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B. Rouben McMaster University 4D03/6D03 Nuclear Reactor Analysis 2016 Sept.-Dec. Assignment 2 Assigned 2016/09/21 Due by 2016/09/28 11:00 am 5 Problems; 4 problems will be marked. Total for marked problems: 35 marks 1. [10 marks] At t = 0, a certain reactor is in steady state with reactivity = 0. An incident happens where the reactivity starts to increase linearly at the rate of 4 mk/s, for 3 s. But the protection systems identify the problem and actuate a shutdown system (SDS) at t = 0.8 s. The SDS enters the reactor core at t = 1.5 s. Beyond that time the SDS inserts a total reactivity of -72 mk at a linear rate over the next 1.6 s. Tabulate keff and the reactivity at 0.2-s intervals, from t = 0 to t = 4 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 thermal neutrons escape from 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 2.44 fast neutrons For the fuel, /f = 1.0735 30 fast neutrons are produced by fast fission 48 thermal neutrons are absorbed in other than the fuel 7 fast neutrons escape from the reactor 99 non-thermal neutrons are absorbed 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 2.43 neutrons Number of thermal fissions = 1000/2.44 = 409.84 (bottom box) (b) Using the number of thermal fissions & value of Thermal captures in fuel = 409.84* = 409.84*1.0735 = 439.96 (2nd box from bottom) (c) Using (a) & (b) Thermal absorptions in fuel = 409.84+439.96 = 849.80 (3rd box from bottom) (d) Using (c) & number of thermal neutrons absorbed in other than fuel total number of thermal neutrons absorbed anywhere = 849.80+48 = 897.80 (5th box from bottom) (e) Using number of neutrons from thermal fissions and fast fissions total number of fission (fast) neutrons = 1000+30 = 1030 (3rd box from top) (f) Using (e) & fast leakage Number of fast neutrons surviving fast leakage = 1030 - 7 = 1023 (5th box from top) (g) Using (f) & absorption of non-thermal neutrons Number of neutrons surviving to thermal energies = 1023 - 99 = 924 (7th box from top) (h) Thermal leakage number of thermal neutrons surviving to thermal energies – number of thermal neutrons absorbed = (g)-(d) = 924 – 897.8 = 26.20 3. [5 marks] In an infinite uniform non-multiplying medium where the diffusion length and diffusion coefficient are L and D, there are four infinite plane sources of neutrons. Two of them are at z = a and z = b. The other two are at y = c and x = d respectively. The source strengths are Sa, Sb, Sc and Sd (per cm2 of plane) respectively. Write a single equation for the total flux at any point (x, y, z) in space not on any of the planes. Solution: The answer can be written very simply as a sin gle formula ( for any po int ( x, y, 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 y c / L d e x d / L 2D 2D 2D 2D 4. [10 marks total; 4 for (a), 6 for (b)] Consider 2 point sources located in an infinite, homogeneous, non-multiplying medium. The point sources are on the x-y plane at positions (y = -8 cm, x = 0) and (y = +8 cm, x = 0). The sources are of equal strength S = 4*107 s-1. The medium has L = 5 cm and D = 2 cm. (a) Derive the mathematical expression and calculate the numerical value of the flux at the point (x = -16 cm, y = 0 cm) on the x-y plane. (b) Calculate the magnitude and direction of the total neutron current at the same point. Solution: S e r / L (a ) The flux at any location P in space from a po int source is , where r is the 4D r dis tan ce between P and the po int source . Here D 2 cm, L 5 cm, S 4 * 10 7 s 1 for both sources. The dis tan ce of P ( x 16 cm, y 0 cm) from each source is 16 2 8 2 17.889 cm. The total flux at P is the sum of the fluxes from the 2 sources. These 2 fluxes are equal. P 2 * 4 * 10 7 s 1 e 17.889cm / 5cm * 4 * 2 cm 17.889 cm 4.972 * 10 3 cm 2 .s 1 (b) The current from a po int source to P S 1 1 e r / L rˆ, where rˆ is the unit vector 4 L r r from the source to P. The total current at P is the sum of the currents from the 2 po int sources. Since P is at the same dis tan ce from each of the 2 po int sources , and the sources are of equal strength, then the magnitudes of the 2 currents are equal. By symmetry, the y components of the 2 current " arrows" will cancel one another, and the x components will add up, .i.e., the total current will be in the negative x direction! 16 The x components are calculated u sin g the cos ine of the angle to the x axis, i.e., 0.894 17.889 e 17.889 / 5 4 * 10 7 s 1 1 1 * i.e., J 2 * 0.894 * * (iˆ) 4 * 3.1416 5 cm 17.889 cm 17.889 cm 2.276 * 10 3 cm 2 .s 1 iˆ, where iˆ is the unit vector along the positive x direction. 5. [This problem will not be marked. You should still do it.] 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 z s J z s explicitly that J z s 0 implies that 0 (5) 4 2 Then ( your job is almost finished ), u sin g the Fick ' s Law relationsh ip between 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