Adjoint flux for multi-group diffusion equation for criticality problem?

In summary, Taitae25 is a new neutronics expert who is trying to calculate the adjoint flux for a critical system. She understands how to derive the adjoint operator for the diffusion equation, but is confused about how to calculate the adjoint flux for a critical system. The adjoint flux is important for the diffusion equation, but it is non-conservative. She can solve for the adjoint flux using power iteration, or solve for the regular flux, determine what the criticality (keff) is, and then solve for the linear system for the adjoint flux.
  • #1
taitae25
10
0
Hi,

I'm new to the entire neutronics field. I've learned about adjoints as a physics student in undergrad and I'm doing nuclear engineering for my graduate studies. I understand how to derive the adjoint operator for the diffusion equation, but I'm a bit confused as to how to calculate the adjoint flux for a critical system.

I'm tryring to calculate the adjoint flux numerically. The regular flux is simple to calculate using power iteration, but what does criticality mean for adjoint flux? Can I use power iteration for adjoint flux calculation? If adjoint flux gives a sense of importance of the flux, then I can use this to perform sensitivity analysis on the criticality (keff). But I also learned that the adjoint operator for the transport equation is non-conservative. In that case, how can I calculate adjoint flux numerically?

Can I still solve for it using power iteration? or do I simply solve for the regular flux, determine what the criticality (keff) is and then solve for the linear system for the adjoint flux? i.e. For my two group diffusion equation, with only down scatter, would my adjoint diffusion equation read as follows? (assuming prompt fission neutron only appears in the lowest energy group (group 0,fast neutrons)).

-D [tex] _0[/tex][tex]\frac{\partial^2 \phi_0}{\partial x^2} + (\Sigma_{a,0} + \Sigma_{s,0->1})\phi_0 - \Sigma_{s,0->1}\phi_1 = (\nu_{0} \chi_{0}\Sigma_{f,0}\phi_0 + \nu_{1} \chi_{0}\Sigma_{f,1}\phi_1)/k [/tex]

-D [tex] _1[/tex][tex]\frac{\partial^2 \phi_1}{\partial x^2} + (\Sigma_{a,1})\phi_1 = 0[/tex]

And just solve for the coupled linear system?

Thank you very much.
 
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  • #2
Sorry, I just can't get the LaTeX script to work but there should be a [tex]\chi_{0}[/tex] right next to th [tex]\nu_{0}[/tex] and [tex]\nu_{1}[/tex] respectively, for the fast group diffusion equation ([tex]\phi_{0}[/tex]). Please also note that this is my adjoint flux equaitons. So with the "k" known from the normal flux calculation, hence a constant, can I just solve the linear system once and solve for the adjoint flux?

Thanks again in advance.
 
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  • #3
It is easier to think of in terms of matrix operators. The equation can be rewritten in terms of a multiplication operator, M, and a fission operator F where:

[tex]M = \nabla D \nabla + \Sigma_a[/tex]

[tex]F = \nu \Sigma_f \Chi[/tex]


These operators can then be written in matrix form for a two-group equation. The adjoint of these operators is just the transpose of their normal form.

1)Re-write the equations in matrix form
2)Re-write the equations after transposing the operators
3)Solve for the flux in the transposed equations, which gives the adjoint flux
 
  • #4
Hi joek856,

First, thanks a lot for the reply. I see for how to actually calculate the adjoint flux. That's a neat compact, concise way to put it. So with that given, does it mean that I have to solve this then using something similar to power iteration? Do I still iterate on the fission neutron source for the adjoint flux then ? I guess the keff for the adjoint flux doesn't mean anything does it? After this, I should be able to get moving.
 
  • #5
the keff for the adjoint flux will be the same as for the normal flux, 1 in a critical reactor. I am unfamiliar with the power method for solving these equations, but if you can solve the equations for regular flux, the same methodology will apply for the equations with the adjoint operators in place of the normal operators.
 
  • #6
Okay, I'll give it a shot. Thanks a lot.
 
  • #7
Joek856,

Just wanted to say thanks. My adjoint flux is calculated correctly and I'm able to study uncertain quantification and sensitivity analysis using the adjoint method. Greatly appreciated.

-taitae25
 

1. What is the purpose of using adjoint flux in multi-group diffusion equations for criticality problems?

The adjoint flux is used to calculate the sensitivity of a criticality problem to changes in neutron flux. This is important for optimizing nuclear reactors and other systems that rely on controlled nuclear reactions.

2. How is the adjoint flux calculated in multi-group diffusion equations for criticality problems?

The adjoint flux is calculated by solving the adjoint diffusion equation, which is derived from the original diffusion equation by taking the adjoint of the operator. This involves inverting the diffusion coefficient matrix and solving for the adjoint flux.

3. Why is the adjoint flux important in criticality problems?

The adjoint flux provides valuable information about the sensitivity of a criticality problem to changes in the neutron flux. This allows for more efficient optimization and control of nuclear reactors and other systems that rely on nuclear reactions.

4. Can the adjoint flux be used for other types of problems besides criticality problems?

Yes, the adjoint flux can be used for other problems involving neutron transport, such as shielding and source multiplication calculations. It can also be applied to other types of differential equations, not just diffusion equations.

5. Are there any limitations to using the adjoint flux in multi-group diffusion equations for criticality problems?

One limitation is that the adjoint flux assumes a linear relationship between the neutron flux and the response of the system. This may not always hold true for highly non-linear systems. Additionally, the adjoint flux may not accurately capture the effects of strong neutron absorption or scattering.

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