Laws of Thermal Radiation

Abstract

All bodies emit radiation to their surroundings through electromagnetic waves due to the conversion of the internal energy of the body into radiation. Reflectivity, absorptivity and transmissivity of bodies are defined. Planck’s law for the spectral distribution of the emissive power of a blackbody is presented in Sect. 10.4, which has been used to determine Stefan–Boltzmann’s equation of the total emissive power of a blackbody. Wein’s displacement law is given in the next section. Concept of gray body has been presented. Kirchhoff’s Law has been stated and proved. Equality of emissivity and absorptivity has been established. Lambert’s cosine law has been presented and utilized to determine the intensity of radiation in terms of total emissive power of a blackbody.

Appendices

Review Questions

• 10.1 What is difference between thermal radiation and other types of electromagnetic radiation?

• 10.2 Define absorptivity, transmissivity and reflectivity of a surface.

• 10.3 The Planck’s law governing the change in emissive power of a blackbody with the wavelength is given by

  $$E_{b\lambda } = \frac{{c_{1} \lambda^{ - 5} }}{{\exp (c_{2} /\lambda T) - 1}}$$

  Using the above equation show that the total emissive power is given by

  $$E_{b} = \sigma T^{4} .$$

  where σ is the Stefan Boltzmann constant.

• 10.4 Distinguish between black, real and gray bodies giving suitable examples.

• 10.5 What is the Wein’s displacement law?

• 10.6 State and prove the Kirchhoff’s law of thermal radiation.

• 10.7 State the Lambert’s cosine law.

• 10.8 Define intensity of radiation and prove that for a diffuse-blackbody, the intensity of normal radiation is 1/π times of the total emissive power of the body.

• 10.9 Distinguish between specular and diffuse surfaces.

Problems

• 10.1 A photon has an energy of 0.3 × 10⁻¹² J. Calculate its frequency and wavelength in vacuum.

  [Ans. ν=E/h = 0.3 × 10⁻¹²/6.6236 × 10⁻³⁴ = 4.53 × 10²⁰  Hz; λ = c_o/ν = 2.998 × 10⁸/(4.53 × 10²⁰) = 6.62 × 10⁻¹³ m.]

• 10.2 Radiant energy with an intensity of 1000 W/m² is incident normal to a flat surface whose absorptivity is 2.5 times the transmissivity and 2 times of the reflectivity. Determine the energy transmitted, absorbed and reflected.

  [Ans. α + ρ + τ = 1, substitution gives α + α/2.5 + α/2 = 1, hence, α = 1/1.9; Q_A = 1000/1.9 = 526.32 W/m²; Q_T = Q_A/2.5 = 210.53 W/m²; Q_R = Q_A/2 = 263.16 W/m².]

• 10.3 The temperature of a black surface of 0.5 m² area is 727°C. Calculate: (a) the total amount of energy emission, (b) the intensity of normal radiation, (c) the intensity of radiation at an angle of 60^o and (d) the wavelength of maximum monochromatic emissive power .

  [Ans. E_b = AσT⁴ = 0.5 × 5.669 × 10⁻⁸ (727 + 273)⁴ = 28345 W. I_n = E_b/π = σT⁴/π = 18045 W/(m² sr); I_ϕ = I_n cos60^o = 9022.5 W/(m² sr); λ_max = 2897.6/1000 = 2.9 μm.]

• 10.4 Figure 10.17 shows the variation of reflectivity with λ for an opaque surface. The irradiation G (radiation impinging) on the surface from a source is approximated as given below. Determine the energy absorbed.

  Fig. 10.17

  

  Problem 10.4

  $$\begin{array}{*{20}c} {0 \le \lambda < 0.4\quad G = 20{\text{ W}}/{\text{m}}^{2} } & {0.4 \le \lambda < 1.0\quad G = 150{\text{ W}}/{\text{m}}^{2} } \\ {1.0 \le \lambda < 2.0\quad G = 100{\text{ W}}/{\text{m}}^{2} } & {2.0 \le \lambda < 3.0\quad G = 20{\text{ W}}/{\text{m}}^{2} } \\ {3.0 \le \lambda < \infty \quad G = 30{\text{ W}}/{\text{m}}^{2} } & {} \\ \end{array}$$

  [Ans. Using the relation absorptivity = (1 – ρ), we have G_absorbed = (1 – 0) × 20 + (1 – 0.5) × 150 + (1 – 0.8) × 100 + (1 – 0.9) × 20 + (1 – 0) × 30 = 147 W/m².]

• 10.5 For a spectrally selective, diffuse surface the spectral distribution of absorptivity and reflectivity are shown in Fig. 10.18. Determine spectral transmissivity, τ_λ.

  Fig. 10.18

  

  Problem 10.5

  [Ans. $\tau_{\lambda } = 1 - \alpha_{\lambda } - \rho_{\lambda } .$ Hence, for $0 \le \lambda \le 2$ μm, $\tau_{\lambda } = 1 - 0.1 - 0.05 = 0.85$ and for $\lambda > 2$ μm, $\tau_{\lambda } = 1 - 1 - 0 = 0,$ i.e. the surface is opaque for radiation of λ > 2 μm.]

• 10.6 A window glass transmits 90% of the radiation between wavelengths 0.3–3 μm. It is practically opaque to all other wavelengths. Determine the percentage of radiant heat flux transmitted when it is coming from a blackbody source at 2000 K.

  [Ans. λ₁T = 0.3 × 2000 = 600 μm K, λ₂T = 3 × 2000 = 6000 μm K; $F_{{0 - \lambda_{1} T}} =$ 0.112 × 10⁻⁶, $F_{{0 - \lambda_{2} T}} =$ 0.73777 from Table 10.3, $F_{{\lambda_{2} T - \lambda_{1} T}} \approx$ 0.73777; Flux transmitted = 0.90 × 0.73777 × 100 = 66.4%.]

• 10.7 If the incident solar radiation on the Earth is 1390 W/m² for a mean distance r_se = 1.5 × 10¹¹ m, determine the radiation on planet Mercury if its orbital radius is 5.8 × 10¹⁰ m.

  [Ans. The radiation incident on a surface is inversely proportional to the distance squared (the inverse square law). Hence, $G_{\text{Murcury}} = G_{\text{Earth}} \times \left( {r_{\text{se }}/ r_{\text{sm}} } \right)^{2} = 1390\times\left({1.5 \times\ 10^{11} /5.8 \times 10^{10} } \right)^{2} = 9297\,{\text{W}}/{\text{m}}^{2}$.]

• 10.8 Radiation flux G_o enters into a cavity through a small opening. The absorptivity of the cavity surface is less than 1. Determine the flux after n reflections.

  [Ans. On first reflection G_o reduces to G₁ = (1 – α) G_o; Similarly, on the second reflection, G₂ = (1 – α)² G_o, …, after n reflection, $G_{\text{n}} = (1 - \alpha )^{\text{n}} G_{\text{o}}$; As n increases, $1 - (1 - \alpha )^{n} \to 1$, i.e. the radiation will be completely absorbed.]

• 10.9 A large plate (ε₁ = 0.8) emits 300 W/m². Another plate (ε₂ = 0.4) of the same surface area emits 150 W/m². If these plates are brought very close and parallel to each other, determine the net heat exchange per unit area of the plates.

  [Ans. For a gray surface E = εE_b. Hence, $E_{1} = \varepsilon_{1} E_{{{\text{b}}1}} = \varepsilon_{1} \sigma T_{1}^{4}$, and $E_{2} = \varepsilon_{2} E_{{{\text{b}}2}} = \varepsilon_{2} \sigma T_{2}^{4}$; Ratio E₁/E₂ gives T₁ = T₂ hence q₁₂ = 0.]