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H A&S 220c Energy and Environment: Life Under the Pale Sun 2 Dec. 2004. Reviewing for the final. The final is on Monday Dec. 13 (10.30-12.20 in the classroom, MGH 238). It is closed-book, but you may bring a single sheet of paper with notes on both sides (not too microscopic please). I handed out a 10-page review, which was my ‘essay’ about the content of the course. It did not include many numbers or equations, but tried to show some of the relationships between the four areas of interest in the course. The final ‘quiz’ (I like the word better than ‘exam’) will start with some short questions (1.), then as in the midterm quiz, will have some questions from the science core (2.) and finally (3.) some short essay questions. You will choose a few of many questions in each section. I have written a miniature lecture below on ‘box-models’. This was part of our reading in The Spherical Cow. Please read this (a related question will be on the quiz). Some practice questions are below. Others will be given in class Thurs. Dec 9. Some of these will appear on the final. The review questions for the mid-term (posted on the website on 3 Nov 2004) are still relevant). Typographical error in the REVIEW: p.2, line 7 ‘directly’ should read ‘inversely’. 1. Short Answers, 2 or 3 lines (will do 2 on final) 1.1 What was Operation Cat Drop of the British Royal Air Force? (Hawken/Lovins) 1.3 Where is Curitiba and what is its significance? (Hawkin/Lovins) 1.4 What differences between the oil deposits in Saudia Arabia and the US, which has economic importance [Roberts]. 1.5 What is the significance of the specific heat capacity of a gas or liquid? 1.6 What are the principal animals hunted by natives of Greenland? 1.7 Explain why the flame of a candle has different colors in it. 1.8 Breaking apart water molecules and forming hydrogen gas and oxygen gas requires a net input of energy equal to 118 Kcal/mol, because the H-O bond in water is stronger than the H-H bond in hydrogen gas. Electrolysis is the use of electricity to do this. What kinds of forces are involved in this chemical energy? 1.9 Water is a great solvent: many solids will dissolve in it, meaning that their rigid bonds between atoms are broken as it interacts with the water molecule. What property of H2O molecule is particularly important for this? 2. Questions from the science core (will do 2 on final). 2.1 Why is the Arctic, and Greenland in particular, very sensitive to global warming (the rising temperature of the Earth’s atmosphere and oceans)? 2.2 Compare the efficiencies of car and bicycle as people-transporters. An efficient bicyclist can ride 20 mph (32 km/hour) for long stretches, burning perhaps 750 Kcal (food calories) per hour. [1 Kcal = 4184 Joules ] A typical car does 25 miles per gallon of gasoline, which is 10.65 km/liter. The energy content of gasoline is about 45 MJ/kg. 2.3 Greenland is about 2500 km long and 1000 km wide. The ice cap peaks at 3 km above sea level, and holds about 10% of the Earth’s fresh water (Antarctic ice holds about 70% of the fresh water). If Greenland were to melt, roughly how much liquid water (in m3) would be released, and how much would global sea level rise, on average? The area of the oceans is about 3 x 1014 m2. [radius of the Earth, R = 6400 km, area of oceans = 70% of 4π R2]. What if Antarctica were to melt too? 2.2 The new Airbus 330-220 burns about 8.5 tonnes of fuel per hour at its cruising speed of 900 km/hr. If that fuel contains 40 x 106 Joules/kg of energy, how many watts of power does this rate of burning represent? 1 tonne = 103 kg. How does this fuel use compare with each of the 330 passengers driving a car that uses 10 kg/hour of gasoline, yet is going at only 100 km/hr? 2.3 In the ocean circulation water rich with oxygen sinks from the surface at 107 m3/sec, flows round the world and slowly rises back to the surface again. If the volume of the world’s oceans is 1018 m3, how many years does it take, on average, for the water to return to the sea surface to ‘breathe’ again? 2.4 If the sun heats the top 10m of the ocean in spring, with a power of 200 watts/m3, how many degrees warmer will it be after 3 months? The specific heat capacity of water is about 4000 J/kg 0C and the density of water is about 103 kg/m3. 2.5 The solar heating of the Earth changes with the seasons, in part because the Earth is closest to the sun in January, closer by about 3% than in July. What change in mean temperature of the Earth would this cause if the simple greenhouse model of the Earth were accurate? [Spherical Cow] 2.6 Why do we dam rivers to generate hydroelectric power? To think about this calculate the energy flow (in watts) of a stream of water falling from a height h, of 100m from a reservoir, with a flow rate Fw (m3/sec) ….(the water has negligible kinetic energy in the reservoir). Is this a useful power source for a river with flow rate Fw = 103 m3/sec? [The potential energy per m3 of water, due to the gravity force is ρgh where ρ is the density of the water, 103 kg/m3 and g = 9.8 m/sec2 is the acceleration due to gravity.] 3. Essays. Choose one from these topics and write a short essay…typically 2 pages of handwriting. 3.1. What do Lovins/Hawkens mean by “the present industrial system is, practically speaking, a couch potato”. Would Lomborg agree? 3.2 If global energy use of 4 x 1020 Joules/year were cut by 90% and distributed equally among the 6 billion citizens, how much ‘poorer’ would Americans be? Recall that energy cost is about 6% of Gross Domestic Product. What positive changes by our political and industrial leaders could help ease this transition? [This is quite a deep question, if followed through.] 3.3 What are some of the hidden costs of oil production and use, which make the price per barrel an imperfect measure? What would Lomborg say about these costs? What would Hawken/Lovins’ say about these costs? 3.4 With solar, wind, biomass and hydropower supplying only about 7% of our global energy use, write a paragraph describing a Utopian (positive, idealistic) image of life with just this supply of energy and no fossil fuels. Give at least one quantitative calculation (for example, based on your personal energy profile). What population migrations might result from this change in energy supply? 3.5 Environmentalists argue that building taller smokestacks cleaned up London’s polluted air and yet ‘exported the pollution to fall as acid rain in Scandanavia. Lomborg (The Skeptical Environmentalist) argues that saving 64,000 lives per year by improving London’s air is more important than saving Scandanavian trees. What do you think? 3.6 “…the sense among governments was that oil was too important to be left in private hands.” [Roberts, The End of Oil]. Discuss with detailed examples. =============================================== A Small Lecture on Steady State Box Models. Steady-state box models appear in Spherical Cow, p36. I want to take a little extra time to discuss this important idea. Here we have a lake with a river flowing in one end and out the other. A factory dumps an amount, Fp, 160 kg of pollutant each day into the lake (that is 0.16 tonnes). If the pollutant mixes evenly round the lake, so that its concentration (call it C, in kg/m3) is the same everywhere, what is that concentration? This is a key question for environmental protection. We are given the river flow-through rate, Fw = 8 x 104 m3/day the volume of the lake, Mw = 4 x 107 m3. The sketch on p37 is important in visualizing the problem This harks back to the discussion of ‘transport’ or movement of energy in a river. We saw in lectures that the rate at which energy moves past a point in a river is given by concentration of energy x V x A. Here it is the same thing, for the concentration, C, of pollutant. transport of pollutant = (concentration of pollutant C in kg/m3 ) x (velocity of river V in m/sec) x (cross section area of river, A in m2 ) A is equal to width x depth if the river has uniform depth. Note the units work out to be kg/sec which makes good sense for the transport. {This was derived by realizing that river water moves a distance Vt in a time interval t, and the volume moving past an observation point is thus VtA. The amount per second passing this point is VA, which is what we called Fw above . The total amount of pollutant in that volume is CVA} The rate at which pollutant enters the lake is Fp = 160 kg/day and in a steady situation, this must leave the lake at the same rate. So Fp = CVA = CFw because the river through-flow is the velocity times the cross-section area. Divide through to give the answer, C = Fp/Fw with units (kg/sec) / (m3/sec) = kg/m3. Putting in the numbers, C = 160/8x104 = 2 x 10-3 kg/m3. Because water has mass 3 10 kg/m3 (1 tonne per cubic meter) this could be expressed as 2 x 10-6 (kg pollutant)/(kg water) or ‘two parts per million’ pollution. Note that the answer is independent of the size of the lake (Mw), which seems surprising. More on this below. (Notice that Harte derives the same answer in a more casual way, using the idea of residence time of water in the lake.) The non-steady problem and exponential decay. The problem could be made more accurate in a number of ways. What about the size of the lake? It seems important, yet it didn’t affect the steady-state concentration C. To investigate this, imagine that the pollutant inflow is started suddenly: how does the concentration build up with time? This is a very interesting problem involving exponential growth rate. It does involve a differential equation and is described on p 111. Don’t bother to read this now. Without all the detail, you can visualize that the concentration C will grow from zero to its steady- state value (given above). We suspect it might take about as long as the ‘residence time’ to become steady. The input is Fp, a constant, yet the outflow is CFw, which varies as C varies. The equation is rate of change of C = (Fp – CFw)/Mw We saw in discussing population growth that an exponential curve is one whose slope is proportional to its height. Here the ‘slope’ is the rate of change of C and the height is C itself. The result is that the solution to our equation is an exponential, C = (Fp/Fw)(1 - exp(-Fw t/Mw) where exp(α) means e to the power α It is plotted in the figure below. This answers the question above…the time it takes to reach steady state does depend on the lake volume; this occurs when the exponential curve dies way to nothing. The residence time Mw/Fw is an approximation to this time, because when t = Mw/Fw the exponential term is exp(-1) which is ‘on its way down’ to zero. Notice that at the start time, t=0, C is zero as it should be, and at large time when the exponential dies to zero, C = Fp/Fw also as it should be. Notice also, when we say ‘slope is proportional to height’ for this plot the height is measured from the line C/Co = 1. Graph: concentration of a pollutant dumped into lake, as a function of time.

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