EART 121 PROBLEM SET #3 SOLUTIONS

1. Imperfect Greenhouse Gas Layer [20 pts] In class, we derived a model of greenhouse warming of the Earth assuming a “perfect greenhouse gas layer.” We will now instead figure out how close to perfect this layer is in order to maintain the Earth at an average surface temperature of 288 K. (a) Draw a diagram to show the different fluxes of energy (sun in, sun out…) if we assume now that the greenhouse layer has an absorptivity of some fraction f (i.e. before, our perfect greenhouse layer used f = 1, which led to an overly warm Earth). Make sure to account for Kirchhoff’s Law [emissivity at any wavelength = absorptivity at that same wavelength]! (b) There are two unknowns in this model, the temperature of the greenhouse gas layer TGHG (which isn’t 255 K anymore) and f. Solve for both of these quantities using energy balances. Sketch into the diagram of (a) where those energy balances are being performed. (c) We’ll take this a bit further: from energy balances, derive an equation that relates the surface temperature of the Earth, Te, to f (with no other variables, only constants). Now, differentiate this equation IMPLICITLY (look in your calculus book if you don’t remember how) to show that:

( )2

18

1f

ee T

df

dT

−=

(d) As we add GHGs to the atmosphere, we know that f will increase. Very approximately, over the last decade f has increased by 0.03 (corresponding to a 20 ppm increase in CO2). Using the above equation from (c), estimate the change in surface temperature due to this change in the GHG layer. SOLUTION: See attached. Please ignore that the solution says “Problem #2”.

Problem #2 EART 121: Solutions to Problem Set #3

1

Question 2: Atmospheric circulation and energy transport The tropics are net exporters of energy – they receive more from the sun than they emit as IR. Regions outside the tropics, particularly at the poles, are the opposite – they receive less from the sun than they emit. This imbalance is made up by atmospheric and oceanic transport of energy from the tropics to the poles. This imbalance is also the root of the global circulation patterns. Let’s estimate the global horizontal transport of energy. We’ll make a number of assumptions:

1. Earth’s albedo α is the same at all latitudes and is 0.30 2. The emissivity of the GHG layer f is the same at all latitudes and is the value from Question #1(b) above. 3. The sun is fixed in a position directly overhead the equator. 4. The horizontal energy transport is dominated by the atmosphere (in reality, it’s about 60% atmosphere and 40% ocean). Assumptions 1, 2 and 3 are all simplifications that aren’t true in reality, but simplify the analysis. Remember: Our eventual goal is to calculate the amount of energy that is redistributed by the

global circulation as a function of latitude θ.

The solar intensity is given by: πθθ /)cos()( 0SS = where S0 is the solar constant.

PART 1: Temperature discrepancies and the radiative energy flux deficit [20 pts] (a) Because sunlight changes with latitude, we can calculate an expression for the average

surface temperature based only on radiation as a function of latitude, Tr(θ).

(b) The actual observed average surface temperature Ta(θ) can be described by:

baTa += θθ cos)(, where a = 58.6 K and b = 241.4 K

Use a spreadsheet to generate values of Tr and Ta as a function of θ (for θ from 1 to 77° latitude in 2° increments). Plot both values. Explain in words (just a few sentences) what you see in this plot.

(c) Using the GHG model, use Ta(θ) to calculate the total outgoing IR radiation flux, FIR(θ). Assume that the energy budget of the GHG layer is at steady state. What are the units of FIR?

The net energy flux is, therefore, Fnet (θ) = (1–α)S(θ)– FIR (θ), where positive Fnet is defined as net incoming radiation). The first term is the net input of sunlight. The second term is the total IR emitted.

PART 2: Horizontal transport [20 pts]

(a) Show that the surface area contained in a narrow latitude range, dθ, can be given by:

θθπ dRdA ⋅⋅= cos2 2

where R is the radius of the Earth [hint: it’s just geometry.] We’ll approximate this equation as:

θθπ ∆⋅⋅=∆ cos2 2RA

Important: what are the units of θ in this equation?

(b) Find an expression for the net radiative power gain/loss in any given latitude band, ∆P(θ,

∆θ) in units of Watts. This is different from the net energy flux because the units are different. (c) Continue the spreadsheet you started earlier. We’ll take the middle latitude of each band as representative of the entire band (so the first band is 0 to 2°S, centered on 1°S). Here are

some important columns that you can already fill out: net solar flux; IR output flux F(θ); area

of band ∆A, net radiative power ∆P(θ). (i) Now do a calculation starting with the band closest to the equator. Calculate the net horizontal power transport into/out each latitude band for steady state to be maintained. We will assume this energy is sent to the next higher latitude band (i.e. excess energy in the 4 to 6° band is send to the 6 to 8° band). For the first band, 0 to 2°, assume a symmetric boundary so the energy flow across the equator is zero. (ii) Once you have completed this column, calculate the total horizontal energy flow leaving

each latitude βανδ Η(θ). This is the sum of the two terms: (1) the energy coming into the band from the lower band and (2) the amount of energy that band needs to maintain steady state. Plot results from (i) and (ii) versus latitude using a computer program. (d) Now, compare your plot to the one provided (from Hartmann, Physical Climatology). ** Make sure your spreadsheet has column headings that make sense, and that every column has not just a header, but also the UNITS of that column. Also make sure the number of significant figures is appropriate. **

300

280

260

240

220

Tem

pera

ture

(K)

806040200Latitude (degrees)

Tactual (Ta)

Trad (Tr)

2.6x1015

2.4

2.2

2.0

1.8

1.6

1.4

1.2

1.0

0.8

0.6

0.4

0.2

0.0

H, e

nerg

y ou

tput

flow

from

eac

h la

titud

e (W

)

80706050403020100Latitude (degrees)

200x1012

150

100

50

0

-50

-100

-150

Pnet , horizontal transport needed to m

aintain each band at steady state (W

)