Chapter 12, Problem 12.1P

To determine

The concentration of induced Na and Cl radioactivity in the cooling water after a single passage through the reactor core if the mean thermal flux is 10¹¹neutrons per cm²/s and the mean temperature in the core is 80^o C.

Answer to Problem 12.1P

  ${}^{24}Na:92\text{Bq/L}$

  $C{l}^{36}:3.2\times {10}^{-5}\frac{\text{Bq}}{\text{L}}$

  $S^{35}:0.47\frac{\text{Bq}}{\text{L}}$

  $C{l}^{38}:431\frac{\text{Bq}}{\text{L}}$

Explanation of Solution

Given info:

Water boiler reactor core at a rate is $4\text{L/min}$

Coiled stainless steel tube’s inside diameter is $6.4\text{ mm}$

Coiled stainless steel tube’s length is $213\text{ cm}$

The concentration of Na and Cl in the water is 5atoms each per million molecules ${\text{H}}_{\text{2}}\text{O}$

Mean thermal flux is ${10}^{11}$ neutrons per $\frac{{\text{cm}}^{\text{2}}}{\text{s}}$

The mean temperature in the core is ${\text{80}}^{\text{o}}\text{C}$

Formula used:

Activity, $A=\left(\lambda {\mathrm{s}}^{-1}\right)\times n$

Distance $=$ velocity $\times$ time

  $\begin{array}{l}\text{velocity}\text{=}\frac{\text{νolume}\text{flow}\text{rate}}{\text{area}}\\ \text{distance}\text{=}\frac{\text{νolume}\text{flow}\text{rate}}{\text{area}}\times t\end{array}$

The number of radioactive atoms/L produced by thermal neutron irradiation of N target atoms/L during an irradiation time t seconds,

  $n=N\frac{\text{targets}}{\text{L}}\times \sigma \frac{{\text{cm}}^{\text{2}}}{\text{target}}\times \varphi \frac{\text{neutrons}}{{\text{cm}}^{\text{2}}\text{×s}}\times t\text{s}$

Calculation:

Since the irradiation time of one passage through the core is much less than the

The half-life of the activated isotopes, the resulting activity is:

  $A=\left(\lambda {\mathrm{s}}^{-1}\right)\times n$

To correct for the cross-section at a different temperature, the following equation is applied to the cross-sections (Etherington):

  $\sigma ={\sigma }_o\frac{1}{1.128}\sqrt{\frac{293K}{\left(273+80\right)K}}={\sigma }_o\times 0.808$

The following reactions and their associated parameters are used for this problem.

Also listed is the number of target atoms per liter for each element, which were calculated as shown below.

  

Calculate the number of atoms of each nuclide present per ${\text{cm}}^{\text{3}}$

  ${\text{H}}_{\text{2}}\text{O}$ , with a density of $0.97{\text{g/cm}}^{\text{3}}(CRC)$ at ${80}^{\text{o}}\text{C}$ , and a MW of $18\text{ g/mole}$ ;

  $\begin{array}{l}=\frac{6.03\times {10}^{23}\text{molecules}}{\text{mo1e}}\times \frac{\text{1mole}}{18\text{g}}\times \frac{970\text{g}}{\text{L}}\\ =3.25\times {10}^{25}\frac{\text{molecules}{\text{H}}_{\text{2}}\text{O}}{\text{L}}\end{array}$

Since there are 5 atoms Na and Cl for every million molecules of water;

  $\begin{array}{l}=3.25\times {10}^{25}\frac{\text{molecules}{\text{H}}_{\text{2}}\text{O}}{\text{L}}\times \frac{5{\text{atoms}}^{\text{23}}\text{Na}}{{10}^6\text{molecules}{\text{H}}_{\text{2}}\text{O}}\\ =1.625\times {10}^{20}\frac{\text{atoms}{\text{Na}}^{\text{23}}\text{and}\text{Cl}}{\text{L}}\end{array}$

There are two isotopes of Cl: $75.4\%$

  ${}^{\text{35}}\text{Cl}$ and $24.6\%$

  ${}^{\text{37}}\text{Cl}$ , whose concentrations are

  $0.754\times 1.625\times {10}^{20}\frac{\text{atoms}{}^{35}\text{Cl}}{\text{L}}=1.23\times {10}^{20}\frac{\text{atoms}{}^{35}\text{Cl}}{\text{L}}$

and

  $0.246\times 1.625\times {10}^{20}\frac{\text{atoms}{}^{37}\text{Cl}}{\text{L}}=4.0\times {10}^{19}\frac{\text{atoms}{}^{37}\text{Cl}}{\text{L}}$

Calculate the irradiation time that is, the time that the water spends in the reactor

  $\text{distance}\text{=}\text{velocity}\text{×}\text{time}$

  $\text{velocity}\text{=}\frac{\text{νolumetric}\text{flow}\text{rate}}{\text{area}}$

  $\text{distance}\text{=}\frac{\text{νolumetric}\text{flow}\text{rate}}{\text{area}}\text{×t}$

  $213\text{cm}=\frac{4000\frac{{\text{cm}}^{\text{3}}}{\text{min}}}{\frac{\pi }{4}{(0.64\text{cm})}^2\times 60\frac{\text{s}}{\text{min}}}\times \text{t}$

  $t=1.03$ s

For the case of $N{a}^{24}$ , the induced atomic concentration after a single passage through the reactor core:

  $n=N\frac{\text{targets}}{\text{L}}\times \sigma \frac{{\text{cm}}^{\text{2}}}{\text{target}}\times \varphi \frac{\text{neutrons}}{{\text{cm}}^{\text{2}}\text{.sec}}\times t\sec$

  $n=1.625\times {10}^{20}\frac{\text{targets}{}^{\text{23}}\text{Na}}{\text{L}}\times 0.43\times {10}^{-24}\frac{{\text{cm}}^{\text{2}}}{{\text{atom}}^{\text{23}}\text{Na}}\times {10}^{11}\frac{\text{neutrons}}{{\text{cm}}^{\text{2}}\text{sec}}\times 1.03\sec$

  $n=7.2\times {10}^6\text{atoms}{}^{24}\text{Na/L}$

And, the activity is

Activity $=\lambda \times n$

Activity $=\left(1.28\times {10}^{-5}{\sec }^{-1}\right)\times \left(7.2\times {10}^6\right)=92\text{Bq/L}$

By similar calculations, we find that for the other activations:

  $C{l}^{36}:n=7.13\times {10}^{-14}{\sec }^{-1}\times 1.23\times {10}^{20}\times 35.2\times {10}^{-24}\times {10}^{11}\times 1.03=3.2\times {10}^{-5}\frac{Bq}{L}$

  $S^{35}:\text{n}=9.22\times {10}^{-8}\times 1.23\times {10}^{20}\times 0.40\times {10}^{-24}\times {10}^{11}\times 1.03=0.47\frac{Bq}{L}$

  $C{l}^{38}:n=3.08\times {10}^4\times 4\times {10}^{19}\times 0.34\times {10}^{-24}\times {10}^{11}\times 1.03=431\frac{Bq}{L}$

Conclusion:

The concentration of induced

  ${}^{24}Na:92\text{Bq/L}$

  $C{l}^{36}:3.2\times {10}^{-5}\frac{Bq}{L}$

  $S^{35}:0.47\frac{Bq}{L}$

  $C{l}^{38}:431\frac{Bq}{L}$