Radiation heat transfer basics

Black and grey bodies, Stefan–Boltzmann and Wien's laws, Kirchhoff's law, view factors, radiation exchange between surfaces and radiation shields.

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Why it matters

Above a few hundred degrees Celsius, radiation dominates heat transfer. It controls heating in furnaces and heat-treatment ovens, heat loss from ladles, molten-metal surfaces and hot castings, the thermal load on operators near a furnace, and the readings of infrared pyrometers used to measure process temperatures. Radiation shields and low-emissivity coatings are standard tools for controlling it.

Key ideas

Nature of thermal radiation. All matter above absolute zero emits electromagnetic radiation because of its temperature. Thermal radiation lies roughly in the 0.1–100 μm wavelength band (part ultraviolet, all visible, and infrared). It needs no medium; it travels best through a vacuum or transparent gases. Unlike conduction and convection, it depends on the fourth power of absolute temperature, which is why it dominates at high temperature.

Black body. An ideal surface that absorbs all incident radiation at every wavelength and direction, and therefore also emits the maximum possible radiation at any temperature (a perfect emitter). Its total emissive power is E_b = σT⁴ (Stefan–Boltzmann law), with σ = 5.67 × 10⁻⁸ W/m²·K⁴. A small hole in a large cavity is a near-perfect black body.

Spectral distribution. Planck's law gives how black-body emission is spread over wavelength. The peak shifts to shorter wavelengths as temperature rises — Wien's displacement law: λ_max·T = 2898 μm·K. The Sun (about 5800 K) peaks near 0.5 μm (visible); a body at room temperature peaks near 10 μm (infrared). This is why steel glows red, then orange, then white as it is heated.

Real surfaces. For incident radiation, absorptivity α + reflectivity ρ + transmissivity τ = 1; for an opaque solid τ = 0, so α + ρ = 1. Emissivity ε = E/E_b (0 to 1). Polished metals have low ε (0.02–0.1); oxidised metals, refractories, paints and most non-metals are high (0.8–0.95). Values depend on surface condition and temperature — take them from a data book.

Kirchhoff's law. For a surface in thermal equilibrium, ε = α at each wavelength. A grey body has ε independent of wavelength, so for a grey surface ε = α overall. Most engineering calculations assume diffuse, grey surfaces.

View (shape, configuration) factor. F₁₂ is the fraction of radiation leaving surface 1 that strikes surface 2 directly. It depends only on geometry. Rules:

  • Summation: ΣF₁ⱼ = 1 for an enclosure.
  • Reciprocity: A₁F₁₂ = A₂F₂₁.
  • A flat or convex surface cannot see itself: F₁₁ = 0. A concave surface can (F₁₁ > 0).
  • For a body completely enclosed by another, F₁₂ = 1.

Radiation exchange. Treat it as a network: each grey surface has a surface resistance (1 − ε)/(εA), and each pair a space resistance 1/(A₁F₁₂). Important results:

  • Small body (area A₁, ε₁) in large surroundings: Q = ε₁σA₁(T₁⁴ − T₂⁴).
  • Two large parallel plates: q = σ(T₁⁴ − T₂⁴)/(1/ε₁ + 1/ε₂ − 1).
  • Concentric cylinders or spheres: Q = σA₁(T₁⁴ − T₂⁴)/[1/ε₁ + (A₁/A₂)(1/ε₂ − 1)].

Radiation shields. A thin, highly reflective (low-ε) sheet placed between two surfaces adds two surface resistances and a space resistance, cutting the exchange sharply. With N shields of the same emissivity as two parallel plates, the heat flow drops to 1/(N + 1) of the unshielded value. Shields are used in furnaces, cryogenic tanks, thermocouple protection and fire-fighting suits.

Radiation heat transfer coefficient. For combined convection and radiation, radiation may be expressed as h_r = εσ(T₁² + T₂²)(T₁ + T₂), so that Q_rad = h_r·A·(T₁ − T₂).

Formulas

E_b = σ·T⁴

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

E = ε·σ·T⁴

  • Grey-body emission (W/m²).

λ_max·T = 2898 μm·K

  • Wien's displacement law.

α + ρ + τ = 1; opaque α + ρ = 1; Kirchhoff ε = α

  • Surface properties (dimensionless).

A₁·F₁₂ = A₂·F₂₁; Σ F₁ⱼ = 1

  • View-factor rules.

Q = ε₁·σ·A₁·(T₁⁴ − T₂⁴)

  • Small body in large enclosure (W).

q = σ·(T₁⁴ − T₂⁴)/(1/ε₁ + 1/ε₂ − 1)

  • Large parallel plates (W/m²).

q_N = q_0/(N + 1)

  • N shields, all emissivities equal.

h_r = ε·σ·(T₁² + T₂²)·(T₁ + T₂)

  • Radiation coefficient (W/m²·K).

Worked examples

Example 1 (standard): radiation loss from a steam pipe. Given: an uninsulated pipe, 100 mm outer diameter and 1 m long, has surface temperature 200 °C and ε = 0.8. It is in a large room whose walls are at 25 °C. Find the radiation loss and the equivalent radiation coefficient.

  1. A = π·D·L = π × 0.1 × 1 = 0.3142 m²
  2. Absolute temperatures: T₁ = 473.15 K, T₂ = 298.15 K.
  3. Pipe is small compared with the room: Q = ε·σ·A·(T₁⁴ − T₂⁴) = 0.8 × 5.67 × 10⁻⁸ × 0.3142 × (473.15⁴ − 298.15⁴)
  4. Q = 601.6 W
  5. h_r = Q/(A·ΔT) = 601.6/(0.3142 × 175) = 10.9 W/m²·K Answer: Q ≈ 602 W per metre of pipe; h_r ≈ 10.9 W/m²·K — comparable to natural convection, so radiation cannot be ignored even at 200 °C.

Example 2 (GATE level): parallel plates with a shield. Given: two large parallel plates at 800 K (ε₁ = 0.8) and 400 K (ε₂ = 0.6). Find the net radiation per m². Then a thin shield with ε = 0.05 on both sides is placed between them; find the new heat flux and the percentage reduction.

  1. σ(T₁⁴ − T₂⁴) = 5.67 × 10⁻⁸ × (800⁴ − 400⁴) = 21 772.8 W/m²
  2. Without shield: R = 1/0.8 + 1/0.6 − 1 = 1.9167, so q = 21 772.8/1.9167 = 11 360 W/m².
  3. With shield: R = (1/0.8 + 1/0.05 − 1) + (1/0.05 + 1/0.6 − 1) = 20.25 + 20.667 = 40.917
  4. q = 21 772.8/40.917 = 532 W/m²
  5. Reduction = 1 − 532/11 360 = 0.953 Answer: q ≈ 11.4 kW/m² without the shield; ≈ 532 W/m² with it — a 95% reduction

Example 3 (quick): Wien's law. A furnace at 1000 K radiates most strongly at λ_max = 2898/1000 = 2.9 μm (infrared), whereas the Sun at 5800 K peaks at about 0.5 μm.

Common mistakes

  • Using °C in T⁴ — always convert to kelvin first.
  • Writing ε(T₁ − T₂)⁴ instead of ε(T₁⁴ − T₂⁴).
  • Using the small-body formula for two parallel plates or for comparable-size enclosures.
  • Assuming a white-painted surface has low infrared emissivity; most paints are near 0.9 in the infrared whatever their colour.
  • Forgetting that a concave surface has a non-zero self-view factor.
  • Applying ε = α between surfaces at very different temperatures (e.g. solar absorptivity vs infrared emissivity) — valid only for grey surfaces.

For GATE PI

Expect Stefan–Boltzmann and Wien's-law numericals, view-factor algebra with reciprocity and summation (including enclosures and hemispheres), net exchange between parallel plates and concentric cylinders or spheres, effect of radiation shields, and conceptual questions on black, grey and real surfaces and Kirchhoff's law. Practise the resistance-network method; it handles every two-surface problem.

Quick check

  1. By what factor does black-body emission rise if T rises from 300 K to 600 K?
  2. What is F₁₁ for a flat plate?
  3. A sphere sits inside a larger sphere. What is F from the inner to the outer sphere?
  4. With 3 shields of equal emissivity between two plates of the same emissivity, by what factor does heat flow drop?

Answers: 1. 16 2. 0 3. 1 4. To one-quarter (1/(N + 1) = 1/4)

Interview questions

All Thermal and Fluids Engineering interview questions

Try answering each one aloud before you open it.

  1. 1.What is radiation heat transfer?Concept

    Radiation heat transfer is the process by which heat energy is emitted or absorbed in the form of electromagnetic waves. Unlike conduction and convection, radiation does not require a medium to transfer heat and can occur in a vacuum.

  2. 2.Explain the Stefan-Boltzmann Law.Concept

    The Stefan-Boltzmann Law states that the total energy radiated per unit surface area of a black body is directly proportional to the fourth power of the black body's absolute temperature. Mathematically, it is expressed as E = σT⁴, where E is the emissive power, σ is the Stefan-Boltzmann constant, and T is the absolute temperature in Kelvin.

  3. 3.What is emissivity and how does it affect radiation heat transfer?Concept

    Emissivity is a measure of a material's ability to emit thermal radiation compared to a perfect black body. It ranges from 0 to 1, where 1 represents a perfect black body. Higher emissivity means the material is more effective at emitting radiation, thus affecting the rate of heat transfer.

  4. 4.Why is radiation heat transfer important in industrial furnaces?Application

    Radiant exchange scales with T₁⁴ − T₂⁴, while convection scales roughly with T₁ − T₂, so at furnace temperatures (above about 800 °C) radiation from the flame, hot gases and refractory walls carries most of the heat to the charge. That is why furnace design focuses on wall emissivity, the view factor between walls and charge, and gas radiation from CO₂ and H₂O. The same T⁴ dependence makes radiation the main heat loss through openings and from hot ladles.

  5. 5.What happens to radiation heat transfer if the surface temperature of an object doubles?Application

    If the surface temperature of an object doubles, the radiation heat transfer increases by a factor of 16. This is because the Stefan-Boltzmann Law states that radiation is proportional to the fourth power of temperature (E = σT⁴), so (2T)⁴ = 16T⁴.

  6. 6.How does a vacuum affect heat transfer between two surfaces?Application

    A vacuum removes conduction and convection, because both need a medium, so radiation becomes the only mode. It does not make radiation itself any stronger: the exchange still depends on emissivities, view factor and T₁⁴ − T₂⁴. This is why vacuum flasks and cryogenic tanks combine a vacuum gap with low-emissivity silvered or multilayer shield surfaces — the vacuum kills conduction and convection, the low emissivity cuts radiation.

  7. 7.Calculate the radiation emitted by a black body at 500 K with an area of 2 m². Use σ = 5.67 × 10⁻⁸ W/m²·K⁴.Numerical

    Use Q = σ·A·T⁴. T⁴ = 500⁴ = 6.25 × 10¹⁰ K⁴, so Q = 5.67 × 10⁻⁸ × 2 × 6.25 × 10¹⁰ = 7087.5 W, about 7.09 kW. This is the emitted power; the net loss would subtract what the body absorbs from its surroundings.

  8. 8.What is the role of view factors in radiation heat transfer?Concept

    View factors, also known as configuration factors, quantify the fraction of radiation leaving one surface that directly reaches another surface. They are crucial in calculating radiation exchange between surfaces, especially in complex geometries, as they account for the orientation and distance between surfaces.

  9. 9.Explain how radiation shields work and where they are used.Application

    A shield is a thin sheet of low-emissivity (highly reflective) material placed between two surfaces. It adds two large surface resistances (1 − ε)/(εA) and an extra space resistance to the radiation network, so the net exchange falls sharply. For large parallel plates, N shields with the same emissivity as the plates cut the heat flow to 1/(N + 1) of the unshielded value; low-ε shields cut it far more. They are used in cryogenic tanks (multilayer insulation), furnaces, thermocouple radiation shields and protective suits.

  10. 10.A surface with an emissivity of 0.8 is at 400 K. Calculate the radiation it emits per unit area. Use σ = 5.67 × 10⁻⁸ W/m²·K⁴.Numerical

    E = ε·σ·T⁴ = 0.8 × 5.67 × 10⁻⁸ × 400⁴. 400⁴ = 2.56 × 10¹⁰ K⁴, so E = 0.8 × 5.67 × 10⁻⁸ × 2.56 × 10¹⁰ = 1161 W/m². This is emission only; the net flux to surroundings at T₂ would be εσ(T₁⁴ − T₂⁴).

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