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    Calculates locally in your browser · participating-gas radiation
    Heat transfer / combustion

    Radiation from a participating gas

    A flame is not a gray surface. CO₂ and H₂O absorb and emit in narrow bands, so the gas is transparent at some wavelengths and nearly opaque at others — which is why a furnace calculation cannot use a single emissivity. The usual engineering treatment is the weighted sum of gray gases: a handful of fictitious gray gases fitted to the band data.

    pL0.50Band modelLowerWall ε0.01
    Gas emissivity
    0.46
    at pL = 0.50 atm·m
    Heat flux
    1.1 kW/m²
    2.5 kW/m² from the gray approximation
    Absorptivity
    0.81
    evaluated at the wall temperature
    Effective emissivity
    0.01
    gas and wall in series
    The gas, the wall and the bands
    A participating gas is not a gray surface: emissivity depends on path length and temperature.
    1.1 kW/m²
    The gas, the wall and the bands
    QuantityValueNote
    Gas temperature1200 °C1473 K
    Wall temperature400 °C673 K
    Partial pressure of CO₂ + H₂O0.250 atmtypically 0.2–0.3 for flue gas
    Mean beam length2.00 m3.6 V/A for a box
    pL0.500the scaling group
    Gas emissivity0.461WSGG at the gas temperature
    Gas absorptivity0.808at the wall temperature
    Wall emissivity0.01refractory or tube
    Effective emissivity0.0101/(1/ε_g + 1/ε_w − 1)
    Heat flux1.14 kW/m²band model
    Gray approximation2.52 kW/m²single emissivity
    Difference-55%band model lower

    At pL = 0.50 atm·m the gas emits with an emissivity of 0.46, and the band model gives 1.1 kW/m² against 2.5 kW/m² from a single-gray-gas approximation — -55% different. The discrepancy is why furnace models use band data rather than one emissivity.

    Inspect the model
    Choose a quick result view, a scaled comparison, or the complete diagnostic set.
    Emissivity against beam length
    Saturates as the path becomes optically thick — a large furnace behaves like a blackbody, a small one does not.
    Emissivity against beam length0.0000.1340.2680.4030.5370.6710.5 m0.5 m1 m1 m2 m2 m4 m4 m8 m8 mε
    Band model against gray
    The gap is why a furnace model needs band data.
    Band model against gray0.0000.5961.191.792.382.981.14Band modelBand model2.52GrayGraykW/m²
    Canonical equations
    The exact relationships implemented by this engine.

    ε_g = Σ a_i(T)(1 − e^(−b_i pL)) · Three gray gases.

    q = ε_w σ(ε_g T_g⁴ − α_g T_w⁴) · Hottel form.

    α_g = ε_g(T_w)(T_g/T_w)^0.5 · Temperature correction.

    L = 3.6 V/A · Rectangular box.

    engineering methodology

    A clear path from inputs to results.

    A flame is not a gray surface. CO₂ and H₂O absorb and emit in narrow bands, so the gas is transparent at some wavelengths and nearly opaque at others — which is why a furnace calculation cannot use a single emissivity.

    Step-by-step methodWorked exampleAssumptions explained

    How to use

    1. Enter the gas and wall temperatures.
    2. Set the partial pressure of CO₂ and H₂O.
    3. Add the mean beam length and the wall emissivity.
    4. Read the gas emissivity, the absorptivity and the flux.

    The formula

    εg=∑iai(T) (1−e−bipL)\varepsilon_g = \sum_i a_i(T)\,\left(1 - e^{-b_i pL}\right)

    Reference: Three-term weighted-sum-of-gray-gases fit; flux in the Hottel form.

    Worked example

    Flue gas at 1,200 °C against a 400 °C wall

    Inputs
    • Gas 1,200 °C, wall 400 °C.
    • 25% CO₂ and H₂O by volume, 2 m beam length.
    • Wall emissivity 0.8.
    Calculation path
    1. 1

      The pressure-path-length product is 0.25 × 2 = 0.5 atm·m.

    2. 2

      At 1,473 K the WSGG weights give a gas emissivity of 0.46.

    3. 3

      At the wall temperature the absorptivity is higher, 0.81, because the bands narrow as the gas cools.

    4. 4

      The flux is 0.8 × σ × (0.46 × 1473⁴ − 0.81 × 673⁴) = 91 kW/m².

    5. 5

      A single-gray-gas approximation gives a different answer, which is why band models exist.

    Result

    ε 0.46 · 91 kW/m² · absorptivity 0.81.

    Interpreting the result

    The emissivity falls with temperature and rises with the path length, so a large furnace is more radiative per unit area than a small one at the same temperature. That is the opposite of a gray surface, whose emissivity does not depend on size at all.

    Assumptions & limitations

    • A three-term WSGG fit for a 1 atm flue-gas mixture.
    • Gray wall with a stated emissivity.
    • No soot, ash or luminous flame contribution.
    • Uniform gas temperature and a mean beam length.

    Frequently asked questions

    Why not use a single emissivity?

    Because the gas is not gray. The error from a single-gray-gas approximation is typically 10–30%, and it goes the wrong way at high temperature.

    Why is absorptivity different from emissivity?

    Because the gas absorbs at the wall temperature, where the bands are narrower. The correction factor (T_g/T_w)^0.5 accounts for it.

    Does this work for a boiler furnace?

    Yes, as the radiation term. A full furnace calculation couples it with the heat balance and the gas flow, which is what a zone model or CFD does.

    What is the mean beam length?

    The effective path length for radiation, 3.6 V/A for a rectangular box. A large furnace has a long one, and that raises the emissivity.

    What about a luminous flame?

    Soot particles radiate continuously and add a large gray component. A WSGG model alone underestimates a sooty flame, sometimes badly.

    Methodology last reviewed 2026-10-22. Results are informational and not financial or engineering advice — see our disclaimer.