Carbon Dioxide from Burning a Fuel (Factor per Mass of Fuel)

When a fuel burns, the carbon in it becomes carbon dioxide, and the arithmetic is only a ratio of molar masses. Every kilogram of carbon in the fuel becomes 44/12 kilograms of carbon dioxide, the ratio of the molar mass of CO2 to that of carbon, so the emission factor is $EF = w_C\,\frac{44}{12}$, the carbon content $w_C$ times 3.667. The emitted mass is then the factor times the mass of fuel burned, $m_{CO_2} = EF\,m_{fuel}$.

The factors for ships' fuels come from Table 21 of the International Maritime Organization's Fourth Greenhouse Gas Study (2020), taken from the 2018 EEDI Guidelines, in grams of CO2 per gram of fuel: 3.114 for heavy fuel oil (carbon content 0.8493) and for low-sulphur heavy fuel oil, 3.206 for marine diesel and gas oil (0.8744), 2.750 for LNG (0.7500) and 1.375 for methanol (0.3750). Methanol's factor is the lowest because so little of its mass is carbon. The factor is a number of kilograms of CO2 per kilogram of fuel, so it works in any mass unit: a tonne of heavy fuel oil makes 3.114 tonnes of carbon dioxide. For a fuel sold by the litre or the gallon, the mass is the volume times the density.

The worked examples are the table's own figures: 1 t of heavy fuel oil gives 3.114 t of CO2, 1 t of marine diesel gives 3.206 t, 100 t of LNG gives 275 t and 10 t of methanol gives 13.75 t. A larger case comes from the ship fuel cubic law page: 31.4868805 t of heavy fuel oil burned in a day at 3.114 makes 98.0501 t of CO2. The page also runs backwards, giving the fuel burned from the CO2 and the factor, or the factor from the CO2 and the fuel.

This is the whole of the combustion step and no more. Unburned carbon, methane escaping from the engine (the “slip” of an LNG engine) and the CO2 of making and delivering the fuel are not in it. For a flared gas, where the combustion efficiency and the CO2 already in the gas matter, the flare gas CO2 page keeps a carbon balance instead.