- #1

- 6

- 0

In diffusion equation ,if we have a infinite slab of moderator with thickness ±a and the sources is

s(x)= cos(x)

i think the first boundary

flux( ±a ) = 0

what will be the second boundary condition ??

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter victor77
- Start date

- #1

- 6

- 0

In diffusion equation ,if we have a infinite slab of moderator with thickness ±a and the sources is

s(x)= cos(x)

i think the first boundary

flux( ±a ) = 0

what will be the second boundary condition ??

- #2

- 25

- 2

To get the first one, you will have to apply the boundary condition of the total flux decay at the boundary; which you have already mentioned in your question. This will lead you to numerous possible solutions/harmonics.

To get the second one, you have to know the power of the assumed reactor.

- #3

- 6

- 0

To get the first one, you will have to apply the boundary condition of the total flux decay at the boundary; which you have already mentioned in your question. This will lead you to numerous possible solutions/harmonics.

To get the second one, you have to know the power of the assumed reactor.

this is for reactor but in my case it is a moderator

- #4

- 25

- 2

- #5

- 6

- 0

i do not have the power, i am thinking to use laplace transformtion

if the boundary condition

flux( ±a ) = 0

Derivative flux(0)=0

- #6

- 25

- 2

However, my first reply was the general scheme for the solution of the problem, but seems I underestimated your example as it seems that it will get more complicated with this cosine-shaped source.

As far as I know the general scheme for your solution will be as follows..

Solve using the diffusion equation using eigen functions solution (the way I mentioned in my first reply) after you apply your boundary condition (most cases, you can just expect the shape of the flux instead of solving the differential equation)..then take advantage of the orthogonality of the eigen set of functions that you have by putting your source term in an eigen functions summation as well as your flux (that you just got assuming no Source) term, and insert both of these terms in the diffusion equation. Now you are left only with the coefficient to be calculated (the constant that you were to get it by knowing the power), and you can get it. This is the over view of the solution, and obviously it may be not so helpful to just read it, so I advise you to read section 5-10 (especially subsection: distributed sources in a finite medium) from John R. Lamarsh, Introduction to Nuclear Reactor Theory.

- #7

- 25

- 2

- #8

- 6

- 0

i read the lamarsh(intro to nuclear reactor ) book but in lamarsh(intro to nuclear enginnering ) problem 34 he use laplace transformtion

that why i am confused now ,,anyway thank for helping

- #9

- 25

- 2

Problem 34, which chapter?

- #10

- 6

- 0

5Problem 34, which chapter?

Share:

- Replies
- 5

- Views
- 2K

- Replies
- 3

- Views
- 1K

- Replies
- 1

- Views
- 1K

- Replies
- 4

- Views
- 5K

- Replies
- 1

- Views
- 2K

- Replies
- 1

- Views
- 3K

- Replies
- 4

- Views
- 5K

- Replies
- 1

- Views
- 3K

- Replies
- 6

- Views
- 6K

- Replies
- 1

- Views
- 5K