ML20247F590
| ML20247F590 | |
| Person / Time | |
|---|---|
| Issue date: | 05/11/1998 |
| From: | Advisory Committee on Reactor Safeguards |
| To: | |
| References | |
| ACRS-T-3036, NUDOCS 9805190441 | |
| Download: ML20247F590 (245) | |
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GRlG Nh- AcRSF3o36 OFFICIAL TRANSCRIPT OF PROCEEDINGS L
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NUCLEAR REGULATORY COMMISSION L .
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Title:
! SUBCOMMITTEE ON--
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, DATE: Monday,May 11,1998 PAGES:1 - 132 144 - 207 o-249 277 289 - 294 9805190441 980511 PDR ACRS T-3036 PDR ,
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I DISCLAIMER
-UNITED STATES NUCLEAR REGULATORY COMMISSION'S ADVISORY COMMITTEE ON REACTOR SAFEGUARDS
, MAY 11, 1998 The contents of this transcript of the proceeding of the United States Nuclear Regulatory Commission Advisory' 1 A
() Committee on Reactor Safeguards', taken on-May 11,'1998, as reported herein, is a record of the discussions recorded at the meeting held on the..above'date.
This transcript had not been reviewed, corrected 3
~
and edited and it may contain inaccuracies.
i i
l
l 1
1 UNITED STATES NUCLEAR REGULATORY COMMISSION I e 2 ADVISORY COMMITTEE ON REACTOR SAFEGUARDS 3
4 ***
5 SUBCOMMITTEE ON 6 THERMAL-HYDRAULIC AND SEVERE-ACCIDENT 7 PHENOMENA 8
9 10 U.S. Nuclear Regulatory Commission 11 Two White Flint North '
12 11545 Rockville Pike 13 Rockville, Maryland 20852-2738 14
() 15 Monday, May 11, 1998 16 1
L 17 The Committee met pursuant to notice at 8:30 a.m. 1 1
18 19 MEMBERS PRESENT:
l
l l! 22 MARIO H. FONTANA, Member, ACRS 23 24 l
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2 1 PROCEEDINGS 2
d(' - [8:32 a.m.]
3_ CHAIRMAN KRESS: The meeting will now please come 4 to order. This is a meeting of the ACRS Subcommittee on 5 Thermal-Hydraulics and the Severe Accident Phenomena.
6 'I am Thomas Kress, the Chairman of the 7 Subcommittee. ACRS members in attendance are Mario Fontana 8 and Robert Seale. Graham Wallis would have been here but I 9 understand he is slightly ill today. ACRS consultants in 10 attendance are Ivan Catton, Virgil Schrock and Novak Zuber.
11 The purpose of this meeting-is for the 12 Subcommittee to continue its review of the Westinghouse Test 13 and Analysis Program being conducted in support of the AP600 14 design certification. For this meeting, the Subcommittee A) t, s 15 will review the issues and concerns pertaining to the AP600 16 reactor coolant system that were cited in our Committee's 17 February 19th, 1998 interim report on the design -- AP600
'18 design certification review.
19 The Subcommittee will gather information and 20 analyze relevant issues and facts and formulate proposed 21 positions and actions, as appropriate, for deliberation by 22_ the full Committee.
23 Paul Boehnert is the Cognizant ACRS Staff Engineer l 1
24 for this meeting.
1 25 Portions of this meeting will be closed to the ANN RILEY & ASSOCIATES, LTD.
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3 1 public to discuss Westinghouse Electric Company proprietary s
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3 Rules for participation in today's meeting have 4 been announced as part of the notice of this meeting, 5 previously published in the Federal Register on April 22nd, 6 1998.
7 The transcript of the meeting is being kept, so it 8 is requested that speakers first identify themselves, come 9 to a microphone, identify yourselves and speak with 10 sufficient clarity and volume so that you can readily be 11 heard.
l 12 We have received no written comments or requests 13 for time to make oral statements from members of the public.
14 If you will recall, in our interim report in February we 10
( ,) 15 took what I thought was a consolidation of all the ACRS 16 unanswered concerns with respect to the Test and Analysis 17 Program on both the RCS and the containment, and put them 18 all together in one place and sent them off with an interim 1
19 letter to the EDO, I guess. And today I think we are going 20 to hear some responses from Westinghouse to those questions, 21 or at least give us directions or road maps on where we can 22 find the answers.
23 So with that, I will ask if any member, l
1 24 Subcommittee members, or consultants want to make any 25 comments before we start?
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1 DR. CATTON: I took a look into my concerns about
() 2 3
DNBR. I could comment on them at this time if you wish, or I could just give you a report, written report.
4 CHAIRMAN KRESS: I don't know how to decide until 5 I hear what'your concern is. Why don't you give me an idea 6 -- why don't you just give me an idea of what the concern is 7 now?
8 DR. CATTON: Well, I don't think the base --
9 CHAIRMAN KRESS: I will want a written report.
L 10 DR. CATTON: I have it right here.
11 CHAIRMAN KRESS: Okay.
12 DR. CATTON: I have it right here. Why don't I 13 just read you the conclusions?
14 CHAIRMAN KRESS: Okay. That would be good.
b(j .15 DR. CATTON: There are several problems with what 16 has been done. The CHF correlation was changed using 17 Westar, which is a code I had never heard of before, under 18 conditions where cross-flow is minimal. There is no 1
19 reference to Westar in the SSAR, but.THINK 4 is referenced.
l 20 None of the documents show how cross-flow and 21 channel-to-channel mixing comparisons, with each other or 22 with data. No basis or even reasoned arguments are given 23 for increasing the channel-to-channel mixing by 50 percent.
24 And I don't believe the NRC conditions on the use of the i 25 revised thermal design procedure are met. !
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1 Now, the staff says that the AP600 core design is
, 'i 2 similar to that of operating Westinghouse plants, therefore, N' J 3 therefore, and everything is fine. I don't know whether I l
4 am missing documentation, or whether indeed it just slipped 5 through the crack. Because the requirements are clear. I 6 think I have them here.
l 7 The staff position that was written in the SER 8 when they blessed this new method says that sensitivity 9 factors for a particular plant and their ranges of 10 applicability should be included in the SSAR or reload 11 submittal. (2) Any changes in DNB correlation, THINK 4 12 correlations, or parameter values listed in Table 3.1 of 13 WCAP 11397, outside of previously acceptable ranges, require 14 re-evaluation of the sensitivity factors and the use of
(,
15 equation 2.3 of the topical report. Equation 2.3 is just a 16 trivial thing.
17 If the sensitivity factors are changed as a result 18 of correlation changes, or changes in the application of the 19 use of the THINK code, the use of an uncertainty allowance 20 for application of equation 2.3 must be re-evaluated and so 21 on. And I don't think that was done.
22 Now, I don't know how much the correlations 23 change. I don't know what this acceptable range of change 24 is. But I don't think THINK 4 was used in the process, 25 unless there is a document missing.
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6 1 CHAIRMAN KRESS: Well, I will look forward to
() 2 reading your whole report.
3 DR. CATTON: I also have a couple of comments on 4 LOFTRAN. Would you like them now or do you want the report?
5 Either way.
6 CHAIRMAN KRESS: Why don't we wait for the report.
7 DR. CATTON: Do you want to --
8 CHAIRMAN KRESS: And if the time comes up during 9 the meeting, if a place comes up during the meeting that 10 seems appropriate to bring that up, you can bring it --
11 DR. CATTON: I don't know where it will come up 12 because you have already passed Chapter 15.
13 CHAIRMAN KRESS: Okay. This is part of Chapter 14 15.
() 15 DR. CATTON: So it probably won't come up again.
16 CHAIRMAN KRESS: Well, why don't you make them now 17 then?
18 DR. CATTON: Why don't I which?
19 CHAIRMAN KRESS: Why don't you make your comments 20 now on LOFTRAN?
21 DR. CATTON: With LOFTRAN, V&V was done with 22 SPES 2, and on the surface, it looks okay. But when you 23 start to dig, you find that for a given heat load, which I 24 assume was measured when they did the SPES 2 testing, they 25 have to distribute the heat load somehow on the apparatus.
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7 1 As they shift it, if you say that 80 percent is the steam
() 2 generator, 20 percent is the RCS, you get one set of 3 results. As you shift it more and more to the RCS, things 4 change, and the changes are really dramatic. At 80/20, you 5 pretty much fill up the pressurizer. When you move it over 6 to 65/35, the level in the pressurizer is significantly 7 below the measured values.
8 And when you read all this, I just wonder how you 9 can say anything about its use on the AP600? As a matter of 10 fact, with what was in the V&V document, I don't think you 11 can. So in my view, LOFTRAN really hasn't been V&V'ed 12 against anything. Now, I thought it was going to be V&V'ed 13 against the OSU data, but apparently it is not.
14 CHAIRMAN KRESS: Could you clarify that a little
() 15 for me, Ivan? This heat distribution, is that --
that's a 16 boundary condition, initial condition on the test?
17 DR. CATTON: Again, I am not even really sure how 18 they apply it.
19 CHAIRMAN KRESS: Yeah.
20 DR. CATTON: You can't tell from the document.
21 But we all knew before they went into SPES 2, that for 22 anything beyond a large break LOCA or something, it was 23 going to be iffy because of the --
24 CHAIRMAN KRESS: Because of the heat losses.
25 DR. CATTON: -- high area to volume ratio.
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8 l' CHAIRMAN KRESS: Of the heat losses.
() 2 3
DR, CATTON:
CHAIRMAN KRESS:
The problem.
So they had to up the core heat 4 to take care of that?
5 DR. CATTON: They have to do something in order to 6 make use of the data. Well, they did something to make use 7 of the data for LOFTRAN. And what they did is the first 8 runs in the preliminary V&v document, they said, well, gee, 9? 80 percent of this heat loss from the primary fluid is to 10 the steam generator.
11 CHAIRMAN KRESS: I see. That's the distribution 12 you are talking about.
13 DR. CATTON: Yes.
14 CHAIRMAN KRESS: With the heac loss.
( 15' DR. CATTON: They take this amount of heat that 16 .they have got to deal with and they split it up, and they 17 shift the split, and they just adjusted it.
, And you could 18 adjust it to get whatever you want. So, given that you 19 adjust it to match SPES 2, I think what you have is a code 20 V&V'ed for analysis of SPES, not anything else. Because the 21 heat loss has too big an impact.
22 When they change it from 80/20 to 65/35, the 23 pressurizer level changes dramatically from being above the i
24 data to being significantly below the data. i
-25 CHAIRMAN KRESS: I gather what you are saying is l
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9 1 that'if you validate the LOFTRAN against SPES under these 2 conditions, what you are validating is the heat lost b
3 equations, as opposed -- because they dominate as opposed to 4 the other parts of LOFTRAN? i 5 DR. CATTON: Well, I am not sure what you are 6 validating. Because'it is so sensitive to the heat loss 7 distribution, you can get whatever result you need. So now 8- if you move to something else, you don't know whether you 9 have covered up other problems or what you have done.
10 CHAIRMAN KRESS: Well, it would be worth thinking 11 about. My initial reaction is, as long as the code can 12 predict what you have, and if you still have your PIRT chart 13- telling you'you are looking at the right phenomena, and the 14 right phenomena are sufficiently strong enough that it shows 15' up in the test, that you may not-be -- it may not be a 16_ problem. I'll have to --
17- DR. CATTON: Well, whenever you can go into --
18 whenever you have this unknown parameter in whatever you 19' doing, --
20 CHAIRMAN KRESS: Yeah. Well, that --
21 DR. CATTON: -- and you can turn it and make the 22 results come down to whatever you want, I think you have a 23 problem, no matter whatever else you have.
24 -CHAIRMAN KRESS: Oh, you are tuning this in the 25: code, you are saying?
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i 10 1 DR. CATTON: Well, they started out picking 80/20,
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v' 2 the base case became 65/35. Why isn't the base case 80/20?
- 3. CHAIRMAN KRESS: I see. Now, I think I understand 4 the question a little more.
5 DR. CATTON: And this makes it -- this make the j 6 code --
7 CHAIRMAN KRESS: It looks like it is tuned to SPES 8 if you do it that.way.
9 DR. CATTON: Right. And what do you do now when l
10 you want to make predictions of LOFTRAN? Or is this heat l 11 load so low it.doesn't matter?
12 Well, you don't know where you are at, because you 13 don't know whether you have covered other kinds of 14 difficulties or not. And what leads me to that is that the
) 15 shapes of a lot of the curves aren't quite right. You know, 16 pressure drops too step, this is too high. You know, 17 nothing really. If you had a nice circumstance where the 18 parameter, whatever it is, is a function of time, was very 19 similar to the other value, then you could say, well, you l 1
20 know, maybe everything else is kind of right. But when it '
21 is not, and you sort of just get this thing so it wiggles 22 back and forth across the data, well, I don't buy that. I 23 don't think that's proper.
24 CHAIRMAN KRESS: Do you guys understand what his 25 problem so that you might address it at one time? I don't l ANN RILEY & ASSOCIATES, LTD.
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11
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1 know if you are prepared to do that.
4 3- 2 MR. McINTYRE: We don't -- we didn't travel with
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3 our LOFTRAN expert this time. But Paul forwarded I think a 4 synopsis of the conclusions, or sentences. If there is a 5' whole report on this, that would obviously be useful.
6 DR. CATTON: Brian, that was before I had your 7 documents.
8 MR. McINTYRE: You did get the --
9 DR. CATTON: I just amplified a little bit on my 10 reaction to the meeting that was on the 31st of March in 11 that e-mail to Paul. I got your documents and tried to go
-1:2 .through them reasonably careful-in a weekend.
13 It had a severe impact on me. It was Mother's Day 14 Sunday.
(O j. 15 DR. SCHROCK: I was going to say, you mean like 16 leaving there at Noon yesterday.
17 DR. CATTON: But I did go through it. I 18' MR. McINTYRE: Did you have the DNB reports?
19 DR. CATTON: I had three and, let's see -- oh, I 20 passed all my reports down there. Why don't you give me 21 back one of-them, Paul?
22 I had WCAP 11397-P-A, which is the revised thermal 23 design procedure. That is where these conditions that I 24 read into the record were listed. There were about eight or ,
1 25 'nine but only the three are relevant.
l !
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12 1 The second was WCAP 12330-A, and that is improved
() 2
'3-THINK modeling and really you just did it better, although I felt that the comparisons with the other codes were 4 incomplete. All they compared was the vertical mass flow in 5 a flow channel -- really should have looked at cross flow 6 'and mixing but they didn't. That is neither here nor there, 7 but it wasn't done.
8 And then I looked at WCAP 14371, which is AP600 9 low flow critical heat flux test data and analysis. Well, 10 they tuned WESTAR -- they developed a correlation for 11 WESTAR, not for THINK 4. Now you may have another document 12 where you develop the correlation for THINK 4, and then 13 follow through on thece three requirements that the Staff 14 laid down on you in what, 1989, or something, but I don't
() 15 have that documentation.
16 CHAIRMAN KRESS: Did you have other comments, 17 Ivan, before we move on?
. 18 DR. CATTON: I think that's enough.
l- -19 CHAIRMAN KRESS: Novak, did you want to say ;
20 anything?
1 21 DR. ZUBER: I have four short comments.
22 First, I would like to compliment Westinghouse for 23 sending us that roadmap. I think it was useful, but my :
24 first try was completely a failure. I was looking where is L
25 the momentum flux and I think it is not there.
1
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13 1 This will be the first question really to address.
I V
3 2 The second thing is on scaling, the Stage 4 and 5, 3 and I.will not say what my concerns.are. After you have the 4 presentation I will let you know my concerns.
5 Third, I would like I hope that Westinghouse 6 presentations will be as candid as Mike's are usually, so 7 you should serve as an example for your colleagues as far as 8 candid presentation.
9 I am looking forward to hearing them. Thank you.
10 DR. SCHROCK: I just would like to say that I 11 thought the roadmap idea was good, but like Novak's first 12 try, I didn't find it all that useful. It could have been 13 made a lot more useful with not very much more effort. ,
14 The quality of the writing and some of the new A( ,) 15 material that'we have is still really poor for technical 16 documentation and I don't understand why it is that somebody 17 at Westinghouse doesn't read the material before it is 18 passed along.
19 There are words missing, there are words 20 misspelled, there are grammatical errors, technical errors 21 still -- it just is not good quality engineering 22 documentation and I will give you some examples as we go l
23.
i l along, but I was hoping that this issue or this list of 10 l l 24 issues could be resolved, that we would hear some clear, l
l 25; concise answers to them, but instead what we have is a very j
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14 1 diffuse set of documents that tend to obscure as much as
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[ ';- 2 they tend to enlighten, and it is not helpful. y 3 That-is my reaction I have so far. .I hope as I 4 hear the presentations that that view will be changed, but 5 as of just this time, having simply read as much as I could 6 of these documents, I do not find it really that helpful in 7 clarifying answers to these 10 issues.that are before us 8- today.
9 CHAIRMAN KRESS: Okay. I'll ask any of.the.
10 subcommittee members, will you make any comments?
. 11 DR. SEALE: None.
12- CHAIRMAN KRESS: Okay. With that, let's turn it 13 over to Westinghouse and the floor is yours.
14 MR. PIPLICA: Thank you. My name is Gene Piplica.
(O,) 15 I was involved in the testing program for the AP600 and I am 16 here today to give you an introduction to the material that 17 we are going to present on the AP600 Test and Analysis 18 Program.
19 We are going to focus on the 10 issues that apply
. 20 to the passive core cooling systems, known by the acronym l l
21 PXS in Westinghouse language. l 22 We are passing out all the presentation material 23 in a booklet. What I am showing you is the earlier !
24 material.
25 CRAIRMAN KRESS: By the way, I appreciate having ANN RILEY & ASSOCIATES, LTD.
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-n 15 1 in a booklet like that. It makes it easy for me not to lose
/ Y 2 it*
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3 MR. PIPLICA: Okay, good.
4 DR. BOEHNERT: There is some proprietary material 5 in here, right, Gene?
6 MR. PIPLICA: Yes. It is marked. Some of it is 7 proprietary and it is marked 8 Paul was so kind to prepare the agenda based on n 9 our input and he added and we labelled the 10 items that 10 were in the letter from February 19th, and these numbers 11 show up on the agenda so we have organized the material to 12 address these 10 issues.
13 Now the issues themselves in the order they are
'14 didn't lend theaselves to the presentation so we are not b)
(, 15- going to addrass them in this order but rather you will be 16 able to see'on the agenda which item we are addressing.
17 We are hoping that will clarify what we are 18 talking about, and wa are relying on-the material in the 19 documentation that you have and I understand the comment
'20 about how diffuse the documentation is because it was 21 generated over a five-year period. It tried to incorporate 22 as much as it could the feedback from the Staff and from the 23' subcommittee meetings we have had over this long period of 24 time.
25 Unfortunately, we not always able to go back and 1'
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16 1 redocument everything as we might'best have, so that leads <
[m<s')
2 to -- some of the confusion I believe is the fact that this 1
3 has all been done over such a broad period of time and j 4 involves such a large test program and so many computer 5 codes and reviews that the staff has done.
6 Now as I mentioned, we are here to respond to 1 1 those 10 issues today, but we are also looking to conclude 7 i 8 your review of the test and analysis program for the core t
)
l 9 cooling systems. We believe that with the SER out and the 10 staff's review being complete of the Test and Analysis 11 Program that we have provided sufficient information for the l 12 certification program, and we are hoping today that we can 13 address adequately these 10 issues and bring this series of 14 meetings to a successful conclusion.
[~'T
( ,1 15 DR. ZUBER: Do you mean this is the last meeting 16 on this subject?
17 DR. CATTON: They hope.
18 DR. ZUBER: You hope?
19 MR. PIPLICA: That is our goal.
20 To do this, we plan to present the material in 21 three phases. I was going to give a overview of the core 22 cooling systems in the Test Program and Mr. Brown is going 23 to give an overview of the scaling approach for the core 24 cooling systems, and then we would get into the specific 25 issue of multi-loop scaling of ADS-4 to IRWST transition,
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17 1 which'we added to our PIRT scaling report.
[)
% _,1 2 Then Mr. Young and his associates will talk about 3 NOTRUMP, the LOCA code. Now unfortunately we are not 4 talking about LOFTRAN, the non-LOCA code, today, and we 5 don't have our LOFTR'NA or our fuel experts here to deal with 6 the issues that Dr. Catton has brought up.
7 In that regard, the fuel that we are using for 8 AP600 is commercial fuel, Vantage-5H, and the licensing 9 process that has been applied to that fuel for commercial 10 plants is applicable to the AP600, so we relied on our 11 Fuels Division --
12 DR. CATTON: Excuse me. You keep saying that, yet 13 your Fuels Division went and made measurements and CHF is 14 different and they corrected their code.
' (~h t
) 15 MR. PIPLICA: That's right. All I want to say is 16 that I am not familiar with the details of the process they 17 used to do that.
18 DR. CATTON: They even call it -- well, the title 19 of the document says "AP600 Fuel."
I 20 MR. PIPLICA: Yes.
21 DR. CATTON: So there is something different.
l 22 The CHF was different. I i
23 MR. PIPLICA: Yes. The difference is in the size 24 of the core. We have a larger core --
l 25 MR. McINTYRE: Dr. Catton, the fuel is the same.
x l 4
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18 j 1 The difference --
1 i' 'h l
) 2 DR. CATTON: Fine. I hear you. But your written i LJ 3 word says differently.
4 MR. McINTYRE: The difference is the plant because 5 of the canned motor pumps, they coast down differently.
6 That was the extension of the DNB correlation.
7 MR. PIPLICA: And I am going to touch on that, if 8 I may.
9 DR. CATTON: Okay.
10 MR. PIPLICA: Now our objective in going over the 11 core cooling systems was not to reiterate what we had done 12 before but just to base the meeting and to hit the 13 highlights of the systems that are related to the issues we 14 are going to talk about today, so I am just going to take 15
,/-
( ,) 15 or 20 minutes and quickly go through the core cooling 16 features for the AP600.
17 We were going to do this mainly for Dr. Wallis' 18 benefit. Unfortunately, he is not here. But I still think 19 it will be beneficial. We have passive decay heat removal, l 20 it is performed by the passive RHR. This is a natural
'21 circulation heat exchanger.
.22 DR. ZUBER: May I make a comment?
23 MR. PIPLICA: Yes.
24 DR. ZUBER: I saw a big difference between last l 25 meeting and this one. First, you read Moody's book, because
/m ;
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l 1 19 1 on-every second page you reference to Moody's book. And
[I ). 2 since Graham became a member of the ACRS, his name also 3 appears in your -- conspicuously in your write-up.
4 MR. PIPLICA: In our defense, his name was in 5 there in-the beginning, he was in there before.
6 DR. ZUBER: So if this is by accident or 7 intentional, I won't ask, but I am just putting it on the 8 record.
9 MR. PIPLICA: Yeah. Okay. .All right.
.10 MR. BOEHNERT: Novak, you know, I have seen your 11 name in there too.
12 DR. ZUBER: Without success.
! 13 MR. PIPLICA: And, of course, we have the passive 14 safety injection systems, the pressurized accumulators, the
- .O i _) 15 core makeup tanks,-the gravity drain, and they reside at RCS E16 pressure at the beginning of any transient. We have the 17 IRWST, which sits at containment pressure prior to the )
18 transient. And then we have the automatic reactor coolant I 19 ~ system depressurization system. And we will be discussing 20 all of these features in the meeting today. ;
.21 Now, this is a picture of the power generatic.
22 system, the NSSS, without the core cooling features 23 attached, but yet we still call this passive safety 24 features. And the reason is we have a fat, lazy core. We l 25 have a large than normal core in order to get the linear
[j
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20 1 . power density lower in the AP600. This is an advantageous
() 2 3~
feature ~for this plant, and it is important with respect to the DNB issue. i l
4 We have two cold legs connected to each -- on each 5 side of canned motor pumps. Now, these canned motor pumps 6 have a high inertia' rotor attached to it, and the reason 7 they need that is the system coasts down much more rapidly 8 'than it does with a shaft sealed pump that we have in our 9 commercial plants. For that reason, we in the AP600 10 program, in concert, or in conjunction with our Fuel 11 Division people in Columbia -- that's Columbia, South 12 Carolina, we performed DNB testing-of the vantage 5H fuel at 13 Columbia University in New York City. The whole purpose was 14 to verify that the DNBR correlation that vnt use for this O 15
\ ,) fuel was still applicable in the range of conditions that it 16 is going.to operate in, and extend the correlation so that 17 low flow conditions could'be compared. And that was why 18 there is AP600 written on those documents.
19 DR. CATTON: But there's still the 50 percent-20 increase in the channel-to-channel mixing that I can find no 21 trace of anywhere except in your SSAR.
22- MR. PIPLICA: Okay. That I can't address. But --
23 DR. CATTON: And that could make a big difference 24 on the~CHF.
25 CHAIRMAN KRESS: Would that have to do with the I
[)
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21 1 fact that the core is fatter and larger?
2 DR. CATTON:
( I don't think -- I think this just --
3 I don't know what it --
see, when they do the testing, they 4 do a small bundle. So things like cross-flow don't make 5 much of a difference.
6 When you get into a big core, the channel to 7 channel mixing can play an important role in cooling down 8 the fluid that is in the hot channel, and they have 9 increased the magnitude by 50 percent, and I can't find a 10 record or a trace or anything.
11 To me, that is a significant change, unless you 12 show otherwise by sensitivity, which is what you are 13 supposed to do.
14 MR. PIPLICA: On those issues we have to go back (O,)- id and. speak to our thermal hydraulic people in the Fuel 16 Division.
17 DR. CATTON: I don't have a current copy of 18 Chapter 15.
19 Maybe you took it out of there. I don't know --
20 and I asked for one at the last meeting.
21 MR. PIPLICA: Paul, 22 DR. CATTON: It's not Paul. It's Noel. Not
'23 Paul's' fault.
24 MR. HUFFMAN: This is Bill Huffman of the NRC 25 Staff.
l
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1 22 1 DR.'CATTON: Do you know about that 50 percent 2 ' increase?
3 MR. HUFFMAN: No, I don't, but I think that maybe 4 you might focus on Chapter 4 of the SAR where they talk l 5 about the DNBR correlation and also our Staff's-SER. I 6 think it is Section 4.4, where we discuss about the low flow l 7 DNBR correlation evaluation, not in Chapter 15.
8 DR. BOEHNERT: Are you talking about Chapter 15 of '
10 DR. CATTON: I am confused by all of the --
11 DR. BOEHNERT: The fuel analysis is Chapter 4.
12 DR. CATTON: I took Chapter 15 and the references 13 and I raised a question. I got documents from Westinghouse 14 that were to address them. They used WESTAR with whatever 15 they did with this correlation, and that is not appropriate.
16 Now I am told that I should look at Chapter 4.
17 Well, I need Chapter 4, I guess.
18 MR. HUFFMAN: Chapter 4 is the fuels analysis and 19 the associated --
20 DR. CATTON: Well, it sounds to me like the fuel 21 analysis people don't talk to the Chapter 15 people.
22 MR. McINTYRE: That is perhaps a little harsh.
)
23 DR. CATTON: Well, I don't know. I mean what I am i 24 saying is strictly based on what has been put on the table.
25 MR. McINTYRE: We will somehow endeavor to get you
'[
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_ _ _ _ _ . _ _ _ _ _ _ . _ _ _ _ _ _ _ _ _ _ _ _ . _ _ _ _ _ _ _ _ _ _ . . . _ _ _ _ . _ _ _ _ _ . _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ ___ J
23 1 a copy.
-(
q- 2 DR. CATTON:
4 Get me a new Chapter 4 and a new
.%)
3 Chapter 15.
4' CHAIRMAN KRESS: Paul is looking into that. He 5 will.
6 DR. CATTON:. Okay.
7 DR. BOEHNERT: But what'you are saying is the fuel 8 has the same dimension, the same pitch to diameter, the same 9 spacers --
10 MR. McINTYRE: The same grid, the same nozzles.
11 DR. BOEHNERT: -- you just have more assemblies to 12 get a lower power density. Is that all it is? How do you 13 get a big fat core?
14 MR. McINTYRE: It has more assemblies than a
.n
'I( ,/
)
15 traditional two-loop 600 megawatt plant would have.
- 16. We do have cores of that same fuel design which i
17 are larger. They are in four-loop.
18 DR. CATTON: But unless you have done this kind-of 19 analysis for those plants, it doesn't make any difference, 20 does it?
l l
21 MR. McINTYRE: No. )
22 ER. CATTON: And your fuels people probably did" 23 that, so maybe if you could get the documents that report on 24- what they did, it would be helpful.
1 25 MR. McINTYRE: I think Paul went out to look for a 1
[
~%#)
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24 1 Chapter 4. If not, Westinghouse can send you another --
() 2 3
DR. CATTON: Well, I suspect Chapter 4 isn't going to do it, although I don't know. There is probably some 4 report somewhere within your Fuels Group that addresses 5 these issues, and they should be by reference to AP600 to 6 close it.
7 MR. McINTYRE: Gene and I aren't the people to 8 answer this. We'll look into it over the next couple of 9 days.
10 DR. CATTON: Okay. I think the biggest issue 11 though, if it is the same fuel, is the 50 percent increase 12 in the mixing.
13 MR. McINTYRE: All right. The other feature I 14 ' wanted to point out was we have a large pressurizer, larger
' (_,1 15 than we would for this size of plant and power rating and we 16 also have a serpentine surge line, which we will be 17 discussing today.
18 DR. CATTON: You're not going to drop that on me, 19 Paul?
l 20 DR. BOEHNERT: I am.
21 DR. CATTON: What I would like to do is maybe to 22 get it mailed to me.
- 23 MR. PIPLICA: This three-dimensional overhead show j 24 the relationship of the passive safety features that we are 25 going to be talking about today.
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25 1 We.are going to be discussing ADS 1-2-3, these two 2- flow paths into the IRWST through the sparger.
'3 We are going to be talking about the, as I 4 mentioned before, from the pressurizer and its connection to 5 the hot leg.
6 We are going to be talking about Stage 4, which is 7 a connection'from the hot leg to two equal flow paths on 8 each side of the. reactor vessel, and we are going to be l 9 talking about return through the sump recirculation screen 10 and injection through the DVI line.
11 Those are all areas that are going to be discussed 12 to address the 10 issues.
'13 Of course, in the scaling an important feature 14_ that we are going to be talking about is the multi-loop 15 scaling analysis of what occurs when you have. venting
.16 through Stages 1, 2 and 3 through the pressurizer. You also 17 .have venting off the hot leg through fourth stage. At the 18 same time you have injection from the IRWST and spill out 19 .the break.
20 'The core is creating steam that has to be vented l
21 -through these flow paths and we need to show that -- the 22 purpose of all this is to show that going from the
.23 transition phase, from when fourth stage opens until you get !
24 IRWST injection that what we have done at OSU is properly 25 scaled and provides the data for the appropriate phenomena l
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26 1 that we need to validate NOTRUMP.
2 DR. ZUBER: Okay.
( This is a question. You should 3 go into detail.
l 4 MR. PIPLICA: And we are. Mr. Brown is going to 1
5 go into great detail about his scaling of this transitional 6 period.
7 DR. ZUBER: Something I found out, I read 8 something, that this is really very atypical in OSU.
9 MR. PIPLICA: Atypical?
10 DR. ZUBER: Atypical. It is not well-scaled.
11 MR. PIPLICA: It is not well-scaled.
12 DR. ZUBER: I am bringing it up so you can think 13 about it.
14 MR. PIPLICA: Okay, well., that is not the
() 15 conclusion that we came to. We felt that it was adequately 16 scaled, but we'll discuss it.
17 DR. ZUBER: Then you have to define what is 18 adequate.
19 MR. PIPLICA: Right. Now the recirculation phase, 20 which occurs when the compartments have flooded up and you 21 have recirculation back through the screen, had been scaled 22 on its own and we discussed it briefly in the previous 23 meeting that we had, so today we are going to be focusing on 24 this transitional period in the scaling analysis.
25 DR. SCHROCK: Before you take that away, the one O ANN RILEY & ASSOCIATES, LTD.
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l 27 1 labelled " Injection" there -- one of the similar schematics 2 in the report, corrections to, I think, WCAP 14727, has 3 pressurizer half-filled with liquid and it is trying to 4 justify presumptions about the status of the system for that 5 time domain.
6 It says that the ADS 1, 2 and 3 are open but not 7 participating or some such words. Now you show the 8 pressurizer as a liquid seal on a system which is otherwise 9 largely filled with vapor and then you argue that the system 10 is -- that it has these flows which don't need to be 11 considered.
12 On the face of it, ridiculous, and so --
13 MR. PIPLICA: Your comment is well-founded.
14 DR. SCHROCK: Okay.
15 MR. PIPLICA: And we are going to specifically 16 address that when we present this and show you that that 17 figure should not have been drawn that way and that the 18 actual scaling analysis that was done --
19 DR. SCHROCK: Okay. Well, this was my lead 20 comment. You see, we are here to resolve, after many hours 21 of dealing with these things, here it is boiled down to 10 22 issues for today, and what we have is a bunch of documents 23 that represent poor engineering documentation of technical 24 issues, and y:2 want us to accept it on that basis.
25 It's hard to do.
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28 1 MR. PIPLICA: Okay. I hope that when we have the
!o)
LJ 2 discussion on this that we can clarify this and get this 3 straightened out.
4 DR. ZUBER: Let me say something. Decisions 5 should not be made on discussions. Decisions should be made 6 on documents, because documents stay and we can say I refer 7 to this document -- I made this decision on that.
8 After this meeting, it is hearsay, what you said 9 at this meeting, what I said. It's really not a document.
10 That is the reason I support what Virgil said.
11 Whatever you do, document it so you can defend it 12 technically and we can defend you technically in turn, or 13 criticize you, but it has to be made on documents, not --
14 MR. PIPLICA: I hear you.
g i
) 15 DR. ZUBER: -- not on actual explaining today. We 16 have been explaining stories for the last five years.
17 MR. PIPLICA: Understood.
l 18 DR. CATTON: Seven.
19 MR. PIPLICA: There will be discussion on Stages j 20 1, 2, 3, especially on the ADS test, and the configuration
! 21 is that we have three parallel flow paths off of a header l l
22 from the pressurizer.
23 We have a control valve which controls the flow, 24 and we have an isolation valve. The isolation valve is a 25 motor-operated gate valve and the control valve is a globe (p\ 'j ANN RILEY & ASSOCIATES, LTD.
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29 1 valve, and these were tested and we are going to discuss 2 some of those tests today.
.M.
[ )' .
3 We are also going to have a lot of discussion on 4 the fourth stage,.and this is what it looks like, for your 5 information. It is coming -- we have two of these, one on 6 each hot leg and a vertical T comes off the hot leg and it 7 splits into two parallel paths, and these are located in the 8 subcompartments next to the reactor vessel and it quickly 9 reduces to a 10 inch line here.
10 It goes through a motor operated gate valve 11 through a short length of pipe, through a -- can't think of 12 the name now -- squib valve -- and it discharges directly 13 into the containment.
14 DR. CATTON: And if you remember our primary k 15- concern was where you had your hand a few minutes ago, where 16 the words hot leg are.
17 MR. PIPLICA: Right here.
18 DR. CATTON: Yes.
l- 19 MR. PIPLICA: 'Okay.
20 DR. CATTON: It's how the fluid gets into the l 21 pipe.
22 MR. PIPLICA: Right -- so there will be a lot of l
i 23 discussion when we get into the NOTRUMP arena of how the 24 system performs and some investigations that have been done 25_ with respect to it.
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30 1 I was going to touch on the five tests that we're
[~)
L/
2 interested in today, just to benchmark us, ground us, if you 3 will, on what those tests are.
4 The important thing here is that we have two 5 integral tests, SPES and OSU and we are going to talk about 6 three separate effects tests today.
7 The test program consists of the core makeup tank 8 test that was completed in '94. We did some PRHR heat 9 exchanger tests using three single tubes, not a tube bundle, 10 and we used that to develop a correlation that was used in 11 the LOFTRAN code.
12 We have the ADS tests that were performed in Italy 13 at the VAPORE facility and we did those in two phases.
14 Phase A was steam-only blowdown tests through simulated
() 15 piping and Phase B for ADS was done with the actual valves 16 in place.
17 Then of course we had our long-term cooling test 18 at Oregon State University, which is still undergoing 19 testing under a program that is directed by the Research 20- Branch of the NRC, and we did our DNB tests that I mentioned l
21 earlier at Columbia University in order to extend the DNBR 22 correlation to the lower flows that will be experienced by 23 the AP600.
24 We are not going to talk much about the Phase A of 25 the ADS tests, but the main thrust of that testing was to
'[
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31 1 look at IRWST effects on sparger performance, lowering and
() 3 2 raising the sparger.
sparger was an important part of the test, and that was used The water level with respect to the 4 as a prelude to the Phase B-1 tests.
5 CHAIRMAN KRESS: On that Phase A --
6 MR. PIPLICA: Yes.
7 CHAIRMAN KRESS: -- for the second part of the 8 bullet, the hydrodynamic loads, was the tank itself 9 prototypical size or --
10 MR. PIPLICA: No. The tank was cylindrical and it 11 was well anchored in the ground.
12 CHAIRMAN KRESS: The sparger in the middle of it?
13 MR. PIPLICA: And the sparger was in the middle.
14 In fact, I have a picture of it.
/
(%)
s 15 CHAIRMAN KRESS: You measured the loads on the 16 wall? I see.
17 MR. PIPLICA: So here you can nc+ the tank --
18 CHAIRMAN KRESS: So the sparger was prototypic.
19 MR. PIPLICA: The sparger was prototypic and we 20 raised and lowered the water level and the water temperature 21 to simulate different conditions and we had instrumentation 22 racks where we had pressure transducers -- high response, 23 high frequency transducers to pick up the loads that were 24 generated by the blowdown.
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32 1 back --
[
~
2- MR. PIPLICA: They did analysis to do that. We
- 3 took the data, analyzed-it, to back out what.the forcing 4 function would be as'a result of the blowdown loads.
5- That~ forcing function was assumed to'be a-point 6- load --
7' CHAIRMAN KRESS: That was.my'next question.
8 MR. PIPLICA: And then it was used to analyze the 9 performance of the'AP600 plant.
10 CHAIRMAN KRESS: Using the actual geometry of.the 11 AP600?-
12 MR. PIPLICA: Right. '
13- CHAIRMAN KRESS: Okay.
.14 MR. PIPLICA: -And we had a single sparger in the k test but in the plant you have two and both point loads were
~
15 L 16 . superimposed to get the worst condition and that was the p 17 frequency response and the loadings that were generated on
'18' the IRWST itself --
19 CHAIRMAN KRESS: Superimposed in this case means 20 .
you calculated-them separately and added the loads at the --
21 MR. PIPLICA: Right. You assume they are both 22 working-in the same frequency ~or response, which gives you a 23 higher load -- a conservative approach.
i
- 24 CHAIRMAN KRESS: -Thank you.
25' DR. SCHROCK: I remember reviewing that when we O
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33 1 made a visit to the plant about six or seven years ago and I
,m '
(x.y) 2 was not at all convinced about the analysis, especially the 3 assumption of the point source in the analysis.
4 I guess it was left in my mind as kind of an 5 unresolved issue and I haven't really dug out of later 6 documentation how the Staff has dealt with that. Maybe they 7 will comment more about that.
8 MR. PIPLICA: Yes. There has been significant 9 work done since then on the analysis, methodology and 10 results and the Staff review has been quite heavy in this 11 area, especially since it is such an important load that is 12 superimposed on the plant.
13 Okay. Just to show you what the valve piping 14 package looked like, this is the thermal loop seal that
.fh
( ,) 15 comes up and goes to what I call the donut and the four-inch 16 motor operated valve in Stage 1 is located here.
17 Now here you have coming this direction you have 18 Stage 2, and we have the motor operated gate valve and here 19 we had a flange and we simulated what a globe valve would i 20 look like in this position and then later in Phase B-2 we l
21 actually inserted a globe valve into that position and 22 performed additional testing, and confirmed that this l
23 reduced section of piping did give you response 24 characteristics that were similar to what a globe valve 25 would be.
i l
(s)
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34 l
1 This simulated the L over D and the resistance of
[ 2 the globe' valve in this position, and then of course here is G) 3
( Stage 3, and the Stage.3 motor operated gate valve is back 4 here.
l l 5 CHAIRMAN KRESS: I'm interested in how you --
l-f 6 .before ycu had the data on the globe valve -- to back out an 7 equivalent L over D, how you estimate an L over D per globe 8 valve?
9 .MR. PIPLICA: The systems engineers did that.
10 CHAIRMAN KRESS: They were able to do that?
11 MR. PIPLICA: By working with the vendors and --
12 DR. KRESS: Okay.
13 MR. PIPLICA: -- and getting a range, a possible 14 range of what the vendors based on their operating (O,/ 15 experience.
16 CHAIRMAN KRESS: So it was based on operating 17 experience?
18 MR. PIPLICA: Right, and then we tried to pick one 1
l 19 that would be conservative and'we-had two, because we wanted 20 to maximize loads in the IRWST, simulated IRWST, and then we 21 wanted to maximize flows, so those are contradictory goals, 22 so we actually ran tests where we changed this piece to give l 23 us that range of conditions.
24 Then when we did the actual testing with the globe 25 valves, we used globe valves from two separate manufacturers
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35 1 to give us that broad range, because globe valves can have IT 2 quite a range in L over D and resistance, so we have a range V-3 of conditions that we tested.
4 DR. CATTON: And then it's very different when it 5 is two phased than when it is single phased.
6 MR. PIPLICA: Yes, yes.
7 CHAIRMAN KRESS: Do any of the characteristics of
.8 the globe valve show up in the ITAAC or the --
9 MR. PIPLICA: Oh, yes. Yes.
.10 CHAIRMAN KRESS: So the specifications will be --
11' MR. PIPLICA: The Staff has made sure that we 12 ITAAC these extensively.
13 There are significant ITAAC tests that are done on 14 these valves.
( 15 Okay. I think I have talked enough about ADS that 16 the two slides that I have in your package. The only reason 17 I want to show the schematic _is in case we need'to go back 18 at it and look at it when-we are discussing the ADS tests, 19 but we used the supply source to generate both' single phase 20- and two-phase flow conditions at'the inlet to the valve 21 piping package, and then we actually used the valves in the 22 Phase B-2 tests to initiate the blowdowns.
,23 We actually staged them open as they would be
[ 24 operated in the plant. I l 25- We are going to talk today about, a little bit !
r I
\s Tl, ANN-RILEY & ASSOCIATES, LTD. j Court Reporters 4 1250 I Street, N.W., Suite 300 l Washington, D.C. 20005 L (202) 842-0034 L _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ . . _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ . _ _ _ _ _ _ _
36 1 about how the data was used to assess the computer code i 2 NOTRUMP results, to compare to NOTRUMP and how the 3 validation -- Mike, Mr. Young, is going to talk a little bit 4 about the data reduction and answer some of the questions 5 you had from the previous meeting with respect to the ADS 6 tests. l 7 CHAIRMAN'KRESS: My understanding is that that 8 question revolved around whether or not to use Fanno-like 9 relationships or something else.
10 MR. PIPLICA: Yes, and Mike is going to talk about 11 how we went back and looked at that again and what the l 12 conclusions were that he has for you.
13 CHAIRMAN KRESS: Okay.
14 MR. PIPLICA: Now today we did the PRHR tests and i'
/ 15 I am just going to quickly show that facility. This tank 16 simulated the heighth of the IRWST and the piping that was 17 used for the primary system was at simulated actual pressure 18 and temperature conditions and we had three tubes that we 19 tested and we had a baffle inside the tank so that we could 20 control what the conditions were at these three tubes to 21 simulate -- and we could move this baffle closer and farther 22 away to simulate multiple rows.
23 We used this and we extensively tested this, and 24 came up with a correlation that we use in LOFTRAN, and that 25 has been delineated in the Appendices of the LOFTRAN V&V
'b)
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37 1 report of how that correlation was used, but we are not
(
% ))
2 going to be discussing the PRHR test itself today. That was 3 not an issue for this meeting.
4 And the next test was the core makeup tank test 5 that we performed at our Waltz Mill facility. Just to give 6 you'an idea of the relative size, this tank was a 10 foot 7 high tank two feet in diameter. The actual plant shown on a 8 later overhead -- this shows you some of the scalability, if 9 we are going from a 20 foot by 12 foot diameter tank to a 10 10 foot by 2 foot diameter tank, but we simulated primary 11 conditions, full pressure, full temperature, and we have the 12 elevation head was set up in the test to match that in the 13 plant, so that our source of steam and water in the 14 steam / water reservoir was at the same elevation as the f
(3,) 15 source of steam and water would be in the plant relative to 16 the core makeup tank, so that the gravity draining was not 17 an issue.
18 We tested both the ability to recirculate and heat
- 19. the tank and the ability to drain the tank.
20 CHAIRMAN KRESS: Was the tank in the test -- was 21 it insulated?
22 MR. PIPLICA: Well, the lines were insulated as 23 they were in the plant. Whatever was in the plant was done 24 in the test. The tank, as you can see from the picture, was 25 not insulated. The tank is not insulated in the plant.
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38 1 CHAIRMAN KRESS: Well, did it have the same sort
[ i 2 of wall thickness as the --
V 3 MR. PIPLICA: No. The wall thickness for the tank 4 in the test was appropriate for the pressure that the test 5 was expected to operate at.
6 CHAIRMAN KRESS: So it was a lot thinner.
7 MR. PIPLICA: Yes, it was a lot thinner.
8 CHAIRMAN KRESS: A lot less --
I 9 MR. PIPLICA: It's six to eight inches in the 10 plant, and it's two-and-a-half inches in the test. These 11 differences were scaled. A scaling report was issued for 12 the CMT test, and a NOTRUMP comparison to the test results 13 was also issued.
14 CHAIRMAN KRESS: When you added the diffuser, did j'%
\, ) 15 you check it in this test?
16 MR. PIPLICA: Yes. The diffuser was installed in 17 the facility and the methodology it was used to design that 18 diffuser is the same methodology that's applied for the 19 physical plant diffuser. And the same approach was used for 20 a diffuser for both SPES and OSU. So that the CMT test, the 21 SPES test, and the OSU test all had the same design approach !
22 applied to the diffuser.
23 The SPES test. I want to jump to that now.
l 24 There's a schematic in the overhead package, and we had a 25 full-height test, full pressure, full temperature I i ANN RILEY & ASSOCIATES, LTD.
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r- -
l 39 h
1 conditions. Thermal power was five megawatts. It's 1/395 l[Ql
\
2 volumetric scale. We did not approach this test using the I 3 hierarchical two-tiered scaling approach. It was designed l s4 prior to our entrance into that technology. It was simply 5 part of volume scaling. And for the loop piping it was done 6 based on resistance.
7 DR. CATTON: How did characterize the heat loss?
8' MR. PIPLICA: The heat loss was characterized by 9 running preoperational tests where we brought the entire 10 plant to a constant' condition and recorded where the power 11 was being lost, what the temperatures were on the boundaries 12 of the plant as best we could. We looked at what the heat ;
i 13 load was being dumped to maintain a constant power level.
, . 14 So what we did is we ran the plant at different levels of l r~
( 15 temperature for a steady-state period'of time and recorded 16 the data. And then, using that, we estimated what the heat 17 losses were for the various components.
18 The first estimate that we came out with was the 19 80/20 number. Okay? But when they did the LOFTRAN
- 20 preliminary V&V report, they noticed that we were 21 underestimating the amount of heat loss that was being 22 generated or lost to the environment in the annular !
23 .downcomer region and in the upper head. So additional
) l L 24 analysis.was performed, and it was estimated that there was
-25 a significant heat loss-that we didn't account for, nor did
.[]
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40 1 we pick up inLour preoperational tests. And that was the
[ h' w) 2 rationale'for going to the 65/35 split.
3' DR; CYPTON: I guess when.you talk 80/20 or 65/35 4 it's of a fixed amount.
5 MR. PIPLICA: Right.
.6 DR. CATTON: So-if you missed a major heat loss, 7 it seems to me the total amount should go up, and the amount 8 that.is lost in the steam generator --
9 MR. PIPLICA: But you know your power input, and 10 you know the power that's being -- that's coming out of the
'11: plant.is being condensed. So the difference at a 12 given steady-state condition is the loss to the environment, i
13 and it's just a matter of distribution.
14 That's how it was done. And that was not -- the
( 15 details of that analysis was not in the V&V report, but it 16 was rather documented in a substantial calc note back at 17 Westinghouse. So you weren't -- you didn't have access to 18 that information. And I believe that's the -- that's what 19 happened. And that was done, let's see, about four years 20 ago. And that's my recollection of the approach that was
- 21. used.
22 DR. CATTON: The thing that makes that a little 23 shaky is the fact that 80/20 overpredicts and 65/35 grossly 24 underpredicts.
l 25 MR. PIPLICA: But --
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41 1 DR. CATTON: So you still don't have it right.
2 MR. PIPLICA: 'Well --
3 DR. CATTON: An overprediction is a problem.
4 MR. PIPLICA: We had a methodology'that we used to 5 estimate the heat losses. We stuck to it. And we concluded 6 that this was our best shot at the heat losses. And we put 7 that into the code, and that's what came out. So, yes, we 8 could have said well, what if we go halfway between? But we 9 didn't have a basis for doing that, so we couldn't.
10 DR. CATTON: Did you do an error analysis?
11 MR. PIPLICA: Instrumentation error; yes.
12 DR. CATTON: I mean, a heat-balance error-13 analysis.
14 MR. PIPLICA: On SPES? Yes, there is an error r'
i ~ 15 analysis that's in the SPES test analysis' report, but the 16 heat loads were estimated, and of the total errors, the heat 17 load errors were -- it's a matter of distribution as' opposed 18' to total heat loss. And it wasn't done on the distribution.
19 DR. CATTON: You know it turns out it's really
- 20. important.
21~ MR. PIPLICA: Oh, yes.
22 DR. CATTON: And your analysis shows that.
23 MR. PIPLICA: Yes.
24 DR.-CATTON: And I don't know how important it is 25 to NOTRUMP, because I haven't really gone through the L
i
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42 1 NOTRUMP V&V document, the LOPTRAN part of it -- or I mean
] l 2 the SPES -- in much detail.
3 MR. PIPLICA: Well.--
4 DR. CATTON: But for the small break, it could 5 . well have an impact-on.what:you're doing with NOTRUMP as 6 well.
l 7 MR. PIPLICA: The --
8 DR. CATTON: Maybe you should --
9- MR. PIPLICA: The heat loss is much more important 10 in the non-LOCA transients because you're always longer in a 11' high pressure'. The LOCA transient actually -- the problem 12 .actually goes the other way as you proceed. For example, 13 :when you start'out at the initiation of the break, your heat 14 losses to the environment are excessive because of your --
(k 15' the metal surface area through your water volume is so 16 askew. .But as you' lose water, then-that heat load'from that 17- . metal begins to go back into what water is left, and we 18 .actually have to -- the correction becomes a different
- 19 direction. Instead of it being an uncertainty with the heat 20 loss to the environment, it becomes an uncertainty with the 21 heat loss --'the heat input back into the~ coolant.
-22 DR.'CATTON: So that could warp your whole l
1 23 . analysis.
24 MR. PIPLICA: Well, or it could balance.
[ 25- [ Laughter.)
' .1
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43
- 11. DR. ZUBER: Compensating errors.
( :2 MR. PIPLICA: I can't address that. But, yes, we 3- recognize that that is an issue that you must deal with in 4 .all tests of'this nature. You have tall, skinny components, 5 which is why we;ran a shoot for the LOCA transients.
6 CHAIRMAN-KRESS: Now back to my point'on this.
7 That only impacts trying to interpret the results directly 8 in terms of how AP600 would behave; right?
9 MR. PIPLICA: That's correct.
10- CHAIRMAN-KRESS: But you could still use it to 11 validate a code if you don't artificially put in the splits 12 into the1 code calculation.
13 MR. PIPLICA: That's right.
14 DR. CATTON: I think that would be right.
O Q 15 CHAIRMAN KRESS: .Yes.
16 DR. CATTON: If the split wasn't so important.
17 CHAIRMAN KRESS: You think it's overrides the 18 other phenomena you' re- tryir.g to look at.
19 DR. CATTON: Yes.
'20 CHAIRMAN KRESS: And I think that that's well 21 worth'--'that's well. worth looking isn't.
22 DR. CATTON: .They show.that in their V&V, that you l L23 .can: override most anything just by shifting.
L24 CHAIRMAN KRESS: .I think what that says to me is 25 'that the heat transfer characteristics of a SPES with high ANN RILEY & ASSOCIATES, LTD.
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L__________________ _ _ ___
44 1 surface to volume are such that the controlling phenomena
() 2 are not things you want them to be, is what I'm saying.
3 DR. CATTON: That's right.
4 CHAIRMAN KRESS: And I don't know if that's right 5 or not, but it sounds like an important issue.
I 6 DR. CATTON: You know, it's not that you can't 7 deal with some of these things, but codes like LOFTRAN 8 don't. As a result, you have some artificial way of trying 9 to handle it. You clearly don't need to worry about it on a 10 full-size plant.
11 CHAIRMAN KRESS: Well, does LOFTRAN do like RELAP 12 5 and calculate heat transfer to walls and use -- k 13 DR. CATTON: I don't think RELAP 5 is any better.
14 It's just a problem. These codes are built for a different
() 15 purpose. They're not built for high area to volume ratios 16 where you've got to worry about the heat losses to the 17 environment around you so much.
18 CHAIRMAN KRESS: Okay.
19 DR. CATTON: I mean, for large-break LOCA they're 20 fine.
21 CHAIRMAN KRESS: Yes, because --
22 DR. CATTON: And that was the intent of the SPES-2 23 facility -- the SPES facility. And at the outset that was 24 the intent --
it was the high-pressure end of AP600 behavior 25 that was to be tested with SPES.
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l 45 1 CHAIRMAN KRESS: In the large-break LOCA you're
'[T v'
2 saying that the blowdown constant is so big compared to the 3 heat transfer time constant that it's not as important.
4 DR. CATTON: And there are a lot of other things 5 that swamp the heat transfer to the outside effects.
6 CHAIRMAN KRESS: Okay.
7 DR. ZUBER: Let me ask you now, where do we stand 8 on this problem?
9 MR. PIPLICA: Well --
10 DR. ZUBER: The point is, since you want to close 11 this -- I mean, to close the whole problem today, I mean, in 12 that two days, where do we stand on this one?
13 MR. PIPLICA: We came here today to talk about 14 ECCS injection and NOTRUMP. We're not prepared to discuss
( ,) 15 the LOFTRAN or the non-LOCA transients today.
16 MR. McINTYRE: Gene, if Paul will share the 17 report -- again, this is a case, we don't travel with the 18 full contingent of Westinghouse people to answer all 19 questions. There have been like 150 people that have worked 20 on it, and we can't obviously travel with all those people 21 all the time.
22 If Paul can give us the reports, we can look for
.23 getting you an answer sometime in the next 36-48 hours.
24 And, you know, we can get them faxed back. I think I 25 understand what your concern is, and I think, you know, Tom lf)
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l 46 l l
c 1~
is right, Ivan, that when you start looking at small l
- '[' \v} - 2; facilities that'you-can get overwhelmed by the boundary
[ 3 conditions.. 'But if Paul gives us those we'll send them back
>4 and try to get you an answer. l 1
5- DR. CATTON: But for NOTRUMP SPES-2 doesn't play 6 such a significant role, does it?
7 MR. McINTYRE: For NOTRUMP -- LOFTRAN?
8 DR. CATTON: Let's not talk about LOFTRAN for the
'9 moment, just NOTRUMP. How big a role in the V&V does SPES
{
10 play? l 11 MR. YOUNG: We used SPES to. assess NOTRUMP's
- L:2 performance in the early portion of the transient prior to 13 ADS 103 and then in the early blowdown period. Following 14 that, then OSU became more important. But that's how we 15 used SPES in the NOTRUMP assessment.
16 DR. CATTON: And then there is some overlap with 17 OSU when you take the scaling into consideration.
18 MR. YOUNG: In periods?
19 DR. CATTON: Yes.
- 20. MR. YOUNG: Yes, there is. l 1
21 DR. CATTON: So it's not as important for NOTRUMP l 22 as it is for LOFTRAN is what I gather, because you didn't do !
23 any V&V against OSU for LOFTRAN, or if you did, it's not a 24- part of the document I've got.
25 MR. YOUNG: I can't answer that.
(
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i 1 MR. PIPLICA: OSU was not used for any validation 2 purposes for LOFTRAN.
3 DR. CATTON: So you only used SPES?
4 MR. PIPLICA: Only SPES, and CMT.
5 DR. CATTON: I don't have as much problem rith 6 CMT.
7 MR. YOUNG: The CMT test facility was used.
8 MR. SCHROCK: If Westinghouse is going to get 9 answers in 48 hours5.555556e-4 days <br />0.0133 hours <br />7.936508e-5 weeks <br />1.8264e-5 months <br />, I'll miss the answer. I would like to 10 request that they be sent to me, too.
11 MR. BOEHNERT: Certainly.
12 MR. PIELICA: Okay. And to give you a relative 13 idea of the SPES facility, this is the CMT tank. It is 14 20-foot high. It's diameter, unfortunately, though, is only
() 15 approximately eight inches. I'm showing you the relative 16 size of the facilities here. And then the last, of course, 17 are the OSU facility, which most of the discussion today, if 18 not all, from an integral test approach will be addressed, 19 and it included all of the injection systems, including 20 tanks to represent the lower containment compartments so 21 that we could do the long-term cooling tests at OSU. The 22 entire system was modelled and it was configured relatively 23 speaking in the same configuration as the plant, and the 24 lines were routed in the same configuration that existed in 25 the. plant design at the time the OSU test was performed.
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48 L
1 Of course, as you know, we used the -- Dr. Reyes (L 2. performed a scaling for the facility, and Mr. Brown is going 3 to discuss additional scaling that was performed for the OSU 4 , test today. i 1 5 Just to give you an idea of the relative size, l
- 6. it's hard to see, but the reactor vessel is here behind Dr.
7 Reyes standing in front of the facility. This is the IRWST 8 'here at your left. The core makeup tanks, which would be 9 over here, are not shown, but they are a five-foot-high 10 tank,.approximately two foot in diameter. So we have a 11 range of scales for the integral test facilities.
12 This was a low-pressure test, as you know, 13 initiated at 400 psi and 400 degrees fahrenheit, but it was 14 able to do a simulated blowdown of the AP600 and do all of
/O V 15 the transitions into ADS 1, 2, 3 actuation, ADS 4, IRWST 16 injection and sump injection.
17 Over 30 tests were run by Westinghouse, and I 18 think about the same number of tests have been run by the 19 NRC staff to assess the performance of passive systems. l l 20 MR. SCHROCK: So you say it's able to do a
'21 simulated blowdown. It can't of course, do that from l
22 ' initial condition. '
23 MR. PIPLICA: No, it can't. It does it from 400 24 ~ psi. Right. j 25 MR. SCHROCK: Stretching to say you can simulate I
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49 1 it. You have constructed arguments to justify the fact that 2
( ). you're able somehow to reproduce conditions that would exist
-3 in blowdown of a full pressure system at some midpoint in 4 time.
5 MR. PIPLICA: Yes. That's --
6 MR. SCHROCK: That's a rather'different statement.
7 MR. PIPLICA: I understand. I understand the 8 distinction and you are correct. Okay.
9 I think that's all I need to present. I know I 10 took a little longer than I expected. But'at this time, I i
11 .would like to have Mr. Brown come up and give you first an 12 overview of the scaling approach and then get into the 13 details of the 80S4 to IRWST transition scaling that we 14 performed at.your request.
15 Thank you.
16 MR. BROWN: Again, this was going to be a little 17 bit probably more for Dr. Wallis' benefit to briefly review 18- the approach that we took to PIRT scaling. There's only a 19 few pages here.
20 As you know, we were trying to provide a 21 comprehensive identification, assessment.and ranking of the-22 important phenomenon in AP600 as one major objective, and 23 then to demonstrate that the AP600 integral and separate 24 effects test captured the highly ranked phenomenon and that 25 the test data was adequate for code validation. Those are i-1, 0 ANN RILEY & ASSOCIATES, LTD.
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50 E 1 the two primary purposes.
?
l(n)
! ./
2 To accomplish this, we divided the LOCA transient l
3 into phases, developed the PIRTs for each phase, performed 4 top-down system-level kind of scaling to point towards which l
'5 things from the bottom-up that we had to look up, and the l 6 bottom-up, of course, was really the area where we scaled 7 the facilities when we got down to geometry and 8 thermodynamic parameters and all that kind of thing.
9 The next Elide shows the division that we chose of 10 blowdown natural circulation, ADS, IRWST, and long-term 11 cooling or recirculation. This is where we're going to 12 focus next in a few minutes today, which you guys asked for 13 on the transition between ADS 4 occurring very late here in 14 this ADS operation into IRWST injection.
/O 15 j DR. ZUBER: Let me ask you, this is the only case 16 where you really looked different loops scaling and 17 interacting?
18 MR. BROWN: We --
19 DR. ZUBER: Because this is only one, one instance 20 in this whole transient where you have interacting loops.
21 MR. BROWN: We did look at that also in che early 22 ADS period. I did a multi-path, if you will, here, and 23 that's what was presented I think back in December of --
24 late last year. We had done this back here and a lot of 25 paths interacting right here, and we have now done that l/C ANN RILEY & ASSOCIATES, LTD.
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l 51
, 1 here.
(J');
~
2 3-By the' time we get out to here, there really L aren't very_many paths interacting. Everything is. kind of 4 gone. So there's really.not much in the way of looking at 5 as far as multiple paths so. So really, we have now done
'6 this area and this area.
7 A brief summary of some of the past meetings that 8 we've had -- in '96, in December, Dr. Hochreiter presented 9 the PIRTs. We presented a lot of separate-effects scaling, 10 and Gene just presented on ADS and CMT. We did a first 11- approach at the integral effects test for natural l
, 12 circulation, the early ADS, IRWST and sump, but that was a 13 'more simplified version. We went back in December of '97, I
' 14 then we added the bottom up as well as the multi-paths for 15 ADS operation.
- 16 .Today, we're looking at the multi-path or 17 multi-loop, if you will,- for the late ADS and IRWST 18 transition period, which is the area or period in which we 19 pose the greatest threat to core uncovery.
- 20. So with~that, I'will get into the main event today 21- .for me, and that's the multi-loop for the ADS.to IRWST
.22' transition.
- :23 As Dr. Zuber always likes to get quickly to the 24' end, I'll try to present'a slide which is what I concluded 25 from all this, which is on page 2 from the top down ANN RILEY &. ASSOCIATES, LTD.
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L 52 t 1 multi-path scaling.
/-
~\
f 2 I. scaled conservation of mass, pressure equation
. %) .
3- and conservation of momentum. I found that OSU was fairly well scaled foi ADS 4:and IRWST,-which were the most 4
5 significant-mass flows that'I found in the system'at that 6 time.
7 The SPES ratio.was notably larger versus OSU,'and
~
8 there is an area of dispersion-in SPES. SPES 2.has an. ADS 4 9 Jarea of dispersinn. It's too big. And this really_shows up
'10 in the pressure' equation. SPES distorts the ADS power to 11 volume scaling'due'to the very large area.
12 OSU was pretty well scaled with respect to the 13 area and the volume, certainly with respect to SPES. The 14 only other thing that I found as well is that the PRHR-
) 15l energy removal appeared to be more important in OSU than it 16 did in SPES and what we-might predict for AP600.
17 DR. ZUBER: Let me say what is my concern, is I 18 really the effect'of the inventory in the vessel. I mean, 19 this is'the bottom line. I mean, you may scale everything.
20 How does it affect the inventory in the vessel? -Because 'j 1
21 this is what you are really after.
- 22 ; MR. BROWN
- Uh-huh.
23- DR. ZUBER: So.in your presentation, please focus 24- on that question.
25 MR. BROWN: The conservation of momentum equations i
~
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l 53 1 -- we'didn't -- I did not scale those because of the
'( ) 2 distortion in the area, the volume with SPES. I did not 3 bother doing that, .and I proceed with OSU and I found that 1
4 the gravity. head of IRWST and the ADS 4 non-recoverable 1
5 l acceleration loss were important and significant as well as 6 'the DVI ADS 4 resistances. I found the pressurizer gravity 7 had to be more of a medium importance.
8 Now, to backtrack and go through this top-down
- 9 level scaling, I!11 start with the conservation of mass,.
10 which, as we said, was important because of the mass 11 inventory situation, and this was the model that I used. I i
12 have an IRWST injection through the DBI lines and then the 13 mass out the ADS -- out of the vessel, and it's focusing on 14 the vessel.
) 15 DR. CATTON: When you talk about conservation of
.16 mass, you've got several places that that mass could be. It 17 could be in the IRWST, it could.be in the pressurizer, it 18 could be in the vessel, it could be in the CMT, whatever.
19 MR. BROWN: Yes.
20 DR. CATTON: This is kind of a limited picture.
21 MR. BROWN: Well, this is the picture of interest.
22- -This is where, as Dr. Zuber said, this is where we're 23' focusing on, is the vessel. ;
l 24 DR ' CATTON:
. What I'm trying to lead you to is a r
- 25. ' clear description or clear reason why you ignored the line
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54 1 from the pressurizer to the IRWST.
2 MR. BROWN: This -- this --
3 DR. CATTON: The other line through the 4 pressurizer.
5 MR. BROWN: The other line through the 6 pressurizer. You're talking about the surge line?
7 DR. CATTON: Surge line, pressurizer, because 8 there's liquid hold-up in the pressurizer, and I can't 7
9 figure out how you treat that from what you have put down.
10 I don't want you to do it and I don't want to disrupt your 11 chain.
12 MR. BROWN: Okay.
13 DR. CATTON: I just want to let you know what I'm 14 interested in so that when you get to it, we can emphasize 15 it a little bit.
16 MR. BROWN: Okay. Well, I focused on the vessel 17 because that's what I thought was --
18 DR. Z"BER: Yes. What's in the vessel -- the 19 vessel is the end result of all these interactions.
20 MR. BROWN: Right.
21 DR. ZUBER: See, the liquid can go -- going into 22 the vessel, it can go into the pressurizer, depending on the 23 pressure, and this is, for example, the case for SPES.
24 Instead of going into the vessel, it goes into the 25 pressurizer.
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55 1 MR. BROWN: Right. Well, certainly the IRWST has
() 3 2 no choice but to either inject that or not.
DR. CATTON: But it's a quasi-steady process and 4 what determines what is where is pressure balances, pressure l 5 balances and levels, and you know how much mass you've got 6 in the system. It's simple.
7 MR. BROWN: Next is the status of the flow paths.
8 The ADS 4 flow paths are open and they're discharging intc 9 containment. The ADS 1, 2, 3 paths are open and they're 10 ineffective due to pressure refill and drain.
11 DR. CATTON: But they're not ineffective. You 12 have some steam flow through them.
13 MR. BROWN: Some small amount of steam flow, yes.
14 DR. CATTON: That small amount of steam flow is
( 15 what determines how much water you've got in the 16 pressurizer. Without that steam flow, you probably would 17 have none.
18 MR. BROWN: That is considered in the conservation 19 of momentum equations.
20 DR. CATTON: I didn't see it and I don't think it 21 is, because you need to know the pressure above the liquid 22 in the pressurizer.
23 MR. BROWN: That's right. That's right.
24 DR. CATTON: How do you know it unless you guess 25 it?
l l
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56 1 MR.. BROWN: I have treated that as a boundary
) 2 condition from the IRWST --
3 DR. CATTON: Where did you get it?
4 MR. BROWN: I know what the initial IRWST head is; 5 therefore, I know what the pressure is in the sparger and 6 the discharge line, and I can treat that as a boundary 7 condition to the_ top of the pressurizer.
8- DR. CATTON: But you have steam flow, so you have 9 pressure --
W 10 MR. BROWN: That steam flow is -- and we have seen 11 this in both SPES and OSU -- is very, very, very small, 12 barely measurable.
13 DR. CATTON: Well, your assumption --
14 MR. BROWN: That's not an assumption. That's from
("%)
g 15 the test. I mean, I even know that --
16 DR. CATTON: That you can ignore it is an
- 17. assumption, and this is not what was found by others who 18 have looked at this same problem.
i 19 MR. BROWN: It's only --
{
20 MR. SCHROCK: That's exactly what prompted my 21 earlier question. You can argue until you're blue in the 22 face that it's unimportant. Show some numbers to prove 23 that, okay? Don't tell us it's unimportant. We think it's 24 important, you think it's not important. Give us a l 25 quantitative assessment of it; prove your case.
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57 1 MR. BROWN: Okay. I have not done that here. I 2 have done.that in my own calculations and I have
(
3 approximately 2 or 3 percent of the total mass of the system 4 that's being injected in that vessel is going out through '
5 ADS stages 1, 2, 3 total, and when you look at the momentum 6 from the resistance through ADS 1, 2, 3, it is roughly two 7 to three order of magnitude -- two to three orders of 8 magnitude less than the IRWST head. It's nothing. Very, 9 very small amount. I 10 MR. SCHROCK: Well, you're comparing the wrong 11 things there.
12 MR. BROWN: How so?
1 13 MR. SCHROCK: The level of the liquid in the 14 pressurizer is controlled by the pressure -- pressurizer I
() 15 together with other factors. Pressure in the pressurizer is 1
l 16 controlled by what's happening and its connections through 17- the. vapor with the IRWST, and you're arguing that that 18 connection is of zero consequence.
19 MR. BROWN: It's very small, that's correct.
- 20- DR. ZUBER
- Let me --
l l 21- MR. SCHROCK: Show the numbers in that context.
l 22 We're not talking-about-what's happening at the other end of 23 its opening; we're-talking about what's happening between 24 the pressurizer and the IRWST.
25 MR. BROWN: That's where I'm talking about, right. l l
l
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58 1 MR. SCHROCK: And that influences the level --
\
[Y
\-
2 MR. BROWN: I'm saying I contend it's small.
3 CHAIRMAN KRESS: You're saying you can neglect 4 momentum terms and --
5 MR. BROWN: Yes.
6 CHAIRMAN KRESS: -- friction terms and just use 7 the pressure at the end point.
8 MR. BROWN: Yes.
9 CHAIRMAN KRESS: Which seems to me like a 10 reasonable argument --
11 MR. BROWN: Right.
12 CHAIRMAN KRESS: -- if his numbers are right.
13 MR. BROWN: Right.
14 -DR. ZUBER: Well, we can always go and believe his
/
( )\ : 15 numbers, which we don't see it here. I have seen other 16 analysis which really says that SPES really, instead of 17- having the water going down into the core, goes into the 18 pressurizer. That's one thing. So you said it's 19 negligible; yet, the water goes there. It's a distortion in 20 SPES.
21 You have another distortion in OSU which you 22 neglected. This is the CMT.
23 See, the things which other people find 24 distortion, and really very important distortions, you l 25 neglected. You.said you have some calculation back home.
I l
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59 1 CHAIRMAN KRESS: The CMTs are probably emptied by
] 2' this time.
3 MR. BROWN: They are nearly empty. They're at 4 their 20 percent level whenever the signal is given to ADS 5 4. That head is nowhere near what the IRWST is --
6 DR. ZUBER: The thing is --
7 MR. BROWN: -- and the pressurizer.
8 DR. ZUBER: From the very beginning, as you start 9 under, you go back into the stage 5. CMT's --
b
'10 MR. BROWN: We don't have stage 5. I don't know 11 what you mean by stage 5.
12 DR. ZUBER: I mean the ISW. I mean the last 13 stage.
14 MR. BROWN: You mean stage 4?
(O) 15 CHAIRMAN KRESS: He means the natural circulation ,
16 in IRWST.
17 MR. BROWN: You're talking sump injection now?
18 DR. ZUBER: The last, last stage. You have the 19 thing going from CMT. It's distorted in OSU, the behavior, 20 and you neglect it. !
l 21 MR. BROWN: Because the head associated with that l 22 at the time that it goes off is very small. It's nearly 23 empty, and that's by -- that's the design of the plant.
24 DR. ZUBER: Okay. Put yourself in my position. I 25- have seen analysis which says this is important, and l
I l
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l 60 1 distortion in SPES and distortion in OSU, and you said it's 2' not important, you have calculations. Now, what am I going
(
3 to conclude? I have to conclude what I see.
4 MR.~ BROWN: I hear your message. I hear your 5 message. I understand. I'm saying I have done a 6 calculation and the momentum effect due to the ADS 1, 2, 3 7 is very small once ADS 4 opens up.
8 DR. CATTON: When you talk about momentum effect, 9 what do you mean?
10 MR. BROWN: Pressure drop. The pressure drop from 11 the. top of.the pressurizer to above the. water that's held up 11 2 in the pressure to the ADS discharge, which is in the IRWST-13 is extremely small, and therefore, I have applied the back 14 pressure with the water level above the IRWST, which in the
() 15. real plant is approximately ten feet of water, and applied 16 that pressure back upstream to the top of the pressurizer.
17 DR. CATTON: I'll just sort of emphasize what l
l 18 Novak said in that we have another analysis that disagrees 19 with your conclusion about the lack of importance.
i l 20 DR. ZUBER: Okay. Now, let me say something more, i
21 and this is for you, Tom.
22- The thing I have seen, an analysis of three 23 facilities, ROSA, SPES and OSU. All are distorted, but at
- 24 least one is in the correct range. So I think this is a 25 . good result. So out of three facilities, we have one which j ANN RILEY & ASSOCIATES, LTD.
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61 1 has good results. So if you have a code which addresses b'
f 2 3
these facilities which are scaled, we have confidence in the code'. There is only one event which is not scaled in many 4 facility, which is just this event here, and it's distorted 5 in SPES and it is distorted in OSU, and this is the reason 6 I'm asking you, really --
7 MR. BROWN: Is the CMT -- in the analysis you've 8 seen, is the CMT important? I mean, forgetting about 9 whether it's distorted, is it important? Because if it not, 10 then the distortion is immaterial.
11 DR. CATTON: Well, if it's distorted, that may be 12 the reason it's not important. So it's hard to say.
13 MR. BROWN: I don't think it's so.hard to say when 14 I start at 20 percent level. I know what the maximum head 15 can be.
16 DR. CATTON: Why don't you put numbers on the 17 table and compare them with the other analysis and decide
~18 who's right.
19 DR. ZUBER: See, we have something in writing in 20 black and white -- provide us the numbers and we will see 21 how it comes out.
l 22 MR. BROWN: I got the message.
23 HDR . ZUBER: But-this is the only event which has l
24- distortion in all three facilities, and now, of course, '
l 25 they're not that good that we can put our faith in the code )
i
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62 l' calculations.
- 2 -MR. BROWN
- The conservation of mass equation N._./
3 looked at the mass of the vessel, the IRWST in and the ADS 4 4 out, and I -- scaling that based on the initial conditions
- 5 we can get from this ADS 4 period, I ended up with'a. mass 6 flow ratio and a time constant associated with filling or 7 draining the vessel -- draining the vessel.
8 The change in the nass of the vessel with respect 9 to the ADS 4, discharging the mass out of the system was the 10 time constant.I got, and I got ADS 4 to IRWST mass flow 11 scaling ratio.
12 The reference parameters that I used to try to 13 evaluate these were based on the -- I used the volume 14 associated'with that to drain the vessel from the bottom of
/~
(T) 15 the upper head to the-top of the fuel region, and the 16- scaling factors for SPES and OSU were applied to that.
17 I used a homogeneous equilibrium model for the ADS 18 4. These are the critical mass fluxes based on those. The 19 areas for each of the ADS 4's you see here with a single 20 stage ADS 4 failure. This represents --
l L 21 MR. SCHROCK: As I read your new documentation,
'22 your G/C values are still erroneous. That's owing to the fact that you evaluated derivative at constant enthalpy 23 241 rather than at constant entropy. I guess you still haven't 25 . understood that part of the problem.
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63 1 MR. BROWN: Thank you for your comment.
() 2 3
The scaling ratios that I got for the mass flows indicate again that SPES and OSU were within a factor of 4 two, and the time constant was not -- was not one, but, as 5 you see, roughly .7 and .8 for OSU versus AP600.
.6 Again, the ADS 4 area in SPES and OSU were a bit 7 different relative to the plant, and therefore, you do see 8 some difference between the two.
9 I'll move on to the pressure equation. I drew the
- 10. control volume as the primary system with PRHR energy, core 11 energy, IRWST energy in, ADS 4 energy out, and again, the 12 ADS 1, 2, 3 --
L 13 DR' CATTON:
Now, you set somewhat arbitrarily --
14 MR. BROWN: I wouldn't say arbitrarily.
) 15 DR. CATTON: Somewhat. You didn't -- you claim 16 they are calculations. That's okay. But you set the 17 pressure at the top of the pressurizer.
L 18 MR. BROWN: Right. Based on --
19 DR. CATTON: And you set it to whatever the head l 20 is?
( 21 MR. BROWN: Right, in the --
22 DR. CATTON: That elevation in the IRWST?
23 MR. BROWN: Uh-huh.
24 DR. CATTON: How do you determine what the level
-25 is in the pressurizer? Do you have a level?
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l l
l 64 l 1 MR. BROWN: Yes.
(9
'w) 2 DR. CATTON: How do you determine it?
3 MR. E!CWN: Well, that's a hard one.
4 DR. ZUBER: See, this is exactly what -- I 1
l 5 MR. BROWN: That's a hard one.
6 DR. CATTON: You bet it is, and that's kind of the 7 crux of the whole matter. You've got --
8 MR. BROWN: That's a hard one.
9 DR. CATTON: You've got two paths and what goes on 10 with which path determines what the level is going to be in 11 --
t 12 MR. BROWN: Yes. And that's one of the problems 13 of trying to -- which I don't like -- of'trying to scale 14 these in between phases, is it makes it difficult to
(
n.) 15 determine sometimes what are the initial conditions of some 16 of these components, especially like the pressurizer.
17 DR. ZUBER: How do you address it? Your codes 18 cannot do the calculation too well.
19 MR. BROWN: That's right.
20 DR. ZUBER: See, the point is --
21 MR. BROWN: That's right, and that's why the only 22 thing I could do was to look at sensitivities of, well, gee, 23 the most it could probably be is full, and what does that 24 give me? And then look at things in between. That's all I 25 really could do, and that's what I have done in a case of i
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65 1 things like the pressurizer where I don't know its level
- %['")\ 2 very well initially, is to try to look at the boundary l
3 conditions'that it could take.
4 Things like the IRWST or the CMT where I know what l
l 5 the level is based on the design of the plant or the test l 6 facility at the time, the things that are tied into a level 7 specifically at an instant in time, I know what they are.
8 The pressurizer is a tough one. I don't know what that is.
9 All I can do is range it between the minimum and the 10 ' maximum.
l 11 DR. ZUBER: In SPES, the liquid goes -- raises in 12 the pressurizer, is rising. In OSU, it did just the 13 opposite.
14 MR. BROWN: The liquid rises in the -- there's
-( ) 15- liquid in the pressurizer in both test facilities.
16 DR. ZUBER: But it. rises during the transient in 17 SPES and it drains in OSU It's a different behavior, !
18 because they have distortions, and this is something which 19 does not appear in your analysis.
20 MR. BROWN: Well, when we open up these, they all 21 have a change in level and then there's a draining period.
22 MR. McINTYRE: Excuse me, Novak. Could you 23 clarify that.a little bit? Because what we see is an l '24 initial swell of mixture into the pressurizer and then a 25 drain.
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66 1
-1 - MR '. BROWN: In both cases.
- v[~} 2 DR. ZUBER: The point is the other analysis I have
[3 seen,;the pressurizer drains-in OSU. In SPES, .the-liquid l 4 rises.
=5 MR. BROWN: .They both eventually drain.
6 DR. ZUBER: . Eventually. Eventually,'yes, but the 7' question is when and what's happening during that time in
-8 the' core? We want to know where is the liquid, what happens t
9 to11t, and what happensLin the core.
p 10 DR. CATTON: If you plot the level in the l 11 pressurizer as a function of time, and on the same graph, 12 you plot the level in the core and IRWST, three curves on 13 the same diagram, what do you see? I mean, the core goes 14 through a minimum, the pressurizer does funny things, IRWST
- fg
() 15 behaves pretty uniformly, but somehow, it's the interaction 16 of those heads that leads to that minimum. The question 17- then is, okay, how do you get to AP600 because your codes 18 can' t really do the -job? Hopefully, a scaling analysis 19 would help come to grips with that, but if you leave a part
.20 of it out, you have to -- you're going to be in doubt.
.21 There's phenomena we cannot calculate, at least we 22" don't, and yet you're going to make an extrapolation.
23 CHAIRMAN KRESS: ~What's the total volume of the l
24 -pressurizer compared to the core?
- 25. DR. CATTON: It's one of head, not volume.. That's
{
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67 1 .the question, 2 CHAIRMAN KRESS:
i
{f I know, but the importance of 3 this would be the volume ratio.
4 MR. BROWN: I don't know. Any of you guys?
'5 MR. McINTYRE: For the plant or for OSU? Either 6 one?- Both?
7 CHAIRMAN KRESS: Well, either one or both.
8 MR. McINTYRE: We can get the numbers here in a 9 minute.
10 . CHAIRMAN KRESS: I think whether or not the
-11 pressurizer has a big.importance on the level will depend on 12 the volume -- core volume, 13 DR. CATTON: The amount of water in the 14 pressurizer, all these things are tied together in O) q, 15 -determining how much steam goes where.
16' CHAIRMAN KRESS: Right.
17 DR. CATTON: And then you've got the complicated 18 process where the water wants to lay down in the bottom of 19 the hot leg, but the steam flow keeps dragging it up. If 20 more steam flow goes to the pressurizer, the behavior is 21 going to be different than if more goes to ADS 4.
- 22 CHAIRMAN KRESS: .Yes, but if it doesn't take much 23 liquid to fill it up compared to changing level of the core, 24' then it's not really important. If it takes a lot of 25 liquid, then it's important.
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68 l' DR. CATTON: Well, where it impacts is the flow 2 rate from'IRWST to the core.
3 CHAIRMAN KRESS: Ye, but, you know, once again, if
- 4. the volume is small, it can't have much --
5 DR. CATTON: But it's pressure. It's pressure.
6 If I increase the pressure in the core, I get less flow from 7 IRWST. So it's the heads, not so much the volumes.
8 MR. BROWN: I guess a conclusion I have -- I have 9 found that even by filling the pressurizer up, that is'still 10 a very -- that is still a modest player --
11 CHAIRMAN KRESS: In the head.
12 MR. BROWN: -- in the total head relative to IRWST 13 head and to ADS 4 resistance. Even filling the thing up to 14 the top, picking a boundary condition.
l
(~ .
( )) 15 CHAIRMAN KRESS: There is no heat source in that 16 pressurizer other than the stored energy?
17 MR. BROWN: Right. Right. As a matter of fact,-
16 for a while, you actually have some cold water-from the 19 accumulators and things, you know, discharging in there for 20 a while, too.
21 CHAIRMAN KRESS: I would like to know what those 22 volumes are. It would help me.
23 MR. BROWN: Okay.
-24 Again, I have the same component status with the l
25 exception of I have the PRHR active, which can remove some O ANN RILEY & ASSOCIATES, LTD.
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L-
69 1 systems'or energy from the primary system.
[
\J {
2 The rate of pressure' change equation I went to 3 Moody's text.
4 DR. ZUBER: I think his number of weeks has-5 increased since last February.
6 MR. BROWN: Yes.
7 DR. CATTON: A good sign.
8 MR. BROWN: He has a couple of versions in there.
9 One is for an ideal gas and one is for a two phase 10 mixture, and I have used the one for the two phase mixture.
11' I thought it was more appropriate here for the primary 12 system.
13 Anyway, I have the IRWST, the ADS-4, the core.
14 energy, PRHR energy, and when I non-dimensionalize all
/
- j
( 15 that --
- 16. CHAIRMAN KRESS: By the way, was this also used in 17 the containment or do you know?
18 MR. BROWN: I_used the ideal gas version in the 19 containment.
20 oCHAIRMAN KRESS: You didn't use this one in it?
21 MR. BROWN: No. You can get the ideal gas one
- 22. very easily from this one, yes.
23 It's a simplification of that one and I have used 24' -that one for containment.
25- CHAIRMAN KRESS: Thank you.
1 1
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l
_ _ _ _ _ _ _ - _ - _ _ _ _ _ _ _ _ - _ _ . _ _ _ _ . = _ _ _ _ _ _ _ - - _
70 1 MR. BROWN: Okay. I normalized on the ADS-4
(} 2 energy. I divided all the terms by the ADS-4 energy since I 3 think this is what is driving everything or causing 4 everything to being depressurized during this time and to 5 change the pressure, and when I do that I end up with a 6 depressurization system time cost and IRWST energy scaling, 7 core scaling group, a ratio and a PRHR pcwer, as well as a 8 primary system response scaling group, which is --
9 CHAIRMAN KRESS: Now this denominator is what?
10 Virgil says it should have been constant entropy developed 11 and --
12 DR. SCHROCK: Yes. I am a little curious about 13 your terse thank you for my comment. I guess does that mean 14 that you disbelieve the contention or you have evaluated it (q
s
,j 15 and found a different conclusion? I don't know 11 6 I guess we should discuss it a little bit some 17 time. If now is the right time, I don't know.
18 MR. BROWN: I had a follow-on discussion that was
'19 to discuss the issues on the homogeneous equilibrium model 20 which follows the scaling presentation, so I would rather 21 wait and talk about it that. l 22 DR. SCHROCK: Fine, 23 MR. BROWN: Anyway, I ended up with these scaling l 24 groups.
25 Here is a table of parameters that I used and
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l )
L l
l 71 l 1 fortunately I noticed in my package I don't have Slide 17,
( 2 which shows the results, but if you look at page 17, we have l 3 the time constant ratio and again the SPES is distorted 4 because of the ADS-4 area. It doesn't scale well on time.
5 Obviously, it should have been something 6 one-to-one and it wasn't. You have a very large area.
7 OSU early on in ADS, earlier in time, you'll see 8 the different time constant,'which you all.know that 9 facility runs at a different speed during the early part of f 10 its depressurization and midstream of this event. However, L 11 as we approach the -- as we start-to getting very low in 12 pressure, if you remember the curve I had earlier on the 13 whole LOCA transient, that by the time we get to ADS-4 l
14 things are approaching a quasi-steady situation and the
() 15 pressure is reduced to the point where it is approaching 16 that of the plant as well, so it is interesting to me that i 17 the time scaling seems to come into.line, which is what I 18 would expect when we get down to ADS-4, whereas earlier, 19 when ADS-3 was open at very high pressure, that was not 20 quite the case.
21 We were running at a different speed, which is l
l 22 well documented in Dr. Reyes's scaling report for OSU.
l 23 When I look down at the scaling groups, for core 24 power we have ratios again between a half and two, which we l: 25 believe is good scaling. The PRHR -- we don't believe it is (A ANN RILEY & ASSOCIATES, LTD.
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.72
.1' significant in AP600 or in SPES. However, it does appear to 2 be more.significant in OSU and it is not well scaled in that 3 regard but we don't think it is an important phenomenon 4 'because it is not important in AP600 and in SPES, but it is 5 there, and it does have a little impact -- so it is --
I 6 DR. SCHROCK: -How do you judge that a' scaling
? ratio of 2 is good scaling?
8 MR. BROWN: That is just -- that is the criteria 9- that we'used-and have been working with for the past few 10 years.
11 DR.-SCHROCK: I don't know. If you take the core 12 _ power, for example, it says that core power, one is correct, 13 the other one is off by a factor of two.
14 It is surely going to influence the course of the-I ,
15 transient. What does it mean that it is'" good" scaling?
16 If-you have a scaling ratio of 2, it seems somehow 17 .to need to interpret the scaled facility performance in the i 18 context of that scaling ratio -- so have you done that?
19 MR. BROWN: All I'll say is it'is our position 20 that a' scaling ratio of a factor of two is acceptably or 21 well-scaled.
22 DR. ZUBER: Okay. Tell me physically how do -- I
-23 am looking at this graph, 17, and you'have the first pi 24 group. You have q in the core divided by the entropy
- 25 flowing out.
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J 73 )
.1 MR. BROWN: Yes.
! 2 EDR . ZUBER: And.AP has .44 so it means you are
- . (V) l 3. removing more energy out than --
g4 MR. BROWN: Yes, out of AP-4.
l 5 DR. ZUBER: Okay. Now in SPES it is just the 6 opposite. You are removing less but you in OSU are removing 7 twice as much energy, I think.
8 Do you think this has an effect on the rest of the j
9 transient?
l 10 You know, the point is -- let me -- let me just 4
11 say where I come from. When I look at scaling it gives me a 12 feeling for physics, what is happening and if I can't see a 13 ' difference in numbers I can understand maybe this is what 14 .the physics is. 1
- ) 15- To say this is two or three doesn't mean anything 16 .unless you really provide some more words to explain the l ~17 physics. This is thing that Richard was asking.
18 DR. SCHROCK: Ideally the scaling.comes first. We 19 have said that before, and then the thing is built with some
- 20 anticipation that it is going to provide results that you )
21 will be able to interpret in relationship to the system that 22 you are trying to represent for that scaled facility, but
- j. 23 you wouldn't take the output from your scaled facility 24 without an understanding of what your scaling analysis has !
25 taught you.
l l
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i L 74 I
- 1. What I'have. asked you is did you have a means of
() 2 interpreting results from your scaled facility in each case 3 that takes into account the scaling ratios, and you gave an l'
4 answer which is unrelated to the question, so let me pose it
.5 again.
6 Can you answer the question?
7 MR. BROWN: Well, I guess maybe I should -- what 8 are you looking for, I guess, specifically?
9 DR. SCHROCK: I am looking for how do you i
10 ' interpret results from the tests in the context of your l 11 anticipated performance of the AP600 given the fact that you 12 have a factor of two between these scaling parameters in the 13 two facilities, and similarly for OSU it happens in this'
, , -~ 14 particular instance the difference is in'the opposite
( ,) 15 direction.
16 You cover a wide range of grounds with the two 17 scaled facilities, but you haven't addressed the issue that 18 I have raised, which is how do you interpret the performance 19 of each of the scaled facilities in terms of your need to 20 tell us what the AP600 is going to do.
21 MR. BROWN: .I think we, from our criteria of --
22 which I stated before, on the scaling results we make sure 23 that any phencmenon in the PIRT of course that is highly
~24 ranked, that we make sure that if we have each of those and
.25 what we call a well-scaled range then it simply means that
()
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75 1- the guyc with the code can use the data for code validation.
I\ 2 That's it. We don't do anything more with the k/
- 3. scaling. At that point it has served its purpoi.-
4 DR. SCHROCK: It has not --
5 MR. BROWN: Saying that the code is sufficient for 6 purposes of --
7 DR. CATTON: Let me get a little' closer to the J s
8 real issue.
9 The issue is what happens to the level in the 10 core.
11 The code can't predict it well, so Westinghouse 12 has said gee, what we would like to do would be to set some
- 13. level somewhere for AP600 and we will figure out what that 14 level is by doing something to the OSU calculation.
j ) 15 Okay, now, the question then is when I look at
'16 SPES the behavior in the core is not the same. There is a 17 dip. It drops down and comes back up, goes through a.
18 minimum. None of the facilities is scaled right -- exact 19 scaling is impossible.
l 20 We all know that. That is why when you say "2" I ,
l 21 don't get as unsettled as Virgil providing you --
22 DR. SCHROCK: I don't mind the "2" -- two is good. ,
23' It's the question of what you do with it after you have got !
i 24 :it. I 25 DR. CATTON: Well, providing you can tell me why L
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76
! 1 it goes through that minimum, and what in the configuration
[ [ 2 causes it to go through that minimum -- a couple more l 3 sentences'and then you can --
l 4 Now another person addressed this problem and was 5 actually able with some very simple equations to make 6 predictions that show that minimum. He can explain why it l 7 goes through a minimum. I have never asked him, but he L
8 probably could come to some conclusion about AP600 and that 1
( 9 minimum, and I think where I have been trying.to see you go 10 is to get into that kind of a position, so you can say I 11 know what causes this minimum -- it is -- it is either worse 12 or better on AP600, therefore we will do thus-and-so.
- 13. We are not getting there.
L 14 MR. BROWN: You are trying to do with the scaling (O) 15 what we do with codes.
i 16 DR. CATTON: You can't do it with the code because 17 the code does not have the right kind of physics in it.
18 MR. BROWN: That's what you think, right.
19 DR. CATTON: No. You admitted it right here -- i I
not you but your colleagues have admitted the modeling is
~
L 20 ;
21 l wrong. !
I-22 Okay, now given that the modeling for that is l 23 wrong, what do you do?
l 24 Do you understand what I am saying? i i
25' MR. BROWN: Yeah.
- ~
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77 1 DR. CATTON: And we could talk about what is wrong
() 2 3
with the code if you want.
MR. BROWN: No. That is not my -- I don't want to 4 do that.
L 5 DR. CATTON: Well, you are being a little bit too i 6 focused on just the scaling. This is an interdisciplinary 7 question, particularly now.
-8 MR. BROWN: Well, all I am saying is that we
! 9 ask -- you asked what we do with the scaling and we
! 10 determine that we feel that the highly ranked phenomenon is
! 11 well scaled. That tells these guys with the code go ahead 12 and use the data from that facility -- during a particular 13 phase you can use the code -- you can use the data to l 14 validate the code.
l(-(j- 15 That is what we do. That is what I said in the i 16 beginning with the perspective. We didn't have a number l 17 three that says now try to, you know, try to get more out of 18 it than that. We don't.
l
'19 DR. CATTON: So I guess it is all in your --
20 DR. ZUBER: Wait, wait. I have questions for him.
21 MR. BROWN: But you asked why is it? Because 22 simply if you -- if I look at the ratios I see simply that l
l 23 we get all this mass and energy out ADS-4. I mean that is i
j 24 why we go for the minimum, until the IRWST comes on to i
25 counteract it. I mean that-is simply the reason. That is 1
l
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L 78 i
1- 'the way the plant operates.
e A.
i- 2" We don't have anything else. When we have lost
-(_ /
- 3 the accumulator has-already gone, the CMT is essentially 4 drained, the pressurizer volume is almost nothing. We are 5 down to two things -- this IRWST and this ADS-4 going out, 6 and initially the ADS-4 is huge and that is why you see, 7 that'is why when I went back and I ratioed the core power, 8 -IRWST energy and so on, these are 10, 20, 30, 40 percent 9- relative to the energy and the mass of the ADS-4 initia'ly.
10 That's why. That is what is happening.. Right.
11 We have got a big hole going out there, and that is where it 12 is all going. That is what causes the minimum.
13 When I go through'the calculations, I see that 14 'from that. Ultimately what happens is of course we do get
-t O) 15 the IRWST on and that is what saves the day. That is the
- 06 only thing that refills that core.
17 As a matter of fact, later when I look in the 18 momentum scaling to justify use of ignoring the inertial 19 terms,-I went through looking at the time constants on page
- 20 21 and the time constant associated with, fortunately, with' l l
21 draining and filling the vessel, and when I drain it, and I 22 fill it, you know, I am talking 100 or some seconds, and 23 fortunately'I am able to get enough IRWST injection to turn 24 that thing around -- otherwise you're right -- pffht. ;
25 DR. CATTON: I think we would all agree you can f.
l
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79 1 neglect inertial terms in this phase.
2 MR. BROWN: Yes.
[%)')
3 DR. CATTON: That comes right out of the scaling.
4 MR. BROWN: Right.
5 DR. CATTON: But the' minimum -- what is the 6 minimum a function of?
7 MR. BROWN: Well, as I said, in my mind in looking 8- at this, I concluded it is simply the ADS-4 area versus the 9 IRWST injection. That is what it all comes down to, and 10 that is why I showed that on the vessel as well.
11 DR. SCHROCK: When you talk about justification of 12 steady state treatment of phenomena that are in general 13 transient I think you kind of missed the point here in the 14 discussion about inertial effects.
(a) 15 In a steady state flow there can be inertial 16 effects. The issue is the process that you are looking at 17 one of steady state or is it one in which transient is of 18 any significance.
l 19 Don't answer that. Just --
20 MR. BROWN: Yes, you're right. I mean --
21 DR. SCHROCK: -- there are places in steady flows 1
22 .that inertia is extremely important. I 23 So I don't know about your time constant for l
24 inertia. I don't know exactly how you view it physically. !
i 25 What does it mean?
I i
/%
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j
80 1- MR. BROWN: Do I move on to momentum?
[)
A_/
2 DR. SCHROCK: Page 21. Just tell me physically 3 what does that equation on page 21 mean.
4 MR. BROWN: It means that I looked at the smallest 5 diameter, longest --
6 DR. SCHROCK: It doesn't contain any terms which 7 is an inertia term. It is a change in --
8 MR. BROWN: Well, the L over A of the DVI line to 9 me is the most limiting inertial portion of the entire 10 system.
11 It is a very small di. ar, very long line, and 12 it is very, of course, as you just said a few minutes ago, 13 .it's very key and important in order to be able to recover 14 the Kohler mass inventory.
lO) _,
15 So when I look at the time it takes to establish 16 that flow by the gravity head of the IRWST going from zero 17 to something in the case of OSU -- say one pound a second -- )
18 this is the time it takes me, a few tenths of a second to I 19 establish that flow.
20 When I compare that against anything associated
)
21 with drsining or filling, that volume in the upper head here H 22 down to the top of the core, we are talking a hundred some 1
23 seconds, so in my opinion the impact of including a term I J
)
.24 would' call the inertial term on the left hand side of a 25 conservation to mass equation associated with time is very L
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.f' 81 1 small relative to us filling cnr draining this core, and
/~N l2 therefore I'think I can neglect it.
\- >
13 HDR. SCHROCK: So Delta W, which never appears in l 4 your nomenclature, represents the difference betweencthe
^
5 stagnanticondition --
6' MR. BROWN: Ye s . :
7 DR. SCHROCK: -- and a steady flow condition.
8- MR. BROWN: Yes. The flow established in the DVI 9 line; Right, that is correct.-Yes.
10 DR. ZUBER: Let me ask a question on this graph.
11 MR. BROWN: Yes.
12' DR. ZUBER: What causes the
~
.The first pi~ group.
13 difference between'the two facilities and do you have less 14 power or more,or less -- more power and less enthalpy flow
!( ) 15L out -- what is the difference from SPES and from'OSU?
16 MR. BROWN: Yes, there are some' differences,in
.17 ' power. -Remember that the, power curves because some of the 18, heat losses and things in the facilities were a bit 19: different than they were for the plant.
20' I would have to go back and look at the scaling 21 parts to see that, but there are some. things done to try to 22 -compensate from some of the heat losses and things in there.
~
, 23 DR.'ZUBER: So the difference is'mostly due to the 24: q and not due-to the enthalpy out?
25' MR. BROWN: I believe.so, yes.
!I .
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82 1 CHAIRMAN KRESS: The question on the importance of
( 2 how big the ratios are of the scaling parameters. It seems 3 to me like it revolves around whether on not your phenomena 4 changes or you are dealing with the same phenomena. Do you 5 have a step change in flow regime or do you have some in 6 terms -- with respect to the code itself do you have the 7 e change, are you outside the bounds of your correlations?
8 Did you look at it from that viewpoint at all?
9 MR. BROWN: You are getting into more of a --
10 CHAIRMAN KRESS: Why your criteria of 2 is good, 11 for example.
12 MR. BROWN: When you ask that question, I think 13 you are getting more into what we do, what we presented 14 before about things like bottom up where we were looking at (I 15 correlations or where we look a Rayleigy number or Reynolds 16 number --
17 CHAIRMAN KRESS: Yes, that would be where it would 18 show, at the bottoms up.
19 MR. BROWN: Yes, we have done that and presented 20 that previously in the bottoms up and that is where we look 21 at that kind of thing, to make sure are we in the same range 22 of Reynolds number, Rayleigy number o:: whatever, and look at 23 that.
24 That is where that -- but again from the top down, 25 we are looking more at mass and energies in and out from n/
\-
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l l
83 1 different components in the system and trying to determine
() 2 3
is there anything that specifically is askew there, and as I said, I said one thing I happened to notice, that in the 4
case of OSU that the PRHR seems to be more effective than in 5 SPES during that timeframe.
6 DR. SCHROCK: You do have in this new 7 documentation some pretty big numbers for your ratios on 8 some items --
9 MR. BROWN: Yes.
10 DR. SCHROCK: I guess that will come out later.
11 MR. BROWN: Yes, you're right. There are some 12 that are --
13 DR. SCHROCK: Have some of those been judged 14 unacceptable?
() 15 MR. BROWN: Again, only if it was an important 16 phenomenon and in a PIRT and if it is a high ratio beyond a 17 factor of two, then that is a problematic distortion, yes.
18 If it is within that and it is a high phenomenon, 19 then no, and if it is high but it is not -- it is only 20 ranked a medium or a low in the PIRT then even though it may 21 be distorted, we don't worry about it in the codes.
22 DR. SCHROCK: So it is flexible depending upon the 23 combination of the scaling and the PIRT?
24 MR. BROWN: Yes, the two have to go together.
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84 1 with this.
) 2 We do, as I said, there are examples in which this L 3. obviously -- you could look at this as a distortion, but 4 when we go back to AP600 and especially the plant we are 5 looking at, it is not an important phenomenon, so I don't 6 get too excited about that.
7 I mean if it was, you know, if it was something
'8 that was core power or IRWST that would be a different 9 story.
10 DR. ZUBER: Okay. Let me ask you, you understand 11 ' correctly, you are' putting twice as much energy in OSU than 12 in the plant.
13 MR. BROWN: Yes. Relative to this, yes, the 14 blowdown. Yes.
m
() 15 DR. SCHROCK: In SPES you are putting in half as 16 much and you are losing a lot more.
17 MR. BROWN: Yes. Yes.
18 DR. CATTON: Mike, what level is it that you want 19 to set? It's in the pressurizer, isn't it, that you want to 20 change?
21- MR. BROWN: No. It's the IRWST.
22 DR. CATTON: It's the IRWST.
23 MR. YOUNG: Our penalties apply to the IRWST, yes.
24 DR. ZUBER: Let me just ask a question. Why did 25 you increase the power in OSU -ith respect to AP600?
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l 85 1 MR. BROWN: I don't know offhand, to be honest
[ 2 with you. I don't know. I would have to go back and read 3 the scaling report. I don't know why. I don't remember why 4 it was taken out for awhile, but there is a slightly 5 different curve in it. I don't remember what the reason 6 was.
7 MR. PIPLICA: Dr. Zuber --
8 DR. ZUBER: Yes?
9 MR. PIPLICA: -- that power increase at OSU was 10 early in the transient.to compensate for the fact that we 11 didn't have the full decay heat at the initiation of the 12 transient, but during this phase of the transient, the power 13 was scaled to decay heat.
14 MR. BROWN: It was scaled a lot closer.
15 DR. ZUBER: Well, still if it was scaled, still it 16 is a factor of two there -- 9.92 versus .44 is two.
17 MR. PIPLICA: Yes, and I am wondering if it is not 18 because of the ADS-4 rather than the core power.
L 19 DR. ZUBER: This was my question, and I see the 20 discrepancy. Is it due to the enthalpy outflow or is it due
~
21 to the power? This was my first question.
22 Then you said it is due to the power -- then my l
23 question was then why did you increase the power in OSU 24 vis-a-vis the plant.
25 MR. PIPLICA: But the power. levels that were set
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86 1 in OSU were a result of the scaling that Dr. Reyes did and
[' 2 'what we are looking at is trying to match up the same phase 3 of the transient so the decay -- you know, I am not sure if 4 we are --
5 MR. BROWN: Yes, it's a little toughe~r because we
- 6 are in between --
7 MR. PIPLICA: Because of the time scaling.
8 MR. BROWN: But remember, we start off in OSU and 9 we are kind of in the middle there and so it is -- anyway, 10 there was some compensation for that early-on.
11 DR. ZUBER: Okay, okay.
12 MR. BROWN: Now to the momentum model.
13 IRWST path through the DVI into the core, we have 14 two ADS paths as well as a PRHR path, and then a p)'
q, 15 pressurizer, with a surge line path into the hot leg and a 16 path through ADS-1, 2, 3. However, as I mentioned earlier, 17 I did apply a pressure boundary condition here, and my 18 calculation, which I have not presented, showed me that I 19 'could neglect the pressure drop from here to the IRWST and 20 apply the boundary condition and pressure from the sparger.
21 DR. CATTON: That pipe that's at the top of the 22 pressurizer --
23 MR. BROWN: Um-hum.
24 DR. CATTON: It goes into the bottom of the IRWST 25 somewhere, doesn't it?
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87 1 MR. BROWN: It comes into the top, and then
() 2 3
there's a sparger. It comes in here --
DR. CATTON: And the sparger sits on a platform.
4 MR. BROWN: There's a pipe -- right off bottom 5 there's a structural that supports the sparger. And the 6 plant is roughly ten feet of water above that.
7 DR. CATTON: So you calculate the -- you 8 back-calculate to the P-14.
9 MR. BROWN: Right. This static head right here >
10 applies that --
11 DR. CATTON: So as that head changes, the 12 pressure --
13 MR. BROWN: Right. Right.
14 DR. CATTON: What you're ignoring is the pressure
() 15 drop due to the steam flow through the line.
16 MR. BROWN: Yes, my calculation that you don't see 17 shows that this pressure drop is two to three orders of l 18 magnitudes less than this thing and most everything else in 19 there, and so I put a boundary condition on here.
20 DR. SCHROCK: When I commented during Mr.
21 Piplica's presentation, he indicated that this level in the 22 pressurizer was not something that should have been included 23 in this figure. At least this is the figure I was referring 24 to. Maybe he was referring to a different one. Is this 25' what you mean to show us, or is it not?
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88 1 MR. PIPLICA: What I said was not the level but l () 2 the outlet piping that represents ADS-1, 2, 3, the figure in 4
3 WCAP-14727, that is not shown. It was inadvertently closed
- l. 4 off.
5 MR. BROWN: It was shown like this in the WCAP.
6 DR. SCHROCK: I see. I thought it showed the same 7 thing you have here. I'd have to look again.
l 8 MR. BROWN: Yes.
9 DR. SCHROCK: Well, showing it this way then, 10 what's in the pipe below here. You've got -- in the surge 11 lir o you've got -- is that supposed to be filled with liquid 12 here?
H13 MR. BROWN: Yes, we have a liquid in the l
14 pressurizer --
( 15 DR. SCHROCK: And where is the liquid level in the 16 vessel? Above the hot leg?
17 MR. BROWN: Yes, I mean it is moving around there.
18 DR. SCHROCK: Well, I think you have vapor going 19 to ADS-4 most of the time.
20 MR. BROWN: Right. I think it's somewhere around 21- the hot leg here.
l 22 DR. SCHROCK: You see, when I looked at what you 23 were arguing in the report, it seemed ridiculous to me, as I l 24 said before, because I couldn't imagine that you've got 25 vapor down here in -- as a minimum a stratified flow in that l-i-
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89 1 hot leg, and then you're holding. liquid up in the-2
.( ) pressurizer, but what happens in that line is of no 3 significance. I'm referring to ADS-1 through 3 lines. It 4 has to be of significance.
5- MR. BROWN: Of significance in that yes, there is 6 a small amount of --
7 DR. SCHROCK: With respect to the quantity of 8 liquid held up in the pressurizer.
9 MR. BROWN: Yes. All I'm saying is that given 10 that that allows the steam through, we have ---just for 11 calculation purposes here I have treated this as a pressure 12 boundary condition rather than actually trying to include 13 that from there on in the model. We do allow steam into the 14_ surge line here. In this line we do have steam. This is O
. ( ,) 15 not solid water down to here.
16 DR. ZUBER: So if you leave steam, the pressure 17 then is increasing or the level is coming down.
18 MR. BROWN: Yes, this -- this I've got a liquid 19 level in the pressurizer and I have a two-phase mixture back i
20 to the core in here. Unfortunately this doesn't show that i 21 as well. It's not -- this isn't all water-solid here.
22 DR. SCHROCK: I guess what I'm trying to tell you 23 is that you give this to another competent engineer who's 24 not privy to all the background. He'd look at this and he'd 25 say, I think, this isn't a very good starting point for the 1
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90 1 arguments that are going to follow. Maybe it's reasonable 2
.( and maybe it isn't, but don't present a case which I think 3 is convincing to a competent observer.
i
! 4 DR. ZUBER: Let me get -- essentially you are l
5 closing the top when you're doing these calculations.
I 6 MR. BROWN: Yes.
l 7 DR. ZUBER: Okay. So you have some steam 8 coming --
- 9. MR. BROWN: Yes.
10 DR. ZUBER: And coming up --
i 11 MR. BROWN: Yes.
12 DR. ZUBER: The pressure therefore in that. region 13 14 is increasing. If it's closed and the steam is coming, 14 it has to increase.
': p .
() 15 MR. BROWN: Remember, I'm not doing'a transient I
16 calculation, Dr. Zuber, here.
t 17 DR. ZUBER: I don't know -- let me say, look, I 18 try_to understand what is the physics in your reasoning.
19 MR. BROWN: Yes.
20 DR. ZUBER: You have steam coming up, going 21 through the interface, accumulating in region 4. The top is 22 closed. The pressure then in that region has'to increase, ,
i L 23 This would then push the liquid out of the pressurizer l 1
l 24 faster than'you would predict -- then in a real case -- in a 25 real case the liquid can hang up there because the steam j
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l 91 l'
s 1 .will separate and go through that line and going to the 2 -IRWST. Is that correct? I don't know whether you call it d/ \
3 steady-state transient. This is what's happening.
4 MR. BROWN: This is not a NOTRUMP or a code 5- calculation.
6 DR. ZUBER: Look, I don't . teed the code. I mean, 7 I'm just trying.to see the physics. I mean, if you close 8 the top and you have steam coming, therefore the pressure 9 there increases. It will push the liquid down. Should it?
10 MR. BROWN: Yes.
11 DR. ZUBER: Okay. So the liquid then will go back 12 into the core.
13 MR. BROWN: Yes.
14 DR. ZUBER: So.that's not conservative.
r fi )\ 15 MR. BROWN: Well, I say the liquid will go back 16- down the surge line. I don't want to say where necessarily, 17 and I didn't say hard-wire that into anything I did.
18 DR. ZUBER: Yes.
19 MR. BROWN: Yes, it'll go. It could go here, it 20- could go here, it could go here ;
21 DR. ZUBER: This is the problem I have. I don't 22 know where the liquid is going. And once you close it l
23 there, you bring another question which I don't know the i 24 answer. I can just give you a problem. Because then the 25 liquid is - .if the steam separates and goes through that l
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92 1 line to the-IRWST, the liquid.in the pressurizer -- well, (f 2 the liquid-in the pressurizer can hang up and not drain'into 3 the hot leg.
4 MR.~ BROWN: But it has multiple paths here which 5 to go,.and we allow'that in our model.
6 DR. ZUBER: . Well, you -- but you -- you are.right' 7 in.a nonphysical situation, because you're really forcing 8 the. liquid to go faster'down, because you are increasing the
'9 pressure in region 14.
10 MR. BROWN: I have done the calculation at the 11 initial instant for the boundary condition that I know with 12 a' full IRWST, whatever'its backpressure is, that's the
- l'3 - initial pressure, I've done a calculation at that instant.
14 I did.not carry on a transient calculation.
'15 DR. ZUBER: 'Well, look, we are dealing with a 16 transient. We'are looking where the flow is. It can be in 17 this volume, in that volume, it can --
1 L 18' MR. BROWN: Right.
19 D10 ZUBER: And.it's not a transient. It 20 -circulates. Depending where the pressure is, it will then
- 21 flow according to the -- it's a multipath problem.
J 22.: MR. BROWN: Yes.
23' DR. ZUBER: And this is the reason were are
- 24. interested in the introduction of loops.
1 25- MR. BROWN: Yes. !
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93 1 DR. ZUBER: Now you artificially close one loop s
2 and do some argument. Well, the argument may be sound or (C) 3 not, but it is an artificial situation, and I don't have the 4 calculations in front of me. I can see that the physics is 5 wrong. And I have another report which says -- draws 6 different conclusions. And I am perplexed.
7 DR. CATTON: See, there's two things that set the 8 level in the pressurizer. One is the pressure drop through 9 the line from 14 into the IRWST.
-11 DR. CATTON: And the other is the two-phase 12 countercurrent flow that's taking place in that long surge 1
13 line then up through the pressurizer. They all play a role.
14 And the balance between the pressure drop at 10.
/N
.( ) 15 MR. BROWN: Um-hum.
16 DR. CATTON: And that's -- and that's really the 17 crux of the whole thing. If the level in the pressurizer --
18 well, it's not just the level in the pressurizer, it's the 19 combination. {
20 DR. ZUBER: It's the pressure, yes, very much.
21 DR. CATTON: Because the --
22 MR. BROWN: Yes. And we allow -- this pressure is 23 sitting here and is related to the level in the IRWST and 24 this backpressure.
25 CRAIRMAN KRESS: In essence you're saying --
i n
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94 l 1 MR. BROWN: So it's not really closed off like I
~
l,-)h 2 think you think it is.
i \_
3 CHAIRMAN KRESS: Yes, it's not really closed.
1
- 4 MR. BROWN
- That's why I'm very frustrated. It's 5 not really closed off.
6 CHAIRMAN KRESS: Because you have a boundary 7 condition.
8 MR. BROWN: It's really not. I l 9 CHAIRMAN KRESS: But what I heard is you have some 10 sort of in mind an equation for DL-DT for the core, where L 11 is the core. And it has all of these pressures and areas on 12 one side and DL-DT on this side. And you're saying you 13 don't solve that equation for all time --
14 MR. BROWN: Right.
(O) 15 CHAIRMAN KRESS: Because it changes with time.
16 What you do is you put -- you look at it at some point in i
17 time and maybe -- maybe you're calculating at that point in 18 time, it's a snapshot, and you may change that level in the 19 pressurizer and redo the calculation, and you show for both 20 of those cases there may be end points to the transient, 21 that the -- you end up with the important parameters that 22 need to be scaled at those end points.
23 MR. BROWN: Yes.
24 CHAIRMAN KRESS: So you don't really make a 25 calculation of the transients. You're just looking at the i (o)
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95 t
1 DL-DT equation at particular points in time that.you may or j 2 may not have the right values because you have:to input some 3 levels and you just---
4 MR. BROWN: And I've frozen those levels --
l 5 CHAIRMAN KRESS: You freeze those --
6 MR. BROWN: And I froze them at the beginning of 7' the ADS 4 --
l 8 CHAIRMAN KRESS: And in terms of the actual l- 9 derivation of the DL-DT equation you didn't close that line L
10 off, you just simply -- you just said let it be what it 11 would be.
12 MR. BROWN: Yes.
13 CHAIRMAN KRESS: And --
14 MR. BROWN: And it just turns out that to me my A
(,,/ 15 experience is with the calculation I did that that i
16 resistance is very small even allowing the steam -- allow 17 the steam to go through, the resistance was small, and to 18 simplify the' calculation I apply the boundary condition 19 directly tied to the backpressure in the IRWST.
! 20 I have a pressure here which is related to this 1
21 pressure, plus whatever this, you know, and the resistance 22 . associated with this part is so small I have now neglected 23 that and I have the back pressure with the IRWST in it.
g 24 That's this.
1:
l 25 DR. CATTON: Do you know what the steam flow is
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l l; 96 1 through ADS-1 through 3?
'I i 2 MR.: BROWN; Very small.
d 3 DR. CATTON: Do you know what it is?
4 MR. BROWN: It's about -- in OSU it is about 2 or 5 3, percent of.the total mass injected into the core, and it 6- is approximately -- we have estimates of maybe roughly a 7 quarter of the. total. steam flow generated in the vessel, at
.8 most'.
9 DR. CATTON: The problem that I am having is that 10 we have the other. analysis. Many of us have ccpies of it.
11 CHAIRMAN KRESS: Yeah. I think we need to know
.12 what that is.
13 DR. CATTON: And when you look at that analysis, 14 this is Banerjee's analysis done for research. And when you 15 look at it --
16 CHAIRMAN KRESS: Have we reviewed that as a 17 subject?
18 DR. CATTON: You bet we did, we reviewed it in Los 19 Angeles. The Subcommittee did.
20 CHAIRMAN KRESS: Yes. That's the one you are 21- talking about.
22 DR. CATTON: Yes. And when you look at that --
23 and Westinghouse people were there. When you look at the 24' analysis that he did, he did the scaling analysis and many p 25 parts of it are not a whole lot different than what we see l
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t 97 1 here, except that he took a little bit more freedom with
~
\
[O 2 keeping it simple. Whereas, when you do it, you put r
3 everything in'because you are afraid of what Virgil is going 4 to do to you.
5 MR. BROWN: You guys always say if I miss one 6 thing, say you forgot that. So I put it all in there.
7 (Laughter.]
8 DR. CATTON: I am just . ying to lighten it up a 9 little bit.
l 10 Now, when you look at that analysis, he went one 11 step further. He said, okay, I am going to see what happens 12 if I input initial conditions from data for each of the 13 facilities and use my scaled equations, and see if I can't 14 predict the behavior of the three facilities. And, indeed,
() 15 'he'does. Okay.
16 Now, what that tells me is that at some level, in 17 any event,'these things are predictable. That doesn't tell
.18 me anything about the code, because the code is going to be 19 used in some broader sense, of which this is a small piece.
20 But at least this piece, apparently there is enough r
21 understanding.
22 Now, he also at the end of it -- now, keep in mind 23 he predicted the three facilities. At the end, he says, 24 gee, what are the most important things? CMT, pressurizer, 25 IRWST. And when he says pressurizer, he is referring to ;
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98 1 that flow. He even comments that, although the steam flow 2 is very low, the pressure drop is important.
i i V) So there is a
[ 3 disagreement in the calculated numbers on the table.
1 You are talking about'a pressure drop 4 MR. BROWN:
5 through this portion ao opposed to --
l 6 DR. CATTON: No , no, the top part. The steam 7 line, the steam flow part. So there is a disagreement with )
8 respect to the calculated numbers.
9 On the one hand though, we have seen the results i 10 compared against experiment. And in your case, we haven't i
11 seen anything.
l 12 MR. BROWN: Right.
13 DR. ZUBER: Okay. Now, --
14 DR. CATTON: I tend to come down on the side of
( 'T i
s ,
) 15 the person who presented the numbers.
16 DR. ZUBER: Okay. Now, let me -- one that you did 17 on the simplified accuracy and also another study scaling 18 much more complicated, more rigorous than yours, and it 19 shows'all -- that is really, has distortions. The biggest 20 distortion in all three facilities, and this is the reason I 21 was pushing for this analysis. Now, you do it for an 22 artificial case. I have no arguments. But it is an 23 artificial case. I mean you just forget what is going up 24 and you balance it.
25 ' CHAIRMAN KRESS: Is this a good time to take a
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99 1 break, do you think?
'2 MR. BROWN: Okay.
3 CHAIRMAN KRESS: Did you have a -- okay. Let's 4 take a break for 15 minutes.
l
-5 [ Recess.] ,
6 CHAIRMAN KRESS: Let's start again, please.
Okay. l 7 MR.~ BROWN: Do we.want to move on from the 1
i 8 sketch or -- we have: beat that one. pretty good.
9 CHAIRMAN KRESS: Yeah. I-think we have beaten j l 10 that one to death.
l 11 MR. BROWN: I think one of the things to keep in 12 mind that I thought of at the break was in the model that we
- 13 used, that we have a two phase mixture through the surge 14 line in the hot leg. So, you know, this is not a solid 15 flood of water all the way. So we have the:effect of-16 flooding, if you will, steam in that line, but all I have
[ 17 done is I have not -- I have allowed that steam to get 18 through to here, but when it came to actually applying the 19 boundary, you know, to handling it up at --
!' 20. CHAIRMAN KRESS: You had --
21 MR. BROWN: Yeah, I just, I applied it as a 22 boundary condition because it was small. And, again, while l .
- l. 23 .I have presented'it here, however, we have used that model 24 and actually got good agreement with OSU.
'25 This is the same as before with the major V
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l 100 i
i components, with the exception of we have the passive RHR
! %./
[) 2 on, but that we are allowing -- I am allowing steam to be i k
3 drawn into the PHRR where it is condensed and drains back l 4 into the cold leg.
l 5 I have already presented the next slide, which is I 6 my justification for neglecting the inertial transient terms
{
7 in the conservation of momentum equations so that I can work 8 with a steady state set.
9 DR. SCHROCK: Again, I think you need to think a l
l 10 little more about what it is you are trying to do to justify 11 steady state treatment. When we use quasi-steady state I 12 calculations, what we are essentially doing is taking our 13 usual coefficients as being the same as they would be for 14 the same steady flow situation, the same as the p
, ( ,) 15 instantaneous flow. The friction factor, for example, would 16 be the same as in a steady flow of the same level as the 17 instantaneous flow in that transient. Then it is fine to 18 neglect the transient term in an equation.
19 What you are doing is -- maybe it is overkill on 20 the one hand, but it is not physically examining the issue 21 of is this flow one that can be reasonably treated as 22 quasi-steady. Instead, what you are saying, I'll compare 23 time scales to the time that it would take to accelerate a 24 slug of liquid from rest to that steady flow rate in a l 25 constant area pipe or channel. That is not testing really.
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101 !
I 1 You could accomplish that acceleration via a quasi-steady
/ } 2 process in which the friction factor would be changing as
(_/
3 the Reynolds number builds up. It would.be a quasi-steady 4 flow and you would get a good prediction if it is a slow 5 acceleration.
6 MR. BROWN: A spatial acceleration change?
7 DR. SCHROCK: Yeah. I 8 MR. BROWN: Yes.
9 DR. SCHROCK: And maybe that would be okay for an 10 acceleration at that level of head.
11 MR. BROWN: I am just arguing that the time change 12 of the acceleration of momentum equation, that that term I ;
13 can neglect, not the spatial one. I have included the 14 spatial acceleration term. So it is the change in momentum n
'() 15 of time is what I have neglected. That's so I have a 16- quasi-steady set of equations.
17 DR. CATTON: I thought you said earlier that you 18 neglected the inertial terms. No?
19 MR. BROWN: I said the term associated D -- D rho
'20 V square DT. That term is what I have neglected. The 21 change in momentum with time.
22 CHAIRMAN KRESS: The way you handled the spatial 23 is you only looked at the components that have constant 24 area.
25 MR. BROWN: I look at the ins and outs of the l
I f)
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102 1 components, yes.
/ 2 CHAIRMAN KRESS:
\_) Yes.
3 MR. BROWN: I include that. And that's what you 1
4 will see in the next. I put this in the context of a set of 5 six momentum equations through each of the paths, if you 6 will. I.have. retained the head terms, the resistance terms 7 and a-spatial acceleration term that you see, the mixture 8~ density that I use there is a homogenous form for the 9 density.
10 IRWST gravity head I found to be the most 11 . dominant, importantEt4erm head-wise, from-the IRWST down to 12 the DVI, to the vessel. And so I have divided everything, 13 normalized everything based on that. So all the heads or 14 pressure drops, resistances, are all relative to the IRWST p)
(_ 15 gravity head.
16 DR. ZUBER: Let me ask you, in your scaling, did 17 you ever draw kind of loops like electrical loops, nodes, 18 resistances, something that you can then relay the numbers 19 and really explain in a --
20 MR. BROWN: I used to and I gave up.
21 DR. ZUBER: Why?
~22 MR. BROWN: That was -- well, I don't know, I 23 . guess -- I guess it wasn't -- I don't know. I kind of like
- 24 ' this a little better.'
25' DR. ZUBER: Of course you have to do that. But in
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103
~ 1> ' explaining what'we were just discussing, the pressurizer.
i 2' MR. BROWN: Yeah, you could do that.
3 DR. ZUBER: ~It can be much easier for you to 4 explain and explain your scaling. These-are the reservoirs, 5 the volumes is'at the resistances, these are the loops,-this 6 is:how they interact. And you have really a --
i
,7 - MR. BROWN: It is a better physical picture. I 8 mean. I do that sometimes, you know, :ba my own sketching, but 9- I guess in terms of actually building the modal-and solving
- 10. it, I have read the whole thing out in a set of equations..
11! Since.the tool I'usually use for solving them is MathKit, so 12 it doesn't -- I tend to like to write them'this way,'because 13 that way when I enter them, I can enter them as I have
~
14- . written them. It is pretty user-friendly like that.
()
15: CHAIRMAN.KRESS: The subscript numbers correspond
' 16 ~ to the numbers on your diagram?
17 MR. BROWN: Yes. If you look in the diagram, you
.18 -will see -- right. These -- right, they refer te the-19 diagram. That's.right..
20 CHAIRMAN KRESS: So your normalizing parameters, 21' the gravity head, the IRWST and the reactor vessel.
'22 MR. BROWN: Yes. Yes. If you -- right. If you 231 look at: the, diagram, you will'see the IRWST, that it goes D24 .down throught'he DVI and down into the vessel, or the 25 mixture, the level in the core, that's the gravity head, !
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104 1 that's right. And that's -- I found that to be the most
() 2 3
dominant term of any single other element in the system at this time. Even.over the pressurizer, even with the 4 sensitivity of the pressurizer.being at full head. .
5 And the real reason is simply because, especially 6 when you allow-a flooding type of phenomenon, or two phase ,
l 7 mixture in the hot leg, in the surge line,-that that is '
8 really the reason why it doesn't become~ dominant. Because '
9 if you had a solid column of water all the way back, then I 10 believe the pressurizer'could in fact be the most dominant 11 gravity head in the system. But it is not and I don't see- ,
.12 that in the test.
13 I see water in the pressurizer itself, a solid I 14 head of water in there, essentially, but it is a two phase
.r .
15 mixture through the surge line because of that small amount 16 of steam that is going to through ADS 1, 2, 3 and out from 17 the hot leg. And it is.that two phase mixture which 18 significantly reduces the overall density of that line, ~
or 119- that path, I should say. And so, therefore, that's why I 20 found it to be more of medium importance relative to the 21 IRWST. l 22 So if.you really had a solid column of water from
'23 the pressurizer'to the IRWST, then I think you kind of L
l 24 really would have a well balanced slinky back and forth here 25 between these two -- other than the volume would be a huge
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105
- 1. mismatch. As you mentioned earlier, Dr. Kress, that it is a
() 2
. huge amount of water', and the IRWST, very, very little 3' compared in~the pressurizer, so it is not -- you may have
'4 the head there, but you don't have the capacitance, if you 5 .w ill, of all that volume of water there to really help --
- 6' help with that.
7 CHAIRMAN KRESS: The power doesn't show up 8 anywhere in momentum.
~9 MR. BROWN: No, not in the momentum.- I had that 10 --
11 CHAIRMAN KRESS: This affects the pressure.
12 MR. BROWN: Yes, I had that in the pressure 13 equation.
14 CHAIRMAN KRESS: It's in the pressure equation.
) 15 MR. BROWN: Yes. Yes.
16 CHAIRMAN KRESS: -So somehow, in order to look at 17 the importance of' things, you have'to look at both_of them 18 together.
19 MR. BROWN: Yes. That's why -- that's why I did 20 scale the pressure equation along with the momentum 21 equation, set of momentum. equations here as well. That's 22 right. And, as I said, I found that the IRWST gravity head, 23 and that's why I normalized upon that, is I found that to be 24 the most dominant.
F 25 Anyway, unfortunately, it took two pages to shot' l[
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106 1 all those equations shown off on slides 23 and 24.
' .[%j j 2 DR. SCHRCCK: I had a comment on that 23.
3 MR. BROWN: Okay.
4- DR. SCHROCK: It, again, may be viewed as 5 nitpicking, but I think you lose your credibility when you 6 present stuff that is' nonsense. This lower case g subscript 7 c is a non-existent thing in the world of physics. And here 8 you are non-dimensionalizing it. It is already a 9 non-dimensional number. Okay. It is a fiction of people's 10 imagination from the standpoint of physics. It is nothing.
11' I mean you are better off not ever putting in your 12 equations. But if you are going to insist that it is your 13 right to solve equations using inconsistent units, then you 14 can save the day and get the right answer by inserting g (j 15 sub c.
16 But g sub c, as you will find correctly described 17 in many textbooks nowadays, is in fact a dimensionless 18 correction factor. It has units, but it is a dimensionless 19 correction factor. And the units that it has depends upon 1 i
20 which crime you have committed in selecting the units that l 21 you are using in your equation.
22- So, but to put it in your scaling analysis, you l
23- know, just boggles my mind that you would do that.
l l'
24 MR. BROWN: I understand.
i 25 DR. SEALE: Some old habits die by fire. l k ANN RILEY & ASSOCIATES, LTD.
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107 l
'l MR. BROWN: I understand your point.
( 2 DR. CATTON: -Virgil, maybe they plan to fly it in l 3 low orbit. You would have to have the ratio g over g sub c.
4 DR. SCHROCK: Ivan, I have just explained that g
! 5 sub c doesn't exist in the physical world. Whether --
6 DR. SEALE: At any altitude.
7 DR. SCHROCK: Whether here or out there.
8 MR. BROWN: On the slide 25 and 26 I have shown 9 the numerical results from all this. And the ones, what I 10 would call of most importance or higher importance that I 11 found, at the top of the list is the IRWST head, which'I '
12 . mentioned before is what I normalized all the' equations
- 13. upon.
14- Obviously, the scaling is just relative to like j
I '
) 15 OSU is the quarter-height scaling, so it is actually, when 16 you go into the bottom-up on just the ratio of heads you 17 have a factor of four.
18 CHAIRMAN KRESS: I don't understand that first row 19 up there.
20 MR. BROWN: Pardon me?
21 CHAIRMAN KRESS: That first row, I don't 22 understand. Is that --
23 MR. BROWN: It just says that it is normalized on 24 itself. It doesn't tell you anything more than that.
i
- 25. CHAIRMAN KRESS: Okay.
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108 1 MR. BROWN: It.just --
1 E() 2 CHAIRMAN KRESS: Okay. I understand it then.
3 MR. BROWN: Yeah. That's all.
4 The next one was the spatial acceleration loss i
5 associated with the ADS-4. This, too, indicates on the j 6 pressurizer side. The'next one is the two phase pressure 7 drop resistance associated with the PRHR, that was 8 important. The DVI resistance, then ADS-4, two phase or 9 resistance associated with -- again the ADS-4 on the 10 pressurizer side. And then here at the bottom of that group 11' I had the pressurizer head. That was the order in which I.
12 came up with.
13 The other terms that were in the equations, sort l 14 of a second tier of terms between .1 and .2 are the other l (/"'sj 15 things associated with the ADS-4 on the other side. It's l
16 gravity and acceleration and resistance. The other terms-17 that are not shown were found to be even much than 10 1
18 percent, so I didn't include those. l 19 And so what I concluded from all that was that the l
20 gravity head was most important and was preserved in OSU. i 21 The ADS-4 and DVI resistances were important and they were
. 22 well scaled. .The' ADS-4 non-recoverable acceleration loss 23 ~was also significant in OSU. And the pressurizer gravity- ,
24 head was -- ILclassified that more as of medium importance l-
{ 25 and was well scaled in OSU.
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109 1 Comments. I think one of the things that I find
( 2 is that the IRWST, as designed, once the -- once ADS-4 opens 3 up, ADS-1, 2, 3, essentially becomes, from a -- it does 4 certainly allow a small amount of steam through the line up 5' through the pressurizer, which Mike Young and Kenji will 6 talk about later in more detail, up through the surge line.
7 It certainly is a mechanism for holding up water. However, 8 the pressure drop associated with that in the surge line, as 9 well as the pressure drop downstream of the pressurizer 10 going to the IRWST, I found to be very small. So its effect 11 is-important by the fact that we have the water held up i
12 there. So I have accounted for that in that sense. That's 13 what -- that's why we do have a static head there.
I 14 Of course, if we in fact were able to drain all O)
(, 15 this fluid, you know, it would be perhaps a different 16 scenario. But once the ADS-4 opens up, it is essentially --
17 I mean that's where all the mass is going out of the system 18 and, essentially, IRWST is the only dominant thing going on.
19 The CMT is -- at best, it's at 20 percent level. It is 20 ,nowhere near the IRWST initial elevation. So I find it a 21 little bit hard'to believe in this particular time frame 22 that the CMT is dominant. I believe the IRWST is dominant, 23 as well as the ADS-4 resistance, anything associated with 24 that are really the two biggest players in this.
,25 And, as I said before, because of the problems ANN RILEY & ASSOCIATES, LTD.
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e 110 1 with SPES and the ADS-4 area, the situation is too distorted
() 2 3
to use that m".', therefore, we ar'e using OSU and we find that it is adequate to use that for a validation with 4 NOTRUMP.
5 DR. SEALE: Okay.
6 MR. BROWN: We can go on to the next topic then?
7 CHAIRMAN KRESS: Yeah, let's go to the next one.
8 I think we are ready.
9 MR. BROWN: Okay. The next topic is Applicability 10 of Equation --
11 DR. ZUBER: Let me ask you. I like to kind of put 12 a periodic end of each presentation. Is there any action 13 you are going to take on this concerning our comments?
14 MR. BROWN: According to the way we -- that we at
) 15 Westinghouse were working with this is -- the conclusion I 16 have is that the data from OSU is adequate to use for code 17 validation during this period. So I don't see any more 18 action that 7. need to take, at this point in scaling.
19 DR. ZUBER: Okay. Well, then I make -- I shall 20 make my final comments at the end of the meeting.
21 MR. BROWN: Okay.
22 DR. ZUBER: I disagree with you.
23 DR. CATTON: Maybe for Mike's benefit, I would 24 like to kind of make a statement of where I think we are at. I 1
25 The codes, and I know you don't care about the codes, but l
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1 that's all right. Both NOTRUMP and RELAP 5 do not have the 2 proper physics to treat the steam flow from the core to the 3 pressurizer and the containment through ADS-4. As a result, 4 we must resort to other means to predict the minimum core 5 level during the transition to IRWST flow -- and flow to the 1
6 vessel. Westinghouse would like to penalize the IRWST 7 level. They would like to scale the penalty from what was 8 needed to match OSU behavior. The question then becomes how 9 to scale the penalty from OSU to a full scale AP600. To 10 just change it or linearly reale it is arbitrary.
11 And that puts it all in Mike's lap.
12 DR. ZUBER: Okay. Now, let me then make one 13 comment. I didn't know that you were going to. I feel 14 comfortable with the scaling, not yours. I have seen other 15 analysis with all aspects of the transient. Because of the 16 three facilities, at least one is properly scaled. At least 17 we have, for each phase we have one f acili'.y which performs 18 well, and we can use that facility to validate that code.
19 Except this way we discussed, all three facilities are not 20 well-scaled, they have distortions.
21 Now, we have also the analysis of Banerjee who 22 addresses also these aspects. I was critical in my report 23 to you, Tom, about RES. In this case, they did a very good 24 job. They produced this work by Wolfgang and the work by 25 Banerj ee . So we have two pieces of analysis, which this is l
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112 1 the reason Ivan and I are critical. Unless you really
,j 2 address this particular issue in a technically defensible 3 way, I don't -- I cannot consider that this issue has been 4 closed. The codes cannot do it. We have tools how to 5 analyze it and I think you should address it.
6 MR. BROWN: And your process for addressing it 7 would be?
8 DR. ZUBER: I told you we have two pieces of 9 evidence, it showed that this -- that these three facilities 10 are distorted, and for this particular --
11 MR. BROWN: In important phenomenon, they are 12 distorted?
13 DR. ZUBER: Distorted for in this transient.
14 CHAIRMAN KRESS: The issue that has to be
(-)w .15 s addressed is basically Ivan's. 'How do you justify what 16 penalty to take with the IRWST level? And it bears on these 17 subjects. How do you go about making that judgment?
18 MR. BROWN: I see this connection. I don't see 19 the connection --
- 20 DR. CATTON: Well, I'll make the connection for 21 you if I can. When you say, when you talk about the 22 phenomena, sure. The phenomena probably is in all of the 23 facilities. I wouldn't really try to argue with you about 24 that.
25 If your code was capable of dealing with the f
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113 1 phenomena properly, and for a couple of the facilities you
) .2 were able to show that, we wouldn't have to have this 3 discussion. The reason we need is because the code can not.
4 And given that it can not, you have to go back and scale 5 these phenomena that are relevant or important in 6- . establishing what this penalty is going to be, and then say 7 something about how you get from that to AP600. Because it 8 can't be done in the code. All you can do in the code is to 9 penalize the code enough to account for it.
10 Did I make myself clear?
11 MR. BROWN: I take it that, and I conclude here 12 that the IRWST is the most important significant part in all 13 this, all that phenomena. In fact, that is what these guys 14- have chosen to adjust in the code. So I think that is
()
'15 proper. They have-taken the most important phenomenon, the 16 scaling has indicated that it is the most important 17 phenomenon,'and they are going to take a penalty on that 18 most important phenomenon. They are not taking something --
19 something that is insignificant.
20 DR. CATTON: But, you see, the question is how big l
21 should the penalty be?
22 MR. BROWN: Okay. That's another issue then.
c23 DR. CATTON: No , it is not another issue. They 24 are all together. You want to use -- you have got -- you
, 25 can't just arbitrarily say one foot or whatever, you pick a l
[~
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114
- 1. number.
2 MR. BROWN: I agree, it should not be arbitrary.
3 Yes.
4 DR. CATTON: Now, where are you going to get the 5 number? You can go to OSU and say,. gee, if we penalize the 6 IRWST, level by.X inches or feet, the results look good. And 7 now it is.one-quarter. scale, so I will take four times that 8- for AP600. That you cannot do. Unless you show that that 9 is how -- how that particular thing scales.
10 DR. ZUBER: And in that case you really have to 11 take the interaction, you can not just assume a pressure
- 12. boundary. You have to take all the loops together. And 13 this is -- you did partly -- partly address the problem, but 14 not completely to satisfaction. And we have another A
() 15 analysis which showed the distortions.
16 So unless -- I don't see this issue closed today 17- unless this is really resolved and addressed, addressed and 18 resolved.
MR. PIPLICA: Can we I mean look at that other 20 analysis so we can comment on it?
21 DR. ZUBER: Well, let me ask you --
22 DR. CATTON: That's Westinghouse's problem and you
- 23. need Brian. And he has made the decision that you shan't.
241 MR. PIPLICA: The point is we don't have it.
- 25 CHAIRMAN KRESS: I think the staff over here could b\/ ANN RILEY & ASSOCIATES, LTD.
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115 1 speak to that. We are talking about Banerjee and Wolfgang's
.([ 2 3
-- Wolf's scaling analysis.
. haven't seen them.
Are they available? You 4 DR. CATTON: See, it has been held proprietary 5 within research. The staff doesn't even have it.
6 CHAIRMAN KRESS: Oh, the staff hasn't seen it.
7 DR. CATTON: That's really kind of dumb if you ask 8- me.
9 CHAIRMAN KRESS: It's a little hard for me to 10 comment.
11 DR. ZUBER: 'Well, I did comment. I did comment.
12 CHAIRMAN KRESS: I have some comments here.
13 DR. ZUBER: That is another, another, another.
14- DR. CATTON: See, in this case, I thought research f
15 was supported to support the efforts of NRR, and I thought 16 this work was being done so that the staff could better 17 evaluate AP600. Now, we find that one of the most important 18 pieces of information --
19 DR. ZUBER: Two pieces. Two pieces.
20 CHAIRMAN KRESS: The only thing they got was the i 21 bill?
! 22 DR. CATTON: I think so. Wolfgang's wor.k is a 23 little bit more complex than this problem needs, I think.
24 But Banerjee's was good.
25 CHAIRMAN KRESS: Well, the question is have is,
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116 1 ' can the issue be evaluated based on just Banerjee's, because
'2 it shows the difference in results also, and do we have to
(J) s.
3 go to Wolfgang if'that's the one that's proprietary.
4 DR. CATTON: They're both proprietary.
5 CHAIRMAN KRESS: Oh , they're both being. held.
.6 DR. CATTON: Right.
7 CHAIRMAN KRESS: I didn't know NRC held things --
8 DR. ZUBER: I would advise actually it's up to NRC 9 to take those two reports and their analysis and make a 10 decision what to do. .I think what they have done we have 11 two pieces of evidence, good work, which NRC didn't make use 12 of in evaluating their results.
4 1
13 MR. LANDRY: Ralph Landry from the staff NRR. '
14 Since this came up this morning, I've managed to l3,O) 15 get the last two copies that exist of the reports that 16 Banerjee did. More copies are being made. They're supposed l 17 to be done by the middle of this week for NRR. We're having l
l 18 copies made for our use so that our people can review the l
19- material too.
20 Wolfgang's report-I haven't seen. I don't know 21 who has copies of that.
l
[ 22- DR. CATTON: Is Wolfgang's report out yet, Paul?
23 MR .. BOEHNERT: No , I don't believe so. I believe l l i 24 it's in research.
25 DR. CATTON: It's in research in draft form, I ANN RILEY & ASSOCIATES, LTD.
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117 1 guess. I'd like to see it too.
( ~2 MR. BOERNERT: Well, I haven't been able to get i 3 ahold of it.
4 MR. LANDRY: From the perspective of NRR, we would L
5 prefer to review material that is in final form, not review
-6 draft materials, because draft materials can change. We i
7 don't like to make licensing judgments on the basis of draft 8 materials.
! 9 DR. CATTON: That's a discussion you and research 10 need to have.
11 MR. LANDRY: Somewhere this report did not come-12- through the channels, and we need to get that fixed.
l 13.. DR. ZUBER: Didn't go through the channels.
14 DR. CATTON: Wolfgang's work was done quite some Q(% 15- time ago.
- 16 MR. BOEHNERT: They both were. It was done about-17 the same time.
r 18 DR. CATTON: And -- well, Wolfgang was a little 19 slower.in getting his act together.
20: MR. BOEHNERT: 'Okay. Well, that may have been.
21- DR. CATTON: He stopped-to publish a paper on 22 nuclear energy and_ nuclear engineering design.
-23 MR. BOEHNERT: I think you guys even referred to
- 24 the. paper he published.
L 25 MR. BROWN: Yes.
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118 1 DR. CATTON: But,it was generic.
-n
.~ t 2 2 MR. BROWN: .Yes, it was generic.
O 3 DR. CATTON: Where his report to research was --
4 DR. ZUBER: This is specific. And the reason is I 5 feel comfortable about all phases of the transient, because 6 at least one facility is very scared. So the Code can use 7 those data, except this phase here. All three facilities 8 are distorted.
9 MR. BROWN: You guys seem to be looking more for a 10 prediction of core level, though, right? That's what-I'm 11' gathering, and that just seems'to be a.little more than~I 12 think what we expected or were looking for in our objectives 13 for scaling. And so that's --
.14 DR. CATTON: Well, were looking for ---
'15 ,MR. BROWN: I hear what you're saying --
16 DR. CATTON: Looking for the basis of the penalty 17 that you're going to take.
f 18 MR. BROWN: Of the penalty; right. Right..
19 DR. CATTON: And now, I mean, that's the top-down
,_ 20 statement of what we would like to see.
21 CHAIRMAN KRESS: 'Yes. I mean --
22' DR.'CATTON: And now you can keep cutting it and 23 you could get into the scaling. There's all kind of ways L
[ 24 you could address that.
.25 MR. BROWN: Are you disagreeing -- I guess are you l
1
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1- disagreeing with the results here? You're saying that these j '
2- are in conflict with results you've seen otherwise, so you 3 say --
4 DR. CATTON: No.
5 DR. ZUBER: I do. I do.
6 'DR. CATTON: They're incomplete because you left 7 out the steam flow from the pressurizer to ADS. Now --
8 DR. ZUBER: Let me say, they're in conflict with
)(
9 the results of Wolfgang. /
10 MR. BROWN: Okay. With Wolfgang. How about 11 Banerj ee?
12 DR. CATTON: And with Banerjee. He makes a clear 13' statement.
14' PEl. BROWN: Do those two agree also?
j, s) 15 DR. CATTON: They agree on what's important.
16 MR. BROWN: Okay.
17 CHAIRMAN'KRESS: So you need to resolve the 18 conflict, why are they different and which is correct. j 19 DR. ZUBER: Tom, it's NRR --
20 CHAIRMAN KRESS: NRR has to. Right. And then as 21 to the question as to the penalty, do you guys have a 22 suggestion on how one could use the scaling analysis to 23 arrive at-a justification for that?
24' DR. CATTON: I do. I think that they should
.25 follow the path that Banerjee took, which was to try to l
l l
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120 2 do -- with the scaled equations try to do some simple 2 . calculations.
}} You want to find out what is important'and 3 .
how it impacts.
! 4 CHAIRMAN KRESS: Actually treat the scaled 5 equations as a model, i 6 DR. CATTON: Yes, and you avoid a lot of 1
7 difficulty that they have to face with their code, because 8 you just pluck out this little piece of it and look at it.
9 You don't try to do the whole problem.
10 CHAIRMAN KRESS: I'm still a little perplexed, 11 because what I see is scaled equations on pressure, scaled 12 equations on momentum, scaled equations on energy, and what 13 I'm really interested in is what is the level. How does 14 that translate into the level.
rx 15
( , ) DR. CATTON: Well, pressure can translate pretty 16 directly for this kind of problem.
17 CHAIRMAN KRESS: You think you can look at the 18 pressure equation and actually see how to penalize the IRWST 19 to get a proper level --
20 DR. CATTON: What you do is if you do it in terms 1 21 of mass balance, I mean, how many pounds of water do I have 1
(
22 in the core, if I know the area that directly translates 23 into --
24 CHAIRMAN KRESS: You go to conservation of the l 25 mass equation.
1
(
1
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121
. 1 IMt. CATTON: The conservation of mass or-mass 2 flux, however you want to treat it.
. ( ). But if you know the 3 mass in the various volumes, you know the levels.
4 DR. ZUBER: This is addressed in Wolfgang.
5 CHAIRMAN KRESS: Is that addressed?
! 6 DR. CATTON: He actually has a DL by DT.
7 CHAIRMAN KRESS: It looks to me like that would be 8 of great value if even that draft report could be released.
9 DR. CATTON: Well, that's Brian is the one who has 10 been maintaining the Westinghouse line, which is the
-11 proprietary nature of the data. And apparently --
i 12 CHAIRMAN KRESS: Wolfgang's report was done for 13 research.
14 MR. BOEHNERT: 'Both were done for research.
15 MR. LEVIN: This is Alan Levin from the staff.
16' The issue here is the proprietary nature of the data from 17 the confirmatory testing, and as this committee heard last 18 year, and essentially sided with Westinghouse, that there 19 was a commercial advantage and that it should be held 20 proprietary. There's nothing we can do about it until
- 21. Westinghouse decides it's releasable.
22 CHAIRMAN KRESS: From releasing a draft copy to 23 -Westinghouse?
24 MR. LEVIN: Yes. We cannot selectively release 25 things to the public. Westinghouse is part of the public.
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122 1 It doesn't matter whether it's their data or not, j 2 CHAIRMAN KRESS: That's the problem.
3 MR. LEVIN: Yes.
4 CHAIRMAN KRESS: Thank you, A1.
.5 MR. BOEHNERT: Yes. Catch-22.
6 DR. CATTON: And we -- the Committee agreed with 7 ~ Brian, in part probably because of the way the question was 8 .phrastd to us.
9 MR. LEVIN: And I'll note'that where we have asked 10 Westinghouse specifically to review -- to predict ROSA
-11 behavior, which'is the PRHR stuff, we received agreement 12 from them a priori to place that material in the PDR. .So 13 that material -- allithat data is open literature now. It's 14 nonproprietary. And that's what we've got to do anytime we
, r~)( ,e 15 want to release something related to ROSA or anything else 16 confirmatory into!the public domain,-we.have to get an 17 agreement from Westinghouse not to_ hold it proprietary.
18 CHAIRMAN KRESS: This shouldn't be an issue for 19 the Banerjee report, should it?
20 DR. CATTON: Yes, because it was supported by 21 research also.
22 CHAIRMAN KRESS: Yes, but it had to do only with 23 the SPES --
'24 DR. CATTON: SPES -- it has to do with SPES, OSU, 25 and ROSA.
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l 123 1
1 CHAIRMAN KRESS: Oh, it has those in there too.
()
r 2 MR. LEVIN: If it has any proprietary information 3 in it, that's it.
4 CHAIRMAN KRESS: Thank you. That clears things up
! 5 for me a little.
6 DR. CATTON: You could give it to Westinghouse and 7' have them cleanse it, then give it back to Westinghouse.
8 It's a problem.
9 MR. BROWN: So what you saw that you liked was 10 using the scaled equations to try to predict the facility to 11 give you a handle on what kind of level penalty you think
- 12 you should take.
13 DR. CATTON: You want to know what the important 14 parameter is and how it scales, because then -- whatever it 15 is.
17 DR. CATTON: Well, how do you scale the penalty.
18 That's the question.
19 MR. BROWN: I guess you presented that before.
.20 Right, Mike?
L 21 DR. ZUBER: You cannot with the Code.
22 MR. YOUNG: The relationship between NOTRUMP and 23 the scaling analysis I hope will get a little bit clearer as l
l 24- we go along here.
25 MR. BROWN: Okay. Equation 363 of WCAP-14727.
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124 1 This is in direct response to your comment the last time,
/D 2 Professor Schrock, on the homogeneous equilibrium equation.
V 3 DR. SCHROCK: What I said this morning, it's 4 'that -- still are not doing it correctly. The derivative of 5 quality with respect to pressure should be taken at constant 6 entropy, not at constant enthalpy. When you read 7 Wallis's --
8 MR. BROWN: Of quality with respect -- you said 9 quality with respect to pressure, you said.
10 DR. SCHROCK: Yes.
11 MR. BROWN: Are you referring to down here?
12 DR. SCHROCK: Right. Look at your equation that 13 you're getting G sub C, and the one at the bottom, same 14- derivative. Bututhat needs to be at constant entropy, not f%
( ,) 15 at constant enthalpy.
16 Wallis explains a little bit vaguely, 17 unfortunately, the thing. It's true he wrote in the book --
18 MR. BROWN: Because that's precisely how it comes 19 out of the text. That's why I'm asking.
20 DR. SCHROCK: Yes, but I had told you before in 21 our previous discussion of this that that was not correct, 22 so I guess either you didn't hear it or you didn't agree 23 with me or whatever. But it is not correct, and L 24 MR. BROWN: That's what I'm saying. Are you 25 saying the equation in Wallis's book was incorrect?
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125
~
1 DR. SCHROCK: I think you have to read the words
.( ) 2 .that go with the equation more carefully, and you'll 3 understand that you've jumped too quickly to pick what he 4 put down in the book.
5 I think that the context for the equation was not l 6' exclusively. determination of critical mass flux. That he 7 was pointing out the analogy of this set of equations to the 8 comparable equations in the classical gas dynamics, in which 9 you have these so-called influence coefficients which act on 10 the dimensionless derivatives with respect to area, with 11 respect to whatever.
L 12 And it's true that in both of the two equations in 13 the book -- I think the right one was 2-44, which is one you 14 have here now is 363, and the one that you used
(). 15 previously -- no, only in this one.because the term only
- 16. appears in.this one. It has the derivative at constant 17 enthalpy is the way it's written.
18 But in my report to the subcommittee and in the 19 discussion at two previous --
20 MR. BROWN: Okay. I didn't pick that up.
21 DR. SCHROCK: Subcommittee meetings I pointed out 22 that in fact the correct way to do it is at constant
-23 entropy. Constant enthalpy never has any real meaning.
24 Constant stagnation enthalpy is correct for the application 25 of that to Fanno flow, in which case the term involving I
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126
! 1 derivative of area becomes zero. But the static enthalpy is
} ). -2 never in any instance the correct thing to be used.
- 3. 'MR.. BROWN: Okay. Well, that's what I did, and I 4 was not aware of that then. You may have said-it and I 5 didn't -- it didn't sink in then. I missed that then 6 before.
7 CHAIRMAN KRESS: In cases of low friction and 8 little heat exchange between the walls and the fluid, the 9 two derivatives approach each other?
10 DR. SCHROCK: No , they don't. I mean, if you look 11 at the temperature entropy diagram that contains the lines 12 of constant enthalpy, you'll see that in very limited 13 regions the two may be pretty coincident, but in general it 14 can be grossly in error.
(G
,/ 15 CHAIRMAN KRESS: This is strictly a-fluid property
'16 equation. ;
17 DR. SCHROCK: Right. Strictly --
l 18 CHAIRMAN KRESS: Doesn't have anything to do I i
19 with --
1 20 DR. SCHROCK: Well, in terms of calculating '
21 critical mass flux, I mean, the analogy to the gas dynamics 22 is that what's in the bracket and the denominator is 1 minus I L 23 the mach number squared. That contains the ratio of the
~
24 speed to the sound speed. The sound speed depends upon the 25 derivative at constant entropy, not at constant enthalpy, lN ANN RILEY & ASSOCIATES, LTD.
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127 1 MR. BROWN: Okay. Well, then there's no point in l
() 2' 3
discussing the rest of that other than to say that with I
respect to this equation, what we had done since then is we 4 had actually just gone to the homogeneous equilibrium 5 curves, a family of solutions based on inlet conditions such 6 as published in Leahy and Moody, and so we didn't actually 7 fool witn the equation anyway in the calculations or in the B scaling report.. We present it, and I wasn't aware of the 9 constant entropy --
10 DR. CATTON: Moody's book?
11 MR. BROWN: Leahy and Moody;.yes. Yes, yes. Just L 12 used the homogeneous curves. I just went back to the 13 curves, and that's what I used.
) 14 CHAIRMAN KRESS: Virgil, for my education --
15 excuse me, I'm trapped here -- let me ask the question
! 16 another way. This was a critical flow equation for -- down l
17 a pipe. And --
18 DR. SCHROCK: Well, it's more than the pipe. I 1
i 19 (
mean, the equation has area change in it as well. 1 l 20 CHAIRMAN KRESS: Has area changes in it also.
21 DR. SCHROCK: He's written a general equation.
22 CHAIRMAN KRESS: And it's an equation for though 23 the change in pressure with position along such a pipe.
24 DR.-SCHROCK: Right, right, right.
25 CHAIRMAN KRESS: Now what I was asking you, and l l
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4 I 128.
1 let'me rephrase it to be sure I understand your answer --
'h G 2 DR. SCHROCK: Okay.
L 3- CHAIRMAN KRESS: Was that if that process along j 4 the pipe is one where losses due to friction and exchanges 5 of energy between the fluid and the pipe itself are small, 6 then you could describe that as being a constant enthalpy 7 ' process, as opposed to constant --
8 DR. SCHROCK: Well, no, for'the straight pipe the l 9 only thing that forces an expansion of the fluid is the 10 friction. It drives a loss, the frictional loss of 11 pressure, and the loss in pressure allows the 1
12 compressibility of the vapor to increase the specific
-13 volume.
14 CHAIRMAN KRESS: I understand.
.(A.f 15 DR. SCHROCK: That's the only thing that does 16 that.
17' CHAIRMAN KRESS: I understand.
18 DR. SCHROCK: Now in the application of this to 19 the whole of their ADS lines, as I also said previously, you 'j 20 can think of it as a simple sequence of processes. You have 21 .something near an isentropic entrance. You can take into ,
-22 account the permanent losses in that using loss coefficient.
23 That'll get you from an upstream reservoir into the pipe.
24 You then have a Fanno flow in the constant area l
'25 ' channel until you get to'the end of the. pipe, and there you if%- ) ANN RILEY &' ASSOCIATES, LTD.
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129 1 have another constriction. And that one is reasonably
-2 . represented as an isentropic process. The frictional aspect l 3 of the flow through.your final constriction at the end is 4 certainly negligible. So you have a sequence of those
~5 processes that you could take into account.
6 But the other aspect of what you've done here and-
'7 also I'd commented on previously is that you' don't know i
8 until you've solved that equation from the upstream 9 reservoir to the point of choking what the pressure is at I 10 the point of choking. And what you've done in your NOTRUMP 11 code and what you I think probably have done in getting the 12 numbers that.you put in this scaling report purportedly as 13 the HEM prediction of critical flow for that situation is to 14 use a fixed critical pressure ratio of .58.
15 Well, if you look at the data available on )
16 critical flow measurements, you see that you do not have a 17 fixed ratio. The critical pressure ratio is not universally 1 l 18 .58. It varies quite: considerably, depending on the 1
location in the two-phase flow domain. So you need that to 20 be incorporated into the thing as well. You've calculated a !
l l
21 critical mass flux in some way that you describe simply by i l'
22 saying it's HEM. That's not enough. You have to say how l 23 you get it'from HEM.
l 24 You constructed a model that you solved along the 25- length of the pipe. That's one way to do it. Another way i i
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130 1 to do it if you have the capability of predicting what the
() 2 pressure is at the point of choking, then you can calculate 3 what the sound speed is at the point of choking as well, and 4 you can get it from that. You have to say what it is you're 5 doing to get the number that you put in the report, and 6 you've not said that. I don't know, Tom, have I answered --
7 CHAIRMAN KRESS: That helped me a lot, Virgil.
8 DR. CATTON: I'd just like to ask Virgil a 9 question.
10 In reading about LOFTRAN they made a special 11 version to deal with steam generator tube rupture, and they 12 use Zaludek's brake flow model, 1964. I don't remember very 13 much about that brake flow model.
14 DR. SCHROCK: Zaludek's report is very hard to
() 15 come by. Unfortunately it was never published in the 16 refereed literature, but it was used very early on by code 17 developers, and it's deeply embedded in all of the codes, 18 the NRC codes as well as company codes. It's a little hard 19 to justify at this point, I think, given the fact that it 20 was really never published. I think it was good work, but, 21 my God, we're talking about stuff that was done in --
22 DR. CATTON: Thirty-four years ago.
23 DR. SCHROCK: Yes. If Zaludek we alive today, I 24 don't think he would tell you that he would like the 25 situation as it exists. I mean, it's --
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l 131 l 1- DR. CATTON: He was old when I knew him.
l-
[( 2 He was a very. good piano player. When he was a-
- 3. teenager he stuck his fingers into the chain on a 4 motorcycle. Ended his piano playing.
5 DR. SCHROCK: To wrap this one up, I think what 6 you need to do is recalculate your numbers and you need to 7 state in the documentation the method for sting the 8' numbers, and I can't tell you how much change there is going 9 to be in the numbers for the conditions that you have in i
10 this particular application, but it is again an example.of 11 what I have said on a number of these things.
12 The documentation isn't doing what you are saying-13 the documentation is doing and I think you are left in a 14 weak position because of that.
O
!, j- 15 MR. BROWN: That's all I have, Dr. Kress, unless 16 you have any other comments from anyone.
17 DR. SEALE: Clear the room.
18 CHAIRMAN KRESS: Where are we on the agenda then? l 19 Item C-3.
20 This is a closed part of the meeting, with Mike 21 Young, and so at this point we will have to ask people who 22' are not supposed to be in here to leave and I guess it's up 23 to Westinghouse to verify that -- and this is now a closed ~
24 part of the meeting.
25 [Whereupon, at 11:47 a.m., the meeting proceeded 3 ANN RILEY & ASSOCIATES, LTD.
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20 21' 22
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144 1 AFTERNOO.N SESSION
- '\ 2 OPEN SESSION lU 3 [1:05 p.m.)
4 CHAIRMAN KRESS: We'll start back again. Mike, j 5 the floor is yours. .We are doing pretty good on the agenda, l
l 6 I think.
7 DR. CATTON: Tom, I would like to correct 8 something I said earlier.
9 CHAIRMAN KRESS: Okay.
10 DR. CATTON: My comments about the friction losses l
11 in the line from the pressurizer to the IRWST are incorrect.
.12 Near as I can tell, in looking through Banerjee's report, he 13 made the same assumption that Westinghouse made, which is 14 that the pressure can be set by the IRWST.
y-~ \
(f 15 CHAIRMAN KRESS: Okay. Appreciate that 16 correction.
17 DR. CATTON: I am going to look again through this I 18 to be sure. But on the surface, that is what it looks like.
19 DR. SCHROCK: But that doesn't make it right.
20 [ Laughter.)
21 MR. YOUNG: Okay. Let me go through just a quick overview of what we wanted to talk about this afternoon and 23 tomorrow morning. If I go through the ten issues --
24 CHAIRMAN KRESS: I don't think our consultants I i
25 would agree that we could strike out Item 2 yet.
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- l. 145 i
1 MR. YOUNG: I should have used the marker here.
'[ ) 2 DR. ZUBER: And Item 7.
'V 3 DR. SCHROCK: There are two parts to it. One is 4 the constant entropy and other one is how you get the state 5 at the point of choking in order to invoke that equation.
l 6 Because that equation is dependent upon the state of the 1 7 fluid at the point of choking.
8 CHAIRMAN KRESS: That's Item 9, isn't it, Virgil?
i 9 DR. SCHROCK: That's Item 9. Okay, well, they go 10 together.
j 11 CHAIRMAN KRESS: Yeah. They go together. !
i 12 MR. YOUNG: Okay. This was how we calculate the l 13 quality just upstream of the --
14 DR. SCHROCK: Actually, 9 has more in it than just
! /~N 15 that one point.
l! ,) But the point that I was making is that j 16 your calculation of critical mass flux gives numbers that 17 you put in these reports, has two parts to it that you need i
18 to explain further. One, you need to correct -- get the 19 derivative at constant entropy instead of constant enthalpy. !
20 The other one is that you have to show how you know what the 21 state is at the point of choking. That comes, I believe, l 22 from your code assumption that critical pressure ratio is 23 universally .58, which is not valid.
24 If you have some other way of getting it, then 25 explain that. But from what I have heard in previous
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146 i
t 1 meetings, I think that must be what you have done there, II) 2 MR. YOUNG: All right. Let me --
' \_)
'3 DR. SCHROCK: Mr. Brown did the calculation. I i 4 guess he ought to be able to tell you how he got the numbers
'5 'that are in his table.
j 6I MR. YOUNG: All right. Those are -- it is l
7 completely separate of NOTRUMP. That is simply application 8 of HEM. Is that correct, Bill?
9- MR. BROWN: Yes.
10 MR. YOUNG: Using what as the upstream condition, i 11- the reservoir condition?
12 MR. BROWN: Using the system pressure that we have
- l. 13 at the initial condition. And then, as you have said 14 before, Professor Schrock, we.first -- in the scaling
- g Q-15' report, not in NOTRUMP, but for scaling purposes, we have j 16 used, you.know, critical pressure ratios of the order of a-3 17 half, as you have said, to get the choke point. i l
18 HDR . SCHROCK: What do you mean of the order of a i 1 50 half? You had to use a number. .58 is what you have got in 20 NOTRUMP. Do you have another number you used some other 1 21 time?
22 MR. BROWN: Again, keep NOTRUMP separate from what 23- we did in scaling. All I am saying'is in scaling I used 24 roughly a half of critical pressure ratio relative to my
~
i 25 upstream or stagnation pressure to estimate what the
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147 1 critical mass flux is out through the critical flow path.
(,n) 2 But as far as NOTRUMP is concerned, that's a v
3 separate -- that's a separate issue. Mike could address 4 that, but I -- that's not what I have done.
~5 DR. SCHROCK: Okay. Well, see, my point of view
~
6 is this, that if you say in your documentation, what you 7 have done in your scaling report, that you have used the 8 homogeneous equilibrium model to get a number that you have 9 put in the report, and you expect the reader to understand 10 .what is the homogeneous equilibrium model, so-called, then, 11 clearly, you have made a mistake, because people who are 12 knowledgeable on that would not accept what you really did 13 in order to get your number as being the correct evaluation 14 for that critical mass flux, according to the stated model.
A-(_,), 15 Okay.
16 And while your number isn't grossly in error, my 17 objection to it is that your documentation just sort of i 18 flaunts good engineering practice. If you don't calculate 19 it, according to what is a well known procedure, then don't i 20 say that is what you do. j 21 CHAIRMAN KRESS: Do they need to fix the 22 documentation or --
23 DR. SCHROCK: They need to fix the number that he 24 'has in his table, and they need also to explain what they do 25 -with regard to the application of the homogeneous l
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l, 148 1J . equilibrium model.
l 42 ~ You have not solved the equation that you have r
'3; taken from-Wallis' --
!: 4 MR.. BROWN: No.
l ~5 DR. SCHROCK: -- as the reference to lead to an 6 evaluation'of critical mass flux. What you have done l
! 7 'instead is to'go to the bottom line,' but the denominator in t.
8 that equation, at'the point of choking, goes to zero owing.
'9; to the: fact that.the term'in brackets physically represents
~
i 10 1.minus the square of the MACH number. When the MACH number l
111 .becomes unity, that becomes zero. And now, if you calculate 112~ the1 critical mass' flux in from the fact that the MACH number L
113 is equal to unity,.it means that the critical. velocity;is 14 the sound speed, then you are evaluating the sound speed at j-
. y/.
15 the state at the point of choking, not at some upstream
! 16- condition. .So'you have then.the obligation:of providing-the 17 link between'the upstream stagnation state and the state at-l
!' , 18 the point.of choking. Discuss that.
- 19- MR. BROWN: In the scaling report itself, what.you 20 saw in the table versus the equation 3-63, after I went
- 21 ' .' i through 3-63, for the homogeneous' equilibrium model, I l 22 ' simply went to a tabular graphic form such as published in l 23 Lahey;& Moody to actually.obtain the values for the critical 24 . mass flux.at that point and did not actually use the
$25 ' equation.in this_ latest revision.
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149 1 DR. ZUBER: How did you know then the upstream 2 condition?
3 DR. CCHROCK: In that case he doesn't need it 4 because they -- I am not sure which reference you used, but l
l 5 there are such graphs that have been published that give it 6 in terms of the upstream condition.
7 MR. BROWN: Right. Correct. And that's -- yes.
8 DR. SCHROCK: And where the calculation -- but, on 9 the other hand, that is for an isentropic flow, it is not 10 for a fanno flow, not --
11 MR. BROWN: Right. And that is what has been done 12 in the -- this update, this latest update that is presented 13 today is directly from Lahey & Moody's text for the 14 hor.ogeneous equilibrium curves, based on upstream
() 15 conditions.
16 MR. BOEHNERT: Excuse me, Mike. Mr. Chairman, I 17 have to make a housekeeping chore here. I have to let them 18 know we are in open session since after lunch. Okay. Good.
19 Sorry about that. Thank you.
20 DR. ZUBER: Item 7. I don't consider, as far as I 21 am concerned it is not a closed issue. It is still open 72 issue.
23 MR. YOUNG: Okay. Let me define this as things we 24 have talked about.
'25 DR. CATTON: 5, 6, 7, 8 are the issues. I mean it
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1 l 150 1 is really 5 is the issue. 6, 7 and 8 are subsidiary.
l-
'[ T 2 MR. YOUNG: 6 is --
I %.-l i L
l 3 DR. CATTON: 6 is the flow through the surge line. l
' 1 l 4 7 is that you have enough information. And 8 --
5 MR. YOUNG: That's -- these -- I'll show you why I 6 think these fold into the scaling analysis and how that as 7 related to our application of a level penalty. All right, i 8 But let's assume right now that these issues 9 aren't closed just because they are crossed out. Let's see 10 where we end up with --
11 DR. CATTON: Crossed out means we have talked 12 about them.
13 MR. YOUNG: Crossed out means we have talked about 14 them.
IN
( ,) 15 DR. ZUBER: What I would like to know, since you i l
16- want to put everything to rest, I would like to know where .
17 we stand, not what we talked about.
- 18 MR. YOUNG: Well, clearly, --
19 DR. ZUBER: Let me say it the way I see then. You
- 20. may, okay, or not. We have two analyses on scaling, one by 21 Wolfgang, one by Banerjee. I think Wolfgang has many 22 questions concerns the distortions, especially in that l 23 particular period of time, which need to be addressed. And i 24 I think it may be advisable that ACRS ask either Wolfgang or l 25 Banerjee to present their results and present their l
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151 p'
[ 1 . conclusions, have the input from NRR. At least we know what 2 one side _of the group thinks.
[']
N.
3 CHAIRMAN KRESS: We ought to be able to do that r
4 even though they have this proprietary issue. We can do it 5 in closed session.
l 6 DR. ZUBER: Because if he is right, we should not 7 . hold him back, Westinghouse. If'they are right, then you 8 had better address it.
9 MR. YOUNG: I' wanted to just quickly return to how 10 we talked about these things, these subjects before. As far
! 11 as NOTRUMP comparisons to test data, there is a -- we have 12 in several of the previous meetings talked about the poor 13 -prediction of ADS-1 to 3 flow and the break flow, and some 14 of the OSU tests. And that led to the question ofc well, is
,~.,
i 15 there.a compensating error that we need to watch out for 16 here. So we are going to show some mora comparisons and I 17 think -- I am going to try to summarize for you here where I 18 hope we will be going this afternoon.
19 Yes, there are compensating effects, but we think 20 we understand pretty clearly now why what is happening'here 21' is happening. I think the more important question is this 22 early draining of the pressurizer. And that leads to then 23 early actuation of IRWST and the OSU test predictions. And 24 that has been discussed in previous meetings and led to.the 1
25 issues that you see there. )
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152 1- As far as the NOTRUMP modeling of important
[^}
V 2 phenomena,.this is obviously of importance. We have talked 3 about this quite a bit. Entrainment and treatment of l 4 . flooding in the surge line, as I see it, are all I guess l 5 what I would call lower level or detailed models that are 6 needed for that, essentially-to figure out whether the OSU i- 7 is scaled properly.within this time period.
8 This one is a little bit new to me. We talked 9 about this in July, and I plan to go back over what we did 10 to validate this model that we have for horizontal flow.
11 And then, finally, application of IRWST level 12 penalty, address NOTRUMP deficiencies. Let me talk a little 13 bit about what I saw out of the scaling analysis. And to me 14 the most important thing was the ranking that that analysis 0)
(,, 15 provided for the importance of each of these components.
16 And it shows clearly that ADS-4 resistance, the resistance 17 to flow out of that valve is really the dominant term 18 ' limiting'or delaying the onset of IRWST. You have other 19 ones below that and, in fact, the pressurizer level is i 20 there, but it is somewhat lower in importance.
21 .And I think what we-are. going to show are some 22 calculations that-indicate that if, in fact, you got the l 23- flow resistance through ADS-4 correct, everything else seems _j 24 to fall into place. All right. Now, we will have to wait 25- :until tomorrow morning-before we get those types of l
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i r i
1 153
-1 calculations.
L
() '2~
But_that's what the scaling analysis told me, was L 3 that we should have beenffocusing on'the resistance through 4' ADS-4Eall along. And that, of course, goes back to properly L 5 calculating the pressure drop through that line.
6- DR. ZUBER: Mike, I like your. presentations. No , .
7' really, I --
J8' MR.. YOUNG: But I want'to.make the point thatI am 9; 'not sure.that the message of the scaling analysisEcame.out-10 the way --' at least I saw it when I first, you know, read 11? th'e results.
121 CHAIRMAN KRESS: Mike, let me ask you about that.
13 'If,-indeed, the flow out the ADS-4 is, say, dominant,1and -
that flow is driven by resistance in that line, and the head d ~ 15L ~ of the IRWST down to that sort of level, isn't the
-a 16 relationship'between!the flow and the head a square. root 17' relationship?- And doesn't that -- doesn't question you use 18L of:a linear penalty?
19- MR.< YOUNG: I will talk about how the level 20 ~ penalty can be used to address that effect. All right.
21' DR. SCHROCK: Mike, as I was reading this new 122 ' documentation about this, it recalled to mind an aspect of 23 -- interpretation from Wallis' description there about 1
-24 . flashing. He used the term flashing in a more general 25 sense, he meant phase change. And in the region where you
-l l
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l 154 l
1 are working on this ADS-4 problem, there wouldn't be
] ) 2 flashing, there would'be condensation. It's like expansion 3- through nozzles and steam turbines, if you begin with 4 superheated steam,.you may well'get through the expansion 5- without any condensation even though equilibrium would take 6 you into the 2-phase domain.
7' So you have another non-equilibrium problem which
~
8 is raised in the vicinity of saturated vapor boundary just 9 as you do in the vicinity of saturated liquid boundary. The 10: problem for ADS-4 that we have never discussed, I' don't know 11 whether.it is a serious one or not a serious one, but the 12 frictional effect in the pipe between the pressurizer and 13 the valve creates superheating. To follow a constant 14 enthalpy line, it's like the principle of the old throttling
.f~)
( ,J. 15 calorimeter, you go away from the 2-phase region into the
- 16. superheat region as you throttle the mixture, as you have 17 pressure loss due to friction.
18 So you are going out into the superheat region.
19 Now, when you get to the valve, you are expanding ideally 20 isentropically through the valve, you cross the saturation 21 boundary again. It's following a saturation -- or constant 22 entropy line, you are going down into 2-phase once again.
223. So now there is an issue, does the superheat that 24 is generated succeed in getting rid of all the nuclei for 25 condensation that would enable equilibrium to persist once ANN RILEY & ASSOCIATES, LTD.
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155 1 you are expanding isentropically within what ought to be a
/~ \ 2 ~ 2-phase region. It probably ought to be looked at.
L O.
! 3 MR. YOUNG: All right.
4- DR. SCHROCK: Did you follow what I said?
5' MR. YOUNG: Yeah, I think I did. For essentially 6 a fanno process, where-you do have friction, you would 7 expect, ' essentially, quality to increase with link.
l 8 DR. SCHROCK: Well, by your arguments it is pure t
9 vapor arriving there. It was pure vapor leaving the 10 pressurizer. And if it was saturated steam leaving the '
11 pressurizer, then, in fact, it is inevitably going to 4
}
12 superheat in the pipeline downstream, f1 13 MR. YOUNG: All right. But we are talking about a 14 very small pressure gradient, it's a tenth of a psi, l f, % ,)- 15 basically.
! 16 DR. SCHROCK: Tenth of a psi pressure loss?
l 17 MR. YOUNG: Through ADS-1 to 3, during this time l
18 period.
19 DR. SCHROCK: I am not talking 1 to 3, I am 20 talking 4.
21 MR. YOUNG: Oh, 4 is -- I'm sorry, you said 22 pressurizer, so I thought you meant ADS-1 to 3.
23 DR. SCHROCK: Well, I said the wrong place when I 24 said pressurizer. You're right. I 25 MR. YOUNG: Okay.
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156 1 DR. SCHROCK: Excuse me.
(~'J 2 MR. YOUNG: Okay. So ADS-4 --
N_--
3 DR. SCHROCK: So if it.is wet steam going in 4 .there, then it is not an issue. If it is -- but, on the
- 5. other hand, if you are calculating that it is -- no, I think 6 it is probably not an issue. Because I was thinking in 7 terms of pressurizer and not in terms of the hot leg.
8 MR. YOUNG: Yeah, I think --
9 DR. SEALE: ADS-1 gets it all.
10 DR. SCHROCK: ADS-1 through 3 is where it would be 11 a problem if it is a problem.
12 MR. YOUNG: Right. And then, finally, noting 13 differences between test and AP600, we will talk about those 14 as well.
) 15 Again, we went through the scaling analysis. I am 16 hoping to bring that.more into focus or relevance with 17 regard to NOTRUMP as we walk through these. And then we 18 will talk about-the NOTRUMP calculations. I will be 19 referring to sensitivity studies to demonstrate certain 20 things here. That will be this part of the presentation.
21 Okay. If we move on to the momentum flux issue.
22 Again, let me just summarize what we presented in the last 23 meeting. This is'obviously well known. The NOTRUMP model l i
24 for this piping ignores the momentum flux terms. We made 25 estimates of the.effect of not having those terms on the lN ANN RILEY & ASSOCIATES, LTD.
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157 1 prediction of pressured up and flow rate and found that for ADS-1 to 3 it was a relatively weak -- by weak, I mean a 10
~
2 percent effect, at most, due to the substantially smaller 4 area at the valves.
5 In ADS-4 it is clearly more significant, so I am 6 going to focus on ADS-4. Maybe if we get through that we 7 can go back and talk about what this 10 percent effect might 8 mean for ADS-1 to 3.
9 .Now, we did go through our application of the 10 critical flow model and I think we concluded that it was 11 reasonable. I will return to that again after looking at a 12 detailed model to see how NOTRUMP stacks up. The NOTRUMP 13 model is clearly much simplified compared to this more 14 detailed model I will show you.
(a); 15 DR. SCHROCK: Are you saying that your view of 16 what happened at the last meeting was we agreed that the 17 critical flow model was reasonable?
18 MR. YOUNG: I heard Dr. Catton say it was 19 reasonable.
20 DR. CATTON: I did?
21- (Laughter.)
22 MR.' YOUNG: I suspected that no one would remember 23 that so I did bring in some of the old --
24 DR. CATTON: You brought some of the transcript?
25 MR. YOUNG: No, I brought in the old everheads.
l
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u_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __ _ _ -. ]
158
- 1 We'can go over that again, all right, just to refresh our b'i LJ 2 . memory.
3 DR.'CATTON: The problem is everything is 4 relative. If you talked about the critical flow right after
5 something that was really bad, I might have said it was
-6 reasonable.
7- MR. YOUNG: Good it retrospect.
8 DR. ZUBER: Let me ask you, out of curiosity, I 9 forget. Why did you go back to the technology of 1971 or 10' '72. You know, momentum fluxes were not present in RELAP 11 3-B and then we had that big flop in '73 and then we started 12 to develop RELAP 4, eventually RELAP 5. But momentum' fluxes 13 were important immediately because for the large break LOCA 14 and the blowdown.
15 MR. YOUNG: Well, --
16 DR. ZUBER: No, really, you went back, you turned 17 the clock back to '71, almost 27 years.
18 MR. YOUNG: Well, we had trouble even back then 19 dealing with momentum flux,.its proper application in those
.20 codes that-had very large node sizes. You have to be a 21L little bit careful, especially.when you are talking about 22 flow direction changes and so on. This was an off-shoot of
- 23. a snall break LOCA where we concluded there that velocities 24 were very-low, but we forgot about ADS-4. You know, 25 clearly,'they are not there.
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l 159 1 DR. ZUBER: Yeah.
l (-s
%)
i 2 MR. YOUNG: So you are right, we dropped -- well, 3 we dropped back to NOTRUMP, which we had shown for regular 4 plants,. momentum flux is not an important term.
l 5 Okay. The detailed model essentially involves 6- solving the momentum and energy equations under the 7 assumptions you see there. It just shows steady state 8 equilibrium, homogeneous and adiabatic conditions. I had 9 shown previously that if you introduce slip into the 10 calculation, that would tend to predict-a smaller preesure 11 gradient along the pipe. So it appeared as if a homogeneous 12 assumption would give you a relatively high pressure !
13 gradient as a result of acceleration through this pipe.
14 Taking these two equations, and after some
) 15 manipulation, you end up with --
16 DR. ZUBER: You see, the point is if you have no 17 phase change, but if you have phase change also, due to 18 flashing or heat addition, these du -- dz really becomes 19 important. I mean as you accelerate the flow because it 20 expands.
21 MR. YOUNG: Even if you don't have phase change.
22 That's right.
23 DR. ZUBER: Even if you -- but, particularly, if 24 you have phase change, then it becomes really important.
25 MR. YOUNG: That's right. '
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160 1 DR. ZUBER: And this is with flashing or 2 something.
3 MR. YOUNG: And I think there is a plot where we 4 can see that effect becoming increasingly more important.
5 But, anyway, after rearrangement, you can 6 integrate these two along the pipe network then to calculate 7 the pressure distribution from the hot leg up to the valves.
8 DR. SCHROCK: The eqration in the middle here is 9 this, equation -- no, in the middle.
10 MR. YOUNG: This.
11 DR. SCHROCK: Right there. That's 3-63 from 12 14727, except that here it's one that had been presented in 13 the older version and now we have heard that that has been 14 modified to include the phase change term. Phase change
() 15 term is missing here.
16 MR. YOUNG: No, this is the phase change term 17 here. What I include in this calculation is the effect of 18 the quality change.
19 DR. ZUBER: Okay. What is B?
20 MR. YOUNG: Well, B is this term. I just --
I 21 simplified this so I could squeeze it in here.
22 DR. SCHROCK: Could you get Mr. Brown's viewgraph 23 that has equation 6 -- 3-63, and lay it on the other -- you 24 don't have another --
25 MR. YOUNG: No. Yes, there's another term that O ANN RILEY & ASSOCIATES, LTD.
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161 l-1 has the partial equality.with respect to pressure at I 1 - j
[ j 2 constant'enthalpy. Is that the one j
4 v
3 DR. SCHROCK: It is not here. '
4 MR. YOUNG: Well, this is not an isentropic 5 process here. This is --
- 6 DR. SCHROCK
- Neither was that.
7- MR. YOUNG: All right.
8 DR. SCHROCK: What we are talking about here now 9 is just the general momentum equation for'a general channel.
10 It may or may not have area change, it surely has friction, 11 et cetera.
12 MR. YOUNG: Yes.
13 DR. SCHROCK: Gravitational effects. And so now 14 what I am pointing out is your denominator is not -- it does
[^)\
.. 15 not account for flashing.
16 MR. YOUNG: If I -- the quality as it moves down 17- the pipe is changing.
18 DR. SCHROCK: . Mike, you are dealing with exactly
- 19. the same equation as the one that he used in the scaling 20 report. Now, they can't both be right and be different.
21- :They are different.
22 MR. YOUNG: Well, I started from these. All
]
23 .right. I didn't start from Wallis' textbook, I started from 24 these. And you simply take this, expand the differential.
I 25 This enthalpy is hf plus quality times Hfg. And it is just li
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162 L
p 1 simply a rearrangement of these two equations.
i j-%) 2 L ;V Now, there may be another step that relates this 3 to.the Wallis equation, I have not done that.
4 DR. SCHROCK: Well, this term in the denominator I 5' is different, the bracketed term is not the same as the.one 6 that Mr. Brown'showed, and it is not correct.
7 MR.. YOUNG: Well, I don't agree with that, right
! 8 off the bat, but I can't tell you why.-- at this point, why 9- this term should be'different.
10' DR. ZUBER: Why don't you put both? I mean we 11 have both these' equations. You can explain'what is the-
?2- difference.
13- MR. YOUNG: Well, maybe the thing to do is to go i
14 back. We can compare the two equations and come back. We l (r T) 15 have tomorrow morning to look at this one more time.
l 16 DR. SCHROCK: See, your bottom equation here --
1
-17 the equation on the bottom says the partial derivative'of V 18 with respect to P is equal to -- now, you have terms that 19- are ordinary derivatives, which is certainly correct.for the 20 saturation properties,.they depend only on pressure. You i 21- don't say in that equation how you view that partial 22 derivative. What does it represent?
23 MR. YOUNG: This?
24 , DR . SCHROCK: Right.
25 MR.1 YOUNG: Well, it's partial because the quality
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i 163 1 . derivative'is not completed yet.
[\_t)- 2 DR. SCHROCK: Oh, no. No, no, no. There's-3 nothing. The quality is.a~ point function. It's not 4 anything having to do with the path that is being followed
~
5 for that differentiation. That shouldn't be the partial 6 derivative. How did you. introduce it into the equation as a 7 partial derivative?. All your previous equations are L 8 ordinary derivatives?
9' -MR. YOUNG: 'Right. This term here, which is equal 10 to this, is this term right there. All right. Now, the 11 quality is changing, an incremental distance the quality 12 will be changing. This is simply the energy equation here.
13 And that quality then, the effect of that on the momentum 14 equation is contained in this term.
, 15 We can go back and try to relate it to the Wallis I
.16 equation.
-17 DR. SCHROCK: Well, I don't think we need to go 18 back. Before we leave the meeting, I think we can resolve !
19 it. ,
20 .MR. YOUNG: Okay.
21 DR. SCHROCK: It's unclear to me why you would 22 .want to present the same equation two different ways when it 23 all relates to the same issue here. Issue No. 2 is the
- 24 applicability of equation 3-63. Admittedly, it is 25 specifically to the scaling, but I am amazed that you limit I
l . .
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h 164 1 it to that. I mean the equation is the equation, you are 2 using it in other ways now, talking about NOTRUMP.
l 3 MR. YOUNG: _I simply decided to solve the energy 4 and momentum equations this way, that's all.
l 5- DR.'SCHROCK: If you do it this way correctly, you i 6 get the equation ~which is given in Wallis's book as 2-44 --
l 7 MR. YOUNG: That may well be the case. I haven't 1
8 done that comparison yet. (
- 9 Now I decided to try to independently check the i 10 other simplifications that were done in the NOTRUMP 11 calculation, one of which was to use a fairly old two-phase 12 friction multiplier, Martinelli-Nelson. There are obviously 13 more recent models and correlations. This one is applicable 14 to not just friction conditions but with various values of C q
i h 15 it could also be applied to fittings and bends and so on.
16 So the two-phase frictional --
17 DR. ZUBER: Is this Chisholm?
18 MR. YOUNG: Yes, this.is Chisholm, Dewey Chisholm.
'19 And the. frictional expression is in HTFS, but it's 20 the same --
21 DR. ZUBER: I'm sorry to interrupt you, but since l l
22 we're on this subject, what do you use in COBRA / TRAC?
23 MR. YOUNG: For a two-phrase friction multiplier?
24 DR. ZUBER: Yes. How do you model? How do you !
i 25 model a frictional pressure drop in COBRA / TRAC? i l'
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165 1 Are we going to discuss COBRA / TRAC this meeting?
l ) 2 MR. YOUNG: No, we weren't planning to. But it's k m 3 not -- it's a simpler formulation. It's actually derived 4 from something that Wallis has. I think I talked about this 5 some years ago. It's a significantly more simple 6 formulation.
7 DR. ZUBER: But do you' partition --
8 MR. YOUNG: Depending on the flow regime. If, for 9 example, you're an annular flow, all right, and the 10 frictional force is applied to the liquid field, all right, 11 if you're in dispersed droplet vapor flow, then the force 12 from the wall is applied to the vapor.
13 DR. ZUBER: What do you do in slug flow?
14 MR. YOUNG: Slug flow, it's liquid. We don't have
() 15 a transition. We just assume it's liquid unless it's 16 dispersed vapor flow.
17 DR. ZUBER: Okay.
18 MR. YOUNG: And there are other equations, again 19 trying to take the best from what's available. For pressure 20 changes through a pipe as it passes a tee, there's a change 21 in the momentum as the tee draws some vapor mass off, and 22 there's also a loss term.
23 For these remaining equations and figures, 24 basically 3 represents the branch pipe; 1 is the main pipe 25 upstream of the tee; and 2 is the main pipe downstream of 1
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l l- 166 )
1 the T.
,/ 't ' 2 DR. ZUBER: And X-1 and X-2? 1 V-3 MR. YOUNG: X-1 is the quality upstream of the 3
4 tee, X-2 is the quality downstream.
5 DR. ZUBER: Okay.
6 MR. YOUNG: And then the other thing I wanted to 7 try to see if there's anything important here is the flow 8 split. The ADS piping -- and let me just quickly throw that 9 . figure up here so we can know what we're talking about --
10 .looks like this.
11 This is the hot leg, and this is a typical. setup, 12 a bend with a fairly large pipe goes to a tee, horizontal, 13 where the flow splits. You have one ADS pipe and then a 14 second ADS pipe coming off horizontally and then going like 15 that.
16 So the question that might be asked is well, if 17- you have phase separation there, what might happen to the
'18 total flow rate, and is.that something we have to worry l 19 about. And again there.are some correlations available that 20 basically relate the branch quality to the upstream quality 21 in terms of the mass velocities.
22 Finally, at the end of this. pipe in which we saw l 23 the momentum and energy equations, and as you'll see we'll j 24 _ subdivide it into some detail, we applied HEM then at the 25 point in which you had the ADS squib valve or the valve in l
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167 L
1 which the critical flow' conditions were likely to occur. '
s LlL) 2 And we use then the total enthalpy and entropy at that last 3 cell just upstream of.where the valve is and then applied
- 4. HEM to that. So at that point we assumed adiabatic 5 frictionless flow. Up to that point then it was adiabatic
'6 but not frictionless.
7 We tried to bench-mark the model to the 8 incompressible total loss coefficient,~and the way I did ;
l i'
9 that was simply to run at a very low pressure drop so that 10 the only terms that were important in the momentum equation 11 were. essentially the irrecoverable loss terms. Now because 12 of some of the differences in the treatment of losses in the 13 tee and -- I think it was mostly the Tee -- we had to adjust 14 the total resistance, the irrecoverable resistance, by about
-( ) '15 ten percent to get agreement with this incompressible' loss 16 coefficient'that we use essentially in NOTRUMP, which is i
17 derived from essentially upper-bound calculations. And then 18 we calculated flow through ADS-4 as a function of pressure 19 and quality.
{
20 Now let me walk through what this calculation 21- produces. First of all for integrating the equations this 22 just gives you an idea of how the area changes within ADS 4, 23 and you can see for one thing that the piping that's assumed l 24 in the calculation of a resistance that we use in NOTRUMP is !
25 very long. It's 45 feet. And that's basically an i
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168 1 upper-bound type.of configuration that one would expect.
() 2 3
The typical configuration you saw there was about maybe 10 or 15 feet. So you have a piping that is this -- this is l
L 4 square feet -- this area. You then have a tee that draws 5 off one branch. This pipe continues to a reducer, and that l
6 goes to the other branch then, and that continues to two 7 gate valves on each branch, and finally the squib valves at.
8 the end of the piping.
9 There are about 400 cells or so in this 10 calculation, and more in the gate valve.
11 DR. SCHROCK: The gate valve -- that shape that 12 you're showing there --
.13 MR. YOUNG: Well, this is a --
14 DR. SCHROCK: Two restrictions.
-']
/
( ,/ 15 MR. YOUNG: Yes. Unfortunately this overlays the 16 two branches.
17 One of them is on 1 and one is on DR. SCHROCK:
18 the other?
19 MR. YOUNG: Yes.
20 1 DR. SCHROCK: Okay. I see what you're saying.
21 MR. YOUNG: So if we look first at the static ;
22' pressure in the pipe for a hot-leg pressure of about 50 psi 23 and 100-percent steam quality, you start out with a fairly l 24 substantial pressure drop already as you accelerate into the j i
25 . piping, and then in this region, there's a further pressure ]
i~ ;
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!. 169 1 drop, and that's due to friction and the bend. You'll.see l.
, ~.
.2 1
( }- the velocities here are pretty high.
L 3- There's a pressure drop, then a recovery passing-4 the tee, and then you can.see the pressure dropping, and L 5 these steps here are-the various elbows that are assumed in 6 the model. But the pressure drops from about 50 to perhaps l 7. 35 -- 35 psi before it gets to the squib valve. That's a 8 fairly significant' pressure drop,-all right? And the 9 question then is how much of this is friction and how much 10 .is all the other terms, and we can look at that a'little bit
.11 further.
12 Finally the squib valves are down here, so there's 13 still critical flow at-the squib valves. That's where the 14 minimum area still limits the flow.
O
( ,f 15 DR. ZUBER: May I just interrupt. Sorry. How do 16 you do it in RELAP? How do you model that, or did you 17 model?
18 I'm sorry, but you know this is the same problem, 19 what tool do we have to address this problem, RELAP 20 versus -- how do you model this problem in the RELAP?
21 MR. KELLY: Okay. I'm a little rusty-on this, 22 because I haven't looked at -- this is Joe Kelly from 23 -research. I haven't looked at the RELAP calculations in the 24 -AP600 in a while, so I'm a little rusty. I was not very i 25- familiar with the plant model. I was much more familiar l'
[' )
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I 170 1 with the way we model the experimental facilities.
'fgj ) 2 And if you're talking about the critical flow in 3 the experimental facilities, they were always across orifice 4 plates?
5 DR. SCHROCK: Could you say that again? I don't 6 understand.
7 MR. KELLY: In the experimental facilities, their 8- critical flow for ADS 4, that was always modeled as being an 9 orifice plate, and those are the plant input decks that I 10 would be most comfortable describing to you. Anything else, 11 I would have to go and get an answer and come back.
12 DR. ZUBER: My concern is really how do you model 13 the pressure drop and the problems associated with ADS 4, and 14 sometimes maybe in the future we should really look into it.
/-
(x) v 15 MR. KELLY: In the input decks they did model the '
16 piping as it was -- well, for the experimental facilities as 17 it was built, you know, including loss coefficients, not 18 only the frictional, you know, L over D's for the pipe, but 19 also for the elbows and so forth.
l 20 DR. ZUBER: Okay. Thank you, i 21 DR. SCHROCK: This illustrates that your reduction 22 in area at the valve is only about 15 percent for the ADS 4.
23 .In the previous one.that showed the area it shows that your 24 area of reduction's only about 15 percent. Then this graph l
25 is indicating that the pressure drop across the
~
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171 1 critical-flow device is.in-fact very small. It's a little
~
( '
- 2 hard to read from -- but in 35 it's not more than 2 and L/
3 probably less than 2, 2 psi out of 35.
4- MR. YOUNG: You're talking about the gate valves.
5 These?
6 DR. SCHROCK: Right.
7 MR. YOUNG: Yes.
8- DR. SCHROCK: .Yes. And if I understood your 9 explanation correctly, what you've done is to calculate the 10 friction flow in the constant-area pipe up to the entrance 11 Eto the gate valves, and from that back-calculate the 12 stagnation properties at that point, and then use critical 13 -flow tables or calculate critical flow as a function of the 14 stagnation properties without solving.the isentropic flow h-3 ,) 15 .through the valve. Is that right?
16 MR. YOUNG: No. What --
17 DR. SCHROCK: Not right.
18 MR. YOUNG: Right. The application of HEM, and
'19 ' it's a direct application of that model, is applied here.
20 fki this is the cell at which you take the enthalpy and 21 entropy, and then assume an isentropic process down to the 22 squib valve. Okay? Here we're applying the momentum and 23 energy equations, as you saw them. ;
L !
24 Now if this area were in fact substantially 25' larger, all right, then the model would predict thac you ANN RILEY & ASSOCIATES, LTD.
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172 1 would eventually get choke flow here.
s~
-/ h 2 DR. SCHROCK: I guess I don't understand what-
.V-l 3- squib. valves means. Isn't that a separate. valve.that you T4 use to trigger the experiment?
5 MR. YOUNG: No , no - -
l 6' DR. SCHROCK: Something that exists in the plant?
l 7 MR. YOUNG: This is the valve that basically 8 controls the flow. process. The gate valve-is what, just l ;9 .something to" isolate the.--
10 MR. BROWN: Yes -- Bill Brown. The gate valves 11 upstream,. Professor Schrock, are just there for double 12 isolation. They're the ones that --
- 13. .IMt. SCHROCK: So the critical flow is generated at 14 the squib valve.
); 15' MR. BROWN: Yes, that's correct.
16 DR. SCHROCK: And.it's a much greater area of
~
17 reduction.
18 MR. BROWN: Yes.
19- DR. SCHROCK: Than this 15 percent. The gate 20 valve. puts a 15-percent reduction on it, and that_-- through 21- 'the gate valve you're still using your model based on these L 22' equations that you presented.
h 23- MR.-YOUNG: Right.
'24- DR. SCHROCK: Then when you get to the squib 25 valve, what do.you do?
l
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173 1 MR. YOUNG: Assume an isentropic process from that 2 point on.
3 DR. SCHROCK: Assume an isentropic process.
4 MR. YOUNG: And then maximize the mass velocity to 5 get-those.
6 DR. SCHROCK: And how do you calculate the 7 critical flow for'the isentropic process?
8 MR. YOUNG: Energy equation and constant entropy.
9 DR. SCHROCK: How do you apply it in a practical 10 sense? I mean, one'can do the calculation stepwise lower 11 and lower pressures and calculate the mach number at each.
12 station and find where the mach number becomes unity, that's 13 one approach. Another is to say I know the critical 14 pressure ratio and immediately jump to the calculation of
'() 15 change in kinetic energy associated with that pressure 16 change.
17 MR. YOUNG: Right.
18 DR. SCHROCK: So which is it that you do?
19 MR. YOUNG: What I did was I took this equation.
.20 This can be converted to mass flow rate and a density term, 21 so the density term becomes a function of thermodynamic 22 conditions, T&H, if you will, or P&S, sorry. So you have 23 essentially an equation that you can solve for mass flow I
24
. rate or mass velocity as a function of the conditions P&S --
25 DR. SCHROCK: Right. So you can go down --
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- p. 1 l-p 174 1 MR. YOUNG: Right, you just march through --
'+
[( 2' DR. SCHROCK: . Calculating mass flux and'where does 3c it maximize- .
4 MR. YOUNG: Right, g
5 DR. SCHROCK: That's what you've done.
6 MR. YOUNG: -Right.
7 DR. SCHROCK: .Okay.
8 MR. YOUNG: Okay. Let's see where=I~-- oh, this 1
.)
SF is the fluid velocity, again for 100-percent quality --
10 DR. SCHROCK: I guess one last thing that's 11 confusing on these graphs, Mike, .this very heavy line that l 12 looks like a V on each of these on the previous graph here 13 it's inverted --
14 MR. YOUNG: Um-hum.
) 15 DR. SCHROCK: Are those hand-drawn afterwards, or 16 is that something-that your graphics is producing? What l
17: makes it go up again, take branch No. 1, for example. Now j 18 there*s a dip in it. It looks like it goes way up again and 19 then emerges out of that dark line on the previous graph. ,
L 20 MR. YOUNG: Previous graph which was what? Can 21 you remind me what that was, what was the quantity?
22 Pressure, or --
F 23 DR. SCHROCK: Pressure. Pressure. Do you see 24 what'I mean?. Each of -- each of those traces has these
- 25. little dashes describing it, and you get out here to gate p
i i
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175
- 1 - ' valves, and now what's it mean? Why is it suddenly dark.
[' ) 2. MR. YOUNG: Why does it suddenly drop?
< X_ /
3 DR. SCHROCK: And why does it overshoot?
'4 MR. YOUNG: Okay. What you see here, I'm sorry 5 that this was the only.way I could figure out how to plot 6 .this, you see the static pressure ~as it passes through the l 7 ' gate valve. You. accelerate, then decelerate,.the pressure I
8 recovers. I had to apply the loss coefficient-from the gate '
- 9. valve somewhere.
10: DR. SCHROCK: But you drew those on there with a 11 l pen?.
12 MR. YOUNG: No, no, no. That's -- that's 13 substantially more noting detail than the gate valve. So 14 those are-points where the pressure was calculated O
\, ,). 15 specifically.
16 DR. SCHROCK: What I don't understand is for'each 17 of those traces then why does the gate valve produce a i
181 pressure profile which is not--- which is overshooting on !
19 the recovery?. Why does it overshoot on the recovery?
i 20- MR. YOUNG: Well, no, it basically recovers. {
21- Here's -- there's this branch, you. drop, and you go back up.
22 DR. SCHROCK: Okay. And then it drops stepwise J 23 down to --
24- MR. YOUNG: Right.
25- DR. SCHROCK: -Some more stuff. Why?
1 (N\~)
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176 1 MR. YOUNG: That's the application of the loss 2 coefficient of a gate valve. I could have spread it through
[~')T L j 3 the gate valve. I just applied it at the end. So at the 4- end of a gate valve there's a sudden drop in pressure 5 because of the loss to the gate valve.
1 6 DR. SCHROCK: I understand it.
7 MR. YOUNG: This was velocity for 100 percent 8 quality. There is pretty significant velocity from the hot 9 leg to the first pipe prior to the branches, and this is 10 where -- remember, we first talked about velocities on the 11 order of 800 or 900 feet per second. So this is pretty 12 significant and results in a fairly large pressure drop 13 through this first part of the piping. The rest is down 14 here, but for 100 percent quality, velocities at the float i
.s,) 15 are still on the order of 1200 or 1300 feet per second.
16 If we look at a 20 percent quality case, this is 17 what we get, a somewhat similar pressure gradient, and you 18 can see the effect of the separation. I'll show you quality 19 plots for each branch in a second. But now we're talking 20 about 20 percent quality, so that there is a more 21 substantial effect of flashing down the pipe as the pressure 22 drops.
23 MR. SCHROCK: Your figure that -- the 50 percent 24 -- 50 psi and 100 percent quality --
25 MR. YOUNG: Yes.
[]
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177 1 MR. SCHROCK: -- if you look at the pressures that
() 2 3
you have there, these are absolute, they are plotted.
That's not a mistake, is it? Is plotting absolute?
4 MR. YOUNG: As for pressure, yes.
5 MR. SCHROCK: Okay. Then at the squib valve, you 6 have a pressure loss from about 34 -- maybe I've 7 overestima"4d that; it might be closer to 33 -- and it drops 8 suddenly to 22 where your arrow shows squib valves down 9 below. That gives you a critical pressure ratio on the 10 order of .66. So I just wanted to point that out to you to 11 underscore the comment I made before.
12 Your numerical calculation applying HEM has shown 13 you for that case the critical pressure ratio in the area of 14 .66. I'm surprised you would find such a high number for
() 15 that location, but thac's what your numbers give. Anyway, 16 it's a far cry from .58.
17 MR. YOUNG: That's true. The .58 assumption, 18 applying that for all conditions, is not -- I think I have a 19 critical pressure plot here somewhere, and maybe we can talk 20 about that some more.
21 DR. CATTON: Nhy does the pressure in branch 1 22 jump up?
23 MR. YOUNG: What you have is a pressure recovery, 24 because --
25 DR. CATTON: Oh, from the flow -- okay.
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178 1 MR. YOUNG: Right. The flow is being reduced.
/ T 2 Fluid velocities. They are all lower because you V
3 have essentially two-phase flow and you're choking at a 4 lower velocity there, but basically the same. But again, 5 the separation in the two bri.nches is the result of this l 6 type of separation in quality between branch 1 and branch 2.
l 7 Branch 2 is the one that makes a turn, so that's where the 8 separation would occur and you would end up with more water l 9 then flowing down the -- essentially the branch that belongs
'10 to the main pipe that had the straight shot.
11 MR. SCHROCK: In your calculatfsn for this model, 12 what have you done to get that phase separation at the 13 branch?
14 MR. YOUNG: Applied the correlation I showed you,
'()
15 which takes the branch quality as a function of the upstream 16 quality and the mass velocity ratio.
17 MR. SCHROCK: You showed me in the previous 18 meeting or today? j 19 MR. YOUNG: No , I think I -- it was in one of 20 these overheads. It's this right here. It's a fairly 21 involved function. I have the paper here somewhere if you 22 want to look at it.
23 MR. SCHROCK: Is that the same as the model in 24 NOTRUMP or is it something else?
25 MR. YOUNG: No. NOTRUMP doesn't have a model like I
I b
k)
%/
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179 1 this, all~right? What I was trying to do'was to see whether
.['Y in' fact the process of separation in these two branches 2
g) '
3 would lead >to-something like significantly reduced flow or 4 something else like that.
5 MR. SCHROCK: This is'from Lahey? Is that --
6 MR. YOUNG: I think it's due to Seeger.
7 This just shows how qualities -- again, as'a 8 function of the entrance quality, what are the two branch 9 qualities. As the pressure increases, the difference
-10 actually is reduced. So as the pressure increases, you're 11 going like this. I think the mass velocity ratio tends 12 towards one as the pressure increases, and we've got choke 13 conditions at the squib valve.
14' Finally, this is the ratio of -- not the' ratio, fy 15- sorry -- this'is the-throat pressure. So,-I mean, this 16 would be the range, then, of critical pressures with respect 17' to the hot leg pressure, and so there is some variation, no 18 question about it, although it does tend to be at about 50
.19 percent or so in most cases.
20 Now, in-the next overheads, I want to show how I l
21 took this model and tried to assess how NOTRUMP.would be 22 behaving, whether it's performing poorly throughout these 23 conditions relative to this model or whether it's performing .
24' well for whatever reason during some parts of the transient.
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180 1 This shows mass flow rate versus hot leg pressure at various 2 qualities. You can see here the comparison with the more
[Jk 3 simple Crane handbook estimate for compressibility at least 4 at 100 percent quality is pretty good at low pressure. Once 5 you get choking at.the squib valve, which is about here, 6- they tend to diverge.
7 You can generate a function of mass flow rate out-8- ADS for -- as a function of hot leg pressure and quality, 9' all right, and then converting that into a function, 10 calculate the flow rate, then using NOTRUMP calculated 11 pressure in the hot leg and NOTRUMP calculated flow quality, 12 and you can see that --
so what we have here is'the 13 prediction -- well, the way I look at this is this is what 14- the model would have predicted if we had put it into
() 15 NOTRUMP, and assuming, of course, there are no feedback 16 effects with the rest of the system.
17 But while we have critical flow in the ADS 4 squib 18' valve, the two agree fairly well, and I think this points to 19 the fact that our application of critical flow model --
l 20 remember the way we did that was you take the pressure drop 21 and calculate the total pressure downstream up to where the 22 ADS valve is and then use that in the critical flow model, 23 and it seems as if where you do the irreversible loss 1
24 calculation and where you do the density or acceleration 25 calculation is not all that important as long as you do it O ANll RILEY & ASSOCIATES, LTD.
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181 1 somewhere, okay?
'~'I 2 Now, as soon as you go to post-critical flow where L- /
3 we. don't have essentially the acceleration term anymore, l 4 anywhere in NOTRUMP, that's where we get a fairly 5 significant. difference.
6 How to assess this in relationship to predictions 7 of OSU and our predictions of AP600. What we chose to do 8 was to use this model as a way to estimate to what extent 9 one would have to increase the resistance through this ADS 4 10 valve to eliminate this discrepancy, okay? So the question 11 is, how much higher does this ADS 4' resistance have to be so 12 that.I get fairly good agreement with this more detailed 13 model.
14 bUt. SCHROCK: I guess I have difficulty with the 15 conclusion of these calculations without studying the 16 details of them more thoroughly. If I interpret right --
17- correctly, what you're saying here is that for the critical 18 flow, that the ADS 4 is still okay, that you've gone through 19 a model calculation now which is more detailed and you think 20 that what you were doing previously, with the neglect of 21 . momentum flux terms in NOTRUMP, produced essentially the 22 same bottom line critical mass flow.
23 MR. YOUNG: Right.
24 MR. SCHROCK: I have a hard time accepting that, i 25 MR. YOUNG: Yes. Maybe -- let me return just ANN RILEY & ASSOCIATES, LTD.
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182 1 quickly --
L 2 MR. SCHROCK: Yes.
}/)
x-I guess the bottom line, Mike, 3 is that I.wouldn't say that I am convinced without studying 4 in detail what went into the production of this comparison.
5 MR. YOUNG: Okay. I understand that.
6 DR. ZUBER: The only thing I can conclude, that-7 had .you done this' correctly :b1 the very beginning, you would 8 have saved all the headache that you have now.
9 MR. YOUNG: As.you will see in some predictions of 10 OSU, yes, things are falling into place once you get this 11 part fixed.
12 MR. SCHROCK: Mike, let me just make one further 13 comment about what I just said and the reason why I feel 14 that way about it. What you're basically arguing here is
- ( j 15 now we've gone back and we have done a more detailed and 16 more credible calculation according to the HEM model, and we 17L find that the critical mass flux that we predict this way is 18 basically the same as the critical mass flux we produced in 19 a situation where the'model was admittedly erroneous. That 20 means somehow a compensation in error, somehow a 21 ~ compensation that led to the correct answer previously. I 22 don't see that'that's possible.
23 MR. YOUNG: Well, again, let me just -- I wanted !
24 to -- this is a slide I showed at the last meeting, all i
25 right. The point I was trying to make was that what NOTRUMP I
e
((~}[
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183 1- is~doing is' essentially' calculating this-total pressure, all
() :21 3:
right, from -- and P sub zero zero was the hot leg pressure,
'okay? And all NOTRUMP does is takes that pressure, 4 calculates a' loss using a loss coefficient through the ADS
'5. " valve and results in a: pressure then-which'is this total 6 pressure.P zero plus the dynamic head term.' This is what
'7 NOTRUMP uses as input to the HEM, all right, as a reservoir 8 . condition.
9 MR. SCHROCK: Well, it depends on more thanijust 10 pressure and flux state which requires two dynamic 11 variables. You can do it in terms of pressure and quality,
-12 :ba terms of pressure and entropy, whatever you like; but the 13 fact is'that the entropy is increased in that long pipe, and 14 now'all you've done is calculate a stagnation pressure.
'A s ,) 15 What is the state, then, that you are employing'in the 16 NOTRUMP model to get the critical mass ~ flux? The upstream 1" quality? What is the second thermodynamic property --
18 MR.. YOUNG: The second term is the quality.
MR. SCHROCK: Where? At this location or 20 upstream?
21; MR. YOUNG: It's the quality at the location where 22 .this' pressure has been --
23- DR. CATTON: It's in Rhobar, isn't it?
24 MR. YOUNG: Yeah, the Rhobar. -All right.
t l 25 Now, again, perhaps -- well, almost certainly, I
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184 l' this is simply just somewhat fortunate in that the quality, (n) 2 even with a detailed model, does not change very much in 3 this --
4 MR. SCHROCK: See, Ivan wanted to help you, but 5 I'm not convinced that you answered his question thoroughly.
6 I mean, you do one or the other of these things or you do 7 something to choose the thermodynamic variables that you 8 input to your critical flow model.
9 MR. YOUNG: Right.
10 MR._SCHROCK: And I think you've got to look back 11 and.see what, in fact, is that. Is it a local value of 12 . quality calculated at the end of the pipe, or is it the 13 quality at the upstream head?
14 MR. YOUNG: It's calculated -- the quality is (s
() 15 calculated downstream of where this loss is applied.
16 MR. SCHROCK: See, because previously, as we went 17 through the evaluation of what NOTRUMP was doing, you were 1 18 neglecting any change in stagnation variables up to the 19 restriction. You were just arguing that the difference 20' between the stagnation properties and those in the upstream 21 reservoirs in fact are negligible. That you showed is not 22 the case when you found a 40 percent difference in your j l
23' comparison for ADS 4 on that long pipe. j 24- MR. YOUNG: Yes, but I --
-25 MR. SCHROCK: So I, you know, I think you've got !
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185 1 to be complete in your description of what you're doing V[~'\
2< here, and I couldn't make a final comment about it one way
'3 or the other until I know what.you've actually done.
4 MR. YOUNG: In:NOTRUMP, you're talking about now.
5 MR. SCHROCK: I'm talking about'in this 6 calculation now, leading to this figure that shows that, in 7 fact, the calculated flow rate is not significantly 8 different according to the way you used to calculate it 9 using NOTRUMP with the neglect of momentum flux terms and 10 the way you calculate it now, taking this into' account in a
^11 correct way.
12 MR. YOUNG: All right. I'm not sure what you're 13 telling me. Do you need to look at it further or --
14 MR. SCHROCK: I'm telling you you haven't
,O
( ,) '15 presented sufficient information about how you've done this 16 new calculation to make a case that I can believe that 17 produces this result.
18 MR. YOUNG: Okay.
19 MR. SCHROCK: I just find it incredible that the 20 calculation that you made previously could produce the same 21 result as the correct calculation.
22 MR. YOUNG: Well, I think I -- well, let me go 23 back to this one, then. I think I showed at least a partial 24 reason, and this is where I thought --
25 MR. SCHROCK: You're not doing the same thing you O .
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186
'l did before.
MR. YOUNG: Pardon me?
y/ O. .- 2 3- MR. SCHROCK: Here, you're not doing the same
'4 thing you did before. Previously, you calculate the 5 critical flow from your critical flow model neglecting _the 6 change in the state.that's imposed by'the Fanno part of the 7 ' process. Now you're in a way taking that into account. .
8 MR. YOUNG: Right. For various reasons --
9 MR. SCHROCK: _So am I misinterpreting what it is i'
10 you've compared here?
- 11 MR. YOUNG: No. What I'm -- let me try to explain 12 what I think -- why I think the calculation is relatively 13 similar or is really not very different when you have choke 14 conditions at the ADS-4 valve.
O)
(_, - 15 One thing in our favor basically is that even with 16 the detailed model, the quality, the mass quality of the 17 mixture as it flows down the pipe does not change very much.
-18 There's significant flashing, but the mass term or the 19 effect on the -- the mass of the water on the mixture is not
~20 that strong. So the quality itself does not change much by 21- the time you get to the ADS-4 valve.
22 This presentation was designed to show that if you 23 cal'culate the total pressure, all right -- remember, what I~
i 24 said was you could apply the momentum equation to calculate 25 critical flow. Usually, that's assuming frictionless flow, i
l[)' \/
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187 1 so there's a zero there, And basically, what NOTRUMP is
() 2 31 doing is calculating the total pressure, this entire differential here, taking into account a frictional loss, 4 and that seemed to be' roughly equivalent to a more exact
'5 solution of-the critical flow model, all right?
6 So the way I saw it was, using the total pressure 7 as one reservoir condition, using whatever quality NOTRUMP 8 calculates just upstream of the ADS valve is not too much of 9 an error, because qualities don't change much even with an 10 ' exact solution, results in a critical flow that's not too 11 far off.
12 MR. SCHROCK: The words here following equation' 13 14, the NOTRUMP procedure is to calculate the total 14 pressure, you're referring to what is labelled in this graph (h
'Q 15 the NOTRUMP procedure or in the code NOTRUMP?
16 MR. YOUNG: In the code. No, it's -- that's what
! 17 is done.in the code.
18 MR. SCHROCK: Well, that's not what you said in 19 the previous meeting.
20 MR. YOUNG: That is what I said. I presented this 21 overhead at the last meeting. I distinctly remember Ivan 22 saying, they're doing it right. It might be by accident, t.
23' but they're doing it right.
L' H24 [ Laughter.)
25 DR. _ATTON: Don't blame me.
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188 l
l 1 MR. YOUNG: -Well, let me try to just finish.
() '2- There is no question that there may be compensating errors l
l 3 leading to a similar result here. What struck me about this
.4' comparison, though, was that at'the point where we took off 5 the critical flow calculation, which meant that any L 6 acceleration terms now anywhere in the pipe were gone from 7 the NOTRUMP calculation, that suddenly is where we saw the 8 difference.
l 9 CHAIRMAN KRESS: What causes those fluctuations?
10 MR. YOUNG: Well, these are changes in quality.
11 The level in the hot leg is moving up and down as we're 12 moving along in time here. It's a very long time scale 13 here.
14 CHAIRMAN KRESS: Those two curves don't really
) 15' look identical to me.
16 MR. YOUNG: No , they're not identical. They're 17 similar, but not identical.
18 CHAIRMAN KRESS: Like the model gave you higher 19 values during that period and then lower ones during the 20 post-critical.
'21' MR. YOUNG: Well, yes. No, I'm saying here,
-22 they're quite similar.
23 CHAIRMAN KRESS: They're close enough.
24 MR. YOUNG: Right. Here, they're not. And this 25 is where we have to think about what to do with the NOTRUMP l'
l l
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1 'calcul'ation.
1 7
2- DR. ZUBER:
o v
( ) .If I understand correctly, in the 3 region where you would' expect the fluxes to be important, l; 4 you don't.see any difference; and in the region where you 5 don't~ expect the momentum fluxes to be important, you see a 6 big difference. Is this --
7 MR. YOUNG: Not quite. Not quite. What I --
8' DR. ZUBER: This is the way I see this.
9 MR. YOUNG: Right. No. But let me just try to 10 rephrase what you said. In the region where ---in this 11 region, where fluxes are definitely important, all.right.
12 NOTRUMP is indirectly calculating the effect of those fluxes 13 by using HEM. HEM is simply an acceleration down to a float 14 condition. Here, fluxes are still important, but NOTRUMP'is
'f3
(_,) - 15 no longer calculating critical flow. So the only thing it 16 has now is delta P equals K will be squared over two, okay?
.i
- 17. So-as long as there is some calculation somewhere in the -l l
- UB path of the=effect of'the expansion, even if it's, you know, 19 isotropic as opposed to isonthalpic, you know, that leads to 20 a reasonable prediction here, or a prediction that doesn't
{
- 21. .have gross error.
22 You'll see that the effect of acceleration is 23 important practically all the way down to atmospheric 24 pressure. I mean, the vapor density is very sensitive to 25 pressure, okay?
i
[9
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190 l 1 DR. ZUBER: I guess had you taken the momentum i
/~'T 2 fluxes in that left region, plus the homogeneous, you would 1O
'3 expect a good' agreement or not?
1 4 MR. YOUNG: Taking momentum flux in NOTRUMP?
5 DR. ZUBER: Yes, in addition to calculating your i 6 quality.
L l 7 MR. YOUNG: Well, if we had momentum flux in 8 NOTRUMP, we would have had to -- we have to be careful about 9 what pressure to use in the HEM model.
l 10 DR. ZUBER: But then, if I may rephrase it, you 11 are changing it in order to obtain agreement on the quality, l
12 and then by adjusting this, you obtain a relative agreement l 13 in that region.
14 MR. YOUNG: No. I just took -- all I did was I
() 15 took the hot leg pressure, I did take the quality, the flow
-16 quality that NOTRUMP is calculating entering the pipe, and 17 took those as boundary conditions through the detailed i
18 model, okay, and then simply compared that with a prediction 19 from NOTRUMP of the flow rate. So I wasn't doing any /
20 adjusting to get agreement here, okay? Maybe the reason is 21 it really needs to be more fully explained, all right, in 22 terms of what the NOTRUMP model is, because they are i 23 substantially different.
24 MR. SCHROCK: You indicated that these 25 oscillations are due to oscillation in the quality coming
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191 l'
into the pipe. -Does that-represent this range of qualities
[D l %-)
2- that you did the calculation on, roughly 20 percent to 100
.3 percent?
4 MR. YOUNG: Twenty to 100 percenti yes.
5 MR. SCHROCK: What do you do if the quality is low 6' enough so_the HEM is no valid? I mean, you have adopted for 7 your computation Henry Fauske in the low quality range, and y 8 then the HEM in the higher quality range. Have you no 9 situation in which you ought to be using Henry Fauske 10 instead of the HEM?
11 MR. YOUNG: Well, Henry Fauske is applied in 12 NOTRUMP if qualities fall below ten percent. That -- Henry 13 Fauske is not in this more detailed model. And based on the 14 conditions we looked at, it doesn't appear as if you fall i (( ,) 15 below 20 percent very much at all during the transient.
16 MR. SCHROCK: That's a huge quality.
17 MR. YOUNG: This is what I.did to try to assess 18 the impact of the NOTRUMP model' efficiency -- estimate -- to 19 estimate the increase in the flow resistance needed to 20 account for the lack of momentum flux. Again, I used the 21 detailed model. You can calculate an effective loss i
22 coefficient essentially. It's now a function of pressure l 23 because of the compressibility effect, and then examine l
24 AP600 and OSU predictions with a modified loss coefficient.
l 25 This is -- if you simply take --
i i
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192
~
1 DR. ZUBER: What do you mean examine AP600?
I l i[~) ~ 2 1MR . YOUNG: I'll show you in a minute what I did.
i %.J l
i 3 The first step was to take -- essentially calculate an l 4 effective loss coefficient, which is simply the pressure 5 drop, total flow rate through the ADS valve, and the average 6 density basically, mixture density in the hot leg, which f 7 essentially would be coming from the flow quality.
8 DR. ZUBER: This is for what? For OSU or for --
9 MR. YOUNG: This is for AP600, 10 DR. ZUBER: Okay. Okay.
11- MR. YOUNG: And what you see is at 14.7 psi in 12 which acceleration effects you might expect are not very 13 strong, flow rates.are very low, this would be the effect of 14 .the two-phase multiplier effect. This is 100 percent 1%
(_,) 15 quality and this is 20 percent quality.
16 Then as you move it to a higher' pressure, hot leg
-17 pressure, then you see the change in the effective loss 18 coefficient, and you see a change right around here, which 19 is where the squib valve finally becomes choked, okay?
20 CHAIRMAN KRESS: That pressure is not the same 21 pressure as the one you've got upstairs in the equation?
22 MR. YOUNG: This is the loss coefficient. !
23 CHAIRMAN KRESS: In the equation, PHL is not -- 1 24 MR. YOUNG: PHL is hot leg pressure.
25 CHAIRMAN KRESS: Yes. Plus 14.7.
l
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l l 193 1
i l
MR .- YOUNG: Well, no. The delta P here would be li,,h L 2 -- yeah. This delta P would be almost zero at this point.
!V-
! 3 CHAIRMAN KRESS: Okay. That was confusing me for
- 4 a minute. Thanks.
5 MR. YOUNG: What we can then do is take the loss 6 coefficient and normalize it to the value at 15 psi for each 7 pressure. In other words, you have a quality of 20 percent, 8 let's say, at 15 or 14.7 -- sorry -- 15 psia, you have a 9 certain loss coefficient and you normalize, then, all the 10 values for higher pressures to that for that quality, and 11 you get something like this.
12 I think what's significant about this is that at 13 pressures of about 20 psi, you are already getting a pretty 14 significant increase in the loss coefficient or resistance 73
(,) 15 to flow through the pipe because of density changes.
16 The other thing I think that this shows is that 17 from here to about here, all of the qualities are basically 18 behaving the same, which means that what you're seeing is 19 the effect of the vapor density, the change in vapor density 20 due to pressure, and then as you get to higher and higher 21 pressure gradients, you start to see the flashing effect, 22 okay, the spreading out of the qualities. So low quality 23 -gives you essentially an effectively higher resistance 24 because you start to evaporate additional vapor from the 25 water.
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194 1 But what this is saying, then, is that in the l j
() 2 range where you a7:e approaching IRWST flow, IRWST injection,
- 3 where an AP600, the system pressure would be between, oh, 25 i 4 to 30 psi, it'looks like there should be a resistance 5 increase of about 60 percent, all right? There's no 6- critical flow here yet; it's just because of the 7 compressibility of the mixture as it flows.through the pipe.
8 For OSU similarly, there ought to be an increase 9 --
10 DR. ZUBER: This is compensating for the gradient
'11 momentum fluxes acceleration; is that correct?
12 MR. YOUNG: Right. This is the effect of the 13 acceleration converted --
14 DR. ZUBER: And you convert it now in a friction 1
- . j 3
)
i 15 of --
16 MR. YOUNG: Right.
17 DR. ZUBER: -- flux coefficient.
l 18 MR. YOUNG: Which is something we can put into 19 NOTRUMP and see what happens.
20 DR. ZUBER: I see. I see. I see.
21 MR. YOUNG: So what you'll see is calculations in 22 which for OSU, we increase the ADS-4 resistance by about 40 23 percent and AP600 by about 60 percent, j 24 DR. ZUBER: Why the difference?
i l 25 MR. YOUNG: Because AP600 system pressure is a
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195 i
1 llittle higher,.okay? That's because all the levels are a nO 2 little bit higher.
l.V' 3 DR. ZUBER: Oh, okay. But the pointHis that you 4 really; assume without saying.is that the quality, everything 5' else is prototypical in OSU, so.that you can take the 60 6- percent, the 40 percent, apply it, then extrapolate to --
i.
7 .this is your basis for -- assuming that everything is well 8 scaled in OSU, then you calculate what -- the correction for 9 not having that acceleration, you are compensated with the 1 l 10 friction, and then you will tie it to OSU, and then OSU has 11- different pressure than AP600 and you want another 20 psi.
I 12 So the entire argument, it very well rests on.the 13 case that OSU is well scaled for this particular process.
14 -MR. YOUNG: I still have to make that argument, E!x_j 15 yes.
l 16 DR. ZUBER: No, no, no, I mean, this is the basis. l l
17 I mean, that's the basis.
f l .18 MR. YOUNG: Yes. But the emphasis has changed a 19 little bit, and you'll see when we do the calculations that j 20 it wasn't so much not predicting the pressurizer level for 21 whatever reason, all right, it was mispredicting the ADS-4 l
i 22 resistance, because when we do that right, we are going to l j
i l
23 predict the draining of the pressurizer right, as you will l 24 see.
25 But this allows us to identify to what extent we l
[\
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196 1 need to increase or compensate for the lack of momentum flux 2 .in NOTRUMP for both OSU and AP600, okay, and this is how we 3 did that.
- 4. DR. CATTON: Basically what you're doing is what 4 5 they do in flow and forced medium. You have a friction
)
6 coefficient that is like a constant divided by velocity plus 7 a constant, and the second constant deals with the inertial 8 effect and the first is the basic friction. That's kind of 9 what you're doing here.
10 MR. YOUNG: Yeah. And that makes this a function 11 of --
12 DR. CATTON: But normally, they don't do it this 13 way. They would write -- you look at what is.the effective, 14 normalized, or however you want.to phrase it, friction as a 15 function of the velocity throughLthe system and then -- l 16 MR. YOUNG: Yes.
17- DR. CATTON: -- you use the correlation of its 18 --and it makes more sense, it's more physical that way.
19 This -- I have a lot of trouble following the logic in doing 20 it this way,. although I understand the words you use.
21 MR. YOUNG: This seemed like a convenient way to l 22 -- I mean, it does collapse these points down here.
- 23. DR. CATTON: Well, when you get to very low flow, i
l -24 the friction. factor is different, and that's what you're i i 25 doing. When you:get down to zero delta P, you've got l i
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197 i
1 . essentially zero flow.
Friction is like one over velocity.
f'~) 2 MR. YOUNG: So.that all you have is the V
3 recoverable loss and no -- virtually no --
.j 4 DR. CATTON: This is friction. That's right. The j i
5- second term has to do with the recoverable part )
6 MR. YOUNG: Right.
l
-7 DR. CATTON: -- and the inertial effects.
8 MR. YOUNG: Okay. There probably are different 9 ways one could do this.
10 DR. CATTON: Well, there's a different way that-11 most people do it.
12 MR. YOUNG: Okay.
13 CHAIRMAN KRESS: Now, you could have either put 14 this correction factor into NOTRUMP or you could take a
'( ). 15 penalty on the IRWST level.
16 MR. YOUNG: Right.
17 CHAIRMAN KRESS: And these two are related.
18 MR. YOUNG: They're related, and I'll try to 19 explain how. It might make more sense after we look at;i.he !
20 NOTRUMP runs that we did to do that. "
21 CHAIRMAN KRESS: Okay.
.22 DR. CATTON: If I change the mixture ratio in the 23 hot leg, that will change this curve, won't it?
L 24 MR. . YOUNG: The quality?
L 25 DR. CATTON: If I change the ratio of vapor to
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198 4
1 liquid, or does the normalization take care of that?
i .
2 MR. YOUNG: Yes. This is the range of 20 to 100
[J)
\_ .
t 3 percent quality, and it tends to collapse it down by the~
4 normalization. ,
5 DR. CATTON: Okay.
6 MR. YOUNG: Okay.
7 DR. CATTON: So let me see. So you take care of 8 the stagnation conditions by your normalization. You take 9 care of the acceleration recovery terms by this graph.
10 MR. YOUNG: Right.
11 DR. CATTON: Okay. I still would have liked it l 12 the other way. I would have understood it easier.
13 MR. YOUNG: Okay.
14 DR. ZUBER: Mike, I see you are trying to save i) 15. something.
16 MR. YOUNG: That's what I'm trying to do.
17 DR. ZUBER: I would feel much more comfortable if 18 I could really be convinced that OSU is well scaled for this i
19 particular case, because if it isn't, then you really don't 20 have a case. You cannot really make a case.
21 DR. CATTON: That's right.
22 MR. YOUNG: But again, let me go back to what I 23 said,-which.was the scaling analysis showed us that ADS-4 24 resistance is really the dominant effect here, all right.
25 And we were working around that as one -- our misprediction I
b'
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199 1 of ADS-4 pressure drop as one possible source of our
/~~
() '2 misprediction of the pressurizer drain rate. We are looking 3 at entrainment. What I think we found is that our 4 conclusion, our NOTRUMP calculations support exactly what 5 .the scaling analysis found, okay, that ADS-4 resistance. If
'6 you can get that right, then everything else falls into 7 place. And in fact, how well scaled the pressurizer is when 8 it's draining is not quite as crucial as -- at least I 9 thought in the past. I thought that was where all the 10 importance is. But this appears to be really a crux of the 11 entire issue.
12 Now, this is the run that we did. You'll see more 13 details about all --
14 DR. ZUBER: Let me again interrupt.
/~~'N
() '15 MR. YOUNG: Sure.
16 DR. ZUBER: Was this the conclusion from'RELAP 17 analysis? Was this your conclusion from the RELAP that the 18 resistance in the ADS-4 was the most important factor?
19 MR. KELLY: Joe Kelly from Research.
L 20 Dr. Zuber, yes. If you can remember back to the 21 meeting we had out in L.A. --
22 DR. ZUBER: I wasn't there, so there is nothing to 23' remember.
24 MR KELLY: Okay. Then I'll jog Professor 25 Catton's memory. ,
i
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200 1
(- I gave a presentation where I tried to describe f
d 2 the phenomenology behind the oscillations that were observed 3 in the ROSA test, and I ended up relating that directly to 4 the ADS-4 loss coefficient. Basically, if you can vent the i
j 5 vapor out ADS-4, it's not going to take the ADS-1, 2, 3 l 6 path, and so it won't hold the liquid up in the pressurizer.
! 7 JBut if you clamp down and don't let it go out ADS-4, then it 8 has to go out through the pressurizer and then it holds the l '9 liquid up.
10 DR. ZUBER: Okay.
l 11 MR. KELLY: So I agree with that.
l 12 MR. YOUNG: This is the NOTRUMP run in which we 13 increased the resistance of the ADS-4 piping 60 percent, and l .
'14 we do have now pretty good agreement with the detailed l (O)
- 15 model.
16 We will show more results of this in terms of the l 17 effect on the total inventory and so on in Andy's i
18 presentation.
19 DR. SCHROCK: And the basis of 56 percent increase 20 in K is simply that it produces a better correspondence or l
21 is there another reason for that?
22 MR. YOUNG: Well, what we had estimated from this 23 curve was that the increase in resistance for AP600 should 24 be about 60 percent. We missed, we had a slight error in
,. 25 the actual calculation. It should have been 60. We made it r
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l
- j 201 1- 56 percent in the calculation, but this says that we should
([ 2 increase the resistance about 60 percent for AP600. Okay?
3 This also says that we should be increasing the l 4 resistance about~40 percent --
5- DR. SCHROCK: That's not so much increasing 6 resistance as it is'taking into account the compressibility
-7 effect.
8 DR. ZUBER: Yes, right.
9 MR. YOUNG: Right. Offsetting the lack of a 10 compressibility term.
11 DR. CATTON: That sounds a lot better than
- 12. increasing the resistance.
13 DR. SCHROCK: You are making an arbitrary 14 adjustment in something. You're invoking the right physics.
() 15 MR. YOUNG: Right -- very indirectly, yes.
16 DR. CATTON: Does this mean you won't want to take 17 the penalty, you won't need to?
18 MR. YOUNG: No , this means what -- what I am going 19 to try to show is that applying a level penalty is roughly
'20 equivalent to doing this -- to essentially increasing ADS-4 21 resistance, okay, and basically it :bs -- I mean you have a 22 pressure trying to force water into the system. That is the 23 IRWST level. Then you have a pressure preventing that, and 24- that is the ADS-4 pressure drop.
25 DR. SCHROCK: So you.get one wrong, so you will
..h
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202 1" adjust the other? That should be out of the circuit
() 2 3
analysis. You-should show, be able to show --
MR. YOUNG: If I keep this ratio fixed, I'm okay.
4 CHAIRMAN KRESS: Yes. It's just Delta H over C --
5 DR. CATTON: But he can do it easily, just a 6- little circuit analysis and he'll wind up with a ratio of 7 the resistances.
8 CHAIRMAN KRESS: But there is no circuit. It just 9 goes straight -- because only ADS flow is important. It's 10 just Delta H over C.
11 DR. CATTON: -Well, it is the one coming in and the 12 other going out.
, 13 MR. YOUNG: Right. Well, I am going to try to 14 show how basically they are equivalent' steps, okay, but
,(Q ,j :05 again I'll just do that after we have seen some of the 16 :NOTRUMP calculations.
17 So'that is basically what we ended up with on 18 momentum flux and the detailed model I think did show some 19 valuable'information, and tends to confirm the conclusion --
20 well,. supports the conclusion that.they made in the scaling 21 analysis that you need to be -- certainly take into account 22 all the terms that could lead to higher pressure drop 123 through that piping.
L 24 CHAIRMAN KRESS: In the spirit of Novak liking to 25' know -- put~a period at the end of each of these -- I am ANN RILEY & ASSOCIATES, LTD.
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203 I wondering what the consultants think about this as a way to i /~m
.( [ 2 address both the momentum flux term and later on the penalty
(' 3' on-the IRWST.
l 4 Any thoughts on that right now, Ivan? Or --
5 DR. CATTON: Go ahead, Novak.
6 DR. ZUBER: I don't see what they can do at this 7 point in time with the momentum flux. I mean that is a sin 8 committed years ago. You have.to live with the 9 consequences.
10 You have to find a remedy. I would feel much more 11 comfortable were I assured.that OSU was well scaled for this 12 thing here, for this particular case -- for the resistances.
13 I would go then back and address that whole 14 problem, the ADS-4, and the beginning of the IRWST. This is
! ,r~
t (_, 15 the question which Ivan wrote up and this is the question I 16 have greatest concern.
17 This is the only part on the transient.really 18 which concerns me at this point in time. If you can put.
1 19 this to rest, I can say. fine, you can go ahead, and to do
-20 this I would advise to have a meeting with Banerjee and l 21 Rogen or somebody --
22 CllAIRMAN KRESS: I think I have asked Paul to try !
l 23 to put that on our agenda.
l
! 24 DR. ZUBER: And then we have to discuss it and see 25 who is right, who is wrong, whether they are both right or l
b
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204 I
1 both wrong, g
,.[v)' 2 There is no sense holding them back if they are 3 right, and there is no letting them loose if they are wrong, 4 so I think until we resolve that question, I think as far as 5 I am concerned, this is still an open question, but as far 6 as the momentum flux, you have to live with the sins.
7 CHAIRMAN KRESS: As far as the scaling question, l 1
8 what you are saying is he's derived this correction factor !
1 9 independent of scaling but to show that it works he has to !
l 10 compare it to something and that's SPES -- l 1
11 DR. ZUBER: Yes, he shows it works but -- j J
12 CHAIRMAN KRESS: If it works for SPES --
13 DR. ZUBER: If SPES is okay, then you have reason 14 to say okay, this'may apply also --
r'
(%) 15 CHAIRMAN KRESS: Then you can say it works for -- l 16 DR. ZUBER: Hopefully it works for the AP600, but 17 if it doesn't work for SPES --
I mean for OSU, then --
18 CHAIRMAN KRESS: Then you have got a problem.
19 DR. ZUBER: -- then you have a problem.
20 CHAIRMAN KRESS: Yes, so first show it works for 21 OSU and then show OSU is well-scaled for this part.
22 MR. YOUNG: Right. Two steps there.
23 CHAIRMAt KRESS: Two steps. Okay. Ivan, did you I
24 want to --
25 DR. CATTON: I essentially agree with that and I
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205 1 think that what would really be helpful is if Westinghouse
() 2 3
would just do a little bit of conversion of the scaling analysis so that it is put in terms of the core level, 4 because that is really where we are at -- IRWST level, core 5 level -- and then we can actually compare all these things.
6 DR. ZUBER: And maybe to help us, what a different 7 education before the computers -- have a circuit analysis.
8 It's really very instructive.
9 DR. CATTON: Yes. You see, the little circuit 10 analysis will -- should come out with the ratio of the 11 resistances if your arguments are correct -- or a ratio of 12 the voltages or the heighth of the pool. Then you can 13 argue, gee, that this thing is the same "if" --
14 MR. YOUNG: Height is voltage and --
() 15 DR. CATTON: Well, essentially your heights 16 becomes the voltage or the potential. Then it is simple.
17 It's clean and what should drop out of it are the parameters 18 that are important in making the step between the facilities 19 and you could look at them.
20 DR. ZUBER: And the lesson to be learned, if you 21 ever design a new reactor, and you'want to develop a code, 22 put all that good physics in it. Don't commit a sin in the 23 very beginning.
24 CHAIRMAN KRESS: Is that a message to the Research 25 staff?
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206 1 DR. CATTON: I get a different message. The>
(} 2 message I get-is " KISS" -- you know, that was the phrase 3 that EPRI coined a long time ago, which is " Keep It Simple, 4 Stupid."
l 5 MR. YOUNG: To me, the scaling analysis is what 6 tells the story. That tells you ADS-4 is important.
7 Therefore you have got to pay a lot more attention to it.
8 DR. ZUBER: Maybe he is trying to slip me up.
9 DR. CATTON: But you still have to make the leap 10 of faith from the experimental regs that you have.
11 MR. YOUNG: Yes, that 's true.
12 DR. CATTON: And the scaling analysis helps you do 13- that also, and also it will tell you if something ain't 10 quite right. It will tall you also that if you look at S
( ,) 15 these facilities that maybe you bracketed the process, so if 16 you can predict all three, you've got it made.
17 DR. ZUBER: See, the thing is all three 18 facilities, at least one is well-scaled for designs except 19 this phase here, and if you can put this to rest, you can i i
20 then -- you are satisfied. At least I am satisfied.
21 MR. YOUNG: Okay.
22 CHAIRMAN KRESS: Thank you. You are still on, 23 apparently.
24 DR. BOEHNERT: We have to go into closed session 25 now.
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207 1 CHAIRMAN KRESS: Would you like to have a short 2 break before we go into closed session?
! 3 .MR. YOUNG: Why don't we do it? Yes.
4 CHAIRMAN KRESS: Let's take a 15-minute break.
5 [ Recess.]
6 (Whereupon, at 3:01 p.m., the hearing proceeded in 7 in-camera session.]
8 9
10 11 12 13 14
[
d 15 16 17 18 19 20 21 22 23 24 25 I
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l 249 !
i 1 OPEN SESSION l '[d N.
3- 2 [4 :04 p.m.]
3 MR. TAKEUCHI: My name is --
4 DR. ZUBER: May I -- again, since we are -- we 5 have the problem with the code here, I would like to raise i
6 the question with RELAP.5. How do you calculate this with 7 RELAP 5? What kind of success do you have, or failure?
8 MR.. KELLY: Are we talking about the horizontal .,
9 stratification?
-10 DR. ZUBER: ADS-4.
11 MR. KELLY: Our conclusion was that for all the i 12 design basis scenarios we looked at, it was very much a 13 self-correcting process, pretty much as Mike Young has 14 ' indicated. If, for example, you are over-predicting the O'
( ,/ 15 entrainment into ADS-4, so that the two-phase pressure drop 16 across the valve goes high, then that in effect blocks off 17 the system. As' system pressure begins to rise, that shuts 18 off the IRWST flow. The system starts to boil down a 19 little. When it boils down enough so that it clears the i 20 entrainment, the vapor can thin out, the pressure falls, and 21 the IRWST flow comes in.
22 DR. ZUBER: And you can predict that with RELAP?
23 MR. KELLY: You cannot predict exactly the right
- 24. level in the hot leg with RELAP.
25 DR. ZUBER: What about the level in the core?
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l 250 1 MR. KELLY: Within -- the mixture level is up in
"() 2 the upper plenum, up somewhere in the hot leg, and it will 3 vary by some fraction of the hot leg depending upon how good 4 or bad your models are, but that's all. It's only within 5 that. amount. So the core is well submerged because the 6 level's up in the hot leg. So for any of the design basis 7 scenarios, there was no problem at all, and you had to go to 8 . massive failures of ADS-1, 2 and 3 and several of the ADS-4 9 valves before what was in entrainment became important.
I 10 DR. ZUBER: Okay. Thank you.
11 CHAIRMAN KRESS: Is there a model for entrainment 12 out of T in RELAP 5?
13 MR. KELLY: Joe Kelly again from Research. Yes, 14 there is a model in RELAP. It's based -- it was developed
() 15 by the RELAP 5 team on this database. It's a correlation 16- very similar to what was shown, and the -- basically in 17 these calculations, it turns on and off. The water level 18 comes up enough so you get the entrainment, then it shuts l
19 -the vapor shuts off and so on. I 20 CHAIRMAN KRESS: Okay. Thank you, Joe.
21 MR. SCHROCK: It was developed by Keith Ardron, 22 actually, as a composite of the KFK and UCB results. That's 23 what they put in RELAP almost ten years ago.
i
~24- CHAIRMAN KRESS: Thank you.
~25 MR. TAKEUCHI: May I start?
I 1
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l l
251 1 My name is Ken Takeuchi, a fellow engineer at
() 2 3
Westinghouse Electric Company, and I was asked to find out where the most limiting counter-current flow limit, or CCFL, 4 would take place.inside the pressurizer surge line.
5 The conclusion is that limiting CCFL will take 6 place in the vertical section of the surge line, and so my i
j 7 presentation is to explain how I got this conclusion.
~
8 The circumstances of a small break loss of coolant 9 accident is illustrated by the OSU test SB18, and the 10 typical -- after the blowdown the natural circulation, the 11 ADS valves 1 and 3 are opened up for 100 seconds, and the 12 pressurizer is refilled and drained, and at about 1,000 13 seconds, ADS-4 is activated. As a result, the draining --
14 slow draining lasted for about 80 -- 800 seconds. In the
() 15 meanwhile, IRWST is activated at 1200 seconds.
-16 The time period of interest is from 1,000 to-1,800 17 seconds. The system pressure is approximately 20 to 35 l 18 psia, and I have used the 35. psia for a scoping study and 19 the 20 psia for verification. Fluids are saturated and ]
20 liquid drain rate is .15 pounds per second, which is R 21 approximately one-tenth of the normal drain rate, so that 22 the CCFL must be taking place inside the pressurizer surge 23 line.
24' This is a schematic of the pressurizer surge line L 25 of OSU, quite similar to AP600. Here, there is a vertical l
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252
'l section and this one is connected to the slightly inclined
() '2-
- 3
- red. pipe'through elbow, and this horizontal line is connected to another slightly inclined straight pipe.through 4 the horizontal elbows here. So'there are various competing 5- sections-in that for the CCFL.
L 6 .The method of analysis for the vertical section is
.7 to apply;the correlation and to generalize the' correlation l
l 8' expressed in terms of the Kutateladze numbers. For the
'9' ' straight inclined, slightly inclined pipes, we use a 10, Taitel-Dukler phase' transition theory. For the elbows, we 11 apply the Banrjee centrifugal force treatment.
12 First, the CCFL in the vertical pipe -- or in the L 13 horizontal pipe is compared with the vertical CCFL followed 14 by a scaling study. CCFL in the elbow are compared with the
()
E15 finclined straight. pipe. CCFL in the. vertical elbow will be
- 16. compared with vertical CCFL. Jmd finally, the test data of 17- OSU SB18 will be compared with the prediction.
- 18. What is -- the correlation for the vertical pipe.
l19 is expressed in terms of dimension -- that is for the steam 20 and liquid phases. Jg, for instance, is equal to the
- 21; superficial velocity of the steam times the square root of 22 -steam density divided by acceleration gravity, pipe 23; ' diameter, D, and the density difference, Rho.
24- C ranges from .75 to 1.0, and we will talk about 25 the scaling study-later. On the other hand, Taitel-Dukler O ANN RILEY & ASSOCIATES, LTD.
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253 1 ' theory'is to. apply the Kelvin-Helmholtz critical condition 2 to a steady-state stratified flow, and this theory is 3 verified comparing the traditional against the test data for 4 phase. transition from stratified flow, cold current 5 stratified flow to annual dispersed or intermediate phases.
6 We apply this one to the counter-current stratified flow to 7 predict the flooding.
.8 Curve in Hammel's condition can be expressed in 9- terms of Wallis non-dimensional constant j (G) star in this 1'
L10 expression, where A sub L is the water level in the 11 stratified flow, alpha is the void fraction and theta is the 12 inclination of the. pipe. And.C-2 is equal to 1 minus A sub 13 .L divided by pipe diameter. This one is proposed by 14 , Taitel-Dukler for circular pipe.
15 This equation 2 is actually equivalent to this 16 expression. And what this one means is illustrated by the
.17 figure below. Suppose --
18 DR. ZUBER: Wait, wait, let me ask you1something.
19' -MR. TAKEUCHI: -Yeah.
20 DR. ZUBER: Alpha is dimensionless.
21 MR. TAKEUCHI: Alpha?
' 22 - DR. ZUBER: Yeah. Okay. That's a void fraction.
23 MR. TAKEUCHI: That alpha is void fraction.
24 DR. ZUBER: Okay. This, the. diameter actually is 25 the height, the ratio is dimensionless. Where do you get
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_ _ _ _ _ _ _ _ _ _ _ _ - _ _ - _ _ - _ - _ _ _ _ _ _O
l 254 1 the velocity?.
! O 2 MR. TAKEUCHI: j (G) star.
(:V 3 DR.tZUBER: Where do you get -- you don't have the 4 ' gravity term here?-
- 5. MR. TAKEUCHI: Gravity term'is cosine -- cosine g 6 times cosine theta. This one is gravity effect.
'7 DR. ZUBER: Is there a mistake? Is this 8 correction there?
9 MR. TAKEUCHI: No , this one is correct expression.
10 MR. YOUNG: I think the G term is subsumed into 11 j (G) star.
12 DR. ZUBER: I see. Okay.
13 MR. TAKEUCHI: Is it okay, can I go ahead?
14 DR. ZUBER: Go ahead, (p) 15 MR. TAKEUCHI: Suppose that the Wave is formed by 16 -- as a perturbation. Then' steam flow increases over -- i 17' dynamic head increases at the narrowed part. That decreases 18 the starting pressure and--- increases amplitude of the 19 wave. On the other hand, gravity is going to; suppress the 20 wave amplitude. If the dynamic pressure, the winds, the 21 amplitude keeps increasing and the stratified flow changes 22 to other flow regime.
23 And'later in my discussion, in addition to the 24' gravity force, self-extension and centrifugal force will be i
25 taken into account qualitatively.
I
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L l i
255 I 1 One more. Equation 2, right on side becomes only 2 a function of void fraction.
(u_)3 1
3 DR. SCHROCK: How does it eliminate d and d alpha 4 da Hl?
5 MR. TAKEUCHI: H1 is water level so it is a 6 function of void fraction. It is a stratified flow.
7 DR. SCHROCK: Okay. But how is d eliminated?
-8 MR. TAKEUCHI: d and H1 becomes a dimensionless 9 diameter and that becomes a void fraction. Diameter and H1
'10 is related to the void fraction. This quantity.
i 11 So this is equation 2 and this indicates that this i 12 portion is a zone of stable stratified flow.
13 Interfacial force ba?ance equations can be i l
14 obtained from the steady state momentum equations for liquid f
(3 j 15 and steam phases. Here the taos are the wall and the 16 interfacial sheers and the S's are the surface, interfacial-17 and the wall surfaces. Taking difference of these two 18 equations, we have interfacial force balance equations.
19 With taking the rather realistic sheer forces, we 20 find' equation.(5) is a function of superficial velocities
- 21. for liquid and steam phases and void fraction. And we have-22 seen that the critical condition is a function of steam 23 velocity and the void fraction. So by eliminating void 24 fraction from these two equations, we have a critical curve.
1 25 And this curve'is compared with the Wallis flooding curve in l t
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y 256 L
1- the next slide.
l
['
s.
2 Curve No. 2 is obtained from Taitel-Dukler theory 3 for inclination 2.5 degree. Curve 3 is for inclination 7.7 l :4- degree and Curve No. 4 is 13 degree inclinations. Curve 1, l
5 there are two straight lines for the Curve 1. Is Wallis l
6 flooding correlation. This one is for flooding curve, 7 flooding constant is .75, and this is for flooding constant 8 1.0. In other words, this is sort of'the range of the 9' flooding correlations by Wallis.
10 And the same calculation is performed for AP600, 11 larger diameter. And you can see that both results are 12 about the same.
13 DR. SCHROCK: Your Wallis lines are both M equal 14 to unity?
S
- s ,) 15 MR TAKEUCHI: Yeah, both. Both, Wallis curve is 16 the same for both.
17 And another point of interest is that as 18 inclination increases, those curves become higher and 19- higher. And for a given steam flow rate, this much of the 20- liquid --
21 DR. SEALE: It's kind of hard to see when you 22 ~ stand right there.
23' MR. TAKEUCHI: Oh, sorry. Suppose steam flow rate
'24 of this much is given. According to the Wallis flooding 25 curve, liquid of about this much can downflow through the
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~ - __- _ _______ _____-
- 257 1- '~ vertical pipe. According to the Taitel-Dukler, in the
[~I
, y/
2 slightly inclined pipe, this much water can flow without
-3 disturbing the. stability of stratified flow. So this 4 indicates --
5 CHAIRMAN KRESS: Four is a bigger incline that 3?
6 MR. TAKEUCHI: Bigger flow can be larger, much 7 larger than that flow can go through the vertical pipe.
8 CHAIRMAN KRESS: Curve 4 has a bigger angle that 9 Curve 3?
10 MR. TAKEUCHI: That's right, yes.
11 CHAIRMAN KRESS: Why doesn't it approach the 12 vertical? The bigger the angle, why don't you -- it is 13 measured from the vertical?
14 MR. TAKEUCHI: That's right.
V(^N 15 CHAIRMAN KRESS: Okay. So it is --
16 MR. TAKEUCHI: This one is going away.
17 CHAIRMAN KRESS: It's getting shallower and 18 shallower?
19 MR. TAKEUCHI.: No , the other way around. It's 20 getting closer but - .the inclination is increasing and this 21 one becomes farther away from the vertical. But eventually 22 it'will come back and that one I will talk about.
23 CHAIRMAN KRESS: It will turn around eventually, 24 is what you are saying?
25 MR. TAKEUCHI: That one I will talk about later.
l l
i .'
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258 1 But the thing to be noticed here is that as the 2- curve is higher, more liquid can go through, so that it is 3 less restrictive. So Wallis flooding curve is much 4 restrictive than slightly inclined pipes, straight pipe.
5 DR. SCHROCK: In Wallis' description of the I 6 process, there's the influence of end conditions that he 7 talks about, you know, the influence of end conditions.
8 MR. TAKEUCHI: End conditions. That one appears 9 --
10 DR. SCHROCK: And the end conditions in various 11 experiments were somewhat different for the vertical pipe 12 and the way the fluid is introduced at the bottom and at the 13 top can influence value of M and also the value of C. None 14 of those experiments had such asymmetric geometry as you 15 have for the vertical section in this application, where it 16 is fed through an elbow going from the horizontal flow, and 17 then into the pipe. So I wonder if it isn't a little 18 dangerous to presume that the Wallis correlations encompass 19 enough range of end conditions to lead to the conclusion l
20 that you are going to make on this, that Wallis flooding is 21 the controlling --
22 MR. TAKEUCHI: Yes.
-23 DR. SCHROCK: It may be that flooding in the l 24 vertical section is controlling, but it may be that we don't 25 know what the flooding process is in the vertical section i
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259 1 when it is fed by this elbow from a nearly horizontal pipe.
Y)
.L/
2 MR. TAKEUCHI: The liquid is fed from the top.
3 Steam is coming from the bottom.
4 DR. SCHROCK: That's right. But that has --
5 MR. TAKEUCHI: End condition important for this 6 equation, maybe the steam --
7 DR. SCHROCK: It is the steam that is holding --
8 it is the steam that is holding the liquid back. And so it 9 is at the bottom of the channel where the critical situation 1 10 exists is in side orifices and BWRs and so forth, that's the 11 location where the process is really governing.
12 I suspect that the process that governs the 13 counter-current flow limitation here is also at the junction 14 between that elbow and the vertical section. But you need
() 15 data to establish what that relationship is. You are not 16 going to get it very reliably by putting together the 17 correlations from vertical pipes with symmetric entrances at i
18 both ends. I I
19 -MR. TAKEUCHI: Could you raise that question at '
20 the very end of the presentation?
21 DR. SCHROCK: Sure, l
22 MR. TAKEUCHI: The end -- end effect is included 23 within this uncertainty of the flooding. If the end effect 24 is not -- the end can be smooth end or not very smooth end.
25 But flooding correlation can be enclosed within this range.
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f -___- _- _ - _ _ _ _ -.
260 1 DR. SCHROCK: Well, I am raising a question as to
() 2 how sound that argument is, and that is based on the fact 3 that none of the experiments leading to Wallis' bounding of 4 this kind of performance, and it is dependent upon the end 5 conditions, went as far as introducing the vapor or the gas 6 phase into the bottom of the tube through an elbow.
7 MR. TAKEUCHI: That part -- but you are aware that 8 the Wallis range are is taking into account that end 9 effects within the limitations, not --
10 DR. SCHROCK: I know very well what the 11 experiments are that it is based on. I have studied all 12 those.
13 MR. TAKEUCHI: Yes.
14 DR. SCHROCK: And what I am telling you is that
() 15 none of them look like an elbow at the entrance.
16 MR. YOUNG: Excuse me, Virgil. I though the end 17 effect for those tests was essentially at the top. Wouldn't 18 it be where the pressurizer connects with the vertical pipe?
19 I thought that's where the end effect was, since that is 20 where you introduce the water.
21 DR. SCHROCK: No, it is both ends, in general, and 22 it may well be shown that one end or the other is 23 controlling the situation in a given geometry, or in a given 24 experiment. But the entrance into the channel is frequently 25 the location where the flooding relationship controls what O ANN RILEY & ASSOCIATES, LTD.
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261 1 the-correlation is giving you.
- l. , x) 2 MR. YOUNG: Entrance from the point of view'of the
. ;.J 3 l'iquid?
4' DR. SCHROCK: At the bottom. The entrance of the 5 gas into the thing.
6 MR. YOUNG: .Well, Kenji was going to look at the.
7 elbow and make some conclusions, so maybe we should wait for 8 that.
9 DR. SCHROCK: Yeah. Well, I think what I am 10 -saying is only that it points up the need for a simple 11 experiment in which you essentially mach up the geometry of 12 the elbow going into a vertical pipe and mach up the 13 . geometry at,the base of the pressurizer. That wouldn't be 14 an expensive experiment-to do and you could have had a (n,)- .
15 correlation for it that would either meet your needs'or f
J 16 verify the presumptions that you have made here.
17- MR. TAKEUCHI: Yes. We didn't have that 18- experiment in hand, but we are trying to do the best we can
-19 do.
R20 Where am I? Another. program of Wallis correlation 21 is that that one is applicable.only for small pipes like 22 less than - diameter less than 2.5 inches. However, we
'23 have generalized Wallis' flooding correlations by 1
24 : introducing the dimensionless diameter D star in this !
25' ' equation, and dimensionless - ' flux correlation L star in ,
i l
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262 1 such a way that the Kutateladze liquid holding up condition L[
d' 2 'is going to be satisfied. The result is equation 2.4. In
,3 fact, for small diameter, equation 2.4 is radius to the i 4 flooding correlation by Wallis, and for large diameter, 5 equation 2.4 is. radius to equation 2 -- 6 in terms of the 6 Kutateladze number.
7 If you plug in K L star is equal to zero you can 8 see that K D star is code 3.2, that is Kutateladze record 9' holding up condition.
10 Yes?
11 DR. SEALE: He is going to recharge you there.
12 DR. SCHROCK: Could I make one further comment to 13 -l try to clarify the problem I see for your flooding i 14 correlation? Introduce the liquid at the top from the
/\
( ,). 15 pressurizer. There is going to be a tendency for the liquid 16 to fall as a film on all of the wall.
17 MR. TAKEUCHI: Yes.
18 DR. SCHROCK: Even if it all came in one side, 19 with.significant height, it would tend to spread around.
20 MR. TAKEUCHI: Yes.
21 DR. .SCHROCK: Okay. When it gets to the. bottom, 22 the gas is coming in a stratified flow, it has to break the l
23 liquid to get in.
24 MR'. TAKEUCHI: Yes.
25 DR. SCHROCK: Okay. So it has the capability now ANN RILEY & ASSOCIATES, LTD.
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263
, 1 to stop.the liquid and push it back in a way that is not r
j 2 represented in the correlations that Wallis gave you. None 3 of those other experiments had this kind of breaking through 4 Lat the bottom like that'I think.
5 DR. CATTON: There is.also the possibility of a 6 hydraulic jump at the bottom.
7 DR SCHROCK: Yeah.
8 MR. TAKEUCHI: But that one is not -- if you can 9 suspect, you can suspect that way, but at the moment we 10 . don't have that-test, so.
11 .And this, in the summary of. scaling study, the 12 previous. equation, could that generalize the flooding 13 . correlation is applied. Wallis flooding curve is.here, and 14 the OSU surge line is larger, 3.5 inches is larger than f'
'15 -Wallis flood correlation can be applied to it, and it 16 becomes like this, for AP600 it becomes more lower than'the 17 Wallis curve, so as the pipe diameter increases, the 18 flooding becomes more and more restricted in the vertical 19 pipe.
- 20. On the other hand, for the slightly inclined pipe 21 --
22 CHAIRMAN KRESS: Isn't that a little
~
23 counter-intuitive to you?
24 MR. TAKEUCHI: Excuse me?
25 CHAIRMAN KRESS: Isn't'that a little-() ANN RILEY & ASSOCIATES, LTD.
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264 l' ,
il counter-intuitive?
~ 2'- MR. TAKEUCHI: Counter-intuitive?
-3 -CHAIRMAN KRESS: That the bigger the pipe is, the 4 -more' restrictive the flooding is? Is that counter-intuitive
)
l' I5. to you,' Virgil?
6 .DR. CATTON: No.
7' CHAIRMAN.KRESS: It's counter-intuitive to"me.
8: -DR. CATTON: Tom asked the question, so'I am just
- 9 --
10 DR.'ZUBER: To me, it'is. Can you explain really 11 why is that so?
- 12. MR. TAKEUCHI: Why is that so? Well,'the actual-
'13 ' condition is - this one is equal to 3.2. This is 14 Kutateladze for-steam.
DR. .ZUBER: That is independent of diameter?
i 16 MR. TAKEUCHI: That's right. Depends on the ---
17 7
-faced agent, but independent -- becomes independent of pipe 11 8 - diameter. As pipe line becomes larger. But'Ioam drawing 19- 'this curve on -this Wallis square : root of j (G) star 20 ' coordinate' system..
. 21- DR. ZUBER: Kutateladze would not even come on 22 that graph.
1 92 3 - MR.-TAKEUCHI: Huh?'
24 DR. ZUBER: Kutateladze's number would not even 25' .come on that graph..
l
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- 265 l-l
'l MR. TAKEUCHI: You see that they are related by
'2 this expression. Kutateladze number becomes constant as the
(
L 3 . pipe diameter becomes larger. But Wallis, the correlation 4 depends on the diameter. And I am just floating the 5 constant value in the square root of j (G) star coordinate 6 system. Actually, steam velocity. The steam velocity to 7 hold up~the liquid becomes independent of diameter. When 8 the diameter is more, it depends on diameter. When it 9 becomes -- diameter becomes larger --
10 DR. ZUBER: Okay. Then why you can say it is more 11 restrictive in increasing diameter? Because this is an l 12 artifice, this is the way.you brought it, you came to this 13 conclusion.
14 MR. TAKEUCHI: The restrictive coming from how
- 15 much liquid can go through.
16 DR. CATTON: You are not answering his question.
17 MR. TAKEUCHI: Yeah.
18 DR. ZUBER: It is independent of diameter. The 19 Kutateladze's number is independent on diameter. If I take 20 a large' diameter or a little bit smaller, it doesn't make
- '21 'any difference. There is no restriction on the diameter.
22 DR. CATTON: Once the diameter'is large enough.
23 D'R . ZUBER: Large enough.
MR. TAKEUCHI: Steam flow rate is independent of 25 . diameter. But a liquid flow rate can be larger. l 1
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266 1 MR. YOUNG: I think, just to try to clarify, 4
i :2 really what we are saying is that beyond some diameter, the
,%/
l 3 flooding performance becomes independent of diameter.
I
! 4 DR. ZUBER: We agree.
5 DR. SCHROCK: We would agree with that.
! 6 MR. YOUNG: I think what'-- the way he is saying l
L 7 this is that the Kutateladze type scaling is more 8 restrictive than J-star. scaling is J-star scaling were to be t
9 applied to larger diameter pipes. With J-star scaling, the 10' larger the pipe, the more downflow you would allow for a 11 given steam upflow. So what he is talking about here is the 12 more restrictive scaling that comes out of the K-star and 13 Kutateladze number.
14 DR. CATTON: So what you are saying is that gj 15 -NOTRUMP is more restrictive.
16 MR. YOUNG: Since it uses the Kutateladze number.
17 Yes.
18 MR. TAKEUCHI: I think the superficial velocity 19 may be the same, but the total amount of the water coming 20 down is larger.
21 For CCFL in the horizontal elbow is compared to 22- the CCFL in the inclined straight pipe. For this let me 23 pick -- select a convenient set'of steam and liquid flow 24 rate. Steam flow rate.is the flow of steam just to hold 25 that liquid in the vertical section of the pipe. So from
,[ \
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267 1 this Kg star islequal to 3.2, the Wallis number for this
() 2' condition is obtained for OSU and the AP600. And the liquid ,
3 phase of flow rate is selected in such a way that flooding 4 just takes place in the inclined pipe with this steam 5 upflow.
6 Applying these values to figures 3 and 4 for OSU 1
7- tests, 0.71 is applied and we get the steam flow rate .575.
8 That's where this number comes from.
9 In the case of the AP600 this number is applied to
,10 here and you've got 1.08 here.
11 Now we need another thing, that's the void 12 fraction. That is obtained from the critical conditions, 13 applying steam flow rate for OSU, .534, and we get void
'14 fraction .8.
g ,/ 15 And applying 0.26 of the AP600 values, and we get 16 void fraction .63.
'17 So we have with those conditions applied steam 18 velocity and.the liquid velocity is obtained~like this.
19 The location of liquid inside the elbow is 20 investigated by Banerjee, and this is a cross-sectional view 1
- 21 of the 61bnw with curvature R. So actually this pipe is 1
22 formed in this fashion. And the location of liquid in this ]
23 cross-section is indicated by annular delta. And tangents 24- of delta is the ratio of centrifugal forces and gravity j 25 forces. ;
-l f
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268
- 1 If I apply the previously obtained liquid and the
() 2 3
steam velocities to this formula, I get large negative angles for both OSU and the AP600.
4- As long as the delta is negative, the centrifugal 5 force exerted through the liquid combined with the gravity 6 force makes a larger body force than the gravity force by 7 itself.
8 So this one is going to stabilize the liquid phase 9 more than gravity force alone. Therefore, it is concluded i
10 that the stability in the elbow is higher than the straight 11 pipe of the sections.
12 Next subject is a CCFL in a vertical elbow, that i
13 would be compared to the vertical CCFL. Now for this let me j 14 think of a series of straight pipes tangent to the elbow and
( ) 15 increase the angle of inclination and see what happens.
I 16 First, gravity will do two things. One component l 17 of the gravity that is orthogonal to the length of the pipe 18 is going to stabilize -- stratify the flow. Another 19 component -- that one is along the pipe length, that one l 20 will accelerate the fluid. So once the inclination of the 21 pipe is increased, stability of that stratified flow is not 22 much influenced, but liquid starts running faster.
23 Therefore, CCFL becomes less restricted. That's the reason l 24 we saw that the flooding curve became higher, higher with 25 -inclination.
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269
( 1 However, when the inclination becomes larger than b)
\.
2 60 degrees or so, the effect -- the component that we used 3 for stabilizing the stratified flow becomes significantly i
4 less so that we get this kind of results.
5 Curve 2, this one is for 45 degree inclination, 45 ;
l 6 degree. Curve 3 is for 70 degree. And curve 4 is 85 7 degrees. It's coming backwards. And with inclination it 8 gets more and more restrictive.
9' When the angle becomes 90 degrees, the component, 10 that's when the stabilizer stratified flow becomes zero, and {
11 surface tension takes over the stabilizing role.
12 All those things considered, CCFL in the vertical 13 section of the pipe is the most restricted.
14 And here --
. d- 15 DR. SCHROCK: Excuse me --
16 +
MR. TAKEUCHI: Yes.
17 DR. SCHROCK: In your graph 4-1.
18 MR. TAKEUCHI: 4-1. Yes. j l
19_ DR. SCHROCK: I see you don't draw the lines all l 20 the way back to the intercept for the inclined pipe, and 21 what you say.is right, you know, for large values of j f 22 down --
23 MR. TAKEUCHI: Yes.
24 DR. SCHROCK: What you've said is right; for small 25 . values isn't the opposite true?
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l 270 1 MR. TAKEUCHI: Small values. I missed the small.
2- Which one'is --
(% )i 3 DR. SCHROCK: Well, you have curves. numbered 2, 3,
- 4. and 4 --
5 MR. TAKEUCHI: Yes.
15 DR. SCHROCK: Which do not extend to the intercept 7 for the axis.
8 MR. TAKEUCHI: That was -- the only reason was 9 ~that -- numerical matter. But you can --
10 DR. SCHROCK: Well, the point I'm making --
11 MR..TAKEUCHI: It is kept separate like this.
12 DR. SCHROCK: Let me make it again. If you look 13 at line No. 4 --
14 MR. TAKEUCHI: Yes.
) 15 DR. SCHROCK: It cuts across the other.two. So 16 the argument that you're making is valid in part of the 17 range, but the opposite seems to be true for the remaining 18 part.
-19 MR. TAKEUCHI: Yes, that is true, because that one 20 .is a type of theory I have applied, and it has only gravity 21 force. So up to 90 degrees this may collapse down to here 22 because up to 90 degrees gravity is not going to stabilize l 23' the stratified flow. However, surface tension is going to i
L 24 stabilize the flow.
25 DR. SCHROCK: Well then would you clarify the
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271 1 meaning of the angle. The angle is measured from the i
Ll9- \ ,/'
2 vertical down.
3 MR. TAKEUCHI: Oh, Oh.
- 4 DR. SCHROCK: Or the angle is measured from the L.
L 5 horizontal up.
6 MR. TAKEUCHI: Angle is the horizontal up.
7 DR. SCHROCK: So if it's horizontal up, it 8 means --
9 MR. TAKEUCHI: That one-is --
10 DR. SCHROCK: 4 should-be approaching i finally.
11' Line No. 4 when you go from 85 to 90 degrees --
12 MR. TAKEUCHI: Yes.
13 DR. SCHROCK: Ought to become line No. 1.
14 MR. TAKEUCHI: Yes.
' 15. DR. SCHROCK: But it's crossed over it.
16 MR. TAKEUCHI: Crossed over it because this one 17 does not take into account the effect of surface tension.
18 :In this calculation it takes only gravity effect in it. You 19 can see'in that Kutateladze number that surface' tension is-20 going to'be critical values for that --
21' DR. SCHROCK: But'you're plotting it as Wallis 22 . numbers, and there is no surface tension.
'23 MR. TAKEUCHI: The pipe size is so small. Wallis
'24 7 correlation is empirical correlation for smaller pipes.
lMi MR. YOUNG: Virgil, what Kenji's trying to show is e ,.
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=-_____-__
272 1 that as the inclination increases towards vertical, the t
2 lines' start to collapse back down toward the Wallis lines,
)
3 but they won't become the Wallis lines because the stability 4 criterion does not include the surface tension.
- 5. DR. SCHROCK: Well, he's attributing it to surface 6 tension effect, and the coordinates that he's using here are 7 the Wallis parameters in which there is no surface-tension 8 effect. So it's unclear why it's plotted in this context if. H 9 surface tension is a parameter of,importance.
10 You know, you're presenting the figure in order to 11 convince us that something is going in a certain trend, but 12 when I point out to you that the. correlation that you're 13 showing line No. 4, which ought to be coming in to become 14 line No. 1 as you get to finally a vertical pipe, you argue f3
(_) -15 then well, the reason that it doesn't do that is on account l 16 of it doesn't have surface tension. There isn't any surface 17' tension in anything up there that's evident. It's in Wallis 18 parameters, 19 MR. TAKEUCHI: But you see that flooding curve 20 with surface tension is included here in the OSU and the 21 AP600 flooding curves, and they're lower than the worst 22 flooding curve.
23 MR. YOUNG: The surface tension effect comes about 24 by d star there, and it's just a way of putting everything 25 on the same coordinate system. The traditional way is to ANN RILEY & ASSOCIATES, LTD.
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273 1 use k star as the flooding limit, but here he just used j
( 2 star, and there's a d star there that relates j star to k 3 ' star. So that the surface tension effect essentially is in 4 there for those vertical flooding lines.
5 DR. SCHROCK: These are just Wallis's lines. You 6 could compare it to some others, your OSU line and your 7 AP600 line.
8 MR. YOUNG: That's true. Virgil is right there.
9 The AP600Lline should be lower than those two there because 10 of the larger diameter.
11 MR. TAKEUCHI: They are lower on this, yes.
12 MR. YOUNG: Okay.
13 MR. TAKEUCHI: However, those lines -- however, 14 this line even becomes lower than those AP600'and the OSU
(-
l
- s. ) 15 curve too as inclination grows much higher because surface 16 tension is not included in this calculation.
17 DR. SCHROCK: Am I convinced? No.
'18 MR. TAKEUCHI: No?
19 DR. SCHROCK: I have heard what you have said.
20 MR. TAKEUCHI: Finally, the test data is analyzed.
21 The data, liquid downflow rate, is .125 pound per second.
.22' This-number is converted to the Taitel-Dukler number for the 23 liquid phase, 0.253. This number is applied.to the 24 generalized flooding curve and computed by Kg-stars, square 25 root of it'becomes 1.~44. This number is converted to the l'
A
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l 274 1 steam flow rate to the 0.117 pound per second.
{) 2 This number is to be compared with the steam flow l
1 3 rate in the ADS valve 1 to 3 -- they are all together -- and 4 that becomes .05 to .08 pound per second. Those are the 5 data.
6 Those two has to be compared and if I plot tnia 7 one in the flooding curve, we have the OSU data here, and 8 the prediction is shown by a circle.
9 I think their agreement is fairly good to support 10 the conclusion that the liquid holding up is taking place in 11 the vertical section of the surge line.
12 DR. SCHROCK: Which correlation -- this is your 13 OSU correlation with the intercept at 179 on Kg scale?
14 MR. TAKEUCHI: Yes.
k 15 DR. SCHROCK: 131 on K1 scale.
16 MR. TAKEUCHI: That's right.
17 DR. SCHROCK: And your lower than that by about 25 18 percent or 30 percent, something in that range?
19 MR. TAKEUCHI: That's right and then there are 20 some end effects included. This may have some range of the 21 flooding curve.
22 In the case of Wallis, the range of uncertainties 23 is 1 to .75, if I multiply .75 to this number and draw 24 another line here the this one will be fairly close to the 25 flooding curve -- we said this one is below the flooding O ANN RILEY & ASSOCIATES, LTD.
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275 1 curve also, It is not here.
{ 7 DR. SCHROCK: That's deemed good enough, I guess.
3 Well, the point that I was making about the liquid being 4 broken by the steam that is entering at the bottom would be 5 ' expected to produce the kind of comparison that you show 6 here between the data and.what you expected the curve to be.
7 MR. TAKEUCHI: That was quite interesting a test 8 to see the results of it.
9 DR. .SCHROCK: So you have data enough from OSU to 10 do several such determinations and plot a flooding line 11 . based on OSU performance, haven't you?
12 MR. YOUNG: Yes, we have. There is a range of 13 steam flows in OSU that we could look at and a range of F.4 drain rates. That's true. I mean this is one point.
!/ \
l ( ,/- 15 DR. SCHROCK: Well, I would think that if you 16 obtained a correlation from the OSU data that applies to 17 this specific geometry with the' elbow entrance at the l
18 bottom, then you could translate that into a correlation for 19 the AP600 geometry.
20' Have you missed an opportunity there? ]
I 21 MR. YOUNG: Probably one of several, I guess. I i
22 [ Laughter.] l L
23- MR. YOUNG: Yes, but again let's look at how the f 24 prediction for drain rate is now, with NOTRUMP. Maybe
- 25. there's another time to discuss this again.
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276 1 MR. TAKEUCHI: So the conclusion is that limiting
( ) 2 CCFL in the pressurizer surge line is taking place in the 3 vertical section of the surge line, and the vertical 4 flooding model used by NOTRUMP is applicable in this 5 situation.
6 Of course, the unknown factor is the entrance of 7 the steam coming in. The way it is coming into the vertical 8 section is an unknown factor.
9 DR. SCHROCK: Well, my impression is that there is 10 too much emphasis on reaching the bottom line condition that 11 NOTRUMP is applicable. NOTRUMP isn't applicable and there 12 are enough data now from the AP600 tests to do something 13 better than that, and it would be in your interest to do 14 that, I would think, but I would say what you have shown
- 15 here probably indicates that the use of NOTRUMP is not going 16 .to create a major problem for you in this area, although it 17 is tending, I believe, to produce results that are 18 conservative and that it might be in your interest to 19 sharpen it up a bit by extracting from the OSU data what you 20 can'in the way of a flooding curve that could be argued for l 21 AP600.
22 MR. TAKEUCHI: Thank you very much.
- 23. DR. BOEHNERT: We are going into closed session j 24 now.
25 [Whereupon, at 5:00 p.m., the meeting proceeded in l i
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. Pr '
tJ. 2 3:
4 5 -.
6
- 7. -
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- 10 11-
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- 17
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l 289 l
1 OPEN SESSION l
[)
\_-
2 [5:16.p.m.]
3 CHAIRMAN KRESS: We are back into open session.
l 4 Do you guys want to hold your comments for 5 tc.aorrow or do you prefer -- that way you can think-about 6 them overnight? Or would you prefer to make some now?
7 IMt. CATTON: Leave it for tomorrow? Novak wants 8 to'go home.
9 CHAIRMAN KRESS: Okay. Why don't we hold them --
-10 DR. ZUBER: No, no -- there are other things I 11 would like to do -- to read something before I make my final 12 comments.
13 CHAIRMAN KRESS: Okay. I think that would be 14 perfectly fine. With that then --
() 15 DR. SCHROCK: I had commented that I thought their 16 NOTRUMP calculation was conservative but I think I had that 17 backwards. Conservative that you are letting more liquid 18 down out of the pressurizer by your NOTRUMP calculation than 19 occurs in the actual situation.
20 I think that this result is the opposite of that, 21- isn't-it? That you are holding me more -- so it is not 22 conservative.
23 . CHAIRMAN KRESS: I am not sure which way is 24 conservative -- we have a three-way conversation here.
25 DR. CATTON: Okay, I'm sorry. I'm sorry.
L %
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l 1 l l-290
'l CHAIRMAN KRESS: Okay.
() 2 3 you like.
DR. CATTON: But I could make a brief comment, if 4 From what I have heard today it seems to me that 5 the key parameter is this ADS-4 resistance and it does two 6 things.
7 As you increase the resistance, you will delay the 8 onset of IRWST injection and you will deepen the minimum l 9 that it goes.through because -- I could go into more detail 10 if you want -- so to me I think what needs to be done is to 11 'somehow argue what the maximum ADS-4 line resistance is 12 going to be.
13 It is important at the. outset between the period 14 when ADS-4 starts and you get the pressure.down to where C\ 15
( ,/ IRWSt is going to initiate, it is important there because 16 that is what the term is, when it will initiate.
-17 Also it is probably a little more difficult to l 18 describe because the levels are higher and this l
19 intermittency is probably more of a headache.
u
-20 Once the IRWST is flowing, I think the comments 21 that.were made by Mike and some others about the balance are 22 absolutely right, and you -- if the flow is not enough, 23 through the ads-4, the pressure is going to go up. If the 24 pressure goes up, the level drops and you clear the line, so 25 you are going to get a cycle, so as long as the frequency I
])
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291
-1 isn't too'-- between on and off isn't tuo great, it's just a V
L coffee pot, butLthose initial stages I think are very 3 important and somehow getting a grasp of what the maximum L4 value of the ADS-4 resistance is-for the AP600 is an 5 important. step in closing all of this.
6 I think that Novak will have something to say 7- tomorrow about the scaling and what you can do with OSU with 8- respect to this. question, 9' But I think we came a long way today. I think we
.10 ' sort of have honed in on exactly where the difficulties are.
11 DR. ZUBER: And I think you understand what you 12 are really driving at.
'13 CHAIRMAN KRESS: Pardon?
14 DR. ZUBER: I think you understand our concerns.
-15 There'is one more thing, Tom. If you schedule a 16 meeting with RES --
17' CHAIRMAN KRESS: I am planning on doing that i 18 . definitely.
)
19 DR. ZUBER: What they should also address, what is l20 their program at OSU.
21 CHAIRMAN.KRESS: Their program?
22 DR. ZUBER: Their program, because they are 23 ' running experiments. We have to ask them questions -- what 24 kind of instrumentation, why did they put that 25L instrumentation, what information they want to get it from?
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292 1 You see, the needs -- because they short or something --
/ ,)
2 what were the needs and what instruments did they provide
\ _./
3 and when do they expect some results?
4 Essentially, hopefully we could put this program 5 within a short time to rest -- hopefully.
6 DR. CATTON: You could make one or two runs with 7 OSU with different orifice plates in the ADS-4 line and very 8 quickly establish whether -- at least the conclusions that I 9 have come to, whether they are correct.
10 DR. ZUBER: One could' plan a few experiments just 11 to put the thing to rest, if you can.
12 CHAIRMAN KRESS: It may be asking a lot to plan 13 some more experiments.
14 LR. CATTON: But, you see, Jose I think has been f%
( ,/ 15 given the charter to look into thic question.
16 CHAIRMAN KRESS: I see.
17 DR. CATTON: But when --
18 CHAIRMAN KRESS: Well, they may be prepared to run 19 it right away.
20 DR. CATTON: See, I am not sure he is going to run 21 the right experiments.
l 22 CHAIRMAN KRESS: We can ask. This is the type of I
23 thing we can ask Research.
I 24 DR. CATTON: My recollection is he was going to i 25 run the experiments that would enable him to better model
<~
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293 1_ 'the liquid takeup part of the process.-- the water spout and l'- l,)- 2 that part of it were I think at this point -- at the time
- 3 when I' heard about that I thought, gee, that's pretty neat, L
4 that will take care of everything.
l 5 Now I am beginning to think that a better
-6 experiment at least to get to.the bottom of the issues j 7 associated with AP600 would be to have run two'or three very 8 simple runs and just change the flow resistance in ADS-4 --
9 and then decide whether or not we'want to understand it.
10 CHAIRMAN KRESS: We will look into seeing when we 11 can schedule it. I think it would be a very worthwhile 12- meeting.
13 DR. CATTON: You have to talk to the microphone.
14 MR. BROWN: I think they have run some I think
/~T
( ,/ . 15 with multiple ADS, additional ADS-4 failures which isn't 16 exactly what you are saying, but you will get -- but.it is 17 sort of equivalent. You will obviously get overall ADS-4 18 increased resistance and I knew that they have run some 19 additional cases like that.
20 DR. CATTON: They may already have the data then.
1 21 MR. BROWN: Yes, but I don't think they have run I 22 some with large or small orifices but they have run some. I 23 think they have additional ADS-4 failures.
24 DR. CATTON: I don't care how they increase the 25 flow resistance as long as you know what it is.
s
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294 1, MR. BROWN: I think that is part of the lesson of what the intent was, which wasn't carried out in SPES.
{) 2 --
3 There were some other reasons why it was larger size but 4 actually I think between SPES and OSU when you compare some 5 of those two, unfortunately it was a bit too large, but you 6 kind of do see some effect -- the effect of the resistance 7 in which case the pressurizer drained very, very rapidly in 8 SPES, where in OSU you see a much more extended draining 9- period in there, so you do see some effect of the resistance 10 there also.
11 CHAIRMAN KRESS: Okay. With that then, I am going 12 to declare a recess until tomorrow.
13 (Whereupon, at 5:21 p.m., the meeting was 14 recessed, to reconvene at 8:30 a.m., Tuesday, May 12, 1998.)
() 15 16 17 ,
18 19 y 20 i I L 21 22 23 24.
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