Chapter 5 Subcritical Multiplication

Ryan Schow

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.

Primary -n

NEUTRON HIERARCHY

NEUTRONS

FISSION SOURCE

PROMPT DELAYED INTRINSIC INSTALLED

-n S.F. Secondary

PHOTO-NEUTRON REACTION

n H H 1

0

1

1

2

1

nHHH 1

0

1

1

*2

1

2

1

0

0

ALPHA-NEUTRON REACTION

nNeOHe 1

0

21

10

18

8

4

2

nNeO 1

0

21

10

18

8

4

2

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

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

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

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

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

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

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

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

EQUATION

N)k-(1S eff0

eff

o

k1

SN

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

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

EQUATION

teff0 N)k-(1S

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

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.

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

COUNT RATE RATIO

2

2eff

2

1

1eff

1

2

1

k1

1S

k1

1S

CR

CR

h

h

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

COUNT RATE RATIO

2eff21eff1 k1CRk1CR

On NRC Equation sheet

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

COUNTS VS keff DURING STARTUP

keff

400

200

100

300

.9 .925 .95 .975 1

"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

"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

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.

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

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

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!

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

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.

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.

INVERSE MULTIPLICATION

Where:

CRn = some count rate at a condition “n”

Cro = initial count rate

o

n

CR

CRM

RELATIONSHIP OF keff TO 1/M

2

1

eff

eff

1

2

k1

k1

CR

CR

M

11keff

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

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

1/M PLOT

100 200 300 400 500

0.2

0.4

0.6

0.8

1.0

CONTROL ROD NOTCHES WITHDRAWN

1 /

M

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

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

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

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

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