EnginePage 5 of 7
Sulphur content of a blended tank: a mass balance
Why it matters Burning fuel over the sulphur limit breaks MARPOL and brings fines or detention; the blend decides where this tank may be burned.
Words on this page
- Sulphur content
- how much of the fuel's weight is sulphur, in per cent.
- Remainder
- the old fuel still in the tank before the new fuel goes in.
- Delivery
- the new fuel the supplier pumps into the tank.
- Delivery note
- the supplier's paper that states the fuel's quantity and its sulphur.
- Parcel
- an amount of fuel taken as one lot.
- Mass balance
- counting the real tonnes of sulphur from each part, then sharing them over the whole tank.
- MARPOL Annex VI
- the international rules on air pollution from ships, the sulphur limits among them.
- ECA
- emission control area: sea areas with a stricter sulphur limit.
- Viscosity
- how thick the fuel is, or how easily it flows.
An everyday picture
Pour a small glass of very sweet tea into a big jug of tea with hardly any sugar.
The mix does not taste halfway between the two. It tastes much closer to the big jug.
To know how sweet it is, count the sugar in each; add it up and divide by all the tea.
In the ship, sugar is sulphur. The small glass is the fuel left in the tank; the jug is the new delivery.
The rule
- Percentages cannot be added; quantities of sulphur can. So turn each part into tonnes of sulphur: mass × sulphur % ÷ 100.
- Add the sulphur from the remainder and from the delivery. Then divide by the total mass in the tank.
- The remainder counts even when it is small; here it is the higher-sulphur part. The blend sits nearer the bigger parcel.
- Compare the blend with the limit where the ship is going. MARPOL Annex VI: 0.50 % outside, 0.10 % inside an ECA. Sulphur blends by mass; viscosity does not, and has to be measured.
Formula
- Sulphur of the blend (%) = (m₁ × S₁ + m₂ × S₂) ÷ (m₁ + m₂)
- m₁, S₁ = what remains in the tank; m₂, S₂ = the delivery
Worked example
| Fuel remaining in the tank | 62 t |
| Sulphur content of what is remaining | 0.42 % |
| Quantity to be delivered into the same tank | 340 t |
| Sulphur content on the delivery note | 0.09 % |
- Sulphur in the remainder = 62 × 0.42 ÷ 100 = 0.2604 tWe find how many tonnes of sulphur the old fuel holds.
- Sulphur in the delivery = 340 × 0.09 ÷ 100 = 0.306 tWe find the same for the new fuel.
- Blend = (0.2604 + 0.306) ÷ (62 + 340) × 100 = 0.5664 ÷ 402 × 100 = 0.141 %We add the sulphur and share it over all the fuel in the tank.
Answer0.141 %
Common mistake
Averaging 0.42 % and 0.09 %, as if the tank held as much old fuel as new.
There are 62 t of the one and 340 t of the other. The plain average, 0.255 %, counts them as equal. It shows the tank far nearer a limit than it is; the mass balance gives 0.141 %.
Note for mariners Fuels do not always mix as evenly as tea, and two fuels can be incompatible; test a sample before relying on the figure.