Net Radiation Exchange Between Surfaces

Q˙=εσA(T14−T24)\dot{Q} = \varepsilon \sigma A (T_1^4 - T_2^4)

Worked example: eps 0.85, 1.2 m2, 500 K to 300 K → 3146 W — press Try an example to run it live, then adjust anything.

Enter your known values, leave one input blank, and solves for the missing one. Tap a variable’s symbol to see what it means, with a typical value. Try different units for next level excitement!

Here the solver did the work — could you?

Radiation exchange →

UniversityThermodynamics & Heat Transfer

Test your skills in the Exam Room: new numbers every attempt — free lessons for students, no sign-up, just pure learning.

See your Report Card
Compete with your friends
share your results
Learning zone

Net Radiation Exchange Between Surfaces explained

T1T2QεA

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 formula

Q˙=εσA(T14−T24)\dot{Q} = \varepsilon \sigma A (T_1^4 - T_2^4)
Where
  • Q˙\dot{Q}= Net radiant heat rate (W)
  • ε\varepsilon= Emissivity
  • AA= Surface area (m²)
  • T1T_1= Surface temperature (°C)
  • T2T_2= Surroundings temperature (°C)