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CHAPTER

 

 

 

Thermal radiation

8

 

8.1  Introduction

Any body above a temperature of zero Kelvin emits radiation. This is a fundamental law of nature and is known as Prevost’s law. Radiation occurs due to molecular activity that arises as a consequence of temperature. The molecular activity is due to the translational, rotational, and vibrational energy of matter. Radiation ceases only at absolute zero temperature. Hence, it stands to reason that any body, or anything that exists for that matter, has to and will emit radiation. It is quite intuitive to guess that radiation is a function of temperature and that it shows a monotonic increase with temperature. The challenge now is to characterize this radiation and find ways to quantify it as a mode of heat transfer in engineering problems, when several surfaces at different temperatures and radiation characteristics, whatever they may be, are present.

The theoretical framework for radiation comes from two key game changers in modern physics: (1) electromagnetic theory and (2) quantum theory.

According to electromagnetic theory, the radiation emitted by any surface travels in the form of waves with the velocity of light. In such a case, radiation can be characterized by velocity, frequency, and wavelength.

These are related by

c = νλ

(8.1)

In Eq. (8.1), c is the velocity in m/s, ν is the frequency in Hz, and λ is the wavelength in µm (because meter (m) is typically too big a unit for wavelength in the study of thermal radiation).

Quantum theory, on the other hand, proposed that the radiation emitted from any surface is in the form of discrete packets called quanta.

E = hν

(8.2)

Here h is the Planck’s constant that has a value 6.627 × 1034 J s and E is the energy in J.

Table 8.1 shows a very basic classification of radiation based on the wavelength of the emitted radiation.

Radiation that occurs in the wavelength 0 λ 100 m is frequently referred to as thermal radiation. In the study of radiation heat transfer, this wavelength range is of primary interest to us. Radiation from a surface or volume can be emitted in all

Heat Transfer Engineering. http://dx.doi.org/10.1016/B978-0-12-818503-2.00008-3

233

Copyright © 2021 Elsevier Inc. All rights reserved.

234 CHAPTER 8  Thermal radiation

Table 8.1  A very basic classification of radiation.

Wavelength, λ, µm

Type of radiation

<0.4

Ultraviolet

0.4–0.7

Visible

>0.7

Infrared

 

 

directions. Hence, we need to understand some basic ideas in geometry to proceed further in our study of radiation.

8.2  Concepts and definitions in radiation

The elemental plane angle (see Fig. 8.1A) dα is defined as

dα =

dL

(8.3)

r

 

 

In Eq. (8.3), dα is in radians.

As opposed to this, the solid angle subtended by an elemental area dAn associated with the hemisphere is given by

dω

 

=

dAn

steradians

(8.4)

n

 

 

 

r2

 

A schematic representation of solid angle is shown in Fig. 8.1B.

The solid angle subtended by an elemental area dAn about a point on another elemental area dA1 in the spherical coordinate system is given by

dAn

= (r sinθ ) dφ.rdθ

(8.5)

dA

 

= r2 sinθ dθ dφ

(8.6)

n

 

where r,θ,φ are the radius, zenith angle, and azimuthal angle, respectively.

FIGURE 8.1

Schematic representation of (A) a plane angle and (B) a solid angle.

8.3 Black body and laws of black body radiation 235

FIGURE 8.2

Solid angle subtended by an elemental area dAn about a point dA1 in the spherical coordinate system.

The idea of a solid angle is better understood from Fig. 8.2, which shows both the cartesian and the spherical coordinate systems.

Example 8.1: Evaluate the total solid angle for the positive hemisphere.

Solution:

The elemental solid angle dwn is given by

dω

 

=

dAn

sinθdθdφ

(8.7)

n

 

 

 

r2

 

 

 

 

 

π

 

Integrating from 0 to π/2 for θ and 0 to 2π for φ, we have ωn = 02π 02 sinθ dθ dφ

 

ωn

= 2π steradian

(8.8)

Hence, the solid angle associated with the hemisphere is 2π sr and that for a sphere is 4π sr, where sr is steradian.

8.3  Black body and laws of black body radiation

The study of a black body and its characteristics is central to the study of radiation heat transfer. A black body is a benchmark against which all radiative surfaces are compared.

236CHAPTER 8  Thermal radiation

8.3.1  Black body

A black body is defined as one that absorbs all the incident radiation, regardless of the direction and the wavelength.

In view of the above definition, it is evident that

1.A black body is an ideal absorber.

2.The black body is a perfect emitter. This logically follows from the definition of a black body and can be proved by a thought experiment involving a black body that is in equilibrium with its surroundings.

3.Black body radiation is independent of position (i.e., homogeneous) and direction (i.e., isotropic).

4.Consequent upon #3, the radiation from a black body is a function of only temperature.

Additionally, if all the incident radiation is absorbed and nothing is reflected, then the body appears black. However, what is visually black need not be radiatively black. Even so, all radiatively black bodies have to be visually black.

8.3.2  Spectral directional intensity

The spectral directional intensity of emission Iλ,e from a black body is given by

Iλ ,e (λ,θ,φ,T) =

dQ

(8.9)

dAcosθdωdλ

 

 

From the above mathematical definition, it is clear that the spectral direction intensity of emission Iλ,e is the rate at which the radiant energy emitted by a surface (dQ) in the direction of (θ,φ) per unit area of the surface normal to this direction (dAcosθ) per unit solid angle dw about (θ,φ) per unit wavelength dλ about λ.

dQ = Iλ ,e (λ,θ,φ,T)dA1cosθdωdλ

(8.10)

Let dQ = dQ

(8.11)

λ

 

dλ

 

and dqλ

=

dQλ

(8.12)

dA1

 

 

 

dqλ = Iλ,e (λ,θ,φ,T )cosθdω

(8.13)

For a black body, Ib,λ (λ,θ,φ) ≠ f (θ,φ).

Therefore, the hemispherical spectral black body emissive power of a black body at a temperature T in the wavelength interval dλ about λ is given by

π

 

Eb (λ,T) = Ib,λ (λ,T)02π 02 cosθ sinθ dθ dφ

(8.14)

8.3 Black body and laws of black body radiation 237

Eb (λ,T) = Ib,λ (λ,T) × 2π ×

1

(8.15)

 

2

 

Eb (λ, T) = π Ib,λ (λ, T)

 

(8.16)

The total hemispherical, emissive power from a black body defined as Eb(T) is

Eb (T ) = 0Eb,λ (λ,T ) dλ

(8.17)

π

 

Eb (T ) = 002π 02 Ib,λ (λ,T )cosθ sinθ dθ dφ dλ

(8.18)

Eq. (8.18) is a powerful equation that connects is known, Eb(T) can be determined.

Ib,λ (λ, T ) to Eb(T). Hence, if Ib,λ (λ, T)

8.3.3  Planck’s distribution

After many unsuccessful attempts by many researchers to get the correct distribution, Planck proposed the following distribution for the spectral intensity of emission from a black body at a wavelength λ and temperature T as

 

 

'

5

 

Ib,λ (λ,T) =

C1λ

 

 

(8.19)

e(C2

/ λT ) 1

 

 

where C1' and C2 are the first and the second radiation constants respectively and are given by

C ' = 2hc2

= 1.192 × 108 Wµm4

1

0

m2sr

C2 = hck0 = 1.4380 × 104 µmK

k = 1.38 × 1023 J/K (Boltzmann’s constant) h = 6.627 × 1034 Js (Planck’s constant)

c0 = 3 × 108 m/s (velocityof radiation in vacuum)

The hemispherical, spectral emissive power Eb,λ (λ, T) given by π Ib,λ (λ, T ) with Ib,λ (λ, T ) from the above distribution is plotted against λ for a few temperatures and is shown in Fig. 8.3. In this case, the constant C'1 gets changed to C1, where C1 =πC'1.

From the plots, the following are evident.

1.For every temperature T, Eb,λ continuously varies with λ.

2.At a given λ, Eb,λ is higher at a higher temperature.

3.For every temperature T, there exists a wavelength, λ at which Eb,λ has a peak.

4.This peak shifts to the left side at higher temperatures, which means that at higher temperatures, the peak occurs at lower wavelengths.

238 CHAPTER 8  Thermal radiation

FIGURE 8.3

Planck’s distribution for a few representative temperatures.

The following two laws emerge from the foundations laid by Planck’s distribution:

1.Wien’s displacement law

2.Stefan-Boltzmann law

8.3.4  Wien’s displacement law

Let us now try to determine where the peak Ib,λ occurs at a given temperature T. For maxima or minima of Ib,λ, the condition given below has to be satisfied.

 

 

 

Ib,λ

 

= 0

 

 

 

 

 

λ

 

 

 

 

 

 

 

 

C1' λ 5

= 0

 

 

 

 

 

 

 

 

 

 

C

 

/ λT

 

λ

(e

2

 

 

− 1)

 

λmaxT = 2898 mK

(8.20)

(8.21)

(8.22)

(Details of the full solution to Eq. (8.21) are presented in Example 8.3).

8.3 Black body and laws of black body radiation 239

8.3.5  Stefan-Boltzmann law

Eb (T ) =

Eb,λ dλ = σT 4

(8.23)

 

λ =0

In fact the, Stefan Boltzmann law was obtained before the discovery of Planck’s distribution using the thermodynamic route with the constant σ that has a numerical value of 5.67 × 108 W/m2K4 (known as the Stefan-Boltzmann constant), coming from a matching of experimental data of Eb(T) against T4. Hence, it was an absolute necessity for any proposal of Ib,λ (λT ) to satisfy the Stefan Boltzmann law when π Ib,λ (λ, T ) is integrated from λ = 0 to λ = .

Example 8.2: Consider the Planck’s distribution for the spectral, hemispherical emissive power of a black body given by

Eb,λ =

C1λ5

 

C

 

 

2

1

 

e

λT

where C1 =πC'1

1.In the limit of λCT2 0, what form will the Planck’s distribution take?

2.Compute the % error between the values of Eb,λ (Planck’s) with the Eb,λ calculated

from the simplified expression obtained in (1) for (a)

C2

= 1, (b)

C

2

= 0.1, and

λT

 

λT

(c)

C2

= 0.01. Comment on the results.

 

 

 

 

 

 

 

λT

 

 

 

 

 

 

 

 

 

 

 

 

3.In the limit of λCT2 → ∞, what form will the Planck’s distribution take?

4.Compute the % error between the values of Eb,λ(Planck’s), with the Eb,λ calculated

from the simplified expression obtained in (3) for (a)

C2

 

= 1, (b)

C

2

= 10, and

λT

 

 

λT

(c)

C2

= 50. Comment on the results.

 

 

 

 

 

 

 

 

 

 

 

 

 

λT

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Solution:

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Eb,λ =

C1λ5

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

C

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2

 

1

 

 

 

 

 

 

 

 

 

 

 

 

 

 

C2

 

e

λT

 

 

 

 

 

 

 

 

 

 

 

 

 

1. When

0

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

λT

 

 

 

 

x2

 

x3

 

 

 

 

 

 

 

 

 

 

 

 

We have ex = 1+ x +

+

+…..

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2!

3!

 

 

 

 

 

 

 

 

 

 

 

 

 

Here, x =

C2

; for x 0, ex = 1+ x, e

C2

= 1+

C2

 

 

 

 

 

 

 

λT

 

 

 

 

 

 

 

 

λT

 

 

 

 

 

 

 

λT

 

 

 

 

 

 

 

 

 

 

Cλ5

Eb.λ = C1 2

λT

240 CHAPTER 8  Thermal radiation

This is known as the Rayleigh-Jeans distribution; it holds good for longer wavelengths and it fails in the asymptotic limit of λ 0.

2. % error =

Eb,λ (Plancks) − Eb,λ (Rayleigh Jeans)

100

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Eb,λ (Plancks)

 

 

 

 

 

 

 

 

 

 

 

 

C λ 5

 

C λ 5

 

 

 

λT

 

 

 

 

 

 

 

C2

 

 

C

 

 

 

 

 

 

 

1

 

 

 

 

1

 

 

 

 

 

C2

 

 

 

 

(

 

(

 

 

)

 

)

 

 

2

 

 

 

 

e

 

 

−1

 

e

λT

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

=

 

 

 

 

 

−1

 

λT

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

100

= 1

 

 

 

 

 

 

100

 

 

 

 

 

 

C λ 5

 

 

 

 

 

C2

 

 

 

 

 

 

 

 

 

1

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

C2

 

 

 

 

 

 

 

 

 

λT

 

 

 

 

 

 

 

 

(e(

 

)−1)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

λT

 

 

 

 

 

 

 

 

 

 

(a)λCT2 = 1

%error = 71.82%

(b)λCT2 = 0.1

%error = 5.17%

(c)λCT2 = 0.01

%error = 0.502%

As the value of C2 decreases, the simplified expression approaches Planck’s

λT

values of Eb,λ. Hence, the Rayleigh-Jeans approximation, works well for longer wavelengths (and so low values of λCT2 )

3. When λCT2 → ∞

Eb,λ =

C1λ5

 

C

 

 

2

1

 

e

λT

C2

eλT 1

C λ5

Eb,λ = 1C2

eλT

The above equation is known as the Wien’s distribution; while this works well for shorter λs, at higher λs it tends to saturate.

4.

=

Eb,λ (Planck) − Eb,λ (Wien)

100

 

 

 

 

 

 

 

 

 

% error

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Eb,λ (Planck)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

C1λ

5

 

C1λ

5

 

 

 

C2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

C2

 

 

 

 

 

 

 

C2

 

 

 

eλT

−1

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

1−

 

 

 

 

 

 

 

C2

 

 

=

eλT −1

 

 

 

 

eλT

100

=

 

 

 

 

100 = e

100

 

 

 

 

 

 

 

 

 

λT

 

 

 

 

5

 

 

 

 

 

 

C

 

 

 

 

 

 

 

 

 

C1λ

 

 

 

 

 

 

 

 

 

 

 

 

 

2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

e

λT

 

 

 

 

 

 

 

 

 

 

 

 

C

2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

e

λT

−1

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

8.3 Black body and laws of black body radiation 241

(a)λCT2 = 1

%error = 36.79%

(b)λCT2 = 10

%error = 0.0045%

(c)λCT2 = 50

%error = 1.93 × 1020%

As the value of

 

C2

 

increases, the simplified expression approaches Planck’s

 

λT

values of Eb,λ.

 

 

 

 

 

 

 

 

 

 

 

 

Example 8.3:

 

 

 

 

 

 

 

Consider

 

C1λ 5

 

.

 

C2

 

 

 

e

λT

1 σ T 5

 

 

1.What are the units of the above quantity?

2.Determine the value of λT at which the value of this function is the maximum.

3.Hence determine the maximum value of this function.

4.What potential use do you see in introducing such a function?

Solution:

Consider,

 

 

 

 

 

 

 

 

 

 

 

C1λ 5

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

C2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

eλT

1 σ T 5

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

1. Let us first work out the units of the above quantity.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

hc

 

4

 

 

 

 

C1 = 2π hc02 = 3.742 × 108 Wµm4/m2 , C2

=

 

0

= 1.439

× 10

µmK

 

 

k

 

 

Hence, the units of given quantity are

W m4 /m2 × ( m)5

=

1

 

 

W /m2 /K4 × K5

 

µmK

2.

 

 

 

C1λ5

 

 

 

 

 

 

 

 

 

 

 

 

 

 

d

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

C

 

)σ T

 

 

 

 

 

 

 

 

 

 

 

 

 

(

λT

5

 

 

 

 

 

 

 

 

 

 

 

 

 

e

 

2

1

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

= 0

 

 

 

 

 

 

 

 

 

 

 

 

 

 

d(λT)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

242 CHAPTER 8  Thermal radiation

 

 

 

 

 

 

 

 

 

 

 

 

 

C1

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

d

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

(

 

C2

 

 

 

)σ

 

 

 

 

 

 

5

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

(λT)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

e

λT

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

1

 

 

 

 

 

 

 

 

 

 

 

 

= 0

 

 

 

 

 

 

eλT

1 + (λT )5

d(λT)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

5

(λT )4

 

eλT

C2

 

 

 

 

 

 

 

 

C2

 

 

 

 

 

 

 

 

 

 

 

 

 

C2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

λT

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

= 0

 

 

 

 

 

 

(λT )5

 

 

C2

 

 

 

2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

eλT

1

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

e

C2

1

C2

e

C2

= 0

 

 

 

 

 

 

 

 

 

5

λT

λT

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

λT

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

C2

e

C2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

λT

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

λT

 

 

 

= 5

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

C2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

1

− eλT

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Let

 

C2

= x

xex

 

 

 

 

= 5

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

ex

1

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

λT

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

x = 5

ex 1

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

e

x

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Writing this in algorithmic form and solving by the successive substitution method details of which are given in Table 8.2, we have

xi +1

= 5

exi 1

 

 

 

x

 

 

 

e i

 

In the above expression, i refers to the iteration number.

x* = λCT2 = 4.965

Table 8.2  A successive substitution method for example 8.3.

Iteration

xi

xi+1

Residual (xi+1xi)2

1

3

4.75

3.06

2

4.75

4.95

0.044

3

4.95

4.964

1.96 × 10−4

4

4.964

4.965

1.96 × 10−6