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 table 6. 3 so f ? 0. 484 We will use 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. 022188sec;t d,Be =2. 0124e-3sec;t d,C ? 8. 772e ? 3sec 7. 5 One? delayed? neutron group reactivity equation; ?? ?lp 1 ? ?lp ? ? ? where ? ? 0. 0065; ? ? 0. 1sec? 1 1 ? ?lp ? ? ? For lp ? 0. 0sec For lp ? 0. 0001sec For lp ? 0. 001sec Note:In this question examine the figure 7. and see that to give a constant period value ,say 1 sec,you should give much more reactivity as p. neutron lifet ime increases. 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 neutrons:p? 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 delayed

Now you can compare 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 small so small reactivity assumption can be used as, 1 1 T= ? ?i t i ?? ? ? 0. 0848(from table 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 using 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 t

P0? 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 Insert 7. 56 into 7. 57 and plot reactivity vs rod radius Using 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( ) ?

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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 protected] 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 First 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 convergence ….. 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 absorption 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 Note:When 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 between 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 & ? 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 below; ?? ?? 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 & N0 are equal to equilibrium concentration for 157 Eu and 157Gd . ? No fission & 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 Maximum 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 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 ]

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