prob.viiii-4 - fifim: a PROBLEM 9—4 SOLUTION Zane 1...

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Unformatted text preview: fifim: a PROBLEM 9—4 SOLUTION Zane 1 Zane 2 {lflfl assemblies} [3i] assemblies} Assembly 1 Assembly 2 _' e e x” Drifieing Eleek _ Since each zene prnduees Sfl‘fin ef the teta] reactor power, eaeh zune produces the afiiepewer, i.e., Q1 = and assuming eaeh assemny pmduees the same power. .01 where Nlr'meriT] = = NgrflgefiT; HI = lflfl, N3 = 30. and M} = fiT: {as given in Lhe problem statement] 1:i11 = [LE 1h; A1511: e11— = lUUfi11+EUfi12 = 1UUIi'11 +3fl[1.25fi1|} = seem :_:I ' I 1 3-Whfim the McAdam-rs relatjen can he used to express I: {LE f1 = H.13qflufl‘t1‘2 = “.134 u Hear Dereher 15. [993 Page 1 ef 3 pages thlem 9-4 Sulufiun - and am can he re~axpraaaed as: an = [1.392 p0-3V§'5D§1'2L3p1 EL; = 4m and AP: = flaa = 145E“ N’mz' = 2.n24Eaa mm3 (1} .— na fl -1-?- 3 = LEM = W F vi D: H 3.134%] 0.134(4111} {Eng up the, expression fur pmsaurc drnp across assemblies in Zone 1, where H11: orifice: flak inflame; cunuact'mn, fmuan. and expansion 105535: 1 '2 a .I V U a... mu 9"“: b+fh%(P—2fi)+faB—:(F%l] m where the: McAdam rclatinn can trace again tn: uatrl la expraaa f: a v 41.2 {a = {1134 may“; = [ma-qlab SDE and {3} 43.1 a, = [1.134 [Reflfl'z = U.134[Pvfiflh) and min = PAth when; an?) m —- _.|l ' - = J A5] f‘ 4 , rah III;I 2m marefare Vb = m a = {11.513015 kgthIhflfiflfl a] = 35; Egg migs (4) 2m:- paDE am (am kgrmi} 1mg 13% ulisfituting figs. [3} and (4} into {E} yieids: a kcp . =_' " an = _.2i _ + aaaLbPaa (RENE-jg )Ia DEW}; :— as kc? W + —_EJI_ + QDQZLapfifivg .SDiI .Epflfi Du: D2 {'23 ('5) N [ha = Gala: =:~ pawn = 0.3mm whcrcAfi= a: marefan, Va = ELEV: {1r} Page 2 01'3 pagas T‘mblem 9‘4- Slflu liun - DIWELapfi,EVg.SDe-ILEHG.E d, nflggL m0.5{fl3}1.3v%-3Dilflufli hail-[111;me Eq, (1) Ihat pU-H v5” nil-4 “1 = 2.024314, 1111:1451. Harm of Eq. (5) Lemmas 1.2416EU5L3. :Esfihs'fimfiflg [1143. given vnlLIBS [:ka = [15, k“ = Ll]. p = EU'U kgfmi 3.: = ZE-fl‘lhl‘fifif IE-IE, 1L3: 114 m, and L3 = 3.6 In intn Eq. [5} and assuming £131: £1111: 3, I {4.5} {344 144413] 145505 Mm? = —J-‘—— 2 + {1091 {414 m] {SUD kgi'm3P‘a {EH-n4 Harm’IP‘E 0% { 1 .0} {3m kqufl-filgfllf + —2L+ I.24:3ED.5 {3.4 m} 7.45ED5 Mm? = + W + W = 4.454315 Di”: Dfil‘ 13% 'Suflving fur Db gives: Db = “11215 m I11" 2.15 cm Page 3 flffi pages ...
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This note was uploaded on 02/02/2012 for the course PHY 4049 taught by Professor Acosta during the Fall '08 term at University of Florida.

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prob.viiii-4 - fifim: a PROBLEM 9—4 SOLUTION Zane 1...

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