Radiation Definitions and laws Heat transfer by conduction and convection required the existence of a material medium, either a solid or a fluid. However, it is not required in heat transfer by radiation. Radiation can travel through an empty space at the speed of light in the form of an electromagnetic wave. As shown in the electromagnetic spectrum in Fig , thermal radiation covers the range of wavelength from 0.1~100 μm Absorptivity Thermal radiation impinging on the surface of an opaque solid is either absorbed or reflected. The absorptivity is defined as the fraction of the incident radiation that is absorbed:

46 [2.7.1] Where q (a) is the energy absorbed per unit area per unit time and q (i) is the energy impinging per unit area per unit time. Let us define q (a) and q (i) such that q (a)d  and q (i)d represent respectively the absorbed and incident energies per unit area per unit time in the wavelength range to  +d. The monochromatic absorptivity  is defined as : [2.7.2] For any real body α λ < 1 and depend on. A graybody is a hypothetical one for which α λ < 1 but independent of and temperature. The limiting case of α λ = 1 for all. and temperature is known as a blackbody. In other words, a blackbody absorbs all the incident radiation Emissivity The emissivity of a surface is defined as [2.7.3] Where q (e) and q b (e) are the energies emitted per unit area per unit time by a real body and a blackbody, respectively. Table lists the emissivities of some Material. These averaged value have been used widely even though the emissivity actually depends on the wavelength and the angle of emission.

47 Let us define q (e) and q b (e) such that q (e) d  and q b (e) d represent respectively the energies emitted per unit area per unit time in the wavelength range to  +d of a real body and blackbody. The monochromatic emissivity  is defined as : [2.7.4]  =1 for blackbody and <1 for a real body.

Kirchhoff’s law Consider the body enclosed in the cavity shown in Fig Suppose the body is at the same temperature as the wall of the cavity, that is, the two are at thermodynamic equilibrium with each other. Since there is no net heat transfer from the cavity wall to The body, the energy emitted by the body must be equal to the energy absorbed. [2.7.5] Where q (e) is the energy emitted per unit area of the body per unit time, A the surface area of the body, q (i) the energy impinging per unit area of the body per unit time, and  the absorptivity of the body. If the body is replaced by a blackbody, Eq.[2.7.5] becomes [2.7.6] Dividing Eq. [2.7.5] by Eq.[2.7.6], we get [2.7.7] Substituting Eq. [2.7.7] into Eq.[2.7.3], we have [2.7.8]

Plank’s distribution law Plank derived the following equation for the energy emitted by a blackbody as a function of the wavelength and temperature: [2.7.10] Where T = temperature in K c = speed of light h = Plank’s constant k = Boltzmann’s constant This is Kirchhoff’s law, which states that for a system in thermodynamic equilibrium the emissivity and absorptivity are the same. Following a similar procedure, we can show that [2.7.9]

The solid angle Stefan-Boltzmann law The Plank distribution law can be integrated over wavelengths from zero to infinity to determine the total emissive energy of a blackbody: [2.7.11] or [2.7.12] Where  (Stefan-Boltzmann constant) = x Wm -2 K -4. Consider a hemisphere of radius r surrounding a differential area dA 1 at the center. Fig a shows only one-quarter of the hemisphere. On the hemisphere [2.7.13]

51 The solid angle that intersects dA 2 on the hemisphere, shown in Fig.2.7-4b, is defined as: [2.7.14] The energy leaving dA 1 in the direction given by the angle θ is I b dA 1 cosθ, where I b is the blackbody intensity. The radiation arriving at some area dA 2 at a distance r from A 1 would be Where dA 2 is constructed normal to the radius vector. The dA 2 /r 2 represents the solid angle subtended by the area dA 2 which can be expressed as so that J. P. Holman, “Heat Transfer”, 1977, p.285 r1r1 r2r2 dθ r 1 dθ r 2 dθ

52 This yields the following equation: [2.7-18] and Radiation between blackbodies Consider two black surface A 1 and A 2 shown in Fig Define the view factors: F 12 = fraction of energy leaving surface 1 that reaches surface 2 F 21 = fraction of energy leaving surface 2 that reaches surface 1

53 As such, the energy leaving surface 1 that reaches surface 2 per unit time is Similarly, the energy leaving surface 2 that reaches surface 1 per unit time is Therefore, the net energy exchange rate is as follows: If both surfaces are at the same temperature T, there is no net energy exchange and Q 12 =0. Then substituting Eq.[2.7-12] into Eq. [2.7-22], we have so that Which is called the reciprocity relationship. Substituting Eqs. [2.7-12] and [2.7-24] into [2.7-22], we get [2.7-20] [2.7-21] [2.7-22] [2.7-23] [2.7-24] [2.7-25]

54 Let us now proceed to determine the view factors F 12 and F 21 between the two black surfaces. From Eq. [2.7-19] [2.7-26] where dΩ 12 is the solid angle subtended by dA 2 as seen from dA 1. Let r 12 be the distance between the centers of dA 1 and dA 2. Consider the hemisphere of radius r 12 and centered at dA 1. The projection of dA 2 on the hemisphere is cos  2 dA 2. Therefore, from Eq.[2.7-14] [2.7-27] Substituting Eq.[2.7-27] into Eq. [2.7-26] And so Similarly it can be shown that [2.7-28] [2.7-29] [2.7-30]

55 Substituting Eq. [2.7-29] and Eq. [2.7-30] respectively into Eqs. [2.7-20] and [2.7-21], we get Eqs. [2.7-31] and [2.7-32] [2.7-31] [2.7-32] and The integration in Eq. [2.7-31] and [2.7-32] is often difficult and needs to be done numerically. Analytical equations are available for a number of special cases. Some examples are given below:

56 For large (infinite) parallel plates, long (infinite) concentric cylinders and concentric spheres F 12 = 1, as shown in Fig

57 For the two identical, parallel directly opposed rectangles shown in Fig where X and Y are defined in the figure. [2.7-33]

58 For the two parallel concentric circular disks shown in Fig , [2.7-34] where X, R 1, and R 2 are defined in the figure. For the sphere and the disk shown in Fig , [2.7-35]

59 Example Consider the radiation from the small area d A1 to the flat disk A 2, as shown in the Fig. The element of area dA 2 is chosen as the circular ring of radius x. Calculate F dA1  A2.

Radiation between graybodies In radiation heat transfer between blackbodies, all the radiant energy that strikes a surface is absorbed. In radiation heat transfer between nonblackbodies, the energy striking a surface will not all be absorbed; part will be reflected. Since the energy emitted and the energy reflected by a nonblack surface both contribute to the total energy leaving the surface (J), we can write [2.7-36] Where J is called radiosity, is the total energy leaving a surface per unit area per unit time and , called the reflectivity, is the fraction of the incident energy reflected. From the definition for J, we see that J - q (i) = net energy leaving a surface per unit area per unit time and q (i) - J= net energy received at a surface per unit area per unit time For an opaque material the incident energy is either reflected or absorbed:  +  =1 [2.7-37]

61 If Kirchhoff’s law can be applied, that is,  =  according to Eq. [2.7-8], Eq.[2.7-37] becomes  +  = 1 [2.7-38] From Eqs. [2.7-36] and [2.7-38] [2.7-39] Let us consider radiation heat transfer between two gray surface A 1 and A 2. Since the net energy transfer from A 1 to A 2 (Q 12 ) equals either the net energy leaving A 1 or the net energy received at A 2, we can write, with the help of Eq. [2.7-39]: and [2.7-40] [2.7-41]

62 Similar to Eqs. [2.7-20] and [2.7-21] for two black surfaces, we can write the following expressions for two gray surfaces in terms of the view factors: [2.7-42] [2.7-43] and Since A 1 F 12 =A 2 F 21 according to Eq. [2.7-24] [2.7-44] From Eqs. [2.7-40], [2.7-44], and [2.7-41] [2.7-45] [2.7-46] [2.7-47] For blackbody[2.7-20]

63 Adding these equations, we have [2.7-48] [2.7-49] [2.7-50] Substituting Eq. [2.7-12] ( ) into Eq. [2.7-48], we have  For blackbody Compared to Eq.[2.7-48a], therefore, [2.7-48a] For A 1 =A 2

64 Example Radiation shields Given: Hot surface T 1,  1, cold surface T 2,  2, radiation shield T 3,  3 Surfaces 1 & 2: two infinite parallel gray surface with the area of A Find: The radiation heat transfer between two parallel gray surfaces Since A 1 =A 2 =A 3, and the surfaces are infinite and parallel F 12 =F 32 =1 (by definition, F 13 = fraction of energy leaving surface 1 and reaches surface 2) From [2.7-50] Substituting above two equations into Eq.[2.7-49], we get

65 Since Q 13 =Q 32 =Q 132, we have Adding the two equations, we have and

66 The ratio of radiation heat transfer with a shield to that without one is If  1 =  2 =  3 =  Without a radiation shield, from Eqs [2.7-49] and [2.7-50]

67 Example: Two very large parallel planes with emissivities 0.3 and 0.8 exchange heat. Find the percentage reduction in heat transfer when a polished-aluminum radiation shield (  =0.04) is placed between them. Sol: The heat transfer without the shield is given by The radiation network for the problem with the shield is placed, based on Eq. [2.7-50], is given The total resistance is (24) + 2(1) = The heat transfer is reduced by 93.6% E’ b1 E’ b2 E’ b3

68 Summary 1. Emission and absorption of a black body and a gray body (black body) T1T1 T 2, A 2 2. Radiation between black bodies (Q 12 : The net energy exchange rate ) (black body) T 1, A 1 (gray body) T1T1 and

69 Summary (gray body) T 2, A 2 3. Radiation between gray bodies (gray body) T 1, A 1