Radiation laws: Stefan-Boltzmann, Kirchhoff and Wien

Why it matters

Radiation dominates heat transfer in furnaces, fired heaters, kilns and reformers, and it is a large part of the loss from any hot uninsulated pipe or vessel. It also causes measurement errors: a thermocouple in a hot gas duct "sees" the cooler walls and reads low. The black-body laws — Planck, Wien and Stefan–Boltzmann — and Kirchhoff's law for real surfaces are the starting point for every radiation calculation.

Key ideas

Thermal radiation. Electromagnetic energy emitted by all matter above 0 K because of its temperature, mainly in the range about 0.1–100 μm (part of the UV, the visible 0.4–0.7 μm, and the infrared). It needs no medium and travels at the speed of light. Unlike conduction and convection, net exchange depends on the difference of the fourth powers of absolute temperatures.

Absorptivity, reflectivity, transmissivity. Incident radiation (irradiation) is partly absorbed, reflected and transmitted: α + ρ + τ = 1. Opaque solids have τ = 0, so α + ρ = 1.

Black body. An ideal surface that absorbs all incident radiation of every wavelength and direction (α = 1), and for a given temperature emits the maximum possible energy at every wavelength. It is a diffuse emitter. A small hole in a large isothermal cavity is the practical approximation.

Planck's law. The spectral emissive power of a black body E_bλ(λ, T) rises from zero, peaks, and falls again. As T rises the whole curve rises and the peak moves to shorter wavelengths.

Wien's displacement law. The peak lies at λ_max·T = 2898 μm·K. The sun (≈ 5800 K) peaks at about 0.50 μm, in the visible; a furnace wall at 1200 K peaks at 2.4 μm; a room-temperature surface (300 K) at about 9.7 μm, in the far infrared. This is why glass, transparent to visible light but opaque beyond about 3 μm, traps solar energy (the greenhouse effect).

Stefan–Boltzmann law. Integrating Planck's law over all wavelengths gives total black-body emissive power E_b = σT⁴, with σ = 5.67 × 10⁻⁸ W/m²·K⁴ and T in kelvin. Doubling absolute temperature multiplies emission 16 times.

Real surfaces and emissivity. Emissivity ε = E/E_b (0 to 1) compares a real surface with a black body at the same temperature. It depends on material, surface finish, temperature, wavelength and direction: polished metals 0.02–0.1, oxidised metals 0.3–0.8, paints, refractories, water and skin 0.85–0.97. A gray body has ε independent of wavelength; most engineering calculations assume diffuse gray surfaces.

Kirchhoff's law. For a surface in thermal equilibrium with its surroundings, its emissivity equals its absorptivity at each wavelength and direction: ε_λ = α_λ. The total form ε = α holds for gray surfaces (or when the incident radiation comes from a black body at the same temperature as the surface). Good absorbers are good emitters. Solar collectors use selective surfaces: high α in the short-wavelength solar band and low ε in the long-wavelength infrared band where the surface itself emits — allowed because the two bands are different wavelengths.

Net exchange with large surroundings. A small gray body of area A₁ inside large surroundings at T_sur: q = ε₁·σ·A₁·(T₁⁴ − T_sur⁴). This can be written with a radiation coefficient h_r so it can be added to a convective h.

Formulas

E_b = σ·T⁴ Black-body emissive power (W/m²); σ = 5.67 × 10⁻⁸ W/m²·K⁴; T absolute (K).

E = ε·σ·T⁴ Real (gray) surface; ε emissivity (–).

E_bλ = C₁/{λ⁵·[exp(C₂/(λ·T)) − 1]} Planck's law (W/m²·μm); C₁ = 3.742 × 10⁸ W·μm⁴/m², C₂ = 1.439 × 10⁴ μm·K, λ in μm.

λ_max·T = 2898 μm·K Wien's displacement law.

α + ρ + τ = 1; opaque: α + ρ = 1 Radiative properties.

ε_λ = α_λ; gray surface: ε = α Kirchhoff's law.

q = ε₁·σ·A₁·(T₁⁴ − T_sur⁴) Small gray body in large enclosure (W).

h_r = ε·σ·(T₁² + T_sur²)·(T₁ + T_sur) Radiation heat transfer coefficient (W/m²·K), so q = h_r·A₁·(T₁ − T_sur).

Worked examples

Example 1 (standard). An uninsulated steam pipe of 100 mm outside diameter has a surface temperature of 150 °C and emissivity 0.8. The surroundings are at 25 °C. Find the radiation loss per metre and the equivalent radiation coefficient. Also find the wavelength of peak emission from the pipe.

Given: T₁ = 423.15 K, T_sur = 298.15 K, ε = 0.8, A₁ = π × 0.1 × 1 = 0.3142 m².

1. q = ε·σ·A₁·(T₁⁴ − T_sur⁴) = 0.8 × 5.67 × 10⁻⁸ × 0.3142 × (3.206 × 10¹⁰ − 0.790 × 10¹⁰).
2. q = 0.8 × 5.67 × 10⁻⁸ × 0.3142 × 2.416 × 10¹⁰ = 344 W per metre.
3. h_r = ε·σ·(T₁² + T_sur²)·(T₁ + T_sur) = 0.8 × 5.67 × 10⁻⁸ × (179 056 + 88 893) × 721.3 = 8.77 W/m²·K (check: 8.77 × 0.3142 × 125 = 344 W).
4. λ_max = 2898/423.15 = 6.85 μm (infrared).

Answer: q ≈ 344 W/m; h_r ≈ 8.8 W/m²·K, comparable to or larger than the free-convection coefficient; λ_max ≈ 6.9 μm.

Example 2 (GATE level) — radiation error of a thermocouple. A thermocouple bead (ε = 0.6) in a gas duct reads 650 K. The duct walls are at 500 K, and the gas-to-bead convective coefficient is 80 W/m²·K. Neglecting conduction along the leads, find the true gas temperature.

1. Steady energy balance on the bead (per unit area): convection gained = radiation lost to the walls.
2. h·(T_g − T_tc) = ε·σ·(T_tc⁴ − T_w⁴).
3. T_tc⁴ − T_w⁴ = 650⁴ − 500⁴ = 1.785 × 10¹¹ − 0.625 × 10¹¹ = 1.160 × 10¹¹ K⁴.
4. Radiation loss = 0.6 × 5.67 × 10⁻⁸ × 1.160 × 10¹¹ = 3947 W/m².
5. T_g = T_tc + 3947/80 = 650 + 49.3 = 699.3 K.

Answer: true gas temperature ≈ 699 K; the thermocouple reads about 49 K low. A radiation shield, a smaller (higher-h) bead or a lower-emissivity bead reduces the error.

Common mistakes

• Using °C in σT⁴ or Wien's law; radiation formulas need kelvin.
• Writing σ·ε·(T₁ − T₂)⁴ instead of σ·ε·(T₁⁴ − T₂⁴).
• Applying Kirchhoff's total ε = α to a non-gray surface irradiated by a source at a very different temperature (e.g. solar on a selective surface).
• Confusing emissive power (emitted) with irradiation (incident) and radiosity (leaving: emitted plus reflected).
• Forgetting that the Wien constant is in μm·K (2898) or m·K (2.898 × 10⁻³).

For GATE CH

Expect direct Stefan–Boltzmann and Wien calculations, ratio questions (emission when T doubles), net exchange of a small body with large surroundings, combined convection–radiation losses, thermocouple radiation errors, and conceptual statements about black bodies, gray bodies and Kirchhoff's law. Practise keeping four significant figures in T⁴ terms, since differences of large numbers lose accuracy quickly.

Quick check

1. By what factor does black-body emission rise if T goes from 300 K to 600 K?
2. Peak wavelength of a 1000 K black body?
3. An opaque surface has ρ = 0.3. What is α? If gray, what is ε?
4. Which law justifies ε_λ = α_λ?
5. E_b at 1000 K?

Answers: 1. 16. 2. 2.90 μm. 3. α = 0.7, ε = 0.7. 4. Kirchhoff's law. 5. 56.7 kW/m².