Introduction To Health Physics
Introduction To Health Physics
5th Edition
ISBN: 9780071835275
Author: Johnson, Thomas E. (thomas Edward), Cember, Herman.
Publisher: Mcgraw-hill Education,
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Chapter 12, Problem 12.8P

(a)

To determine

To Calculate: The uranium concentration for 10% enriched uranyl sulfate UO2SO4 .

(a)

Expert Solution
Check Mark

Answer to Problem 12.8P

  151.4gUO2SO4LH2O

Explanation of Solution

Given:

For criticality,

  η=ε=p=f=1k=1

Formula used:

The mean number of fission neutrons per neutron absorbed in uranium is calculated as

  η=σf5σa5+( N 8 N 5 )×σa8v

Where,

  σf5=549b fission cross-section 235 U

  σf5=650b absorption cross-section 235 U

  σf5=2.8b absorption cross-section 238 U

  N8N5=9010=9 molar ratio of U-8 to U-5 atoms in 10% enriched U

  v=2.5 for 235U= average number of neutrons per fission

The dependency of f on the composition of the aqueous UO2SO4 solution is

  f=σa5M5+σa8M8σa5M5+σa8M8+σaH2OMH2O+σaO2SO4MaO2SO4

Where,

  M5= moles U-5 per mole water

  M8= moles U-8 per mole water

  MaO2SO4= moles OSO4 per mole water

  σaH2O= absorption cross-section of H2O=0.664b

  σaO2SO4= absorption cross-section of O2SO4=0.491b

  MH2O= mole water per mole water = 1

Calculation:

For criticality,

  k=ηεpf=1

Eta (η) , the mean number of fission neutrons per neutron absorbed in uranium, is calculated as

  η=σf5σa5+( N 8 N 5 )×σa8v

  η=54.9650+(90)×2.8×2.5=2.03

The fast fission factor, ε , can be assumed to be 1, since this is a homogeneous assembly. Additionally, p , the resonance escape probability, is also assumed to be

1, since the molar ratio of the moderator to fuel is very high.

The value of f , the thermal utilization factor that will lead to criticality, k=1 is

  k=ηεpf

  1=2.03×1×1×ff=0.4926

If M = the number of moles of UO2SO4 per mole H2O to attain criticality, then for

10% enrichment

  (U5)/U=0.1(U8)/U=0.9

the number of moles per mole of H2O for each of

the components of the UO2SO4 solution is given in the table below.

    Component

      σab

    moles

      σa × moles

      H2O

      0.664

      1

      0.664

      O2SO4

      0.491

      M

      0.491M

      U-5

      650

      0.1M

      65M

      U-8

      2.8

      0.9M

      2.52M

The dependency of f on the composition of the aqueous UO2SO4 solution is

  f=σa5M5+σa8M8σa5M5+σa8M8+σaH2OMH2O+σaO2SO4MaO2SO4

  f=65M+2.52M65M+252M+0.664+0.491M=0.4296

M =0.00745 mole 10% enriched UO2SO4 per mole of water for criticality.

The molecular weight of the enriched uranium is

  (235×0.1)+(238×0.9)=237.7g/mol of enriched uranium and the molecular weight of the UO2SO4 solution is

  237.7+(2×16)+32+(4×16)=365.7g/moleUO2SO4

The concentration of 10% enriched UO2SO4 required to attain criticality:

  365.7g1mol UO2 SO4×7.45×103mol UO2 SO4molH2O×1molH2O18g×103g1L=151.4g UO2 SO4LH2O

Conclusion:

The uranium concentration is 151.4gUO2SO4LH2O of uranyl sulfate UO2SO4 aqueous solution that can go critical if the uranium is enriched to 10%.

(b)

To determine

To Calculate: The uranium concentration for 90% enriched UO2SO4 .

(b)

Expert Solution
Check Mark

Answer to Problem 12.8P

  20.64gUO2SO4LH2O

Explanation of Solution

Given:

For criticality,

  η=ε=p=f=1k=1

Formula used:

For criticality,

  k=ηεpf=1

Eta (η) , the mean number of fission neutrons per neutron absorbed in uranium, is calculated as

  η=σf5σa5+( N 8 N 5 )×σa8v

Where,

  σf5=549b fission cross-section 235 U

  σf5=650b absorption cross-section 235 U

  σf5=2.8b absorption cross-section 238 U

The dependency of f on the composition of the aqueous UO2SO4 solution is

  f=σa5M5+σa8M8σa5M5+σa8M8+σaH2OMH2O+σaO2SO4MaO2SO4

Where,

  M5= moles U-5 per mole water

  M8= moles U-8 per mole water

  MaO2SO4= moles OSO4 per mole water

  σaH2O= absorption cross-section of H2O=0.664b

  σaO2SO4= absorption cross-section of O2SO4=0.491b

  MH2O= mole water per mole water = 1

Calculation:

The calculations for 90% enriched UO2SO4 are made in a similar manner,

using the values for η and for f appropriate to 90% enrichment

  η=σf5σa5+( N 8 N 5 )×σa8v=549650+( 10 90)×2.8×2.5=2.11

The value of f needed for criticality (k=1) is

  1=ηεpf

  1=2.11×1×1×f

Since, f=σa5M5+σa8M8σa5M5+σa8M8+σaH2OMH2O+σaO2SO4MaO2SO4 the above equation become

  1=2.11×σa5M5+σa8M8σa5M5+σa8M8+σaH2OMH2O+σaO2SO4MaO2SO4

By letting M = the number of moles of 90% enriched UO2SO4

, we obtain the following values to substitute into the equation above:

    Component

      σab

    moles

      σa × moles

    H2O

      0.664

      1

      0.664

    O2SO4

      0.491

      M

      0.491M

    U-5

      650

      0.9M

      585

    U-8

      2.8

      0.1M

      0.28M

Substituting these values into the equation,

  1=2.11×585M+0.28M585M+0.28M+0.664M+0.491M

  M=1.023×103 mole UO2SO4

The molecular weight of 90% enriched uranium is (235×0.9)+(238×0.1)=235.3 and the molecular weight of 90% enriched UO2SO4 is 235.3+(2×16)+32+(4×16)=363.3

The concentration of 90% enriched UO2SO4 required in the aqueous solution to

attain criticality is

  363.3g1mol UO2 SO4×1.023×103mol UO2 SO4molH2O×1molH2O18g×103g1L=20.64g UO2 SO4LH2O

Conclusion:

The uranium concentration is 20.64gUO2SO4LH2O of uranyl sulfate UO2SO4 aqueous solution that can go critical if the uranium is enriched to 10%.

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