Massachusetts Institute of Technology Physics Department Physics 8.21 Fall 2012 Physics of Energy November 13, 2012 Quiz 2 Instructions Problem Points 1 20 points 2 20 points 3 20 points 4 20 points 5 20 points 6 (+ 20 points ) Total 100 + 20 points You must do problems 1-5. You may do problem 6 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 [20 pts] Analyzing an engine Problem 2 [20 pts] Nuclear properties in a fictional world Problem 3 [20 pts] Energy density in 238 U Problem 4 [20 pts] Short solar problems Problem 5 [20 pts] Insolation and orbit Problem 6 [+20 pts] A change of phase 1
MIT 8.21 Physics of Energy Fall 2012 2 Mass excesses of light proton, neutron and the lightest nuclei Mc 2 (MeV) Mass excess (MeV/c 2 ) Mass excess (µu) Proton 938.272 6.778 7276 Neutron 939.566 8.071 8665 Hydrogen 938.783 7.289 7825 1 dalton 931.494 0 0 Electron 0.511 0.511 549 4 2 He 3728.401 2.425 2603 Selections from the National Nuclear Data Tables Wallet Cards Periodic Table
MIT 8.21 Physics of Energy Fall 2012 3 Some nuclear data
MIT 8.21 Physics of Energy Fall 2012 4 Problems Problem 1. [20 points] Analyzing an engine An ideal-gas engine works on n-moles of a gas. The engine s cycle is shown in the pv -plane in the figure below. Note that the heat capacity of the gas (at constant pressure) is C p = 5 2 nr. Figure 1: (a) Describe in words (e.g. adiabatic compression, isothermal expansion, etc.) the three steps, [12], [23], and [31], in this cycle (b) Define the compression ratio r V 1 /V 2 and show that the ratio of pressures p 2 /p 1 and the ratio of temperatures T + /T can be written in terms of r. (Remember this is an ideal gas.) (c) How much work is done in each step of the cycle? Be careful about signs; express in terms of pressures and volumes (not temperatures). (d) In which step(s) does this engine absorb heat from a heat reservoir? How much? (e) Define the efficiency of the engine in terms of the quantities you have already calculated and express it as a function of r alone. (f) Suppose r = 2. How does this engine s efficiency compare with the Carnot limit for an engine working between the same maximum and minimum temperatures? Why do you suppose this engine is so inefficient? (a) [12] Isothermal compression at T from V 1 to V 2 [23] Isobaric (constant pressure) heating from T to T +, where T + is chosen so the volume increases back to V 1.
MIT 8.21 Physics of Energy Fall 2012 5 [31] Isometric (constant volume) cooling at volume V 1 from T + back to T. This returns the system back to its starting configuration. (b) For an ideal gas pv = nrt. Since T 1 = T 2, p 1 V 1 = p 2 V 2, so p 2 /p 1 = r. Likewise nrt + = p 2 V 1 while nrt = p 2 V 2, so T + /T = r. (c) Step [23], isobaric expansion. The engine does work: p is constant, so W [23] = p 2 (V 1 V 2 ). Step [12], isothermal compression. Work is done on the engine: W [12] = V 2 V 1 pdv = p 1 V 1 ln(v 2 /V 1 ). W [12] is negative, reflecting that work was done on the engine. Step [31] Constant volume: no work is done. (d) Step [23], isobaric expansion. ([31] is isometric cooling so heat is expelled, and [12] is isothermal compression, where heat is also expelled.) Q [23] = C p (T + T ). (e) The efficiency is work done/heat absorbed, so η = W [23] + W [12] = p 2(V 1 V 2 ) p 1 V 1 ln(v 1 /V 2 ) = 2 Q [23] C p (T + T ) 5 ( 1 ln r ) r 1 (f) For r = 2, η = (2/5)(1 ln 2) = 0.123. The Carnot efficiency is η c = (T + T )/T + = (r 1)/r, which is 0.5 in this case. The engine is inefficient because so much of the heat is added at temperatures far below T +.
MIT 8.21 Physics of Energy Fall 2012 6 Problem 2. [20 points] Nuclear properties in an fictional world Consider a world in which protons have electric charge + 1 2 e and neutrons have electric charge 1 2 e. The masses of neutrons and protons and the strong interactions of neutrons and protons do not change. As in our world, the neutron β-decays into a proton, an electron and an anti-neutrino: n p + e + ν e. (a) Write down the semi-empirical mass formula (SEMF) for the binding energy B of a nucleus as a function of the number of protons, Z, the number of neutrons, N, and the mass number A = Z + N in this fictional world. (b) For the most stable nucleus with a given value of A, what does the SEMF predict to be the ratio N/Z? (c) Are there any (electrically) charged nuclei in this world? Explain your answer consider both even and odd A. (d) Ignoring the pairing energy, what does the SEMF predict for the binding energy per nucleon, b(a), for the most stable nucleus as a function of A. (e) Can nuclear fission (either spontaneous or slow neutron induced) occur in this world? Explain. (f) Do you think light nuclei can fuse in this world? Is it easier or harder than in our world? Explain. (a) Only change is that Z 1 2 (Z N) in Coulomb energy. B(Z, N) = ɛ V A ɛ S A 2/3 (Z N) 2 (N Z) 2 ɛ C 4A 1/3 ɛ sym + η(z, N) A A 1/2 (1) (b) At fixed A, this clearly has its minimum at Z = N, so Z/N = 1. (c) Yes, all odd-a nuclei are charged, and there are electrically charged nuclei with even-a because the pairing energy allows several different nuclei to be stable, when both Z and N are even. (d) Since N = Z for the most stable isotope, b(a) = 1 A B(A/2, A/2) = ɛ V ɛ S /A 1/3. (e) No, there is no nuclear fission in this world. Nuclei just keep becoming more stable as A increases. There is no Coulomb term to destabilize them. (f) Yes, the mechanism for fusion: reducing the surface to volume ratio, still holds. Furthermore many light nuclei are electrically neutral, so there is no Coulomb barrier to fusion, making it much easier than in our world.
MIT 8.21 Physics of Energy Fall 2012 7 Problem 3. [20 points] Energy density in 238 U 238 U is not fissile, but the fissile nucleus 239 Pu can be bred from it by neutron capture in a reactor. This problem asks you to compare the energy content of natural uranium when (a) it is used only for its 235 U content and (b) it is used as a source of 239 Pu. Natural uranium contains 0.72% 235 U. The rest is 238 U. Some relevant nuclear data is given at the beginning of the exam. (a) On average, when 235 U fissions, it releases approximately 195 MeV of useful energy. (This includes energy promptly released plus subsequent radioactivity, but excludes neutrinos that escape to outer space.) Compute the energy density of natural uranium (J/kg) viewing it as a source of 235 U fission fuel. Natural uranium is 0.72% 235 U, so 1 kg of natural uranium contains 7.2 g of 235 U. This is equal to 0.0306 mol or 1.85 10 22 atoms of 235 U. At 195 MeV/atom, the total energy density then is (1.85 10 22 atoms 235 U/kg U)(195 MeV) = 577 GJ/kg natural U. (b) 238 U can absorb a thermal neutron and undergo a sequence of decays that lead to 239 Pu. Write down formulas for this reaction and for the subsequent decays. Compute the total energy released in this process. (i) 238 U + n 239 U + γ Q = ( 238 U) + (n) ( 239 U) = 47.310 + 8.071 50.575 MeV = 4.806 MeV (2) (ii) 239 U 239 Np + + e + ν e (iii) 239 Np 239 Pu + + e + ν e Q = ( 239 U) ( 239 Np) = 50.575 49.3137 MeV = 1.2613 MeV (3) Q = ( 239 Np) ( 239 Pu) = 49.3137 48.5912 MeV = 0.7225 MeV (4) Only the total energy was requested. It is obtained by adding the three: Q tot = 6.790 MeV (c) The reaction you described in part (b) occurs in a breeder reactor, creating 239 Pu that can subsequently fission. On average when 239 Pu fissions, it releases approximately 200 MeV of useful energy. Combine this with the result of part (b) to compute the energy density of natural uranium (J/kg) viewing it as a source of 239 Pu fission fuel. The large neutron flux inside something like a reactor could convert the 238 U into 239 Pu. Each 239 Pu fission releases around 200 MeV. So, in 1 kg of natural uranium, there are 992.8 g of 238 U, or 2.51 10 24 atoms. If all of this is converted to 239 Pu, then fission of 239 Pu can release (200 MeV/atom)(2.51 10 24 atoms/kg) = 80.5 TJ/kg natural U. If the plutonium is created inside a reactor, where the energy released from neutron capture and subsequent decays can be used, then we instead get around 205 to 207 MeV per uranium atom (some of the energy in the decays is carried off by the neutrino), so a total of around 83 TJ/kg natural U can be extracted from plutonium production and fission.
MIT 8.21 Physics of Energy Fall 2012 8 Problem 4. [20 points] Short solar problems (a) Solar concentrators on Mars Consider an ideal focusing concentrator. Increasing the concentration causes the receiver temperature to increase up to a point. Beyond certain maximum concentration, the temperature remains constant. What is the maximum concentration that can be used on Mars for a 3-D case? The radius of the orbit of Mars about the Sun, is approximately 2.3 10 8 km. The acceptance angle θ a should be chosen to be equal to the half-angle subtended by the sun as seen from Mars. That gives θ a = tan 1 (R sun /R orbit ) = tan 1 (6.96 10 8 /2.3 10 11 ) = 0.18 degrees. The maximum concentration is thus C 1/ sin 2 θ a = 99915 10 5. (b) Laser power delivery A startup company proposes to develop technology which delivers power using laser beams. The lasers shine from outer space, delivering monochromatic light onto solar cells mounted on the Earth. Suppose a 10 W/m 2 beam of laser light is normally incident on a 2 m 2 silicon PV cell. The PV cell converts each incident photon to an outgoing electron with 20% probability (this is known as its external quantum efficiency ). What electrical current (in amperes) can be generated by the PV cell, if (a) the light has wavelength 980 nm, and (b) the light has wavelength 1.55 µm? (These are technologically important wavelengths, though practical delivery from outer space would have to take into consideration the absorption window of water. 1 ) (a) The energy of a 980nm photon is E = hc/λ = 2.03 10 19 J. In electron volts, this is E/q = 1.27 ev. Because this is larger than E g = 1.1 V, photons of this energy are absorbed by the PV. The 20W incident power is 20/2.03 10 19 = 9.85 10 19 photons/sec. 20% of this gives an electron current of 1.97 10 19 electrons/sec which, multiplied by 1.602 10 19 Coulombs, gives 3.16 Amperes. (b) The energy of a 1.55 µm photon is about 0.8 ev. Because this is less than E g, no photons are absorbed, and thus no photocurrent results. (c) Photodiode efficiency Everything else being the same, the efficiency of a photodiode rises: (a) when the operating temperature rises (b) when the operating temperature falls 1 980 nm is the wavelength at which erbium doped fiber lasers are pumped; this is the technology which makes undersea fiber-optic transmission cables feasible. 1.55 µm is the wavelength used for virtually all modern fiber-optic telecommunications, because photons of this energy suffer the minimum dispersion absorption in glass fibers.
MIT 8.21 Physics of Energy Fall 2012 9 (c) when the light power density rises (d) when the light power density falls Indicate which of the above are true. (b) and (c) are true. Problem 5. [20 points] Insolation and orbit (a) Solar insolation What is the insolation (W/m 2 ) on a flat and level surface, at latitude 45 degrees North, at 10:00 local time on November 13, 2012 (i.e. 34 weeks after the spring equinox)? Assume that the insolation on a surface facing the sun is 1000 W/ m 2. The two relevant equations are sin δ = sin ɛ sin α and cos β = sin λ sin δ+cos λ cos δ cos ωt, where δ is the declination, ɛ is the tilt, α the day of year measured from the spring equinox, β is the angle of insolation, and t is the time of day measured from solar noon. For α = 2π7 34/365, and using ɛ = 23.45, we find that δ = sin 1 (sin ɛ sin α) = 0.331 = 9.5. The insolation we wish to compute is given by ωt = 2(2π)/24 = π/6, the above value of δ, and λ = 45 = π/4 plugged into: P = 1000 cos β (5) = 1000(sin λ sin δ + cos λ cos δ cos ωt) (6) = 349 W/m 2. (7) (b) Lost in space You have landed on an unknown planet and, must determine both your latitude λ and the angle, ɛ 0, the inclination of the planet s axis of rotation referred to its orbital plane. To accomplish such a determination, all you have is a ruler and plenty of time. You erect a vertical pole, exactly 5 m tall and observe the length of the shadow at noon as it varies throughout the year. The shortest length is 1.34 m and the longest is 18.66 m. Give λ and ɛ 0 (both in degrees). At noon, t = 0 and thus β = λ δ. The longest length shadow is obtained at the winter solstice, when α = π/2, at which point δ = ɛ 0 ; similarly, the shortest shadow is when δ = ɛ 0. We thus have two equations and two unknowns: 1 18.66 tan 5 1 1.34 tan 5 Solving these gives us λ π/4 = 45, and ɛ 0 = 30. = λ + ɛ 0 (8) = λ ɛ 0 (9)
MIT 8.21 Physics of Energy Fall 2012 10 Problem 6. [+20 points extra credit] A change of phase? (a) A 1 m 3 container contains 10 kg of water at NTP. 5 10 6 calories of heat are added to the water in the container (the pressure is kept fixed at 1 atm). What is the final temperature of the contents of the container? What (mass) fraction of the water is in the form of vapor? 5 10 6 calories is about 21 MJ. The energy needed to bring the water to its boiling point is C v (100 20)K 10 kg 3.3 MJ. The remaining 17.7 MJ is less than the 23 MJ needed to boil the water. The final temperature is thus 100 C. (b) A solar concentrator of a parabolic trough design is used in a desert location (with insolation 1 000 W/m 2 ) to harness solar power. The system can focus 100 m 2 of sunlight with concentration C = 100 onto a long central pipe containing a working fluid. Explain briefly why each of the following kinds of engines is good or bad (or inappropriate) for converting the resulting heat into mechanical power. Which is best? Otto Cycle Engine Stirling Cycle Engine Rankine Cycle Engine Brayton Cycle Engine Otto: inappropriate, as it consumes a fuel for its operation Stirling: possible to use, but not as efficient as Rankine cycle under realistic circumstances; low power density Rankine: best, taking advantage of phase change energy (eg steam); high power density. One drawback is that the steam must be cooled to condense it back to water this may require cooling water. Brayton: inappropriate, as it consumes a fuel for its operation (c) Consider the Stirling, Rankine, and Brayton cycles. Which of these engine cycles is most appropriate to use for the following energy sources, and why? Cheap source of blocks of ice, in a warm climate Kerosene Burning wood Uranium reactor undergoing controlled fission reaction Ice: Stirling cycle, because of low T compared with ambient temperature of environment. Kerosene (jet fuel): Brayton cycle, because it is an open cycle Burning wood: either Stirling or Rankine cycles, because the heat from the wood can be used to generate steam Uranium reactor: Rankine cycle