Nuclear Reactor Physics I Final Exam Solutions

.11 Nuclear Reactor Physics I Final Exam Solutions Author: Lulu Li Professor: Kord Smith May 5, 01

Prof. Smith wants to stress a couple of concepts that get people confused: Square cylinder means a cylinder whose height is the same as its diameter. Extrapolation distance in #6: be careful that d extra = D g is energy dependent so it s best to use R critical + D g in the expression for buckling, for instance, ( ) Bg π = (1) R critical + D g then solve for R critical. If you solve the two energy group using the same B then subtract D out of the critical radius it wouldn t be accurate because the extrapolation distance depends on energy. Energy-independent critical buckling: assume B to be the same for two groups, write out the -group diffusion equations, and solve for B. Units: it is recommended to put units with your solution, and also to carry the units through computation to check for consistency. For instance #30 it is less likely to make mistakes if we carry the units for α, β, ρ around. Fission yield vs. fission fraction. See #14. Background: All problems in this exam use the neutronics data supplied in Table 1 and two-group diffusion theory.

General Knowledge 1. What are the average energies at which prompt and delayed fission neutrons are emitted? Answer: MeV for prompt neutrons, 0.4 MeV for delayed neutrons.. What are the units of real and adjoint flux? Answer: Real flux has a unit of neutrons per cm per second. Adjoint flux is typically unitless. 3. What is the difference between true and effective downscatter in two-group cross sections? Answer: True down-scattering cross section Σ s,1 measures the probability of scattering from group 1 to group. Effective down-scattering cross section ˆΣ s,1 = Σ s,1 Σ s, 1 is like combining flux-weighted up-scattering and down-scattering into one down-scattering term. 4. What phenomenon is responsible for the 1/v tail of thermal scattering cross sections for isotopes with constant elastic cross sections at higher energies? Answer: Thermal motion/thermal vibration of the interaction material. 5. What is the mean number of isotropic elastic scattering collisions with Livermorium, Lv 93 116, required to slow neutrons from 1 MeV ( to 1.0) ev? A 1 Answer: A = 93, α = = 0.986, ξ = 1 + α ln α A + 1 1 α = 6.8 10 3, n = ln(e /E 1 ) = 08.6. ξ Hence it requires 09 collisions. 6. What is the average spacing (in ev) of U 38 resolved capture resonances? Answer: 5 ev. 7. For Einsteinium Es 53 in the energe range from 11.10 ev to 99.653 ev, what is the ratio of group absorption cross section to resonance integral? ( ) E Answer: Recall that RI eff and σ g are related through: RI eff = σ g ln, hence the ratio is, σ g RI eff = 1 ( ) = 0.456. E ln E 1 8. What is the definition and units of dilution cross section for resonance absorbers? Answer: Dilusion cross section is defined as, E 1 σ d = N mr m N r () where N m, r m are the number density and cross section of the moderating material, and N r is the number density of the resonant material. σ d has the unit of cm. 9. What is the approximate flux disadvantage factor at 0.5 ev for a 17x17 PWR lattice? Answer: Notice flux disadvantage factor seem to be defined two ways: The moderator flux over the fuel flux (as in Reuss book, e.g. Fig. 9.4) in which case the flux disadvantage factor would be about 1.05 for standard fuel, 1.1 for MOX fuel. The fuel flux over the moderator flux as in Kord s lectures, in which case the flux disadvantage factor would be around 0.9 Hence results within 0.9 to 1.1 receive credits for this problem. 10. What is the approximate Doppler temperature coefficient in pcm/k in an LWR? Answer: -3 pcm/k (it is important that the Doppler coefficient is negative), see slide 17, Lecture 10. 3

11. What are the two primary sources for the production of Xe 135? Answer: Iodine decay (major source), fission/burnup (minor source). 1. What is resonance escape probability for a material with a resonance integral of 100 barns, a dilution cross section of 000 barns, and a mean logarithmic energy decrement of 0.333? Answer: Recall the expression for resonance escape probability: ( p exp RI ) ( ) eff 100 barns = exp = 0.861 (3) ξσ d 0.333 000 barns 13. What is Dancoff factor for unclad PWR fuel rods with radius 0.5cm in a voided uniform infinite lattice with a pitch of 1.99 cm? Answer: When the moderator is void, that is the opacity is zero, hence Dancoff factor is 1.0. 14. What is the difference between delayed neutron fraction and delayed neutron yield? Answer: Absolute delayed neutron yield is the number of delayed neutrons per fission. Relative delayed neutron yield is the number of delayed neutrons per fission for an isotope divided by number of delayed neutrons per fission for all isotopes. Delayed neutron fraction is the absolute yield divided by ν. 15. What are the approximate delayed neutron fractions for U 35, U 38, and Pu 39? Answer: U 35 : 0.00665. U 38 : 0.01650. Pu 39 : 0.005. 16. What is an approximate expression for the extrapolation distance in diffusion theory for one-group bare homogeneous reactor? Answer: = 3Σ tr 3 λ tr = D. 17. What is an expression for the mean cosine of the scattering angle in the lab system for an isotope with atomic mass A, if scattering is isotropic in the CM system? Answer: In 3D it would be cos θ = 0, cos φ = 3A. 18. For dense matrices of size N, what is the order of numerical operations required to find the full matrix inverse, as N becomes large? Answer: N 3. 19. If an infinite-repeating lattice calculation that is used to produce flux form functions and discontinuity factors has assembly-averaged flux of 0.9 10 13 and an surface-averaged flux of 1.1 10 13, what is the assembly discontinuity factor (ADF)? Answer: Recall that ADF = surface-averaged flux asembly-averaged flux 1.1 1013 = = 1.. 0.9 1013 0. If two neighboring nodes have ADFs of 1.30 and 1.5 in the thermal group, what is the percentage discontinuity in homogeneous thermal flux at the shared interface when the nodal diffusion equations are solved using the NEM method? Answer: Recall f n φ HOM n = φ HET n difference in the homogeneous thermal flux is, = φ HET n+1 = f n+1 φ HOM n+1, then φhom n+1 φ HOM n 1.30 1.5 (1.30 + 1.5)/ = 3.9%. = f n f n+1. Thus the percentage 4

Exam 1 Material 1. If paraffin (C 5 H 5 ) is used as the moderator and coolant for a new district-heating reactor, what is the macroscopic elastic scattering cross section of paraffine at 10.0 ev and a density of 0.8 gm/cc? Answer: First we calculate the molecular number density, N molecule = ρn A M = 0.8 0.60 35 = 0.00137 10 4 1/cm 3 (4) Then we know the number density for Carbon atoms are N C = 5N molecule, N H = 5N molecule. Recall those elastic scattering cross section plots we had in the beginning of the semester (or consulting KAERI s table of nuclides), σ C = 4.7 barns, σ H = 0 barns, then Σ = σ C N C + σ H N H = 5 0.00137 4.7 + 5 0.00137 0 = 1.584 cm 1 (5) If you use 5 barns for σ C, the final answer should be 1.59cm 1.. Using Bell s approximation, estimate the ratio of escape cross section of an isolated 0.5 cm radius rod to the fuel escape cross section in a uniform hexagonal lattice with a Dancoff factor of 0.6? (1 C)b Answer: Essentially we are asked to approximate. We are already given the Dancoff factor (1 C) + Cb C = 0.6. Recall the plot of Bell s factor for various geometries, we approximate b as 1.1 (as in the examples in lectures), then σ e (1 C) + Cb 0.4 + 0.6b (1 C)b (1 C)+Cb σ = = =.4 (6) (1 C)b 0.4b e 5

Exam Material 3. Compute the finite-medium energy-independent critical buckling and fast-to-thermal flux ratio for Fuel 1. Answer: This is essentially #10 in Exam. We start by writing the -group diffusion equations with B being the same for the two groups, From Eq. 8, we can write, Plug into Eq. 7 and divide every term by we get, D 1 B + (Σ a1 + Σ s,1 ) = νσ f1 + νσ f (7) D B + Σ a = Σ s,1 (8) = D B + Σ a Σ s,1 (9) (D 1 B + Σ a1 + Σ s,1 νσ f1 ) D B + Σ a Σ s,1 = νσ f (10) Solving this quadratic equation, we get B = 0.008876. Then plug into Eq. 9, we get, = D B + Σ a Σ s,1 = 4.31 (11) 4. Using the infinite medium spectrum from #3, compute the effective one-group cros sections for Fuel 1 by preserving reaction and leakage rates. Σφ Answer: Recall collapsing cross sections is essentially Σ =. Thus we do, φ νσ f = ν φ 1Σ 1 f1 + ν Σ f = 0.03515 (1) + 1 Σ a = Σ a1 φ1 + Σ a = 0.0384 (13) + 1 D = D 1 φ1 + D = 1.7387 (14) + 1 6

5. Using one-group cross sections from #4, what is the one-group, critical radius of a bare square cylinder (i.e., height = diameter) using diffusion theory extrapolation distances for all surfaces? Answer: There are two slightly different ways to do it. Using collapsed one-group cross sections, we can write the material buckling as, B material = νσ f Σ a D = 0.008876 (15) The geometrical buckling with extrapolated distance D is, B geometrical = ( ) ( ).405 π R + D + R + 4 D (16) Setting the two bucklings to be equal, we can solve for the critical radius to be R = 7.94cm. Alternatively, we can recognize that we have already solved for B in #3 and get 0.008876 (which agrees with the previous method using collapsed one-group cross sections, because we are preserving leakage by collapsing D). Taking advantage of the square cylinder geometry and the assumption of one-group, if we define R = R + D where R is the critical radius we are actually solving for, R is the distance including the extrapolated distance, we can easily solving for R through and get R = 30.49cm, R = R D = 7.94cm. ( ).405 ( π ) B = R + R (17) 7

6. What is the two-group, critical radius of a bare square cylinder of Fuel 1 using diffusion theory extrapolation distances for each energy group and all surfaces? Answer: We absolutely cannot use the nd method in #5 anymore because the extrapolation distance D which is different for the two energy groups. Instead, we have to define a bunch of transver buckling terms (similar to #1 in Exam ). This problem can be break into the following steps: (a) Defining transverse buckling terms: there are two transverse leakage directions, the radial direction and the z-direction, (.405 Bg,R = R + D g ) ( Bg,H = π (R + D g ) ) (18) where the subscript g means energy group (thus 1 or ), and subscript R, H designate Radial direction or Height direction. (b) Then we can write our two-group diffusion equations: (D 1 B 1,R + D 1 B 1,H + Σ a1 + Σ s1 ) = νσ f1 + νσ f (19) (D B,R + D B,H + Σ a ) = Σ s1 (0) (c) Solving the above system. We do the typical thing: write φ1 from Eq. 0, plug into Eq. 19 and divide both side by, = D B,R + D B,H + Σ a Σ s1 (1) D 1 B1,R + D 1 B1,H νσ f Σ s1 + Σ a1 + Σ s1 = νσ f1 + D B,R + D B,H + Σ a () We obtained a one equation one unknown (R), thus can solve for the critical radius R = 7.668cm. 8

7. Los Alamos scientists have created a system to monitor sub-criticality of spent fuel pools by comparing detector responses from two identical point sources, one source/detector pair placed in the pool and another source/detector pair placed in the fuel, as depicted in the following figure. What is the required distance, L, between the water pool source and detector such that the signal will be exactly equal to the fuel source and detector signal as the k of the fuel reaches the regulatory limit of 0.95? Answer: Recall in lecture we solved the point source in an infinite non-multiplying medium problem to have φ(r) = S 0 4πD e r/l r φ(r) = S 0 4πD. This problem is essentially the same set-up, thus we can write, e B r r, where B = Σ F a νσ F f D F = 0.0707 Fuel region Σ P a D P = 0.1414 Pool region (3) where νσ F f = k Σ F a. A quick review of how we obtained the above flux expression, Start by writing the diffusion equation using spherical coordinate systems, The solution is in the form of, d φ dr + dφ r dr B φ = S 0 δ( r ) (4) D φ(r) = A e B r r + B e B r r (5) BC1: lim = 0, thus B = 0. r BC: lim r 0 4πr J(r) = S 0, thus φ(r) = S 0 e B r 4πD r (6) Coming back to Eq. 3, the question is basically asking for, S 0 e 0.1414r 4πD r = S 0 e 0.0707 00 4πD 00 (7) Thus r = 104.58cm. 9

New Material 8. When a control rod worth 37 cents is removed from a critical reactor (with beta-effective of 0.007), what is the ratio of power after and before the reactivity insertion (e.g., the prompt jump)? Answer: We use prompt jump approximation, ρ = $0.37 = 0.37β, T (0 + ) = β T 0 β ρ = β = 1.59 (8) β 0.37β 9. If a fully-enriched, uranium-fueld, thermal-spectrum reactor has a stable period of 60 seconds after a control rod is moved following long-term stable operation at power of 100 watts, how much reactivity (in dollars) was inserted? (note: use the eight group delayed neutron data from Lecture 17, slide 1). Answer: Given β i, λ i etc, we can calculate: β = 6.65 10 3, ω = 1 T = 1/60s 1 ; Λ = 1 vνσ f 1.87 10 5 s. Plug everything in the In-hour equation, ρ = ωλ + 8 i=1 β i ω ω + λ i = 9.60 10 4 = $0.1444 (9) You can also omit the ωλ part as it is very small and get ρ = 9.59 10 4 = $0.1443. 30. If a room temperature critical reactor undergoes a $5.0 reactivity excursion from near zero power, what is the approximate final core-averaged fuel temperature (K)? Note, assume prompt neutron lifetime of 10 5 sec, beta-effective of 0.007, specific heat of the fuel of 1 J/gm-K, and Doppler coefficient of 3.0 pcm/c Answer: Be careful about the units! First we convert all the units into k : ρ β = 4β = 0.08 k, α = 3 10 5 k. Thus, Tfuel end (ρ β) = + Tfuel 0 = 0.08 k α 3 10 5 + 93 = 159.67K (30) k 31. What is the fast-to-thermal adjoint flux ratio in an un-reflected critical slab reactor of Fuel 1 material? Answer: Keep in mind that un-reflected suggests there are leakage terms. We have solved earlier in #3 that B = 0.008876, which would remains the same for the forward problem and the adjoint problem. We write the two-group diffusion equations for the adjoint problem, From Eq. 31, we can write (D 1 B + Σ a1 + Σ s1 νσ f1 )φ 1 = Σ s1 φ (31) νσ f φ 1 = (D B + Σ a )φ (3) φ 1 φ = Σ s1 D 1 B + Σ a1 + Σ s1 νσ f1 = 0.655 (33) From Eq. 3, we can write φ 1 φ = D B + Σ a νσ f = 0.655 (34) Notice the fast-to-thermal adjoint flux ratio can be solved from either equation and they come out to be the same. 10

3. A CASMO-5 calculation for a fuel assembly, core baffle, and homogeneous reflector (as depicted in the following figure) was performed to produce cross sections and discontinuity factors for a nodal diffusion model of the equivalent baffle/reflector nodes. Answer: First of all, we only need to worry about fast group for this problem. Secondly, since we are asked for the DF of the left-face of the homogenized baffle/reflector node, and we are giving information at the right face of the fuel node, we only need to worry about the baffle/reflector node. (a) We expand the flux using 4th order polynomial, ( (ξ) = φ + a 1 ξ + a (3ξ 1/4) + a 3 (ξ 3 1/4ξ) + a 4 ξ 4 3 10 ξ + 1 ) 80 (35) (b) There are two other expressions that we need: the net current expression is, J 1 (ξ) = D 1 h 1 d (ξ) dξ (36) and the diffusion equation is, D 1 h d (ξ) dξ + Σ a1 (ξ) + Σ s1 (ξ) (37) (c) There are four unknowns, a 1, a, a 3, a 4, and we need 4 constraints: J( 1/) = 0.55701. J(1/) = 0. Multiply diffusion equation by first order polynomial (ξ), integrate over -1/ to 1/, set it to zero. Multiply diffusion equation by second order polynomial (3ξ 1/4), integrate over -1/ to 1/, set it to zero. (d) Solve the above 4-equations-4-unknowns system by hand or by Mathematica and get the four coefficients to be: a 1 =.701 a = 1.3739 a 3 = 4.77 a 4 = 3.587 (38) Plug in ξ = 1/, we get the homogeneous flux at the interface to be ( 1/) =.89466, hence DF= 3.45.89466 = 1.1919. 11

33. Using the SANM nodal model, compute the thermal group discontinuity factor for the left face of the homogenized baffle/reflector node described in #3. Answer: The thermal flux is in the form of 1, (ξ) = a cosh ( κ ( ξ 1 )) + b sinh ( ( κ ξ 1 )) ( + (ξ = 1 ) Σs1 (39) Σ a with its net current expression as, J (ξ) = D h dφ(ξ) dξ = D ( ( ( h κ a sinh κ ξ 1 )) ( ( + b cosh κ ξ 1 ))) (40) There are two unknowns a, b, and we have two constraints: We can solve for a, b to be, J ( 1/) = 0.05509, J (1/) = 0 (41) 0.05509 a = = 0.00470603, b = 0 (4) D h κ sinh( κ) Plug in ξ = 1/, we get the homogeneous flux at the interface to be ( 1/) = 5.43808, hence DF = 0.63 5.43808 = 0.11585. 1 use the ξ 1/ expressions if you are going to solve it by hand, because then the boundary condition of J(1/) = 0 would makes the coefficient of sinh goes to zero. If you use Mathematica, then the shifting term is not necessary, you can just let (ξ) = a sinh(κξ) + b cosh(κξ) + φ(ξ) Σ s1 ; the coefficients would come out to be different than the ξ 1/ case, but the final results would be all the same as Σ a tested using Mathematica. 1