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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
4D
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/(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