Question 10.13: Consider uranium fuel in a thermal reactor with an initial e......

We begin with Eq. (10.38):

C R(t)=\frac{{\sigma_{\gamma}^{28}}N^{28}(0)}{{\sigma_{a}^{25}}N^{25}(t)+{\sigma_{a}^{49}}N^{49}(t)}

Part a. At the beginning of life: N^{49}(0)=0 The enrichment is

\tilde{e}=\frac{N^{25}(0)}{N^{25}(0)+N^{28}(0)}

or

\frac{N^{28}(0)}{N^{25}(0)}=\frac{1}{\tilde{e}}-1=\frac{1}{0.04}-1=24

Using thermal cross section data from Table 3.2:

C R(0)={\frac{\sigma_{\gamma}^{28}N^{28}(0)}{\sigma_{a}^{25}N^{25}(0)}}={\frac{2.42}{591}}24.0=0.098

Part b: Substituting Eqs. (10.31) and (10.37) into the conversion ratio:

CR(t) = \frac {\sigma_{\gamma}^{\,28}N^{\,28}(0)}{{{\sigma}}_{a}^{\,25}N^{\,25}(0)\exp[-\sigma_{a}^{\,25}\Phi(t)]+\sigma_{\gamma}^{\,28}N^{\,28}(0)\bigl\{1-\exp[-\sigma_{a}^{\,49}\Phi(t)]\bigr\}}

C R(t)={\frac{1}{\exp[-\sigma_{a}^{25}\Phi(t)]+C R(0)\left\{1-\exp[-\sigma_{a}^{49}\Phi(t)]\right\}}}C R(0)

For 50% burnup, from Eq. (10.31):

N^{25}(t)/N^{25}(0)=\exp[-\sigma_{a}^{25}\Phi(t)]=0.50

Thus

\Phi(t)=-\frac{1}{\sigma_{a}^{25}}\ln(0.50)=-\frac{1}{591\cdot10^{-24}}\ln(0.50)=1.13\cdot10^{21}\,\mathrm{n/cm^{2}}

\sigma_{a}^{\,25}\Phi(t)=-\ln(0.50)=0.693

and

\sigma_{a}^{49}\Phi(t)=-\frac{\sigma_{a}^{49}}{\sigma_{a}^{25}}[\ln(0.50)]=\frac{973}{591}0.693=1.14

Thus

C R(t)={\frac{1}{\exp(-0.5)+0.098\,\{1-\exp(-1.14)\}}}0.098=0.129

Part c.

\frac{P^{\prime \prime \prime 49}}{P^{\prime \prime \prime}}=\frac{P^{\prime \prime \prime 49}}{P^{\prime \prime \prime25}+P^{\prime \prime \prime49}}=\frac{\gamma\sigma_{f}^{49}N^{49}(t)\phi}{\gamma\sigma_{f}^{25}N^{25}(t)\phi+\gamma\sigma_{f}^{49}N^{49}(t)\phi}

Using Eqs. (10.31) and (10.37)

From Table 3.2

(\sigma_{f}^{49}/\sigma_{f}^{25})(\sigma_{\gamma}^{28}/\sigma_{a}^{49})=(698/505)(2.42/973)=3.44+10^{-3}

and using the results from part b:

{\frac{P^{\prime \prime \prime49}}{P^{\prime \prime \prime}}}={\frac{3.44\cdot10^{-3}\cdot24\left\{1-0.320\right\}}{0.50+3.44\cdot10^{-3}\cdot24\left\{1-0.320\right\}}}=0.101\;\;\;\mathrm{or}\;\;10.1 \%