ML20141G289

Exam Rept 50-461/OL-85-02 on 851021-25 & 28-31.Exam Results: All Candidates Passed Written Exam & 6 Operators & 13 Senior Operator Candidates Passed Oral/Operating Exam

ML20141G289
Person / Time
Site:	Clinton
Issue date:	01/05/1986
From:	Lang T, Mcmillen J, Plettner E NRC OFFICE OF INSPECTION & ENFORCEMENT (IE REGION III)
To:
Shared Package
ML20141G278	List: ML20141G274 ML20141G289
References
50-461-0L-85-02, 50-461-L-85-2, NUDOCS 8601100117
Download: ML20141G289 (71)

Text

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U.S. NUCLEAR REGULATORY COMMISSION REGION III Report No. 50/461/0L-85-02 Docket No. 50-461 License No.

Licensee: Illinois Power Company 500 South 27th Street Decatur, IL 62525 Facility Name: Clinton Power Station Examination Administered At: C11rt- Station Examination Conducted: October 21-25 and October 28-31, 1985 Examiner (s): Ln /fd e

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Dhte' Approved By: i ef perating Licensing Section

// [d 04te' Examination Summary Examination administered during the weeks of October 21 and 28, 1985 (Report No. 50-461/0L-85-02)

Examinations were administered to 14 senior operator candidates and 8 reactor operator candidates.

Results: All candidates passed the written examinations and six operators and 13 senior operator candidates passed the oral / operating examination.

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4 DETAILS

1. Examiners T. Lang, Region III E. Plettner Region III J. I. McMillen, Region III W. Cliff, PNL

2. Examination Review Meeting Copies of the examinations and answer keys were given to the facility personnel for review at the conclusion of the written examination.

Facility personnel gave their comments to the examiners on October 24, 1985. These comments are enclosed as Attachment 1 to this report.

Resolution of these comments is Attachment 2 to this report.

3. Exit Meeting On October 31, 1985 an exit meeting was held. The following personnel were present at this meeting:

R. F. Scheller, Director-Training, IP J. C. Wemlinger, Supervisor Training, IP J. W. Wilson, Clinton Plant Manager, IP F. A. Spangenbert, Manager-Licensing and Safety, IP J. S. Perry, Manager, Nuclear Program Coordination, IP E. Plettner, USNRC, Region III J. McMillen, USNRC, Region III The facility was informed of those persons who had clearly passed the oral / operating examinations.

Attachments:

1. Facility Comments Reactor Operator Examination

2. Resolution of Facility Comments 2

ATTACHMENT 1 FACILITY. COMMENTS i REACTOR OPERATOR EXAMINATION SECTION 1 COMMENTS 1.03 Add to answer key:

d. Equilibrium Samarium does not change as a function of power.

e. Time to reach equilibrium Samarium or equilibrium shutdown Samarium is long and therefore does not create power transients,

f. The core has sufficient excess reactivity to overcome the effects of Samarium.

Reference:

Standard Nuclear Principles 1.07 Should be d. (The CPS test bank is incorrect).

Reference:

Standard Fluid Mechanical Equation 1.12 Add to answer key:

d. Spontaneous fission of CM-242 and CM-244 in highly exposed fuel assemblies.

e. Neutrons liberated by cosmic ray interactions with nuclei.

Reference:

Standard Nuclear Principles 1.13 b. A rod withdrawal in a high power region can cause a large power increase in the adjacent fuel. This power may cause a PCIOMR and/or thermal limit violation. Answer which refer to PCIOMR or thermal limit violations should be considered correct.

Reference:

5tandard Nuclear Principles SECTION 2 COMMENTS 2.01 The key incorrectly gives the drywell pressure setpoint of 2 psig.

The drywell pressure setpoint is 1.68 psig, and the allowable value is 1.88 psig. These answers should be considered correct.

Reference:

Technical Specifications Table 3.3.2-2 l

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2.09 a. Although large centrifugal pumps should be started against a shutoff head, they should not be operated against a shutaff head to avoid damage from overheating. Therefore the answer should be true.

Reference:

Standard Operating Practices 2.10 This question may be interpreted two ways and the answer depends on the interpretation.

First: Note that the Service Water system (WS) normally supplies all Shutdown Service Water (SX) flow under non-accident conditions, and that under accident conditions or loss of WS flow the SX flowpath isolates from WS and the SX pumps provide component flow to a, b, and d. The normal source of flow to components c and e is the Component Cooling system (CC) which is backed up by SX. With this interpretation the correct answer is a, b, c, d, and e.

Second: The flowpath to the components a, b, d from the SX system does not change when flow shifts from WS to SX, but individual component valves operate on c and e to shift from CC to SX. This interpretation supports a correct answer of c and e.

It is recommended that either answer be accepted (i.e., a, b, c, d, e or c, e).

SECTION 3 COMMENTS 3.01 Answer (f) Under and Over Voltage This answer is not listed in the reference given on the key.

In the RTS lesson plan "RR" (Section RR4, C.4.b.3) the alarm is shown as Under Voltage /Over Current. This means Under Voltage or Over Current..

f. The following should be added to the answer key:

g. " Return line filter" high dp.

h. " Pressure filter" (discharge filter) high dp.

1. " Actuator Drain" (leakage greater than 0.5 gpm).

J. Pump motor stop.

k. Fan motor stop.

2

1. HPU shutdown.

Reference:

RTS Lesson Plan "RR" 3.03 b. RTS lesson plan " Generator and Auxiliaries", (Section G and A4, C.1.d.3) states that on a loss of stator water cooling the Turbine trips on a runback if:

(1) Generator load above 80% after 2 minutes.

(2) Generator load above 25% after 3.5 minutes.

These % load levels should be accepted as well as the corresponding current values now in the key.

Reference:

RTS Lesson Plan "G and A" 3.06 The last portion of the answer states " makeup with feed pumps" however, the question did not ask for response of the Feedwater system. Failure to discuss the Feedwater system should t.ot be considered a deficiency.

Change 60% turbine load to 65% in the key, and change 40% to 35% in the key. (35% is the Turbine Bypass Valve capacity).

Reference:

RTS Lesson Plans "EH" and Mt, GS, T0" 3.07 There is a typographical error on the key: "DMS Computer" should be "PMS" Computer."

Reference:

RTS Lesson Plan "IP" 3.09 b. The setpoint value was not asked for and should not be required for full credit.

3.11 Setpoints were not asked for, and failure to include them should not be a deficiency.

Level 2 or High Drywell Pressure is given in the off normal procedure 4001.01. The change to Level 1 is a recent setpoint change which has not yet been incorporated in all documents.

Reference:

CPS 10N4001-01 " Reactor Coolant Leakage" SECTIOP 4 COMMENTS 4.02 A drawing of the '.anel operating area (P0A) or list of control room panels comprising the POA should also be an acceptable answer.

Reference:

CPS 1401.05. " Main Control Room-Conduct of Personnel" 3

4.03 As per Clinton procedure master index dated October 1, 1985, this Reference Procedure (1401.06) has been cancelled. Therefore, the question should be disqualified or the answer given as per old procedure as stated on the key.

Reference:

Clinton Procedure Master Index 4.05 The immediate action taken as per 4008.01 " Loss of Reactor Coolant Flow" should also be an acceptable response.

Reference:

CPS 10N4008.01 " Loss of Reactor Coolant Flow" 4.07 The answer "a" is stated in the question so the student probably would not give it as a response. Either the two responses "b" and "c" should be given full credit or any reasonable response accepted as the third answer.

Reference:

-CPS 1024.15 Exposure Control 4.12 4401.01 Section 6 states that ECCS shall not be secured unless i misoperation is verified by two independent sources, adequate core cooling is assured, or directed to do so by the E0P. This should also be an acceptable answer.

Reference:

CPS 4401.01 " Level Control Emergency" l

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FACILITY COMMENTS

SENIOR OPERATOR EXAM i

SECTION 5 COMMENTS l

5.03 b. The key should also state that Alpha D will increase core delta K/K.

Reference:

Standard Nuclear Principles 5.08 Strictly speaking, the answer should be decrease even if only i slightly.

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Reference:

Standard Nuclear Principles l 5.10 a. "MAPLHGR" should be APLHGR limit"

Reference:

General Electric S.N.E. Manual l

SECTION 6 COMMENTS 6.01 (a) The answer incorrectly explains the response of the SSTS for the NSPS Bus. The correct answer is as follows:

l The SSTS will automatically transfer from the normal to the j alternate power source on a loss of normal power (undervoltage of 105 VAC). Restoring power to the normal source must be done manually.

Reference:

RTS Lesson Plan "IP" f 6.03 (a) The question did not ask for setpoints so the answer should not

< be considered deficient if they are not included.

6.04 (a) The answer is Group 7 not Group 6.

Reference:

RTS Lesson Plan "PC" l 6.05 a. The setpoints were not asked for so the answer should not de l deficient if not included. Also note that 40% power is the l Technical Specification value for the RPT-E0C trip bypass.

Reference:

Technical Specification 3.3.4.2 l

6.06 Although not given in the key, no action occurs for a loss of stator

cooling at 10% power.

The setpoints were not asked for so the answer should not be considered deficient if they are not included.

Reference:

RTS Lesson Plan "G and A" i

6.07 (c) The source of hood spray is from the condensate booster pump discharge, not from the cordensate pump discharge.

Reference:

RTS Lesson Plan "C0/FW" 6.08 (b) The trips listed in the key are correct for the Division 1 and Division 2 Diesel Generators. The Division 3 Diesel Generator trips that are still in effect are only Overspeed and Generator Differential Current.

Reference:

RTS Lesson Plan "DG" 6.10 A series of four (4) restricting orifices in the charging in the charging line also limits the runout flow of the pump to 200 GPM to prevent the pump from tripping on overcurrent. This is in addition to the design features listed in the key.

Either answer should be accepted.

Reference:

CPS Lesson Plan " Control Rod Hydraulics" SECTION 7 COMMENTS 7.01 Item "d" should be Recirculation loop " temperatures" not " flow."

7.03 Part "a" is correct but b, c, d, e, and f answers are for Primary Containment Integrity vice Drywell Integrity.

The answer should read:

b. All drywell equipment hatches are closed and sealed.

c. The drywell airlock is operable pursuant to Specification 3.6.2.3.

d. The drywell leakage rates are within the limits of Specification 3.6.2.2.

e. The suppression pool is operable pursuant to Specification 3.6.3.1.

f. The sealing mechanism associated with each drywell penetration; e.g., welds, bellows or 0 " rings," is operable.

Reference:

3020.01, Revision 1 7.05 According to 4402.01, Revision No. 3.

a. Suppression Pool greater than 95 F 2

r - - - - _ - - - - - - - - - - - - _ _

b. Drywell Temperature greater than 135 F

c. Suppression Pool level less than 18'11"

d. Suppression Pool level greater than 19'5"

e. Containment Temperature greater than 120

f. Drywell pressure greater than 1.68 psig.

It is possible to get either of the answers depending on which revision is used for answer key.

7.08 a. The terminology "SRV Acoustic Monitor Alarm" should be accepted for No. 2 in the key.

Reference:

CPS 10N4009.01 " Inadvertent Opening of SRV" SECTION 8 COMMENTS 8.02 a. This was deleted by the examiners. There is no low condenser vacuum scram.

b. The question is misleading. It is interpreted as asking for the basis of both the MSIV isolation closure and the scram.

CPS Technical Specification Pages B2-8 and B3/4 4-5 states that the low steam line pressure isolation limits the^ amount of fission product release for certain postulated events. It states that the MSIV closure scram anticipates the pressure and flux transients which could follow MSIV closure and thereby protects the reactor vessel pressure and fuel thermal / hydraulic safety limits.

c. The Technical Specifications basis (Page B2-9) does not assume a subsequent failure of the BPVs so this should be deleted from the answer.

Reference:

Technical Specification, Pages B2-8 B3/4.4-5, B2-9) 8.03 If an operator is qualified in an area, he is by definition Auxiliary Operator qualified or above. Therefore the key should accept either " qualified in the area" or " Auxiliary Operator qualified in the area."

Reference:

CPS 1402.04 " Operations Department Watch Standing Organization and Qualifications.

8.06 a. The answer in the key is correct per CPS 1401.02 however, the Control Room Supervisor position has been deleted. There are now 3 positions, (not 4) who can limit the number of persons in 3

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'the control room. They are Shift Supervisor, Assistant Shift

~ Supervisor, and Control Room Operator. The examiner should accept either these 3 positions or the answer given in the key,

Reference:

CPS 1401.01 " Operation Department Organizations, Responsibility and Minimum Qualifications

b. The examiner should also accept a diagram of the Panel Operating Area (P0A) or a list of panels making up the P0A.

Reference:

CPS 1401.05 " Main Control Room-Conduct of Personnel.

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ATTACMENT 2 RESOLUTION OF FACILITY COMENTS 1.03 The suggested answers are general statements that express the answers given in the key. No change was made to the answer key and credit was given for answers that indicated an understanding of the concept.

1.07 Comment accepted, answer key corrected.

1.12 Coment accepted, answer key corrected.

1.13.b. Answer key already contains reference to thermal limits. No change necessary. Credit given for answers that demonstrate understanding of concept.

2.01 Comment accepted.

2.09 Disagree with coment Question says pumps should never be operated against a shutoff head. Since they can be started against a shutoff head the answer is false.

2.10 No reference was given therefore question was graded per the answer key. Only three candidates received less than full credit and the result of the examination would not be affected by the proposed answer. Question will be clarified if used in the future.

3.01 Coments accepted. Answer f was changed. The other suggested answers were used during grading but were not added to the answer key.

3.03 Coment accepted. Credit given for either answer.

3.06 Correction to answer key concerning turbine load and steam flow accepted. Disagree with the statement concerning " makeup with feed pumps". The question asks for final steady state condition which implies a reans of makeup to the reactor.

3.07 Comment accepted.

3.09.b. Coment accepted.

3.11 Coment accepted.

4.02 Coment accepted.

4.03 Question graded per answer key. Deleting question would not change overall results of examination.

4.05 Disagree. Candidate must list the imediate action steps.

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4.07 . Comment accepted. Point value redistribu'ted.

4.12: Comment ' accepted.: Question graded according to suggested answer. Key will be changed in question bank.

5.03 Comment accepted..

5.08 Reviewers opinion. Credit was given for answers that demonstrated a.

knowledge of.the concepts' involved. In particular that core orificing -

-is an important feature of plant design.

'5.10.a.

Connent" accepted.

6.01 '. Comment accepted.

6.03- Comment accepted.

,6.04.a. Coment . accepted.

.6.05.a. Comment accepted.

6.06 . Disagree. Question requests initiation signals which implies setpoints.

6.07.c . Comment accepted.

'6.08.b. Connent accepted.- If candidate specified the Division, full credit was granted.

6.10 Comment accepted. Credit given for either inswer.

7.01 Comment' accepted.

7.03- Examiner decided question was confusing and therefore deleted it.

Point value for section~ reduced.

7.05 Comments accepted.

7.08 a. Comment accepted.

8.02- a. -Question' deleted. Point value for section reduced.

b. Question'was apparently clear to' candidates. All but three received full credit.

c. Coment accepted.

8.03 Connent accepted.

8.06 . a. and b. Connent accepted.

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MASTER COPY U. S. NUCLEAR REGULATORY COMMISSION REACTOR OPERATOR LICENSE EXAMINATION FACILITY: clinton REACTOR TYPE: nun-cr 6 DATE ADMINISTERED: ncenher 21.1985 EXAMINER: v_ plettner APPLICANT:

INSTRUCTIONS TO APPLICANT:

U3 ceparate paper for the answers. Write answers on one side only. Staple question ch::t on top of the answer sheets. Points for each question are indicated in parzntheses after the question. The passing grade requires at least 70% in each catcgory and a final grade of at least 80%.

% of Cat gory  % of Applicant's Category V'lue Total Score Value 25 95 1. Principles of Nuclear Power Plant Operations, Thermodynamics, Heat Transfer and Fluid Flow 25 25 2. Plant Design Including Safety and Emergency Systems 25 2s 3. Instruments and Controls 25 . 25 4. Procedures - Normal, Abnormal.

Emergency and Radiological Control 100 100 TOTALS Final Grade  %

All work done on this exam is my own, I have neither given nor received aid.

Applicant's Signature

cAIEeggy_1 PRINCIPLES OF NUCLEAR POWER PLANT OPERATION, THERMODYNAMICS HEAT TRANSFER AND FLUID FLOW 1.01 Explain or define the following terms:

a. Delayed neutron fraction Beta (0.5)

b. Reactor period (0.5)

.c. Subcritical multiplication (0.5) 1.02 Describe the problems and hazards involved in starting (1.0) up a reactor without an adagate neutron source.

,1.03 Give three reasons why samarium is not considered a (1.5) problem during reactor operations.

1.04 Refering to the Boiling Heat Transfer curve Fig 1.0 match the region marked "A" through "E" on the curve with the appropriate made of heat transfer below:

1. Nucleate boiling (0.5)

• 2. Subcooled boiling (0.5)

3. Film boiling '

(0.5)

4. Transition boiling (0.5)

5. Forced convection (single phase flow) (0.5) 1.05 What is the purpose of:

a. Deep rodc (two required) (3.0)

b. Shallow rods (one required) (0.5) 1

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1.06 The tabulation below illustrates REACTIVITY COEFFICIENT VARIATIONG due to increases in several core parameters. For each condition (a-f) listed below, INDICATE how the VALUE of that coefficient varies if the indicated parameters are INCREASED. (Answer with more negative, unchanged, or less negative). (3.0)

CORE PARAMETER MODERATOR CORE FUEL CORE TEMP VOIDING TEMP AGE COEFFICIENT VOID COEFFICIENT (A) (D)

MODERATOR TEMP

. COEFFICIENT (C) (D)

FUEL TEMPERATURE COEFFICIENT (E) (F) 1.07 A motor driven centrifugal pump is operating at rated flow. (1.0)

You' start closing down the discharge valve. Which of the following statements best describes the parameter changes that will occur with this action?

a. Flow remains constant, discharge pressure remains constant, motor amps increase, net positive suction head increases.

b. Flow decreases, discharge pressure increases motor anps increases, not positive suction head increases.

c. Flow decreases discharge pressure increases motor amps decreases, net positive suction head decreases.

d. Flow decreases, discharge pressure increases, motor empt decreases; net positive suction head increases.

1.00 Boiling w+ter reactors are designed to have "under moderated (1.0) cores". Which statement bgst describes under moderated?

a. The ratio of moderator to fuel is such that the temperature and void coufficient will both be the same (both positive or both negative).

b. The ratio of moderator / fuel is such that increasing moderator density increaues K eff.

c. The ratio of moderator t fuel is such that the amount of under moderation increases during care life.

d. The ratio of fuel to moder ator is such thal increasing ,

moderator density will decrease M eff.

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.1.09 The reactor is on a 75 second period. Moderator temperature is (2.0) 153 degrees F. With NO OPERATOR ACTION, what will the moderetor temperature be when the reactor it again on an infinite period.

Note 1. Assume the moderator temperature coefficient is -1.5 times 10 to the minus forth power. delta !. per k per degree F. Use beta =.007 and lamda = .100

2. Show all work.

1.10 Which of the thermal limits protects the fuel from clad (1.0) rupture due to plastic deformation?

a. MAFLHGR b. MCPR c. LHGR d. DEA LOCA 1.11 What is water hammer and how does it affect a piping systen? (1.0) 1.12 What are three (3) sources of neutronc other than installed (3.0) sources when the reactor is shutdown? (Briefly describe each) 1.13 The reactor is started up after a refueling outage. Rods are pulled to the 100% line and power is then increased to 100% with recirculation flow. After approximately 20 hours2.314815e-4 days <br />0.00556 hours <br />3.306878e-5 weeks <br />7.61e-6 months <br />, reactor power has decreased to about 98% with no operatior action.

a. Whct is the primary cause for this reduction in power? (0.5)

b. Driefly enplan why control rod withdrawals are not (1.0) recommended at a high power levels.

1.14 Concerning the core thermal limits:

a. For each condition (1-4) given below , Indicate whethcr it will cause and INCREASE, DECREASE,or have NO EFFECT on the CRITICAL POWER RATIO.

1. Local peal:ing f actor INGREAggg. (0.5)

2. DEGRE@@E in inlet subcooling (0.5)

3. Dj[RE6SE in reactor pressure (0. 5)

4. Axial power peak shifts from Bgttgm to Tgg of channel (0.5) 3 f

b. .With regard to MAPRAT:

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1. What is the. relationship between MAPRAT and MAPLHGR? (0.5)

2. Is a MAPRAT of 1.05 acceptable? (.25)

3. What physical consequences could occti- if the MAPRAT (.75) limit is exceeded?

End of Category 1 4

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CAIEEOBY_2 PLANT DESIGN INCLUDING SAFETY AND EMERGENCY SYSTEMS 2.01 What sin signals will automatically start the standby gas (3.0) treatment system.

2.02 With regard to the Low Pressure Core Spray (LPCS) System:

a. INDICATE whether the following statements are True or False:

1. When a manual override is performed on a pump or valve, automatic pump restart or valve opening is disabled until the indications are reset. (0.5)

2. Manually starting the LPCS pump automatically establishes adequate Standby Service Water flow path. (0.5)

b. On en AUTO initiation LIST 4 of the 5 actions that occur in the LPCS system. (1.0) 2.03 The ADS system has two-position control switches P601 and P642. (3.0)

a. What are the three positions on the switch and briefly explain what occurs in the system for each of the positions.

2.04 a. Explain how the Master Manual controller in the Recirc (1.0)

Flow control system pcrforms it intended funtion.

(Hint the inputs and outputs used in Auto) b' . What happens when the flux controller is shifted to manual? (0.5) 2.05 What two types of air are used to operate the flow control (1.0) val ve in the Control Rod Hydraulic System?

2.06 An automatic RCIC initiation has occurred. Subsequently, RCIC injection was automatically terminated due to high reactor water level,

a. What component in the RCIC system f unctioned to automatically

-terminate the injection? (0.5)

b. Assuming no operator action, how will RCIC respond to a subsequent decreasing water level? (1.0)

c. If a RCIC " Turbine Test" had been in progress when the initial automatic initiation signal had been received, how would the system have responded? (1.0)

d. If, following the initiation, the RCIC turbine had tripped cn overspeed, could it be reset from the Control Room?

Enplain (1.0) 5

-2.07 What five indications should the operator use to verify that Standby liquid control is operating? (2.5) 2.08 List five automatic signals that will cause a Group 6 (RCIC) (2.5) i sol ati on to occur.

2.09-Concerning the HPCS system: True or False

a. The HPCS pump shall Jgggy: be operated against a shutoff (1.0) head.

b. Stopping the HPCS pump with an initiation signal prevents (1.0) starting the pump until the signel is reset.

c. HPCS injection val ve F004 adte closes when level 8 in (1.0) reached and reopens at level 3.

d. The HFCS pump shall be stopp$d under all conditions when (1.0) room, winding, or bearing temp high conditions are reached.

2.10 The shutdown service water system is a source of backup cooling (1.0) for which of the following components / systems?

a. Switchgear heat removal condensing unit

b. Contrcl room HVAC chiller

c. Fuel pool cooling system heat exchanger

d. Diesel generato heat exchangers

e. Reacto- recirc pump seal and motor bearings hear exchanger

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2.11 What are two (2), reasons for having main steam line flow (1.0) restrictors? s End of Category 2 l4 d

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CATEGORY 3 INSTRUMENTS AND CONTROLS 3.01 What alarms and setpoints if applicable are associated with the (3.0)

HPU units in the Recirc Flow Control System (si:. required) 3.02 a. Explain briefly the principles of the operation of an SRM (2.0)

(f i ssi on chamber). You may want to draw a figure to help clarify your explanation.

b. How are gammas compensated for in an SRM or are they? (1.0) 3.03 a. What two conditions will cause an automatic main turbine (1.0) runback? (include setpoints)

6. Two time delay relays are associated with the main turbine (1.0) runback. What conditions, if NOT met, will cause these time delay relays to send a trip signal to the turbine?

3.04 What are two functions of the Traversing In Core Probe System? (2.0':

3.05 a. Why does the Red Pattern Controller require control rods to (1.0) be pulled in a " checker board pattern"?

b. With regard to the Rod Pattern Controller, what process (0.5) is.used to determine the low power setpoint?

3.06 Describe how the EHC pressure control and logic system would (3.0$

respond if while operating at 100% power, the Load Limit slowly failed to nero. Take your discussion to a final steady state condition (i . e. Reactor scr am) Assume no operator action and all systems are in normal full power lineup and Retire flow control is in master manual.

3.07 What are four loads supplied by the UPS Inverter Bus "A"? (2.0; 3.08 What are the three modes of operation of the safety / relief (1.5',

valves?

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3.09 One of the flow units associated with the APRM's is inoperative.

a. What automatic action will result from the INOP signal? (0.5)

b. Other than an INOP what signal associated with the flow (0.5) units will cause the same automatic action to occur as in part "a" above.

c. What are two conditions which will cause an INOP trip of a (0.5) unit? s 3.10 Assume the FEEDWATER LEVEL CONTROL SYSTEM is being operated in 3-ELEMENT control using reactor LEVEL DETECTOR CHANNEL 'A'.

Reactor power is at 85%, STEADY STATE. For each of the instrument or control signal failures listed below, STATE HOW REACTOR LEVEL WILL INITIALLY hESPOND (i n cr ease , decr ease , or remains constant) and BRIEFLY EXPLAIN WHY in terms of WHAT is happening in the Feedwater Control System IMMEDIATELY AFTER THE FAILURE.

(FOR EXAMFLE, your answers should include the following detail, "Causes reactor. level to decrease due to a steam flow / feed flow error signal, steam flow _ feed flow, resulting in a signal to increase the speed of the reactor feedpumps(s),"

IF APPLICABLE.)

a. B Steam Loop FLOW signal fails HIGH (1.0)

b. Channel A REACTOR LEVEL detector signal fails High (1.0)

c. LOSS OF CONTROL SIGNAL to " A" Reactor Feed Pump Speed (1.0)

Controller 3.11 What will cause inboard and outboard valves to auto i sol at e (1.0) in the Instrument air system? (2 required) 3.12 List three of the s i :' automatic trips on the service air (1.5) compressor. Setpoints not required.

End of Category 7 9

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• CATEGORY 4 PROCEDURES - NORMAL, ABNORMAL EMERGENCY AND RADIOLOGICAL CONTROL 4.01 All~drywell' penetrations required to be closed during accidents (2.0) must meet certain criteria. What are these two criteria?

4.02 What does "at the controls" mean in reference to the control (0.5; room operator's responcibility during normal plant operation?

4.03 It is permissible for a person in training to manipulate the (1.0) controls of the reector. Under what two circumstances shall the licensed Reactor Operator assume control of the reactor from the trainee?

4.04 What 6 actions should an operator perform when he receives (3.0) notification of.a fire?

4.05 What are you required to do on the trip of a recirculation pump? (3.0)

(Assume.locp is not to be i sol ated )

4.06 What are four immediate actions you would perform when the (2.0) control room must be evacuated?

4.07 NRC dose limits state that a whole body exposure of up to 3 (3.0)

REM per quarter is permitted provided three criteria are meet. What are these three criteria?

4.08 Individuals at Clinton Power Station sign a Blanket Radiation (0.5) work permit.(BWRP) How long is the BWRP in effect?

4.09 Match the following to its dose rate per hour: (1.5)

~i. High Radiation area a. 500 mrem

2. Restricted High Radiation area b. 100 mrem

3. Radiation area c. 50 mrem

d. 1000 mrem e 5 mrem

f. 10 mrem 9

4.10 What are the thrte immediate operator actions when a high alarm /

fail alarm is received on any criticality monitor? (1.5)

Note: Ac knowl edge the al arm is not one of the answers.

'4.11 a. What automatic action should occur when the setpoint of (0.5) 1 ppm i s reached on the chlorine detector / monitor?

b. What are the two possible responces of Main Control Room (1.0)

Personnel.

4.12 What criteria shall be satisfied before you can stop HPCS (2.0) injection flow following automatic initiation?

4.13 Inadver tent opening of one SRV will result in a relatively small plant transient that should not cause a rapid deterioration of conditions; however, timely actions are required.

a. In your own word, describe Four symptoms you would see. (1.0)

b. What are the immediate operator actionc for this condition? (2.5)

End of Category 4 L

1 10

Ti1Ma 2. Prspxrtiss et S:turated Steam cnd Snur ted htu(Pressure)

.)

Jese. Temp. Volume, It3 /lbm Enthalpy, Bru/lbm Entropy, Bru/lbm x R Energy, Bru/lba hde F Nater Evap. Steam Water . Evap. Steam Water Evap. Ste2m Water Steam

's 'fe 's i le E

e 8g 8 #

te 8 I 8 S.0 sel.se e.01e39 2.26es 2.2e13 355.5 442.s

. e.0 :et.96 s.01838 2.2912 2.3095 119e.3 e.5434 1.0016 1.545e ase.s 1113 7 .*

354.6 e=3.6 119e.2 - 0 5=2e 1 0035 1 5463 1113*'

53.e 3ee.12 e. ele 36 2.3139 2 3322 253.7

..e 379.26 e. ele 35 2.3370 2.3554 e==.4 1198 1 e.5417 1.e054 1 5471 353 's 1113 5 353.

20 370.%e e. ele 34 2.3606 2.3790 352.s e=5 1 1197 9 e.5406 1 007e 1 54ee 352 1 1113 4 251 9 495.9 1197.e .e.5395 1.ee94 1 5949 351 2 1113 2 X 0.0 377.53 4. ele 33 2.se47 .

2.tese 250.9 au.7 1197 6 -

344.3 37%.65 e.01832 2.4093 2.4276 e.53e* 1.8113 1 549e 35e.3 1113 1 350.o a=7 5 1197.5 s.5273 1.0133 1.5507 349.= 1113.s Be6 0 375.17 e. ele 31 2.4394 2.4527 249.1 see.3 1197 3 3C4.9 37*.e4 e.close 2.460s "2.47e3 e.5262 1 0153 1 5516 ses.4 1112 9 3es.1 649 1 1197.2 e.5351 1.e174 1.5525 347.5 1112.s X2.0 373.96 e. ele 2e 2.4862 2 5045 247 2 449.9 1197.e e.5339 1.e194 1 553, 346 5 1812J seo.S 373.04 e.01e27 2.5129 2.5312 246 2 850.7 1196 9 110.0 372.n e.ote2s 2.5=92 2.55e5 e.532s 3.0215 1 5543

• 30 6 us2.5 173.o 371.24 e.01e25 2.56e1 2.5464 24 % 2 e51 5 1196 7 e.5316 1.o226 1 5552 2.=.6 m24 244 2 852 3 1156 5 e.53e5 1.0257 1 5542 343 6 1112 3 174.0 370.31 e.01e24 2.5964 2 4149 243 2 e53 1 1196 4 369*37 e.01823 a.5293 1.e279 1.5511 3n2 7 1112 2 172.0 2.625e 2 6tte 242 2 453 9 1196.2 s.52e1 1.0200 1 5541 391 7 3133**

370.0 368.42 8. ele 21 2.6556 2.6738 2gt.2 454.s 115.0 367.g7 s.elete 2.6e61 2.7093 240 2 1196.e e.5269 1.e322 1 5591 340 7 1111.9 e55.6 1195.s e.5256 1 0244 1 5601 339*7 1111*8 166.0 366 50 e.01e19 2.7173 2 7355 229 2 45o.5 1895 7 8.5244 15b.0 345.53 e.01018 2.7493 2.7674 1 0367 1.5611 230 6 1111 6 338 2 857 3 1195 5 e.5232 1 0389 1 5621 337 6 1111.5 132.3 364.5g e.elet? 2.7820 2.e001 227 1 e58 2 1195 3 e.5219 1.0412 1 5631 336 6 1111 3 310.0 3e3.55 0 01815 2.e155 2 8336 336.1 e59 0 1195.1 e.5206 158 9 362 55 0 01e19 2.849e 2 8679 235 0 1 5935' 1 5641 335 5 1111 2 859.9 1194.9 0.5194 1 0453 1 5652 329 5 1111.e 356.e 361.53 e.01813 2.8849 2.9031 233.9 860.e 1194.7 a.5181 1.04e2 154 0 340 51 e*01e12 2.921e 2 9391 332 8 1 5662 333** 1118*'

e61 6 1194 5 a.5168 1 0506 1 541 332.3 1110.7 152.s 359.46 e.01818 2.9579 2.9760 , 231 8 e62 5 1194 3 c.5154 1 56e,3 1 053e 331 2 111e.6 150.e 3*e.43 e. ele 09 2.9958 3.e139 230 6 843 4 1194.1 0 5141 1.e554 338 1 ille.*

gg3.g 357.91 e.0180s 3.0151 3 0322 1 5695 330 1 e63.1 1194.s e.5134 1 0566 1 570s 229.6 1118*3 388 3 357.24 e.01004 3.0347 3 0528 229 5 364 3 1193.1 s.5127 Ag7.5 356.0g e.01007 3.0545 3 0726 229 0 1.0579 1 5706 229.0 111e*3 864 0 1193.e 0 5120 1.e591 1.5712 32a.5 1118 2 146.8 356.31 e.eleOS

• 3.0746 3 8927 22s.4 e65.2 8193.6 e.5114- 327,9 113e.1 1.e604 1 5717 115.e 355.77 e*Ste06 3.0950 3.1130 227.8 e65 7 1193.5 e.5107

' 1%.O 355.23 e.01805 3.1156 3.3337 227 3 1.eet6 1 5723 327.4 111L*8 en.2 1193.4 e.510e 1 0629 1 5729 226.8 11M*9 133.e 35g.69 e.01e05 3.1365 3.1546 326 7 au.6 1193.3 22s.2 11e9 8 -

354.1% e.01004 3.1577 3 1757 e.5e93 1 0642 1.573e 142.s 326 1 e67.1 '1193 2 e.See6 1.0655 1.57:se 325 6 1189 7 141.0 353 59 e.stt03 3.1792 3 1972 225 5 e67 5 1193.1 0.5e79 325.1 11e9 7 1.8664 1 5796 110.0 353.e, e. ele 03 3.2010 3 2150 225.0 see.e 1193.e e.5e71 1. Seat 324 5 Hes.4 339.3 352.te e.01802 3.223e 3 2411 1 5752 324.4 e6e.5 1142.4 e.Sett 1.8699 223 9 1189 5 130.0 351.92 e.e1001 3.2454 3 2639 223.5 e63.9 1192.7 31 575,e e.818e1 3.26e1 8.5857 1.e707 576 323.3 1109*4 137.0 351 26 3 2361 223 2 869.4 1192 6 e.5050 1.0720 1.577e 222 7 1189 3 33%.3 350.75 0 01000 3.2912 3 3091 222 6 e69.9 1192 5 e.5e03 1 0733 1 5776 222.1 1109 2 135.s 350 23 0.01799 3.3145 3.3325 322.0 878.9 1192.4 8.5835 1.8707 221 5 1109 1 124.0 349.65 9.e1799 3.3382 3.3562 1.5742 321.4 870.3 1192 2 e.5028 1.0760 1 5733 320.9 1109.e 133,e 349.0e e.0179e 3.3622 3.3802 320.e 871 3 1192.1 0 5e20 1.e774 1 579, 320.3 110s.9 132 0 348 50 8.e1797 3.3466 3 4046 320 2 571 8 1492.8

! 131.0 397.92 e*01797 3.4113 3.9293 e.5e13 1.8788 1 1800 319.7 1198 8 219.6 872 3 1191.9 e.5005 1 0801 3 5007 31'*1 1188*#

130.0

• 347.33 0 01796 3.4364 3.4544 219.0 372 8 1191 7 e.e994 1 0e15 129.0

• 346.7% e*e1795 3.4619 3.4799 1.5813 318*5 M 88 '

210.3 37J.3 1191 6 e.4990 1.e829 1 5019 317.9 1100 5 12a.e 346.15 0.01794 3.4878 3.5057 317 7 873.0 1191 5 8.49e2 127.0 3m5 55 e.81794 3.5141 1.ee=3 1.5e26 317 3 11ee.t 3.5320 217 1 379 3 1191 3 e.gt75 1.0e5e 1 5822 316 7 1100 3 325.s 3e=.95 e.01793 3.5407 3.55e6 316.9 374.s 1191 2 e.t967 1 8472 1 5e39 316.e 1108 2 125.0 3*=.35 e.01792 3.5678 3 5e57 215.e 875.3 1191 1 e.4959 1.eee6 315 4 1108 1 3g3.7e 3.5953 1.5845 124.0 e.e1792 3.6132 315 2 375 0 1190 9 8.9951 1.0901 1.5e!2 31**8 II'8**

123.9 393 13 e.01791 3.6232 3.6411 314.5 876 3 1190 8 e.4943 1 0915 1 5858 31".1 1107 9 122.0 392 11 0 01790 3.6516 3.u15 313.9 876 0 1890.7 e.4935 1 8930 1.5865 313 5 1137.e 121.9 341.89 .e.81798 3.6e09 3 6983 213 2 e77 3 1190.5 1.0945 312.s 1147 7 8.4927 1 5e72 320.0 391.27 e.81739 3.7097 3.7275 312 6 SM.e 1890.4 212 2 1187.6 111.0 290.6% e.017ee 3.1399 e.4919 1.H60 1 5879 3.7573 211 9 373 3 1190.2 e.4911 1 0175 1 5835 311*5 1187*S

• 113.3 240.01 e.01787 3.7697 3.7375 311 3 378.0 1890.1 0.4903 1.e690 318*9 1187**

117.0 329 37 c.01737 3. ele 3 1.5992 3.5004 310.6 879 3 1149.9 8.ne94 1 1005 1.5899 310 2 1107 3 116.0 328.13 e*eDe6 3.8316 3 8495 209.9 479.9 1189.8 8.4886 1 10n 1 5906 3M.5 11 6 I 115.0 328 00 0 01785 3.0634 3 8813 209 3 800.4 11e9 6 e. set? 1 1036 1 5913 308 9 1187.e

! 114.9 331.41 e.017e5 3.8957 3.9136 200 4 800.9 11e9.5 e.te69 1.1052 1.5121 30s.2 1106.9 i

112.0 326.78 e.017et 3.9206 3 9444 207.9 801.4 11e9.3 0.4460 1 1067 1.5924 307.5 1106 8 i

112.0 326.12 0 01783 3.9620 3 9795 207.2 se2.0 1149.2 1 5915 3e6.8 1186 7 111 0 335.66 e.4452 1 1083

, 0 01752 3.9960 4.0138 206 5 332 5 1149.e 0 4443 1 1099 1 59*2 30* a II88*'

110 0 324.19 e.01722 4.0306 4.0tet 205.8 833.1 1168.9 e. gest 1 1115 1.595e 205.4 He6.5 i

129.0 314.11 e.01781 4.065e 4.0837 205 1 se3.6 1148.7 204.7 1194 3 10e.e e.01780 e.4e26 1.1132 1 5957 333.** 4.1017 4 1195 209.4 see.1 11ee.5 e.ge17 1.11ge 1 59e5 20s.O 1106 2 107.0 322.71 e.01779 4.1382 4.1510 203 7 444.7 Itge.4 203 3 I m *1 r 109 0 322.C6

e. gest 1 1165 1.gg73 e.0!M9 4.1753 4.1931 203 0 ses.2 11ee.2 e.9799 1.1101 1 19e0 382 6 11e6.0

, m.0 m.n 4.n 32 4.uG9 202a en.e e.o, 1.u9e m9 nao u.. e uo.. .e.e 0 mme, 4a5n 4.u15 m .5 est., site.,.

im. e.,7,0. 1. m 5

1. Sus 1.Su5 2a a 1 m,.e m.u .0me 4a9 0 4 30 o 200 0 .. 9 11 o .7 8.on 1.i m 1.e m m. 1;; M,-

a .0 m.a m .o in.e m .5, e.e m 6 e.e m 5 4.u!,

4.m 4.3* 7 4.m5 ma so.5 m .1 1 m .5 ima e.o62 s.. m 1.u 9 11m

. .o i t

1. 0i9 299.7MSly-219a i m m.o ... m . 4.4 n. e.. tie m .5 . . . . ima .. 3 1.n e. 1.an m.= ne3a

~'

Tr.ble 2. Pr:psrties cf Satursted Steam and Stturatzd Wetsr (Pressura)

Fress. Temp. Volume, k'/mm Er.thalpy, Btu /lbm Entmpy,8tu/Em x R Energy, Btu /lba pdn F Water Evap. Steam Water Evap. Steam Water Evap. Steam Water Steam

'I fe 's f fe e 8g a,g s, ag s,

,see.e 6,t.sa . .so7, .ut-e ',

17ea.c 41% *7 e.e2463 e.1973e

..u. 1 4,s., - se3.e 11 ,2.3 e.ee , e. 66, 1 38,, 4. i.79.s e.22897 646 1 5e7.5 1153 6 173e.0 617.9e s.eatst e*.aeese o.a253e e.e395 e.=183 1 209e ese.e lose.s 643 7 511.a 11w.9 17te.o 616.33 e. saws e*.as**a e.22ee7 441 3 514.9 3156.a e.esio e.47e* 1 311e 435 7 lost.s 172s.o 414 73 e. eau 7 e".aeost e.232*3 63e.9 s. essa e.47es 1 2137 433.e seea.s Ste.5 1157.4 e.e331 e.este 1 3157 631 1 1843**

1793.e 413.13 e.satze e*.2tt7e e.ases?

14es.e 611 51 e.e242e e.2155e e.2397e 636 5 saa.a 1154 6 e.ases e. net? 1.3174 eae.e 1844 4 ees.e7 63=.e sas.e 1:39.8 e. east 1610.0 e.ea=11 e.21945 e.2=3S7 431 6 549 5 e.49e9 1.3194 436 5 1885 3 164e.0 ese.aa e.82*e3 e"32341 a.2=74*

1161.s e.e265 8.4954 1 32ts tae.a tese.a eas.1 533.1 1162.2 e.sans e.4992

.1329 e 6e6.55 e.sa39s e"2274s e.2514e 626.7 536.1 1 3233 eat.e see7 1 1:43 4 e.sa21 e.5834 1.32Ss e19 5 tees.e Stie.e es*.e7 e.ea3e7 e'.23tS9 e.255*5 424.a 544.3 35e0.0 6es.17 e.e237e s*.23Se t 116m.5 a.st99 e.se76 1.327e 437 1 sees.9 e.25960 e21 7 ses.9 11es.7 136e.e ' 681 45 e.e2379 e*.atete e.263a4 6t9.2 547 6 e.e176 e.51ts 1 32 % 414.e 1889 8 13se.c 599.72 e.easta e*.24*S6 1166.e s.eass e.stee 1.231. saa.* 1898 4 e.26ste e16.7 SSA.a 1167.9 1330.e s97.97 e.e2354 e*.24909 e.27263 614 3 554.s s.st31 e.sae3 1.3333 ele.e 1891.5 1169.e o. esse e.5245 1 33s3 6e7.e 1e92 3 1tos.o S96 2e e.e2346 o*.25372 e.27719 611 1 SSe.g 117e.1 14ee.e S99.41 . e.4233e s*.2S897 e.28te6 e. sees e.52ee 1 3373 6es.2 1993 1 tus.o 392.61 e.02331 6cs.1 542.s 117:.2 e.seet e.5332 3 3393 eaa.7 1099.8 e'.26334 e.28665 set.6 565.6 1172.2 atte.c 59e.7e e.e2323 e*.26833 e.29156 eeg.e e.sess e.5375 1 3=t3 ese.3 Sete.e Sea.53 569.a 3173.3 e.se e.se19 1 3e23 5.9 7. e 1895 6 a:20.e e.sa315 o'.37345 e.2966e not.g 37a.9 117g.3 e.79 ,te e e.Se63 1.3es3 595.3 1e96 3 Stee.o 547.07 e.caser e".27e71 e.3017e 59e.e 576 5 13se.e ses.te e.e23ee e*.2stle 1113.3 e.1966 s.sse7 1.3et, 39e.9 se97 1 e.3e71e S96 2 see.1 1176.3 1360.0 ses.28 e.02292 e*.2e965 e.31256 e.79ea s.5552 1.3499 598 3 1897.9 553.6 ses.1 1117.3 e.791e e.SS97 13te.e set.as e.oraes e*.29539 e.3 tate 590.9 ser.e 11?e.3 1.351s se7.e 1898 6 132s.o 579.te e.82277 e*.30119 s.32394 e.7e93 e.5692 1.3535 ses.3 1999.*

See.3 591.s 1179 3 e.7ste s.56es 2.3556 56a.7 31ee.1 1500.0 577.42 . e.e2269 e'.3e722 e.32991 58s.6 594.6 11ee.2-129s.0 576.43 a.e2265 e13te29 e.33295 e.7ses e.5733 3 3sn see.1 11e8.9 .

see.2 596 5 stee.7 8.7sst e.5757 1 35e7 57s.e 31st.a late.e 575.nz e.e2262 e*.31341 e.336e 3 se2.9 59e.3 ma7s.e 57*.n2 e.saats s*.316Se o.33916 5e1.5 tiet.a e.78te e.57ee 1.359e 377 3 11e1 6 late.0 57J.40 a.e2259 6ee.1 stat.4 e.7ees e. sees 2 36es 576.a 11e1 9 e'.319ee e.34234 See.2 6e1.9 11e2 1 e.7793 e.5826 1 3619 574 9 Ste2*3 1253.s 572 38 e.e225e o*.323e6 e.34S56 57s.e

• saws.o 371.36 e.02247 e*.32637 e.3teeg ses.e 11e2 6 e.77ee e.Sese 1.363e 573 6 Ste2 6 577.4 6e5 6 3:e3.o e.7767 e.5e7=

Aase.o sie.Ja e.e22*3 e'.32973 e.35216 576.e 1 36ee 572.3 anos.e 122s.3 569.2e e.e2239 e'.33314 e.3SSS4 57e.6 687.4._11e3 5 e.175* e.5e97 1 3651 57s.9 3183.3 1a1s.0 See.2m e.e2236 609.3 1:ss.9 e.77e1 e.5921 1 3s62 369 6 1183 7 e*33661 e.3Se97 573.a 611 1 11ee.* e.772e e.5995 1 3673 ses.a Stes.e 1200 0 567.19 e.e2232 e'.34e13 e.36245 571.9 613.e 1153.0 546 13 e.8222e stee.t e.77tt e.5969 1 36es . 566.9 38e9.3 e*.34371 e.36S99 570.5 Alte.e 565.06 a.e2225 e*.3473* e.36958 569.e ele.e ^ 11es.3 e.77e1 e.5993 1 3694 565.s Stes.7 117s.e 563.99 e.e2221 616 6 11e5.7 e.76es e.6et7 1.370s 564 2 stes.e

' e*.35103 0 37324 567.6 ele.S Ste6 1 Alte.e S62 91 e.e2217 e*.35478 e.3745S e.7670 e.6092 1 3786 342.s lle%3 ,

566 2 620 3 11e6.6 e.1461 e.6e66 1 3727 3e1.4 1185.6 1159.4 541.82 e.e2214 e".35459 8 3ee73 564.s 622 2 site.o

• 560.7J e.0221e e*.36247 1137.0 ' e.76e7 e.6091 1.3734 540 1 Stes.e 9 384S7 563.3 624 1 11*e.e S*S.63 0.022e6 4136641 e.3e697 Ste7.4 s.7634 s.4115 1.3799 SSe.7 18e6 3 1120.0 S*8.52 S.822e3 561.9 62S.9 ste7.8 s.742e e.elte 1376e 557 3 11e6 6 e'.37e41 0.39244 See.S 627.8 1198.2 s.76ee 1810.e 537.44 e.92199 e".37449 e.3949 e SS9.e 0.4365 1.3771 SS5*9 slee*'

629.6 11ee.1 e.7592 S.6190 1.3783 S54 5 1807 2 1190.0 Sth.2e e.e219$ e*.37863 c.40eSe 557.S 631 5 119e.e

• 555.14 e.e2192 1189.1 e.7578 e.6236 1.3794 553 1 18e7.5 4

e138205 8.4e476 556 1 633.4 11e1.5 e.756e lee 3.0 * *. S* * .0 0 e.ettee e'.38719 e.44902 554 6 e.4241 1 3ee5 551.7 Ste7.e te78.0 SS2.e4 e.02184 e;39tSe 0 41335 635.3 11e9.9 e.75Se e.6266 1.3817 55e.2 Stes.1 553 1 637.1 185s.3 e.7536 4.6292 548*e l 3064.0 551 7e 0 02181 e*.39599 9.91775 SSI.6 639.0 119e.7 1 382e 1188**

0.7522 9 431e 1 3ees Se7.g 11ee.7 gese.e Ste.53 e.02177 e*.4es47 e.42224 550 1 me.9 i lose.O 549.36 0.02174 1191.s e.1507 e.6344 1.2eSt 545 9 11e9.e e* meso? e.42641 240 6 642 0 1151 4 se30.0 Ste.te e.e217e e*.40976 e.43146 e.?t93 e.637e 2 3843 544 5 1189 3 547 1 644 7 1151.s e.747s 3 3ere 543.e 1809.4 1 182e.0 546.99 e.e2:46 e'.414Se e.4362e 545.6 6e6 4 1192.2 e.6396 1910.0 545.79 ,0 02163 S.7463 e.6423 3.see6 341 5 1:e9.9 e*.41991 e.44te3 544 1 693.5 1192 6 0.794 9 S9e*# 311e*1 e.6449 1.3898 1ect.3 544.5e 0 02159 e*.42434 e.4459S 142.6 650.4 953.0 143.36 e.c2 59 e;42942 1192.9 e.7434 0.6476 1.39te 538 6 111e.4 e.45097 541.s 652 3 1153 3 8 7e19 9es.O S42.1% e.c21*2 e*.43tS7 0.456e9 539.S e.6593 f.3922 537 1 11:4*T 970.e S40.99 659 2 1193.7 e.70e9 e.653e 1 393e 535.6 1111.s

• e.0214e e*.439e2 e.46130 137.9 456 1 1154.s j 56e.0 S39.65 e.e2145 e.7309 e.6SS7 1 39e6 534.0 1811 2 e*.4451e e.46662 536 3 654.s 1154.4 e.7373 e.6See 532 5 ,1111 5 1 3958 453.8 538.39 0.82141 c*.4Seu e.4720$ 534.1 668.0 448.0 S37.13 0.02137 1354.7 S.73Se 0 6432 3.3978 531 8 1111*T e".45621 e.47759 533 2 661 9 1855 1 0.7342 i 930.3 S35.e5 e.02134 e*.46tSe e.44329 131.6 e.6640 1 3982 529 4 11:2.s 923 0 536.56 e.0213e 643.s 1155 4 e.7327 e.6660 3 3915 527.9 1112 2 e*.447 70 e.ne191 530 0 665 8 1155 7 e.7311

, 950 9 533.2e e.e2127 e.47343 8.4949e S28 3 e.6696 1.eee7 526 3 1112 5 667.7 1154.1 e.7295 e.4724 1.4419 S29*e 1112 7 t

' toe.S S31.95 0 02123 e*.41968 s.50e91 526 7 M97 1194.4 e$e.0 S20.63 3.e2119 e*.4eSe6 S.Se706 S25 1 e.7279 8.6753 14e32 523.2 1813.e ete.e S29.38 671 6 1154.7 e.7263 e.6782 1.4045 521.4 1113 2 S.e2416 e'.4921e e.S1333 123 4 673 4 1897.e 0.7247 07e.0 227.96 0.02112 8 49e63 e.St975 521.s e.6stl 1.4eS7 520.e 3113.4 est.3 Ste.te 475 6 1117.3 e.723e e.6ege 1.447s 51e** II13*I e.02149 e.SeS22 0 52631 320 1 477.6 1897.7 8.7214 e.6e49 3.4043 516.7 1813.9 ele.0 525.2s e.e2teS sk51157 e.513e2 Ste.4 679.5 ete.e 533.86 8.02341 e*.Stege

!!9e.e e.7397 e.6e99 1.4096 515 1 1814 1 e.S39ee 516.7 est.5 1194 2 e.7 tee St3** SII'*3 e34.0 S22.46 . e.ete94 0*.52513 S.S4tt9 Slbe e.6529 1.4149 e20.e 538.06 6e3.5 1898.5 e.7343 e.6959 1.4122 St!*e 183'*5 e.e2054 e*.S331* e.SS648 313.3 6e5.5 1158.e 0.7 tg6 ele.S S19.44 S.02e91 e.4990 3.gt24 518.1 Sit **e 8.S4e52 e.S6143 511 6 6e7.4 1859 1 0.7329 Ste** 11II**

s.702e 1.9449 v

v= Cycle effi: ten:y = ;:.e t.:. i f ..a s/t

• out)/(Energy in) 2 - -

w = eg s = Vgt + 1/2 at A = Ne$(1-e )

E = m:2 ---

.it-2 A = An KE = 1/2 mv , a = (Vf - V )/t g ,. A = A,e ,

PE = mgh .

yg = Y, + a t w= e/t 1 = Ln2/t 1/2 = 0.693/t 3f7 V2 1/2 eff = [(t p )(In))

1

.P -

d9I N# [It1/2)

• IIO)3

.E = 931 t.m -Ex I=Ieg o = ..cptt Q = UAt1 . I = l o

e"*

Psr = v fan I = 1,10-*/IV' TVL = 1.3/u 5

P = P,10 "#II) HVL = -0.693/u t

P = Pn e /T ,

SUR = 26.06/T , SCR = 5/(1 - K,ff)

CR, = S/(1 - Kgf,)

~

SUR = 25c/t* + ( E - p )T CR)(1 - Kdf1) = CR 2 II - k dQ)

T = (t*/p) + [(8 -p)/193 M = 1/(1 - Kgf) = CR)/CR g i T = 1/(p - 8) M = (1 - Kggg)/(1 - Kdf1)

T = (a - p)/(10) SDM = (1 - Kgf)/Kgf

, = (K,ff-1)/K ,ff = d ,ff/K di t= = 10 seronds

-I A = 0.1 seconds e = [(1"/(T Kgf)] + [I,f f (1 / + 18 )]

. j =1d P = (z V)/(3 x 1010) .

I1)d)d) 2 =2 Id g2 2

2 I = pH R/hr = (0.5 CE)/d (meters)

Water Parameters Miscellaneous Conversions 1 gal. = B.345 lbm.. I curie = 3.7 x 10 10 aps 1 gal. = 3.78 liters 1 kg = 2.21 lbm 1 ft3 = 7.48 gal. I hp = 2.54 x 10 Btu /nr Density = 62.4 lem/f t3 1 ms = 3.41 x 10 Etu/hr Density = ) gm/cm3 lin = 2.54 cm Heat of vaporization = 970 Btu /lom *f = 9/5*C + 32 Heat of fus. ion = 144 Btu /lbm *C = 5/9 (*F-32)

• - 1 Atm = 14.7 psi =, 29.9 in. Hg.

~ '

MASTER COPY CATEGORY 1 ANSWERS 1.01 a. The ratio of delayed neutrons to the total neutrons. (0.5)

b. Time in seconds required for power to change by a factor (0.5) of "e".

c. The process by which neutron level remains at a constant (0,5) value (or increases due to positive reactivity addition) due to having source r eutrons ir, the vicinity of the fuel with Keff less than or. a .

Reference CrS Reactor theory page 51,55,43 1.02 Without the source, monitoring the rise in reactor pcwer is (0.5) difficult and you could establish a super critical condition without realizing it, and could continue pulling rods, establishing an excessively short period (.25) which could damage the core when power overshoots in the power range (.25)

Reference CPS Reacter theory page 95 1.03 a. Relatively low cross section for absorbtion (0.5)

b. low fission yield (0.5)

c. long half-life (0.5)

Reference CPS Reactor theory page 93 1.04 1. c (0.5)

2. b (0.5)

3. e (0.5)

4. d (0.5)

5. a (0.5)

Reference CFS Thermal sciences 9-6 1.05 a. gross power distribution (0.5) radial power distribution (0.5)

b. ax i .a1 flux shaping (0.5)

Reference CFS Thermal sciences 10-24,25 1

L

1.06 a. More negative (0.5)

b. More negative (0.5)

c. More negative (0.5)

d. less negative (0.5)

e. more negative (0.5) f less negative (0,5)

Reference Standard nuclear theory 1.07 d. (1.0)

Reference CPS question bank 1.17 1.08 b. ( 1. 0 )

Reference CPS question bank 1.21 1.09 Reactivity of core rho = beta /1 +1amda t rho = .007/1 +

( .1 x 75) rho = .00082 delta k/k (1.0) delta temp of mod = rho core / alpha temp delta temp of mod = .00082 delta k/k divided by -(-1.5 :: 10 -4 delta L/k/F delta temp of mod e 5.49 F (0.5) final temp = 153 F + 5.49 F = 158.49 F (0.5)

Reference Standard nuclear principles 1.10 c. (1.0)

Reference CPS Thermal science 10-10 1.11 Dissipation of fluid momentum oy sudden flow stoppage charact- (1.0) erized by a pressure shock wave which can result in damaged pipes.

Reference CPS Thermal science 14-17

/,0 1.12 a. Spontaneous fission uranium 238 and uranium 235 with (O r5) uranium 238 being the largest contributer. or Cuvism 24L

b. Photo-neutron source naturally occuring deuterium and ( S) fission of decay gammas react to form hydrogen and a neutron.

l.0

c. Alpha-neutron reactions oxygen 16 with en alpha particle (Dr5) to produce a neutron Reference CPS Reactor Theory page 95,96 2

t

~

t 1.13 a. Ar the reactor operates at power, Xenon builds into (0.5) equilibrium, adding negative reactivity, causing power to decrease.

b. A rod withdrawal from a high power region will cause a (1.0) power increase in the adjacent fuel rods because of being closer to thermal limit and therefore cause damage.

Reference GE Reactor theory 1.14 a. a Decreases (0.5) b Decreases (0.5) c Decreases (0,5) d Decreases (0,5)

b. a MAF R AT is the ratio of MAPHLHOR to MAPLHGR or the (0.5)

KAPLHGR (ACTUAL) to MAPLHGR (LIMIT CONDITION) b NO (.25) c The clad temperature can e::ceed 2200 degrees F. (.75) during a DBA LOCA.

Referent GE HTLFF page 16617, CPS Thermal science 10-16 End of Category 1 3

CATAGORY 2 ANSWERS 2.01 1 High drywell pressure 2 psig (0.5) 2 Level 2 (0.5) 3 Hi gh radiation in the containment fuel transfer floor (0.5) 4 High radiation in the containment exhaust (0.5) 5 High radiation in the fuel building exhaust, Fuel floor (0.5) 6 High Radiation Levels in the continuouc containment purge (0.5) exhaust

Reference:

CPS License Review Manual VG3 page 1 2.02 a. l.- True (0.5)

2. False (0,5)

b. 1. LPCS pump start

2. LFCS room cooler start

3. F012 Test return valve close

4. FOO5 Injection valve open

5. F0ll Minimum flow valve open (4 required G 0.25 each)

Reference CPS Lesson Plan LPCS 2.03 Off: (.25) Disables pressure relief function 3 position control (0.5) switch on P601 and P642 (both switch on P601 and P642 must be in off)

Off disables pressure relief function (switch on (0.5)

P601 controls solencid 'a' and switch on P642 controls solenoid 'b'). Both switches must be in off to (.25) disable the associated valve.

Auto: (.25) Valvo actuation controlled by reactor pressure (0.5) setpoints (pressure relief function).

'Opent (.25) Opens the SRV (0.5) i Reference RTS Lesson plan ADS 3-2 4

i- -

2.04 e. With both loop controllers and the flux controller in automatic, the FCV's are controlled with the master controller. (.25) The master controller " raise" and

" lower" slide switch is used to adjust the power demand signal. (.25) The power demand signal is compared with actual thermal power and then sent to the flux controller.

(.25) In this mode the system will automatically adjust for small changes in reactor power due to reactor pressure or feedwater temperature fluctuations. (.25)

b. The master controller automatically shifts to manual (0.5)

Reference CPS Lesson Flan Recirc Flow control page 14 2.05 a. Actuator air (0.5)

b. Control air (0.5)

Reference RTS Leucon plan Control Rod Hydr aulic System 2.06 a. The Steam Supply Valve (F045) (0.5)

b. When Icvol decreaces to tfe inatiation level, the FOSS (1.0) valve will reopen.

c. The turbine test circuitry would be automatically bypassed (1.0) and the flow controller would control normally

d. No, must be reset locally. (1.0)

Reference:

CPS Lecson Plan RCIC 2.07 a. loss of squib valve continuity (0,5)

b. injection check valven indicate open (0.5)

c. pump run indication (run light and pump discharge pressure normal) (0.5)

d. tant level decreasing (get tank hi/ low level GLC tant alcrm) (0.5)

c. power decreasing 40.5)

Reference RTS lesson plan SC D

2.00 a. High steam flow to RHR/RCIC

b. Turbine e:thause diaphragm pressure

c. Low RCIC steam line pressure

d. RCIC area high temperature

e. RCIC area high differential temperature

f. Main steam tunnel high temperature

g. Main steam tunnel high dif f urential temperature

h. RHR area high temperature

i. RHR area high differential temperature Any five at (0.C) e ., c h Reference CPS exam bank 2.40 2.09 a. False (1.0)

b. True (1.0)

c. False (1.0)

d. False (1.0)

Reference RTS Icsnon plan HP3-7,8 2.10 c. (0.5)

e. (0.5)

Reference RTS Icsson plan SXO-7 2.11 a. limit loss of coolant from the vos. col during MSIV closure,

b. limit radiological releacau outside the drywell prior to MSIV closure

c. provide flow signals f or MGIV closure Any two each (0.9)

Referenco RTS leucon plan MG3-1 End of Category 2 6

CATEGORY 3 ANSWERS 3.01 a. tent Icw 70 gal (C.5)

b. toni empty 60 gal (0.5)

c. oil warm 145 F (0.5)

d. oil hot 150 F (0.5)

e. Iow pressure 1650H (0.5) th/bf8 e,g, , m f

f. underd and over = ' ' : :7 (0.5)

Reference CFC Retirc Flow Control 1.0-6.26 page 6 3.02 a. There is an inner electrode and ond outer electrode with the outer electrode coated with uranium 235. (0.5) The neutronu penetrates the coating, fiunioning a uranium 235 atom couning fiction products to retail into the chamber . (0.5) The f i t.u i on products otrip electrons from the argon gat ioni:ing the ges. (0.5) The bios voltage will cause the electronc and ions to be collected by the electrodes. (0.5)

(M)lo

b. Gamma pulous, are much smaller than neutron pulcus and this difference in pulse ni o allows the discriminator to eliminate the gamma pultos while passing neutron pulces.

Referenct Enem Duestion Bent 2.00 a. 1. Lone of Ltator coolant prennure correcponding tc ( l' . 5 )

427 gpm.

2. High stator c ool .an t temperature (01 dec. C.) (0.5)

b. 1. Generator output current not roduced to the "no ( L' 5 )

liquid cooling"turrent vclue (9071 .757.

ampo) in 2 7 minutes. (0.5)

2. Generetnr output current not reduced below the check current value (30,297 ampo) in 2 minutuu.

J0 Yo Rwfurence: CPC L s:Luon Pl an Main Gener ator G ?. A 4 7

3.04 a. Frovi de a t. l g n cl treportional to the auial neutron flux (1.0) distribution at cel ected small axial intervals over the region of the core where LPRM system detector assemblies ore 1ocated. This nignal i t. of high precinion to al1ow ruliable calibration of LFRM gains.

b. Provide accurate indication of the flun measurement e5ich (1.0) allows pointwise or continuous measurement of the aniel neutron flun distribution.

Reference CFS Exam bant 3.09 3.05 c. The RFC requirec this pattern te limit the reactivity (1.0) incerted should a rod that it suppound to be fully uithdrawn but is in fact fully inter ted and Leparcted from itt a muchanism and de op out. of the core. This limits the fuc1 energy to 203 cal /gm.

b. Turbinc fir st stage chell prcosure. h Ref er ence CPS Enam bani: 2.13 3.06 Control valvet will start to clore ord the bypass valves will utart la open, of try the small close bias cignal i t, overcome.

At appr,'pimately fE4 turbine load and bypass velves at 100".

open, pf% Lteam flow, the header precsure will ctart to increso. (1.0)

As the Icod 1imit as further reduced reactor procturu cnd power will start to increawe. The reactor will scram on high preocure or htgh power to flow. (1.0) The final steady state conditionu will be the reatter shutdown with prenture being maintained with the bype:t valvus and mal eup wi th feed pumps (1.0)

Feference R1B I c o t.on plan EH 2-0 3.07 MC cortputer DC5 dioplay generator DCC ccN uter cron,-ti c to UF C bun "E" Fire protection panel panoic p679,600,070 GCTARC cobinent 1H10-612 Inot. Fanel 1paCOJ Rh power supply on Hi!p640 Aue building MCC 11 any four each (O.5)

Referenco RTL Lonton plan IP7-4 0

2.03 a. telf-ectuation on high pressure (0.5'

b. Pilot actuation on ADS cignal or pressur e relief function (0.5)

c. Manuti opening (0.5)

Reference:

CPG Leccon Plan ADS IIIA 3.09 a. A rod block will be generated (0.5)

b. High flow 1 1 1 */. (0.5)

c. Module unplugged. (O rJ)

Mode switch out of operate. ( 0 J5)

Reference Tech Spec page 3/4-3-57 3.10 4. Causes reactor level to INCEASE (0.25) duc to the Level Control System having a STEAM FLOW / FEED FLOW ERROR , STEAM FLOW > FEED FLOW (0.375) resulting in a SIGNAL to I f CRE ACE the SPEED OF THE REACTOR FEED FUMPS. (0.375)

b. Cauces reactor level to DECREASE (0.25) due to the Level Contro 1 System having a LEVEL ,RROR, E with NO compencating FLOW ERROR (0.375) resulting in a SIGNAL to DECREASE the SPEED OF THE REACTOR FEED PUMPS (0.375)

c. Reactor level should REMAIN CONSTANT (0.25) becaucc the "A" FEED FUMP Turbine Control Unit (0.375) will lock the pumps at the speed at the time of fcilure. (0.375)

Reference CPS Lennon Plan Feedwater Control (0,5) 7.11 c. 1.88 ptig or N.$h ow Pr.,s.,. ,

( 0. 5 ;'

Roacto- ventel level 1 Reference RTG 1esson plan SA/IA4-2 0.12 low oil proccure abnormal oil t e.np c r e t ur e high dicharge air tnmperoture low CCW precouru overcurrent high vibration any throu each (0.5)

Reference RTS lennon plan SA/FA4-1 End of Category 5 9

CATEGORY 4 ANSWERS 4.01 a. Capable of being closed by an operable drywell automatic (1.0) isolation system.

b. Closed by at 1 cast one manual valve, blind flange or (1.0) deactivated automatic valve secured in its closed position.

Reference CPS Tech spec 1.10 a 4.02 At least one operator must remain within the area required (0.5) by the procedure.

Reference CPS No 1401.05 Rev.3, sec 8.5.2 4.03 a. An unplanned reactor trip (0.5)

b. An unschedualed or unexplained change in power (0.5)

Reference CPS No 1401.06 Rev.4, sec E.3.2 4.04 1 Notify the Shift Supervisor (0.5) 2 Sound station fire alarm (0.5) 3 Announco location, type, and si:c of fire (0.5) 4 Dispatch the fire brigade to the appropriate assembly area (0.5) 5 Start a Diesel Fire Pump if not already running (0.5) 6 Initiate / verify initiation of appropriate fire suppression (0.5) system.

Reference:

CPS No. 1893.07, DISCOVERY OF A FIRE CPS No. 1990.03, OPERATOR ACTION UPON NOTIFICATION OF A FIRE 4.05 Shut the Recirculation Pump Discharge Valve, 1933-F)67A(B), (1.0) to allow time for the pump to stop and prevent reverse rotation.

Ghut the idle loop's Racirc. FCV, 1B33-F060A(B), to minimum (1.0) ,

p or+ 1 t i on . ,

If the loop is 02t to be isolated, reopen the Recirculation (1.0)

Pump Discharge Valvo af ter appro:timately five (5) minute.

Reference:

CPS No. 0302.01 10 l

F 4.06 c. Manually scram the reactor and observe all control rods (1.0) f ully inserted.

b. Sound the Containment Evacuation Alarm (0.5)

c. Report to the Remote Shutdown Panel (0.5)

References CPS No. .4003.01 3.1 - 3.3 4.07 a_ Ou r t e 1 y e vm-" ch1 be 1i-itcd tc 7 cc H re-)

b. Records of previous occupational exposure to radiation (FHEr are complete. /e (

s. f

c. Accumulated dose to the whole body shall not evcced 5 (n-18) ( k.dn Reference CPS No 1024.15 sec e.1.1.b 4.00 One calendar quarter (0.5)

Reference CPS No. 1905.11 sec 6.5 4.09 1. b (0.5)

2. d (0.5)

3. e (0.5)

Reference CPS No. 1905.20 -

4.10 a. Immediately notify the shift / assistant shift supervisor (0.5)

b. Immediately notify the Radiation Protection Office (0.5)

c. Evacuate the affected crea of non-ensential Perconnel 'O.5)

Reference CPS 4979.09 sec 3.1 4.11 c. Control room HVAC should shift to the chlorine recponce modc (0.5)

b. Don a self contained breathing apparatus (0.5)

Cvacuate the area (0.5)

Reference CPS 4969.01 sec 2.0.3.3 4.12 a. positively verified that the initiation conditionc are no (1.0) longer present.

b. HPCG flow is not needed to satisfy existing plant cooling (1.0) requirements.

Reference CPS No. 3309.01 sec 4.0 11

+- , . ..

t' 4.13 a. ADS or Saf ety relief Valve leaking annunciator alarms.

SRV moni toring systen trouble annunciator al arms. ,

SRV temperature recorder output increasing f or one or more SRV's.

SRV position indicates open.

Decrease in generator output.

Unexplained increase in reactor water icvel followed by level decrease.

Increasing supprection pool temperature.

. Increasing hotwell makeup.

Any four at (.25) .cach

b. Sound the containment evacuation alarm (0.5)

Determine which SRV has lifted (0.5)

Attempt to close the ERV by placing its' associated control (0.5) switch on panel F601 in the open position and then in the off. position.

Attempt to close the SRV by placing its associated contrcl (0,5) switch on panel P642 in the open position and then in the off pocition.

IF the SRV cannot be shut ~within two minutes then place the (0.5) mode switch in SHUTDOWN and refer to opporiate porcedures.

Reference CPS No. 4009.01 End of Category 4 12

XSE copy

~ ~~ ~

M U.S. NUCLEAR REGULATORY COMISSION SENIOR REACTOR OPERATOR LICENSE EXAMINATION FACILITY: Clinton REACTOR TYPE: BWR DATE ADMINISTERED: 10/21/85 EXAMINER: T. Lano APPLICANT:

INSTRUCTIONS TO APPLICANT:

Use separate paper for the answers. Write answers on one side only. Staple question sheet on top of the answer sheets. Points for each question are indicated in parentheses after the question. The passing grade requires at least 70% in each category and a final grade of at least 80%.

% of Category  % of Applicant's Category Value Total Score Value , Category 25 25 5. Theory of Nuclear Power Plant Operation, Fluids, and Thermodynamics 25 25 6. Plant Systems Design, Control, and Instrumentation

1. 81 25 7. Procedures - Normal, Abnormal, Emergency, and Radiological Control EM 25 8. Administrative Procedures, Conditions, and Limitations 100' 9f 100 TOTALS Final Grade  %

All work done on this exam is my own, I have neither given nor received aid, f [. .P.!.!c s Sy f

( A w, ( . . L , a L L, n d

)

I Theory of Nuclear Power Plant Operations, Fluids, and Thermodynamics 5.01 Define the following terms .

a. Neutron Lifetime (0.5)

b. Neutron Generation Time (0.5)

c. Thermal Diffusion Time (0.5) 5.02 Prior to startup (all rods full in), the SRM count rate is 10 cps and K effective is 0.%

a. If control rods are pulled to give a Ak of +0.035, what count rate on the SRMs should be expected when the period becomes infinite? (1.0)

b. If additional control rods are pulled to give a Ak of +0.003, would the time required to reach an infinite period be greater or less than the time in part "a"? Give the reason for your answer. (1.0) 5.03 a. List the three (3) reactivity coefficients in a BWR at 1005 power and give approximate values for each.

Assume 80L core conditions. (1.5)

- b. What effect (Increase, Decrease, or No Effect) do each of the coefficients have on total core reactivity following a safety / relief valve failing open? Briefly explain why the dominant coefficient effects reactivity in the manner you indicate. (2.0) 5.04 In which of the following situations is the center control rod worth greater? Explain your answer. (2.0)

Situation #1 All control rods are fully inserted and the center control rod is then fully withdrawn.

i Situation #2 '

All control rods are fully withdrawn and the center control rod is then fully inserted.

5.05 If equilibrium xenon is obtained (the reactor has been and the operated at constant reactor power power is doubled, willfor themany new hours)Ilibrium equ ,

xenon concentration be twice at great? Explain your answer. (2.0)

5.06 On the attached temperature-entropy (Ts) diagram of Figure 4.2. Label each of the three regions.

A B C (1.5) 5.07 Does the centerline temperature of the fuel pin located

. closest to a control rod change when the control rod is moved when operating near full power? Explain your answer using general concepts of heat transfer. (2.0) 5.08 You increase core power by pulling control rods around the center fuel bundle. Assuming that recirculation flow is kept constant, would the flow through the center bundle increase, decrease, or stay the same?

Explain your answer. (2.0)

' 5.09 Explain the effects of increasing the following uore parameters on steady state critical power.

s. Core flow f.

b. Inlet subcooling I,.

c. Reactor pressure *

(.

5.10 Your reactor operator infoms you that MPRAT is 1.02.

a. Is the MPRAT, as stated, conservative? Explain your answer. (1.0)

b. In regards to MPRAT, which of the following statements are True and which are False?

(1) MPRAT maintained within limits ensures that ,

transition boiling will not occur in 99% of the fuel bundles. (0.5)

(2) Maintaining MPRAT limits ensures that peak clad temperature will not reach 2200'F during a LOCA. (0,5) 2

5.11 A reactor startup is in progress. The reactor engineer has estimated that the reactor should go critical at Notch 28 (

on a particular control rod if the reactor operator reaches that notch at 0800. How would each of the following conditions or events affect actual critical rod position (more rod withdrawal, less rod withdrawal, or no significant affect)? Assume that if no conditions changed the reactor would have been critical at Notch 28 of the indicated rod,

s. The reactor has been shutdown from extended full power operation for 35 hours4.050926e-4 days <br />0.00972 hours <br />5.787037e-5 weeks <br />1.33175e-5 months <br /> and due to circumstances beyond your control, you will not have the indicated rod at Notch 28 until approximately 0900. (0.5)

b. Shutdown cooling is terminated. (0,5)

c. Moderator temperature is gradually decreasing. (0,5) 5.12 In discussing core thernal hydraulics, both steam quality and void fraction are considered. ,

s. Explain the difference between these two variables. (1.0) i

b. Reactor power is increased from 80% to 90% of full power by increasing recirculation flow. Did the average void fraction increase, decrease, or remainthesame(comparethevaluesaftersteady state has been reached). Briefly explain your answer. (1.0)

END OF SECT!DN l

3

i Plant Systems Design, Control, and Instrumentation i j

6.01 The NSPS bus can be transferred either by a manual bypass or via the 55T5.

' a. How does the SSTS furetion in a situation in which

! normal power is lost for 10 minutes then returns.  ;

include in your answer all modes of operation of

. the $$T5. (2.0) l

b. Other than the NSPS bus, what other bus has a 55T57 (0,5) 6.02 In regards to the Fire Protection System, how do the j following systems operate?

a. Wet Pipe Sprinkler System .

b. Automatic Deluge System .

c. Automatic Preaction System .

6.03 a. What are six (6) initiating signals which will automatically start 58GT7 (1.5)

b. Why should the use of S8GT be avoided when there is heavy welding taking place in the secondary  !

containment? (0.5)

1. l 6.04

~

For the following six (6) valves specify which Primary Containment Group they are assigned.

! a. ACIC Exhaust Vacuum Breaker Outboard Isolation Valve

{ b. Drywell Purge Supply Outboard Isolation Valve l0.5)1 0.5,

c. RNR S to Radweste Secondary Isolation Valve 0.5) i j d. Shutdown Cooling Outboard Suction Isolation Valve 0.5)

e. Mn Steam Outboard A Drn Isolation 0.5p

f. Recirculation System Sample Isolation 0.5) {

! 6.05 The recirculation pumps have an ATWS trip as well as an RPT trip.

a. What are the initiating signals for each trip? (1.0)

- b. What is the purpose of each trip? (2.0) 1 f6.06 What protection is there for the Turbine Generator when a loss of stator cooling occurs? (Assumethe loss occurs at 101 power and again at 100% power.)

Include initiation signals in your answer. (2.0) l l

6.07 a. under what conditions is hood spray provided and why. (0.5)

b. What conditions will automatically open the hood spray valve? (0.5)

c. What is the source of water for the hood spray system? (0.5) ,

6.08 a. If the Diesel Generator is supplytr.g the bus in parallel with off-site source and a LOCA initiation occurs, what automatic actions will take place in regards to the Diesel Generator operations. Include in your answer any actions you must perform in order to ensure a safety power source is available. (2.0)

b. What three Diesel Generator trips are still in effect, even if a LOCA signal is present? (1.5) 6.09Whataresix(6)loadssuppliedbyTBCCW? (3.0) 6.10 What design feature prevents the control rod drive pump from going into runout conditions after a s: ram? (1.5)

END OF SECTION

{

2

., r ,,

,.8e, "A s

j f s  :

N Vroced'uris., Nnrual, h ormal, Emergency,

? A ahe hJiologidal Control l ., s

+. g . s, x f 7.01 What are four items rc' wired to be logged once the ,

reactor has gone critical? (2.0) 7.02 According to your procedure for "Heatup a xi Pressurization" there is a method of determining control rod coupling integrity. Briefly explain how and when this is done. (2.0) t 7.03 What six (6) conditions must exist in order for Drywell Integrity to d irt? (3.0) h,04 Ad.ording to yo8((rocedure " Primary andkr Secondary Containment fengr1ty 'Verificatf~ ca" what does the term verify mean? '

(1.0) 7.05 What Are the six (6). entry cos.Mn'ons for " Containment Control"?s + *

(3.0) 7.06 ~According Yo your procedure 4003.01 " Remote Shutdown" what are three steps or actions which must be performed 4

prior tievacuating the control room? (1.5) 1

7' ing to your procedure for " Lost of Control Air"

- I .07 Acco the issnediate actions require you to manually scram the reactor if three conditions occur. What are these three conditions? (2.0) y 7.08 a. What areJ four (4) indications which would indicate an iliadvertent opening of a safety / relief valve? (2.0)

- b. What' Ution' must be taken if a safety relief valve wili not close? (0.5) 7.09 What are five entry condition,,s  ?'

forf*your " Level Control -

Emergency" procedure? (3.0)

'7.10 What criteria or conditions must se met in ordhr to manually override or secure a l'CCS system? (2.0) 7.11 What are f.hree crite91a used to determine if the reactor can be maintained in a shutdown condition with control rods?t ,

(1.5)

~ ,-

l 1 #

?

, '7 s 5 W

_ - , . . . . _ _ _ _ _ . _ . . .__,____,._,,._,,_,,.._%.,j,,..,..,,,__.,__,._,_,.,.,_,__,___,__,,,,_._.y. _, _ _ _ _ ,_ _ , _ _ _ , . . _ , , . , . , _ . , , . . _ _ . ,

' 7.12 According to your procedure on " Secondary Containment /

Radiation Release Control - Emergency" if an area temperature or radiation level is at or above its alann point then all systems discharging into the area must be insolated unless they perfonn any of three functions.

What are the three functions? (1.5)

END OF SECTION 2

Administrative Procedures, Conditions, and Limitations 8.01 Define the following terms according to your Technical 4 Specifications.

a. Limiting control rod pattern (1.0)

- b. Channel check ,

(1.0)  !

8.02 According to Technical Specifications, what is the, basis for the following: ,

. O n t = ' e "; r- %= t$

- b. Main Steamline isolation Valve Closure isolation and scram (2.0)

c. Turbine Control Valve Fast Closure Scram (1.0) 8.03 In order to remove a safety-related system from service the outage must have an independent verification.

a. Can anyone verify the system is removed from service?

Explain your answer. (1.0)

~ b. If a verifier finds an error in a valve lineup, what action must be taken in accordance with your procedures? Be specific in your answer. (2.0) 8.04 a. What is the difference between an unrestricted CR0 i

and a restricted CRO? Explain each in your answer. (2.0)

b. How can a restricted CR0 become an unrestricted CRO? (1.0) 8.05 Who must sign a TCF for " Management Staff Member"?

Include any specific requirements in your answer. (2.0) 8.06 In regards to conduct of personnel in the main control room *

a. Who has the authority to limit the number of persons in the control room? Four required for full credit. (1.0)

b. 10 CFR 50.54 states, "An operator or senior operator licensed pursuant to Part 55 shall be at the controls at all times during operation of the facility."

What must an operator do to meet this requirement? (2.0) 8.07 What action is required by Technical Specification when you ~

have'more than one control rod scram accumulator inoperable? (2.0)  ;

E D

~

s

8.08 What is the restrictions in regards to recirculation loop flow mismatch under the following conditions?

a. Core Flow 80%

b.

(0.5)

Core Flow 10% (0.5) 8.09 An individual shall be designated to assume the control room command function when the Shift Supervisor is absent.

a. What restrictions and/or requirements are there for the individual who takes over when the unit is operating at 100% power? (1.0)

b. How would the above change if the individual took over when in condition 4? (1.0) 8.10 What requirements must be met in order for Drywell Integrity to exist?

(3.0)

=

END OF TEST l

l 2

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NUCLEAR POER PLANT T>WtMAL SCgENCES See c ear Steam Power Plant September, 1994 Figure 4.2 Steam Cycle Temperatr e Entropy Diagram l

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1 Page 4-4

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I 1

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Toble 1. Satureted Steam: Temperature Table-Contheued E

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'Critetal temperatver E

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l l

Table 2: 5.tutsted Steam: Pie..ure Table G

3. etat Delsme Intnespy Att Press lea. 8.t Set Ret SM S Entrep?

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• l I

10 CFit 20 Appendix B -

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Table 2. KEUTRON FLUX DOSE EQUIVALENTS *

~

Ecutron Number of Neutrons per squarc

-ncrgy centimeter equivalent to a dose Average flux to deliver

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'~~ Energy (MEV) h*ater -

Concre_t_e_ -.. It c ..

Pb 0.5 . 0.090 0.21 0.63 1.0 1.7 0.067 0.15 0.44 1.5 0.057 0.77 0.13 0.40 0.57 2.0 0.048 0.11 0 .33 0.51 2.5 0.042 0.097 0 .31

~~g 0.49 3,0 10t am

2 O

ANSWERS - Theory of Nuclear Power Plant Operation, Fluids, and Thermodynamics 5.01 a. Neutron Lifetime (1) = the time between neutron birth and absorption or the length of time a neutron exists as a free neutron (either answer will be accepted). (0.5)

. b. Generation Time = the average length of time between successive generations. (0.5)

c. Thermal Diffusion Time = the time between neutron thermalization and absorption. (0.5)

Reference:

Reactor Theory, pgs. 18 and 19 5.02 a. CR N _ Keffo 1-Keff i

CR M _ 1.96 1 .995 aCR _ CR2 = 80 cps (1.0)

IT - .04 555

b. The time to reach an infinite period would be greater, due to the fact that there are more generations, each representing a period of time, required to reach equilibrium. (1.0)

Reference:

Reactor Theory, pgs. 42 and 43 5.03 a. 1. Moderatgr temperature coefficient alpha T equals

-1 x 10 4 per degree change in temperature.

< 2. Moderator void coefficient alpha V equals -1 x 10 3

per % change in voids.

3. Fuel tem _perature coefficient alpha D equals

-1 x 10

• per degree change in fuel temperature.

b. Alpha T increases core delta k/k i~ Alpha V decreases core delta k/k Dominant effect - relief valve opening results in decreasing reactor pressure which increased voids and decreases moderator density resulting in more neutrons leaking out of the core and reducing power.

7

Reference:

Reactor Theory, pas. 68, 74, 76

5.04 Generally speaking, a rod will experience its greatest total worth when it is fully withdrawn and all other rods remain inserted (situation #1). This is due to ,

the high flux created at the rod, while the average neutron flux remains very low.

I

Reference:

Reactor Theory, p. 83 5.05 No. The production rate is directly proportional to power level, but removal rate is proportional to xenon concentration

. and it contains a power dependent term, thermal neutron flux.

Since flux is directly proportional to power level, the burnout term becomes more significant. This results in an equilibrium xenon value which is higher than the original value, but not twice as high.

Reference:

Reactor Theory, p. 85 5.06 A = subcooled liquid region

• B = liquid and vapor region C = superheated vapor region-

Reference:

Thermal Sciences, p. 4-4 '

5.07 Yes.' As the rod is moved, neutron flux will change. Heat  ;

flux from fission is directly proportional to neutron flux. i The water temperature next to the fuel rod will change only '

slightly if at all. (The only change would be heating up the subcooled water to saturation conditions in the lower part of the fuel.) Heat transfer is directly proportional to delta T. 'As water temperature stays approximately '

constant, centerline temperature must change as flux changes.

Reference:

Heat Transfer, pgs. 5-37 5.08 As fuel temperature is increased (due to Control Rod Pull)

i. more voiding is created. Therefore, sore back pressure, which would ;aean less flow. However, because of core orificing back prcssure is less significant and flow through the bundle is almost the same. (2.0)

Reference:

Thermal Sciences, pgs. 9 .16 5.09 a. Increasing core flow will cause critical power to increase c,Je to the increased heat removal capability.

(1.0)

b. Increasing subcooling causes an increase in critical power. The inlet enthalpy will be reduced and the

heat which can be removed will increase. (1.0)'

2 i

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-..,.-._.7T- -

c. Increasing reactor pressure reduces the energy to be at critical quality therefore critical power will be lower. (1.0)

Reference:

General Thermodynamic Theory and Thermal Sciences, p. 10-10 5.10 a. The MAPRAT of 1.02 is not conservative. With a MAPRAT greater than one, it means that the MAPLHGR has been exceeded because:

MAPRAT = *') A et N'wCe g3 ny f*

b. 1. False M7ME - (0.5)

2. True (0.5)

Reference:

Thermal Sciences 5.11 a. Less rod withdrawal due to xenon (0.5)

b. More withdrawal due to heatup (0.5)

c. Less withdrawal due to cooldown (0.5)

Reference:

Reactor Theory 5.12 a. Void fraction is the ratio of volume of steam to volume of fluid (steam und water).

Quality is the ratio of mass of steam to mass of fluid (steam and water).

b. Slight decrease. In order for net reactivity to be zero (Rx critical), void fraction must decrease a slight amount to compensate for the increased negative reactivity added to Doppler as fuel temperature increased.

Reference:

Thermal Sciences END OF SECTION 3

ANSWERS - Plant Systems Design, Control and Instrumentation 6.01 a. SSTS will automatically transfer from the normal to the alternate power source on loss of normal power (undervoltage of 105VAC). A local switch allows the SSTS to be placed in a normal seeking mode where it will automatically transfer back to normal source if that power is restored. If the SSTS is not in the normal seeking mode, the transfer back to the normal source must be manually initiated. (2.0)

b. UPS (0.5)

Reference:

System IP, IP 4-1 6.02 a. The WPS system is charged with water up to the sprinkler heads. Heat from a fire melts a fusible link in the sprinkler head, allowing water to flow. /*O

b. The Automatic Deluge uses open sprinkler heads. A deluge valve is opened automatically when a heat detector is activated or it can be opened manually at the valve. When the deluge valve is opened, water flows from all the sprinkler heads. I' g

c. The APS system uses closed sprinkler heads, a check valve, and a deluge valve in that order. The system is pressurized with air between check valve and sprinkler heads. If a sprinkler head opens air pressure will drop, which initiates a supervisory trouble alarm in the control room. For wate spray to be initiated, two separate things must happen. Id

1. The deluge valve must be opened, either automatically or manually.

2. Heat from a fire must melt the fusible link in the sprinkler head.

Reference:

System F.P., pgs. 3-4 6.03 a. 1. ' High Drywell Pressure 2 psig

2. ' Level 2

3. High radiation in the containment fuel transfer floor exhaust

4. High radiation in the containment exhaust

5. " High radiation in the fuel building exhaust, fuel floor

6. High radiation levels in the continuous containment purge exhaust duct (0.25 each)

b. The heavy welding will reduce the charcoal beds ability to remove methyl-iodine. (0.5)

Reference:

System VG 3 6.04 a. Group 6 67

b. Group 10

c. Group 2

d. Group 3

e. Group 1

f. Group 5 (0.5 each)

Reference:

System PC

1. o 6.05 a. ATWS = Level 2 or 1150 psig reactor pressure (D?s)

RPT = stop valve closure or Control valve Test Closure with greater than 30% power.

b. ATWS trip designed to insert negative reactivity

, in the event RPS fails to scram the reactors to limit the high power and pressure transient. The purpose of the RPT is to provide additional CPR margin which is lost at the end-of-cycle. Closure of the turbine stop/ control valves results in a significant pressure increase adding positive reactivity to the core. At the end-of-cycle with the control rods almost completely withdrawn, this pressure increase / positive reactivity addition can occur faster than the rods can insert negative reactivity via a scram. Therefore, the recirculation pumps trip to slow speed and subsequent core voiding provides a rapid insertion of negative h. O) reactivity.

Reference:

System RR 6.06 Stator Cooling Runback - Initiation signals stator (inlat) cooling water pressure decreases to that corresponding to 427 gpm flow.

Stator (outlet) cooling water temperature 81'C Generator load above 80% runs back within 2 minute or trips Generator load less than 25% runs back witnin 3.5 minutes or trips 2

Reference:

System G&H 6.7 a. During startup and low load conditions when steam flow is not sufficient to provide for cooling of the last stage buckets and exhaust hood. (0.5)

b. The spray valve will automatically open on high exhaust hood temperature. (0.5)

c. The source of the spray is condensate pump discharge. (0.5)

Reference:

System GS/T0/MT 6.08 a. If the DG is supplying the bus in parallel with an off-site source and a LOCA initiation occurs, the DG output breaker will trip. In order to restore the DG to a standby condition, place the DG breaker control switch to the after trip position (i.e., with a green @(

• OJ flag). y,n n- y

b. perspeed,generatordifferentiapandovercranking.

(1.5)

Reference:

System DG Ti s 6.09 Any six (6) for full credit

1. Condensate Booster Pump Oil Cooler

2. Turbine EHC Fluid Cooler

3. Condensate Pump Motor Oil Coolers

4. Condenser Vacuum Pump

5. Glycol Cooler for Off-Gas System

6. Control rod Drive Pump Oil Cooler

7. Motor Driven RFP Oil Cooler

8. Isolated Phase Bus Duct Coolers

9. Alternator Exciter Coolers

10. Feedwater Process Sample Refrigeration Skid

11. Turbine Building Sample Panel

12. Feedwater Sample Panel

13. Hotwell Sample Panel 3

Reference:

System WT 6.10 Because the charging header connection is downstream of the FCV flow metering element (0.5), water drawn off by the charging header during a scram or accumulator charging will create a high flow signal (0,5) causing the flow control valve to close and divert most of the pump discharge to the charging water header.

This prevents pump run out (0.5).

Reference:

System RD END OF SECTION 4

ANSWERS - Procedures - Normal, Abnormal, Emergency, and Radiological Control 7.01 a. Rod group number, rod position, and sequence (0.5)

b. SRM reading (0.5)

c. Period (0.5)

, d. Recirculation loop flow (0.5)

Reference:

CPS No. 3001.01 7.02 Rod coupling should he monitoring by observing discernible nuclear instrument response during rod withdrawal. Rod coupling for each fully withdrawn control rod should be observed by monitoring control rod drive water stall flow without receiving the rod over-travel annunciator. (2.0)

Reference:

CPS No. 3002.01 7.03 a.' 11 drywell penetrations required to be closed during a ident conditions are either: capable of beir.g clo d by an operable drywell automatic isolation syste or closed by at least one manual valve, ind flange, r deactivated automatic valve secur in the closed po tion. (0.5)

b. All containee equipment hatches are losed and sealed. (0.5)

/ c. Each containment ai ck is i ompliance with requirements of speci ' ati 3.6.1.3 (operable). (0.5)

d. The containment leaka ra s are within the-limits of specifica on 3.6.- (within Technical

( Specification li its). (0.5)

e. The suppress' n pool is in complia with the requireme s of specification 3.6.31 erable). (0.5)

f. The aling mechanism associated with eac rimary c ainment penetration; e.g., welds, bello or

-rings, is operable. (0.5)

Reference:

CPS No. 3020 01 7.04 Verify: As used in this procedure, verify means to prove to be true by observation or by demonstration. (1.0)

Reference:

CPS No. 3020.01

7.05 a. Suppression pool temperature .-

above 95*F ,/ ,

b. Drywell temperature above 150*F /35

c. Suppression pool level below 18' 11"

d. Suppression pool level above 19' 5"

e. Containment temperature above 122*F l20' F

f. Drywell pressure above 2 psig f, j g (0.5 each)

Reference:

CPS No. 4402.01 7.06 a. Leave the Mode Switch is Run to ensure the main steam isolation valves (MSIVs) shut at 850 psi steamline pressure. (0.5)

b. Manually scram the reactor; observe all control rods fully inserted. (0.5)

c. Sound the Containment Evacuation Alarm. (0.5)

Reference:

CPS No. 4003.01 Immediate Actions 7.07 a. The air pressure drops to 60 psig and (0.5) the air pressure cannot be restored or (0.5)

b. The SDV level increases and results in a rod block g (0.5)

c. The control rod begins to drift (0.5)

Reference:

CPS No. 4004.01 7.08 a. Any four (4) for full credit. (0.5 each)

1. ADS or Safety Relief Valve leaking annunciator alarms.

^

2. SRV Monitoring System Trouble annunciator alarms. c'L86VA

3. Safety Relief Valve (SRV) temperature recorder output increasing for one or more SRVs.

4. SRV position indicates open.

5. Decrease in generator output.

6. Unexplained increase in reactor water level followed by a level decrease.

7. Increase in suppression pool temperatures.

2

8. Ir. crease in hotwell makeup.

Any reasonable answer accepted.

b. If the SRV cannot be shut within 2 minutes, the mode switch must be placed in shutdown. (0.5)

Reference:

CPS No. 4009.01 7.09 a. Reactor water level below Level 3 (+8.9") (0.5)

b. Drywell pressure above 1.68 psig (0.5)

c. RPV pressure above 1064.7 psig (0.5)

d. A condition which requires MSIV isolation (0.5)

e. A condition exists which requires a reactor scram and any of the following exist:

Reactor power > 3% (0.25)

Reactor power cannot be determined (0.25) aference: CPS No. 4401.01 7.10 Do not secure or place an ECCS in manual override unless by at least two, independent indicators (0.5)

1. Misoperation in the automatic mode is confirmed (0.5)

2. Adequate core cooling is assure g (0.5)

3. Unless otherwise directed to do so by the procedure (4401.01) (0.5)

Reference:

CPS No. 4401.01 7.11 The following criteria may be used to determine if the reactor can be maintained in a shutdown condition with control rods:

a. All control rods are inserted (0.5)

b. No more than 8 control rods are not fully inserted and rods out are at least 2 cells apart (0.5)

c. Nuclear engineer has determined that the existing rod position cannot result in criticality in cold conditions. (0.5)

Reference:

CPS No. 4403.01 3

7.12 a. Systems required to shut down the reactor (0.5)

b. Systems required to assure adequate core cooling (0.5)

c. Systems required to suppress a fire (0.5)

Reference:

CPS No. 10N4406.015 END OF SECTION 4

ANSWERS - Administrative Procedures, Conditions, and Limitations 8.01 a. A ~ ' 'n Control Rod P Pfrstra tern q

which resu s e being on a thermal hydraJue imit, i.c. , opera ing vn e list-ing V Me for APLHGR, LHGR, or MCPR. (1.0)

b. A Channel Check shall be the qualitative essessment of channel behavior during operation by observation.

This determination shall include, where possible, comparison of the channel indication and/or status with other indications and/or status derived from i independent instrument channels measuring the same parameter. (1.0)

Reference:

T.S. Definitions 8.02 a. ss of condenser vacuum occurs when the condenlier can longer handle the heat input. Loss of condense uum initiates a closur 4f the turbine stop valves an the heat input to t rbine bypass,va ndenser. Closure of the stop s which eliminates h) and bypass valves cause ssure transient, neutron flux rise, and an increase in su eat flux. To prevent the, cladding safety limit from be exceeded if this curs a reactor scram occurs on turbine stop valv osure in the Run mode. (1.0)

b. The inw nro uuro isa13tian af the main steam ett.

825 psig was provided to give protection against rapid reactor depressurization and the rapid cooldown of the vessel. Advantage was taken of the scram feature in the Run mode which occurs when the main steamline isolation valves are closed to provide for reactor

shutdown so that high power operations at low reactor pressure does not occur, thus providing protection for the fuel cladding integrity safety limit. (2.0)

c. The turbine control valve fast closure scram is '

provided to anticipate the rapid increase in pressure and neutron flux resulting from fast closure of the turbine control valves due to a load rejection 4 sub:c';"cnt-fai4uee-of-the- bypss =1=n- (1.0)

Reference:

Technical Specifications 8.03 a. To perform independent verification minimum qualifications shall be Auxiliary Operator qualified on the area in which the function is to be verified. (1.0) i

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

b. If an' independent verifier finds an error, a second independent verifier (preferably the shift / assistant shift supervisor) shall make the verification after it has been corrected. (2.0)

Reference:

CPS No. 1405.01 8.04 a. A Restricted Operator is an operator who is qualified Control Room Operator (CRO) but has limited experience working as a CR0 or who works the Control Room Operator areas in an upgrade capacity only. (1.0)

An Unrestricted Operator is an operator who is an experienced Control Room Operator who has no constraints placed on standing the CR0 watchstation. (1.0)

b. To be considered unrestricted, the restricted operator must be promoted to CR0 and, in the opinion of the employee's Shift Supervisor, capable of standing the A or B area watchstation while a restricted operator stands the other. (1,0)

Reference:

CPS No. 1402.04 8.05 Two members of plant staff one of which shall be the organizational supervisor / designee who is responsible for the affected procedure. The other member of the plant staff shall hold a SRO license on the unit affected. (jo) 4

Reference:

CPS No. 1005-07 8.06 a. The Shift Supervisor, Assistant Shift Supervisor, i

C::t-d S::: ! ;:r & :r, and Control Room Operator each have the authority to limit the number of persons in the control room. (1.0)

b. In order to comply with this paragraph, an operator will be considered to be at the reactor controls

! if he/she is physically within the operating area in front of the unit panels. Q,0)

Reference:

DAP 1401.055 i 8.07 With more than one control rod scram accumulater inoperable, declare the associated control rods g inoperable, and (0.5) l 2

L-

a. If the control rod associated with any inoperable  ;

scram accumulator is withdrawn immediately verify that at least one control rod drive pump is operating (0.5) by inserting at least one withdrawn control rod at least one notch or place the reactor mode switch in the shutdown position. (C. d

b. Insert the inoperable control rods and disarm the associated directional control valves either: (0.5)

1. Electrically or

2. Hydraulically

Reference:

Technical Specifications 8.08 a. 5% (0.5)

b. 10% (0.5)

Reference:

Technical Specification 3/4 4-3 8.09 a. An individual (other than the Shift Technical advisor (0.5)) with a valid Senior Operator License (0.5) shall be designated to assume '

the control room command function. (1.0)

b. In condition 4 either an individual with a valid senior operator's licensee (0.5) a valid operator's license (0,5) can be designated. (1.0)

Reference:

T. S., p. 6-5 8.10 Drywell Integrity shall exist when:

a. All drywell penetrations required to be closed during accident conditions are either:

1. Capable of being closed by an operable drywell automatic isolation system g (0.25)

2. Closed by at least one manual valve, blind flange, or deactivated automatic valve secured in the closed position. (0.25)

b. The drywell equipm nt hatch is closed (0.25).

and sealed (0.25) (0.5)

c. The drywell airlock is operable (0.5)

d. The drywell leakage rates are within limits (0.5)

e. The suppression pool is operable (0.5) 3

. f. The sealing mechanism associated with each drywell penetration; e.g., welds, bellows, or 0-rings is operable (0.5)

Reference:

Technical Specifications END OF SECTION 4