We will first consider the following model for the average global temperature - we will treat the entire earth as a single point (as shown below).
BASIC
program
for an Energy Balance Model) for this project is from the excellent book:
A Climate Modeling Primer, 2nd Edition,
K. McGuffie and A. Henderson-Sellers, Wiley, (1997).
Other references include:
In an energy balance model, the main goal is to account for all heat flows in (P_{Gain}) and out (P_{Loss}) of the system. If these are balanced (P_{Gain}=P_{Loss}), the system will be in a steady state and the system will be at a constant temperature. If the heat flows are not balanced, the temperature of the system will change.
The heat flow to the earth is from the sun. The Solar Constant (S) is the amount of energy arriving (during a 1 second period on a 1 square-meter area oriented perpendicular to the sun's rays) from the sun at the upper atmosphere. The annual average value of the solar constant is S=1370 W/m^{ 2}. This arrives primarily in the form of visible light, with some smaller amounts of infrared and ultraviolet radiation.
A fraction of the sun's radiation is immediately reflected back into space, either from the atmosphere, clouds or the earth's surface. The Albedo (α) of the earth is the fraction of the sun's radiation which is reflected back into space. Thus, the net amount of solar radiation arriving on a 1 m^{ 2} area (perpendicular to sun) on the earth's surface is S(1- α).
From the point of view of the sun, the earth appears to be a disk with a radius R, so the total amount of power absorbed by the whole earth is the product of the arriving solar radiation times the area of a disk the size of the earth: P_{Gain}=πR^{2} S(1-α)
Any object at a temperature T_{K} (in Kelvin) will emit thermal radiation at a rate given by: P_{Loss}=ε σ T_{K}^{4} times it surface area. The factor ε is the emissivity (approximately 1), σ is Stefan's constant, and the total surface area of the spherical earth (4πR^{2}). Recall that a temperature in Kelvin is T_{K}=T_{0}+T where T is the temperature in centigrade and T_{0}=273.15
In the steady state, the incoming radiation must balance the outgoing
radiation. This leads to an energy balance equation for
P_{Gain}=P_{Loss}:
T | The Temperature of the Earth in Centigrade |
S | Solar Constant (1370 W/m^{2}) |
α | Albedo - Fraction of incident solar radiation reflected (about 0.32) |
σ | Stefan's Constant (5.6696E-8 W/m^{2}K^{4}) |
T_{0} | Conversion from Kelvin to Centigrade (273.15) |
Matlab
script to calculate T. To illustrate some of the GUI capabilities
of Matlab, you may use the script
ebmgui.m
(along with
ebmcalc.m
)
which allow you to easily change some of the parameters.
To run this script in matlab, type ebmgui
and enter an initial guess for the temperature in the upper
left, and make sure that the "Blackbody" button in the lower left
is on. Then click the "Calculate" button on the right.
EXERCISE 1: Using the ebmgui
program, calculate T in Centigrade for the earth.
Is this a reasonable value, or would we be mighty cold?
EXERCISE 2: Since the intensity of solar radiation is proportional
to 1/r^{2}, the solar radiation received at any other
planet in the solar system can be calculated. Since Mars
is about 1.5x further from the sun, it will receive
roughly 1/(1.5)^{2}=0.444 times the solar radiation
on Earth. If it has an albedo of 0.16, what is the black-body temperature
for Mars?
Planet | Distance (AU) | Albedo | Ave. Temperature | Atmosphere |
Venus | 0.723 | 0.76 | 425 ^{o}C | 95 Atm, 96% CO_{2} |
Earth | 1.000 | 0.32 | 15 ^{o}C | 1 Atm, N_{2}, O_{2}, Trace H_{2}O, CO_{2} |
Mars | 1.524 | 0.16 | -50 ^{o}C | 0.02 Atm, 95% CO_{2} |
Europa (A Moon of Jupiter) | 5.203 | 0.64 | -145 ^{o}C | Essentially None |
The greenhouse effect is a due to absorption and re-emission of infrared radiation by the earth's atmosphere. As discussed earlier, the earth receives visible light from the sun and emits heat in the form of infrared radiation. The earth's atmosphere is nearly transparent in the visible region of the spectrum, but gases (such as water vapor, CO_{2} and others) in the atmosphere can absorb infrared radiation. When a gas molecule (such as CO_{2}) absorbs some infrared radiation, it ends up in an excited state and eventually it decays to the ground state. In the process, it will re-emit energy in the infrared in a random direction (some of it back towards the earth). The net result is that some of the infrared radiation is "reflected" back towards the earth, thus reducing the loss of heat from the earth.
This is called the greenhouse effect because a similar mechanism occurs in a greenhouse (or a closed car in a parking lot on a sunny day). Visible light passes into the greenhouse and is absorbed by material inside. This material re-emits energy in the infrared, which is reflected back by the glass of the greenhouse (or car). Because of this trapping of heat, it is possible for a sealed greenhouse (or car) to reach a temperature far higher than the outside temperature.
The major contributor to the greenhouse effect in the atmosphere is water vapor (because it is the most abundant contributor). The next leading contributor is CO_{2}, with others including ozone, N_{2}O, methane and CFC's. Human activities are causing changes in the concentrations of these gases, which may lead to increased greenhouse effects.
Since all of our temperatures are within about +/- 20^{o}C,
it is possible to re-write the Stefan-Boltzmann equation using the
binomial expansion: (1+x)^{n} is approximately equal to
(1+nx) if x is much less than 1. Therefore, we can write
T_{K}^{4} = (T_{0}+T)^{4} =
T_{0}^{4}(1 + T/T_{0})^{4}
which is approximately
T_{0}^{4}(1 + 4T/T_{0}).
This allows us to re-write the Stefan-Boltzmann equation as
P_{Loss}=(4πR^{2})
(A + B*T)
where the two constants are A=
ε
σ
T_{0}^{4}=315 W m^{-2} for
a black body, and
B=4ε
σ
T_{0}^{3}=4.6 W m^{-2}
^{o}C^{-1} for a black body.
The greenhouse effect can be included by modifying the values
used for A and B.
In the steady state, the incoming radiation must balance the outgoing
radiation.
This leads to an energy balance equation for
P_{Gain}=P_{Loss}:
T | The Temperature of the Earth in Centigrade |
S | Solar Constant (1370 W/m^{2}) |
α | Albedo - Fraction of incident solar radiation reflected (about 0.32) |
A | Constant Coefficient for Thermal Heat Flow (A=204 W m^{-2}) |
B | Temperature Dependent Coefficient (B=2.17 W m^{-2} ^{o}C^{-1}) |
EXERCISE 3: Calculate T in Centigrade with the addition of the greenhouse effect. How much temperature change is due to the greenhouse effect? Other models use A=202 W m^{-2} and B=1.45 W m^{-2} ^{o}C^{-1} (Budyko:1969) or A=212 W m^{-2} and B=1.6 W m^{-2} ^{o}C^{-1} (Cess: 1976). How sensitive is the global temperature to these A & B parameters? Try changing the albedo to 0.31 or 0.33, or changing the multiplier for the solar constant to about 0.99 or 0.95.
The model described in Equation 2 is the steady state (equilibrium) temperature of the earth. Obviously, our experience indicates that the earth is never in equilibrium during the short term (the local weather changes on the scale of hours). However, if all of the factors in Equation 2 remained the same, the yearly average temperature would be the same.
In reality, due to various natural and man-made effects, the factors in Equation 2 can have long-term changes. A volcanic eruption or large fires can modify the the earth's albedo. Human activity has changed the amount of greenhouse gases in the atmosphere, thus changing the A & B factors. We could re-run the model to find the new equilibrium temperature after changing some or all of these factors.
A large fraction of the thermal energy storage which effects the climate is due to the upper layer of mixed water in the ocean (about the upper 70 meters). We can write the effective heat capacity of a 1 square-meter area of the earth as: C_{E} = f ρ ch, where:
f | Fraction of the Earth Covered by Water (0.7) |
ρ | Density of Sea Water (1023 kg/m^{3}) |
c | Specific Heat of Water (4186 J/kg^{o}C) |
h | Depth of Mixed Layer of Sea Water (70 m) |
Putting these together gives a heat capacity of C_{E}= 2.08 × 10^{8} J/m^{2} ^{o}C. In other words, to change the temperature of 1m^{2} of the earth by 1 ^{o}C will take on the average 2.08 × 10^{8} J of energy. This will tend to moderate the effects of "rapid" changes in any of the parameters governing our global equilibrium temperature. Thus it may take many years for the temperature to converge to a new equilibrium.
As discussed in the previous model of the
hot/cold cans, the
temperature change which results from a difference in P_{Gain}
and P_{Loss} is given by:
T | The Temperature of the Earth in Centigrade |
Δt | Time interval (as measured in Seconds) for each iteration. |
ΔT | Temperature change during the time interval Δt. |
S | Solar Constant (1370 W/m^{2}) |
α | Albedo - Fraction of incident solar radiation reflected (about 0.32) |
A | Constant Coefficient for Thermal Heat Flow (A=204 W m^{-2}) |
B | Temperature Dependent Coefficient (B=2.17 W m^{-2} ^{o}C^{-1}) |
C_{E} | Global Heat Capacity of Earth (C_{E}= 2.08 × 10^{8} J/m^{2} ^{o}C) |
You can develop a M-file to model this case (you may wish to use ploss.m as a starting point):
EXERCISE 4: Create a script file to calculate T in Centigrade as a function of time. What is the steady-state temperature. if there is a 1% or 5% decrease in the solar constant (caused by a volcanic eruption, for example), how much does the global temperature change from the steady-state of Exercise 3? Use an initial temperature from Exercise 3 and make a plot of the temperature as a function of time. How long does it take the earth to reach its new equilibrium?
α_{Ice} | Albedo of Ice (0.6) |
T_{Ice} | Global Temperature at which entire Earth freezes (-10 C) |
α_{Land} | Albedo of Land (0.32) |
T_{Land} | Global Temperature at which entire Earth melts to current state with small polar ice caps) (+10 C) |
Now we can proceed to develop a 1-Dimensional model of the earth which is broken up into 10^{o} latitude regions.
http://physics.gac.edu/~huber/envision/instruct/ebm1doc.html