Massachusetts Institute of Technology Physics Department

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1 Massachusetts Institute of Technology Physics Department Physics 8.21 Fall 2011 Physics of Energy November 8, 2011 Quiz 2 Instructions Problem Points (+ 20) Total 100 You must do problems 1-3. The last part of problem 3 is optional 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 [50 pts] Short problems Problem 2 [25 pts] Nuclear thermal power Problem 3 [25 pts + extra credit] Solar concentrator 1

2 MIT 8.21 Physics of Energy Fall Data on specific nuclides

3 MIT 8.21 Physics of Energy Fall Problems Problem 1. Short problems [50 points total] (a) [8 points] Explain the advantages and disadvantages of the Atkinson cycle. Where is this cycle currently used and how is it implemented? Answer: Advantage: can have greater expansion than compression ratio so efficiency can be greater than Otto cycle with same fuel. Disadvantage: lower power. Currently used in hybrid automobiles. Implemented by keeping valve open during beginning of compression stroke. (2 points for mentioning each of these 4 things) (b) [10 points] p + 3 Temperature T + T p Entropy (b) Explain what happens in each step of the Rankine cycle above, including whether work is done or heat transfer occurs. What is the rationale for the locations of points 1 and 3? Answer: 1 2: Pure liquid water is compressed adiabatically, work is done on the fluid. 2 3: The water is heated, transformed completely to vapor, and the vapor is superheated. Heat is transferred to the fluid. 3 4: The vapor expands, isentropically. As it expands the fluid does work on a turbine. 4 1: The vapor is condensed back into liquid at constant temperature. Heat is dumped from the vapor to the environment. Point 1 is chosen outside the saturation dome so that compression is done on pure liquid; pumps work better on a pure phase and this reduces work since liquid compresses less than mixed liquid-vapor. Point 3 is chosen far enough out that isentropic expansion leaves it almost completely outside the saturation dome. This is done because turbines can t handle mixed-phase vapor without sustaining damage or extra materials engineering.

4 MIT 8.21 Physics of Energy Fall (c) [8 points] What is the purpose of a dedicated breeder reactor? Why would one build breeder reactors instead of conventional fission reactors? Answer: The purpose of a breeder reactor is to use neutrons from fission in a uranium-235 based reactor to produce another fissile material, generally 239 Pu from U-238 or 233 U from 232 Th. One might build breeder reactors to increase the amount of nuclear fuel supplies of U-238 and Th-232 are each roughly 100 times greater than the supply of U-235. (d) [8 points] What important aspect of nuclear fission reactions enables control of a fission reactor on a human time scale? Explain. Answer: Delayed neutrons make possible control with reaction times on the order of seconds or minutes rather than milliseconds. Although the time between links in a nuclear fission chain reaction is on the order of 10 4 s, a few neutrons are released after order 10 s. By tuning the k factor to be just above 1 when the delayed neutrons are included and just under 1 when they are not included, only the delayed neutrons cause the reaction to grow, so that the timescale for growth is 10 s, and dozens of steps in the chain reaction take several minutes, within the time scale of humans and machines controlling reactors to react. (e) [8 points] Germany is aggressively developing solar power as a domestic energy resource. Estimate the insolation on a clear day in Hamburg, Germany (latitude N) at solar noon on October 21. Answer: On October 21 it is one month after the Autumn equinox, so α = 7π/6 = 210, and sin α = 1/2. So we have sin δ = 1 2 sin ɛ = 0.2 and δ = 0.2 = At solar noon the sun is directly overhead at the same longitude and latitude δ so β = λ δ = 65, cos β = 0.42 With typical overhead insolation I 0 = 1000 W/m 2, this gives an insolation in Hamburg on October 21 at noon of I = I 0 cos β = 420 W/m 2. (f) [8 points] You are given a choice of three materials from which to construct a single-junction photovoltaic solar cell. The band gaps in the three materials are 0.2 ev, 1.5 ev, and 4 ev. Which material would you choose to work with? Explain the physics behind why the other materials would not be good choices. Answer: The one with band gap 1.5 ev is the best option. With a band gap of 0.2 ev the energy collected per photon is too small and the collection efficiency will be poor. With a band gap of 4 ev most photons in incoming solar radiation will not excite an electron and most energy will be lost.

5 MIT 8.21 Physics of Energy Fall Problem 2. Potassium thermal energy [25 points] You are a venture capitalist. An entrepreneur comes to you with an idea for a power system based on using thermal energy from the decay of potassium. His plan involves packing potassium chloride (KCl, g/mol, 2g/cm 3 ) into cubic containers of dimension (10 m) 3. The containers are well-insulated, with the walls having a thermal conductance of 0.01 W/m 2 K. Thermal energy from radioactive decay of the potassium builds up in the potassium salt and is used to power a turbine. Do a quantitative analysis of his proposal and decide whether you are interested in investing based on your analysis. Answer: First let s calculate the rate of energy production. The total mass of potassium salt is The number of potassium atoms is then M = 1000 m kg/m 3 = kg. (1) N K = N A M/(74.55 g/mol) = (2) From the table, roughly 0.01% of potassium is radioactive potassium-40. τ 1/2 = years so the decay rate is The half-life of K-40 is 1 τ = ln 2 τ 1/2 = s 1. (3) The number of atoms that decay per second is then N K /τ = % of the time the decay is a β decay to 40 Ca ( = MeV). The rest of the time the decay is to 40 Ar with a similar mass excess, so we ll just use the mass excess from Ca-40. For the purposes of this device both decay products are stable. So net energy release is E 1 = MeV ( MeV) = 1.3 MeV per decay. Total rate of energy release is then P = N K E 1 /τ = 6.2 mw (4) So even without worrying about thermal losses and efficiency of the conversion, this is only a tiny amount of energy. You tell the entrepreneur to go away and you spend no more time or money on this proposal. Note: the thermal conductivity of the walls is not relevant to coming to the right conclusion in this problem. Sometimes you need to sort out what information is relevant from what information is irrelevant.

6 MIT 8.21 Physics of Energy Fall Problem 3. Solar concentrator [25 points + 20 points extra credit] Consider a solar concentrator based on a parabolic mirror of length 5 m and width 2 m. The symmetry axis of the parabola lies in a vertical plane, and the sun is directly overhead. All light incident along the parabola axis is reflected to a cylindrical absorber of total area 1 m 2. Total insolation is 1000 W/m 2. Assume ambient temperature is T = 300 K. (a) [10 points] Compute the temperature of the absorber when it is in radiative equilibrium with the incoming radiation. Answer: The concentration is C = A aperture /A absorber = 10. σt 4 = CI = 10, 000 W/m 2 T = 648 K (5) (b) [15 points] Now assume that the absorber is at T =450 K and that all incoming energy not radiated as thermal radiation from the absorber is used to produce electricity using a Stirling engine. What is the maximum possible efficiency with which electrical energy is produced from thermal energy and what is the maximum power output? Answer: Assuming ambient temperature is T = 300 K, maximum efficiency is Power radiated out is η = η Carnot = T T T = 150 = 33.3% efficiency (6) 450 P radiated = σt 4 = 2325 W (7) So available power is P = 7, 675 W. At 1/3 efficiency, maximum power output is P max = P/3 = 2558 W. (8) (c) [20 points extra credit] Note: this part is not easy to complete. Do not spend significant time on this until you are sure that you have answered all the other questions to the best of your ability. Now assume that the absorber is at a general temperature T. Write algebraic expressions for the maximum possible thermodynamic efficiency and resulting power output. Use this to write an algebraic expression for the temperature T optimal at which power output is maximized assuming optimal efficiency. Estimate the temperature T optimal and corresponding power output (using any method you like within the parameters of the quiz). Answer: Generalizing the previous solution, and denoting T = 300 K, η = T T T (9)

7 MIT 8.21 Physics of Energy Fall and Available power is then (using P total = 10 kw) Power output at maximum efficiency is then P radiated = σt 4. (10) P = P total P radiated = P total σt 4. (11) P max = (P total σt 4 )(1 T /T ) = P total + σt T 3 σt 4 P total T /T (12) This vanishes at T = T and T = T eq = (P total /σ) 1/4 (ambient temperature and radiative equilibrium), and is positive between. Maximum power output is optimized when This vanishes when dp max dt = 3σT T 2 4σT 3 + P total T /T 2 = 0 (13) f(t ) = 3σT T 4 4σT 5 + P total T = 0. (14) It is hard to solve a quintic generally but we know that at T = T = 300 K the derivative of (12) is positive and at T = T eq = 650 K it is negative. So we can use trial and error and linear interpolation to rapidly estimate where the derivative vanishes. Checking some values (all in WK), starting at the points where we know the derivative is positive and negative, and the midpoint, we have f(300) = , f(650) = f(475) = So T is close to 475 K, but slightly larger. Trying f(500) = and doing a rough linear interpolation that the solution should be slightly over 10% of the way from 475 to 500, we have T = 478 K, with maximum power P max = 2620 W At this temperature, radiative losses are σt 4 = 3000 W and efficiency is roughly 37.5%.