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# 1 - l NRE 4214 HW Set 1 — First Law of Thermodynamics...

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Unformatted text preview: l. NRE 4214 HW Set # 1 — First Law of Thermodynamics Consider a four- loop 3300 MWt PWR. Each of the four reactor coolant pumps is rated at 3.5 M'We. The secondary system‘s feedwater, condensate, and heater drain pumps require a total of 14.6 We. The turbineigenerator is rated at 1126 We. Determine the heat rejection ._ rate in the condenser and the secondary cycle efficiency. ' Following a reactor trip of a fourwloop, 3300 MWt,, Westinghouse type PWR, the Operators are required to maintain a combined minimum of 400 gpm of auxiliary feedwater ﬂow {AFW} to the steam generators (SG). The AFW temperanne is 70°F while the SG pressure is 1000 psia. Assuming 100% steam quality at the SG outlet, determine the steady state heat removal capability of the 865 under these conditions. Compare the result with the expected decay heat level at the end of cycle conditions. (Table 3). Explain the rationale for such requirement. Steam is exhausted from a low—pressure turbine at 2 psia and 9 percent moisture content with a mass ﬂow rate of 4.0 x 106 lb tn/hr. The condensate is pumped out of the hot well at 102°}? and 1.5 psia. 3. Determine the heat transfer rate across the condenser tubes. 13. The circulating water enters the tubes of the above condenser at 65°F with a ﬂow rate of 1.35 x 108 1h iii/hr. Determine the simulating water exit temperature. Consider a E‘WR primary system response following a total loss of feedwater ATWS (Anticipated Transient without Scram). A few minutes aﬂer event initiation, the secondary side of the steam generators becomes totally dry so that the heat transfer rate from the primary to the secondary becomes negligible. The primaryr sy'stem temperature and pressure rapidly increase while core power slowly decreases due to the negative moderator temperature coefﬁcient. The primary system pressure increases above the shut—off head of the high pressure safety injection pumps, so that no emergency core cooling ﬂow can be injected into the system. The increase in system pressure is limited by automatic opening of the pressurizer power operated relief vaives (PORVS) and safety valves, thereby releasing a significant amount of primary coolant into the containment. At some instant during the transient the conditions listed in Table 2 prevailed. a. Determine the rate of change of primary coolant speciﬁc internai energy at that instant. b. Assuming the safety valves will be state to maintain the system pressure at 2500 psia, determine the rate of change of primary coolant average temperature confesponding to the result in "a” above. 0. Determine the rate of change of Coolant rnass within the primary system and discharge ﬂow rate corresponding to the result in part "b" above. NRE 4214 —~ HW Set #1 (Contd.) mm Maximum Decay Heat Levels Following Rx. Trip 30 min m 2.0% 1.6% 1.0% 0.7% 0.6% Table 2 Primary System Conditions Following Total Loss of Feedwater ATWS Core Thermal Power 100 MWt Pressurizer Water Level 100% Primary System Pressure 2500 psia Primary Coolant Average Temperature 632°F Pressurizer PORVs and Safety Valves status Open PORVS ani Safety Valves Discharge Flow Rate 1000 gprn PORVs and Safety Valves Liquid Discharge Temperature 656°F Total Prim S stem Volume (ineludin; the Pressurizer' 6,600 ft1 NRE 4214 HW Set # 1 -- First Law of Thermodynamics Solutions 1. Consider a Eur-Inn}; 3300 MW. PW Each of 111:1 four reactor cgolﬁnt "pumps is ratcé at 3 5 MWe.'17h¢ scwndar-y system 5 feedw 1“,; condensate; and heﬁ’tgr drain pumps regain: a tam} of 14 6 111ch The imbine/generatur 15 fateé at 1126 MWe,Determ1ue the: heat rejectian rate in the condenser and-111:2 secondary cycla efﬁciency. . . w Var a \$111513»! 53512.11?" me-‘mgw ax- €416,109“) ““08 \$1131 Lewd? of Wcmmﬁiﬁnﬁwms \$I’a‘1‘cs 1 (W m 1 w: w“- 2501 “M39111; 9? (Ga-1:1 )‘Cagaig 3'" '( w-hei’ 1541559 & ApP‘tuil‘Mj HM Prrsiv Law! ‘95? ‘Hu PIT-{NAVY \$115111»: Gm” 1'» 33m 14th 9 15954: Q 4» 110212111" 3314111111; I Cor: 2” . .. SCa " g APlem‘j HM FWSP Law) 1W H” Seams-lam? Sti\$+£wl NRC?“ .. 14- MM 1 .. w, ' a “ . 33314111111 ... 1anduur‘ W Q35, + 1 WP““P£1 M whirl: ' Qgﬂ?‘ ' -- " \$60“? 1:. 33% 4.14.1 ..... I126» ' " A «,2: 22522.1, 111% (Am) 1' .3 “as a. . .. gawk W3 n [756W‘j z: Wmeb/ QM . _ WECHZ" I4 (“j/(3314)] «3111313 .. Wpum?\$w 1414) MW? 5! 155.3 :55/ (fans) 3 w. ' " I ' , Qialﬁmwt . 113111.114” Qs¢+1wrwrs1wwco§ .— Q 1‘" iurb 2:. 13:4 4.14.1111121 2'; 222922Q MUN: (Amy) it {758 l f]: \Nnek/QM “[001“ I4 “ﬂaw-1):} 1.11m {319“mb 1411141111 33 '51 / LAW) 11 2. Following a reactor trip of a four-loop, 3300 MWt, Westinghouse type PWR, the operators are required to maintain a combined minimum of 400 gpm of auxiliary feedwater ﬂow (AF W) to the steam generators (86). The AFW temperature is 70°F while the SG pressure is 1000 psia. Assuming 100% steam quality at the SG outlet, determine the steady state heat removal capability of the SGs under these conditions. Compare the result with the expected decay heat level at the end of cycle conditions. (Table 1). Explain the rationale for such requirement. Table 1 Maximum Decay Heat Levels Following Rx. Trip 1 sec 1 min 30 min «L111; 8 hrs 48 hrs 4.5% 2.0% 1.6% 1.0% 0.7% 0.6% A . Q56 . For at stench-I Stat... I steaaltj ‘Find' Process ) i3 marina Changes in knife“ c and PaiéWHt—a CMCJ‘TJ ) ‘Hu First- Law wields ‘ Q = Y?“ (hem -' halal ”W"! W‘ = ”‘5. = V": * (400%Pmi7‘3'F) = 0o 4. o.l337 .ﬂ' 9: ' and: 4 mm; 30.2 cams "' °° I" a: 2.0 X10 “om/hr M O I. Q : 2.0xm: [IHZH _ 38.05] 3.4l3Xl0 \I‘r’ . T inplooo pit“ hf @ 70F . “ﬂue“im 'Fac'l'or {3m Btu/l"- +0 th Q =. 97.7 th a: 27. FP (AM; 0 This is c\$uivalen+ 11> elem heal aft-er Sol/minutes OF eveni- 'mzliahm. The Cambinairl‘m of HMS Ade heat removal capab3litnl and HA: .mH—a‘nl water [WUCA‘I'WJ tin 'Hu 35; Should he SuFFicie-A‘i .+o Yamove 430,“! keg-f 'l'hfoltjhaw‘i‘ HM +rans.‘ewl' £ollawntj Y‘cao‘i'or 1'er . 3/6 3. Steam is exhausted from a low-pressure turbine at 2 psia and 9 percent moisture content with a mass ﬂow rate of 4.0 x 106 lb m/hr. The condensate is pumped out of the hot well at 102°F and 1.5 psia. a. ' Determine the heat transfer rate across the condenser tubes. b. The circulating water enters the tubes of the above condenser at 65°F with a ﬂow rate of 1.35 x 108 lb mfhr. Determine the circulating water exit temperature. LP 'lurla‘Ine Eth1 4 .0 mo lb... Aw 2psu‘aJ cu zebulih Condenser '- SleaLoL-1 Shall, Slam-1 “Flwd' process; {Quart AKE’ APE 2 '-- Heat irauSFer mi}. +13 Mu. Cu'vwlai'u‘ﬁ Condzﬂsai'l- W sﬂsfem/ ch’ {5 giugm b3: 4.0x") lbw/Kw . “92°F, 1-59qu QM“ = 7“ (kinki- -- he“) = 4.09:.10" D024 .2 -7o.o] Q”: 3.82 Kit;El Btu/bur (Ans) r. c.‘«m|a+.{a water : Sieadut sf-aLn.’ steak-f ‘H‘ﬂ-O Praia»; Eaves-f AKEJ APE '2 . . [.35x103lbw/hv chz m (heﬁl-"hu'nbi\ @655 F a s q” 3.32sz = l-35'xl0 [hem —33.os7] he“ & 5.1.35 Btu/11am Tag a: 613.3 °F CAM.) 135x o‘lun 4/é 4. Consider a PWR primary system response following a total loss of feedwater ATWS (Anticipated Transient without Scram). A few minutes after event initiation, the secondary side of the steam generators becomes totally dry so that the heat transfer rate from the primary to the secondary becomes negligible. The primary system temperature and pressure rapidly increase while core power slowly decreases due to the negative moderator temperature coefﬁcient. The primary system pressure increases above the shut-off head of the high pressure safety injection pumps, so that no emergency core cooling ﬂow can be injected into the system. The increase in system pressure is limited by automatic opening of the pressurizer power operated relief valves (PORVs) and safety valves, thereby releasing a signiﬁcant amount of primary coolant-into the containment. At some instant during the transient the conditions listed in Table 2 prevailed. a. Determine the rate of change of primary coolant speciﬁc internal energy at that instant. _ b. Assuming the safety valves will be able to maintain the system pressure at 2500 psia, determine the rate of change of primary coolant average temperature corresponding to the result in "a" above. c. Determine the rate of change of coolant mass within the primary system and discharge flow rate corresponding to the result in part "b" above. Table 2 Primary System Conditions Following Total Loss of Feedwater ATWS Core Thermal Power 100 MWt Pressurizer Water Level 100% Primary System Pressure 2500 psia Primary Coolant Average Temperature 632°F Pressmizer PORVs and Safety Valves status Open PORVs and Safety Valves Discharge Flow Rate 1000 gprn PORVs and Safety Valves Liquid Discharge Temperature 656°F Total Primary System Volume (including the Pressurizer) 6,600 ﬁ3 Ykihlel- = Zero woo 3pm @ asa'F CL" Q“; ._ 2: r0 QCDYC = [Do HUJt 5/4 .D‘Isdnar‘ae mama 'Flnzd' Y‘all! g m . = ”\$44 = 1000*043368 ___ 8L6? lbm/s 3”" 71);”? so ,5 0.02728 . . C a); 4; 656-.F o Primaru.‘ ,SL‘5+EM undem+pr3 - f @ .6. ‘3 MSI-‘SIBwL: 535*““ =~_____éf.—°-—?—-—— = 2.599 KID lb,“ ‘1 . C 1:544} @asa‘F : . S Oat-I 4 *0-02539 _ Usage.» G459 ._ 2 ° :78 _ \$94.2 Bin/"aw LG“: h€@ 6937:}- t mucus?“ '9!ch 'FruM (”of 91’.) JC.» 33+“ _ﬂ= 2 via- _ n‘a. =o_81.b7-— Hallo... dt 'Inlel» m gawk em ——2 '3— . Embrﬁa E1>MHM (un1¥wm Shh) MV'FWM 'Flmr ; m-cafgﬂalc Ghana: u'ux K51: PE) W—Jdu / '=' lye.“ H T?“ + ”5%“ (-13% M dqsysk... __ ' ' . dt _ Get-ore "" Mex-«4 hex-9+ — “5.153... 3% dM - 31 UL L“ SWUShh/Lhnj +W cTi — —. me“ (cszu ua-ta' CE nth ) du M . M ’3?“ “ (Deere "' may} Chad- "' usqsl'w‘) - M alumna... ___ too * 348K104 _. aha (101.4—4543) cit 3600 .ca ‘F he“ kfcaése. -— +9.0“: X104 Btu/s —' 4 - dusgsku- ._ 4. 9.046210 = °_34‘2 BW/5-\bm(AMS-) . .. dt 25%wa . For 2? t: 250? [>514 @ TMS = 4,32 F umﬂw = \$54.2 aha/lb.“ 5 6 ° 2555* Ill-1140.02 (. Q T 5: G36: F usqg,“ ‘72..“ —- _____:7___’_g___._..——— : 54on BIN/LL... (all N 660‘; —- 65—4-2 t “59 BW/.F° ”PM .nu— T y 63!. __ (,32 p=25caop£m =20” ' Elly-[ﬁtn- dT u -. (Bu Shun _ 3' g———-- ° ddéqs‘m ' {ii /3? + 31‘ at _ = 0.2.2. °F/\$ec c: 792. ”Why at [.53 (Am .) O AsSUL-M-Ivtj Hm: Prlﬁar-T SLtskM +0 he Tiahi" 0.. _— I- 5—- ...— o A; _. _ .— Pt .._—- :é - -- 5 =- d“ - ”"7“0 1: My (772) ”0m /5 (Aux) ‘3 ’- 0-0253? 63(- -632 3 9 DIM/Luge mi... = —/U'C_i_i% = 0.0253741524 =3.8él-F+/5 N ‘7338—P‘M Aug. _ _5.475K!O§‘bm/hr = —-152.l ...
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1 - l NRE 4214 HW Set 1 — First Law of Thermodynamics...

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