Lecture 3a: Surface Energy Balance Instructor: Prof. Johnny Luo http://www.sci.ccny.cuny.edu/~luo
Surface Energy Balance 1. Factors affecting surface energy balance 2. Surface heat storage 3. Surface radiative heating and cooling 4. Atmospheric boundary layer (ABL) 5. Variations in surface energy balance 6. Surface energy flux over the oceans
Surface Energy Balance How do we define the surface of the Earth? To study surface energy balance, we have to include some 1) atmosphere, 2) ocean, and 3) land (e.g. a few m of soil) Surface energy balance determines the surface climate which obviously has practical values.
E s t = R s LE SH ΔF eo W m -2 Ø SH: conduction (sensible heat) Ø LE: evaporation (latent heat) Ø R s : net radiation Ø de s /dt: storage TOA energy balance
Outlines 1. Factors affecting surface energy balance 2. Surface heat storage 3. Surface radiative heating and cooling 4. Planetary boundary layer (PBL) 5. Variations in surface energy balance 6. Surface energy flux over the oceans
How much total heat energy is stored in land, atmosphere and ocean? E s = C e T e - C e is effective heat capacity (J m -2 K -1 ): the amount of energy required to change temperature by a unit degree; m -2 is here because we consider a unit area column, - T e is the effective temperature of whatever material (land, atmosphere, ocean) of interest. mass per unit area Atmosphere: C e (atm) = C p x M = C p x P s /g = 1004 J K -1 kg -1 100,000 Pa / (9.8 m s -2 ) = 1.02 x 10 7 J K -1 m -2 1 m 2 1 pa = 1 N m -2 = 1 kg m s -2 m -2 C p is the specific heat capacity of air: the amount of energy required to change 1 kg of air by 1 degree (J m -2 K -1 kg -1 ):.
Atmosphere: C e (atm) = C p x M = C p x P s /g = 1004 J K -1 kg -1 x 100,000 Pa / (9.8 m s -2 ) = 1.02 x 10 7 J K -1 m -2 ρ w x d w gives mass per unit area Ocean: C e (ocn) = C w x M = C w x ρ w x d w = 4218 J K -1 kg -1 x 1,000 kg m -3 x 4,000 m = 1.69 x 10 10 J K -1 m -2 The difference b/w atmosphere and ocean in surface energy budget: the whole atmosphere is involved, while only the top 70 m of the ocean is involved. Ocean: C e (ocn) = C w x M = C w x ρ w x d w = 4218 J K -1 kg -1 x 1,000 kg m -3 x 70 m = 2.96 x 10 8 J K -1 m -2 Ocean heat capacity is ~30 times that of the atmosphere.
Atmosphere: C e (atm) = C p x M = C p x P s /g = 1004 J K -1 kg -1 x 100,000 Pa / (9.8 m s -2 ) = 1.02 x 10 7 J K -1 m -2 Ocean: C e (ocn) = C w x M = C w x ρ w x d w = 4218 J K -1 kg -1 x 1,000 kg m -3 x 70 m = 2.96 x 10 8 J K -1 m -2 Soil inorganic material Specific heat C p (J kg -1 K -1 ) Density ρ (kg m -3 ) C p x ρ (J m -3 K -1 ) 733 2600 1.9 10 6 Soil organic material 1921 1300 2.5 10 6 water 4182 1000 4.2 10 6 air 1004 1.2 1.2 10 3 So, what matters is the depth of the material involved. For atmosphere, d ~ 10,000 m, for ocean, d ~ 70 m. What about land?
Heat storage of the land surface: even thinner than the ocean, only a very small depth of land participates in energy transfer. For typical soil, only the top 10 cm participates in diurnal heat transfer and the top 1.5 m in annual heat transfer. Why only a thin layer of land participates? Unlike ocean and atmosphere, land is made of solid material, which can t move up and down to transport heat (convection). It depends solely on conduction - the most inefficient way of transferring heat. Vertical heat flux (w m -2 ) by conduction: F = K T dt dz Negative sign means down gradient transport
Vertical heat flux (w m -2 ) by conduction: F = K T dt dz land If you solve this equation with a forcing with period of τ (e.g., diurnal cycle of solar radiation), you will see the heat penetration depth h T follows: Change in heat storage is determined by heat flux convergence: C s T t = z F h T = = z (K T dt dz ) = K T K T /C s τ 2 T z 2
Diurnal variation of soil temperature: what s the penetration depth for surface energy transfer?
Outlines 1. Factors affecting surface energy balance 2. Surface heat storage 3. Surface radiative heating and cooling 4. Planetary boundary layer (PBL) 5. Variations in surface energy balance 6. Surface energy flux over the oceans
R s = S S + F F solar IR Ø SH: conduction (sensible heat) Ø LE: evaporation (latent heat) Ø R s : net radiation Ø de s /dt: storage
R s = S S + F F
S S = S (1 α s ) The key parameter here is the surface albedo. Surface albedo, α s, varies wildly, depending on surface type & conditions. Take albedo of water surface as an example
0.7 µm Surface albedo, α s, is also a function of wavelength
R s = S S + F F F = (1 ε)f + εσt 4 ε is the emissivity of the surface and (1-ε) is reflectivity F F = F [(1 ε)f +εσt 4 ] = ε(f σt 4 ) F and T are measurable using ground-based instruments. 1) ε is close to 1 (surface close to blackbody) 2) Because of strong GHE, the two terms in the parentheses are close so the surface can t cool itself much through.
R s = S S + F F