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( ) ?

T then to find the constants you must 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 blade 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 Multiply the n. current density by the area of the blades in the cell… –What is the area of the blades in the cell: From fig 7. 9,assume unit depth into the page so the cross sectional area of one of four blades, A=(l-a) ? 1 Divide by the total number of neutrons thermalizing per second in the cell —What 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 average 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. 5$@ x ? H a) For x ? 3H / 4 ? 75cm 1 ?x ? ? Sin(2? x / H ) ? ? ?? (3H / 4) ? ?0. 4545$ ? H 2? ? so the positive reactivity insertion is -0. 4545$-(-0. 5$)=0. 04545$ ?? ( x) ? ?? ( H ) ? b) The rate 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 let’s 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 ? p@ 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 ]