom of sulfur (32 kg) combines with one O2 molecule (32 kg) to form SO2, a 1:1 mass ratio, meaning 1 kg of sulfur needs 1 kg of oxygen. Scaled to air: 1 ÷ 0.232 ≈ 4.32 - the formula's sulfur coefficient, following exactly the same derivation logic as carbon and hydrogen.

Applying This Understanding to the Formula

Recognizing that every constant in this formula traces back to nothing more exotic than atomic weights and air's roughly 23.2%-oxygen-by-mass composition makes the formula far less of a black box - and confirms it will hold for any hydrocarbon-and-sulfur fuel's ultimate analysis, not just the specific worked example on the calculator page, since the underlying atomic chemistry it's built from doesn't change from one fuel to another.