# Nuclear Fuel Cycle Cross Rate

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```					Reaction Rates

Dr. Kathryn Higley
NE 113
Reaction Rates - Overview

What they are
How they are used
Example Calculations
Learning Objectives

Calculate:
The fission rate of 235U based on the power
level of a reactor
The burnup rate of nuclear fuel
Explain the process of
Fission
Reaction Rate (R)

Units: interactions/cm3•s
Applications
Absorption of neutrons yields activation
products through reactions such as:
(n,), (n,), (n,2n), (n,p),
Fissioning of a nuclear target
Reaction Rate (R)

Basic Equation:
atoms/cm3

R=N        neutrons/cm2•s

cm2

interactions/cm3•s
N: Atom Density

Units: atoms/cm3
Atom density can be calculated from:
, Elemental density (g/cm3)
A, Atomic weight (g/mol)
No, Avogadro’s number (6.023 x 1023
atoms/mol)

No 
N
A
Flux,   
Neutron flux
Units neutrons/cm2•s
Given for
Specific neutron energies
Specific reactor power levels
: Microscopic Cross Section
Units: cm2
Describes the theoretical “size” that an atom
presents, like a target, to be hit by an
approaching neutron

Frequently reported in “barns”
1 barn = 10-24 cm2
Examples

Fission Rate
Rf=Nf

Absorption Rate
Ra=Na
Example:        235U

Some nuclei can either absorb a neutron and
undergo fission:

Where the reaction cross section is listed as f,
or
Graphic is from the Student’s page of the NRC
website: www.NRC.gov
Example:           235U,   continued

Where the reaction cross section is listed as :

U  n U  
235
92
1
0
236
92
Reactor Power

How to estimate fission rate from reactor
power:
Determine power level
Symbol P
Specified in megawatts (MW)
Determine “recoverable energy” per fission
Symbol ER
Define fission rate as fissions/d
Reactor Power, continued

Use dimensional analysis:
fissions
day
 106 J   1 fission   1 MeV 
 MW  s    E MeV    1.6 1013 J   86,400 s / d 
 P( MW )                     
          R                        
23 P
 5.4 10
ER
Practical Applications

The fission rate of the OSU Triga Reactor (OSTR)
Power level: P=1 MW
Energy per fission: ER (for 235U)  200 MeV

OSTR fission rate: 5.4x1023 (P/ER)
= 5.4x1023 (1 MW / 200 MeV)
= 2.7 x 1021 fissions/d
Practical Applications, cont’d

Commercial Reactors
Power level: P=1100 MWe (electrical)
P (MWe) = PMWth
 is the fraction of thermal energy converted to
electrical energy (typically 30 - 33%)

Commercial fission rate: 5.4x1023 (P/ER)

= 5.4x1023 (1100 MW / 0.33)/(200 MeV)
= 8.9 x 1024 fissions/d
Practical Applications, cont’d

What is the fuel consumption rate?
Assume
Each fission consumes 1 atom of 235U
Fission rate = burn-up rate
Burn-up rate  5.4x1023 (P/ER) atoms per day

How is this converted to mass?
Practical Applications, cont’d

Use dimensional analysis
Burn-up rate  5.4x1023 (P/ER) atoms per day of    235U

Need to convert # atoms to total mass

Multiply by atomic weight and divide by Avogadro’s
number:
= 5.4x1023 (P/ER) A/N0
= 5.4x1023 (P/ER) (235 g   235U/mol)/(6.023   x 1023atoms
per mol)
= 0.895 PA/ER, in g/d
Practical Applications, cont’d

For   235U   burn-up, fill in the numbers:
Burn up (g/d) = 0.895 PA/ER
A = 235
ER = 200
(0.895)(P)(235)/200 = 1.05 P (g/d)

For a reactor at 1 MW the burn up rate is:
1.05 (1) = 1.05 g/d
Practical Applications, cont’d

How much fuel is burned in the OSTR each day?

Burn-up = 1.05 P (g/d)
P=1 MW
How long does the reactor operate each day?
Burn up = 1.05 (g/d) =1.05 (8/24) = 0.35 g
How much does it burn in a year?
Assume 100 days/y operation
35 g/year are consumed
Practical Applications, cont’d

How long will the OSTR fuel last?

The reactor has 12.5 kg of fissionable fuel
8 kg is needed for a critical mass
Burnable excess= 12.5 - 8 = 4.5 kg
4.5 kg •(103g/kg) / (35 g/y) = 128.6 y
Practical Applications, cont’d

How long will the fuel last in a power reactor?

Remember: Burn-up = 1.05 P (g/d)
Typical power level:
1100 MWth
Conversion Efficiency of ~33%
P=1100/0.33=3300
Burn-up = 1.05 (3300) (g/d) = 3500 g/d
3500 g/d(10-3kg/g)  3.5 kg/d
3.5 kg/d (300 d/y operation) = 1000 kg/y
Consumption of Fissile
Material

235U can capture a neutron and:
fission, or

What is the true consumption rate?
Burnup = 0.895 PA/ER (g/d) for fission
losses?

Use :
the ratio of the fission and capture cross-
sections
 / f
For   235U,

=0.175
fission occurs six times more often than
capture
cont’d

The consumption rate of fissile materials equals:
0.895 (1+ ) PA/ER
For   235U,   this becomes
=0.175
0.895 (1+ 0.175) PA/ER
0.895 (1.175) P (235/200)

Or, 1.24 g/d per MWth

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Description: Nuclear Fuel Cycle Cross Rate