Heat Transfer · Chapter 11
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Heat Transfer · Chapter Eleven

Fundamentals of Thermal Radiation

Every surface above absolute zero emits energy as electromagnetic waves. Radiation needs no medium, travels at light speed, and scales with the fourth power of temperature — making it the dominant mechanism at high temperatures.

This chapter introduces the electromagnetic nature of thermal radiation, the idealized blackbody and its laws, the concept of radiation intensity, and the real-surface properties — emissivity, absorptivity, reflectivity, transmissivity — that govern radiative exchange.

Learning objectives

  • Explain thermal radiation as spectral and directional electromagnetic transport.
  • Calculate blackbody emissive power and peak wavelength using absolute temperature.
  • Relate intensity, solid angle, emissive power, irradiation, and radiosity.
  • Distinguish spectral, total, directional, and hemispherical surface properties.
  • Apply energy partition and Kirchhoff's law only under appropriate conditions.
  • Audit radiation calculations using bounds, units, and limiting behavior.
11–1

Introduction

Radiation transports energy by electromagnetic waves and needs no material medium. All matter above \(0\) K emits, but net radiative heat transfer requires an imbalance between what a surface emits and what it absorbs. Conduction and convection may occur simultaneously at the same boundary.

Radiation as Electromagnetic Waves

Thermal agitation of charged particles produces a spectrum commonly important from about \(0.1\) to \(100\,\mu{\rm m}\). Vacuum does not suppress it; intervening gases may be transparent or participating depending on molecular species and wavelength.

Physical meaning

Emission is a one-way property of a surface state. Net exchange is an energy balance involving irradiation from the environment; \(E\) alone is never the net heat loss.

Problem-solving strategy

  1. Identify whether the target is emitted power, incident power, a spectral band, or net exchange.
  2. Convert every temperature in a fourth-power relation to kelvins.
  3. Choose blackbody, gray, spectral, diffuse, or directional modeling from the available property data.
  4. Use blackbody fraction tables with \(\lambda T\) when a wavelength band is specified.
  5. Audit \(0\le\alpha,\rho,\tau,\varepsilon\le1\), energy partition, units, and hot-to-cold direction.
11–2

Thermal Radiation

The spectrum depends strongly on temperature: increasing \(T\) raises total emission and shifts the distribution toward shorter wavelength. Spectral emissive power \(E_\lambda\) has units W/(m²·µm) when wavelength is measured in micrometres; integration over wavelength yields \(E\) in W/m².

Spectral Nature

A total property is a spectrum-weighted average, not an arithmetic average. Likewise, a hemispherical quantity integrates directional intensity over the outward hemisphere with a projected-area factor \(\cos\theta\).

11–3

Blackbody Radiation

A blackbody absorbs all incident radiation and emits the maximum possible energy at each wavelength and direction. It is a reference idealization; a cavity with a small aperture can approximate it. Planck's law requires \(\lambda\) and its constants to use a consistent unit convention.

Planck's Law and the Stefan–Boltzmann Law

Planck's law gives the spectral distribution of blackbody emission as a function of wavelength and temperature. Integrating it over all wavelengths gives the Stefan–Boltzmann law: total blackbody emissive power rises with the fourth power of absolute temperature.

\[ E_{b\lambda}(T) = \frac{C_1}{\lambda^5\left[\exp(C_2/\lambda T)-1\right]}, \qquad E_b = \sigma T^4 \]

Wien's Displacement Law

The wavelength at which blackbody emission peaks is inversely proportional to absolute temperature: hotter bodies radiate most strongly at shorter wavelengths. This is why a heated metal glows red, then white, as it gets hotter.

\[ (\lambda T)_{\max\text{-power}} = 2897.8\ \mu\text{m·K} \]

Blackbody radiation in a band (fraction function)
\[ f_{0\to\lambda} = \frac{\int_0^{\lambda} E_{b\lambda}\,d\lambda}{\sigma T^4}, \qquad f_{\lambda_1\to\lambda_2} = f_{0\to\lambda_2} - f_{0\to\lambda_1} \]
EXAMPLE 1Calibration load from a furnace aperture
Problem. A \(0.0150\) m² furnace aperture behaves as a blackbody at \(950\) K. Find its total emitted power and peak wavelength for specifying an infrared calibration sensor.

Strategy

The aperture is explicitly black and isothermal, so use Stefan–Boltzmann for total emission and Wien for spectral peak.

Givens

  • \(A=0.0150\) m², \(T=950\) K, \(\sigma=5.670\times10^{-8}\) W/(m²·K&sup4;).

Physical/model interpretation

The requested power is outward emission, not net exchange with the sensor or room.

Numbered derivation
1. \(E_b=\sigma T^4\).
2. \(E_b=(5.670\times10^{-8})(950)^4=4.618\times10^4\) W/m².
3. \(\dot Q_e=AE_b=(0.0150)(4.618\times10^4)=693\) W.
4. \(\lambda_{\max}=2897.8/950=3.05\,\mu{\rm m}\).

Unit/sign/limit checks

Units: \(E_bA\) gives W and \((\mu{\rm m\,K})/{\rm K}\) gives \(\mu{\rm m}\).

Signs: emission is reported positive outward; net loss would be smaller in warm surroundings.

Limit: increasing \(T\) raises power as \(T^4\) and shortens peak wavelength, consistent with both laws.

Final answer: \(\boxed{\dot Q_e=693\ {\rm W},\ \lambda_{\max}=3.05\,\mu{\rm m}}\).

Engineering insight

A detector centered far from \(3\,\mu{\rm m}\) may miss much of the signal; actual calibration also requires aperture view factor and spectral detector response.

11–4

Radiation Intensity

Intensity \(I\) resolves radiation by direction and has units W/(m²·sr), based on area projected normal to the ray. The solid-angle element is \(d\omega=\sin\theta\,d\theta\,d\phi\), and the \(\cos\theta\) projection prevents grazing rays from contributing as strongly.

Emissive power from intensity (diffuse surface)
\[ E = \int_{\text{hemisphere}} I\cos\theta\,d\omega = \pi I \quad (\text{diffuse emitter}) \]
A diffuse surface radiates equally in all directions, so total emissive power is \(\pi\) times the intensity.

Solid Angle and Irradiation

The solid angle \(\omega\) (steradians) is the radiation analogue of a planar angle. Irradiation \(G\) is the radiation flux arriving at a surface from all directions; radiosity \(J\) is the total leaving it (emitted plus reflected).

EXAMPLE 2Power emitted into an inspection cone
Problem. A diffuse \(0.0800\) m² hot target has constant intensity \(I=1200\) W/(m²·sr). Determine the power emitted within a cone of half-angle \(30^\circ\) about the surface normal.

Strategy

This is a directional subset, so integrate \(I\cos\theta\,d\omega\) over the cone rather than using the full-hemisphere relation \(E=\pi I\).

Givens

  • \(A=0.0800\) m², \(I=1200\) W/(m²·sr), \(0\le\theta\le30^\circ\), \(0\le\phi\le2\pi\).

Physical/model interpretation

Diffuse means \(I\) is direction-independent, but projected area still introduces \(\cos\theta\).

Numbered derivation
1. \(\dot Q=A\int_0^{2\pi}\int_0^{\theta_c}I\cos\theta\sin\theta\,d\theta\,d\phi\).
2. Integration gives \(\dot Q=AI\pi\sin^2\theta_c\).
3. The cone emissive flux is \(\pi(1200)\sin^230^\circ=942\) W/m².
4. \(\dot Q=(0.0800)(942)=75.4\) W.

Unit/sign/limit checks

Units: steradian is dimensionless, leaving W after multiplication by area.

Signs: \(\cos\theta\ge0\) in the outward cone, so power is positive.

Limit: at \(\theta_c=90^\circ\), the expression becomes \(AI\pi\); this cone captures \(\sin^230^\circ=25\%\) of it.

Final answer: \(\boxed{\dot Q_{\rm cone}=75.4\ {\rm W}}\).

Engineering insight

Optical collection depends on projected solid angle, not merely the cone's geometric solid-angle fraction.

11–5

Radiative Properties

Real surfaces emit less than a blackbody and partition irradiation into absorption, reflection, and transmission. These properties can depend on wavelength, direction, temperature, finish, oxidation, and contamination; tabulated values require a matching condition.

Emissivity, Absorptivity, Reflectivity, Transmissivity

Emissivity \(\varepsilon=E/E_b\). Incident radiation satisfies \(\alpha+\rho+\tau=1\); opaque engineering surfaces have \(\tau=0\), but glass and thin films may not. A gray model assumes spectral independence over the important bands.

Kirchhoff's Law

At thermal equilibrium, matching spectral and directional emissivity and absorptivity are equal. Equality of total values requires compatible source and surface spectra; it can fail as a shortcut when solar irradiation and low-temperature emission occupy different bands.

\[ \varepsilon(T) = \alpha(T) \]

11–6

Atmospheric and Solar Radiation

The sun is often approximated as a \(5780\) K source. Atmospheric attenuation makes ground-level irradiance directional, spectral, weather-dependent, and generally below the extraterrestrial solar constant. Solar absorptivity \(\alpha_s\) and long-wave emissivity \(\varepsilon\) must therefore be kept distinct.

Solar constant
\[ G_s \approx 1373\ \text{W/m}^2 \]
Course Note

The key to solar design is that absorptivity depends on the source temperature. A surface can have low absorptivity for the sun's short waves but high emissivity for its own long-wave radiation — this is exactly how selective coatings on solar collectors and spacecraft radiators work. Don't assume \(\alpha = \varepsilon\) when the source is the sun.

EXAMPLE 3Radiative balance of a selective roof coupon
Problem. A coated roof coupon at \(330\) K receives \(850\) W/m² solar irradiation. Its measured solar absorptivity is \(0.25\) and long-wave emissivity is \(0.85\). With sky surroundings idealized at \(300\) K, find the net radiative heat gain per area, neglecting convection.

Strategy

Separate short-wave absorption from long-wave exchange; do not invoke \(\alpha_s=\varepsilon\) across dissimilar spectra.

Givens

  • \(G_s=850\) W/m², \(\alpha_s=0.25\), \(\varepsilon=0.85\), \(T_s=330\) K, \(T_{\rm sky}=300\) K.

Physical/model interpretation

Positive flux is into the coupon. The sky is a large long-wave surrounding; atmospheric spectral detail and convection are excluded.

Numbered derivation
1. Absorbed solar flux: \(q''_{\rm sol}=\alpha_sG_s=(0.25)(850)=212.5\) W/m².
2. Long-wave loss: \(q''_{\rm LW}=\varepsilon\sigma(T_s^4-T_{\rm sky}^4)\).
3. \(q''_{\rm LW}=0.85(5.670\times10^{-8})(330^4-300^4)=181.2\) W/m².
4. \(q''_{\rm net}=212.5-181.2=31.3\) W/m² inward.

Unit/sign/limit checks

Units: \(\alpha_sG_s\) retains W/m² because absorptivity is dimensionless, while \(\varepsilon\sigma(T_s^4-T_{\rm sky}^4)\) gives \((1)[{\rm W/(m^2K^4)}]({\rm K^4})={\rm W/m^2}\); therefore their difference is dimensionally valid.

Signs: the hotter coupon emits a long-wave loss; solar absorption is inward.

Limit: with \(G_s=0\), net radiation is outward; with \(T_s=T_{\rm sky}\), the long-wave term vanishes.

Final answer: \(\boxed{q''_{\rm net}=31.3\ {\rm W/m^2\ inward}}\).

Engineering insight

The small remainder is sensitive to sky temperature and convection; a full roof-temperature prediction needs both plus conduction and weather data.

Chapter recap

  • Blackbody laws set the upper emission scale and spectral peak.
  • Intensity carries directional information; emissive power is its projected hemispherical integral.
  • Real-surface properties are bounded, spectrum-dependent energy fractions.
  • Solar and thermal bands usually require separate properties.

Before moving on

You should be able to use kelvins in radiation laws, distinguish emission from net exchange, integrate diffuse intensity, partition irradiation, and state when gray and Kirchhoff simplifications apply.