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Kinetics With Delayed Neutrons. B. Rouben McMaster University EP 6D03 Nuclear Reactor Analysis (Reactor Physics) 2009 Jan.-Apr. Prompt and Delayed Fractions. We now have to distinguish between the prompt and delayed neutrons . Let’s write:. Prompt and Delayed Sources.

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Kinetics with delayed neutrons

Kinetics With Delayed Neutrons

B. Rouben

McMaster University

EP 6D03

Nuclear Reactor Analysis

(Reactor Physics)

2009 Jan.-Apr.


Prompt and delayed fractions
Prompt and Delayed Fractions

  • We now have to distinguish between the prompt and delayed neutrons. Let’s write:


Prompt and delayed sources
Prompt and Delayed Sources

  • Without delayed neutrons, we had gotten to the following evolution equation for the neutron density:

  • Now, we will need to separate the neutron source into a prompt part,

    and a delayed part.


Prompt and delayed sources1
Prompt and Delayed Sources

  • The delayed source comes from the decay of delayed-neutron precursors. We saw last time that there are many delayed-neutron precursors. However, we will for now assume only 1 delayed-neutron precursor group. We will write its concentration as C.

  • By the radioactive-decay law, the decay rate of the precursor is C, where  is the decay constant of the precursor.

  • There is 1 delayed neutron born from the decay of 1 precursor nuclide. Therefore the production rate of the precursor must be equal to the fission rate times d:


Prompt and delayed sources2
Prompt and Delayed Sources

  • Therefore the equation for the evolution of the precursor concentration is (production rate – decay rate)

  • and the new equation for the evolution of the neutron density is

  • The above coupled equations are the point-kinetics equations in 1 energy group, with 1 delayed-neutron-precursor group.


Point kinetics equations with 1 precursor
Point-Kinetics Equations with 1 Precursor

  • At the end of the previous presentation, we had just reached the point-kinetics equations with 1 delayed-neutron-precursor group in the following form:

  • Equation for the evolution of the precursor:

    (we hadn’t previously shown explicitly the dependence of C and on space and time - we now added this for completeness)

  • Equation for the evolution of the neutron density:


Point kinetics equations with 1 precursor cont
Point-Kinetics Equations with 1 Precursor (cont.)

  • Simplify Eq. (2) by factorizing from first 3 terms:


Point kinetics equations with 1 precursor cont1
Point-Kinetics Equations with 1 Precursor (cont.)

  • So let’s rewrite the final point-kinetics equations for the neutron density and precursor concentration in terms of time only (write equation for n first, as per convention):

  • We can also rewrite the equations in terms of , using

    , but this will involve :


Solution of point kinetics equation
Solution of Point-Kinetics Equation

  • Let’s try to solve the point-kinetics equations, (6) & (7).

  • Note that there are 2 time constants (  and 1/) which enter into the equations, so we may expect that the time evolution will somehow involve these 2 time constants.

  • We have a system of two differential equations in 1st-order in 2 variables, n and C. The usual form of solution tried for linear differential equations is the exponential form, so let’s try:

  • Substituting this form into Eqs. (6) & (7) yields, after division of both sides by the common factor et, found in every term:


Solution of point kinetics equation1
Solution of Point-Kinetics Equation

  • These are in fact 2 homogeneous equations in 2 unknowns:

  • We know that there will be a non-trivial solution only if the determinant is zero, i.e.,

  • There are 2 ways to attack this equation:

    • One is to realize that this is a quadratic equation, whose roots can be found. We’ll come back to this method later.

    • The other method is graphical. In this approach, the form of the equation usually seen is obtained by writing  in terms of l*. Remember from Eq. (19) in the previous presentation:


Inhour equation
Inhour Equation

  • Eq. (15) then becomes:

  • This form of the equation is called the Inhour equation (for 1 delayed-neutron-precursor group).

  • The Inhour equation is not easy to solve, as the left-hand side is a discontinuous function which goes to  at 2 values of  (at G+1 values of  if we had done the analysis with many delayed-neutron-precursor groups).

  • The way to visualise the solutions is to plot the left-hand side in terms of , and see where it crosses a horizontal line at height .


Inhour equation1
Inhour Equation

  • To plot Eq. (16), we need to note the following features:


Inhour equation cont
Inhour Equation (cont.)

  • The plots of the l.h.s. and r.h.s. of Eq. (16) are then as follows: [From Lamarsh]

  • There are always 2

    solutions for :

  • When  < 0, both

     values are negative

  • When  > 0, one

     value is positive, and

    the other is negative.

  • When  = 0, one

     value is 0, the other is

    is negative.


General solution for 1 precursor group
General Solution for 1 Precursor Group

  • Thus there are always 2 values of  as solutions, and consequently the general solution for the neutron density and the precursor concentration is a sum of 2 exponentials:

  • By convention, let’s agree that 1 is the algebraically larger solution, i.e., the rightmost one.

  • Note that l* is very small (~1 ms), therefore the vertical line at 1/l* is very much to the left of the line at  (i.e., the figure is far from being to scale).

  • Thus, although 2 is algebraically smaller (i.e., it is the one on the left), in absolute magnitude it is much larger than 1:


General solution for 1 precursor group1
General Solution for 1 Precursor Group

  • Physically, l* is the time constant at which the neutron density (or power) would evolve without delayed neutrons (as we showed before), whereas 1/ is the time constant corresponding to the delayed neutrons.

  • Because of the coupled nature of the kinetics equations, the time constants for the evolution of the neutron density (and power) are not equal to l* or 1/, but at least one of the time constants is intermediate between these values.


General solution for 1 precursor group cont
General Solution for 1 Precursor Group (cont.)

  • Summarizing, therefore:

  • In the general solution, then, the exponential term in 2 will first die away (quite quickly), and the term in 1 will remain as the asymptotic time evolution of the neutron density:

    • increasing exponentially if  > 0, decreasing exponentially if

       < 0, and tending to a constant value if  = 0.


Stable period
Stable Period

  • Since the asymptotic evolution of the neutron density (or the flux, or power) will be proportional to the exponential , then we can write

    which will be apositive period (increasing flux) if 1>0, and a negative period if 1<0.


Kinetics with 6 delayed neutron precursor groups
Kinetics with 6 Delayed-Neutron-Precursor Groups

  • If we retain 6 delayed-neutron precursor groups, the analysis will be similar, but the Inhour equation will have 7 roots instead of just 2. The graphic representation of the Inhour is: [from Duderstadt & Hamilton]


Kinetics with g delayed neutron precursor groups
Kinetics with G Delayed-Neutron-Precursor Groups

  • Generalizing to any number (G) of delayed-neutron precursors, the point-kinetics equations will be:

  • and in this case the inhour equation will be

    which will have (G+2) branches and (G+1) roots.

  • If <0 all roots will be negative, but

  • If >0 one root will be positive (1), and all other roots will be negative


Kinetics with g delayed neutron precursor groups1
Kinetics with G Delayed-Neutron-Precursor Groups

  • With G delayed-neutron-precursor groups, then, the general solution will be a sum of (G+1) exponentials:,


General solution
General Solution

  • By convention, we denote 1 the algebraically largest root (i.e., the rightmost one on the graph)

  • 1 has the sign of .

  • Since all other  values are negative (and more negative than 1 if 1 <0), the exponential in 1 will survive longer than all the others.

  • Therefore, the eventual (asymptotic) form for n and Cg is exp(1t), i.e., the power will eventually grow or drop with a stable (or asymptotic) period .


General solution cont
General Solution (cont.)

  • In summary, for the asymptotic time dependence:

  • For  not too large and positive (i.e., except for positive reactivity insertions at prompt criticality or above):

  • , i.e., things evolve much more slowly than without delayed neutrons


Situation at steady state
Situation at Steady State

  • The point-kinetics equations apply even in steady state, with =0.

  • The relationship between the precursor concentrations and the neutron density can be obtained by setting the time derivatives to 0 in the point-kinetics equations. For G precursor groups at steady state (subscript ss):

  • From Eq. (26) we get

  • Summing Eq. (27) over all g yields back Eq. (25), since


Prompt jump or drop approximation
Prompt-Jump (or Drop) Approximation

  • We have seen that the time evolution of the flux/power is a sum of exponentials, all with negative exponent, except for a single one if  > 0.

  • In either case, whether  > 0 or <0. the exponential with the (algebraically) largest exponent, 1, will quickly be the surviving one, the other exponentials dying away more quickly.

  • In a sudden insertion of reactivity (whether > 0 or < 0) in a reactor running steadily, there will therefore be a prompt jump or prompt drop, during which these dying exponentials die away, after which the flux/power will be on an exponential behaviour with a stable period 1/1 (see figure).


Prompt jump
Prompt Jump

  • Illustration of prompt jump – prompt drop is similar:


Estimating the prompt jump drop
Estimating the Prompt Jump/Drop

  • An estimate of the size of the prompt jump (or drop) can be obtained most easily by making the following approximations:

  • 1) The delayed-neutron-precursor concentrations cannot change substantially during the prompt jump, therefore the precursor concentration at the start of the stable period can be taken equal to the steady-state value.

  • 2) The derivative of the time evolution at the start of the stable period is much smaller than the derivative during the prompt jump, therefore the derivative at the start of the stable period can be approximated by zero. cont’d


Estimating the prompt jump drop cont
Estimating the Prompt Jump/Drop (cont.)

  • Let’s do this estimate with one precursor group, for simplicity:


Prompt jump invalidity
Prompt-Jump Invalidity

  • What happens if  >  ? We can see from Eq. (30) that the r.h.s. would be negative, i.e. n would be < 0!

  • This is a symptom that the prompt-jump approximation breaks down when the value of (positive)  comes close to the value of .

  • The reason that it breaks down is that as  increases and crosses the value of , the value of 1 becomes very large (the exponential increases extremely rapidly), and the approximation of setting the derivative of n to zero simply breaks down.

  • Note that this is not the case for any negative value of : The prompt-drop approximation does not break down.


Prompt criticality
Prompt Criticality

  • The condition  =  corresponds to:

  • This means that keff = or >1 even if we ignore the delayed neutrons ( ), i.e., the reactor is critical on prompt neutrons alone. This condition is called prompt criticality.

  • The delayed neutrons then no longer play a crucial role, and when  increases beyond (prompt supercriticality),very very short reactor periods (< 1 s, or even much smaller, depending on the magnitude of ) develop.

  • Thus, it is advisable to avoid prompt criticality.

  • The next slide shows how the period changes above prompt criticality for various prompt-neutron lifetimes.



Alternate treatment for 1 precursor group
Alternate Treatment for 1 Precursor Group

  • Now that we have analyzed the kinetics by means of the Inhour equation, we will return to investigate the alternate solution for the  values, for 1 precursor group. We return to Eq. (15):

  • This is in fact a quadratic equation in :

  • We can write the exact two solutions for :


Alternate treatment for 1 precursor group cont
Alternate Treatment for 1 Precursor Group (cont.)

  • We can see that 1 and 2 are respectively the solutions with the plus sign and the minus sign in front of the square root.

  • Under these conditions, then, we can see that


Alternate treatment for 1 precursor group cont1
Alternate Treatment for 1 Precursor Group (cont.)

  • As for 1:

  • The general solution is


Alternate treatment for 1 precursor group cont2
Alternate Treatment for 1 Precursor Group (cont.)

  • Each C is related to its corresponding n as per Eq. (12):

  • Therefore Eqs. (35) & (36) become


Alternate treatment for 1 precursor group cont3
Alternate Treatment for 1 Precursor Group (cont.)

  • So Eqs. (35) & (36)’ become, at t = 0:


Alternate treatment for 1 precursor group cont4
Alternate Treatment for 1 Precursor Group (cont.)

  • Using the value of 1 from Eq. (34),


Summary
Summary

  • This presentation has focused on presenting the principles and concepts of delayed-neutron influence on the kinetics of a reactor core.

  • Most of the analysis was based on a 1-delayed-neutron-precursor-group model. Some of the analysis was generalized to 6 or G precursor groups.

  • However, remember that all the analysis was based on point kinetics, in which the flux shape did not change.

  • In reality, for a more accurate analysis of what happens in neutronic transients, spatial kinetics is required. We will not cover this here.