Net Radiation Exchange Between Surfaces

Q˙=εσA(T14T24)\dot{Q} = \varepsilon \sigma A (T_1^4 - T_2^4)

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Every surface both emits and absorbs radiation, and the net exchange with large surroundings goes as the difference of fourth powers, not as a simple ΔT. That non-linearity is why radiation is negligible in a chilled-water pipe and dominant in a furnace. A 1.2 m² oxidised steel panel (ε = 0.85) at 500 K facing a 300 K room radiates 0.85 × 5.670 × 10⁻⁸ × 1.2 × (6.25 × 10¹⁰ − 8.1 × 10⁹) = 3146 W — with a convective coefficient of 10 W/(m²·K) it would shed only about 2400 W by convection, so more than half the heat leaves as light you cannot see.

Emissivity is where field work goes wrong. Polished aluminium is 0.04, mill-finish steel 0.2–0.3, the same steel after a summer outdoors 0.7–0.85, and almost every paint, oxide, brick and organic surface is 0.85–0.95 regardless of colour — white paint is as good a radiator in the infrared as black. That is the trap in infrared thermography: point a camera set for ε = 0.95 at a shiny bus bar and it reports the temperature of whatever the bar is reflecting, which is usually you. Tape a square of matte tape on the target and read that instead. The formula assumes the surroundings are large enough to behave as a black enclosure; two comparable surfaces facing each other need view factors and a full radiosity network.

Net Radiation Exchange Between Surfaces
Q˙=εσA(T14T24)\dot{Q} = \varepsilon \sigma A (T_1^4 - T_2^4)
Where
  • Q˙\dot{Q}= Net radiant heat rate
  • ε\varepsilon= Emissivity
  • AA= Surface area
  • T1T_1= Surface temperature
  • T2T_2= Surroundings temperature