Chapter 12, Problem 12.8P

(a)

To determine

To Calculate: The uranium concentration for 10% enriched uranyl sulfate ${\text{UO}}_{\text{2}}{\text{SO}}_{\text{4}}$ .

Answer to Problem 12.8P

  $151.4\frac{\text{g}{\text{UO}}_{\text{2}}{\text{SO}}_{\text{4}}}{\text{L}{\text{H}}_{\text{2}}\text{O}}$

Explanation of Solution

Given:

For criticality,

  $\begin{array}{l}\eta =\epsilon =p=f=1\\ k_{\infty }=1\end{array}$

Formula used:

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

  $\eta =\frac{{\sigma }_{f5}}{{\sigma }_{a5}+\left(\frac{N_8}{N_5}\right)\times {\sigma }_{a8}}v$

Where,

  ${\sigma }_{f5}=549b$ fission cross-section 235 U

  ${\sigma }_{f5}=650b$ absorption cross-section 235 U

  ${\sigma }_{f5}=2.8b$ absorption cross-section 238 U

  $\frac{N_8}{N_5}=\frac{90}{10}=9$ 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 ${\text{UO}}_{\text{2}}{\text{SO}}_{\text{4}}$ solution is

  $f=\frac{{\sigma }_{a5}{M}_5+{\sigma }_{a8}{M}_8}{{\sigma }_{a5}{M}_5+{\sigma }_{a8}{M}_8+{\sigma }_{a{H}_{2}O}{M}_{H_{2}O}+{\sigma }_{a{O}_{2}S{O}_4}{M}_{a{O}_{2}S{O}_4}}$

Where,

  $M_5=$ moles U-5 per mole water

  $M_8=$ moles U-8 per mole water

  $M_{a{O}_{2}S{O}_4}=$ moles OSO₄ per mole water

  ${\sigma }_{a{H}_{2}O}=$ absorption cross-section of $H_{2}O=0.664\text{b}$

  ${\sigma }_{a{O}_{2}S{O}_4}=$ absorption cross-section of $O_{2}S{O}_4=0.491\text{b}$

  $M_{H_{2}O}=$ mole water per mole water = 1

Calculation:

For criticality,

  $k=\eta \epsilon pf=1$

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

  $\eta =\frac{{\sigma }_{f5}}{{\sigma }_{a5}+\left(\frac{N_8}{N_5}\right)\times {\sigma }_{a8}}v$

  $\eta =\frac{54.9}{650+\left(90\right)\times 2.8}\times 2.5=2.03$

The fast fission factor, $\epsilon$ , 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, $k_{\infty }=1$ is

  $k_{\infty }=\eta \epsilon pf$

  $\begin{array}{l}1=2.03\times 1\times 1\times f\\ f=0.4926\end{array}$

If M = the number of moles of $U{O}_{2}S{O}_4$ per mole $H_{2}O$ to attain criticality, then for

10% enrichment

  $\begin{array}{l}\left(U-5\right)/U=0.1\\ \left(U-8\right)/U=0.9\end{array}$

the number of moles per mole of $H_{2}O$ for each of

the components of the $U{O}_{2}S{O}_4$ solution is given in the table below.

Component	${\sigma }_{a}b$	moles	${\sigma }_a$ × moles
${\text{H}}_{\text{2}}\text{O}$	$0.664$	$1$	$0.664$
${\text{O}}_{\text{2}}{\text{SO}}_{\text{4}}$	$0.491$	$M$	$0.491M$
$\text{U-5}$	$650$	$0.1M$	$65M$
$\text{U-8}$	$2.8$	$0.9M$	$2.52M$

The dependency of $f$ on the composition of the aqueous ${\text{UO}}_{\text{2}}{\text{SO}}_{\text{4}}$ solution is

  $f=\frac{{\sigma }_{a5}{M}_5+{\sigma }_{a8}{M}_8}{{\sigma }_{a5}{M}_5+{\sigma }_{a8}{M}_8+{\sigma }_{a{H}_{2}O}{M}_{H_{2}O}+{\sigma }_{a{O}_{2}S{O}_4}{M}_{a{O}_{2}S{O}_4}}$

  $f=\frac{65M+2.52M}{65M+252M+0.664+0.491M}=0.4296$

M =0.00745 mole 10% enriched $U{O}_{2}S{O}_4$ per mole of water for criticality.

The molecular weight of the enriched uranium is

  $\left(235\times 0.1\right)+\left(238\times 0.9\right)=237.7\text{g/mol}$ of enriched uranium and the molecular weight of the ${\text{UO}}_{\text{2}}{\text{SO}}_{\text{4}}$ solution is

  $\text{237}\text{.7+}\left(2\times 16\right)+32+\left(4\times 16\right)=365.7\text{g/mole}{\text{UO}}_2{\text{SO}}_{\text{4}}$

The concentration of 10% enriched ${\text{UO}}_{2}S{O}_4$ required to attain criticality:

  $\begin{array}{l}365.7\frac{g}{1\text{mol}{\text{UO}}_{\text{2}}{\text{SO}}_{\text{4}}}\times 7.45\times {10}^{-3}\frac{\text{mol}{\text{UO}}_{\text{2}}{\text{SO}}_{\text{4}}}{\text{mol}{\text{H}}_{\text{2}}\text{O}}\times \frac{1\text{mol}{\text{H}}_{\text{2}}O}{18g}\times {10}^3\frac{g}{1\text{L}}\\ =151.4\frac{\text{g}{\text{UO}}_{\text{2}}{\text{SO}}_{\text{4}}}{\text{L}{\text{H}}_{\text{2}}\text{O}}\end{array}$

Conclusion:

The uranium concentration is $151.4\frac{\text{g}{\text{UO}}_{\text{2}}{\text{SO}}_{\text{4}}}{\text{L}{\text{H}}_{\text{2}}\text{O}}$ of uranyl sulfate ${\text{UO}}_{\text{2}}{\text{SO}}_{\text{4}}$ aqueous solution that can go critical if the uranium is enriched to 10%.

(b)

To determine

To Calculate: The uranium concentration for 90% enriched ${\text{UO}}_{\text{2}}{\text{SO}}_{\text{4}}$ .

Answer to Problem 12.8P

  $20.64\frac{\text{g}{\text{UO}}_{\text{2}}{\text{SO}}_{\text{4}}}{\text{L}{\text{H}}_{\text{2}}\text{O}}$

Explanation of Solution

Given:

For criticality,

  $\begin{array}{l}\eta =\epsilon =p=f=1\\ k_{\infty }=1\end{array}$

Formula used:

For criticality,

  $k=\eta \epsilon pf=1$

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

  $\eta =\frac{{\sigma }_{f5}}{{\sigma }_{a5}+\left(\frac{N_8}{N_5}\right)\times {\sigma }_{a8}}v$

Where,

  ${\sigma }_{f5}=549b$ fission cross-section 235 U

  ${\sigma }_{f5}=650b$ absorption cross-section 235 U

  ${\sigma }_{f5}=2.8b$ absorption cross-section 238 U

The dependency of $f$ on the composition of the aqueous ${\text{UO}}_{\text{2}}{\text{SO}}_{\text{4}}$ solution is

  $f=\frac{{\sigma }_{a5}{M}_5+{\sigma }_{a8}{M}_8}{{\sigma }_{a5}{M}_5+{\sigma }_{a8}{M}_8+{\sigma }_{a{H}_{2}O}{M}_{H_{2}O}+{\sigma }_{a{O}_{2}S{O}_4}{M}_{a{O}_{2}S{O}_4}}$

Where,

  $M_5=$ moles U-5 per mole water

  $M_8=$ moles U-8 per mole water

  $M_{a{O}_{2}S{O}_4}=$ moles $OS{O}_4$ per mole water

  ${\sigma }_{a{H}_{2}O}=$ absorption cross-section of $H_{2}O=0.664\text{b}$

  ${\sigma }_{a{O}_{2}S{O}_4}=$ absorption cross-section of $O_{2}S{O}_4=0.491\text{b}$

  $M_{H_{2}O}=$ mole water per mole water = 1

Calculation:

The calculations for 90% enriched ${\text{UO}}_{\text{2}}{\text{SO}}_{\text{4}}$ are made in a similar manner,

using the values for $\eta$ and for $f$ appropriate to 90% enrichment

  $\eta =\frac{{\sigma }_{f5}}{{\sigma }_{a5}+\left(\frac{N_8}{N_5}\right)\times {\sigma }_{a8}}v=\frac{549}{650+\left(\frac{10}{90}\right)\times 2.8}\times 2.5=2.11$

The value of $f$ needed for criticality $\left(k_{\infty }=1\right)$ is

  $1=\eta \epsilon pf$

  $1=2.11\times 1\times 1\times f$

Since, $f=\frac{{\sigma }_{a5}{M}_5+{\sigma }_{a8}{M}_8}{{\sigma }_{a5}{M}_5+{\sigma }_{a8}{M}_8+{\sigma }_{a{H}_{2}O}{M}_{H_{2}O}+{\sigma }_{a{O}_{2}S{O}_4}{M}_{a{O}_{2}S{O}_4}}$ the above equation become

  $1=2.11\times \frac{{\sigma }_{a5}{M}_5+{\sigma }_{a8}{M}_8}{{\sigma }_{a5}{M}_5+{\sigma }_{a8}{M}_8+{\sigma }_{a{H}_{2}O}{M}_{H_{2}O}+{\sigma }_{a{O}_{2}S{O}_4}{M}_{a{O}_{2}S{O}_4}}$

By letting M = the number of moles of 90% enriched ${\text{UO}}_{\text{2}}{\text{SO}}_{\text{4}}$

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

Component	${\sigma }_{a}b$	moles	${\sigma }_a$ × 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,

  $1=2.11\times \frac{585M+0.28M}{585M+0.28M+0.664M+0.491M}$

  $M=1.023\times {10}^{-3}$ mole ${\text{UO}}_{\text{2}}{\text{SO}}_{\text{4}}$

The molecular weight of 90% enriched uranium is $\left(235\times 0.9\right)+\left(238\times 0.1\right)=235.3$ and the molecular weight of 90% enriched ${\text{UO}}_{\text{2}}{\text{SO}}_{\text{4}}$ is $235.3+\left(2\times 16\right)+32+\left(4\times 16\right)=363.3$

The concentration of 90% enriched $U{O}_{2}S{O}_4$ required in the aqueous solution to

attain criticality is

  $\begin{array}{l}363.3\frac{g}{1\text{mol}{\text{UO}}_{\text{2}}{\text{SO}}_{\text{4}}}\times 1.023\times {10}^{-3}\frac{\text{mol}{\text{UO}}_{\text{2}}{\text{SO}}_{\text{4}}}{\text{mol}{\text{H}}_{\text{2}}\text{O}}\times \frac{1\text{mol}{\text{H}}_{\text{2}}O}{18g}\times {10}^3\frac{g}{1\text{L}}\\ =20.64\frac{\text{g}{\text{UO}}_{\text{2}}{\text{SO}}_{\text{4}}}{\text{L}{\text{H}}_{\text{2}}\text{O}}\end{array}$

Conclusion:

The uranium concentration is $20.64\frac{\text{g}{\text{UO}}_{\text{2}}{\text{SO}}_{\text{4}}}{\text{L}{\text{H}}_{\text{2}}\text{O}}$ of uranyl sulfate ${\text{UO}}_{\text{2}}{\text{SO}}_{\text{4}}$ aqueous solution that can go critical if the uranium is enriched to 10%.