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Purdue ME 606: Radiation intensity, Surface properties, Kirchoff's Law (Lecture 3)

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Radiation Intensity

A method for examining and measuring radiation in Cartesian coordinates is through a quantity known as intensity.

Intensity is defined as:

I(\theta,\phi) = \frac{energy~per~unit~time}{(projected~area~of~emitter)(solid~angle)} ~ =\frac{dq''}{cos\theta~d\omega} ~ (eq. 1).

From this the heat flux is defined as:

dq''=I(\theta,\phi)cos(\theta)d\omega ~[\frac{W}{m^2}] ~(eq. 2).

The cosθ term results from the emitting surface needing to be normal to the direction of the emitted radiation.

To solve for the total heat flux, eq. 2 must be integrated over the angles of interest. Before this can be done a solid angle dω must be defined as,

d\omega = \frac{dA_s}{r^2} ~(eq. 3).

Here dAs is a differentially small area normal to the θ and φ directions at a radius, r, away from the emitting surface.

Figure 1 shows the concept of a solid angle [1]

For a hemisphere it can be shown that dAn = r2sinθdθdφ. Integrating over the surface of a hemisphere and sphere the resulting solid angles are 2π and 4π respectively.

The heat flux emitted over a hemisphere is found by combining eqs. 2 and 3 and using the definition of a solid angle over a sphere,

q'' = \int_0^{2\pi} \int_0^\frac{\pi}{2} I(\theta , \phi) cos\theta sin\theta d\theta d\phi ~ \frac{W}{m^2}~(eq. 4).

Note that if the heat flux is being calculated in non-spherical coordinates (e.g. heat transfer between two flat plates), eq. 4 does not hold because the solid angle would change. Instead the solid angle for the geometry would need to be calculated (i.e. area normal to the emitting surface divided by the distance squared)and then used in eqs. 2 and 3.

It should be noted that while intensity has been defined relative to an emitting surface, it can also be used to described radiation incident on a surface. In this case the normal area is the area of the object intercepting the radiation and the solid angle is formed by the cone above the object.

Surface Properties

Diffuse (Lambertian) Surface

A surface is diffuse if the intensity is emitted uniformly in all directions (theta and phi).

From equation (4) when a surface emits diffusely the intensity becomes independent of direction and the emissive power becomes,

E = q'' = I\int_0^{2\pi}\int_0^{\frac{\pi}{2}}cos\theta sin\theta d\theta d\phi = I\pi (eq. 5)

A blackbody is always a diffuse emitter Eb = πIb(T).

It is important to note that a diffuse emitter does not emit uniform energy in all directions, although the intensity is uniform. Recall that intensity is the rate of energy released per unit area normal (of the emitting surface) to the emitting direction. Consequently, the energy will be the greatest when the emitting surface is normal to the emitting direction and zero, when the emitting surface is parallel to the emitting direction.

Reflection, Absorption, and Transmission

When incident radiation reaches a surface the radiation maybe reflected (ρ), absorbed (α), and transmitted (τ). These properties can be spectrally independent or dependent. The sum of what is reflected, absorbed, and transmitted equals 1, α + ρ + τ = 1.

For opaque surface, \tau=0, ~1=\rho+\alpha.

For a perfect mirror surface, \rho=1, ~0=\tau=\alpha.

For a blackbody surface, \alpha=1, ~0=\rho=\tau.

Emissivity

Eλ(T)=Spectral emissive power with units of W/m^3 (it is m^3 because of the spectral dependence). E(T)=Emissive power with units of W/m^2.

The spectral emissive power of a blackbody with a constant refractive index is defined as,

E_{b\lambda}(T,\lambda)=\frac{2\pi hc^2}{n^2\lambda^5(e^{hc/n\lambda kT}-1)},

where k is the Boltzmann constant (1.3807×10^-27 J/K), h is Planck’s constant (6.626×10^-34 Js),n is the index of refraction, and C is the speed of light in a vacuum(2.998×10^8 m/s).

Integrating the blackbody spectral emissive power for all spectra simplifies to the Stefan-Boltzmann constant (σ) multiplied by the temperature (in Kelvin) to the fourth,

E_b = \int_0^\lambda E_{b\lambda} d\lambda ~ = \sigma T^4,

where σ equals 5.67×10^-8 (W/m^2K^4).

The spectral emissivity of an object is defined as,

\epsilon_\lambda = \frac{E_\lambda(T)}{E_{b\lambda}(T)}.

The total blackbody emissivity is,

\epsilon = \frac{E(T)}{E_b(T)} = \frac{\int_0^\inf E_\lambda(T)d\lambda}{\int_0^\inf E_{b\lambda}(T)d\lambda}=\frac{\int_0^\inf \epsilon_\lambda E_{b\lambda}(T)d\lambda}{\int_0^\inf E_{b\lambda}(T)d\lambda} .

Kirchoff’s Law

  1. Consider a closed cavity with uniform temperature. The radiation field within the cavity equals G(T).
  2. A small disk introduced into the cavity with arbitary surface properties
    • disk absorbs
      • \alpha A_{disk}G(T) ~=~\alpha A_{disk}E_b(T).
    • The rate that energy leaves the disk equals,
      • εAdiskEb(T).
    • Because the disk is at thermal equilibrium,
      • \alpha E_b(T)~=~ \epsilon E_b(T),
    • therefore
      • α= ε.

This holds when the incident or radiated intensity is from a blackbody.

  1. This thought experiment can be repeated to show that,
    • αλ= ελ

when the radiation or irradition is diffuse

and
    • αλθ= ελθ

holds for all conditions.

Last modified on 14 Oct, 2008