Diesel Generator Efficiency Calculator

Work out diesel generator efficiency instantly with clear inputs, formula shown and shareable results.

About Diesel Generator Efficiency Calculator

Generator efficiency is electrical output divided by fuel energy input. Fuel mass per hour times net calorific value gives the input in kilowatts. A well-loaded modern diesel generator reaches 38-43% at full load, falling steeply below about 50% load, which is why oversized sets are expensive to run.

Formulas

Thermal efficiency

Input (kW) = fuel (kg/h) x CV (MJ/kg) x 1000 / 3600; efficiency = output / input

Specific fuel consumption

SFC (g/kWh) = fuel (kg/h) x 1000 / output (kW)

Worked example

With the default values below, Diesel Generator Efficiency Calculator returns Thermal Efficiency of 38.603%. Change any input above and every figure here is recalculated in your browser.

Inputs used

Electrical Outputminimum 0.01 kW: 500 kW
Fuel Consumptionminimum 0 L/hour: 130 L/hour
Fuel Densityminimum 0.01 kg/m3: 840 kg/m3
Net Calorific Valueminimum 0.01 MJ/kg: 42.7 MJ/kg

Results

Thermal Efficiency: 38.603%
Specific Fuel Consumption: 218.40
Litres Per kWh: 0.2600
Fuel Energy Input: 1,295.23

Step by step

1Values used
2Thermal efficiency
3Specific fuel consumption
4Thermal Efficiency= 38.603
5Specific Fuel Consumption= 218.40
6Litres Per kWh= 0.2600
7Fuel Energy Input= 1,295.23

Assumptions and limits

The result is derived only from the 4 inputs listed above. Anything not among them is not part of the calculation.
In particular, this calculator does not take inflation, tax, fees or charges or compounding frequency into account.
Figures are rounded for display to 4 decimal places, so totals built from several rounded parts can differ slightly from the unrounded arithmetic.

How these calculators are built and tested is described on the Methodology page. If a figure looks wrong, please report it.

Frequently Asked Questions

What SFC should a good set achieve?

195-215 g/kWh at full load for a modern medium-speed diesel. Above 240 g/kWh indicates part-load operation, poor maintenance or a fuel quality problem.

Why is part-load operation so inefficient?

Friction and auxiliary losses are nearly constant while output falls, so their share of fuel rises. Below 30% load, wet stacking and cylinder glazing also become a maintenance risk.