G α Absorbed irradiation
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2 Thermal energy emitted y matter as a result of virational and rotational movements of molecules, atoms and electrons. The energy is transported y electromagnetic waves (or photons). adiation reuires no medium for its propagation, therefore, can take place also in vacuum. ll matters emit radiation as long as they have a finite (greater than asolute zero) temperature. The rate at which radiation energy is emitted is usually uantified y the modified Stefan- Bolzmann law: where the emissivity, ε, is a property of the surface characterizing how effectively the surface radiates compared to a "lackody" (0<ε< ). E/ (W/m ) is the surface emissive power. σ is the Stefan-Boltzmann constant (s5.67x0-8 W/(m K 4 )). T is the asolute surface temp. (in K)
3 G α αg sored irradiation
4 Electromagnetic radiation spectrum Thermal radiation spectrum range: 0. to 00 mm It includes some ultraviolet (UV) radiation and all visile ( mm) and infrared radiation (I).
5
6 The Planck Distriution The Planck law descries theoretical spectral distriution for the emissive power of a lack ody. It can e written as where C.74x0 8 (W.μm 4 /m ) and C.49x0 4 (μm.k) are two constants. The Planck distriution is shown in the following figure as a function of wavelength for different ody temperatures.
7 Spectral lackody emissive power
8 Planck Distriution t given wavelength, the emissive power increases with increasing temperature s the temperature increases,more emissive energy appear at shorter wavelengths For low temperature (>800 K), all radiant energy falls in the infrared region and is not visile to the human eyes. That is why only very high temperature ojects, such as molten iron, can glow. Sun can e approximated as a lackody at 5800 K
9 Solar Irradiation
10 ngles and rc Length We are well accustomed to thinking of an angle as a two dimensional oject. It may e used to find an arc length: α L L r α Solid ngle We generalize the idea of an angle and an arc length to three dimensions and define a solid angle, Ω, which like the standard angle has no dimensions. The solid angle, when multiplied y the radius suared will have dimensions of length suared, or area, and will have the magnitude of the encompassed area. r r dω
11 Projected rea The area, d, as seen from the prospective of a viewer, situated at an angle θ from the normal to the surface, will appear somewhat smaller, as cos θ d. This smaller area is termed the projected area. projected cos θ normal d θ d cos θ Intensity The ideal intensity, I, may now e defined as the energy emitted from an ideal ody, per unit projected area, per unit time, per unit solid angle. I d cosθ d dω
12 Spherical Geometry Since any surface will emit radiation outward in all directions aove the surface, the spherical coordinate system provides a convenient tool for analysis. The three asic coordinates shown are, φ, and θ, representing the radial, azimuthal and zenith directions. d θ sin θ d In general d will correspond to the emitting φ surface or the source. The surface d will correspond to the receiving surface or the target. Note that the area proscried on the hemisphere, d, may e written as: d [( sin θ ) dϕ] [ d ] d θ sin θ dϕ dθ ] Δφ ecalling the definition of the solid angle, d dω we find that: dω sin θ dθ dφ
13 eal Surfaces Thus far we have spoken of ideal surfaces, i.e. those that emit energy according to the Stefan-Boltzman law: E σ T as 4 eal surfaces have emissive powers, E, which are somewhat less than that otained theoretically y Boltzman. To account for this reduction, we introduce the emissivity, ε. E ε E Emissive power from any real surface is given y: E ε σ T as 4 eceiving Properties Targets receive radiation in one of three ways; they asorption, reflection or transmission. sorptivity, α, the fraction of incident radiation asored. eflectivity, ρ, the fraction of incident radiation reflected. Transmissivity, τ, the fraction of incident radiation transmitted. eflected adiation sored adiation Transmitted adiation Incident adiation, G
14 We see, from Conservation of Energy, that: α + ρ + τ In this course, we will deal with only opaue surfaces, τ 0, so that: α + ρ elationship Between sorptivity,α, and Emissivity,ε Consider two flat, infinite planes, surface and surface B, oth emitting radiation toward one another. Surface B is assumed to e an ideal emitter, i.e. ε B.0. Surface will emit radiation according to the Stefan-Boltzman law as: E ε σ T 4 and will receive radiation as: G α σ T B 4 The net heat flow from surface will e: ε σ T 4 - α σ T B 4 Surface, T Surface B, T B Now suppose that the two surfaces are at exactly the same temperature. The heat flow must e zero according to the nd law. If follows then that: α ε
15 Thermodynamic properties of the material, α and ε may depend on temperature. In general, this will e the case as radiative properties will depend on wavelength, λ. The wave length of radiation will, in turn, depend on the temperature of the source of radiation. The emissivity, ε, of surface will depend on the material of which surface is composed, i.e. aluminum, rass, steel, etc. and on the temperature of surface. The asorptivity, α, of surface will depend on the material of which surface is composed, i.e. aluminum, rass, steel, etc. and on the temperature of surface B. Black Surfaces Within the visual and of radiation, any material, which asors all visile light, appears as lack. Extending this concept to the much roader thermal and, we speak of surfaces with α as also eing lack or thermally lack. It follows that for such a surface, ε and the surface will ehave as an ideal emitter. The terms ideal surface and lack surface are used interchangealy.
16 Diffuse Surface: efers to directional independence of the intensity associated with emitted,reflected,or incident radiation. Grey Surface: surface for which the spectral asorptivity and the emissivity are independent of wavelength over the spectral regions of surface irradiation and emission.
17 elationship Between Emissive Power and Intensity By definition of the two terms, emissive power for an ideal surface, E, and intensity for an ideal surface, I. E I hemisphere cos θ dω eplacing the solid angle y its euivalent in spherical angles: E π 0 π 0 I cos θ sin θ dθ dϕ Integrate once, holding I constant: E π I cos θ sin θ dθ π 0 Integrate a second time. (Note that the derivative of sin θ is cos θ dθ.) E π sin θ π I π I 0 E π I
18 I d cos θ d dω Next we will project the receiving surface onto the hemisphere surrounding the source. First find the projected area of surface d, d cos θ. (θ is the angle etween the normal to surface and the position vector,.) Then find the solid angle, Ω, which encompasses this area. d I cos θ d dω d cos θ d To otain the entire heat transferred from a finite area, d, to a finite area, d, we integrate over oth surfaces: I cos θ d cosθ d
19 Total energy emitted from surface : emitted E π I View Factors-Integral Method Define the view factor, F -, as the fraction of energy emitted from surface, which directly strikes surface. F emitted I cos θ d π I cosθ d F cos θ cosθ π d d dr d j D Example Consider a diffuse circular disk of diameter D and area j and a plane diffuse surface of area i << j. The surfaces are parallel, and i is located at a distance L from the center of j. Otain an expression for the view factor F ij. j d i θ i θ j d i L
20 cos cos d d F π θ θ Since d is a differential area cos cos d F π θ θ ( ) dr r L F π π 4 dr r L F Let ρ L + r. Then ρ dρ r dr. 4 d L F ρ ρ ρ 0 D L L L F + ρ ρ D L D L D L L F D + +
21 Enclosures In order that we might apply conservation of energy to the radiation process, we must account for all energy leaving a surface. We imagine that the surrounding surfaces act as an enclosure aout the heat source which receive all emitted energy. For an N surfaced enclosure, we can then see that: eciprocity i F i j j F N j F i, j This relationship is known as the Conservation ule. j i Example: Consider two concentric spheres shown to the right. ll radiation leaving the outside of surface will strike surface. Part of the radiant energy leaving the inside surface of oject will strike surface, part will return to surface. Find F,. pply reciprocity. D F, F, F, F, D This relationship is known as eciprocity.
22 ssociative ule Consider the set of surfaces shown to the right: Clearly, from conservation of energy, the fraction of energy leaving surface i and striking the comined surface j+k will eual the fraction of energy emitted from i and striking j plus the fraction leaving surface i and striking k. F i ( j+ k) Fi j + F i k This relationship is known as the ssociative ule. k j i adiosity adiosity,, is defined as the total energy leaving a surface per unit area and per unit time. ε E + ρ G ε E ρ G G
23 Net Exchange Between Surfaces Consider the two surfaces shown. adiation will travel from surface i to surface j and will also travel from j to i. i j i i Fi j i j likewise, j i j j Fj j The net heat transfer is then: j i (net) i i Fi j - j j Fj j From reciprocity we note that F F so that j i (net) i i Fi j - j i Fi j i Fi j (i j)
24 Net Energy Leaving a Surface The net energy leaving a surface will e the difference etween the energy leaving a surface and the energy received y a surface: [ε E α G] Comine this relationship with the definition of adiosity to eliminate G. ε E + ρ G G [ - ε E]/ρ {ε E α [ - ε E]/ρ} ssume opaue surfaces so that α + ρ ρ α, and sustitute for ρ. {ε E α [ - ε E]/( α)} Put the euation over a common denominator: ( ) E E E + α α ε α ε α α ε α assume that α ε ( ) E E ε ε ε ε ε
25 Electrical nalogy for adiation We may develop an electrical analogy for radiation, similar to that produced for conduction. The two analogies should not e mixed: they have different dimensions on the potential differences, resistance and current flows. Euivalent Current Euivalent esistance Potential Difference Ohms Law I ΔV Net Energy Leaving Surface? ε E - ε Net Exchange Between Surfaces i? j F
26 Example: Consider a grate fed oiler. Coal is fed at the ottom, moves across the grate as it urns and radiates to the walls and top of the furnace. The walls are cooled y flowing water through tues placed inside of the walls. Saturated water is introduced at the ottom of the walls and leaves at the top at a uality of aout 70%. fter the vapour is separated from the water, it is circulated through the superheat tues at the top of the oiler. Since the steam is undergoing a sensile heat addition, its temperature will rise. It is common practice to sudivide the super-heater tues into sections, each having nearly uniform temperature. In our case we will use only one superheat section using an average temperature for the entire region. Coal Chute SH section Surface Superheat Tues Surface Burning Coal Surface Water Walls
27 Energy will leave the coal ed, Surface, as descried y the euation for the net energy leaving a surface. We draw the euivalent electrical network as seen to the right: The heat leaving from the surface of the coal may proceed to either the water walls or to the super-heater section. That part of the circuit is represented y a potential difference etween adiosity: F ε ε E σ T 4 F F E F F E It should e noted that surfaces and will also radiate to one another.
28 It remains to evaluate the net heat flow leaving (entering) nodes and. ε ε E F F ε ε E F E ε ε Insulated surfaces. In steady state heat transfer, a surface cannot receive net energy if it is insulated. Because the energy cannot e stored y a surface in steady state, all energy must e re-radiated ack into the enclosure. Insulated surfaces are often termed as re-radiating surfaces. Black surfaces: lack, or ideal surface, will have no surface resistance: ε ε 0 In this case the nodal adiosity and emissive power will e eual.
29 Large surfaces: Surfaces having a large surface area will ehave as lack surfaces, irrespective of the actual surface properties: Consider the case of an oject,, placed inside a large enclosure,. The system will consist of two ojects, so we proceed to construct a circuit with two radiosity nodes. Now we ground oth adiosity nodes through a surface resistance. ε ε ε ε /( F ) 0 (-ε )/(ε ) /( F ) (-ε )/(ε ) E σt 4 E σt 4
30 Since is large, 0. The view factor, F (-ε )/(ε ) /( F ) E σt 4 E σt 4 Sum the series resistances: Series (-ε )/(ε ) + / /(ε ) Ohm s law: i ΔV/ or y analogy: ΔE / Series ε σ(t 4 T 4 ) eturning for a moment to the coal grate furnace, let us assume that we know (a) the total heat eing produced y the coal ed, () the temperatures of the water walls and (c) the temperature of the super heater sections.
31 pply Kirchoff s law aout node, for the coal ed: Similarly, for node : E nd for node : E The three euations must e solved simultaneously. Since they are each linear in, matrix methods may e used: E E
32 Surface : Find the coal ed temperature, given the heat flow: σ σ T T E Surface : Find the water wall heat input, given the water wall temperature: 4 T E σ Surface : (Similar to surface ) Find the water wall heat input, given the water wall temperature: 4 T E σ
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