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8 - NRE 4124 HW Set 8 Two-Phased Fluid Mechanics 1 Problem...

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Unformatted text preview: NRE 4124 HW Set # 8 - Two-Phased Fluid Mechanics 1. Problem 11-5 of Todreas and Kazimi, Vol. 1, pp. 520. 2. Problem 11—6 of Todreas and Kazimi, Vol. 1, pp. 520. 3. A 6-fi-high boiling-water channel operates at 700 psia average, 23.1°F subcooling, and 6 % exit quality. The voids in the upper part of the channel, however, cause strong neutron- flux depression there, so that the axial flux distribution is represented by "' C ~iizZH . 7E =e sm— tp H where C is a (9113an 2 = 0 indicates the channel entrance, and H is the channel height. Find (a) the height 2 at which the flux is maximum, (13) the value of C in terms of the maximum flux, and (c) the non-boiling and boiling heights. 4. A S-fi-high boiling-water channel has an equivalent diameter of 0.5 in. Water enters the channel at the rate of 2 x 105 lbmlhr at 10 fps (forced circulation), 22°F subcooled. Five Mw(t) of heat are added sinusoidally in the channel. The slip ratio is 1.8, and the average pressure is 900 psia. Neglecting the extrapolation lengths, calculate the friction and acceleration pressure drops. Consider the cladding surface to correspond to smoothdrawn tubing. NRE 4214 HW Set # 8 ~— Two Phase Fluid Mechanics Solutions - 1.. friabléifiu-‘s'bmiafi a. mam. ' no; ,5 b) Banko‘ff’!’ an? :carkela.+rm:_7' « C a: a K @ .. ' ' “'C'E'b Huff) ? ': “HEN, Op 3775' K ’2 a'Tl-I- a 09:3er (psi) - (“€443 :-'K"= ¢.1l.q-o.o.o¢1*,.72¢e 14-504- ,: 0.814 5‘05?- 9.63] l I AW5 . Dd” F‘Qx “9:25 lint? 'Cmrfc‘ffilhoin. flan? Ca 2: V03 C) ‘0“- ‘I . I _. ‘ (”II—43) Co (t) 4- “3;: CDI: {3 L'Jr ("E—13b] - I[n-4r} 2/13 D.‘ b = LEE-5) .-.- %) = 9.743 0-4!.) '7 o 4 7 ” ‘ 0.793 .. C 5 _77s ______ _ .___ . 8 D 6 [‘4' (0.775 1) 1 ID ’4' n VVJ. = (1-x) V” (115m _ 94 Far Cflwv,‘ (21..., “=0 , V” .,. [:53 0'3 (61-60] (Tab—f; u-z) V106;- 'Fw TW'ngc’ —> (7:17.13 *{53 A“ / v,» = g1Wf41. (74°F = 9-185 Ml!) ‘FW mzo VJJt=VoO=°lgs _'.. d —_-,-, _____.__].——————_——"‘ "=— 0-703 Au5_ LOSCJ- + o'lgS‘XB? J :I e .115 6-15 ,1 £933 .3 I Sign... ! 59b" A L 2. Problem 11-45 ofTodIcas and Kazimi, Vol. I. 1311.521}. ' ' Fin“ 511:4: Slut». {and fiheua 5 “law: “1" Q = G‘Chfihm— LEE-‘43 I H 315-...“ ham= hgglmfh = new: Bram...“ ‘4 " la 4mm 9:..th fin-.1" @T = . ’- _ fl . F H “(HF 63:95:. Hm ' L ., WEI-w: 1;... ___ '69:; 3 unz. me = 3-73Mu‘1hnghr (um 9’... 41? .1) (fiPfirM)m MW ‘— (flvvfi‘dBIfli-L tr- + (&P -L )Iri'ryi: :‘( D‘P PIIi'l'|"-)||":---i-:.r (b?a'mu\'1‘ '- <A PWM)r-'Jeg (EAL-t etrur) ofl'H '. lilac :7 .535. 2 8+“ Kiel“ 3. Eng saq-?—F q 13 1 fig, "-'—' D can Hs/Ibm - I? : 4:..‘15 lbw/93.3 (. alt. ens." (I‘M) E? "I' {x Q} 0': 5 9" ' I = l = b L‘Ié HELP-4 1-3 a": mun-zit. l+ T file; '1' a 1*..QUS‘L -' em“ = C1—*¢-H¢-)« —L—.—_—. .1. ImL‘lL-y ._—'—-—— = :5.” IL»; 9.01115... qu-Sf— fri £5- : $.49- 4,5 3* {5425' 32.2 t] .33 f L PP:1.)VI‘H{ d I; ) Iii-H- $31.: Hg P51 Fifi“,- : W? = 19-1 ‘lf—fi'“ CF: = [l+x(%flu)] ELM—.39. HEM { 3. IE: :: 2.593 T? '— ”(b P ulna : 3.23.5: F5: 7' GEE; (bPfif: ){flfl-r la ( Pfi£¢)fisw '. 25L; ' l H Pin. in H E Vt - H LMlaJr (”a-flaw Few «1 Is ~ Fr 3% A +rIn-E MA firm EEC-Chm ”*9 “CE-4r! 5w“ ( f3 = FE: (RE) ava‘ MR1 Mriitr' L‘Atph: -—"-—» “Fifi“: 33'9ch [hf-1; xfikl‘ +..l—-— Gem’- = L11 1311' arts . A fi-fiphigh boiling-water chant-1e] uperates at THE psia average. 23.]"F aubeeeliug, and 6 “/2. exit quality. The voids in the upper part ef the ehmmel, hewever, eeuse strung neutron-flux depressien there, so that the axial fiux distribution is represemed by — m . its =Cem sm— m H where C is a constant, z= D indicate: the channel emu-mace, and H is the channel height. Find (a) the height 2 at which the flux is maximum, ['b) the value ef C in terms of the maximum flux, and (e) the uen-heiiing and hailing heights. VIE» 7'3 4. A E-fi-high boiling-water eharme] has an equivalent diameter of [1.5 in. Water enters the channel at the rate of 2 x 105 Ihmihr at IU firs (forced circulation}, 22°F subcooied. Five Mutt) of heat are added sinusoidaliy in the channel. The slip ratio is 1.8, and the average pressure is Qflfl psia. Neglecting the extrapolation lengths, eaieuiate the friction and acceleration pressure drops. Consider the cladding sinfaoe to correspond to smootlvdrawn tubing. I Tm -= Tween e SioF :- L‘h'fi- 4%.? Bhfldibm ass-gs: mater»? trams. r. v. C9 E HU- -.: ”*3” = 1.14- mo 11:.“th Fr .W €._ ‘ 12:3?” _ 3 1 12;: W + (0.031?) 4, e5ee‘ll __ than 3 U " 9-5—51.) ear-r55? a.u1i 2.3 enema} ‘2 L115 uh: 11:: {1.55. U; = mar_f’_..;.._ .. 16 21+ fr} .. ef- Ea - ofilbr *1“: 3x15:- __3.. fl, = a sill-E? lu'" REE. ; F; U5: be e: 3 ”(.103 _._.:.. $.52 a an}: Pr ""1 ID '53me HA! hm: 1914.44 ‘FY'I‘L‘h‘ma-E [MA-m1, clfl? m 10$);qu lube—3U) we sun—tel Clea-{dc HE v.49 Smraj 1‘63me ”Share X. L; mwd 1’13 Una-n1 “nth!“ Mat-L 2. f l . a" Refi'fm 1'. m1 5; 5:— 15 , H5: = 15.53 JI: 2.5’ £23335" 5 143—: has" "£1; 335241 5-5:" 5 Ham = 1.2:” (lbw-n59“? 71.1. f ALIAGLMM]; J'LM-s T15)“ I q"? '52-!- L D ._..—.. 2.9:: 592-5 9432.4 {.5} 3.75 512.; 5-5933 23 Sun 595.: a-DV'TS' 3-! For rdaqlmii +84 WMLMWLJ W»... MME-Lf/ _ T? HE fu "(&Pfrrn) H31: (#3:: (#9: '3': £57; 1') -.- 33p aflmq-zfib:§_3 *fla-Lfl!) . £337.: 210:. unzgsllullm- : 5.173 19;. FW Ycaaq‘m 11'- HI I“ a 1-7; = {;_ym.=i2 —l.Si-°-n}1°‘mwifi-Wfl *LLcfl-Lf} Zeta-all: I: 32.2 11+? cur/n. .2. #- “13.2.?! ,. 1-! 5.1:“: r32.1;t-IH-‘+ . T? T? . (AF- ) (11p (5 TP) “r? H' = 4' P - + 5 . 5:, H3 )HB: If- H311 ( P41. )Hgm .—. at"?! + 5.15:9 +afi$3 = 1-292, P5,: . ‘11? ”(13%, 3 H = Q: PS: j”; + @an-LM, -: p.3rL-I- 1.93; = -1éw ...
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