Doubling Time $($cont$\text{'}$d$)$

$DT=\frac{1}{[f_C({}_a{)}_{f}f(-2-L)]}$

Compound Doubling Time

$D{T}_C=\frac{0\mathrm{.}6931}{ln[1+(\frac{1}{DT})]}$

Breeder Reactor

$h=(\frac{S_f}{S_a})V$

$S_a=$ absorption microsccopic cross section

$V=$ neutrons produced per fussion

$h=$ number of necutions per absooption

Comversion Ratio$*(CR)=\frac{\text{f}\text{i}\text{s}\text{s}\text{i}\text{l}\text{e}\text{a}\text{t}\text{o}\text{m}\text{s}\text{p}\text{r}\text{o}\text{d}\text{u}\text{c}\text{e}\text{d}}{\text{f}\text{i}\text{s}\text{s}\text{i}\text{l}\text{e}\text{a}\text{t}\text{o}\text{m}\text{s}\text{c}\text{o}\text{n}\text{s}\text{u}\text{m}\text{e}\text{d}}$

$L=$ losses $CR=h-1-L$

$C_{\text{m}\text{a}\text{x}}$ is obtained if $L$ is zero or

$*C_{\text{m}\text{a}\text{x}}=h-1$

Bineeding Gain $(BG)$

$**BG=CR-1=-2-L$

$**B{G}_{\text{m}\text{a}\text{x}}=C_{\text{m}\text{a}\text{x}}-1=-2$

Doubling Time $($DT$)$

$N_b=N_{f}BG=N_f(-2-L)$

$N_f=F_c(N_0{)}_f({}_a{)}_{f}f(T)$

where: $,f=$ average reactor neutron flux, neutrons $\frac{?}{c{m}^2}-S$

$N_b=$ number density of new fissionoble nuclei produced in the brever

reactor, nudei $1c{m}^3.$

$N_f=$ number of original fissiomable fuel nuclei consumed

$($No$)$f number of fissionable fuel per nudei present in the core

and tide up in the fuel cycle, nudei $\frac{?}{c{m}^3}$

$F_C=$ fraction of $(N_0{)}_f$ that is in the cone, dimensionless

$({}_a{)}_f=$ microscopic absonption nuclear cross section of fuel, $c{m}^2$

USING THIS FORMULAS

ANSWER THIS QUESTIONS WITH COMPLETE SOLUTIONS

A fast breeder reactor using a plutonium$-$uranium mixed oxide fuel has a breeding ratio of $1\mathrm{.}20$ and a

compound doubling time of $10$ years. Assume that fuel in the core is $75\%$ of the reactor fuel inventory