Lecture%2011%20Climate%20Physics-slides

# Lecture%2011%20Climate%20Physics-slides - Today Energy...

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Today Energy Balance Model Greenhouse effect Does everybody get my weekly e-mail? Alternative: Also posted as an announcement on vista. Midterm help sessions: Friday: 3:00 – 5:30 p.m. in Henn 301 Monday 1:00 – 3:30 p.m. in Hebb 12

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Step 2: Geometry of our System Why is the incoming radiation directed? Why is the emitted radiation in all directions? Incoming solar radiation Outgoing thermal radiation due to surface temperature
Step 2 Physics Principles The temperature on the surface of the Earth is constant if P in = P out . Power in: Earth is heated by radiation from the Sun: The amount of available power per m 2 is given by the solar constant S . Power out: Earth loses heat due to its surface temperature according to Stefan’s law : 4 E E out T A e t Q P

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Shadow of Earth on an imaginary sheet is a disk with the radius of the Earth Step 2: Assumptions and Relationships Assumptions: Atmosphere transmits Sun’s radiation 100%. Earth emits and absorbs IR radiation very well (e = 1). Known facts: Earth is a rotating sphere. Intensity of incoming sunlight: S = 1367 W/m 2 30% of sunlight is reflected by atmosphere and surface (albedo), 70% is absorbed by the surface. Earth is so far away from Sun that rays are parallel: The effective area that is “seen” by the rays is a disk with the radius of the Earth
Step 3: Plan a Solution We want to solve P in = P out Need expressions for P in and P out . • P in = ? • P out = ? Our final step will be to isolate the surface temperature of Earth.

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Step 3 Question 1 We want to set up an energy balance equation for the Earth: P in = P out . We know that the solar constant S is the amount of radiation at the location of Earth. The expression for the total incoming power P in should be A. P in = S B. P in = S · A disk (= S R Earth 2 ) C. P in = S · A half sphere (= S 2 R Earth 2 ) D. P in = S · A sphere (= S 4 R Earth 2 ) E. P in = S · V sphere ( =S 4/3 R Earth 3 )
Example from Roland B. Stull, Meteorology for Scientist and Engineers Radiation Absorbed by Earth Sun is far away: Rays hitting Earth are (almost) parallel. Area that intercepts the Sun’s rays: A = R Earth 2

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Solar Radiation Annual average power per m 2 on Earth is the solar constant S = 1367 W/m 2 . Total power of sunlight on Earth: Multiply solar constant with cross sectional area A (disk). • P in = S A
Solar Radiation Annual average power per m 2 on Earth is the solar constant S = 1367 W/m 2 . Total power of sunlight on Earth: Multiply solar constant with cross sectional area A (disk). • P in = S A • P in = (1 - A ) S π R Earth 2 What’ this?

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Step 3 Question 2 We want to set up an energy balance equation for the Earth: P in = P out . The expression for the total power emitted by Earth P out should be A. P out = S π R Earth 2 T Earth 4 B. P out = e σ π R Earth 2 T Earth 4 C. P out = S · 4 R Earth 2 D. P out = e σ 4 π R Earth 2 T Earth 4 E. P out = e σ 4/3 π R Earth 3 T Earth 4
Step 4: Solve the Problem Simple model: P in = P out P in = (1- A ) S π R Earth 2 P out = 4 π R Earth 2 e σ T Earth 4 (e = 1.0)

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