PHYS/ENVS 3070 Homework Assignment #9: Problem #1. In the year 2020, it is conceivable that 20 GigaWatts of electricity could be continuously generated in a single western state by coal burning power plants. (5 points total) Part I: At 38% efficiency for generating electric power, how many tons of coal would be burned per year? If we want 20 GigaWatts = 20 x 109 Watts or Joules/second for one year (or 3 x 107 seconds), we need (20 x 109 Joules/second) x (3 x 107 seconds) = 6 x 1017 Joules of electrical energy. If our generating power plant is only 38% efficiency, then we need much more input fuel energy of 6 x 1017 / 0.38 = 1.6 x 1018 Joules. One ton of coal has an energy of 2.81 x 1010 Joules (according to the textbook front cover inset). Therefore we can calculate the number of tons of coal needed. tonstonJJTonCoalEnergyedEnergyNeedTonsCoal71018106.5/1081.2106.1/×=××== Thus, we need 56 million tons of coal each year for this power plant. Part II: If the sulfur content of the coal averages 1%, how many tons of SO2 would be released per year? If sulfur is 1%, then 5.6 x 105 tons of sulfur is contained in the coal. However, given the atomic weight of sulfur (32) and oxygen (16), the mass of SO2 is half sulfur and half oxygen (two atoms in each molecule). Thus the 5.6 x 105 tons of sulfur will turn into 1.2 x 106 of 1.2 million tons of SO2. Part III: If a standard of 0.1 pounds of particulates per million BTU is adhered to, how many tons of particulates would be released per year? Since this limit on particulates is given in terms of BTU, we need to convert the coal energy of 1.6 x 1018 Joules into BTU. Using the conversion in the front cover of the book, we get 1.5 x 1015 BTU. The wording of this problem could imply it is per BTU of electrical energy instead of fuel energy, and so either number is fine. If we take 1.5 x 1015 BTU / 1 x 106 BTU = 1.5 x 109 million BTU. So there is 1.5 billion million BTU of energy. If for every million BTU there is 0.1 pounds ofparticulates, then there are 0.15 billion pounds of particulates that can be emitted each year. A ton is 2000 pounds, and so this is 75,000 tons of particulates. Part IV: How many tons of CO2 would be released per year, assuming that the coal is 100% carbon? This problem is the same as a previous homework problem. Since we have 56 million tons of coal and CO2 is made of carbon (mass 12) and two oxygen (mass 16), for every ton of carbon or coal, there is about three times as much CO2 produced. This is thus 168 million tons of CO2. Problem #2. In the above problem, assume that the amount of SO2 released per year is 1.18 x 106 tons. If a thermal inversion persists for a 24-hour period over the whole state and the released SO2 can rise no higher than 1000 meters, what concentrations of SO2 (in units of micrograms per cubic meter) will be present in the air? Assume that the state is 500 km x 500 km in area. (5 points total) We are keeping all the 1.18 million tons of SO2 in a very large box. The box with dimensions 500,000 meters x 500,000 meters x 1000 meters (note converting all distances to meters), has a volume = 2.5 x 1014 m3 or cubic meters. Thus, we can just divide the mass by the volume to get the answer. However, we want to convert tons in units of micrograms or 10-6 grams. One ton = 2000 pounds and 1 pound = 454 grams and 1 gram = 1 x 106 micrograms. cubicmetermicrogramsmetermicrogramsSOgrammicrogramspoundgramstonpoundsmtonsmetermicrogramsSO/102.4)(21101145412000105.21018.1)(233631463×=⎟⎟⎠⎞⎜⎜⎝⎛××⎟⎟⎠⎞⎜⎜⎝⎛×⎟⎠⎞⎜⎝⎛×⎟⎟⎠⎞⎜⎜⎝⎛××= In the above problem, the wording is somewhat ambiguous. It makes more sense to say you only have the emissions from the 24-hour period in the box, not from the whole year. This would then result in a SO2 density that is a factor of 1/365 smaller. How does this level compare to standards set from the clean air legislation? In the textbook in Table 9.3 are listed the National Ambient Air Quality Standards which for SO2 is 365 micrograms per cubic meter for a 24 hour period or 1300 micrograms per cubic meter for a three hour period. In either case the abovecalculated number is above these limits. You can also find more information on clean air legislation from other resources. Problem #3. The pH of normal rain is _______ and it contains _______________. a) 4.5 H2SO4 b) 7.0 no acid at all c) 5.6 carbonic acid d) 7.0 carbonic acid e) 6.5 HNO3 f) 7.5 H2SO4 (1 point) C is the correct answer. As detailed in Section 9.10 of the textbook "pure rainwater has a pH of around 5.6, somewhat lower than the pH of 7 for a neutral solution. This slight acidity comes about mainly from the formation of carbonic acid (H2CO3)." Problem #4. Pick the set of answers which best characterize the major fuel, time of occurrence, principle constituents, and principle effect of photochemical smog (e.g. in Los Angeles): a) petroleum, early morning, (CO, SO2), bronchial irritation b) petroleum, midday, (O3, NO, NO2, CO), eye irritation c) coal, midday, (O3, NO, NO2, CO), eye irritation d) petroleum, early morning, (O3, NO, NO2, CO), eye irritation (1 point) B is the best answer. See details in Section 9.6 of the textbook. Many emissions start early in the morning, but the photochemical smog takes some time to develop. Problem #5. The most environmentally harmful emissions from automobiles are ____________. a) CO, SO2, and particulates b) hydrocarbons, CO2, and SO2 c) hydrocarbons, CO, and NOx d) CO, particulates, and NOx e) CO2, SO2 and NOx f) CO, SO2, and NOx g) SO2 and NOx h) radioactivity and SO2 (1 point) C is the best answer. See details in Section 9.7 of the textbook. In particular, Table 9.4 is quite useful in considering the various sources of pollutants including transportation.Problem #6. If there are 0.28 ppm of SO2 in the air at STP (standard temperature and pressure), how many SO2 molecules are there is 1 cubic meter? Note that the atomic weight of sulfur is 32 and of oxygen is 16. (2 points) Example 9.2 of the textbook is a good example for this problem. Air at standard temperature and pressure (STP) has Avogadro's number 6 x 1023 molecules per mole, which occupies 0.0224 cubic meters. Thus, we can calculate the number of molecules total in one cubic meter as 1 cubic meter x (6 x 1023 molecules / 0.0224 cubic meters) = 2.7 x 1025
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