Feedback Reactivity Effects due to Fission Product Poisons

Transcription

1 SUPPLEMENT TO CHAPTER 11 OF REACTOR PHYSCS FUNDAMENTALS This supplement reviews the text mteril rom slightly dierent point o view. t lso extends the mthemticl description outlined in the text. You should be milir with the text mteril beore studying this supplement. Feedbck Rectivity Eects due to Fission Product Poisons Buildup o odine nd non Bulk non Trnsientnon rium nd Other Fission Products John L. Groh

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3 non & odine * TnkModel or non nd odine Production nd Removl The igure opposite cn be used to construct most o the reltionships described in the text. Numericl vlues in the igure re typicl CANDU vlues, but they do dier little rom sttion to sttion. Tble 1 gives up to dte vlues. Numericl estimtes in this supplement generlly use the old vlues, or consistency with numbers in the text. TABLE 1 PARAMETERS l = s 1 (G.E. Nucler Chrt 1996) l = s 1 (G.E. Nucler Chrt 1996) s = 3.5 x 10 6 b= cm 2 (New Trnsent vlue is cm 2 ) g = 6.3% (New Trnsent vlue ª 6.4 % or equilibrium uel & 6.3% or U235 issions.) g = 0.3% (New Trnsent vlue ª 0.6 % or equilibrium uel & 0.24% or U235 issions.) S ª 0.1 cm 1 (resh CANDU uel) S ª cm 1 (equilibrium uelling) is burnup dependent F.P. = ncm 2 s 1 (uel lux t ull power/equilibrium uelling: BNGSB predictor) F.P. = ncm 2 s 1 is convenient vlue or clcultion, nd close enough. time constnts or inl = ull power lux (or equivlent hl lives multiply by ln2 =0.693): s + l Ê Á Ë ˆ 1 inl ª 491.min utes [ s ( l l )] 1 inl ª 53 7 utes (hl time 34 minutes). min (hl time 37 minutes) 1/l = 569 minutes (hl lie 6.6 hours) 1/l =790 minutes (hl lie 9.1 hours) 1/(ll) =2032 min = 33.9 hrs (hl time 23.5 hours) To convert rom number concentrtion to mk worth o non135, tke 1 mk ª toms * Clculte the rction o mss 135 ission rgments tht re non nd the rction tht re iodine. 6.6% o issions produce either non135 or iodine135. Direct productionby ission is 0.003/0.066 = 4.5% non nd 0.063/0.066 = 95.5% iodine The newest numbers in the tble give bout 4% non,96% iodine or resh uel nd 8.6% non, 91.4% iodine or equilibrium uel.

4 * Equilibrium Stedy Stte Conditions or non nd odine Show tht the %oproduction o non once equilibrium is chieved is lmost 95% rom iodine decy nd 5% direct ission production. Once equilibrium is chieved, iodine decy =iodine production so the sme per cents pply s the ission rctions bove. * Show tht the removl o non t norml ull power lux conditions is more thn 90% by burnout nd lmost 10% by decy. Clculte rtio o burnout to totl removl: [Ns/(Ns + ln)] = [1 + (l/s)] 1 l/s =2.09x10 5 /3.5 x cm 2 s 1 =6x10 12 cm 2 s 1 nd tke ª1x10 14 [1 + (l/s)] 1 = (94%) (text vlues o 90%/10% use much lower lux) * Develop ormul or equilibrium iodine concentrtion nd show tht equilibrium iodine concentrtion is proportionl to stedy stte lux. Equte odine tnk inlow = outlow t equilibrium to get the text eqution. Solve or N to get eqution 11.3 in the text; N eq = g S /l * Develop ormul or equilibrium non concentrtion nd show tht the non Lod t equilibrium is nerly lux independent or high lux rector Equte the two inlow terms in the non tnk to the two outlow terms to get the text eqution Solve or N to get eqution 11.4 in the text nd rerrnge: N ( eq ) = l + ( g + g ) ( g + g ) ( g + g ) S s = S S = Ê ˆ l Ê l ˆ s s Á1+ Á1+ Ë s Ë s For >> l/s this eqution is the sme s the high lux eqution in the text. * Wht power level constitutes "high lux"? l/s =2.09x10 5 /3.5 x cm 2 s 1 =6x10 12 cm 2 s 1 ª 6% ull power lux. High power is much greter thn this, perhps 60% F.P. or higher. Above thispower equilibrium non rectivity worth is lwys close to 28 mk. 4

5 * Determine the reltive concentrtions o iodine nd non nd use equilibrium non mk worth =28 mk bsorption to clculte the reserve o non stored s iodine (odine Lod). N N ( eq) ( eq) = ( ) ( g l ) + s g + g l ª 0.95 ( )/2.93) = 12 Equilibrium non is worth 28 mk, so the rectivity bnk o iodine is ª mk = 336 mk worth. Lower lux gives smller number. (N eq µ) * Buildup o odine nd non to Equilibrium Describe the time behviour o odine s it builds up. Describe the time behviour o non during its buildup. At ny instnt, the rte o chnge o iodine concentrtion is given by the dierence between inlow nd outlow to the iodine tnk. Similrly the rte o chnge o non is given bythe inlow nd outlow dierence in the non tnk. These dierences give the dierentil equtions 11.1 nd 11.2 in the text. These cn be solved or the initil buildup o iodine nd non. NTAL BULDUP The time dependent expression or the initil buildup o 135 in resh core is: t N() t = N eq( e l ( ) 1 ) (1) N ( eq ) = g S (2) l The time dependent expression or the initil buildup o 135 in resh core is more complicted. t cn be written s two terms: È l N l t ( eq) t ( ) N t N ( e ) N ( e ) [ ] t l s eq eq ( ) ( e ) l l () = ( ) 1 Í ( ) 1 1 (3) ÎÍ s N l l ( eq) ( g + g ) N ( eq ) = s Ê l Á1 + Ë s S ˆ (4) Equtions 1 nd 3 re grphed in the text s igures 11.2 nd 11.3, respectively. 5

6 The ormuls bove clculte concentrtion in number density (toms/cm 3 ). A conversion rom number density to mk worth requires tht the lux nd mk worth be speciied. The ollowing give results tht gree, within roundo, with Ontrio Hydro s CANDU rector non Predictor equtions used by the Bruce Sttion opertions st. 1. = F.P. P where power, P, vries such tht 0 < P < 1 nd F.P. ª cm 2 s mkrequires non135 toms per cm 3. Using these vlues nd prmeters rom tble 1 into (2) nd (4) gives; N (eq) = 322 kworth o iodine in reserve. N ( eq) mk P = P = 28mk Ê 006. ˆ Á Ë P (5) This reltion or equilibrium non is grphed in igure 11.4 in the text. BREF DSCUSSON OF THE EQUATONS The irst term in (3), tken by itsel, shows the sme time dependence or non buildup s or 135 buildup in (1). The second long messy term is smll correction or 135 holdup. non comes mostly rom iodine decy so must wit or iodine to be produced. During the irst ew iodine hl lives the holdup term reduces non concentrtion little. The holdup term is zero or t = 0 nd disppers gin ter ew iodine hl lives. The squre brcket is numericlly bout 0.09 or ull power lux nd the pek vlue o the product o time dependent brckets is bout t 2.2 hours, so the mximum reduction is pproximtely 28 mk ª 1.8 mk t 2.2 hours. 6

7 TRP FROM FULL POWER EQULBRUM CONDTONS The eqution or the time dependent non concentrtion ter trip rom equilibrium stedy stte is simpler thn the one or 135 buildup in reshly uelled core: l t l l t l t N () t = N( eq) e + N( eq) { e e } (6) l l l The numericl equivlent, in mk, is t e { e e } t l t l t () = By tking time derivtive o N (t) nd equting to zero one cn solve or the time rom the trip to the pek o the trnsient t pek = l 1 È l Í 1 ln l Îl l l È lní1 + ÎÍ ( l l ) l N N ( eq) ( eq) (7) Using Tble 1 nd equtions (5), eqution (7) cn be written numericlly: t pek (hours) = ln[ /(0.94 P+0.06)] This gives t pek = 11.1 hrs 0.8 hrs = 10.3 hours. For very high lux rectors the time to the pek pproches 11.1 hours. For trips rom lower power it tkes less time to rech the pek. The size o the pek t 10.3 hours ter trip rom ull power is, using (6): 28 mk [2.93/( )] 322mk { } = = 151 mk Asimplernumericlmethodcnbeusedto estimte the pek size. At the pek, non decy = iodine decy, so l (N eq e l t )=l N pek,giving,using eqution (2): N pek l = Ê e Ë Á g S ˆ l l proportionl to preshutdown lux. t pek l This explicitly shows the pek size is Numericlly, N pek =(l /l ) 322 mk 2 (10.3/6.6) 7

8 The estimte o the non pek size rom this gives N ner 152 mk. 8

9 * non Trnsient Behviour ter Triprom Full Power. time nd decision nd ction time poison override The net non production rte immeditely ter trip rom ull power is, (l 322 mk l 28 mk) = ( ) 10 3 mk/s ª 0.5 mk/min. (ln gives the number o decys per second). Buildup continues t this irly stedy pce immeditely ollowing the trip. The rector control system (djuster rod removl)cndd + 15 mk o rectivity to oset the loss. Approximtely 30 minutes ter the trip the non lod will hve incresed by 15 mk (30 min 0.5 mk/min), mking recovery impossible. This time intervl is clled the poison override time. Since removing djuster rods tkes tlest5 to 10minutes, thetrip must be reset within bout 20 minutes, the decision nd ction time. n most cses investigtion o the cuse o the trip tkeslonger thnthis. Most trips result in ull rector shutdown withrestrt in bout 1_ dys, ter the iodine is gone nd non decreses rom the pek to ner its equilibrium vlue. * rium Use the tnk model to ind equilibrium smrium concentrtion. Show it is independent o lux. The sme methods cn be used to nlyze Promethium nd rium buildup to equilibrium nd ter trip. Here we will, but simply present some results. PROMETHUM & SAMARUM EQUATONS Promethium nd rium obey the sme set o dierentil equtions s odine nd non, only the prmeters re dierent. The dierentil equtions nd initil conditions in resh core re identicl or 149 nd 135, so the equtions or initil buildup o promethium re: t () = eq ( ) l ( ) 1 N t N e with (S1) 9

10 ( ) = g S l N eq (S2) 10

11 TABLE 2 l Æl = s 1 (G.E. Nucler Chrt 1989) l Æl = 0 s Æs = 4.2 x 10 4 b= cm 2 (New Trnsent vlue is cm 2 ) g Æg = 1.13% (Nuc. Theory notes) g Æg = 0 S ª 0.1 cm 1 (resh CANDU uel) S ª cm 1 (equilibrium uelling) is burnup dependent F.P. = ncm 2 s 1 (uel lux t ull power/equilibrium uelling: BNGSB predictor) time constnts or inl = ull power lux (or equivlent hl lives multiply by ln2 =0.693): Ê Á Ë s ˆ 1 inl ª 72. 6hrs = 3. 0dys (hl time 2.1 dys) [ s l ] 1 inl ª 1375hours = 57. 3dys (hl time 39.7 dys) 1/l = 76hours = 3.2 dys (hl lie 53 hours or 2.2 dys) Here is the generl solution or rium149 buildup: s ( l ) ( ) ( ) s l s 1 l Ï g S 149 Ô s ()= t Ì( 1 e ) e e s Ô Ó t t t Ô Ô (S3) When <<l/s (very low lux) the buildup depends on the slow burnout rte. g S s t ( 1 ) 149 ()= t e s When >>l/s (very high lux) the buildup depends on the decy rte g S l t ( 1 ) 149 ()= t e s Neither o these lux conditions is correct or CANDU t high power. 11

12 For l /s = c = / = cm 2 s 1 ( vlue pproching ull power lux or typicl CANDU) both the numertor nd denomintor o the second term in S3 re zero. n other words, typicl CANDU t ull power hs conditions such tht the burn up constnt ( = s 1 ). is bout the sme s the decy constnt ( s 1 ) When = l/s, the eqution requires L Hôpitl s Rule to evlute it, giving l t { 1 ( 1 l ) } g S 149 ()= t + t e s n this eqution, by tril nd error, builds up to 97% o its equilibrium vlue or lt =5.307, giving buildup time o 282 hours or bout dys, pproching the 11 dy buildup time in very high lux nd bout hl the time required or buildup in 10% F.P. lux. As long s the lux is ner c,i.e. in situtions where roundo errors in (S3) give poor numericl results, the eqution with l = s cn be used. EQULBRUM SAMARUM Setting equilibrium smrium149 production, 100% rom promethium decy, equl to equilibrium 149 removl, 100% by burnup, yields: l N = g = N s S so ( ) ( eq) ( eq) S N ( eq ) = g s (S4) Equilibrium smrium is the sme or ny lux, but the time to get there chnges The rtio o bsorption rtes or non nd rium re N s /N s = (s N /s N )=g /(g +g ) [1+l /s ] =(1.13/6.6) [ / ]= = Full power equilibrium bsorption in 135 is worth 28 mk, 12

13 so is worth = 5.1 mk, independent o power level. This estimte depends minly on how ccurtely the yields re known, nd is nerly insensitive to the estimte o the ull power lux. 13

14 SAMARUM BULDUP AFTER A TRP The dierentil equtions re simple to solve or lux dropping to zero. n ct the solution is obvious, since there is no decy o smrium, the promethium equilibrium bnk decys to dd to the smrium lredy present: È s Ne () t = N( eq) Í1+ 1e Î l l ( t ) (S5) With lªsthis becomes, numericlly, (t) = 5.1[2e l t ] The time or most o the increse to show up, bout 5 hl lives, is 265 hours = 11 dys. The inl vlue is, N pek È = N( eq) Í1 + s Î l (S6) = 5.1 mk [ ] or = Thus the shutdown lod or CANDU ter trip is bout double the norml lod OTHER FSSON PRODUCTS (& Plutonium) Severl other ission products, minly 151 nd Rh105, together with Pu 239, contribute smll but mesurble trnsient rectivity eects ter shutdown nd on restrt. The eects on opertion re mostly negligible. They cn ect the required uelling rte in the irst ew weeks ter rector restrt ollowing long mintennce shutdown. 14