Boiler Efficiency Calculator
Estimate boiler efficiency by the direct (input–output) method: the useful heat carried away in the steam divided by the chemical energy supplied in the fuel. Enter the steam and feedwater conditions, the steam and fuel flows and the fuel's calorific value to get efficiency plus the heat and fuel-energy rates in kilowatts.
Enter Values
How to use this calculator
- Enter the steam mass flow ms and fuel mass flow mfuel using the same time basis (both kg/h here).
- Enter the steam enthalpy hs and feedwater enthalpy hf in kJ/kg — hs must be larger, since the boiler adds heat.
- Enter the fuel calorific value CV in kJ/kg, then read the efficiency along with useful heat and fuel input in kW.
How it works
The direct method compares energy out with energy in: η = ms·(hs − hf) / (mfuel·CV) × 100 %. The numerator is the heat the boiler adds to each kilogram of water times the steam flow; the denominator is the fuel flow times its calorific value. Dividing each per-hour energy by 3600 converts kJ/h into kW, giving the useful heat rate to the steam and the fuel energy input rate for a physical feel for the duty.
Worked example
Worked example. With ms = 1000 kg/h, hs = 2800 kJ/kg, hf = 400 kJ/kg, mfuel = 100 kg/h and CV = 40 000 kJ/kg: η = 1000×2400 / (100×40 000) × 100 = 60 %. Useful heat = 1000×2400 / 3600 = 666.67 kW and fuel input = 100×40 000 / 3600 = 1,111.11 kW.
Common mistakes
- Mixing time bases — steam flow per hour with fuel flow per second, for instance. Both flows must share the same time unit.
- Swapping hs and hf. The steam enthalpy must exceed the feedwater enthalpy, otherwise the boiler would be removing heat.
- Using a gross calorific value with an efficiency expected on a net basis (or vice versa); be consistent about which CV you enter.
Frequently asked questions
What is the difference between the direct and indirect methods?
The direct method used here divides useful heat by fuel energy — quick and needing few measurements. The indirect method instead adds up individual losses (flue gas, radiation, blowdown, unburnt fuel) and subtracts them from 100 %, which pinpoints where efficiency is being lost but requires flue-gas analysis and more data.
Why convert the energy rates to kilowatts?
The mass flows are per hour, so ms·(hs − hf) and mfuel·CV come out in kJ/h. Dividing by 3600 (seconds per hour) converts kJ/h to kJ/s, which is kW — a more intuitive rating for the boiler's thermal duty.
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- Carnot Efficiency Calculator
- Steam Quality Calculator
- Thermal Resistance Calculator
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Tip: Enter any known values to calculate the remaining results.
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