The Layer Model Approximation to the Greenhouse Effect
Designed by Prof. David Archer, University of Chicago
This lab exercises your understanding of "layer models" of the
greenhouse effect. These are also called "isothermal slab models," or
"glass atmosphere models." They have serious shortcomings in their
neglect of the way thermodynamics and convection alters the vertical
temperature structure of real atmospheres, but they are still useful
in understanding the basic way the greenhouse effect operates.
Consider a spherical planet with no atmosphere. Energy pours
in as visible light from the sun, and energy is lost by emission of
infrared light from the surface of the earth.
The amount of energy coming in to the planet is given by
Flux In = (1) r^{2} L
where is the albedo, defined
as the proportion of the incident light which gets reflected back to
space. Snow has a high albedo, and black velvet paintings of Elvis
have generally a low albedo (unless they have sparkles on them). L is
the intensity of sunshine, which at the mean radius of the earth's
orbit equals 1380 W/m^{2}. The factor r^{2} reflects the fact that the
sphere intercepts a "disk" of radiation of radius r.
For outgoing energy, the process is blackbody radiation. We know from
the StephanBoltzmann Radiation law that for a blackbody the rate at
which energy is lost by outgoing infrared light can be calculated as a
simple function of temperature, as
Flux Out = Area_{sphere} T_{e}^{4}
where is the StefanBoltzmann
constant, and T_{e} is temperature of the surface of the
earth. The area of a sphere is 4
r^{2}. Of course we're assuming here that the surface of the
planet is all of uniform temperature, but this kind of cavalier
cowboystyle assumption making is what thought experiments are for.
If we set the incoming and outgoing energy fluxes equal to each other,
we end up with
(1)
r^{2} L = 4 r^{2}
T_{e}^{4}
or, rearranging,
L (1) / 4 = T_{e}^{4}
which gives us a relation between the solar constant L, the albedo of
the planet , and the temperature at
the surface of the earth T_{e}.
Now let's clothe our planet with an atmosphere. We'll do it in a
simplified "idealized" way to start with, using a thought experiment
known as the "layer model". The idea is to wrap the earth in a layer
which has the properties of optical transparency in the visible and
complete absorbtion in the infrared.
This complicates things a bit, as we have to calculate the energy
balance of both the earth and its atmospheric wrapper. The picture
looks like this:
The energy coming in from the sun is the same as before, and it makes
it through the atmospheric "layer" unchanged. Infrared light is
emitted from the ground (IR_{e}), using the StefanBoltzmann
law as before. This energy (IR_{e}) is completely absorbed by
the atmospheric layer, which itself radiates energy, both upward
(IR_{a,up}) and downward (IR_{a,down}). (In case you
are wondering: There are just two directions, rather than a whole
sphere's worth of directions, because the world extends infinitely far
in both directions, and is completely uniform. So some light shines
off at angles other than plain up and down, but what goes out of the
picture by flowing sideways is balanced by what comes into the picture
from nextdoor. If you weren't wondering about the two directions,
don't worry about it.)
The crux of the whole thing, the whole idea of the greenhouse
effect, is based on the assumption that some of the IR from the
atmospheric layer shines downward. OK here's how it works. We
know that the energy going into the layer has to balance the energy
leaving the layer.
We know that half of the energy leaving the layer makes it to space.
Therefore, the temperature of the layer is determined by the energy
balance as we used it before
Solar in = IR_{a,up}
which implies
L (1) / 4 = T_{a}^{4}
But the energy flowing upward from the atmosphere (IR_{a,up})
is only half of the energy going in to the atmospheric layer, which
comes from the surface of the earth (IR_{e}). That is to
say,
IR_{e} = 2 IR_{a,up}
which means that
T_{e}^{4} = 2 T_{a}^{4}
or
T_{e} = 2^{1/4} T_{a}
Now, notice that the temperature of the top layer in this calculation
(T_{a}) is the same as the temperature of the bare earth in
the Case I calculation (T_{e}). We can therefore see that the
planetary surface is warmed up by the greenhouse atmosphere than the
noatmosphere case by a factor of the fourth root of 2, Or about 20%.
This is the greenhouse effect.
 The basic calculation
The basic greenhouse effect calculation for a 1layer blackbody
atmosphere model is so fundamental that you should practically be able
to do it in your sleep. So, to start out this lab, make sure that you
can begin with a blank sheet of paper, and show (without referring to
the notes) that a simple 1layer blackbody atmosphere that is
transparent to solar radiation increases the surface temperature by a
factor of 2^{1/4} above what the temperature would be on a
noatmosphere planet. Try a minor modification of the derivation by
assuming that the surface has nonzero albedo , reflecting a proportion of solar radiation back upwards. Would
your answer change if it were the atmosphere rather than the surface
that were doing the reflecting?
 Variant 1: 1 layer model with solar absorption in the atmosphere
Modify the 1layer model by assuming that the atmosphere isn't
transparent, but instead absorbs a fraction of the incoming solar radiation. You may
still assume that the surface albedo is zero.
The energy budget of the atmospheric layer is now:
L / 4 +
T_{e}^{4} =
2T_{a}^{4}
What is the corresponding energy budget of the surface? Solve for the
temperatures and discuss how they depend on . How do your results
change if the surface has a nonzero albedo instead of being
completely absorbing?
 Variant 2: A 2layer model
In this exercise, you can go back to assuming that the atmosphere is
transparent in the visible spectrum. Now, instead of representing the
atmosphere as a single blackbody layer, we represent it as two layers
which have different temperatures. They are coupled to each other and
to the ground only by radiation (i.e. no heat transfer by
convection).
Write down an energy budget for each layer and for the ground, and
determine all the temperatures as a function of L. How does the
ground temperature compare with that of the 1layer case?
Example: The energy budget for the middle layer is: T_{e}^{4} + T_{2}^{4 } = 2 T_{1}^{4}
Explain this equation, and write a similar budget for the ground and
for the top layer.
If you know that the absorbed solar radiation is (1) L, can you determine the temperature of the
top layer without knowing what the other layers are doing? Explain
why or why not. You may assume that the system is in equilibrium.
How do your results change if you assume that convection couples the
two atmospheric layers so strongly that their temperatures are always
identical to each other?
 Variant 3: Nuclear Winter
Suppose that some kind of catastrophe injected a layer of black soot into the
upper atmosphere. The soot layer is perfectly absorbing both in the IR
and the solar parts of the spectrum. The soot layer is so high up that it is
essentially thermally isolated from the ground (i.e. there is no energy
exchange between it and the ground except by radiation). Approximate the
atmosphere as an IRabsorbing, solartransparent blackbody. What is the
effect of the soot layer on the surface temperature? On the atmospheric
temperature?
