Kinetics With Delayed Neutrons

1
Kinetics With Delayed Neutrons

• B. Rouben
• McMaster University
• EP 4D03/6D03
• Nuclear Reactor Analysis
• 2008 Sept-Dec

2
Prompt and Delayed Fractions

• We now have to distinguish between the prompt and
  delayed neutrons. Lets write
• 
  
  
• 
  
  

3
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.

4
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

5
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.

6
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 hadnt 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

7
Point-Kinetics Equations with 1 Precursor (cont.)

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

8
Point-Kinetics Equations with 1 Precursor (cont.)

• So lets 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

9
Solution of Point-Kinetics Equation

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

10
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. Well 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

11
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 G1 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 ?.

12
Inhour Equation

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

13
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 ? lt 0, both
• ? values are negative
• When ? gt 0, one
• ? value is positive, and
• the other is negative.
• When ? 0, one
• ? value is 0, the other is
• is negative.

14
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, lets 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

15
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.
  

16
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 ? gt 0, decreasing
  exponentially if
• ? lt 0, and tending to a constant value if ?
  0.

17
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 a positive period (increasing
  flux) if ?1gt0, and a negative period if ?1lt0.

18
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

19
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 (G2) branches and (G1)
  roots.
• If ?lt0 all roots will be negative, but
• If ?gt0 one root will be positive (?1), and all
  other roots will be negative

20
Kinetics with G Delayed-Neutron-Precursor Groups

• With G delayed-neutron-precursor groups, then,
  the general solution will be a sum of (G1)
  exponentials,

21
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 lt0), 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 .

22
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

23
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

24
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 ? gt
  0.
• In either case, whether ? gt 0 or lt0. 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 gt 0
  or lt 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).
  

25
Prompt Jump

• Illustration of prompt jump prompt drop is
  similar

26
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.
  contd

27
Estimating the Prompt Jump/Drop (cont.)

• Lets do this estimate with one precursor group,
  for simplicity

28
Prompt-Jump Invalidity

• What happens if ? gt ? ? We can see from Eq. (30)
  that the r.h.s. would be negative, i.e. n would
  be lt 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.

29
Prompt Criticality

• The condition ? ? corresponds to
• This means that keff or gt1 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 (lt 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.

30
Reactor Period vs. Reactivity for Various
Prompt-Neutron Lifetimes
31
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 ?

32
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

33
Alternate Treatment for 1 Precursor Group (cont.)

• As for ?1
• The general solution is

34
Alternate Treatment for 1 Precursor Group (cont.)

• Each C is related to its corresponding n as per
  Eq. (12)
• Therefore Eqs. (35) (36) become

35
Alternate Treatment for 1 Precursor Group (cont.)

• So Eqs. (35) (36) become, at t 0

36
Alternate Treatment for 1 Precursor Group (cont.)

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

37
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.

38

• END