PROBLEM SET 8 ECE 4130: Introduction to Nuclear Science

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1 PROBLEM SET 8 ECE 4: Introduction to Nucler Science Chirg Bhrdwj 6 October 6. Problem 6-. Let us denote by F the fuel nd by S the coolnt in the mixture. Then, Σ F nd Σ S represent the bsorption crosssection for the fuel nd coolnt, respectively, while Σ represents the nucler cross-section of the mixture of the two i.e. Σ Σ F + Σ S. In our cse, the fuel F is plutonium enriched to w/o while the coolnt S is liquid sodium. We know tht for ech constitutent element X of the mixture, the tom density N X is given by N X w Xρ MX, where w X is the reltive mss frction of X in the mixture, MX is the tomic weight of X, nd ρ is the density of the substnce from which the element is derived, nmely the mixture. Furthermore, becuse this produces units of mol, we cn multiply by Avogdro s number N cm A to produce units of toms i.e. units of tom density: cm N X w Xρ MX N A. The bsorption cross-section for n tom of ech constituent element X cn be found in stndrd tble, so we cn compute the mcroscopic bsorption cross-section tht ech constituent element of the mixture contributes, reclling tht the mcroscopic bsorptio cross-section Σ is given by Σ Nσ, where N is the tom density nd σ is the nominl one-group bsorption cross-section for X in fst rector. We proceed with finding the reltive contributions for ech constituent element X in the mixture: Σ X N X σ X w Xρ MX N Aσ X. We then clculte this vlue for ech of the constituent elements of the mixture, nmely sodium nd plutonium, consulting Tble 6. for the nominl one-group constnts σ N nd σ Pu : ΣX N w g Nρ MN N Aσ N.97. cm.9898 g mol ΣX Pu w g Puρ MPu N Aσ Pu.. cm 9.5 g mol toms mol 4 toms mol.8 b. b 4 cm. 5 cm b 4 cm.6 4 cm b Now, we find the fuel utiliztion f using the reltion f ΣF Σ : f ΣF Σ Σ F Σ F + Σ S + ΣS Σ F + ΣN Σ Pu +. 5 cm.6 4 cm.89 We lso compute the infinite multipliction fctor k using the reltion k η f, where η is nominl one-group number of neutrons relesed in fission per plutonium neutron bsorbed by fissile nucleus in the fst rector: k η f where we used η Pu from Tble 6.. Since k >, we see tht the rector is supercriticl.

2 b Problem 6-9. i. We first find the mteril buckling of the rector defined in Problem 6- using B k L k Σ D D k, where D is the nominl one-group diffusion coefficient in the rector, pproximtely given by D λ tr Σ tr Σ Σ tr Σtr X X N X σtr X X where ech X is constituent element of the mixture. In our cse, we cn use the vlues from Problem 6- nd Tble 6. to find D if X rnges over sodium nd plutonium: D [.54 4 cm. 4 cm cm cm.95 cm. ] We now use this in combintion with the results from Problem 6- to find the mteril buckling: Σ B D k.8 4 cm cm..95 cm The results of Section 6. give us tht the first-order geometric buckling of bre sphericl rector is given by B R, where R is the criticl rdius. Then, we cn rerrnge this s R B nd solve for the extended rdius using the mteril buckling in plce of the first-order geometric buckling since they re equl: Then, we cn find the criticl rdius: R B B cm cm R R d R.D 46 cm..95 cm 98 cm ii. As the end of Section 6. shows, the mximum flux in the bre sphericl rector is given by φ mx P 4E R Σ f R, where P is the therml power level t which the rector opertes, E R is the recoverble energy of the rector in joules per fission, nd Σ f is the nominl one-group mcroscopic fission cross-section of the moderting mixture. We first find the mcroscopic fission cross-section using the results from Problem 6- nd the vlues from Tble 6.: Σ f Σ N f + Σ Pu f N N σ N f + N Pu σ Pu f.54 4 cm 4 cm + Assuming recoverble energy of MeV per fission, we get: cm.85 4 cm.4 4 cm. 5 MW φ mx.6 4 MeV neutrons J.4 ev 4 cm 98 cm cm sec iii. The probbility tht fission neutron will escpe or lek from the rector is given by the inverse reltion: P E P L % k.,

3 . Problem 6-5. We consider the cse of criticl bre cylindricl rector of rdius R nd height H, s shown below. We denote by the term rdius of r" ny position within the cylinder tht hs n L -distnce of r wy from the centrl z-xis: Figure : A criticl bre cyndricl rector bse t the origin of rdius R nd height H t centrl-xis L -distnce r. We wish to derive the constnt fctor A tht chrcterizes the flux of such rector t the specified rdius r using the formentioned Lebesgue mesure s our metric of distnce. We begin with the one-group rector eqution φ + B φ, where B represents the buckling of the rector. Since we re exmining cylindricl rector structure, it would mke the most sense to trnsform the one-group rector eqution into cylindricl coordintes i.e. trnsformtion into circulr polr coordintes in the xy-plne with the z-direction left untouched: φ r + φ r r + φ z + B φ. However, we recll tht our boundry conditions remin intct, nmely tht the flux φr, z must stisfy certin constrints with respect to its two prmeters. One such condition is tht the flux vnishes beyond the boundry of the modertor in the rector. In other words, φ R, z for ny height z bove the bse of the cylinder. The other condition is tht the net flux through ny point in the circulr surfce through the center of the cylinder vnishes, s regrdless of how the flux is distributed, the net flow through the center cncels out by symmetry. Since the center is t height /, we hve tht the flux through ny rdius r wy from the centrl xis in the z / plne is zero. In other words, φr, / for ny rdius r wy from the centrl xis. We cn now propose using the method of seprtion of vribles tht φ is of the form φr, z ρrzz, where ρ nd Z re independent functions. Then, the differentil eqution is d ρ dr Z + Z dρ r dr + ρ d Z dz + B ρz. Dividing through by ρz, we hve the equivlent differentil eqution: d ρ ρ dr + dρ ρr dr + d Z Z dz + B. We now seprte out the vribles so tht the eqution reduces to d ρ ρ dr + dρ ρr dr + B d Z Z dz. However, the LHS is purely function of r, nd the RHS is purely function of z. Since we sid tht r nd z re independent prmeters, it must follow tht both sides re equl to some constnt, which we will cll λ. Then, we end up getting two seprte equtions:

4 d ρ ρ dr + dρ ρr dr + B λ Z d Z dz λ. We will solve the second eqution first. We first rewrite the second eqution s the eigenvlue problem: d Z + λz. dz We ssume solution of the form Z Z e kz for constnt k. Then, we cn rewrite the differentil eqution s k + λz e kz, nd since we don t wnt Z for non-trivil solution, we insted rrive t the chrcteristic k + λ by the zero-fctor property. We then brek into cse-by-cse nlysis bsed on the vlue of λ: Cse. λ >. In this cse, to ssert tht λ >, we set λ θ for some θ. Then, the solutions to the chrcteristic eqution re k ±θ. Thus, there re two solutions to the differentil eqution, nmely Z e θz nd Z e θz. However, by the superposition theorem, ny liner combintion of these two fundmentl solutions is lso solution. Thus, we cn sy tht the solution to the differentil eqution is Z c Z e θz + c Z e θz w e θz + w e θz for some other constnts w nd w. Cse. λ. This is the simplest cse. Here, the only solution to the chrcteristic eqution is k. Then, we hve Z Z. Cse. λ <. In this cse, to ssert tht λ <, we set λ θ for some θ. Then, the solutions to the chrcteristic eqution re k ±θi. Thus, there re two solutions to the differentil eqution, nmely Z θz nd Z sin θz. We gin ppel to the superposition theorem to rrive t the following liner combintion: Z c Z θz + c Z sin θz w θz + w sin θz. Now, we pply our boundry conditions to ll three cses. Since φr, / holds for ll rdii r, this implies tht it is condition on z lone, i.e. we see tht Z / is the true condition. Symmetry bout the cylinder lso tells us tht Z Z. We now brek into our cse-by-cse nlysis: Cse. λ >. In this cse, the second boundry condition gives us tht w + w w e θ w e θ. We cn rerrnge this nd collect like terms to rrive t the following: Solving for w, we get the following: w + e θ + w + e θ. w + eθ + e θ w. We rewrite the originl solution now s the following: w e θz + eθ + e θ e θz. We now use the first boundry condition, which gives us tht w e θ + e θ + e θ e θ There re two possibilities: w nd the term in prentheses is. We first void the trivil solution: 4.

5 e θ + eθ + e θ. This leds to e θ + + e θ, which is true regrdless of θ. Thus, it must be the cse tht w for the boundry condition to hold. This tells us tht w s well, which leds to the trivil solution Z, so we rule out this cse. Cse. λ. The second boundry condition leds to Z Z, nd the first boundry condition directly gives us tht Z. Thus, Z, which is the trivil solution, so we rule out this cse, too. Cse. λ <. We pply the second boundry condition to get tht w w θ w sin θ. Seting corresponding prts equl, we see tht θ nd sin θ. Both conditions yield θ. Thus, we see tht the solution is of the form We pply the first boundry condition to get tht w / w z + w sin z. + w sin / which reduces to w. It follows tht our solutions re ll of the form w z. Thus, the solution to this prt of the differentil eqution is Zz w z. We now solve the other eqution... Rerrnging, we hve tht, d ρ ρ dr + dρ ρr dr + B λ θ. d ρ dr + dρ B r dr + + ρ. Let us denote by B B + the effective buckling of the rector i.e. B B + θ. Then, we hve tht d ρ dr + dρ r dr + B ρ. We solve this differentil eqution by ssuming series solution form: ρr n r n. We proceed: n n nn n r n + n n n r n + B n n r n. We notice tht some of these sum indices re redundnt, so we cn rewrite this compctly: n n + n + n+ r n + r + n + n+ r n + B n n n r n. Becuse we require tht ll exponents re to the positive power, we cnnot hve the /r term, s it would not be cnceled out by ny term in the series nd the RHS is homogeneous. Thus, this term cnnot exist, i.e. : n We end up with the following: n + n + n+ r n + n 5 n + n+ r n + B n n r n.

6 n + + n + n+ + B n r n. n Since the RHS is homogeneous, we need every term to cncel out, i.e. we need the following recurrence to hold: B n+ n + n. Since, we see tht 5 s well, so every odd term cncels out. We re left with only the even terms. Let us sy tht n is m for some m Z. Then, n + is m +, nd we hve the following recurrence: B m+ m + m, B which we cn rewrite more netly s m m m. Thus, we cn define the entire set of coefficients in terms of some bse coefficient indices re now over the m-rnge. Tht is: m m B m 4 m m!. Then, the entire solution is simply ρr + m r m. We compute this below: m ρr + m B r m m 4 m m! + J B r J B r, where J is Bessel s function of order zero. We leve out the first becuse its vlue is infinitesiml compred to the sum of infinite terms in the series. We now combine our solutions to the generl form: z z φr, z ρrzz w J B r AJ B r for some other constnt A. As the textbook gives, B.45, so our solution is of the form R z φr, z ρrzz AJ.45r R The only thing left to do is to determine the constnt A. We find the power in the rector to determine the constnt: P E R Σ f φr, zdv E R Σ f V R AE R Σ f dθ 4 R AE R Σ f.45 φr, zrdrdθdz z.45 R dz uj u du 4 R AE R Σ f.45 [uj u].45 4 R AE R Σ f.45 Then, we cn rerrnge to solve for A:.45r rj dr R. J R AE R Σ.45 f.864 R AE R Σ f A P /.864P.864 R E R Σ f R.69P E R Σ VE R Σ f f 6

7 b Problem 6-. The result is the sme s in Problem 6-5, s we hd ssumed power of P wtts in the first plce, s long s it is true tht R R nd H, i.e. d.. Problem 6-5. We cn find the tomic density of 5 9 U using conversion fctors: N U g L mol.67 4 toms L cm 5 g mol 9 toms.56 cm Similrly, we find the moleculr density of H O using conversion fctors. In the solution, we know tht there is L of wter per L of solution i.e. the liquid solution is ll wter with U dissolved inside, so we get the following: N w L cm L g L L cm cm g mol.67 4 toms toms cm g mol cm b We cn then find the fuel utiliztion f using the method of Problem 6-: f + Σw Σ U + N wσ w N U g σ U.4 + toms/cm cm toms/cm cm using the vlues for σ given in Exmple 6., prt nd g C.978 from Tble.. c The therml diffusion re without modertor L TM is given by L TM D D w Σ Σ U + Σ w D w N U g σ U + N w σ w.6 cm.56 9 toms/cm cm +.4 toms/cm cm 8.5 cm where we used vlues from Tble., Tble 5., nd Tble 6.. We cn now find the true therml diffusion re: Then, the therml diffusion length is simply L T f L TM cm 4.6 cm L T d We lstly find the infinite multipliction fctor s follows: L T.4 cm k η U f where we used η U t C from Tble 6.. Since k <, we see tht the rector is subcriticl. 4. Problem 6-9. We recll tht in n infinite homogeneous mixture of fuel in modertor, the infinite multipliction fctor is given by k η T f, where η T is the nominl one-group constnt for fst rector. We rerrnge this s f k. However, we know tht η T f ΣF Σ Σ F Σ F + Σ S, which we cn rewrite s f Section 6.6. Solving for Z:, where Z is the one-group rtio introduced in + ΣS + /Z Σ F + /Z k η T f k Z η T k 7

8 T We recll from Eqution 5.59 tht Σ F g T Σ F E T t some other temperture T, where T 9.6 K is room temperture nd E.5 ev. In our cse, we wnt the vlue t T T, so we expnd everything: Σ F gf T Σ F T E T g N F σ F. However, we know from the previous pge tht Z Σ F /Σ S, so we cn eqully rewrite this s Σ F Now, we cn solve for N F in terms of the other quntitites: N F ZΣ S g σ F g N F σ F ZΣ S. k Σ S η F k g σ F where Σ S is the mcroscopic bsorption cross-section of the modertor the solution. We know tht this rector is criticl, so k. The other fctors re modertor-dependent, so we do individul clcultions. We find N F for 5 U in vrious modertors. All vlues re from Tble., Tble 5., Tble 6., nd Tbles II./. Modertor: H O. Note the correction fctor of / for Σ w to offset the outer multipliction of /. N U Σ w η U g σ U /. cm cm 9 toms mol 5 g cm. cm.67 4 toms mol L. g L, Modertor: D O. We ssume tht it contins H O t 5 w/o. N U Σ D η U g σ U 9. 5 cm cm 7 toms mol 5 g cm.47 cm.67 4 toms mol L.57 g L Modertor: grphite C. N U Σ C η U g.65 7 toms.78 cm.48 g L σ U.4 4 cm cm mol.67 4 toms 5 g mol cm L 8

9 b We find N F for 9 Pu in vrious modertors. All vlues re from Tble., Tble 5., Tble 6., nd Tbles II./. Modertor: H O. Note the correction fctor of / for Σ w to offset the outer multipliction of /. N Pu Σ w η Pu g σ Pu /. cm cm 9 toms mol 9 g cm.96 cm.67 4 toms mol L 7.78 g L Modertor: D O. We ssume tht it contins H O t 5 w/o. N Pu Σ D η Pu g σ Pu 9. 5 cm cm 6 toms mol 9 g cm 9.7 cm.67 4 toms mol L.68 g L Modertor: grphite C. N Pu Σ C η Pu g σ Pu.4 4 cm cm 7 toms mol 9 g cm.9 cm.67 4 toms mol L.949 g L 5. Problem 6-. We consider the cse of criticl bre cubicl rector of side length, s shown below. Figure : A criticl bre cubicl rector bottom-left corner t the origin of side length. We begin with the one-group rector eqution φ + B φ, where B represents the buckling of the rector. Since we re working with Crtesin coordinte system, it mkes most sense to write the eqution in them: 9

10 φ x + φ y + φ z + B φ. We now consider boundry conditions on the flux in this rector. Just s in Problem 6-5, the flux t the midpoint of ech fce of the cube must vnish. Tht is, we hve the following: φ, y, z φ x,, z φ x, y,. Similrly, symmetry bout the cube gives us the following three constrints: φ, y, z φ, y, z φ x,, z φ x,, z φ x, y, φ x, y,. We cn now propose using the method of seprtion of vribles tht φ is of the form φx, y, z Rx, yzz, where R nd Z re independent functions. We now brek prt B into its x-, y-, nd z-components, i.e. B B x + B y + B z for the buckling x-component B x, y-component B y, nd z-component B z. Then, we get: R x Z + R y Z + R d Z dz + B RZ. Dividing through by RZ nd rerrnging, we get the following: R R x + R R y + B x + By d Z Z dz B z. Since R nd Z re both independent functions, the only wy the LHS nd RHS re equl is if they re both equl to some other constnt, which we we will cll m. Then, we end up getting two seprte equtions: R R x + R y + Bx + By m d Z Z dz + B z m. We first rewrite the second eqution s d Z dz + B z + mz or d Z + tz for some other constnt t. We then dz brek up the first eqution, which cn be rewritten s follows: R x + R y + B x + By mr. We gin use seprtion of vribles, ssuming tht Rx, y is of the form Rx, y XxYy, where X nd Y re gin independent functions. Then, the differentil eqution becomes d X dx Y + X d Y dy + B x + B y mxy. Dividing through by XY nd rerrnging, we get the following: d X X dx + B x m d Y Y dy B y. Agin, both sides re equl to some other constnt k. Then, we end up getting two more equtions: d X dx + B x kx d Y dy + B y + k my.

11 We cn rewrite these s d X dx + rx d Y dy + sy for some other constnts r nd s. Note tht r + s + t B x + B y + B z B. We now convert ll of our boundry conditions into equivlent conditions for X, Y, nd Z: X Y Z X X Y Y Z Z. These differentil equtions with these boundry conditions re exctly identicl to those tht we explored in Problem 6-5 erlier. Thus, we lredy hve our generl solution there is no point repeting the sme nlysis: We thus know our overll solution: φx, y, z XxYyZz w x which we cn rewrite s Xx w x Yy w y Zz w z. w y φx, y, z A x w z y z w w w x for some other constnt A. We notice tht the flux is symmetric in ll directions, s expected. y z, b Problem 6-4. We use the method of Problem 6-5 gin, noting tht if the rector is operting t power of P wtts, this power is relted to the flux φ vi the full rel-volume integrl: P E R Σ f φx, y, zdv V AE R Σ f x dx z dy ] [ AE R Σ f p dp [ / AE R Σ f u du [ AE R Σ f 8 [ AE R Σ f / sin u / ] ] u du z dz ] 8 AER Σ f sin Thus, we cn solve for the constnt A, which gives us A P 8E R Σ f sin /

12 c Problem 6-5. i. In this cse, we wish to solve for the criticl boundries, given the composition. In prticulr, we re given tht the rtio β n F /n S is. 5, where n F is the tomic composition of the urnium-5 fuel nd n S is the tomic composition of the modertor in the solution/mixture, which is grphite. Now, we turn to Section 6.6 nd exmine Cse, which describes the cse where the tomic density composition is specified: B k MT, where MT L T + τ T nd τ T is the neutron ge. Also, for cubicl rector of side length, we know tht s described lter in Section 6.6. Thus, we cn rewrite this s B L T + τ T k L TM f + τ T. k As we sw in Problem 6-9, we cn write k s k η T f, where f computtion, we relize tht we cn rewrite f s + /Z nd Z ΣF Σ S. Before doing ny f + /Z Z +, L TM Z + + τ T which mkes our work slightly esier, s now. However, since k η k T f, we see tht Thus, we cn rewrite s Now, we write n eqution for Z: Z ΣF gf TΣ F T E T Σ S T ΣS E T k η T f η TZ Z Z + η T. Z + gf TΣ F E Σ S E L TM + τ TZ + Using T 5 C 5.5 K nd E.5 ev, we get: where we used vlues from Tble. nd Tble II.. η T Z gf TN F σ F E N S Σ S E 687. Z cm.4 4 cm.9, We now find L TM t T insted of T using the djustment formul from Chpter 5: L T ρ, T L T ρ ρ T, T. ρ T. N F g F Tσ F E N S σ S βg F T σf E E σ S E. However, since β, we see tht n F n S, i.e. the mixture is lmost purely grphite. Thus, the density of the mixture is pproximtely tht of pure grphite, or ρ ρ. Then: L TM ρ, T L TM ρ T 5.5 K, T 5 cm T 9.6 K 467 cm, where we used vlues from Tble 5.. Then, we cn finlly solve for : Using the extended tble on Pge of Effective Cross-Section Vlues for Well-Moderted Thennl Rector Spectr", C.H. Westcott.

13 467 cm + 68 cm cm where we used vlues from Tble 5. nd Tble 6.. ii. The criticl mss is bsed on the full rel-volume, so we must first compute from : d.d 4 cm cm 46 cm, where we used D from Tble 5.. We cn now find the tomic density of U s follows: N U N F βn S βn C. 5 4 toms 9 toms.8 cm.8 cm, where we used vlues from Tble II.. We cn represent this quntity s mss density insted: 9 toms mol 5 g.8 cm g toms mol cm. Finlly, we find the criticl mss: m U ρv ρ iii. We recll from Problems 6- nd 6-4 tht the flux is given by. 4 g kg cm 46 cm.9 kg g φx, y, z P 8E R Σ f sin / x y z. We cn mximize this by choosing x, y, nd z so tht ech of the ine terms is mximized. This occurs principlly when x y z, so ech ine term is unity. Then, we cn simply compute the rest. We first find Σ f for the urnium-5: Σ U f T gu f TN Uσ U T f T T, where we hve T 5 C 5.5 K nd T 9.6 K. Then: Σ U f toms cm cm Finlly, we find the mximum flux ssuming tht recoverble energy of MeV: 9.6 K 5.5 K.9 4 cm. φ mx φ,, kw.6 8 MeV 9 J 46 cm.9 ev 4 cm 4 cm sin 4 cm which evlutes to φ mx 6. 9 neutrons cm sec

13: Diffusion in 2 Energy Groups

13: Diffusion in 2 Energy Groups 3: Diffusion in Energy Groups B. Rouben McMster University Course EP 4D3/6D3 Nucler Rector Anlysis (Rector Physics) 5 Sept.-Dec. 5 September Contents We study the diffusion eqution in two energy groups