METO 621Upper half-range intensitySlide 3Slide 4Formal solution including Scattering and EmissionFormal solution including scattering and emissionSlide 7Radiative Heating RateSlide 9Slide 10Separation into diffuse and direct(Solar) componentsDiffuse and direct componentsSlide 13Slide 14Slide 15Slide 16Slide 17Slide 18Slide 19METO 621Lesson 10Upper half-range intensity• For the upper half-range intensity we use the integrating factor e-( )μτνμτννμτνμτμττ///)(1−−++−+−=⎟⎟⎠⎞⎜⎜⎝⎛−= eBeIddIeIdd• In this case we are dealing with upgoing beams and we integrate from the bottom to the top.Upper half-range intensity( )or)'(')'('),*,(),,0(''*0/'/'0*/*/'0*∫∫∫−−−++−+=−=−=ττμττμττμτννμτνττμττμτφμτφμττBedBedeIIeIddd€ Iν+(0,μ,φ) = Iν+(τ*,μ,φ)e−τ*/ μ+dτ'μ0τ*∫Bν(τ')e−τ'/ μUpper half-range intensity•To find the intensity at an interior point , integrate from * to  and obtain€ Iν+(τ,μ,φ) = Iν+(τ*,μ,φ)e−(τ*−τ)/ μ+dτ'μττ*∫Bν(τ')e−(τ'−τ)/ μ• What happens when  approaches zero. This is where the line of sight traverses an infinite distance parallel to the slab. Here€ Iτ±(τ,μ = 0,φ) = Bν(τ)Formal solution including Scattering and Emission• Note that the source is now due to thermal emission and multiple scattering€ S(τ,ˆ Ω ) = 1− a(τ)[ ]B(τ) +a(τ)4πdω' p(τ,ˆ Ω ',ˆ Ω )I(τ,ˆ Ω ')4 π∫• The independent variable is the extinction optical depth, the sum of the absorption and scattering optical depths. We can write)ˆ,()ˆ,()ˆ,(Ω+Ω−=Ωττττμ SIddIFormal solution including scattering and emission• The method of using an integrating factor can be applied as before€ I τ(P2),ˆ Ω [ ]=I τ(P1),ˆ Ω [ ]e−τ(P1,P 2)+ dt Sτ(P1)τ(P2)∫(t,ˆ Ω )e−τ(P P2)• In slab geometry the solutions become€ I−(τ,μ,φ) = I−(0,μ,φ)e−τ / μ+dτ'μ0τ∫S(τ',μ,φ)e−(τ−τ')/ μFormal solution including scattering and emission€ I+(τ,μ,φ) = I+(τ*,μ,φ)e−(τ*−τ)/ μ+dτ'μττ*∫S (τ')e−(τ'−τ)/ μ• where€ I±(τ,μ = 0,φ) = S (τ,μ = 0,φ)€ S(τ,μ,φ) = 1− a[ ]B(τ)+a4πdφ' dμ' p(μ',φ';μ,φ)I+(τ,μ',φ')01∫02π∫+a4πdφ' dμ' p(−μ',φ';μ,φ)I−(τ,μ',φ')01∫02π∫• andRadiative Heating Rate• The differential change of energy over the distance ds along a beam is € ∂(d4E) = dIνdA dt dν dω• If we divide this expression by dsdA, (the unit volume, dV), and also ddt then we get the time rate of change in radiative energy per unit volume per unit frequency, due to a change in intensity for beams within the solid angle d.Radiative Heating Rate• Since there is (generally) incoming radiation from all directions, the total change in energy per unit frequency per unit time per unit volume is∫∫∇⋅Ω=πνπνωω44)ˆ( IddsdId• The spectral heating rate H is )ˆ(4∫∇⋅Ω−=πννωΗ IdRadiative Heating Rate• The net radiative heating rate H is€ Η =− dν0∞∫dω(ˆ Ω ⋅∇Iν )4 π∫• In a slab geometry the radiative heating rate is written € H = − dνδFνδz0∞∫= −2π dν cosθd(cosθ)∂Iν∂z−1+1∫0∞∫where Fν= Fν+− Fν− is the radiative flux in the z directionSeparation into diffuse and direct(Solar) components•Two distinctly different components of the shortwave radiation field. The solar component:€ IS−(τ,θ,φ) = FSe−τ / μ0δ(ˆ Ω −ˆ Ω 0)= FSe−τ / μ0δ(μ − μ0)δ(φ− φ0)• We have two sources to consider, the Sun and the rest of the medium€ I−(τ,μ,φ) = Id−(τ,μ,φ) + IS−(τ,μ,φ)Diffuse and direct components•Assume (1) the lower surface is black, (2) no thermal radiation from the surface, the we can write the half range intensities as € −μdI−(τ,ˆ Ω )dτ= I−(τ,ˆ Ω ) − (1− a)B −a4πdω' p(+ˆ Ω ',−ˆ Ω )I+(τ,ˆ Ω ')+∫ −a4πdω' p(−ˆ Ω ',−ˆ Ω )I−(τ,ˆ Ω ')+∫Diffuse and direct components•And for the +ve direction € −μdI+(τ,ˆ Ω )dτ= I+(τ,ˆ Ω ) − (1− a)B −a4πdω' p(+ˆ Ω ',+ˆ Ω )I+(τ,ˆ Ω ')+∫ −a4πdω' p(−ˆ Ω ',+ˆ Ω )I−(τ,ˆ Ω ')+∫Diffuse and direct components∫∫∫−−−+−−−−−−ΩΩ−Ω−−ΩΩ−Ω+−ΩΩ−Ω−−−−Ω+Ω=Ω−Ω−)'ˆ,()ˆ,'ˆ('4)'ˆ,()ˆ,'ˆ('4)'ˆ,()ˆ,'ˆ('4)1()ˆ,()ˆ,()ˆ,()ˆ,(τωπτωπτωπττττμττμddSSdSdIpdaIpdaIpdaBaIIddIddINow substitute the sum of the direct and diffuse componentsDiffuse and direct componentsgets one hence/ and beamsolar direct theis But μτdIdIISSS−−−−=€ −μdId−(τ,ˆ Ω )dτ= Id−(τ,ˆ Ω ) − (1− a)B − S*(τ,−ˆ Ω ) −a4πdω' p(+ˆ Ω ',−ˆ Ω )Id+(τ,ˆ Ω ')+∫ −a4πdω' p(−ˆ Ω ',−ˆ Ω )Id−(τ,ˆ Ω ')−∫Diffuse and direct components•where€ S*(τ,−ˆ Ω ) =a4πdω' p(−ˆ Ω ',−ˆ Ω )FS−∫e−τ / μ0δ(ˆ Ω −ˆ Ω 0)=a4πp(−ˆ Ω 0,−ˆ Ω )FSe−τ / μ0• One can repeat this procedure for the upward componentDiffuse and direct components€ −μdId+(τ,ˆ Ω )dτ= Id+(τ,ˆ Ω ) − (1− a)B − S*(τ,+ˆ Ω ) −a4πdω' p(+ˆ Ω ',+ˆ Ω )Id+(τ,ˆ Ω ')+∫ −a4πdω' p(−ˆ Ω ',+ˆ Ω )Id−(τ,ˆ Ω ')−∫Diffuse and direct components€ S*(τ,+ˆ Ω ) =a4πdω' p(−ˆ Ω ',+ˆ Ω )FS−∫e−τ / μ0δ(ˆ Ω −ˆ Ω 0)=a4πp(−ˆ Ω 0,+ˆ Ω )FSe−τ / μ0Diffuse and direct components•If we combine the half-range intensities we get€ udI(τ,u,φ)dτ= I(τ,u,φ) −a4πdφ'02π∫du', p(u'.φ';u,φ)I(τ,u',φ')−1+1∫ − (1− a)B − S*(τ,u,φ)• Where u is cosand not