# New Flux in a Sphere of moderator of Radius R

1. Mar 7, 2006

### Lucky mkhonza

I have a problem where I must show that neutron flux in the sphere, for a sphere of moderator of radius R. This is one of the problems (5.14.a) in chapter 5 of "introduction to Nuclear Engineering" by J. R. Lamarsh. I would like to include the equation in this request but I'm unable to paste it in.

Anyone can tell me how do I go about solving this problem, I would appreciate it.

2. Mar 7, 2006

### Astronuc

Staff Emeritus
Is that the 3rd edition of Lamarsh. I have the 1st edition, but I just ordered the third. The problem numbers are different.

If the problem is a point source then see this thread,

If the problem is a distributed source,

please show your work, i.e. the governing differential equation, i.e. the diffusion equation for spherical geometry, and the boundary conditions.

Then we can assist you.

BTW, this is a homework problem and actually belongs in the Engineering Homework section.

3. Mar 7, 2006

### theCandyman

Lucky mkhonza, you can use Latex to post equations. Read abou it here: https://www.physicsforums.com/showthread.php?t=8997.

I will save you the time and post the problem.

A sphere of moderator of radius R contains uniformly distributed sources emitting S neutrons/cc- sec.

Show that the flux in the sphere is given by
$$\phi = \frac{S}{\Sigma_a}[1 - \frac{R+d}{r}\frac{sinh(\frac{r}{L})}{sinh(\frac{R+d}{L})}]$$

You can see my other thread, but since it deals with a single source, the problem is going to have a different differential equation and hence, a different solution for the flux.

EDIT: S is divided by the macroscopic absorption cross section, not the microscopic absorption cross section. I cannot get a capital sigma in latex.

EDIT 2: I fixed the sigma. Thanks, Grogs.

Last edited: Mar 7, 2006
4. Mar 7, 2006

### Grogs

In the 2nd edition, it's problem 5-16a. I can't speak for the 1st though.

Try this:

$$\Sigma$$

@Lucky mkhonza:

This problem is going to look a lot like the infinite reactor with a point source. The differences are:

1. You'll have a source term in the diffusion equation, so it will be inhomogeneous.

2. Your boundary conditions will be different.

The diffusion equation will take the form:

$$\nabla^2\phi - \frac{1}{L^2}\phi = -\frac{S}{D}$$

where S is a constant in this case.

This is subject to the following Boundary Conditions:

$$\lim_{\substack{r\rightarrow 0}} \phi(r) < \infty$$

$$\phi(R+d) = 0$$

You've got an inhomogeneous equation here. There are a few different ways to solve it, but I find the variation of constants easiest myself. The general solution will take the form of the homogeneous portion (the solution for a point source problem) plus the inhomogeneous part. The inhomogeneous part will take the form of your source term, or in this case, a constant. Solve for your inhomogeneous constant by substituting it back into the diffusion equation as $\phi$ then proceed to apply BC's and solve as you would any other problem.

Last edited: Mar 7, 2006
5. Mar 8, 2006

### Lucky mkhonza

Thanks for all your replies...let met try

Here is the Diffusion Equation (in spherical coordinates) for the problem

$$\nabla^2\phi - \frac{1}{L^2}\phi = -\frac{S}{D}$$

general solution

$$\phi = A \frac{e^{-\frac{r}{L}}}{r} + B \frac{e^{\frac{r}{L}}}{r} + C$$

A, B (homogeneous solution) and C (non-homogeneous solution) are constants.

Substituting the above equation into Diffusion Equation gives C.

$$C = \frac{SL^2}{D}$$

therefore: $$\phi (R+d) = A \frac{e^{-\frac{R+d}{L}}}{R+d} + B \frac{e^{\frac{R+d}{L}}}{R+d} + \frac{SL^2}{D} = 0$$

Applying boundary conditions

$$B= -Ae^{-\frac{2(R+d)}{L}} - \frac{SL^2(R+d)}{D}e^{-\frac{(R+d)}{L}}$$

Substituting B

$$\phi = \frac{A}{r}e^{-\frac{r}{L}} - \frac{A}{r}e^{\frac{r}{L} - \frac{2(R+d)}{L}} + \frac{SL^2(R+d)}{Dr}e^{\frac{r}{L} - \frac{(R+D)}{L}} + \frac{SL^2}{D}$$

From here I used Fick's Law to solve for A but I get a mess of terms which at the end I don't get the correct answer.

I will appreciate if anyone comments on my work above.

Last edited: Mar 8, 2006
6. Mar 8, 2006

### theCandyman

I believe your particular solution will also have a r term in it. I calculated the particular solution as follows.

$$\phi_p = Er + F$$
Guess the solution.

$$\phi_p = \frac{Er}{L^2} - \frac{F}{L^2} = -\frac{Sr}{D}$$
Solving for the constants.

$$E = \frac{L^2S}{D}$$
E is solved, there is no constant in the answer so F must equal zero.

$$\phi(r) = A\frac{e^{-\frac{r}{L}}}{r} + B\frac{e^{\frac{r}{L}}}{r} + Er$$
The flux for a multisource sphere.

Last edited: Mar 8, 2006
7. Mar 8, 2006

### Lucky mkhonza

I followed your suggestion but I'm getting a different E than yours. I believe you used the spherical coordinates

$$E = \frac{L^2Sr}{D(2L^2 - r^2)}$$

I will carry on to see if I could get the correct answer.

Thanks to all......

8. Mar 8, 2006

### Grogs

Try this form for a general solution:

$$\phi = A \frac{\sinh{-\frac{r}{L}}}{r} + B \frac{\cosh{\frac{r}{L}}}{r} + C$$

I was taught that if the inhomogeneous part is a polynomial, your solution should be of the same order (0 in this case,) so I agree with the 'C' you posted above. You can simplify that to:

$$C = \frac{SL^2}{D} = \frac{S}{\Sigma_a}$$

because

$$L^2 = \frac{D}{\Sigma_a}$$

With sinh and cosh, when you apply the $\lim_{\substack{r\rightarrow 0}} \phi(r) < \infty$ BC, you'll get B=0, so you're left with:

$$\phi = A \frac{\sinh{-\frac{r}{L}}}{r} + \frac{S}{\Sigma_a}$$

Solve for A using the extrapolated boundary condition, and you should be able to get the answer you're looking for after a little bit of algebra.

9. Mar 8, 2006

### theCandyman

If the nonhomogenous part of the equation does not have an r term in it, it will disappear out of the equation after you apply Fick's law though. Is the current at the center of the sphere zero?

10. Mar 8, 2006

### Grogs

It should be. You've got a uniformly distributed source, so if you pick an infinitely small sphere, i.e., a point, you have no source there. At the center of the sphere, you've got the same amount of material -> the same source in every direction, so there should be no net current.

If you substitute a constant into the diffusion equation:

$$\nabla^2 C- \frac{1}{L^2} C = -\frac{S}{D}$$

the first term is zero and you end up with:

$$-\frac{C}{L^2} = -\frac{S}{D}$$

which makes sense. If you have an r in your source (inhomogeneous) term, the flux will actually increase as you move away from the center of the sphere, which makes no sense for a uniformly distributed source.

EDIT: OK, I see the problem. In this equation:

$$\phi_p = \frac{Er}{L^2} - \frac{F}{L^2} = -\frac{Sr}{D}$$

you shouldn't have the 'r' in your source term. That's a linearly dependent source term, not a uniform source term, which should just be a constant. If the right-hand side is $-\frac{S}{D}$ you end up with E = 0 instead of F.

Last edited: Mar 8, 2006
11. Mar 8, 2006

### Astronuc

Staff Emeritus
Assuming 'uniformly' distributed source, uniform BC and symmetry, yes, current is zero at center.

12. Mar 9, 2006

### Lucky mkhonza

Aha...you guys rock.

$$\phi = A\frac{\sinh({-\frac{r}{L})}}{r} + \frac{S}{\Sigma_a}$$

Solving for A using the extrapolated boundary condition, this is what I get.

$$A = -\frac{S}{\Sigma_a} \frac{(R+d)}{ \sinh (-\frac{R+d}{L})}$$

Substituting A

$$\phi = \frac{S}{\Sigma_a}[1 - \frac{R+d}{r}\frac{sinh(-\frac{r}{L})}{sinh(-\frac{R +d}{L})}]$$

using the following relations of Sinh

$$\sinh({-\frac{r}{L}}) = -\sinh({\frac{r}{L}})$$

similarly

$$\sinh({-\frac{R+d}{L}}) = -\sinh ({\frac{R+d}{L}})$$

I get the answer I have been looking for.

$$\phi = \frac{S}{\Sigma_a}[1 - \frac{R+d}{r}\frac{sinh(\frac{r}{L})}{sinh(\frac{R +d}{L})}]$$

Thank you so much guys...you have made my day.

Last edited: Mar 9, 2006
13. Mar 9, 2006

### Grogs

Arrgh! I just noticed that minus sign in the SINH term. It shouldn't be there. I cut and pasted the latex from the exponential form and forgot to take it out. It doesn't change the answer in this case, but in a non-symmetrical problem, I think it would. That general form should read:

$$\phi = A \frac{\sinh{\frac{r}{L}}}{r} + B \frac{\cosh{\frac{r}{L}}}{r} + C$$

14. Mar 9, 2006

### Lucky mkhonza

Will it make any difference if someone used your first suggestion besides considering non-symmetrical problem? that's using..

$$\phi = A \frac{\sinh{-\frac{r}{L}}}{r} + B \frac{\cosh{\frac{r}{L}}}{r} + C$$

Obviously using

$$\phi = A \frac{\sinh{\frac{r}{L}}}{r} + B \frac{\cosh{\frac{r}{L}}}{r} + C$$

one get directly to the solution

$$\phi = \frac{S}{\Sigma_a}[1 - \frac{R+d}{r}\frac{sinh(\frac{r}{L})}{sinh(\frac{R +d}{L})}]$$

Thanks Grogs

Another thing...How do I include someone's QUOTE on my reply?

Last edited: Mar 9, 2006
15. Mar 9, 2006

### Astronuc

Staff Emeritus
Hit the quote button below the particular post you wish to quote, but it will not catch other quotes in that post.

Alternatively, copy the text, paste it and put [Q=person] at beginning and [/Q] following text, where Q=quote.

Also and , where S=sup for superscript and S=sub for subscript

16. Mar 9, 2006

### Grogs

Well, like I said, it shouldn't make a difference on this problem. My only concern is that, if it really does make a difference on *some* problems - and I honestly can't say for certain if it does, your instructor might take off a point or two because the equation is technically wrong. I'm probably in a similar class to what you're in right now (I'm taking reactor theory) and I've got a pretty picky instructor, so I try to avoid all those little errors so I don't get lots of little deductions on my HW problems.

I guess a better way of putting it is I know the
$$\phi = A \frac{\sinh{\frac{r}{L}}}{r} + B \frac{\cosh{\frac{r}{L}}}{r} + C$$
form works, so that's what I use.