Stack Heat Loss (Siegert)
Calculate the heat loss through the exhaust gases by the Siegert formula, loss = K × (T_gas − T_air) ÷ CO₂, from the fuel factor K (~0.5 for natural gas, ~0.6 for oil), the gas and combustion air temperatures (°C) and the CO₂ percentage in the gases. The result, in %, is the largest energy loss of a boiler or furnace — the heat escaping hot through the stack. Lowering the gas temperature (with economizers and preheaters) and adjusting the excess air (which dilutes CO₂) minimizes this loss. The combustion efficiency is approximately 100% minus this loss. Enter the K factor, the temperatures and the CO₂.
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Stack heat loss (Siegert formula)
In a boiler or furnace, the largest single energy loss is almost always the heat that escapes hot up the stack with the exhaust gases — energy released by combustion but never transferred to the process. The Siegert formula estimates that loss in a practical way: loss = K × (T_gas − T_air) ÷ CO₂, where K is an empirical fuel factor (about 0.5 for natural gas, 0.6 for fuel oils, 0.7 for coal), (T_gas − T_air) is the difference between the flue gas temperature in the stack and the combustion air temperature (°C), and CO₂ is the percentage of carbon dioxide in the gases. The result, in %, is the fraction of the heating value carried away by the stack. The formula reveals the two levers that cut this loss. First, lower the flue gas temperature: the cooler the gases leave, the less heat they carry off — hence the use of economizers (exchangers that preheat the feedwater with the exhaust gases) and air preheaters (which recover heat to warm the combustion air), bringing stack temperature down from 250-300 °C to 120-150 °C. (There is a floor: below the acid dew point of the gases, water vapour and SO₃ condense and corrode the stack.) Second, maximize CO₂ (which sits in the denominator): high CO₂ means little excess air (concentrated gases, little useless air diluting them) — so trimming excess air to the minimum required (O₂ control) raises CO₂ and cuts the loss. Combustion efficiency is roughly 100% minus this stack loss (plus small losses from unburnt fuel and radiation). Flue gas analysers that read temperature, CO₂ and O₂ compute this loss in real time and guide the tuning — one of the largest industrial energy saving opportunities. Enter the K factor, the temperatures and the CO₂.
Related Tools
Acid Dew Point of Flue Gas
Computes the temperature at which sulphuric acid starts to condense on the cold surfaces of a boiler, using the Verhoff and Banchero correlation: the reciprocal of the absolute dew point temperature is a combination of the logarithms of the partial pressures of water vapour and sulphur trioxide in the flue gas, plus the product of those two logarithms. The result is the thermal floor of the design — keeping the stack, the economiser and the air preheater above it is what prevents the acid corrosion that eats steel in a few weeks, and it is why heavy fuel oil boilers throw away up the stack heat they could otherwise recover. SO₃ is what rules here, not humidity, and the reason is the range each one spans: water vapour barely leaves the 5% to 15% band in a flue gas, which accounts for 11 °C end to end, while SO₃ varies by orders of magnitude with the sulphur in the fuel — going from 1 to 10 ppm alone raises the dew point by almost 22 °C. The Verhoff and Banchero correlation was adopted, with partial pressures in millimetres of mercury at atmospheric pressure, as it is the one most used in boiler design, in its original form with the interaction term between the two logarithms; the later Okkes correlation returns 1 to 8 °C lower for the same composition, so treat the value as a reference and not as an exact limit. Enter the water vapour content and the SO₃ content of the flue gas.
Excess Air (from Flue Gas)
Calculate the excess air of a combustion from the oxygen in dry flue gas, EA = O₂ ÷ (20.9 − O₂) × 100%, from the measured O₂ percentage in the stack. The result, in %, shows how much air was supplied beyond stoichiometric — measured by the leftover oxygen in the exhaust gases. Some excess air (10-30%) is needed to ensure complete combustion (avoid CO and soot), but too much wastes energy heating useless air that leaves hot through the stack. Gas analyzers measure O₂ and compute the excess air to optimize combustion efficiency. Enter the O₂ percentage in the gases.
Boiler Efficiency
Calculate the thermal efficiency of a boiler by the direct method, η = (m_steam × Δh) ÷ (m_fuel × LHV) × 100%, comparing the useful heat absorbed by the water/steam (steam flow × enthalpy gain) with the energy released by burning the fuel (fuel flow × lower heating value). The result, in %, shows how much fuel energy actually reached the steam; the rest is lost in flue gases, blowdown, radiation and unburnt fuel. Well-run industrial boilers reach 80–90%. Enter the steam flow, enthalpy gain, fuel flow and LHV.
CO₂ Volume Produced
Calculate the CO₂ volume produced in complete combustion of a hydrocarbon C_xH_y, V_CO₂ = x × 22.4 ÷ M, from the number of carbon atoms x and the fuel molar mass M (g/mol). The result, in Nm³ of CO₂ per kg of fuel, is the carbon dioxide generated by complete burning — information for emission inventories, exhaust system sizing and gas analysis. Each carbon atom in the fuel becomes one CO₂ molecule. Methane produces ~1.4 Nm³/kg. Fuels with more carbon per unit mass (coal, heavy oils) produce more CO₂. Enter x and the fuel molar mass.
Natural Chimney Draft
Calculate the natural draft (depression) of a chimney, ΔP = 353 × h × (1/T_air − 1/T_gas), from the chimney height h (m) and the absolute temperatures of the outside air and the hot gases (K). The result, in pascals, is the pressure difference that 'pulls' the gases up and the combustion air into the burner, generated by the density difference between the hot (light) gases and the cold (dense) air — the chimney effect. Taller chimneys and hotter gases generate more draft. It is the basis of natural-draft furnace and boiler chimney design; insufficient draft requires fans (forced draft). Enter the height and the air and gas temperatures.
Steam Loss Through a Trap or Orifice (Napier)
Computes the saturated steam flow escaping through a trap stuck open or a leak hole, using the Napier formula for critical flow: ṁ = C_d · 0.5244 · A · P_abs, with the orifice area in mm² and the absolute line pressure in bar, giving kg/h. Above roughly 1.9 bar absolute the flow is choked, and from there on the rate depends only on the UPSTREAM pressure, not on downstream back pressure — which is why the loss grows linearly with line pressure and with the square of the hole diameter, and why a high-pressure line loses disproportionately more through the same defect. The constant 0.5244 kg/(h·mm²·bar) is the exact conversion of the published imperial form, ṁ[lb/h] = 51.43 · A[in²] · P[psia], and with C_d = 1 it reproduces the isentropic choked flow to within 1 %. A hole of just 3 mm at 8 bar, with a discharge coefficient of 0.7, lets 20.8 kg/h escape — over 180 tonnes of steam per year of continuous operation, the central economic argument of any steam trap maintenance programme. Enter the discharge coefficient, the orifice diameter and the absolute line pressure.
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