ML20199D913

Draft Peach Bottom Containment Event Tree Documentation

ML20199D913
Person / Time
Site:	Peach Bottom
Issue date:	02/07/1986
From:	Amos C, Griesmeyer J, Haskin F SANDIA NATIONAL LABORATORIES, TECHNADYNE ENGINEERING CONSULTANTS, INC.
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9 DRAFT Peach Bottom C.E.T. Documentation C. N. Amos (Technadyno Engineering Consultants, Inc.)

J. M. Griesmeyer and F. E. Haskin (Sandia National Laboratories)

, February 7 , 1986 This document describes the computerized containment event tree for the Peach Bottom Atomic Power Station (PBAPS) which has been developed at Sandia National Laboratory in conjunction with the Severe Accident Risk Reduction Program (SARRP) . SARRP has three principle objectives. The first objective is to characterize evaluate the offsite risks for commercial Light Water Reactors (LWRs) using state-of-the-art methodology. The second objective is to is to characterize the uncertainty associated with these risk

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profiles. The third objective is to evaluate selected means of reducing offsite risk through design or operational modifications.

The program is sponsored by The United States Nuclear Regulator Commission (NRC). SARRP results are expected to be used in support of NUREG-ll50; which is scheduled to be published by the NRC staff 4

during the summer of 1986.

Containment event trees (CETs) are used in SARRP to track accident progression and containment failure modes and assess their probability. Such trees are being developed for sorry, Grand Gulf, Sequoyah and LaSalle as well as for the PBAPS. The PBAPS tree is the most complex tree developed to date. The CETs are processed by the EVNTRE computer code which was written to allow the construction of trees with sufficient complexity to adequately model accident progression.

Each question represents a branch point of the tree which may have several (up to ten) alternate branches. The majority of questions have only two. The probability of a given branch can be dependent on the path through the tree to that point. To determine the appropriate set of probabilties, the path is compared against boolean expressions involving the occurence or absence of particular branches at previous questions on the tree. The probabilities associated with the first case for which there is a match are used. For most questions, the user supplies the case dependent branch point probabilities for each question. The code i

does, however, have the capability to internally calculate branch point probabilities. This capability is utilized, for example, in the calculation of containment failure probability. The code carries the values of " parameters", which are specified in a case 8603240407 860311 PDR P

ADOCK 05000277 PDR

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Paga 2 DRAFT Panch Bottco C.E.T. Documantation l

dependent manner through the CET. Parameters, such as the pressure of containment at the start of core degradation, the pressure rise

due to hydrogen generation, and the pressure rise at vessel breach, can be summed and compared to a user input containment fragility curve

(which could also be case dependent -- reflecting temperature effects for example). This comparison determines the branch point

probability for the gaestion (such a question is limited to'two

branches).

i The EVNTRE code allows six types of questions:

i i Type 1 User supplies branch point probabilities

- independent of the branches taken in previous

! questions.

Type 2_ User supplies branch point probabilities that are dependent on previous branches (two or more j " cases" are given).

Type 3 User supplies branch point probabilities and

specifies the value of one or more parameters (value dependent on branch taken).

l Type 4 User supplies branch point probabilities and

! specifies the value of one or more parameters a (value dependent on branch taken) dependent on previous branches (two or more cases given).

Type 5 User specifies a reference value to which the sum i of specified parameters is to be compared. The method of comparison is also specified. Branch j point probabilities are calculated.

Type 6 User specifies a reference value to which the sum i of specified parameters is to be compared in a

case dependent manner. The method of comparison

! is also specified. Branch point. probabilities are calculated.

I i The code marches through the CET and evaluates the probability of each.possible pathway (combination of events). Typically, the user will specify a cut-off probability such that if the probability.of a pathway drops below the cut-off after a given branch then further exploration of that pathway is abandoned. The cut-off probability (conditional on the initiating event) is typically specified at 1 E-8.

Marching through the event tree and finding the probability of each pathway yields many possible outcomes (1000 to 50000 depending upon the sequence and quanitification) which are grouped into bins that represent unique initial and boundary conditions for source term analysis. The characteristics considered in binning the outcomes are specified by the user. It earlier PRAs, binning could be as simple as early, intermediate, and late containment failure.

The EVNTRE code allows many more binning " dimensions". For the

, Paga 3 l DRAFT P0cch Botton C.E.T. Documsntation PBAPS tree, for instance, the binning dimensions may include:

containment failure time, extent of suppression pool bypass, extent i of reactor building bypass, timing of pool bypass; along with four or more additional key characteristics. .

For SARRP the quantification of the containment event tree i pathways has been done with three different evaluations of the

! branch point probabilities for each question. The probabilities t

have been evaluated based on an optimistic, a central, and a

, pessimistic estimate (termed the OCP approach). The central i estimate reflects the authors' best estimate of the probability based on their understanding of current thinking amoung researchers

in the LKR severe accident area. The optimistic evaluation gives what the authors judge to be the greatest reasonable probability to

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the branch of a given question which is expected to result in the.

lowest offsite consequences. The pessimistic evaluation gives what

the authors judge to be the greatest reasonable probability to the branch of a given question which is expected to result in the greatest offsite consequences. The tree is quantified three times

once assuming all the optimistic evaluations, once with all the

, central evaluations, and once with all the pessimistic evaluations. Each quantification has been termed a "walkthrough".

This approach is intended to give a feel for the range of possible

, results. However, it has some recognized deficiencies which limit its usefulness.

Uncertainty in results will thus also be studied using a limited application of Latin Hypercube Sampling (a Monte-Carlo type j, of approach). This study will make use of the EVNTRE code l capability to sample. Sampling involves simultaneously varying

, sets of individual (case dependent) branch probabilities, while leaving the structure of the tree untouched. For each sample observation, the values for the probabilities are randomly selected from within their range. The resulting distributions of the i outputs of interest provides information regarding uncertainties, i and exlporation of the relationships between the inputs and those l outputs provides sensitivity information.

The PBAPS CET consists of 92 questions with associated logic i and quantification. This document provides a detailed desciption i

of the PBAPS CET. The description begins with a summary of the major sections of the tree. A question by question and case by case description of the structure and logic follows. Where j appropriate, the rationale used in the quantification of the branch

point probabilities and code parameters is discussed. The actual l values of these quantities has not been included in this report j since some have not yet been determined.

Questions #1 through #16 define the intitial and boundary l conditions for the accident progression analysis (i.e. the plant damage state). The branches taken for these questions are l determined by inspection of the cut sets developed under the Accident Sequence Evaluation Program (ASEP) for the accident sequence under consideration. The ASEP cut sets for the various accident sequences are grouped according to the initial and I e

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Paga 4 DRAFT Panch Botton C.E.T. Docum:ntation '

boundary conditions that they imply for evaluation of the CET.

J The remaining questions on the tree (#17 through #92) have 4

branch point probability evaluations which are dependent on the branches taken in previous questions and which normally do not change as a function of the cut set (s) being considered. Some will i

be dependent on the cut sets for sequences in which the determination of whether the core melts depends on phenomena 4 occurring during accident progression.

! The after the questions used to establish initial and boundary conditions for the analysis, the questions are divided into groups i which cover distinct time regimes in the accident. Questions #17 >

l through #21 complete the sequence definition by considering events

• and actions whicn are implied by the ASEP results but not specifically quantified. Questions #22 through #31 consider system

, availability and the integrity of the plant so as to determine if

the accident can be recovered prior to core melt. . Recovery prior 4

to core melt is considered in Question #32. Questions #33 through 4

1. 59 determine the accident progression, system availability and plant integrity up until the time of vessel breach. Question #60 allows for accident recovery prior to vessel breach. The_ remaining questions determine accident progression at vessel breach (Questions #61 through #71) and after vessel breach (Questions #72 through #92).

As mentioned above, each. path through the tree is binned 4

according to its implied initial and boundary conditions for the sourcre term calcuations. At this time, only preliminary selection of binning dimensions has been completed and thus binning of containment pathways is not discussed further in this document.

A detailed description of the event tree structure (questions

{ and cases) follows.

Question # 1 : What is the initiating event?

Defines the initiating event. ~ These events are those i initiators identified by ASEP as being the most probable i contributors to overall risk (ie frequency > IE-8).

This is a Type 1 question where branch point probabilities are input and are independent of the branches taken in previous l questions.

There are 7 branches:

i 1 . A: Initiator is a large break loss of coolant accident.

2 . S1: Initiator is a small break which is in the size  !

range such that High Pressure Coolant Injection (HPCI) cannot maintain level and yet the Automatic Depressurization System (ADS) must be used to allow low pressure coolant injection.

P2g3 5 DRAFT Peach Bottom C.E.T. Documentation 3 .

S2/3: Initiator is a small break wich is small enough that HPCI provides adequate injection to prevent loss of RPV level to the' point that ADS is

. actuated. (ie level is maintained above Level 1) .

(ASEP frequency < 1 E-8) 4 . TLOSP: Initiator is a transient which is either induced by or results in the loss of offsite power.

5 . TPCS: Initiator is a transient induced by turbine trip or closure of the Main Steam Isolation Valve (s)

(MSIV). The Power Conversion System is assumed to be isolated from the Reactor Pressure Vessel (RPV).

6 . TC: Initiator is some transient followed by a failure of the Reactor Protection System (RPS) such that no scram occurs.

7 . IORV: The initiator is the Inadvertent Opening of a Relief Valve by the operator.

Question # 2 : What is the initial break location?

Specifies the pathway of affluent steam (or hydrogen / fission products) from the RPV to the containment. This question is not referenced in the current tree. For some plants and for certain cut sets this question could be used in determining success critera. It is included here for the sake of completeness.

~

This is a Type 1 question whete branch point probabilities are input and are independent of the branches taken in previous questions.

There are 5 branches

1 . RPV: Break is in piping connected to the RPV but not in the Recirculation Lines, Main Steam Lines (MSL) of Feedwater (FW) lines.

2 . Rcirc: Break is in the recirculation line.

l 3 . MSL: Break is in the Main Steam Line.

l 4 . FW: Break is in the Feedwater line.

5 . SRV: There is no break. Steam leaves the RPV only j through the Safety / Relief Valves.

Question # 3 : What is the level of pre-existing leakage or isolation

P;gs 6 DRAFT Peach Botton C.E.T. Documentation failure?

BWR operating history indicates some instances where plants were operating with leakage to the environment or the reactor building in excess of the technical specification limits. This excessive leakage was usually a result of a failure to properly close primary containment penetrations. It is believed, and recent data substantiates, that this is not a concern with inerted containments. Inerted containments carry an overpressure of nitrogen with respect to ambient pressure which would result in rapid detection of any large leakage. Leakage in excess of technical specifications is thus deemed impossible. This question is not referenced in the current version of the tree. This is equivalent to assuming that branch 5 (no excessive leakage) is always taken (probability of 1.000).

This is a Type 1 question where branch point probabilities are input and are independent of the branches taken in previous questions.

There are 5 branches:

1. E1L2D: There is pre-existing leakage from the drywell large enough to remove decay heat from the containment ten minutes after shut-down.

2 . ElLlD: There is pre-existing leakage from the drywell in excess of techical specification limits but not sufficiently large to remove all the deacay heat generated.

3 . ell 2W: There is pre-existing leakage from the wetwell large enough to remove decay heat from the containment ten minutes after shutdown.

4 . ell 1W: There is pre-existing leakage from the watwell in excess of technical specification limits but not sufficiently large to remove all the decay heat generated.

5 . ElLO: There is no leakage from the containment in excess of technical specifications. ( < 0.5% of the containment free volume per day).

Question # 4 : Is there a loss of offsite power?

Determines if there is power available to AC powered equipment which is not connected to the emergency busses. Probability of LOSP is unity if intiating event is TLOSP.

This is a Type 1 question where branch point probabilities are input and are independent of the branches taken in previous questions.

P go 7 DRAFT Petch Bottom C.E.T. Documentation There are 2 branches:

1 . El-LOP: There is a loss of offsite power.

2 . EinLOP: Offsite power is available.

Question f 5 : Is there a Station Blackout (Diesel generators fail)?

Determines if there is power on the emergency busses. A station blackout means that there is no offsite power and the diesel generators are not supplying power to the emergency busses (ie. neither of the two dienel generators is running). The branch taken is determined from the ASEP cut set (or sets) of interest.

.This is a Type 1 question where branch point probabilities are input and are independent of the branches taken in previous questions.

There are 2 branches:

1 . SB: There is a station blackout. No AC power is available even on the emergency busses.

2 . nSB: There is not a station blackout. AC power is available at least from the emergency busses.

Question # 6 : Is DC power not available?

This question determines the availability of the station battery. The station battery supplies DC power to the following critical equipment: 1) Vessel instrumentation (level, pressure, etc.), 2) Pilot valves on the Safety / Relief Valves (S/RVs) (allow automatic depressurization and relief function), 3) Vessel injection from the High Pressure Coolant Injection (HPCI) and Reactor Core Isolation Cooling (RCIC) systems, and 4)

Diesel-Generator starters. ASEP has identified common mode failure of the station batteries as being a significant contributor to the frequency of station blackout events. This question is thus highly significant in the determination of the plant damage state.

This is a Type 1 question where branch point probabilities are input and are independent of the branches taken in previous questions.

There are 2 branches:

1 . ElfDC: The station batteries are failed. There is no DC power.

2 . El-DC: Power is available on the DC busses.

I

Paga 8 DRAFT POcch Bottca C.E.T. Docuncntation Question # 7 : For TC, does SLC fail to inject?

This question applies only to sequences involving failure of the Reactor Protection System (RPS) scram or Alternate Rod Insertion (ARI) functions (ie the initiator is an Anticipated Transient Without Scram -- ATWS ).. In such sequences, the Standby Liquid Control (SLC) provides a means to achieve plant shut-down.

SLC is manually actuated and consists of two pumps (only one at a time may operate) which inject enriched boric acid into the RPV.

The branch taken is determined by ASEP cut sets. It must be noted that these cut sets are for SLC operation within two minutes of the transient event. The assumption that such prompt action is necessary to achieve shutdown is quite conservative. ASEP is at this time revising its assessment of SLC failure.

This is a Type 1 question where branch point probabilities are input and are independent of the branches taken in previous questions.

There are 3 branches:

1 . ElfSLC: SLC is not initiated within two minutes of the transient.

2 . El-SLC: SLC is initiated within two minutes of the transient.

3 . na: The event is not a TC, the RPS performed correctly.

Question f 8 : Does a S/RV stick open early?

This question determines whether there is a failure of a least one Safety / Relief valve to re-seat after operation. ( P failure).

j Branch point probabilities are determined by ASEP and are dependent on the cut set (or sets) of interest.

This is a Type 1 question where branch point probabilities are input and are independent of the branches taken in previous questions.

There are 2 branches:

1 . El-SORV: There is at least one stuck open relief valve.

2 . EinSORV: There are no stuck open relief valves.

l l

Question # 9 : Do the HPCI and RCIC systems fail to inject?

l

j. .

DRAFT Pe2ch Bottcc C.E.T. Docum:ntation i

! Determines whether the High Pressure Coolant Injection (HPCI)

! and/or Reactor Core Isolation Cooling (RCIC) systems are operating i initially. Both systems are driven by steam turbines which draw i steam from the Main Steam Lines. Turbine exhaust is dumped to the 2

suppression pool. Suction for both systems is initially from the i Condensate Storage Tank (CST). Suction will switch to the i suppression pool on a low CST level signal. Both systems require i DC power to fuction. Both systems will isolate and stop injecting i on high containment pressure (> 40 psig) or low RPV pressure (<

l 150 psig). The branch taken is dependent on the ASEP cut sets j being considered.

4 l This is m' Type 1 question where branch point probabilities are j input and are independent of the branches taken in previous

questions.

There are 3 branches:

1 . ElfMPC

Both High Pressure Cooling systems (HPCI & RCIC) i have failed. '

j 2 . ElrHPC: Neither HPCI or RCIC is currently providing water

injection to the RPV. However, at least one of j these systems can be recovered.

4 l

3 . El-HPC: Either HPCI or RCIC (or most likely both) is providing high pressure make-up water to the vessel.

i l Question # 10 : Does the CRD hydraulic system inject?

i i Determines the availability of the Control Rod Drive (CRD)

! hydre.ulic system for injecting water into the vessel. The CRD I system provides cooling flow to the control rod drives during

! normal operation. Normal flow is approximately 70 gpm. Suction for CRD flow is drawn from the CST. The CRD pump head can be up to 1800 psi. Using both CRD pumps, plant operators can manipulate the system to attain flows up to 200 gpm; with the RPV depressurized, over 300 gym is possible. The branch taken is determined by the l ASEP cut sets of interest. It should be noted that the CRD pumps are normally powered off the station bus (offsite AC). At Peach Bottom, emergency power (Diesels) may be used.

This is a Type 1 question where branch point probabilities are input and are independent of the branches taken in previous questions.

There are 3 branches 1 . ElfCRD: The CRD cooling system is failed.

2 . ElrCRD: The CRD cooling system is not currently operating i.___._.__ . _ _ _ . _ . _ . _ . _ . _ _ _ _ _ _ _ _ . . _ _ _ _ . _ _ . _ . _ . _ . _ _ , _ _

a . .

Pago 10

{ DRAFT POnch Bottco C.E.T. Documentation 1

but is recoverable. This situation could arise in i '

a loss of offsite power.

3 . El-CRD: The CRD system is providing high pressure make-up

water to the RPV.

Question # 11 : Do the LPCS and LPCI systems fail?

l Determines whether Low Pressure cooling systems are available to inject water to the RPV. There are two low pressure ECC systems:

J (1) the Low Pressure Core Spray (LPCS) and Low Pressure Coolant l Injection (LPCI) . Two of the LPCI pumps are also the Residual Heat Removal (RHR) system pumps and thus provide shut-down cooling, s

suppression pool cooling, and containment (drywell) spray. Each system (LPCS & LPCI) consists of four pumps divided such that two

pumps from each system are powered by a separate electrical division. Thus there are four low pressure pumps powered by each diesel. Each pump has a run-out flow in excess of 5000 gpm.

Suction is normally drawn from the suppression pool, but can be taken from the CST in the case of LPCS. High pressure interlocks prevent the injection valves from opening with the RPV above 450 psi.

This is a Type 1 question where branch point probabilities are input and are independent of the branches taken in previous j questions.

, There are 4 branches: .

I 1 . ElfLPC: Both LPCS & LPCI are failed.

2 . ElrLPC: Both LPCS & LPCI are not available for injection j to the vessel. However, at least one of these I systems is recoverable. This situation could arise in a station blackout.

3 . ElaLPC: Either LPCS or LPCI (or both) are available (ie the pumps are running) but there is no injection to the vessel (injection valves are closed either due to high RPV pressure or system failure).

l 4 . El-LPC: Either LPCS or LPCI (or both) are providing water injection to the vessel.

Question # 12 : Do both the SPC and CS modes of RHR fail?

Suppression Pool Cooling (SPC) and the Containment Spray System (CSS) are two modes of the Residual Heat Removal (RHR) system.

LPCI (see above) is also a mode of RHR. In either the SPC or CSS modes of operation the RHR system can remove heat from the suppressien pcci by passing water from the pool through heat

_ --- . = -. - . - . _- . . . .

, Pags 11 DRAFT P32ch Dottcm C.E.T. Docu2Ontation l exchangers (with service water on the shell side). In spray mode, water is sprayed into the drywell. Spray water can return to the suppression pool once the water level in the drywell reaches the

bottom of the vent pipes (approx. 2 ft deep). Both SPC and CSS

must be manually initiated. Following a LOCA signal (low RPV level

, & high drywell pressure) automatic interlocks prevent actuation of

! SPC or CSS for ten minutes so as to assure LPCI availability to I provide water to the RPV. The branch taken is determined by the l ASEP cut sets which are of interest.

1 I This is a Type 1 question where branch point probabilities are j input and are independent of the branches taken in previous j questions.

There are 3 branches:

1 . ElfRHR: Both SPC and CSS are failed.

2 . ElrRHR: Both SPC and CSS are not currently available but i

are recoverable. This condition could occur in a i

station blackout.

3 . El-RHR: Either SPC or CSS (or most likely both) are available.

l l

Question # 13 : Does the condensate system fail?

Determines the availability of the condensate system. Since the feedwater pumps are turbine driven,' closure of the Main Steam 1 Isolation Valves (MSIVs) results in the loss of feedwater (ie the

ability to inject condensate at high pressure). However, condensate and condensate booster pumps are motor driven and thus

.)

the condensate system can be available whenever offsite power is available. Injection is through the feedwater lines to the feedwater sparger inside the vessel. The condensate system can I thus deliver water when vessel pressures are up to 600 psi. Since

! this is a high pressure system there are no pressure interlocks and

! isolation is done manually if at all. Suction is taken from the condenser hot well. The branch taken is determined by the ASEP cut sets of interest.

This is a Type 1 question where branch point probabilities are l input and are independent of the branches taken in previous j questions.

i L There are 3 branches:

l 1 . ElfCOND: The condensate system is failed.

2 . ElrCOND: ,

The condensate system is unavailable but can be recovered. Such a condition would result from a

! loss of offsite power (for example).

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. . i Pago 12 DRAFT POcch Bottc3 C.E.T. DocunOntation '

l 3 . ElaCOND: The condensate system is available for injection to the RPV but is not necessarily in use.

Question # 14 : Does HPSW fail in a mode which would preclude injection?

Determines the availability of the High Pressure Service Water system for providing injection to the vessel through a cross-tie with the PHR (shut-down cooling mode) system. HPSW normally provides cooling flow for the RHR heat exchangers and drywell coolers. The system has four pumps, two powered by each emergency electical division. Suction is taken essentially from the river.

The branch taken is determined by the ASEP cut sets of interest.

This is a Type 1 question where branch point probabilities are input and are independent of the branches taken in previous questions.

There are 3 branches:

! l . ElfHPSW: HPSW is not available for injection to the vessel.

2 . ElrHPSW: HPSW is unavailable but may be recovered. Loss of offsite power is the most likely cause of this condition.

3 . ElaHPSW: HPSW is available. Operator action is required to align this system for injection to the vessel.

Question # 15 : Is there a failure which precludes drywell cooler operation?

Determines the availability of drywell coolers for containment heat removal. The coolers are sized to remove the drywell heat load during normal operation. This load is approximately 1 megawatt. While the drywell coolers are inadequate to remove decay heat, their operation could impact accident progression. The branch taken is determined by the ASEP cut sets of interest. This question is not referenced in the current tree.

This is a Type 1 question where branch point probabilities are input and are independent of the branches taken in previous questions.

There are 3 branches:

! l . ElfDWC: The drywell coolers are failed.

2 . ElrDWC: The drywell coolers are unavailable but may be recovered.

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J _ Pngs 13 DRAFT Peach Botton C.E.T. DocunOntation l

1 3 . ElaDWC: The drywell coolers are available.

4 Question # 16 : Is ADS blocked or not called upon?  ;

The Automatic Depropressurization System is designed to lower the RPV pressure in the event that the high pressure systems are  ;

not maintaining adequate vessel level. This situation may occur during a small break LOCA. ADS logic requires three indications to initiate: (1) low RPV level (Level 1), (2) at least one LPCS or i LPCI pump -running (discharge pressure indication) , and (3) the ADS i timer (actuated on Level 1 signal) is timed out. The Peach Bottom ADS timer runs for 8 minutes. In an ATWS event the Emergency Operating Procedures (EOPs) dictate that ADS should be inhibited.

The reason an 8 minute timer is incorporated is to allow time for the operators to inhibit the ADS. Thus, no ADS is interpreted as success for ATWS. ADS may also be actuated manually. However, manual depressurization is more likely to occur through manual -

3 actuation of the S/RVs in relief mode. ADS relief valves (7) have their own air accumulators and require only DC power to the solenoid valve on the accumulator to hold the associated valve t open. The branch taken is determined by the ASEP cut sets of  !

interest. For this question, manual depressurization under a

circumstances which would-have lead to ADS will be taken to be the j same as ADS.

i This is a Type 1 question where branch point probabilities are input and are independent of the branches taken in previous questions.

There are 2 branches:

1 . ElnADS: There has been no automatic depressurization of the vessel.

2 . El-ADS: Automatic (or equivalent manual) depressurization 3 of the vessel has occured.

i Question # 17 : Does the RPV remain pressurized?

Determines the RPV pressure once the plant damage state has been established. The question accounts for primary system breaks as well as depressurization as described in Question #16. Emergency Operating Procedures (EOPs) prescribe vessel depressurization in circumstances when ADS would not occur. In particular, when the suppression pool temperature rises above the pre-determined limit with respect to RPV pressure. This limit is the Hot Condensation Temperature Limit (HCTL) which is designed to avoid condensation oscillation dynamic loads in the pool which are postulated to occur  :

when steam is condensed in a hot suppression pool. During  !

isolation events when'no ECC injection (or RCIC) is available, I

i DRAFT P02ch Botton C.E.T. DocunOntation depressurization is included in the procedure to prolong core cooling and provide for the use of the condensate system or HPSW for injection to the vessel.

This is a Type 2 question where branch point probabilities are input and are dependent on branches taken in previous questions.

' There are 2 branches:

1 . E2nDeP: The vessel is not depressurized prior to core damage (if any).

2 . E2-DeP: The vessel is depressurized prior to core damage (if any).

Case i 1 :

In this case the Reactor Pressure Vessel (RPV) has been depressurized either through a large break or through the use of ADS. The RPV is thus certainly depressurized.

Case i 2 :

In this case, at least one S/RV is stuck open. For the

' dominant cut sets SPC is failed and low pressure systems are available. The suppression pool is heating. The open S/RV and the action of HPCI will be lowering the vessel pressure. The Emergency Operating Procedures (EOPs) instruct the operator to depressurize

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the vessel to stay within the HCTL limit (described above).

However, there may be some reluctance to follow this procedure since opening additional S/RVs may be perceived to exacerbate the existing problem of an S/RV which is not closed. This possible reluctance is considered in quantifying branch point probabilities 1

for this case.

Case i 3 :

In this case there is a station blackout with initial high

pressure injection (HPCI / RCIC) . The emergency procedures would lead the operators to depressurize the vessel to maintain the vessel pressure below that which is required by the HCTL curve.

Consideration is given to the fact that coolant injection comes 1

only from turbine driven systems. There may thus be concern over

! the loss of high pressure systems should the vessel be depressurized. It is recognized that depressurization to 150-200 psi will not cause isolation of HPCI or RCIC.

Case i 4 :

In this case there is a station blackout and DC pcwer has failed. There is thus no power to operate.the ADS system or use the relief mode of.the S/RVs. The RPV must remain at the safety valve setpoint (close to operating pressure).

- - - - - - - - -- --c -----

DRAFT Pocch Botton C.E.T. Docussntation l

Case # 5 :

This case includes transient initiated sequences with loss of all injection. Such sequences include blackout with loss of both HPCI and RCIC. The availability of condensate and/or High Pressure Service Water may differentiate the non-blackout sequences.

However, the EOPs would dictate depressurization in either case.

It may be true that the operator would be more likely to depressurize knowing condensate or HPSW were available. It is noted that if these systems were available that the sequence is i

unlikely to lead to core melt. This being the casa quantification assumes the operators pay no attention to the availability of other injection sources. This approach is consistent with the use of symptom- based emergency procedures.

Case # 6 :

In this case there was a transient initiator and high pressure injection is provided by HPCI, RCIC or both. Failure of RHR is implied since with containment heat removal available such a sequence is not core vulnerable. Since the two high pressure make-up systems will eventually fail without containment heat removal, low pressure systems provide the only means of eventual recovery. However, the operator should be following the emergency procedures and depressurize the vessel to remain within the HCTL curve. Thus, even pessimistically, there is only a negligible possibility that the vessel would remain pressurized.

Case # 7 :

This case is similar to the previous one except that CRD flow is available. This case includes the ASEP TW sequences in which high pressure injection is relied upon in the initial stages.

Long-term CRD flow can provide adequate cooling. Since the operator could chose to rely on CRD flow and ignore the HCTL limit

! (or forget that the high pressure systems will fail without l containment heat removal) some slightly larger probabilility for the operator failing to depressurize is given.

Case f 8 :

This case includes all ATWS sequences (TC) in which ADS has been correctly inhibited. ASEP has made an assesment of the j . probability of controlled depressurization in a TC sequence given I that ADS was inhibited. These probabilities ar's used in the C.E.T.

Case # 9 :

This case covers transient initiated sequences with no high.

pressure injection but with low pressure systems available.

Depressurization is certain for these cases. Probably the operators would manually depressurize (not waiting for the ADS timer). If there were no manual depressurization ADS would occur.

This case does not arise in any of the ASEP cut sets.

Pago 16 DRAFT Peach Botton C.E.T. Documentation Question # 18 : For TC is inadequate level maintained with LPI?

This question provides ASEP results for TC sequences to the tree. All six TC sequences defined by ASEP are considered. Branch probabilities are calculated from ASEP results. The question pertains only to the use of Low Pressure Injection since the turbine driven high pressure systems (HPCI & RCIC) are inherently incapable of maintaining level. Throughout this analysis it is assumed that the HPCI and RCIC pumps cannot pump water at a temperature above 100 C for prolonged periods due to failure of the pump seals. Net Positive Suction Head (NPSH) is not at issue as long as the containment remains intact..

This'is a Type 2 question where branch point probabilities are input and are dependent on branches taken in previous questions.

There are 3 branches:

1 . TL: Operator attempts to control level fail and the I

water level is allowed to fall sufficiently that j adequate core cooling is not maintained.

2 . TAF/OSC: Operator attempts to' control level are at least partially successful. Either the level is maintained at Top of Active Fuel (TAF) as specified by the EOPs or the level oscillates but still provides adequate cooling.

3 . na: The sequence is not TC.

case # 1 :

l The operator failed to inhibit ADS. As the vessel pressure decreased past the interlock setpoint injection of 20,000 gpm of cold water would begin. Large level and power oscillations would result. Oak RidgeNational Laboratory (ORNL) analyzed this sequence for the Brown's Ferry Plant (which is very similar to Peach Bottom) and found that containment pressurization will continue at a rapid rate but that core cooling would be maintained as long as water injection continued. The sequence.is thus considered to be core vulnerable. This is sequence #61 on the ASEP event tree.

Case # 2 :

In this case ADS is inhibited and SLC is successful. Reactor power is decreasing due to poison injection. ASEP sequences #49 and #50 are covered by this case. Sequence 49 includes a failure to maintain adequate cooling with the low pressure systems.

Sequence 50 includes a failure to depressurize the RPV, preventing injection-by these systems, and thus adequate core cooling. In either-sequence, water level is not maintained. If it were maintained this would constitute a success path. The probability

Paga 17 7

DRAFT Panch Bottom C.E.T. Documsntation that core damage is avoided when SLC is successful is already factored into the TC sequence frequency input to the tree.

Case # 3 :

In this case ADS is inhibited but SLC was not initiated.

I Vessel pressure was lowered by manual action. This case covers ASEP sequences 58 and 59. In the former sequence there is RPV level control but the containment integrity is still threatened so the sequence is core vulnerable. The latter sequence involves the operator lowering the level too low. Branch point probabilities in the central estimate were assessed directly from the ASEP event trees. The optimistic and pessimistic estimates reflect a 4

reasonable range of uncertianty in these results.

Case # 4 :

In this casa no depressurization occurs. The operator is attempting to control level with only high pressure systems which will eventually fail as a result of the high temperatures produced in the~ suppression pool (from which these systems will take suction). The RPV water level is thus sure to fall too low. This sequence is identified on the ASEP ATWS tree as #60.

Case f 5.:

This case covers sequences which are not TC. Therefore, this ,

question does not apply.

, Question # 19 : What type of seqdence is this (summary of plant damage)?

Determines the sequence specified by the cut set (or sets) i being considered. This is a summary question included only to

~

simplify the logic in the questions that follow. Thus, all branch probabilities are zero or unity.

This~is a Type 2 question where branch point probabilities are input and are dependent on branches taken in previous questions.

There are 10 branches:

1 . AE : The sequence is a large break LOCA with a loss of all injection.

2 . AW: The sequence is a large break LOCA with a loss of j the containment heat removal functions of RHR.

3 . SlE: The sequence is a small break with a loss of all injection.

4 . S1W: The sequence is a small break with a loss of the containment heat removal functions of RHR.

5 .TQUV/FTB: The sequence is a transient initiated loss of all I _ - _ - _ - .

Pcga.18 DRAFT Poach Botton C.E.T. Docuncntation' RPV make-up water. Included are station blackouts in which both HPCI and RCIC are inoperable.

6 . TB: The sequence is a station blackout in which there

' is initialy injection to the vessel. (The sequence may have been intitiated by a plant transient).

7 . TW: The sequence is a transient initiated event in which core cooling is maintained but the containment heat removal functions of RHR are lost.

8 . TC-fDeP: The sequence is a transient followed by a failure to scram (ie ATWS). The operator does not depressurize.

9 . TC-TL: The sequence is a transient followed by a failure to scram (ie ATWS) . The operator depressurizes, or fails to inhibit ADS, and allows the level to fall to low to maintain adequate core cooling.

10 . TC-CV: The sequence is a transient followed by a failure to scram (ie ATWS). The operator controls level at TAF or the level-is~ allowed to oscillate. The CV i

designation indicates that these.areicore vulnerable sequences, not necessarily melts.

Case # 1 :

High pressure systems are turbine driven and are thus unavailable with RPV pressure relieved through the large break in the primary system. Low pressure systems have failed. Loss of all injection following a large break defines AE.

Case # 2 :

A large break loss-of-coolant acciant occurs; followed by a failure of containment heat removal. This defines AW.

Case # 3 :

The initiator is a small break. High Pressure Injection is terminated by depressurization of the vessel. Low pressure injection has failed so there is a total loss of all make-up flow.

This defines SlE.

Case # 4 :

A small break loss of cooant accident (in the size range which l requires ADS) occurs. Containment heat removal subsequently fails. This defines S1W.

l

l Pcga 19 .

DRAFT Poach Botton C.E.T. Documsntation  !

Case # 5 : 1 This case covers all transient initiated events with no injection. This situation defines TQUV. This case includes station blackout with no high pressure systems available (termed Fast 4

station blackout -- FTB).

Case # 6 :

A station blackout is a station blackout (profound, no?).

Initial operation of at least one of the high pressure systems is implied. Since core cooling can be maintained by the turbine driven systems for approximately six hours (battery failure prevents'further operation) such sequences are termed long term station blackouts in the tree. Sometimes the term " station blackout" is also used, without qualification, to mean'the same type of sequence.

Case # 7 :

In this case there is a transient initiator followed by a failure of the containment heat removal functions of RHR. This defines TW. For this case IORV is included as a transient. It  !

i should be noted that the large majority of these sequences are expected to be recovered.

Case # 8.:

In this case the initiator is ATWS and the operator inhipits ADS and later fails to depressurize. This defines the TC - failure to depressurize sequence. This is ASEP sequence #50.

1 Case # 9 : D In this case there is an ATWS event. The operator depressurizes-the RPV and unsuccessfully attempts to control level with the low pressure systems. This defines TC - Failure to maintain adequate level sequence. This includes ASEP sequences #59 and #60.

Case # 10 :

~

In this case there is an ATWS and core cooling is at least temporarily maintained using the low pressure pumps. The containment continues to pressurize. This defines the TC - core vulnerable sequence. For reference, this is ASEP sequence #58.

Case # 11 :

! The program should never reach this point. All cases are defined above.

Question #-20 : Is there an LP system break induced by power cycling (ATWS)?

e --w, , - - ,w, -

w--,. ,-r- -- ,--.,,,_n-. n.___...- _ _ - . ,- , - - . . - - ,

1 Paga 20 DRAFT Peach Botton C.E.T. Documentation '

4 Determines if a break in the low pressure piping (of LPCS or LPCI) occurs as a result of possible power spikes during an ATWS.

This is conceivable since the injection of cold water to the core i with little or no reactivity control may occur. This is especially

true when the operator is attempting to manually control injection

with large capacity systems. Power spikes could be sufficiently

severe or rapid that a. check valve may fail resulting in exposure  ;

of the LP piping to high pressure. It is noted that BWR low '

pressure piping is expected to withstand full RPV pressure and has

! withstood static testing at these pressures. For completeness, allowance has been made for these postulated dynamic failures.

This is.a Type 2 question where branch point probabilities are input and are dependent on branches taken in_ previous questions.

a There are 2 branches:

1 . PCyBk: A break occurs in the LPCS or LPCI piping as a result of an ATWS power spike.

2 . nPCyBk: No break is induced in low pressure piping as a result of ATWS.

Case # 1 :

A large power spike is judged to be likely if ADS were to occur during an ATWS event. This judgement assumes that no steps were previously haken to inhibit Iow pressure injection. Following ADS both the LPCS and the LPCI would inject. The voids in the reactor-i core would collapse.- With the core filled with cold water, an over-moderated condition would result. The core power is expected to increase to a' level which would be substantially larger than rated power. Fuel rods would heat rapidly, voids would be formed, i resulting in the reactor power decreasing. However, with the void-formation, there would be an large increase in the vessel pressure. This occurrence is judged to have the largest potential for inducing a pipe break in the low pressure systems.

Case # 2 :

This is a TC in which the operator has at limited success in-contolling level. There is some opportunity for cycling of the power level to be induced if (or when) the operator depressurizes the vessel and attempts to use the low pressure, high capacity, systems to control the RPV level. However, since in the sequences considered here the result of this attempt is that injection is inadequate there is judged to be little chance of significant power cycles and even less chance of a break being induced.

l Case # 3 :

In this case the operator is maintaining adequate core cooling at a reduced level. Manual control of the ECCS pump flowrates has

Page 21 DRAFT POsch Bottom C.E.T. Documsntation ,

i l

been taken. It is thus conciavable that a power spike could occur' ,

from a sudden increase in water injection. However, it is I pressumed for these sequences that the operator follows the l Emergency Operating Procedures. The logical opportunity for a power spike comes as the vessel is depressurized and injection is transferred-to the large capacity low pressure pumps. With slow depressurization and adequate throttling of the flow as the injection valves are opened no problems should result. If the depressurization were too rapid and if the pumps were allowed to begin injection at full flow, a power spike may result as water from the suppression pool (with high.subcooling and thus a good

' moderator) is introduced to the core. With mixing of saturated water already in the vessel any power increase is not likely to be severe enough to cause repressurization of the vessel. A break in the low pressure piping from repressurization thus induced is even i less likely.

Case # 4 :

i This case covers all sequences which are not TC. Power cycles cannot occur if the Reactor Protection System (RPS) functiuons.

Since there are no power cycles, a power cycle break is impossible.

I Question # 21 : Do low pressure ECC systems continue /begin to inject?

This question determines whether low pressure systems are used to provide cooling flow to the core. The status of the low pressure systems (LPCS and LPCI) is also reflected for reference by future quastions. This status will be unchanged from that reflected in Question ill above unless the systems were previously available and are now being used for injection.

This is a Type 2 question where~ branch point probabilities are input and are dependent on branches taken in previous questions.

There are 4 branches:

1 . E2fLPC: Low pressure ECCS (both LPCS and LPCI) have failed.

2 . E2rLPC: Neither LPCS nor LPCI are available. However, at least one of these systems is recoverable.

3 . E2aLPC: Either LPCS or LPCI (or both) are available (the pumps are running) but are not providing make-up i water to the RPV. (Most likely because the RPV pressure is too high).

4 . E2-LPC: Either LPCS or LPCI (or both) are providing make-up water to the RPV.

i l Case # 1 :

Pago 22 DRAFT Patch Botton C.E.T. Docum3ntation In this case low pressure systems were initially recoverable

, and nothing has happend to alter this status. They remain recoverable.

Case # 2 :

In this case the vessel is not depressurized. Even though the low pressure systems may be availabls, there can be no injection since the injection valves are closed. This case includes the TC sequence identified by ASEP in which the operators' fail to  :

depressurize the vessel. This case is singled out since it specifically represents a branch point on the ASEP tree and does not reflect any judgement on the part of the authors of the containment Event Tree with respect to branch point probabilities. ,

However, since the case requires that the RPV remain pressurized, it is certain that there will be no low pressure injection.

Case # 3':

Low pressure systems were both failed initially; they remain so l since they cannot be recovered, or inadequate level is maintained with LPCS and LPSI in an ATWS.

Case # 4 :

This case covers all sequences in which the vessel.is depressurized and at least one of the low pressure systems is available. Since low pressure injection is automatic for such circumstances, injection is deemed certain.

Question #'22 : Do containment sprays fail before core degradation?

This question determines the availability of containment spray mode of the RHR system. This includes the availability of spray water-only, not the ability to use the sprays for the removal of heat from containment; the latter is considered in questions pertaining to the RHR system. For instance, if Emergency Service Water is failed the RHR heat exchangers are useless but spray water

' can still be directed to the drywell and this would profoundly influence both the accident progression and source terms. Use of HPSW for containment spray has been included. The Peach Bottom P&ID for LPCI indicate that valve alignment to spray service water is possible. While this mode would not remove energy from the containment, the containment temperature and pressure would be lowered.

This is a Type 2 question where branch point probabilities are

< input and are dependent on branches taken in previous questions.

i There are 3 branches:

1 . E3fCS: . Containment sprays are failed.

2 . E3rCS: Containment sprays are unavailable but may be recovered.

Paga 23 DRAFT Peach Bottom C.E.T. Documentation p

3 . E3-CS: Containment sprays are used to spray the drywell.

Case # 1 : -

In this case the accident initiator is either a primary system break or a transient followed by the failure of the Reactor Protection System (ie an ATWS) . For such sequences the drywell pressure is rapidly elevated. The containment spray system is intended to provide rapid reduction in the drywell pressure. RHR availability is assumed to imply that spray mode can be used.

That i

the operator will use spray under these conditions of high containment pressure is judged to be almost certain.

Case # 2 : <

In this case the accident initiators considered are the same as those in the previous case of this question. The plant damage state is different in that RHR is not available so the use of spray will not lower the total heat within the containment. However, either the RHR pumps (LPCI) or High Pressure Service Water is available for use in containment spray. (There is a slight inaccuracy here in that the tree references the availability of LPCS and LPCI together so LPCI failed and LPCS'available could be the plant damage state for a sequence included in this case. Since none of the ASEP core melt or core vulnerable cut sets contain this damage state, the event tree results are not impacted.) As noted

above, High Pressure Service Water can be used for containment spray and this will have the effect of lowering the containment temperature and (at least temporarily) the pressure. Since containment spray is not available in the usual sense and because the use of HPSW requires more unusual operator actions, a lower probability for successful spray is judged to apply in this case.

Case # 3 :

i In this case the accident initiators considered are the same as the two previous cases of this question. The plant damage state in this case has the RHR and HPSW systems to be unavailable; both are recoverable. With at least some of the pumps which can be used for spray recoverable, the sprays themselves are recoverable.

Case # 4 :

This case covers sequences in which the initiating event is a transient and there is little presurization of the containment because the suppression pool will prevent significant increases-in the containment pressure. With no pressurization of the containment, there will be no call for containment spray so the operator ~will not use it. This case also covers TW sequences; the small fraction which result in core damage. This inclusion, which l implies no containment spray prior to core damage, assumes that if spray were possible then injection to the vessel would be possible

Pogo 24 DRAFT Pacch Botten C.E.T. Documentation and core damage would be averted.

Question # 23 : Do S/RV-tailpipe vacuum breaker (s) stick open?

This question determines whether one or more of the vacuum breakers on the S/RV tailpipes has stuck open. A stuck open

, tailpipe vacuum breaker would provide a pathway for suppression  ;

4 pool bypass since gasses released from the vessel down the tailpipe would pass directly into the drywell.. Tailpipe vacuum breakers 4 ,

i will open after the associated S/RV discharges steam through the tailpipe into the suppression pool. When the steam in the tailpipe condenses on the pipe walls a vacuum is formed. The vacuum breaker prevents the tailpipe from drawing suppression pool water into it. '

A stuck open tailpipe vacuum breaker is only significant if it is the vacuum breaker on the tailpipe for an S/RV which is expected to be open after core damage occurra. Thus the cases below consider which vacuum breakers are challenged by the sequence during the boildown phase of the accident. The question only reflects significant vacuun breakers being stuck open (ie ones which will result in fission products bypassing the suppression pool).

l This is a Type 2 question where branch point probabilities are

! input and are dependent on branches taken in previous questions. ,

(

There are 2 branches:

1 . oSRVBkr: There is a least one S/RV tailpipe vacuum breaker stuck open.

2 . cSRVBkr: There are no S/RV tailpipe vacuum breakers stuck open.

Case # 1 :

This case covers transients followed by loss of all injection (TQUV) with no depressurization and no. stuck open relief valves.

The vessel will boil down to the point at which core damage occurs in about an hour. There will be relatively few challenges to the S/RVs during this short boil- off time. Also, for most of these i

sequences, the operators will depressurize the vessel and thus the S/RV challenged during the high pressure boil-off will not be the one which is open during core degradation. (During boil-off the low- low set valve will cycle, the operators are quite unlikely to 1

open this particular valve in relief mode.)

case # 2 :

! This case covers long term isolation transients. These are similar to those covered in the previous case except that high pressure systems provide initial core cooling. There are thus many '

more challenges to the tailpipe vacuum breakers. With the vessel at high pressure (with no SORV), the low-low set S/RV will cycle.

This is the same valve through which fission products could be i

..._m_,,..__ .,.._,_..m,-. , _ . _ . . . , , _ .._~.__._,.,_......__.-._____.,-,__,.___,.m_. , _ _ _ _ , _ _ _ - . _ , _ - _ _ - -

i . .

Paga 25 i

DRAFT Patch Botton C.E.T. Documentation released if core damage followed, thus a stuck open vacuum breaker could be.significant. Quantification for this case thus reflects a judgement of the probability that a tailpipe vacuum breaker will i stick open after approximately six hours of vessel-boil-off to the -

pool. This time period represents between one and two hundred challenges.

F Case # 3 :

This case includes TC sequences with a failure to depressurize the vessel. The power level is such that one or more S/RVs will be held open in relief mode if depressurization does not occur. The valve which would cycle, thereby challenging the tailpipe vacuum breaker, is not the low-low set valve and thus would not be the valve through which fission products pass. A smaller' probability of this bypass is thus judged valid. Only.for a very short period, 4

late in the boil-off, following injection failure, would the i low-low set' valve tailpipe vacuum breaker be challenged. The number of challenges should certainly be less than 50. ,

Case # 4 :

This case covers sequences with primary system breaks and/or early vessel depressurization. For such cases the release pathway is not through an S/RV which would have been cycling earlier. Thus there.can be no bypass through any stuck open tailpipe vacuum breaker.

i Question # 24 : In the containment vented before core degradation?

Determines whether the operator successfully vents containment prior to core damage. The EOPs instruct containment venting when

containment pressure exceeds 60 psig. There are three likely vent pathways: The first is a 2" line (with a 1" valve) which connects-i

to the torrus airspace and is part of the vent and purge system.

The second consists of two 18" lines, also connected to the torrus airspace and also part of the vent and purge system. The third is the Integrated Leak Rate Test (ILRT) line which connects to the torrus and vents outside the reactor building. This latter line has~four valves in series and is terminated by a blind flange. The other pathways can be opened from the control room provided AC power is available. The procedure for opening the 18" lines calls for local action to deflate the valve seals prior to opening the valve. In addition there are lines connected to the drywell which could be used for venting. However, since the procedures call for opening the wetwell lines first, and significant time would be required for these actions, the possibility of drywell venting is judged to be negligibly small and is not considered in the containment event tree. Successful venting is defined to mean that a depressurization of the containment occurs. This definition is particulary significant for TC sequences in which opening an 18" line is unlikely to allow enough steam to escape the containment to result in the containment pressure being lowered.

This is a Type 2 question where branch point probabilities are

~

_-_L-_--- . - - _ _ _ - - _ .. -_ __

DRAFT P0cch Botton C.E.T. Docussntation input and are dependent on branches taken in previous questions.

There are 2 branches:

1 . E3nVENT: The containment is not successfully vented prior to core damage.

2 . E3-VENT: The containment is successfully vented prior to core damage.

Case # 1 :

This case includes station blackout (High Pressure Injection continues until battery depletion). The containment venting pressure of 60 psig will not be reached prior to core damage. It

! is also true that the operator has no indication of containment pressure due to the loss of AC power which is used for the pressure transmitters. (A mechanical pressure gauge could be rigged to give this indication but procedures to do this are not specific.) 'It is

, judged that there is some probability of venting. This possibility arises because there is uncertainty as to the containment pressure, due to the loss of indication, and because the operating staff would be aware that it is a blackout and may anticipate future problems with venting by venting before core damage occurs. (A subsequent question considers venting after core damage has occurred.)

Case # 2 :

This case covers break initiators followed by a subsequent loss of containment heat removal. This sequence is slow to progress.

Venting (in accorance with EOPs) would occur approximately 8 to 10

. hours after the break occured. The probability of venting for these accidents has been assesed by ASEP. The optimistic and central estimates reflect what is judged to be an appropriate range of this value. It is noted that with a primary system break there will be alarms indicating high radiation in the containment. These alarms could bring about some reluctance to vent the containment.

Case # 3 :

In this case-there is a loss of containment heat removal following a transient'avent. Transient initiated events are even slower progressing than those which are break-initiated. It requires on the order of a day to reach the venting threshold.

There will probably be no radiation alarms in this. sequence. The operators will have plenty of time to prepara for venting. Public officials can have ample notification to consider measures which may be needed to protect or reassure the surrounding population.

For these sequences, venting is the only measure which is expected to avert core damage, thus it seems all but certain that venting would occur.

DRAFT Poach Bottom C.E.T. Docum3ntation Case # 4 :

i In this case the sequence is TC. For an ATWS sequence f successful venting is defined to mean arresting the rise in containment pressure. This is doubtful with only one 18 inch vent line if the core power level is 18%. However, there are large uncertainties in the power level calculation; this is especially true when the vessel is at low pressure, as it would be for these sequences. Quantification of the probabilities for venting in this case reflects this uncertianty. current calculations probably

-overestimate the power in a depressurized vessel since there is-increased voiding due to the larger change in the density

, steam-water heated at low pressure. Sequences involving a power cycle break in the low pressure systems containment pressure are excluded from this case since the containment is bypassed and no venting will occur.

Case # 5 :

This case catches those sequences in which the containment challenge, if any,'comes after core degradation. Blackouts, in which pressure indication is lost, are not included but have been considered in a previous case of this question. Since the pressure has not reached the venting threshold, it is judged to be certain that the operators will not decide to vent. To do so would violate the EOPs.

Question # 25 : Does the RPV repressurize before core damage?

The S/RVs or ADS valves are held open by air pressure. The valves stay open only as long as the air pressure in the accumulators (containment isolation isolates the accumulators from plant air) exceeds the drywell pressure. If the containment is allowed to pressurize to within 20-25 psi of the air pressure level (ie containment at approx. 110 psi) the valves will reclose and the RPV will begin to repressurize.

This is a Type 2 question where branch point probabilities are input and are dependent on branches taken in previous questions.

There are 2 branches:

1 . rep: The Reactor Pressure Vessel (RPV) repressurizes prior to core damage.

2 . nReP: The Reactor Pressure Vessel (RPV) does not repressurize prior to core damage.

Case # 1 :

This casa excludes: TB (station blackout), TW (transient followed by loss of RHR), and TC-CV (ATWS with successful level control). These are the only sequences for which repressurization

~

l Pcga 28 i DRAFT Pcach Battcm C.E.T. Documsntation prior to core damage is possible. This case includes sequences initiated by breaks and transient intitaited sequences where there is a loss of adequate RPV water level before containment pressurization is significant. Repressurization does not occur '

before core damage for these sequences.

Case # 2 :

j In this case there is a stuck open relief valve. Containment i pressurization cannot ressat this valve so regardless of whether other valves are opened in relief mode and can thus be reclosed, there is judged to be a negligible probability that the Reactor Pressure Vessel could repressurize.

Case # 3 :

In this case an ATWS power cycle has induced a primary system break in the reactor building. With a break in the primary system, repressurization is impossible.

Case # 4 :

In this case the sequence is a TC and the operator is successfully controlling the RPV water level using the low pressure systems. Venting is successful and the containment pressure

! decreases. It is certain that the S/RVs will not reclose.

Case # 5 :

In this case the sequence is a long term blackout (initial high

, pressure injection).- The vessel was depressurized by operation action to stay within the HCTL. Containment pressurization will be~

-insufficient ~to cause S/RV reclosure regardless of any action taken to vent. Repressurization will occur before core damage if the station batteries fail prior to the water make-up pumps (HPCI &

RCIC) failing as a result of the pumps overheating. If the batteries fail, injection ceases and the S/RVs reclose. If the pumps fail first, the vessel will remain depressurized. Even i

though there are methods available to the operators for prolonging j

t battery life, battery failure is the more likely cause of repressurization. Quantification of the probabilities for this L case thus reflects the probability of battery failure for blackout i sequences with initial high pressure make-up.

Case # 6 :

In this case the sequence was transient initiated and depressurization occurred prior to loss of injection capability.

i The containment continued to pressurize and was not vented. The l

S/RVs are certain to reclose as a result of the continued increase in the containment pressure.

case # 7

l This case covers sequences in which the vessel was not l

..,_.-e ,a. , , , - - ,

- , - - - . - - - - - - - - - , - - - - - ~ , - - - - - - - ~ ~ - ' - - - - - - - - ' - - - - - - - - - - - - - - - ' - - - - - - - - ^ ^ ^ - ' - - ' - - - - -

~. . - - -_ .. __- _-_ . .

. . 1 DRAFT Pscch Botton C.E.T. Documsntation

. 1 depressurized.- Since it was not depressurized it cannot repressurize. TW sequences in which the containment were successfully vented are also covered by this case.- Venting will preclude repressurization.

Question # 26 : Is there a repressurization break of low pressure ,

systems?

Determines whether a break occurs in the low pressure piping of the LPCS or LPCI systems as a result of the RPV repressurization.

Since the low pressure piping on a BWR can probably withstand .'

l primary system pressure, and since the check valves should function to. protect the " low pressure" piping, a break induced as a result of the vessel repressurizing is not considered-to be of significant probability. This question is included primarily for completeness since earlier BWR risk assessments have show "V" sequences to be significant risk contributors. While ASEP did not identify interfacing system LOCAs as significant initiating events, they may be considered more likely if the vessel is repressurized with the low pressure ECCS operating. In this case action of the check valve, exclusive of pressure interlocks closing the injection valves, is needed to protect the upstream piping from full RPV pressure. However, even if these valves were not to close, it is judged extremely unlikely that the piping would fail.

i This is a Type 2 question where branch point probabilities are input and are dependent on branches taken in previous questions.

Thera are 2 branches:

1 . RePBk: A break in the low pressure ECCS piping results from RPV repressurization.

2 . nRePBk: There are no breaks in the low pressure ECCS piping resulting from vessel repressurization.

Case # 1 :

This case includes all sequences in which there is a repressurization of the primary system prior to core damage. RPV pressure thus increases past the shut-off point of low pressure injection. Only in such sequences is a repressurization break possible. Quantification reflects the probability that the check valve will fail to reclose and that the piping has some defect which results in a break occurring. The piping has sufficient design strength to withstand the pressure.

Case # 2 :

1 This case includes all sequences where there is no repressurization. Therefore, a repressurization break is impossible.

Paga 30 DRAFT Patch Bottos C.E.T. DocunOntation Question # 27 : Does containment fail before core degradation?

This question determines whether or not the containment fails prior to core degradation. This.is possible only when injection to the vessel is maintained but heat removal from the containment is inadequate.

This is a Type 2 question where branch point probabilities are input and are dependent on branches taken in previous questions.

There are 2 branches:

1 . E3-CFP: The containment fails as a result of overpressurization prior to core degradation.

2 . E3nCFP: The containment is not failed prior to core degradation.

Case # 1 :

This case covers sequences with an induced break in the low pressure ECOS piping. With such a break, containment is bypassed.

Since there is no challenge to the containment, overpressure

, failure is impossible.

Case # 2 :

In this case the containment is successfully vented. By i

definition,this precludes containment failure due to overpressure.

!, For this case failure is then impossible.

Case # 3 :

l l

This case includes all sequences in which there is no challenge l to containment prior to core degradation. Included are all l

sequences without initial injection to the vessel, blackouts, and ATWS events in which injection is insufficient to maintain adequate.

cooling even though the systems are operable (failure to depressurize or level maintained too low sequences). For all these sequences core damage will occur before the pressure rise in the containment is sufficient to challenge its integrity. Loss of containment heat removal (W) sequences and the core vulnerable TC sequences are specifically excluded.

Case # 4 :

In this case the sequence is an ATWS with adequate core cooling being provided by the low pressure systems. The power level is estimated to be approximately 184. Venting was not attempted or was unsuccessful and the containment pressurizes rapidly. As the containment continues to pressurize, there are two possibilities.

Either the containment will fail or the S/RVs will reclose due to the elevated containment pressure early enough to prevent

Paga 31 1 DRAFT POsch Botten C.E.T. Docuncntetion l l

pressurization of the containment to the point of failure. In the latter case core degradation will occur in an intact containment.

This case thus essentially determines if containment feils after injection is lost but before the level in the RPV boils down to the point at which core damage will occur Case # 5 :

This case includes transient initiated sequences with loss of containment heat removal. High pressure systems will fail on high containment pressure (this case includes no venting) or high pool temperature, the RPV will repressurize eliminating low pressure injection. CRD flow is also not available so there is no injection and core damage will occur in an intact containment. Thus, for this case, containment failure prior to core degradation is impossible.

Case # 6 :

This case covers the remaining sequences for which injection continues in the absence of containment venting and containment heat removal. Subsequent failure of the containment prior to core damage is certain.

Question # 28 : What is the containment leakage level before core degradation? .

This question determines the level (or rate) of leakage from the containment prior to core degradation. Two distinct levels of leakage are allowed. Level 2 corresponds to a small leak which provides essentially no reduction in the containment pressure (leak rate approximately equal to vessel steaming rate). Level 3 corresponds to a rupture of the containment (the containment depressurizes).

This is a Type 2 question where branch point probabilities are input and are dependent on branches taken in previous questions.

There are 3 branches:

l . E3nCL: There is no leakage from the containment i

significantly in excess of technical j specifications prior to core damage.

2 . E3-CL2: There is leakage from the containment. prior to core damage. However, the leakage is insufficient to result in depressurization of the containment.

3 . E3-CL3: There is a failure of containment prior to core damage of a size sufficient to cause depressurization of the containment.

Case # 1 :

1

~-- _ _ . - - - - - -_. _ . _ -

Pago 32 1

DRAFT P0cch Botton C.E.T. Docum0ntation j

In this case a. break has been induced in the low pressure system piping. The containment is effectively bypassed.

Containment pressure will remain at approximately its current level so long as the RPV remains intact. However, leakage of steam from the vessel is sufficiently large that, with respect to impact on the reactor building,.this is considered to be Level 3 leakage.

This classification is consistent with subsequent references to this question.

Case # 2 :

In this case the containment has been successfully vented.

With the containment vented, leat ge from the containment is, by ~

the definition of successful vencing, Level 3.

Case # 3 :

This case is an ATWS in which centainment failure occurs prior to core damage. With the high pressurization rate assumed to occur for ATWS, a large size failure of containment is' judged to be more likely. The IDCOR analyses assume a leak type failure of containment when the drywell temperature is elevated but this will not be so for ATWS events with early containment failure resulting from steam pressurization. In addition, the pool will be hot and will. flash when the containment pressure begins to drop. This added energy release will have the effect of exacerbating the

' containment leak size.

j Case # 4 :

In this case the containment fails due to a gradual overpressurization with steam. The containment shell will be heated in this process. Based on intuition and the results of a single Sandia experiment on the overpressure failure of a steel building (not similar to a containment and not at elevated temperature), a large failure is judged to be more likely.

Consideration is given to the IDCOR analyses which indicate leak type failures are more likely when the temperature of the steel containment shall is elevated. As in the previous case, however, pool flashing will occur which is judged to make a larger leakage rate more likely. It is also noted that while the containment shell temperatures will be elevated, they are not at the levels typical of the IDCOR analyses after vessel breach (> 550 C) when the leak type failures seem more plausible.

Case f 5 :

In this case the containment is not bypassed, vented, or failed. The probability of significant leakage is thus judged to be negligible.

Question # 29 : Does AC power remain lost during core degradation?

This question accounts for the delayed recovery of offsite L - -- ,.,. - - - _ - - - . . . . - , - . . - - - , - . - - - - - - - - - . - - . - - - - - - . - - - - - - - - - - -

Paga 33 DRAFT POcch Ectton C.E.T. Documentation a

power. ASEP accounts for recovery which occurs prior to core damage in estimating the sequence frequency but later recovery is also possible. Late recovery of AC may prevent vessel melt-through and otherwise mitigate accident consequences.

This is a Type 2 question where branch point probabilities are-input and are dependent on branches taken in previous questions.

4 There are 2 branches:

1 . E4fAC: AC power is not recovered prior to vessel breach.

2 . E4-AC: AC power is recovered during core degradation. '

Case # 1 :

j This case includes the so-called fast station blackouts in which high pressure systems fail to inject. For these sequences recovery during core degradation is eqivalent to the probability of power being restored between 1 and 2 hours2.314815e-5 days <br />5.555556e-4 hours <br />3.306878e-6 weeks <br />7.61e-7 months <br /> after the loss. The central estimate of the probability was made based on available

. data for non-recovery of offsite power as a function of time. The 4

optimistic and pessimistic estimates were made assuming a plus or minus 300% error in the non- recovery probability.

Case # 2 :

j This case includes blackouts in which there is early high

pressure coolant make-up. AC power would have to be restored between 6 and 8 hours9.259259e-5 days <br />0.00222 hours <br />1.322751e-5 weeks <br />3.044e-6 months <br /> after the blackout.

~

The recovery probability was estimated in the same manner as the previous case and

, represents the probability of offsite power recovery by.the time 8

- hours has elapsed given that it was not recovered 6 hours6.944444e-5 days <br />0.00167 hours <br />9.920635e-6 weeks <br />2.283e-6 months <br /> after the offsite power was lost.

case # 3 : .

For the remaining sequences some form of AC power is available

~

l l (either emergency or offsite) thus AC power remains available with a probaility of unity.

Question # 30 : What is the status of CRD?

Steam flooding of the reactor building is a possible consequence of containment failure or of venting. Steam flooding can impact the operability of control rod drive flow by causing shorts or other malfunctions in the pumps motor controls. This question reflects the possible impact of the steam flooding i

phenomenon. Also included is the impact of delayed recovery of AC power.

! This is a Type 2 question where branch point probabilities are input and are dependent.on branches taken in previous questions.

s, ,e-- , , - - - - - - ,w.n.,,.w-,, ---,,,n ,.-------~.------w..~--.--r , - - - - - - - - - . - , - - - - - - - - - - - - - - - - - - - - - - . - - - - - -

Page 34 DRAFT Peach Bottom C.E.T. Documentation

~

There are 3 branches:

1 . E34fCRD: The control rod drive cooling flow is failed.

2 . E34rCRD: The control rod drive cooling flow is unavailable, 4 but may be recovered.

3 . E34-CRD: There is flow from the control rod drive cooling system into the vessel.

4 Case # 1 : ,

In this case there is leakage from the containment large enough to cause the containment to depressurize. For these early failure / vent sequences the pool will have been superheated relative to atmocpheric pressure and large volumes of steam woould be released. Alternatively, a break has been induced in~the primary _

system. Either case has the highest possible potential for damage to the CRD hydraulic system which in this case was, until steam could damage it, was either cperating or was going to be recovered i

as offsite pcwer was restored. In either case the CRD pump motors and motor controls would be energized in a high steam environment.

While these systems are qualified to withstand post-LOCA and steamline break conditions, the conditions resulting from a total breach of containment would be more severe. A significant probability of pump failure must thus be recognized.

Case # 2 :

l This case is similar to Case #1. It is. judged that the probability of failure of CRD flow due to steam flooding is independent of whether or not the CRD pumps are operating at the l time of containment failure. In this case the CRD pumps are not

currently injecting but are recoverable. Thus this case is similar to Case #1 of this question except that if steam flooding.does not cause failure the CRD flow remains recoverable rather than remaining operational.

Case # 3 :

In this case CRD pumps have been providing injection all along

. or have been recovered by the recovery of AC power. There is no i significant steam leakage to the reactor building. Thus the pumps are considered certain to remain operational.

I Case # 4 :

( In this case there is also no significant leakage to the containment to impact the status of the CRD cooling system. Since the system was unavailable but recoverable, it remains in this J condition.

._.-,--~.e. -- _. _ _ _ _ _ _ _

Paga 35 DRAFT Pasch Bottom C.E.T. Documsntation Case # 5 :

This case includes only those sequences in which the CRD cooling system was previously failed. This being the case, it is certain to remain failed.

Question # 31 : What is the status of HPSW?

This question deteremines the availability of High Pressure Service Water for injection to the vessel through the. cross-tie with the LPCI system. Operator action is required to open the valves on the cross-tie line. In addition, LPCI injection valves must be opened (shut-down cooling mode). The HPSW pumps require offsite power. The HPSW pumps are not in the reactor building, and thus could not fall victim to induced failure by steam flooding.

However, the motor operated valves on the cross-tie are in the reactor building and are therefore vulnerable. Since these valves are normally closed failure of the valves would result in HPSW being unavailable for injection to the RPV.

This is a Type 2 question where branch point probabilities are input and are dependent ~on branches taken in previous questions.

There are 3 branches:

1 . E34fMSW: High Pressure Service Water has failed in a manner which precludes injection to the vessel.

2 . E34rHSW: High Pressure Service Water is unavailable for injection to the vessel but is recoverable.

3 . E34aHSW: High Pressure Service Water is available to supply low pressure injection to the RPV.

Case # 1 :

In this case, as in Case #1 of the previous question, there is a high level of steam flooding in the reactor building. Prior to the loss of containment integrity, either HPSW was available or it was going to be recovered through the restoration of AC power.

Unlike safety related systems, the HPSW cross-tie is not qualified for severe conditions in the reactor building. qualified for such environments. Thus, the probability that the HPSW cross- tie will be inoperable, and thus HPSW will be unavailable (for either vessel injection or containment spray), is judged to be significant.

Case # 2 :

This case is similar to Case #1 of this question except that

, service water was unavailable prior to containment failure. While l the condition was recoverable, steam flooding of the reactor building could induce failure. Quantification of the branch probabilities for this case is exactly as in case #1 with the l

L -

Pago 36 DRAFT Pacch Bsttom C.E.T. Documsntation exception that even if there.is no induced failure, HPSW is only recoverable rather than judged to be injecting to the vessel.

Case # 3 :

1 In this case, either HPSW is available and thus remains so, or HPSW was recoverable and AC power has been recovered. making HPSW once more available. Any leakage which may be occurring from the containment to the reactor building is judged to be insufficient to impact valve operability.

Case # 4 :

, In this case HPSW was initially recoverable. Leakage from the containment is sufficiently low that it is judged that there would be no impact on valve operability. It is thus certain that HPSW will remain recoverable.

Case # 5 :

This case catches all sequences where High Pressure Service

' Water is initially failed. This being the case it certainly i

remains failed.

Question # 32 : Does the core melt? .

This question resolves core vulnerable sequences into two groups: 1) Those which and in safe stable conditions and 2) those in which core damage occurs. All sequences defined by ASEP to be core melts are properly identified as such by this question.

This is a Type 2 question where branch point probabilities are input and are dependent on branches taken in previous questions.

There are 2 branches:

I 1 . MELT: At least some core damage occurs.

2 . nMELT: Core vulnerable sequences assigned to this branch result in stable shut-downs with no core damage.

l 1

Case # 1 :-

~This case covers sequences initiated by a primary system break l in which the containment heat removal system fails. However, the i

condensate system is available and could inject to the vessel.

Operator action is required to align the condensate system and

re-start the condensate pumps. It is assumed that given that the operator does this, core damage will be avoided. (Even if no l venting occurs'the containment would flood.) Qantification of the

probabilities for this case is based on the fact that the condensate system is one with which the operator is very familiar.

l Condensate would, more than likely, be the system the operator i

~_- - - _-..- - .- . ._. .

l Paga 37 DRAFT P ach Botton C.E.T. Documentation i

l would turn to. l Case # 2 :

This case is similar to case #1 of this question. The  ;

difference being that instead of condensate being available for '

, injection to the vessel, it is High Pressure Service Water. This case is quantified in exactly the same manner. It is recognized that use of the HPSW cross-tie is not a familiar action for the operator. However, in these sequences, containment pressurization is sufficiently. slow that the operator.would have on the order of ten hours to align HPSW. It is highly unlikely that the operator

! would allow the core to melt with any injection system which is identified in the procedures being available, given this much time. ,

i Case # 3 :

l This case is again similar to cases 1 & 2 of this. question. In this case however neither condensate or.High Pressure Service Water is available for injection. The only system that is available long term is the CRD cooling system. For these sequences the containment must be vented or eventually failed. It is assumed i that this will cause loss of Net Positive Suction Head (NPSH) for the ECCS pumps. Calculations by, Oak Ridge National Laboratory _

(ORNL) for Brown's Ferry (similar to Peach Bottom) indicate that CRD flow can remove decay heat from the core 6-8 hours after shut-down. These calculations were performed for a transient initiator. It was also assumed that the operators took action to optimize the flow from tha CRD pumps. . There is some uncertainty that CRD flow could be adequate when there is a primary system break. This is particulary true is the elevation of the break is below Top of Active Fuel (TAF). This uncertainty is accounted for in the evaluation of the success probabilities for this case.

1 Case # 4 :

In this case the initiating event is a transient. Referring to the comments in case #3 of this question, the Oak Ridge calculations indicate that CRD flow will provide adequate cooling for long term transients. This being the case, the probability of 1

core melt is negligible for a TW sequence with CRD flow available, regardless of RPV pressure.

Case # 5 :

In this case the sequence was initiated by a transient and containment heat removal subsequently failed (TW). The RPV was

~

subsequently depressurized. Repressurization of the RPV has either not occurred ~(ie the containment was vented) or the vessel did repressurize and now is again at low pressure either becase a break

occurred in the low pressure ECCS piping or because the containment l failed thus allowing the S/RVs to reopen. The condensate system is

, available. As in case #1 of this question, all that is required is i for the operator to align the system and re-start the pumps. The success probabilities are determined in exactly the same manner as

--.,_,,,p-,,._,.,,- -, , . , , + - -

--e ------

.. ._ _- . = - . _ .. _ - - . . .-

Page 38

] DRAFT POcch Botton C.E.T. Documsntation

1 for Case #1 above.

Case # 6 :

. This case is similar to Case #5 of this question except that High Pressure Service Water is available for injection rather than

, the condensate system. Quantification is identical.

Case # 7

This case is similar to case #5 of this question. The vessel pressure is low as described in that case. However, the sequence was intitiated by an ATWS (it is TC - core vulnerable) and therefore the operator has far less time than in the TW sequence to restore injection after the containment fails or is vented. This sequence only requires on the order of an hour to reach the point where injection will be lost due to the loss of NPSH for the ECCS pumps. The TW sequence takes about a day and a half. Once injection is lost.for this sequence, there are only a few minutes available in which to avert core damage by restoring injection.

For the TW sequence the operators have approximately two_ hours to restore injection after the ECCS is lost. However, this case has condensate available. Since the operators would be very likely to be trying to restore condensate with the ECCS running, quantification of the success probability is the same as for TW.

Again, it is judged highly unlikely that the operators would allow the core to melt with the condensate system available.

Case # 8 :

.This case is similar to Case #7 of this question with the exception that it is the High Pressure Service Water on which the operator must rely rather than the condensate system. Because the

! operator would be less familiar with using the High Pressure i Service Water for injection than'with using condensate and because

! there are more valves which must be opened or closed, it is judged that success is less likely than when condensate is available.

Case # 9 :

This case catches all the ASEP " core melt" sequences. In other

, words the previous cases include all sequences identified as being core vulnerable. For this case the core melt probability must thus be unity to maintain consistency with ASEP.

Question # 33 : What is containment pressure before core degradation?

This question provides an assesment of the containment pressure at the onset of core damage. These estimates are based on j thermal-hydraulic calculations performed using BWR-LTAS (written at ORNL by M. Harrington) or on calculations performed for the Containment Loads Working Group (CLWG) or the Containment Performance Working Group (CPWG). In some cases Source Term Code Package (STCP) calculations may also have been used. The containment pressure depends on both the sequenco definition and

,-----.-----,-vrw-ym  :. .--,w. -

,w-v,,-e-,--ww-,

4 . .

Pago 39 DRAFT Panch Bottcm C.E.T. Documsntation the plant damage state as enumerated by the cases described below.

This is a Type 4 question which is used to input parameter values to the code. Branch point probabilities are also input. '

They are dependent on the branches taken in previous questions, as are the parameter values.

There are 1 branches:

1 . PBasel: This is not a branch as in the previous questions. This is more of a parameter name and it is the value (in psig) of the containment pressure prior to core damage (pressure due to steam + heating effects). This " branch" is not referenced below.

Case # 1 : ,

This case is for a large break with loss of injection. The containment pressure is relatively low. It will be less than the peak calculated for a Design Basis Accident (DBA) LOCA since there is no injection to the vessel in AE. Vessel injection would transfer more heat to the containment. The central estimate for the base pressure is that found in BMI-2104 for the AE sequence (Figure 6.8). Optimistic and pessimistic estimates represent a reasonable range of uncertainty in the BMI evaluation.

Case # 2 :

This case is an S1E with delayed loss of injection. High

' pressure systems are able to maintain adequate cooling for 2 to 3 hours3.472222e-5 days <br />8.333333e-4 hours <br />4.960317e-6 weeks <br />1.1415e-6 months <br />. For this case the containment pressure is elevated relative to the case with'no initial injection. Quantification is based on hand calculations of the added pressure rise due to 2.5 hours5.787037e-5 days <br />0.00139 hours <br />8.267196e-6 weeks <br />1.9025e-6 months <br /> of

decay heat added to the containment subsequent to the break. The

' optimistic value for the pressure assumes a break area equivalent to a 2" line and takes credit for a large mass of heat sinks. The central estimate is merely the mean of the optimistic and '

l pessimistic values. The pessimistic value assumes'a break area roughly equivalent to a 6" line and assumes heat sinks similar to l those used in a typical FSAR containment calculation.

Case # 3 :

I This case is that of a small break with no injection.

Containment pressure will be lower than for the large break (AE) case. The actual pressure is somewhat dependent on break size.

Variation amoung the three walkthroughs accounts for this.

Optimistically, the pressure in the containment could range form that associated with a high pressure boildown of the vessel to almost that associated with a large break. An intermediate value l is selected. The central estimate is merely the mean of the i

optimistic and pessimistic values. The pessimistic estimate is

Pega 40 DRAFT Peach Botton C.E.T. Documentation approximately that associated with the large break in case #1 of '

this question (actually a few percent lower) .

Case i 4 :

This case icludes transient initiated sequences in which no initial injection occurs (TQUV or fast station blackout). The containment pressure rise associated with the boildown of the RPV can be calculated with a fair degree of accuracy. The central .

estimate is taken Irom calculations for the Brown's Ferry plant  !

(almost identical to Peach Bottom) performed at Oak Ridge National Laboratory (ORNL) as part of the SASA program. The optimistic and

- pessimistic estimates represent what are judged to be reasonable variations about the mean to account for calculational .

uncertainties (eg. heat capacity of structures) and variations between the two plants. The variation is 15%.

Case # 5 :

This case includes station blackout sequences with early injection from high pressure systems. Containment pressure at the onset of core damage is thus higher than for the previous case.

Because injection delays core damage more heat is transferred to the containment. Quantification of this case was done of the basis of BWR- LTAS calculations. ~ A larger variation around this central estimate (20%) was used since there is greater uncertainty in a long term calculation (>6 hours).

Case # 6 :

This case includes those sequences in which there was a power cycle break. That break could occur at almost any time during the Tc sequence thus the containment pressure at the time of its occurence is indeterminate. Once the break occured, the containment pressure would remain essentially constant. Thus in quantifying the pressure the approximate range of possibilities is covered by the three walkthroughs. . Owing to the low probability of this event the quantification here is expected to have a negligible l impact on risk. Optimistically, the containment pressure could be

[

as low as a few psi above normal operating conditions. The central estimate is the mean of the optimistic and pessimistic estimates.

Pessimistically, the containment pressure could be up to approximately the threshold for repressurization of the vessel.

Once repressurization occured and low pressure injection ceased there would be little (if any) chance of power cycles.

l Case # 7 :

1 This case includes repressurization breaks. Quantification rs*lects a reasonable variation around the central estimate of the containment pressure at which vessel repressurization occurs. This threshold pressure is determined on the basis of the estimated performance of the S/RV accumulators. The containment pressure for a repressurization break would be approximately the pressure at which the accumulator pressure could no longer maintain the valves w e ,, ,- y,---- .-y --,m_-,,,. ,._yr,..,,,m,_m,,,_.,,.,e m.-,,.--m.,-,,._w.-----,-.,,,--..,._.-.,.,....,,,,_,.--_..,..._-.~_--..,,._-o,__

Paga 41 DRAFT P0ach Bottca C.E.T. Docum0ntation open. Once the break occurred there would be little further containment pressurization. The central estimate thus reflects the i

best estimate of the containment pressure at which relief mode of i the S/RVs will fail. The optimistic and pessimistic values reflect a 10% estimated uncertainty in this value.

Case f 8 :

This case is a TW in which the containment was not vented and there is no CRD flow. The vessel repressurizes and high pressure systems are failed due to the high containment pressure. Core i

degradation thus occurs in an intact containment. The central estimate of containment pressure is the pressure at which the S/RVs are expected to reclose plus the pressure rise due to vessel boil-down. The optimistic and pessimistic values represent a l' plus-or-minus 15% uncertinty in this evaluation.

i Case # 9 :

In this case the sequence is an ATWS (TC) and adequate core cooling was lost prior to containment failure (ie the containment is still intact). Because' the containment pressurizes very rapidly in an ATWS there is considerable uncertainty as to the exact pressure. The most likely scenario is that injection is lost at the time the containment pressure reches the S/RV reclosure point.

/

This liklihood is reflected in the quantification of this case.

The central estimate of containment pressure is approximately 10%

higher than this value.

I Case # 10 :

For this case the containment has failed but the leakage rate is insufficient to result in depressurization of the containment.

l The containment pressure is thus at approximately the ultimate <

capability of the containment as the central estimate. The l optimistic and central estimates represent resonable variations l around the assessed mean containment failure pressure.

Case # 11 :

This case catches all sequences in which the containment is either vented or has failed such that the containment is ,

depressurized. With this depressurization the containment pressure is anticipated to be slightly higher than atmospheric. Th'is applies for all walkthroughs. The exact numerical value used here does~not impact evaluation of the event tree since the significant fact is that the containment is vented or failed.

Question # 34 : What is the status of containment sprays?

This question determines the availability of the drywell sprays. Drywell spray is a mode of the RHR. In spray mode, the RHR pumps draw suction from the pool and return the water through the RHR heat exchangers (cooled by emergency service water if available) to spray headers in the drywell at the approximate.

~

l Pags 42 DRAFT Poach Bottom'C.E.T. Documsntation i

elevation of the vessel head. Because spray water can significantly impact fission product transport, with a dependence 1 on spray water temperature which is'small compared to other j uncertainties in fission product transport calculations, the presence or absence of the heat removal function of the spray is not considered here but is left to be included in questions regarding the status of RHR. Thus, if both spray and RHR were shown to be in use (for example) then this could mean that the operator was using spray mode for containment heat removal. (It is also possible that one RHR loop is in spray mode and the other is in pool cooling mode. However, for the purposes of the event tree the difference is unimportant.)

This is a Type 2 question where branch point probabilities are input and are dependent on branches taken in previous questions.

There are 3 branches:

1 . E4fCS: Containment spray mode of the RHR system is failed.

2 . E4rCS: Containment spray mode of the RHR system is unavailable but may be recovered.

-3 . E4-CS: Containment syrays are actuated.

i

, Case # 1 :

In this case containment sprays have failed previously. They thus remain failed.

Case i 2 :

In this case containment sprays would operate if AC power were available but there has been no recovery of AC power. Thus, the sprays remain in the recoverable condition.

i l Case # 3 :

l In this case either High Pressure Service Water or the LPCI pumps are available to operate the sprays. Since core damage is incipient, the containment pressure and/or temperature will be above the technical specification limit. Thus the operator may use l sprays to reduce temperature and/or pressure.While the Emergency

! Operating Procedures would indicate the use of sprays, this is a l manual action and the operators may be so concerned with the vessel j status that containment-spray will not be actuated. Quantification l of the spray probability thus reflects the possibility that the operators concentrate on restoring vessel injection.

Case i 4 :

l This case includes sequences in which there are no water

Pago 43 DRAFT Pacch Botton C.E.T. Docum0ntation ,

sources to operate the sprays. The sprays are thus failed.

Question # 35 : What is the H2 generation and pressure rise during CD?

This question determines.the amount of hydrogen generated during core degradation. Three possible levels of hydrogen production are considered. The question quantifies the level of hydrogen production in terms of the pressure rise (in psi) in the primary containment which results from hydrogen generation. In each case the optimistic walkthrough corresponds to low hydrogen production (Branch 1), the central walkthrough to medium hydrogen production (Branch 2), and the pessimistic walkthrough to high

. hydrogen production (Branch 3).

This is a Type 4 question which is used to input parameter values to the code. Branch point probabilities are also input.

They are dependent on the branches taken in previous questions, as are the parameter values.

There are 3 branches:

1 . H2 Low: Hydrogen production in-vessel is in the range predicted by IDCOR's MAAP code for BWR's.

2 . H2Med: Hydrogen production in-vessel is in the range predicted by MARCON (ie 30-40% of the active fuel j cladding).

1 3 . H2Hi: Hydrogen production in-vessel is in the range predicted by MARCH 2 (ie 80+% of the active fuel cladding).

Case # 1 :

In this case, hydrogen production, regardless of magnitude, cannot pressurize the primary containment beacuse its integrity has been previously violated. The pressure rise is thus zero for all walkthroughs.

Case # 2 :

This case includes sequences initiated by primary system breaks in the drywell. Only a portion of the gas escaping from the vessel is cooled / condensed by passing through the suppression pool. The fraction passing through the pool is dependent on the break size.

l A relatively small break will result in almost complete bypass of L the suppression pool ( a 2" line break results in approximately 85%

flow bypass assuming ADS valves are all open). Bypass results in very hot ( > 500 C) gasses being released to the drywell. Without the cooling affored by the pool, the pressure rise resulting from hydrogen release is an estimated 50% higher than for a transient.

It is also true that BMI-2104 calculations indicate higher hydrogen

~

production for the AE sequence than for the more slowly developing

Paga 44 DRAFT-Peach Botton C.E.T. Documentation transients. Since there is some doubt as to the physical basis of this result, it is reflected only in the pessimistic walkthrough.

Case # 3 :

1 This case includes station blackouts with delayed restoration of AC power. Exception is taken to the general rule for this question that the optimistic walkthrough takes Branch 1, the central walkthrough Branch 2, and the pessimistic Branch 3. This is due to the fact that the hydrogen production resulting from late recovery sequences is highly uncertain. Few analyses are available and it is not clear that those which have been done employ suitable models. Thus in the optimistic walkthrough the probability that

, hydrogen production will be of a medium level is indeterminate

! (with zero probability of high hydrogen production). In the i

central estimate, the probability of high hydrogen production is indeterminate (with zero probability of low hydrogen production) .

Passimistically, hydrogen production is certain to be high. The pressure rise associated with hydrogen production is that associated with transients which are not recovered (Case #5 of this question) plus the pressure rise in the containment due to depositing the heat contained in the core in the suppression pool The additional pressure rise is calculated using various values for i

the mean core temperature such that the three walkthroughs reflect the range of reasonable possibility i

Case # 4 :

In this case the sequence is transient initiated but a S/RV l vacuum breaker has failed to reclose allowing ~ partial bypass of the

! suppression pool. Hydrogen production in-vessel is identical to that during any transient. However, the containment pressure rise is increased due to the suppression pool bypass as discussed in case #2 of this question for small primary system breaks.

Case # 5 :

This case includes all transients with no pool bypass.

Hydrogen prodution calculations for such cases are numerous. It is considered that the differences between ATWS, short term TQUV, intermediate term TB, and long term TW sequences are not sufficiently large so as to warrant treatment as separate cases.

Question # 36 : Does H2 discharge to SP induce WW failure?

When core degradation occurs with RPV at high pressure the the low-low set S/RV opens peri'odically in relief mode. Once the valve opens, pressure within the RPV drops approximately 50 psi before the valve will re-close. Assuming the volume available to gas (steam + hydrogen) within the vessel is 12,000 cubic feet and that the gas mixture behaves as ideal, if the mean RPV pressure is 1070 psia and the mean gas temperature is 2000 R then each valve pop releases nearly 11,000 SCF of gas to the pool. A fifty percent volume fraction of hydrogen is not unlikely (10% by mass). Since S/RV is open for less than 2.5 seconds (0.1 ft opening,-2100 ft/sec

Pcgs 45 DRAFT Psach Bottom C.E.T. Documtntation choked velocity), even if all the steam were condensed, a 5000 cubic foot bubble could be formed at the quencher (equivalent'of a

~10 ft radius). This bubble has the potential to do over 350 thousand foot-pounds of work (0.5 megajoules) this is the energy of a small car travelling 10 mph. Clearly, this magnitude of energy release has the potential for damaging the wetwell structure. Both 4

the vent pipes and vent header could be damaged. Either would

, result in a bypass of the suppression pool. Also, the'torrus itself could be damaged. If the torrus were damaged, at best the containment could be vented in the wetwell airspace, at worst, the suppression pool could be drained. This question. considers the possibility that the torrus ire damaged resulting in leakage from the containment. The following question (# 37) considers the possibility of iduced bypass.

This is a Type 2 question where branch point probabilities are input and are dependent on branches taken in previous questions.

There are 3 branches:

1 . E4nHWF: There is no structural damage to the watwell resulting from the release of hydrogen under high pressure.

2 . E4-HWF2: The release of hydrogen from the vessel results in a Level 2 leak in the torrus. The leakage rate is insufficient to result in significant I

depressurization of the containment.

3 . E4-HWF3: There is structural damage to the torrus as a result of high pressure hydrogen release. A

rupture of the torrus has resulted and the

! containment depressurizes.

i Case # 1 :

! In this case the RPV was depressurized early in the sequence and did not repressurize prior to core damage. Core degradation thus occurs at low pressure and there is no potential for stuctural ~

[ damage in the torrus due to hydrogen release. This would also 4

apply if the S/RV vacuum breaker were stuck open since in this case only a fraction of the hydrogen would reach the pool.

Case # 2 :

n In this case the level of hydrogen production is sufficiently high that in vessel concentrations may reach or exceed the 50% by volume. These high levels result in the greatest probability of structural damage to the wetwell. Because the behavior of such large gas bubbles is not well understood, the probability of damage l to the wetwell structure is can only be surmised. It is conciavable that no bubble will be formed and the pool will be

" tunnelled". It is thus possible that there will not be any i

. - . . _ _ . . . . _ _ _ _ _ _ _ - _ _ - - _ - - - - . , _ _ , - - - _ _ - - - - - - , - - _ _ - - . _ - _ _ ~ _ - _ --__ _

Paga 46 DRAFT Peach-Bottom C.E.T. Documentation significant damage. At least, not to the extent that the steel containment shell is breached.

Case # 3 :

In this case there is only moderate hydrogen production. There is considerable uncertainty as to the maximuu concentration of hydrogen within the vessel. Low concentrations (<10% by volume) would present far less threat to' the terzus since only small bubbles would be formed. In quantifying the probailities of wetwell failure in this case consideration is made both of the possibility that high hydrogen concentrations are not built up within the vessel and of the possibility that even if large gas bubbles were formed in the suppression pool, they would not impact .

the torrus with sufficient energy to cause a breach.

Case # 4 :

This case applies to sequences in which hydrogen production rates are low. With these low rates (typified by MAAP calculations) in-vessel hydogan concentrations are on the order of a few percent. .The possibility of iduced damage in the wetwell is thus considered to be negligible.

Question # 37 : Does H2 discharge lead to SP bypass?

This question considers the possibility of suppression pool bypass induced by the release of hydrogen from the vessel. Bypass could be induced by two mechanisms: 1) Hydrogen discharged to the pool will raise the watwell pressure above that in the drywell thus causing the vacuum breakers to open. 2) Structural damage to the vents and/or the vent header could occur (described in Question #36 above). Vacuum breakers, installed in the vent-header in the wetwell, prevent the pressure difference from the wetwell to the drywell~from becoming large enough to push water from the pool onto the drywell floor. The vacuum breakers are essentially large check valves. Like any check valves there is the potential that they will stick open. There are recorded incidents in which vacuum breakers on Mark I BWRs have failed to re-close. Failure of one of the vacuum breakers to re-close would result in a bypass of the suppression pool subsequent to vessel failure (should core damage not be arrested). Release of hydrogen under high pressure is more likely to cause pool bypass than the structural damage to the torrus described in the previous question. This is because pool water swelling upward as.a result of hydrogen bubble formation would certainly strike the vent header before impacting on the inside of the torrus.

This is a Type 2 question where branch point probabilities are input and are dependent on branches taken in previous questions.

There are 2 branches:

1 . H2-SPB: At least one of the Wetwell-to-Drywell vacuum breakers does not fully re-close. Alternatively

t Paga 47 DRAFT Poach Botten C.E.T. Docuar.ntation 1

there has been structural damage to the drywell vent system.

2 . H2nSPD: There is no early (prior to containment failure) bypass of the suppression pool.

Case # l':

This case includes primary system breaks in the drywell. Since the majority of the gasses affluent from the RPV are discharged to the drywell the vaccum breakers are not challenged. There is also i no high pressure hydrogen discharge to the pool. This being the case, a stuck open vacuum breaker is impossible.

Case # 2 :

This case includes transient (or IORV) initiated sequences.

The-vacuum breakers will be challenged. In this case the RPV.is depressurized so pressurization of the watwell is gradual. The vacuum breakers are expected to open periodically. There is are large dynamic loads in the torrus since hydrogen release is at low pressure. Thus, induced bypass is not a significant consideration.

j Case # 3 : -

1 In this case there has been sufficient structural damage to the wetwell (torrus) that a leak or rupture has developed. between the pool surface and the upper part of the torrus, it is assumed that i

damage to the torrus implies (with high liklihood) that a bypass of i the suppression pool resulting from damage to the drywell vent 4

system has occurred.

Case # 4 :~

l This case (and those following) covers those sequences during which there is discharge of hydrogen to the suppression pool under high pressure. In this case hydrogen production is high and each discharge would raise the watwell pressure approximately 2 psi (assuming 50% in-vessel hydrogen concentration).

~

, In addition,

hydrogen dischare will result in the large dynamic loads described I

in Question #36 above. These loads are substatially placed on the vents and vent header.. The breaking off of a vent pipe or a crack in the vent header would be equivalent to a stuck-open drywell vacuum breaker insofar as the suppression pool.would be bypassed i for future discharge from the drywell.

i I

Case # 5 :

l This case is similar to the previous case (# 3) except that the

hydrogen production rate in-vessel is assumed to be moderate. The drywell vaccum breakers are still challenged many times before vessel breach (assuming that it is breached). There is thus a significant chance that a vacuum breaker will stick open. The

= _ - - -.-_ - - - _ - ___.-- - ---__- - - _ - --. _ _ , . - - . . - - - _ _ _ _ - -

Pcgo 48

. DRAFT Peach Bottom C.E.T. Documentation possibility of dynamic loads on the vent pipes and vent headers is lower than in the previous case since the in-vessel concentration of hydrogen is expected to be lower.

Case #'6 :

This case covers sequences with low hydrogen production but with a pressurized vessel. Challenges to the drywell vacuum breakers will be few and it is anticipated that the dynamic loads will be low since no large bubbles will be formed in the pool. A small but significant probability of pool bypass is considered to be appropriate.

Question # 38

Is the vent threshold reached during core degradation?

This question instructs the event tree code to calculate the sum of the containment pressure at the onset of core damage (Parameter #1) and the pressure rise due to in-vessel hydrogen production (Parameter #2). This sum is compared to the vent threshold pressure of 60 psig stated in the Peach Bottom Emergency Operation Procedures. If the sum is greater that 60 psig branch 1 is taken if not branch 2 is taken.

l This is a Type 6 question in which branch point probabilities i are calculated by the code. The input is a list of parameters to be summed and values of reference parameters to which the sum is compared. The list, the reference parameters, and the method of comparison is dependent on the branches.taken in previous

' questions.

There are 2 branches:

i i

1 . E4-Vth: The containment pressure does exceed the vent threshold pressure during core degradatiori.

~

2 . E4nVth: The containment pressure does not exceed the vent threshold pressure of 60 psig during core degradation.

Case # 1 :

This case includes all sequences in which there has been no failure of the wetwell (torrus) induced by the discharge of hydrogen form the vessel under-high pressure. Containment failure modes other than this (described in Question #36) are accounted for in the base pressure used in the evaluation of the pressure at the i

and of core degradation (Pbasal defined in Question #33 above).

l Case # 2 :

This case includes all sequences in which there has been a rupture of the wetwell (torrus) as a result of hydrogen being discharged form the vessel under high pressure. The containment is

DRAFT Peach Botton C.E.T. Documentation I

I thus failed.and the vent threshold will never be reached. A dummy l

value of the vent threshold (200 psi rather than 60) is used as the l l parameter in this question to insure this case always results in  !

Branch 2-(vent threshold not reached) being taken. '

Question # 39 : Does containment venting not occur during core degradation?

4 This question determines whether or not the containment is l ' vented during core degradation (prior to vessel melt-through if core damage is not arrested in - vessel). Venting is instructed by the emergency procedures once the containment reaches 60 psig.

There is no qualification of this procedure to account for radiation levels in the containment. Venting would be carried out as described under Question #24 above.

This is a Type 2 question where branch point probabilities are input and are dependent cns branches taken in previous questions.

There are 2 branches:

l 1 . E4nVENT: The containment is not vented during core degradation (ie prior to vessel melt- through). .

. 2 . E4-VENT: The containment is vented during core degradation; 1

prior to vessel melt- through.

t i

Case # 1 :

4 In this case the sequence is a TW. The venting threshold was exceeded long before core damage occurred. . If the operator did not vent'before,now that there is high radiation in the containment, 4

the probability that he vent now is judged to be negligible.

Case # 2 :

! In this case the vent threshold is exceeded and there is AC i

power available to operate the valves on the lines required for l' venting. However, there is radiation in the containment and while the EOPs instruct venting there are at least two causes for

reluctance. First, while the 18" lines can be opened from the control room the EOPs direct deflating the valve seals which must 4

be done local to the valve. This procedure would thus result in an operator being exposed to a high radiation field as flow began through the line following deflation of the seal. Second,the plant owners and government officials would think twice (at least) before allowing the operators vent radioactive gasses (and possibly particulates depending on their faith in pool scrubbing) to the environment. Quantification of this case takes into account this i relucance and the probability of venting used presumes a decision 4

to sacrifice the valve seals and ignore possible offsite (and

.onsite) consequences.

For the central estimate the probability of venting is thus indeterminate. Reluctance factors are assumed to

Paga 50 DRAFT Pasch Botten C.E.T. Documsntation balance the requirements of the procedures.

Case # 3 :

This case includes sequences in which AC power is recovered during core degradation. The venting threshold has also been exceeded. Thiscase is similar to the previous one and is quantified identically.

l

-Case # 4 :

This case is a station blackout and there is no power available to permit venting of the containment by remote actuation of the valves from the control room. Based on the EOPs, the operators would be instructed to open the 6" ILRT line. Opening this line would require personnel to go into the reactor building to the valves located on the floor above the torrus room. These valves I

could be opened manually and it is judged that, until the last of the four valves was opened, radiation levels would be tolerable.

However, as the last valve was opened highly radioactive gas would pass through this pipe adjacent to where the personnel was standing. It seems likely that the personnel involved would be at least seriously injured. Even though this case includes only

, sequences in which the vent threshold pressure has been exceeded, it is far from certain that the operators would be aware of this fact. As discussed under Question #24 above, containment pressure

indication requires AC power. Thus, for this case, venting is considered very unlikely. It should be noted that venting prior to

, core degradation, anticipating the above mentioned problems, has i been considered.

i Case # 5 :

This case catches all sequences for which the venting threshold is not reached during core degradation. If the threshold isanot reached, the EOPs do not instruct venting. Thus the probability of venting is assumed to be negligible.

Question # 40 : Does the RPV repressurize during core degradation?

This question is analogous to Question #25 above. However, in this question the code is instructed to calculate whether the containment pressure exceeds the level above which the S/RV accumulators can not hold the valves open. The containment pressure at the onset of core damage (Parameter #1) is summed with I the pressure rise due to hydrogen generation (Parameter #2).- If the l

sum exceeds the threshold value, Branch 1 is taken and .~

repressurization is assumed to occur.

a i This is a Type 6 question in which branch point probabirities are calculated by the code. The input is a list of parameters to be summed and values of reference parameters to which the sum is compared. The list, the reference parameters, and the method of comparison is dependent on the branches taken in previous a questions. m i .

. - . _ _ - -. -= . _ . . _ _

Paga 51 DRAFT Poach Bottca C.E.T. Docum2ntation There are 2 branches:

1 . E4-rep: The RPV repressurizes as a result of containment j pressurization during core. degradation.

2 . E4nReP: The RPV does not repressurize during core degradation.

Case # 1 :

In this case the initiating event is a transient, there are no stuck open relief valves, and the containment has not failed or been previously vented. Thus repressurization is possible and the

. code proceedes with the calculation described above.

Case # 2 :

This case is the complement of case #1 and includes all sequences in which the RPV could not repressurize. In this case, Parameter #2 is excluded from the sum which assures that repressurization will not be indicated.

Question # 41 : Is there a repressurization break during core degradation?

This question is analogous to Question #26 above. Refer to the description there r This is a Type 2 question where branch point probabilities are input and are dependent on branches taken in previous questions.

l There are 2 branches:

~

1 . E4-RePB: There is a break in one of the low pressure systems as a result of RPV repressurization during core degradation.

2 . E4nRePB: There is no. break in the low pressure systems as a result of RPV repressurization during core i

degradation.

l l

Case # 1 :

In this case the vessel repressurizes during core degradation.

Quantification of the probability of repressurization is identical to Question #26 above.

Case # 2 :

This case covers all sequences in which there is no

, - . , - - - - , _ - - - - - - - - , , - - . - _ - . . . - . ,,.- r,-. . - - ._ _ . -.,. - - --., ,- -_- --..., , - - - - - - - - - - - - - , .

a .

• Paga 52

, DRAFT P3ach Bottca C.E.T. Documsntction repressurization of the RPV and thus a repressurization break is impossible.

i Question # 42 : Does the containment fail by pressure during core degradation?

This question instructs the code to calculate the probability

~

of containment failure during core degradation. The containment

fragility is assumend to be normally distributed. The mean failure pressure is estimated based on optimistic, central, and pessimistic assumptions. The code calculates the sum of the containment pressure at the onset of core degradation (Parameter ~#1) and the pressure rise due to hydrogen generation (Parameter #2). Based on ,

the fragility, the probability of containment failure (of Branch 1 4

being taken) is then calculated.

This is a Type 6 question in which branch point probabilities l are calculated by the code. The input is a list of parameters to i be summed and values of reference parameters to which the sum is compared. The list, the reference parameters, and the method of comparison is dependent on the branches taken in' previous questions.

There are 2 branches:

1 . E4-CFP:' The containment fails during core degradation.

i 2 . E4nCFP: There is no failure of the containment during core degradation.

i Case # 1 :

In this case the containment was not vented and there was no repressurization break resulting in containment bypass during core degradation. Failure due to overpressure is a possibility.

Case # 2 :

In this case the containment has been vented or has previously failed. The case also includes sequences in which there has been a repressurization break.

Question # 43 : What is the level of containment leakage before VB7 This question determines the level (or rate) of leakage from the containment at the time of vessel breach. A full description of the failure level definitions is given in Question #28 above.

This is a Type 2 question where branch point probabilities are input and are dependent on branches taken in previous questions.

There are 3 branches: ,

-v- ,y~----%o- --.-,w ,w---,-, w-- --,.-,+m--em,-v--,-+y--,..r-,.vv,-,--w.ww-+,w '

t 1 .. .

Pcg2 53 I DRAFT Pach Bottcc C.E.T. Docum0ntation

1'. E4nCL: There is no leakage from the containment, j significantly in excess of technical.  ;

specifications, prior to vessel breach. i

-2 . E4-CL2: There is leakage from the containment which is i

significantly in excess of technical 4 specifications prior to vessel breach but it is insufficient to cause the containment to

depressurize.

i

3 . E4-CL3: There is a containment breach prior to containment

failure which is sufficiently large to cause the  ;

containment to depressurize. ,

i Case # 1 :

l In this case the containment previously ruptured, the rupture will certainly persist.

p Case # 2 :

! In this case the containment is coupletely bypassed through a i

primary system break in the reactor building. The terminology here is somewhat confusing since depressurization of (leakage from) the containment would not begin until vessel breach. However, the main i purpose of this question is to determine the impact of leakage rate 4

of the reactor building thermal-hydraulics and on fission product transport. For sequences with containment bypass,-containment behavior prior to vessel breach does not have a significant impact I on the consequence evaluation so this slight ambiguity is acceptable. Subsequent questions properly distiguish containment bypass sequences (which are expected to have negligible impact on risk due to their low frequency) from sequences with a rupture in the containment. Since a break in low pressure systems is expected to be large (pumps seal failures are regarded to be negligible leaks) this case is assumed to be certain to have level 3 leakage.

case # 3

In this case the containment fails during core degradation as a

result of the overpressure. Until recently, it was generally i

believed that the overpressure failure of a steel containment would be a rupture resulting in de pressurization. However, IDCOR has sponsored work which apparently suggests that high temperatures

! within the containment would result in a leakage failure rather than a rupture. Such a failure could still be sufficiently large to depressurize containment. In the quantification of this question allowance has been made for the possibility of leakage j failures.

{ case # 4 :

I In this case the containment has been successfully vented.

a i .

, . - ~ _ . . _ _ . . _ - . . . _ _ , _ _ . . , , . . . . . _ _ _ . . _ . _ . . . , . _ _ . . - - - _ , - . . _ , , . _ , . . - - - . _ - - - - , , . - , - - - .._...____m___,..-- m-., ,_ . -

Pago 54 DRAFT Patch Bottom C.E.T. Docum:ntation Successful venting has been defined to mean that the containment depressurizes. Thus this case is certain to result in level three leakage in all walkthroughs.

Case # 5 :

In this case a leak has developed earlier but the leakage rate was insufficient to depressurize the containment. There has been no further significant threat to the containment. Thus it is considered certain that the leak will remain as only a leak.

Case # 6 :

This case catches all sequences where there has been no previous leakage and no significant containment challenge.

The containment is thus assumed certain to retain its integrity.

Question # 44 : What is the location of containment leakage before VB?

This question determines the location of containment leakage and is used later to evaluate the extent of suppression pool bypass. As such, this is only a summary question.

This is a Type 2 question where branch point probabilities are input and are dependent on branches taken in previous questions.

There are 3 branches:

1 . E4nCL: There is no leakage from within the containment in excess of technical specifications.

2 . E4-CLD: The leakage from within the containment comes from the drywell (or from within the RPV).

3 . E4-CLW: The leakage from within the containment comes from the wetwell.

Case # 1 :

In this case there is a complete bypass of the containment due L

to a primary system break in the reactor building. Subsequent to vessel breach this would be very similar to a drywell failure.

Thus it is classified as such.

Case # 2 :

l In this case there has been a rupture failure of the l containment. Accounting for the IDCOR position that leakage from the containment occurs from the drywell, given that the failure is a rupture, there should be some possibility that it rupture occured in the watwell. This possibility is accounted for in the

! quantification of this question. Otherwise, it is the general concensu's of experts in this area that the drywell has lower

PLg2 55 DRAFT Poach Botten C.E.T. Docunantation I

pressure capabilility in a Mark I containment than does the wetwell. Therefore the probabilities are weighted heavily toward drywell failure.

Case # 3 : .

In this case the discharge of hydrogen from the vessel under high pressure resulted in a leak or rupture of the wetwell (torrus). The failure location is thus in the wetwell for this sequence (by definition).

case # 4

In this case the overpressure failure of the containment resulted in leakage only. As stated earlier (Case #2 of this question) -it is generally agreed that the weakest structural point in nthe containment is in the drywell. In addition, the IDCOR position on leakage favors a drywell failure location. ~This being the case, the probability of a watwell leak, given that there was

no previous damage to the torrus, is considered negligible.

Case # 5 :

In this case the containment has been vented. Since the Emergency Operating Procedures give priority to wetwell vent lines the possibility that a drywell vent line would be opened is judged to be negligible. Leakage from the containment is thus from the

, wetwell (torrus).

J j Case # 6 :

4 This case catches all sequences in which there has been no i breach of the containment. Thus leakage is from neither the wetwell or drywell and "no significant leakage" is certain for all

- walkthroughs.

Question # 45 : Is the RPV depressurized before vessel breach?

This question determines the RPV pressure at the time of vessel-breach.

4 This is a Type 2 question where branch point probabilities are input and are dependent on branches taken in previous questions.

! There are 2 branches:

1 . E5nDeP: The RPV is at high pressure (approx. 1100 psig) at.

I vessel breach.

2 . E5-DeP: The RPV is depressurized at the time of vessel I

breach.

Case # 1 :

i

._ . _ . . ~_ --- . . . . - .

Pago 56

, DRAFT Peach Botton C.E.T. Documentation i

In this case there has been an induced break in the primary

. system. Such breaks are assumed to always result in depressurization of the vessel.

Case # 2 :

In this case the RPV was previously depressurized, but.

n repressurized'during core degradation. Allowance is made for the fact that failure of containment or restoration of AC power will result in the RPV once again depressurizing in spite of prior repressurization. If the RPV repressurized, and nothing has changed that would result in depressurization, the condition is '

certainly that of being pressurized.

Case # 3 :

In this case the vessel was not depressurized prior to core damage. Given this is the case there is little (if any) justification to believe it would be depressurized prior to vessel breach. It is noted that induced primary system breaks are excluded from this case. (These sequences fall into Case #1 of j- this question).

! Case # 4 :

M.

This case catches all sequences in which the vessel has remained depressurized since early in the sequence. Nothing is

' assumed to change in this respect.

Question # 46 : What is the status of CRD?

As noted in a previous question (# 30) there is the possibility j that steam flooding of the reactor building, resulting from

, containment failure, could impact the operability of the CRD I pumps. In this question the possibility of hydrogen burns in the reactor building impacting CRD pump operability must also be considered since hydrogen would be released from the containment, subsequent to core damage, if its integrity was violated. The treatment of hydrogen burns in this question is not at all rigorous l (in terms of consistency with subsequent questions relating to hydrogen burns in the reactor building) but they are not expected '

! to be a significant contributor to CRD pump failure.

I This is a Type 2 question where branch point probabilities are input and are dependent on branches taken in previous questions.

I j There are 3 branches:

1 . E5fCRD: The CRD pumps are not (or are no longer) available to supply make-up water to the vessel.

! 2 . E5rCRD: The CRD pumps are unavailable but may be

] recovered.

i 1

- . . , - - . ---~~m,-mm_.. _,- - . . . - . - - - . - . . . - - . ~ ~ -

i.

Pago 57 DRAFT POcch Bottca C.E.T. Docuanntstien 4

3 . E5-CRD: The CRD pumps are injecting water;to the vessel.

, Case # 1 :

In this case there has been a rupture of the containment resulting from overprassurization. Large quantities of steam and/or hydrogen would be released into the reactor building. The CRD pumps are operating at the time of the failure. Some probability of failure is thus. allowed.

Case # 2 :

This case is like the previous one except that CRD pumps are unavailable. Their recoverability can be impacted by the containment leakage which has occurred. This impact is reflected

by this case.

Case # 3 :

In this case there was previously flow from the CRD system to 1 the vessel. The containment has not ruptured or did rupture prior to core degradation.(with no effect on the CRD pumps). What

} leakage there may be from the containment is judged to have a negligible impact on the CRD pumps.

Case # 4 :

This case is similar to the previous one except that it is the recoverability which is being examined rather than the operating status.

Case f 5 :

This case covers sequences in which the CRD pumps were previously failed. They are certain to remain so.

! Question i~47 : 'What is the status of HPSW?

This question treats the status of HPSW much as the question i above treated the status of CRD flow. There are two significant differences. First, the HPSW pumps will not be impacted by steam flooding of the reactor building or by hydrogen burns. As was noted in Question #31 (above), the HPSW pumps are outside the the

reactor building and only valve operability could be impacted.

Thus, if HPSW were already aligned to the LPCI system there would be a negligible probability of any impact. If HPSW were not in i

service, conditions in the containment could impact its i operability. The second difference is in the branch definitions, i failure is taken to include failure of the operator to use the system.

This is a Type 2 question where branch point probabilities are l

input and are dependent on branches ~taken in previous questi-  ;.

. , - - , . , , - - - , . . - - , - - - - ~ - - - , . - - , , - - , - - - , - .

• ^

Pago 58 DRAFT Pacch Botten C.E.T. Docu entation l

There are 3 branches:

1 . E5fMPSW

The High Pressure Service Water system cannot be (or is not due to operator error) aligned to l provide water to the LPCI system.  ;

1 2 . E5rHPSW: The High Pressure Service Water system remains i

! unavailable but may be recovered.

l 3 . E5-HPSW: The High Pressure Service Water system is j supplying water to the LPCI system.

! i

Case # 1

'

l In this case High Pressure Service Water is available but the

! operator has not previously aligned it to the LPCI system. There

! is steam flooding and hydrogen in the reactor building. Two l

factors are considered in the quantification. First, the probability that the reactor building environment will cause one or more of the required valves to fail in the closed position. This

is quantified exactly as in the subsequent case of this question with regard to failure of the HPSW system given that it was ,

! previously recoverable. The second factor is the probability that i i the operator will take the actions required to align.the system.

2 This is quantified in the subsequent case of this question which 3

includes sequences in which there is no containment rupture and the

HPSW system is available.

! Case # 2 : '

In this case there has been a rupture of the containment and the HPSW system was previously in an unavailable but recoverable  ;

i condition. Quantification of this question reflects the  ;

probability that the resultant severe reactor building conditions will cause at least one of the motor operated valves required to

! align HPSW to the LPCI system to fail in the closed position.

4 Case # 3 :

. In this case there has been no rupture of containment and the HPSW system is available. Quantifict. tion of this case reflects a judgement of the probability that the operator will align the HPSW l system to supply water to the LPCI system. Account must be taken of the fact that this was not done prior to core damage.

Case # 4 :

In this case there was no containment rupture during core degradation and the HPSW system was recoverable. It is assumed that the HPSW will remain in the recoverable condition. This case

) includes station blackout sequences in which venting occurred.

1 Such venting occurs through the ILRT line which bypasses the l

t _. _ ___ __- _ _ _ _ _ _ _

___ - . _ _ - -. __ - - .- . - - . - _ - . - ~

t . .-

Paga 59 '

l DRAFT Pacch Bottcm C.E.T. Documentation reactor building (no steam flooding).

i

! Case f 5 :

4 The case catches all sequences in which the HPSW system was i previously failed.- It is thus certain to remain so.

Question # 48 : Is MSIV leakage control or condenser vacuum maintained?

This question determines the availability of the MSIV leakage control system (MSIV-LCS). The purpose of the MSIV-LCS is.to direct-i fission products which may leak from the MSIVs after they have closed to the SGTS system. If condenser vacuum is maintained the same end is achieved. The MSIV-LCS consists of two (inboard and outboard) safety systems. They are powered by the emergency i busses. Actuation is by manual action from the control room.

L Interlocks prevent actuation if the Main Steam Line pressure is I high (not a concern here since MSIVs are assumed closed). This-

manual actuation is a normal part of the post-LOCA procedure. This l question is not referenced in the current tree but is included j since this may be of interest in source term binning. For Peach

Bottom, possible influence on risk is probably minor owing to the  !

large releases associated with the dominant core melt sequences.

This is a Type 2 question where branch point probabilities are input and are dependent on branches taken in previous questions.

There are 2 branches:

1 . MSIVnLC: There is no MSIV leakage control.

2 . MSIV-LC: The MSIV leakage control system is operating.

I i

Case # 1 :

i In this case AC power is available. The operation of the MSIV 1 Leakage Control System depends only on action by the operator to

start the system. There is some possibility that this would not be i done andthat the condenser vacuum pumps were not running. It is noted that actuation of this system is a routine part of operator

, response to a LOCA.

Case # 2 :

t j This case covers sequences in which there is no AC power. It

! is certain that the MSIV Leakage Control System would be inoperable

! and that there would be no condenser vacuum.

i Question # 49 : Is the suppression pool drained before VB?

This question determines the availability of water in the torrus to provide for pressure suppression and pool scrubbing.

Given that one allows for wetwell failures'one must consider

Paga 60 DRAFT P:Ech Botton C.E.T. DocunOntation

~

failures which may occur below the pool surface. Failure occuring more that about 2 meters below the water level would drain the pool sufficiently that flow comming through the drywell vents would not lun impacted by the pool (ie, the pool would be bypassed by vent flows).

This is a Type 2 question where branch point probabilities are input and are dependent ~on branches taken in previous questions.

.There are 2 branches:

1 . E5-SPD: The suppression pool is drained to a distance below the normal level that results in pool. bypass for at least drywell vent flows. ,

2 . E5nSPD: The suppression pool maintains at least normal operating level.

Case # l':

In this case the torrus has ruptured. In the central estimate i

i the failure is assumed to be equally likely either above or below the water line. There may be a slight tendency toward the area below the water line since this supports the water's mass. This is

judged insignificant. The optimistic and pessimistic walkthroughs i

reflect a reasonable range of the uncertainty.

. Case # 2 :

In this case the release of hydrogen to the pool under high pressure has resulted in a large breach in the torrus. In the i central estimate it is judged that this failure is likely to be

above the pool level. This is because the hydrogen bubbles would

! tend to lift the pool, throwing water against the upper part of the i torrus. The fall back under gravity would be far less energetic.

i The energy released in this case is discussed under Question

1. 36above.

Case # 3 :

l

! In this case there is no large failure of the torrus. Leaks

} are assumed to be of insufficient size to result in a significant

loss of pool inventory. Thus the pool is certain to maintain an adequate inventory of water.

i l Question # 50 : Will the SP flash following containment vent or rupture?

i This question determines whether or not the suppression pool is

! saturated. (Saturation is defined to mean that the bulk temperature of the pool is equal tho the saturation temperature which corresponds to the watwell airspace pressure.) This is distinct from the pool being in equilibrium with the airspace

-~ . . . .- ..

d I

l

, Pago 61  ;

DRAFT Pcach Botton C.E.T. Docu=Ontation '

s (partial pressure of steam in the airspace equal to the saturation pressure which corresponds to the pool surface temperature). The significance of the pool being saturated is three-fold: First, the pool flashes at containment failure resulting in the release of large quantities of steam to the reactor building (approximately

, 360 metric tons if vented at 60 psig, approximately 540 metric tons if failure occurs at 135 psig). Second, in a saturated pool, all the energy deposited in the pool is used to convert water into steam; none of the energy is stored as sensible heat. Thus the saturated pool absorbing a unit mass unit volume of steam initially at 537 C in a low pressure vessel releases 1.8 volumes to the continment. If the vessel were at high pressure 48 volumes would be released to the containment. Third, the ECCS pumps (and RCIC) draw suction from the pool in BWRs such as Peach Bottom. These pumps cannot pump water which is at or near saturation since the pumps will cavitate. Thus, if the pool saturates on containment failure there is no ECCS. NOTE: The pool cannot be saturated if the containment is intact but can be if the containment integrity is not maintained.

This is a Type 2 question where branch point probabilities are input and are dependent on branches taken in previous questions.

There are 2 branches:

4 1 . SatSP: The suppression pool will flash and saturate following containment venting or a rupture.

2 . SubcSP: The pool temperature is subcooled relative to

, atmospheric pressure (temperature below 100 C).

l Case # 1 :

In this case the the high pressure systems have supplied I

I make-up water to the vessel for sufficiently long that enough heat has been deposited in the pool to raise the temperature to 100 C.

The RHR system has not been used to remove the heat. This occurs in TW sequences and in blackouts where there is no battery failure. Under these conditions the pool will certainly flash when the containment is depressurized.

Case # 2 :

r In this case low pressure ECCS was operated but the RHR system was not used to remove the heat deposited in the pool. Since the low pressure systems did inject they are certain to run at least until the pool reaches 100 C. This situation arises in all the loss l of containment heat removal (W) sequences in which there was no early high pressure injection (AW, SlW and some TW).

Case # 3 :

In this case the sequence is an ATWS. Pool heat-up is so rapid

j Pago 62 DAAFT P31ch Botto2 C.E.T. Docum:ntation l that 100 C is reached in no more than 25 minutes. The pool in thus certain to flash and saturate when containment integrity is lost.

Case # 4 :

i

' This case catches all sequences in which the pool c'annot saturate until after vessel breach (if at all). This includes most

station blackout sequences, AE, S1E, and TQUV.

Question # 51 : What is the level of suppression pool bypass before VB?

This question determines the extent to which flows from the primary system into the containment or the reactor building (or to i the environment) bypass the suppression pool. Pool bypass determines the pressure suppression ability of the containment as a well as the fission product removal'(due to pool scrubbing) along the release pathway.

This is a Type 2 question where branch point probabilities are i input and are dependent on branches taken in previous questions.

There are 3 branches:

)

i 1 . E5nSPB: The suppression pool is not bypassed. Gas flow l from the vessel to the containment is directed toward the pool.'

~

2 . E5-SPB2: There is some significant amount of pool bypass.

Less that 50% of the flow leaving the vessel will bypass the pool.

, 3 . E5-SPB3: The suppression pool is essentially bypassed. No i more than 50% of the gas affluent from the vessel

can pass through'it.

case # 1 :

i In this case there is a primary system break in the drywell and i the containment is failed in the drywell. Thus there is a direct i

pathway from the vessel to the reactor building which most of the l gas affluent from the vessel will take. Suppression pool bypass is

! certain.

Case # 2 :

i i

In this case the containment is completely bypassed since there is a primary system break outside containment. The pool is also

certainly bypassed.

j Case # 3 :

l In this case there is a primary system break in the drywell.

l The pool level has been lowered sufficiently that the bottom of the

Paga 63 DRAFT POnch Bottoa C.E.T. Docu22ntation vent pipes are uncovered. Gas flows from the vessel thus pass from the drywell directly into the airspace above the suppression pool.

The pool is completely bypassed.

Case # 4 :

In this case the pool level is lowered significantly but the primary release pathway from the vessel is through the S/RV discharge quenchers which are near~the bottom of the torrus. The pool would have to be nearly empty for total bypass to occur.

Qualtification thus allows for partial bypass. This should be interpreted as a temporal transition from no bypass at the start of core damage to perhaps total bypass at vessel breach.

Case i 5 :

In this case there is a primary system break and either the wetwell-to-drywell vacuum breakers are stuck open or there is leakage from the drywell vent system due to damage induced by the release of bdrogen from the vessel under high pressure (Question

1. 37 above). Either occurrence implies a complete bypass of the suppression pool.

Case # 6 :

In this case there is an S/RV tailpipe vacuum breaker stuck open. Thus there will be at least a partial bypass of the suppression pool.

Case # 7 :

This case catches all sequences for which there is no opportunity for suppression pool bypass prior to vessel breach.

Question f 52 : What is level of RB breach / bypass before VB without burn?

This question determines the extent to which fission products could be retained in the reactor building (assuming they are released from the primary containment). There is an implicit assumption that fission product retention varies inversely with the extent to which the reactor building is bypassed or with the flow rate of aerosol through it. This assumption maybe in conflict with the IDCOR model which appears to rely heavily on aerosol impaction for fission product removal. This question does not consider the impact of possible hydrogen burns in the reactor building. In order to correctly represent the sequence progression, the impact of hydrogen burns (if any) are dteremined in Question #56, below.

This is a Type 2 question where branch point probabilities are r input and are dependent on branches taken in previous questions.

There are 3 branches:

1 . E5nRBB: There is no breach or bypass of the reactor 1

1 _ _ _ - _ . - _ _ -. - - .-. -- - - - - - - - .

I DRAFT POcch Bottos C.E.T. Documintetion 4

i building. What flow may exist to the reactor building could be handled by the Standby Gas Treatment System (SGTS) if it is operational.

2 . E5-RBB2: There is leakage from the reactor building but a significant fraction of the leakage flow from the primary containment passes through the building.

3 . E5-RBB3: The reactor building is essentially bypassed by

! leakage flow from the primary containment. Either the flow path bypasses the building or the through flowrate is large.

Case # 1 :

In this case there is a primary system break in the reactor building. During core degradation leakage flow to the reactor-building is small (compared to that resulting fro.m a rupture in the containment) but large enough to cause the reactor building blowout pannels to lift. However, the low pressure ECCS piping (assumed to be the release pathway) is low in the reactor building . Thus, the released gasses must pass completelt through the building in order to be released.

Case # 2 :.

In this case the sequence is a station blackout. Venting 4

occurred without AC power. The vent line which must have been used is the 6" ILRT line (de cribed under Question #24). This line is made of piping and passes through the reactor building at the next level above the torrus exhausting outside the building. With this line is open, the reactor building is bypassed completely.

Case # 3 :

In this case there is a rupture of the containment in the drywell. This will certainly cause a breach in the reactor building but there is only a small probability that total bypass of

! the reactor building will occur.

Case # 4 : .

In this case the containment is succesfully vented. It is assumed that the 6" ILRT line which would be used in a blackout (see previous case #2) would not be employed for non-blackout sequences. Thus the reactor building would not be completely bypassed as in that case. However, opening one or both of the 18" ventlines would cause rapid pressurization of the reactor building and loss of its integrity. A partial bypass is thus certain.

Case # 5 :

In this case the containment has ruptured in the wetwell.

-.n ~ , , - - - , - - - , - ,v---n.-.,-- - - . . ,n, .- -----r--~~-m,-- - --- - - - - - - - - - - - - - - - - - - , - + - - - - - - - - - - - - - - - - - -

• Paga 65 DRAFT POcch Botton C.E.T. Docum ntation Because the torrus is low down in the reactor building and the blowout pannels are at the top, the release pathway involves the entir building. Thus there is only a partial bypass and complete bypass is excluded.

Case # 6 :

In this case there is leakage from the containment but the leakage rate is low compared to that associated with a rupture or venting. There is significant probability that the Standby Gas Treatment System could handle such flow rates. Structural heat <

sinks may absorb enough heat from gasses leaving the primary  :

containment to prevent the hot gas from closing the fire dampers in the SGTS system. Quantification thus allows for the possibility of no breach or bypass. i Case # 7 :

This case catches sequences for which there is no significant .

leakage from the primary containment prior to vessel breach. This  !

being the case, the probability of reactor building breach is negligible.

Question # 53 : Are fire systems actuated before VB without burn?

This question determines whether the fire sprays in the reactor buiding-are triggered. Triggering due to hydrogen burns is considered in a subsequent question. This question considers only the impact of hot steam being released to the building prior.to vessel breach. Fire sprays will start automatically when the building temperature is raised significantly above normal operating limits. The sprays in Peach Bottom are limited to the area above the cable trays thus the impact on fission product retention is not as great as it would be if the spray was more global (like the containment spray). The fire sprays are-independent of plant AC power since the pumps are driven by dedicated fire system diesels.

This is a Type 2 question where branch point probabilities are input and are dependent on branches taken in previous questions.

There are 2 branches:

1 . EFireS: Fire sprays are actuated in the reactor building prior to vessel breach.

2 . nEFireS: Fire sprays are not actuated in the reactor building as a result of steam being released from the containment prior to vessel breach.

Case # 1 :

In this case the containment leaks or has ruptured. The reactor building is not completely bypassed so there is significant

Pcgo 66 DRAFT Poach Bottcm C.E.T. Docum3ntation

~

~ hold-up of the gasses released from the containment within the i reactor building. Also, either the suppression pool is bypassed o it is saturated. These suppression pool conditions imply that

either the gas coming into the reactor building is hot (no cooling effect from the pool) or that there are large amounts of steam .

released because the saturated pool has no ability to condense j

steam and will actually release more in the process of absorbing the heat transported to it. Such conditions are judged to be certain to trigger the fire sprays.

Case # 2 :

This case is the complement of case one. Sequences with the i' following character istics are included: 1) The containment is intact and no significant quantity of gas is released to the

• reactor building. 2) The reactor building is bypassed so there is i

no accumulation of released gasses within the building. 3} The

suppression pool is not bypassed and is subcooled. In the.latter l instance any gasses released to the reactor building will be i relatively cool. If any of the above three conditions are met the

! reactor building will not heat up and the fire sprays will not be l triggered.

4 Question # 54 : Does SGTS fail before VB without burn?

I This question determines the availability of the Standby Gas Treatment System (SGTS). The SGTS is a safety system designed to limit the leakage of particulate fission produts from the reactor building. When activated, SGTS draws air (or whatever gas is l present) from the reactor building and exausts it through charcol filters up the stack. At Peach Bottom, the SGTS lines contain fire dampers which are intended to close if there is a fire in the reactor building. Dampers are secured open by fusible links.

These links fail at approximately 160 F causing the dampers to close (by the action of gravity). Thus, when the reactor building temperature is raised, the fire dampers close and SGTS fails. The SGTS draws power from the emergency busses and will not operate in a station blackout.

This is a Type 2 question where branch point probabilities are input and are dependent on branches taken in previous questions.

There are 3 branches:

1 . E5fSGTS: The Standby Gas Treatment System is failed.

2 . E5rSGTS: The Standby Gas Treatment System is unavailable but may be recovered.

3 . E5-SGTS: The_ Standby Gas Treatment System is operating.

Case # 1 :

Pag 3 67

DRAFT Peach Botton C.E.T. Documentation In this case the reactor building temperature has not been raised sufficiently high to actuate the fire sprays (discussed in j the previous question). The containment is not vented (which may cause SGTS to isolate -- see Case #2 of this question). Lack o fire sprays implies that the SGTS dampers will be open (high temperatures in the reactor building would both start the fire f

sprays and close the fire dampers). Only sequences with AC power are included in this case so all all the conditions are met for SGTS.to be operating. It is thus certain to be operating.

Case # 2 :

In this case the primary containment is vented to the SGTS (or to the reactor building if the duct work on the vent lines cannot .

withstand the pressure). With hot gasses channeled to the SGTS it may cause the dampers to close. Whether or not the dampers close depends on the suppression pool temperature. The pool temperature will probably be in a range which may cause the dampers to fail so some uncertainty is reflected in the quantification. Sequences in which the pool is definitely hot enough to cause the dampers to close are not included in this case by virtue of the fact that fire

systems would be operating in this instance. As in the previous-i case, fire systems not operating is used in the tree as a surrogate

[ for reactor building temperatures below the temperature at which

. the fire dampers close.

4 Case # 3 :

This case includes station blackouts in which the reactor building temperature is still low enough not to result in actuation of the fire systems. The SGTS thus is unavailable due to the lack of AC power. However, the dampers remain open so the system could be recovered.

Case # 4 :

This case covers all sequences in which the fire sprays start 4 due to high 4eactor building temperature. It is thus expected that i the fire dampers will close; failing SGTS.

)

Question # 55 : Does hydrogen burn in RB before VB7 This question determines whether or not there are hydrogen burns in the reactor building. This becomes possible if there is a

loss of containment integrity and the reactor core is damaged.

Even though the primary containment is inerted, the reactor building contains sufficient air that even if mixed with all the

, nitrogen escaping from the primary containment hydrogen combustion could still be supported. Hydrogen burning is assumed to occur

)l even in the absence of electrical power since ignition is very

easily achived. However, it is also assumed that if large quantities of steam are released to the reactor building, the building will be steam-inerted.

This is a Type 2 question where branch point probabilities are l

Pago 68 DRAFT Petch Botten C.E.T. DocunOntation i

1 input and are dependent on branches taken in previous questions.  ;

There are 2 branches:  !

i l 1 . HBbVB: Hydrogen burns in the reactor building prior to vessel breach.

2 . nHBbVB: There is no hydrogen burn in the reactor building  !

prior to vessel breach.

Case # 1 : '

In this case ther is a primary system break in the reactor building. Very hot hydrogen and superheated steam will be released directly into.the air of the reactor building. Combustion is judged to be certain.

Case # 2 :

In this case.there is a station blackout, AC power has not been

! restored, and yet the containment is vented. Venting without AC power can only be accomplished using the 6" ILRT line. This line '

bypasses the reactor building (see Question #24 above). With the reactor building totally bypassed in this manner, no significant amount of hydrogen can leak into it and thus there will be no i

• combustion.

Case # 3 :

In this case, hydrogen production has been assumed to be low (on the. order of that predicted by the MAAP code for BWRs). With only a few hundred pounds released from the in-vessel reaction of.

l Zircaloy, combustible concentrations will not accumulate in the reactor building.

Case # 4 :

In this case there is some leakage of steam, hydrogen, and nitrogen from the primary containment. However, the leakage is not sufficient to depressurize the primary containment (Level 2).

Because the affluent mixture is initially inert and is expected to -

disburse, significant hydrogen burns are not expected. This is equivalent to assuming that combustible concentrations (>4-54 by volume) will not form.

Case # 5 :

In this case the reactor building is substatially bypassed.

This implies that the residence time for affluent gas is short and that there is little time for combustible mixtures to accumulate (the gas leaving the containment is inert). Thus, the probability of a hydrogen burn is considered to be indeterminate.

Paga 69 l

, DRAFT Peach totton C.E.T. Documentation i

Case # 6 :

In this case there is a large leak from the primary' containment into the reactor building. The Standby Gas Treatment System is i operating. The suppression pool does not flash, so large

quantities of steam were not released. There is a possibility that SGTS will help prevent combustible hydrogen concentrations from accumulating in the reactor building. This is reflected in the  ;

quantification.

Case # 7 :

i In-this case the're is a large leak from the primary containment but the Standby Gas Treatment System is not operating. There is a possibility that steam inerting will occur as a result of

, suppression pool flashing.- However, condensation of steam on the

, structures in the building and on the building walls would work i against inerting. With the SGTS not running there is then considerable potential for combustible concentrations to accumulate. The probability of burns is thus fairly high.

Case # 8 :

This case covers sequences in which there is no significant 3

I leakage from the primary containment. There is thus little, if any, hydrogen in the reactor building which could burn.

Question # 56 : What is level of reactor building (RB) breach / bypass

} before VB?

l This question determines the impact of possible hydrogen burning on the integrity of the secondary containment. If hydrogen burning occurs in the reactor building rapid pressure increases

~

would occur which could alter gas flow pathways through the building by damaging doors and/or walls or causing blow-out pannels to fail. The burns would also tant to incrase'the level of bypass by shortening the residence time of the aerosol-carrying gasses (less effective retention). l This is a Type 2 question where branch point probabilities are input and are dependent on branches taken in previous questions.  !

There are 3 branches:

1 . nERBB: There is no loss of the reactor building integrity prior to vessel breach.

2 . ERBB 2: There is significant leakage from the reactor buiding prior to vessel breach. However, the majority of the gas flow passes through the building with some opportunity for fission product deposition.

3 . ERBB 3: The leakage from the reactor building prior to S vessel breach is very large or gas escaping the

Pago 70 DRAFT PO2ch Bott03 C.E.T. DocunOntation primary containment bypasses the building completely.

Case # 1 :

In this case large leakage from, or bypass of, the reactor building began prior to the opportunity for hydrogen burning. The building thus remains in this condition.

Case # 2 :

In this case a hydrogen burn occurs in the reactor building.

Prior to the burn the building either has negligible leakage or a partial bypass. It is judged to be certain that hydrogen burning would cause a failure of the reactor building pressure boundary, thus creating at least a partial bypass. Even if the building was leaking prior to the burn, there is a significant probability that hydrogen burning would exacerbate the bypass by doing further structural damage or by significantly reducing the gas residence time.

Case # 3 :

In this case the reactor building initially had some leakage but no hydrogen burns occur which could exacerbate it. The partial bypass is certain to. persist.

Case # 4 :

This case covers sequences in which the reactor building was intact and remains so after the possibility of hydrogen burns is considered since none are found to occur. (Either because primary containment is intact or there was just no burn.)

Question # 57 : Are fire systems actuated before VB?

This question determines the impact of possible hydrogen burning on the operation of the reactor building fire sprays. The fire spray system is discussed above (Question #53).

This is a Type 2 question where branch point probabilities are input and are dependent on branches taken in previous questions.

There are 2 branches:

1 . EFireS: The reactor building fire sprays are operating prior to vessel breach.

2 . nEFireS: The reactor building fire sprays are not operating prior to vessel breach.

Paga 71 l DRAFT PC,0ch Botton C.E.T. Documentction

]

Case # 1 :

In this case the fire sprays were not operating previously and no hyrogen burn occured in the secondary containment which would

cause them to start. Thus, fire sprays are certainly not operating. ,

l l Case # 2 :

This case covers sequences in which either the fire sprays came -

on regardless of any hydrogen burn in the reactor building or

. sequences in which there was a hydrogen burn. Hydrogen burns are

certain to activata .the. fire . system since they .are designed to come

• on if there is a fire in the reactor building.

Question # 58 : Does SGTS operate until VB? j This question determines the impact of possible hydrogen burns j in the reactor building on the Standby Gas Treatment System. As noted above, fire dampers are expected to isolate,the SGTS should '

the temperature in the reactor building be elevated significantly.

This feature is intended to protect the SGTS from fire.

This is a Type 2 question where branch point probabilities are  !

input and are dependent on branches taken in previous questions.

i There are 3 branches:

1 I 1 . EfSGTS: The Standby Gas Treatment Sytem is failed l (isolated).

2 . ErSGTS: The Standby Gas Treatment System is unavailable but may be recovered.

E-SGTS: The Standby Gas Treatment System is operating

~

3 .

prior to vessel breach.

Case.#.1 : 3 In this case the SGTS was operating prior to the possibility of a hydrogen burn. The fire sprays have not started (ie. there was no burn). Since fire sprays and SGTS isolation are triggered by the same thing (high reactor building temperature the SGTS must contiue to run.

Case # 2 :

In this case SGTS was not running prior to the possibility of a hydrogen burn but it was recoverable (ie the fire dampers remain

~open). As with the previous case the fire sprays are not on so the fire dampers.will.ra== h _open. EGTS thus remains recoverable.

Case # 3 :

-.-n.. , - , - - , , , < - - - - - - , . -

--,-----,-,e-

Paga 72 DRAFT P3cch Bottem C.E.T. Documentation This case covers sequences in which the SGTS was previously failed or in which a hydrogen burn occurred. The burn is assumed to both actuate the fire sprays and isolate the SGTS.

Question # 59 : What is containment pressure after CF or vent during

meltdown?

This question determines the containment pressure at the tire of vessel breach for those sequences in which the containment has been vented, or has failed, during-core degradation. This. question is thus irrelevant for sequences in.which either the containment has remained intact or in which it failed prior to core degradation. Containment pressure at the time of vessel breach for the lat.ter group of sequences is given by Question #33, above. For those sequences with containment failure during vessel breach, this question inputs the containment pressure (in psig) to the event

, tree code as Parameter #3.

! This is a Type 4 question which is used to input parameter values to the code. Branch point probabilities are also input.

They are dependent on the branches taken in previous questions, as are the parameter values.

There are 1 branches:

1 . PBase2: This question actually has no branches. The value of Phase 2, the containment failure at vessel breach for sequences in which the containment failed or was vented during core degradation, is merely determined for future refernce by the tree.

Case # 1 : M In this case the containment has ruptured or been vented. It is thus depressurized. The pressure will be only slightly greater

than atmospheric so Parameter #3 is assigned an appropriate value.

Case # 2 :.

This case covers sequences in which the containment is-intact or has a leak which is of a size which is insufficient to cause depressurization. For sequences in which the containment is intact after core degradation the parameter evaluated by this question

(#3) is not referenced. Thus, this case considers sequences in which there was a leakage failure (Level 2 as described in Question

1. 28 & #43 above) during core degradation. The containment pressure is thus slightly below the failure pressure of the containment.

Question # 60 : What.is the nature of meltdown?

This question determines whether the core melt proceeds by the

' _ Pcga 73 DRAFT Panch Botton C.E.T. Documsntation  !

- i 1

1 gradual flow of molten corium into the lower plenum or by catostrophic failure of the core supports resulting in complete collapse of the core. Despite numerous analyses and various models there is still no widespread agreement amoung. experts as to how core degradation will progress. This question relsolves the disagreement into two distict possibilities. A slump type of melt was that envisioned in WASH-1400 and propagated ino the current

~

MARCH 2 code (some variations on the original slump concept are modelled). The slump model assumes that as the core melts the debris collects on the core plate until the core plate fails (due to heating and/or melting). Failure of the core plate is assumed l to result in the entire reactor core " slumping" downward. This model is more applicable to PWRs than BWRs since the BWR core plate provides latteral support only. The core is supported by the control rod guide tubes in four fuel bundla groups. However, the source term code package models assume a slump type failure for

, BWRs. Other models for BWRs, most notably MELRPI (developed by R.

' Lahey at al for the NRC Severe Accident Sequence Analysis program) and the IDCOR program MAAP model, assume that molten fuel can accumulate on the core plate (or fuel bundle tie plates) and at some point will locally melt the core plate allowing flow of the molten material down the guide tubes into the lower plenum. This flow type melt is more gradual than that assumed in the slump I

model.

This is a Type 2 question where branch point probabilities are input and are dependent on branches taken in previous questions.

There are 3 branches:

1 . Slump:. The entire core slumps into the lower plenum prior to vessel breach.

2 . Flow: Molten material flows from the core region into-the lower plenum. vessel breach may occur with r some part of the core intact. i 3 . nMelt: Core degradation did not' occur.  !

Case # 1 :

This case includes sequences in which adequate injection was provided by operator action prior to significant core damage.

There is thus no core melt.

Case # 2 :

In this case either some means of water injection was restored during core degradation or CRD flow has continued to operate but~

was insufficient to prevent core melt. The injection of water into the vessel is expected to make slump type melts even less likely than with no injection. This is because water accumlation in the lower plenum will result in more cooling on the core support i

DRAFT POnch Botton C.E.T. Dscumsntation structure, lessening its vulnerability (if any) to catastrophic failure. In addition, late injection of water may well retard or prevent the melting process resulting in a greater tendency for i fuel material to stay in the core region. Arresting the core melt would prevent a slump completely (and also vessel failure --

Question #61 below). t case # 3 :

This case covers sequences in which there is no delayed restoration of water injection. Quantification of this question is l judged to reflect current thought in the reactor safety community as to whether BWR core melts are more likely to be flow type melts or slump type melts. Flow type melts are considered likely in the central estimate.

Question # 61 : What is the mode of vessel breach?

This question determines the nature of the failure of the vessel.

This is a Type 2 question where branch point probabilities are input and are dependent on branches taken in previous questions.

There are 4 branches:

1 . Alpha: There is a steam explosion in vessel with debris forming a missile which subsequently lifts the upper vessel head resulting in subsequent failure of the drywell.

2 . SE-BH: An in-vessel steam explosion causes catostrophic failure of the vessel bottom head.

3 . Mthru: The vessel botta head melts through as a result of the accumulation of hot core debris on its inner surface.

4 . nBreach: Core melt is arrested in-vessel.

Case # 1 :

This case covers sequences with no core damage. There can be no vessel breach.

Case # 2 :

In this case it is assumed that a slump type of core melt occurs. This (according to some experts) gives rise to the possibility that a steam explosion could occur. Only the pessimistic walkthrough allows this as a significant possibility.

Case # 3 :

Pcga 75 DRAFT Panch Botton C.E.T. Documsntation

This case includes sequences in which injection is restored during core degradation. There is thus some possibility that

vessel breach may be avoided. This depends on whether the core debris in the vessel is in a coolable geometry. Just what this mean is not clear and expert opinion varies. Clearly there will be a sensitivity to the extent of core damage at the time injection begins. The IDCOR position is that less than 20% of the core must be melted in order for there to be recovery. Since melt proceeds rapidly once it has begun, there seems little justification for a high probability of preventing vessel failure. It is thus judged that the probability of preventing vessel failure with late injection is indeterminate.

Case # 4 :

This case covers sequences in which a flew-type melt has been assumed. Steam explosions are excluded on the basis that there is insufficint mechanical energy available to trigger an explosion which has enough energy to fail the vessel. These sequences are thus assumed to result in a vessel melt-through.

i Question # 62 : Is the RPV blowndown before vessel breach?

! This question determines the pressure within the Reactor

~ Pressure Vessel (RPV) at the time of vessel breach. Note that Question #45 above has previously determined the RPV pressure during core degradation. The purpose of this question is to allow

.for the possibility that the melt-through of the vessel (determined by Question #61) occurs in such a way that the gas pressure is relieved before a significant amout of corium is ejected. This could occur if the RPV pressure boundary is broken first in a control rod guide tube at a location which is both above the debris accumulated in the bottom head (if any) and below debris accumulated at the elevation of the core plate. . Failure of an

. instrument tube penetration before a large amount of debris has collected on the bottom head of the vessel could have the same result.

This is a Type 2 question where branch point probabilities are input and are dependent on branches taken in previous questions.

There are 2 branchec:

-1 . nEBlowD: The Reactor Pressure Vessel is at high pressure when vessel breach occurs.

2 . EBlowD: The Reactor Pressure Vessel is depressurized at the time of vessel breach.

Case # 1 :

In this case core degradation occurs with the RPV at low

- - - =. .- , - _- . __ - _- - ._. ._ . - . _.

Page 76 DRAFT Poach Botten C.E.T. Docum0ntation pressure. The pressure is thus low at vessel breach regardless of the failure mode.

Case # 2 :

In this case the core melt is assumed to be of the flow type (described in Question #60 above). The vessel bottom head fails by melting-through. In a flow melt, corium will flow down the control rod guide tubes and will accumulate around the stub tubes where the control rod drive enters the vessel. A leak at this point in the pressure boundary could occur before a significant quantity of core debris had accumulated in the lower plenum. This leak would have the potential for depressurizing the RPV prior to a gross melt through of the head. A less likely possibility is that corium flows through the fuel support casting and down the inside of uthe contol rod guide tube. 'A melt through of the pressure boundary inside the guide tube could then occur; isolated from any large quantities of debris which may have accumulated. The latter is less likely since the inside of the guide tube is probably full of water.

Case # 3 :

This case covers sequences in which the vessel is a high pressure during core degradation and there is a slump type melt so there is a reduced probability of blowdown prior to a large failure of the bottom head.

.i, Case # 4 :

This case covers sequences in which there is no breach of the vessel or in which vessel failure results from an in-vessel steam explosion.

Question # 63 : Does direct heating occur?

This question determines whether there is direct heating of the containment. If the vessel fails at high pressure, very hot core debris can be blown into the containment atmosphere. If the containment contains significant oxygen the debris may burn as zirconium, and uranium oxide oxidize.- In the inert Peach Bottom containment the sensible heat of the debris may be trasferred to the containment atmosphere with relatively high efficiency. This results in rapid pressurization of the containment. This phenomenon is direct heating.

This is a Type 2 question where branch point probabilities are input and are dependent on branches taken in previous questions.

There are 2 branches:

1 . DH : ' Direct heating occurs at vessel breach.

2 . nDH: There is no direct heating of the primary containment.

f a r-- - - - . . , ,w, n , , n--- -,.,-,v,---------.-,,,,---,.------,..-,--,---,,.n.,-,--.,---,,-----,--,--.-.~.r---

,n,

DRAFT POcch Bottom C.E.T. Documsntation Case i 1 :

In this case the vessel pressure is high at the time of. vessel breach. The melt progression is assumed to be a core slump so that there are large quantities of core debris in the vessel bottom head. These are the conditions which are assumed to allow direct heating to occur. However, the confined geometry of the BWR pedestal and the restricted opening to the drywell (2 meters by 1 meter) will serve to mitigate the dispersal of debris into the drywell airspace. (It is also noted that the pipes and associated whip restraints, as well.as pumps and equipment in the drywell, will also tend to hamper debris dispersal). Thus, even if the conditions at in the RPV at vessel breach favor direct heating, the probability that it will occur is judged to be indeterminate.

Case # 2 :

This case covers sequences in which either there has been a flow type melt (so little debris has accumulated on the bottom

, head) or the vessel is depressurized at vessel breach. Either of these conditions is assumed to prevent direct heating.

Question # 64 : Does an ex-vessel steam explosion occur?

This que'stion determines whether a steam explosion occurs as a l result of corium ejection from the Reactor Pressure Vessel. A steam l explosion could result from the hot melt falling into water in the

sumps or water which has accumulated on the drywell floor. For the purposes of this question, only steam explosions of sufficient magnitude ~to cause a significant. increase in containment pressure are consisdered. Small steam explosions, which may result in small incoherent pressure spikes, are more probable but are not included since they have a negligible impact on containment integrity.

~

This is a Type 2 question where branch point probabilities are input and are dependent on branches taken in previous questions..

There are 2 branches:

1 . ExSE: There is a steam explosion in the containment at vessel breach.

2 . nExSE: Thesre is no steam explosion in the containment.

Case f.l.:

l l In this case a slump type melt has occurred. Slump type core melts are.a prerequisite for steam explosions. The slump melt allows for the possibility of a large quantitity of hot molten material to fall from the vessel into the water beneath at the time

I Paga 78 DRAFT Peach Bottcm C.E.T. Documsntation i

l of vessel breach. If a sufficiently large mass of corium falls j into a water pool with enough velocity to induce fragmentation of the molten material a steam explosion is likely. Three factors I mitigate the probability of a steam explosion: First, the amount of l material which is molten when the vessel bottom head fails may not 1 be very large or may mix with solid debris prior to reaching the water in the bottom of the pedestal. Second, the control rod drive l mechanisms and their supporting structural beams will ihibit the fall-of molten material and a significant quantity my refreeze on the steel surfaces. Third, apart from water in the sumps, there can be at most 2 feet of water accumulated on the pedestal floor.

This water could be displaced by the falling material and mixing of -

the water and corium could be poor. However, it should be noted that the pedestal would tend to contain the water and while it ,

might be temporarily displaced eventual mixing is likely. The l above factors were considered quantifiying the probability of an ex-vessel steam explosion in this case.

Case # 2 :

1 This case covers sequences in which either there is a flow melt i or in which core melt is arrested in-vessel. In either case, there is no possibility of a steam explosion. If the vessel remains i intact, this is clearly true. In the case of a flow melt corium leaves the vessel more-or-less as the fuel melts. This is a relatively slow rate. The corium is also close to the liquidus temperature and should readily refreeze on the control rod drive structure below the vessel. With small quantities of solid, or almost solid, debris falling into the pedestal water (if any) there

! is a negligible probability that a steam explosion, of a magnitude l which could impact the containment integrity, would occur.

l Question # 65 : What is pressure rise due to blowdown (effective VS and

! RCdry)? <

l I This is the first of two questions ( #65 and #67 ) which addresses the pressure rise in the containment resulting from

. vessel melt-through. .Each of the questions assigns a value to a separate pressure rise parameter. The sum of these three parameters is the total pressure rise (in psi) resulting.from vessel breach. This question quantifies the pressure rise which

. results from the blowdown of the vessel. The blowdown of the vessel from 1100 psi, assuming 50% steam by volume, results in approximately a 20 psi rise in the containment pressure.

Uncertainties in this value arise from uncertainties in the bulk gas temperature in vessel, the hydrogen contentent, and the final taperature in the containment (considering PV work). The impact of possible direct heating is also considered in this question.

Quantification of the pressure rise due to direct heating is accomplished with a simple model'but the uncertainties are large.

These uncertainties are principally, the amount of debris which participates in direct heating, the extent of distribution into the ,

containment atmosphere, and the degree to which the debris undergoes exothermic reactions with the atmosphere.

PCga 79 DRAFT Poach Botton C.E.T. Docum:ntation

,' This is a Type 4 question which is used to input parameter values to the code. Branch point probabilities are also input.

They are dependent on the branches taken in previous questions, as are the parameter values.

There are 1 branches:

1 . DPblow: This question has no actual branches. The purpose of this question is to evaluate Dpblow, the pressure rise in the containment due to blowdown

'of the RPV. The question is not reference by subsequent questions, only the evaluated parameter

(#4) is used.

Case i 1 :

In this case there is direct heating. Implicit in this, the vessel must be pressurized at vessel breach. The central estimate

. of the pressure rise is the calculated value for direct heating, using the assumptions given.below, plus the estimated 20 psi pressure rise resulting from the blowdown of hot gas. The optimistic a pessimistic variations represent what is judged to be a reasonable uncertainty in this. number.

Case # 2 :

In this case the vessel is at high pressure but there is no direct heating. Quantification of the pressure rise indicates a reasonable uncertainty around the central estimate of pressure rise due to blowdown of the vessel.

Case # 3 :

This case covers sequences in which the vessel is depressurized at the time of vessel breach. There is thus no significant pressure rise in the containment resulting from the equilibration i of the pressures inside and outside the vessel.

4 Question # 66 : Does the cavity have more than initial sump water?

This question determines whether there is more water in the base of the reactor pedestal than that contained in the sumps. The pressure rise which results from the core debris contacting that excess water is also evaulated'in Question #67 (the next question). The water which may accumulate on the drywell floor as a result of condensation of steam passed back to the drywell through the drywell vacuum breakers is considered negligible.. This question only considers water from primary system breaks or drywell sprays.

This is a Type 2 question where branch point probabilities are input and are dependent on branches taken in previous questions.

Paga 80 DRAFT P0ach.Botton C.E.T. Documsntation There are 2 branches:

1 . RCwet: The reactor pedestal floor is covered by up to 2 feet of water.

2 . RCdry: There is no water in the reactor pedestal beyond

! what is contained in the drywell sumps.

Case f 1 :

In this case the accident initiator is a primary system break -

in the drywell. With this leakage from the RPV there will be much more water on the drwell floor than would be present due to condensation in sequences where steam discharge from the RPV is passed to the suppression pool through the S/RVs.

Case # 2 :

In this case the drywell sprays were running prior to vessel '

breach. The drywell will be flooded up to the bottom of the vent 3 pipes (2 ft' of water) . Thus the pedestal floor is considered to be

" wet".

Case # 3 :

This case covers sequences in which there is no added water in the pedestal. These are transient initiated sequences with no drwell spray operation.

Question # 67 : Is SP bypassed or vents not cleared at VB (effect. VS)?

-- DPfVS This question is the second of the two which are used to assess the containment pressure rise at vessel breach. The pressure rise indicated in the previous question (# 65) was evaluated assuming that pressure suppression occurs. This question provides for adding a pseudo-pressure rise which accounts for suppression pool bypass, pool. temperatures close to the saturation point, and dynamic loads resulting from steam explosions and direct heating.

Thus, the parameter evaluated below is a correction to the previously evaluated pressure increases to account for any lack of effectiveness in the pressure suppression system. . The question has branches and evaluates the pressure rise parameter. Thus there can be " cases within a case". In other words the value that the pressure rise parameter (#5) as'sumes is dependent on which question branch is taken in that case. Thus the probability that the vents have time to clear in a specific case will impact the prediced pressure rise.

This is a Type 4 question which is used to input parameter values to the code. Branch point probabilities are also input..

They are dependent on the branches taken in previous questions, as are the parameter values.

l Paga 81 i DRAFT Ponch Bottom C.E.T. Documsntation There are 3 branches:

1 . Vcir: The drywell vents clear (no large dynamic load) ,

and the pressure suppression system works as designed.

2 . nVcir: There is a large, rapid increase in the drywell pressure and the drywell vents do no clear with sufficient rapidity.

3 . Bypass: The suppression pool is bypassed and there is no vapor suppression function. '

case # 1 :

In this case there is an ex-vessel steam explosion but no direct heating. The suppression pool is not bypassed but its temperature is high enough that the effeciveness of vapor suppression is reduced. Because a steam explosion would produce a large dynamic load, there is some probability that the drywell vents will not clear. The resulting pressure rise in the drywell would be large since the steam generated by the explosion would not be condensed. The value assigned to the pressure rise parameter, given that the vents do not clear, is large enough to give a significant probability of containment failure. If the vents do

. clear, the pressure rise is that associated with transferring the energy of the explosion into a hot suppression pool. This pressure rise may or may not be. sufficient.to fail containment. The probability nof containment failure is explicitly evaluated in Question #68 below.

Case # 2 :

This case is similar to the previous one with the exception that the suppression pool has a substantial amount of subcooling.

Pressure suppression will thus be effective if the vents clear.

The probability that the vents do clear is evaluated exactly as in the previous case. When the vents do not clear the pressure rise is evaluated such that there is a significant probability that the containment will fail. On the other hand, if the vents do clear, the pressure rise is taken to be negligible.

Case # 3 :

In this case, direct heating occurs, there is no ex-vessel steam explosion, the suppression pool is not bypassed but the water in it is hot. Direct heating can result in a very rapid increase in the containment pressure. Quantification of the probabilities

for this case allows for the possibility that the drywell will pressurize too rapidly for the drywell vents to clear. If the vents to not clear the pressure rise will be large enough to assure containment failure. If the vents do clear, the pool r2tains the

Page 82 DRAFT Poach Botton C.E.T. Documentation I

energy released but steam and some quantity of hydrogen (from core debris oxidation), will be generated resulting in significant containment pressurization.

l Case # 4 :

This case is similar to the previous one except that the suppression pool is significantly subcooled. Provided that the vents clear, there is only a moderate increase in containment pressure resulting- from the generation of hydrogen. As in the previous case, if the vents do not clear, containment failure is assured.

Case # 5 :

In this case, both direct heating and a steam explosion occur at vessel breach. Such an occurrence is extremely unlikely; as indicated by previous quantification of the tree. The suppression pool is also hot so the effectiveness of pressure suppression is reduced. With both a steam explosion and direct heating the drywell will pressurize very rapidly. The probability of the drywell. vents not clearing will.tMt greatest under these

, circumstances. As in the previous case, if the vents do not clear, the pressure rise is assigned a value so large that containment failure is assured. Given that the vents clear the pressure rise is that associated with suppression pool heat-up and the generation of hydrogen as a result of direct heating.

Case # 6 :

This case.is similar to the previous one with the exception that ths suppression pool is subcooled and the steam released to, or generated in, the drywell is largely condensed. There is thus a negligible pressure rise from this steam provided that the drywell vents clear. The probability that the vents clear.is evaluated-exactly as in the previous case.

Case # 7 :

In this case there is a significant amount of water on the drwell floor. The vessel pressure was high at the time of failure so there was blowdown to the containment (the debris was n'ot necessarily discharged under high pressure). The suppression pool i temperature is high but it has not been bypassed. Interaction of i

.the dedris with the water on the drywell floor will produce steam and hydrogen in addition to what is released from the vessel. The effectiveness of pressure suppression is reduced by the hot pool.

It is certain that the vents will clear (the system is designed for very rapid pressurization of the drywell). The pressure rise parameter for this case accounts for two effects: 1) The additional steam generated from the suppression pool in absorbing the steam and hydrogen released by the blowdown of the vessel (relative to a subcooled pool). 2) The generation of steam and hydrogen resulting

{ from the core debris interacting with water on the drywell floor.

i e

Paga 83 f

DRAFT Peach Bottom C.E.T. Documentation I

Case # 8 :

This case is similar to the previous one with the exception that the vessel was depressurized during core degradation. There is thus no blowdown of the vessel to be taken into account. The pressure rise for this case is just that due to the suppression pool (hot) absorbing the energy transferred ~to it by the steam and hydrogen generated in the interaction of the core debris with water on the drywell floor plus that due to the hydrogen being added to the containment.

Case i 9 : ,

In this case there is no water on the pedestal floor above what is contained in the sumps. The suppression pool is hot. The vessel pressure may either be high or low. The suppression pool being hot reduces the effectiveness of vapor suppression and thus there is a small added pressure rise from that given for vessel blowdown in the question above (#65). This pressure rise corresponds to that resulting from pool heat-up as a consequence of condensing the steam which was released from the vessel at high pressure. There is a slight inaccuracy in this case in that if the vessel is depressurized at vessel breach the pressure rise is 4

expected to be lower that that given in this case. However, the i

difference is less than 10% of the total failure pressure so there is no impact on the results predicted by the event tree code.

Case # 10 :

l In this case the suppression pool is subcooled and not ,

4 bypassed. There is also no ex-vessel' steam explosion or direct heating (such sequences are considered in previous cases). Because vapor suppression is effective ther is no additional pressure rise for this case. (The vents are certain to clear). All the appropriate pressure rise is given by Question #65 above.

Case # 11 :

In this case the suppression pool is at least partially bypassed but the containment has previously ruptured. With the containment ruptured there can be no pressure rise (definition of rupture). The branch of this question indicating suppression pool bypass is taken in this caseand in all the succeeding cases of this

- question. All the non-bypass sequences are covered by the previous ten cases.

Case # 12 :

• In this case there is a significant quantity of water on the drwell floor, the vessel was pressurized prior to vessel failure and the suppression pool is bypassed. (A partial bypass is assumed equivalent to a complete bypass for the puposes of evaluating containment pressure rise.) The pressure rise parameter thus accounts for the increase in containment pressure if vessel blowdown occurs with no pressure suppression over that occurring

Paga 84 DRAFT Patch Botto3 C.E.T. Docum:ntation with normal pressure suppression (the assumption made in Question (65). In addition, the pressure rise due to steam and hydrogen generation from the core debris interacting with water on the drywell floor is also taken into account.

Case # 13 :-

This case is similar to the previous case (#12) with the exception that the RPV was depressurized at vessel breach. Since there is little pressure rise associated with vessel breach with the vessel depressurized, even with the suppression pool bypassed, the pressure rise parameter for this case essentially accounts for the pressure rise associated with the interaction between core debris and the water on the drywell floor. To pressure suppression occurs.

Case f 14 :

P -In this case the vessel is depressurized at vessel breach, there is no steam explosion and the pedestal floor is dry (ther may be water in the sumps however). The suppression pool is bypassed but there is no significant increase in the vessel pressure because the release of energy from the vessel at the time of breach was negligible.

4 Case f.15 :

i This case covers only sequences in which the pedestal floor is dry, the suppression pool is bypassed and there is a steam explosion in the containment at the time of vessel breach.

Containment failure is certain in such sequences. The pressure rise paramenter is consequently set high enough to assure this will be predicted in a subsequent question. The RPV pressure at the time of breach is irrelevant for this case.

Question # 68 : Does pressurization fail containment at VB?

This question instructs the Event Tree Code to determine the  !

l probability of containment failure immediately after (at) vessel

[ breach. To do this the code calulates the containment pressure l using parameter values supplied by previous questions. The i

-calculated pressure is used to evaluate the probability of failure based on an input containment fragility. This question is analogous to Question #42 abopve;.the same fragility is assumed.

l This is a Type 6 question in which branch point probabilities I

are calculated by the code. The input is a list of parameters to

, be summed and values of reference parameters to which the sum is

compared. The list, the reference parameters, and the method of comparison is dependent on the branches taken in previous questions.

I

There are 2 branches

I

! l . IP: The containment fails at vessel breach.

1 i

Paga 69 DRAFT POnch Bottom C.E.T. Documsntation Case'# 6 : l In this case there is a large leak from the primary containment into the reactor building. The Standby Gas Treatment System is operating. The suppression pool does not flash, so large quantities of steam were not released. There is a possibility that SGTS will help prevent combustible hydrogen concentrations from accumulating in the reactor building. This is reflected in the quantification.

Case i 7 :

In this case there is a large leak from the primary containment but the Standby Gas Treatment System is not operating. There is a possibility that steam inerting wi,11 occur as a result of suppression pool flashing. However, condensation of steam.on the

, structures in the building and on the building walls would work against-inerting. With the SGTS not running there is then l considerable potential for combustible concentrations'to accumulate. The probability of burns is thus fairly high.

Case f 8 :

This case covers sequences in which there is no significant leakage from the primary containment.. There is thus little, if any, hydrogen in the reactor building which could burn.

Question # 56 : What is level of reactor building (RB) breach / bypass '

before VB?

This question determines the impact of possible hydrogen

! burning on the integrity of the secondary containment. If hydrogen burning occurs in the reactor building rapid pressure increases would occur which could alter gas flow pathways through the i building by damaging doors and/or walls or causing blow-out pannels to fail. The burns would also tent to incrase the level of_ bypass by shortening the residence time of the aerosol-carrying gasses (less effective retention).

! This is a Type 2 question where branch point probabilities are

input and are dependent on branches taken in previous questions.

There are 3 branches:

1 . nERBB: There is no loss of the reactor building integrity prior to vessel breach.

! 2 . ERBB 2: There is significant leakage from the reactor

! buiding prior to vessel breach. However, the

majority of the gas flow passes through the building with some opportunity for fission product deposition.

3 . ERBB 3: The leakage from the reactor building prior to vessel breach is very large or gas escaping the

Paga 70 DRAFT P0cch Botto:a C.E.T. Docum:ntation primary containment bypasses the building completely.-

Case i 1 :

In this case large leakage from, or bypass of, the reactor building began prior to the opportunity for hydrogen burning. The building thus remains in this condition.

Case l'2 :

i -

In this case a-hyam n burn occurs in t.he'-reactor building.

Prior to the burn the building either has negligible leakage or a partial bypass. It is judged to be certain that hydrogen burning would cause a failure of the reactor building pressure boundary, thus creating at least a partial bypass. Even if the building was leaking prior to the burn, there is a significant probability that hydrogen burning would exacerbate the bypass by doing further structural damage or by significantly reducing the gas residence 4

time.

Case # 3 :

In this case the reactor building initially had some leakage but no hydrogen burns occur which could exacerbate it. The partial bypass is certain to persist.

j Case # 4 :

1 This case covers sequences in which the reactor building was intact and remains so after the possibility of hydrogen burns is considered since none are found to occur. (Either because primary

, containment is intact or there was just no burn.)

Question f 57 : Are fire systems actuated before VB7 l This question determines the impact of possible hydrogen I burning on the operation of the. reactor building fire sprays. The

[ fire spray system:1s discussed:above (Question 153).

This is a Type 2 question where branch point probabilities are input and are dependent on branches taken in previous questions.

l There are 2 branches:

l . EFireS: The reactor building fire sprays are operating prior to vessel breach.

! '2 . nEFireS: The recctor building fire sprays are not operating prior to vessel breach.

M

- - . , - , . - _ - - - - - . - , - . , . - . , - - - - - . . . - . ~ -- - ..- -- - - ~ - -

w -- - m. s - _ .- - . -

c Paga 71 DRAFT POcch Bottcm C.E.T. Docussntation i

Case i 1 :

In this case the fire sprays were not operating previously and no hyrogen burn occured in the secondary containment which would cause them to start. Thus, fire sprays are certainly not operating.

Case # 2 :

This case covers sequences in which either the fire sprays came on regardless of any hydrogen burn in the reactor. building or sequences in which there was a hydrogen burn'. Hydrogen burns are certain to activate the fire system since they are designed to come on if there is a fire in the reactor building.

Question f 58 : Does SdTS operate until VB7 This question determines the impact of possible hydrogen burns in the reactor building on the Standby Gas Treatment System. As noted above, fire dampers are expected to isolate the SGTS should the temperature in the reactor building be elevated significantly.

This feature is intended to protect the SGTS from fire.

This is a Type 2 question where branch point probabilities are input and are dependent on branches taken in previous questions.

There are 3 branches:

1 . EfSGTS: The Standby Gas Treatment Sytem is failed (isolated).

2 . ErSGTS: The Standby Gas Treatment System is unavailable i but may be recovered.

3 . E-SGTS: The Standby Gas Treatment System is operating prior to vessel breach.

Case # 1 :

In this case the SGTS was operating prior to the possibility of a hydrogen burn. The fire sprays have not started (ie. there was no burn). Since fire sprays and SGTS isolation are triggered by the same thing (high reactor building temperature the SGTS must contiue to run.

Case # 2 :

In this case SGTS was not running prior to the possibility of a hydrogen burn but it was recoverable (ie the fire dampers remain open). As with the previous case the fire sprays are not on so the fire dampers will remain open. SGTS thus remains recoverable.

Case # 3 :

i

'Pcgo 72

. DRAFT Panch Bottom C.E.T. Documsntation This case covers sequences in which the SGTS was previously failed or in which a hydrogen burn occurred. The burn is assumed to both actuate the fire sprays and isolate the SGTS.

Question ~# 59 : What is containment pressure after CF or vent during meltdown?

This question determines the containment pressure at the time -

of vessel breach for those. sequences in which the containment has been vented, or has failed, during core degradation. This question is thus irrelevant for sequences in which either the containment has remained intact or in which it failed prior to core degradation. Containment pressure at the time of vessel breach for

.t the latter group of sequences is given by Question #33, above. For those sequences with containment failure during vessel breach, this question inputs the containment pressure (in psig) to the event tree code as Parameter #3.

j This is a Type 4 question which is used to input parameter values to the code. Branch point probabilities are also input.

They are dependent on the branches taken in previous questions, as are the parameter values.

There are 1 branches:

i 1 . PBase2: This question actually has no branches. The value of Pbase2, the containment failure at vessel breach for sequences in which the containment failed or was vented during core degradation, is

' merely determined for future refernce by the tree.

i Case # 1 :

In this case the containment has ruptured or been vented. It is thus depressurized. The pressure will be only slightly greater than atmospheric so Parameter #3 is assigned an appropriate value.

Case # 2 :.

i This case covers sequences in which the containment is intact or has a leak which is of a size which is insufficient to cause

depressurization. For sequences in which the containment is intact

, after core degradation the parameter evaluated by.this question

(#3) is not referenced. Thus, this case considers sequences in 1 which there was a leakage failure (Level 2 as described in Question L #28 & #43 above) during core degradation. The containment pressure is thus slightly below the failure pressure of the containment.

Question # 60 : What is the nature of meltdown?

i This question determines whether the core melt proceeds by the i

d Pega 73 DRAFT Poach bottcm C.E.T. Docum ntation gradual flow of molten corium into the lower plenum or by catostrophic failure of the core supports resulting in complete collapse of the core. Despite numerous analyses and various models there is still no widespread agreement amoung experte as to how core degradation will progress. This question relsolves the disagreement into two distict possibilities. A slump type of melt was that envisioned in WASH-1400 and propagated ino the current MARCH 2 code (some variations on the original slump concept are  ;

modelled). The slump model assumes that as the core melts the debris collects on the core plate until the core plate fails (due to heating and/or melting). Failure of the core plate is assumed to result in the entire reactor core " slumping" downward. This model is more applicable to PWRs than BWRs since the BWR core plate provides latteral support only. The core is supported.by the control rod guide tubes in four fuel bundle groups. However, the source term code package models assume a slump type. failure for

. BWRs. Other models for BWRs, most notably MELRPI (developed by R.

I Lahey at al for the NRC Severe Accident Sequence Analysis program)

and the IDCOR program MAAP model, assume that molten fuel can

. accumulate on the core plate (or fuel bundle tie plates) and at some point will locally melt the core plate allowing flow of the

{ molten material down the guide tubes into the lower plenum. This flow t1pe melt is more gradual than that assumed in the slump model.

This is a Type 2 question where branch point probabilities are input and are dependent on branches taken in previous questions.

There are 3 branches:

1 . Slump: The entire core slumps into the icwer plenum prior to vessel breach.

2 . Flow: Molten material flows from the core region into the lower plenum. Vessel breach may occur with some part of the core intact.

3 . nMelt: Core degradation did not occur.

Case # 1 :

l This case includes sequences in which adequate injection was I provided by operator action prior to significant core damage.

There is thus no core melt.

Case # 2 :

In this case either some means of water injection was restored during core degradation or CRD flow has continued to operate but was insufficient to. prevent core melt. The injection of water into the vessel is expected.to make slump type melts even less likely than with no injection. This is because water accumlation in the lower plenum will result in more cooling on the core support l

1

Paga 74 DRAFT P0cch Bottem C.E.T. Documsntation structure, lessening its vulnerability (if any) to catastrophic failure. In addition, late injection of water may well retard.or

. prevent the melting process resulting in a greater tendency for fuel material to stay in the core region. Arresting the core melt 1

would prevent a slump completely (and also vessel failure --

Question #61 below).

Case # 3 :

This case covers sequences in which there is no delayed restoration of water injection. Quantification of this question is judged to reflect current thought in the reactor safety community as to whether BWR core melts are.more likely to be flow type melts or slump type melts. Flow type _1 salts are considered likely in the central estimate.

Question ~# 61 : What is the mode of vessel breach?

This question determines the nature of the failure of the vessel.

This is a Type 2 question where branch point probabilities are input and are dependent on branches taken in previous questions.

There are 4 branches:

1 . Alpha: There is a steam explosion in ves'sel with debris forming a missile which subsequently lifts the

. upper vessel head resulting in subsequent failure of the drywell.

2 . SE-BH: An in-vessel steam explosion causes catostrophic failure of the vessel bottom head.

3 . Mthru: The vessel botta head melts through as a result of the accumulation of hot core debris on its inner surface.

4 . nBreach: Core melt is arrested in-vessel.

l l Case # l':

This case. covers sequences with no core damage. There can be no vessel breach.

Case# 2 :

In this case it is assumed that a slump type of core melt occurs. This (according to some experts) gives rise to the possibility that a steam explosion could occur. Only the i pessimistic walkthrough-allows this as a significant possibility.

l Case # 3 : '

I

DRAFT Panch Bottom C.E.T. Documsntation This case includes sequences in which injection is restored during core degradation. There is thus some possibility that vessel breach may be avoided. This depends on whether the core debris in the vessel is in a coolable geometry. Just what this mean is not clear and expert opinion varies. Clearly there will be a sensitivity to the extent of core damage at the time injection begins. The IDCOR position is that less than 20% of the core must be melted in order for there to be recovery. Since melt proceeds rapidly once it has begun, there seems little justification for a high probability of preventing vessel failure. It is thus judged that the probability of preventing vessel failure with late-injection is indeterminate.

Case # 4 :

This case covers sequences in which a flow-type melt has been assumed. Steam explosions are excluded on the basis that there is insufficint mechanical energy available to trigger an explosion which has enough energy to fail the vessel. These sequences are thus assumed to result in a vessel melt-through.

1 Question f 62 : Is the RPV blowndown before vessel breach?

This question determines the pressure'withi~n the Reactor Pressure Vessel (RPV) 'at the time of vessel breach. Note that Question #45 above has previously determined the RPV pressure during core degradation. The purpose of this question is to allow for the possibility that the melt-through of the vessel (determined by Question #61) occurs in such a way that the gas pressure is relieved before a significant amout of corium is ejected. This could occur if the RPV pressure boundary is broken first in a l control rod guide tube at a location which is both above.the debris accumulated in the bottom head (if any) and below debris j accumulated at the elevation of the core plate. Failure of an instrument tube penetration before a large amount of debris has collected on the bottom head of the vessel could have the same result.

l This is a Type 2 question where branch point probabilities are input and are dependent on branches taken in previous questions.

There are 2 branches:

1 . nEBlowD: The Reactor Pressure Vessel is at high pressure when vessel breach occurs.

2 . EBlowD: The Reactor Pressure Vessel is depressurized at the time of vessel breach.

Case #- l :

In this case core, degradation occurs with the RPV at low

DRAFT Pasch Battcm C.E.T. Documsntation pressure. The pressure is thus low at vessel breach regardless of the failure mode.

Case # 2 :  ;

In this case the core melt is assumed to be of the flow type (described in Question #60 above). The vessel bottom head fails by melting-through. In a flow melt, corium will flow down the control rod guide tubes and will accumulate around the stub tubes where the control rod drive enters the vessel. A leak at this point in the pressure boundary could occur before a significant quantity of core debris had accumulated in the lower plenum. This leak would have the potential for depressurizing the RPV prior to a gross melt through of the head. A less likely possibility is that corium

, flows through the fuel support casting and down the inside of uthe l contol rod guide tube. A melt through of the pressure boundary inside the guide tube could then occur; isolated from any large quantities of debris which may have accumulated. The latter is less likely since the inside of the guide tube is probably. full of water. .

Case # 3 :

This casa covers sequences in which the vessel is a high pressure during core degradation and there is a slump type melt so there is a reduced probability of blowdown prior to a large failure

of the bottom head.

Case # 4 :

This case covers sequences in which there is no breach of the vessel or in which vessel failure results froa an in-vessel steam explosion.

Question # 63 : Does direct heating occur?

This question determines whether there is direct heating of the containment. If the vessel fails at high pressure, very hot core debris can be blown into the containment atmosphere. If the containment contains significant oxygen the debris may burn aus zirconium, and uranium oxide cxidize. In the inert Peach Bottom containment the sensible heat of.the debris may be trasferred to the containment atmosphere'with relatively high efficiency. This results in rapid pressurization of the containment. This phenomenon is direct heating.

This is a Type 2 question where branch point probabilities are input and are dependent on branches taken in previous questions.

There are 2 branches:.

1. DH : Direct heating occurs at vessel breach.

2 . nDH: There is no direct heating of the primary containment.

DRAFT Poach Bottom C.E.T. Documsntation I

i l

Case # 1 :

In this case the vessel pressure is high at the time of vessel breach. The melt progression is assumed to be a core slump so that there are large quantities of core debris in the vessel bottom head. These are the conditions which are assumed to allow direct 3 heating to occur. However, the confined geometry of the~BWR pedestal and the restricted opening to the drywell (2 meters by 1 meter) will serve to mitigate the dispersal of debris into the drywell airspace. (It is also noted that the pipes and associated whip restraints, as well as pumps and equipment in the drywell, will also tend to hamper debris dispersal). Thus, even if the

, conditions at in the RPV at vessel breach favor direct heating, the probability that it will occur is judged to be indeterminate.. l Case i 2 :

i This case covers sequences in which either there has been a flow type melt (so little debris has accumulated on the bottom

head) or the vessel is depressurized at vessel breach. Either of these conditions is assumed to prevent direct heating.

Question # 64 : Does an ex-vessel steam explos5on occur?

i

' This question determines whether a steam explosion occurs as a result of corium ejection from the Reactor Pressure. Vessel. A steam explosion could result from the hot melt falling ~into water in the sumps in water which has accumulated on the drywell floor. For the purposes of this~ question, only steam explosions of sufficient I

magnitude to cause a significant increase in containment pressure are consisdered. Small steam explosions, which may result in small incoherent pressure spikes, are more probable but are not included since they have a negligible impact on containment integrity.

This is a Type 2 question where branch point probabilities are input and'are dependent on branches taken in previous questions.

l There are 2 branches:

i 1 . ExSE: There is a steam explosion in the containment at vessel breach.

2 . nExSE: Thesre is no steam explosion in the containment.

Case # 1 :

In this case.a slump type melt has occurred. Slump type core melts are a prerequisite for steam explosions. The slump melt allows for the possibility of a large quantitity of hot molten material to' fall from the vessel into the water beneath at the time

Paga 78  !

DRAFT Poach Bottos C.E.T. Documsntation l l

of vessel breach. If a sufficiently large mass of corium falls into a water pool with enough velocity to induce fragmentation of the molten material a steam explosion is likely. Three factors mitigate the probability of a steam explosion: First, the amount of material which is molten when the vessel bottom head fails may not be very large or may mix with solid debris prior to reaching the water in the bottom of the pedestal. Second, the control rod drive mechanisms and their supporting structural beams will ihibit the fall of molten material and a significant quantity my refreeze on the steel surfaces. Third, apart from water in the sumps, there can be at most 2 feet of water accumulated on the pedestal floor.

This water could be displaced by the falling material and mixing of the water and corium could be poor. However, it should be noted that the pedestal would tend to contain the water and while it might be temporarily displaced eventual mixing is likely. The -

above factors were considered quantifiying the probability of an ex-vessel steam explosion in this case.

Case # 2 : -

This case covers sequences in which either there is a flow melt or in which core melt is arrested in-vessel. In either case, there is no possibility of a steam explosion. If the vessel remains intact, this is clearly true. In the case of a flow melt corium leaves the vessel more-or-less as the fuel melts. This is a relatively slow rate. The corium is also close to the liquidus temperature and should.readily refreeze on the control rod drive structure below the vessel. With small quantities of solid, or almost solid, debris falling into the pedestal water (if any) there is a negligible probability that a steam explosion, of a magnitude

, which could impact the containment integrity, would occur.

Question # 65 : What is pressure rise due to blowdown (effective VS and l RCdry)?

This is the first of two questions ( #65 and #67 ) which addresses the pressure rise in the containment resulting from vessel melt-through. Each of the questions assigns a value to a separate pressure rise parameter. The sum of these three parameters is the total pressure rise (in psi) resulting from

vessel breach. This question quantifies the pressure rise which results from the blowdown of the vessel. The blowdown of the vessel from 1100 psi, assuming 50% steam by volune, results in approxim~ately a 20 psi rise in the containment pressure.

Uncertainties in this value arise from uncertainties in the bulk gas temperature in vessel, the hydrogen contentent, and the final taperature in the containment (considering PV work). The impact of possible direct heating is also considered in this question.

Quantification of the pressure rise due to direct heating is accomplished with a simple model but the uncertainties are large.

These uncertainties are principally, the amount of debris which participates in direct heating, the extent of distribution into the containment atmosphere, and the degree to which the debris undergoes exothermic reactions with the atmosphere.

DRAFT Pscch Botton C.E.T. Documsntation This is a Type 4 question which is used to input parameter

~ values to the. code. Branch point probabilities are also input.

They are dependent on the branches taken in previous questions, as are the parameter values.

There are 1 branches:

1 . DPblow: This question has no actual branches. The purpose of this question is to evaluate Dpblow, the pressure rise in the containment due to blowdown of the RPV. The question is not reference by subsequent questions, only the evaluated parameter

(#4) is used.

Case # 1 :

In this case there is direct heating. Implicit in this, the vessel must be pressurized at vessel breach. The central estimate of the pressure rise is the calculated value for direct heating, using the assumptions given below, plus the estimated 20 psi pressure-rise resulting from the blowdown of hot gas. The optimistic a pessimistic variations represent what-is judged to be a reasonable uncertainty in this number.

Case # 2 :

In this case the vessel is at high pressure but there is no direct heating. Quantification of the pressure rise indicates a reasonable uncertainty around the central estimate of pressure rise due to blowdown of the vessel.

l Case # 3 : .

This case covers sequences in which the vessel is depressurized

, at the time of vessel breach. There is thus no significant l pressure rise in the containment resulting from the equilibration i of the pressures inside and outside the vessel.

l Question # 66 : Does the cavity have more than initial sump water?

i This question determines whether there is more water in the base of the reactor pedestal than.that contained in the sumps. The pressure rise which results from the core debris contacting that excess water is also evaulated in Question #67 (the next question). The water which may accumulate on the drywell floor as j a result of condensation of steam passed back to the drywell through the drywell vacuum breakers is considered negligible. This

question only considers water from primary system breaks or drywell sprays.

This is a Type 2 question where branch point probabilities are input and are dependent on branches taken in previous questions.

-c..- - - -..

Pcgo 80 DRAFT Pach Botten C.E.T. Documsntction There are 2 branches:

1 . RCwat: The reactor pedestal floor is covered by up to 2 J

feet of water.

2 . RCdry: There is no water in the reactor pedestal beyond what is contained in the drywell sumps.

Case # 1 :

In this case the accident initiator is a primary system break in the drywell. With this leakage from the RPV there will be much' more water on the drwell floor than would be present due to condensation in sequences where steam discharge from the RPV is passed to the suppression pool through the S/RVs.

Case # 2 :

In this case the drywell sprays were running prior to vessel breach. The drywell will be flooded up to the bottom of the vent pipes (2 ft of water) . Thus the pedestal floor is considered to be

" wet".

Case # 3 :

. This case covers sequences in which there is no added water in the pedestal. These are transient initiated sequences.with no drwell spray operation.

4 Question # 67 : Is SP bypassed or vents not cleared at VB (effect. VS)?

DPfVS

~

This question ~is the second of the two which are used to assess the containment pressure rise at vessel breach. The pressure rise

. indicated in the previous question (# 65) was evaluated assuming that pressure suppression occurs. This question provides for adding a pseudo-pressure rise which accounts for suppression pool bypass, pool temperatures close to the saturation point, and dynamic loads resulting from steam e>Tiasions and direct heating.

Thus, the parameter evaluated belov is a correction to the previously evaluated pressure indtrase s to account for any lack of effectiveness in the pressure alpyren non system. The question has branches and evaluates the pressare riaa parameter. Thus there can be " cases within a case". In other words the value that the

, pressure rise parameter (#5) assumes is dependent on which question

, branch is taken in that case. Thus the probability that the vents have time to clear in a specific case will impact the prediced pressure rise.

This is a Type 4 question which is used to input parameter values to the code. Branch point probabilities are also input.

They are dependent on the branches taken in previous questions, as are the parameter values.

l 1 1 DRAFT Patch Bsttsu C.E.T. Docutontation l

There are 3 branches:

1 . Vcir: The drywell vents clear (no large dynamic load) and the pressure suppression system works.as j

, designed.

2 . nVcir: There is a large, rapid increase in the drywell pressure and the drywell vents do no clear with sufficient rapidity.

3 . Bypass: The suppression pool is bypassed and there is no vapor suppression function.

Case # 1 :

In this case there l's an ex-vessel steam explosion but no direct heating. The suppression pool is not bypassed but its temperature is high enough that the effeciveness of vapor suppression is reduced. Because a steam explosion would produce a i

large dynamic load, there is some probability that the drywell vents will not clear. The resulting pressure rise in the drywell would be large since the steam generated by the explosion would not be condensed. The value assigned to the pressure rise parameter, given that the vents do not clear, is large enough to give a significant probability of containment failure. If the vents do clear, the pressure rise is that associated with transferring the energy of the explosion into a hot suppression pool. This pressure rise may or may not be sufficient to fail containment. The probability nof containment failure is explicitly evaluated in Question #68 below.

Case # 2 :

This case is similar to the previous one with the exception

.that the suppression pool has a substantial amount of subcooling.

Pressure suppression will thus be effective if the vents clear.

, The probability that the vents do clear is evaluated exactly as in the previous case. When the vents do not clear the pressure rise is evaluated such that there is a significant probability that the

, containment will fail. On the other hand, if the vents do clear,

! the pressure rise is taken to be negligible.

Case # 3 :

In this case, direct heating occurs, there is no ex-vessel steam explosion, the suppression pool is not bypassed but the water in it is hot. Direct heating can result in a very rapid increase in the containment pressure. Quantification of the probabilities for this case allows for the possibility that the drywell will pressurize too rapidly for the drywell vents to clear. If the vents to not clear the pressure rise will be large enough to assure containment failure. 'If the vents do clear, the pool retains the

PEga 82 >

DRAFT P0ach Botton C.E.T. Documentation energy released but steam and some quantity of hydrogen (from core debris oxidation), will be generated resulting in significant containment pressurization.

Case # 4 :

This case is similar to the previous one except that the suppression pool is significantly subcooled. Provided that the vents clear, there is only a moderate increase in containment pressure resulting from the generation of hydrogen. As in the i

previous case, if the vents do not clear, containment failure is assured.

i Case # 5 :-

In this case, both' direct heating and a steam explosion occur i at vessel breach. Such an occurrence is extremely unlikely; as indicated by previous quantification of the tree. The suppression pool is also hot so the effectiveness of pressure suppression is reduced. With both a steam explosion'and direct heating the drywell will pressurize very rapidly. The probability of the drywell vents not clearing will be greatest under these circumstances. As in the previous case, if the vents do not clear, the pressure rise is assigned a value so large that containment failure is assured. Given that the vents clear the pressure rise is that associated with suppression pool heat-up and the generation of hydrogen as a result of direct heating.

Case # 6 :

This case is similar to the previous one with the exception that the suppression pool is subcooled and the steam released to, or generated in, the drywell is largely condensed. There is thus a

', negligible pressure rise from this steam provided that the drywell vents clear. The probability that the vents clear is evaluated exactly as in the previous case.

Case # 7 :

l In this case there is a significant amount of water on the

! drwell floor. The vessel pressure was high at the time of failure i

so there was blowdown to the containment (the debris was not necessarily discharged under high pressure). The suppression pool temperature is high but it has not been bypassed. Interaction of the dedris with the water on the drywell floor will produce steam i and hydrogen in addition to what is released from the vessel. The i

' effectiveness of pressure suppression is reduced by the hot pool.

It is certain that the vents will clear (the system is designed for very rapid pressurization of the drywell). The pressure rise parameter for this case accounts for two effects: 1) The additional steam generated from the suppression pool in absorbing the steam and hydrogen released by the blowdown of the vessel (relative to a subcooled pool). 2) The generation of steam and hydrogen resulting from the core debris interacting with water on the drywell floor.

4 Pega 83 DRAFT P0ach Botto2 C.E.T. Documentation l

i l

Case # 8 : i This case is similar to the previous one with the exception that the vessel was depressurized during core degradation. There l is thus no blowdown of the vessel to be taken into account. The l pressure rise for this case is just that due to the suppression  :

pool (hot) absorbing the energy transferred to it by the steam and hydrogen generated in the interaction of the core debris with water on the drywell floor plus that due to the hydrogen being added to the containment.

Case # 9 :

In this case there is no water on the pedestal floor above what I is contained in the sumps. The suppression pool is hot. The vessel pressure may either be high or low. The suppression pool being hot reduces the effectiveness of vapor suppression and thus there is a small added pressure rise from that given for vessel blowdown'in the question above (#65). This pressure rise corresponds to that resulting from pool heat-up as a consequence of condensing-the steam which was released from the vessel at high pressure. There is a slight inaccuracy in this case in that if the vessel is depressurized at vessel breach the pressure rise is expected to be lower that that given in this case. However, the

, difference is less than 10% of the total failure pressure so there is no impact on the results predicted by the event tree code.

3 Case # 10 :

I In this case the suppression pool is subcooled and not bypassed. There is also no ex-vessel steam explosion or direct

heating (such sequences are considered in previous cases). Because vapor suppression is effective ther is no additional pressure rise l for this case. (The vents are certain to clear). All the appropriate pressure rise is given by Question #65 above.

Case # 11 :

In this case the suppression pool is at least partially bypassed but the containment has previously ruptured. With the containment ruptured there can be no pressure rise (definition of rupture). The branch of this question indicating suppression pool bypass is taken in this caseand in all the succeeding cases of this

! question. All the non-bypass sequences are covered by the previous ten cases.

Case # 12 :

• In this case there is a significant quantity of water on the

-drwell floor, the vessel was pressurized prior to vessel failure and the suppression pool is bypassed. (A partial bypass is assumed l equivalent to a complete bypass for the puposes of evaluating j containment pressure rise.) The pressure rise parameter thus accounts for the increase in containment pressure if vessel L

blowdown occurs with no pressure suppression over that occurring I

Paga 84 DRAFT P0ach Botten C.E.T. Documentation

, ~.

with normal pressure suppression (the assumption made in Question

1. 65), In addition, the pressure rise due to steam and hydrogen generation from the core debris interacting with water on the drywell floor is also taken into account.

! Case # 13 :

This case is similar to the previous case (#12) with the exception that the RPV was depressurized at vessel breach. Since there is little pressure rise associated with vessel breach with the vessel depressurized, even with the suppression pool bypassed, the pressure rise parameter for this casa essentially accounts for i'

the pressure rise associated with the interaction between core debris and the water on the drywell floor. To pressure suppression .

4 occurs.

Case # 14 :

In this case the vessel is depressurized at vessel breach, there is no steam explosion and the pedestal floor is dry (ther may be water in the sumps however) . The suppression pool is bypassed but there is no significant increase in the vessel pressure because-the release of energy from the vessel at the time of breach was negligible.

. Case # 15 :

This case covers only sequences in which the pedestal floor is i . dry, the suppression pool is bypassed and there is a steam explosion in the containment at the time of vessel breach.

Containment-failure is certain in such sequences. The pressure rise paramenter is consequently set high enough to assure this will i be predicted in a subsequent question. The RPV pressure at the

' time of breach is irrelevant for this case.

Question #'68 : Does pressurization fail containment at VB?

This question instructs the Event Tree Code to determine the probability of containment failure immediately after (at) vessel breach. To do this the code calulates the containment pressure using parameter values supplied by previous questions. The calculated pressure is used to evaluate the probability of failure based on an input containment fragility. This question is

analogous to Question #42 abopve; the same fragility is assumed.

This is a Type 6 question in which branch point probabilities are calculated by the code. The input is a list of parameters to be summed and values of reference parameters to which the sum is compared. The list, the reference parameters, and the method of comparison is dependent on'the branches taken in previous questions.

There are 2 branches:

1 . IP: The containment fails at vessel breach. ,

Paga 85 DRAFT Patch B tton C.E.T. Docuz0ntation i

2 . NIP: The containment retains its integrity following '

vessel breach.

i 1

Case # 1 : ,

1 In this case the containment has not been vented and did not fail during core degradation. The pressure in the containment immediately prior to vessel breach is "Pbasel" which is evaluated in Question #33 plus the pressure rise due to hydrogen production' in-vessel (Question #35). The containment pressure after vessel breach is evaluated by summing the above defined pressure rise parameters (Questions #65 and #67)~to this initial value.

Case # 2 :

4 In this case the vessel does not melt-through so the probability of containment failure is assessed based on the I

containment pressure evaluated by summing the pressure at the onset -

of core degradation (Question #33) and the pressure rise due to hydrogen production (Question #35). The result is the containment pressure when the accident is recovered (assuming that it did not fail).

Case # 3 :

l.

[ In this case the containment was previously failed or has been i vented. The pressure after vessel breach is thus evaluated using the base pressure given by Question #59 which accounts for containment failure or venting during core degradation. The pressure rise in containment following vessel breach is added (Questions #65 and #67). This allows for what may previously been Level 2 leakage (defined under Question #43) to be exacerbated to a rupture.

Question # 69 : Does direct melt-structure attack fail containment at VB?

This question determines whether containment failure occurs as a result of core debris contacting the steel containment shell at the point of embedment. Core debris may flow from the vessel pedestal and spread over the drywell floor to the point at which the steel containment shell-is embedded in the concete of the drywell floor and basemat. Two failure modes can be conceived: The debris maybe sufficiently hot that the carbon steel of the shell is melted. Or, the combined effects of high containment pressure and local heating of the steel shell may result in a creep-rupture type failure. IDCOR (Chicago Bridge and Iron) has evaluated the latter mode and concluded that the shell is. adequately restrained to preclude this mode of failure. It should be noted however that this evaluation was carried out assuming a 1500 degree Fahrenheit

, debris temperature, whereas the MAAP code apparently predicts a l 1500 C temperature. At 1500 C considerable softening (to the point

l Pags 86 DRAFT P cch Bottom C.E.T. Documentation of melting -- approx. 1450 C ) would be expected. A cooler crust might still form on the debris preventing a melt-through. However, melt-through of the containment is clearly possible (especially assuming the CORCON-VANESA model for core-concrete interaction, which can, under some conditions, predict very high debris temperature > 2000 C ). The factor most likely to limit containment failure by melt-structue attack is the failure of the corium to flow out as far as'the containment shell. This containment failure mode is neglected if the containment failed at vessel breach as a result of overpressure. If this was the case.the containment melt-through will no significantly impact the source term (drwell failure is almost certain anyway).

This is a Type 2 question where branch point-probabilities are input and are dependent on branches taken in previous questions.

There are 2 branches:

1 . IM: The containment is fails as a result of melt-through at the point of embedment.

2 . NIM: The containment is not failed as a result of melt-structure attack.

Case # 1 :

In this case the core debris was dispersed at vessel breach as a result of the high pressure.in the RPV (slump-type melt is i

assumed). Such dispersal is judged to decrease the probability of containment failure due to melt-through. The fact that dispersal l 'may result in some cooling of the debris (and limit self heating) is considered.

, Case'#. 2 :

In this case the drywell floor is dry and core debris falls from a depressurized vessel. Depending on the temperature of the debris it may spread over the drywell floor and contact the-containment shell at the point of embedment. Even if the debris falling from the vessel is close to the solidus temperature it may reheat and spread out as core-concrete interaction (CCI) begins.

This case has the higest probability of melt-structure attack.

Case # 3 :

In this case the core debris has been dynamically disperced ss a result of an in-vessel steam explosion which failed the bottom head. This dispersal decreases the probability that the core debris will contact the containment shell. A slump type melt in vessel is also required for this kind of interaction to occur since there must be a sufficient supply of molten corium at the time of vessel breach. The probability of drywell melt-through is quantified exactly as in Case #1 of this question.

. .Pagn 87 DRAFT P3Ech Bottom C.E.T. Docum:ntetion i

Case # 4 :

This case covers sequences in which the containment fails immediately after vessel breach as a result of pressurization or in which the drywell floor has excess water. In the former case the  ;

nelt-through is not significant since there will be considerably greater resistance to gas flow along the leakage pathway opened by failure at the embedment. In the latter case the accumulated water is expected to prevent hot dedris from contacting the containment shall thus; preventing melt-through.

Question f 70 : What is the containment leakage level after VB?

This question determines the level of containment leakage

-immediately after vessel ~ breach. Given that the containment is no

longer intact, two levels of leakage are considered. Level 2 leakage is a failure which results in only slow depressurization of the. containment (on the order of 2 hours2.314815e-5 days <br />5.555556e-4 hours <br />3.306878e-6 weeks <br />7.61e-7 months <br /> to reach atmospheric pressure). Level 2 leakage corresponds to a leakage are on the order of 0.1 square feet. Level 3 leakage is a rupture of the containment (several square feet of area) which results in containment depressurization is several minutes.

This is a Type 2 question where branch point probabilities are input and are dependent on branches taken in previous questions ~.

There are 3 branches:

1 . nICL: The containment retains its integrity immediately following vessel breach.

2 . ICL2: The containment to leaks subsequent to vessel i

breach. The leakage rate is results in very slow containment depressurization.

3 . ICL3: The containment ruptures, resulting in depressurization, subsequent to vessel breach.

Case # 1 :

In this case there is either an alpha failure (WASH-1400 type in-vessel steam explosion), or the core debris melts the containment at the point of embedment, or there is an overpressure failure at the time of vessel breach. The alpha mode failure has very rapid depressurization; the drywell head is assumed to be lifted. Alpha mode is certainly Level 3 leakage. It is assumed that if the drywell is melted at the point of embedment that the leakage area is large enough to result in rapid depressurization.

Containment failure at the time of vessel breach is also assumed to result in containment rupture since the pressurization is rapid and typically peak pressures are well in excess of the assessed ultimate capability of the containment. Sequences in which the

! Pcga 88 DRAFT Poach Bottom C E.T. Documentation containment previously ruptured are also included in this case.

The rupture is certain to remain.

Case # 2 :

The containment has previously developed a leak. ~ Since,there-has been no exacerbation of this leak (ie subsequent containment failure) the leak continues.

' case # 3 :

! This case covers sequences in which there has been no significant threat to the containment integrity up to the time the core debris is released from the vessel. (Sequences in which there l

is no vessel breach are also included). There is thus no containment failure.

Question # 71 : What is the location of containment failure after VB?

This question determines the location in the. containment from

, which leakage (if any) occurs. This question is subsequently used 1

to evaluate the extent of suppression pool and/or reactor building bypass. Questions #28 and #43 above are analogous.

This is a Type 2 question where branch point probabilities are input and are dependent on branches taken in previous questions.

j There are 3 branches:

1 . nICL: There is no leakage from the containment up to the i

' time at which core debris is released from the vessel (or core melt is arrested).

4 2 . ICLD: Containment leakage is from the drywell.

3 . ICLW: Containment leakage is from the wetwell (torrus),

i l Case # 1 : s In this case containment failure occurs either at vessel breach due to an Alpha Mode steam explosion in-vessel, or as a result of melt-structure attack (Question #69). These failures are by i

i definition in the drywell. Sequences in which the drywell was failed prior to vessel breach are also included in this case since

these must also have drywell failures.

Case # 2 :

In this case containment over-pressure failure resulted from either direct heating or a steam explosion. Either occurrence would place a large dynamic and thermal load on the drywell making this the most probable location for containment failure.

Paga 89 DRAFT Panch Bottom C.E.T. Documentation i

J Case # 3 :

In this case the-containment failed due to over-pressurization at vessel breach. In the absence of large dynamic loads to the i drywell there is some possibility that failure may occur in the watwell.

Case # 4 :

In this case the containment failed in the wetwell prior to vessel breach. This failure remains. It should be noted that if the previous wetwell failure was Level 2 (see Question #43) then the possibility that there is subsequent failure in the drywell at the time of vessel breach is allowed by the previous case of this question.

Case f 5 :

In this case the containment remains intact following vessel-breach. There-is no leakage significantly in excess of tecnical specifications.

Question # 72 : Is AC power not available late?

This question allows consideration of offsite power being restored subsequent to vessel breach. Quantification of the probabilities is done using the same data as was used in the ASEP quantification of the. blackout sequence frequencies. National data for the time for recovery of offsite power was used.

This is a Type 2 question where branch point probabilities are 4

input and are dependent on branches taken in previous questions.

There are 2 branches:

I 1. LfAC: AC power is not recovered during the sequence i until after most of the fission product release 4

has occurred.

l 2 . L-AC: AC power is recovered subsequent to vessel breach i and in time to impact the core-concrete interaction.

! ~ Case # 1 :

In this case there was initial operation of the high pressure make-up systems and thus vessel breach occurs approximately 8-9 4

hours after the loss of offsite power. The probability for AC power avaialability thus approximates the probability of of offsite power being available at 10 hours1.157407e-4 days <br />0.00278 hours <br />1.653439e-5 weeks <br />3.805e-6 months <br /> (somewhat arbitrary) following interruption, given that it was not available at 9 hours1.041667e-4 days <br />0.0025 hours <br />1.488095e-5 weeks <br />3.4245e-6 months <br /> following

interruption.

i 4

s-,---, --,_v-----.---.,--.-y,- w,-- , v -vv e- . . - - - .- . . - - . - . , - . - - - , , - , - - - , - - - - - -_ _ _ _ _ _ - - - - - - - - - - _ _ - - - - - - - - . * - -

DRAFT Psach Bottom C.E.T. Documsntction Case # 2 :

This case includes the so-called fast station blackout sequences in which there is no early high pressure injection.

Vessel breach would occur'approximately-2 to three hours following vessel' breach. The probaility of AC power recovery is taken to be i the probability that it was recovered after 3 hours3.472222e-5 days <br />8.333333e-4 hours <br />4.960317e-6 weeks <br />1.1415e-6 months <br /> given that it was not recovered after 2 hours2.314815e-5 days <br />5.555556e-4 hours <br />3.306878e-6 weeks <br />7.61e-7 months <br />.

f-Case # 3 :

This case covers sequences in which AC power was not lost or was recovered prior to vessel breach. It is thus certain that AC power remains available.

Question # 73 : Is RHR not operating late?

.This question determines the availability of containment heat removal late in the sequence. Containment heat removal can be accomplished using either the spray or suppression pool cooling

modes of RHR operation. Heat removal from the containment by i venting or as a result of containment failure is considered in' other questions.

. This is a Type 2 question where branch point probabilities are inp.ut and are dependent on branches taken in previous questions.

There are 2 branches:

l i 1 . nLRHR:

There is no heat removal from the containment (using the~RHR system) subsequent to vessel breach.

2 . LRHR: The RHR system is used to remove heat from the l containment subsequent to vessel breach.

Case # 1 :

In this case either: the RHR system has previously failed, or there is no AC power available to operate the system, or the I suppression pool is drained so there is no water to pump. Thus, l there is certainly no containment heat removal.

l l

Case # 2 :

l In this case the RHR is not available due to the lack of AC

, power. Power is restored afeter vessel breach. The RHR system I cannot operate unless the LPCI pumps have sufficient Net Positive Suction Head (NPSH) thus either the containment must be intact or the suppression pool must be subcooled in order for RHR to be operable.The RHR system is manually initiated. Quantification of the probability that the RHR system is operated for this case thus r

considers the probability that the operator will start the system.

l t

- . - - , - - - - . , . , , , , - - - . - - - - , - - - - - - , - - - - - - ------r--

DRAFT Paach Botton C.E.T.- Documentation

, Operation is far from certain since the plants condition has gone far beyond anything that plant personnel are trained to handle.

Case # 3 :

In this casa either the RHR has been operationg up until the time of vessel breach or it was recovered subsequent to vessel 4

breach as a result of AC power being restored. However, the containment has failed at the suppression pool subsequently saturated The RHR pumps do not have adequate NPSH so the system is failed.

! Case # 4 :

l This case covers all sequences in wh'ich the RHR system was previously operating. The containment has remained intact so RHR continues.

Question # 74 : Do containment sprays not operate late?

This question determines whether the containment sprays are operating subsequent to vessel breach. Operation of the l containment sprays will profoundly influence the release of fission products to the environment.. In particular, the operation of sprays late in the sequence (during core-concrete interaction) will significantly reduce, and possibly prevent, the release of non-volatile fission product species.

This is a Type 2 question where branch point probabilities are input and are dependent on branches taken in previous questions.

f j There are 2 branches:

~

1 . nLCS: The containment (drywell) sprays are not operating i

subsequent to vessel breach.

! 2 . LCS: The containment (drywell) sprays are operating subsequent to vessel breach.

l Case # 1 :

! In this case either the sprays failed earlier in the sequence (perhaps initially) or there is no AC power available to power the l pumps. Either way, there is no late containment spray.

[ Case # 2 :

In this case the sprays were previously unavailable but AC power is recovered subsequent to vessel breach. In addition, i either the containment is intact or the suppression pool is j subcooled so if the operators al'ign the RHR' system for spray mode 3

and start the pumps there will be sufficient NPSH for the pumps to provide spray flow. (The suppression pool-is not drained.)

Paga 92 DRAFT POsch Bottcm C.E.T. Documentation 4

, Quantification of this case thus reflects the probability that plant personnel will initiate sprays late in the sequence.

Case # 3 :

I In this case the High Pressure Service Water System (HPSW) was recoverable and AC power has been restored. HPSW can be used for containment spray and the quantification of the probabilitiy of spray in this case reflects a judgement as to whether the plant operators will use HPSW for drywell spray. Late in the accident this is the best use of HPSW as far as mitigating accident consequences.

-Case # 4 :

i In this case containment sprays were operating up until the time of vessel breach using the RHR pumps. The containment has

failed and either the suppression pool is saturated (the pumps loose NPSH) or the suppression pool is drained (no water to pump).

j In either case, drywell spray does not continue.

Case # 5 :

This case covers sequences in which the containment sprays were operating at the time of vessel breach. In these sequences either:

the containment remained intact, the suppression pool was subcooled, or spray water was being delivered by the High Pressure Service Water (HPSW) system. With any (or all) .of these conditions met, the sprays will continue to operate.

Question # 75 : Is' water not supplied to the debris late?

This question determines whether water is delivered to the core debris in the drywell subsequent to vessel breach. This is significant since water has the potential to quench the debris, i halting the core-concrete interaction (CCI) and preventing the further release of fission products. Even if CCI is not prevented,

fission product releases will be mitigated.

This is a Type 2 question where branch point probabilities are input and are dependent on branches taken in previous questions.

There are 2 branches:

1 . nLDBWat: No water is supplied to the core debris on the drywell floor.

2 . LDBWat: Water is supplied to the core debris.

)

c Case # 1 :

I In this case the core debris is quenched in-vessel and there

] was no breach of the vessel. This implies that water was supplied i

Paga 93 DRAFT POcch Botton C.E.T. Docum:ntction to the core debris.

Case i 2~:

In this case the containment (drywell) sprays are operating.

There is-thus at least 5000 gpm of water being pumped into the drywell. This water will certainly reach the core debris.

Case # 3 :

In_this case the condensate system is available to pump water into the vessel. Since the vessel bottom head is breached, if the condensate pumps were started the water would reach the debris on the drwell floor. In the quantification of this case it was considered that the codensate system is the system of choice for i supplying water to the core. However, for these severely degraded conditions it is not certain that the operators would avail themselves of this option.

Case # 4 :

In this case there was water injection to the vessel (either

, CRD flow or High Pressure Service Water) but the core debris was not quenched in vessel. The vessel breach was due to a melt-through and there was no subsequent (immediate) failure of the containment. With no steam explosion or containment failure there is no threat to either of these injection systems. Injection to the vessel continues and since the vessel is breached this water is l supplied to the debris on the drywell floor.

Case # 5 :

In this case either the CRD flow was recovered (AC power has been restored after vessel breach) or CRD flow was being injected to the vessel at the time of vessel breach. For sequences in which CRD flow was operating previously, either the vessel failed as the result of a steam explosion or the containment failed as a result of the vessel breach (other sequences in which there was CRD flow were considered in the previous case of this question). The steam explosion and/or containment rupture presents the possibility that the CRD pumps are damaged or the flow of water through the CRD system is otherwise prevented. For sequences in which AC power is recovered subsequent to vessel breach operator action is required to restore CRD flow. The probability of CRD flow failure resulting from a steam explosion or containment rupture and the probability that the operator will restore CRD flow after the vessel has failed are quantified similarly.

Case # 6 :

This case is similar to the previous one except that it is the High Pressure Service Water (HPSW) system which is being considered as a source of cooling water to the debris rather than the CRD cooling water system. The HPSW system is considered to have a similar potential for' damage and, like the CRD system, operator

j Pago 94

DRAFT Psoch Botton C.E.T.'Documsntation action would be-required to use it for injection to the vessel wore AC power to be recovered late in the sequence.

Case # 7.:

i In this case-the RHR system is operational subsequent to vessel

breach. Plant operators could use this system to deliver water to i the core debris through use of the LPCI or shut-down cooling modes. Operator action would be required to switch to one of thess l- modes from the containment cooling mode of operation. It is not

, certain that plant operating personnel would perceive this to be an advantageous action. However, a reasonably high probability is

assigned. ..

Case # 8 :

I This case covers sequences in which there are no systems  ;

i available for the injection of water into either the vessel or the i drywell. Therefore, no water can be supplied to the core debris.

Question # 76

Is the debris not coolable without wa'ter? ,

i i

This question determines whether the core debris, which has

fallen or been ejected from the vessel, is distributed such that it i

' gives up sufficient heat to the drywell to prevent core-concrete interaction from being sustained. The heat loss mechanisms would be: direct radiation to the drywell structures, convection, and conduction to the surfaces on which the debris has been deposited.

I l This is a Type 2 question where branch point probabilities are input and are dependent on branches taken in previous questions.

There are 2 branches:

1 . nDBCwoW: The core debris'cannot be cooled without water.

l 2 . DBCwoW: The debris is distributed in the drywell such that it is self-cooling (no water needed).

Case # 1 :

l In this case debris has been dispersed in the drywell as a result of a steam exposion (in-vessel or ex-vessel) or direct

heating has occurred. If there has been direct heating, the debris will necessarily have been dispersed. Dispersal does not assure coolability, but with any of these events the dispersal is expected l to be widespread %Athin the drywell such that there is a significant probability that core-concrete interaction cannot be

! sustained.

Case # 2 : ,

1 i

This case covers sequences in which there has been no energetic j

Pcg2 95 DRAFT Peach Bottom C.E.T. Documentation dispersal of the core debris. The debris is thus pooled on the drywell floor. In this geometry, cooling by covection and radiant tranfer to the drywell structures is not expected to be adequate to remove the decay heat plus the heat generated as a result of rections in the melt (ie not coolable without water).

Question # 77 : Is the debris coolable with water?

This question determines whether not the core debris can be cooled with water. The phenomenological uncertainty with regard to whether the heat transfer to water covering a pool of molten corium which is reacting with concrete is adequate to remove the heat generated within the melt. Recent experiments performed at Sandia National Laboratory indicate that the addition of water to the melt will~not prevent core concrete interaction from continuing.

However, analyses performed by IDCOR indicate that the addition of water will result in quenching of the melt.

This is a Type 2 question where branch point probabilities are input and are dependent on branches taken in previous-questions.

There are' 2 branches:

1 . nDBC: The core debris in the drywell is not coolable.

2 . DBCwW: The core debris in the drywell can be quenched' (further core-concrete interaction prevented) by supplying water to it.

Case # 1 :

In this case either the debris was dispersed sufficiently at vessel breach that it can be cooled even without water (see the previous question) or the Reactor Pressure Vessel was not breached. If the latter is true, it is only because the debris was I

coolable with water that the vessel was not breached. Thus, in l

i either instance, the debris can certainly be cooled by supplying water.to it.

Case # 2 :

In this case, as in case #1 of the previous question, the debris was energetically dispersed at the time of vessel breach.

Sequences in which this dispersal wa sufficient to allow cooling without water were considered under the previous case of this question. However, given that the debris cannot be cooled without water, there is still a significant probability that it can be cooled by the addition of water. A judgement of this probability is reflected by the quantification of this case.

Case # 3 :

In this case there was a slump type melt in-vessel. The core

P2go 96 DRAFT Pcach Botten C.E.T. Documsntation t

debris was dumped on the pedestal floor en masse when the vessel was breached. A significant fraction of the debris is expected to ,

i -

be molten, thus there is expected to be some flowing of the melt out from the pedestal area (through the aanway) onto the drywell floor. With the debris in one lump with a substantial fraction being molten,.the probability of it being coolable is judged to be at a minimum.

Case # 4 :

This case covers sequences in which there was a flow type melt. Malted debris leaving the vessel is expected to be closer to the solidus temperature than.for a slump type of in-vessel melt.

For this type of melt there is a greater chance that the debris t will (at least initially) be retained in the pedestal area.However, ,

whether the geometry of the debris resulting from this type of core melt is more or less' easily cooled is not known. There is some reason to ascribe a greater probability to coolability since the debris temperature is lower.

i Question # 78 : Is the debris not cooled?

l This question sumarizes the results of the three previous questions with respect to whether core-concrete interaction (CCI)

l. is prevented or mitigated by cooling the core debris.

This is a Type 2 question where branch point probabilities are input and are dependent on branches taken in previous questions.

There are 2 branches:

1 . nCoolDB: The' core debris in the drywell is not cooled sufficiently to remove the decay heat plus the heat generated by core-concrete interaction.

2 . CoolDB: The core debris in the drywell.is cooled sufficiently to prevent further core- concrete interaction.

Case # 1 :

In this case either the debris recieved adequate cooling j

in-vessel that vessel breach was prevented or the debris was sufficiently dispersed at vessel breach that cooling is assured.

Case # 2 :

) In this case the debris in the drywell can be cooled if water i is supplied to it and some method of delivering water flow to the drywell has been established. On the basis of the way the previous

] questions are defined, the debris is certainly cooled.

l Case # 3 :

s

e .

' Paga 97 DRAFT Patch Botton C.E.T. Documsntation This esse covers sequences in which either the debris is not ccolable or there is no water supplied to it. Either way, the debris is not cooled and core concrete interaction will proceed.

1 Question # 79 : Does the containment fail late due to temperature?

3

' This question determines whether there is a failure of drywell penetration seals as a result of elevated temperature. High temperature failures of the drywell penetration seals was the assumed mode of failure in the IDCOR Task 23. 1 report on Peach Bottom. Loss of drywell integrity due to high temperatures can occur even if the containment has previously failed in the wetwell

! or been vented. Thus, such failures may significantly impact fission product release to the environment.

This is a Type 2 question where branch point probabilities are input and are dependent on branches taken in previous questions.

There are 2 branches:

1 . LTCF: There is a failure of one or more drywell j penetration seals, due to elevated temperatures in 1

the drywell, subsequent to vessel breach.

4

2 . nLTCF

There are no drywell penetration seal failures

resulting from high drywell temperatures.

1 Case # 1 :

In this care the core debris is distributed in the drywell and is giving up its heat (decay power) to the drywall structures and equipment. No water is being supplied to the drywell so there is 1 no effective mechanism for removing this heat. The drwell temperature will continue to rise until heat conduction through the

! containment wall reaches equilibrium with the decay heat being l generated. This would'not occur until the temperature in the  ;

i drywell, and in the walls, was well above what the seals could withstand. Temperature induced failure is thus certain. (Note that the structure of the tree allows overpressure failure to occur prior to temperature failures).

Case # 2 :

In this case there is debris on the drywell floor and core-concrete interaction is occurring. Heat generated in the melt is either transferred to the concrete, carried away by the gasses I

being generated (most of this portion will be absorbed by the pool and/or released from the containment), or goes into the drywell 1

equipment and structures. The magnitude of this latter fraction (and also the extent of exothermic reactions in the melt) largely determines the drywell temperature. The IDCOR Task 23.1 assessment was that the drywell temperature would rise sufficiently to cause I

1

~ __ _ _ _ . _ _ _ _ _ _ _ . _ . . _ _ _ _ _ - . _ . _ _ . . _ -._ _

i .

DRAFT Porch Bottea C.E.T. Documsntation loss of the seal integrity. However, there is sufficient uncertainty in the CCI phenomena that such a consequence is not certain.

Case # 3 :

This case covers sequences in which water is supplied to the drywell. Whether or not the debris is cooled or quenched by the water, the boil-off of this water will provide an effective means of transferring the heat which is being generated drywell to the suppression pool and eventually out of the containment or directly out-of the containment. The presence of water in the drywell will prevent temperatures from rising to levels at which seal integrity is threated to the point at which there is a significant probability of failure.

Question # 80 : Is the containment not vented late?

This question determines whether the containment is vented from the wetwell airspace after vessel breach.

This is a Type 2 question where branch point probabilities are input and are dependent on branches taken in previous questions.

There are 2 branches:

1 . LnVent: The operator does not (or is not able to) vent the containment after vessel breach. (Note: The containment may have been previously vented.)

2 . L-Vent: The plant operators successfully vent the containment only after the vessel has been breached.

l

! Case # 1 :

l l In this case either the containment has already lost its integrity and the pressure can no longer reach the level at which the procedures require venting or, there is no AC power available.

Venting is judged not to be a credible occurence, without AC power, once the RPV has been breached. Such a venting operation would l require manual action in the reactor building. Once core debris r

was on the drywell floor, radiation levels in the reactor building would be extremely high. Even if an operator were to consider the measured radiation levels acceptable, the action of venting would expose him to certain lethal radiation as radioactive material I flowed out the vent line.

1 Case # 3 :

In this case the accident has essentially been arrested. Decay l heat is being removed from the containment via the RHR heat exchangers and core-concrete interaction has been terminated.

Pcga 99 DRAFT Patch Bottca C.E.T. Docum0ntation Since the containment pressure would:be decreasing there is no necessity for venting.

Case # 3 :

This case covers sequences in which the containment pressure contines to rise subsequent to vessel breach. While the containment pressure will certainly exceed the venting threshold, venting is considered to be less likely at this point in the accident than it would have been at earlier time. Radiation levels in the containment would be extreme. In addition, there has been time to notify local government officials of the plant situation.

Under these circumstances there will be considerable reluctance to intentionally vent the containment regardless of what the emergency operating procedures dictate.

Question f 81 : Does the containment fail late?

This question determines whether containment ' integrity is lost as a result of core-concrete interaction, or containment heat-up, after vessel breach.

This is a Type 2 question where branch point probabilities are input ~and are dependent on branches taken in previous questions. l There are 3. branches:

1 . LPCF: The containment fails as a result of overpressurization due to core-concrete interaction (or the containment is vented).

2 . LTCF: The containment fails as a result of drywell peneteration seal failure after vessel breach.

3 . nLCF: Containment integrity is maintained throughout the accident.

case # 1 :

f.

! In this case the drywell penetration seals have failed.

) Containment integrity is thus lost as a result of high l temperatures.

(

i Case # 2 :

! In this case water is being supplied to the core debris on the l drywell floor which is thereby successfully cooled. The

containment heat removal (or RHR) system is operating. The heat i generated by core debris in the drwell will be transferrred to the <

suppression pool by the water flow through the drywell vents back l, into the suppression pool. Under these conditions the containment l pressure will be decreasing and the containment integrity will not j be further threatend.

l _ . _ _

Paga 100 l DRAFT Pacch Buttom C.E.T. Documentation i

1

' case # 3-:

In this case there is water being supplied to the core debris on.the drywell floor and the debris is thereby successfully cooled. However, there is no heat removal from the containment so

! the containment will continue to pressurize. Eventual overpressure failure is thus certain.

case # 4 :

In this case the debris is not cooled but containment heat removal is operating. The containment will fail as a result of the i

overpressure from noncondensable gasses (carbon dioxide, hydrogen,

carbon monoxide) produced by ccI.

! case # 5 :

In this case water'is being supplied to the core debris on the drywell floor but core-concrete interaction contines. The containment is still intact after vessel breach but pressurization i

continues. The presence of water in the drywell will keep drywell j temperatures below levels at which seal failures due to temperature

{ are a significant concern. Containment failure as a result of ov

, arpressurization by the combination of steam and/or non-condensable gases is certain.

case # 6 :

t In this casa no water is supplied to the core debris in the drywell and core- concrete interaction continues. ~ The containment j is intact subsequent to vessel breach but prssurization continues due to the heat and non-condensable gas generation. Drywell

! thermal failures were previously. considered ( Question #79 and Case

1. 1 of this question) so the only type of failure considered here is

failure due to overpressure, which is certain to occur.

i case # 7 :

This case covers sequences in which the containment has previously been failed or vented. The containment pressure is not rising and secondary failure of the drwell pensteration seals due to high drywell temperatures has be previously considered.

(Question.#79 and Case #1 of this question).

Question # 82 : What is the level of late containment leakage?

This question determines the level of leakage from the containment in the time period after the beginning of core-concrete interation, or the time of containment failure (whichever is earlier).

This is a Type 2 question where branch point probabilities are input and are dependent on branches taken in previous questions.

Paga 101 DRAFT P= ch BottO:a C.E.T. Docum:ntation 1

There are 3 branches:

1 . nLCL: The containment retains its integrity over the entire course of the accident. Leakage rates are not significantly higher than technical specification limits.

2 . LCL2: The containment develops a leak but the leakrate is sufficient to cause only slow ( > 2 hours2.314815e-5 days <br />5.555556e-4 hours <br />3.306878e-6 weeks <br />7.61e-7 months <br />) depressurization of the containment.

3 . LCL3: The containment is ruptured and the interior pressure is approximately atmosphereic pressure.

Case # 1 :

In this case the containment was previously ruptured and thus remains ruptured.

Case # 2 :

In this case there has been drywell failure as the result of excessive thermal loading on the drywell seals. Once such a seal begins to leak, with the drywell pressure high, the seal is expected to be erroded and the leak rate would increase to a level which would be sufficient to cause the containment to depressurize. In the quantification of this case some allowance is made for the possibility that the leak rate would remain small.

Case # 3 :

1 In this case the containment fails as a result of 1

overpressurization during the core-concrete interaction phase of the accident. With elevated drywell temperatures, IDCOR has suggested that containment leakage rather that rupture is the most likely mode of failure. Some allowance for the possibility of leaks (Level 2) has been made in the quantification of l probabilities for this case.

( Case # 4 :

i In this case the containment has been vented late in the sequence. Venting will result in containment depressurization and is thus a Level 3 leak, defined to be equivalent to a rupture of the containment.

I Case # 5 :

In this case the containment was leaking immediately after I

vessel breach and subsequent progression of the accident has not placed additional loading on the containment. Leakage is considered to continue and the containment pressure will continue

to gradually decline.

l

Paga 102 ,

DRAFT POcch Letton C.E.T. Docum:ntation

~_

case # 6 :

This case includes sequences in which the accident is recovered either prior to, or shortly after vessel breach and the containment had not previously failed. Since there is recovery, by definition, the containment ~ integrity is maintained. The containment leakage t rate is not significantly higher than technical specification limits.

Question # 83 : What is the location of late containment leakage?

This question determines the location of containment leakage.

The leakage location profoundly influences fission product retention along the release pathway to the environment. ,

This is a Type 2 question where branch point probabilities are '

input and are dependent on branches taken in previous questions.

There are 3 branches:

1 . nLCL: There is no leakage from the containment in excess of technical specification limits.

2 . LCLD: Containment leakage subsequent to vessel breach is from the drywell.

3 . LCLW: Containment leakage subsequent to vessel breach is I from the watwell (torrus) .

Case # 1 :

In this case excessive thermal loading in the drywell has resulted in failure of the penetration seals. With a failure in the drywell, the dominant leakage pathway would be from the drywell unless there had previously been a rupture in the watwell and the suppression pool was bypassed. With watwell rupture and pool bypass, if the drywell leakage were only Level 2 there would be less resistance on the wetwell pathway and this would be dominant.

These latter circumstances are specifically excluded from this case.

Case # 2 :

In this case the containment was successfully vented after the vessel was breached. There are no penetration failures in the-drywell as a result of thermal loading. The dominant pathway for gas flow from the containment is thus through the wetwell.

Case # 3 :

Immediately after vessel breach, leakage from the containment was through the drywell. Even if the previous leakage were small

_-. . = - _ - - - .. . - _ ..

DRAFT PO7,ch Botton C.E.T. DocunOntation (Level 2) and~the cont'ainment subsequently ruptured, it is judged that the rupture would occur in the drywell since the containment

! shell had previously been weakend there.

case # 4

?

.In this case there was leakage from the wetwell immediately after vessel breach. Provided there has been no subsequent leakage of the drywell penetration seals as a result of thermal loading, the dominant leakage pathway will continue to be through the i wetwell.

Case # 5 :

In this case. containment failure occured subsequent to vessel i breach as a result of overpressurization. However, the failure is not a rupture (it is Level 2). IDCOR results suggest leak type

! failure are dominant when the temperature of the containment shell l is elevated. Since temperatures are very high (> 400 C) only in the drywell, this suggests the dominant location for these Level 2 i

failures will be the drywell.

Case # 6 :

In this case the containment has failed as a result of.

! pressurization during core-concrete interaction. The failure was a i

rupture (Level 3). Allowance is made for the possibility that j failure would occur in the wetwell. Analyses performed to date do, i

.however, usually assume drywell failure (except WASH-1400).

f Case # 7 :

i' The case covers sequences in which the containment integrity is maintained throughout the accident. There is thus no leakage.

l Question # 84 : Is the suppression pool drained late?

This question determines whether the suppression pool still contains enough water to cover the drywell vents and therefore

maintain fission product scrubbing for sequences in which the l dominant release pathway is through the wetwell. If the i

containment has failed in the wetwell, this allows the possibility that the failure has occured below the pool water level. Partial i (or complete) draining of the pool could then occur. The level i

need only be lowered 3-4 feet before there will be essentially no i scrubbing of the gas affluent from the containment.

This is a Type 2 question where branch point probabilities are l input and are dependent on branches taken in previous questions.

i There are 2 branches:

l 1 . LSPD: The suppression pool is drained below the level of the drywell vents during the period following the 4

deposition of core debris into the drywell.

1

DRAFT POsch Bottc3 C.E.T. Docun0ntation 2 . nLSPD: The suppression pool is not drained. The drywell vents remain submerged in the suppression pool.

Case i 1 :

In this case the pool level was previously drained down and the drywell vents were uncovered. The pool level will remain too low.

Case # 2 :

In this case the containment has failed in the wetwell-either at the time of vessel breach or subsequent to that time. There is no evidence to suggest whether failure is more likely above or below the water line.

Case # 3 :

This case covers both sequences in which containment failure occurred prior to vessel breach with pool level was maintained and sequences with later failures in the drywell. The pool is'not drained in these sequences.

Question f 85 : What is the level of late suppression pool bypass?

! This question detremines the extent to which the suppression pool is bypassed. Suppression pool bypass can profoundly influence l the quantity of fission products released to the reactor building since pool scrubbing will not be effective.

This is a Type 2 question where branch point probabilities are input'and are dependent on branches taken in previous questions.

There are 3 branches:

1 . nLSPB: The dominant release pathway for fission products to the reactor building is through the suppression pool.

2 . LSPB2: There is a partial bypass of the suppression pool. A significant fraction of the gas flow from the containment is not scrubbed.

3 . LSPB3: Gas flow from the containment completely bypasses the suppression pool. There is no fission product scrubbing.

Case # 1 : ,

In this case the vessel has been breached and either the dominant release pathway from the containment is through the

. _~

Pag 3 105 DRAFT P22ch Botton C.E.T. Docuar,ntation I

i drywell or the suppression pool has been drained or otherwise bypassed. Suppression pool bypass is complete.

) Case # 2 :

In this case there has been an induced primary system break in the reactor building. The suppression pool is thus certainly bypassed.

Case # 3 :

In this case the vessel was not breached. However, there was a

previous partial bypass (level 2) of the suppression pool. This  ;

l bypass remains.

Case # 4 :

In this case the vessel is not breached but the suppression pool has previously been completely bypassed. The complete bypass remains.

Case # 5 :

l 1

This case covers sequences in which the containment either, remains intact, has been vented, or in which failure of the containment occurs in the wetwell above the suppression pool water level. Pool scrubbing will retain a significant quantity of fission products in the pool during these sequences; there is no pool bypass.

! Question f 86 : What is the level of late RB-bypass without a burn?

~

This question determines the extent to which fission products released from the containment subsequent to vessel breach could be

! retained in the reactor building. There is an implicit assumption that fission product retention in the building is proportional to the gas residence time. The extent to which the gas bypasses the

{ building (the length of the release pathway) and the flowrate l through the building are thus considered. This basic assumption ,

may be in conflict with the IDCOR model which appears to rely heavily on impaction for fission product retention in the reactor building.

i This is a Type 2 question where branch point probabilities are input and are dependent on' branches taken in previous questions.

4 j There are 3 branches:

i 1 . nLRBB: There is no breach or bypass of the reactor

building. What leakage there may be from the l containment could be handles by the Standby Gas Treatment System (SGTS) assuming that it were j operational.

j 2 . LRBB2: There is leakage from the reactor building but a j

i

- - - - - - - - . -,, ,.-_-,,_.,.-n---- - - - . - _ - . - - - - , - - . , ,--,_,--,,-__--,-,.n-. ---.,--n,----c,, -n,- , - - - . - , -. - - -.

V W Pags 106 DRAFT POnch Bottos C.E.T. Docum:ntation 4

significant fraction of that flow passes through the building on the pathway from the primary 3 ,

containment.

i 3 . LRBB3: The reactor building is essentially bypassed and j

there is negligible retention of fission pr,oducts within it.

Case # 1 :

In this case the reactor building was bypassed prior to vessel o i breach. The building will thus remain bypassad.

case # 2

,

i In this case vessel breach was the result of an Alphd mode-

{ steam explosion. By definition, all containment (including the j reactor building) is completely bypassed. ,

case .# 3

In this case the drywell ruptured at vessel breach. Sudden pressurization of the reactor' building would have occurred causing the blowout panels to give way. The extent to which the reactor building was bypassed would depend on the drywell failure locttion. Failure of the drywell head would result in complete bypass. The more likely failure would be lower down in the i

drywell. This would result in a partial bypass of the reactor building. Fission products released at the time of containment

rupture would likely be swept out of the reactor building. Fission products released later would be far more likely to be retained

, since the flowrate of gas through the reactor building would be lower. This case thus assumes a'substatial probability of only partial bypass of the reactor building.

l Case # 4 :

, In this caso the wetwell was ruptured at vessel breach. As in

the previous case the reactor building blowout panels would fail.

However, with a failure in the wetwell the reactor building could not be completely bypassed unless some breach in the reactor building wall, other than the blowout panels, was made.This latter failure mode is considered tc.have a negligible' probability.

Case # 5 :

In this case the steel shall of the drywell was melted through at the point of embedmont (see question #69). This failure location is low in the reactor building so only partial bypass of the

building is expected.

4 Case # 6 :

l-4

, . 'I t

i Paga 107 DRAFT Pasch Botton C.E.T. Documsntation {

In this case there is a drywell rupture as the result ~of slow pressurization of the containment following vessel breach. As in l Case #3 of this question the extent to which the reactor building

is bypassed depends on the drywell failure location. The more likely failure locations would not result in significant bypass of j the reactor building.

Case # 7 :

1 In this case there has been a rupture of the wetwell (torrus)

as a result of slow pressurization of the containment following vessel breach. As in Case #4 of this question, failures in the wetwell will most likely not result in a large bypass of the j reactor building.

I

case # 8

In this case the containment has a Level 2 failure and the leak rate to the reactor. building is relatively small (compared to a

rupture type failure). There is no significant possibility that the reactor building would be bypassed but the leakage flow is

anticipated to be high enough that there is a'significant release

of fission products from the reactor building.

Case # 9

In this case the reactor building developed significant leakage prior to vessel breach. This leakage has not been exacerbated so j the level of reactor building bypass remains the same.

Case # 10 :

This case covers sequences in which the containment retains its integrity. There is thus no significant leakage from the reactor building.

l Question # 87 : Are fire systems operating late without a late HB?

This question determines whether the fire sprays in the reaction building are running at the time fission products which are rolesed subsequent to vessel breach enter the reactor building. Operation of the fire sprays will increase fission product retention in the reactor building. However, because the Peach Bottom fire sprays are located only above the cable trays, the increased retention may not be significant. The fire spray system is described above under the discussion of Question #87.

This is a Type 2 question where branch point probabilities are input and are dependent on branches taken in previous questions.

There are 2 branches:

1 . LFireS: The reactor building fire sprays are running subsequent to vessel breach.

DRAFT POsch Bottc3 C.E.T. Docum:ntation 2 . nLFireS: The reactor building fire sprays are not started as a result of a late failure of the containment.

Case # 1 :

In this case the accident is arrested prior to containment failure. There is no release to the reactor building and thus the fire spray system would not be actuated.

Case # 2 :

In this case the fire spray system was actuated prior to vessel breach. The fire sprays will continue to run. '

case # 3 :

In this case the reactor building is essentially bypassed.

With no accumuation of hot steam or gasses in the areas of the reactor building which house the cable trays, the fire sprays would not be triggered.

. Case # 4 :

, In this case there is a leak from the drywell as a result of slow pressurization following vessel breach. The most likely location for a leak from the drywell is from the head closure (see Chicago Bridge & Iron analysis performed for IDCOR). Leakage from the head closure would pass up into the refueling area. Leakage of hot gas into this region would not result in the fire sprays being triggered since there are no cables there to be protected.

Case # 5 :

In this case the release pathway from the containment to the reactor building is through a subcooled suppression pool. With only warm gasses being passed into the reactor building, the fire sprays would not be triggered.

Case # 6 :

This case covers all sequences in which there is release of hot gas from the containment to the reactor building. The building is not completely bypassed and so there will be significant heating of areas of the building which are protected by fire sprays. This heating will cause the fire sprays to be triggered.

Question # 88 : Does standby gas treatment not work late without late HB?

This question determines whether the Standby Gas Treatment System-(SGTS) is operating during the period in which fission products, released from the containment after vessel breach , are in the reactor building. The impact of possible burning of

. s DRAFT Patch Bottca C.E.T. DocunOntation combustible gasses in the reactor building on SGTS operation is considered in a subsequent question. SGTS characteristics have been described previously (Question #54).

1 This is a Type 2 question where branch point probabilities are

input and are dependent on branches taken in previous questions.

There are 2 branches:

, 1 . nLSGTS: The Standby Gal; Treatment System does not operate

subsequent to 5issel breach.

t 2 . LSGTS: The Standby Gas Treatment System is operating after the containment has failed, following vessel l breach. '

I case # 1 :

In this case the SGTS has not previously failed and there is AC power available to operate it. The fire dampers in the SGTS duct work are still held open by the fusible links. Since SGTS 4

operation is automatic (assuming the system is operable) it is considered certain to run. (Note: It is assumed that conditions in the rector building which would result in the fire sprays being triggered would also cause the isolation of SGTS by failing the fusible links on the fire dampers.)

i Case # 2 :

This case covers sequences in which the SGTS has failed (or the fire dampers have closed rendering the system inoperable) or in

.which there is no Ac power.

Question # 89 : Does H2 burn in the reactor building after vessel

! breach?

This question determines whether hydrogen-(and/or possibly carbon monoxide which is also released by the corium-concrete interaction) burns in the reactor building following vessel breach. Late combustion of hydrogen released earlier is not considered (if a burn was going to occur it would have already) .

Combustion in the reactor building can impact fission product retention by increasing the level of reactor building bypass and actuating the fire protection systems.

This is a Type 2 question where branch point probabilities are input and are dependent on branches taken in previous questions.

There are 2 branches:

1 . LHB: Combustible gases burn in the reactor building following vessel breach.

l l

l l.-. . - . . - . - - ---. .- ---- . - . - - . - - - - - - - . - - - - - - - - - - - - --

. e l Page 110 i DRAFT Pasch Botto3 C.E.T. Docunontation 2 . nLHB: There is no burning of combustible gasses in the reactor building following vessel breach.

Case # 1 :

In this case the containment failed prior to vessel breach and the vessel was depressurized when it failed. There was thus no large influx of hydrogen into the reactor building after the vessel was breached and what may have been present before is assumed to have either burned or been swept out. The core debris is cooled so there is no further production of combustible gas. Under these conditions there will be'no combustion ~in the reactor building.

Case # 2 :

In this case there is significant leakage of combustible gasses from the containment subsequent to vessel breach. The Standby Gas Treatment System is operating however, and depending on the rate at which combustible gasses are produced, they may or may not accumulate to flammable concentrations in the reactor building.

The SGTS will reduce the chance of combution by removing some portion of the gas from the reactor building an lessening the probability that there will be flammable concentrations.

Case # 3 :

In this case there.is significant leakage from the containment and the Standby Gas. Treatment System is not operating. Without SGTS, combustible gasses may accumulate to flammable levels depending on the production rate inside the containment. The IDCOR model predicts that flammable concentrations are not attained in the reactor building. However, the Source Term Code Package models predict combustible concentrations are readily attained.

~

It would seem that combustion is not certain but even from an optimistic standpoint, some consideration to theis event must be given.

Case # 4 :

This case covers sequences in which the containment retains its integrity throughout the accident. No significant quantities of combustible gas are released to the reactor building so ther is no chance of combustion.

I Question # 90 : What is the level of late RB bypass?

This question considers the impact of possible combustible gas burns in the reactor building on its capability for fission product

retention. Burning within the building will increase the

(

probability of bypass, and also the level of bypass. Doorways and hatches could be blown out and leakage through the building walls l could be increased. In addition, burns will accelerate the flow of gas through the building, decreasing the residence time.

- ~

Pago 111 DRAFT Patch Botton C.E.T. Documentation This is a Type 2 question ~where branch point probabilities are input and are dependent on branches taken in previous questions.

There are 3 branches:

1 . nLRBB: The reactor building retains its integrity throughout the accident.

2 . LRBB2: There is a partial bypass of the reactor building but a significant fraction of the fission products released from the containment are retained.

3 . LRBB3: The reactor building is essentially bypassed late in the accident. There is no significant retention of fission products released from the containment during the core-concrete interaction

phase of the accident.

Case # 1 :

The reactor building was bypassed even without the effects of later combustible gas burns. The building is certain to remain bypassed.

Case # 2 :

In this case ther is a hydrogen burn in the reactor building and there is a Level 2 leak in the drywell. SGTS is not operating. As discussed previously, such a leak is probably from the drywell head closure. With the leak pathway being directly to the refueling floor level this is certainly where combustion occurred; immediately adjacent to the blowout panels. Under the assumption that the burn blew out a substantial portion of the panels, the bypass of the building would be essentially complete.

Case # 3

l l

In this case, there was a combustible gas burn in the reactor

' building. Prior to the burn the building had significant leakage.

There is some probability that the burn exacerbated the leakage sufficiently that the bypass becomes essentially complete.

Case # 4 :

In this case there was previously significant leakage from the reactor building. No burn occurred. The reactor building remains only partially bypassed.

Case # 5 :

In this case the containment remains intact throughout the accident and so does the reactor building.

DRAFT Patch Botten C.E.T. Docum:ntation Question # 91 : Are fire systems operating late?

This question determines the impact of possible hydrogen burns in the reactor building on the operability of the fire sprays. The description of Question #53 above includes a description of the fire spray system.

This is a Type 2 question where branch point probabilities are input and are dependent on branches taken in previous questions.

There are 2 branches:

i 1 . LFireS: The reactor building fire sprays are running

, during at least some portion of the period following vessel breach.

2 . nLFireS: The fire sprays are not actuated during the accident.

, Case # 1 :

In this case the fire sprays were not actuated prior to vessel breach. Reactor building temperature remained below the threshold at which the fire sprays are triggered after vessel breach and no

.\ combustible gas burns occurred. The fire spray will thus never be ij actuated.

! Case # 2 :

In this case the fire sprays were not actuated prior to the hydrogen burn which occurs following vessel breach. However, because the drywell failure location is assumed to be in the drywell head closure seal the burn is assumed to occur above the refueling floor (described in the discussion above). A burn in

! this location will not cause the fire sprays to come on.

Case # 3 :

This case covers sequences in which the fire sprays were operating prior to vessel breach or were activated subsequent to vessel breach either by the hot gasses released from the containment or by a subsequent hydrogen burn.

Question # 92 : Does standby gas treatment work late?

This question determines whether the Standby Gas Treatment System (SGTS) continues to operate throughout the accident.

Hydrogen burns in the containment (except if they occur above the refueling floor) will cause SGTS to isolate since the fusible . links in.the duct work will fail, resulting in the closure of the fire dampers.

This is a Type 2 question where branch point probabilities are

1 Pago 113 DRAFT Porch Bottc2 C.E.T. Docun ntation input and are dependent on branches taken in previous questions.

There are 2 branchest 1 . nLSGTS: The Standby Gas Treatment System (SGTS) does not operate during the latter porion of the accident following vessel breach.

2 . LSGTS: The Standby Gas Treatment System (SGTS) continues to operate throughout the accident.

Case i 1 :

In this case the SGTS was operating after vessel breach and no hydrogen burns occurred in the reactor building. The fire dampers will thus remain open and SGTS is certain to continue to operate.

Case # 2 :

This case covers sequence in which the SGTS has previously failed or fails due to combus'tible gas burning following vessel breach. .

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