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Principles of naval engineering - Historic Naval Ships Association

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PRINCIPLES OF NAVAL ENGINEERINGAs stated previously, naturally radioactiveisotopes decay bythe emission <strong>of</strong> alpha particles,beta particles, gamma rays, or a combinationthere<strong>of</strong>. In the case <strong>of</strong> induced nuclear reactionsthere are many other phenomena which mayoccur, including fission and the emission <strong>of</strong>neutrons, positrons, nutrions, and other forms<strong>of</strong> energy.^CONSERVATION OF MASS-AND ENERGYThe conservation <strong>of</strong> energy is discussed inchapter 8 <strong>of</strong> this text. It now becomes necessaryto consider mass and energy as two phases <strong>of</strong> thesame principle. In so doing, the law <strong>of</strong> conservationbecomes:(mass + energy) before =(mass + energy) after.Fundamental to the above and to the entiresubject <strong>of</strong> nuclear power is Einstein's massenergyequation where the following relationholds:wheremCE = energy in ergs,M = mass in grams,C = velocity <strong>of</strong> light (3 X 10^° cm/sec)Mass and energy are not conserved separatelybut can be converted into each other.Several units and conversion factors whichhave become conventional to the field <strong>of</strong> nuclear<strong>engineering</strong> are listed below.1 ev (electron-volt = the energy acquired byan electron as it movesthrough a potential difference<strong>of</strong> 1 volt1 Mev (million electronvolts)= lO^ev= 1.52 X 10-l6BtuFor detailed information on nuclear particles, referto Samuel Glasstone, Sourcebook on Atomic Energy(2d ed.; Princeton: D. Van Nostrand Company, Inc.,1958).1 amu 1.49 X 10 erg1.66 X 10 24gm931 Mev1.415 X 1013 BtuNUCLEAR ENERGY SOURCE1 amu (atomic mass unit = 1/16 <strong>of</strong> an oxygenatom (by definition)It was previously stated that the atomicmass number is the total number <strong>of</strong> nucleonswithin the nucleus. It can also be said thatthe atomic mass number is the nearest integer(as found by experiment) to the actual mass <strong>of</strong>an isotope. In nuclear equations, the entiremass must be accounted for; therefore the actualmass must be considered.The atomic mass <strong>of</strong> any isotope is somewhatless than indicated by the sum <strong>of</strong> the individualmasses <strong>of</strong> the protons, neutrons, and orbitalelectrons which are the components <strong>of</strong> thatisotope. This difference is termed mass defect:it is equivalent to the binding energy <strong>of</strong> thenucleus. Binding energy may be defined as theamount <strong>of</strong> energy which was released when anucleus was formed from its component parts.The binding energy <strong>of</strong> any isotope may befound, as in the following example <strong>of</strong> copper(ggCu"^) which contains 34 neutrons, 29 protons,and 29 electrons. Using the values givenin figure 24-2 we find:34 X 1.00894 = 34.30496 amu29 X 1.00785 = 29.21982 amu29 X 0.00055 = 0.01595 amuTotal <strong>of</strong> component massesLess actual mass <strong>of</strong> atomMass defectConverting to energy, we find:931 Mev/amu x 0.61093Mev, or 560.8 + 63 == 63.54073 amu= 62.9298 amu= 0.61093 amuamu = 568.775838.9 Mev/nucleonThe relationship between mass number andthe average binding energy per nucleon isshown in figure 24-5.Since binding energy was released when anucleus was formed from its component parts,it is necessary to add energy to separate anucleus. In the fissioning <strong>of</strong> uranium 235, theadditional energy is supplied by bombarding thefissionable fuel with neutrons. The fissionablematerial absorbs a neutron and is converted618

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