Basic Radiative models/Earth’s climate system analysis Pt. 1

I thought I would work backwards a bit and go over some of the basics of radiative transfer/balance in the climate system, and how daily weather systems work. The post is rather broad, and (hopefully) easy-to-read version of what factors determine the Earth’s climate.

The ultimate driver of climate and weather systems on Earth is the sun. Over 99.9% of the energy the Earth receives is from the sun, with negligible fractions coming from underground geothermal sources or tidal energy from gravitational influences by the sun and moon. For a given area facing the sun at the top-of-atmosphere the amount of energy intercepted each second of a surface area of one square metre is ~1,370 watts (S). This is at a distance of 1 astronomical unit (the average distance from earth to sun) and may vary slightly over cycles of changes in the eccentricity of Earth’s orbit over 100,000 year cycles.

The energy above (S = solar constant) averaged over the entire planet is

S (pi R^2)/ (4 pi R^2)

cross out some terms, and this = S/4

where pi R^2 is the Earth’s cross-sectional area with radius R, and “4” comes from the energy being distributed over the globe. So, the net incoming solar radiation is about 342 W/m-2, defined as the solar radiation coming in at the top-of-atmosphere (TOA). However, the global and annual *net* absorbed shortwave irradiance is less, because some of that is reflected back to space:

A= bC

where C = the 342 W/m-2, and A = the actual ~237 W/m-2. This corresponds to b = ~0.69 (the coalbedo) and albedo (the fraction of reflected solar radiation on a scale from 0 to 1) is 0.31. Albedo is calculated by the reflected radiation/incident radiation. Anyone knows wearing a black shirt on a summer day will make you hotter than a white shirt, because a black shirt has a low albedo close to zero, and so absorbs most solar radiation, whereas a white shirt has a high albedo, near one, and thus reflects quite a bit of incident radiation. The Earth, on average, reflects about 31% of the solar radiation coming in although this varies significantly by surface (ice sheets, sand, land, ocean, etc all vary significantly in their albedo). In other words, about 107 W/m-2 of the initial C goes right back to space, so the net solar radiation coming in is about 237 W/-2 (Note* the math will not be perfect here because of different sources which round off or say “about” so expect discrepancies of one or two W/m-2 if you do the math yourself).

To balance the incoming energy the Earth must radiate the same amount. Just think if the Earth continues to absorb radiation without emitting radiation, it would continue to heat up without bound and would have boiled over). Fortunately, the Earth emits radiation back to space at E= εσT^4 (Stefan-Boltzmann law). That is, the total energy flux emitted by an object across all wavelengths is proportional to the fourth power of the temperature. σ represents Stefan-Boltzmann constant, or about 5.67 x 10^-8 (in units J s^1 m^2 K^4). Temperature must be in Kelvin (welcome to thermodynamics), which is the same as Celsius only offset so that absolute zero (which you can see would be the only temperature at which an object no longer emits radiation) is at 0 K. We’ll return to this in a second.

From Wein’s displacement law, we can calculate the wavelength of most intense radiation of an object with a certain temperature. To calculate this, use

λmax= 2897/ T(K) where λmax is the wavelength of energy of which an object prefers to emit most of its energy, 2897 is a constant when λmax is in µm (micrometers) and T(K) is the temperature of the object, in Kelvin. This is illustrated in Figure 1

fig10.gif

The figure illustrates the wavelengths of peak (or preferential emittance) as a function of temperature, and the two examples are the Sun at nearly 6000 K, and the Earth (which is actually 288 K as I’ll discuss below). The sun’s preferential emissions is at 0.5 µm, which is in the visible light spectrum, while the Earth emits at longer wavelengths (wavelength of peak emittance is inversely proportional to temperature), notably in the infrared part of the electromagnetic spectrum. (Apply Stefan-Boltzmann and you’ll see the vast difference in energy output by the two bodies!).

Back to the Earth emitting radiation, as I promised we’d return to, we now know that the Earth emits primarily in the infrared part of the light spectrum. Solar radiation comes in, and our atmosphere is rather transparent to it, meaning it comes in nearly freely. Of that which is not reflected back to space (from clouds, aerosols, gases, etc), only about 67 W/m-2 is absorbed by the atmosphere. About 168 W/m-2 is absorbed by our surface.

To emit the ~240 W/m-2 necessary to maintain radiative equilibrium, the temperature would have to be a chilly 254 K (-19 C, or less than -2 degrees F). This is obviously not done at the surface, which is much warmer, but rather up in the atmosphere approximately 5 km above the surface (temperature decrease with altitude following the appropriate adiabatic lapse rate). Lower pressures correspond to higher altitude. As warm air rises, it expands and cools (by adiabatic expansion) and some of this cooling is offset by the heat released by condensation of water vapor, so temperatures go down at about 6.5 degrees C/km of altitude. Temperatures decreasing at this rate with height will be said to follow the moist adiabat.

We can now calculate the temperature of the Earth at equilibrium where the solar irradiance coming in, 240 W/m-2 , is roughly equal to the outgoing longwave radiation (OLR) at the TOA.  Ignoring heat from radioactive decay (which is a good approximation on the terrestrial planets with a well defined surface) and no atmosphere (discussed in Part 2), the planet’s temperature will be set by the radiative balance at the top of the atmosphere and will not be able to go above the Stefan-Boltzmann values accounting for  incoming and relevant emissivity and reflectivity parameters.  The expected temperature for the Earth is 255 K (or about -18 degrees C or approximately -0.4 degrees F). Seems like my calculations produced a planet a bit too cold for life!

To be continued…

Parts on the Greenhouse Effect (a bit more radiative transfer, TOA energy balance, surface energy balance, and greenhouse gas effect) and then part 3 on atmospheric and oceanic circulation.

3 responses to “Basic Radiative models/Earth’s climate system analysis Pt. 1

  1. Pingback: Basic Radiative models/Earth’s climate system analysis Pt. 2 « Climate Change

  2. Pingback: Basic Radiative models/Earth’s climate system analysis Pt. 3 « Climate Change

  3. Pingback: blog

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s