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It's part of my homework to find an analytical solution to a slab reactor with a reflector. The setup of the core is two slabs of equal thickness (-a to 0 and another 0 to +a) but different values for the diffusion coefficent, macroscopic absorption cross sections, and [tex]\nu \Sigma_f[/tex]. There is a reflector on each side and each has the same thickness (-b to -a and +a to +b) and material properties.

As boundary conditions, I have that the flux is zero at the extrapolated distances from the reflector ([tex]\phi(X_{ex})=0[/tex], where [tex]X_{ex}[/tex] is the extrapolated distance). To link each part of the system together I have found that

[tex]\phi_{r1}(-a) = \phi_{c1}(-a)[/tex]

[tex]\phi_{r2}(a) = \phi_{c2}(a)[/tex]

[tex]\phi_{c1}(0) = \phi_{c2}(0)[/tex]

Using these conditions and solving the general time-independent diffusion equation for each section,I got

[tex]\phi_{r1}(x) = A_1 sinh(\frac{x+X_{ex}}{L_r})[/tex]

[tex]\phi_{r2}(x) = A_4 sinh(\frac{X_{ex}-x}{L_r})[/tex]

for each reflector, where [tex]L_r[/tex] is the neutron diffusion length. In the core, I got

[tex]\phi_{c1}(x) = A_2 sin(B_{c1}x)+C_2cos(B_{c1}x)[/tex]

[tex]\phi_{c2}(x) = A_3 sin(B_{c2}x)+C_3cos(B_{c2}x)[/tex]

where [tex]B = \frac{\nu\Sigma_f - \Sigma_a}{D}[/tex] (these three variables are defined in the problem).

I then take the equations for flux of each half of the core and since they are equal at x=0, I find that [tex]C_2=C_3[/tex].

Here's where I am somewhat stuck. I said the reflectors were symmetric about 0 so then

[tex]\phi_{r1}(-a)=\phi_{r2}(a)[/tex]

and using the earlier mentioned continuity equations I get

[tex]A_2sin(B_{c1}*(-a))+C_2cos(B_{c1}*(-a))=A_3sin(B_{c2}*(a))+C_3cos(B_{c2}*(-a))[/tex]

which is one equation with three unknowns (since [tex]C_2=C_3[/tex]).

I'd like a little input into if I have all the conditions that link the sectors together and if my equations for the core are correct. I can't think of a reason not to have the [tex]Asin(B_cx)[/tex] term, but I can see that the [tex]Ccos(B_cx)[/tex] is necessary to prevent zero flux at x=0.

As boundary conditions, I have that the flux is zero at the extrapolated distances from the reflector ([tex]\phi(X_{ex})=0[/tex], where [tex]X_{ex}[/tex] is the extrapolated distance). To link each part of the system together I have found that

[tex]\phi_{r1}(-a) = \phi_{c1}(-a)[/tex]

[tex]\phi_{r2}(a) = \phi_{c2}(a)[/tex]

[tex]\phi_{c1}(0) = \phi_{c2}(0)[/tex]

Using these conditions and solving the general time-independent diffusion equation for each section,I got

[tex]\phi_{r1}(x) = A_1 sinh(\frac{x+X_{ex}}{L_r})[/tex]

[tex]\phi_{r2}(x) = A_4 sinh(\frac{X_{ex}-x}{L_r})[/tex]

for each reflector, where [tex]L_r[/tex] is the neutron diffusion length. In the core, I got

[tex]\phi_{c1}(x) = A_2 sin(B_{c1}x)+C_2cos(B_{c1}x)[/tex]

[tex]\phi_{c2}(x) = A_3 sin(B_{c2}x)+C_3cos(B_{c2}x)[/tex]

where [tex]B = \frac{\nu\Sigma_f - \Sigma_a}{D}[/tex] (these three variables are defined in the problem).

I then take the equations for flux of each half of the core and since they are equal at x=0, I find that [tex]C_2=C_3[/tex].

Here's where I am somewhat stuck. I said the reflectors were symmetric about 0 so then

[tex]\phi_{r1}(-a)=\phi_{r2}(a)[/tex]

and using the earlier mentioned continuity equations I get

[tex]A_2sin(B_{c1}*(-a))+C_2cos(B_{c1}*(-a))=A_3sin(B_{c2}*(a))+C_3cos(B_{c2}*(-a))[/tex]

which is one equation with three unknowns (since [tex]C_2=C_3[/tex]).

I'd like a little input into if I have all the conditions that link the sectors together and if my equations for the core are correct. I can't think of a reason not to have the [tex]Asin(B_cx)[/tex] term, but I can see that the [tex]Ccos(B_cx)[/tex] is necessary to prevent zero flux at x=0.

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