Worked example at 75 % load
The set delivers 75 % of its 16 kW rating. The part-load multiplier at 75 % is 1.03, so each kWh costs slightly more fuel than at full load.
P = 20 kVA × 0.8 × 0.75 = 12 kW
SFC = 0.24 L/kWh × 1.03 = 0.2472 L/kWh
F = 12 kW × 0.2472 L/kWh = 2.966 L/h
F = 2.966 / 3.785411784 = 0.7836 US gal/h
24 h: 2.966 × 24 = 71.19 L
Tank: 50 L / 2.966 L/h = 16.9 h
Curve and assumptions
- Full-load specific fuel consumption of 0.24 L/kWh (about 200 g/kWh at 0.835 kg/L), a value within the range quoted for modern diesel sets. It is not taken from any manufacturer's datasheet.
- Generic part-load multipliers 1.4 at 25 %, 1.12 at 50 %, 1.03 at 75 %, 1 at 100 %. At 25 % load the set uses about 40 % more fuel per kWh than at full load.
- Load in kW at a power factor of 0.8, the rating basis of ISO 8528-1. A load at a better power factor draws more kW from the same kVA and burns more fuel.
- A steady load for the whole period, with no warm-up or idling time.
What changes the numbers
- The engine itself: two sets of the same kVA from different makers, or of different ages, can differ noticeably. The datasheet consumption at 25, 50, 75 and 100 % load is the figure to use once the model is known.
- A varying load: average the consumption over the hours spent at each load level rather than using the average load in one step, because the curve is not linear.
- Site and condition: altitude, hot intake air, clogged filters, injector wear and low coolant temperature at light load all raise consumption.
- Fuel itself: diesel density changes with temperature and grade, so the same mass of fuel occupies a slightly different volume.