Radiation Heat Transfer

Thermal radiation is electromagnetic radiation emitted by a body because of its temperature, in the range about $0.1$ to $100\ \mu$m (ultraviolet, visible and infrared).

Planck's law

It gives the spectral emissive power of a black body at wavelength $\lambda$ and absolute temperature $T$:

$$E_{b\lambda} = \frac{C_1\,\lambda^{-5}}{\exp\left(C_2/\lambda T\right)-1}$$

$C_1 = 3.742\times10^8\ \text{W}\,\mu\text{m}^4/\text{m}^2$, $C_2 = 14\,388\ \mu\text{m K}$. At every wavelength $E_{b\lambda}$ rises with $T$; each curve has a peak.

Wien's displacement law

Setting $dE_{b\lambda}/d\lambda = 0$ gives the wavelength at the peak:

$$\lambda_{max}T = 2898\ \mu\text{m K}$$

The peak moves to shorter wavelengths as $T$ rises (a heated iron goes from dull red to white). The sun (5800 K) peaks near $0.5\ \mu$m, a room-temperature body (300 K) near $9.7\ \mu$m.

Stefan-Boltzmann law

Integrating Planck's law over all wavelengths gives the total emissive power of a black body:

$$E_b = \int_0^\infty E_{b\lambda}\,d\lambda = \sigma T^4,\qquad \sigma = 5.67\times10^{-8}\ \text{W/m}^2\text{K}^4$$

For a real (grey) surface $E = \varepsilon\sigma T^4$.

Kirchhoff's law

At thermal equilibrium the emissivity of a surface equals its absorptivity, $\varepsilon_\lambda = \alpha_\lambda$. For a grey (diffuse) surface $\varepsilon = \alpha$ at all conditions. A good absorber is a good emitter.

Lambert's cosine law

The radiation intensity $I$ of a diffuse surface is the same in all directions, so emissive power in direction $\theta$ from the normal varies as $\cos\theta$, and total emissive power $E = \pi I$.

Frequency, wavelength and temperature

Wavelength and frequency are related by $\lambda = c/\nu$ ($c = 3\times10^8$ m/s in vacuum). A hotter body emits at higher frequency (shorter wavelength) for its peak, according to Wien's law, and also emits more energy at all frequencies according to the $T^4$ law.

Radiation properties

When radiation of total amount $G$ (irradiation) strikes a surface, part is absorbed, part reflected and part transmitted:

• Absorptivity $\alpha$ = fraction absorbed.
• Reflectivity $\rho$ = fraction reflected.
• Transmissivity $\tau$ = fraction transmitted.

$$\alpha + \rho + \tau = 1$$

For an opaque solid $\tau = 0$, so $\alpha + \rho = 1$. For a black body $\alpha = 1$; for a white body $\rho = 1$; for a transparent body $\tau = 1$.

Types of bodies

Emissivity $\varepsilon = E/E_b$ is the ratio of the emissive power of a surface to that of a black body at the same temperature.

Kirchhoff's law (proof)

Place a small body of area $A$, emissive power $E$ and absorptivity $\alpha$ inside a black enclosure at the same temperature $T$. The enclosure sends radiation $E_b$ per unit area on the body, of which the body absorbs $\alpha E_b A$. It emits $EA$. At thermal equilibrium (no net heat flow, since temperatures are equal):

$$EA = \alpha E_b A\;\Rightarrow\;\frac{E}{E_b} = \alpha \;\Rightarrow\; \varepsilon = \alpha$$

Thus at the same temperature the emissivity of a body equals its absorptivity.

(a) Peak wavelength (Wien's law)

$$\lambda_{max} = \frac{2898}{T} = \frac{2898}{1500} = 1.932\ \mu\text{m}$$

This is in the near-infrared; only a small tail lies in the visible range.

(b) Black-body emissive power (Stefan-Boltzmann law)

$$E_b = \sigma T^4 = 5.67\times10^{-8}\times(1500)^4 = 2.870\times10^{5}\ \text{W/m}^2$$

(c) Grey body radiation

$$\begin{aligned} E &= \varepsilon\sigma T^4 = 0.8\times2.870\times10^5 = 2.296\times10^5\ \text{W/m}^2\\ Q &= E\,A = 2.296\times10^5\times0.5 = 1.148\times10^{5}\ \text{W} \end{aligned}$$

Comparison with the sun

$$\lambda_{max,sun} = \frac{2898}{5800} = 0.4997\ \mu\text{m}$$

The sun peaks in the visible (green), at about 0.5 $\mu$m, about one quarter of the wavelength of the 1500 K surface.

Answer: $\lambda_{max} = 1.93\ \mu$m; $E_b = 287\ \text{kW/m}^2$; $Q = 114.8$ kW; sun peaks at $0.50\ \mu$m.

Setup

Two large parallel plates, so that all radiation leaving one reaches the other: $F_{12} = F_{21} = 1$. Plate 1 at $T_1$ ($\varepsilon_1$), plate 2 at $T_2$ ($\varepsilon_2$), $T_1 > T_2$, both opaque grey surfaces ($\rho = 1 - \varepsilon$).

Radiosity $J$ is the total radiation leaving a surface (emitted plus reflected). The net heat leaving a surface:

$$Q = \frac{E_b - J}{(1-\varepsilon)/(\varepsilon A)}$$

where $(1-\varepsilon)/\varepsilon A$ is the surface resistance. The exchange between two surfaces uses the space resistance $1/(A_1F_{12})$.

Network

 Eb1   J1              J2    Eb2
  o--/\/\--o----/\/\----o--/\/\--o
  (1-e1)/   1/(A1F12)   (1-e2)/
   (e1 A1)              (e2 A2)

Total resistance (series):

$$R = \frac{1-\varepsilon_1}{\varepsilon_1A} + \frac{1}{AF_{12}} + \frac{1-\varepsilon_2}{\varepsilon_2A}$$

With $A_1 = A_2 = A$, $F_{12} = 1$:

$$Q_{12} = \frac{E_{b1}-E_{b2}}{R} = \frac{\sigma\left(T_1^4 - T_2^4\right)A}{\dfrac{1}{\varepsilon_1}+\dfrac{1}{\varepsilon_2}-1}$$

Result

Per unit area:

$$\frac{Q_{12}}{A} = \frac{\sigma\left(T_1^4 - T_2^4\right)}{\dfrac{1}{\varepsilon_1}+\dfrac{1}{\varepsilon_2}-1} = \varepsilon_{eff}\,\sigma\left(T_1^4 - T_2^4\right)$$

with $\varepsilon_{eff} = \left(\dfrac{1}{\varepsilon_1}+\dfrac{1}{\varepsilon_2}-1\right)^{-1}$. For black plates $\varepsilon_{eff} = 1$ and the result becomes $\sigma(T_1^4-T_2^4)$.

For two parallel grey plates $q = \dfrac{\sigma(T_1^4 - T_2^4)}{\frac{1}{\varepsilon_1}+\frac{1}{\varepsilon_2}-1}$.

$$\sigma(T_1^4-T_2^4) = 5.67\times10^{-8}\,(800^4 - 500^4) = 5.67\times10^{-8}(4.096\times10^{11}-6.25\times10^{10}) = 19\,681\ \text{W/m}^2$$

(a) Without shield

$$q_{12} = \frac{19\,681}{\dfrac{1}{0.8}+\dfrac{1}{0.5}-1} = \frac{19\,681}{2.25} = 8747\ \text{W/m}^2$$

(b) With shield (emissivity $\varepsilon_3 = 0.1$ on both sides)

The shield adds two surface resistances and one space resistance, so the total resistance is the sum of the plate-to-shield gap and the shield-to-plate gap:

$$\left(\frac{1}{\varepsilon_1}+\frac{1}{\varepsilon_3}-1\right)+\left(\frac{1}{\varepsilon_3}+\frac{1}{\varepsilon_2}-1\right) = (1.25+10-1)+(10+2-1) = 10.25 + 11 = 21.25$$

 Plate 1    Shield    Plate 2
 800 K   |  T3  |    500 K
 e=0.8   | e=0.1|    e=0.5

$$q_{with} = \frac{19\,681}{21.25} = 926\ \text{W/m}^2$$

(c) Reduction

$$\frac{8747 - 926}{8747}\times100 = 89.4\%$$

(d) Shield temperature

Heat from plate 1 to shield equals $q_{with}$:

$$q_{with} = \frac{\sigma(T_1^4-T_3^4)}{10.25}\;\Rightarrow\;T_3^4 = 4.096\times10^{11}-\frac{926\times10.25}{5.67\times10^{-8}} = 2.422\times10^{11}$$ $$T_3 = 701.5\ \text{K}$$

Check with the other side: $T_3^4 - T_2^4 = 2.422\times10^{11}-6.25\times10^{10} = 1.797\times10^{11}$; $\sigma\times1.797\times10^{11}/11 = 926$ W/m$^2$. Correct.

Answer: (a) 8747 W/m$^2$; (b) 926 W/m$^2$; (c) 89.4 % reduction; (d) 701.5 K.

(a) Shape factor and algebra

The shape factor $F_{ij}$ is the fraction of radiation leaving surface $i$ that strikes surface $j$ directly. It depends only on geometry and orientation.

• Reciprocity: $A_iF_{ij} = A_jF_{ji}$.
• Summation rule: for an enclosure of $N$ surfaces, $\sum_{j=1}^{N}F_{ij} = 1$.
• Superposition: if surface $j$ is made of parts $j_1, j_2$, then $F_{i(j_1+j_2)} = F_{ij_1}+F_{ij_2}$ and $A_iF_{i\to j}$ adds similarly.
• For a plane or convex surface, $F_{ii} = 0$; for a concave surface $F_{ii} > 0$.

(b) Concentric spheres

      .------------.
    /    .------.    \
   |    |  1     |  2 |
   |    |  T1    |  T2|
    \    '------'    /
      '------------'

Areas: $A_1 = 4\pi(0.1)^2 = 0.1257$ m$^2$, $A_2 = 4\pi(0.2)^2 = 0.5027$ m$^2$.

Shape factors. All radiation from the inner (convex) sphere reaches the outer one, so $F_{12} = 1$ and $F_{11} = 0$. By reciprocity:

$$F_{21} = \frac{A_1}{A_2}F_{12} = \frac{0.1257}{0.5027} = 0.25$$

By the summation rule for surface 2: $F_{22} = 1 - F_{21} = 0.75$.

Heat exchange. For a small body enclosed by a large one:

$$Q_{12} = \frac{\sigma A_1\left(T_1^4 - T_2^4\right)}{\dfrac{1}{\varepsilon_1}+\dfrac{A_1}{A_2}\left(\dfrac{1}{\varepsilon_2}-1\right)}$$ $$\begin{aligned} \sigma A_1(T_1^4-T_2^4) &= 5.67\times10^{-8}(0.1257)(600^4 - 300^4) = 865.7\ \text{W}\\ \text{denominator} &= \frac{1}{0.6}+0.25\left(\frac{1}{0.4}-1\right) = 1.667 + 0.375 = 2.042\\ Q_{12} &= \frac{865.7}{2.042} = 424\ \text{W} \end{aligned}$$

Answer: $F_{12} = 1$, $F_{21} = 0.25$, $F_{22} = 0.75$; $Q_{12} \approx 424$ W from the inner to the outer sphere.