Chapter 12: Radiation Heat Transfer, Study notes of Law

Download Chapter 12: Radiation Heat Transfer and more Study notes Law in PDF only on Docsity! Chapter 12, E&CE 309, Spring 2005. 1 Majid Bahrami Chapter 12: Radiation Heat Transfer Radiation differs from Conduction and Convection heat t transfer mechanisms, in the sense that it does not require the presence of a material medium to occur. Energy transfer by radiation occurs at the speed of light and suffers no attenuation in vacuum. Radiation can occur between two bodies separated by a medium colder than both bodies. According to Maxwell theory, energy transfer takes place via electromagnetic waves in radiation. Electromagnetic waves transport energy like other waves and travel at the speed of light. Electromagnetic waves are characterized by their frequency ν (Hz) and wavelength λ (µm), where: λ = c / ν where c is the speed of light in that medium; in a vacuum c0 = 2.99 x 108 m / s. Note that the frequency and wavelength are inversely proportional. The speed of light in a medium is related to the speed of light in a vacuum, c = c0 / n where n is the index of refraction of the medium, n = 1 for air and n = 1.5 for water. Note that the frequency of an electromagnetic wave depends only on the source and is independent of the medium. The frequency of an electromagnetic wave can range from a few cycles to millions of cycles and higher per second. Einstein postulated another theory for electromagnetic radiation. Based on this theory, electromagnetic radiation is the propagation of a collection of discrete packets of energy called photons. In this view, each photon of frequency ν is considered to have energy of e = hν = hc / λ where h = 6.625 x 10-34 J.s is the Planck’s constant. Note that in Einstein’s theory h and c are constants, thus the energy of a photon is inversely proportional to its wavelength. Therefore, shorter wavelength radiation possesses more powerful photon energies (X-ray and gamma rays are highly destructive). Chapter 12, E&CE 309, Spring 2005. 2 Majid Bahrami Fig. 12-1: Electromagnetic spectrum. Electromagnetic radiation covers a wide range of wavelength, from 10-10 µm for cosmic rays to 1010 µm for electrical power waves. As shown in Fig. 12-1, thermal radiation wave is a narrow band on the electromagnetic wave spectrum. Thermal radiation emission is a direct result of vibrational and rotational motions of molecules, atoms, and electrons of a substance. Temperature is a measure of these activities. Thus, the rate of thermal radiation emission increases with increasing temperature. What we call light is the visible portion of the electromagnetic spectrum which lies within the thermal radiation band. Thermal radiation is a volumetric phenomenon. However, for opaque solids such as metals, radiation is considered to be a surface phenomenon, since the radiation emitted by the interior region never reach the surface. Note that the radiation characteristics of surfaces can be changed completely by applying thin layers of coatings on them. Blackbody Radiation A blackbody is defined as a perfect emitter and absorber of radiation. At a specified temperature and wavelength, no surface can emit more energy than a blackbody. A blackbody is a diffuse emitter which means it emits radiation uniformly in all direction. Also a blackbody absorbs all incident radiation regardless of wavelength and direction. Chapter 12, E&CE 309, Spring 2005. 5 Majid Bahrami Example 12-1 The temperature of the filament of a light bulb is 2500 K. Assuming the filament to be a blackbody, determine the fraction of the radiant energy emitted by the filament that falls in the visible range. Also determine the wavelength at which the emission of radiation from the filament peaks. Solution The visible range of the electromagnetic spectrum extends from 0.4 to 0.76 micro meter. Using Table 12-2: ( ) ( ) 05271.0 053035.0.1900250076.0 000321.0.100025004.0 12 22 11 =− =→== =→== λλ λ λ µµλ µµλ ff fKmKmT fKmKmT which means only about 5% of the radiation emitted by the filament of the light bulb falls in the visible range. The remaining 95% appears in the infrared region or the “invisible light”. Radiation Properties A blackbody can serve as a convenient reference in describing the emission and absorption characteristics of real surfaces. Emissivity The emissivity of a surface is defined as the ratio of the radiation emitted by the surface to the radiation emitted by a blackbody at the same temperature. Thus, 10 ≤≤ ε Emissivity is a measure of how closely a surface approximate a blackbody, εblackbody = 1. The emissivity of a surface is not a constant; it is a function of temperature of the surface and wavelength and the direction of the emitted radiation, ε = ε (T, λ, θ) where θ is the angle between the direction and the normal of the surface. The total emissivity of a surface is the average emissivity of a surface over all direction and wavelengths: ( ) ( ) ( ) ( ) ( ) ( ) 4 4 TTTE T TE TE TET b σε σ ε =→== Spectral emissivity is defined in a similar manner: ( ) ( ) ( )TE TE T bλ λ λε = where Eλ(T) is the spectral emissive power of the real surface. As shown, the radiation emission from a real surface differs from the Planck’s distribution. Chapter 12, E&CE 309, Spring 2005. 6 Majid Bahrami Fig. 12-4: Comparison of the emissive power of a real surface and a blackbody. To make the radiation calculations easier, we define the following approximations: Diffuse surface: is a surface which its properties are independent of direction. Gray surface: is a surface which its properties are independent from wavelength. Therefore, the emissivity of a gray, diffuse surface is the total hemispherical (or simply the total) emissivity of that surface. A gray surface should emit as much as radiation as the real surface it represents at the same temperature: ( ) ( ) ( ) 4 0 T dTET T b σ λε ε λλ∫ ∞ = Emissivity is a strong function of temperature, see Fig. 12-20 Cengel book. Absorptivity, Reflectivity, and Transmissivity The radiation energy incident on a surface per unit area per unit time is called irradiation, G. Absorptivity α: is the fraction of irradiation absorbed by the surface. Reflectivity ρ: is the fraction of irradiation reflected by the surface. Transmissivity τ: is the fraction of irradiation transmitted through the surface. Radiosity J: total radiation energy streaming from a surface, per unit area per unit time. It is the summation of the reflected and the emitted radiation. Chapter 12, E&CE 309, Spring 2005. 7 Majid Bahrami 10 radiationincident radiation dtransmitte:vitytransmissi 10 radiationincident radiation reflected:tyreflectivi 10 radiationincident radiation absorbed:tyabsorptivi ≤≤== ≤≤== ≤≤== ττ ρρ αα G G G G G G tr ref abs Applying the first law of thermodynamics, the sum of the absorbed, reflected, and the transmitted radiation radiations must be equal to the incident radiation: Gabs + Gref + Gtr = G Divide by G: α + ρ + τ = 1 Fig. 12-5: The absorption, reflection, and transmission of irradiation by a semi- transparent material. For opaque surfaces τ = 0 and thus: α + ρ = 1. The above definitions are for total hemi-spherical properties (over all direction and all frequencies). We can also define these properties in terms of their spectral counterparts: Incident radiation G, W/m2 Reflected ρG Absorbed αG Transmitted τG Semi-transparent material Emitted radiation ε Ebλ Radiosity, J (Reflected + Emitted radiation) Chapter 12, E&CE 309, Spring 2005. 10 Majid Bahrami Tsky = 285 K for warm cloudy sky Using Kirchhoff’s law we can write α = ε since the temperature of the sky is on the order of the room temperature. The View Factor Radiation heat transfer between surfaces depends on the orientation of the surfaces relative to each other as well as their radiation properties and temperatures. View factor (or shape factor) is a purely geometrical parameter that accounts for the effects of orientation on radiation between surfaces. In view factor calculations, we assume uniform radiation in all directions throughout the surface, i.e., surfaces are isothermal and diffuse. Also the medium between two surfaces does not absorb, emit, or scatter radiation. Fi→j or Fij = the fraction of the radiation leaving surface i that strikes surface j directly. Note the following: The view factor ranges between zero and one. Fij = 0 indicates that two surfaces do not see each other directly. Fij = 1 indicates that the surface j completely surrounds surface i. The radiation that strikes a surface does not need to be absorbed by that surface. Fii is the fraction of radiation leaving surface i that strikes itself directly. Fii = 0 for plane or convex surfaces, and Fii ≠ 0 for concave surfaces. Fig. 12-7: View factor between surface and itself. Calculating view factors between surfaces are usually very complex and difficult to perform. View factors for selected geometries are given in Table 12-4 and !2-5 and Figs. 12-41 to 12-44 in Cengel book. Plane surface, Fii = 0 Convex surface, Fii = 0 Concave surface, Fii ≠ 0 Chapter 12, E&CE 309, Spring 2005. 11 Majid Bahrami View Factor Relations Radiation analysis of an enclosure consisting of N surfaces requires the calculations of N2 view factors. However, all of these calculations are not necessary. Once a sufficient number of view factors are available, the rest of them can be found using the following relations for view factors. The Reciprocity Rule The view factor Fij is not equal to Fji unless the areas of the two surfaces are equal. It can be shown that: Ai Fij =Aj Fji The Summation Rule In radiation analysis, we usually form an enclosure. The conservation of energy principle requires that the entire radiation leaving any surface i of an enclosure be intercepted by the surfaces of enclosure. Therefore, 1 1 =∑ = N j ijF The summation rule can be applied to each surface of an enclosure by varying i from 1 to N (number of surfaces). Thus the summation rule gives N equations. Also reciprocity rule gives 0.5 N (N-1) additional equations. Therefore, the total number of view factors that need to be evaluated directly for an N-surface enclosure becomes ( ) ( )1 2 11 2 12 −=⎥⎦ ⎤ ⎢⎣ ⎡ −+− NNNNNN Example 12-2 Determine the view factors F12 and F21 for the following geometries: 1) Sphere of diameter D inside a cubical box of length L = D. L = D 1 L D L 2 3 A1 A2 A1 A2 A3 A1 A3 A2 Chapter 12, E&CE 309, Spring 2005. 12 Majid Bahrami 2) Diagonal partition within a long square duct. 3) End and side of a circular tube of equal length and diameter, L = D. Assumptions: Diffuse surfaces. Solution: 1) sphere within a cube: By inspection, F12 = 1 By reciprocity and summation: 6 11 6 1 6 222221 2 2 12 2 1 21 π ππ −=→=+ =×== FFF L DF A AF 2) Partition within a square duct: From summation rule, F11 + F12 + F13 = 1 where F11 = 0 By symmetry F12 = F13 Thus, F12 = 0.5. From reciprocity: 71.05.02 12 2 1 21 =×== L LF A AF 3) Circular tube: from Fig. 12-43, with r2 / L = 0.5 and L / r1 = 2, F13 ≈ 0.17. From summation rule, F11 + F12 + F13 = 1 with F11 = 0, F12 = 1 - F13 = 0.83 From reciprocity, 21.083.04/2 12 2 1 21 =×== DL DF A AF π π The Superposition Rule The view factor from a surface i to a surface j is equal to the sum of the view factors from surface i to the parts of surface j. Chapter 12, E&CE 309, Spring 2005. 15 Majid Bahrami ( ) ( )WTTFAQQ jiij N j i N j iji 44 11 −== ∑∑ == •• σ Using the sign convention, a negative heat transfer rate indicates that the radiation heat transfer is to surface i (heat gain). Now, we can extend this analysis to non-black surfaces. It is common to assume that the surfaces are opaque, diffuse, and gray. Also, surfaces are considered to be isothermal. Also the fluid inside the cavity is not participating in the radiation. Radiosity J is the total radiation energy streaming from a surface, per unit area per unit time. It is the summation of the reflected and the emitted radiation. For a surface i that is gray and opaque (εi = αi and αi + ρi = 1), the Radiosity can be expressed as ( ) ( ) ( )blackbody afor /1 4 2 ibiii iibiii iibiii TEJ mWGEJ GEJ σε εε ρε == −+= += Note that the radiosity of a blackbody is equal to its emissive power. Using an energy balance, the net rate of radiation heat transfer from a surface i of surface area Ai can be expressed as ( ) ( ) ( )ibi i ii i biii iii iiii JE AEJ JAQ WGJAQ − − =⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ − − −= −= • • ε ε ε ε 11 In electrical analogy to Ohm’s law, a thermal resistance can be defined as ii i i i ibi i A R R JE Q ε ε− = − = • 1 where Ri is called the surface resistance to radiation. Fig. 12-10: Surface resistance to radiation. Q• i Ji Ebi Surface i Chapter 12, E&CE 309, Spring 2005. 16 Majid Bahrami Note that the surface resistance to radiation for a blackbody is zero. For insulated or adiabatic surfaces, the net heat transfer through them is zero. In this cases, the surface is called reradiating surface. There is no net heat transfer to a reradiating surface. Net Radiation between Two Surfaces Consider two diffuse, gray, and opaque surfaces of arbitrary shape maintained at uniform temperatures. The net rate of radiation heat transfer from surface i to surface j can be expressed ( ) ( ) ( )WJJFAQ WFJAFJAQ jiijiij jijjijiiij −= −= • • yreciprocit Applying In analogy with Ohm’s law, a resistance can be defined as iji ij ij ji ij FA R R JJ Q 1 = − = • where Rij is called the space resistance to radiation. Fig. 12-11: Electrical network, surface and space resistances. In an N-surface enclosure, the conservation of energy principle requires that the net heat transfer from surface i to be equal to the sum of the net heat transfers from i to each of the N surfaces of the enclosure. ( )W R JJ QQ N j ij ji N j iji ∑∑ == •• − == 11 We have already derived a relationship for the net radiation from a surface Q• ij Jj Ebi Surface i Surface j Ebj Rij Ji Rj Ri Chapter 12, E&CE 309, Spring 2005. 17 Majid Bahrami ( )W R JE Q i ibi i − = • Combining these two relationships gives: ( )W R JJ R JE N j ij ji i ibi ∑ = − = − 1 Method of Solving Radiation Problem In radiation problems, either the temperature or the net rate of heat transfer must be given for each of the surfaces to obtain a unique solution for the unknown surface temperature and heat transfer rates. We use the network method which is based on the electrical network analogy. The following steps should be taken: 1. Form an enclosure; consider fictitious surface(s) for openings, room, etc. 2. Draw a surface resistance associated with each surface of the enclosure 3. Connect the surface resistances with space resistances 4. Solve the radiations problem (radiosities) by treating it as an electrical network problem. Note that this method is not practical for enclosures with more than 4 surfaces. Example 12-4: Hot Plates in Room Two parallel plates 0.5 by 1.0 m are spaced 0.5 m apart. One plate is maintained at 1000°C and the other at 500°C. The emissivities of the plates are 0.2 and 0.5, respectively. The plates are located in a very large room, the walls of which are maintained at 27°C. The plates exchange heat with each other and with the room, but only the plate surfaces facing each other are to be considered in the analysis. Find the net heat transfer rate to each plate and the room; neglect other modes of heat transfer, i.e., conduction and convection. Assumptions: Diffuse, gray, and opaque surfaces and steady-state heat transfer. Solution: This is a three-body problem, the two plates and room. The radiation network is shown below.