Massachusetts Institute of Technology Physics Department Physics 8.21 Fall 2011 Physics of Energy October 4, 2011 Quiz 1 Instructions Problem Points 1 30 2 45 3 25 4 (+ 20) Total 100 You must do problems 1-3. You may do problem 4 for extra credit Do all problems in the white exam booklets Remember to write your name clearly on every exam booklet you use Open book, open notes, open 8.21 Energy Information Card Do not use red ink No cell phones or computers Calculators are allowed The problems have different point values as marked You may need to make assumptions to solve some problems. Explain your assumptions Problem summary Problem 1 [30 pts] Energy quantities Problem 2 [36 pts] Short problems Problem 3 [34 pts] Stirling engine Problem 4 [extra credit] Carbon sequestration 1
MIT 8.21 Physics of Energy Fall 2011 2 Problems Problem 1. Energy quantities [30 points (5 points each)] How much energy (in joules) is contained in each of the following? (a) 1000 liters of ethanol E = 84 MJ/gallon 1 3.785 l/gallon 1000 l = 22.19 GJ. (b) 1000 liters of water in a reservoir 50 m above the base of a dam E = mgh = 1000 kg 9.8 m/s 2 50 m = 490 kj (c) 1000 liters of water heated from 10 C to 20 C E = 1000 kg 4.186 kj/kgk 10 K = 42 MJ (d) 1000 liters of water vaporized at atmospheric pressure, 100 C 1000 kg 2.26 MJ/kg = 2.26 GJ (e) A 1000 Farad capacitor charged to 10 V E = 1 2 CV 2 = 50 kj (f) 1000 photons of visible (red) light with wavelength 700 nm E = 1000 2π c/λ = 3 10 16 J
MIT 8.21 Physics of Energy Fall 2011 3 Problem 2. Short problems [45 points (9 points each)] (a) Arrange the following 4 types of heat engine in order of practically realizable efficiency: Atkinson, Diesel, Otto, Stirling. Explain. From highest to lowest efficiency: 1. Stirling (realizes maximum Carnot efficiency). Note that in some situations Stirling engines are highly efficient but that in other contexts there are practical challenges in implementing Stirling engines; credit was given if Stirling was placed lower in the list with a clear explanation of why. 2. Diesel (realizes highest compression ratio because fuel cannot knock or combust prematurely since it is not injected until compression is complete) 3. Atkinson (realizes higher expansion ratio than Otto cycle and therefore hire efficiency) 4. Otto (b) A truck with length 20, width 8, height 7 and mass 6 tonnes, is driving at 70 mph along a flat highway. Estimate the rate of energy loss to air resistance. Use this to estimate the number of miles/gallon achieved by the vehicle, assuming a typical diesel engine. Area is approximately A = 5.2 m 2. Mass is irrelevant. We can estimate the truck drag coefficient at c d = 0.5 (higher than car which has cd = 1/3). The velocity of 70 mph corresponds to v = 70 4/9 m/s = 31 m/s. The rate of energy loss to air resistance is P = 1 2 c daρv 3 = 47 kw. Note that ρ = 1.2 kg/m 3 should be air density, not the density of the truck! Work done over one mile W mile = 3600 s 70 miles 47 kw = 2.4MJ. Assuming a diesel engine at 45% efficiency, with 1/2 of mechanical energy going to air resistance and 140 MJ/gallon (see card), each gallon gives 32 MJ, so roughly 15 miles/gallon. (c) A hot spring carrying water at 50 C at a flow rate of 2 l/s emerges from a rock near your remote home. You consider installing a Stirling engine to convert the thermal energy to electrical energy. What is the absolute maximum power you could get theoretically, assuming ambient temperature of 20 C? Do you think this is practical? A Stirling engine would get (ideally) Carnot efficiency, η = (T + T )/T + = 10% efficiency. The energy difference of water between 50 C and 20 C is Q s = 30K 4.2 kj kgk = 126kJ/kg, (1)
MIT 8.21 Physics of Energy Fall 2011 4 and flow rate is f = 2 kg/s, so the maximum power possible would be P η Q s f = 0.1 252 kj/s = 25 kw. In practice, the temperature differential is so small that it would be hard to get this much power. (d) Why do long-distance transmission lines operate at high voltage? Assuming that the load is substantial, the power lost over power of the load is roughly so the larger the voltage the smaller the loss. P lost P load = RP load V 2 RMS (e) A house upgrades 16 windows each of area 1.5 m 2 from single pane windows with R-value R = 1 to double-pane windows with R-value R = 3 (U.S. units). Assuming 120 days of cold outdoor temperatures with average difference between indoor and outdoor temperature of 15 C, compute the energy saved over the course of a year s heating season. With the original R = 1 windows, in SI units R = 1/5.68 Km 2 /W. We then have power lost P = qa = A T/R = 2 kw Replacing with R = 3 drops this by a factor of 3. So change in rate of energy loss is ( P = P 1 1 ) = 2 3 3 P = 1.33kW. (2) The total energy saved is E = 120 24 3600 s 1.33 kw = 13 GJ. Roughly 13 days energy use for a typical American in energy savings.
MIT 8.21 Physics of Energy Fall 2011 5 Problem 3. Stirling engine [25 points] Consider an industrial-strength Stirling engine. Maximum and minimum volume of the chamber containing the working fluid are V 1 = 10 l, V 2 = 1 l, respectively. High temperature is 1000 K, low temperature is 300 K. Minimum pressure at maximum volume is p 1 = 1 atm. (a) [4 points] Sketch (in your white exam booklet) the idealized thermodynamic cycle in the V -p plane. Label the 4 points at the beginning of each step in the cycle. p 3 3 p 2 2 p p 4 p 1 4 1 V 2 = V 3 V 1 = V 4 V (b) [3 points] Name the thermodynamic process associated with each step in the cycle. (see next part) (c) [3 points] For each step, state whether work is done, and indicate direction of heat flow. 1 2: isothermal compression, work done on fluid, heat goes out of fluid into environment 2 3: isometric heating, heat Q h flows out of regenerator into fluid 3 4: isothermal expansion, work done by fluid, heat from high-temperature source into fluid 4 1: isometric cooling, heat Q c = Q h flows out of fluid into regenerator (d) [3 points] Compute the pressure p and volume V at all the (labeled) points in the cycle. The volume at points 1 through 4 are V 1 = V 4 = 10 l, V 2 = V 3 = 1 l. Let s now turn to the pressures, we have that p 1 = 1 atm. Along an isotherm pv is constant so p 2 = p 1 V 1 V 2 = 10atm.
MIT 8.21 Physics of Energy Fall 2011 6 Along the isometric transforamtion 2 3, the volume is constant V 2 = NK B T 2 p 2 = NK B T 3 p 3 and analogously p 3 = p 2 T 3 T 2 = 33.3atm p 4 = p 1 T 4 T 1 = 3.3atm. so (e) [4 points] Compute work done in each step. Work from 1 2 is W 1 2 = V2 V 1 N K B T 1 dv V = p 1V 1 ln V 1 /V 2 = 10 3 (ln 10) J = 2300 J Work from 3 4 is same multiplied by T 4 /T 1 = 10/3: Work done per cycle is W 3 4 = p 4 V 4 ln V 4 /V 3 = T 4 T 1 p 1 V 1 ln V 1 /V 2 = 7.7 kj W = W 3 4 W 1 2 = 5.3kJ. (f) [4 points] Compute net heat input The net heat input is same as work done during isothermal expansion (cancelling the heat to and from the regenerator, Q in = 7.7 kj) (g) [4 points] Use the last two answers to compute overall efficiency and compare to your expectation. The efficiency is given by η = W Q in = 5.3 kj 7.7 kj = 0.7 just as expected from Carnot efficiency T/T + = 0.7.
MIT 8.21 Physics of Energy Fall 2011 7 Problem 4. Carbon capture [up to 20 points extra credit] The level of carbon dioxide in the atmosphere was around 280 parts per million (ppm) by volume in preindustrial times. Currently the level is around 390 ppm. Carbon dioxide in the atmosphere traps outgoing infrared radiation from Earth, warming the planet. It has been proposed that removing carbon dioxide from the atmosphere may eventually be necessary to mitigate climate change. This problem explores the role of entropy in making direct carbon capture from the atmosphere a challenging operation. (a) Estimate the change in entropy if one mole (44 g) of CO 2 is separated out from the appropriate volume of air. [hint: the entropy of the air changes very little; this is somewhat like the reverse of the entropy change in free expansion.] Change in entropy for each molecule is k B ln(10 6 /390), so total change in entropy is S = N A k B ln(10 6 /390) = 65 J/K. (b) Assume that the entropy associated with the mixing of CO 2 into the atmosphere is reduced by capturing the CO 2 as above. Use the second law of thermodynamics to argue that there is a minimum energy that must be used in the capture process. Estimate the minimum energy needed to capture one kg of CO 2 (that might be released from a coal plant, for example, when 1.1 kg of coal is burned, releasing perhaps 33 MJ of energy of which 30%, or 10 MJ of useful energy can be extracted) The entropy of the environment must be increased by S = 1 kg 44 g 65 J/K = 1500 J/K for one kilogram of CO 2 to be captured. This requires an energy transfer to the environment of E = 300 K 1500 J/K = 450 kj. This is almost 5% of the energy extracted from the coal that gave rise to the CO 2 in the first place. Any realistic process will take several times this amount of energy. As we will discuss later in the course, capturing carbon near the power plant point source is easier since the density in flue gases is higher than in the atmosphere.