4 - N RE 4214 HW Set # 4 — Thermal Analysis of Fuel...

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Unformatted text preview: N RE 4214 HW Set # 4 — Thermal Analysis of Fuel Elements Solutions 1. Problem #8-3 ofTodreas and Kazimi, Vol.1, p. 340. 44/1; +¢MPeW CT ...'T'c) you‘ll be Prolwa-Ma—l +0 “Ha... sum 9‘: my“! mstsl'aaces cfi-Lu 'Fue' Mal-eer 4.44.6! citadel-5L3! So +ha.’r: A... S (Tm-n 1.2:)sz _ z :00)- “1' if; (TM“TC) DC a S W“ hi- c" 2 law "EC Hm Tablet. 8-»! g. 3-2. ‘3? +1.1 'l-wc} 1‘00). = 3 ‘9 LII/1M K kUc = 2.?» W/W‘ K a .. [A] m K 21:61:19.3” l3 / T 0.01:: 0.002 ‘ A U0: : z‘ofls'b + ‘5 2-. 4'19 Ami ATUC 9.0!!) 9 5:32 W41 2. Problem #8-4 of Todreas and Kazimi, Vol. I, p. 340 8*?! +0 carved fir fuel Parasihd‘. ______&§_§.__—-—- no.“ , “.236 FL“ ' __,__o;3§__..—. {7,035 ‘4. 05,005.05) . k N ‘16 k IFo-‘B'BfiTW-k“ "' 0'8 fins; CW4 c: 2.I5 ud/w' K “(coHSIM|'_':""“"""“"It c,- O‘ 3‘“, kcoqfii uncradu-‘l ' c: 2.42 Wu K ’ l AT = . (3—113) I AT :1 I .. aw '1 2n Rah: (8 m I ARM = ‘5 .6. (Re, /I2c,c) (3—115 In) . 2. 1r kc For uncraoLeJ Fuel W 3 9.26. 44"") LA .c...o we. a: éofl K “Em = 211.“? (‘1’- s. W ATE!” = 44 x'o 2L211 [5.zaxt53—o.36X153—- 23F”? 11194-30" = 399 K ., Tf : 355.9 + 308 z eéan 'C I E] LTM ._. (963.9 + 14-47 a! For a. Parabolic +eMPcPa+que Profile [:4 q Luthhalm‘cafl f-uel elemeu‘l‘Z T 3 TM+ Rio _ ~36 Ms z ,3 lSS'TC. C9" Upland: of a Parablaid 2: 1‘5: K 194-414 4’54 x [4332”) For 'Crackcd 'F'M3l : RcPea‘hfi ‘HM abode PIOCeaL-APQ M Lanai/Doc: LO/qu/ ufi' fit}: - _ 92. LAJI'HA -2*ISWK "g 3. Problem #8-6 of Todreas and Kazimi, Vol. I, p. 341. For 'Hfu. Gem-+4:ch am : 0; T""'T’“" 15?.‘i4k;+"|¢;+kc L». U) M g. =3 Lbs —-D)/2 Subsfihi'u‘tj for vmr3m FMamC‘I‘dV‘S m +Lu- RHSJ E734!) hlfclc's 1 o ’i‘m—fl‘M = 2030.3 c. 0, l C J.- Tmo—TM .. cl) sl T; + In} L2) EM I a 5"- Z? 9.. + L (3) “VT”: 1515LE+I§ kc '~ Where, 51 + 5z. 3 D (‘1‘) 9 4% €15: S ‘6 5' (5') 1: = 1”» m M % s i g. C7 SubS'i'l‘l‘u‘l'ifj "PW \J’MFA—v: PMRW‘lCVx I‘d-4. (2345.80) M4 solufj Eggs. 2/3, W 4 ) Magi. ._ St "-3 my“ 52’ _-_- 2-8 MM GM 0 TM°_ TM 2 Zoaé\g C They-2F”!!! / "I J— A‘so/ Tj—Tw» '3 [ kc. + 4. Problem #8—? of Todreas and Kazimi, Vol. I, p. 342. ’ l AT = ’ buzz/:2 w 7) ——-)2nke ' Aqud :7 1’ ,9,“ (Rs/22) T x 2"" kc A " ._ 1 FJ». — ‘1) 211 R3 L For a; arabola‘c +em emrwu. chsb'FE ' vi Hap Fuel Mach-Braid “hm TMJ '- T‘D — "' (Em-TH) = Ji-AT‘Fuslm-J T -T a ’ .__|__._ Meg/2,) Ema/2;) I M3 '- CE E‘B‘nk‘c + z'nkj % 2.1112c fangs“ 12M «357.5 .. ‘2; i I 4.4(489/4“) m $111135” 211 r a 277 "Va 5. Consider a typical PWR fuel pellet, 9.4 mm in diameter, made of sintered U02 with 95% of the theoretical density. At that location in the core, the linear heat rate is 39.4 KWlm. The fuel material surface temperature is 610°C. Because of self-shielding effects, the thermal neutron flux, and hence the volumetric heat generation rate in the fuel pellet, varies radially according to the relation: q'”(r)=qg’ 1000‘), where K = 0.213 mm'1 (1: is a function of enrichment, pellet diameter, and density). a. Determine the maximum fuel temperature. (Include the variation of q'” with r and the dependence of thermal conductivity on temperature; assume kf to be given by Eq. 8—16a of Todreas and Kazimi.) b. How would the result in (a) change if the radial dependence of the heat generation rate is ignored? (Assume the linear heat rate to remain unchanged.) c. Repeat parts (a) and (b) above assuming the themal conductivity to remain constant and equal to its value at the mean fuel temperature. The flavors/tuna cii'FFgrcwi-lu ecbmhfi; _. ‘l a 9LT "’ _ . "” ”’ -.= ta...) + w -o More were“) In:a¢a.i'i:j.( um? 31' 1 I J lsz -= 1:; [100cm -- lofixrfl a) T‘ x} R (Oi-rs. a) . I 9’ Tim. linear lA-cad' rate : (B = . ZTTY‘ AV ’4': b L ab, 1 2.1 in 2 110m) (2) 3C (2) [n+0 0) I w r-_-.o ) M .. 15‘“ / Jha7:__:§______ [IOCKM-I] _ (3) T zirxa 1.00.) . 5 Fur “sis CM: 36: 0.2% mui' , l2: 4:"? WW. I-XR .-= [.00 ) 1° OCR)=|~ZU- I IICKRV 0-9932 9/01 T-FM . 01 T _._. 3Q“? «ire-26¢ _ __ . Ila ZTY #I.D$o.gész ‘- z'qg’ Ts Ecb.L8—Iéa) 0F HM +e>c+: 39.24. .43 i T :: + 14C) W LJZS’E ,uo (T+273) . T a]. T ~13 (T 4- -‘ fledT -: 332+ (hi—452A) + 1531-4- XJo +3.73) 0 —8.€0é3 filo - (+3 FOY T7: T5 = GIG C Té l2 dT = 35-374- W/Cm :- 3.53'7 kW/w T o 7‘de = 44.99 w/m D . 5%.(43 am he mm +0 W HM fouwué, TQM TCC) I500 chao moo [800 Jth (LU/w) $0.94!; 63.24; 4.51549 675330 JV"? ‘3'“ I It AT: 1. = 3439-4 kw/m T; 4"" 1? h d?" :: 35337Q- + 31-394 : Q; 123 kCTm_T$): 2.95! (5.) W Ta 54 Ha HM.er emoluméka a} HH m, EAL/t +em1nemiur‘e ?‘ : "rem-*1} ' 2. Trio-J'- g< Erna?“ So lud'l‘m : _ AS$ULM ‘ ¥ :7. Ugo-C. =-> k = 2.4045- Kiel w/Cu-L'C .‘. wild} (Tm—Ts) : 23-5} 2 = “33 ‘C 2.945 xfi: ." T __ 0 ! _FM.. [743 ci Aus- __ 0 (mm: T = tl7L.C;c -— .. I k (TM‘TS) '-'-' 1 = 3.1394 [cud/M U3) _.. 4'" .... -2 A‘E-‘Suuu... T 7 [2,50 ‘5 :- :- k :2~531XID w/Qw‘c Gag . ((9) Tim .ngT, = l23"! °c :1»: 18491 AM. at. $312302 .._‘ I: cow-e MWMW 6L ...
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4 - N RE 4214 HW Set # 4 — Thermal Analysis of Fuel...

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