Download Global Warming Fall 2013 Building Up the nth

Global Warming
Fall 2013
Building Up the nth -Layer Model
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2. In this question we are deriving the equation Tg = (n + 1) 4 TBE by solving for Tg
when n = 0, n = 1, n =2, and n = 3 and then generalizing for any n.
Zero-Layer Model (n = 0):
Energy Balance Equation: Ein = Eout
L(1−α)
= σTg4
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Whenever we see the incoming solar flux balanced by a blackbody flux, then we will designate
the temperature of that blackbody as the ’Bare Earth’ temperature (i.e. TBE ).
Therefore: Tg = TBE ;
L(1−α)
4
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= σTBE
Given a solar luminosity, L, of 1370 W m−2 and an albedo, α of 0.3, Tg = TBE ≈ 255K.
One-Layer Model (n = 1):
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Energy Balance Equations: Ein = Eout
4
Top of Atmosphere (TOA): L(1−α)
= σTa1
4
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Atm Layer: σTg4 = 2 σTa1
4
= σTg4
Ground: L(1−α)
+ σTa1
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Right off the bat we should notice that there is an extra energy flux into the ground as
compared the the Zero-Layer (i.e. ’Bare Earth’) Model. So long as Ta1 >0, Tg >TBE !
Also, we should notice that the Top of Atmosphere (TOA) energy balance equation is the
same as the Zero-Layer Model energy balance equation. Thus, Ta1 = TBE .
4
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Using the above realizations: σTg4 = 2 σTa1
−→ σTg4 = 2 σTBE
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Therefore: Tg = (2) 4 TBE .
Given the solar luminosity and albedo from the Zero-Layer Model section, Tg ≈ 303 K.
Two-Layer Model (n = 2):
Energy Balance Equations: Ein = Eout
4
Top of Atmosphere (TOA): L(1−α)
= σTa2
4
4
4
Atm2 Layer: σTa1 = 2 σTa2
4
4
Atm1 Layer: σTg4 + σTa2
= 2σTa1
4
Ground: L(1−α)
+ σTa1
= σTg4
4
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As in the One-Layer Model, the equation for the TOA is the same as the Bare-Earth Model.
Therefore, the Ta2 = TBE . It is ALWAYS the case in these simple models where the atmosphere is simulated as a blackbody IR pane of glass that the temperature of the top-most
layer (be it an atmospheric layer when n ≥ 1 or the surface layer when n = 0) will have the
’Bare Earth’ temperature (i.e. TBE ).
Therefore, we can replace any Ta2 in the above equations with TBE :
4
= σTBE
Top of Atmosphere (TOA): L(1−α)
4
4
4
= 2 σTBE
Atm2 Layer: σTa1
4
4
4
Atm1 Layer: σTg + σTBE = 2σTa1
Now we can use the Atm1 Layer equation and the Atm2 Layer equation to solve for the
ground temperature in terms of the ’Bare Earth’ temperature.
4
4
σTa1
= 2 σTBE
4
4
4
4
σTg4 + σTBE
= 2 (2 σTBE
) −→ σTg4 + σTBE
= 4 σTBE
4
4
= 4 TBE
Tg4 + TBE
4
4
Tg = 3 TBE
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Therefore: Tg = (3) 4 TBE
Given the solar luminosity and albedo from the Zero-Layer Model section, Tg ≈ 335 K.
Three-Layer Model (n = 3):
Energy Balance Equations: Ein = Eout
4
= σTa3
Top of Atmosphere (TOA): L(1−α)
4
4
4
= 2 σTa3
Atm3 Layer: σTa2
4
4
4
Atm2 Layer: σTa1 + σTa3 = 2σTa2
4
4
= 2σTa1
Atm1 Layer: σTg4 + σTa2
4
Ground: L(1−α)
+ σTa1
= σTg4
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As in the other cases, the top most layer of our model (i.e. Atm3 (a3)) has the temperature of the surface in the ’Bare Earth’ model. Therefore, Ta3 = TBE .
Substituting TBE for Ta3 :
4
Top of Atmosphere (TOA): L(1−α)
= σTBE
4
4
4
Atm3 Layer: σTa2 = 2 σTBE
4
4
4
Atm2 Layer: σTa1
+ σTBE
= 2σTa2
Substituting the relationship for Ta2 in terms of TBE in the equation for Atm3 Layer in
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for Ta2 in the equation for Atm2 layer, we find the following relationship:
4
4
4
σTa1
+ σTBE
= 2 (2 σTBE
)
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Therefore: Ta1 = (3) 4 TBE , which was the ground temperature for the Two-Layer Model.
Using the Equation for Atm1 Layer and the relationships for Ta2 and Ta1 in terms of TBE ,
we find the following relationship for Tg :
4
4
)
= 2 (3 σTBE
σTg4 + 2 σTBE
4
4
4
Tg + 2 TBE = 6 TBE
1
Therefore, Tg = (4) 4 TBE
Given the solar luminosity and albedo from the Zero-Layer Model section, Tg ≈ 360 K.
nth -Layer Model:
We could continue drawing pictures, writing out equations, and figuring out the ground
temperature for any number of panes of glass. However, we should recognize a pattern. Let’s
look again at the results from above:
4
# of layers
0
1
2
3
Tg
(1) TBE
1
(2) 4 TBE
1
(3) 4 TBE
1
(4) 4 TBE
1
4
1
Note: (1) 4 = 1.
When the number of layers is 0, the number raised to the 14 power is 1. When the number of layers is 1, the number raised to the 14 power is 2, etc. In general, the power raised to
the 41 power is one larger than the number of layers in the model!
Therefore, a generalized equation for the ground temperature given n number of atmospheric
layers is:
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Tg = (n + 1) 4 TBE
where TBE is the ’Bare Earth’ temperature. Given a solar luminosity, L, of 1370 W m−2 and
an albedo, α, of 0.3, TBE ≈ 255 K.
2. In this question we are exploring the governing energy balance equations and the equatorial surface temperature for an atmospheric-less planet without heat transport.
First, let’s identify what we know from the prompt.
1) L = 1370 W m−2 . From the prompt we know that the star is Sun-like and that the
distance of the planet from the star is roughly the same as the distance between the Earth
and the Sun (i.e. 1 A.U.). The dominant things that affect L are the star’s luminosity, which
is dependent on the temperature of the star, the radius of the star, and the distance between
the star and the planet. Since it is a Sun-like star and the distance is about that of the
Earth’s, it is safe to assume that L is the same for this planet as for the Earth.
2) α = 0.3
3) The planet does not have heat transport
The first two points are important on being able to get an answer, but the crux of this
problem is the third point.
In all of the layer models to date, the area receiving incoming stellar radiation was different than the area reradiating IR away from the planet. This was because we assumed
that the ENTIRE planet was radiating at the same amount, and thus the ENTIRE surface
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of the planet was at the same temperature. This is of course a false assumption–the tropics
are warmer than the middle of Antarctica. However, this assumption actually is pretty close
to true; the equator-to-pole temperature difference is about 30 - 50 ◦ C, which we will see in
this problem is really not that great. Also, this assumption allows us to build a simple model
that can actually provide robust information about the climate system.
In making this assumption regarding the homogeneity of the planetary temperature we are
implicitly assuming that our planet is REALY REALY REALY effective at distributing heat
around the globe. In this problem, we are assuming that our planet does not transport any
heat. Therefore, if a 1 m2 patch of surface absorbs some incoming stellar flux, then that
same square meter surface must reradiate that flux away. Thus the surface energy balance
equation will be the following:
F luxin Ain = F luxout Aout & Ain = Aout
Therefore: F luxin = F luxout
For an atmospheric-less planet, the F luxout is just σTg4 .
The DOMINANT incoming energy for the Earth is solar radiation. Similarly, the incoming energy for this hypothetical planet will just be the incoming stellar radiation, I.
I(1 − α) = σTg4
Thus, for a given incoming radiation flux, I, and albedo, α, we can determine the temperature of a given square meter patch of ground.
If we assume that this this planet has zero obliquity, then the incoming radiation will be
at a maximum at the equator at high noon and sinusoidally tend towards zero at the poles.
That maximum stellar flux will be the solar constant at the given orbital distance, which we
said we can assume to be 1366 W m−2 . Also, we can assume the albedo of the surface to be 0.3.
Therefore, at high noon on the equator, the surface temperature, Tg would be ≈ 360 K.
That is HOT! Water boils at 373 K.
At night, there is no incoming stellar flux. Therefore the value of L at night would be 0
W m−2 . Thus, the surface temperature at night, Tg , would be 0 K.
This situation is very similar to the moon. However, most materials have some sort of heat
storage capacity–think about a blacktop after the sun goes down; it is usually still warmer
than the air since it absorbs and stores some of the sunlight during the day and slowly cools
off towards the air temperature over the course of the evening/night. The takeaway: the
diurnal cycle on a basically airless body is DRASTIC!
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