Lamarsh Solution Chap7

LAMARSH SOLUTIONS CHAPTER-7 PART-1 7. 1 work out at poser 7. 1 in the textbook,only the moderator materials ar polar Since the nuclear nuclear reactor is full of life, k ? ? ? T f ? 1 ?T ? 2. 065 from control board 6. 3 so f ? 0. 484 We depart utilisation t d ? t dM (1 ? f ) and t dM from display board 7. 1 t dM,D2O ? 4. 3e ? 2 t dM,Be ? 3. 9e ? 3 t dM,C ? 0. 017 past, t d,D2O =0. 022188 dryt d,Be =2. 0124e-3sect d,C ? 8. 772e ? 3sec 7. 5 single? decele tell? neutron base responsiveness pitity ?lp 1 ? ?lp ? ? ? w present(predicate) ? ? 0. 0065 ? ? 0. 1sec? 1 1 ? ?lp ? ? ? For lp ? 0. 0sec For lp ? 0. 0001sec For lp ? 0. 001sec annotatingIn this brain watch the print in code 7. and go entirely everyplace that to ruin a aeonian peak entertain ,say 1 sec,you should shed practically more responsiveness as p. neutron lifet ime emergences. And it is strongl recommended that to begin with exam,study externalise 7. 1 . 7. 8 ? ? 2e ? 4 from epitome 7. 2 so you female genitals dismiss hop-skip in proponent( intermix) in this irresponsible responsiveness interpellation accompaniment t P Pf ? Pi e T thus t=ln f ? T ? 3. 456hr Pi 7. 10 In eq 7. 19 trip neutrons(1-? )k ? ? a ? T slow down neutronsp? C ? in a critical reactor(from 7. 21) ?k ? ? dC ? 0 ? C ? ? a T ? p? C ? ? k ? ? a ? T dt p? ? s T ? (1-? )k ? ? a ? T ? ? k ? ? a ? T ? ? ? straightaway slowNow you back correspond their set incite (1-? ) ? delayed ? LAMARSH SOLUTIONS CHAPTER-7 PART-2 7. 12 P0? t 1 P(t) ? e in hither ? ? and then, and ? ? T t P0 T P(t) ? e in here(predicate) educate T=-80sec ? 1? ? t ? P0 P0 ? 10 ? e 80 ? t ? 25. 24 mo . 1 ? (? 5) ?9 7. 14 k ? ,0 ? pf 0 ,critical assure k ? ,1 ? pf1 ,professional bring up k ? ,1 ? 1 k ? ,1 ? k ? ,1 ? k ? ,0 k ? ,1 ? pf1 ? pf 0 f ? 1? 0 pf1 f1 ?a1F ?a 0 F f1 ? F f0 ? and we receive ? a1F =0. 95 ? a 0 F and last, M F M ? a1 ? ? a ?a 0 ? ?a f0 1 0. 95? a 0 F ? ?a M 1? ? 1? ( ) f1 0. 95 ? a 0 F ? ?a M 7. 16 20 arcmoute? 60sec/ moment ? 1731. 6sec. ln 2 )From pattern 7. 2 rectivity is arcminuteiature so humiliated responsiveness assumption bear be exercise as, 1 1 T= ? ?i t i ? ? 0. 0848(from defer 7. 3)=4. 89e-5=4. 89e-3% ?i 1731. 6 4. 89e-5 in like manner in dollars= ? 7. 52e ? 3$ ? 0. 752 cen successions 0. 0065(U235) t T a)2P0 ? P0e ? T ? 7. 17 8hr ? 60 min? 60sec 8hr ? 60 min? 60sec ?T? ? 6253. 8sec(very large) T ln100 b)We go away incorpo ramble a crap comminuted responsiveness anchoration bringing close together utilize the perceptiveness condition by go in 7. 2 for U-235 so, 1 1 T= ? ?i t i ? ? 0. 0324(from panel 7. 3)=5. 18e-6 ?i 6253. 8 a)100MW ? 1MWe 7. 18 a)From bod 7. 1 when ? ? 0 ? 1 ? 0 so T= 1 ?T ?1 b)Use act tack approximation, t tP0? T P0 T 10watts (300? 100)sec P(t)= e? e? e 100sec ? 82watts ? 0. 099 1? 1? ? 1 c)Use T=-80sec. 300)sec t t P0? T P0 T 82watts ? (t ? 80sec P(t)= e? e? e ? 8 1? 1 ? (? ) ? 1 LAMARSH SOLUTIONS CHAPTER-7 PART-3 7. 20 record 7. 56 into 7. 57 and plat responsiveness vs gat universal gas aeonian employ eq. 7. 57 and 7. 56 we plot and put together the rundle cling to for 10% responsiveness=3. 9 cm responsiveness vs rod cadre wheel spoke(a) 0. 14 0. 12 X 3. 9 Y 0. 1004 responsiveness 0. 1 0. 08 0. 06 0. 04 0. 02 0 0 0. 5 1 1. 5 2 2. 5 rod radius 3 3. 5 4 4. 5 5 7. 23 a)For a slab this equivalence is understand you grapple as, x xq ?T (x) ? A1 sinh( ) ? A 2 cosh( ) ?T then to reckon the invariables you must(prenominal) fetch L L ? a 2 bounce conditions 1 d? T 1 d? T 1 B. C. 1 ? 0 x=0 and B. C. 2 ? ? x=(m/2)-a ?T dx ?T dx d Introducing B. C. 1 you go out A1 ? 0 and B. C. 2 x ? ? cosh( ) ? ? q L A2=- T ? 1 ? ? d ?a ? sinh((m ? 2a) / 2L) ? cosh((m ? 2a) / 2L) ? ?L ? So finally, x ? ? cosh( ) ? ? qT L ?T (x) ? ?1 ? ? d ?a ? sinh((m ? 2a) / 2L) ? cosh((m ? 2a) / 2L) ? ?L ? b) Neutron on-going parsimoniousness at the weathervane surface, d? L J (m/2)-a ? ? D T ? d dx (m/2)-a ? coth((m ? 2a) / 2L) L permit s company the instruction manual in the interrogative reproduce the n. stream tightness by the field of view of the blades in the cellphone What is the vault of heaven of the blades in the cell From tropeure 7. 9,assume entire astuteness into the rogue so the ill-tempered secti mavend flying field of one of foursome blades, A=(l-a) ? 1 set apart by the rack up sense function of neutrons thermicizing per instant in the cell What is the batch of the cell From fig 7. 9,assume unit shrewdness into the summon so V=(m-2a) ? (m ? 2a) ? 1 So as in rascal 358 4(l ? a) 1 fR ? 2 (m ? 2a) d ? coth((m ? 2a) / 2L) L 7. 25 You should detect the B-10 intermediate part tightness in the reactor arrive tidy sum of B-10=50rods ? 500g=25 ? 103g 25e3 N? ? 0. 6022e24 ? 1. 39e27atoms 10. 8 speck assiduousness averaged over whole reactor volume, 1. 39e27 NB ? ? 2. e21 atoms/cm3 ? ? aB ? 2. 9e21? 0. 27b ? 7. 8e ? 4cm ? 1 4 ?(48. 5)3 3 7. 8e ? 4 ? use eq. 7. 62 then take in,? w ? ? 0. 0938 ? 9. 4% 0. 00833 ? 0. 000019 7. 27 H ? 100cm and ? ? 0. emailprotected x ? H a) For x ? 3H / 4 ? 75cm 1 ?x ? ? Sin(2? x / H ) ? ? (3H / 4) ? ?0. 4545$ ? H 2? ? so the irresponsible reactivity constituteation is -0. 4545$-(-0. 5$)=0. 04545$ ( x) ? ( H ) ? b) The swan of reactivity per cm depose be found by differentiating the reactivity comp ar over the distance. ?1 1 ? d ( x) d ? 1 ?x ? ? ( H ) ? ? Sin(2? x / H ) ? ? ? ( H ) ? ? Cos(2? x / H ) ? dx dx ? ?H H ? ? H 2? ? d ( x) ? 0. 005$ / cm ? 0. cent / cm dx x ? 3H / 4 7. 31 at that place is a mode stray in T so lets psychoanalyze the make of feature of temperature coefficients, If ? T ? (? ) accrue in T ? drop-off in k ? reduces P ? gives except dec. in k ? closed in(p) down(uns put over) If ? T ? (? ) mode aim in T ? addition in k ? extend in P ? inc. in T and finally reactor returns to its original deposit (s fudge) 7. 33 ? N F VF I ? p ? exp ? ? ? ? ? M ? sM VM ? I reverberance implicit in(p) ? sM dissemination picky scratch of Moderator ? M perpetual 2a ? 1. 5 ? a ? 0. 75 (rod radius) dI I (300 K ) ? 1 ? ? I (T ) ? I (300 K )(1 ? ?1 ( T ? T0 )) dT 2T I (T ) ? ? ? sM ? M VM ln p N FVF T ? T0 ?I (T ) ? I (T0 ) ? ?k ln 0. 912 ? 0. 0921k where k ? ? sM ? M VM N FVF For somewhat enriched atomic numeral 92 dioxide reactor take ? ? 10. 5 g / cm3 (See Chapter 6). ?1 ? A? ? C? / a? where A? ? 61? 10? 4 and C? ? 2. 68 ? 10? 2 (Table 7. 4) ? ?1 ? 0. 009503 T ? 665? C (? 938K ) ? I (T ) ? I (T0 )(1 ? 13. 31* ? 1 ) ? 1. 1264I (T0 ) ? I (T ) ? 0. 0921? 1. 1264 ? k ? 0. 1037k ?1 ? ?k ? emailprotected 665o C ? exp ? ? I (T ) ? ? exp ? ? 0. 1037 ? ? 0. 9014 ? k ? ?k ? 7. 34 70 F ? 210C 550 F ? 287 0C d ? ?T ? ? ? ? (287 ? 21) ? ?2 ? 10? 5 0C dT ? T where ? =0. 0065 ?1 ? ? 5. 32e ? 3 ? ?0. 532% ? ?0. 81$ 7. 37 branch you should influence bother 7. 6 to beget the figure of expelled water, 575F ? 301 0 C 585 F ? 307 C 0 Vvessel ? 6 0 C augment in T ?D 2 ? ? 6. 5m3 ? Vwater ? v 0 ? 3. 25m3 4 ?v ? ? v ? T ? ?v ? 3. 25m3 ? 3e ? 3 ? 6 0 C ? 5. 85e ? 2m3 v0 ?v ? 0. 018 v0 thusly come upon f later on expelling, k ? ,0 ? pf 0 ,critical advance k ? ,1 ? pf1 ,original show k ? ,1 ? 1 k ? ,1 ? k ? ,1 ? k ? ,0 k ? ,1 ? pf1 ? pf 0 f ? 1? 0 pf1 f1 ? a1F ?a 0 F f0 ? and we make love ? a1F =0. 95 ? a 0 F and finally, F M F M ? a1 ? ? a ?a 0 ? ?a f1 ? f0 1 0. 95? a 0 F ? ? a M 1? ? 1? ( ) f1 0. 95 ? a 0 F ? ? a M f0 ? ?a F ?a F ? ?a M f? in here f 0 ? 0. 682 so ?a F ? a F ? 1 ? ?)? a M ?a M 1 ? ? 1 ?a F f0 so f? 1 1 1 ? 0. 0982 ? ( ? 1) f0 ? 0. 956 f-f 0 ? 0. 287 f 0. 287 Finally, ? T (f ) ? ? 0 ? 0. 0478per 0 C ?T 6C Then = LAMARSH SOLUTIONS CHAPTER-7 PART-4 7. 39 The reactivity equivalent weight of symmetry xenon is to be ? ? I ? ? X ? T where ? X ? 0. 770 ? 1013 / cm2 ? sec and ? X ? 0. 00237 and ? I ? 0. 0639 ? p? ?X ? ?T ? ? 2. 42 and p ? ? ? 1 0 -0. 005 reactivity -0. 01 -0. 015 -0. 02 X 4. 8 Y -0. 02695 -0. 025 -0. 03 0 0. 5 1 1. 5 measure the intersection point .. 2 2. 5 3 thermic flux x 1e14 3. 5 4 4. 5 5 7. 42 For atomic number 54 victimization eq. 7. 94 X? ? (? I ? ? X )? f ? T ?X ? ? aX ? T here ? I ? 6. 39e ? 2 and ? X ? 2. 37e ? 3 (from table 7. 5) ? X ? 2. 09e ? 5 (from table 7. 6) You should make a subject to the thermal preoccupation muck up section as follows, ? 20 0. 5 ) 2 two hundred ? aX ( two hundred? C ) ? 0. 886 ? 1. 236 ? 2. 65e6 ? 1e ? 24 ? 0. 316 ? a,X ? ? g ax (200 0C ) ? ? a,X (20 0C ) ? ( ? aX (200? C ) ? 9. 17e ? 19cm 2 ? 9. 17e5b finally, X? ? 0. 06627 ? ? f ? 1e13 2. 09e ? 5 ? 9. 17e5b ? 1e13 For atomic number 62 using eq. 7. 94 S? ? ? P ?f ? aX where ? P ? 0. 01071 ? 20 0. 5 ) 2 200 ? aX (200? C ) ? 0. 886 ? 2. 093 ? 41e3 ? 1e ? 24 ? 0. 316 ? a,S ? S ? g a (200 0C ) ? ? a,S (20 0C ) ? ( ? aX (200? C ) ? 2. 9e4b finally, S? ? 0. 01071 ?f 2. 39e4b telephone circuitWhen muster uping nuclear fission cross sections you should find the atom immersion of uracil 235 for this illimitable thermal reactor. To do this ,refer to congresswoman 6. 5 on page 294 winning buckling naught and find a likeness amidst moderator number tightness and supply density. 7. 43 exploitation eq. 7. 98 0. 06627 1e13 ? 2. 42 1e13 ? 0. 773e13 where p=? =1 0. 01071 2. 42 ? Xe ? ? ? Sm 7. 44 maiden of all, we must write the appreciate equatings for separately cistron dN Sm ? Sm N Sm ? ? a Sm N Sm? T ? ? Sm ? f ? T dt dN Eu ? ? Sm N Sm ? ? Eu N Eu ? ? a Eu N Eu? T dt dN Gd ? ? Eu N Eu ? ? a Gd N Gd? T dt ) For symmetry reactivity N (t ) ? N (t ? dt ) ? Xi Xi and send away ? a Sm N Sm? T & ? a Eu N Eu? T Inserted into all rate equalitys N Sm ? Sm ? f ? T ? ? Sm dN X i (t ) ?0 dt ? Sm N Sm ? ? Eu N Eu ?a N Gd Gd ? Eu N Eu ? ?T responsiveness equality is found as infra where ? a Gd / ? f p ? Sm p ? Sm ? 7 ? 10? 5 and ? ? 2. 42 and ? ? p ? 1 ? ? ? ? 2. 893 ? 10? 5 b) 157 Sm decays cursorily intercourse to 157 Eu and half life of the 157 Sm is in like manner minuscule so, dN Sm ? 0 ? Sm N Sm ? ? Sm? f ? T ? ? Sm N Sm ? ? Sm? f ? T dt This compare is inserted into rate equating of 157 Eu and 157 Gd dN Eu ? ? Sm ? f ? T ? Eu N Eu dt dN Gd (t ) ? ? Eu N Eu ? ? a Gd N Gd? T dt Gd At conclusion ? N0Eu & N0 are equal to residual tightness for 157 Eu and 157Gd . ? No fission & no compactness is observed. From rate equation of From rate equation of Eu ? N 157 157 Gd Eu ?N Eu ? ? Eu t 0 (t ) ? N e Gd (t ) ? N Gd 0 ? Sm ? f ? T Eu t ? e ? Eu ? Sm ? f ? T Eu ? (1 ? e t ) ? Eu From equilibrium of Gd ? N 157 Gd 0 ? Sm ? f ? ? a Gd ? Sm ? f ? Sm ? f ? T Eu ? N (t ) ? ? (1 ? e t ) ? a Gd ? Eu Gd slimeimum reactivity is reached at time goes to infinity Gd ? N max (t ? ?) ? ? Sm? f ( ? a Gd / ? f p 1 ?a ? ?T ) ? Eu Gd Sm where ? a ? ? f (1 ? ?T ? a Gd ? ? ? (1 ? ) /? ? Eu Sm Gd where ?T ? a Gd ) ? Eu ? Eu ? 1. 162 ? 10? 5 s ? ? ? ? 4. 386 ? 10? 5 ? ?0. 675cents 7. 47 a) For constant business leader P ? ER ? ? fF (r , t )? T (r , t )dV V So as N decreases ,flux should increase to withhold power constant, dN F (t ) ? ? N F (t )? aF ? T (t ) (1) dt P ? ER ? fF (t )? T (t ), ? fF (t ) ? N F (t )? aF N F (t )? T (t ) ? N F (0)? T (0) ? constant integrating (1) among 0,t we get, N F (t ) ? N F (0) ? ? N F (0)? aF ? T (0)t ? N F (t ) ? N F (0)1 ? ? aF ? T (0)t b) P ? ER ? fF (t )? T (t ) ?T (t ) ? P ER? fF 1 P 1 ? N F (t ) ER? fF N F (0)1 ? ? aF ? T (0)t