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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
  Radiative capture
  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
They can undergo radiative capture:




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?
  About 1/3 of the 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
  undergo radiative capture


What is the true consumption rate?
  Burnup = 0.895 PA/ER (g/d) for fission
  How do we adjust for radiative capture
   losses?
Radiative Capture Losses

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

  =0.175
  fission occurs six times more often than
   capture
Radiative Capture Losses,
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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posted:12/19/2010
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Description: Nuclear Fuel Cycle Cross Rate