Radiation from a sooty flame
A clean flue gas radiates only in the CO₂ and H₂O bands. A flame with soot radiates continuously across the whole spectrum, and a pulverised-coal flame adds ash particles on top. Both are grey emitters that add to the band absorption rather than replacing it, so the coefficients are added and one exponential is taken.
| Term | Value | Note |
|---|---|---|
| Gas temperature | 1200 °C | 1473 K |
| Mean beam length | 2.00 m | as entered |
| CO₂ + H₂O partial pressure | 0.250 atm | flue gas is 0.2–0.3 |
| Gas emissivity (WSGG) | 0.461 | bands only |
| Gas absorption coefficient | 0.309 /m | backed out of the emissivity |
| Soot volume fraction | 0.20 ppm | k = 3.6 f_v / λ, λ = 1.4 μm |
| Soot coefficient | 0.51 /m | continuous spectrum |
| Ash loading | 0.10 g/m³ | d_p 10.0 μm |
| Ash coefficient | 0.008 /m | 1.5 c / (ρ d) |
| Total coefficient | 0.83 /m | sum of the three |
| Emissivity | 0.810 | 1 − e^(−kL) |
| Flux | 163.77 kW/m² | Hottel form |
Soot and ash raise the emissivity from 0.46 to 0.81, a 78% increase in flux. Soot dominates: 62% of the absorption.
k_s = 3.6 f_v / λ · λ ≈ 1.4 μm.
k_a = 1.5 c / (ρ_p d_p) · Mass concentration.
k = k_gas + k_s + k_a · Coefficients add.
ε = 1 − e^(−kL) · One exponential for all three.
A clear path from inputs to results.
A clean flue gas radiates only in the CO₂ and H₂O bands. A flame with soot radiates continuously across the whole spectrum, and a pulverised-coal flame adds ash particles on top.
How to use
- Enter the gas and wall temperatures and the beam length.
- Set the soot volume fraction and the ash loading.
- Add the ash particle diameter.
- Read the total emissivity and the flux against the clean-gas case.
The formula
Reference: Soot k = 3.6 f_v/λ (Beér); ash k = 1.5 c/(ρ d); gas from the WSGG model.
Worked example
A gas with 1 ppm of soot
- 1,200 °C gas, 2 m beam, 25% CO₂ + H₂O.
- 1 ppm soot, 0.1 g/m³ ash at 10 μm.
- 1
The clean-gas emissivity is 0.46, which corresponds to an absorption coefficient of 0.31 /m.
- 2
Soot at 1 ppm gives 3.6 × 10⁻⁶/1.4 × 10⁻⁶ = 2.57 /m — eight times the gas.
- 3
Ash adds only 0.008 /m, because the particles are large and the loading is low.
- 4
The total coefficient is 2.89 /m, so the emissivity is 1 − e^(−5.77) = 0.997.
- 5
The flux rises from 91 to 196 kW/m² — the flame is effectively opaque.
ε 0.997 · 196 kW/m² · soot dominates.
Interpreting the result
A little soot goes a long way. One part per million of soot volume already makes the flame nearly opaque, which is why a luminous flame transfers far more heat than a clean gas flame at the same temperature — and why soot is both a heat-transfer asset in a furnace and a pollutant in a chimney.
Assumptions & limitations
- Rayleigh-limit soot absorption at a mean wavelength of 1.4 μm.
- Geometric-optics ash absorption with spherical particles.
- Gas contribution from the three-term WSGG model.
- No scattering, and a uniform particle concentration.
Frequently asked questions
Why add coefficients rather than emissivities?
Because absorption is additive and emissivity is not. Converting the gas to a coefficient, adding the particulates, and taking one exponential is the only combination that behaves correctly at both limits.
Does ash matter as much?
Rarely in a gas flame, but it does in pulverised coal, where the loading is much higher. The particles are large, so they absorb less per unit mass than soot.
What about scattering?
Soot scatters little at these wavelengths, but ash can. A scattering correction needs the albedo, which is beyond a hand calculation.
What soot loading is realistic?
0.1 to 5 ppm by volume in a luminous hydrocarbon flame. Above about 5 ppm the flame is opaque and more soot only changes the colour.
Does this replace the clean-gas page?
No — it extends it. With the soot and ash set to zero it reduces exactly to the WSGG result.
Methodology last reviewed 2026-10-22. Results are informational and not financial or engineering advice — see our disclaimer.