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

taitae25

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.

taitae25

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.

joek856

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

taitae25

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 doesnt mean anything does it? After this, I should be able to get moving.

joek856

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.

taitae25

Okay, I'll give it a shot. Thanks a lot.

taitae25

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