Presentation is loading. Please wait.

Presentation is loading. Please wait.

0100020003000 Energy Consumption Fossil Fuel Contribution to Global Energy Demand Year.

Similar presentations


Presentation on theme: "0100020003000 Energy Consumption Fossil Fuel Contribution to Global Energy Demand Year."— Presentation transcript:

1 Energy Consumption Fossil Fuel Contribution to Global Energy Demand Year

2 The constraints of limited/vanishing fossils fuels in the face of an exploding population

3

4 …together with undeveloped or under -developed new technologies The constraints of limited/ vanishing fossils fuels

5 Nuclear will renew interest in nuclear power

6 volume term grows as nuclei added surface term corrects for surface nuclei Coulomb term repulsion due to protons symmetry term drives number of protons ≈ neutrons pairing energy term increased stability for paired nuclei

7

8

9 The fusion of 2 protons actually involves positron decay which converts one of them into a neutron. Q=0.42 MeV The sun’s longevity is due to the fact that the accompanying positron identifies this as a weak, i.e., rare process. It is only slowly consuming its hydrogen. Since this allowed the evolution of life on earth - the weakness of the weak interaction is of significant to us! The fusing of 2 protons might seem to exemplify the basic process, but there is no 2-proton state! This is, however, the first step of hydrogen burning in the sun.

10 ttriton nucleus of tritium 3 H d deuteronnucleus of deuterium 2 H  alpha nucleus of helium 4 He The lightest nuclides And their binding energies Nuclide Number of bonds Total Binding energy(MeV) per particleper bond 2 H 3 H 4 He

11 The Deuteron, 2 H Charge+1.6  C = 1e mass MeV = u spin1 magnetic moment   N total angular momentum, J1 Does not exist in an excited state ( of higher ℓ ). What does this tell us about the nuclear force? What does this tell us about the orbital angular momentum?

12 Odd – Odd Isotopes IsotopeSpin, I 

13 The total wave function describing 2 identical fermions must be anti-symmetric to particle exchange. Space Part Spin Part Any part will contribute an independent symmetry of its own.

14 B A B r1r1 r2r2 Particle Exchange: A r2r2 r1r1 Parity Operation: A B r1r1 r2r2

15 B A B r1r1 r2r2 Particle Exchange: A r2r2 r1r1 Parity Operation: A B r1r1 r2r2 r A B r1r1 r2r2 r 1 r 2 r r r 1 r 2 r 1 r 2

16 The Spherical Harmonics Y ℓ,m ( ,  ) ℓ = 0 ℓ = 1 ℓ = 2 ℓ = 3

17 The total wave function describing 2 identical fermions must be anti-symmetric to particle exchange. The space part of an ℓ = 0 state has no change in sign when particle positions are exchanged! so the spin part of this wavefunction must be anti-symmetric. Which would allow only a total spin of 0.

18 Q=2.23 MeV The simplest fusion event to consider: forming a deuteron from a proton and neutron. but doesn’t exemplify the typical fusion reaction very well: Q=0.42 MeV Since is suppressed by its (weak) cross section

19 In the fusion of light elements into medium nuclei the coulomb barrier has to be penetrated from the outside. In fission an elongated neck develops, prolonging scission. In fusion approaching nuclei are flattened by coulomb repulsion, delaying contact. Because of this repulsion between nuclei, fusion is not a naturally occurring process on earth. In order to bring nuclei close enough together for fusion to occur requires kinetic energy.

20 Q=23.8 MeV but this energy release is larger than that required to remove a proton or neutron from 4 He! Q=0.42 MeV The sun can make deuterium only through the weak (slow) process: Once there is sufficient deuterium, could consider: Q=3.3 MeV So there are more likely products of deuteron fusion (d-d): Q=4.0 MeV

21 liberating ~1 MeV/nucleon can be performed in the laboratory with a beam of deuterons on a deuterium target. Assuming R  1.5 fm, the electrostatic potential:  0.5 MeV.

22 A (typical)  A beam (even if every deuteron struck a deuterium nucleus and fused) would produce the net rate of energy:

23 Could a hot deuterium gas have enough thermal kinetic energy to overcome the coulomb barrier and allow nuclei to fuse? With 0.25 MeV thermal KE, 2 deuterium atoms meeting in a head-on collision would have the 0.5 MeV KE needed. 1 c of heavy water D 2 O has the potential of liberating 5  J even if extracted over the course of 24 hours 50 MW! Ordinary water ~0.015% D 2 O Fusion energy available in 1 liter, ordinary water chemical energy available in 300 liters of gasoline =

24 What temperature would provide a mean kinetic energy of 0.5 MeV? By comparison, the temperature of the surface of the sun  6000 K.

25 Q=17.6 MeV A fusion reaction involving the heavier hydrogen isotope, tritium is This deuterium-tritium or D-T reaction is currently the reaction of choice in fusion reactor designs. In the center of momentum frame, the 4 He and n share the final energy with equal but opposite momenta: T n =14.1 MeV

26 From the mass difference: this should release 26.7 MeV of energy. Energy production in stars Stars form out of gas clouds which collapse under mutual gravitational attraction. The gravitational potential energy is converted into kinetic energy of the gas molecules - the gas heats up (and its density increases). About 99.9% of all nuclei in the Universe are hydrogen or helium. 1 H is dominant, accounting for 92.5% of the nuclei. Only 7.4% are 2 He. He nuclei are ~4  more massive than hydrogen. So by mass of the Universe is 24% helium. In stars the 1 st fusion processes which can occur must involving protons. The first step in the process is of turning 4 protons into an .


Download ppt "0100020003000 Energy Consumption Fossil Fuel Contribution to Global Energy Demand Year."

Similar presentations


Ads by Google