Radiation: black body, view factors and radiosity networks

Why it matters

Radiation needs no medium and grows with the fourth power of absolute temperature, so it dominates in furnaces, reflow ovens, engine exhausts, spacecraft and vacuum equipment, and it is a large share of the cooling of any enclosure or heat sink in still air. Infrared thermometers and thermal cameras work on the same laws, so emissivity errors become measurement errors.

Key ideas

Thermal radiation is electromagnetic radiation emitted because of temperature, mostly at wavelengths of about 0.1–100 μm. All surfaces above 0 K emit; the net exchange is what matters for heat transfer.

Black body. An ideal surface that absorbs all incident radiation at every wavelength and direction, and therefore emits the maximum possible radiation at each temperature and wavelength (it is a perfect absorber and emitter, and a diffuse emitter). Its spectrum is given by Planck's law; its peak moves to shorter wavelengths as temperature rises (Wien's law); its total emissive power is σT⁴ (Stefan–Boltzmann law).

Real surfaces. Emissivity ε = E/E_b (0 to 1). Incident radiation G is absorbed, reflected or transmitted: α + ρ + τ = 1; for an opaque surface τ = 0, so α + ρ = 1. Kirchhoff's law: at the same wavelength and temperature, ε_λ = α_λ. A gray surface has ε independent of wavelength, so ε = α. Polished metals have low ε (0.03–0.1); oxidised metals, paints of any colour, and anodised aluminium are high (0.8–0.95) in the infrared. White paint is a good emitter but a poor absorber of sunlight, because α for solar wavelengths differs from ε at room temperature.

View factor F_ij. The fraction of radiation leaving surface i (diffusely) that strikes surface j directly. It depends only on geometry, not on temperature or emissivity. Rules:

• Summation: Σ_j F_ij = 1 in an enclosure.
• Reciprocity: A_i F_ij = A_j F_ji.
• A flat or convex surface cannot see itself: F_ii = 0. A concave surface can.
• If surface 1 is completely enclosed by surface 2, F₁₂ = 1.

Radiosity and the network method. Radiosity J is all radiation leaving a surface per unit area: emitted plus reflected, J = εE_b + ρG. For a gray, diffuse, opaque surface the net heat leaving is Q = (E_b − J)/[(1 − ε)/(εA)] — a surface resistance. Between two surfaces, Q = (J_i − J_j)/(1/(A_iF_ij)) — a space resistance. Draw the network like an electrical circuit with E_b as source potentials. A black surface has zero surface resistance, so J = E_b. A re-radiating (insulated) surface has net Q = 0, and its J floats.

Radiation shields. A thin, low-ε sheet between two surfaces adds two surface resistances and one space resistance. N shields of the same ε as the surfaces cut the exchange to 1/(N + 1) of the unshielded value.

Formulas

E_b = σ T⁴ — σ = 5.67 × 10⁻⁸ W/m²·K⁴, T in kelvin, E_b (W/m²). λ_max T = 2898 μm·K — Wien's displacement law. E = ε σ T⁴ — gray surface. α + ρ + τ = 1; opaque: α + ρ = 1; gray (Kirchhoff): ε = α. Σ F_ij = 1, A_i F_ij = A_j F_ji J = ε E_b + (1 − ε) G — opaque gray surface (W/m²). R_surface = (1 − ε) / (ε A), R_space = 1 / (A_i F_ij) — m⁻². Q₁₂ = σ (T₁⁴ − T₂⁴) / [ (1 − ε₁)/(ε₁A₁) + 1/(A₁F₁₂) + (1 − ε₂)/(ε₂A₂) ] — two-surface enclosure. q = σ (T₁⁴ − T₂⁴) / (1/ε₁ + 1/ε₂ − 1) — large parallel plates (W/m²). Q = ε₁ σ A₁ (T₁⁴ − T₂⁴) — small body (A₁ ≪ A₂) in a large enclosure. Q = σ A₁ (T₁⁴ − T₂⁴) / [1/ε₁ + (A₁/A₂)(1/ε₂ − 1)] — concentric cylinders or spheres (1 inside 2). h_r = ε σ (T_s² + T_sur²)(T_s + T_sur) — radiation coefficient, so that q = h_r (T_s − T_sur).

Worked examples

Example 1 (standard, parallel plates and a shield). Two large parallel plates are at 800 K (ε₁ = 0.8) and 500 K (ε₂ = 0.6). Find the net exchange per m². Then find it with one thin shield of ε = 0.05 on both sides placed between them.

1. σ(T₁⁴ − T₂⁴) = 5.67 × 10⁻⁸ × (800⁴ − 500⁴) = 19 681 W/m².
2. Without shield: 1/ε₁ + 1/ε₂ − 1 = 1.25 + 1.667 − 1 = 1.917; q = 19 681 / 1.917 = 10 268 W/m².
3. With the shield there are two gaps in series: (1/ε₁ + 1/ε_s − 1) + (1/ε_s + 1/ε₂ − 1) = (1.25 + 20 − 1) + (20 + 1.667 − 1) = 20.25 + 20.667 = 40.917.
4. q = 19 681 / 40.917 = 481 W/m², a 95 % reduction.

q ≈ 10.3 kW/m² without the shield; ≈ 0.48 kW/m² with it

Example 2 (GATE level, view factors and a pipe in a room). (a) A flat disc of radius R is covered by a hemispherical dome of the same radius. Find F₂₁ and F₂₂ for the dome (surface 2). (b) A horizontal steam pipe, 0.1 m outer diameter and 1 m long, at 400 K with ε = 0.8 sits in a large room whose walls are at 300 K. Find the radiation loss.

1. (a) The flat disc cannot see itself, and all its radiation hits the dome: F₁₂ = 1.
2. Reciprocity: F₂₁ = A₁F₁₂ / A₂ = πR² × 1 / (2πR²) = 0.5.
3. Summation: F₂₂ = 1 − 0.5 = 0.5 (the concave dome sees itself).
4. (b) Small body in large enclosure: A₁ = π × 0.1 × 1 = 0.3142 m².
5. Q = ε σ A₁ (T₁⁴ − T₂⁴) = 0.8 × 5.67 × 10⁻⁸ × 0.3142 × (400⁴ − 300⁴) = 0.8 × 5.67 × 10⁻⁸ × 0.3142 × 1.75 × 10¹⁰ = 249 W.

F₂₁ = 0.5, F₂₂ = 0.5; Q ≈ 249 W per metre of pipe

Common mistakes

• Using °C in σT⁴. Always convert to kelvin before raising to the fourth power.
• Taking (T₁ − T₂)⁴ instead of T₁⁴ − T₂⁴.
• Assuming F₁₂ = F₂₁; they are equal only when A₁ = A₂.
• Giving a convex or flat surface a non-zero self view factor.
• Forgetting that the parallel-plate formula 1/ε₁ + 1/ε₂ − 1 applies only to large plates (F₁₂ = 1, A₁ = A₂).
• Treating colour as emissivity: white paint has a high infrared emissivity.

For GATE ME

Expect view-factor algebra (summation and reciprocity for spheres, hemispheres, cylinders, cubes), net exchange between parallel plates or concentric cylinders and spheres, the effect of radiation shields, Wien and Stefan–Boltzmann calculations, and conceptual questions on Kirchhoff's law, gray bodies and radiosity. Practise building the resistance network for two- and three-surface enclosures, including one re-radiating surface.

Quick check

1. By what factor does black-body emissive power rise when absolute temperature doubles?
2. A sphere of area A₁ is inside a cube of area A₂. What is F₂₁?
3. For an opaque gray surface with ε = 0.3, what is its reflectivity?
4. One shield of the same ε as two parallel plates reduces radiation by what fraction?
5. At what wavelength does a 1000 K black body emit most strongly?

Answers: 1. 16. 2. A₁/A₂. 3. 0.7. 4. Half (to 1/2). 5. 2.898 μm.