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    Heat transfer / radiation

    Radiation between surfaces

    Radiation is the only heat transfer that does not need a medium, and it goes as the fourth power of absolute temperature — which is why it dominates at high temperature and is negligible at room temperature unless the areas are large. The two-surface enclosure formula reduces the problem to three resistances in series: two surface resistances and one space resistance.

    σT⁴2842 W/m²Shields0h_rad11.0
    Radiative flux
    1983 W/m²
    1983 W over 1.00 m²
    Blackbody flux
    2842 W/m²
    what a perfect emitter would radiate
    Convective flux
    1800 W/m²
    at h = 10
    Dominant mode
    Radiation
    radiation carries more heat here
    Three resistances in series
    Two surface resistances and one space resistance — that is the whole enclosure formula.
    1983 W/m²
    Three resistances in series
    TermValueNote
    Surface 1 temperature200 °C473.1 K
    Surface 2 temperature20 °C293.1 K
    Blackbody emissive power2842 W/m²σT₁⁴
    Emissivity 1 / 20.90 / 0.900 to 1
    Surface resistance 10.1111 1/m²(1 − ε₁)/ε₁A₁
    Space resistance1.0000 1/m²1/(A₁F₁₂)
    Surface resistance 20.1111 1/m²(1 − ε₂)/ε₂A₂
    Exchange, unshielded1983 Wthe enclosure formula
    Exchange, shielded1983 W0 shields
    Flux1983 W/m²per unit of surface 1
    Radiative coefficient11.01 W/m²·Kflux ÷ ΔT
    Convective flux1800 W/m²h ΔT

    Radiation dominates at this temperature, which is why insulation alone is not enough on a hot surface: a low-emissivity surface finish matters as much as thickness.

    Inspect the model
    Choose a quick result view, a scaled comparison, or the complete diagnostic set.
    Flux against surface temperature
    The fourth power is why radiation is negligible at room temperature and dominant at furnace temperature.
    Flux against surface temperature0.0001.371e+42.741e+44.112e+45.483e+46.853e+450 °C50 °C100 °C100 °C200 °C200 °C300 °C300 °C500 °C500 °C800 °C800 °CW/m²
    Radiation against convection
    The crossover is usually somewhere between 150 and 250 °C for a natural-convection surface.
    Radiation against convection0.000467.88935.771403.651871.532339.42RadiationRadiationConvectionConvectionW/m²
    Canonical equations
    The exact relationships implemented by this engine.

    E_b = σT⁴ · σ = 5.670 × 10⁻⁸ W/m²·K⁴.

    q = σ(T₁⁴ − T₂⁴)/ΣR · Two surface and one space resistance.

    (1 − ε)/(εA) · Zero for a blackbody.

    q_n = q/(n + 1) · For equal emissivities.

    engineering methodology

    A clear path from inputs to results.

    Radiation is the only heat transfer that does not need a medium, and it goes as the fourth power of absolute temperature — which is why it dominates at high temperature and is negligible at room temperature unless the areas are large.

    Step-by-step methodWorked exampleAssumptions explained

    How to use

    1. Enter the two surface temperatures and their emissivities.
    2. Set the areas and the view factor.
    3. Add any radiation shields.
    4. Compare the radiative flux with convection.

    The formula

    q=σ(T14−T24)1−ε1ε1A1+1A1F12+1−ε2ε2A2q = \frac{\sigma (T_1^4 - T_2^4)}{\dfrac{1-\varepsilon_1}{\varepsilon_1 A_1} + \dfrac{1}{A_1 F_{12}} + \dfrac{1-\varepsilon_2}{\varepsilon_2 A_2}}

    Reference: Stefan-Boltzmann law with the gray-diffuse two-surface enclosure; σ from CODATA 2018.

    Worked example

    A 200 °C surface facing 20 °C

    Inputs
    • 200 °C and 20 °C, both with emissivity 0.9.
    • 1 m² each, view factor 1.
    Calculation path
    1. 1

      The blackbody emissive power at 200 °C is 2,838 W/m².

    2. 2

      The two surface resistances are 0.111 each and the space resistance is 1, giving 1.222 in total.

    3. 3

      The net exchange is 2,422/1.222 = 1,981 W over the square metre.

    4. 4

      Convection at 10 W/m²·K and 180 K gives 1,800 W/m², so radiation is the larger term.

    5. 5

      Adding two shields cuts the exchange to a third, 660 W.

    Result

    1,981 W/m² · radiation dominates · two shields give 660 W.

    Interpreting the result

    The shield result is the one worth remembering: each shield adds two surface resistances and no space resistance, so n shields divide the exchange by n + 1. It is the cheapest way to reduce radiative loss, and it is why a vacuum flask works.

    Assumptions & limitations

    • Gray-diffuse surfaces with constant properties.
    • A two-surface enclosure with the areas and view factor you supply.
    • No participating gas between the surfaces.
    • Steady state, with no conduction in the surfaces.

    Frequently asked questions

    When does radiation dominate?

    Above roughly 200 °C for a surface in air, or at any temperature where the areas are large and the emissivity is high. Below 100 °C it is usually comparable to or smaller than natural convection.

    Why do shields work so well?

    Because each one adds two surface resistances and no space resistance. With equal emissivities, n shields divide the exchange by n + 1.

    What about a fire or a furnace?

    The gas becomes participating, absorbing and emitting in bands, and a gray-surface model is no longer valid. That needs a band model or a CFD solution.

    What is a view factor?

    The fraction of the radiation leaving one surface that reaches the other. For infinite parallel plates it is 1; for a small object in a large room it is effectively 1 as well, but the enclosure formula then reduces to εσA(T₁⁴ − T₂⁴).

    Does emissivity matter as much as temperature?

    At high temperature, more. A low-emissivity coating on a hot surface can cut the radiative flux by 90%, which is worth more than adding insulation thickness.

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