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8.4 Properties of real surfaces 253

2. Hemispherical total emissive power

Eb = εσ T 4

= 0.7224 × 5.67 ×10−8 ×10004

Eb = 40.964 kW/m2

3.Fraction of emissive power in the region 0 ≤ λ ≤ 6 m

λ1T = 4000 µmK : F(0 → λ1 ) = 0.4805

λ2T = 6000 µmK : F(0 → λ1 ) = 0.7374

ε

= 0.8 × 0.4805 + 0.6 × (0.7374 − 0.4805) 0.7224

Fraction = 0.7396

4.For the emissivity of a gray body at 1200 K to have the same hemispherical, total emissive power as this body as calculated in (2).

Eb = 40.964 × 103 W/m2 = εσ Tb4 here, Tb = 1200 K

ε= 40.964

5.67× 10−8 × 12004

ε= 0.348 × 103

Though we have given an example with an unrealistic, piecewise constant, spectral emissivity distribution, the same technique can be used for more complex, continuous emissivity distributions as well.

Hence, a smooth distibution can be considered to be made up of a large number of infinitesimal “boxes” using constant emissivities.

8.5 Kirchoff’s law 255

Eq. (8.56) is a key engineering result that relates the absorptivity and reflectivity of an opaque surface. Please note that the reflectivity, absorptivity and transmissivity are dimensionless quantities, having the range 0 to 1.

8.4.3  Spectral directional absorptivity

The quantity α appearing in Eq. (8.56) is an average quantity over an hemisphere and over the wavelength 0 to ∞. Now, we need to get down to the most basic definition for a particular angle (θ, φ) and a wavelength interval dλ about λ. This is denoted as spectral, directional absorptivity. The spectral directional absorptivity gives the ratio of radiation absorbed to that incident in a given direction and wavelength.

Mathematically, it is given by

For a gray diffuse surface α′λ = α. This is a major engineering simplification and is often not a bad assumption.

The relation between emissivity (ε) and absorptivity (α) cannot be obtained by theory because they come from completely different processes. Absorption is the capacity of the body to absorb, whereas emission is related to translation, rotation, and vibration of molecules. So, any relation between α and ε should come empirically from the experiments. Kirchoff was the first scientist who performed those experiments and proposed the famous Kirchoff’s law that relates emissivity and absorptivity.

8.5  Kirchoff’s law

Kirchoff’s law states that, the spectral directional absorptivity is equal to the spectral directional emissivity.

For a diffuse surface,

256CHAPTER 8  Thermal radiation

For a gray and diffuse surface,

Eq. (8.63) is one of the most important simplifications in radiation heat transfer. For an opaque and diffuse surface, if we know ε, then from the above equation we know α then and from Eq. (8.56) we know r. Hence we know every property that is required to perform radiation calculations involving surface to surface radiation heat transfer.

8.6  Net radiative heat transfer from a surface

Consider a thin, opaque surface exposed to an incident radiation of G (W/m2) as shown in Fig. 8.10.

At steady state, from energy balance, the net radiation from a surface is given as

The radiosity (usually denoted by J) of a surface is the leaving radiation from a surface, consisting of emission and reflection from a surface. The incoming radiation is also known as irradiation and is denoted by G.

FIGURE 8.10

Depiction of net radiation from an opaque surface.

8.6 Net radiative heat transfer from a surface 257

For an opaque and diffuse surface, 1− ρ = α = αλ

where ‘α’ is the total hemispherical absorptivity or simply absorptivity and is given by

In Eq. (8.71), the numerator represents the radiation absorbed by the surface in the wavelength interval 0 to ∞. The denominator shows the total incident radiation in the same wavelength range. It is straight forward so that, αl = αλ for a gray surface for which αλ≠ f(λ).

Example 8.6: For an opaque diffuse surface ελ varies with λ as shown in Fig. 8.11. Plot αλ vs. λ and ρλ vs. λ for this surface.

Solution:

From Kirchoff’s law, αλ = ελ, and for an opaque and diffuse surface,

ρλ = 1 − αλ

Hence αλ and ρλ will vary with λ as shown in Fig. 8.12.

FIGURE 8.11

Spectral emissivity distribution for example 8.6.

258 CHAPTER 8  Thermal radiation

FIGURE 8.12

Spectral distribution of (A) absorptivity and (B) reflectivity for example 8.6.

Example 8.7: Consider an opaque surface whose spectral hemispherical absorptivity varies with wavelength, as shown in Fig. 8.13(A). The spectral distribution of incident radiation is also shown in Fig. 8.13(B).

1.Determine the hemispherical total absorptivity of the surface.

2.If this surface is diffuse and is at 1200 K, what is its total hemispherical emissivity?

3.Determine the net radiation heat transfer from the surface.

FIGURE 8.13

Variation of (A) hemispherical spectral absorptivity and (B) spectral irradiance with wavelength.

8.6 Net radiative heat transfer from a surface 259

Solution:

1. Hemispherical total absorptivity of the surface.

α = ∫λ∞=0 αλ Gλi dλ

∫λ∞=0 Gλi dλ

Total irradiance or irradiation G in W/m2 is given by

G= 126000 W/m2

α= 0 × ∫01.8 Gλi dλ + 0.7 × ∫1.83.4 Gλi dλ + 0.7 × ∫3.45 Gλi dλ

126000 126000 126000

α= 0.48

2.Total hemispherical emissivity. For a diffuse surface, αλ = ελ

λ1 = 1.8 m, λ1T = 2160 mK

From the F function chart, we get F0→λ,T = 0.094

ε= 0.6342

3.Net radiation heat transfer from the surface.

Q = εσT 4 + ρG − G = εσT 4 − αG = 0.6342 × 5.67 × 10−8 × 12004 − 0.48 × 12600

Q = 14.09 kW/m2

Example 8.8: Solar flux of 975 W/m2 is incident on the top surface of a plate whose solar absorptivity is 0.9 and emissivity is 0.1. The air and surroundings are at 30 °C, and the convection heat transfer coefficient between the plate and air is 25 W/m2 K. Assuming that the bottom side of the plate is perfectly insulated, determine the steady state temperature of the plate. (Refer to Fig. 8.14)

8.7 Radiation heat transfer between surfaces 261

8.7  Radiation heat transfer between surfaces

There are two ways of dealing with radiation heat transfer between multiple surfaces.

1.Network method

2.Enclosure theory

The network method is hard to scale up for a large number of surfaces. In this book, our focus is on the enclosure theory. The enclosure theory was developed by Prof. E. M. Sparrow and his colleagues at the University of Minnesota, Minneapolis, in the United States in the early 1960s and was originally developed for gray diffuse evacuated surfaces. Let us now see how this method works.

Consider an N surface evacuated enclosure as shown in Fig. 8.15

According to the enclosure theory, if we account for all the radiation coming out of every surface (Ji) and find out what is the radiation falling on that surface (Gi) from the other surfaces, then the net radiative heat transfer from any surface is given by Eq. (8.73).

In Eq. (8.73), Ji is called the radiosity or outgoing radiation and Gi is called irradiation or incoming radiation both in W/m2 as already mentioned.

For example, consider a three-surface enclosure with the top surface exposed to atmosphere, as shown in Fig. 8.16A This problem can be handled using the enclosure theory by closing the open cavity with an imaginary surface having an emissivity (ε) = 1 at temperature T∞. Similarly, a single surface too can be handled by closing the surface with a hemispherical black body at temperature T∞ as shown in Fig. 8.16B.

This way, even open surfaces can be handled effectively. So the enclosure theory can be applied from N = 1 to N → ∞, where N is the number of surfaces in the enclosure.

FIGURE 8.15

A typical N surface evacuated enclosure.