Radiation Fundamentals

Outline • • • • Review last topic Basic ideas of heat exchangers Overall heat transfer coefficient Log-mean temperature difference method • Effectiveness –NTU method • Practical considerations Radiation Fundamentals 2 Electromagnetic Radiation • Wavelength, λ, ranges from 10-9 to 1010 μm is distance between wave peaks • EM waves travel at speed of light = 299,792,458 m/s (in a vacuum) • Frequency, ν = c/λ, units of Hz = s-1 • Radian frequency ω = 2πν, units s-1 • For ν = 60 Hz = 60 s-1, λ = (299,792,458 m/s) / (60 s-1) ≈ 5x106 m = 5x1012 μm • Radiation heat transfer by electromagnetic radiation – Part of much larger spectrum – Thermal radiation transfers heat without contact • Use of fire or electric resistance heating are best examples • Thermal radiation lies in infrared and visible part of spectrum (with some in ultraviolet) Figure 12-3 from Çengel, Heat and Mass Transfer EM Radiation Properties 3 Black-body Radiation 4 Black-Body Radiation II • Basic black body equation: Eb = σT4 • Perfect emitter – no surface can emit more radiation than a black body • Diffuse emitter – radiation is uniform in all directions • Perfect absorber – all radiation striking a black body is absorbed – Eb is total black-body radiation energy flux W/m2 or Btu/hr·ft2 – σ is the Stefan-Boltzmann constant • σ = 5.670x10-8 W/m2·K4 • σ = 0.1714x10-8 Btu/hr·ft2·R4 – Must use absolute temperature • Radiation flux varies with wavelength – Ebλ is flux at given wavelength, λ 5 6 1 Radiation Basics Ultraviolet Infrared Stefan-Boltzmann Constant Spectral Ebλ • Found experimentally, but later analysis relates σ to other fundamental constants • Energy (W/m2) emitted varies with wavelength and temperature • Maximum point occurs where λT = 2897.8 μm·K • T increase shifts peak shift to lower λ – σ = 2π5k4/(15h3c2) – k = Boltzmann’s constant = 1.38065x10-23 J/K (molecular gas constant) = Ru/NAvagadro – h = Planck’s constant = 6.62607x10-34 J·s • First notion of quantum mechanics that energy associated with a wave, ε = hν = hc/λ – c = 299,792,458 m/s = speed of light in a vacuum 7 8 Figure 12-9 from Çengel, Heat and Mass Transfer Spectral Black-body Energy Total Energy • Ebλdλ = black-body emissive power in a wavelength range dλ about λ • Total energy is integral over all wavelengths – Typical units for Ebλ are W/m2·μm or Btu/hr·ft2·μm Ebλ dλ = ( C1 C2 λ T ) ∞ dλ ∫ Eb = Ebλ dλ = σT 4 λ e −1 • C1 = = 3.74177 W·μm4/m2 • C2 = hc/k = 14387.8 μm/K 5 0 • Recall that σ = 2π5k4/(15h3c2) 2πhc2 – h = Planck’s constant, c = speed of light in vacuum, k = Boltzmann’s constant Figure 12-11 from Çengel, Heat and Mass Transfer 9 Integral Proof Integral Proof II • Show IEbλdλ = σT4 on this and next chart • Get single variable z and integrate – First get common variable C2/λT = C2/y λ =∞ Eb = λ =∞ ∫ Ebλ dλ = ∫ λ5 (eC λ =0 2 λ =0 Eb T4 y =∞ = ∫ y =0 ( λT =∞ C1 C1 y 5 e C2 y λT ) −1 ) dλ = T ∫ 4 −1 C25 C25 λT = 0 C2 / y = 0 ∫ dy = C2 / y = ∞ Eb C1 (λT )5 (eC2 λT − 1) ( C1 y 5 e C2 y ) −1 C25 C25 d (λT ) C2 y ⇒ dz = − C2 y 2 dy = − C2 (C2 z ) 2 dy ⇒ dy = − dy 4 C2 / y = 0 = ∫ C2 / y = ∞ z =0 C1z 5 − ez −1 z =∞ ∫ ( 5 ( y e ) C1 C2 y 1 C2 C25 z 2 ) C25 −1 C25 dz = C1 z =0 dy = z =∞ C z5 1 ∫ (e z1− 1) C25 dy z =∞ z3 C π4 ∫ (e z − 1) dz = C214 15 C24 z =0 – Standard integral found from Matlab command int(‘z^3/(exp(z)-1)’,0,inf) C2 z T = • Define z = C2/y and get dy in terms of dz z= 10 2 11 dz Eb = C1 π 4 4 2πhc 2 π 4 4 2π5 k 4 4 T = T = T = σT 4 C24 15 15h 3c 2 (hc k )4 15 12 2 Radiation Basics Partial Black-body Power Radiation Tables Black body radiation between λ = 0 and λ = λ1 is Eb,0-λ1 λ 1 ∫ Eb,0−λ1 = Ebλ dλ • Can show that fλ is function of λT fλ = λ 1 σT 4 1 λ C1 ∫ Ebλ dλ = σT 4 ∫ λ5 (eC 0 2 λT 0 ) −1 dλ = 1 σ λT ∫ (λT )5 (eC1 λT − 1) d (λT ) C 2 0 0 Fraction of total radiation (σT4) between λ = 0 and any given λ is fλ fλ = λ 1 σT 4 ∫ Ebλ dλ' 0 • Radiation tables give fλ versus λT – See table 12-2, page 672 in text – Extract from this table shown at right 13 14 Figure 12-13 from Çengel, Heat and Mass Transfer Δλ Sample Problem • Radiation in finite band, Δλ f λ1 −λ 2 = 1 σT 4 1 σT 4 λ2 ∫ Ebλ dλ − λ2 ∫ Ebλ dλ = λ1 1 λ1 ∫ Ebλ dλ 4 T σ 0 0 = f (λ 2T ) − f (λ1T ) • A conventional light bulb has a filament temperature of 4000oF. Find the fraction of visible radiation from this filament, if it is a black body. • Given: T = 4000oF and visible region • Find: Fraction of total radiation in region • Missing information: Visible region is between 0.4 μm and 0.76 μm • Conversion: 4000oF = 4460 R = 2478 K 15 16 Figure 12-14 from Çengel, Heat and Mass Transfer Sample Problem Solution Radiation Exchange • Compute λT at λ1 and λ2 and find corresponding fλ values in Table 12.2 – – – – • In general, radiation leaving a surface can vary in direction λ1T = (0.4 μm)(2478 K) = 991 μm·K λ1T = (0.79 μm)(2478 K) = 1883 μm·K f(λ1T) = 0.000289 (interpolation in table) f(λ2T) = 0.04980 (interpolation in table) – Ideal of diffuse radiation is uniform in all directions – Need coordinate system for radiation leaving a surface – Look at hemisphere on top of surface and use spherical coordinate system – I(θ,φ) is radiation intensity in direction (θ,φ) • Fraction in visible range = 0.04980 – 0.000289 = 0.0495 or about 5% in visible range for conventional lighting • See chart after next for diagram 17 18 3 Radiation Basics Review Spherical Coordinates • • • Figure 2-3 from Çengel, Heat and Mass Transfer Radiation Intensity Use angular coordinates φ and θ φ is polar angle in x-y plane θ is azithmuthal angle with z axis • Radiation intensity, I, is radiation in a particular direction, (θ,φ) • Intensity depends on area at outer surface of cone • Measure of this surface is solid angle 19 20 dω = sinθ dθdφ Solid Angle II Solid Angle • Similar to radian angular measure in 2D θ Figure 12-16 from Çengel, Heat and Mass Transfer • Arc length, l = rθ so θ = l/r • Differential arc length dl = rdθ so dθ = dl/r • Partial surface area, S = r2ω so ω = S/r2 (Total area = 4πr2) • Units for ω are called steradians (sr) • Just as there are 2π radians in a circle, there are 4π steradians in a sphere • Differential area, dS = dω/r2 Figure 12-18 from Çengel, Heat and Mass Transfer 21 22 Figure 12-17 from Çengel, Heat and Mass Transfer Radiation Intensity II Radiation Intensity III • Radiation intensity, I, is radiation energy in a particular direction, dSnormal = (θ,φ) per unit area dAcosθ normal to the direction, Solid per unit solid angle, ω angle ω • Normal area is projection of dA normal to direction = dAcosθ Figure 12-16 from Çengel, Heat and Mass Transfer 23 • Emitted intensity, Ie, is radiation energy, dQe, in a particular direction, dSnormal = (θ,φ) per unit area dAcosθ normal to the direction, Solid per unit solid angle, ω angle ω dQ& e I e (θ, φ) = dA cos θdω Figure 12-16 from Çengel, dQ& e Heat and Mass Transfer = dA cos θ sin θ dθdφ 24 4 Radiation Basics For integration over hemisphere: 0 ≤ φ ≤ 2π and 0 ≤ θ ≤ π/2 Emissive Power • Radiation flux for emitted radiation (energy per unit area of surface) dE = dQ& e = I e (θ, φ) cos θ sin θdφ dA cos θdω 2π π / 2 E= ∫ dE = ∫ ∫ I e (θ, φ)cos θ sin θdθdφ hemi − sphere For constant Ie, E = πIe Figure 12-18 from Çengel, 25 Heat and Mass Transfer • Ii = incident intensity – function of direction • G = total radiation impinging on surface G= 2π π / 2 hemi − sphere ∫ ∫ Ii (θ, φ)cos θ sin θdθdφ • Emissivity, ε, is ratio of actual emissive power to black body emissive power – May be defined on a directional and wavelength basis, ελ,θ(λ,θ,φ,T) = Iλ,e(λ,θ,φ,T)/Ibλ(λ,T), called spectral, directional emissivity – Total directional emissivity, average over all wavelengths, εθ(θ,φ,T) = Ie(θ,φ,T)/Ib(T) – Spectral hemispherical emissivity average over directions, ελ(λ,T) = Iλ(λ,T)/Ibλ(λ,T) – Total hemispheric emissivity = E(T)/Eb(T) – Called “spectral” quantities • Previous quantities are then integrals over all wavelengths ∫ 0 ∞ ∫ 0 28 Emissivity • Previous discussions of I, E, G, and J have not considered wavelength • Can define Ie,λ, Ii,λ, and Ie+r,λ I i = I i,λ dλ Figure 12-21 from Çengel, Heat and Mass Transfer 27 Spectral Quantities ∞ J= Iemitted + Ireflected ∫ dG = φ=0 θ=0 I e = I e,λ dλ 26 Radiosity, J, total intensity leaving surface (sum of directly emitted plus reflected) Radiosity Irradiation G Figure 12-20 from Çengel, Heat and Mass Transfer φ = 0 θ =0 ∞ ∫ I e + r = I e + r ,λ dλ 0 29 30 5 Radiation Basics Emissivity Assumptions εθ • Diffuse surface – emissivity does not depend on direction • Gray surface – emissivity does not depend on wavelength • Gray, diffuse surface – emissivity is the does not depend on direction or wavelength • In general, total directional emissivity is constant over most of surface – Simplest surface to handle and often used in radiation calculations 31 32 Figure 12-26 from Çengel, Heat and Mass Transfer 33 34 Properties Figure 12-31 from Çengel, Heat and Mass Transfer • When radiation, G, hits a surface a fraction ρG is reflected; another fraction, αG is absorbed, a third fraction τG is transmitted • Energy balance: ρ + α + τ = 1 35 Properties II • Fractions on previous chart are properties – Reflectivity, ρ – Absorptivity, α – Transmissivity, τ Figure 12-31 from Çengel, Heat and Mass Transfer • Energy balance: ρ+α+τ=1 36 6 Radiation Basics α Data Properties III • As with emissivity, α, ρ, and τ may be defined on a spectral and directional basis – Can also take averages over wavelength, direction or both as with emissivity – Simplest case is no dependence on either wavelength or direction – Reflectivity may be diffuse or have angle of reflection equal angle of incidence 37 • Solar radiation has effective source temperature of about 5800 K Figure 12-33 from Çengel, Heat and Mass Transfer Kirchoff’s Law Effect of Temperature • Absorptivity equals emissivity (at the same temperature) • True only for values in a given direction and wavelength • Assuming total hemispherical values of α and ε are the same simplifies radiation heat transfer calculations, but is not always a good assumption • Emissivity, ε, depends on surface temperature • Absorptivity, α, depends on source temperature (e.g. Tsun ≈ 5800 K) • For surfaces exposed to solar radiation 39 From Çengel, Heat and Mass Transfer 38 – high α and low ε will keep surface warm – low α and high ε will keep surface cool – Does not violate Kirchoff’s law since source and surface temperatures differ 40 41 7