Exam1_F11_solns

Exam1_F11_solns - KW NEEP 411 Exam #1 October 215‘, 2011...

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Unformatted text preview: KW NEEP 411 Exam #1 October 215‘, 2011 ( 2/5) l. A UOg-fuelled LWR has a heterogeneous cylindrical core containing 60,000 fuel rods. At full power, the maximum thermal neutron flux is 5 x 1013 n/cmZ-s. The fuel enrichment is 3%. The average moderator temperature is 300 °C. The fuel pellets are 8 mm in diameter. The core is 3.5 m high and 3.5 m in diameter. Neglecting extrapolation lengths, calculate the total thermal power produced in this reactor in MWth. "ME ML/kthmP sen/demo Tom WMM. mega mm mm: Mamas; 0F: (WNWGFMWUS FVeL Q/ho 35‘ 63WMS€0 Br gem aw 01—30113 0?. (4'3053. usmor w mm, Q = olmmAL—l " L _§ J‘ 3 a“ _ z: “I: 1‘ {0.00%} =§.oz7xmm‘ “2019,0th " AS” 4 431. (4: if..« "I @h — Gé'jc 1000 G. —; [31) He\/ 2 = u 0’; L3 0 10 3 1C a; ,_ A3. VWCC; - (a 1x (0103\[0i8’X’ISM/Ol) /cm Nfi: '" 14% “CM 235' 20 r; A/ - 7 045’110 CM \ ‘H A ‘12 ,2 ’ _ 2580!? ’3 ijbb r15 ’ 0L, Z T 573 3 ~24 7.. m '3 z : max/rev 704mm W numb 0M WOW] PM 3 w M M ~13 M1? to c — ZZEUCHDIY W/cw.s >< “AND A >< 3 {40¢ w‘ MM) 3 z 3729 A .A’ ~ MM) 6316 = 0.2% (amuseva strut»)ng a» 1 Vi \ok 213% NW ...more workspace for Problem 1... ( 303 2. The steady heat conduction equation in cylindrical geometry is 1% d1”) q"'(r) k An actual U02 fuel rod experiences some “self—shielding” that is, it is a sufficiently strong absorber that the neutron flux experiences a small depression as neutrons move into the fuel. The result of this is that the heat generation rate is not really uniform. Instead, it can be approximated as a parabolic shape: HI m “I L 2 q (r)—qo[1+a[RH Here is the power density in the center of the fliel rod and a is a dimensionless parameter that measures the strength of the self—shielding. ( a») (a) Assuming the rod’s outer surface temperature is known and equal to TS, find an expression for the temperature profile in the rod. ( ’5) (b) What is the temperature difference across the rod if it has a diameter of 1 cm, is 300 MW/m3, a = 0.2 and the rod’s thermal conductivity is 2 W/m-°C? (T) g (c) How much larger or smaller is this than the case without self—shielding, where the power density throughout the rod is uniform and equal to (l ~I— a) ? n, 1 \ pk all" A h f; (x r Ar Var) if] V V H/ r3 __, ér(r:0320 r - gfl(r4fl : ~Z0 r+c<“21 gus‘ air at k R O 0 H, l o< oLl‘ V a ,. r A 1 ~ Z): (I + 4 at} ‘1 air a \c 3 "’ r (39’? Q”D(:+§:Ev)flirjmtre 5 AT — b 2— 4 e r \7 (:1 Z O< r " 2 “up: o<r4\)1\:_°(7+,7; _ D .——\ + "-n.‘ e—H r k- \‘i h : Ha 7' "I Z Z t. newer/wows) 1 mg} kg (Wis : {a} wmmr sax» 5H’(9’L0m)(r) 0 A , -,~ \(16 C. \L T5 T1295 mm- mm saF-SH+-ym»w(r IS SMA’LLLAZ, ET A mm ,, N; If} ML \25 7., WWI/WY. LL (3D 3. In our pressurized thermal shock problem, the heating from neutron attenuation is more distributed because not every neutron interaction is an absorption reaction. Scattering interactions result in deeper penetration than is conveyed in a simple exponential decay. To model this, let the heat source be represented as follows: d2T dx2 = j]:— (m) emf—ADC) Here the presence of the linear term, ,ux, results in a less severe attenuation as we move from the inside (x = 0) to the outside (x = t) of the vessel. (7,0) (a) If the downcomer water temperature is T w and the heat transfer coefficient is h, find an expression for the temperature profile throughout a vessel of thickness t. As with our previous approach, assume the outside of the vessel (x = t) is insulated. [0) (b) Find an expression for the average temperature through the vessel. (H (c) For t z 20 cm, k = 20 W/m-°C, h = 50000 W/m2-0C, # = 0.2 cm“, = 0.2 MW/m3, and steel material properties E = 200 GPa, v: 0.3 and a = 8 x 10“6 C4, what is the thermally—induced hoop stress on the inner vessel surface? You may find the following integrals useful: Ix exp(—ax)dx = — 3E exp(—ax) ~ ~12— exp(—ax) a a 2 No? V JXZ CXp(—aX)dx: x eXp( LIX) exp( ax) 23€Xp< (1)0] ULM)®)_.. o (a) ’ fl 7 TD 5tMvL1f—‘T W INTFCAW} \T‘ \3 Human. n USE wre- wnztmzw Sussiwww ’U': 41% r -\ N a “I ..'V“ -V" ’ @235 O'qxe'o’de: _“gveai1r—“Zi[-ve_e]+c\ JAG wIk 41K Ill Z “fix * x 4;: :1 D Lax e + a 41 3 + 6‘ 05¢ «’11 __ C is DWWMW W 0: :o M‘ xat (Mum SUD-woe) + d’“ (u .— O A; ...m0re workspace for Problem 3... (u _ ~ #426 Tm: 21': gveercV” (WfiJr‘C firm H; at“! '\r~..’\/“ Jr “"161+C _\ Z" i~1f *C ~c “CHMPQC’ V Z - path 5. "' “fix ” x "7* TU\ " Q “ 2)“ iqx e + 26 fl +— (‘1X3(\+’7+\ c 1 2' 412’?” g 39’? woe-V GYWKM‘W 0"] f'wwws i’M’e’ F69. “LT/AX {W W \fi'LLlSIWéSS "NWT \3 ./.\. more workspace fog Problem 3... L0 To FWD mm mm m: HEW W bxmww BEWWU "m; Lam/L [25W WW WVW/W/w, um 'F—W "MI? Vmckus wry “MK II/ ’ TOW“ 1 T Jr 12L Lb (“th6 “j x:o\—~ " *7 “1 W —\ 4 W I": (9.2xloé" “O (“my VQAAM uric“, 47:02.. CM : ZOM/ k’zow QC) Zo \ M3 -. a ,o- m V C “Howdy ma “(109703 -T « fir-75 4 4 LZDW 3 W \_,/'\/\——- ,. 2- {1 o.ZZ7\ ~l.>032. 0'13; 1 W 6 —).033/% Tbcrauw-fi: 1 “L73’C, 3 ’c " o .. gx (T~f\ fl ZUOXIO M?“ (3))“ 06:8(Z7SC> \_ (“LIX/VIPA. 0; : " “f 0’) W \——U (7/0.) 4. A slab—like nuclear fuel element is sandwiched between cladding plates of different thermal conductivity. Power density is uniform in the nuclear element, and there is negligible power produced in the cladding. Water flows on either side of the cladding. The water temperature on either side is the same, but the heat transfer coefficients are different. The geometry of the problem is indicated below: kF kL kR t E t hL hR Tw Tw I» x (to) (a) Outline the procedure for finding the temperature profile in this element, including the fuel and the cladding. This should include the shapes of the temperature profiles in both the cladding and the nuclear fuel, and any boundary or interfacial conditions you would use to find any undetermined coefficients from the heat conduction equation. (w) (b) For the case of kp << kL, kR with [Q < [CR and h < m, sketch the expected Shape of the temperature distribution in this element. (60 one PriSSWa? or: mrwr Gmmmw m we CLJrDDzoG Awe WE » I mi wth mm GLWWW to W FUEL Mews we um UU A , WWL/fiw/e’ WOW/ES w M cmonoe MAFWBOLLC SlHrS’LSU It) We WEL- . : TLLM : C\ 4/ CZX W Z / Tam 63 -l« C4x TPLM = ~Z% + C§X+Cb L WWW W St! our/uer umrfiugm’ mth MUST Be“ ofimwéD W T I ,. . A .. 3 A ; R [(L if; {X :4) : ML LTLUL—(A - Tool / [LW—Jfl \Pix 1") \4 “‘1 (1:0 2 L if (“’63 ; Tmuzellrlgxef‘fl) L. 032 F fix A otil l \v‘ - kl ti : \LF ‘ M M __l \C (Mai ({:Z++i): M ((:Z+*Jl—-Tw\ E... g; s?— , 7 ...more workspace for Problem 4... (M 0mm NE mems mm, “MW” \8 W WWI/W, Man‘ch "m "M? L5H- 30 “NYE mon w e’Wetm) “17> 8? 5mm rd I (Pm/me 3mm LET/r OF Mum WK — «m “MT .3 ...
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Exam1_F11_solns - KW NEEP 411 Exam #1 October 215‘, 2011...

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