Flux Shape in Various Reactor Geometries in One Energy Group

Transcription

1 Flux Shape in Vaious eacto Geometies in One Enegy Goup. ouben McMaste Univesity Couse EP 4D03/6D03 Nuclea eacto Analysis (eacto Physics) 015 Sept.-Dec. 015 Septembe 1

2 Contents We deive the 1-goup lux shape in the citical homogeneous ininite-slab eacto and ininitecylinde eacto + Execises o othe geometies. eeence: Dudestadt & Hamilton Section 5 III 015 Septembe

3 Diusion Equation We deived the time-independent neuton-balance equation in 1 enegy goup o a inite, homogeneous eacto: D 0 (1) a We showed that we could intoduce the concept o geometical buckling (the negative o the lux cuvatue), and ewite the equation as 0 () whee has to satisy the citicality condition 1 D k 1 L (3) a 015 Septembe 3

4 Solving the Flux-Shape Equation Eq. () is the equation to be solved o the lux shape. We will study solutions o this shape equation o vaious geometies, and will stat with the case o an ininite cylindical eacto. The thing to emembe is that the solution must satisy the diusion bounday condition, i.e. lux = 0 at the extapolated oute suace o the eacto. While Eq. () can in geneal have a lage multitude o solutions, we will see that the addition o the bounday condition makes Eq. () an eigenvalue poblem, i.e., a situation whee only distinct, sepaated solutions exist. 015 Septembe 4

5 Solving the Flux-Shape Equation We will now apply the eigenvalue equation to the ininite-slab geomety and the ininitecylinde geomety, and solve o the geometical buckling and the lux shape. As concluded beoe, we will ind that the cuvatue o the 1-goup lux in a homogeneous eacto is always negative. 015 Septembe 5

6 Execise Given that the cuvatue o the 1-goup lux in a homogeneous eacto is negative, whee do you think the maximum lux would have to be in a egula-shaped eacto? Explain. 015 Septembe 6

7 Maximum Flux In a 1-goup homogeneous eacto o egula shape, the maximum lux must necessaily be at the cente o the eacto. We can explain that by eductio ad absudum. Suppose the lux is a maximum not at the cente o the eacto; i we daw a staight line om the maximum point to the cente o the eacto, then by symmety, thee would have to be anothe maximum on the same line on the othe side o the cente. Theeoe the cente o the eacto would be at a local minimum along that staight line, which would imply that the lux does not have negative cuvatue along that staight line! 015 Septembe 7

8 Ininite-Slab Case Let us study the simple case o a slab eacto o width a in the x diection, and ininite in the y and z diections. Eq. () then educes to its 1-dimensional vesion, in the x diection d dx x x 0 (4) We can without loss o geneality place the slab symmetically about x = 0, in the inteval [-a/, a/]. cont d 015 Septembe 8

9 Ininite-Slab eacto Geomety x = -a/ x = a/ x = a ex / x = Septembe 9

10 We also know that: Ininite Slab (cont d) The lux must be symmetic about x = 0, and The lux must be 0 at the extapolated boundaies, which we can call a ex / Eq. (4) has the well-known solutions sin(x) and cos(x). Theeoe the most geneal solution to Eq. (4) may be witten as x A x A sinx cos 1 (5) cont d 015 Septembe 10

11 Ininite Slab (cont d) Howeve, symmety about x = 0 ules out the sin(x) component. Thus the eacto lux must be x A (6) 1 cos x We can detemine om the bounday condition at a ex : aex A1 cos 0 (7) Now emembe that the cos unction has zeoes only at odd multiples o /. Theeoe must satisy: a ex n, i. e., n a ex with n odd (8) cont d 015 Septembe 11

12 Ininite Slab (cont d) It looks as i thee is an ininite numbe o values o. While that is tue mathematically, the only physically possible value o the lux in the citical eacto is the one with the lowest value o, i.e. o n =1: a, i. e. the buckling ex a ex (9) [Why? What do the othe solutions look like?] See sketch in next slide cont d 015 Septembe 1

13 Ininite Slab (cont d) phi with = pi/a phi with = 3pi/a Septembe 13

14 Ininite Slab (cont d) We can conclude that Eq. (9) is the only physical value o, om the act that the solutions with n = 3, 5, 7, all eatue egions o negative in the eacto, and that is not physical. Also to be noted om Eq. (9) is that the buckling inceases as the dimensions o the eacto (hee a ex ) decease [as had also been concluded ealie] the cuvatue needs to be geate to oce the lux to 0 at a close bounday! cont d 015 Septembe 14

15 Ininite-Slab Case (cont d) The 1-goup lux in the ininite-slab eacto can then be witten x x A cos (10) 1 a ex The absolute value o the lux, which is elated to the constant A 1, is undetemined at this point. This is because Eq. (3) is homogeneous theeoe any multiple o a solution is itsel a solution. The physical signiicance o this is that the eacto can unction at any powe level. Theeoe, to detemine A 1, we must tie the lux down to some quantitative given data e.g., the desied total eacto powe. cont d 015 Septembe 15

16 Ininite-Slab Case (cont d) ecause the slab is ininite, so is the total powe. ut we can use the powe geneated pe unit aea o the slab as nomalization. Let this be P W/cm. I we call E the ecoveable enegy pe ission in joules (and we know that this is ~00 MeV =3.*10-11 J), then we can wite P E a / a / a / a / x 1 dx a x dx E A cos( ) (11) And doing the integation will allow us to ind A 1 : a ex x aex a P E sin 1 1 sin A E A aex a / aex P A1 (13) i.e., a ae sin aex So that inally we can wite the absolute lux as P x x cos (14) a sin aex ae a a / 015 Septembe ex 16 ex (1)

17 Additional Notes In solving this poblem, we ound that the eacto equation had vey speciic solutions, with only speciic, distinct values possible o. Actually, thee is a geneal point to be made hee: Equations such as Eq. (3), holding ove a cetain space and with a bounday condition, i.e., the diusion equation o the eacto, all in the categoy o eigenvalue poblems, which have distinct solutions [eigenunctions] - hee the lux distibution -, with coesponding distinct eigenvalues (hee the buckling ). Although in this poblem we ound only 1 physically possible eigenunction o the steady-state eacto (the undamental solution), the othe eigenunctions ae peectly good mathematical solutions, which do have meaning. While these highe eigenunctions (which have lage values o ) cannot singly epesent the tue lux in the eacto, they can exist as incemental time-dependent petubations to the undamental lux, petubations which will die away in time as the lux settles into its undamental solution. 015 Septembe 17

18 Case o Ininite-Cylinde eacto Ininite height adius 015 Septembe 18

19 Ininite-Cylinde eacto Fo a homogeneous bae ininite cylinde, the lux is a unction o the adial dimension only. All axial and azimuthal positions ae equivalent, by symmety. 0 We wite the eigenvalue equation in cylindical co-odinates, but in the vaiable only, in which the divegence is 1 d d d (15) The 1-goup diusion equation then becomes 1 d d( ) 0 (16) d d d y evaluating the deivative explicitly, we can ewite Eq. (16) as d d 1 d d 0 (17) cont d 015 Septembe 19

20 Ininite-Cylinde eacto (cont d) We may be completely stumped by Eq. (17), but luckily ou mathematician iend ecognized it as a special case o an equation well known to mathematicians, essel s equation (m is a constant): d 1 d m 0 (18) d d Eq. (17) coesponds to m = 0, o which this equation has solutions, the odinay essel unctions o the 1 st and nd kind, J 0 () and Y 0 () espectively. These unctions ae well known to mathematicians (see sketch on next slide)! [It sue helps having mathematicians as iends, isn t it, even i Nobel didn t like them!] cont d 015 Septembe 0

21 Ininite-Cylinde eacto (cont d) I sketch the unctions J 0 (x) and Y 0 (x) below: Although, mathematically speaking, the geneal solution o Eq. (17) is a combination o J 0 (x) and Y 0 (x), Y 0 (x) tends to - as x tends to 0 and is theeoe not physically acceptable o a lux. cont d 015 Septembe 1

22 Ininite-Cylinde eacto (cont d) The only acceptable solution o the lux in a bae, homogeneous ininite cylinde is then AJ 0 (19 ) The lux must go to 0 at the extapolated adial bounday ex. Theeoe we must have J0 ex 0 (0) The igue in the pevious slide shows that J 0 (x) has seveal zeoes, labelled [the 1st is at x 1 =.405, the nd at x 5.6] ut because, physically, the lux cannot have egions o negative values, o the ininite cylinde can be given only by 1 x1.405 ex ex (1) Theeoe the buckling o the ininite cylinde is.405 ex () 015 Septembe

23 Ininite-Cylinde eacto (cont d) The 1-goup lux shape in the ininite homogeneous cylindical eacto is then ex AJ (3) As beoe, the absolute magnitude o the lux (i.e., the constant A) can be detemined only om some quantitative inomation about the lux, o example the powe pe unit axial dimension o the cylinde. I we denote that powe density P, and the enegy eleased in ission, we can wite: P d AE J d (4) E dv AE J Septembe 3 0

24 Ininite-Cylinde eacto (cont d) The integal on the essel unction may look obidding, but it can be evaluated om known elationships between vaious essel unctions. I ll just give the inal esult hee without deivation. I we ignoe the extapolation distance, which gives o the 1-goup lux the inal equation A 0.738P E J (6) 0.738P E (5) 015 Septembe 4

25 Execise/Assignment: Apply to Othe Shapes Execise: Apply the eigenvalue Eq. (3) to the ollowing geometies to ind the geometical buckling and the lux shape: Paallelepiped Finite cylinde Sphee Note: in the cases o the paallelepiped and o the inite cylinde, invoke sepaability, i.e., wite the solution as a poduct o unctions in the appopiate dimensions, each with its own diectional bucklings, which add to the total buckling. 015 Septembe 5

26 Paallelepiped eacto y +b/ z c/ -c/ -b/ x -a/ +a/ 015 Septembe 6

27 Paallelepiped eacto Fo a paallelepiped, the eigenvalue equation becomes d d d (7) dx dy dz To solve this, we get the billiant idea o tying a sepaable om o, i.e. x, y, z xgyhz (8) Substituting this om into Eq. (7) gives dx and i we divide both sides by (x)g(y)h(z): x d gy d hz g( y) h( z) ( x) h( z) ( x) g( y) dy dz d 1 x d dx x 1 g y d g dy y 1 h z d h dz z xgyhz (30) 0 (9) cont d 015 Septembe 7

28 Paallelepiped eacto (cont d) Since the 3 tems on the let-hand side ae unctions o x only, y only, and z only espectively, and the sum is a constant, then each tem must be a constant: 1 x d dx x x 1, g y d g dy y (31) and the 3 patial bucklings must add to the total buckling: x y z (3) We solve each pat o Eq. (31) in the same way as o the ininite slab, and, as in that case, we ind the solution to be a pue cosine unction: cont d 015 Septembe 8 y 1, h z d h dz z z

29 Paallelepiped eacto (cont d) I the paallelepiped has sides a, b, c in the x, y, and z diections, then x A x cos xx, g y Ay cos y y, hz Az cos zz (33) with the patial s as o the ininite slab: x, y, a b ex ex since the lux must go to 0 at the extapolated boundaies. (34) [Any odd multiple o each o these values is mathematically possible as well, but as beoe, only the lowest value is physically possible, as the highe values give egions o negative lux.] The total buckling o x y z 015 Septembe aex bex cex 9 z the c ex paallelepiped is

30 Paallelepiped eacto (cont d) We can inally wite the complete solution o the lux shape by substituting Eq. (33) into Eq. (8), getting the om x y z x, y, z Acos cos cos a (35) ex bex cex As in the case o the ininite slab, the lux amplitude A can be detemined only by anchoing the lux to some measued o desied quantity. Usually this is the total powe P, o pehaps the powe at some point in the eacto. I we use P, we can integate Eq. (35) ove the volume to ind (neglecting the extapolation distance) 3 P A 8 abce (36) 015 Septembe 30

31 015 Septembe 31 Paallelepiped eacto Diectional bucklings: Total buckling: Flux shape: I total powe = P, and neglecting extapolation distance: (37),, ex z ex y ex x c b a (38) z y x ) (39 cos cos cos,, ex ex ex c z b y a x A z y x (40) 8 3 abce P A

32 Finite-Cylinde eacto Finite height +H/ adius -H/ 015 Septembe 3

33 Finite-Cylinde eacto Fo a homogeneous bae inite cylinde, the lux is a unction o and also o the axial dimension z. All azimuthal positions ae equivalent, by symmety. In the divegence we must theeoe add the nd deivative in z d 1 d d The 1-goup diusion equation then becomes d 1 d d d d d d dz dz 0 (41) (4) cont d 015 Septembe 33

34 Finite-Cylinde eacto Just as we did o the paallelepiped eacto, we ty a solution in sepaable om:, z Z z (43) Substituting this into Eq. () gives d 1 d d Z z Z z Z z d [Function () not to be conused with adius ] and i we divide both sides by ()Z(z): 1 d d 1 d d d 1 Z z Z z (44) (45) ecause the tems in and Z ae sepaated, and thei sum is a constant, they must each be equal to a constant: 1 d d 1 d d dz (46) cont d 015 Septembe 34 d Z dz 1 Z z z d Z dz z & z

35 Finite-Cylinde eacto (cont d) whee the diectional bucklings must add to, so that:. 405 z ex The sepaate equations ae: d d d d 1 (47) (48) and d Z z Zz 0 (49) dz Eq. (48) is the essel equation o the ist kind, ode 0 (as beoe), and so the adial solution is J 0 () as o the ininite cylinde, and Eq. (49) gives cos(z) axially, as o the paallelepiped. So, in all.405 z (50) 0, z AJ 0 cos ex H ex H ex 015 Septembe 35

36 Finite-Cylinde eacto Diectional and total bucklings:.405, z, z ex H ex Flux shape: (51),.405 z 0 ex H ex z AJ cos (5) 015 Septembe 36

37 Spheical eacto Fo a homogeneous bae sphee, the lux is a unction o the adial dimension only. All latitudinal and longitudinal (azimuthal) positions ae equivalent, by symmety. We wite the eigenvalue equation 0 in spheical co-odinates, but in the vaiable only since the othe dimensions don t ente; the divegence is then 1 d d (53) d d The 1-goup diusion equation then becomes 1 d d( ) d d 0 (54) cont d 015 Septembe 37

38 Spheical eacto (cont d) To solve Eq. (54), we hit upon the teiic idea o tying o () a om such as (55) whee () is an (as yet) unknown unction, to be detemined (again, not to be conused with adius o sphee).. When we substitute the om (55) into Eq. (54), we get d d 1 d d 1 d d d d d d d d d d d (56) Septembe d 38 This educes magically to (57) cont d 0 0 0

39 Spheical eacto (cont d) And we know the geneal solution o this equation! Asin Ccos om which we then get sin cos A C (58) (59) ut we can ule out the cos tem, because the lux must be inite eveywhee in the eacto, and cos (60) as 0 The sin tem is O.K., because it emains inite at the oigin: y L Hôpital s ule, sin cos (61) 1 as 0 cont d 015 Septembe 39

40 Spheical eacto (cont d) Theeoe the lux shape in the bae homogeneous spheical eacto can inally be witten sin A 1 sin ex A whee, o the same easons as in the othe geometies, must take the lowest value allowed, 1 ex (6) (63) to guaantee that thee will not be egions o negative lux in the eacto. 015 Septembe 40

41 Flux Amplitude o a Spheical eacto We can integate Eq. (66) to evaluate the lux amplitude A o a given total eacto powe P. I E is the enegy eleased pe ission, and neglecting the extapolation distance: sin / P E *4 d 4AE d 0 4AE 4AE 4AE A 4 P E 0 sin d 0 cos ( and 0 0 sin 0 int egating cos d 0 4A by E pats ) 015 Septembe 41

42 Spheical eacto In summay then o a sphee o extapolated adius ex : uckling: (65) ex whee, o the same easons as in the othe geometies, must take the lowest value allowed, to guaantee that thee will not be egions o negative lux in the eacto. and lux shape: sin ex A (66) 015 Septembe 4

43 Summay We can obtain the solution o the 1-goup lux shape in bae homogeneous eactos o vaious geometies. In each case we detemine diectional bucklings (i applicable) and the total buckling, in tems o the dimensions o the eacto. The buckling(s) must take the lowest mathematical values allowed, to ensue that the lux solution is physical eveywhee in the eacto. 015 Septembe 43

44 END 015 Septembe 44