Download Assignment-2 Solutions

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
4D 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/(3tr) = tr/3, then d = 2tr/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   4A  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  4B
 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