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HW1-12-13a

# HW1-12-13a - 1-4 Fundamental Concepts Chap 1 12 Dry air at...

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Unformatted text preview: 1-4 Fundamental Concepts Chap. 1 12. Dry air at normal temperature and pressurehas a mass density of 0.0012 gj cm3 with a mass fraction of oxygenof 0.23. What is the atom density (atomjcm3) of 180? Solution: From Eq. (1.5), the atom density of oxygen is N(O) = wopNa - 0.23 x 0.0012 x (6.022 x 1023) A(O) 15.994 = 1.04 X 1019atomsfcm3. From Table A.4 isotopic abundanceof 180 in elemental oxygen if /18 = 0.2 atom-%. Thus the atom density of 180 is N(IS0} = nsN(O) = 0.002 x 1.04 x 1019= 2.08 X 1016 atomsfcm3. A reactor is fueled with 4 kg uranium enriched to 20 atom-percent in 235U. The remainder of the fuel is 238U.The fuel has a mass density of 19.2 gfcm3. (a) What is the mass of 235Uin the reactor? (b) What are the atom densities of 235Uand 238Uin the fuel? Solution: (a) Let ms and ms be the m85Sin kg of an atom of 23SU nd 23SU, nd let a a ns and ns be the total number of atoms of 23SU nd 23SUin the uranium a m85SMu = 4 kg. For 20% enrichment, ns = 4ns, so that Mu = nsms+ nama= nsms+ 4nsmg= nsms nsms Here nsms = Ms is the massof 23SUin the uranium mass Mu. From this result we obtain using ms/ms ~ 235/238 -1 =0.7919 kg. The mass of 23SUMs = Mu -Ms = 3.208 kg. Julv 24. 2002 1+~~ 13. ...
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