Heat Transfer · Chapter 10 of 10 · Advanced
Thermal Radiation
Every surface glows. Heat crosses empty space as electromagnetic waves, and the rate climbs with the fourth power of absolute temperature.
Readiness check
From Chapter 1, where radiation was introduced as the third mode. Tick only what you can do closed-notes.
- State that radiation needs no medium.
- Work in absolute temperature (kelvin).
- Raise a number to the fourth power confidently.
- Recall the convection rate law q = hA ΔT.
- Combine two parallel heat paths from a surface.
The core idea
Every surface emits electromagnetic energy at a rate set by the fourth power of its absolute temperature; net radiant exchange is the difference between what it emits and what it absorbs.
Eb = σT⁴σ = 5.67×10⁻⁸ W/m²·K⁴q = εσA(Ts⁴ − Tsur⁴)Radiation needs no medium, so it carries heat across a vacuum (the Sun to the Earth) and dominates at high temperature. Because of the fourth-power law, doubling absolute temperature multiplies emission sixteenfold, which is why radiation is negligible near room temperature but governs furnaces, filaments, and spacecraft.
The skills, taught in order
Radiation closes the course by returning to the third mode from Chapter 1, now with the laws to quantify it. Six skills take you from the blackbody ideal to a real surface losing heat by radiation and convection together.
10.1 Why radiation is different
Conduction and convection need matter; radiation does not. Energy leaves a surface as electromagnetic waves and crosses a vacuum freely. The rate depends on the fourth power of absolute temperature, not on a temperature difference, so it is steeply nonlinear.
10.2 Blackbody and the Stefan-Boltzmann law
A blackbody is the perfect emitter: Eb = σT⁴ with σ = 5.67×10⁻⁸ W/m²·K⁴. Its spectrum peaks at a wavelength given by Wien's law, λmaxT = 2898 μm·K, which shifts to shorter (visible) wavelengths as a body heats, the reason hot metal glows red then white.
10.3 Real surfaces: emissivity and absorptivity
A real surface emits a fraction ε of the blackbody value and absorbs a fraction α of incident radiation. For a gray surface in thermal equilibrium Kirchhoff's law gives ε = α. Emissivity ranges from about 0.03 for polished metal to 0.95 for matte black paint.
| Surface | Emissivity ε (approx.) | Behaviour |
|---|---|---|
| Polished aluminium | 0.03 to 0.05 | poor emitter; good radiation shield |
| Oxidised metal | 0.6 to 0.8 | emits strongly once tarnished |
| Matte paint, most non-metals | 0.85 to 0.95 | near-blackbody emitters |
| Human skin, water | ≈ 0.95 | nearly black in the infrared |
10.4 Surface-to-surroundings exchange
The practical workhorse: a small surface inside large surroundings loses q = εσA(Ts⁴ − Tsur⁴). Both temperatures are absolute, and the surroundings absorb everything (behaving as a blackbody enclosure), which is why only the surface emissivity appears.
10.5 View factors
When two finite surfaces exchange radiation, only part of what one emits reaches the other. The view factor Fij is that fraction. It obeys reciprocity AiFij = AjFji and, in an enclosure, the summation rule ΣFij = 1. For two large parallel plates each sees only the other, so F12 = 1.
10.6 Combined radiation and convection
A real surface usually loses heat both ways at once. Defining a radiation coefficient hrad = εσ(Ts+Tsur)(Ts²+Tsur²) lets you add it to the convection coefficient and treat the total loss as (hconv+hrad)A ΔT, closing the loop with Chapter 1.
Engineering connection: furnaces, incandescent and infrared heaters, spacecraft thermal control, building heat loss, and the multilayer insulation that exploits low-emissivity shields.
Worked example 1: a hot panel losing heat to a room
A 0.5 m² panel at 200 °C with emissivity 0.85 faces a large room whose walls are at 25 °C. Find the net radiant heat loss, and express it as an equivalent radiation coefficient.
- ProblemFind the net radiant loss from the panel in Figure 1, and the equivalent hrad.
- Given / findA = 0.5 m², ε = 0.85, Ts = 200 °C = 473 K, Tsur = 25 °C = 298 K. Find q and hrad.
- AssumptionsGray diffuse surface, large surroundings (so they act as a blackbody enclosure), steady state.
- ModelApply the surface-to-surroundings law, then define hrad = q/[A(Ts−Tsur)].
- Equationsq = εσA(Ts⁴ − Tsur⁴) hrad = q/[A(Ts − Tsur)]
- SolveTs⁴ = 473⁴ = 5.01×10¹⁰, Tsur⁴ = 298⁴ = 7.89×10⁹, difference = 4.22×10¹⁰ K⁴. q = 0.85 × 5.67×10⁻⁸ × 0.5 × 4.22×10¹⁰ = 1016 W ≈ 1.0 kW. Then hrad = 1016/(0.5 × 175) = 11.6 W/m²·K.
- CheckUsing Celsius (200⁴ − 25⁴) would have been meaningless; the kelvin values are essential. The radiation coefficient sits in the natural-convection range, so at this temperature radiation and free convection lose comparable heat.
- ConclusionA kilowatt leaves by radiation alone. Lowering emissivity (a polished or foil surface, ε ≈ 0.05) would cut this loss roughly seventeenfold, the principle behind reflective insulation and emergency blankets.
Worked example 2: exchange between two parallel plates
A furnace wall at 800 K (ε₁ = 0.8) faces a parallel workpiece plate at 400 K (ε₂ = 0.5), close enough that each sees only the other (F₁₂ = 1). Find the net radiant flux exchanged per unit area.
- ProblemFind the net radiant flux between the two parallel gray plates in Figure 2.
- Given / findT₁ = 800 K, ε₁ = 0.8; T₂ = 400 K, ε₂ = 0.5; F₁₂ = 1. Find q/A.
- AssumptionsLarge parallel plates (negligible edge loss), gray diffuse surfaces, no participating medium between them.
- ModelUse the two-surface gray enclosure result for infinite parallel plates, where the emissivity term combines the two surface resistances.
- Equationsq/A = σ(T₁⁴ − T₂⁴) / (1/ε₁ + 1/ε₂ − 1)
- SolveT₁⁴ = 800⁴ = 4.10×10¹¹, T₂⁴ = 400⁴ = 2.56×10¹⁰, difference = 3.84×10¹¹. Denominator = 1/0.8 + 1/0.5 − 1 = 1.25 + 2.0 − 1 = 2.25. q/A = 5.67×10⁻⁸ × 3.84×10¹¹ / 2.25 = 9680 W/m² ≈ 9.7 kW/m².
- CheckIf both plates were black (ε = 1), the denominator would be 1 and the flux would rise to σ(T₁⁴−T₂⁴) = 21.8 kW/m². The gray surfaces cut the exchange to 44% of the blackbody value, as their emissivities should.
- ConclusionInserting a single low-emissivity shield (ε ≈ 0.05) between the plates would add two large resistances and slash the flux by more than an order of magnitude, which is exactly how radiation shields and multilayer insulation work.
Misconceptions and diagnostics
| Mistake | Symptom | Diagnostic question | Correction |
|---|---|---|---|
| Using Celsius in T⁴ | Wildly wrong heat rate | "Are my temperatures in kelvin?" | The Stefan-Boltzmann law requires absolute temperature; convert first. |
| Forgetting the surroundings term | Emission counted but not absorption | "Did I subtract Tsur⁴?" | Net exchange is the difference of fourth powers, not σTs⁴ alone. |
| Ignoring radiation at high T | Underpredicted loss from hot surfaces | "Is hrad comparable to hconv?" | Above a few hundred °C radiation often rivals or beats convection; include it. |
| Confusing ε and α | Different values used arbitrarily | "Is the surface gray and near equilibrium?" | For a gray surface Kirchhoff's law gives ε = α at the same temperature. |
Practice ladder
Find the blackbody emissive power of a surface at 1000 K.
Show answer
Eb = σT⁴ = 5.67×10⁻⁸ × 1000⁴ = 5.67×10⁻⁸ × 10¹² = 56 700 W/m² ≈ 56.7 kW/m². A glowing surface radiates tens of kilowatts per square metre.
The Worked Example 1 panel also loses heat by natural convection at hconv ≈ 8 W/m²·K. Using hrad = 11.6, what fraction of the total loss is radiation?
Show answer
Radiation fraction = 11.6/(11.6+8) = 11.6/19.6 = 59%. At 200 °C radiation is already the larger of the two paths, and its share grows quickly as the surface gets hotter because of the T⁴ law.
For the Worked Example 2 plates, a thin radiation shield of emissivity 0.05 is placed between them. Estimate the new flux and explain the size of the drop.
Show answer
The shield adds two surfaces of ε = 0.05. The series resistance roughly becomes (1/0.8 + 1/0.05 − 1) + (1/0.05 + 1/0.5 − 1) ≈ 20.25 + 21 = 41.25 versus the original 2.25, so the flux falls by about 18 times, to roughly 0.5 kW/m². One low-emissivity sheet dominates the whole network.
Find a radiation-driven design (a wood stove, a halogen heater, a thermos flask, a survival blanket). Identify the temperatures and emissivities at play and explain the design choice in terms of the T⁴ law or emissivity control.
What good work looks like
Absolute temperatures used, the surface-to-surroundings or two-surface law applied, emissivity values justified, and the design explained as either exploiting or suppressing radiant exchange.
Working with AI, and proving it yourself
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Portfolio task
Write a one-page radiation note on a real hot surface: estimate its temperature and emissivity, compute the net radiant loss, and compare it with the convective loss to decide which mode dominates.
Retrieval and spaced review
Closed notes. Answer out loud, then reveal.
1. State the Stefan-Boltzmann law and the value of σ.
Eb = σT⁴ with σ = 5.67×10⁻⁸ W/m²·K⁴, T in kelvin.
2. Write the net loss from a surface to large surroundings.
q = εσA(Ts⁴ − Tsur⁴), both temperatures absolute.
3. What does Kirchhoff's law say for a gray surface?
Emissivity equals absorptivity, ε = α, at the same temperature.
4. Define the view factor and give its reciprocity rule.
Fij is the fraction of radiation leaving i that reaches j; reciprocity is AiFij = AjFji.
5. Why does radiation dominate at high temperature?
Because it scales as T⁴, while conduction and convection scale roughly linearly with ΔT, so radiation overtakes them as T rises.
Textbook mapping
| Item | Mapping |
|---|---|
| Primary source | Lienhard and Lienhard, A Heat Transfer Textbook (6th ed), Chapter 10 (radiative heat transfer) |
| Cross-reference | Incropera, Ch. 12 and 13 (radiation, exchange between surfaces) · Çengel and Ghajar, Ch. 12 and 13 |
| Core topics | 10.1 Nature of radiation · 10.2 Stefan-Boltzmann · 10.3 Emissivity · 10.4 Surroundings exchange · 10.5 View factors · 10.6 Combined modes |
| Engineering connection | Furnaces, infrared heaters, spacecraft thermal control, and low-emissivity insulation. |
| Read next | You have completed the Heat Transfer course. Return to the course hub to review. |