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Chapter 5 Subcritical Multiplication
Ryan Schow
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OBJECTIVES 1. Define subcritical multiplication.
2. Describe primary and secondary neutron sources,
critical loading experiment, and addition of a neutron
source to a critical reactor.
3. Plot inverse multiplication versus keff or fuel mass and
explain the possible curve shapes.
4. Solve problems involving subcritical count rates, core
reactivity, and rod worth.
4. Explain how doubling count rate during a startup can
be used by the operator to predict criticality.
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Primary -n
NEUTRON HIERARCHY
NEUTRONS
FISSION SOURCE
PROMPT DELAYED INTRINSIC INSTALLED
-n S.F. Secondary
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PHOTO-NEUTRON REACTION
n H H 1
0
1
1
2
1
nHHH 1
0
1
1
*2
1
2
1
0
0
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ALPHA-NEUTRON REACTION
nNeOHe 1
0
21
10
18
8
4
2
nNeO 1
0
21
10
18
8
4
2
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INTRINSIC NEUTRON SOURCE RELATIVE STRENGHTS
Immediately following a reactor shutdown:
1. Photo-Neutron Sources
2. Alpha-Neutron (Transuranic) Sources
3. Spontaneous Fission Sources
Several weeks following shutdown:
1. Alpha-Neutron (Transuranic) Sources
2. Spontaneous Fission Sources
3. Photo-Neutron Sources
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INSTALLED NEUTRON SOURCES
nBeBe 1
0
8
4
9
4
e Te Sb 0
1-
124
52124
51
Sb n Sb 124
51
1
0
123
51
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SUBCRITICAL MULTIPLICATION
31 73 139 200 100 274 300
500
1000
1500
2000
666
GENERATIONS
keff = 0.80
keff = 0.85
keff = 0.90
keff = 0.95
NU
MB
ER
OF
NE
UT
RO
NS
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SUBCRITICAL MULTIPLICATION WITH keff = 0.80
Generation Neutrons from
Fission
Source
Neutrons
Total Net
Gain
0
1
-
n
3
4
2
0
80
-
400
195
236
144
100
180
-
500
295
336
244
-
80
-
0
51
41
64
100
100
-
100
100
100
100
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SUBCRITICAL MULTIPLICATION WITH keff = 0.85
Generation Neutrons from
Fission
Source
Neutrons
Total Net
Gain
0
1
-
n
3
4
2
400
425
-
566
464
479
446
500
525
-
666
564
579
546
-
25
-
0
18
15
21
100
100
-
100
100
100
100
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SUBCRITICAL MULTIPLICATION WITH keff = 0.90
Generation Neutrons from
Fission
Source
Neutrons
Total Net
Gain
0
1
-
n
3
4
2
566
600
-
900
657
681
630
666
700
-
1,000
757
781
730
-
34
-
0
27
24
30
100
100
-
100
100
100
100
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NEUTRON POPULATION
Nn = number of neutrons present after n generations
So = source strength in neutrons per generation
keff = effective multiplication factor
n = number of generations
Nt = the total number of neutrons present at equilibrium
Where:
eff
n
effon
k1
k1SN
eff
otk1
1SN
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Assume keff becomes insignificant when keff is equal to
0.001. Determine the number of generations required for
keffn to equal 0.001 when keff is 0.85 and 0.95:
effkln
001.0lnn
(Take natural log of both sides) 001.0lnklnn eff
001.0kn
eff
When keff = 0.85 sgeneration4385.0ln
001.0lnn
sgeneration13595.0ln
001.0lnn When keff = 0.95
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EQUATION
N)k-(1S eff0
eff
o
k1
SN
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Neutron Life
Cycle
Neutrons
at Beginning of Cycle
Source
Strength
Neutron from Fission
at End of Cycle
Neutrons
lost in cycle
(1-keff)N
ROLE OF SOURCE NEUTRONS
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Assume that Reactor A has a keff = 0.8 and a neutron source
strength of 100 neutrons/ generation. Reactor B, which also
has a keff = 0.8, has a neutron source strength of 200
neutrons/generation. Of the two reactors described, which one
has the larger equilibrium neutron population?
Solution:
000,18.01
1200N
Bt
500
8.01
1100N
At
eff
otk1
1SN
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EQUATION
teff0 N)k-(1S
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A nuclear power plant that has been operating at rated power
for two months experiences a reactor scram. Two months after
the reactor scram, with all control rods still fully inserted, a
stable count rate of 20 cps is indicated on the source range
nuclear instruments.
The majority of the source range detector output is being
caused by the interaction of ____________ with the detector.
A. intrinsic source neutrons
B. fission gammas from previous power operation
C. fission neutrons from subcritical multiplication
D. delayed fission neutrons from previous power operation
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Solution:
This question is intended to mislead the student into thinking
that the different neutron source must be considered. After a
reactor trip, the keff of the core is still greater than 0.5. This
means that more than half of the neutrons in the core, and
therefore those interacting with the Source Range NI’s will
have come from fission of fuel.
C. is the correct answer.
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CR = neutron count rate indicated on the detector meter
Nt = actual neutron count in the core
h = detector efficiency
COUNT RATE
h
k1
1SCR
eff
o
h tNCR
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COUNT RATE RATIO
2
2eff
2
1
1eff
1
2
1
k1
1S
k1
1S
CR
CR
h
h
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COUNT RATE RATIO (cont’d.)
2ef f
1ef f
2
1
k1
1
k1
1
CR
CR
1
k1
k1
1
CR
CR 2ef f
1ef f2
1
1ef f
2ef f
2
1
k1
k1
CR
CR
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COUNT RATE RATIO
2eff21eff1 k1CRk1CR
On NRC Equation sheet
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Given an initial count rate of 100 cps and a keff = 0.9,
the operator pulls control rods and establishes a final
count rate of 200 cps. Determine the new keff.
Solution:
95.0k 2eff
05.01k 2eff
2effk1
9.01
100
200
2eff
1eff
1
2
k1
k1
CR
CR
9.01100)k1(200 2eff
200
10k1 2eff
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COUNTS VS keff DURING STARTUP
keff
400
200
100
300
.9 .925 .95 .975 1
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"HALVING the DISTANCE to CRITICALITY
DOUBLES the COUNTS"
When reactivity is added to a subcritical reactor in an
amount equal to the amount associated with
1/2 (1-keff), the count rate will double.
THUMB RULES
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"CRITICALITY in 5 to 7 DOUBLINGS"
When the initial count rate at the beginning of a
startup has doubled 5 - 7 times, the reactor
will be at or near critical.
THUMB RULES
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THUMB RULES
“FIXED REACTIVITY ADDITIONS
versus
COUNT RATE DOUBLING"
When enough reactivity is added to the reactor
to double the count rate,
if the same amount of reactivity is added to the
reactor again, the reactor will be supercritical.
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A reactor startup is initiated with an initial count rate of
100 cps and keff of 0.9.
After pulling the control rod bank a total of 200 steps,
the count rate has doubled to 200 counts.
Assume each step of control rod motion adds the same
amount of reactivity.
What would happen if the rods were pulled an additional
200 steps?
Solution: Reactivity associated with keff = 0.9
k/k1111.09.0
0.19.0o
eff
eff
k
1k
o
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After pulling the control rod bank a total of 200 steps:
Recall the thumb rule “HALVING the distance to
CRITICALITY DOUBLES the COUNTS” can also be seen as
“doubling the count rate will half the distance to criticality.”
Therefore we can now say that since the count rate was
doubled we are halfway to criticality.
Halfway between 0.9 and 1.0 is 0.95.
Reactivity associated with keff = 0.95
k/k0526.095.0
0.195.0eff
eff
k
1k
1
Reactivity added:
= 1 – o = -0.0526 - (-0.1111)
= 0.0585k/k
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After the rods were pulled an additional 200 steps:
Second rod pull adds the same amount of
reactivity, 0.0585 k/k.
= 2 – 1
2 = + 1
2 = 0.0585 + (-0.0526) = 0.0059k/k
1
1kef f
006.1k/k0059.01
1kef f
The reactor is supercritical!
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A reactor startup is being commenced with initial source
(startup) range count rate stable at 20 cps. After a period
of control rod withdrawal, count rate stabilizes at 80 cps.
If the total reactivity added by the above control rod
withdrawal is 4.5 %∆k/k, how much additional positive
reactivity must be inserted to make the reactor critical?
A. 1.5 %∆k/k
B. 2.0 %∆k/k
C. 2.5 %∆k/k
D. 3.0 %∆k/k
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Solution:
The easiest way to solve this problem is to use the
thumb rule regarding count doublings. We see that
there have been 2 doublings, from 20 cps to 40 cps and
then to 80 cps.
We do not know the amount of reactivity that was initially
necessary to bring the reactor critical, so we will call that
“x”.
The first doubling must have added one-half of x. The
second doubling added half again as much reactivity
(x/4). Thus the total reactivity added to that point is 3x/4.
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This means that 4.5 %∆k/k is ¾ of x.
Therefore, quick algebra shows us that x must have
been 4/3 times 4.5% or 6.0 %∆k/k.
So, we needed 6.0 %∆k/k but we have already added
4.5 %∆k/k, so the net remaining to be added is:
ANSWER = 1.5 %∆k/k.
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INVERSE MULTIPLICATION
Where:
CRn = some count rate at a condition “n”
Cro = initial count rate
o
n
CR
CRM
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RELATIONSHIP OF keff TO 1/M
2
1
eff
eff
1
2
k1
k1
CR
CR
M
11keff
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INVERSE MULTIPLICATION PLOT
So = source count rate
Nt = neutrons per second counted at a
particular time
Where:
1
2
CR
CRM
2
1
CR
CR
M
1
0.1sec/n100
sec/n100
N
S
M
1
t
o
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1/M PLOT FOR A FUEL LOADING
1.0
.9
.8
.7
.6
.5
.4
.3
.2
.1
0
0 10 20 30 40 50
A
B
C
D
E
F
G
INVERSE MULTIPLICATION PLOT
1/M
NUMBER OF FUEL ASSEMBLIES LOADED
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1/M PLOT
100 200 300 400 500
0.2
0.4
0.6
0.8
1.0
CONTROL ROD NOTCHES WITHDRAWN
1 /
M
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1/M PLOT DURING FUELING
OCCURS WHEN:
1. FUEL LOADED TOWARD DETECTOR
2. DETECTOR INITIALLY TOO FAR FROM CORE
3. DETECTOR TOO CLOSE TO SOURCE
1.0
0.8
0.6
0.4
0.2
0.0
0 2 4 6 8 10
INCORRECT DETECTOR LOCATION
FUEL ASSEMBLIES LOADED
1/M
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1/M PLOT FOR DETECTOR TOO FAR FROM SOURCE
1.0
0.8
0.6
0.4
0.2
0.0
0 2 4 6 8 10
DETECTOR TOO FAR AWAY FROM SOURCE
FUEL ASSEMBLIES LOADED
1/M
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1/M PLOT FOR APPROACH TO CRITICALITY
1.0
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
ROD HEIGHT (STEPS)
ICR
R
0 40 80 120 160 200 228 A
0 40 80 120 160 200 228 B
0 40 80 120 160 200 228 C
0 40 80 120 160 200 228 D
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A reactor is taken critical. The data for the approach to
criticality is shown below. Using this data, create a 1/M (or
ICRR) plot. After you have plotted your graph, refer to the
graphical solution to compare the results.
Rod Position Count Rate 1/M (ICRR)
0 400 cps
50 500 cps
100 890 cps
130 1,290 cps
160 1,905 cps
180 3,333 cps
200 8,000 cps
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EXAMPLE (cont’d) 1.0
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
0 50 100 150 200 230
ROD POSITION
—
0.8
0.45
0.31
0.21
0.12
0.05
400
500
890
1,290
1,905
3,333
8,000
0
50
100
130
160
180
200
ROD
POSITION CPS
1/M
OR
ICCR