B. Rouben
McMaster University
4D03/6D03 Nuclear Reactor Analysis
2015 Sept.-Dec.
Assignment 9
Assigned 2015/11/18
Due by 2015/11/25 11 am
3 problems, total 25 marks.
1. [15 marks] Run the Exkin program 6 times with different inputs, as follows:
[Note: I posted the executable code Exkin2.exe (file “EXKIN Program” on the course
website) and a sample input Exkin.txt (“EXKIN Sample Input” on the website). You need to
copy both files to your computer. To run the code, first ensure that you modify the input file
according to the problem at hand, and then click on Exkin2.exe. Each time you run the
program, you will get output files EXKINOUT.TXT and EXOUT.TXT. You can ignore
EXOUT.TXT entirely. But you must in any case remove the files or change their names
before running Exkin again.]
(a) Case 1. Insert +4.8 mk linearly from 0 s to 1.2 s (i.e., reactivity =0.4 mk at 0.1.s, 0.8
mk at 0.2. s, etc., until 1.2 s), then leave the reactivity at 4.8 mk. This may be typical
of a loss-of-coolant accident without action of a shutdown system. What are the
maximum power over the 3-s interval, and the integrated power at 3 s (in FullPower-Seconds, FPS)?
(b) Case 2. Same as Case 1, but starting at 1.1 s insert (in addition to the loss-of-coolant
reactivity) a shutdown system reactivity of -60 mk linearly over 1.5 s, i.e., an
additional -4 mk every 0.1 s. So the total reactivity will be +0.8 mk at 1.2 s, -3.2 mk
at 1.3 s, -7.2 mk at 1.3 s, etc., until -55.2 mk at 2.6 s, and then constant to 3 s. What
is the maximum power over the 3-s interval, and the integrated power at 3 s? How
much has the shutdown system helped in reducing these values?
(c) Case 3. Same as Case 2, but the shutdown system (still worth -60 mk) starts to act
0.1 s earlier, i.e., it starts at 1.0 s. So the net reactivity will be +0.4 mk at 1.1 s, -3.2
mk at 1.2 s, etc., … , and maximum negative reactivity = -55.2 mk at 2.5 s, then
constant to 3 s. How much did the 0.1 s help in maximum power and integrated
power?
(d) Case 4. Same as Case 2, but the shutdown system starts acting at 0.9 s. How much
did this extra 0.1 s help?
(e) Case 5. Same as Case 2, but the shutdown system starts acting only at 1.2 s. How
much worse is this than Case 2?
(f) Same as Case 2, but the full shutdown-system reactivity is -75 mk and its insertion
time is as in Case 2. Compare the results to those of Case 3 and of Case 4. What is
more effective: having 15 negative mk more in shutdown-system reactivity, or
inserting the shutdown system earlier by 0.1 s or 0.2 s?
2. [5 marks] A certain CANDU reactor has a fuel-temperature coefficient of -.007 mk/C
and a full-core void reactivity of +14 mk. At steady state the average fuel temperature
is 695 C and the average coolant density is 0.82 g.cm-3. What will be the reactivity if,
in a certain accident, the fuel temperature goes to 1,300 C and the coolant density
goes to 0.40 g.cm-3? How much B would have to be injected “instantaneously” to
bring the reactivity to -10 mk?
3. [5 marks] A steady-state critical reactor can be thought of as a subcritical assembly
with a reactivity of - ( the “reactivity” of the delayed neutrons), with the steadystate flux being the subcritically multiplied “external” source represented by the
delayed neutrons.
Derive the prompt-neutron jump (or drop) formula by starting with this model and
showing what the flux will be (in relation to the original flux) if an additional step
reactivity  is inserted.