524B-Assignment1 Solutions
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CE/ABE/ENSCI 524B – Assignment #1 Solutions Global Atmospheric Change 8.1 Suppose the ratio of (18π ⁄16π ) in standard ocean water is 0.00200. If another sample of ocean water has an (18π ⁄16π ) ratio of 0.00199, what would be the value of πΏ18π ? Would the sample correspond to a warmer or colder climate? The principle used is that less ice on Earth leads to more dilution with lighter water and therefore a warmer climate. (18π ⁄16π )π πππππ 0.00199 οΌ πΏ18π = [(18 π ⁄16π )π π‘ππππππ − 1] × 103 = [0.00200 − 1] × 103 = −5. οΌ The sample with a ratio of 0.00199 , would correspond to a warmer climate. A lower (18π ⁄16π ) ratio means a lower amount of 18π . A lower amount of 18π in ocean water means warmer temperatures. 8.2 A relationship between the mean annual surface temperature of Greenland and the value of πΏ18π of the snow pack on the Greenland ice sheet is given by π(β) = 1.5πΏ18π (0⁄00) + 20.4 An ice core sample dating back to the last glaciations has a value of πΏ18π equal to −35. What would the estimated surface temperature have been at that time? οΌ π(β) = 1.5πΏ18π (0⁄00) + 20.4 = 1.5(−35) + 20.4 = −32.1β. 8.5 The solar flux π arriving at the outer edge of the atmosphere varies by ±3.3 πππππππ‘ as the earth moves in its orbit (reaching its greatest value in early January). By how many degrees would the effective temperature of the earth vary as a result? οΌ The solar flux π, arriving at the outer edge of the atmosphere, is represented by the solar π constant of 1,370 π2 . For this problem we are assuming that the solar flux varies by π ±3.3 πππππππ‘ as the earth moves in its orbit. So, π = 1,370 ± 45.21 π2 . οΌ We also know that the energy absorbed by Earth is πππ 2 (1 − πΌ) and the energy radiated back to space by earth is π4ππ 2 ππ 4 , assuming that the earth is a blackbody and behaves according to the Stefan-Boltzmann Law. οΌ As we know that the energy absorbed by Earth equals the energy radiated back to space, we can set the two equations equal and solve for the effective blackbody temperature. 1 πππ 2 (1 2 4 − πΌ) = π4ππ ππ . This leads us to ππ = π(1−πΌ) 4 [ 4π ] . π οΌ The average albedo for Earth constant, πΌ = 0.31. We also know that π = 5.67 × 10−8 π2 πΎ4 , where π is the Stefan-Boltzmann Constant. We also know, from previously, that π οΌ π = 1,370 ± 45.21 π2 . Using these quantities and solving for ππ , we get ππ = 1 π 1,370±45.21 2 (1−0.31) 4 π [ π 4(5.67×10−8 2 4 ) π πΎ ] = 254 ± 2 πΎ. The earth would vary from 252 πΎ to 256 πΎ Any atmospheric heat retention has been ignored, but the 4K variation is correct. 8.8 π In Figure 8.13, the average rate at which energy is used to evaporate water is given as 78 π2 . Using 2465 ππ½/ππ as the latent heat of vaporization of water, along with the surface area of the earth, which is about 5.1 × 1014 π2 , estimate the total world annual precipitation in π3 /π¦π (which is equal to the total water evaporated). Averaged over the globe, what is the average annual precipitation in meters of water? o Assuming average conditions everywhere and also that there is enough water for evaporation everywhere, we can estimate what the total world annual precipitation in π3 /π¦ would be. This is not realistic, of course. π o Using the average rate at which energy is used to evaporate water, 78 π2 , and the surface area of the earth, we can solve for the total power used to evaporate water over the surface of the earth. 78 π π2 5.1 × 1014 π2 = 3.978 × 1016 π = 3.978×1016 π½ π = 3.978×1013 ππ½ π . o Using the total power used to evaporate water over the surface of the earth, we can solve for the total energy used to evaporate water over the surface of the earth in the course of a year. 3.978×1013 ππ½ 3600 π 24 β 365 π π β π π¦ = 1.25×1021 ππ½ π¦ . o Using the latent heat of vaporization of water, we can solve for the total mass water evaporated from the total energy used to evaporate water per year. 5.08×1017 ππ π¦ 1.25×1021 ππ½ ππ π¦π 2465 ππ½ = . o Using the density of water, we can solve for the total volume of water evaporated from the total mass of water evaporated = 5.09×1017 ππ π¦ π3 = 1000 ππ 5.09×1014 π3 π¦ , a rough estimate of the total volume of precipitation per year, worldwide is therefore 5.09 × 1014 π3 . o By dividing by the surface area of the earth, we can find the average annual precipitation in meters of water. 5.09×1014 π3 5.1×1014 π2 = 0.998 π, say 1m. A better approach would have been to accept water for evaporation only available on 70% of the Earth’s surface, but that precipitation is spread over all of Earth. That would result in an average rainfall of 0.7m per year on average, which is close to reality. 8.14 The radiative forcing as a function of concentration for π2 π is modeled as a square root dependence: ΔπΉ = π2 (√πΆ − √πΆ0 ) Assuming it has been in that region since preindustrial times when the concentration was 275 πππ, find an appropriate π2 if the current concentration is 311 πππ and the forcing is π estimated to be 0.14 π2 . Estimate the added radiative forcing in 2100 if it reaches a concentration of 417 πππ. (a) Find an appropriate π2 if the current concentration is 311 πππ οΌ Using the preindustrial concentration for nitrous oxide, listed on page 481, we know that π2 = ( ΔπΉ √πΆ−√πΆ0 =( ) 0.14 π π−2 √311−√275) = 0.133. (b) Estimate the added radiative forcing in 2100 if it reaches a concentration of 417 πππ οΌ Using the π2 that we calculated for nitrous oxide, we can solve for the added radiative forcing, if the concentration reaches 417 πππ. The added forcing that results from an increase in concentration from 311 πππ in 2010 to 417 πππ in 2100 is ΔπΉ = π2 (√πΆ − π √πΆ0 ) = 0.133(√417 − √311) = 0.371 π2 . 8.20 Compute the global warming potential for a greenhouse gas having atmospheric lifetime π = 42 π¦ππππ and relative forcing per unit mass that is 1630 times that of πΆπ2 over the following time periods: οΌ To calculate the Global Warming Potential (GWP) of a gas we need to know the radiative forcing of the gas, πΉπ , the exponential decay of that gas, π π (π‘), and the time period for cumulative effects, π, and use them in the equation π πΊπππ = ∫0 πΉπ β π π (π‘) ππ‘ π ∫0 πΉπΆπ2 β π πΆπ2 (π‘) ππ‘ . οΌ In this problem, we know the relative forcing as compared to carbon and we know the atmospheric lifetime, which gives us the exponential decay function of the gas π π (π‘) = π‘ π 0 π −42 . Assuming an average lifetime of 100 years for CO2, and a decay function of π‘ π πΆπ2 (π‘) = π 0 π −100 The GWP for three different time periods are as follows. (a) 20 years π a. πΊπππ = b. πΊπππ = π ∫0 πΉπΆπ2 βπ πΆπ2 (π‘) ππ‘ 1630(−26.09+42) (−81.87+100) = 20 − 1630πΉπΆπ2 ∫0 π 0 π 42 ππ‘ π‘ 20 − πΉπΆπ2 ∫0 π 0 π 100 0 − − 1630[(−42π 42 )−(−42π 42 )] π‘ 20 ∫0 πΉπ βπ π (π‘) ππ‘ = ππ‘ 20 0 − − [(−100π 100 )−(−100π 100 )] = 1630(−26.09+42) . (−81.87+100) = 1431 (b) 100 years a. πΊπππ = b. πΊπππ = π ∫0 πΉπ βπ π (π‘) ππ‘ π ∫0 πΉπΆπ2 βπ πΆπ2 (π‘) ππ‘ 1630(−3.88+42) (−36.7+100) = π‘ 100 − 1630πΉπΆπ2 ∫0 π 0 π 42 ππ‘ π‘ 100 − πΉπΆπ2 ∫0 π 0 π 100 ππ‘ 100 0 − − 1630[(−42π 42 )−(−42π 42 )] = = 100 0 − − [(−100π 100 )−(−100π 100 )] = 983 (c) 500 years π a. πΊπππ = b. πΊπππ = ∫0 πΉπ βπ π (π‘) ππ‘ π ∫0 πΉπΆπ2 βπ πΆπ2 (π‘) ππ‘ 500 = 1630(−0.0002838+42) (−0.6738+100) 1630πΉπΆπ2 ∫0 500 πΉπΆπ2 ∫0 = 689 500 π‘ − π 0 π 100 ππ‘ 0 − − 1630[(−42π 42 )−(−42π 42 )] π‘ − π 0 π 42 ππ‘ = 500 0 − − [(−100π 100 )−(−100π 100 )] . 1630(−3.88+42) . (−36.7+100)
