Homework Problem #3In 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 setupFirst 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".
2. Enter your data
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 read composition pu 1 1 293 94239 100 end h2o 2 1 293 end end composition read celldata 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
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
=csas1 parm=bonamiThe 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.
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.)
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),
pu 1 1 293 94239 100 end
This line is the description of material 1:
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 of the composition description.
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:
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"
(Of course, the outer radius of each layer must be larger than the
This closes out the zone description, the cell data, and the
input deck itself.
Running the first caseOkay. 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 editorif 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 deckNow 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 caseFor 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
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
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
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:
Okay, let's run the new problem.
8. Put these changes into the deck, save it as "pt2.in"
9. Make the material and radius changes to the deck.
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:
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