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B. Rouben McMaster University 4D03/6D03 Nuclear Reactor Analysis 2015 Sept.-Dec. Assignment 2 Assigned 2015/09/16 Due by 2015/09/23 11:00 am 6 Problems; 4 problems will be marked Total for marked problems: 36 marks 1. [8 marks] A reactor is critical up to t = 0, at which instant a perturbation occurs and the reactor multiplication constant starts to increase. The perturbation inserts reactivity linearly with time, at the rate of 4 mk/s up to t = 1.5 s, and at the rate of 3 mk/s thereafter. To counter the supercriticality, the reactor shutdown system (SDS, consisting of vertical shutoff rods) is actuated automatically at 0.9 s. It takes 0.3 s for the SDS to enter the reactor and a further 2 s to come to full insertion. The SDS has a total reactivity of -72 mk, which it inserts linearly with time during the 2 s of its descent within the reactor. Tabulate keff and the reactivity at 0.1-s intervals, from t = 0 to t = 3 s. Show keff to 2 decimal places and in mk, to a tenth of a mk. Plot vs. t. Solution: In separate file. 2. [10 marks] This is an exercise on quantitative aspects of the neutron cycle. Refer to the figure below which pertains to a critical reactor. Refer also to the notes in the figure. You are asked to calculate how many epithermal and fast neutrons are absorbed in the reactor per unit time. Remember that the two main things that can happen when a neutron is absorbed in the fuel are capture (where the neutron is absorbed and a gamma ray is emitted) and fission (where the neutron is absorbed and fission is induced). So the ratio of capture to fission, /f, is an important parameter. To solve the problem, use the data given and find the right sequence (up and/or down) for filling numbers into the boxes. Note that numbers need not be exact integers. In each box, retain non-integer numbers to 3 decimal places. 1000 fast neutrons Fast Fissions Fast Leakage Epithermal and Fast Absorption 1000 fast neutrons are born from thermal fissions A thermal fission releases 2.41 fast neutrons 20 thermal neutrons leak out For the fuel, /f = 1.12 28 fast neutrons are born from fast fission 39 thermal neutrons are absorbed in other than the fuel Fast leakage = 40% of thermal leakage Thermal Leakage Thermal Absorptions in Fuel Non-Fuel Thermal Absorption Thermal Captures Thermal Fissions Solution: (a) 1000 fast neutrons are born from fission & a thermal fission releases 2.41 neutrons Number of thermal fissions = 1000/2.41 = 414.938 (bottom box) (b) Using the number of thermal fissions & value of Thermal captures in fuel = 414.938* = 414.938*1.12 = 464.730 (2nd box from bottom) (c) Using (a) & (b) Thermal absorptions in fuel = 414.938+464.730 = 879.668 (3rd box from bottom) (d) Using (c) & number of thermal neutrons absorbed in other than fuel total number of thermal neutrons absorbed anywhere = 879.668+39 = 918.668 (5th box from bottom) (e) Using (d) and thermal leakage total number of fast neutrons which are thermalized = 918.668+20 = 938.668 (6th box from bottom) (f) From the clues, total number of fast neutrons born from all fissions = 1000 + 28 = 1028 (3rd box from top) (g) Fast leakage = 40%*20 = 8. Using this and (f) Number of neutrons surviving fast leakage = 1028 - 8 = 1020 (5th box from top) (h) Using (g) and (e) Result Sought, Number of epithermal and fast neutrons absorbed = 1020 – 938.668 = 81.332 3. [8 marks] Calculate the neutron flux due to U-238 spontaneous fission in a universe uniformly filled with U-238. Use the following data (not guaranteed to be totally accurate): Atomic mass M = 238 g.mol-1 Avogadro’s constant = 6.022*1023 mol-1. Density = 19.1 g.cm-3 Half-life 1/2= 4.468*109 yr Number of days in a year = 365.25 U-238 decay mode: Overwhelmingly -decay, with branching ratio for spontaneous fission = 5.5*10-5% Number of neutrons released per spontaneous fission = 2.4. Microscopic absorption cross section a = 2.73 b Compare this to the order of magnitude of the flux in the moderator of a CANDU reactor at full power (FP): ~1014 cm-2.s-1 Solution: First we calculate the U - 238 nuclide density N : N A0 6.022 * 10 23 mol 1 * 19.1 g.cm 3 4.833 *10 22 cm 3 M 238 g.mol 1 Overall U - 238 decay constant ln 2 1/2 0.693 4.916 *10 18 s 1 1 4.468 *10 yr * 365.25 * 24 * 3600 s. yr Now we will need th U - 238 decay constant for spontaneou s fission (SF) SF : 9 SF * Branching Ratio for SF 4.916 * 10 18 s 1 * 5.5 *10 7 2.704 * 10 24 s 1 We can now calculate the fission source S. If we assume 2.4 neutrons released per spontaneou s fission, we get S SF N * 2.4 2.704 *10 24 s 1 * 4.833 * 10 22 cm 3 * 2.4 3.137 *10 1 cm 3 .s 1 We also need the macroscopi c absorption cross section a : a N a 4.833 * 10 22 cm 3 * 2.73 *10 24 cm 2 1.319 * 10 1 cm 1 The neutron flux due to U - 238 SF and the absorption cross section is 3.137 *10 1 cm 3 .s 1 S 3.106 cm 2 .s 1 1 1 a 1.319 *10 cm This is about 14 orders of magnitude lower than the neutron flux in the moderator of a CANDU reactor at Full Power! 4. [Not to be marked] In an infinite uniform non-multiplying medium where the diffusion length and diffusion coefficient are L and D, there are four plane sources of neutrons parallel to the x–y plane. The plane sources are located at z = a, z = b, z = c, and z = d respectively, where the signs of a, b, c and d can be anything, but where you know that d > c > b > a. The source strengths are Sa, Sb, Sc and Sd (per cm2 of plane) respectively. So it is clear that source Sd is definitely in the “highest” z location, and source Sa is in the “lowest” z location. There are 5 regions of space which do not include the planes. Find the flux at any point z in each of these regions. Solution: The answer can be written very simply as a sin gle formula ( for any po int z not on the source planes ) : S L S L S L S L z a e z a / L b e z b / L c e z c / L d e z d / L 2D 2D 2D 2D 5. [10 marks total; 4 for (a), 6 for (b)] An infinite plane source of strength S = 2*106 cm-2.s-1 is located on the x-y plane in a uniform infinite non-multiplying medium with L = 6 cm and D = 2 cm. There is also a point source of strength Sp = 5*1010 s-1 at (x, y, z) = ((0, 0, 10 cm). Consider point P located halfway between the point source and the plane source. Calculate: (a) the neutron flux at P (b) the current (magnitude and direction) at P Solution: (a ) P is located at z 5 cm, at equal dis tan ce from the plane source and the po int source . We can write the total flux at P as the sum of the fluxes from the 2 sources. S p e r / L SL z / L i.e., P *e * 2D 4D r 8 2 1 2 *10 cm .s * 6 cm 5 cm / 6 cm 5 *1010 s 1 e 5 cm / 6 cm *e * 2 * 2 cm 4 * 3.1416 * 2 cm 5 cm (1.304 * 10 6 1.729 *10 8 ) cm 2 .s 1 1.742 *10 8 cm 2 .s 1 (b) The total current at P is the sum of the 2 currents from the 2 po int sources. If kˆ is the unit vector along the positive z direction , then S z / L ˆ S p 1 1 e r / L J P e k kˆ 2 4 L r r 2 *10 6 cm 2 .s 1 5 cm / 6 cm ˆ 5 *1010 cm 1 1 1 e 5cm / 6 cm ˆ * *e k * k 2 4 * 3.1416 6cm 5 cm 5cm 4.346 *10 5 1.268 *10 8 cm 2 .s 1 kˆ 1.264 *10 7 cm 2 .s 1 kˆ The neutron current at P is in the negative z direction. 6. [This problem will not be marked. You should still do it. You are guided.] In this problem you will essentially derive the boundary condition with vacuum and the extrapolation distance. You are guided through the solution. One of the assumptions in deriving diffusion theory from transport theory is that the neutron flux has a “weak” angular dependence; by weak dependence is meant that the angular flux is only linearly anisotropic. In concrete terms, let us simplify to plane geometry for the moment, so that the angular flux () is dependent only on the (polar angle) from the z-axis and not on the azimuthal angle in the x-y plane. Also, the anisotropy of the angular flux implies that the total current will be only in the z direction, i.e., the magnitude J of the current is equal to the z component of the current. The figure below shows a “1-d” reactor on the left-hand side of the page and vacuum on the right-hand side, the boundary with the vacuum being at zs (I took the z-axis as horizontal in this example). The linearly anisotropic dependence is then expressed by saying that the angular flux is only linearly dependent on ( cos ), i.e., we can write ˆ A B (1) where A and B are constants (which we will determine). [In Eq. (1) I dropped the z variable for simplicity for now.] (a) By using the definitions for the total flux and the z-component of the current J (equal to the current’s magnitude) (see equations 2 and 4 in learning module 3), evaluate the integrals to write and J in terms of A and B, and thereby show explicitly that A 4 i.e., and B 3 J 4 4 3J 4 (2) (3) (b) Now let’s look at the boundary [The generaliza tion to a polar axis in any direction is (but you are not asked to show this ) : ˆ 3 J ˆ condition at the reactor surface (z = zs). (4)] 4 4 The transport-theory boundary condition is that the current is zero at any incoming angle, i.e., J(zs, < 0) = 0. Since in diffusion theory we eliminate angles from the treatment, the corresponding condition in diffusion theory is that the total incoming current is zero, i.e. J-(zs) = 0 (since incoming corresponds to - in the above figure). By evaluating the integral for J-(zs), i.e., the number of neutrons passing from vacuum into the reactor, corresponding to angles in the range /2 to , show explicitly that J z s 0 (5) 4 2 Then ( your job is almost finished ), u sin g the Fick ' s Law relationsh ip between J z s 0 implies that zs the total current and total flux, show that Eq.(5) gives z s D d z s 4 2 dz 0, i.e., 1 d 1 zs dz 2D (6) “Geometrically”, Eq. (6) means that if the flux is extrapolated linearly beyond the surface with the vacuum, the flux would appear to go to zero at a distance 2D beyond the boundary (see the Figure below). This is the “extrapolation distance” d. Since D = 1/(3tr) = tr/3, then d = 2tr/3. Note: A more accurate transport-theory treatment gives: Extrapolation distance d = 0.7104 tr = 2.1312 D (7) Solution: 2 1 1 (a) d d 2 A B d 2 A B 2 4A A ( S .1) 2 4 0 1 1 1 2 1 1 ˆ zˆd 2 A B d J J d 1 1 1 0 1 1 1 1 A B A B 4B 2 A 2 B 3 2 [ ] 3 3 2 1 2 3 2 3 3J B ( S .2) 4 3J We have derived Eq.(3) 4 4 2 0 0 (b) J z s d z d 2 A B d 0 1 1 0 1 A B J 1 2 A 2 B 3 2 0 (u sin g S .1 and S .2) ( S .3) 3 2 3 2 4 2 1 J zs 0 J zs 1 d z s dz zs 2 zs D 1 2D d dz zs zs 2 (U sin g Fick ' s Law) This is Eq. (6) QED