solar Recall that the solar irradiance ie the solar con stant is S o 1366 7 Wm

# Solar recall that the solar irradiance ie the solar

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solar Recall that the solar irradiance (i.e., the solar con- stant) is S o ≈ 1366 ± 7 W·m –2 (equivalent to 1.11 K·m s –1 in kinematic form after dividing by ρ · C p ) at the top of the atmosphere. Some of this radiation is at- tenuated between the top of the atmosphere and the surface (Fig. 2.14). Also, the sine law (eq. 2.19) must be used to find the component of downwelling solar flux K that is perpendicular to the surface. The re- sult for daytime is K S T o r ↓ = − · ·sin( ) Ψ (2.34) where T r is a net sky transmissivity . A negative sign is incorporated into eq. (2.34) because K is a downward flux. Eq. (2.6) can be used to find sin( Ψ ). At night, the downwelling solar flux is zero. Net transmissivity depends on path length through the atmosphere, atmospheric absorption characteristics, and cloudiness. One approximation for the net transmissivity of solar radiation is T r H M L = + ( . . sin )( . )( . )( . ) 0 6 0 2 1 0 4 1 0 7 1 0 4 Ψ σ σ σ (2.35) where cloud-cover fractions for high, middle, and low clouds are σ H , σ M , and σ L , respectively. These cloud fractions vary between 0 and 1, and the transmissivity also varies between 0 and 1. Of the sunlight reaching the surface, a portion might be reflected: K A K ↑ = − · (2.36) where the surface albedo is A . Figure 2.14 Fate of sunlight en route to the Earth’s surface. B U N P T Q I F S F 4 , o TVOMJHIU DMPVET : QBUI &BSUI Sample Application Downwelling sunlight shines through an atmo- sphere with 0.8 net sky transmissivity, and hits a ground surface of albedo 0.5, at a time when sin( Ψ ) = 0.3. The surface emits 400 W m –2 IR upward into the atmosphere, and absorbs 350 W m –2 IR coming down from the atmosphere. Find the net radiative flux. Find the Answer Given: Tr = 0.8, S o = 1366 W m –2 , A = 0.5. I = 400 W m –2 , I = 350 W m –2 Find: F * = ? W m –2 Use eq. (2.34): K =–(1366 W m –2 )·(0.8)· (0.3) = –381 W m –2 Use eq. (2.36): K = –(0.5)·(–381 W m –2 )= 164 W m –2 Use eq. (2.33): F * =(–381)+(164)+(–350)+(400) W m –2 F * = –167 W m –2 Check : Units OK. Magnitude and sign OK. Exposition : Negative sign means net inflow to the surface, such as would cause daytime warming. Figure 2.13 Typical diurnal variation of radiative fluxes at the surface. Fluxes are positive upward. 11 17 23 29 35 -PDBM!6JNF!)I* 511 1 m511 m911 'MVY!)80N 3 * + m + o , m , o ' . R. STULL PRACTICAL METEOROLOGY 45 longwave (ir) Upward emission of IR radiation from the Earth’s surface can be found from the Stefan-Boltzmann re- lationship: I e T IR SB ↑= · · σ 4 (2.37) where e IR is the surface emissivity in the IR por- tion of the spectrum ( e IR = 0.9 to 0.99 for most sur- faces), and σ SB is the Stefan-Boltzmann constant (= 5.67x10 –8 W·m –2 ·K –4 ). However, downward IR radiation from the at- mosphere is much more difficult to calculate. As an alternative, sometimes a net longwave flux is de- fined by I I I * = ↓ + (2.38) One approximation for this flux is I b H M L * ·( . . . ) = 1 0 1 0 3 0 6 σ σ σ (2.39) where parameter b = 98.5 W·m –2 , or b = 0.08 K·m·s –1 in kinematic units.  #### You've reached the end of your free preview.

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