Question 4.10: Using the data from problem [4.6], vary the coolant/fuel vol......

From Eq. (4.49), and the enrichment is {\tilde{e}}=0.03

\eta_{T}=\frac{\nu\Sigma_{f}^{f}}{\Sigma_{a}^{f}} =\!\frac{\tilde{e}N\nu^{25}{\sigma}_{f}^{25}}{\tilde{e}N\sigma_{a}^{25}+(1-\tilde{e})N\sigma_{a}^{28}+2N\sigma_{a}^{o}} =\frac{{\widetilde{e}}\nu^{25}\sigma_{f}^{25}}{{\widetilde{e}}\sigma_{a}^{25}+(1-{\widetilde e})\sigma_{a}^{28}+2\sigma_{a}^{o}}

Using data from Table 3.2 and oxygen data from Table E-3

\eta_{T}=\frac{0.03\cdot2.43\cdot505}{0.03\cdot591+(1-0.03)2.42+2\cdot0.0002}=1.82

From Eq. (4.40)

p=\exp\left[-\left(\frac{V_{f}}{V_{m}}\right)\frac{N_{f}}{{\xi}\Sigma_{s}}I\right]

For the fuel A = 238 + 2.16 = 270,

N_{f}=\rho N_{o}/\,A=11.0\cdot0.6023\cdot10^{24}\,/\,270=0.0245\cdot10^{24}\ \mathrm{gm/cm}^{3}

From Table 4.3 for UO2

I=4.45+26.6{\sqrt{4/{\rho{D}}}}=4.45+26.6{\sqrt{4/(11\cdot1.1)}}=19.74\,b

where we have increased the diameter by 10%: Taking \xi \Sigma_s = 1.28 for water from Table 3.1, we obtain

p=\exp\left[-\left(\frac{V_{m}}{V_{f}}\right)^{-1}\frac{0.0245\cdot10^{24}}{1.28}19.74\cdot10^{-24}\right]

or

p=\exp\Bigl[-0.378(V_{m}/{V_{f}})^{-1}\Bigr]

From Eq. (4.55)

f=\frac{1}{1+\varsigma(V_{m}/V_{f})(N_{m}/N_{f})(\overline{{{\sigma}}}_{a T}^{m}/\overline{{{\sigma}}}_{a T}^{f})}

N_{m}{N}_{f}=\frac{\rho_{m}N_{o}/{A}_{m}}{\rho_{f}N_{o}\left/A_{f}\right.}=\frac{\rho_{m}A_{f}}{\rho_{f}A_{m}}=\frac{1.0\cdot270}{11\cdot18}=1.364

With the thermal disadvantage factor ς = 1.16 (from problem [4.6]) and taking the absorption cross sections from Table E-3:

f={\frac{1}{1+1.16\cdot(V_{m}/V_{f})\cdot1.364{(0.5896/6.540)}}}

or

f=\frac{1}{1+0.1426({ V}_{m}/{V}_{f})}

Finally, with ε = 1.24(form problem [4.6], we combine the above equations to obtain:

The plots of p, ƒ and k_{\infty} are as follows:

We obtain a maximum value of k_{\infty}  of  k_{\infty}|_{\text{Max}} = 1.46

At a volume ratio of V_m / V_f = 1.82