Engineering & Technology · Heat Transfer & Thermal Management

Thermal radiation, surfaces and radiative exchange

A surface emits thermal radiation even when it is at the same temperature as its surroundings. Equal temperature does not stop emission; it can make incoming and outgoing radiation balance. Cooling predictions therefore need net exchange, not just the surface’s emission alone.

Thermodynamics & Energy Systems provides state, energy, heat and boundary vocabulary. Material Structure, Processing & Properties supports conductivity and property conditions. Symbols, ratios, resistance and exponential-response notation are explained here; calculus is optional. All numerical materials, contacts, operating records, curves and limits are constructed teaching cases. Named physical constants retain their ordinary meanings. No equipment operating procedure is given.

Back to the chapter

After reading, you can

Distinguish emitted radiation from net exchange and identify the surface and geometry assumptions of a radiation model.

Use absolute temperature and a surface model

An ideal blackbody absorbs all incident radiation. Its emitted power per area is Eb=σT⁴, with absolute T in kelvin and σ≈5.67×10⁻⁸ W/(m²·K⁴), the Stefan–Boltzmann constant. The spectrum spans wavelengths and changes with temperature. A visibly dark appearance alone does not establish blackbody behavior at thermal wavelengths.

A real surface has emissivity ε comparing its emission with a blackbody at the same temperature. It can depend on wavelength, direction, temperature and surface condition. A diffuse gray approximation simplifies direction and wavelength dependence. Absorptivity and emissivity relations must use compatible spectral and directional conditions; a single number is not automatically appropriate for both sunlight and long-wave thermal radiation.

Net exchange includes the surroundings

For a small diffuse gray surface completely surrounded by a much larger isothermal enclosure, a useful limited model is Q̇=εσA(Ts⁴−Tsur⁴). The surroundings provide incoming radiation as well as receiving outgoing radiation. The medium is assumed not to absorb or emit significantly. Convection and conduction, if present, require additional terms.

At Ts=Tsur, net radiation is zero even though each side emits. Below the surrounding temperature, the expression becomes negative: net radiation enters the surface. With multiple finite surfaces, geometry matters through view factors, and reflections can make exchange depend on more than one emissivity. The simple enclosure equation cannot be transferred unchanged to every pair of exposed surfaces.

Area0.10 m2, surroundings300 K. Qdot=epsilon*sigma*A*(Ts^4−300^4), sigma5.67e−8 W/(m2 K4). At400 K, epsilon0.8 gives79.38 W and0.2 gives19.845 W. Solid and dashed curves are model values, excluding other heat paths.

Surface condition matters for transfer and measurement

A coating or oxidation can change thermal emission independently of bulk conductivity. Low emissivity may reduce outward radiation from a hot body, but may also affect absorption of incoming radiation. Decide which surroundings, wavelengths and service matter before calling a surface treatment beneficial. Radiation can be important alongside convection even without visible glow.

An infrared instrument detects radiation and converts a signal into an inferred temperature using a measurement model. Surface emissivity, reflected surroundings, spatial resolution and calibration can matter. A bright patch in a thermal image need not represent a proportionally hotter object. A model that neglects reflection and geometry may confuse radiance contrast with temperature contrast.

Subtract incoming radiation before naming the loss

For a constructed gray surface with ε=0.8 and A=0.10 m², Ts=400 K and Tsur=300 K. Net outward radiation is 0.8×0.10×5.67×10⁻⁸×(400⁴−300⁴)=79.38 W. Use absolute temperatures before taking fourth powers.

The emitted term is 116.1216 W and the absorbed enclosure term36.7416 W in this model. Their difference is 79.38 W. Changing ε to 0.2 reduces the net model rate to 19.845 W under the same stated conditions. These numbers are not measured coating performance or a total cooling capacity.

Try the reasoning

Does zero net radiative exchange mean neither body emits?
Show the answer and reasoning
No. Emission and absorption balance at equal temperature.
Why cannot100 °C simply replace373.15 K in a fourth-power law?
Show the answer and reasoning
The radiation law requires absolute temperature; the Celsius zero is an arbitrary offset.
Is the small-surface enclosure equation sufficient for two small surfaces that barely face each other?
Show the answer and reasoning
No. View geometry and exchange with other surroundings must be included.

What to carry forward

Use net radiative exchange, absolute temperatures and an explicit surface/geometry model. Emission alone is not the cooling result.

References

Teaching resources and research supporting this lesson’s explanations and examples.