B. Rouben
McMaster University
4D03/6D03 Nuclear Reactor Analysis
2016 Sept.-Dec.
Assignment 6
Assigned 2016/11/09
Due by 2016/11/16 11:00 am
6 Problems (only 5 to be marked)
Total: 48 marks
1. [Not for marking]
A 235U-fuelled reactor originally operating at a constant power of 0.5 milliwatt
is placed on a constant positive reactor period. It reaches its full power of 90
megawatt within 4 hours and 23 minutes. What was the reactor period?
(b) Starting from full power, the same reactor was then subjected to a reactor
period of -1.5 minutes. When will the power reach 2 W?
Solution:
Powert   P0et / T
(a)
where P0  initial power and T  reactor period ( from definition of period )
(a ) Here P0  0.001 W
When t  4h 23 min  263 min, Power (t )  90MW
 90 M W  0.001 W * e 263 min/ T
T 
263 min
 90 *106 W
ln 
 0.001 W

263 min
 10.43 min
25.223



(b) Here P0  90 MW , T  1.5 min, and P(t )  2 W


t

 2 W  90 M W exp 
  1.5 min 
 2W

 t  1.5 min * ln 
  26.4 min
6
 90 *10 W 
2. [10 marks]
Consider a homogeneous, bare cylindrical reactor with diameter 6.1 m and height
6.6 m. In 1 energy group its average properties are:
 = 2.39, f = 0.00125 cm-1, a = 0.00312 cm-1, D=0.98 cm.
[Note: neglect the extrapolation distance.]
If the 1-group neutron speed is 2,255 m/s, calculate the neutron generation time 
and the prompt-neutron lifetime l*. [You will need to calculate keff for the reactor.]
If there were no delayed neutrons, what would be the reactor period?
Solution:
2
2
 2.405   3.1415 
5
2
B2  

 
  8.483 *10 cm
 305   660 
 f
2.39 * 0.00125
k eff 

 0.9327
2
DB   a 0.98 * 8.483 *10 5  0.00312
  1
1
1
 1
 0.00722
k eff
0.9327
1
1

 0.00148 s
 f  2.39 * 0.00125 cm 1 * 2255 m / s

* 
1
1

 0.00138 s
5
DB   a  0.98 * 8.483 *10  0.00312 cm 1 * 2255 m / s

2



If there were no delayed neutrons, the flux or power would evolve as
t
 0.00722
t
e   e 0.00148  e 48.63t , where t is in sec onds
keff 1
(or , equivalently , it will evolve as e
*
t
e
0.93271
t
0.00138s
 e 48.63t )
Without delayed neutrons, we would therefore have
Re actor period 
1
s  0.0206 s negative period , sin ce   0 .
 48.63
3. [10 pts: 2 each for a) and c), 3 each for b) and d)]
The average following values are given for a certain critical reactor:
p = 2.23
d = 0.014
f = 0.004 cm-1
 = 2.3 km/s
There is one group of delayed-neutron precursors, with  = 0.12 s-1.
This reactor is suddenly subjected to a reactivity insertion of 100 pcm.
a) Calculate the delayed-neutron fraction.
Solution:
   p  d  2.23  0.014  2.244
 d 0.014

 0.00624
 2.244
b) Calculate the approximate exponents in the exponential terms for the evolution of
the power.
Solution:
100 pcm  1 mk  0.001
Delayed  neutron fraction  
1 

0.12 s 1 * 0.001 0.00012 s 1


 0.0229 s 1
   0.00624  0.001
0.00524
Neutron  generation time  
2 
 


1
1

 4.844 *10 4 s
1
1
 f  1 2.244 * 0.004 cm * 230,000 cm.s
0.001  0.00624
 10.82 s 1
4.844 *10 4 s
Note huge difference in order of magnitude between 1 and  2 .
c) Calculate the approximate power at 10 s (taking initial power equal to unity).
Solution:
 1t
  2t
Power t  
e 
e  1.191e 0.0229*10  0.191e 10.82*10
 
 
The 2nd exp onential is totally negligible
 Power 10 s   1.191e 0.229  1.59
d) What would the power at 10 s have been without delayed neutrons?
Solution:
Without delayed neutrons,
Powert   e
t

 Power10 s   e
0.001*10 s
4.844*10 4 s
Hallelujah for delayed neutrons!
 e 20.64  9.2 *108 !!!
4. [8 pts: 4 for a), 4 for b)]
Consider a reactor with 1 delayed-neutron-precursor group, with  = 0.02 s-1.
The reactor shutdown system can insert -80 mk essentially instantaneously.
Following the actuation of the shutdown system, it is found that the reactor is on a
stable period of -52 s.
a) Evaluate the delayed-neutron fraction.
Solution:
1  
The stable period 

 52 s
1


  0.080
0.02 s 1 * ( 0.080)
 52 s
   52 * 0.0016  0.080  0.0832  0.080  0.0032
b) If the reactor started at neutron power of 730 MW, what is the neutron power 5 s
after actuation of the shutdown system?
Solution:
Following the decay of the exp onential with exp onent  2 , the equation for the
(relative) power is
t
Pt 


e  
P0    

P5s 
0.0032


e
P0 0.0032  0.080
0.02 s 1 *(0.080)
*5 s
0.0032 0.080
0.008

0.0032  0.0832
e
0.0832
 0.0384615 e 0.0961538  0.0384615 * 0.908324
 0.0349 [ 3.49%]
 P5s   0.0349 * 730 MW  25.5 MW
5. [8 pts: 4 for a), 4 for b)]
Consider a reactor with 1 delayed-neutron-precursor group, with a delayed-neutron
fraction of 0.0056 and  = 0.05 s-1.
The reactor is running in steady-state critical at a neutron power of 1 (arbitrary units),
when an unknown perturbation suddenly inserts some positive reactivity, and the
neutron power jumps to a value of 1.1.
a) What was the reactivity inserted by the perturbation?
Solution:
U sin g the prompt  jump approximation :

0.0056
 1 .1 
 1 .1
 
0.0056  
0.0056
   0.0056 
 0.000509  0.51 mk
1 .1
b) If the perturbation is not countered, when will the neutron power reach 1.3?
Solution:
The evolution of the power is
P t  

 
e

t
 
 1.1e
0.05 s 1 *( 0.000509)
t(s)
0.0056 0.000509
 1.1e 0.00500t
 The time at which the power reaches 1.3 satisfies
1.3  1.1 e 0.005t
 1.3 
 0.005t  ln  
 1.1 
 0.005t  ln 1.1818
t 
0.167054
s  33.41 s
0.005
6. [12 marks}
A reactor has 1 precursor of delayed neutrons. The precursor has half-life 96 s. The
reactor is in steady state at a power of 1.000 (arbitrary units). Then, starting at t = 0,
an instantaneous reactivity was inserted and kept constant. The power increased and
was found to be 1.823 at t = 5s.
If the reactivity insertion had been only 1/3 of the value it was in the description
above, what value would the power have reached at t = 5 s?
Solution:
The decay cons tan t  of the precursor group is given by  
ln 2
 0.00722 028 s 1
96 s
By t  5s we know we can ignore the  2 term, and we can write
P (t )  P (0) *

 
e

t
 
i.e., P5s   1.823  1 *

 
e
0.00722028s 1 * 
*5 s
 
Note that the right  hand side can be written in terms of sin gle var iable , say x 
0.036101x


1
In terms of x, 1.823 
e 1 x
1 x
This is a transcendental equation and can be solved by trial and error or by Newton' s method.
I found x  0.435934.
Now u sin g 1 / 3 of this value, i.e., x  0.145311, we find P (5s )  1.1772