Introduction
Engine makers declare power and fuel consumption at the ISO reference conditions: 100 kPa, 25 °C air, 30 % humidity and 25 °C charge-air coolant. An engine room in the tropics offers 45 °C air and 36 °C cooling water, and the engine can neither make the declared power nor burn fuel at the declared rate. ISO 3046-1 gives the standard way to move between the two: a ratio of indicated power built from the pressure, temperature and coolant ratios with exponents that depend on the engine type, a power adjustment factor that allows for the friction power which does not scale, and the fuel consumption factor β = k/α. This calculator applies it for the eight formula references with recommended values and adds the calorific-value scaling that every shop-test comparison needs.
It is for marine and power-plant engineers comparing shop tests with site performance, performance analysts normalising SFOC data, and students of internal combustion engines.
Formulas
k = [(p_x − a φ_x p_sx)/(p_r − a φ_r p_sr)]^m (T_r/T_x)^n (T_cr/T_cx)^s ISO 3046-1 (3), T in K
α = k − 0.7 (1 − k)(1/η_m − 1), P_x = α P_r (2), (1)
β = k/α, b_x = β b_r clause 10.4
b_fuel = b_x × 42 700 / LCV equal energy per kWhTable 2 (a, m, n, s): A 1, 1, 0.75, 0 · B 0, 1, 1, 0 · C 0, 0.7, 2, 0 · D 0, 0.7, 1.2, 1 · F and J 0, 0.57, 0.55, 1.75 · G 0, 0.7, 1.2, 1 · I 1, 0.86, 0.55, 0. Reference conditions: 100 kPa, 298 K, 30 %, 298 K.
Assumptions and limitations
- Guideline factors (Table 2 note 2); the manufacturer may declare its own, and low-speed two-stroke engines (E, H) have none.
- β = k/α is the ISO default; electronically controlled engines may behave differently.
- Humidity matters only where a = 1 (A, I); the Magnus vapour pressure is used.
- η_m at the declared power, 0.8 when none is declared; s = 0 when the charge air is jacket-water cooled.
Worked example
A medium-speed four-stroke diesel (reference D) declared at 10 000 kW and 175 g/kWh, mechanical efficiency 0.8, operating at 100 kPa, 45 °C engine-room air and 36 °C charge-air coolant.
- k = (298.15/318.15)^1.2 × (298.15/309.15)^1 = 0.9250 × 0.9644 = 0.8921.
- α = 0.8921 − 0.7 × 0.1079 × 0.25 = 0.8733 → 8 733 kW available, 12.7 % less.
- β = 0.8921/0.8733 = 1.0216 → 178.8 g/kWh, 2.2 % more fuel per kWh.
- Burning a 40 500 kJ/kg residual fuel instead of the 42 700 kJ/kg reference: 178.8 × 42 700/40 500 = 188.5 g/kWh, i.e. 1 646 kg/h at the site power.
For a non-turbocharged engine (reference A) the humidity enters: at 60 % humidity and 45 °C the dry-air pressure is 94.2 kPa against 99.0 kPa at reference, which alone costs 5 % of power.
FAQ
Why does the power fall more than the indicated power?
Friction does not shrink with the air density, so the brake power loses a larger share: that is what the 0.7(1 − k)(1/η_m − 1) term represents.
Which formula reference is my engine?
The manufacturer states it; D (turbocharged, charge-air cooled, medium speed) is the common marine four-stroke case; for two-stroke main engines use the maker's own derating curves.
Why does the SFOC rise?
The fuel per indicated kWh stays the same while the brake kWh fall, so fuel per brake kWh rises by k/α.
What are the reference conditions?
ISO 15550: 100 kPa total pressure, 298 K air, 30 % relative humidity, 298 K charge-air coolant. Manufacturers may declare substitute conditions; enter them in the reference fields.
How does this relate to ISO 19030?
Performance monitoring compares service SFOC with the shop test; both must be on the same fuel energy basis, which the 42 700 kJ/kg scaling provides.
Why is humidity ignored for turbocharged engines?
The standard sets a = 0 for them: the charge air cooler and turbocharger largely compensate; the effect exists but is small.
Related calculators
- Generator load, kVA and fuel — the loads this power serves.
- Engine BMEP, torque and power — the mechanical side of the same engine.
- Fuel net specific energy and CO₂ factor — the LCV from the bunker delivery note.
- Engine room ventilation (ISO 8861) — the air that sets T_x.
- Fuel and voyage cost — consumption at sea.