Radiation pyrometers

Why it matters

Molten steel, glass furnaces, rotating kiln shells, turbine blades and semiconductor wafers are too hot, too fast-moving or too delicate to touch. Radiation pyrometers read their temperature from the radiation they emit. The physics is exact; the practical errors come almost entirely from emissivity and from what lies between the target and the instrument.

Key ideas

Thermal radiation. Every body above absolute zero emits electromagnetic radiation whose spectrum depends on its temperature.

• A black body absorbs all incident radiation and emits the maximum possible at each wavelength. Its spectral distribution is given by Planck's law.
• Wien's displacement law: the wavelength of peak emission is inversely proportional to absolute temperature (λ_max·T = 2898 µm·K). At 1200 K the peak is at 2.4 µm, in the infrared; only at several thousand kelvin does it move into the visible.
• Wien's approximation to Planck's law (valid when λT is much less than C2 ≈ 14 388 µm·K) is used for spectral pyrometers.
• Stefan–Boltzmann law: total power radiated per unit area is σT⁴ for a black body and εσT⁴ for a real ("grey") body. Radiation is a strong function of temperature — the basis of high sensitivity at high temperature.
• Emissivity ε (0 to 1) is the ratio of a real surface's emission to a black body's at the same temperature. It depends on material, surface finish, oxidation, wavelength and angle: polished metals can be 0.05–0.2, oxidised steel about 0.8, most non-metals 0.9 or higher. By Kirchhoff's law, at equilibrium a surface's emissivity equals its absorptivity; for an opaque surface reflectivity = 1 − ε, so low-emissivity targets also reflect radiation from hot surroundings.

Types of pyrometer.

• Total radiation pyrometer: a lens or concave mirror focuses radiation over a broad band onto a thermopile (or bolometer); output ∝ ε(T⁴ − T_d⁴), where T_d is the detector temperature. Calibrated against a black body; reading is independent of distance as long as the target fills the field of view.
• Optical (disappearing-filament) pyrometer: the observer adjusts the current in a calibrated lamp filament until it vanishes against the image of the target, viewed through a red filter (about 0.65 µm). A narrow-band spectral method, used roughly from 700 °C to 3000 °C.
• Spectral (narrow-band) IR pyrometer: a photodetector with a filter measures radiance at one short wavelength. Shorter wavelengths reduce the error caused by an uncertain emissivity.
• Ratio (two-colour) pyrometer: measures radiance at two nearby wavelengths and uses their ratio; if emissivity is the same at both (grey body), it cancels. It also tolerates partial obstruction by dust or a target smaller than the field of view.

Practical points. Set the instrument's emissivity correctly; make sure the target fills the field of view (distance-to-spot ratio); avoid absorption by water vapour, CO₂, smoke and dirty windows (use an atmospheric window wavelength, purge air or sight tubes); beware reflected radiation from furnace walls; response is fast (milliseconds) because there is no thermal contact.

Formulas

λ_max·T = b, b = 2.898 × 10⁻³ m·K (Wien's displacement) λ_max = peak wavelength (m); T = absolute temperature (K).

E = ε·σ·T⁴, σ = 5.67 × 10⁻⁸ W/m²·K⁴ (Stefan–Boltzmann) E = emissive power (W/m²); ε = emissivity.

q = ε·σ·(T⁴ − T0⁴) (net exchange with surroundings at T0, large enclosure)

T = Ta / ε^(1/4) (total pyrometer emissivity correction; Ta = black-body-calibrated reading, K)

1/T = 1/Ta + (λ/C2)·ln ε (spectral pyrometer, Wien approximation) λ = operating wavelength (m); C2 = 1.4388 × 10⁻² m·K; Ta = indicated (brightness) temperature (K).

Worked examples

Example 1 — total radiation pyrometer. A total pyrometer calibrated for ε = 1 reads 1200 K on oxidised steel of emissivity 0.8. Ignore reflected and ambient radiation. Find the true temperature.

1. The instrument sees σ·Ta⁴ = ε·σ·T⁴, so T = Ta/ε^(1/4).
2. ε^(1/4) = 0.8^0.25 = 0.9457.
3. T = 1200/0.9457 = 1268.8 K.

Answer: about 1269 K — the reading was 69 K low.

Example 2 — spectral pyrometer (GATE level). The same target is viewed by a narrow-band pyrometer at λ = 0.65 µm, which reads (brightness temperature) 1200 K with ε set to 1. Using the Wien approximation, find the true temperature and compare with Example 1. Also find the peak wavelength at 1200 K.

1. 1/T = 1/Ta + (λ/C2)·ln ε = 1/1200 + (0.65 × 10⁻⁶ / 1.4388 × 10⁻²)·ln 0.8.
2. = 8.3333 × 10⁻⁴ + 4.5177 × 10⁻⁵ × (−0.2231) = 8.3333 × 10⁻⁴ − 1.0081 × 10⁻⁵ = 8.2325 × 10⁻⁴ K⁻¹.
3. T = 1214.7 K — an error of only about 15 K, against 69 K for the total pyrometer.
4. Peak wavelength: λ_max = 2.898 × 10⁻³ / 1200 = 2.415 × 10⁻⁶ m = 2.42 µm.

Answer: T ≈ 1215 K; λ_max ≈ 2.42 µm. Short-wavelength measurement reduces emissivity error.

Common mistakes

• Using °C in σT⁴ or Wien's law; both need kelvin.
• Thinking a low-emissivity setting error makes the reading high — an uncorrected ε < 1 makes the indicated temperature low.
• Assuming the reading depends on distance; it does not, as long as the target fills the field of view.
• Expecting a ratio pyrometer to fix non-grey targets; it cancels emissivity only if ε is the same at both wavelengths.
• Forgetting reflected radiation from hot surroundings when measuring shiny metal.

For GATE IN

NAT problems use Stefan–Boltzmann (emissive power, temperature from power, emissivity correction for total pyrometers) and Wien's displacement law; some use the Wien approximation for spectral pyrometers. MCQs compare total, optical and ratio pyrometers and test emissivity and Kirchhoff's law. Keep σ, b and C2 at your fingertips.

Quick check

1. What is the peak emission wavelength of a black body at 2898 K?
2. A grey body (ε = 0.9) emits 500 W/m². What is its temperature?
3. Does an uncorrected emissivity below 1 make a pyrometer read high or low?
4. Which pyrometer type largely cancels grey-body emissivity? Answers: 1. 1 µm; 2. (500/(0.9 × 5.67 × 10⁻⁸))^(1/4) = 314.6 K; 3. low; 4. the ratio (two-colour) pyrometer.