Lamarsh Solution Chap7

LAMARSH SOLUTIONS CHAPTER-7 PART-1 7. 1 Look at example 7. 1 in the textbook,only the moderator materials are different Since the reactor is critical, k ? ? ? T f ? 1 ?T ? 2. 065 from circuit board 6. 3 so f ? 0. 484 We forget rehearse t d ? t dM (1 ? f ) and t dM from table 7. 1 t dM,D2O ? 4. 3e ? 2 t dM,Be ? 3. 9e ? 3 t dM,C ? 0. 017 Then, t d,D2O =0. 022188sect d,Be =2. 0124e-3sect d,C ? 8. 772e ? 3sec 7. 5 One? delayed? neutron radical responsiveness equation ?? ?lp 1 ? ?lp ? ? ? where ? ? 0. 0065 ? ? 0. 1sec? 1 1 ? ?lp ? ? ? For lp ? 0. 0sec For lp ? 0. 0001sec For lp ? 0. 001sec pedigreeIn this question exa seconde the turn 7. and see that to give a eonian period value ,say 1 sec,you should give much more responsiveness as p. neutron lifet ime subjoins. And it is strongl recommended that before exam,study figure 7. 1 . 7. 8 ? ? 2e ? 4 from figure 7. 2 so you can ignore jump in power(flux) in this positive reactivity insertion situation t P Pf ? Pi e T then t=ln f ? T ? 3. 456hr Pi 7. 10 In eq 7. 19 prompt neutrons(1-? )k ? ? a ? T delayed 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 ????? ? ?? ? ? ?? prompt delayedNow you can comparison their values prompt (1-? ) ? delayed ? LAMARSH SOLUTIONS CHAPTER-7 PART-2 7. 12 ?? P0? ??? t ?? 1 P(t) ? e in here ? ? then, and ? ? ??? ??? T t P0 T P(t) ? e in here take T=-80sec ? 1? ? t ? P0 P0 ? 10 ? e 80 ? t ? 25. 24 min . 1 ? (? 5) ?9 7. 14 k ? ,0 ? ?? pf 0 ,critical state k ? ,1 ? ?? pf1 ,original state ?? 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 know ? a1F =0. 95 ? a 0 F and finally, 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 min? 60sec/ min ? 1731. 6sec. ln 2 )From fig 7. 2 rectivity is menial so small reactivity assumption can be designd as, 1 1 T= ? ?i t i ?? ? ? 0. 0848(from ta ble 7. 3)=4. 89e-5=4. 89e-3% ?i 1731. 6 4. 89e-5 also in dollars= ? 7. 52e ? 3$ ? 0. 752cents 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 will make small reactivity insertion approximation employ the insight given by figure 7. 2 for U-235 so, 1 1 T= ? ?i t i ?? ? ? 0. 0324(from table 7. 3)=5. 18e-6 ?i 6253. 8 a)100MW ? 1MWe 7. 18 a)From fig 7. 1 when ? ? 0 ? 1 ? 0 so T= 1 ?T?? ?1 b)Use prompt jump 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 cut-in 7. 56 into 7. 57 and plot reactivity vs rod wheel spoke utilize eq. 7. 57 and 7. 56 we plotted and found the radius value for 10% reactivity=3. 9 cm reactivity vs rod radius(a) 0. 14 0. 12 X 3. 9 Y 0. 1004 reactivity 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 equation is solved you know as, x xq ?T (x) ? A1 sinh( ) ? A 2 cosh( ) ?T then to adventure the constants you mustiness introduce L L ? a 2 boundary 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 find 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 current density at the brand surface, d? L J (m/2)-a ? ? D T ? d dx (m/2)-a ? coth((m ? 2a) / 2L) L Let s follow the instructions in the question breed the n. current density by the welkin of the blades in the stall What is the area of the blades in the cell From fig 7. 9,assume unit depth into the rapscallion so the cross sectional area of one of four blades, A=(l-a) ? 1 Divide by the total telephone number of neutrons thermalizing per second in the cell &8212What is the volume of the cell From fig 7. 9,assume unit depth into the page so V=(m-2a) ? (m ? 2a) ? 1 So as in page 358 4(l ? a) 1 fR ? 2 (m ? 2a) d ? coth((m ? 2a) / 2L) L 7. 25 You should find the B-10 middling atom density in the reactor Total mass of B-10=50rods ? 500g=25 ? 103g 25e3 N? ? 0. 6022e24 ? 1. 39e27atoms 10. 8 Atom density 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 find,? w ? ? 0. 0938 ? 9. 4% 0. 00833 ? 0. 000019 7. 27 H ? 100cm and ?? ? ? 0. email&clxprotected x ? H a) For x ? 3H / 4 ? 75cm 1 ?x ? ? Sin(2? x / H ) ? ? ?? (3H / 4) ? ?0. 4545$ ? H 2? ? so the positive reactivity insertion is -0. 4545$-(-0. 5$)=0. 04545$ ?? ( x) ? ?? ( H ) ? b) The come in of reactivity per cm can be found by differentiating the reactivity equation 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 There is a decrease in T so permits examine the effects of sign of temperature coefficients, If ? T ? (? ) decrease in T ? decrease in k ? reduces P ? gives further dec. in k ? shut down(unstable) If ? T ? (? ) decrease in T ? increase in k ? increase in P ? inc. in T and finally reactor returns to its original state (stable) 7. 33 ? N FVF I ? p ? exp ? ? ? ? ? M ? sM VM ? I Resonance Integral ? sM Scattering Cross-Section of Moderator ? M Constant 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 slightly enriched uranium 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 ? email&160protected 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 starting you should solve problem 7. 6 to find the fraction of expelled water, 575F ? 301 0 C 585F ? 307 C 0 Vvessel ? 6 0 C increase 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 Then find f after expelling, k ? ,0 ? ?? pf 0 ,critical state k ? ,1 ? ?? pf1 ,original state ?? 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 know ? 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 of equilibrium 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 Note the converging .. 2 2. 5 3 thermal flux x 1e14 3. 5 4 4. 5 5 7. 42 For Xenon using 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 correction to the thermal submersion cross section as follows, ? 20 0. 5 ) 2 200 ? aX (200? C ) ? 0. 886 ? 1. 236 ? 2. 65e6 ? 1e ? 24 ? 0. 316 ? a,X ? ? g aXe (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 Samarium 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 NoteWhen finding fission cross sections you should find the atom density of uranium 235 for this infinite thermal reactor. To do this ,refer to example 6. 5 on page 294 taking buckling zero and find a relation amongst moderator number density and fuel density. 7. 43 Using eq. 7. 98 0. 06627 1e13 ? 2. 42 1e13 ? 0. 773e13 where p=? =1 0. 01071 ?? 2. 42 ? Xe ? ? ? Sm 7. 44 First of all, we must write the rate equations for each element 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 equilibrium reactivity N (t ) ? N (t ? dt ) ? Xi Xi and ignore ? a Sm N Sm? T &038 ? a Eu N Eu? T Inserted into all rate equations 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 Reactivity equation 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 rapidly relative to 157 Eu and half-life of the 157 Sm is too small so, dN Sm ? 0 ? ?? Sm N Sm ? ? Sm? f ? T ? ? Sm N Sm ? ? Sm? f ? T dt This equation is inserted into rate equation 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 shutdown ? N0Eu &038 N0 are comprise to equilibrium concentration for 157 Eu and 157Gd . ? No fission &038 no absorption 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 utmost reactivity is reached at time goes to infinity Gd ? N grievous bodily harm (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 power P ? ER ? ? fF (r , t )? T (r , t )dV V So as N decreases ,flux should increase to keep 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) between 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