ML20148K320
| ML20148K320 | |
| Person / Time | |
|---|---|
| Issue date: | 02/24/1988 |
| From: | NRC |
| To: | |
| References | |
| NUDOCS 8803310074 | |
| Download: ML20148K320 (110) | |
Text
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O UNITED STATES NUCLEAR REGULATORY CORDdISSION In the Matter of:
BWR MARK I CONTAINMENT PERFORMANCE INFORMATION EXCHANGE WORKSHOP l
CORRECTED COPY l
EVENING SESSION ,
i
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Pages: 162 through 270 Place: Baltimore, Maryland Date: February 24, 1938
.............. .......... er.....me..............r....==..== =i HERITAGE REPORTING CORPORATION O amaw-1220 L Street, N.W., Suite 600 Washington, D.C. 20005 8803310 74 880224 (202) 6M PDR T GMvGE*J E m PDR
E 162 NUCLEAR-REGULATORY COMMISSION O . )
In the Matter of: )
)
BWR MARK I CONTAINMENT )
INFORMATION'EXCHANGS )
WORKSHOP )
)
12th Floor Ballroom The Belvedere Hotel 1 East Chase Street ,
, Baltimore, Maryland Wednesday, February 24, 1988 The above-entitled matter came on for hearing, pursuant to notice, at 7:34 p.m.
APPEARANCES:
On behalf of the Nuclea_ 'equlatory Commission:
DR. ERIC BECKJORD MR. JERRY HULMAN DR. THEMIS SPEIS MR. LEONARD SOFFER DR. WAYNE HOUSTON O
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163 C0NTENTS STATEMENT OF: PAGE:
CHUCK JOHNSON, VICE PRESIDENT, S. LEVY, INC. 164 ;
KEN BERGERON, CONTAINMENT MODELING DIVISION 642a, SANDIA NATIONAL LABORATORIES .170 ROBERT E. HENRY, FAUSKE AND ASSOCIATES 181 MICHAEL CORRADINI, UNIVERSITY OF WISCONSIN 203 DOUG TRUE 225 I
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+ l 164 1 PROCEE' DINGS
/~'N 2 MR. HULMAN:' We still have some people that are out'
(_/ I 3 to dinner, but I'd like to start in now.
4 (Pause) l 5 MR. HULMAN: The agenda shows that we're going to !
6 have a second session on Core Debris Attack,. starting'off with l 7 Chuck Johnson from SLA.
8 Chuck.
9 STATEMENT OF CHUCK JOHNSON, VICE PRESIDENT, S. LEVY, 10 INC. ,
11 DR. JOHNSON: Good evening.
12 I'm glad I get this opportunity. The attenuation may 13 be just a couple of folks,here by tomorrow morning.
14 This is my first real interaction on this subject. I :
1 15 started in on this about six months ago. I'm afraid I'm 16 learning more than you may be learning from me. l l
17 As I watched each of the speakers, I kept saying, l 18 cops, I've got to change something. So, now, I'm getting 19 curious to see what it is I have to say.
20 We started in on the thought that perhaps the shell j 21 of the containment would act as a fin. I guess that's not a 22 new concept to most people. That's where we started from. We 23 said let's put together a little conduction model here and see . l 24 what that fin effect is worth.
O 25 May I have the next slide, please? Next slide, Heritage Rep rting Corporation j (202) 628-4888
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165 1 please. We're having to leave it up there very long, but we r~
g) 2 all recognize what that is. We're going to try and model this 3 situation with and without the water. We're going to try and 4 model it in the transient fashion. We're going to try and 5 model the debris building up transiently against the shell.
6 We're going to try and model this thing with the transiently 7 filling up the debris against the shell.
8 May I have the next slide, ple~ase. The model itself, 9 seen here, is a conduction model. The cubes you see are the 10 body heating to the casing and to the chemical energy. We have 11 conduction coefficients and radiation off the other surface.
12 We have conduction and radiation coefficients off of the seal
() 13 into the containment, wnich is water pressure we now have.
14 Conduction coefficients are the ones shown here.
15 They correspond to nuclear boiling. You see there the huge C to 16 the square term, indicates that there is a rapid temperature 17 difference. l l
18 The inflateds on the next slide. We've got a ;
19 conduction coefficient to simulate the CH 4, simulate the 20 expansion of the steel down in the concrete. The HC 3 is a 21 conduction coefficient which kind of glues together and gives 22 some Swiss data, so that at high temperature, we represent the 23 attack of the concrete in terms of the thermal energy put into 24 the concrete.
25 It comes down on that temperature exponent such as Heritage Reporting Corporation. )
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166 1 lower temperature. We shut off that heat loss into the 1
2 concrete. Let's see. The chemical energy is also temperature I U(~T 3 dependent. At high temperatures, we generate chemical energy 1
4 in the body. As temperature declines down to a thousand, it i l
5 turns on. )
6 The numbers are either standard numbers or ones that 7 we got from Mr. Hodge's folks. For instance, the chemical 1
8 energy was donated by the ORLL, the Oak Ridge people. I think ;
1 9 that's their main contribution here. We're using a low 10 conductivity and we're using our zero zero porosity. j 11 We released some of the constraints, for instance, 1
12 the porosity constraint and some sensitivity.
I 13 Let's have the next slide, please.
1 14 GENERAL AUDIENCE: (Question) l l
15 DR. JOHNSON: Would you put that slide back up there?
16 The question is that F? The F squared. I caught a 17 little heck from two of my peers at home. That means that the i 18 -- we multiplied by temperature squared and on the lower one, 19 that is simply an artifice used to get us through transition 20 from the heat transfer into the concrete at high temperature to 21 the transfer. It's simply a forced-fed between two temperature 22 areas. It doesn't have a physic'al meaning. It just gets it 23 from one temperature to the other.
24 You look puzzled.
25 GENERAL AUDIENCE: (Question)
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167 1 DR. JOHNSON: It is. I'm just trying to show it's.a
( ) 2 function of either the temperature squared, temperature 3 forward, container temperature or something else. It's not a 4 normal transfer coefficient.
5 Let's have the.next slide. . What we cal.1 the forcing 6 function. We were going to predict the -- rather, enforce a 7 debris height as a function of time. The critical organ there 8 should be in inches there. Debris height in inches. It's 9 unlabelled.
10 The base case shown here is taken from the Oak Ridge 11 data. It assumes that the release, that that release'that 12 those folks calculated for the fast black-out case. It assumes
[I) v 13 that the material is spread uniformly over the pedestal and the
- 14 dry well. That may or may not be the case. So, we will run a 15 case which is ten times faster than that to test the 1
1 16 sensitivity.
17 We're using the temperature degree shown there and' 18 we're using an ambient and initial temperature in the system or 19, 318. So, we run transient cases.
20 Let's see the next slide, please. And our low code 21 calculation for temperature profiles. You see the dashed line 22 for isoperms. A little legend to the right here goes over what 23 the issperms are. We go through step by step, time by time, 24 tracking the maximum temperature of the shell.
O 25 In this case, it's on that C contour right down at lieritage Rewrting Corporation (202) 628-4888 e - , , - - ~ - - , - , , - ,, ? -,. r- . . - , ~ - - . - , , , ~.ew
168 l 1 'the second note from the bottom. The maximum temperature here l 1
() 2 is 973. We did this repeatedly. For a particular transient, 3 you get a curve like this. _These are the temperature responses d as the debris builds up and the ambient transient goes through 5 completion for a case with no water, only radiation, the dashed 6 line, and in that case with water above the material.
i 7 Next slide shows the transient. This is a case with 8 water where you see the solid line debris builds up. We stop ,
9 the build-up arbitrarily at four inches and come across at four 10 inch dashed line, which is an asymptote. You see the six-inch 11 line. We let the debris go to six inches. We go out on that 12 six-inch asymptote.
/~N 13 Now, we said we're going to run this case, so we let V
14 the build-up go at a factor of ten faster and I became 15 concerned that perhaps we were building up so fast that we'd 16 get an overview of the temperature. You see, these transients l 17 are showing the temperatures as always monatomic. So, we're 18 sure that the steady state asymptcte is the maximum !
19 temperature. It's not clear that that's always the case. ;
20 So, for the ten times rate build-up, let's see the 21 next slide, we show basically the same results. The ten times 22 case of the four-inches approaching the same asymptote as the 1 23 base case and the build-up to six inches at ten times rate also 24 approaches the same asymptote.
O 25 So, we reach a conclusion that there wasn't any pump fleritage Reporting Corporation (202) 628-4888
169 1 or any surge in temperature.that always rose, at least for
(} 2 these rates of build-up,_ that always rose asymptotically to the 3 steady state value.
4 One intention of this calculation was to investigate 5 the problem of doing a quench or don't you quench to mix of the-6 debris, but these calculations to me indicate that very likely 7 these rates of deposition, that what we've got l~s a process 8 where the debris is quenched first and then reheats, which may 9 wind up at about the same situation but a lot later in time.
10 Let's go to the next. We then proceed to start there-11 some of our parameters. In this case, porosity. The variation 12 we made in the porosity was the reduced density, reduced the 13 heat generation per cubic volume, and we adjusted the CT J
14 conductivity.
15 We did not change or try to take into account the 16 effect porosity might have on the heat transfer to water. You ;
17 see a slight -- well, you see a difference here. It may be 18 illusionary because wa -- at the same height, we get a lower l
19 temperature with higher porosity, but, really, if you had the 1 20 same mass of material, you'd probably have more depth and be 21 back where you started from.
22 Let's see the last case. In this case, we tested the 1
1 23 sensitivity of the affected chemical energy and you see the 24 result there. It has a fairly considerable effect.
25 That's all the cases I have to show. We have been Heritage Reprting Corporation (202) 628-4888
170
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1 asked to give a one-minute summary. I'd say it's pretty
() 2 obvious that what's really important is the depth of the 3 debris, and I also think it's crucial that we know how fast,the 4 debris builds up on the wall and at what temperature it has 5 when you do that because it may make a complete difference of 6 how you -- what conclusion you come to as to when the liner-7 hea'ts up.
8 Water, of course, is very important. Some of the 9 other parameters, like chemistry if you're going to go to very 10 high temperatures.
11 MR. HULMAN: The next speaker is Ken Bergeron from 12 Sandia.
13 While Ken is coming up here, let me tell you there's
(}
14 been a change on the agenda. Instead of Gene Hughes, we will 15 have Doug True from Erin Engineering, following Mike Corradini.
16 Ken.
17 STATEMENT OF KEN BERGERON, CONTAINMENT MODELING 18 DIVISION 642a, SANDIA NATIONAL LABORATORIES 1
19 DR. BERGERON: I'm going to expose you to another l l
20 sensitivity study of the liner or the shell attack question.
21 This was something that's been performed over the past three 22 months as a result of a request by the NUREG.1150 projects at 23 Sandia.
24 I think this exercise differs from some of the ones O 25 that you've seen before because it had a limited purpose, a Heritage Reporting Corporation .
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171 1 purpose essentially of educating.and illuminating issues for a
() 2 small group of people. As a group of experts on the panel 3 associated with multi-core coolant interactions, .they were 4 addressing the uncertainty issues more than NUREG 1150.
5 That made the study a little different in terms of 6 orientation than studies that are oriented towards this group 7 or towards the regulatory issues.
8 Next slide. As you know from history, the issue of 9 shallow-shell melt through was introduced by Greene in the 1984 10 time frame. The first 1150 panel identified as a very important 11 contributor to the uncertainty risk. That study is being done-12 for the second time, for the final 1150, and it is apparent, 13 all the' indications on that will also be an important CJT 14 contributor.
15 But what's also important is that the 1150 program 16 felt there was a need for some kind of a large overview to be 17 .done to assist the panel in making their assessments of 18 uncertainty.
19 To do this, we chose a pure conduction code. The 20 reason for choosing a pure conduction code is not because it's 21 the best tool for the problem. I'll go into that in a minute.
22 But because the issue is uncertainty and the need is for a 23 large number of sensitivity studies and the need is for a tool 24 that people that haven't been immersed in this issue for a long O 25 time can understand. They can get their hands around. In fact,
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4 172 1 could we produce the calculations that they needed to at home?
/~% 2 To supplement the pure conduction code approach, the V
3 1150 program is providing-the more detailed mechanistics of 4 severe accident model results that you've seen in previous 5 discussions today.
6 So, this is not intended to be the only input to the 7 expert panels, but it has capabilities in the sense of well-8 defined models, lots and lots of sensitivity studies that the.
9 more complex and more mysterious codes don't have.
10 Next slide. TAC 2D is a code that many of you may 11 have heard about. It's a standard conduction package. It has 12 many options. It is well validated. I' don't think we have to 13 worry about whether it solves the conduction equation.
14 Probably not.
15 It's in Fortran and it's relatively easy to adapt to 16 special purpose heat transfer models, and I'll describe a 17 couple of the ones that we're developing in this new 18 computation. !
l 19 Before I do that, let me discuss what you really need l 20 for this problem. Aside from the vessel scenario, the' initial j 21 conditions and bonding conditions, given that you've got debris 22 against a liner or shell wall, what do you need. Well, you
.23 need the initial heat conduction. That's clear.
24 You need reasonable heat transfer and that means l 25 radiation heat transfer into an atmusphere which contains Heritage Re p rting Corporation (202) 628-4888
173 1 aerosols and optically active gasses. Decay heating is
(} 2 important. Crust formation may be important. Core ablation is 3 impbrtant. It affects the temperatures and the rate of gas 4 energation. Chemical heating is important, and the conduction 5 process involving sparging of gasses is also important.
6 There is no code that does all these things. That's 7 the starting point of this exercise. The other starting point 8 of this exercise is that these' people on the panel are going to 9 get elicited next week, okay. So, we've got a hard time 10 boundary to provide information to them, to say we can develop 11 a neat code that does all these things, but it's going to take 12 six months or whatever. That's irrelevant to that group 13 because once a snapshot of their opinions has been taken next 14 week, which is what they call elicitation in this business, 15 then they go home and anything that they learn after that point 16 ic irrelevant to the 1150 process. j l
17 So, we chose a conduction code that did not do a'11 18 these things and we had to substitute perimetria treatments for i
l 19 the things that we do.
l 20 Next slide. That's what I just said. Expert 21 panelists need the study soon.
22 The strong point is the two-dimensional conduction 23 problem is done well. We can address these issues of wicking 24 up.the liner. We can address the issues of heat transfer O 25 upward and downward, determining the bulk temperatures and heat Heritage ReDorting Corporation (202) 628-4888
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174 I transfer sideward, determining the temperature drop radially.
(} 2 That's something that simpler codes, like Corcon 3 code, can't do, which is not -- which is two-dimensional, but 4 it's not done.
5 However, we do correlate on three of the' issues that 6 I listed before. Chemical heating, heat transfer to the fluid 7 motion, because it's a conduction code, not a fluid code, and 8 crust formation.
9 To deal with these uncertainties then in the code, we 10 had to use quantities and capabilities in the code as 11 surrogates for the processes that we're modeling. We call them 12 surrogate parameters. We're going to do-sensitivity studies.
13 volumetric heating rates and debris fill-up cohductivity were 14 the way that we address these three untreated issues.
15 Here's the plant that we're concerned with, Peach 16 Bottom. Of course, 1150 is concerned only with Peach Bottom.
17 The area in the rectangle is the area of interest, of course.
18 Next slide. This is how we modeled it. You notice a 19 few things. You don't see the twenty-five degree angle 20 anymore. We deter.T.ined relatively early that it's a fairly 21 small contributor to the issue and it's much easier to run also 22 calculations in the linear arrangement.
23 You'll see that we have provisions in the hatch areas 24 for heat transfer, especially the heat transfer models that O 25 I'll describe in a minute, and you'll also notice that we have Heritage Reprting Corporation (202) 628-4888
175 l' left a gap in the region between the liner and the heat shield
(} 2 wall.
3 There is, by the way, a letter report which has been 4 prepared and transmitted to the expert panel, which describes 5 this calculation in great detail and gives the results. Copies 6 of that will be made available Friday.
7 Here are_a few highlights of what our sensitivity 8 study did. First, the forty-five degree angle has been 9 eliminated and replaced with ninety degrees. It doesn't make 10 much difference to the problem. .
11 We looked at two debris configurations because 12 there's a lot of uncertainty about that. There's the two-layer 13 configuration, oxide of the metal. The classic core code 14 configuration after layer flip has occurred. l l
15 We also looked at a uniform layer configuration, 16 where the metal in the oxide had been mixed together in some l l
17 sort of a slurry because of the high level of mixing. It's l 18 uncertain which is going to be appropriate. Therefore, we're 19 making calculations for both for the experts.
20 We looked at two different conditions above the note.
21 Dry atmosphere, but it is not a transparent gas. It is an 22 aerosol laden atmosphere which is very active and represents a i
23 significant heat sample. And, secondly, a water pool which we 24 chose to be at a 100 degrees C, which is probably low but O 25 doesn't make any difference to the problem.
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l 176.
1 We looked at two options for the gas between the shell and the wall. First, air filled. -We did radiation heat
( )' 2 3 transfer across' that gap. Second, insulated fill. -The reason 4 we chose the second option is because there appears to be some 5 uncertainty at Peach Bottom whether the foam lining between the 6 shell and the shield wall was ever removed or whether it was 7 fully removed. There's ambiguous documentation on that issue. ;
8 It turns out that that makes this analysis more 9 relevant and useful for plants like Brown's Ferry, where the 10 sand area rises a little higher and, in fact, an insulator'made l
11 the better assumption. So, again, there's a sensitivity study. l 12 It's going to be up to the experts to decide what is the best ,
. l 13 interpretation of reality. l
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14 It's kind of nice that when you're doing a l 15 calculation study for one of these 1150 panels, you absolve 16 yourself of all requirements of being an expert because they're 17 the experts, and, so, you say we're just going to do the 18 calculations. We're going to tell you all the assumptions. You 19 get to decide which of our calculations is most probable or 20 not. It's really -- it makes writing reports much easier.
l 21 Next slide. The nodalization. There's lots of 22 those. Specialized heat transfer models were required. For 23 example, upward to the water, we used the horizontal flat plate 24 boiling. Standard boiling curve.
O 25 We were concerned, however, about the possibility of Heritage ReWrting Corporation (202) 628-4888
177 1 enhanced boiling to sparging. This is an issue that was
..() 2 discussed by Dana Powers earlier. So, we ran some' sensitivity 3 studies with a factor of JiX times the boiling curve.
4 On the verti' cal wall, we took an average of 5 horizontal and vertical.because of the fact it's really forty-6 five degrees. The radiation across the shell concrete. gap 7 using reasonable values of certainty and unisivity and we did 8 some sensitivity studies on the unisivity, and we radiated to 9 the gas using a side calculation for the gas unisivity,which is 10 discussed in'the report.
11 Next slide. Now, a point that we emphasize strongly 12 in our report and I want to make again here. We chose two j
13 values for every one of the sensitivity parameters. We chose
(} ,
14 the base case, we chose the variation. We did not intend.those 15 to span the range of uncertainty. We are looking more for the 1
16 sensitivity with respect to parameters than to make statements 17 about uncortainty. l 18 Remember, the panel has responsibility for deciding !
1 i
19 uncertainty ranges. It was not our responsibility to do.so. )
i 20 We felt, however, that we could help them understand how you l 21 translate uncertainty in a parameter to an uncertainty in a 22 bottom line quantity.
23 So, we're really looking for partial derivatives of !
24 bottom line numbers with respect to these parameters. We were a 1
25 little bit -- we didn't spend a great deal of time analyzing l Herita'ge Reporting Corporation (202) 628-4888
178 )
-l 1 -what are the uncertainty ranges. However, we tried to choose '
N 2 numbers that were within the range and that were not NJ .
3 unrealistic.
4 We chose two basic cases and from each of the basic 5 cases, we did water time sensitivity calculations, and for -
6 special purposes, we did a couple of parameter sensitivities.
7 A total of twenty-three cases.
8 Here are the base case and variation for all the
. 9 sensitivity parameters. I don't want to go through the entire 10 table, but I want to point out a couple of the important ones.
11 One of the issues, one of the problems with using a 12 conduction code for this problem is that the debris is not 13 going to be solid. It is not going to be a purely conducted 14 medium. At least a significant portion of the debris is going 15 to be liquid, and that is going to be the dominant heat 16 transfer mechanism.
17 How do you deal with that in the conduction code?
18 Standard answer, you jack up the thermal conductivity because i
19 that's really the only relevant quantities of heat transfer.
20 Now, if you want to represent conductive motioning in l l
21 a conduction code, then you have to multiply the conductivity l 22 by perhaps several orders of magnitude. However, a crust might .
23 form, and if you've got a crust forming, then you want to have 1 24 your conductivity done.
25 So, we had to make some compromise between th'e lleritage Reporting Corporation (202) 628-4888
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179 1 insulating effect of the crust and the enhanced conductivity
() 2 effect of fluid motion. So, this is a challengelof the expert 3 panels to see how our conductivity enhancement might represent 4 this combined situation and it represents a deficiency in the i 5 model.
6 Tom Kress has approached and he described earlier 7 today, I think, may attack this problem in a different way and l l
8 may b'e superior with respect to this crust formation issue.
9 So, we chose two values of enhancement conductivity, 10 a relatively small range, ten and two, times the s'o lid values. j l
11 Chemical heat ratio is the ratio between the energy l 12 generated by oxidation reactions in the metal and the decay
() 13 heat. A range -- the real range'is probably much larger than 14 seven. We chose two values.
15 As I said, we have a boiling enhancement of six in 16 one of our variations and a range of emissivities, initial 17 debris temperatures. These debris temperatures are initially 18 fairly cold. We didn't want to have artificial cases where very 19 hot coriu'm contacted the liner and then melted the liner just l
20 in the cool-down process.
21 Instead, we took a fairly low temperature initially 22 and we allowed it to take the chemical heat to raise the 23 temperature.
24 We chose two values on contact depth. The .19 meters 25 corresponds to fifty percent of the core with a forty percent I
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180 1 porosity, and the .38 meter corresponds to a hundred percent of
(') 2 the core with forty percent porosity.
3 Here is the base case result. One of the things 4 you'll notice.here is that there's a rather broad under-layer o 5 in the sense of a temperature drop on the bulk metal and the 6 bulk debris to the layer edge, and that stands a rather large 7 distance, perhaps thirty centimeters. This is a result of the 8 relatively low conductivity representing the fluid motion.
9 This is, of course, the temperature as a function of 10 the radius. Base case again. Looking at an axio-variation in 11 the dry case. You'll see that the temperature peaks out to a 12 high value in the -- this is the liner temperature. And it is 13 somewhere -- you'll see the front edge of the liner is right in
( ]3 14 here. It's dry.
15 Next slide.
16 MR. HULMAN: Ken, two minutes.
17 DR. BERGERON: Okay. And, again, this is the dry I l
18 case. The metal -- the shelf of the two mil torps that go 1
19 parallel to each other and as you see, the melt temperaturn at l 20 1765 is exceeded at a time which is about seventeen minutes.
21 Next. We had one case in our standard one at a time 22 sensitivities that did not melt through, and that was the case !
l 23 of misconfiguration and enhanced boiling, which resulted in I 24 cool down of debris from the beginning of the problem, and very i 7-V 25 soon after the problem started, the debris was below the Heritage ReW rting Corporation (202) 628-4888
181 1 melting point of steel and, therefore, the shell did not' melt.
() 2 However, the shell temperature reacned a very high temperature 3 as we see'the heat temperature over 1400 on one side, over 1200 4 on another.
S All right. In summary, of twenty-three cases,.all but 6 two resulted in melt through. Time of failure ranged between 7 twenty-nine minutes, that time being determined essentially by 8 the heat-up time of the debris itself. Remember, we'd start at 9 a certain temperature, but the decay and chemical heat raised 10 the bulk temperature. The bulk temperature essentially 11 determined when the later failure would occur.
12 There were two cases of no melt through. However, 13 the temperatures, heat temperatures in those cases exceeded
{}
14 1100 K and certainly one has to address the question of the
.15 loss of strength of steel at these high temperatures even in 16 those cases.
17 That's all.
l 18 MR. HULMAN: The next speaker is Bob Henry from 19 Fauske Associates. I 20 Bob.
21 STATEMENT OF BOB HENRY, FAUSKE AND ASSOCIATES 22 DR. HENRY: Jerry, when I first heard about the l
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23 workshop in the meeting we had in December, I was impressed by 24 the fact that I think people pretty well understand, of course, O 25 the differences in the model, and I thought to be productive at Heritage Re wrting Corporation .
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182 1 this workshop, it would be key to provide some experiments that ,
() 2 would allow us perhaps to get an idea of which of these models 3 or mechanisms or whatever is most appropriate.
4 I certainly agree with George Bankoff in that regard, 5 that these are fairly complex processes, also that they're 6 pretty well defined and what's really needed are some 7 fundamental experiments to drive at where these differe.nces are 8 so that we can sort out which is the best approach.
9 Next slide, please. In the IDCOR analysis, there are 10 two fundamental assumptions that were made in the model and 11 there are points that were made this afternoon by George Green 12 and these are the key ones. They basically come down to the
() 13 same issues Tom Kress was. talking about. What's the debris 14 inventory, what's the debris temperature.
15 Next slide, please. When we assessed the debris 16 inventory for that model, we felt that what we were getting was l
17 a conservative assessment because when we do mechanistic 10 calculations, the most we would get out at the time of vessel i
19 failure is about thirty percent core material. So, we used 20 something like fifty percent. The remaining comes down over an 21 extended period of time and if one wants to look at the true 22 threat to the liner, if you will, it's really governed by this 23 hot material coming out pretty fast. -
24 So, we used this conservative assessment of fifty O 25 percent. l Ileritage Reporting Corporation
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183 1 Next slide, please. The degree temperature, which is
() 2 another assumption, we used 2100 degrees Kelvin, and the reason 3 we used that was because we thought that's about the lowest
'4 ' temperature that we would have and still be flowing, and the 5 reason we did that is because we thought there would be a lot 6 of energy transfer occurring between the materials that came 7 out of the vessel, the structures in the pedestal region, the 8 structures in the dry well re'gion, and the water that may also 9 be in the containment at that point in time.
10 We used this 2100 degree Kelvin just so.you 11 understand why the assumptions -- what the assumptions were, 12 but I really think to have to be productive here, we really
() 13 don't want to sit around and argue model against model, let's la get down to some real fundamental facts.
15 So, to carry that out, since that, we did talk about 16 this in December, we put together some experiments I'd like to i 17 lay out for you here.
I 18 Jeff, if you would. The major elements of the 19 experiments described for you are we have a path.between the l 20 release of the material, the dry well wall simulations, about i
21 one-tenth of linear scale. It's a meter long. The debris l 22 stimulant is iron thermite and, more specifically, Type II 23 Thermite from Orgo Thermite, Incorporated.
24 We used masses instead of fourteen kilograms. That 25 should be 11.4 kilograms and twenty kilograms in the test, and Heritage Reporting Corporation-(202) 628-4888
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184 1 if you use different concrete basements, we used both basalt
() 2 and limestone to get some idea of what the interactions might 3 be.
4 Next one, Jeff. The experiments were performed both L
5 with and without water available before the thermite was 6 released into the simulation of the dry well so that we could 7 get a perspective of the role the water played. There's been a 8 lot of talk today about what kind of role water plays. I think I
9 it's about time we did something to try and sort it out, just 10 how important it might be.
- 1) We have a simulation of the pressure suppression pool 12 used in subsequent quenching process. We have the steam that 13 comes off of the quench. Dry well sprays are simulated with-(} _
14 the recirculation system, so that sprays can continue. If we 15 dry it out, for instance, the original water inventory would be 16 there at the time of vessel failure.
17 We also have a simulated dry well or wet well regions l l
18 where inerted nitrogen is. I chould also mention not only l 19 inerted, but they were pressurized. We did that aheaa of time.
20 We pressurized it to about three atmospheres, 50 psi gauge, and 21 the reason we did it was to try and suppress the pot *ntial for 22 very dynamic interactions at low pressure. We wanted to 23 optimize, if you will, the potential for debris to try and get 24 to the wall. l O 25 So, we minimized thermal interaction, dynamic thermal Heritage Reporting Corporation (207.) 628-4888
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185 1 interaction, by increasing the pressure to begin with. The
_() 2 pedestal floor and walls, the floor is obviously concrete and 3 different kinds have been used, and the walls are painted with 4 a gray epoxy paint at the containment.
5 Melt release from the crucible is initially iron, as 6 I'll show you, at the reaction tamperature followed by aluminum 7 oxide. The estimate we have of the temperature, if we measure 8 it, I'll show you it, gives us a confidence that the 9 temperature of melt we had is releasing into the circulated 10 pedestal dry well regions as it goes to 2400 degrees Kelvin.
11 So, if you factor the test here, which frc , our 12 viewpoint. is highly conservative in that we're promoting things 13 going to the wall, we're using iron, which has a super-heat of
(}
14 abcut 600 degrees Centigrade at this point, and coming into a 15 fairly small system with a channel as you see that forces it 16 down towards the simulated dry well wall.
17 Next slide. This is a schematic of the experiment.
18 You see the test box which is the meter long. It's made out of 19 three-quarter inch carbon steel. It's separated by a doorway 20- between the simulated pedestal region on the right, the dry 21 well region on the left.
22 At the base of the pedestal region is a te.mperature 23 measurement. It's a Tungsten Rhenium Thermal Couple, so we can 24 get an idea of what the material temperature is when it O 25 contacts the base mat. We see the crucible sits above that.
Heritage Reporting Corporation (202) 628-4888
186 1 It's separated from the pedestal region by a plug. This is
() 2 stainless steel. It's a carbon steel plug with a lead insert in 3 it so that when the material finally melts the lead, the 4 material, the molten material comes into the pedestal region.
~
5 The crucible holds the thermite materials ignited by 6 something that basically looks like a sparkler with a small 7 heater just bound around it to ignite the spark and the 8 sparkler ignites the thermite. There is a transducer to 9 measure the pressure in the crucible, as you see, and in the 10 test box itself. In the far end~of the test box, there's a 11 rake of thermal couples. That's the simulated dry well wall so 12 we can measure the thermal couples with the temperature
() 13 transient that'e w have at the wall, as this material comes out 14 and moves down the channel.
15 The circulation spray shown on the spill-over goes i
16 into the circulation sump, then can be pumped back in. The flow
. I 17 meter measures the spray flow rate, and then there's a 18 calibration line to make sure that the recirculation sump 19 doesn't try to act as our quench tank. The quench tank is on 20 the upper left, which you see, and is coupled to the test box 21 with a two-inch line. It has a full-throated two-inch ball 22 valve in it that goes into a sparger to measure the temperature l
73 in the quench tanks so that you can get an-idea.of how much 24 energy is coming into the quench tank, how fast.
O 25 There's also another two-inch line through a 200 psi lieritage Rewrting Corporation
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187 1 gauge rupture disk which provides safety of compression of the 2 box if the temperature goes up as high as that'and it has.
3 I should mention the first test, which l'11 get to 4 that ball valve and the release valve, the low pressure rupture 5 disk simuletion. We did break the rupture disk. So, in that 6 test, which I'll show you, since the quench tank was basically 7 one atmosphere end the test box was about three atmospheres, we 8 did blow down and that had some influence on the overall 9 interaction. I'll discuss that with you when we get to it.
10 But that is the simulation of the quench vessel. You 11 see we also pressurized with nitrogens in the test box. The 12 circulation sump, quench tank were all pressurized prior to ,
13 initiating the test, about three atmospheres over pressure, and 14 then the vent for the first flow is also tied to another very 15 large tank that's off to the right to maintain constant 16 pressure of the crucible before we had ignition. As soon as we 17 had ignition, we closed the vent valve so that it had its own 18 pressurization mechanism.
19 The next slide is a photograph of this, so that you !
20 can see the test box down here at the base, which is a meter -l 21 long, to give you some idea of the dimensions. In the pedestal 22 region, the gas base is thirty centimeters high. A foot into 23 the figure, it's six inches or fifteen centimeters deep.
24 You see the crucible uses 11.4 or twenty kilograms of 25 molten material released into here. You can see the Thompson tieritage Reporting Corporation (202) 628-4880
l 188 1 Renal Thermal Couple down here at the bottom to measure (s-m !
2 temperature. Over at the far end, you can see circulation sump.
3 Thermal couples are embedded inside that brace. You can see 4 bolts to basically reinforce the sides of the box as we have a 5 rectangular pressure vessel here.
6 The two tillage lines, you can see the one on the 7 right, is the one that goes to the quench tank, and the one on 8 the left is the one that has the two-coded psi rupture disk in 9 it, and the other line coming in, the one-inch pipe, is the 10 spray in the circulation system, and you can see the flow make-11 up of the BP cell involvement.
l 12 Next slide, please. This is the inner-crucible, l 13 which holds the thermite mixture. Again, this is from Orgo i
! 14 Thermite. The extension, which is just normal iron, to hold 15 something as much as twenty kilograms, forty-four pounds. When l 16 this ignites, it melts the plug at the bottom and there's a 1
17 small pressure differential between the two that is measured l l
18 and that pushes it 11. l I
19 Next slide. This is looking down at the box itself.
i 20 We say it's three quarters of steel. You can see the l
21 simulation pedestal region on the le#t. The doorway that 22 separates the two. We initially had a sump configuration which l
23 you can see the material would be kept to the pedestal region.
) 24 In order to promote it to go to the wall, we had a channel down 25 the middle which is three inches wide extending the total Heritage Reprting Corporation .
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189 r~' 1 length of the box and 1.25 inches deep.
2 Up_at the top, you can see the plate that shows on 3 this end. It's already been painted. It has baffles for-the 4 recirculation line and the calibration lines, just like you ,
5 would have in the reactor system, of course, and that also.has 6 a thermal coupling. Above that, you see the plates in the other ,
7 -end as well.
8 Next slide,.please. This is a spray header, which is 9 flipped upside down for the picture and comes in from the top.
10 Two two-inch lines on the left are underneath that plate. So, 11 they're also baffled to try to keep material in-the test 12 chamber of the reactor, and the solid left is where the O 13 thermite comes into the simulated pedestal region.
14 Next one, please. That's back to the other one.
15 This is a schematic of the fire plate. It shows the thermal 16 couple positions. You can see the channel in concrete. The l 17 big hole is the recirculation line and the fourth rim couple is 18 used. It's one centimeter for two or one centimeter off the 19 base of the concrete and the next two or two cent. '. meters up, on l 20 the left-hand side, you can see it expand a little bit. Yhe 21 thermal couple wells are drilled within 85,000ths of an inch of 22 the inner-surface so that we can get an idea of the peak 23 response of the wall if raaterial comes in direct contact with 24 it.
25 Next slide, please. Test program that was carried
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190 1 out. The first test would be called test zero. The material 2 stayed in the injector. It was 11.4 kilograms and we just used 3 that to demonstrate what the melt looks like to the second.
4 The first real test, 11.4 kilograms, basalt concrete. Spray 5 condition was it was on before ignition so that the test box 6 was filled up to the recirculation limit and continued to 7 recirculate.
8 The pressure,' as I said earlier, of the test box was 9 sixty-feet PSIA, and the quench tank had a pressure of about 10 one atmosphere and that's the one that had a relief valve 11 rupture disk configuration, so it could load up.
12 The second test was twenty kilograms of basalt just O 13 like test one, just more material. The third test was like the ,
14 first test, except that no water was available. The water was 15 turned on thirty seconds after the melt came into the box, so 16 we could see the influence of water directly from the 17 experimental measurements.
I 18 In the fourth test, twenty kilograms of limestone i
19 with water available to reach the same kind of pressurization '
20 conditions, and the ones that we were fairly considering doing 21 started to fabricate twenty kilograms of basalt with the water 22 coming on three seconds after release, and then doing the same l 23 thing a~t 11.4 and twenty kilograms limestone to get an idea of 24 the influence both with limestone basalt with and without.
I 25 water.
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191 1 Next, slide, piease. Now, we go back-to the pictures.
2 Excuse me, Jeff.
3 This is what the material looks like before it comes 4 out of the injector. You can.see this is in the-crucible.
5 This is the plug with the iron down at the bottom. It's got 6 some of'the injector crucible oxide on the surface and.about' 7 half way up,. you can see.the transition to the oxide, which is 8 a slag or aluminum oxide. This is essentially an FE203 and 9 aluminum thermite slightly enriched with iron and the iron is 10 the first thing that comes out, basically the reaction 11 temperature.
12 Next slide, please. This is the steel plug that ,
O 13 separates the two after the material comes through. It 14 initially starts with a hole. It's lead about one inch and 15 this is now about one and a half inches in diameter. That, of 16 course, is the pressure boundary prior to melting through.
1 17 The next one is a pressure history down at the first l I
18 test. I want to spend just a little bit of time, and I '
19 apologize for the fact that the label in terms of the 20 experiment.
21 The upper curve is the pressure inside the crucible.
22 The lower curve is the presaure in the test box. And this is 23 somewhat of a collapsed scale, but you can see rapid 24 pressurization in a couple of atmospheres when this material 25 comes into the test box, water available, and lasts for about IIeritage Reporting Corporation (202) 628-4888
192 1 two seconds, that pressurization, and you-can see it blew down 2 because this is the one that had the-rupture disk-in it on the 3 label which we subsequently made equal pressurization between 4 the two.
- 5 Now, I want to spend a-little bit of time on that 6 pressurization spike because it turns out'that that 7 pressurization spike is discharging steen into the quench tank, 8 as you can see it increase the temperature of-the quench tank, 9 during that two seconds. That's an energy level that's about 10 sufficient to freeze the material that just came out.
11 That's occurring over a couple of seconds and when we 12 get to the larger inventories, it looks like the material gets
(/ 13 to the wall in about one second. It also turns out, if we want l 14 to analyze that based upon the CHF limitation, it's about 15 somewhere in the ball park of an order of magnitude greater 16 than normal CHF. So, it's a very dynamic process going on l
17 which has extracted a lot of heat from the melt in a very short 18 time.
l 19 I have used a lot of simulate tools in the past, and j 20 I've also looked at film boiling with lead and water, and I 21 guarantee you with lead and water, you won't see this kind of i 22 behavior. You can see the pressurization come down later in 23 time and I think that also has a substantial bearing on what 24 the debris looks like.
25 The next slide is the one with wall temperatures, and l HeritageRe$$rtin24888
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(} 1 you can see the pumps and renal thermal couple, which is the 2 ' upper one. The core temperature is something approaching about 3 1700 degrees Centigrade in this test. Remember, it's.under 4 water to begin~with'and it's also right up against the 5 concrete. So, it just tells us that we've got some very hot 6 stuff now and-it's iron to begin with, of course, and it gives 7 us an idea of when the material came in as well as the 8 pressurization.
l 9 The wall temperature basically saw nothing. As you 10 can see, the mtit didn't-get to the wall in this case. We've 11 got material coming out at about 600 degrees C super-heat which 12 doesn't get to the wall in the case of 11.4 kilograms.
O ' 13 Now, we go to the pictures. This.is a view of the 14 test box overall after test one. You can see the melt in the I 15 right which came in. What you're looking at is basically slag.
16 This is the simulated pedestal region. The doorway is in this 4
17 partition here. Flow is through the doorway. It doesn't get to 18 the wall. What you see is iron in the channel and when we 19 disassemble it, the iron is about roughly a centimeter away 20 from the wall. It's stopped, and this kind of relates to wiiat -
27 Theo was saying today. You've got to be careful how you look
- 22 at these things because you can't pull material any more than 23 when you finally get to the thickness which is dictated by 24 surface tension, but this material isn't wetting the surface, l 25 and that's basically with this inventory runs out before it Heritage Reporting Corporation l (202) 628-4888 1
D 194 1 gets to the wall.
(]
a 2 Now, we're talking about a fairly-sized inventory for 3 such a very small system. Also, you can see that these 4 reinforcement bolts are about four inches up off the top of the 5 concrete. The debris, because it did flow down, you can see 6 the different character when we get to the higher pressure 7 test, is propped up well over t'hese bolts.
8 The next picture is one of the doorway configuration.
9 You can see this material came out and all through the doorway, 10 very fluid when it got there. Aluminum oxide has a melting 11 temperature of about 2050 Centigrade. So, that tells you this 12 material came out. It was in the range of 2300 Kelvin hete
.'~~S .
\"# 13 already. It just gives you the reinforcement that we have 14 temperatures in the range of about 2400 degrees Kelvin.
15 Next one, please. This is getting close to the end.
16 You can see that the water definitely influenced the oxide 17 because you see that what we don't do is have watar there l l
18 initially. The oxide goes all the way down to the end and just I l
l 19 equilibriates, levels out, and that is iron in the channel. !
l 20 Looks like iron came out first, followed by the oxide, which 21 makes sense ithout materials that are configured in the l 22 crucible 23 Next one, please. And this is what it looks like up
() 24 close to the wall. You can see the baffles. This is the 25- recirculation sump on the left, and the iron filled up the IIeritage Reporting Corporation
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('}
V 1 channel but it stopped, and then it's about roughly a 2 centimeter thick when it got just very close to'the wall.
3 The second test then, we used twenty kilograms, which 4 is almost twice the inventory. The pressure history for the 5 second test shown here, PT-1, is the pressure inside of the 6 crucible. PT-3 is in the test box and the artist has taken a 7 little bit of license with the pressure in this crucible. He 8 thinks it equals the test box pressure.
9 You can see again substantial pressurization, and if 10 you look at the steam which is being dumped into the quench 11 tank during these three seconds, which is when this 12 pressurization occurs after the material comes in, again that's i 13 two things. That's the energy, it's a sense required to freeze L i
14 that material. It's also the energy required to vaporize all l
l 15 the water which is in there to twenty kilograms. So, within 2 ,
16 three seconds, we had a very dynamic interaction and you can 17 see that energy being deposited into the quench tank, with that 18 dynamic interaction translate that back to a heat flux and 19 again it's more than an order of magnitude above normal CHF.
20 The dynamics of the process are most important. I 21 I think one of the things this tells us right up front is we l 4
22 should not be carrying out experiments without looking at the l 23 role the water plays. It's very, very important, and if we do
( ) 24 experiments without looking at this dynamic process, it's going 25 to give us very misleading idea of what these processes are and HeritaQe Reporting Corporation (202) 628-4888 J
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196 1 conclusions tha.t one might draw from them.
}
2 Next, which is the wall temperature.-This shows you -
3 the wall temperatures and PC-3 is again the Tungsten Thermal 4 Renal Couple in the. pedestal region. This will record the i
5 temperature that's in excess of 2000 degrees Centigrade,.which 6 again translates to something in excess of around 2300' Kelvin.
7 Remember that this thermal couple is laying on the concrete, so 3
8 it's not going to give you the absolute temperature either.
l
. 9 Measured wall temperatures, TC-4, the material 10 definitely gets there now, and the temperature gets to about i 11 507 degrees Centigrade, and the other wall temperatures, TC-7s 12 run parallel to it, TC-5s are above it, and you can see the TC-0 13 5 obviously not quite as hot. We'll come back to this in a .
I 14 second, i
15 The debris itself, which is shown on the next, the 16 test two, again on the pedestal region, you can see the fluid, 17 and now this test was run at elevated pressure. Three l 18 atmospheres over pressure. You can see none of the materials 19 up around the bolts at all, which is much different than the-I 20 one which was at one atmosphere, and when we start down towards a
i 21 the -- next one, Jeff. Flip it around. Flip it the other way.
I 22 When we get clore to the simulated dry well wall, l
- 23 which is three-quarter inch steel plate, the oxide stops
() 24 roughly around this bolt, but the metal came in contact with it
- 25 to a depth of about four centimeters of pure iron, and let me fleritage Rewrting Corporation (202) 628-4888 ,
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197
() 1 see, I think it's on the next one, let me see. .
2 This is -- the oxide is taken away. Here's the .
3 plate. This is the metal which came up against it. We start 4 off with a temperature of about 2400 degrees Kelvin, and this 5 lu adhe' red to the plate by a second metal. There are obviously 6 other metals that we have in the test because we have lead. We-7 have the thermal couplea which get dissolved, etc.
8 This is high temperature melt that came up against 9 this, knocked this eff with a hammer. There was absolutely no 10 attack of the wall underneath that.
11 Next one, please. This is what the metal looks like 1 .2 at the pedestal region. You can see the pours which came from O 13 the concrete attack, driving the moisture out of the concrete, 14 etc., as it was going through this dynamic process of freezing.
15 Let's see. Which -- next one is a picture. This is a i 16 third test. If you remember, there was 11.4 kilograms. No water ,
i 17 was added till thirty seconds after it was in the test veeJe1. ,
18 The only pressurization that you see, which is a very minor 19 one, comes' from the crucible pressure, which is somewhat above 20 the test box pressure. The material, you can see no where i
21 close to the bolts, all the way down to the end.
22 Remember the other one, we didn't have -- the metal 23 didn't quite get to the end. Lo and behold, you chip away the 24 oxide, the metal didn't get to the end in this one either, 25 wh'ich is again the point that Theo made earlier, these things .
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198 1 can't necessarily flow without,and they'll only get duwn to the 2 surface tension-dominated thickness and that's it.
3 But the oxide went all the way to the wall. So, the 4 kind of temperature profile which we saw, which I don't have S with me,-because we did this one very, very recently, about 6 four days ago, got up to about 300 degrees C peak wall 7 temperature.
8 If I could have that one, yes. So, the summary 9 results of the tests we've done to~date shows that, as I said, 10 the four tests, two of them 11.4, two at twenty kilograms. Of 11 these, number three is the only one that we've done without 12 water. But one, two and three were Basalt 4, with limestone O 13 concrete.
l 14 The peak temperatures that we' measured approached 15 2300 degrees Kelvin and upwards in the pedestal. The peak box 16 pressure shown here and the duration compared three to one, in 17 test one to test three, you can see that the test one 18 definitely gets much more pressurization and lasts much longer j
19 and that's because the steam is being generated.
20 As I was telling you, if you translate that into how 21 much steam is being pushed over by that Delta P, it's 22 essentially the amount of energy required to freeze the 23 material which is in there ao being done at a rate that's about j
() 24 ten times CilF, which means that things are very dynamic. It's
' 25 not just a plain plate having water on top,of it. ;
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{} 1 The peak wall temperature we have, the twenty 2 kilograms, we got to about 507, eleven kilograms with no water, 3 288, twenty kilograms and limestor.e, we got to about 580, and 4 these are all.the various peak. temperatures in a tJnge of maybe :,
5 thirty to fif ty seconds and it was f airly flat when it gets up, 6 this peak value, no wall attack in any of'the experiments done 7 to date.
8 So, how do we interpret this? Next slide, please.
9 If you take test two, which is the one I showed you where metal 10 came out smack against the wall and you look at how you would 11 interpret this and this is exactly the' kind of thing Tom Kress
, 12 was talking about today, it should be 1700 Kelvin up there, not O 13 1800, in the analysis, you can see a perfect conduction profile 14 of the thermal couple which is its lowest in there, the 15 material came up against the wall with no super-heat, that less 16 than a second before that left the super-heat of 600 degrees 17 Centigrade.
18 Again, I make the point that dynamics in the process 19 are most important and the only way you're going to see this is 20 to look at something experimentally that represents the
]
21 dynamics of the process. ;
22 Next slide, please. So, we translate this into the i l
23 reactor system. This is a fairly complicated slide. We're kind l O 24 of running out of time, but let me just mention this. This is l 25 a MAAP calculation for Peach Bottom, which is the reference i Heritage Reporting Corporation .
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'2 The MAAP discharge is a function of time after vessel 3 failure and what it is, when we get about seventy tons of 4 material which is discharged, a lot of the CO2 which is roughly 5 forty tons, we.get about forty tons of zirc and we get about l t
6 eight tons of iron, and if you then look at how much of this t 7 flows out into the dry well region, what's the depth, you get a 8 depth of about three centimeters. ;
l 9 Three centimeters is something we've already
.10 addressed in the experiment we did at higher super-heats than' :
11 de get here. How do we scale this to the reactor system. What 12 do we think it means to the reactor system, and one of the ways !
O 13 of looking at scaling, nex.t one, please, is to basically just 14 look at the potential for transferring heat in and transferring !
i 15 heat out.
16 I'm going to represent this as a convection process.
17 Where it loses all super-heat, it's a conduction process. I 18 can still relate H as some K over growing thermal boundary 19 layer and sense whether zire or iron because we have 1
20 essentially the same kind of parallel, we expect the same kind 4
21 of H, and the flux coming in is this H Delta which is the 22 thickness of debris as mentioned before times super-heat.
23 Again, the point Tom Kress was making earlier today. -
24 Heat fuel minus two fuel melt. The ability to conduct 25 heat out is basically the Delta P from steel melting to Heritage Reporting Corporation (202) 628-4888
201 1 whatever the steel happens to be times conductivity times its (v"]
2 thickness, which is different for our test and for the reactor 3 system, and the debris thickness, too. I'm going to let it 4 conduct in both directions.
5 If I take that ratio, it gives me an idea of what I 6 can get in versus what I can get out. I'm going to look at the 7 ratio, what that ratio is for the reactor system for my test.
8 Next slide, please. So, when we take these ratios, 9 everything falls out except these terms, and I make the 10 assumption that the H's are pretty similar here. .So , it comes 11 down to the ratio of this R value for the reactor versus the 12 test is just the fuel and the debris super-heat divided by the O>
( 13 steel super-heat.
14 Knowing the test, I've got something like 600 degrees 15 Centigrade. The Delta, the thickness of the debris squared in 1
16 both cases, and the thickness of the plate, lower case t. I ,
i 17 would like that to be equal to one so when I carry that out in 18 the test, we've got about 600 degrees C super-heat. The 19 reactor case, over-stated, maybe even less than this, as the 20 Oak Ridge guys were saying today, and I truly believe that once 21 the material fails, the vessel is going to come out at pretty 22 close to saturation of what that material is.
23 Let's take 200 degrees super-heat in the reactor 24 case. Thickness of a containment wall, 0.035, two centimeters 25 in the test and that basically says that when we get something Heritage Reporting Corporation
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i 202 j )
1 at five centimeters, that looks like a twelve centimeter event 2 in the reactor system. .
3 So that the thermal transient of this material toming i
4 to the wall'is greatly dependent upon super-heat and the 5 ability to extract that because that's what's going to govern-6 the super-heat when it gets there.
7 So, in conclusion, what we've-done and seen so far !
8 with these experiments of significant scale, no thermal attack ,
9 on dry well wall in any test, regardless of what kind of ,
10 concrete we use. Water has a substantial influence on the i
11 debris distribution, particularly on the aluminum oxide, but I
12 it's also pooling everything very rapidly on its way to the C:)13 wall, and to just think of it as normal CHF or normal nuclear, l 14 boiling is dead wrong.
l 15 Water provides substantial additional cooling debris, i
l
! 16 also cools the wall, that says by axiol conduction, excuse me, I
17 it's vertical conduction. The thermal transient of the dry
{
18 well wall is dictated by conduction from high temperature t
19 debris in contact with the wall and just by going back and 20 looking at the experiments, we can see the crust formation is !
l 21 what's controlling the heat flux into the wall and that's j 22 what's controlling, therefore, the thermal transient.
23 The last thing I'm going to say is that basically
. 24 what we find is by the time this material goes through one 25 meter, it has virtually -- it has no super-heat, l
i
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203 1 That's all I have.
_( }
i 2 MR. HULMAN: Two housekeeping' remarks. ,
f 3 First, will all the presenters please make sure that 4 their viewgraphs and any of the material that they have brought 5 . with them that they want bound into the record and' copies made
- 6 of, see that it gets to my secretary or John Lane.
7 Secondly, the hotel has asked that I announce that 8 breakfast is available not.only downstairs but a continental i
9 breakfast will be available starting at 7:30.on the 12th floor 10 for a couple of dollars. ,
11 Our next speaker is Mike Corradini from the 1
12 University of Wisconsin.
13 Mike.
, 14 STATEMENT OF MIKE CORRADINI, UNIVERSITY OF WISCONSIN 15 DR. CORRADINI: What I'd like to do is discuss some f 16 analyses we've been doing on the Mark I Containment, in 17 particular with the application of liner analysis.
10 Put the next one up, please.
19 DR. SPEIS: Talk louder, Mike. Get to the 20 microphone.
) 21 DR. CORRADINI: Okay. The presentation outline is 1 22 given here. I'll try to go through it in somewhat quick 23 fashion.
- 24 Next one. The background is, of course, that the 25 Mark I Containment building has some purely unique issues that
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204 1 m'ust be considered. The dry well liner is one. Bob, I guess I
(%./^')
2 would use the word, talking about fuel cool interactions for 3 the last twenty minutes, and I think that has an important 4 consideration.
5 Also, that's very unique about the dry well and has 6 been talked about in many areas is the MCCI that could occur in 7 a very shallow pool and that has some interesting points, and, 8 then, finally, hydrogen control on surface containment.
9 Our objective is to present a methodology for Mark I 10 analysis. The calculations we'll do here and show as a sample 11 calculation only illustrate what one should do and aren't 7,
12 really definiti" in any way. More calculations will be 13 necessary.
14 First of all, the initial conditions. The current 15 calculation I'll show tonight are based on the Peach Bottom ,
16 station black-out sequence, particularly for short-term ADS 17 sequence.
IR I got most of my information thanks to discussions 19 with Cliff Hyman and Steve Hodge over the phone, and most of ,
l 20 the numbers are based on that. I'll come to the end and talk 21 about other sequences.
22 The initial core inventory is different here. The 23 range of temperatures based on the Oak Ridge analysis, again (s) 24 this is a simple point of reference, are in the range of 2300 25 Kelvin and 2650 Kelvin, and I'll get back to that at the end of Heritage Reporting Corporation (202) 628-4888
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{} 1 my discussion about temperatures and consistency in 2 calculations.
3 The mass flow into the pedestal is taken again from 4 Oak Ridge for lack of anything better at this point, and the 5 analysis I'm taking is one of the four which are going to be 6 presented next week as part of the NUREG 1150 work for the BWR 7 Mark I.
8 The clock is essentially the mass versus time where 9 early on, we were depositing about a 120,000 pounds of material 10 into the dry well over about ten minutes, and then after those 11 tan minutes, I deposit the remaining 480,000 pounds in about 12 three hours. I'm looking over there to make sure Oak Ridge O 13 agrees with the way I heard it over the phone. Okay.
14 That breaks down to about ninety-one kilograms a 15 second for the first ten minutes to about twenty kilograms a 16 second for the next three hours, and that gets it up to about 17 seventy-five percent of the core.
18 The one difference I'll point out here and this goes 19 back to the initial core inventory, for the short-term ADS, we 20 have very little zirc oxide produced in core. About fifteen 21 percent of the core is oxidized. Okay. Or the zirconium is 22 oxidized in the core. The remaining eighty-five percent is 23 available to be oxidized on the floor.
24 If we change the sequence to a long-term TC sequence 25 with no ADS, we have a shift and we have a lot of zire oxide in IIeritage Reporting Corporation (202) 628-4888
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(} 1 the core and very little remaining out of the core. So,.I'm 2 particularly looking at a situation where we have a lot of zire j 3 available for oxidization on the floor.
4 The final part of the initial condition is that one f 4
5 has to consider the in-pedestal steel structure, and we tried ,
6 to do it on a relatively rudimentary fashion I'll get to. .That 7 accounts for about eighty metric tons of material and if you 8 look at the drawings and the photographs that were supplied to i 9 me by Greg Kruger from Philadelphia Electric, that's about 465 -
10 , square meters and an average thickness of about two 11 centimeters. So, about a little less than an inch-thick and a lot of it. Okay.
(:)12
,13 The dry well basement, I had a lot of trouble with 14 this one. I talked originally to people, part of NUREG 1150, 15 and they told me that Peach Bottom was limestone CRBR-type ,
l 15 concrete, and I called PECO and Greg told me, after the phone ,
17 conversation, no, it wasn't, it was salacious, and I got over 18 the phone by telefax the construction specs on the concrete in l l
j 19 the basement of Peach Bottom, and they are given here, and if 20 we look at the fine, which should be fine, not five, fine and ;
21 coarse aggregate of about seventy percent is in that form, the ,
i 22 fine aggregate appears to be silica based.
23 So, what we're using in the analysis here is 1
j 24 limestone common sand rather than limestone CRBR based on 25 conversations with Philadelphia Electric. Okay.
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Next slide. So, those are our. initial conditions.
(}-l' 2 Now, let's walk through what would happen if the stuff started a
3 coming out. Well,-when the stuff comes out of the vessel, the :
4 first thing you have to worry about is it's got to go through l
5 the in-pedestal structure and there is a good deal of it. :
6 If you look at it, I said there's about eighty_ metric 7 tons of it, about 465 square meters, and the thickness is about i
8 an inch, average thickness. Well, this accounts for about ten !
9 percent of the volume in the upper. pedestal region. If you look i
10 at the photographs at Peach Bottom, I've not been in it, so I 7 11 can only go by the photographs I've seen, is that it's ;
12 relatively open at the bottom but relatively crowded at the top ,
(:)13 with the' control rod drives, which is a major part of its ;
14 structure, and that implies that if I had a core out of the 15 vessel, the core is going to go through all this structure.
4 16 So, one of two things are going to happen. Either 17 it's going to freeze up there or it's going to lose all that l l
18 heat as it goes through, and, so, my approach in the simple
]
l 19 calculation I've done is say rather let's look at how much can i I
20 freeze and be held up on that structure rather than lose the 21 super-heat. They are essentially equivalent. I'm just losing l
. 22 some of the overall thermal energy of the core.
23 I'll take the large core, which is about a hundred i
24 kilograms a second. If you look at the opening between the end j 25 control rod drive structure, if I had the core at a hundred
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208 1 kilograms a second and one single core stream, that's about 2 five centimeters square, and the holes are something like that 3 or a little bit smaller. So, you can imagine, for instance, 4 these things flashing around and breaking up to smaller streams 5 and the streams would probably be about three to four times 6 smaller than the initial portion. Okay.
7 So, my vision of what's happening if it were to come 8 out is it's' going to come out and break up into smaller 9 streams, maybe two to four times smaller than the original 10 diameter of five centimeters. Qkay.
11 Nov, given that, now I have a set of multiple streams 12 coming down on to the floor. Okay. As it's pouring down, it's O
\- 13 flashing through the structure and I'd be interested in how 14 much it freezes up on it. If I do a very simple freezing 15 calculation, where I have written here three balance equations, i f, essentially the density times the latent hect of the material 17 freezing times the D Delta DT is equal to how much heat is IR going into it, and the heat is controlled essentially by 19 conduction through what is freezing up on the crust and that is 20 equal to the heat-up of the steel.
21 One can do a simple calculation and see how much 22 crust builds up over time and the result of that is one l
23 estimates a frozen melt mass, not a large mass, about five 24 percent of the core, but it turns out to be substantial enough 25 that you ought not to neglect it. You can skip the next one Heritage Reporting Corporation )
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4 1 just for time purposes and we'll go on.
4 2 Okay. So, if I consider the in-pedestal structure, 3 you've got to consider that you only lose about five percent of j 4 the thermal energy or you're going to lose aboutifive percent l l 5 of the core as it freezes up on the structure as it has to go 6 through it.
7 The next thing is FCIs. Bob addressed this with 8 experiments. .I think they're very interesting experiments.
9 Maybe we can discuss the scaling afterwards.
10 What I'd like to do is talk about just a general
) 11 concept of the FCI and how it actually is probably more helpful
]
q 12 than it is harmful under these situations because our next 13 vessel with the large volume.
14 In the dry well, if I have water present, the maximum 15 depth of water I could possibly have is about a three-quarters
- 16 of a mater simply because anything more than that is going to !
- i 17 start spilling over into the suppression pool. Okay. ;
18 So, there is a possibility of an FCI or partial melt i 19 quenching when the stuff comes in through the pedestal i i ,
4 20 structure and then starts pouring down the multiple streams. s 21 Now, if we consider an JCI, one thing that we're j 22 going to have to also consider is hydrogen generation. O'kay.
23 You can't have one without the other. So that I think the )
'4 thing we talk about when we talk about quenching of material 25 and rapid quenching of materials, we're going to have to think 1
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I about what hydrogen is generated from metallic fraction.
2 In the calculation, I'll show you what we did is we 3 assumed that the material coming in breaks up to about two or 4 four times smaller melt streams and that gives you the 5 materials coming into the water about an inch balls or maybe a 6 little bit less than an inch. Okay.
7 At those large sizes, what you find at that shallow 8 depth is you're not going to get a lot of quenching. Okay. You 9 may throw stuff around, but with the material coming in at this 10 low rate, you're not, in these large free volumes, going to
- 1) throw the stuff around enough to disperse it and completely 12 quench it. You may ha partial quench, but not a large 13 quench.
- 14 You can put the next viewgraph up. This is a 15 calculation. The computer model is simply a computer-coded 16 hand calculation, where we look at a series of lump volumes.
17 One for the pedestal, one for the dry well and one for the wet 18 well.
19 The two calculations with lines that aren't colored 20 are without a wet well. That is, we essentially just looked at 21 the pedestal and dry well pressurization and I tried to bound 22 it. On the lcwer one, with the blue, is essentially the rate 23 of pressurization you would get if I had the one-inch balls l
( 24 falling in at a rate of about a hundred kilograms a second.
25 Okay. Which is the melt addition rate that I got from Oak i
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/~N 1 Ridge.
NI 2 On the other hand, if I completely quench what would 3 come in without a wet well, I have the upper line. Okay.
4 Which corresponds to about sixty kilograms of water being 5 vaporized per second. All right. This is without a wet well.
6 If I now say, well, I've got the wet well here, what 7 does that effectively do for me, what it effectively does is 8 the pressure is relatively low with the wet well and the only 9 pressurization you get which is the red line is due to hydrogen 10 generation. .
11 Where I have penciled in this number based on an 12 empirical result, the empirical result is in all the Sandia
'13 experiments with FCI, whether they be explosive or non-14 explosive, you see something on the order of twenty-five to 15 forty percent of the metal fraction being oxidized very rapidly 16 as I have the FCI, and that has to be considered i.. this case 17 because that's the partial fraction that's not going to 18 condense.
1 19 Okay. So, the bottom line for this part of the 20 analysis is that you ought to consider FCI, but if you consider 21 FCI, you have to be very careful of the fact that with the i 22 large volumes we have and under the conditions of the 23 relatively slow pouring in that you have, you don't have a l 24 great potential for stirring the stuff around and moving around 25 and getting a rapid quench. You would probably get a partial i Huritage Reporting Corporation
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212 1 quench but not a large one.
2 Next one. Okay. Now, for the MCCI analysis. The 3 important thing that I've seen since I heard about the liner 4 melt through concern is that I've never seen a calculation done 5 which tries to get the two parts of the boundary condition 6 correct at the same time. !
7 That is, I've not seen any need to do a calculation a
8 that looks at the details of what the dry well really looks 9 like in terms of the cavity, if you want to consider the whole 10 dry well a cavity, and, secondly, a time-dependent mass 11 addition because that's what we have.
i
] 12 ,
In all the calculations that Oak Ridge has presented, 13 they have particularly pointed out the rate of mass addition is '
t 14 relatively slow. Now, you may not believe that, and that we i 1
15 can discuss separately or argue about in terms of the initial 16 conditions, but given the fact that it may come in slowly on 17 the order of minutes to hours, that implies that you have to do j i
18 the calculation with the molten core concrete reaction on the 19 order of minutes to hours, not instantaneous deposition of i
20 sixty-five percent of the core. It's unphysical.
21 so, what we've done is we've tried to look at this by i
4 j 22 looking at a series of factors. Okay. And what we've looked j 23 at is first trying to pin a mass addition where I plot here on fU 24 my little graph the Oak Ridge result. That after ten minutes,
, 25- about fifty-five tons comes in, after three hours, I get all of i .
3 Heritage Re wrting Corporation (202) 628-4888 l'
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213 T' 1 it, which is about 272 tons.
(>}
2 The melt temperature ranges from 2300 Kelvin to 2650 3 Kelvin. Now, I get back to my temperature concerns. In the 4 Oak Ridge analysis, they have the metal coming in first at a 5 relatively low temperature, at about 1800 Kelvin. It then ran 6 up to about 2300 Kelvin as I get a mixture of oxide and metal 7 and eventually it peaks out at 2650 Kelvin.
8 I'm looking at Oak Ridge to make sure I got it 9 represented about right. Start flow goes hiah because I'm 10 getting metals and then oxide.
- 1) The thing that one must be very consistent about is 12 that when the stuff comes out, it's got to be consistent with O 13 material motion and in a lot of analyses, you'll see low 14 temperatures when the material really is going to be solid or 15 the opposite. Okay.
16 So, in the calculations we've done, we've looked at a 17 range of temperatures. The calculation I'll show you, we tend 18 to be conservative and picked the 2650 Kelvin value to make 19 sure that both the oxide is at least above its solidness and 20 the width weight is -- the metal is also liquid.
21 One also should consider mixing, that is, we have 22 relatively shallow pools, we have a possibility of high 23 superficial gas quantities, which means that the two layers may
)
24 not be layers at all, but may well mix because of the i
25 hydrodynamics of the stirring of the gasses. So, that's l l
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'[} J something that ought to be considered.
2 Finally,'the zirconium content, I mentioned in our l t
3 calculat. ions that I'll show today, are based on short-term ADS 4 station black-out, which means the amount of zirc is about .
5 fifteen percent reactor by the time it makes it in, and, t he n',
6 finally, the effect of water, and I'll get to that also.
7 The two key points to remember are the possible rapid P
.i 0 cooling of the pool surface and the ability to spread through 9 the dry well vall. Okay, With that as a proviso, let me show d
10 you what we did. ;
11 This is my cartoon. I didn't have time to draw it 12 earlier, so I drew it today.
O 13 What we've done is we've run a Corcon calculation ,
li l 14 with essentially no modifications. This is Corcon Mode 2, 15 where we have used a facility that nobody seemed to want to use ,
t 16 in Corcon which is arbitrary cavity. So, we have marked up the i 17 sub-region and we've essentially just assigned what the cavity !
1 l
. 10 looks like as if I went out from the sump out to the dry well 1 19 wall. Okay.
2 20 The one unphysical thing about this calculation is 21 that there's a pedestal door, all right. So, it's actually 22 three-dimensional and Corcon only knows two dimensions. So, l
23 what we've done and you see this little break in the elevation i
24 here is we've artificially made the outer anulist of the dry 25 well a little bit steeper to account for the fact that as it
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(} 1 goes out the door, It's got to go around as well as towards the 2 liner. Okay.
3 So that what we're doing is we're filling this up 4 basically like a bath tub. It comes out and Corcon essentially 5 instant'aneously lets it fill up like a bath tub and we look at 6 the Corcon interaction as it's occurring as a function of time.
7 All right. And the rate of mass addition I already mentioned to 8 you. That is, I gave you the curve of how it comes in versus 9 time.
10 And then what we're going to eventually focus on now 11 that we've talked about how much would be held up in the 12 pedestal region because of the structure, how much might be O 13 quenched because of the water there, if there were water there, 14 okay. I'll then eventually go to the liner and talk about what 15 are the key plusses of the liner and the possibility of 16 survival.
17 Next one. Okay. I'm going to show you a series of la calculations with Corcon under the conditions that the melt i l
1 19 temperature is 2650 Kelvin, we have no coolant nor no mixing, 2n okay. The oxidation is fifteen percent of the zirc. We have a 1
21 homogenous addition of the melt. That is, the melt is coming 22 in at the proportions I showed you in the very first viewgraph 73 of initial conditions. Homogeneously is a function of time but 24 with the ramping with the mass time I showed before comes in l 25 about a 120,000 pounds in ten minutes and the remainder of i
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(} 1 480,000 in three hours.
One 2 And what I show you here is two sets of plots.
3 is in short times from 12,000 seconds to 15,000 seconds or 4 about the first hour, and then I'll show you one expanded out .
5 to the whole four hours.
6 First thing is the bulk temperature. Because you 7 have this interesting situation where you're filling up a. bath 8 tub with a funny shape, you start off with a melt relatively 9 high. It fills up the sump and we start rapidly oxidizing the 10 rire at high temperatures. Okay. And what's also interesting 11 is as we till it up, we start diluting the oxide with the 12 concrete materials and both the oxide and the metal fall like a O 13 bandit to very low temperatures as I start filling up my bath ,
14 tub and going out towards the wall. Okay. -
15 Next one. Well, what does that do for erosion? Well, 15 early on, while in the sump, I had some pretty nice erosion. I l' go about five centimeters in a matter of about two minutes. I-18 have very fast erosion. Okay. Partially because of the 19 chemical in zirc and partially because of the high super-heat 20 of the metal. All right.
21 The oxide is relat 2vely near its solid point, and 22 could you go back to that funny -- the cartoon and I'll have to 23 explain these points?
24 What we plotted here is the RZ coordinate at the 25 max.'.n.um axial location and the RZ coordinate at the maximum Heritage Re orting Corporation
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9 217 1~ radial location. So, RNZ and ZNZ is the RZ erosion at the-(}
2 lowest point and RMR and ZMR are the radial and axial erosion 3 at the furthest radial point.
~
4 Okay. So, what you see is in the first hour, you get 5 nothing near the liner because there's just not enough material 6 there to make it to the liner and in the first hour, the 7 material cools down very rapidly. Okay. So, we have a lot oi 0 erosion in the sump region, essentially nothing out by the 9 liner because we just don't have the material there.
10 Next one. And that shows up in the radial plot 11 because nothing is there, nothing erodes.
12 Next one. Okay. Now, the important thing that one O 13 ought to look; at is the superficial gas velocity coming out of 14 the'Corcon calculation, and this is kind of interesting. Even 15 though we have relatively high temperatures, because it cools 16 off very rapidly, the amount of level swell that's predicted is
~
17 relatively small. Only about ten percent void pressure is 18 predicted at most of the calculations.
19 Early on, it's large because the metal temperature is 20 high, but because of the rapid cooling due to the surface area 21 and the fact that I'm essentially diluting it, okay, the bubble 22 rise velocity ends up to be relatively low, about thirty-three 23 centimeters per second, which is on the b~reak point that we may 24 or may not have mixing between the layers. It all depends on 25 exactly what the level -- I'm sorry -- the thickness of the Heritage Reporting Corporation .
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218 1 layers are and the relative densities as to whether-or not we U(~%
2 have mixing between them.
3 Finally, crust. What~I plotted here is the top crust 4 as a function of. time and you can see it starts off because of 5 the fact I have a high temperature and I've chosen it high 6 enough that both materials are liquid, okay, and can flow. I 7 have no crust. I start cooling off and the crust develops.
8 Next one. The radial crust. There is no radial 9 crust in this early time simply because it isn't at the wal1 10 yet. It's still filling up the bath tub. .
11 Next one. Now, the bottom crust also starts off with
_ 12 zero, okay, because we have minor super-heat and then starts U 13 building um to something live the -- the exponent is wrong 14 here, and it should be 4.5 centimeters instead of 4.5 15 millimeters. Okay.
16 Now, same cdlculation, except now instead of for four 17 hours1.967593e-4 days <br />0.00472 hours <br />2.810847e-5 weeks <br />6.4685e-6 months <br />, everything is collapsed and compressed, and the first 10 hour1.157407e-4 days <br />0.00278 hours <br />1.653439e-5 weeks <br />3.805e-6 months <br /> from 12,000 seconds to 15,000 seconds is over there on the 19 left, and the remainder of it is on the right. What yo'. see is 20 now as I start adding the rest of the material in very small --
21 at a very slow rate over a long time, I don't see anything. The 22 material essentially stays relatively cold, okay, and as I add 23 the zirconium, I reactei. Okay. -
24 If 1 now eventually fill up my bath tub so it makes 25 it to the wall, I now get the heat flux at the wall and I Heritaae Rew rting Corporation
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(} l apologize for my scribbling. There were no units on this and I 2 had to go back and look at the calculations. It should be 3 watts times ten.to the seventh. So, you end up with your --
4 wnen I finally make it to the wall by adding the material, I'm 5 getting something like twenty-eight watts per squared 6 centimeter and then, as I build up the crust, the crust grows, 7 the heat flux decreases markedly because I'm totally dominated
~
8 by crust growth.
9 And just to show you there is chemical heat, early on 10 - when we pour in the zire, we get a very high amoun.t of chemical 11 heat, something like a 180 megawatts of power from the zire 12 reaction, but it essentially shuts itself off because of the O 13 surface area and the fact that I'm cooling off and ropidly i
14 reacting it, and essentially at long-time, it's a minimal 15 amount of zirc reaction, a combination of the surface area and 16 the rate of addition.
I 17 okay. That's enough of those guys. Let's now move 18 to liner analysis. Next uord. Thanks. l l
1 Okay. Now, if I take the multi-core concrete 19 20 reaction and I say all right, that's my boundary condition for 21 the liner analysis, what does the liner look like. Well, I'm 22 not going to do a liner analysis. Everybody has done one. So, 23 I'm going to borrow IDCOR's liner analysis. It's a nice one.
'T 24 Okay.
25 It's essentially a thin calculation based on Mike-Heritag~e Re w rting Corporation .
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() 1 Epstein's simple analysis. I ~ think you can do it, too, if I
2 you're familiar with conduction heat transfer. I give it to my 3 students in class. '
q 4 I take a given depth of a melt pool and I treat it as 5 a heat flux boundary condition. Okay. I consider the liner to 6 be approximately as an extended surface or fin and I l 7 essentially use the. quasi-1-D analysis. I essentially borrow 8 Mike Epstein's calculation that was given to us about a month 9 and a half ago by Marty Pleise from Fauske and Associates.
l 10 I'll try to estimate the peak liner temperature 11 steady state as a function of the pool depth and the melt to 12 liner heat flux, and I think those are the tuo key design
^
( 13 parameters or parameters of interest. The heat flux to the l
14 liner from the pool and the depth of the material.
15 For the sample calculation, this is the thermal 16 conductivity of the steel. That's the thickness of the steel. !
17 That's the water heat transfer coefficient if I were to have 18 water, where I've essentially taken a boiling heat transfer l 19 coefficient, a constant times Delta T squared. Delt.a T is T-20 wall minus T-sec. The pool depth can be varied up to twenty 21 centimeters. The melt liner heat flux would be gone from five 22 to a hundred watts per squared centimeter or 50,000 to one r
23 megawatt per squared meter.
24 This is what it looks like. That's good enough. You 25 .can move it over a teensy. Okay. What we have to know is Heritage Reporting Corporation (202) 628-4888
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() 1 essen'ially t the height of the. pool which is L and the water, 2 the heat flux goes in this way into the fin, it goes out that-3 away, and back into the water on top.
4 It's adiabatic on the bottom, adiabatic on the left-5 hand side and infinite in extent. Those two assumptions are 6 simplified so you can do a hand calculation, but for all 7 intents and purposes, they're pretty good. Okay.
8 The key point that I want to make is that this is a 9 nice analysis because you can throw in radiation if you don't 10 believe that there's water there and see the effect of 11 ra'diation because the upper heat flux is just essentially a 12 constant times Delta TQ. You can put in a constant times Delta O -
13 T to the fourth and then pick up radiation. Okay.
14 So, if you put that in the M instead of being three 15 is now four and you can do the same analysis. You break it up 16 so that the peak temperature is down here at the root at T0 of 17 the fin and it's essentially equal to saturation plus the Delta 18 T you get from the heat flux coming in from the debris plus the 19 Delta T that I go from *.he potential level of T-5/T-7 and the 20 two equations are given here.
21 You can skip the pretty picture. We'll just go to 22 the results.
-23 okay. So, what I plotted here is the maximum
('
24 temperature steady state as a function of the corium depth and 25 the heat flux and you can pick your corium depth that you want Heritage Re w rting Corporation (202) 628-4888
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() I to believe and ycur heat flux you want to believe and then read
~2 off essentially the peak temperatures. The heat flux and the 3 corium depth I want to pick, though, are the ones I just 4 calculated by sample calculation, which was relatively, if you 5 want to use the word, conservative, relatively~ boundary. Okay.
6 Some people may not like the word "conservative".
7 At that point, I said that ao soon as it es-entially 8 made it to che liner wall, it was about twenty-eight watts 9 squared centimeter, was the heat flux, and the depth at that 10 time was around five or six centimeters building up to about 11 twenty centimeters when I have the total core in there. Okay.
12 When I have total options, I have seventy-five percent of the O 13 core in, the depth after a surface -- the depth of the pool to 14 the liner is about twenty centimeters.
15 So that if I start out with a small depth, all right, 16 at twenty watts per squared centimeter, you'll see that if I 17 take the radiation line and, unfortunately, I did my analysis 18 with water, I assumed there was water there, and somebody 19 called me up and said no, you can't assume that,' if I assumed 20 -- if I consider the dry case with no water, it's the red line.
21 Okay. And what you find is if I come up to a corium depth of 22 anything large, which it will be after a short amount of time, 23 we have a problem. Okay.
O 24 That is, with a relatively large cooling depth, 25 anywhere from five centimeters to twenty centimeters and Heritage Reporting Corporation (202) 628-4888
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{} 1- considering radiation only, we cannot keep the liner cool 2 enough, okay, given the heat flux. calculation I just told you 3 about. Okay. Twenty-eight watts per squared centimeter and 4 cooling fast.
.5 If, however, I had a water available, the story 6 changes because the water and the boiling phenom-acts as a very
~
7 nice heat sense and I now have the curve that I have drawn in 8 here where I can pick off any of these and if I take the 9 twenty-eight watts per squared centimeter case, all right, and 10 something like five centimeter depth, we're in a region of a 11 peak temperature of around 600 degrees K.
12 So, the point of the analysis is
- hat you can develop O
13 the relatively simple methodology using Corcon without 14 modification, essentially using Corcon as you would -- are 15 provided if you want to use it, and come up with an estimate of 16 what the wall heat flux is, okay, as a function of time, and 17 then can do a simple liner analysis to see, you know, am I'all l
18 right or am I not so all right.
'19 So, my observations. At this point, you really can't 20 have conclusions. My observations are that the MCCI modeling 21 and the Peach Bottom dry well must consider the cavity shape as 22 well as the time-dependent corium mass addition together, not i
23 separately. Okay. You have to consider it together.
24 The point I want to make about initial temperature is 25 we chose relatively high temperatures because we wanted to make ,
1 l
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() I sure that the material made it. The material was supposed to 2 move when it was at a given temperature; that is, we weren't 3 going to be inconsistent with a low temperature and a solid 4 material and depositing it'into those dry wells.
5 So, we chose high temperatures. One must be.very
.I 6 consistent with the temperature you choose and you check the 7 melting temperatures that are there.
8 And, then, finally, Corcon. appears to be a not too l l
9 bad of a usable tool for this sort of stuff, and I think if -
10 you're going to do this calculation, you're going to have to l
'I il look at five very important sensitivities. l 12 One is the mass input versus time. One is the mixing i
() 1,3 of layers because if I mix the layers, the chemistry that's j i
14 involved there may totally change to the benefit or,to the 15 detriment, okay. The heat flux is a function of time at the 16 wall. The melt temperature, the presence of water, I 17 mentioned, is quite important, and the zirconium fraction. We-18 chose a relatively high amount of zirconium metal in the 19 calculation as shown. y 20 And, so, our preliminary analysis suggests that liner I 21 survival might be possible if we enhance dt with the presence 22 of water.
23 I'll stop the're.
24 'MR. HULMAN: Our next speaker is Doug True from Erin.
25 Doug.
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() STATEMENT OF DOUG TRUE, ERIN 2
l 3 MR. TRUE: I have the misfortune today. I have good j 4 fortune to be late and I got to hear everybody's talks before I 5 spoke, but at the same time, finding that a lot of the points I 6 wanted to make in my presentation have been made by other l 7 people which is good, I think, showing that people are moving I 8 in the direction I think is proper.
9 .I want to talk about the necessary considerations in 10 evaluating potential cora debris attack on the Mark I dry well l
11 shelf.
12 To put it in the overall perspective of the Mark I 13 issues and the challenges to containments and how this can be a 14 function of sequence dependencies and quench dependencies.
15 As a background, I guess you probably all heard 16 enough times about all the analyses that have been done. I 17 guess the bottom line of it all is that there have been a bunch 1
18 of different analysis, a bunch of different input assumptions. j 19 There's uncertainty, different opinions on things, but it's a 20 very important issue to Mark I's and that's why we're here 21 discussing it for these four hours today.
22 In a microscopic sense, there are several key 23 parameters that control the evaluation, including rate of 24 debris movement through the wa.1, which is made up of a couple 25 of things, the debris release rate from the vessel, the
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(} 1 _ temperature and fluidity of the melt, temperature debris when 2 it reaches the wall, the type of debris at the wall, whether 3 it's homogenous or mixed la~yers or metals ihat get there first 4 or oxide that gets there later, and then the containment 5 conditions in terms of whether there's water present or not, 6 what the temperatures are, what the heats are for the debris, i
7 and the contributing factors that we talked about are core 8 concrete interaction, ma'terial that comes out first, what the 9 composition is versus time.
10 Now, I want to switch to looking at the perception of 11 the overall risk of Mark I. Accident initiators were not all
_ 12 created equal, and that may be sort of an obvious statement,
%)
13 but it has a big impact on the way you view my opinion.
14 In terms of station black-out which is something that 15 we spent a lot of time on today, even station black-outs aren't 16 clearly created equal. In fact, if you want to look at it even 17 further down from that, even short-term station black-outs 18 aren't created equal.
19 The reason I say that is that short-term station 20 black-outs can be when NUREG 1150 dominant sequence was the 21 loss of electrons, as I call it, there was no power at all 22 available or it can be a situation where you lost AC poter but 23 your DC power lo available and you've lost HPCI and RCIC 24 earlier.
25 Those situations have different sequence Heritage Reporting Corporation (202) 628-4888
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I characteristics and melt progression is going to be different,.
2 vessel failure mode is going to be different and that 3 contributes to different containment situations for liner melt 4 through.
5 The long-term station black-out is much different.
6 It's much further out in time. I think Steve Hodge showed some 7 really interesting plots this morning, debris versus time, the 8 difference between short and long-term station black-outs, and 9 then the other considerations, whenever you're. dealing with 10 station black-out is power recovery, and how that might 11 influence things.
12 In short-term station black-outs, for example, you're O 13 real likely to have power back before the vessel fails. That 14 means the short-term station black-out, you may be looking at a 15 wet case of containment and not a dry.
16 Transients and locas are different case.all together 17 because you have power available in the plant. You may have 18 vessel injection available through either low flow sources, 19 like CRD or SLC, and you may then have low pressure systems 20 available and not be able to depressurize the system in time to 21 prevent core damage, which would make the situation of outside 22 containment wet.
23 ATWS and TW or heat removal cases, they have 24 particular conditions associated with them with containment and 25 the conditions of containment at the time the debris is fleritage Reporting Corporation (202) 628-4888
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{} 1 released are going to be much different than for the.other 2 sequences.
3 Looking at some Oak Ridge results for BWR SAR, short-4 term and long-term station black-outs, I think you'll see some 5 of the differences I was talking about in-terms of the timing-6 of the events.
7 In a short-term station black-out, you can see it
- 8 goes from time zero initiator to maybe core relocatio'n in about 9 seventy-five minutes. In long-term, you're looking at about an 10 order of magnitude of a longer period of time. . The kinetic 11 energy is dropped by quite a bit, and your time from dropping 12 water level to vessel failure is extended quite a bit between 13 the two cases. You have about a 160 minutes or "7-0 minutes of 14 short-term black-out KE in this analysis, you've got about 270 15 minutes or so between top of reactor fuel and vessel failure.
16 That gives you a margin for recovery, changes the complexion of 17 your containment in terms of some openings.
1 18 The zirc, the amount of zirc reaction and pressure l
19 conditions in the containment when you fail the vessel. i 20 I was warned by Steve to make sure I referenced the 21 bottom line there. When debris contacts with the dry well 22 shell, it was calculated by Oak Ridge, but it was a preliminary 23 number and it was based on the models that.they discussed this 24 morning, which they are improving upon, but there is a big l time 25 difference between the vessel failure time and debris contact Heritage Re orting Corporation (20 ) 628-4888 l
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229 1 with the wall.
}
2 If you extend that out to the long-term station 3 black-out, not accounting for the differences in the amount.of 4 zirconium that was r'eacted to produce the core concrete !
5 interaction effects and the lowered KE rate, you get out the 6 time to be much longer than twenty-four hours before the debris 7 might contact the wall. .
8 As I said, those are based on the preliminary models 9 that Oak Ridge developed.
10 So, core damage state, which is leading you down the 11 accident prevention path towards the wall attack, is a function
, 12 of the sequences which lead you to different containment U '13 conditions, the availability of low volume injection sources.
14 Today, those have been largely ignored in the discussions 15 because we've been focusing on short-term station black-outs, 16 but I believe that they are important.
17 The EPGs provide pretty sure direction to use any 18 means of getting water in early containment, CRD, SLC, whatever 19 they have available to get it there, and I believe that most 20 sequences outside of station black-outs where AC power is 21- available, they are going to be an important factor in accident 22 prevention.
l' 23 Vessel pressure is also obviously a function of O, 24 sequence, and the EPG interface in terms of what operators are 25 told to do with venting sprays and depressurization of vessels.
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1 So, within our specific accident sequence that we 2 evaluate, we get to a point where the vessel may be breached.
3 As Paul Hill pointed out this morning, there may be 4 opportunities for recovery in the vessel, but since we're 5 trying to get to the point where the liner is challenged, we'll 6 go beyond that and look at the vessel failure modes.
7 MAAP has one failure mode model. Oak Ridge has a 8 different failure mode model. March has a different failure 9 mode model. Each of those codes generate a different degree of i
10 distribution rate in the containment and those models that are 11 made up of the core melt progression and vessel failure l l
12 directly impact how you get the core concrete interaction, how
( 13 you get to dry well shell interaction, and each of those_is a l
i 14 function of sequence that you're in.
15 The removal path from the debris that's overheated 16 and Steve described his model, a simplified model, the fact 17 that there are a lot of heat removal paths into the CRD tube, 18 into the water above, and I guess the whole point of this 19 basically is that since the instrument tubes aren't cooled, it I
20 looks like they are probably the weak link in the bottom head ;
21 of the BNR RV which is going to lead you to a more prolonged 22 release from the vessel.
23 Discussing the debr.s from the vessel, we have to 24 look at the composition of temperatures and release rates of 25 those, including the eutectics, the slurry, the super-heats, lieritage Reporting Corporation
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() 1 and Mike did a good job in his presentation previous to mine in 2 discussing how those impact the overall dry well shell melting, 3 but those, too, end up being a function somewhat of the 4 sequence as you look at'the impacts of the different systems 5 you have available in vessel and the progression of the 6 accident in vessel followed by the codes and the sequence.
7 If you look at three -- I probably shouldn't use the 8 word "typical", but three analyses that have been used in the 9 analysis -- of three code calculations used in the analysis of 10 dry well shell interaction, we can see you get quite a spectrum 11 of different debris versus temperature profiles that you can- ,
12 use.
O 13 Debris composition versus time, I think everybody has 14 pretty well talked about and the BHR SAR calculations and MAAP 15 calculations. It goes back to the same thing about how the 16 progression is occurring, what's happening in the vessel, how ;
17 much has melted at time of vessel failure, and allsthose things i
i 18 are a function of the accident sequence that you're in and are l 19 going to be a function of the master model that you use.
20 Debris poolability also is important. The Oak Ridge 21 analysis showed the crust formation at the wall can have a 22 significant impact on the penetration of the wall. Those 23 calculations are particularly insightful when you combine them 24 with the time versus debris depth in containment and consider {
25 the fact that you're going to get small, relatively small fleritage Re w rting Corporation (202) 628-4888
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I debris depths first followed by deeper debris depths and it's 2 likely you may have a crust formation early which may~or may 3 not be stable in the long-term, and that has to be integrated 4 into any analysis of the core melt through issue.
5 Any heat transfer models have to look at the entire
- 6 integrated package of an event, from the initiation of the 7 accident to release from the vessel, what that rate is, what 8 the composition is versus time, melting and refreezing 9 analyses, interactions that go on during the processes, and 10 integrated containment conditions, and from listening to 11 today's presentations, it looks like we're heading on that 12 path.
O 13 Mike's analysis is a step towards that. Bob's- '
14 experiment provides an interesting experimental viewpoint in 15 terms of the progression from vessel failure to wall attack.
16 This is another occurrence in the dry well wall air l
17 space, and there are a couple of points I want to make in this.
18 Depending upon the circumstances that you postulate are going- l l
19 on in containment, you can end up with a lot of different 1
20 circumstances when the debris interacts with the wall, and the l 21 failure mode of the wall is immediately obvious.
22 Some of the CBI analyses that were done during IDCOR l
23 and 187425 show that it's likely there may be some creep that 24 occurs in the containment dry well shell which may actually 25 move the dry well liner over against the shield wall whic'h Heritage Re orting Corporation
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{} 1 would limit any release of radioactivity from the. debris into 2 the Torus Room.
3 I'm not saying it.would stop it, but it's going to 4 . reduce the-flow path.
~
5 Secondly, there is.a relatively long gap, about 6 twenty feet, something like that, in the debris depth in the 7 Torus Room that's going to be relatively cool, which may have 8 an impa~ct on source term, and Dana Powers pointed out that it's 9 important in source term and that it's likely if the wall melt 10 does occur, the water, if it's frozen, is going to chase 11 whatever debris release goes out through that path and that can ,
12 impact the source term, too.
l 13 14 15 l
16 j l
17 18 19 20 21 22 23 24
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1 MR. TRUE: Just concluding quickly this 2 presentation, I believe that the accident sequences'are 3 specific, and that t,he property interactions and drywall 4 interactions that you are going to get are going to be sequence 5 specific. They cannot be decoupled from the in vessel 6 phenomena.
7 I think that point has been made pretty clearly by a 8 number of presenters today, and the f act that you have got to 9 start at the beginning and not in the middle of the problem 10 with some debris bed depth at some temperature that has no l 11 calculational basis ahead of time and has not looked at the i 12 transient effects. I
(} 13 14 It has got to be, as I said, integrated, and I think ;
that the Oak Ridge analyses are heading in that direction. -
15 EPRI's calculations are headed that way. Mike obviousl'y is 16 making steps in that direction, too.
17 There are a lot of uncertainties. And I think that 18 FAI experiments may provide a lot of interesting information 19 which may be able use a bench mark for some calculational l
20 models that are developed for this. But I think that the point j 21 that.I am trying to make'is tha't it is sequence specific, and 22 you cannot divorce it from the sequences. And the sequences 23 are quite specific, which takes us from the realm of looking at 24 only station blackout with no water to events that have to be 25 looked at in a varying manner.
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i 1 MR. HULMAN: Our last speaker on core debris attack 2 on the containment shell is Dr. Theofanous.
3 DR. THEOFANOUS: I am relatively new to this problem,
~
4 so I come to'it with the benefit of being new. I consider to 5 be a benefit to this problem. I would like to show you how I-6 began to look at it. Let me have the first one.
7 (Slides shown.)
8 DR. THEOFANOUS: The first impression that I got is 9 the same one that many of you must have gotten today. And that 10 is that it is really a very complex problem, and it has many, 11 many parts to it. So I immediately became overwhelmed with the l
12 complexity of it. A r:! I could not find myself doing a
(} 13 compilation, because at every point I had to stop and decide.
14 I could not be probablistic, because I did not know exactly ,
15 what the story was. I did not want to be conservative or 16 optimistic, because I knew that I had to do that several times. i l
17 And when you do that several times, then you compound I 18 conservatism upon conservatism, or optimism upcn optimism. And 19 then you do not know where you are.
20 So I would like to emphasize today and I would like 21 to show you how I began to dissect the problem into something 22 that I think makes if amenable to some reasonable framework or 23 technical discussion. I i
24 For the purposes of this presentation, I would like-25 to take those variables. First, I would like to consider only O
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r 236 1 one set of sequences. The boiling water reactors, as you know, 2 they have the ADS, and we have a very high likelihood that 3 those reactors are going to-be depressurized.~ If there is any 4 question on that, I feel that they should. And I feel'that 5 they should be made to be depressurized, not only for this 6 reason but for a whole host of other reasons. So I take that 7 for a given. ,
8 The other thing that I think for a given is that we 9 will consider that sprays are operational. But if their 10 availability is not adequate, then it should improve. Again 11 not only from the point of view of this problem but from the 12 point of source data and what have you, it is absolutely
(} 13 14 essential that we have it operational.
so let's do it. '
It is a very big deal, 15 So considering those two variables, I would'like to 16 proceed to show you how I work through this problem. Another 17 way of saying the same thing is that for this particular 18 problem, if any of those two are not satisfied, then I find 19 myself in a space where I do not feel comfortable for a person 20 to be to offer my judgment about what is going to happen, and I 21 do not like to do that.
22 Now the approach, this dissection approach then. We 23 begin with a probablistic framework. Because in the end in 24 order for this result to be useful to anybody, these people 25 like to know what is the likelihood, what is the probability '
Heritage Reporting Corporation (202) 628-4888
237 O 1 for failure of the liner. And it is a complex problem, and it 1
2 has many parts to it. And when you ask an expert to tel1Jyou 3 what the probability is, he has to perform a mental integration 4 of all of those different things that you saw today.
- 5 And I find myself in very great difficulty doing that 6 in my head. So I would rather do it explicitly. So put 7 .everybody on the spot at the site. As you will see now, every 8 one of those items is very important. And then they will have 9 to justify and tell me why the decision or why the judgment is 10 what they say it is.
11 Again for the purpose of emphasis here, again I want l
12 to emphasize methodology. I am going to show you-an
(} 13 illustration of what my judgment is, but I must emphasize that. l 14 It is only a preliminary judgment. I only worked on this 15 problem for about a week, and I do not like to be held up to 16 those judgments.
17 But I would like to illustrate the method. And I-4 18 hope that at the end of this conference that I am going.to look 19 at what everybody else said, and I am going to update those 20 judgments if I feel that it is necessary to do 21 so.
22 So the first thing then is to develop a zero 23 framework which separates the problem into a few essential 24 parts, and I emphasize that. You know, I can break up a 25 problem into millions and millions of pieces. But I would like Heritage Reporting Corporation (202) 628-4888
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1 to have like a minimum set.of parts that kind of interface 2 together to make up this sequence to make up the problem.
l 3 This is really just like doing a PRA. But you are 4 doing a PRA on phenomenonological problems instead on doing it 5 on whct you are accustomed to'in PRAs.
6 Next I would like to proceed with each part 7 independently, and I emphasize that. Because you are going to 8 hear a dispassionate debate on the problem, only if that is all l 9 by itself. When you make the assumptions all the way from the 10 beginning to the end, it tends to push people to extremes, and 11 I think that you have seen some of those extremes today. So I 12 would like to have each part discussed and decided on its own 1
{} 13 14 merits.
Then after I hear the quantification of each one of l
i 15 those parts, and then I wou'.d like to synthesize them, and that j
\
16 is kind of an integral part of the exercise. And then finally l l
and that is an important part hore, after this step is done, 18 there is no stop there, but it is only a beginning there.
19 Because then you are supposed to go into each one of those 20 parts, and then you bring in more and more information as it i
21 becomes available, and you reach the basis upon which those l l
22 judgments were made, and then you synthesize again. I i
23 So here I will provide an illustration. I want to )
24 emphasize that on this part that a workshop like this is very 25 important for that. Because many people work on the same O
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239 l O 1 problem. And you will see now that many of the things that 2 have been said today can be brought to bear with the individual l 3 parts here.
4 Next. So here is the framework. And as part of 5 that, I would like to emphasize that'we have to see this 6 problem right off the bat into two parts. There is a 7 short-term part that is short-term. The moment that the 8 vessels breaches, that the stuff.comes out within a few 9 seconds. That is the short-term. The stuff comes out, and it l
10 is a nuclear solid or whatever and it settles somewhere. And l 11 then there is a long-term part which is whatever was left there l
12 now gradually is being added. All right. !
13 For the purposes of this program, I feel, it is much 14 ;
_ more important to deal with the first part. I think that most i
15 of the concern will be coming out of the first part. Of i i
16 course, both parts have to be done. l l
17 Now to explain what it means is, and I emphacize that 18 this is only really splitting the problem into a few essential 19 parts, the first part is the distribution. Basically, the 20 distribution of the volume of the melt that you expect will 21 come out when the vessel breaches. And I will show you what 22 that means.
23 And then you have a distribution of the temperatures 24 of the melt that comes out when the vessel breaches. And then 25 between the two of them, you have a joint distribution of those O
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1 two things. And that has to be combined then with an 2 independent relationship between the height that will be 3 obtained in the drywall as a function of the volume of the 4 amount and the temperature.
5 It only stands to reason that if I have a very, very 6 high temperafure melt and if the volume is all at the core, 7 then probably it is going to be spread everywhere. Now if we 8 are only going to have a small amount of the melt coming out l 9 with a very high temperature, it stands to reason that it is 10 going to go up to some point and it is going to stop. And at 11 intermediate temperatures , it is going to go to something in 12 between. l c
l
{} 13.
14 So there is.a mechanistic causal relationship between the height that is obtained. And that is the height of the l
15 liner as a function of the V&T, the volume and the temperature. i I
16 Now that can be quantified independently of the other parts, 17 okay. I mean you do that. Maybe some of those heights, and 18 some of those volumes, and some of those temperatures may be 19 never realized with significant probability, but that is 20 contained in the other part.
21 So I find thr.t it really separates out the answers 22 and let people think about technical problems instead of l 23 thinking of where they want to be at after they finish.
24 So then by comparing those two, we could out with the l 25 probability distribution for the height of the liner. And then O
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a 1 we have to have another relationship that tells us what is the 2 failure frequency as a function of the-height.
3 Again that is technical. And I think that most of 4 the discussion today has been oriented to this point. This 5 then is the relationship between the failure frequency and the 6 height. Of course, when the height becomes very high, the 7 failure frequency will also. One hundred percent of the time, 8 you are going to fail. If the height is very, very low like 9 one to two centimeters, then it will be essentially zero. And 10 the other is going to be somewhere in between where there will 11 be some increasing probability of failing the liner, because it i
12 cannot be dissipated. l 13 And then finally thereby combining those two things, 14 cy combining tho.se two, you could out with a probability 15 distribution for the failure frequency. Essentially, that 16 brings you information for a degree of belief that the 17 different failure frequencies would be realized.
18 And a similar thing for the long-term. The long-term 19 exposure you have to worry about, the final equilibrium stage 20 of where the high distribution is and where the failure 21 frequency is. And of course, by combining what the failure 22 frequency is.
23 So now I would go to those different areas, and tell 24 you my preliminary assessment of them. And then I will show 25 you how I will combine them to get a result.
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242 1 Well, first the amount of volume and the temperature.
2 (Continued on next page.)
3 4
5 6
7 8
9 10 11 12 13 14 15 16 17 18 19 20 21 4
22 23 24 25 O
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- h. 1 We have a very unique difference between boiling water reactors 2 and pressurized water reactors. And we really'have to change-l 3 our gears when we think about boiling water react. ors.
4 In boiling water reactors, there is a 100 tons of-5 steel and 90 tons of water. No matter how.the core is in this 6 place, that core is going to quench. Steve Hodge told you that 7 he. expects that-this will be kind of gradual. And because of 8 that, it is going to quench in the water, and then eventually i
9 boil off the water and so on.
10 I claim that even all of the dore melted and fell in 11 there, that if you just move the fractions of the solid to telt 12 in heat capacities and so on, you find out that all of it is 13 going to quench. And what really impressed me_about.that is 14 that I have a full-scale experiment in my laboratory of a 15 section of a boiling water reactor because of some other !
16 problem, because of some mixing.
17 And when you look at that, you see that there is 18 almost no wate'r there. It is just all tubes. There is only a 19 little bit of water in the tubes. So there is no way. No 20 matter how ycu put it there, it is going to quench. So you 21 start off. And this conclusion is supported by the f 22 calculations at Oak Ridge. Even if you do not believe that, I- '
23 am saying that this core is going to quench.
24 Now we start off then. And now, of course, it is )
l 25 going to heat up again. After it dries up, the water is_ going j n
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1 to stsit heat'.ng up. So as it heats up, heat is conducted to ;
2 the head, and that is also heating up. So I did a simple 3 calculation. And forgetting the other functions there, I 4 thought-that it would be worthwhile putting something like that-1 5 up there. l i
6 -That is the ten:perature of the wall' as part of' that 7 hect-up. And what you find when you apply this, the !
8 temperature of the head would be 548 degrees C,-and the l
y 9 temperature of the fuel at the same time is going to about 10 1700 degrees C. It would still lose its structural strength at 11 about 600 to 700 degrees C.
12 What I am saying is that within about an hour to two 13 hours, that head is going to be losing its strength. All 14 right. This is going to be most likely occurring with most of 15 uhe debris solid, most of the debris solid. The vast majority 16 ct the debris will be solid 17 Now that also agrees with what Steve liodge sas i. And 18 again I do not think that you have to go through ail of the 19 details of his model and assumptions, and this and that to come 20 to that conclusion. You come to that conclusion right away 21 just by doing a simple calculation like this.
22 And furthermore, what I am saying here is that we e .
23 should mov eay fr.m this idea that there is a hole and some 24 melt cc- mt. I -
a like to point out that I find it 25 rather sa * - ' ave melt coming out of the vessel head, l
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and the head sitting there looking at you. I think that is 1 ,
2 impossible, and I would-like to bet anyone on that one.- So if ,
3- 'you get melting coming out of that vessel, that head-is going' 4 to come loose.
5 And I would even ask the question that maybe even the 6 support of the vessel might get too hot. I have not looked at 7 it, but there is the possibility that maybe'the vessel support 8 itself is failing and the vessel coupling in this process at 9 this time. ,
10 But this is really when the head is going to come, 11 and I visualize the head coming off in a circumferential kind 12 of a thing and coming down with a lot of the debris actually i 4
{} 13 14 being solid. .
Now in the above consideration here, I diu not even 15 take into account in the teinperature which I cited, I did not >
16 even take into account that the control room tubes are going to i 17 melt. And as they melt, they are going to dribble, as Steve i
18 said, downwards increasing the thermoconductivity of the melt 19 in a downwards direction. I did not take into account 20 that.
21 And I did not take into account also the fact that as 22 the Fteel melted, that is going to moderate the heat-up of the 23 fuel. So both of those things, if you take them together, they 24 are going to emphasize the higher temperature of the wall for 25 the same temperature of the fuel, which again is going to lead Heritt.ge Reporting Corporation (202) 628-4888 i
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1 you to the same kind of conclusion.
2 Now the next observation is that the radial peaking 3 in a boiling water reactor is. pretty flat with 60 percent of 4 the pour within a few percent. And therefore,-I expect that 5 most of the core into the lower plenum, like 60 percent of the 6 core, is relatively coherent. But even if that was to-occur, 7 and even if all of this 60 of the core fell, it_would be 8 quenched.
9 Now the dry-out is going to be from the top to the 10 bottom, of course. And because of the additional haat sinks 11 near the bottom and the additional material that is added up 12 near the top later on, because of additional losses near the 13 top, I expect that the mid-layers of the debris in the lower 14 plenum will heat most rapidly than the other layers.. ;
15 Then this egrees with Oak Ridge. Ana therefore, I 16 expect to have a failure someplace around the circumference, 17 som7 place out here. And this is the support, and this is going 18 off the pictures. And I do not know if it is here or if it is 19 over here, but someplace over here.
20 So having that information, we can make a tirst i 21 attempt at judging this first quantity which is the amount of 4
22 volume released from the veesel where the vessel fails. Now
- 23 that is a way of conveying to you what I feel about how much of I
24 the melt might come out, how much of the debris might come out l l
25 at the time of failure.
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t 247 O 1 As I said, this is preliminary, and somebody raay 2 argue with that. If somebody argues, we can change that. But' 3 unless you actually take a stab ~at quantifying that, you.have 4 no business in trying to decide the likelihood of liner failure ,
5 is.
6 Next. And similarly, the temperature. The !
7 temperatures are based on my calculation. The next thing that
-8 we have to decide is we have to decide this relationship 9 between the melt depth on the liner, in other_words by how-far 10 the liner is submerged, as opposed to the temperature of the 11 debris that is released.
12 Now I would point out that the additional losses in 13 the support structure of the vessel. That was mentioned. If O 14 you go and see this in a reactor, it is overwhelming how much.
) 15 mass is there. I asked some people in the MRC to tell me how 16 mass was there, and I have not gotten an answer yet, but I 17 expect it to be many hundr'eds of tons just by looking at 18 this.
19 And therefore, there is going to be a significant ,
l 20 loss there. Now we can expect how much of the mass is going to 21 the center, and then spread the fraction based on the heat loss e
q 22 on a heat loss calculation. And that is only qualitative at l a
23 this point, bece.use i have not done the calculations yet. But d
24 I actually expect to see a peak in the amount of submergence- as l
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25 a fraction of the temperature nelt. But that is qualitative at Heritage Reporting Corporation ,
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1 this point, but we are in the middle of quantifying that. And 2 I think that we can use a lot of the calculations _that people 3 are doing that are presented here.
4 The next slide. And the way to present this is for 5 different temperatures. And you plot the height versus volume 6 for different temperature ranges. So this is like a two-7 dimensional surface where the height is here, and the 8 temperature is over there, and the volume is over there. So it 9 is like a surface. And what we did for purposes of a 10 presentation here is that we just take cuts. ,
11 (Slides shown.)
12 DR. THEOFAt10US: Again at this point, that is just (g
(~
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13 mechanistic. It is just basically on.the basis of the data 14 that is available and the guess of how much heat losses will be 15 occurring.
16 And the next one. And you will notice that because 17 of the higher temperature that it will spread more. So that is 18 qua.1.itative to the effect that we were measuring bl efore.
19 And by the way, I do believe, and I think that I can 20 demonstrate that if we have all of the water there that. the 21 heat losses are going to be much, much higher as this melt 22 spreads up, much higher than if you have no water. And 23 therefore, I feel that it will be a very significant fact of 24 the water in how the melt actually progresses.
25 But this also has some other very interesting
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( 1 implications, which I wil.1 come to in a minute. Next. Then by ,
i 2 combining all of this up to this point, the point in the )
3 framework that I showed before, you can work with the 4 distributions. This represents a probability of this. 'And ,
5 that really means something to you, because you have used some 6 kind of criteria for defining this probability.
7 And then the next relationship that we .ieed to .
8 establish is what is the further potential of the vessels at 9 depth. And I think that you see the debris bed with water at 10 the top before you really begin to worry.
11 First, you do a simple conduction calculation. And 12 you see the 60 percent. Everybody seems to be drawing.this 3
13 picture, and everybody seems to be doing experiments with .
14 vertical walls. And this is not vertical. This is less than ,
15 45 degrees. You would have a very different conduction ,
16 occurrence over there.
17 So if there is anything over here, if it gets a 18 little more thick of whatever, there will be more and more l
19 tendency for this thing to stay such. And therefore, the l l
20 building of the cast is going to be even more easy and more l j
21 likely to sustain it.
22 Anyway this is a simple calculation by saying how 23 much heat is out of the water here by conduction only from here 24 to nere. All rig:it. At this point, it exceeds something like ;
25 1700 or 1600 degrees heat. So if I combine this estimate.
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1 This is just a lateral kind of transfer of heat.
2 So with those two estimates then, I come up with the 3 idea that we really begin to worry about a liner attack for 4 depth, over about 20 centimeters. And then that leads me to 5 the causal relationship here. Again that is only qualitative.
6 And that is supposed to be the 95 limit, and the other is 7 supposed to be the 5 percent limit qualitative.
8 And I think that based on what I hear today, 9 especially from Tom Kress, I will have to go back and revisit 10 that and see if we can improve it.
11 All right. And then you take that and you combine, 12 and you come out with a frequency. And you get a good physical 13 feel about what is your expectation, your expectation for the 14 liner to fail.
15 Now I think that I have pretty much to the conclusion 16 that the oxygenization effects of this are significant. The 17 oxygenization effects that lead into this, what we referred to 18 before as the high temperature, which is the fit between 19 oxygenization and more oxygenization. And that can become l l
20 under some conditions almost like a run-away reaction. And l l
21 that, of course, 1s.by temperature. All right. l l
22 And now you have a picture where the stuff came out 23 somewhat solid, and now more stuff is coming out. And then it ;
i 24 gradually kind of overflows a little bit, and melts, and j l
25 spreads. It melts and spreads gradually. And I feel that O
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1 the oxygenization effects will be so slow that it will become
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2 involved with everything else. So I would say that is the 3 consideration, and I think that it is an important 4 consideration in assessing this. .
5 Now as you are going to this gradual spreading,_you ,
6 also take different layers, and you cannot accommodate 7 dif'ferent layers. And what is more important is that you have 8- different layers. Because it is perfectly reasonable and 9 violates no law of_ nature to have a two foot or three foot 10 different layer with a crust around it that prevents it, i 11 because it is based on heat losses, prevents it_from spreading ;
12 the other way. ,
13 And I think that everybody who looks over into the 14 drywall forgets about the fact that maybe we are here 15 forgetting tha possibility of power inside the pedestal which 16 gradually eats its way on the pedestal wall, and that could be 17 maybe just as bad. I do not know.
18 Usually when I look at these problems, I do not look i 19 from the point of view of good or bad. But at least we have to 20 look at it from the point of view of what is physically 21 possible. And what I am saying is we have to look at actually j 22 preventing this spreading to the limit.
23 Anyway we have quantified this. When all of the l 24 force has come down, where is the melt inside and what are the 25 different heights. That is very important not only from the O
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252 1 point of view of the liner, but also from the point of view of 2 the pedestal itself. Next.
3 (Continued on next page.)
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1 MR. THEOFANOUS: .(Cont.) And then transmittal by 2 one minute. .Sho-rt of what happened with the water is unlikely 3 I believe at this point, but I think.I don't want to answer it ,
4 yet.
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5 As I said before, I worked with it for a week, but I
- ' 6 think we will be able to prove ultimately and from what I' heard 7 today sounds reasonable.that that might be the case.
8 The outcome of the absence of water, in my opinion, 9 is not going to be very difficulty delayed action. And with 10 the good possibility actually of liner melting.
11 MR. HULMAN: It's time for questions and answers.
12 Let me ask first if all the speakers on containment core 13 attack, core debris attack on the liner will please come up l
14 here and make it a little more convenient. Why-don't you stay !
15 up here, Theo. Bob Henry, the Oak Ridge Boys, the Sandia Boys, ;
16 if all the speakers would please come up. . Raj, make it a i 17 little easier for everybody if you are gathered around, 18 I would like to start the questions off with a 19 question to Raj Sehgal and Bob Henry. .
20 Now, 1 have heard water is good from Bob. I have 4
l 21 heard water is good from Raj. Mike Corradind has said the same I 22 thing potentially. Theo has said the same thing potentially.
23 Where do we get the water ir. a station blackout in
- 24 both the pedestal area and the area outside the pedestal?
25 Raj, can I start with you?
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254 1 MR. SEHGAL: Bob will answer this question.
2 MR. HENRY: I think a lot of things that you have 3 heard today reflect a~ plant-specific analysis, whether it is TB 4 or PRA, whatever. And the particular capabilities of plants 5 are indeed very plant-specific. I think Paul Hill alluded to 6 one of the capabilities at Susquehanna with the dedicated pumps-7 that they have called the cooling pump that can delivery water 8 potentially to drywall sprays as an example. At other plants, 9 again on a plant-specific basis have a capability of blackout' 10 conditions.
11 MR. HULMAN: You are saying that on each plant we 12 would look for the capability to supply water if station
/ 13 blackout were considered a dominant sequence, plant-by-plant 14 basis.
15 MR. HENRY: For station blackout conditions on a i
16 plant-by-plant basis, you would investigate the capabilities of 17 the given plant for additional water sources. I should not :
1 18 only say drywall sprays, but there are plants which can hook up 19 such things as fire water systems to injection.
20 For injections systems, and I haven't seen all plants 21 so I don't know what other capabilities it would have.
a 22 I. guess Bob Cushman has a response to that.
1 23 MR. HULMAN: Bob, do you want to come up. Bob ;
24 Cushman.
i 25 MR. CUSHMAN: This is one of the things that really I i
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i 255 O -1 does indicate a plant-specific approach is absolutely 2 necessary. We happen to be planning a modification. It's 3 $100,000 estimated, fire-dieseled fire pumps into the 4 containment spray.
1 5 Another thing about our plant, and I would like the.
6 NRC to be well aware of this is we have an 1850 megawatt 7 thermal plant, not a 3300. It makes a difference. The power l 8 density ic 80 percent of that at Browns Ferry. That makes a 9 difference.
10 We have an isolation condenser. We can sit there and 11 steam in a station blackout.
12 So, for goodness sakes, when you start making rules, 13 think about individual plants?
14 MR. HULMAN: Any other comments on my question?
15 MR. KREIGER: I'm Greg Kreiger,from Philadelphia 16 Electric, and I would just like to point out some new 17 information.
} 18 In talking with the front-end people writing i
19 NUREG-1150, the station blackout we have all been talking 20 about, the short-term station blackout has now dropped in I
21 probability by a factor of 50.
22 So we are now looking at a long-term station blackout 23 which dces change your perception of getting water in. You 24 have a much longer time. We have operator actions. So, in 4 25 just keeping with the thoughts we heard tonight, the sequence
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f3 256 V 1 'does make a difference, and we definitely could have water in a j 2 long-term station blackout.
3 The water would come.from a number of sources. We 4 have in the past ignored any operator actions. The1 fact that t
5 we do have battery-power in a long-term station blackout, we'do 6 have ADS capabilities. We can easily-refill our condensate 7 storage tank at Peach Bottom to prolong'HPCI emergency 8 injection.
9 So, it is not necessarily getting water on the core 10 debris. The core debris may never get out of the vessel itself 4
11 which is the point.
12 MR. HULMAN: Any questions from the floor, please?
13 Ken Berg,eron.
}
14 DR. BERGERON: I have a question for Boo Henry about .
15 the coring -- I mean the aluminum thermic. All of the 16 transient conduction calculations, or just about all of them 17 about why the melt through showed that some time is needed of e 18 the corium in contact with the liner before the transient 19 affects the initial contact to be lost and for -- to begin. .
l 20 Your experiments are transient bores and nonaluminum ,
21 thermic that spreads and cools very rapidly, the time of 22 contact with the debris -- with the shell is very short. ,
23 Are you asking us to draw the conclusion because l
24 there was no ablation in the shell? In your experiment, there l 25 would be no ablation in the real case where there is volumetric Heritage Reporting Corporation i
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x, 257 1 heating, but with the KE and the - . It's a fairly obvious 2 question and I am sure the thought has occurred to all of us.
3 MR. HULMAN: Bob Henry.
4 MR. HENRY: I guess, Ken, the conclusion I would like 5 you to draw from this is when we take experiments with very 6 large super heat that you bring to the wall in less than a 7 second. And if you were doing the calculations that this moves 8 under the water losing a negligible amount of heat which is -
9 what you would conclude by film boil type of considerations, 10 come in direct contact with the wall just by the contact that a 11 high temperature melt in the wall through an initiated relation 12 just on contact.
(~T 13 What we are finding is that in the short period of
'u) 14 - time that this material moves across the floor underneath 15 water, that there is a very efficient energy transfer mechanism 16 to cool the material so that by the time it gets to the wall, 17 it is essentially at a saturation temperature, freezing
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18 temperature. And when you look at the wall response, that 19 makes good sense.
20 And if it is at that temperature,.it has nc super i l
21 heat. Then there will be no immediate attack on the liner. I 22 This kind of thing that Theo was driving at with really an 23 attack on the material by direct contact of high super heat at 24 highly super heating conditions. The later part is another 25 issue.
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i 258 O 1 The water has to be there long term.as'well. But the ;
2 key to it is when that material comes out and start moving 3 towards the liner in the wall, how much energy is actually 4 moved. What is this temperature?. I guess several people Psve 5 made that point today.
6 And the thing I want you to take away from the 7 experiment-it's not just the normal film boiling CHF kind of 8 heater movements. It is a very dynamic process. It's several l 9 times that.
10 MR. HULMAN: Theo. l 11 MR. THEOFANOUS: I want to make a point this was the l 12 discussion. :
13 ,
I think it is not a very well posed question and I 14 think we should not throw down so easily _ numbers like it loses t 15 its super heat or it doesn't lose its super heat.
16 Whether it los'es or doesn't lose its super heat l 17 really depends on how much melt is there; what is the -- how is I 18 much a layer of depth you have.
19 It's clear that if we had a 1 centimeter layer
{
20 flowing along the floor, that is going to lose super heat in a ;
j 21 very different way than if you have 1 meter of depth going 22 around. I think that we have to be careful, and that's why I 23 brought in this dissection of the problem, and we have got to 24 look at the specifics.
25 If somebody asks you about if it is going to lose l Heritage Reporting Corporation 3 (202) 628-4888 l
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1 I super heat or not you have to ask how much pressure -- and if ,
1 2 the mass is small, I think he is going to loae the super heat.
3 But if it is 500 tons, all at once won't hold it. I think one 4 has a hard time to prove that.
5 MR. HULMAN: Dana, Dana Powers.
6 MR. POWERS: I would like to call attention, first of 7 all, to the point of Dr. Sehgal's presentation, and to direct a 8 question concerning that presentation, 9 Dr. Sehgal recognized that a large amount of mass was 10 being caused -- pressure vessel early in the accident, and then 11 would surge up against the liner, and he performed the surge 12 calculation.
{} 13 14 The question.I have, Dr. Sehgal, is when you have this surge of material up against the liner, you enclose the 15 stagnation heat flux. And I wondered if you had made any 16 estimates on the magnitude of that heat flux and whether just 17 during that surge itself you would threaten the liner 18 integrity.
19 DR. SEHGAL: No, I do have those numbers with me here 20 so I can't answer your question.
21 Basically the velocity determi.ned by the gravity i
22 head, and the gravity head is about 10 cent.imeters from the i
23 pedestal onwards, and that what determines the velocity ;
24 spreading to the liner.
25 There is a surge, but I don't have the number.
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1 MR. HULMAN: Bob Henry. ,
,, 2 MR. HENRY: Let me-just add one thing, because Theo i
- 3 said it exactly the way it should be. You shouldn't just deal- <
4 with the super heat.- I really didn't say it as clearly as I ,
5 should have. .
6 What you should do is translate the energy removal 7 rate into the models which now calculate how fast things are .
8 getting there to see if that influence.s your fundamental 9 conclusions.
10 DR. BERGERON: I agree. Would you advice us -- this 11 is an easy question. l 12 In your document -- ;
8 13 MR. HULMAN: Would you please come up? He can't' hear !
14 you.
15 DR. BERGERON: I agree with your statement. ,
16 The next logical question is, do yca have any 17 estimate of time scale and are able to document these very 18 interesting experiments so tha: you and the rest of the 19 committee can analyze the -- models?
l 20 MR. HENRY: I brought a hundred copies of the j l
, 21 viewgraphs which you saw tonight. You saw that we will try to l
22 do at least three more tests to look at some of these key
- 23 things.
- 24 I would anticipate that these will probably be done J
l j- 25 and written up in about a three-manth time frame from now. But l j Heritage Reporting Corporation (202) 628-4888 i
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261 O 1 you are surely welcome to come and see them at any time, !
i 2 because they were done to try and provide a perspective that I L
- 3 kind of got out of the December meeting when Jerry Adam said we 4 were going to have this workshop and here is the kinds of a 1
! S things that were going to be addressed. People are certainly l
l 6 free to come and see them.
I 7 MR. CHEH: I might say I'am -- my name is Unte Cheh.
1 8 I spent 17 years with.the nuclear industry and national 9 laboratories. I may be a technocrat at NRC.
10 One thing I have read is that I read that most of 11 your recent papers say everyone submitted to the NRC, about 12 3000 pages in the past two months..
13 I tind that you are talking about this syste,m, the 14 temperature pressure without mentioning the system model and 15 the nodalization. It's much depends on nodalization and the i
16 system on how you nodalize and idealize. Nobody mention 17 anything about that.
18 when'we comes to the systems analysis and core i
19 analysis, how to nodalize and how we divided that point to 20 three parts, total parts, five parts because we all have to 1
21 *hink of the cost of the -- nobody mentioned this thing, 22 because one thing what you gentlemen forgot -- the thing is l l
23 when crustability, nobody mentioned this. I brought up this 24 one to the attention of my management. Today here first time 25 comes up. As crust built up, the chemical -- will be the l
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262 1 target. Then that someone here said that 75% wasn't the 2 reaction. That is too arbitrary. So we have got to further 3 the effect of the crust built up here as the second thing.
4 And the third thing, my objective not to embarrass 5 some of you, but someone said that it is one dimensional ,
6 analysis. How it can-be one dimensional analysis? I mean, 7 even a high school kid will know it's not a one dimensional --
-8 to one-fifth. Low -- can be one dimensional analysis because-9 it a configuration is already set. So it has to be at least 10 two dimensional or three dimensional. But someone tried to i
11 deal with it in terms of one dimensional. It was something 12 unbelievable, in my. judgment.
13 And the fourth' thing, the parametric study -- this ;
14 thing go to the Ken Bergeron especially. I see the parametric 15 study, the variation may be 10 percent or 20 percent only. How ;
16 about 50 percent or 100 percent, and to see how parametric )
17 study doing? Say 30, 40 parameters if -- and the changes to I 18 be, then look into further, otherwise -- total -- So, we can 19 converge these things to the source term. That's what we are 20 looking for. So I am just bringing what any findings or why the i
21 view of your discussion here, and also the view of some of your 22 materials so far. So if you have some more discussions, I ;
l 23 would like to have individual basis if you want to, because I l 24 have been working on a LOCA analysis with Combustion 25 Engineering for many years. I
(:) l Heritage Reporting Corporation l (202) 628-4888 l I
263 1 My question is this one. When you talking about 2 temperature pressure 2510, what, is that average temperature in 3 one nodal or an average temperature in the whole as one nodal l 4 or just one small part-pressure or temperature? That goes to 5 Ken Bergeron.
- 6 DR. BERGERON
- I didn't mention what these 7 temperatures were. In these studies, the temperatures.that I ,
8 followed were the highest temperature in each of the layers, in .
9 each layer.
10 MR. HULMAN: In each layer.
11 DR. BERGERON: So there are like four traces and, of 12 course, of the layers.
13 MR. HULMAN: Theo. , ,
14 MR. THEOFANOUS: i just want to make a short comment, j 15 Did you say you were working with three or four? And you don't ;
- 16 like multi-liner analysis?
17 MR. CHEN: Theo, that's one division of analysis for 18 LOCA.
19 MR. HULMAN: Mike? ,
20 DR. CORRADINI: I will answer the question a couple L 21 of way, but I think you are asking a general question. Let me 1
22 try to answer that, too.
l 23 About the liner analysis, the assumption that you can 24 treat it as a fin is only valid insofar as the VO number is L2 5 small. And that implies that the heat transfer coefficient is j Heritage Reporting Corporation (202) 628-4888 ;
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1 going to have to be something.-- the heat transfer coefficient 2 from the pool bulk to the liner is going to have to be 3 something less than about 3,000 watts per meter squared degree 4 K. Because at the conditions I just talked about here, I gave 5 you the thermal -- and thickness, and you back calculate what 6 the VO number has to be.
7 But let me just go a little further and say-because :
8 of the analysis and what you are doing, you have enough 3
9 uncertainty about the heat transfer coefficient. I think if I-10 were to do -- if I were to do anything more than a one
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11 dimensional calculation given the uncertainty of the heat 12 transfer coefficient, personally I would worry, okay, because l
13 you are putting resolution where your initial conditions and 14 your boundary conditions you have very little resolution. So 15 that is my own personal thing.
16 I think where you have to watch about resolution is 17 time resolution, and that is if you are bringing down the ,
18 zirconium relatively quickly and its primarily metal, what we ,
19 found in all our calculations, you have to be very careful i
20 because the zirconlum, if you believe the calculations, if you i
21 believe the calculations, underline "if", emphasize that the !
I 22 rate o'f zirconium burning is extremely fast. And if you don't 23 look at it with enough time resolution, you can get relatively 7
l 24 unphysical results.
25 You can underestimate the chemical heat or O
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1 overestimate the chemical heat.
2 So when you say resolution, I think about spatial as 3 well as temporal.
4 MR. HULMAN: George, George Greene.
5 DR. GREENE: We all know that a little more occurs, 6 and we all know that heat flux occurs. When Bob Henry 7 presented some information here that suggests a more dynamic 8 mode of boiling, and I am very interested in the experiment, 9 haven't see, any of it before, so I am sure it is outstanding.
10 And I would like to ask you, please, to give us some 11 insights as to what the mechanism of the boiling is from the 12 heat that you have reclaimed from the quench tank. You 13 suggested, and I won't hold you to exact numbers, some time of r~3 V
14 heat flux that may be on the order of magnitude greater than 15 the heat flux.
16 And if it is a water entranement into the 17 accelerating or advancing melt front, could you also comment on 18 whether or not that pressure dependent, because your term was ;
I 19 high pressure? '
20 MR. HENRY: I think the basic mechanism which is 21 occurring inside the three-quarter inch steel box,.and we can't 22 really see it, is a dispersal material due to LOCA energetic l 23 interactions which some people may want to call LOCA FCIs. )
24 They may also be generating steam in the concrete, because you 25 are heating that, driving off free water which is dispersing O
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1 material. ;
2 .
So you have an amplification of the available surface 3 area, and you are also dispersing the water at the same time.
4 The heat flux that we get out of this, and this is just an 5 experimental fact, one has to look at it more, is much more in
]
6 line with the fluidization velocity of the water.
7 But that is just three experiments. I certainly 8 wouldn't draw that fundamental conclusion, but I can draw the 9 conclusion that there is an energy extraction very far in 10 excess of -- boiling mechanisms. :
l 11 Is it pressure dependent? It may be, but we had one l 12 of the systems which also depressurized. And if anything, it 13 looked like it did more dispersal, at least on the outside 14 material, and you might anticipate that it had even a more 15 dynamic interaction and therefore we need further amplification 16 in the energy transfer between the two materials, between the 17 coolant and the --
18 MR. HULMAN: Ivan Canton..
19 DR. CANTON: I have heard a whole series of analyses 20 about what happens once the stuff is down below the vessel.
21 And I think that no matter what you do, you probably could sort 22 out and come to some kind of an agreement, l 23 But really it's a three-prong problem. It's how i
24 much, how fast and how high. It seems to me that it is I i
25 incumbent on you people doing the calculations to create that l O
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i 267 O 1 surface so-that we can make a judgment on whether or not you ;
2 are doing your calculations for in-vessel appropriately.- ;
^
3 Anybody have any comments on this? I can't get a 4 measure of whether we are doing it well enough. I suspect we 5 are not, but I can't tell from what you have done.
i 6 DR. THEOFANOUS: Well, maybe I can turn this on. . I 7 think I -- my attempt was exactly to do that, to set the 8 problem in that framework, whereas how fast, how much, how hot ,
9 and to indicate that each of those parts would be actually 10 qualified quite independently, because I guess, no, you can't 11 find independently how much and how high, because that's 12 together with the rest of the part.
13 And I think I give some arguments that show quite 14 well that the -- is quenched. If anybody has objection with 15 that, I think now is the time to say.
16 DR. CANTON: You said this is prel'minary. :
) 17 DR. THEOPANOUS: It's preliminary, of course, yeah. ,
i 18 We can do more than what we have, but we are working on that 19 and we would like to do more of this. I 20 I think it is encouraging that you can also see 21 compilations from -- who have been doing this for a long time.
22 Those calculations give some aspects that make me uncomfortable 23 because a very detailed model. And always when someone tries 24 to put all those things together themselves, it gives some l 25 doubts.
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4 266 i 1 But the reason I presented my thought were I didn't 2 want.those uncertainties to cloud the fact that there is that .
l 3 much heating in the lower head of the boiler water reactor, and ,
4 there are no two ways ~about it. No matter how you put the :
i 5 material from the core slumping into the lower - is going to i
- 6 quench.
7 Now given that, given that the material quenches 8 there, I think you can do a very simple conduction calculation l
9 as I have done with the heat seen at the edge which they 10 present the vessel head, then you find out that you are going 11 to get 'the head very hot before you melt the core. So that 12 tells you some ideas about how much, what is going to fail and 13 what are the temperatures - . The answer is not too much '
14 molting, not too high temperatures - .
15 Do. BERGERON: I want to rise to the defense of the 16 peo>1e that are addressing that question. Theo is one of them 17 and Steve Hodge's people are another.
18 I think there may be a little bit -- first of all, I -
19 agree with your point. Your point is that one must look at the 3
20 in-vessel problem. One has to look at the core problem. One l 21 has to look at the spread problem with the concrete interaction l
- 22 and then the attack, the liner attack problem. !
] 23 We have seen a lot of liner attack calculations to 24 date, mine being one of them. And the answers look very much i
25 the same given the same boundary conditions.
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269 1 Why are there so many calculations? I have a theory.
2 It's because we all love doing something we know how to do.
3 On the other hand, the in-vessel melting pressure 4 problem is a tough one. And the core spread problem in the 5 presence of concrete interactions is a tough one. So you end 6 up with a much smaller number of people, typically braver 7 people, and I think that active work is going on, but the 8 amount of progress in terms of time and dollar invested is not 9 as high.
10 MR. HULMAN: Raj Sehgal.
tl DR. SEHGAL: I have one question for Steve Hodge.
12 This is his model of the melting of the control rod material, 13 and the channel boxes which come down all of a sudden and drop 14 on to the cold plate. This is a 12 feet high core, and the 15 model is based on a 1-foot high experiment. Maybe some kind of 16 experiment or some kind evaluation should be done to be able to 17 extrapolate the results of 12-foot high experiment to a 12-foot 18 high core.
19 DR. HODGE: Yes, I think it was negligent of me 20 before not to point out that in fact for the DF4 experiment, 21 Larry Ott performed the pretest calculations using models that 22 he had developed for use in a full BWR core.
23 Larry Ott was predicting channel box and control 24 blade relocation, early relocation, long before the DF4 25 experiment was ever run. He was predicting it for models for O
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1 rodsL12' and a half feet long.
2 When he did the mock up and did the pretest 3 calculations based on the DF4 experiment, he-of course again 4 predicted it. If.you predicted it for 12.5-foot rods, you 5 would certainly expect to predict it for shorter ones ns well.
6 But the point that I think I didn't make before, and 7 I certainly want to make now is that the code predicted this 8 for fu13. length rods'long before the experiment was ever run.
9 The experiment merely confirmed for the' limited range of the 10- experiment what the code had pre-predicted.
11 MR. HULMAN: It is late. I would like to continue 12 the discussion tomorrow morning. There are several people 13 among our visitors that are not feeling so well, so I would 14 like to cut it off.
15 I do suggest that you remove all of your materials, 16 including your hats and coats. Don't leave them.
17 (Whereupon, at 10:25 'p.m., the meeting was recessed, 18 to resume at 8:30 a.m., Thursday, February 25, 1988.)
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_ 1 CERTIFICATE I 2 3 This is to certi-fy that the attached proceedings before the 4 United States Nuclear Regulatory Commission in the matter of:
5 Names BWR MARK I CONTAINMENT INFORMATION EXCHANGE WORKSHOP 6
7 Docket Number N/A 8 Place: Baltimore, Maryland Date:
February 24, 1988 10 were held as herein appears, and that this is the original 11 transcript thereof for the file of the United States Nuclear 12 Regulatory Commission taken stenographically by me and, 13 thereafter reduced to typewriting by me or under the direction 14 of the court reporting company, and that the transcript is a 15 true and accurate record of the foregoing proceedings.
16 /S/ l W WUANk-~
17 (Signature typed): Andrew M. Emerson 18 Official Reporter 1
19 Heritage Reporting Corporation 20 l 21 22 23 24 25 I
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Heritage Reporting Corporation (202) 628-4888