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Specific fuel oil consumption: fuel per kWh of brake power

Why it matters SFOC shows whether the engine burns more fuel than it should for the work it does; that costs money every day.

Specific fuel oil consumption: the fuel of the hour over the brake power, not the shaft power or the MCRThe flowmeter gives 2016 kg in the hour, 2016000 g. Brake power is taken at the coupling: 11200 kW, 80 % of the 14000 kW MCR. Shaft power, read by the torsionmeter on the shaft line after the shafting losses, is 10640 kW (the question gives both readings). SFOC = 2016000 ÷ 11200 = 180.0 g/kWh.MAIN ENGINEflowmeter: 2016 kg in the hourshaft bearingbrake power — at the couplingshaft power — torsionmeterMCR 14000 kW — a rating, not what she madebrake power 11200 kW (80 % MCR)shaft power 10640 kW (560 kW less, as given)SFOC = 2016000 ÷ 11200 = 180.0 g/kWhPower bars on one scale

Words on this page

Main engine
the big diesel engine that drives the ship.
Specific fuel oil consumption (SFOC)
how many grams of fuel the engine burns for each kilowatt-hour of work.
Brake power
the power the engine delivers at its own output, the coupling.
Coupling
the joint that connects the engine to the shaft line.
Shaft power
the power measured on the shaft line, after the shafting losses.
Torsionmeter
an instrument that finds the power by measuring how much the shaft twists.
Flowmeter
the meter that measures how much fuel flows through the pipe.
MCR
the highest power the engine may give continuously; a rated figure from the maker.
Shop trial
the test run at the factory before the engine goes into the ship.
Calorific value
the heat energy one kilogram of fuel gives when it burns.

An everyday picture

A bakery wants to know how much gas its oven burns for each loaf. The fair count is the loaves as they leave the oven.

Count them at the shop instead, after a few were dropped on the way. Then the oven looks worse than it is.

The engine's brake power is the count at the oven door. The shaft power is the count at the shop.

The rule

  • SFOC is a comparison figure: fuel mass ÷ power developed, in grams per kilowatt-hour. Read the flowmeter and the power over the same hour.
  • Divide by the BRAKE power, the power at the engine's coupling. The shop trial and last week's figures are quoted against it.
  • The torsionmeter's shaft power will not do. It is read on the shaft line after the shafting losses. On a direct-drive slow-speed engine the gap is small, about 1 %; use the two readings the question gives. Nor will the MCR, the rated output she was not making.
  • SFOC cannot see the calorific value of the fuel. A change of bunkers moves it while nothing is wrong with the engine.

Formula

  • SFOC (g/kWh) = fuel burned in the hour (kg) × 1000 ÷ brake power (kW)
  • Brake power = at the engine's coupling; shaft power = on the shaft line, after the shafting losses

Worked example

Fuel burned by the main engine in the hour2016 kg
Brake power developed over the same hour11200 kW
Shaft power measured at the torsionmeter10640 kW
Maximum continuous rating of the engine14000 kW
  1. Fuel in grams = 2016 × 1000 = 2016000 g in the hourThe flowmeter gives kilograms; the figure is wanted in grams, so we multiply by a thousand.
  2. SFOC = 2016000 ÷ 11200 = 180.0 g/kWh (she was at 80 % of MCR)Fuel ÷ brake power = grams for each kilowatt-hour of work.

Answer180 g/kWh

Common mistake

Dividing by the shaft power from the torsionmeter, 10640 kW, instead of the brake power.

The torsionmeter reads the power on the shaft line, after the shafting losses; the question gives the two readings 560 kW apart. So the figure comes out 189.5 g/kWh. The engine looks worse than it is, and the trend no longer matches the shop trial.

Note for mariners Before a shop-trial comparison, SFOC is normally corrected to ISO ambient conditions and the fuel's calorific value.