#### Transcript Ch. 3 Radiation

Ch. 3 Radiation Text: Wallace and Hobbs, Ch. 4, Radiative Transfer p113-152, Liou, An introduction to atmospheric radiation Chs. 3, 4 3.3 Radiative transfer in planetary atmosphere: • 3.3.1 Beer’s law: – quantify radiation attenuation due to scattering and absorption as it passes through a layer gases, in our case, the atmosphere. Based on dI l = -Il rrk l ds (4.17 in text), a beam of radiation passes the earth's atmosphere to a altitude z, can be determined by I l ,z = Il ,¥ ò z k l rrdz, ¥ Il = Il¥e -t l secq where Il¥ incident light at the top of the atmosphere, q : zenith angle of incident beam tl = ò ¥ z k l rrdz, optical depth or thickness, depletion of radiation that a overhead beam (zenith = 0) would experience as it passes through length the atmospheric layer. Tl = e -t l secq : transimissivity of the atmospheric layer Absorptivity : a l = 1 - Tl dIl = -Il kl rrsec qdz • Attenuation is strongest near t~1. small attenuation when t>3, because most of radiation has already lost during the transmission. • The earth’s effective temperature is the temperature approximately at t~1 in the earth’s atmosphere. Fig. 4.23 Vertical profiles of the monochromatic intensity of incident radiation, the rate of absorption of incident radiation per unit height, air density and optical depth, for kλ and r independent of height. • Why? Example: • A beam of radiation passes through a layer of 100 m thick with zenith angle of 600 and 2 types of absorbing gases, each contributes 50% to the total air mass. Gas A has density of 0.1 kg/m3 with mass absorption coefficient of 10-3 for wavelength l. Gas B has density of 0.2 kg/m3 with mass absorption coefficient of 10-2 for the same wavelength. – A) Calculate the optical depth, transmissivity and absorption of the layer at wavelength l. – B) Which gas dominates the optical depth and transmissivity of the layer? a. For the gas A, the mass of absorbing gas that the beam encounter along its slant path length is u a = sec60 o × 0.1 kg/m3 ´ 0.5 ´ 100m = 2 ´ 5kg/m2 = 10kg/m2 t l ,a = kl ,a ua = 10 -3 m2 /kg ´ 10kg/m2 = 10 -2 = 0.01 For gas B : u b = sec60 o × 0.2 kg/m3 ´ 0.5 ´ 100m = 2 ´ 10kg/m2 = 20kg/m2 t l ,b = kl ,b ub = 10 -2 m2 /kg ´ 20kg/m2 = 2 ´ 10 -1 = 0.2 The total t l = 0.21, -(t +t Tl = e l ,a l ,b = 0.81 al = 1 - Tl = 0.19 ) Gas B is the main contributor for optical depth, transimissivity and absorpitivity of the air layer AERONET-sun photometers • World largest observational network for aerosols. Based on Beer’s law, optical depth can be measured by I l = I l¥e-t l secq , ln t l = ln I l ,q1 I lq2 I l ,q1 I lq2 = t l (secq1 - secq 2 ) / (secq1 - secq 2 ) 3.3.2 Absorption and emission of infrared radiation • An beam of infrared radiation will be absorbed and emitted as it passes through a layer of the atmosphere. • The question that governs transfer of infrared radiation through a gaseous medium. Schwarzschild’s equation dI l = -dI l (absorption) + dI l (emission) = -(I l - Bl (T ))kl rrds s1 I l (s1 ) = ò I l ,oe -t l (s1 ,0) + k l rrBl [T (s)]e -t l (s1 ,s)ds monochromatic intensity 0 from 0 to s account for the monochromatic intensity 1 extincition along the way of the radiation emitted by the gas along the path from 0 to s1 account for extincition along the way. Derivation of the Schwarzchild’s law: dIl = [ -Il + Bl (T)]× r × r× kl ds Þ ò t l (s1,s) = - s1 s dIl = -Il + Bl (T) r × r× kl ds rrkl ds' Þ dt l (s1,s) = - rrk l ds dIl = -Il + Bl (T) Þ - dIl e -t l (s1 ,s) = -Il e -t l (s1 ,s)dt l (s1,s) + Bl (T)e -t l (s1 ,s)dt l (s1,s) dt l (s1,s) s1 ò { - dIl e -t l (s1 ,s) + Il e -t l (s1 ,s) dt l (s1,s)} = 0 s1 ò Bl (T)e -t l (s1 ,s)dt l (s1,s) 0 d [ Il e -t l (s1 ,s) s1 ] | = ò B (T)e s1 s l -t l (s1 ,s) dt l (s1,s) 0 Il e -t l (s1 ,s1 ) = Il (0)e -t l (s1 ,0) + s1 ò Bl (T)e t - l dt l (s1,s) (s1 ,s) 0 Il = Il (0)e -t l (s1 ,0) + s1 ò Bl (T)e t 0 - l dt l (s1,s) (s1 ,s) The plane parallel approximation: 53o • • Calculation of 3D radiative transfer is very difficult and expensive. Although radiation transmission has 3-D structure in reality, in most of case, the variations of atmospheric constituents and cloud aerosols layers are primarily a function of height. Thus, radiation transmission can be considered primarily a function of its linear distance. With the plane parallel assumption, the irradiance passes through a layer of atmopshere can be expressed as: Fn-¯(t v ) = ò pI 2 -¯ v (t v , cosq )cosq dw assume I-¯ v is isotropic with azimuthal angle f we have Fn-¯(t v ) = 2p Fn-¯ (t v ) = 2p ò ò 90o 0 -¯ I (t v , cosq )cosq sin q dq let m =cosq v 1 -¯ 1 0 0 I (t v , m )m dm = 2p ò I -¯ v,0 Tn (t v , m ) m dm analogous to Beer's law for flux density v assume Tn (t v , m ) » e - tl m » e-1.66t l where 1 m = 1.66 is a diffusivity factor, equivalent to radiation beam passes through a layer at 53o zenith angle. Fn (t v ) = 2p -¯ ò 1 -¯ 0 - tv m I e m dm = 2p I v,0 -¯ v,0 - e tv m ém2 ù 1 ê ú |0 = p I ë2û -¯ v,0 - e tv m = pI -¯ v (t v , m ) When consider the diffusive factor (3D), a bean of radiation passes through a layer of atmosphere is equivalent to an beam passes through the layer at 53o without diffusive effect under the plane parallel assumption. Plane parallel approximation is not accurate for cumuli clouds. Radiative transfer integrated over all wave numbers: • • • • Radiative flux: the total radiative energy is cumulative radiative energy (irradiance) across all wavelength (number). The transmission of radiative flux is However, to accurately estimate radiative flux, we need to use 10-1 to 10-3 cm-1 wave number interval to calculate extinction for each interval. It is very expensive and slow to do so. To improve computation efficiency, a “correlated k-distribution method” is commonly used. It group lines with similar absorption coefficiencts into one bin and g represent the fractional distribution of the lines with this absorption relative to total lines. Radiative heating/cooling in the atmosphere: • Vertical convergence or divergence causes radiative heating or cooling • Usually we use unit of oC/day. Radiative cooling can reach a few oC/day to 50oC/day at top of dense marine cloud deck. rC p dT ¶F (z) dT 1 ¶ =- R Þ =dt ¶z dt rC p ¶z dT 1 é = ò dt rC p êë 2p = ò 1 -1 [ò ] ù 1 é ¶I v m d w ò ê úû rC p ë 4p ¶z I mdw = 4p v ù ¶Iv dfdm ú where ds = dz/m, m = cosq , dw = secqdqdf û ¶s 2p é 1 ¶Iv ù 2p ò dm = rC p êë -1 ¶s úû rC p [ò 1 -1 rrkn (Iv - Bv )dm ] based on Schwarzchild's equation If we let Iv » 0 i.e., assuming the incidence of longwave from the adjacent layers does not change with height in a thin isothermal atmosphere, we have dT 2p =dt Cp [ ò rk B (z)e 1 0 n v -t v / m ] dm = - p rkn Bv (z)e -t mC p v /m where m = 1.66 as the diffusivity factor Longwave heating rate is mainly determined by Emission and optical depth, thus, heading rate Is much larger in lower troposphere. Radative heating/cooling T trends caused by Radative heating/cooling Of change in greenhouse gases: Why does CO2 warm the troposphere, but cool the stratosphere? Summary-1: • What is the main difference in determining radiative energy extinction between a ray of solar radiation and that of thermal radiation emitted from the earth surface? – Solar radiation experiences scattering and absorption by particles (clouds and aerosols) and also gases (O3 and H2O) when it passes through the atmosphere, whereas terrestrial infrared radiation experiences both absorption and emission. Summary-2: • What physics laws are used to determine the radiative energy absorption and emission extinction in the atmosphere energy? – Beer’s law: Radiative energy loss as it passes through a medium is an exponential function of optical depth of the medium and incident angle of radiation beam. This is used in radiative transfer calculation of the solar radiation in earth’s atmosphere. I l = I l,¥e-t l secq – Schwarzchild’s law: used for longwave radiative transfer, also consider emission of the atmosphere. s1 I l (s1 ) = I l ,oe + ò kl rrBl [T (s)] e-t l (s1,s) ds absorption 0 emission - t l (s1,0) Summary-3: • What are the most commonly used approximation? – Plane parallel approximation: When consider the diffusive factor (3D), a bean of radiation passes through a layer of atmosphere is equivalent to an beam passes through the layer at 53o without diffusive effect Fn-¯ = p In-¯(t n , m ) • Correlated k-distribution: great speed up calculation by group lines with similar absorption coefficiencts into one bin and g represent the fractional distribution of the lines with this absorption relative to total lines. 1 T= Dv ò Dv e -kvm dv = ò 1 -k (g) m 0 e dg Summary-4: • How do we separate the warming due an increase of greenhouses gases from those of solar variability, ozone depletion and volcanic aerosol? – Through finger prints of these radiative agents. • Why does greenhouse gases warm the troposphere, but cool the stratosphere? – In troposphere, greenhouse gases absorb more radiation from the surface than emit to space. Thus, they trap more radiation emitted from the surface. In stratosphere, greenhouse gases absorb radiation in the stratosphere and emit to space. Thus, they cause net loss of radiation in stratosphere.