Jul 2, 2010

Greenhouse effect

A rough numerical estimation

If we consider the Total Solar Irradiance (radiation intensity) reaching Earth as an averaged 1365.5 [Watts/m2] (known as the Solar constant), according to the Stefan-Boltzmann law (that is infered from Planck's law of spectral radiance), a "black-bodied" (absorbs all incoming radiation) Sun's emittance (energy flowing per unit surface area and time) depends on its temperature at a epsilon·rho·T^4 rate, where epsilon=1 for black bodies, and rho states for the Stefan constant of proportionality, whose value is 5.67·10^-8 [W/m2·K4].

Thus, the total energy radiated (radiation power) by the Sun is Wt = Ws · (4·pi·radius^2), being Ws the Stefan-Boltzmann law applied on the surface of Sun. As an isotropic radiation (does not depend on direction), at Earth's distance Wt = We · (4·pi·distance^2), being We the solar irradiance measured on Earth (decreases with distance).

Combining both, the Sun's temperature is derived from Ws = rho·Ts^4 = We · (distance/radius)^2, then Ts = (We/rho)^0.25 · (distance/radius)^0.5 = (1365.5 [W/m2] /5.67·10^-8 [W/m2·K4])^0.25 · (149.5978707M [Km] / 696K [Km])^0.5 = 5775.44 [Kelvin] (actually 5780 K, or 5507º Centigrades), where the Sun-Earth distance is approx. 150 million Kms (that is 1 AU, astronomical unit), and Sun's radius is around 700 thousand Kms.

Wien's displacement law denotes that temperature as a function of the shorter wavelength (lambda) at which radiation is mostly emitted: lambda · T = 0.2898 [cm·K]. The Sun radiates its peak at lambda close to 500 nm (10E-9 m), that corresponds to the yellow. Graphically:

In order to deduce Earth's temperature, we must take into account its albedo, or rate of reflected radiation (what makes Earth visible from space), due to atmosphere (mostly clouds 23%, and air 4%), and Earth's surface (snow, sea, 4%), that may average to 30% as a whole.

Furthermore, Earth's surface average insolation (incoming solar radiation) is approximately Ws/4, averaged. This can be explained by the projection of Earth's "spheric" surface (4·pi·radius^2) as a circle of area pi·radius^2. Because any unit of area, due to Earth's rotation, is not always being insolated. At any time, half the Earth is on the dark side. And the rest depends on the direction of the incident radiation: that is, when they are perpendicular, you get the whole radiation; in any angle, you just get the cosine component (Lambertian cosine law).

The solid angle is then integrated as a function of sin(theta)·d(theta) from 0 to pi/2 (half insolated sphere); do not forget to multiply it by cos(theta), and by d(phi) from 0 to 2·pi (the whole circumference). The latter is just 2·pi; and the former (considering u=sin(theta) and du=cos(theta)·d(theta)) sums 1/2; thus totals 2·pi/2 = pi. So finally get the same 1/4 ratio, as the whole sphere subtends a 4·pi angle.

Then, Earth's temperature might be deducted from the former black body relationship Te = ((1-albedo)·(Ws/4) / rho)^0.25 = ((1-30%)·(1365.5/4) [W/m2] / 5.67·10^-8 [W/m2·K4])^0.25 = 254.79 [Kelvin] = -18.21º [C]. But that's obviously untrue, because of Greenhouse effect.

Incoming solar radiation energy is 95% in the 300 to 3000 nm wavelength range: 10% in the UV (ultraviolet, 300-400 nm); 40% in the visible (400-700 nm), and 50% in the infrared (700-3000 nm). Along with the albedo, not all incoming radiation is transmitted through atmosphere to the ground because of:

- Clouds (3%) and atmosphere (21%) partially absorb it, then converted into heat and causing the emission of their own radiation in all directions. Basically shortwaves are by H2O vapour, CO2 carbon dioxide, methane CH4, nitrous oxide N2O; and ultraviolets mostly by ozone O3, and nitrogen N2 and oxygen O2; also aerosols CFCs contribute. These are the main Greenhouse gases.

- Scattering takes place when small particles and gas molecules diffuse part of the incoming solar radiation in random directions, even back to space; this causes the sky bluish color (Rayleigh's dependance on wavelength^-4) and white clouds (Mie's on wavelength^-1).

Earth emittance is 90% in the range from 3 to 30 microm, with its peak around 10 microm. It corresponds to a black body temperature around 300 K, close to the 288 K average surface temperature.

One step further, if Earth emits energy at a T'e=288 K temperature, then yields an emissivity epsilon = (1-albedo)·(Ws/4) / rho·T'e^4 = (1-30%)·(1365.5/4) [W/m2] / 5.67·10^-8 [W/m2·K4] · 288^4 [K4] = 0.61.

Over half of Earth’s emittance lies in the infrared range from 8 to 14 microm. Without this window, the Earth would become too warm to support life.

Let's build a very simplified model to outline this temperature increase.

If we consider Earth as a black body (thus ideal emissivity = 1), and we close that infrared window, then when thermodynamical equilibrium is again reached, power emitted per unit surface outside is all but the selected window: Pem = Integral[ Radiance(frequency, Temperature) dfreq ] avoiding the infrared range.

The power absorbed per unit surface Pab will be the sum of the known total black body radiance rho·Te^4, plus the Integral[ Radiance(frequency, Temperature) dfreq ] over the window range.

Given that in thermal equilibrium net energy flux must be null, Pem = Pab :

Int[ Rad(freq,Te') dfreq ]{freq2..infinite} + Int[ Rad(freq,Te') dfreq ]{0..freq1} = rho·Te^4 + Int[ Rad(freq,Te') dfreq ]{freq1..freq2}

where Te' is the new equilibrium temperature. We can express this in terms of total radiance :

Int[ Rad(freq,Te') dfreq ]{0..infinite} - Int[ Rad(freq,Te') dfreq ]{freq1..freq2} = rho·Te'^4 - Int[ Rad(freq,Te') dfreq ]{freq1..freq2} = rho·Te^4 + Int[ Rad(freq,Te') dfreq ]{freq1..freq2}

So :

rho·Te'^4 - rho·Te^4 = 2 · Int[ Rad(freq,Te') dfreq ]{freq1..freq2}

Solving the integral :

Int[ Rad(freq,Te') dfreq ]{freq1..freq2} = Int[ (c/4) · (8·pi·freq^2·Kb·Te' / c^3) dfreq ]{freq1..freq2}

here Kb refers to the Boltzmann constant 1.38·10^-23 [J/K].

We are using the Rayleigh-Jeans expression instead of Planck's to afford easier calculations. But take into account that this approximation is only valid for long wavelengths, due to this curve tends to infinite as wavelengths become shorter. Planck's, instead, falls exponentially at short wavelengths.

The c/4 factor is because Planck's (and Rayleigh-Jeans) law stands for density (unit volume) of radiant energy, but what is actually measured is radiant emittance or spectral radiance (per unit surface), which depends as well on speed of outgoing radiation.

Finally :

Int[ (c/4) · (8·pi·freq^2·Kb·Te' / c^3) dfreq ]{freq1..freq2} = 2·pi·(freq2-freq1)^3·Kb·Te' / 3·c^2

Thus :

rho·Te'^4 - rho·Te^4 = 2 · [ 2·pi·(freq2-freq1)^3·Kb·Te' / 3·c^2 ] = 4·pi·(freq2-freq1)^3·Kb·Te' / 3·c^2

Rearranging the latter expression after dividing it by rho·Te^4 :

(Te'/Te)^4 = 1 + 4·pi·(freq2-freq1)^3·Kb·Te' / 3·c^2·rho·Te^4 = 1 + ( (4·pi·(freq2-freq1)^3·Kb / 3·c^2·rho·Te^3) · (Te'/Te) ) = 1 + Constant · (Te'/Te)

because Constant is adimensional, and its value is :

Constant = 4·pi·(freq2-freq1)^3·Kb / 3·c^2·rho·Te^3 = 4·pi·((c/8 [microm])-(c/14 [microm]))^Kb·Te / 3·c^2·((1-albedo)·(Ws/4)) = 4·pi·(3.75·10^13-2.14·10^13)^3·1.38·10^-23 [J/K]·254.79 [K] / 3·(299792458 [m/s])^2·238.9625 [W/m2] = 2.841

Constant's value is "almost" small, so we "can" approximate :

(Te'/Te) = (1 + Constant·(Te'/Te))^(1/4) ~ 1 + (1/4)·Constant·(Te'/Te)

Then :

(1 - (1/4)·Constant) · (Te'/Te) ~ 1 ; Te' ~ (1 - (1/4)·Constant)^-1 · Te = (1 / (1 - (1/4)·Constant)) · Te

Here we will make a new approximation, due to the mentioned trend to infinite for short wavelengths. A geometric series can be  expressed as Sum[ c·r^i ]{i=0..infinite} = c / (1-r), being r the ratio. In our case c = 1 and r = (1/4)·Constant. We take just the first terms of the series and discard from the quadratic term onwards, and :

Te' ~ (1 + (1/4)·Constant) · Te = 435.75 [Kelvin]

That way, Earth's temperature would approximately be increased up to 162.75º Centigrades.

1 comment:

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