prob.xiii-1 - PROBLEM 13—1 SOLUTION 7f“" Heated...

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Unformatted text preview: PROBLEM 13—1 SOLUTION 7f“ " Heated Channel Fewer Limits _ -_- , Operating and Design Cendi‘flem (Example fl-I: Nuei’eer Systems, Vef I, Thrive-:13 and :1: 1:: ' Karim!) Pressure = 15.5 I'vfl’e (Set. Temp -— 344 9T} -- - h;___ mm Temp = 235°C Exit Temp = 324°C Fuel Red Outside Diameter: 9.5 mm Clad Lhiekneee = 13.5"." mm . _. .5313 = 1103 mm Active Fuei Height= 3.66 m Number of fuel pies - 5D,?52 3.;- - Cere flew rate = I'M Mye " {m = average mess flew rate per channel per ealcuietien On page 532= £1.34! kye] | ' I' Hear Ekerwfer mini Mex-me! Preperffes (Take as Cflflfl'mif with tempereexre) I. - {Example 134) e; _ ._ _ k,-—- enema kwfm-“C k“ = 0.01335 kwr‘m-"C ' h —- 34.»: kwfmiee hg = 5.? hermiec '. T It? {Exempie 1.3-1} I; " " ' Cpifiweter} -= 5.453 Lifkgm “— i} The eeelent exit temperature is to remain euheeeled Frem Eq. 13-11, J 1;" _ T {a} T — E11“ L email; (Talc: L=L} *:_'§ _ m “'meP e L “5 ' g Free: the given eendie'en, TM-lt : TmliL) {344.E'DC ': 2_ 7'— -_ " Hence, eelve fer :1"; from Eq. EH 7, Per 3*. = U: ._T'——':' T L _ . . m ;__ . q'n = #mfim: %-fl341 6.61} 1i]J -.1r =43flkwfnl ' L - .. . . . :1: _ :__ binee {1(2) :5 given by eh {I} = ::['fl ~eee[[—zj, the pewer generated frel'n e red 15 as .' - f01|ews:{FerL.,=L) . Rev Febrile“,I ‘11. 1.993 Page 1 Of 3 pages Problem 13-1 Selutien q‘n~L-2 _4eet-3.ee-2 ' 31: IF = [114Tkw era It q : Lyle ureeetzfzfiz — Lets tel-re thi e e as the average premier of me eere reds. Then, total power is q .‘x {it of reds} —' l ISIAHr kw x: SD95: QM = rm J'rfw Hence . I. {if we take the q as the power at the hot channel and etmeider the radial Fewer 5'1- distfibufiem Qmmmu be lower.) a ii) The maximum clad temperature remains below set. temperature. '- _ F rem Eq. [3-22, "1' i " L . rm 1 IE '- "1"~ =T- + ' Sin—+1 + erre— -.-._-_ Le{.z} 1|:l. q {I f L } zflcuh ' 1. 2m Lh ." '- and the max. Tm occurs at zi =—tan'1(__£—-} __" "' '— . fl? mCPrt "I. - 'i. t. e. - e-H- . 15-3415} "i' i.e., z. =J_fifitah']{fi_.fl3—6-—}= 0.65m I; t we rr-fl341-5.E:fl New find [ll-n from Eq. 13‘22 for z = 3,... and Tin-{2t} = 344.9 “C1 frem the given conditions, .- . 344.[ —256 see-(1+ ein'—:r} eee—' It 355 3.1515 ‘5- 1r~fl341~5fifl e.e.5-1e"’ 34.0 Therefnre, 3 q = q'u -—- L = White! :7: Thus, 12mm = .7115 ~ 5995.? = 3954 EH»- Page 2 of 3 pages Problem 13—l Selntien fit] The fuel maximum temperature remains belaw the melting temperature {2491] “{2} From Eq. 13-2? and Eq. 13-213I -v -L W” e E' M; 1 13-21 111"”—rmwfimepx‘m'fu'mfl[mmgh+ ink“ +lrfl415+ml> FED' } 11.:itan'i{_____—L_———-—} (13:1. 13-23} ' 1T - 1 1 Re 1 1 Hum: [——+~—-—1n{——}+ +—-—] Peer Mk: ed eneghg lrdimh and at z '— 21-, we have the max. TEL, Rui=4_[a mmiRg=mi-Ef144.lS—UDSI2=4.14mm . -' ea 3 - = —3 66 ten" {__—_————-}fl _ = flfllfiw J‘ r: nxu34lxfi.6xi3fi?9+1.4T+fi.?4 + e935] Salve fer q'n frem Eq. 13-2? and asking TQL {If} ; CFC. Hill! I T Z,_. , Cl‘e ‘——-——--—-——"“————'14fluu 235 = 45.9takwfm . a.u13 EEG-{l +ELI1mv-5'F} “flu ______—. 9R5 1.4T+ .?-'-1-+ 7 n-HJH-Sfifl “5": are “m + '5 m 9] Then, 2L 2- "nee Qm =q-U.—-5aase=45.ae. “ -5ease=54seMw II fl? Page 3 ef 3 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.xiii-1 - PROBLEM 13—1 SOLUTION 7f“" Heated...

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