prob.iii-5 - .FaeBISept 2 22.031 3 22.312 ENGINEERING OF...

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Unformatted text preview: .FaeBISept 2]. 22.031 3: 22.312 ENGINEERING OF NUCLEAR REACTORS _IIJ1..'! ::: .3- . Problem 3-5 and 3-6N: Due September 221 1993 .I-r'__ ulZiI _-‘i'i"'l:'-: Case I] At shutdown Ihe tieea},r heat power is 4:er to twe contributes: a) ene—halfef the core that has been irradiated fer 13 months I‘ .. ... - ML; I. “+- " ' h] une-heif 0f the cure that has been irradiated far 36 menths Assigning half eere power te eeeh nf the cure halves the deeaja' pewer is new readiiy calculated as: Pm: P+ 1 a g 5.066 (him: (ta1+Tal)_GJLJ.+ .3. D. {166 (“Em-J': (’Eas 4" fag—0| L] ' I ) where tas is the time after shutduwn [l min, 1 hr. 1 41. etc.) and em, tag are the irradiation times of the 13—month—halfand 3&-menth~hait‘. respectively (thus: em=13 months, 1:03:36 mnnthe}. PW I 2.56 4.97 [JESS 1.23 Page i of 5 J EtfiBr'Sflpt 2 I Case 2] ' At steady-state ennditiens and under the assumption sf continues refueling, the rate at Fl ' which the fitel is replaced is pmpartlnnal ta , Where Pu is the rated power 313W TM —‘ Pu Tm. - 1hr is M‘Wth and Tarts: is 315 months. Fer example: an amount of fuel that provided— replaced per each hear at the cycle. By the end of the cycle the entire care will be replace :1. e e' than”:h Emma: "Cassie -———-tF-'l—"'—"l_—————-—+—-—7 W ‘tfl With reference to Fig.1. the rtletrajpr pawer of the fuel in the care can he then calculated as: that“; A!" L r _ 515.. P- P = Jr Pu Elli, EDIE-la. { {it + (fir + (Kayla F t‘) I: Ell This”. I. I I .3 £33“* (£35 + (tailligfl an? CUE, H.135?— Dfiflé .Eu (— as +- whieh pmvides the fellnwing results: _- a...- Time after shutduwn PtMWJ 31.0 32.2 14.1 4.26 (1.?63 I" 1min j 1 hr 1 day ‘F 1. Page! of 5 ...
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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.iii-5 - .FaeBISept 2 22.031 3 22.312 ENGINEERING OF...

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