# Analytical solution to mono-group kinetics with a ramp of reactivity

#### dRic2

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

The system is

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

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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#### rpp

The prompt jump approximation *is* to assume that the precursor population doesn't change, so I'm not sure what the difference is between your 1st method and 2nd method. Aren't both methods considered prompt jump methods?

Irregardless, let's take a look at your math. You stated that the reactivity is zero at $t=0$, and the precursors are a constant value. Therefore, $P(t=0)$ must equal $P_0$ or you've done something wrong.

In both the solution to Method 1 and Method 2, there is still a $t$ in the final solution, even though you've taken the limit $t \rightarrow 0$. If you set this remaining $t$ values equal to zero, you will get the right limit.

This is an interesting problem, let us know if you find the general solution.

#### dRic2

Gold Member
Thanks for the replay.

The prompt jump approximation *is* to assume that the precursor population doesn't change,
No... The prompt jump approximation is very different. If the reactivity is very far from prompt critical ($\rho << \beta$) then we can drop the time derivativa w.r.t $P$, i.e. -as my professor likes to say- we remove the dynamics from the system. The equations then becomes:
$$0 = \frac {\rho - \beta}{\Lambda}P + \lambda C$$
$$\frac {dC}{dt} = \frac {\beta}{\Lambda}P - \lambda C$$

You stated that the reactivity is zero at t=0t=0t=0, and the precursors are a constant value. Therefore, P(t=0)P(t=0)P(t=0) must equal P0P0P_0 or you've done something wrong.
I see no contradiction. For $t=0$ both my solutions yields $P(t=0)=P_0$.

In both the solution to Method 1 and Method 2, there is still a ttt in the final solution, even though you've taken the limit t→0t→0t \rightarrow 0. If you set this remaining ttt values equal to zero, you will get the right limit.
Sorry, I don't understand what you are saying.

Btw, if you do not believe me, instead of taking the limit, I can plot the function and attach an image.

PS: I still do not know which approximation would be the more appropriate, but I feel method 1 is better for small $t$. For very large $t$ it think it is not so realistic to consider the precursors' concentration to be constant, even though both solutions go to zero as t approaches $\infty$.

#### rpp

Thanks for the replay.

No... The prompt jump approximation is very different. If the reactivity is very far from prompt critical ($\rho << \beta$) then we can drop the time derivativa w.r.t $P$, i.e. -as my professor likes to say- we remove the dynamics from the system. The equations then becomes:
$$0 = \frac {\rho - \beta}{\Lambda}P + \lambda C$$
$$\frac {dC}{dt} = \frac {\beta}{\Lambda}P - \lambda C$$
I don't think this is correct. The prompt jump assumes the precursors are constant, not the power. Physically, you are looking for the change in power after a step reactivity change. The prompt neutron population starts to increase/decrease almost instantaneously, but it takes a while for the precursors to "catch up". This initial change in power is the "prompt jump".

It is my understanding that your Method 1 is the prompt jump. It is not clear what assumptions you made in Method 2.

I see no contradiction. For $t=0$ both my solutions yields $P(t=0)=P_0$.

Sorry, I don't understand what you are saying.
Maybe it is I who doesn't understand your equations.

How did you get the second equation in Method 2? It looks like you set the exponent term to 1, but kept the linear term. How do you justify this?

Likewise, for Method 1 you set some of the $t$ terms to zero, but kept other ones. I'm not sure this is justified. How did you come up with the last equation in Method 1 with the form $(1+t)$.

I'm just trying to understand what you did (it's hard to show a detailed derivation in a text box, even with Latex formatting)

#### dRic2

Gold Member
I don't think this is correct. The prompt jump assumes the precursors are constant, not the power.
No. Setting the derivative equal to zero is not a way to say the the power is constant. To avoid more confusion I'll post a page from "Nuclear Reactor Analysis" by Duderstadt & Hamilton.

Likewise, for Method 1 you set some of the ttt terms to zero, but kept other ones. I'm not sure this is justified.
In both cases I was wrong... Sorry... Both my Taylor expansions were wrong! Thank you for spotting my mistake. Now I solved the issue with Method 1 (see below), but still got a problem with Method 2

For method 1 things are a little complicated. The problem is to approximate the integral
$$\int_0^t \exp{- \left( \frac{\gamma t'^2 } 2 + t' \right) \frac {\beta}{\Lambda}} dt'$$
We are lucky it is a gaussian! If we consider a time interval $t << \frac {\Lambda} {\gamma \beta}$ and close to zero, we can see from the graph that the curve is more or less constant and so we can take it outside the integral
$$\exp{- \left( \frac{\gamma t^2 } 2 + t \right) \frac {\beta}{\Lambda}} \int_0^t dt' = \exp{- \left( \frac{\gamma t^2 } 2 + t \right) \frac {\beta}{\Lambda}} t$$
If you put this thing into the final equation of Method 1
$$P(t) = \lambda C_0 t + P_0\exp{- \left( \frac{\gamma t^2 } 2 + t \right)\frac {\beta}{\Lambda}} \approx \lambda C_0 t + P_0 \frac {\beta}{\Lambda} (1 - \frac{\gamma t^2 } 2 )(1 - t)$$
Recalling that $\frac {\beta}{\Lambda} P_0 = \lambda C_0$ we see that the terms containing $t$ cancel out leaving only the higher terms. So in conclusion $P(t) \approx 1 - t^2$ which is physically reasonable.

Method 2.

Here is the correct way to do it:
$$e^{-\lambda t} \approx 1 - \lambda t + o(t)$$
$$(1 + \gamma t)^{\frac{\lambda - \gamma} {\gamma}} \approx 1 + {\frac{\lambda - \gamma} {\gamma}} t + o(t)$$
So
$$P(t) \approx (1 - \lambda t )(1+\frac {\lambda - \gamma} {\gamma}t) = 1 - \lambda t + \frac {\lambda - \gamma} {\gamma}t - \lambda \frac {\lambda - \gamma} {\gamma}t^2 \approx 1+ \frac {-\gamma \lambda + \lambda - \gamma}{\gamma} t$$
If $\gamma < \lambda$ then $\gamma \lambda \approx 0$ because usually $\lambda = 0.008s^{-1}$ (but for very fast transient this value could change very much!) so I still get something that is increasing.

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