NE421 Introduction to Nuclear Criticality Safety

# Homework Problem #3

In this assignment, we are going to use the SCALE sequence CSAS1 to reproduce the curves in Fig. 2-2 in the text (page 10) by calculating the critical mass required for various Pu-239/water mixtures.  We are going to use the CSAS1 sequence because it is deterministic (therefore fast in 1D, as you learned for the midterm).  The remainder of this page is a tutorial that walks you through the first steps of the problem, then sets you loose to complete the assignment.

## Computational environment setup

First of all, we need to establish the computational environment.  I will assume that you have installed SCALE6.2 on the computer you are on.

### 1. Open Fulcrum

I will assume you want to run with the interface Fulcrum that the SCALE group provides. (I personnally prefer to work from a MS-DOS prompt, but I am in a shrinking minority.)

Fulcrum is opened in the Windows system by clicking on the "Start" button (lower left-hand corner of screen), then "SCALE-6.2.1" (could be anywhere, look first in the top section, then search through "All Programs" for it), then click on the choice: "scale SCALE-6.2.1".

Fulcrum serves as both a context-sensitive text editor and an operation environment.

You can keep the files in any directory you choose; I recommend that you create a subdirectory named "NE421" that you can use throughout the course.  But that is up to you.

## Creating the first input deck

### 3. Create "pt1.in".

(You can, of course, use another word processor if you prefer.  If you use a word processor that puts in extra formatting information (like MS Word does), be sure to save the result in ASCII (text) format.  Also, be sure to save the result as "pt1.in" in the subdirectory of your choice.)

Type (or cut and paste) the following lines in the text area of Fulcrum, exactly as printed:

```
=csas1 parm=bonami
fig. 2-2, point 1
v7-56
pu  1 1 293   94239 100 end
h2o 2 1 293   end
end composition
multiregion spherical left_bdy=reflected right_bdy=vacuum cellmix=500 end
1  5
end zone
end celldata
end
```

The input consists of only 13 lines (one of which is unnecessary).  The next part of the tutorial will just walk you through an example, and then you can do the assignment by modifying the deck we will build.
The alternative is to slog your way through the manual.  Although I am going to allow you to short-circuit the manual for now, I still recommend that you find it in the SCALE-6.2.1 directory and learn to use it.

We are going to take a shortcut and I am just going to teach you what each line means, so you can modify this deck to do the assignment.

`=csas1 parm=bonami`
The first line states the sequence that will be used and specifies the resolved resonance processing tool that will be used (Bonami). Most places use the more accurate CENTRM for resonance processiong, but it is much slower and doesn't really add much accuracy for our problems.
`fig. 2-2, point 1`

This is the title of the case.  You should change this every time you modify the deck.

`v7-56`

This gives the name of the cross section library that you want to use.  You have several choices, but in this class we will be mostly using the 56 group library pointed to here, to save a bit of computer time. (The manual says it is mainly for light water reactor work, but it seems to give good answers for our problems.)

`read composition`

This tells SCALE that we are going to begin defining the materials that we will use. Each material is given a number (so we can refer to it later),
and it is best to just number them in order from 1 up. (Unless you organize your work some other way.)

` pu  1 1 293   94239 100   end`

This line is the description of material 1:

• Entry 1 "Pu" is the Standard composition name for plutonium
• Entry 2 "1" is the mixture ID number
• Entry 3 "1" is the volume fraction of the composition in the mixture (1 means 100%.  So this material is all plutonium.)
• Entry 4 "293" is the temperature in Kelvin
• Entry 5 "94239" is the istope's ZA number, which is 1000 * atomic number + atomic weight
• Entry 6 "100" is the weight percent of the isotope in the composition.  (NOTE: Entries 6a and 6b are repeated for each isotope in the composition.  E.g., for a 95% Pu-239/Pu-240 mixture, we would put "94239 95 94240 5".)
• Entry 7 is "END" as required.
` h2o 2 1 293   end`

This is the material card for the water, following the same pattern as the Pu card.  This is the unnecessary card, since we do not use water in this problem; we have it here for use in later problems.

`end composition`

End of the composition description.

`read celldatamultiregion spherical left_bdy=reflected right_bdy=vacuum cellmix=500 end`

This is the beginning of the geometry data.  In this case, we are using Multiregion geometry specification, which is where we put the geometry description in sequence CSAS1:

• Entry 1 specifies that we are using the MULTIREGION option (which is what we will always use for CSAS1)
• Entry 2 "SPHERICAL" is the type of geometry
• Entry 3 "left_bdy=reflected" is the inside (left) boundary condition
• Entry 4 "right_bdy=vacuum" is the outside (right) boundary condition
• Entry 5 "cellmix=500" is not really necessary for this problem (advanced option)
• Entry 6 "end" is a required entry
` 1  5 `

The 1D spherical geometry (which we are using) builds a geometry that is like a jawbreaker--spherical layers of materials radiating out from the center of the sphere.  We build the layers of the sphere, from the center out, by specifying material numbers and outer radii (NOT thickness of the region--a common mistake!).  These line holds the two entries for the first (and only) zone.  The "1" is the material number for the zone and the "5" is the outer radius, in centimeters. If there were more material layers (i.e., zones), each would have its material number and outer radius before the "END ZONE" entry.  (Of course, the outer radius of each layer must be larger than the previous one. )

` end zoneend celldataend`

This closes out the zone description, the cell data, and the input deck itself.

## Running the first case

Okay.  The sample input deck is now made.  Let's run it.

### 4. SAVE THE FILE as "pt1.in" and then pull down the "Run" tab above the text area and choose the first entry: "Run in Background"

It will use "pt1.in" for input and and run it right before your eyes. As the last line printed to the screen says, it put the output in "pt1.out". This is in the same subdirectory where you saved the file.

(NOTE: You will notice that the code gets started well and then "stalls" for a few seconds before continuing. This is the time that it is spending processing the resonance cross sections. The length if this "pause" will get longer and longer as the problems get more complicated. The reason is that the resonance processing step is getting more and more complicated by ORNL as more and more detail is added to this step. In my opinion, this is not worthwhile, but the SCALE group has completely disabled the previous--fast and perfectly adequate-- Nordheim integral treatment. But, alas, there is nothing we can do about this but wait it out.)

### 5. You can look at the output by opening up the "pt1.out" file with your favorite editor

if you want to, but the k-effective value we are after is right there on your screen, after the word "lambda".  There is lots of interesting information in the output deck, but we are going to jump right to the answer.

The k-effective value should be close to 1.01452 (that's what I got).

## Modifying the input deck

Now that we have a deck that works, we will be setting up similar problems by modifying the input deck directly rather than returning to the input processor.  First of all, let us open the input deck and look at it.

### 6. Put in a new guess for the outer radius.

The k-effective, 1.01452, is not close enough to critical for us to declare 5 cm as the critical radius.  For this exercise, I will (arbitrarily) say that we need a value within half a percent of critical (i.e., between 0.995 and 1.005).  So, we need to decrease the radius some amount.

How much?  You have to figure it out by trial-and-error. .)

### 7. Replace the 5 with 4.5 and rerun the problem.

The resulting k-effective should be closer to 1.  I got 0.92635, which is too far off in the other direction.  Interpolating gives about 4.9 (I am eyeballing it!) which gives us a k-effective of 0.99716, which is within our range.

## Running the second case

For the cases beyond the first one, there is a mixture of Pu and water in the sphere.  This mixture is controlled using the two material cards:
` pu          1 1 293   94239 100   end h2o         2 1 293   end`

The first thing we have to do is to get the water into the same mixture as the Pu.  This is done by simply changing the mixture ID on the water card to "1".

` pu          1 1 293   94239 100   end h2o         1 1 293   end`

Now the problem is that both volume fractions are set to 1.00. SCALE interprets this that you want full density Pu and full density water shoved into the same space.  We need, in fact to divide the volume up between the Pu and the water.  The way we do this is to (1) set the the Pu volume fraction to the value needed to get the desired density and then (2) let the water have the rest of the volume.

For example, if we want to create the mixture appropriate for the peak of the curve -- which looks to be at a Pu density of 5 kg/liter (equal, of course, to 5 g/cc), we must dilute the Pu down from its metal density of 19.84 g/cc.  So, the Pu volume fraction should be set to 5/19.84 = 0.252, which leaves 0.748 for the water:

` pu          1 0.252 293   94239 100   end h2o         1 0.748 293   end`

Also, looking at the curve, it looks like a good first guess for the critical radius is 11 cm, so the first layer should become:

` 1            11 `

Okay, let's run the new problem.

### 10. Run the second problem.

If you followed all that, your answer should be about 1.050, which is too high.  Again, we need to perform a trial-and-error search on the critical radius.

## Complete the assignment:

• After you have finished the search on the 5 kg/liter critical radius, repeat the procedure for Pu densities of 0.5, 0.05, and 0.02 kg/liter.
• Now find the critical radius for the peak (not the whole curve) of the BOTTOM curve. (5 kg/liter).  Use a 15 cm water reflector as the region outside the Pu layer. (Be careful to change the water radius when you change the Pu one).
SAVE your input decks.  You will need them again in the next homework.