Boiler combustion efficiency
Boiler efficiency from flue gas oxygen and stack temperature — and the fuel you save by tuning the burner.
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How this calculator works
The flue gas (stack) loss comes from the Siegert formula: loss % = (Tstack − Tair) × (A₂/(21 − O₂) + B), with the published constants for each fuel (natural gas A₂ = 0.66, B = 0.009; fuel oil 0.68 and 0.007; LPG 0.63 and 0.008). Efficiency = 100 − stack loss − other losses, on the net calorific value. Excess air = O₂/(21 − O₂).
The example is a 10 MW gas-fired boiler at 5 % O₂ and 220 °C stack: excess air 31 %, stack loss 9.8 %, efficiency 88.7 %. Tuning the burner to 2.5 % O₂ and cleaning to bring the stack down to 200 °C gives 90.7 % — about USD 57,000 a year in fuel at USD 8/GJ.
Limits: indirect (losses) method for clean, complete combustion; no condensing heat recovery; the Siegert constants are approximate for other fuels. References: Branan, Rules of Thumb for Chemical Engineers (energy conservation); Siegert formula as used in combustion analysers and EN/DIN flue gas loss checks; ASME PTC 4 for full boiler tests.