#### dRic2

Gold Member

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Consider the mono-group model and a ramp in reactivity like ##\rho = -\gamma \beta t##

The system is

$$\frac {dP}{dt} = \frac {\rho - \beta}{\Lambda} P + \lambda C$$

$$ \frac {dC}{dt} = \frac {\beta}{\Lambda} P - \lambda C$$

$$\frac {dP}{dt} = \frac {\rho(t) - \beta}{\Lambda} P + \lambda C_0$$

The solution is:

$$P(t) = \exp{- \left( \frac{\gamma t^2 } 2 + t \right) \frac {\beta}{\Lambda}} \left[ \lambda C_0 \int_0^t \exp{- \left( \frac{\gamma t'^2 } 2 + t' \right) \frac {\beta}{\Lambda}} dt' + P_0 \right]$$

for ##t \rightarrow 0## ##P(t) \approx P_0(1 + t)## !?!? How can the power increase if I insert

the solution is

$$P(t) = P_0 \exp{-\lambda t} (1 + \gamma t)^{-1 + \frac {\lambda}{\gamma}}$$.

Again if ##\gamma < \lambda## then for ##t \rightarrow 0## ##P(t) \approx P_0(1 + \gamma t)^{\frac {\lambda - \gamma} {\gamma}} ## increases!!

I checked my result, the math should be fine.

Any help would be greatly appreciated.

The system is

$$\frac {dP}{dt} = \frac {\rho - \beta}{\Lambda} P + \lambda C$$

$$ \frac {dC}{dt} = \frac {\beta}{\Lambda} P - \lambda C$$

**1st method**: assume that the concentration of precursor doesn't change significantly (and it is equal to ##C_0##). The differential equation to solve is thus:$$\frac {dP}{dt} = \frac {\rho(t) - \beta}{\Lambda} P + \lambda C_0$$

The solution is:

$$P(t) = \exp{- \left( \frac{\gamma t^2 } 2 + t \right) \frac {\beta}{\Lambda}} \left[ \lambda C_0 \int_0^t \exp{- \left( \frac{\gamma t'^2 } 2 + t' \right) \frac {\beta}{\Lambda}} dt' + P_0 \right]$$

for ##t \rightarrow 0## ##P(t) \approx P_0(1 + t)## !?!? How can the power increase if I insert

**negative**reactivity ?!**2nd method**: Prompt Jump (since the reactivity is negative)the solution is

$$P(t) = P_0 \exp{-\lambda t} (1 + \gamma t)^{-1 + \frac {\lambda}{\gamma}}$$.

Again if ##\gamma < \lambda## then for ##t \rightarrow 0## ##P(t) \approx P_0(1 + \gamma t)^{\frac {\lambda - \gamma} {\gamma}} ## increases!!

I checked my result, the math should be fine.

Any help would be greatly appreciated.

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