ML20246M950
| ML20246M950 | |
| Person / Time | |
|---|---|
| Site: | Point Beach  |
| Issue date: | 07/12/1989 |
| From: | Burdick T, Damon D NRC OFFICE OF INSPECTION & ENFORCEMENT (IE REGION III) |
| To: | |
| Shared Package | |
| ML20246M948 | List: |
| References | |
| 50-266-OL-89-03, 50-266-OL-89-3, NUDOCS 8907190264 | |
| Download: ML20246M950 (218) | |
Text
{{#Wiki_filter:, , 1 I U.S. NUCLEAR REGULATORY COMMISSION i REGION III q i Report No. 50-266/0L-89-03 ' f Docket No. 50-266 .;
-(
Licensee: ~ Wisconsin' Electric Power Company 231 West Hichigan Street.- P379 Milwaukee, WI 53201 i Facility Name: Point' Beach Nuclear Plant Examination Administered At: Point Beach Nuclear Plant Examination Conducted: Week of June 26, 1989 s 1 Chief Examiner: . ' mon 7lll / k#) Date Approved By: T. Burdick 7// v[I7 DAte / / Examination Summary Examination administered on June 26-28, 1989 (Report No. 50-266/0L-89-03 )
- to one reactor operator candidate and two senior reactor operator candidates.
Results: The reactor operator candidate failed the examination. 8907190264 890712 PDR ADOCK 05000266 V PDC l
p ;- - - -- -- w m , as 1
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f i.. g REPORTDETAILJ
- i a, . Examiner D.
- :Damon, chief examiner
- 12. Exit Meeting
'~'
.On June 29, the chief examiner met with members of the facility staff to-discuss the examinations. The following persons. attended the meeting:-
G. Maxfield, General Superintendent - Operations WEPC0 J. Reisenbuechler, Superintendent - Operations, WEC R. Bruno,: Superintendent - Training, WEC W. Baines Training,~WEC-F. Flentje Admin Specialist - Regulatory Services, WEC-D. Damon, Examiner, NRC Because of the limited size of the sample, no generic strengths or weaknesses were noted.
.The facility made specific comments on some questions or answers on the-written examamination. These comments, along with their respective NRC-resolution,-follow:
Facility Comment: QUESTION 1.06/4.06 (1.00) The reactor is critical in the intermediate range, BOL, when the reactor. operator. positions control rods to achieve a 1 DPl! startup rate. Assuming no further operator action,.at what level will reactor power stabilize? State all assumptions and show all work. ANSWER 1.06/4.06 (1.00) NOTE: Answer will depend on assumptions made by the candidate regarding the. values of beta-effective,' lambda and power defect. Acceptable ranges: Power defect = 14-16 pcm/% power Lambda = 0.1 j. beta eff = .0052 .0062-Tau =26.06/SUR NOTE: .25 credit for solution rho =(beta eff)/(1+1ambda* tau) NOTE: .25 credit for solution based on candidate assumptions power level = rho /(pcm/% power) NOTE: .5 credit for solution based on candidate assumptions 2
V (J ternate methods of calculation evaluated on a case basis) REFERENCE TRHB 32.4, pp. 17-24 192007K113 192008K110 ..(KA's)
RESPONSE
Acceptable ranges should be expanded based on the attached RODS. Beff range should be expanded to .0068 for BOL Power defect values should be allowed per ROD-7 tables. NRC Resolution Comment accepted. Answer key modified. Facility Comment: OUESTION 1.11/4.11 (1.50) Assume that during the last refueling outage, impulse pressure to the rod control system was calibrated so that Tref reads five degrees higher than.it ' normally would-read. What effect will this have on the following steady-state plant parameters? Limit your answer to I.NCREASE, DECREASE, or REMAIN THE SAME. Assume rod control and EHC are in automatic.
- a. Total Steam Flow
- b. Secondary Efficiency
- c. Reactor Power Answer 1.11/4.11 (1.50)
- a. Remain the same
- b. Increase
- c. Decrease (.5ea)
REFERENCE TRHB 39.2, pp. 7 - 8, 21 Westinghouse Thermo-Hydraulic Principles, pp. 7 - 67 193005kl03 ..(KA's) t
RESPONSE
Consideration for partial credit should be given in instances where differentiation is made between indicated versus actual power. There is also i 3
1 the potential for double jeopardy involved ;n answering parts b and c. Partial credit should be given in instance. where a wrong answer for part b led to an incorrect assumption for part c. NRC Resolution: Since not enough conditions were given in the question to properly limit the answers, parts a and c are deleted. Part b. remains, and the answer remains unchanged. Credit is reduced to 0.5 points. Facility Comment: QUESTION 1 18/4.18 (1.00) A pressurizer level instrument develops a leak in its reference leg. Which ONE of the following statements best describes how and why indicated pressurizer-level will change? ,
- a. Level increases because of less mass in the reference leg, causing DP in the bellows to decrease.
- b. Level increases because of less mass in the reference leg, causing DP in the bellows to increase.
- c. Level decreases because of-less mass in the reference leg, causing DP in the bellows to decrease,
- d. Level decreases because of less mass in the reference leg, causing DP in the bellows to increase.
ANSWER 1.18/4.18 (1.00) a R2FERENCE TRHB 13.6, pp. 5 6 191002K109 ..(KA's)
RESPONSE
PBNP uses two sealed bellows in the reference leg of the pressurizer level instrument. Depending on the leak location and which bellows is referenced, the answer could be either A or B. Due to the configuration of our pressurizer level instrument, either answer A or B should be acceptable. 1 Recognition that indicated level would increase is the important response required. Sinet two bellows (one for sealed reference leg and one for the "DP cell") are u;ed, either A or B should be correct. 4
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h NRC Resolution: 1 The possible answers describe a cause-effect relationship between bellows DP. 4 and indicated level. Since DP across the sealing bellows has no direct effect i on indicated level, assuming that " bellows" meant the sealing bellows.will i have no effect on the answer. The facility pre-exam review examination team requested no changes for this question. In addition, no clarifications were asked for during the conduct of the examination. Answer key remains unchanged. Facility Comment: QUESTION 1.21/4.21 (1.00) A centrifugal pump with a variable speed motor uses 2 KW to develop a discharge-pressure of 50 psi. It is desired to increase the discharge pressure of the-pump to 100 psi by changing the pump speed. How much power will the motor require to develop the increased discharge pressure? Show all work. ANSWER 1.21/4.21 (1.00) Head proportional to speed squared so: if head doubles, (N squared) doubles, or speed increases by a factor of 1.414 (.5) Power proportional to speed cubed so: 1.414 cubed = 2.83 or power increases by a factor of 2.83 (.25) final power = initial power-* increase = 2* 2.83 = 5.66 KW (.25) REFERENCE TRHB 40.2, p. 6 1.9100SK104 ..(KA's) RESPONQ -{ l An error of +/ .1 ks should be allowed since c. calculations are involved and l" rounding off to the nearest tenth should be ac eptcble. NRC Resolution: Comment accepted. Answer key modified. Facility Comment: QUESTION 5.04 (1.50) LER 88-002 describes a situation where electric strikes on fire docrs were 5 i
-. -----_ --- _ _ -- _ ___J
L, y
. designed to fail open on a loss of power. These doors were declared inoperable in accordance with Technical Specifications. What were the two factors that were used as a basis for declaring these doors inoperable?
ANSWER 5.04 (1.50) Fire near~the door could cause a loss of power (.75):and delta"P generated by the fire could cause the door to open (.75) REFERENCE LER 88-002 000067G004 ..(KA's)
RESPONSE
Student may answer that the door was no longer sealed shut by latch, thereby making it' inoperable. This response should also be acceptable for partial credit. NRC Resolution: Comment accepted. The facility requested answer is acceptable for half credit. Facility Comment: QUESTION 2.04/5.05 (1.00) LER 88-010-01-describes a situation where 480 volt safeguards busses 2B04 and 2803 were tied together for maintenance, with the G01 diesel generator red tagged out of service. Describe what Technical Specification (s) was/were violated.. (Specific Tech. Spec. number not required.) ANSWER 2.04/5.05 (1.00) 3 safeguards busses were considered inoperable (violated TS 15.3.7.B.1.e) REFERENCE t.ER 88-010-01 TS 15.3.7.B.1.e
.000056G003 ..(KA's) 1
?
RESPONSE
The fact that with no emergency power supply safeguards components may be inoperable should be ali acceptable answer. An answer that identifies specific components as being " inoperable" should be given full or partial credit. The question asks for a description of the ]
)
6
- _ _ _ _ - _ i
Technical Specification being violated. The answer key states how the Tech Spec was violated, not a description of the Specification.
.NRC Resolution:
Comment partially accepted. Answer key modified to read "one diesel generator was inoperable without the ESF equipment associated with the other diesel generator being operable." Similar wording accepted. An alternate acceptable answer is " Containment Spray, service water, component cooling, RHR, and two charging pumps for Unit'2 are inoperable". Facility Comment: QUESTION 2.06/5.07 (1.00) Choose the ONE correct answer: Assume that rod control is in MANUAL when a dropped rod causes a runback. Pre surizer level will:
- a. decrease because of the drop in Tavg with the dropped rod
- b. decrease, because of a lower setpoint with lower turbine power
- c. remain the same, because of lower Tavg being offset by the runback
- d. increase, because of heatup during the runback ANSWER 5.07 (1.00) d REFERENCE !
TRHB 13.5 TRHB 13.6 J Point Beach Requal Question 043-02 l 000002A106 ..(KA's)
RESPONSE
1 If the candidate assumed that the dropped rod worth was greater than the power- l defect (FSAR assumes cither 500 or 75 pcm) for the 20% runback, then A or B 1 should be correct since Tref will be lowered due to turbine power, steam dumps will maintain Tave within 4 degrees of Tref which will be a lower value than 1 full power Tave which in turn causes pressurizer level to decrease, i Other variables that can affect the answer are: moderator temperature coefficient, rod worth, and power defect. NRC Resolution: Comment accepted. Question deleted. 7
This question was based on a question contained in the facility requalification test bank. It is requested that the facility update or delete the associated question in the requalification test bank. Facility Comment: QUESTION 5.09 (1.50) Assume that you are the SR0 on Unit 2 following a reactor trip and safety injection because of a steam generator tube rupture. While performing actions. in E0P-3, you are informed by the RO that you are in the foldout page red-path for heat sink based on the fact that no feedwater is being supplied to the steam generators. What actions do you take based on this report? ANSWER 5.09 (1.50) Check the status of all CSF status trees (.75), and go co the procedure with the highest priority. (.75) (Alternate: have DTA run status trees and exit to highest priority procedure.) REFERENCE ERG Executive Volume, Generic Issues, Foldout Page Items, p. 6 000038G012 ..(KA's)
RESPONSE
The statement made by the R0 in this question is incorrect. No feedwater is being supplied to the steam generators does not put you into the E0P-3 foldout page red-path for heat sink. Any answer that recognizes this fact and goes into determining if a heat sink red path really exists should be acceptable for full credit. The DTA should commence checking status trees in E0P-0 or when exiting E0P-0. This question states "while performing actions in E0P-3, - ." NRC Resolution Comment accepted. Answer key expanded. Facility Comment: _ QUESTION 2.16/5.19 (2.50) Concerning a faulted steam generator:
- a. List or describe 4 indications that a fault is INSIDE containment.
(1.0)
- b. Explain why feed flow is stopped to a steam generater that is faulted i
upstream of the t1SIV and outside of containment. (1.5) 8
' 5, .If {
.g
- ANSWER 2.16/5'.19 ~ .(2.50).
- a,s ~ containm3nt pressure increase containment temperature increase:
containment sumpflevel increase containment relative humidity increase- (.25 ea)
- b. since the steam leak is unisolable (.5), the SG must boil dry to.stop.the
. leak - ('. 5 ) . Contiaued feed flow would result in continued: steam generation (.5)..
REFERENCE E0P-0 E0P-2 . . , Point Beach-Requal.. Question 031-03 000040K306 ..(KA's)
' RESPONSE The student may answer that the continued steam generator cooldown will cause-
- the RCS to cool down' adding positive reactivity to-the core. The question..is-confusing- since in part "a.", the operator is dealing with a problem '.'INSIDE" containment,.and in part "b" he is referenced "outside"' containment. Equal
- emphasis should be used to' direct'the operator's attention appropriately.
.NRC Resolution:
Comment noted. However, no confusion was evident during the administration
- of the written' examination.
Facility Comment: QUESTION 3.01/6.01 (3.00) Assume that RCS pressure is 2035 psig when the controlling channel for pressurizer pressure fails low.
- 1. ; Describe'the changes that occur for the following items. Limit ycur answer to 1HCREASE, DECREASE, REMAIN'THE SAME, OPEN, SHUT, FULL ON, PARTIALLY ON, FULL OFF, ON, OR OFF.
- a. Actual Pressurizer Pressure b ~. Spray Valves
- c. Proportional Heaters
- d. Backup Heaters LL 9
E 2. . Choose the ONE answer that describes the response of the OT delta'T trip ,. , SETPOINTS. p
- a. The channel associated with the failed pressure instrument will decrease, and all others will not be affected.
- b. All channels'will decrease,
- c. All channels will be unaffected.
- d. The channel associated with the failed pressure instrument will decrease, and all others will increase.
ANSWER 3.01/6.01 (3.00)
- 1. a. increase
- b. shut c.. full on
- d. on (.5 ea)
- 2. d (1.0)
REFERENCE TRHB 10.3, pp. 12 - 14 TRHB 13.3, pp. 4 - 5 012000A205 ..(KA's)
RESPONSE
Part 2, either A or D may be the correct answer depending on what.is assumed to have happened in part 1.a. or if you only take into consideration the instrument channel failure and not the overall plant response. NRC Resolution: i Comment not accepted. The question is formatted to have two separate parts, both associated with the stem of the cuestion. Both parts of'the question require that the candidate be able tc analyze plant response to an instrument failure. Part 2 of th'a question was not meant to limit the res sonse to only the instrument failure; overall plant response must be tacen into consideration. No limiting statements were made in the questions and no questions were asked of the examiner on this issue during the course of the examination. Answer key remains unchanged. Facility Comment: in
l
' QUESTION 3.11/6.11 (2.00)
Assume that all systems are in automatic with the reactor at 100% when a turbine runback occurs. Power stabilizes at 70%. a.- What overall effect will the runback have on the OT delta T setpoint? Limit your answer to INCREASED, DECREASED, or REMAINED THE SAME. (0.5) i
- b. Identify the parameters that affect the OT delta T setpoint, and what effect that each will have on the OT delta T setpoint, following the completed runback.
ANSWER 3.11/6.11 (2.00)
- a. Increased (.5) I
- b. Tavg dropped which caused additional credit.to-the setpoint'(.5) F (delta flux) changed due to large negative delta flux, adding penalties (.5).
Pressurizer pressure will increase, causing' addition credit (.5) (will accept: no change in pressurizer pressure) REFERENCE TRHB 13.3 Point Beach Requal Question 053-02 021000g010 ..(KA's)
RESPONSE
Delta flux may not have an affect on OT delta T if it did not go out of band. This should also be acceptable. NRC Resolution: Comment accepted. Answer key modified. QclityComment: QUESTION 3.13/6.13 (2.09) While reducing load, the " Power Range tnannel Deviation" alarm comes in. l Plant conditions prior to the alarm, and after receiving the a* sara, are as i follow!: ] 4 Paramecer Prior to Alarm After Aiarm N41 100% 85% N42 99.5% 86.5% PM3 100% 85% l I 1 __. _ l
I l. N44 100% 87% Rod G3 215 steps 204 steps Rod C7 216 steps 205 steps : Rod;G11 215 steps 215 steps Rod K7 214 steps 204 steps NTILT 15 1.001 0.989 NTILT 25 0.996 1.007 NTILT 35 1.001 0.989 NTILT 4S 1.001 1.013
- a. What is the cause of the " Power Range Deviation" alarm?
- b. .What actions should be taken to avoid violating Technical Specifications?
ANSWER 3.13/6.13 (2.00)
- a. Rod G11 misaligned (failed to drive in) (1.0)
- b. Rod Gil must be investigated to determine why it is misaligned (.5) and then realigned (.5) (alternate: follow actions of AOP-6B)
REFERENCE Tech Specs 15.3.10.B.3, 15.3.10.D Point Beach Requal Question 039-03 001000K507 ..(KA's)
RESPONSE
Part A, the " Power range deviation" alarm is brought in by a 2% difference in power. Credit should also be given for this response. Part B, since you are approaching a quadrant power tilt conditions, the student may make reference to correct this prior to exceeding this limit. Credit should be given for this response. NRC Resolution Comment for part A sccepted. Part B - Rod realignment is the means used to correct a quadrant power tilt. Full credit will be given if the candidates describe how the tilt is corrected; no credit will be given if the candidates only states that the quadrant power tilt must be corrected. i 12
1 U. S. NUCLEAR REGULATORY COMMISSION
- REACTOR OPERATOR LICENSE EXAMINATION REGION 3
' FACILITY: Point _geach_1_g_2________ f J
l REACTOR TYPE: PWR__-WE_C_2_________________ i DATE ADMINISTERED: 99fg6/26_________________ j CANDIDATE: _________________________ IN!I5gCIJgNg_]Q,CSNglg81El Write answers on one side only. Use separate paper for the answers. sheets. Points for each Staple question sheet on top of the answer The passing question are indicated in parentheses after theand question. grade of at grade requires at least 70% in each category up asixfinal hours after least 80%. Examination papers will be picked (6) ; the examination starts.
% OF CATEGORY % OF CANDIDATE'S CATEGORY
__Y969E_ _lgl@L ___SCg5E___ _y@LgE__ ______________C91 egg 5Y_____________ 23.00 24.47 REACTOR PRINCIPLE 43 (7%) 1. _21:22__ _2;322 ___________ ________ THERMODYNAMICS ( ~, % ) AND COMPONENTS (11%) (FUNDAMENTALS EXAM) 72.00 %9.11 ________ 2. EMERGENCY AND ABNORMAL PLANT
;E3:55__ _52 21 ___________
EVOLUTIONS (27%) 45{14 ________ 3. PLANT SYSTEMS (38%) AND _52 99__ _55i2_7 ___________ PLANT-WIDE GENERIC RESPONSIBILITIES (10%) 94 00 TOTALS _2s 92__ ___________ ________% FINAL GRADE All work done on this examination is my own. I have neither Gi ven nor recei ved sid. Candidate's Signature MASTER COPY wK-___._-__..._____ _ _ . - - - - - - - - - - - _ _ - _ _ _ _ . _ . _ . _ . _ _ . _ . - - _ . __ .__ _ _ h
y'. '. 1 NRC RULES AND GUIDELINES FOR LICENSE EXAMINATIONS During the administration of this examination the f ollowing rules apply:.
?1 '. Cheating on the examination. means an automatic denial of your application-and could result in more severe penalties.
2.,After the examination'has been completed, you must sign the statement.on
~
.the cover sheet indicating that the work is your own and you have This not-must be received or given assistance in completing the examination.
done after you complete the examination.
- 3. Restroom trips are to be limited and only one candidate at a time may leave. You must avoid all contacts with anyone outside the examination room to' avoid even the appearance or possibility of cheating.
4 Use black ink or dark pencil only to f acilitate legible reproductions.
'5.- Print your name in the blank provided in the upper right-hand corner of the examination cover sheet.
I
- 6. Fill' in the date on the cover sheet of the examination (if necessary) .
- 7. You may write your answers on the examination question page or on a
- separate sheet' of paper. USE ONLY THE PAPER PROVIDED AND DO NOT WRITE DN THE BACK SIDE OF THE PAGE. l B. If you write your answers on the examination question page.and you need.
more space to answer.a specific question, use a separate sheet of theDD NOT paper provided and insert it directly after the specific question. WRITE ON THE BACK SIDE OF THE EXAMINATION QUESTION PAGE.
- 9. Print your name in the upper right-hand corner of the first page of each section of your answer sheets whether you use the examination question pages or separate sheets of paper. Initial each page. ,
- 10. Before you turn in your examination, consecutively number each answer sheet, including any additional pages inserted when writing your answers on the examination question page.
- 11. If you are using separate sheets, number each answer as to category and number (i . e. 1.04, 6.10) and skip at least 3 lines between answers to allow space f or grading.
- 12. Write "End of Category " at the end of ycar answers to a category. .
- 13. Start each category on a new page.
14, Write "Last Page" on the last answer sheet. p I
- 15. Use abbreviations only if they are commonly used in f acility literature, Avoid using symbols such as < or > wign4 to avCid a simple, transposi tion error resulting in an incorrect answer. Write it out.
- 16. The point value for each question is indicated in parentheses af ter the f
l question. The amount of blank space on an examination question page is NOT an indication of the depth of answer required. ! I L )
17.. Show 'all celculatione, m2thods, or assumptions unsd to obtain on answsr. ,
- 18. Partial, credit may be given. Therefore, ANSWER ALL PARTS OF THE QUESTION AND DO NOT LEAVE ANY ANSWER BLANK. NOTE: partial credit will NOT be j
given on multiple choice questions.
~
- 19. : Proporti onal . grading will be applied. Any additional wrong inf ormation that,is provided may_ count against.you. For example, if a question is.
L worth one point 'and asks f or four responses, each of which is worth 0.25 - ! points, and you give five responses, each of'your responses will be worth 0.20' points. If one of your five responses is incorrect, 0.20 will be l l deducted and your total credit for that. question will be 0.80 instead of 1.00 even though you got the four correct answers.
- 20. If the intent of a . questi on is unclear, ask questions of the examiner. j onl y. - i
'23. When turning in your examination, assemble the completed examination with In addition, l
ex ami n ati on questi ons, examination aids and answer sheets. I turn in all scrap paper.
- 22. To pass the examination, you must achieve'an overall grade of 80% or greater and at_least 70% in each category.
'23. .There . i s a ti me limi t of (6) hours f or completion of' the examination.
(or- some other time if less that the full examination is taken. )
- 24. When _ you are done and have turned in your examination, leave the examin-ation area as defined by the examiner. If you are f ound in this area
.while the examination is still in progress, your li cense may be denied or revoked.
.. i iI i
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PO92 1 cf 5 SI - DOI 6_ESEEI eCEOCIQB_IBEDBY_EQBdWLOS:- SUR(t)
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_1 . A 1 1 N (N ) H (N ) P 2' 2- 2 2- 2 2 B001eIIQN_eND_GHEd1EIBr_EQBOUL8E R/hr = 6CE/d I, = 10 m CVgg =Cv22-
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G'= DLlutten_Batt 10 C=CO' Volume . . . A=A e A = AN O GQWEBE10NE: 1 gm/cm = 62.4 lbm/ft Density of water (20 C) = 62.4 lbm/ft 1 gal = B.345 lbm
'1 ft = 7.48 gal Avogadro's Number = 6.023 x 10 1 gal = 3.78 liters Heat of Vapor (H O) = 970 Btu /lbm 2
1 lbm = 454 grams Haat of Fusion (ICE) = 144 Btu /lbm
~* grams-e'= 2.72 1 AMU = 1.66-x 10 w = 3.14159 Mass of Neutron = 1.009665 AMU f 1 KW = 738 ft-lbf/sec Mass of Proton = 1.007277. AMU 1 KW = 3413 Btu /cr Mass of Electron = 0.000549 AMU
't HP = 550 ft lbf/ sac. One atmosphere = 14.7 psia = 29.92 in. Hg 1 HP = .746 KW . 'F = 9/5 *C + 22 1 HP = 2545 Btu /hr 'C = 5/9 (* F - 32) i 1 Btu = 778 ft-lbf eR = *F + 460
-6 'K = *C + 273
.1 h.IV = 1.54 x 10 Otu
~I M-sec h = 4.13 x 10 10 fissions /sec '
1 W = 3.12 x 10 2 = 931 MEV/ AMU O g. = 32.2 lbm'-f t/lbf-sec c l '1 inch = 2.54 cm C = 3 x 109 ,f,,c r = 0.1714 x 10 Btu /hr ft R L - - _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ -
.&.....,.s-.,, _. - - - - - - - - = - - - - - -
Paga 4 oL5_
- DGIe_SBEEI-i eVEseGE_IHE858L_CONDUC11M1IL1El K
BatacLal 0.025 Cork- 0.028 Fiber Insulating Board Maple. or Oak Wood - 0.096-
.Guilding Brick _ 0.4
'O.45
.. Wi ndow ' Gl ass .
Conceete O.79=
-17. Carbon Steel 25.00
'11 Chrone-Steel 35.00
. Aluminum. 118.00 Ccpper 223.00-235.00 4 Si1ver Wcter (20 psia; 200' degrees'F) 0.392 Steam (1000 psia, 550 degrees F) 0.046 Uranium Diex1'de. 1.15
' Helium 0.135
-Zircaloy 10.0
' t$1SCELLeNEDUS_INEDBtS8Il0N:
-E = me 2 ,
KE'= 1/2'mv PE = mgh' V '=VO + at-f . Area Volume
' Geometric.Ob.fect
-A = 1/2 bh /////////////////
Tri angl e .
-Square- . A = S* /////////////////
A=LxW /////////////////
' Rectangle Circie. A = wr* /////////////////
l Rectangular Solid A = 2(LxW + LxH + WxH) V=LxWxH Right* Circular Cylinder A= (2 wr#)h + 2(wr*) V = wr#h l Sphere A = 4 wr# V = 4/3 (wr#) Cube ///////////////////////////// V'= S*
1 Paga 5 of 5 Dele _SUEEI diSCELLONEQUS_INEDBdBI1QN_1coattouedi 10 CFR 20 Appendix 8 Table I Table II Gamma Energy Col I Col 11 Col I Col II MEV per Air W3ter Air Water Msterial Hal f-Li f e Disintegration uc/ml uc /ml uc/ml uc/ml
-6 ~0 Ar-41 1.84 h 1.3 Sub 2x10 ----- 4x10 ------
l
~# ~# ~8 -3 Co-60 5.27 y 2.5 S 3x10 1x10 1x10 5x10
~D ~AU ~7 1-131 0.04 d 0.36 S 9x10~ 6x10 1x10 3x10
~D Kr-85 10.72 y 0.04 Sub 1x10 ----- 3x10"# ------
~# ~3 ~ ~
Ni-65 2.52 h O.59 S 9x10 4x10 3x10 " 1x10 *
-A2 ~
~A* -6 Fa-239 2.41x10" y. 0.008 S 2x10 1x10
- 6x10 5x10
~Y ~D 'I ~#
Sr-90 29 y ----- S 1x10 1x10 3x10 3x10 l
~D ~
Xo-135 9.09 h O.25 Sub 4x10 ----- 1x10 * ------
> 2 he Any single which doesradionuclides not decay by alpiawith Tb [ 3x10
~
9x10
~
1x10
~I 3x10
-6 spontaneous fission 2
Neutron Energy (MEV) Neutrons per cm Average flux to deliver equivalent to 1 rem 100 mrem in 40 hours 6 , 670 thermal 970x10 6 280 (neutrons) ! 0.02 400x10 6 30 O.5 43x10 , cm3----ee 10 24x10 17 xs i Linear Absorption Coefficients p (cm~A) Energy (MEV) Water Concrete Iron Lead 0.5 0.090 0.21 0.63 1.7 1.0 0.067 0.15 0.44 0.77 1.5 0.057 0.13 0.40 0.57 ** 2.0 0.048 0.11 0.33 0.51 2.5 0.042 0.097 0.31 0.49 3.0 0.038 0.088 0.30 0.47 l
P0ga 4. 1 1.- REACTOR PRINCIPLES (7%) THERMODYNAMICS 1Z31_egp_CgggggEglS_J1131_J[UNp83ENI8bS_EX801 Y* 4 l . DUESTION 1.01 (2.00) Will the. shutdown margin INCREASE, DECREASE, or REMAIN THE SAME f or the l f oll o.ving condi tions? (Consider each case separately)
- c. Load rejection from 100% to 90% power, control rod position lowers and T-avy =quals programmed T-avg.
- b. The reactor is steady state at 50% power when a small dilution occurs with control rods'in automatic.. .
l )- c. The reactor is steady state ' at 50% power when a small dilution occurs with control rods are in manual.
< d. Reactor power holding at 50%, xenon is increasing.
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Pcga 5~ , LI:__6E99198_[81N{l[LE@_JZ31_IUE@gggyN9919@' { IZ31_6Np_ggg[QNENIS_J1131_1[UNp8DENI@L@_EX801 l I QUESTION 1.02 (1.00)
> An equivalent positive reactivity addition is made'to two critical reactors, one of which is at BOL and the other at EOL. Choose the DNE correct answer that describes the difference in the startup rate on the two roactors.
- c. Larger SUR for the'BOL reactor because Beta Bar ,ef f ective is larger for the BOL reactor.
. b. Larger SUR f or the BOL reactor because Beta'Bar ef f ective is smaller for the BOL reactor.-
ef f ective is larger f or
- c. Larger SUR f or the EDL reactor because Beta Bar the EOL reactor.
Larger SUR for the.EDL reactor because Beta Bar effective is smaller d. for the EOL reactor. , 4 8 .
\
1 (***** CATEGORY 1 CONTINUED ON NEXT PAGE *****) 1
c
; , m
.L1k.~ REACTOR'PRINCIPLEp_jZM ,THERDppVNSDJg!- -Pcg3: 6, 1231.eNp_ Q QDEQNg $1g_j } } Zl_JE((NpSDg $18($_g f @D1 -
a. 32 -
', . 03 - (1.00):
DUESTION
' Choose'.the ONE answer whi ch correctl y. completes _ the bl anks:
The curve of' critical boron concentration vs. core lif e' has an initial
;cteep' drop 1 f ollowed' by a relatively constant concentration f or a short psriodt of time, L af ter which concentration drops linearly with~ burnup.
7The reason for the initial-steep drop i s d ue t o ' _______ ( 1 ) ______ and the:
~
froason. f or the short period of level concentration is due to __(2)._____+
- c. . .(1) buildup of. fission product poisons (2) depletion of: burnable poisons
'b. (1). dep'letion-of' burnable poisons (2). buildup of fi ssion product poisons .
- c. (1) dilution toward initial criticality
-(2) ' depletion of- burnable' poisons .
, d.- '(1) dilution' toward initial criticality (2) buildup of . fission product poisons i.
L. l l p l" l .\ L-. ! 4 i L (***** CATEGORY 1 CONTINUED ON NEXT PAGE *****)
- _ _ _ - . . L. _ .. : .
Pcga 7
- 1. REACTOR PRINCIPLED (77.) THERMODYNAMICS lZ31_999_C90PgygyJg_jligl,jggyp@ggNI@Lg_E38gl QUESTION 1.04 (1.00)
Which DNE of the f ollowing statements is correct concerning various effects on Differential Boron Worth (DBW) over core life?
- a. Boron concentration decrease with core lif e drives DBW more negative, and fission product pc' iso 1 buildup with core lif e drives DBW more negative
- b. Boron concentration decre.:e with core lif e drives DBW 1ess negative, and fission product poison buildup with core lif e drives DBW 1ess negative
- c. Boron concentration decrease with core life drives DBW more negative, and fission product poison buildup with core lif e drives DBW 1ess negative
- d. Boron concentration decrease with core lif e drives DBW 1ess negative, and fission product poison buildup with core lif e drives DBW more negatise .
e l e e 1 (***** CATEGORY 1 CONTINUED DN NEXT PAGE *****) l l l w _ ______- .-
Pcg3 8 1 '-REACTOR PRINCIPLES'(7%) THERMODYNAMICS IZZ1_0Np_CgggpNENIS_J1131_jEyNpSDENI@bS_E38dl QUESTION 1.05 (1.00) _ Which DNE of the following statements is correct concerning the overall of f ect on- Dif f erential Boron Worth (DBW) over core life?-
- c. boron concentration decrease is the dominant factor, which makes DBW more negative with core lif e.
- b. fission . product poison buildup is the dominant f actor, which makes DBW more negative with core lif e,
- c. boron concentration decrease is the dominant f actor, which makes DBW 1ess negative with core lif e.
- d. fission product poison buildup is the dominant f actor, which makes less negative with core life.
- o. boron concentration decreast is offset by fission product poison buildup, maintaining DBW essentially constant over core lif e.
5 5 S . e (***** CATEGORY 1 CONTINUED ON NEXT PAGE *****) u_________-_. ___ -_
1 REACTOR'NRINCIPLES"(72) THERMODYNAMICS'
~-
Peg:2 9 nn_eua_C9eegusu1S_1nn_seougeoguresS_nent ) 1 i
-- DUESTION 1.C6 (1.00) ls UThe reactor-is critical' in the intermediate range, BDL,-when the reactor .f operator- positions control rods to achieve a.1 DPM start-up rate. Assuming:
no f urthe. . operator action, at what l evel will reactor power stabilize? State' all assumptions and show all work. , l l 4 e e e f I i (***** CATEGORY 1 CONTINUED DN NEXT PAGE *****) l t .--a--_.__-.. ______,.__.____________________:__
t -
' P ga'10...
.1. / REACTOR PRINCIPLES (7%). THERMODYNAMICS
- .. -IZ?l;eUp_CggPgyEUIS_jll31.j[gypSDEUIALS_EX8Dl.
( p(: w. lDUESTION. .1.07- ( 1 ~. 50 ) ' a, Why should1 verification of natural circulation conditions- be perf ormed on a ' ' F TRENDING basis versus an instantaneous. set of readings? i e i e O S 4 (***** CATEGORY 1 CONTINUED ON NEXT PAGE *****) Em - _ _
~
Pcgs:11 fl.. REACTDR PRINCIPLES ~(7#/,F THERMODYNAMICS f
.lIZ1_8Np_CgDEgNEDJ@_jll31_j[UNp3 DEN]Skg_E390[
, e LDUESTION 1.00- (1.00)
J Which-DNE of the f ollowing prevents DNB f rom occurring during normal- plant
~
operations?
. o . 0; . Mini mum temperature f or criti cali ty .
- b. ~ Heat-Flux Hot. Channel. Factor ,
c .. ; RCS pressure safety limit
- d. . Enthalpy Rise Hot Channel Factor O. Maximum'KW/ft' limit i
I 1 (***** CATEGDRY 1 CONTINUED ON NEXT PAGE *****f i
)
l 1 ll l 1. l. f
<k Pcgi 12
^
t-r1. . ' REACTOR-PRINCIPLES ~(77) THERMODYNAMICS 12Z1_eUP_990E90ENIE 11121_lEWUP90EUIS(g_E13D1_
.. DUEST ION 1.09. ( 1~. 00)
Whi ch ' ONE ' of. the f oll owing i s NOT characteristic of. f ast neutron' irradiation on reactor vessel metals?
- o. . Increased strength.
b.- Decreased ductility
- c. Nil-ductility Temperature (NDT) decreases.
- d. A change in-the lattice structure.of the metal.
. 2, Increased brittleness 9
4 O OO I (***** CATEGORY 1 CONTINUED DN NEXT PAGE *****)
'Pago 13 l '. REACTOR ' PRINCIPLES - (T/.) THERMODYNAMICS
'zl?31_SUP_C9dE90Eylp_JJ131_jEgyp3DEyISLS_E33D1
. 4 DUESTION 1.10' (2.00)
For the f o13 owing changes . in ' pl ant status,. indicateConsider whether the DNBR in the each change core will INCREASE,.DECPEASE, or. REMAIN THE SAME. The. reactor c parately, and assume all other parameters remain unchanQed. (.5.ea) io initially at 50% power. c.- ' Increase' reactor. power
. b .' Increase.CVCS charging and letdown.
- c. ' Increase pressurizer pressure
- d. , Increase core inlet temperature G .
4 P L (***** CATEGORY 1 CONTINUED ON NEXT PAGE *****) I
- 1. REACTOR PRINCIPLES (7Z)'. THERMODYNAMICS . Pcga-14 L JZZ1_8SP_99dE90ENIE_11131_lE9EP8dEyl36@_E3901 4
DUESTION' - 1.11' +i.0^: (OM Assume that dur.ing the last ref ueling outage, impulse pressure to the rod control system was calibrated so that Tref reads five degrees higher than it normall y woul d read. What effect will this have on the f ollowing steady-state plant parameters? Limit your answer to INCREASE, DECREASE, or. REMAIN THE SAME. Assume rod control and EHC are in automatic.
-: . 'rt s! Etr: "!.m_ beleded
- b. Secondary Efficiency
.:. 9:::t2 "
;. . : --- Ddt4<J O
e S
# e
(***** CATEGORY 1 CONTINUED ON NEXT PAGE *****) l
------_mm_.. m.._mm, ,,
il___BE99198_BBJUg186E!_,1231_IMEED9pyy3DICS Pcg2 15-1.!Z31_eUP_990E90ESI!_.!!!Z1_1EuypeggyIegS_Ef 9D[- d
- - QUESTION 1.12- (1.00)
'The reactor,is.in a condition;with average.. temperature at 557 degrees-F and.
PZR - pressure at .2250 psi g~ pri or to start-up. Calculate- the subcooling margin using the steam tables. 4 9 O O e (***** CATEGORY 1 CONTINUED DN NEXT PAGE *****) "-^-E_- - - . . _ _ _ _ _ _ _ _ _ _ , _ _ _ _ _ _ , _ _ _ _ _ _ _
,7 7,,
. i. . Pqg? 16~
H 1 REACTOR PRINCIPLES: ( 77. ) THERMODYNAMICS. L JZ31_eND_CQUEpNEUIg_JJJZl_jEUND90ENISLg,E3901 4; e QUESTION. 1.13 (1.00) Which DNE of the f ollowing is used for. purposes.other than prevention or rGduction of water hammer?
- o. : maintaining'~ piping systems full of fluid at all times.
- b. starting' centrifugal pumps with the discharge valve shut
- c. draini 7 steam lines with steam traps and drain valves
- d. . ensuring that systems are pressurized prior to operation
- 6. opening valves with a large DP across them slowly e
9 e e 1 CONTINUED ON NEXT PAGE *****) (***** CATEGORY
-- - . _ _ _ _ - _ _ . _ _ _ _ _ j
,r-Paga 17.
' i.- REACTOR PRINCIPLES (7%) THERMODYiGri? CS '
IZ31_eN9_C90E90EyIS_f1131_f[UDp@DEUISh5_Ef8D1
+
s
- DUESTION 1.14 (1.00)
Rsactor Bypass breaker A is racked in and closed f or testing with Reactor Choose-Trip breaker A open. Reactor Bypass breaker B is then racked in.
- the DNE correct response that describes what happens when bypass breaker B
.10 racked in:
i
- c. - bypass breaker A will remain shut and trip break,er B will trip open
-b. bypass breaker A and trip breaker B will remain shut
- c. bypass breaker A will trip open and trip breaker B will remain shut.
- d. both. bypass breakers and trip breaker B will trip open
- 3. bypass breaker A and trip breaker B will trip open 0
4 O . e (***** CATEGORY 1 CONTINUED ON NEXT PAGE *****) 1
1 l Pago 18 1___6E@gIQB_E61 Ngl [6E@_f?31_IdEBDggyN90lg@ IZBl_eNp_QQU[QUEUIS_fi1Zl_fEUUp@DENIS6@_EX801 QUESTION 1.15 (1.00) Actual core exit temperature has remained constant, while containment temperatures have increased from 80 degrees F to 120 degrees F. Which DNE of the f ollowing statements describes how the core exit thermocoup* c readings will react to the containment temperature change?
- o. readings will remain the same because heaters in the thermocouple junction box will offset any changes in containment temperature
- b. readings will remain the same because thermocouple wire is connected-directly to the thermocouple display
- c. readings will decrease because the change in containment temperature will decrease the potential between the two thermocouple legs
- d. readings will decrease because the change in containment temperature will increase the potential between the two thermocouple legs
- o. readings will increase because the change in containment temperature will add to the signal generated by the core exit temperature
- f. readings will increase because the legs of the thermocouple are unevenly af f ected by the containment temperature change
(***** CATEGDRY 1 CONTINUED ON NEXT PAGE *****) ( l l 1
- J
- - - - ~_ _ - - _ - _ _ _ _
Pcgn 19 l 11__8E6E198 E81UElfhEE_l?31_ldgBDppyyeDJCS l?Bl_BUP_G90E99E9I!_ll331_lguypergyIegS_gxeri i 1 1 DUESTIDN 1.16 (1.00) 1 I Sciety valves are used in the main steam systemWhich to protect the the system DNE of the f ollowing piping from damage due to overpressurization. 1 statements concerning the operation of saf ety valves is correct?.
- o. When the activating pressure for a safety valve' returns to the_ lifting set point, a combination of steam and air pressure above the valve disk closes the valve.
b.. As steam pressure increases to the safety set point, the pressure overcomes spring tension on the valve operator, causing the valve to open.
- c. As the disk on a saf ety valve lif ts, less pressure is exerted on .the disk, reducing the upward force on the disk, preventing " valve slamming",
- d. A safety valve is lifted off its seat, then is f orced f ully open by an air-operated piston. .
. )
l (s**** CATEGORY 1 CONTINUED ON NEXT PAGE *****) ( .
I~ 'L coa 20 I
#-!!$![!$p[hhhhhh!!__ffff_!!!0!!$$hf!!g_gxeM1 l
l- DUESTION 1.17 (1.00)-
'Which DNE of the f ollowing statements is TRUE f or the MIXED-BED domineralizers?
- o. Mixed-bed demineralizers are designed to remove only ionic isotopes.
1 L tn Maximum flow rate is limited: to 109 gpm to prevent resin bed l channelling.
- c. An unsaturated mixed-bed domineralizer can decrease RCS boron concentration by approximately 100 ppm.
d.. A mixed-bed demineralized can operate for only two months with 1 */. f ailed f uel. 6 4
)
e . q (***** CATEGORY 1 CONTINUED ON NEXT PAGE *****) i _ _ _ _ _ _ - - - - . l
I Pag 2 21
- 1:__BE39193 fBlypifbEg_jZ31_IdESDgpyy@digg .
.12Zl_eUP_E9dE90EGIg_fl131,fEUUQ9DEUI@65_E39D1 DUESTIDN 1.18 (1.00)
Which
% pressurizer level instrument develops a leak iri its ref ervik:e leg.
DNE of the f ollowing statements best describes how and why .htdicated pressurizer level' will . change?
- a. level increases because of less mass in the ref erence leg, causing DP in the bellows to decrease.
- b. level increases because of less mass in the ref erence leg, causing DP in the bellows to increase.
- c. . level decreases because of less mass in' the ref erence leg, causing DP in the bellows to decrease.
- d. level decreases because of less mass in the reference leg,. causing DP in the hellows to increase.
4 e e (***** CATEGORY 1 CONTINUED ON NEXT PAGE *****) 4
)
j
Pcg2 22 1___BE6919BiEB1991Ebgs_Jzz1_Igggggpyy3rigg
.l?31_eUP_E90E90EUI!_11131_lE99PSDENISkS_gX$rl I s-DUESTION 1.19 (1.00)
I Choose the ONE correct answer: A starting duty-limit (ie. restarting motor, consecutive starts) is placed on the Reactor Coolant Pumps to:
- c. prevent' damage to seals due to low leakof f flow when the pump is of f.
- b. prevent insulation damage in the rotor due to high current
- c. . prevent damage.to the lift oil pump due to repeated starts
- d. prevent damage to the labyrinth seals due to rotor thrust
- a. prevent excessive wear on the anti-rotation pawls and ratchet plate
- f. prevent insulation' damage in the stator due to high current l I
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/
Pcg2 23'
' 1. REACTOR PRINCIPLES (7%) THERMODYNAMICS 12Bl_eU9_c90PpyEylp_jl131_jEgypADEyJ$hp_E39D1 QUESTION 1.20 (1.00)
Which DNE of the- f ollowing describes the Af3TOMATIC f unction of TCV I45' (CVCS- demi ner al i z er inlet divert valve)7
- c. diverts flow around the demineralized when resin -ir," exhausted
- b. diverts flow around the demineralized when fuel elements have failed
- c. diverts flow around the demineralized when letdown temperature rises
' diverts flow around the demineralized when recharging resin beds 4
d. (***** CATEGDRY 1 CONTINUED ON NEXT PAGE *****)
._._.__ .m-. - _ . _ _ _ _ . . _ _ _ _ _ _ _ _ _ . m________m._ _ _ . _ . _ _ _ _ _ _ _ _ . _ _ . . _ _ . _ . _ _._m_ _ . _ _ . _ _ _ _ _ _ _ _ . . - _ _ _ _ _ _ _ _ . _ ~ _ _ _ . . _ __ _ . _ _ _ _ . . _ _ _ . _ . . _ _ _ _ _ . . _ _ _ _ _ _ _ . _ . _ _ . _ . _ . _ _ _ _ . _ _ _ _ _ _ - _ _ _ _ . _
.Pego 24 N il .__. R_ E A_C.T.O_R_ P_R_I N_C_I PL E_S_' _( 7__Z_)_T__H_E_R_M_D_D_Y_N_A_M_I _C_S
.1221_BUP_990E99EyIg_JJ131_JEgypSDgyl% S_g3BD1
. DUESTION 1.21. -( 1. 00) -
A centrifugal pump with a variable speed' motor uses 2 KW to develop-a discharge pressure of 50 psi. It ir. ' desired to increase the discharge
; pressure of'the pump to 100 psi by changing the pump' speed. . 'How much Show power will 'the motor require to develop the increased discharge pressure?
all-work. , 6 s
~
l 8 e l (***** END OF CATEGORY 1 *****) i a
[. Pcg3 25
.j. ,_Et,1,ERGENCY AND' ABNORMAL PLANT; EVOLUTIONS
.- ( 2. -.7 2 )
- i. t p .
4
' QUESTIDN- 12.01 (3.00)-
-As the reactor operator, what immediate actions ' must be taken by you- in
- order.to emergency' borate? Assume'all systems _are.in automatic and the
'RWST in available.
L l> 4 (***** CATEGDRY 2 CONTINUED ON NEXT PAGE *****)
. .. ..E....__1_ _ _ _ _ . _ . . . _ _ _ _ _ _ _ .
2:__EDEB950CL9Np_ggNg6D9kP699]_ gyp (p]JgNg P392 26i d?ZZ1. '
- n. :.
g .-
*n, t
- DUESTION 2.02 '(1.00)
' Fill in the blankt
.__________ is used 14 hours. subsequent'to a large break LOCA t, t;.i...... m 9 to flush. boron"from'the upper regions of the core.
(***** CATEGDRY 2 CONTINUED ON NEXT PAGE *****)
Pego 27; 2 EMERGENCY AND ABNORMAL PLANT EVOLUTIONS. J2231 DUESTION 2.03; (2.00)
;You'are performing EOP-0.2, " Natural Circul ati on Cooldown", followingLa loss of'offsite power, when you notice'that pressurizer level is varying by o large amount,' inconsistent'with the cooldown rate.
1.- Which DNE statement best describes the cause f or the wide variations in pressurizer level in this situation 7
- a. pressurizer' level instruments are inaccurate because of the loss of-offsite power.
- b. pressurizer level is swinging because of contraction of the RCS coolant during the cooldown
- c. pressurizer level is swinging because RCS pressure is varying around the accumulator injection point
- d. pressurizer level is swinging because of voiding in the reactor vessel head area
- 2. Choose the DNE statement that describes the action, level. if any, that would be taken to counteract the variations in pressurizer
- a. nothing can be done until offsite power is restored
- b. depressurize the RCS to facilitate accumulator injection
- c. repressurize the RCS to collapse any voids ,
- d. increase the cooldown rate in order to increase subcooling e e l
l (***** CATEGORY 2 CONTINUED ON NEXT PAGE *****) l
pcg) 23 E2k EMERGENCY AND'A7NDRMAL PLANT _EVDLUTIONS 12Z61
. .l QUESTION 2.04 (1.00)
LER' BB-010-01. describes m' situation where 480 volt saf eguards busses 2BO4 cnd 2B03 were' tied together for maintenance, . with thespecification G01 diesel generator rGd tagged out of service. Describe what technical (s)
- was/were viol ated. (Specific Tech. Spec. number not required.)
b l l (t**** CATEGDRY 2 CONTINUED DN NEXT PAGE *****) I 1 i a._.--__-.-- .- a-.-. . - . _ - - - - - . . . - - - - - - - - .. - _ - - - - - - - - - . _ - - - - - . . - . - , . - - . - . - - .
'T-f PegJ.29.
' 22 LEMERGENCY AND ABNDRMAL' PLANT EVOLUTIONS-
.!22n a
QUESTION 2.05' ( 1. '00 ) ' With Unit.2 at'75%. power, a safety injection.and reactor trip occurred. LThe f ollowing statements summarize key 1 indications that have been noted:
- a. One. steam generator . is 'depressurizing uncontrollably. .
- b. The RCS is losing .. inventory at approximately 200 gpm
. c.. Air ejector-. radiation alarms indicate a tube rupture.
d '. Core Exit Thermocouple are reading about 1300 degrees F. Which statement must be given first priority f or remedial action?
.e a . (***** CATEGORY 2 CONTINUED ON NEXT PAGE *****)
___.__..-__-._--___._--.-.__-2.-- - _ . - . _ - - - _ . - - - _ - . - - . _ - _ _ - . - - - -- . - _ _ _ _ - - - _ - - - _ _ - - . - - _ - - - - -
Paga 30-
- 2. EMEEGENCY AND' ABNORMAL' PLANT ~EVOLUTIDNS Jg7%) !
l 2.06 '(1.00) QUESTION LCh ose the ONE correct answer ! is in MANUAL when a dropped rod causes a runback. Assum gthat rod control Pressuri r level will: 9
- c. decrease, ecause of the drop in Tavg with the dropped rod jl
- b. decrease, beca of a lower setpoint wit h -lower turbine power c.- remain the same, be se of lower Tavg being offset'by the runback
- d. increase, because of hea during.the runback quesb 2.0L Mded
.. (
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7, I- .Pcgo 31' i 21 ' EMERGENCY'AND ABNDRMAL PLANT EVOLUTIONS ( 2 T/,) -
- t. . ;.
QUESTION 2.07 (1.00) Following a transient in which PCV-430.(Pressurizer PORV) has stuck open, cnd.MOV-516 (PORV block valve) has f ailed to close, these readings are obtained in the control room: A B Pcrameter 850 SG Pressure (psig) 850 55 50 SG NR Level (%) 20' AFW f l ow (gpm). 10. -
<530 <530 Tovg (deg!F) 890-
'RCS Pressure-(psig) 850 526 525 TCWR (deg F)- 725 Thermocouple (deg F)- 725 off off RCP's .
17.5 Vassel Level (f t) 18 CTMT Pressure - (psig) . 4.0 Subcooling Meter (deg F) >50 Superheat
'The best pump to start would be: (choose ONE) .
- a. a saf ety injection pump b.. a second charging pump
- c. a reactor ~ coolant pump
- d. a residual heat removal pump O .
(***** CATEGDRY 2 CONTINUED DN NEXT PAGE *****)
1]
'P.cga 32 1
- 2. EhERGENCY AND ABNDRMAL PLANT EVDLUTIONS
( __2_7_'/,)- _ ,
.t
]
QUESTION 2.08 (1.50)
]
Unit l' has sufIndi f ered cated. a total' SG wide' . l oss range of level all AC'from an. finitial s are 190" or "A"condition of
' SG and 195" F 200% power.-
for "B" SG. The turbine-driven AFW can only feed each SG at 75 gpm. mg 7 Explain why. CSP-H.1, " Response to Loss of Secondary Heat Sink", should or should. not be implemented. O e e OG e (***** CATEGORY 2 CDNTINUED ON NEXT PAGE *****) - - _ _ . _ _ . _ _ _ _ . _ _ _ _ _ _ _ _ _ _ _ .__._.__________m____.__._ . _ _ . _ _ _ _ _ _ _ _ _ _ . _ _ . - . . . . _ _ _
Pega 33 2___Edggggggy_9yp_3pyp3036_P($yI_gVp6pIlgyg JZZ31
'4 QUESTION 2.09 ~(1.00)
During a fire f.n the control room with of f-site AC available, reactor power hos been rampeel down to 20% power and it becomes necessary to evacuate the control room. Bef ore leaving the control room, the CVCS system should ber (choose ONE)
- c. aligned with one charging pump.in manual at minimum speed taking suction from the VCT
- b. aligned with one charging pump in manual at minimum speed discharging t o l oop A c ol d l eg
- r. aligned with one charging pump in manual at minir,um speed taking suction from the RWST
- d. aligned with two charging pumps in manual at minimum speed taking suction from the RWST 9
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Pcg2 34-
- 2:__EDEgggNgl,gDp,gpNp3Dgh,3(gNJ,gyg(gligNS 12Z3r QUESTION 2.10 12.00)
Unit 1 is shutting down and at 6% power when intermediate range channel-N36 fails high. Which DNE of the f ollowing statements -best describes how thi s fcilure affects the reactor shutdown and subsequent, operation of the Nuclear Instrumentation system? I
- c. the reactor will trip on high IR flux, and source range NI's will re-energize when N36 reaches the proper setpoint
- b. the reactor will trip on high IR flux, and source range.NI's will have
'to be manually re-energized C. the reactor will not trip, and source range NI's will.re-energize when N36 rear"- the proper setpoint
- d. the react.>r will not trip, and source range NI's will have to be manually re-energized i
I (***** CATEGDRY 2 CONTINUED ON NEXT PAGE *****) d__.____ _ _ _ _ _ _ _ _ _ _ _ . _ _ _ _. _ _ _ . _ _ _ _ _ _ _ _ . _ _ . _ _ _ _ _ _ . _ . __ ._.
Pcg2 35
'2.__gyEBgEypy_ gyp _ggyp3ygk_Pkgy],gyggpIJpyp 32z31 DUEST1DN 2.11 (1.50)
Unit'1 RCS is solid at 340 deg F and 350 psig, . with RCP A running. VCT pressure is 25:psig. A control problem develops with' letdown pressure c ontr ol valve PCV-135 (in both auto-and manual) and RCS pressure drops to 295 psig and then stabilizes. - RCP A #1 seal leakoff flow dropped from O.B gpm to 0.4 gpm, and #1 seal' delta P dr6pped from 325 psig to 270 psig. Why Explain whether or not it is permissible to continue operating RCP A. or why not? 0 0 e e e
- e
(***** CATESDRY 2 CONTINUED ON NEXT PAGE *****) 1
Pcga 36 A ,,EDEgggNgy,9yg,$gygggg(_P($NI_gyg(UI,lgNg _L2zn 4
' QUEST 1DN 2.12 (1.00)
During a power ramp from 30% to 50% on Unit 1, it is discovered th'at water 10 trickling f rom the open pipe end of the containment cooler The service valve will water not oupp)y line drain valve, SW210, at a rate of 0.1 GPM. (choose ONE) close any.further. Continued power operation ist
- c. . allowed, however the containment tooler must be declared inoperable.
- b. not allowed because the containmerit cooler is inoperable.
c.- not- allowed unless a pipe cap is installed to stop the leakage.
- d. allowed with no restrictions.
1 l l (***** CATEGORY 2 CONTINUED ON NEXT PAGE *****) _ _ _ _ _ . _ _ _ _ _ . _ _ _______________u___ _ - - - -
poge:37) h__EDEB9ESCX 9Np,$BNp3dgE_P($NJ_[yp(UllpNp
.L22n
.4' QUEST I DN .- 2.13 (2.00)- -{)
, Unit 2 is.at.98%~ power, with one _ charging pump operating in manual . at mini mum speed. The other charging pumps ARE available. You' note the-
- f ollowing data:
Parameter. t=0 min t=2 min t=4 min. Rx Power (%) 98. 98 98 Tcvg'.(deg F) 571 571 571
- TCWR((deg;F) 543 543 543.
THWR'(deg'F) 600- 400 600 1940 1900 d*ZR Press . (psig) 1985 PZR Level (%) 46 42 38 SG."A" Press-(psig) 700 700 700 SG '.'B" Press (psig) 700 700 700 SF "A" (1b/hr) 3.3E6 3.3E6 3.3E6-
~SF "B" (1b/hr) 3.3E6 3.3E6 3.3E6 SG Atmos. A closed- closed closed
-SG Atmos. B closed closed closed 2RE215-(uc/cc) 1.6E 1.OE+0 1.5E+0 Assume.that 1% PZR level = 60 gallons j Can the . charging system keep up with this transient? Support your ~ answer:
with appropriate calculations. i 9 e 4 (***** CATEGDRY 2 CONTINUED ON NEXT PAGE *****)
- - _ - - _ - _ _ __--_.-.______n
i : 9. . . . Pog2 38 ]~'(2.- . EMERGENCY AND' ABNORMAL PLANT' EVOLUTIONS l (27%)- L.e QUESTION 2.14 (1.50) Aacume'that you are the reactor operator during a reactor startup and you obtain' the f ollowing readings during the startup: N 9x 10E3 cos N-32 9.1 x 10E3 cps N-35 1 x 10E-11 amps N-36 5'x 10E-11 amps c.- What is the cause of the incorrect overlap in the above readings?: ( 0. 5) -
- b. BRIEFLY explain, per Technical Speci ficatione,' why the startup can (1.0) or cannot be continued..
O . e (***** CATEGORY 2 CONTINUED ON NEXT PAGE *****) l l i = - -- - - --- -- _ _ ---- --- -.
Pagy 39 2:__EDE6GEyCy_9yg_9pyg8D96_P6991_Eyg6Ullgy5-12Z31 I GUESTION. _2.15 (1.00) A' fire in the Unit 1 side of the control room has necessitated evacuation.
. Prior to'this, both' units were at 100% equilibri um. Which DNE of the
- f oll owing statements . i s true?
RCP's should be lef t running on Unit 2 and stopped on Unit 1.
<o,.
- b. Each containment will be isolated via' manual containment isolation.
c.' All technical specifications are to be adhered to just as they would be under other operating conditions. Safety Injection is;to.be initiated on Unit 1, then unnecessary d. equipment will be locally secured. l O . (***** CATEGDRY 2 CONTINUED ON NEXT PAGE *****) __ _ _ _ _ . _ _ _ _ _ _ _ _ _ . _ _ __1 _ . _ , . . _
m - = =- ,- . - - Pcg3140 !! 2 1_ EUE8EEUEISOUE.OEUEEDOh,Pb3NJ,[VgbyJJpNS
- ' 122ZL-t
}
- QUESTION 2.16: (2.50)
- l. .
Concerning a faulted steam generator: X 4 ;o. List or describe.4 indications.that a fault i s INSIDE containment. ( 1. O) .
- b. Explain _ why f eed' flow is ~ stopped to a steam generator that is f aulted (2.5) upstream of<the.MSIV and outside of containment.
~
f
-e O
O e s (***** CATEGDRY 2 CDNTINUED ON NEXT PAGE *****) 1 A----- a- - - - E------ ----.--_---a..mene m m mmm .wo n see, e n - e am-* '
...: Paga.41 i2:.__EdEBgENGLSyp_@BUQBd@k & @yI_EyQ6UIl9N@
12231-l ,' QUESTION 2.17 (3.00) BRIEFLY explain why each of Concerning.the Emergency Operating Procedures, (1.0 each) the f ollowing statements are FALSE:
- a. In general, a required task as stated in a procedural step MUST be completed prior to proceeding to the next step..
and'in hot standby. b.- The procedures are applicable only at power
- c. The entry points into the procedures are through EDP-0, ECA O.0 and CSP-S.1.
(***** CATEGDRY 2 CONTINUED ON NEXT PAGE *****)
. _ _ _ _ _ - _ _ _ - _ - - _ - _ _ _ _ _ _ _ _ _ _ _ - _ _ _ - . . = _ _ _ _ _ _ _ _ - - _ - _ _ _ _ _ - _ _ _ _ - - - _ _ _ . _ - _ _ _ _ _ _ _ _ .
m Pcg2 42
.2. ' EMERGENCY AND ABNDRMAL PLANT EVDLUTIONS 12Z31 DUESTION 2.19 (1.00)
Choose the DNE correct answer: Assume;that the reactor has tripped, and one control rod has failed to
. insert into the core.
How i s the actual shutdown margin affected, as compared to the calculated shutdown margin?.
- a. Actual shutdown margin is less than calculated shutdown margin, because of the positive reactivity associated with the stuck rod..
- b. Actual shutdown margin is less than calcul ated shutdown margin, because MTC adds positive reactivity during the cooldown f ollowing the trip,
- c. Actual shutdown margin is the same as calculated shutdown margin, because it is assumed that a rod will stick on a trip.
- d. Actual shutdown margin is the same as calculated shutdown margin, because reactivity added by power defect offsets the stuck rod.
- o. Actual- shutdown margin is greater than calculated shutdown margin, because it is assumed that boration will compensate for a stuck rod,
- f. Actual shutdown margin is greater than calculated shutdown margin, because xenon peaking will offset the stuck rod, e
l (***** CATEGDRY 2 CONTINUED DN NEXT PAGE *****)
Pcgt 02i__EDEggENCy,8yD_9pyggg@(_P6991_EV9691199@ 12z31
< DUESTION' 2.19 '(2.00)
Choose the TWO entry conditions f or ECA-0.0, " Loss Of All AC Power":
- c. all 480 VAC : saf eguards ' busses de-energi zed
.b. all 4160 VAC non-vital busses de-energized
- c. loss of all incoming transmission lines and turbine trip
- d. dual. unit trip with loss of all incoming transmission lines
- o. . response not obtained column from EOP-0,. step'3
- f. loss of 4160 VAC non-vital busses that feed safeguards busses A
4. 4 . m l
)
i (***** END OF CATEGORY 2 *****) l 4 i 1 1
- J
.Pago 44 ANT,,@ygI[D _f}g31:_@yQ,[L991 $19@_@[ygg1C REgggNgigl(1TIEg_J19Z,1 s
- QUESTION 3.01 (3.00)
~
Ahsume that RCS pressure is 2035 psig when the controlling channel 'f or pressurizer pressure fails. low.. Limit your
- 1. Describe the' changes that occur for the f oll owing items.
answer to INCREASE, DECREASE, REMAIN THE SAME, OPEN, SHUT, FULL DN, .
'(.5 ea)
- ~ PARTIALLY DN, FULL DFF, DN, or DFF.
- a. Actual Pressurizer Pressure
- b. Spray Valves
- c. Proportional Heaters
- d. Backup Heaters Choose the DNE answer that describes the response of the DT delta T 2.- (1.0) trip.SETPDINTS.
- a. The channel associated with the failed pressure instrument will decrease, and all others will not be affected.
b. All channels will decrease.
- c. All channels will be unaffected.
- d. The channel associated with the failed pressure instrument will.
decrease, and all others will increase. (***** CATEGORY 3 CONTINUED DN NEXT PAGE *****) s L ----------- - .. _ _ . . ~ _ _ ~ . . . - - . . -..n ~ . - - - - - - -
'P gs.C5l
( 3 .- PdANT= SYSTEMS'(38%)-!AND PLANT-WIDE GENERIC RgpPQNylp1(111gj_flg31 s 1: QUESTION 3.02 (1.00) Which ONE of the following will cause the differential rod worth to DECREASE 7-
- c. Moderator' temperature is increased
. b.: Boron concentration is decreased .
- c. An adjacent burnable poison rod-is depleted-
- d. An- adjacent rod is inserted to the same height s
4
. i
-l l
l
]
(***** CATEGORY 3 CONTINUED ON NEXT PAGE *****) ---~___._____.._______________._____._____.__.__m.__m.mm...., _. . _ . . , _ . .
Pago 46
- 3. PLANT SYSTEMS J3931_gNp_P(gNI;SJpg_@ENgg]C BEggDNpjBJ(JJJgj_jjp31
. QUESTION 3.03 (1.00)
' Which DNE of the f ollowing will best complete the statement?
"During plant cooldown, flow from the RHR system to the RCS is designed to ba ..."
.a. ... constant (around 3100 gpm) and this total flow is maintained by controlling the RHR heat exchanger flow bypass valve (RH-FCV-626) b.- ... constant (around 1550 GPM) and this flow is mainteingd by controlling the RHR heat exchanger outlet valves (RH-HCV 624 & 625)
- c. ... varied (to a flow that meets the desired cooldown rate) and this flow is controlled by the RHR heat exchanger flow bypass valve (RH-FCV-626)
- d. ... varied (to a flow that meets the desired cooldown rate) and this flow is controlled by the RHR heat exchanger outlet valves (RH-HCV-624 & 626) .
(***** CATEGDRY 3 CONT'NUED ON NEXT PAGE *****)
-'^ -
e.ew _* o et._m p e er w e_
5 Pago 47
-;2i_LE69N1_SYSlgdhj}@hl_gNp_g6$N1;WJpE_ GENE 51C BEEE9N@lB161}}Eg_flg31
~
l, QUESTION 3.04 (2. 00). STATE the red path summary. f or. the core cooling. status tree as listed on the EDP-O foldout page. Include setpoints and any coincidence. 4 4 I'
.l
{ (***** CATEGDRY 3 CONTINUED DN NEXT PAGE *****) L. . . . '"-------.--__.____._________m, . _ _ _ _ _ . _ _ _ _ _ _
r Pcg2 48 si__E69NI_Syg}EUS_j3@31_gNp,p(gNI;WJpg_@gNgRJC
;BEgggNgJg}(JJJgj_j1931
,DUESTION 3.05 (1.00)
Assume thatL RCS pressure is stable at 2100 psig f ollowing a reactor trip cnd safety inject ^.on actuation. Which DNE of the f ollowing describes the offect on subcooling margin when a safety injection pump is stopped? L o. subcool)ng will decrease because RCS pressure will now be below the-shutof f nead f or the SI pump
- b. subcooling will decrease because break flow will now be greater than SI.
flow
- c. subcooling will remain constant because pressure is above the shutof f head for the SI pump d
'. subcool i ng . will remain constant because charging pump flow will increase to maintain RCS pressure.
.e
(***** CATEGORY 3 CONTINUED ON NEXT PAGE *****)
Pagh49 2:_ _ Eb eNI_gygIE dg_ J }g31,@Np_ E68NI;MJ pE_ gE Ng 6} C u 5EEE9Ng1916111ES_jlp31 DUESTION 3.06 (1.50). LER BB-005. describes a situation where, during calibration of an NIS channel, a' technical specification violation existed with respect to delta What action must be taken during T channel ' minimum degree of redundancy. NIS channel calibration to ensure that the delta T channel minimumldegree-of redundancy is being met? i.
-(***** CATEGDRY 3 CONTINUED ON NEXT PAGE *****)
- . - - - - - - - - - - - - - - _ - - . - _ _ _ _ _ _ _ s
n . _ _ _ _ _ Pcga 50' 3.' PLANTL' SYSTEMS ( 38'/,) AND PLANT-WIDE' GENERIC j BEEE9NgJp16]IJEg_jlp3) a l DUESTION 3.07. (1.50) 4 j LER 88-002 describes a situation where a Unit 2 reactor trip occurred The trip was caused by a j during a shutdown af ter an end-of-lif e coastdown. f aulty source range instrument while the plant was in hot shutdown with ] Tavg at 540 degrees F. On the Why trip, both main feed regulating valves did the f eed regulating valves go open
~
automatically went full open. in thi s si tuation, despite the f act that Tave was less than 554 degrees F? q v Include any applicable setpoint(s) in your answer. 1 e I
+
f e
.S e
(***** CATEGORY 3 CONTINUED ON NEXT PAGE *****)
Pcga 51 13:__E69NI_SYSIEdg_13@fl_9Np_E(ANI;WlpE_QENE61C BESggN@lg16111ES_fl931 4
'DUESTION 3.08 (1.00)
Assume that the reactor is at 60% power, when N42 begins to fail. Over All the course of the entire shift, N42 drif ts up and eventually pegs high. systems'are in automatic, Which no operator DNE of the f action isbest ollowing taken, and turbine describes the load response rcmains constant. of the rod control system to this f ailure?
- c. Rods will not move because Tavg and turbine powe'r are remaining constant, with no power rate niismatch b, The red control system will cause rods to withdraw, bringing Tavg up in accordance with programmed Tavg consistent with the N42 reading
- c. The rod drop circuitry will cause a turbine runback when the mismatchand rods between N42 and the other power range channels reaches 2.5%,
will step in to reduce Tavg.
- d. The rod control system will cause rods to insert to bring power down, since N42 is driving the auctioneered nuclear power signal high.
O . 4 (***** CATEGORY 3 CONTINUED DN NEXT PAGE *****)
- l Pega-52
- 3. PLANT: SYSTEMS (38%) AND PLANT-WIDE GENERIC BESPQN}}}}(JIJEg_jlpZ.1
.i
. J i
QUESTION 3.09 (1.00) Choose'the DNE correct answer
. Assume that the Unit 1 is operating at 95% when a rod drop occurs, causing
,an NIS spike of -10%_in 1 5 seconds. All systems are in automatic'and no '
operator action is taken. The turbine will runback to 75% power because:- 1
- c. 75% is the pre-determined turbine runback setpoint f or dropped rods b.- turbine runbacks for a dropped rod are always 20%
- c. the turbine runback will stop when Tavg reaches Tref
'd. the turbine load limiter is set at 75%
e O . f l (***** CATEGORY 3 CONTINUED ON NEXT PAGE *****)
nn- _- p . P699 53: L L3; . PLANT: SYSTEMS (38%) AND PLANT-WIDE GENERIC ,, @ES[QN@lBlLillE@, Jig 31 ) l l DUESTION' .3.10- (2.00) A:cume. that Unit 1
- is at .100% power with rod control in MANUAL when a.
l dropped rod occurs, causing an NIS spike of.-5%.in 1 second. For each of f ' . tha. f oll owing l parameters, .woul d the final values be GREATER THAN, . LESS LTHAN, or EQUAL TD the ' final values with rod control in AUTOMATIC?- 1
- c. 'Tavg i
-i
.b. . Steam Flow c.- Pressurizer Level
- d. Generator Load (MWe) ,
l 4 0 4 O S e (***** CATEGORY 3 CONTINUED ON NEXT PAGE *****) e
Pcge 54
- 3. - PLANT SYSTEMS (38%) AND PLANT-WIDE GENERIC BEggpNpJpl61))Ep_jlp31 p
! QUESTION 3.11 (2.00)
Assume that all systems are in automatic with the reactor at 100% when a L turbine' runback occurs. Power stabilizes at 70%.
- c. What-overall'effect will the runback have on the DT delta T setpoint?(0.5)
Limit your answer to INCREASED, DECREASED, or REMAINED THE SAME.
- 6. Identify the parameters that affect the DT delta T setpoint, and what effect that each will have on the DT delta T setpoint, following the completed runback.
e f O O O O O . (***** CATEGDRY 3 CONTINUED ON NEXT PAGE *****) l 1 l i
J Pcg2 55
;:32 PLANT SYSTEMSi(392)AND PLANT-WIDE GENERIC
~
RES[QNSIB16111ES_flg3[
, , im
, -QUESTIONI 3.12 (1.50)
Choose the ONE correct answers Following a ' runback with all' systems in automatic, shutdown margin wills
'c.: decrease, because rods are closer to the insertion limit ti.' remain the same, because boron concentration has not changed
. c .- remain the same, because power def ect is of f set by control rods
<a. increase, because of higher.Tavg on the runback 1
9 E O 1-l. (***** CATEGDRY 3 CONTINUED DN NEXT PAGE *****) _ _ _ _ _ _ _ _ _ __ _ .._.__________-.,____.m___m______..__ _ _
\y, . Pcge.56 L ii__Eb eNI_EY!IEDS_J}BZ.1_9Np_g6991;W1DE_QgNEBIC l
.BEEEDNEIE16111EE_1193L l ,.
4 DUESTION 3.~ 13 (2.00) While reducing load, the " Power Range Channel Deviation" . al arm comes.are i n. as
. Plant conditions prior to the alarm, and af ter receiving. the alarm, f oll ows:-
l^ l Prior to Alarm . After Alarm' l Pcrameter' B5% N41 100% 99.5% B6.5% N42 B5% N43; 100% 100% B7% . N44. ' 204 steps Rod G3. 215 steps 216 steps 205 steps Rod C7 215 steps Rod G11 215 steps 214 steps, 204 steps Rod K7 0.989' NTILT 15 1.001 NTILT 25 0.996 1.007 NTILT 3S' 1.001. 0.999 NTILT 45 1.001 1.013 ( 1 '. 0 ) '
-c. Whatl is the cause of the " Power Range Deviation" alarm?
- b. What' actions should be .taken to avoid violating Technical ( 1. 0 L Specifications?.
O i O . l (***** CATEGDRY 3 CONTINUED DN NEXT PAGE ss***)
)
i Page 57-
- 3. PLANT' SYSTEMS' (38%) AND PLANT-WIDE GENERIC SESEDNSigl(llig@,J1Q31 DUESTION 3.14 (1.30)
When a pressurizer pressure transmitter is removed from service, several
~
bistables must be tripped. Indicate whether or not each of the f ollowing l annunciators in the control room will be illuminated or dark when (.25 the ea) ! bistable tripping is complete.
- a. . Pressurizer Low Pressure SI Channel Alert . ,
1 l
- b. Pressurizer High Pressure Reactor Trip !
- c. , Auto Turbine Runback Overpower delta T
- d. Reactor Cool ant Overtemperature delta T Channel Alert
- e. Auto Turbine Runback Dvertemperature delta T l l
- f. Pressurizer Low Pressure SI Trip 4
*5
(***** CATEGDRY 3 CDNTINUED ON NEXT PAGE *****)
7 Pcgi 38
'3. PLANT SYSTEMS ~(3B%)'AND1 PLANT-WIDE GENERIC Rk&P90&IB1hlIIED 119?I .
DUESTION 3.15 '(1.00) During i d11' power operation, the proportional heaters are observe'd to be on
- cll the time and the backup . heaters are cycling- to maintain RCS pressure at 1985 psig. The f ollowing indications are also noted
- PRT conditions are normal
- Charging ' flow is normal
- A and B spray line - temperatures dif f er by 25 degrees F
- P2R spray valves indicate closed
- No. radiation alarms Which DNE of .the f ollowing is NOT a possible cause f or the above readings?
- e. spray line leak between the spray valve and the pressurizer
- b. . spray valve has excessive seat leakage-
- c. spray bypass valve is throttled open too f ar
- d. spray valve is stuck partially open
- o. spray valve controller malfunction i
4 l I I j (***** CATEGDRY 3 CONTINUED ON NEXT PAGE *****) l l _m_._m.____m__ _ _ _ _ _ _ _ _ _ _ _ _ _ _ . _ _ _ ___
o Pcg7 59 3:__EkeUI_SySIEDg_J38Z,1_9dD,E69y1;WJDE_GENEBIC ' .j L
- reg [Qyp1B16111ES_figf[ 1
{. I'
.. i DUESTION 3.16 ( 1. 00) -
'A 3arge break LOCA occurs on. Unit 1 during' full power operation. At time L t=0, .an FI signal' is- generated; at time t=30 seconds, a containment spray cignal'is generated.. The containment spray system components will operate' Lin the f ollowing sequences. (choose DNE)
- c. t=TO sect ' spray-pumps A and B start; spray discharge valves 860A,1B, C, D open t=2 min 30 sect NaOH' addition valves 836A and B open
- b. t=30 sect spray discharge valves B60A, B, C, D open
.t=40 secs ; spray pumps A and B start
.t=2 min 30 sect ~ NaOH addi tion' valves ' b36A and B open
- c. t=30 sect . spray pumpc A and B start t=40 sect spray discharge va.Yves 860A, B, C, D open t=2 min 30 sect NaOH addition valves B36A and B open d.. t=0 sect-spray pumps A and B start t=30. sect spray. discharge valves B60A, B, C, D open
.t=2fmin: NaDH addition valves B36A and B open O
4 G 9
. 4
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Pcg2 60 2___fbeNI_EX!IEU!_l!!31 899_fkBUI: WIPE _GENgg]C BEEEgyglf]LIIlgg_JJ931 QUESTION 3.17 (1.50) Unit 2 is heating up f ollowing a ref ueling outage when a Saf ety Injection occurs. Given the f ollowing data, STATE why the Saf ety Injection occurred. Include any applicable setpoints in your answer.
. Time RCS Temp. PZR Press. RCS Baron Concentration 450 deg F 1500 psig 2000 ppm 1800 2000 ppm 1900 475 deg F 1600 psig 500 deg F 1800 psig 2000 ppm 2000 2000 ppm 2100 500 deg F 1700 psig S
9 O , (***** CATEGORY 3 CONTINUED ON NEXT PAGE *****)
Pcg7 61'
- 3L PLANT' SYSTEMS (387,) AND PLANT-WIDE GENERIC-BE@EQNSIBILillEg_J1931
'DUESTION 3.18 (1.50)
A ~ controller f ailure has caused 1TCV-130 (component cooling toAssume no non-regenerative heat exchanger control . valve) to go closed. laakage through 1TCV130, and the Letdown Gas Stripper is bypassed. Include any expected alarms and What'effect would this have on CVCS7 Setpoints not required. cutomatic actions that would occur. O . (***** CATEGDRY 3 CONTINUED ON NEXT PAGE *****)
Pitg2 ' 62
-31 PL AN'. SYSTEMS (387,) AND PLANT-WIDE GENERIC.
BESEQU}}B16111Eg_119Z.1 l
- DUESTION 3.19 (2.00) when the f ollowing events
. Assume that Unit 2'is operating at.100% power,
. occur:
- 1433 - "A" steam generator feed reg valve'placed in manual 1442- rod control placed . in manual ' f or replacement of a blown moveable .
gripper fuse on a control bank.D control rod 1445 "A" charging pump is tagged ODS. , 1447 PC-486C = (first. stage pressure) fails high A reactor trip !. 1449 RO attempts .to move rods to .natch Tave to Tref.. l occurs. l Assume that all other systems / components are in their normal 100% power lineups..- Dsscribe the cause of the reactor. trip, and the sequence of events that loads.to the trip. Setpoints are not needed. l e l
)
1 (***** CATEGORY 3 CONTINUED ON NEXT PAGE *****) l I
d,_- Pcge 63 2:__EbeUI_!!!IED!_J3g31_Syp_EheyI;ylpg_ggyg31C
.BEgggyp1g1(JIJEg_jlg31'
[ DUESTION 3.20 (1.00) Which DNE of the f ollowing is NOT true concerning Component Cooling t Tachnical Specifications?
- o. For single unit operation, the reactor shall not be made critical unless the two component cooling pumps and the two component cooling heat exchangers assigned to that unit-are operable.
- b. For single -unit operation, during power operation, a single component cooling pump assigned to that unit may be out of service for a maximum of 24 hours.
c.- For two-unit operation, both reactors shall not be made critical unless three component cooling pumps and three component cooling heat exchangers are operable.
- d. For two-unit operation, during power operation, two heat exchangers may be out of service f or up to 48 hours.
O O . (***** CATEGORY 3 CONTINUED DN NEXT PAGE *****)
Pcgo 64 si__PL9y], gyp][b3,j3931,999,P(@yl;ylpg,@gyEBIC. RESPONSIBILITIES (10Z) } t i I. . DUESTION 3.21 (1.00) Which DNE of the f ollowing is NDT true concerning the reactor makeup cystem? o.- Automatic mode provides dilute beric acid solution, preset to match boron concentration in the RCS.
- b. Dilute mode permits addition of a preset quantity of water at a preset flowrate to the VCT.
- c. Alternate Dilute mode is similar' to Dilute mode. except that water- is supplied directly to the charging pump suction and to the VCT.
- d. Borate mode permits addition of a preset quantity of concentrated boric acid solution at a preset rate to the VCT.
O (***** CATEGORY 3 CDNTINUED ON NEXT PAGE *****)
----.m _ _ - - - - - - - - - - - - _ _ - _ _ _ _ _ _ _ _ _ _ , . _ _ _ _ _ _ _ _ _ _ _
'Pago 65
" 't3:__E69NI_!!!IEDg,J}@31_Byp_E6891; Wipe _GENEBJC: BEgEgygipJb11]Eg_jlp31 l-QUESTION '3.22 (1.50)- P;r. Technical Specification 15.3.14, " Fire Protection System", match'the f ollowing ysituations with the required action: (NOTE: items in the ACTION column-may be used-more than once)
^
ACTION ~ SITUATION
- 2. - penetration fire barrier a. continuous fire watch inoperabim-
- 2. DG sprinkler system inoperable ~ b. periodic fir'e watch inspection-
- 3. fire detection instrument inoperable-(***** CATEGDRY 3 CONTINUED DN NEXT PAGE *****)
v 7-- 4
' k .
Pcqu.'66 's
- 3.1 ' PL ANT: SYSTEMS (387) AND PLANT-WIDE GENERIC 7
R[gPQNS1pl61Ilgg_119Z,1 q.
- GUESTION.- 3.23 ( l'. 00)-
Choose the'one term that_best describes the process of_ determining'an
~ instrument's' accuracy- by . visually comparing the indication to other '(As.it independent' instrument channels measuring the,same parameter.
'i n. defi ned :in -Technical . Speci fications)'
c., Instrument Check' b .- Continuity Check c .- ' Channel- Functional Test
- d. Channel Verification O. Channel Calibration f.- Channel Check e
e e O l
. e 3 CONTINUED DN NEXT PAGE *****)
(***** CATEGDRY .
~Pcg3 67 lei __Eh6El_ETEIEUE.33E31_SyD_E63y1;WJDE_ggyg3]C j
BggEgygJglblIJEg_jlg31 L DUESTION 3.24 (2.00) J Choose the'TWO correct. answers: Under what two. circumstances is it permissible to violate a Technical-Specification Limiting Condition of Operation (LCO)? .
- a. During an emergency to protect public health _and' safety.
~ 6. During a General Area Emergency.
- c. During a Site Area Emergency.
d.- When' f ollowing the Emergency Operating Procedures, O. When f ollowing the Abnormal Operating Procedures.
/. When directed to by the Duty and Call Sup ervi sor ,
- g. When directed to by the Plant Manager
(***** CATEGORY 3 CONTINUED ON NEXT PAGE *****) _--________m____m___._ _ _ _ _ _ . _ _ _ . _ _ ___m.__. ___m__
Page 6B PLANT SYSTEMp_J3g31_gNp,PL3NT Jlpg,g[NgRIC 3. BEgPQNgig161IlES_flp31 s DUESTION- 3.25 (2.00) Par 10 CFR 55:
- a. " Control s" are apparatus and mechani sms, the manipulation of which directly af f ects the reactivity or power level of the reactor. Other than the rod control system, list three systems that are considered (.5 ea)
" controls".
- b. An unlicensed individual may manipulate the controls under the direction and in the presence of a licensed operator or. senior operator 1 under two circumstances. STATE one of these two circumstances. (.5) l l
(***** CATEGORY 3 CONTINUED ON NEXT PAGE *****) m___-__m__.____._._.m_m __ . _ . _ . _ _ _ - _ _ _ . _ _ _ _ . _ . _ . _ . _ . . . .._ m _._.____.-_.m____._____m..__s -
Pcg2 69 3:__E69NI_SYSIEMg_J3Q31_ENQ_g(ENJ:ylgE_gEyEglC'
-BEgggggigibillEg_ fig 31 DUESTION 3.26 (2.00)
Match the f ollowing definitions with the correct terms: L(NOTE: not all the TERMS will be used) DEFINITIONS 1,. Any area, accessible to personnel, in which there exists radiation,
~ -
originating in whole or in part within license material at such levels that a major portion of the body could receive in any one hour a dose in excess of 100 mrem.
- 2. Any area access to which is controlled by the licensee for purposes of protection of individuals f rom exposure to radiation and radioactive materials.
TERMS
- o. Radiation Area
- b. High Radiation Area
- c. Restricted Area
- d. Owner Controlled Area .
(***** CATEGORY 3 CDNTINUED ON NEXT PAGE *****) m
Page 70 2:__EL9NJ_pyp][gg,j}p31,3yp,fg@yJ;yJpg,ggNg3]C , RESPONSIBILITIES (10%) 1 v e QUESTION 3.27 (1.00) f Which DNE of the f ollowing is NOT a method used to control access to high radiation areas per 10 CFR 207
- c. access controlled by a control device which shall cause the level of radiation to be reduced below that at.which indi' "luals might recei ve a dose of 100 mrem in 1 hour upon entry into the ai
- b. access controlled by a control device which shall energize a conspicuous visible or audible alarm signal in such a manner that the individual entering the high radiation area and the licensee or a supervisor of the activity are made aware of the entry maintain the area locked except during p,eriods when access to the c.
area i s required, with positive control over each individual entry.
- d. maintain the area locked with entrance and exit controlled by a key I that is in the direct control of the Health Physics Supervisor or the senior management official on shift .
1 (***** CATEGORY 3 CONTINUED ON NEXT PAGE *****) \ - l L _ __ _
Pcga'71 p 3. . PLANT SYSTEMS(38%)- AND PLANT-WIDE GENERIC L 'BEgPQNSJglLJIJEg_flgZQ (. a..i- ? l . i 'a L -GUESTION 3.28 '( 1. 50) - Match the f ollowing TERMS with the proper DEFINITIONS. TERMS- DEFINITIONS.
- 1. isolated with a. red tag a. ' equipment not.in a condition to.
L perf orm its normal. function
- 2. out-of service
- b. . tags installed and permission
- 3. holdof f clearance granted'to begin work (WE SDD & System Control)
- c. prevents operation that could result.
in danger to personnel or equipment,
.. 1 l
\
I I (***** CATEGORY 3 CONTINUED ON NEXT PAGE *****) I
. J 4
~ ~ - - - - _ . _. ._ __
Pcgp 72 m ::.L_iEb6NJ,5y!] Egg,J35Zd,9ND,fb3NJ-WJDg_ggNgf}C " beg [QNg]!16]IJE),JJQZQ s.
. y_
DUESTION. 3.29 . (1.00). Unit .1 i s at 1007. ' power and Unit 2 is'in a refueling. outage. Fuel i s bring. loaded into Unit 2. - Per technical specifications:
'c)rWhat is- the minimum control ' room licensed personnel staffing (0.5) requirement 7 b)' What is ' the minimum local ref ueling . licensed personnel staffing ( 0. 5 ) .
req.$1rement? O 9 9 e G . (***** END OF CATEGDRY 3 *****) (********** END OF EXAMINATION **********) l x ________ __
- 1. REACTOR PRINCIPLES (7Z) THERMODYNAMICS Pcgp 73
( 77. ) AND COMPONENTS (11%) -(FUNDAMENTALS EXAM) __ . 0, (2.00, MASTER CO2Y (0.5 ea)
- c. remain the same
- b. decrease
- c. decrease
- d. . increase REFERENCE TGehnical Specification 15.1.g,4 192OO2K114 ..(KA's) .-
ANSWER 1.02 (1.00) d REFERENCE TRHB 32.4, pp 9 - 12 192OO3K106 ..(KA's) ANSWER 1.03 (1.00) e REFERENCE W3stinghouse Reactor Core Control for Large PWRs, pp 2-11 to 2-12 192OO7K104 ..(KA's) ANSWER 1.04 (1.00) C s l n'4FERENCE ]
-TRHB 32.5, pp 13 - 14 -
192OO4K111 ..(KA's) (***** CATEGDRY 1 CONTINUED ON NEXT PAGE *****) 1
. l 1
Peg 2 74
- li__BEgglgB_E8]NQlf6g@_JZ31_ldgBDQpyygd}gg .
cl2Z1_BU9_990E9NENIg_J1131_JEyNpgggylg(g_ggggi ANSWER 1.05- (1.00)
- c.
REFERENCE' TRHB 32.5, pp 13 - 14 192OO4K111 ..(KA's) ANSWER 1.06 (1.00) . NOTE: answer will depend on assumptions made by- the candidate regarding tha values of beta-effective, lambda and power defect. Acceptable ranges:
-Power defect =.1^ li r c /% ,,;..;r 72. -
,)o fm h Lccbda = 0.1
'b;ta eff = .0052 .0002 .00GS
'tcu=26.06/SUR NOTE: .25 credi t for sol uti on rho =(beta ef f.1/ (1+1ambdattau) NDTE: .25 credit f or solution based on candidate assumptions power l evel= rho / (pcm/*/. power ) NOTE: .5 credit for solution based on candidate assumptions 3
(c1 ternate methods of calculation evaluated on a case basis) REFERENCE TRHB 32.4, pp 17 - 24 192OOBK113 192OOBK110 ..(KA's) ANSWER 1.07 (1.50) Eliminates effects of instantaneous variations in readings (Will accept: some indications of natural circulation depend on a trend in readings, not on the reading itself) REFERENCE . ERG Executive Volume, Generic Issues, Natural Ci rcul ati on 193OOOK122 ..(KA's) l [ (***** CATEGORY 1 CONTINUED DN NEXT PAGE *****)
. .i-REACTOR PRINCIPLES (77) THERMODYNAMICS 'Pagm 75:
p ilk P (77.) AND COMPONENTS ( 1 17. ) " (FUNDAMENTALS EXAM)
- b. >
'l I
. l. ii ANSWER 1.08 (1.00) d 1 REFERENCE-TRHB 39.3, pg BB W2stinghouse Thermo-Hydraulic Principles, pg-13-36 ,
'193OO9K105 ..(KA's)
ANSWER' 1.09 (1.00) c REFERENCE WWetinghouse Thermal Sciences, Mod D-5, 2.3 193010K105 ..(KA's) ANSWER 1.10 (2.00) (0.5 ea). c.' decrease
- b. remain the same c.- increase d.' decrease REFERENCE TRHB 39.3, pp B2 - 83 W;stinghouse Thermo-Hydraulic-Principles, pp 13-22 to 13-24 193OOOK105 ..(KA's)
' ANSWER 1.11' si. e',
(0.5)
-n u. c. t h: :--- beidel b .~ . increase f at. dr:r:::n Ddde) (.5 ea) ,
(***** CATEGORY 1 CONTINUED ON NEXT PAGE *****)
. . . , ; i Pago-76;
. _. . . . ~,.
J1ar : REACTOR ~PRINCIPLESi(77.)'THERMDDYNAMICSE .
- He dZL1_0N9_G90PDNENI@,11131_1[yND@dENI66@_E36dl IREFERENCE TRHB . 39. 2, - pp =. 7. -- 8, J 21 - .
? WJst'i nghouse 'Thermo-Hydraul i c .Pri nci pl es, pg-7-67~
193OO5K103 '. . (K A 's ):
E> UANSWER' 1.12 >(1.OO)N (654-557)=97 degrees-F-(+/- 2) REFERENCE , ' oteam tables
'193OO3K125- ..(KA's)-
l l'.1'3
-ANSWER (1.00)
.b REFERENCEL
!TRHB'40.2,fpp 52! .58 193OO6K104 ...(KA's)
' ANSWER .1.14 (1.00) lC
' REFERENCE TRHBc13.3, pg 29 .
.19100BK104. ..(KA's)
I
' ANSWER 1.15 - ( 1. 00 )
-REFERENCE LLPO584, 2.4413.d 191002K113 191002K114 ..(KA's)
(***** CATEGORY 1 CONTINUED ON NEXT PAGE *****) ) m .
y. r. f ; Pcg3 77' o fli__BE9CIQ8_E81NClELES_1Z31_ISE85QDyN951CE . . (7%) AND COMPONENTS (11%) (FUNDAMENTALS EXAM) v b T 4 h ' ANSWER- '1.16 -(1.00) l'a l lb,
' REFERENCE.
TRHB 11.1, pg-2
'191001K101. ..(KA's) 1'
,fANSWER' -1.17 (1.00) .
1 c REFERENCE OP-5D, sec E,- precaution 2.3 191007K108 ..(KA's) ANSWER 1.18 (1.00)
.O' REFERENCE
.TRHB'13.6, pp15 - 6 191002K109 ..(KA's)
. ANSWER 1.19 (1.00)
.. f REFERENCE TRHB-10.2, pg 31 19.1005K106 ..(KA's) a ANSWER 1.20 (1.00)
C (***** CATEGORY 1 CONTINUED ON NEXT PAGE *****) l j
r Prg7 78 it.. REACTOR'PRINC]PLES'(77.) THERMODYNAMICS
.l?31_BUP_C9dPgyEUJg_jll31_JEgyp3DEUJ8L@_E39D1
( { _ REFERENCE /
.TRHB 10.6, pg 33 191007K109 ..(KA's) iANSWER 1.'21 (1.00) i hDcd proportional-to speed squared so.
'if head doubles, (N squared) doubles, or speed increases by a factor of 1.414 (.5) power proportional to speed cubed so: ,
14414 cubed = 2.83 or power. increases by a factor of 2.83 (.25)
. final' power = initial power
- increase = 2
- 2.83 = 5.66 kW (.25) ( 0.lAdv)'
REFERENCE TRHB 40.2,.pg 6 191005K104 ..(KA's) l l: I l j ., (***** END OF CATEGDRY 1 *****) L-
EMERGENCY AND ABNDRMAh_P( gT_EVphy))QNS' Peg 3 79L>
-2 122n ANSWER 2. 1 .' 3. OO)f opr>n. charging pump suction from RWST (.5) chut-VCT outlet valve (.5) otcrt all avail abl e charging . pumps ' (.5) op;n HCV-142 (charging flow control valve) for >= 70 gpm f l ow (.5) otcrt bor-ic acid transf er pumps (.5)-
' cptn emergency boration . valve (MOV-350) (.5)
REFERENCE ADP-6E, pg 2 OOOO24G010 ..(KA's) ANSWER 2.02 (1.00) core deluge-REFERENCE JEDP-1, step 27 OOOO11K313 ..(KA's) ANSWER 2.03 (2.00)
- 1. d
- 2. .c (1.0 ea)
REFERENCE ERG Background Document, ES-0.2, pg 45 EDP-0.2, pg 11 j OOOO55K302 ..(KA's) l ANSWER 2.04 (1.00) Cne. M'esd ghwdv was mopewk WMod *k EF 9de %4 AuchaN WM N O
, 3 ..:es .. . u. ... - . - -.. a .. : . :.:; =. ::: :
d iset ge e k k 'rv) OP4 t. (vi ol ated TS 15. 3. 7. B.1. e) Onda'me'*d M (OMj SMe. er- ( O D , % % d e']3 I.h, 1
"Al+tr sMt. anf%k*,
z % e.e ca.o,a w ob s ura 2 4 lod i l (***** CATEGORY 2 CONTINUED ON NEXT PAGE *****) l l l 1 L . . _ . _ _ __-_________-_ - _ _ -
.?te, i, . . . . ,
Pcg:21CO . L 2; <. 2 ; EMERGENCY'AND ABNORMAL PLANT-EVOLUT1DNS. g ---- J iREFERENCE? LER-BB-010 TS115.3.7;B.1.e 000056G003- ..(KA's) EANSWER 2.05 (1.00) di REFERENCE , EOP sta'tus trees-Point Beach'Requal Question 031-03
.OOOO40G011 ..(KA's) sWER 2.06 (1.00)' <
d-o ueh 2.ov ee ad REFERENCE ., TRHB 13.
.TRHB 1 6 Poin each Requal Que on 043-02 OOOO3A106 . . (KA's - ANSWER. 2.07 (1.00)
O REFERENCE CSP-C.1' ] Point Beach.Requal Question 031-03-001 J OOOO74A201 ..(KA's)
. ANSWER 2.08 (1.50)
- During a loss of all AC status trees are to be monitored for information onl y (.75) and CEP's are not to be performed (.75) i I
i (***** CATEGORY 2 CONTINUED ON NEXT PAGE *****) I i
~
p .. ..
- .g. EMERGENCY AND ABNORMAL' PLANT EVOLUTIONS 'Pdga B1-J , 12ZZL y
IF
REFERENCE:
$ LECA-O' cautions S . Point' Beach ~Requal Question 031-01 OOOO55G012- . .- ( K A ' s ) l ANSWER 12.09 (1.00) c REFERENCE . TRHB 10.6 ADP-10A Point Beach Requal Guestion 055-02 OOOO6BA113 ..(KA's) ANSWER, 2.10 -(1.00) b REFERENCE-TRHB.13.1 Logic Sheets 11 and 12 Point Beach Requal Question 053-03 OOOO33A200 ..(KA's) ANSWER 2.11- ( 1.' 50 ) G1 seal delta P is the limiting factor (.5). and is greater than 200 psig (.5) and #1 seal flow is greater than .2 gpm (.5) (so continued operation
" i o ' permi ssi bl e)
REFERENCE 01-1 OP-1A - Point Beach Requal Question 051-01 OOOO27A203 ..(KA's) (***** CATEGORY 2 CONTINUED ON NEXT PAGE *****) j I i am-----_______.__-_______1_ . . _ _ _ . _ _ _ _ _ . _ _ . _ _ _ _
Pcg7 82 2:__EDEQQgyCY_gyD,gpNQQDgh_P(gy},Eyp(yJ}QNg J2231 ANSWER 2.12 (1.00) d REFERENCE Tcch Spec 15.1.D, 15.3.6A Point Beach Requal Question 051-05 OOOO69A201 ..(KA's) ANSWER 2.13 (2.003 Yc;c leak rate = 8% in 4 minutes = 2% = 120 gal / min (.75) charging system has 3 pumps at 60 gpm or 100 gpm capacity (.75) REFERENCE AOP-3A Point Beach Requal Question 051-01-004 OOOO37A212 ..(KA's) ANSWER 2.14 (1.50)
- a. N-35 is overcompensated (.5)
- b. Can continue since tech specs require only 1 operable IR for startup (1.0)
REFERENCE Tcch Spec 15.3.5 TRHB 13.1 Point Beach Requal Question 055-02 OOOO33A211 ..(KA's) ANSWER 2.15 (1.00) b (***** CATEGORY 2 CONTINUED ON NEXT PAGE *****)
mmr -------;--- L2i :EMENGENCY-AND'ABNDRMAL PLANT EVDLUTIONS Pcga C3-
^
(277J; ) L t _ REFERENCE. 1 AOP-10A.
. Pointe Beach ' Requal Question 055-02 OOOO67A213 ..(KA's)
L ANSWER 2.16 (2.50) 1.
- c. containment pressure increase containment temperature increase containment sump level i n c r.e a s e containment relative humidity increase ( . 25 *ea )
- b. since the steam leak is unisol.able (.5),.the SG must boil dry to stop the leak (.5). Continued f eed flow would resultin continued steam generation (.5)
REFERENCE , EDP-O EDP-2 Point Beach Requal Duestion 031-03 OOOO40K306 ..(KA's)
. ANSWER 2.17- (3.00)
- c. Simply starting the step is sufficient.
- b. Some procedures are applicable in hot shutdown and cold shutdown
- c. Entry is not allowed directly into CSP-S.1. (1.0 ea)
(accepts entry into CSP-S.I is through ST-1.or EDP-0) REFERENCE' ERG Executive Volume, Users Guide, pp 5, 20-25 CSP-S.1 entry conditions OOOO29G012 OOOO29G011 .' . (KA's)
. ANSWER 2.18 (1.00)
C (***** CATEGORY 2 CONTINUED DN NEXT PAGE *****)
5 i Pegg: 84
- 2. EMERGENCY AND ABNDRMAL PLANT EVOLUTIONS';
as c .!223.). REFERENCE 4TcchiSpec'15.1.g.4 - OOOOO7 GOO 4 'OOOOO7 GOO 3
. '. (KA's)-
. ANSWER ~ 2.19- -(2.00) l.hi 'e (1.0 ea)'
, REFERENCE,
~
ECA-0.0; entry conditions- . Paint Beach Exam Bank-Ouestion 31-03-09.
.OOOO56G01.1- ..(KA's)
( (***** END OF CATEGDRY 2' * * * * * )
mp - . - --- - - - - -- . _ _ __ 1 i: Pcg? C5'
,31 FLANT' SYSTEMS T (387) AND PLANT-WIDE GENERIC:
.RE@PQN@l@]L111E@_J19ZQ' 1
}. ANSWER 3. Ol' .. (3.00)
't.'o.~ ' increase b.. shut
- c. full on
- d. on~ (.5 ea) l-2, d. ( 1. 0 ) -
REFERENCE TRHB'10.3, pp.12-14 l
-TRHE 13.3,.pp-4-5 J' - 0120004205, ..(KA's)
-ANSWER 3.02 (1.00)
- d
. REFERENCE TRHB.32.5, pg 23
'OO1000K510 ..(KA's)
ANSWER 3.03. (1.00) a
. REFERENCE TRHB 10.7, pg 3 OO5000K402' ..(KA's)
ANSWER 3.04 (2.00) core exit thermocouple greater than 1200 degrees F-(.5) OR core exit TC's greater than 700 degrees F (.5) AND
- vessel level less than 29 ft (.5) with no RCP's running (.5)
(***** CATEGORY 3 CONTINUED ON NEXT PAGE *****)
- n. .
PLANT-SYSTEMS ~ 1387.) AND PLANT-WIDE GENERIC. Pcgo'C6
- 3.
LBESPONgig1(II1Eg_11931 REFERENCE ST-2-(core cooling) EDP-O foldout page . . Oi7000GO15 ..(KA's)- ANSWER 3.05 (1.00) C' REFERENCE , TRHB 10.8, pg 6
'OO6050A102 ..(KA's)
ANSWER 3.06 ( 1. 50 ) - , place . Associated delta T channels ( . 7'5) . in a tripped conditi on. (.75)
~
REFERENCE LER.BB-OO5
-015000 GOO 5 '015000A403 ..(KA's)'
ANSWER 3.07 (1.50)
- Tovg range resistors are installed for coastdown (.5), lowering the low Tavg setpoint (.5) to 539 degrees F (.5)
REFERENCE LER 88-002 012OOOA101' ..(KA's) ANSWER 3.08 (1.00) i l . u l' 4 (***** CATEGORY 3 CONTINUED ON NEXT PAGE *****) i I I I C __ _ __ _ _ _ __ _ _ .__ _ _ _ _ _ _ . . __ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _
Acg2 B7-
,3. . PLANT'PYSTEMS i (37/.) AND PLANT-WIDE GENERIC RESPONSIBILITIES'(10Z) e j.-
REFERENCE
~TRHB:13.8, pp 1-2 OO1000K105 ..(KA's)-
ANSWER 3.09 (1.00) b ,
-REFERENCE TRHB 13.11 Logic.bheet. 16
. Point Beach Requal Question 052-03
.045000K412 ..(KA's)
ANSWER 3.10 (2.00) (.5 ea)
- c. greater thar.
- b. greater than
- c. greater than
- d. equal-to ,
REFERENCE
.TRHB 13.8 TRHB 13.9
. Point Beach Requal Duestion 053-01 _
OO1000K104 . . (KA' s) , ANSWER 3.11 (2.00)
- o. Increased (.5)
- b. Tavg dropped which caused additional credit to the setpoint x addin
(.5) F(delta flux) changed due to large negative delta flud ia 64b ' penalties (.5) [44ndet Futtk M % 6 no cut e4 4 b
-pressurizer pressure will increase, causing additional credit ( . 5)
(will accept no change in pressurizer pressure) (***** CATEGORY 3 CCNTINUED ON NEXT PAGE *****)
.o
- 2 __EE6dl_ EYE.T,gd@_l}@* Pegs 1BB i
f l,999_E6991-ylpg; GENERIC
. RESPONSIBILITIES (10%)-
j
.; REFERENCE-
.TRHB 13.3 i
Point' Beach Requal Question 053-02 012OOOG010 ..(KA's)- E ANSWER 3.12 (1.50) l
.c
. REFERENCE ~ ,
Point. Beach Requal Question 055-01 OO1010A404 ..(KA's) ANSWER 3.13 (2.00)
- c. Rod G11 is misaligned .(f ailed to drive in) (1. 0)[#M1 2% be 8k f*M
- b. Rod'G11 must be investigated to determine why it is misaligned (.5) and then-realigned (.5) (alternate: f ollow actions of AOP-68)
~ f AMi carvak pM y +* ( 5) h *%neh v=Js (s)3 - '
REFERENCE-Tsch Specs 15.3.10.B.3,-15.3.10.D
' Point Beach Requal Question 039-03 OO1000K507 ..(KA's)
ANSWER 3.14 (1.50)
- c. illuminated b ., dark
.c. dark
- d. illuminated
- o. illuminated
- f. dark REFERENCE ARB, Logic Sheets 13, 15, 1B Point Beach Requal Duestion 053-02 012OOOA404 ..(KA's)
(***** CATEGDRY 3 CONTINUED ON NEXT PAGE *****) r -x_____________.___________ _ . _ _ _ _ . . _ . _ _ _ . _ _ . _ _ . . _ . , _ _ _ _ _ _ _ _ . . _ . _ . _ _ _ _ _ . _ _ _ _ _ . _ _ .
l. 3:__E69NI_HYNIEDg_J3g31_99p_gb8MI;WJpE_gENEBJC Pcga 09-J
- BEgggggJpJkJI1Eg_j1931 ANSWER 3.15 (1.00)
.c' REFERENCE TRHB 10.3 Point Beach Requal Duestion 055-02-002 010000G015 ..(KA's)
ANSWER 3.16 (1.00) b-REFERENCE TRHB 10.12 Logics Sheet B,9
. Point Beach Requal Question 051-03 026000K404 ..(KA's)
ANSWER 3.17 (1.50) when pressurizer pressure reached 1775 psig, SI auto-unbl oc k ed ( . 5) The cubsequent pressure reduction to less than 1735 psig (.5) caused an SI on low pressurizer pressure. (.5) REFERENCE OP-1A
-Point Beach Requal Question 053-06 OO6000K405 ..(KA's) q ANSWER 3.18 (1.50) ]
i Ictdown temperature would increase. (.5) "non-regenerative heat exchanger ) Ictdown outlet temperature high" alarm would actuate (.5) (exact title of J clarm NOT required). TCV-145 would divert letdown flow around the d mineralizes.(.5) (***** CATEGORY 3 CONTINUED ON NEXT PAGE *****) l
- u. . .
1 Page 90- _ 3. PLANT SYSTEMS ~(3BX) AND PLANT-WIDE GENERICS [ BEgPQNSIB1(111Eg_11Q31 e
' REFERENCE ,
TRHB 10.6 Figure 10.6.2 Point Beach Requal Question 051-02 OO4010A401 ..(KA's)
' ANSWER 3.17 (2.00) l l
.With.the faulty fuse for a. bank <D rod, this rod will drop when rod' motion
'is demanded (.5). The NI mismatch will result in a (20%) turbine runback.
(.5). With PC-486C failed, steam dumps will not arm or open~(.5). With no cuto rod control or steam dumps, Tavg will rise to the *0T delta T setpoint,- ceusing the reactor.to trip.(.5) NOTE: other causes and . scenarios will be evaluated on a case basi s. REFERENCE TRHB;13.9, pg 4 LTRHB 13.1, pg 9 TRHB 13.3, pp 4-5 015000K103 OO1000A203 OO1000A101 OO1000K105 ...(KA's)
' ANSWER- 3.20 ( 1. 00 ) '
O REFERENCE TRHB 10.9, pp B - 9 Toch Spec 15.3.3.C 000000G005 ..(KA's) ANSWER 3.21 (1.00) d REFERENCE TRHB 10.6, pp 30 - 32 OO4000K106 . . (KA' s ) (***** CATEGORY 3 CONTINUED ON NEXT PAGE *****) u_._-___-m_m_-m____-.-__.. __m. . _ _ _ _ . - _ . _ _ _ _ _ _ _ _ _ __ . _ _ ~ ..___m__ . .
=,- - - -
,r
-Pago 91 "a[(22__P(9NI_gygIEgg[j}g31_9Ng_[(@NI;WJgg_gENgglC
' BEggDNpl31(JIJEp_jlg31' ANSWER' 3.22 (1.50)
-1 . ~ a (2, b
'3.
b (.5 ea) REFERENCE y , Tachnical Specification l15.3.14 194001K116 . . (KA's) .
' ANSWER 3.23 (1.00)- .
~f REFERENCE Tcc'hnical' Speci ficati on 15.1. f 194001A113' ...(KA's)
! ANSWER 3.24 ' '( 2. 00 )
o,~d (1.0 ea) REFERENCE EDP Format and Usage 4 Text pp 12,13 10CFR50.'54x 10CFR50.73 194001A116: ..(KA's)
-ANSWER 3.25 (2.00)
- a. EHC system, reactor water makeup, system, feed system', steam dumps (any 3 & .5 ea)
(other answers evaluated on a case basis)
.b. as part of the training program, or to load / unload f uel (.5 for either) l l
(***** CATEGORY 3 CONTINUED ON NEXT PAGE *****) l
(I,b , ,
; Peg 2192 I
m L3_;lEbeNI_gyg] gyp _j3pf1_ gyp _fgANJ;$Jpg_@gNgRJC L.f lAEgggNgJBJ6111gg_J1931 c, i m,
; REFERENCE-
,n
.. [10,'CFR 55.4 110;CFR 55.13 L '194001A103 194001A109 194001A111 ..(KA's) l ANSWER -3.26 (2.00) .
'1.: b' 2i c1 (1.0 ea) ..
REFERENCE . 10 CFR 20.202 194001K103 ..(KA's)
' ANSWER' 3.27 (1.00) d' REFERENCE 10 CFR 20.203 194001K103 ..(KA's)
ANSWER- 3.28 (1.50)
- 1. c 2 a 3 b (.5 ea)
REFERENCE PBNP.4.13, pp 4 - 5 194001K102 ..(KA's) ANSWER 3.29 (1.00) o) three.(licensed) operators (0.5)
-b) A (licensed) SRD or an SRO li mited to f uel handling (either for 0.5)
(***** CATEGORY 3 CONTINUED ON NEXT PAGE *****)
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. Pcga : 93 .-
53k__EN6dl EYEIEUE .!EE*b1_bdE_EE6dl-WJpg_ggNEBIC'
. RESPONSIBILITIES ( 107. ) -
REFERE'NCE: ). [Tachnical Speci fication 15.6.2 194001A103. ..(KA's) l . s e I
.t'
'l I, (***** END OF CATEGDRY 3 *****) (********** END OF EXAMINATION **********)
t
. v.
TEST CROSS REFERENCE Pcg2 1 09EE1199 _YB695 BEEEEEUEE_ 1.01 2.00 9000743 1.02 1.00 9000746 1.03- 1.00 9000749 1.04 1.00 9000750 1.05 1.00 9000751 1.06 1.00 9000756 1.07 1.50 9000733 5 1.08 1.00 9000740 1.09 1.00 9000741 2.10 2.00 9000742 1.11 1.50 9000744 1.12 1.00 9000748 . 1.13 1.00 9000755 1.14 1.00 9000734 1.15 1.00 9000737 .- 1.16 1.00 9000738 1.17 1.00 9000739 1.18 1.00 9000745 1.19 1.00 9000752 1.20 1.00 9000757 1.21 1.00 9000758 24.00 2.01- 3.00 9000747 , 2.02 1.00 -9000762 2.03 2.00 9000763 2.04 1.00 9000767 2.05 1.00' 9000774 2.06 1.00 9000776 2.07 1.00 9000778 2.08 1.50 9000702 2.09 1.00 9000783 2.10 1.00 9000784 2.11 1.50 9000785 2.12 1.00 9000786 2.13 2.00 9000787 2.14 1.50 9000789 - 2.15 1.00 9000790 2.16 2.50 9000791 2.17 3.00 9000795 2.18 1.00 9000796 2.19 2.00 9000797 29.00 3.01 3.00 9000759 3.02 1.00 9000760 3.03 1.00 9000761 3.04 2.00 9000764 3.05 1.00 9000765 3.06 1.50 9000766 3.07 1.50 9000768 - 3.08 1.00 9000769 3.09 1.00 9000770
%. ..,1-.. t TEST ECRDSS REFERENCE! Pcga 2;
, t VALUE: REFERENCE .
' ;n: ' DUESTIONJ
?
Ur 3.1d- 2;00 9000771 3.11' '2;00 9000772 4
;3.12 1.50 9000773'
!3.13' '2.00- 9000775.
. 3.14 . 1 . 5 0 '. .9000777-
.3.15- 1.00. 9000779
.- 3.16 1.00- 9000700 3.17 -1.50 9000781 3.18 1.50' 9000788-3.39 2.00 9000792 .
3.20' 1.00 9000793
.9000794
, , . 3. 21 t ' 1. 00 ..
3.22 1'.50 .9000729
- 3.23. 1.00 9000730 .
3.24 2.00 9000731- .
,3.25- 2.00 9000732 3.26 2.00- 9000735-t 3!27 .1.00 9000736 ,
~3.2B 1.50' 9000753 3.29 1.00 9000754 43.00
. 96.00 ,
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4
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U. S. NUCLEAR REGULATORY COMMISSION SENIOR REACTOR OPERATOR LICENSE EXAMINATION REGION 3 4 FACILITY: Poi nt _geach _1_g_g________ REACTOR TYPE: PWR-WECg_________________ DATE ADMINISTERED: g9fg6fg6,________________ CANDIDATE: _________________________ , INgI6UgIJgNg_Ig_g@Ngig8IEL 1 Write answers on one side only. l Use separate papar for the answers. Points f or each Stapl e question sheet on top of the answer sheets. The passing question are indicated in parentheses af ter theand question. a final grade of at grade requires at least 70% in each category up six (6) hours after j least 80%. Examination papers will be picked the examination starts. I
% OF ,
CATEGORY % OF CANDIDATE'S CATEGORY ___SCg6E___ _y@6gE__ ______________C8IEGQSy_____________ __YB6WE_ _19196 23.00 32. S !~ ________ 4. REACTOR PRINCIPLES (7%) _i5 22_ _-E2:22 ___________ THERMODYNAMICS (7%) AND COMPONENTS (10%) (FUNDAMENTALS EXAM) 3too 33.33 EMERGENCY AND ABNDRMAL PLANT _ G___; . 0_ _0 _ _ _'2__3 _. _6 _5 ___________ ________.5. g ,, 44.12. PLANT SYSTEMS (30%) AND _d9 99 _ _"2:E2 ___________ ________ 6. PLANT-WIDE GENERIC RESPONSIBILITIES (13%) 101.00 TOTALS j a_ _c_a_. _0_^ __ _ _ _ _ ' _ _ _ _ .________% l FINAL GRADE ( I have neither given j All work done on this examir.ction is my own. i nor received aid, l Candidate's Signature ) i t l MASTER COPY l t < l 1
1 lNRC, RULES'AND'GUIDEL'INES FOR LICENSE EXAMINATIONS l D = During the administration of' this examination the f ollowing rules app 1'ys ' .)
- 1. Cheating'on the examination means an automatic denia11of'your application' ]
and could resul t'- in more severe penalties.
!2.- Af ter: the: examination has been completed,- you must signLth'e statement .on the cover sheet indicating that the. work 1 1s your own and you have nct-i received. or. given assistance in completing the examination. This must be l done af ter you complete the examination.
< I3.; Restroom trips' are to' be limite4 and only one candidate at a time ~ may
- leave. You must. avoid all contacts'with anyone outside the examination
. room to avoid even-the appearance or possibility of cheating. ;
.1 4.L Use bl ack ink or dark pencil only to f acilitate legible reproductions.
- 5. Print :your name 'in the blank provided in the upper right-hand corner. of the examination-cover sheet.
- 6. Fill in the date on the cover sheet of the examination (if necesf ar y)'. - 1 7.fYou.may write your nswers on the examination question page or on a separate sheet of paper. USE ONmY THE PAPER PROVIDED AND DD NOT WRITE DN-THE BACK SIDE OF THE PAGE.
B. If you write lyour answers on the examination question page and you need more. space to answer a specific question, use:a'; separate sheet of theDD NOT-paper:provided and-insert it.directly after the specific question. WRITELON'THE BACK SIDE OF THE EXAMINATION QUESTION PAGE.
- 9. Printfyour name in the upper right-hand corner of the first page of each fsection of your answer sheets whether you use the examination question pages or separate sheets of paper. Initial each page.
- 10. Before you turn in your ex ami nati on , consecutively number each answer sheet," including any additional.pages inserted when writing your answers on the examination question page.
-11. If you are using separate sheets, number each answer as to category and number,ti.e. 1.04, 6.10) and skip at least 3 lines between answers to j allow . space f or grading. l
- 12. Write "End of Category " at the end of your answers to a category.
- 13. Start each category on a new page.
- 14. Write i'Last Page" on the last answer sheet.
- 15. Use abbreviations only if they are commonly used in f acility literature, f Avoid .using symbol s such as < or > signs to avoid a simpla transposition
. error resulting in an incorrect answer. Write it out.
- 16. The point value f or each question is indicated in parentheses af ter the ]
The amount of blank space on an examination question page is question. NOT an indication'of the depth of answer required.
\
.j
- 17. Show all c al c ul ct i ons , m2thods, or cctumptiona u2Gd to obtcin cn cnnwar.
- 18. Partial credit may be given. Therefore, ANSWER ALL PARTS OF THE QUESTION AND DO NOT LEAVE ANY ANSWER BLANK. NOTE: partial credit will NOT be given on multiple choice questions.
- 19. Proporti onal grading will be applied. Any additional wrong information that is provided may count against you. For example, if a question is worth one point and asks for four responses, each of which is worth O.25 points, and you give five responses, each of your responses will be worth 0.20 points. If one of your five responses is incorrect, 0.20 will be deducted and your total credit for that question will be 0.80 instead of 1.00 even th;ngh you got the four correct answers.
- 20. If the intent of a question i s unclear, ask questions of the examiner onl y.
- 21. When turning in your examination, assemble the completed examination with ex ami nati on questions, examination aids and answer sheets. In addition, turn in all scrap paper.
- 22. To pass the examination, you must achieve an overall grade of 80% ce greater and at least 70% in each category.
- 23. There is a time limit of (6) hours for completion of the examination.
(or some other time if less that the full examination is taken.)
- 24. When you are done and have turned in your examination, leave the examin-ation area as defined by the examiner. If you are found in this area while the examination i s still in progress, your license may be denied or revoked.
O O . o
~ - . _ , - _ _
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' - Dele _EBEEI- !
'i IBEBCIQB_IHEDBY_ED8t!ULOS:. j
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-O = r*n c p (AT) y.. .i_0_ ,out q, h
in -hout'real Q (h -h ot i deal in in actual 11 22 supplied 1 2 4:= pAV pAVg3 g = p2 2 2 Av 3
& = aA 4 AP p M = KA oT M = KA o'p J oP' x &n c = KAa t.T . ( i n ) - T ( ut) g, f E th T
" 8.Bx10 AT (in) in ( ---~~- )
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V . L ..CENIBLEUQ8L_EUdE_ LOWS L (N ) ' H (N ) P N 1 1- 1 1 < _1 . _A1 - P (N ) H (N 2
.N 2 2- 2 2 2 .
B0010I1DN_eNQ.CBEd1EISY_EQBdWLeE R/hr = 6CE/d ' I =1 0 e g g =C 22 CV G = Dilutten_3att I=1 0 Ibb 0* Volume 10
~
A=A 0 e . A = AN CONVEBEIDNS: Density of water (20 C) = 62.4 lbm/ft
~
1 gm/cm = 62.4 lbm/ft 1 gal =.S.345 lbm 1ft = 7.40' gal Avogadro *r Number = 6.023 x-10 1l gal = 3.78 liters Heatiof Vapor (H O)2 = 970 Btu /lbm'
< 1 lbm = 454 grams Heat of Fusion (ICE) = 144 Btu /lbm
-24 grams e = 2.72 1, AMU = 1.66 x'10 :
w = 3.14159 Mass of Neutron = 1.008665 AMU 1 KW = 738 ft-lbf/sec Mass of Proton = 1.007277 AMU 1 KW = 3413 Btu /hr Mass of Electron = 0.000549 AMU 1 HP = 550 f t-lbf /sec One . atmosphere = 14.7 psia = 29.92 in. Hg
- 1
- F = 9/5
- C + 32 1 HP = .746 KW .
1 HP = 2545 Btu /hr 'C = 5/9 (* F - 32) i 1 Btu = 77B ft-lbf *R = *F + 460
-16 Btu *K = *C + 273 1 NEV = 1.54 x 10
~
h = 4.13 x 10 M-sec 10 fissions /sec
- 1 W = 3.12 x'10
= 931 MEV/ AMU Sg = 32.2 lbm-f t/lbf-sec c 8 m/sec 1 inch = 2.54 cm C=3x 10
~
r = 0.1714 x 10 Btu /hr ft R
l
' ' . Paga 4 of 5 DOIG SUEEI
!CVE80GE_IHEBd8L CONDUCIIVIIY_1El- l I
.E Untettal O.025 l' Cork Fiber Insulating Board 0.028 .
Mrple.or Dak Wood . 0.096 Cuilding Brick O.4 l' Window Glass O.45 0.79
' Concrete 25.00 1% Carbon. Steel 1% Chrome Steel 35.00
' Aluminum 118.00 '
Crpper 223.00 Cilver 235.00
' Water (20 psia, 200 degrees F) 0.392 i Cteam (1000 psia, 550 degrees F) 0.046 Uranium Dioxide 1.15 l-H211um 0.135 Zircaloy 10.0 UlSCELLONEQUS_INEQBdellON:
E = me 7 KE = 1/2 mv PE = mgh V f =Vp + at Geometric Object Area Volume A = 1/2 kh ///////////////// Tri angle Square . A = S* /////////////// / A=LxW ///////////////// Rectangle Circle A = wr# ///////////////// Rectangular Solid A = 2(LxW + LxH + WxH) V=LxWxH 2 Right* Circular Cylinder A= (2 wr2)h + 2(wr#) V = wr h Sphere A=4wr 2 V = 4/3 (wr#)
// //// / / // / // // // /l.'J//////// V '= 5 #
Cube l
Page 5 of 5-DOIB_EMEEI 01SCELLeNEQUS_INEQBd8IIDW_icoottourdi 10 CFR 20 Appendix B Table I - Table II Gamma -- Energy Col 1 Col 11 Col I Col II MEV per Air Water Air Water M:tarial Half-Life Disintegration uc/ml uc/ml uc/ml uc/ml
~6 ~8 Ar-41 1.84 h 1.5 Sub 2x10 ----- 4x10 .------
~# ~3 ~8 ~3 Co-60 5.27 y 2.5 5 3x10 1x10 1x10 5x10
~Y ~ ~3 ~#
1-131 B.04 d 0.36 S 9x10 6x10 " 1x10 3x10
~3 ~#
Kr-85 10.72 y 0.04 Sub 1x10 ----- 3x10 ------
~# ~# ~8 ~
Ni-6$ 2.52 h 0.59 S 9x10 4x10 3x10 1x10 "
-2z ~
-24 ~6 Pu-239 2.41x10" y 0.008 S 2x10 1x10
- 6x10 -5x10
~ ~ -21 ~#
Sr-90 29 y. ----- S 1x10 ' 1x10 " 3x10 3x10
~ ~#
Xo-135 9.09 h 0.25 Sub 4x10 * ----- 1x10 ------ Any single radionuclides with T > 2 hr ~ ~ -10 -6 which does not decay by alpha br 3x10 ' 9x10 1x10 3x10 spontaneous fission Neutron Energy (MEV) Neutrons per em# Average flux to deliver j equivalent to 1 rem 100 meem in 40 hours thermal 670 970x10'6 400x10 280 (neutrons) 0.02 0.5 , 43x10f , 30 -- g------ - 10 24x10 17 cm x see Linear Absorption Coefficients p (cm-3) Energy (MEV) Water Concrete Iron Lead 0.5 0.090 0.21 0.63 1.7 1.0 0.067 0.15 0.44 0.77 i 1.5 0.057 0.13 0.40 0.57 ** 2.0 0.048 0.11 0.33 0.51 2.5 0.042 0.097 0.31 0.49 ) 3.0 0.038 0.000 0.30 0.47 l l 9 _ _ _m___..__.__________m___.. _ _ . - . _ _ _ _ _ _ . _ _ _
i } Paga 4'
- 4.- REACTOR' PRINCIPLES (7Z) THERMODYNAMICS.
IZ31_9N9_G90E9NENIS_J1931_JEgypSDENISkSiE39D1
- q DUEST10N 4.01 (2.00)
Will the shutdown margin INCREASE, DECREASE, or REMAIN THE SAME;f or the f ollowing conditions? (Consider each case separately) c." Load. rejection from 200% to 90% power, control rod position lowers and
~
T-ave equals programmed T-avg.
~
- b. The reactor is steady state at 50% power when a small dilution occurs with control rods in automatic.
- c. The reactor is steady state at 50% power when a small dilution occurs with control rods are in manual.
d.. Reactor power holding at 50%, xenon is increasing. 4 d e G l = i i i
\
1 (***** CATEGORY 4 CONTINUED ON NEXT PAGE *****) l 1
. l
o l !: - Psg2 5 L Sk__SE@gIQB_E6igglELES,}Z31_ISEgdgpyd@glgg L 1Z31_099_G90E99ENIS_11Q31_1[UBQ@dEgl@(@_Ef@U1 L 1 l L <DUESTION '4.02 (1.00) , An equivalent positive reactivity addition is made to two critical' reactors, one of which is at BOL and the other at EDL. Choose the DNE correct answer that describes the difference in the startup rate on the two reactors. L 'c. Larger SUR for the BDL reactor because' Beta Bar effective is larger fer l the BOL reactor.
- b. Larger SUR f or the BOL reactor because Beta Bar ef f ective is smaller for the BOL reactor,
- c. ' Larger SUR for the EDL reactor because Beta Bar ef f ective is larger f or the EDL reactor.
- d. Larger SUR f or. the EDL reactor because Beta Bar ef f ective is smaller for the EDL reactor.
G 9 O . (***** CATEGORY 4 CONTINUED ON NEXT PAGE *****)
PIga 6
; 'di__BES9196_fBlyglfbgg_JZ31_Jbggyppyygglgg L JZZ1_ed9_E90E9dEyl@_J1931_JEgypBDgyJ3kg_gXgD1 QUESTION 4.03 (1.00) ,
Choose the ONE answer'which correctly completes the blanks: The curve of critical boron concentration vs. core life has an initial steep drop f ollowed by a relatively constant concentration f or a short i period of time, af ter which concentration drops linearly with burnup. The reason f or the initi al steep drop i s due to _______ (1 ) ______ and the j reason ~for the short period of level concentration is due to " ( 2)______.
- o. (1) buildup of fission product poisons (2) depletion of burnable poisons 1
- b. (1) depletion of burnable poisons (2) buildup of fission product poisons
- c. (1) dilution.toward initial criticality (2) depletion of burnable poisons
- d. (1) dilution toward initial criticality (2). buildup of fission product poisons O .
(***** CATEGORY 4 CONTINUED ON NEXT PAGE' *****)
Page 7 4 __5599I95_EB1991 eke!_1231_ISEBDggyNgglgg 1231_eU9_99dE99Eyl@,Ji931,f[UNg8 DEN 16(g,[39D1 DUEST3DN 4.04 (1.00) Which DNE of the f ollowing statements is correct concerning various ef f ects on Differential Boron Worth (DBW) over core life? a.- Boron concentration decrease with core life drives DBW more negative fission product poison buildup with core lif e drives.
~
(decrease), and DBW more negative (decrease).
- b. Boron concentration decrease with core lif e drives DBW 1ess negative (increase), and fission product poison buildup with core lif e drives DBW 1ess negative (i ncrease) .
- c. Boron concentration decrease wi th core li f e drives DBW more negati ve (decrease), and fission product poison buildup with core lif e drives DBW 1ess negative (increase) .
d.. Boron concentration decrease with core lif e drives DBW 1ess negative (i ncrease) , and fission product poison buildup with core lif e drives DBW more negative (decrease). O (***** CATEGORY 4 CONTINUED ON NEXT PAGE *****)
w - - _ _ _ _ _ _ _ _ _ p Pcga 'B.
', 4. REACTOR PRINCIPLES. (77.) THERMODYNAMICS Izz1_eup_C90e99EuIs_J1931_lE999ergulebS_E38d1 QUESTION 4.05 (1.00)
Which DNE of the f ollowing statements is correct concerning the over all of f ect on Dif f erential Boron Worth (DBW) over core life?
- c. boron concentration decrease is the dominant f actor, which makes DBW mor,e negative with core life.
- b. fission product poison buildup is the dominant f actor, which makes DBW more negative with core life.
- c. boron concentration. decrease is the dominant f actor, which makes DBW less negative with core li f e.
- d. fission product poison' buildup is the ' dominant f actor, which makes less negative with core life, o.. boron concentration decrease is of f set by fission product poison buildup, maintaining DBW essentially constant over core lif e.
o O . s (***** CATEGDRY 4 CONTINUED ON NEXT PAGE *****)
y ['
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14 b REACTOR PRINCIPLES (77)-THERMODYNAMICS L 1231_eNp,CQDfgNENIS J1931 j[WUpSDEN]@65,EX9D1 QUESTION- 4.06 (1.00) ,
'The reactor is critical in the intermediate range, BOL, when the reactor operator positions control rods to achieve a 1 DPM start-up rate. Assuming'
~.co f urther. operator action, at what l evel will reactor power stabilize?
State all assumptions and show all work. i t i O . i 1 (***** CATEGDRY 4 CONTINUED DN NEXT PAGE *****) i i
}
- - - -- - . _ - _ _ ~ ~
g . l.;f ip 7s _. j_
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t f k) Pcg3'10 . 2- ' REACTOR PRINCIPLES--(77J THERMODYNAMICS , t J4 'I 1 12Z1 ESP 990E99E9IE 11931_.!E9NDedENISLS_gf$d1 DUEST3DN 4.07 (1.50) . T Why. should verification of natural circulation conditions be. perf ormed ' onL a TRENDING basis versus an instantaneous set of readings? i, M f' (***** CATEGORY 4 CONT 1WUED ON NEXT PAGE *****)
}
f A--------__._--_-___-____-_______
-- .__m____. _ _ _ _ _ . __ U Pcgs 11
.[C.-REACTOR PRINCIPLES (7%)L THERMODYNAMICS L 1Z31.899_C9dPQygyJg_J1931_,((yyp3y[9136S_E3@y1
,'i L i p f/N, DUESTIDN 4.OB- '(1.00) '
' Whi ch .ONE of the f ollowing prevents DNB f rom occurring. during: normal pl ant ~-
operati ons7:
- o. Minimum' temperature for criticality-
- b. Heat. Flux Hot Channel Factor-c.. RCS pressure saf ety limit , .
- d. Enthalpy Ri se . Hot Channel. Factor
- e. Maximum KW/f t limi t 9
e 9
\
j i
\
(***** CATEGORY 4 CONTINUED ON NEXT PAGE *****) , i _ _ - - _ _ _ _ _ _ = _ . .___ _ _ _ _ - _ _ _ . __ - - - . __ _. _
1 V g ,+&++ '
) b i;
Pcg2 ~ 12 T'91__ BESS 193_C3iggl3bEp_JZ31,Idg3ppDfNSDJgp , g, ,,
;1Z31_eU9_99dE995NI!_119Z1_lE99D$dEUIS(S_g39d1 3
'DUESTION 4.09
- . (1.00) ,
Which DNE of the f ollowing is NOT characteristic of f ast neutron-irradiation on reactor vessel metals?
- o.: : Increased strength "b. Decreased ductility
- c. Nil-ductility Temperature (NDT) decreases.
. d .- A change in the lattice structure of the metal.
o.: Increased brittleness 1 O e l (***** CATEGDRY 4 CONTINUED ON NEXT PAGE *****) 4
w' M , '
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Paga 13
;4. REACTDR PRINCIPLES (7Z) THERMODYNAMICS jl
- (7%)- AND CONPONENTS (10X) (FUNDAMENTALS EXAM) i a' ' ,
i i
QUESTION -4.10- (2,00)
For the f ollowing changes in plant status, ' indicate whether the DNBR in the Consider each change i core will INCREASE, DECREASE, or HEMAIN THE SAME. The reactor .i cOperately, and assu:ne all other parameters remain unchanged. (.5 ea) ; 10' ini tiall y at 50% power.
-l
-j o, ' Increase reactor power
- l i
- b. Increase CVCS charging and letdown, c .= . Increase pressurizer pressure
- d. Increase core inlet temperature J
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y a 1 Pcgo 14 do.__BE 99198_EBlyglE6E g,JZ31_ISEgygpyy3Olgg 1 lIZ1_BU9_E90E9NEGIS_J1931_JEgypSygyJ963_EX861
. J QUEST!DN 4.11 ';.500 (0 5) .
' Assume that during the last ref ueling outage, impulse pressure to the rod control system was calibrated so that Tref reads five degrees higher than' it normally would read. What effect will this have on the following oteady-state pl ant parameters? Limit your answer to INCREASE, DECREASE, or REMAIN THE SAME. Assume rod control and EHC are in automatic.
,.. T:i ; L . . . . " I ; w- . DtlM
- b. Secondary Efficiency
- _ e r r + ~- o - r - blei=l 4
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.,7.. .
Pcge?15-A .-. RE ACTOR: PRINCIPLES - (7'/,) THERMODYNAMICS =
,1Z31.9E9_C90E99EyJg_jjp31_j{gNp3DENISLS_E38d1 Je
,Ii DUESTION- - 4.12 . (1.00) . .
-The reactor is. in a condition with average temperature at .557 degrees-F and FZR pressure at 2250.psig prior to start-up.1 Calculate the. subcooling i
' margin l using. the steam tabl es.
i e l (***** CATEGORY 4 CONTINUED DN NEXT PAGE *****)
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t > t ;.
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y -4,. . ,. , N ..
< : PcgG .'16 L l4.- r. 1231_999 := RE ACTOR PRINCI PLES ! (77.) ' THERMDDYNAMI CS .
995E9dENIS 11931_lE9dDSDENIebs_g3eg1 GUESTION 4,13: ' (1.00) l' .Which :DNEl of: the. f allowing 'is used f or purptases other than prevention-'or - rGductionLof-water hammer?'
. o., '. maintaining ' piping. systems f ull of ~ f l ui d at l all . t i mes
- b. starting-centrifugal pumps with the discharge valve-shut .
1 c.- ' draining . steam lines with' steam traps and drain valves
! d . :- ensuring that. systems are pressurized prior to. operation opening . val ves , wi th' a large DP across them slowly.
e.- . t 8 U O G r e .. (***** CATEGDRY 4 CONTINUED ON NEXT PAGE *****)
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l REACTOR PRINCIPLES ( 77,) THERMODYNAMICS Paga 17.
- 4.
IZ31_9Np,CQUggNEN]p_ fig 31,j[yNp@DgN]@h@_E3@D1 l DUESTION 4.14 (1.00) , R2 actor Bypass breaker A is racked in and closed f or testing with Reactor Trip breaker A open. Reactor Bypass breaker B is ", hen racked in. Choose the DNE correct response that describes what happers when bypass breaker.B 10 racked int
- c. bypass breaker A will remain shut and trip breaker B will trip open
- b. bypass breaker A and trip breaker B will remain shut .
.c. bypass breaker A will trip open and trip breaker B will remain shut
- d. both bypass breakers and trip breaker B will trip open
- a. bypass breaker A and trip breaker B will trip open 6
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Pcgo 18
- 4. REACTOR PRINCIPLES (7%) THERMODYNAMICS
( ( _ _N_ F_T __U_ ND_AM ______ E NT AL S_ _ E__X _AM_) __7_%_)_ A_ _N_L ____ . .C_ O_ __ MPO NE_ S_ _( 10%__) DUESTION 4.15 (1.00) . Actual cc,re exit temperature has remained constant, while containment temperatures have increased f rom 80 degrees F to 120 degrees F. Which DNE of' the f ollowing' statements describes how the core exit thermocouple rsadings will react to the containment temperature change?
- a. readings will remain the same because heaters in. the thermocouple junction box will offset any changes in containment temperature
- b. readings will remain the same because thermocouple wire is connected I directly to the thermocouple display
- c. readings will decrease because the change in containment temperature will decrease the potential between the two thermocouple legs
- d. readings will decrease because the change in containment temperature will increase the potential between the two thermocouple legs
- o. readings will increase because the change in containment temperature will add to the signal generated by the core e"it temperature
- f. readings will increase because the legs of the thermocouple e are unevenly affected by the containment temperature change I
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- e Pcg2: 19
- 4. REACTOR PRINCIPLES (7%) THERMODYNAMICS IZ31_8dP_C90E99EGI!_J19Zd_JEUypBDENISLg_E39D1 DUESTION 4.16 (1.00) ,
SSfety valves are used in the main steam system to protect the the system Which DNE of the f ollowing piping from damage due to overpressurization. ctatements concerning the operation of saf ety valves is correct?
- a. When the activating pressure for a saf ety valve returns to the lif ting set point, a combination of steam and air pressure above the valve disk closes the valve.
- b. As steam pressure increases to the safety set point, the pressure overcomes spring tension on the valve operator, causing the valve to open.
i
- c. As the disk on'a safety valve lifts, less pressure is exerted on the i disk, reducing the upward force on the disk, preventing " valve ,
slamming". 1 I d '. A saf ety valve is lif ted of f its seat, then is forced fully open by an air-operated piston. (***** CATEGORY 4 CONTINUED ON NEXT PAGE *****)
x7 r.. - 3 Pcga 20 'l
;fi 4li- REACTOR' PRINCIPLES-(77)'THERMODYNAM3CS E 1 "---liii:595:EDBE59:Esii:liEBI:IE0955BisiE_S_Exeel 'l l
DUESTION- '4.17 (1.00) Which DNE of the f ollowing statements is TRUE 'or the MIXED-BED - Edamineralizers7
- e. ' Mixed-bed domineralizers are designed .to. remove only ionic- isotopes.
'b. Maximum flow rate is limited- to 109 gpm to prevent resin bed .
c h annel l i n g .
~
-c. An unsaturated mixed-bed demineralized can decrease RCS boron' concentration by approximately 100' ppm.
d .' A mixed-bed demineralized can operate for only.two months with 1% f ailed f uel . (***** CATEGORY 4 CONTINUED ON NEXT PAGE *****) r=---_-_-_-_--___-__. -- -_
J Pcgg 21 !4. RE ACTOR PRINCIPLES (T/.) THERMODYNAMICS-JZ31_eyp_Cggggggylg_ fig 31_JguyggDEyl@(@_Ef991 J
~
DUESTION 4.18 (1.00) . A.' pressur i z er level instrument develops a leak in its ref erence leg. - Which DNE.of the .f ollowing statements best describes how and why indicated pressurizer level will change?
.c. level increases because of less mass in the ref erence leg, causing DP in the bellows to decrease.
b.- level increases because of less mass in the ref erence leg, causing DP' in the bellows to increase.
- c. level decreases because of less mass in the ref erence leg, causing DP in the bellows to decrease.
- d. level decreases because of less mass in the reference leg, causing DP in the bellows to increase.
*9
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Pogo 22 d __ bee 9195_BB1991ELEp_jZZQ ,Iyg3DppyN3DJpp , JZZ1_eU9_G90E9EEyIg_jlp31_JEUyp3DgyIghg_g13D1 DUESTION - <4.19 (1.00) , Choose the DNE correct answer: A starting duty limit (i e. restarting motor, consecutive starts) is placed on the Reactor Coolant Pumps to . o.- prevent damage to seals due to low leakof f ' flow when the pump- is of f
- b. -prevent insulation damage in the rotor due to high -current
- c. ' prevent damage to the lift oil pump due to repeated starts
- d. prevent damage to.the labyrinth seals due to rotor thrust
- o. prevent excessive wear on the anti-rotation pawls and ratchet plate
- f. prevent insulation damage in the stator due to high current
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Paga 23
.4. REACTOR PRINCIPLES (7%) THERMODYNAMICS (7%) AND COMPONENTS (10%) (FUNDAMENTALS EXAM)
QUEbTION 4.20 (1.00) . Which DNE of the f ollowing describes. the AUTDMATIC f unction of TCV-tP99 MS (CVCS demineralized inlet divert valve)?
- o. di verts flow around the demineralized when resin is exhausted
- b. diverts flow around the demineralized when f uel elements have failed
- c. ' di vert s f l ow around the demineralized whenLl etdown temperature rises
- d. diverts flow around the demineralized when recharging resin beds 4
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- b
Pags 24 d __BE99195_EB1991ELEp_jZ31_IUggdppyy3DJgp
.3231_eUP_G90E9NEUI!_l1931_JEWU9eDEyISLS_E38D1 QUESTION 4.21 (1.00) .
A. centrifugal pump with a variable speed motor uses 2 KW to develop a discharge pressure of 50 psi. It is desired to increase the discharge pressure of the pump to 100 psi by changing the pump' speed. How much Show power-Lwill the motor require to develop the increased discharge pressure? . all work. . (***** END OF CATEGORY 4 *****)
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7: [. 5 .' 'EMERGENCh'AND ABNDRMAL_ PLANT _EVpLUTigNS'
~
Pcg2-25
* . iggn l-
' DUESTIDN 5.01 -t3.00) .
What immediate actions must be taken Dy' the reactor _ operator in order to amergency borate? Assume all systems .are in automatic. and the RWST is
-cvailable.
,l& s i 4 .
- i e
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Pcga 26 3 5; "EMERGENC !AND.ABN9@AL_P(MT_EV96UTIDNS l
. GE1
. e a
. Q U E S T I O N L. 5.02 (1.00) _
. Fi l l .. i n the blank:
- _______ __ _ - i s ' u s e d 14 hours subsequent to a large break LOCA te ter-i. ;.r t'
'to1 flush boron from the upper regions of the core.-
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.Pcgo 27
- 8:__Ed[Bggggy;99p_9pyp3D9hig[99I_EypbUJJgyg 1223L
-GUESTIDN- 5.03 (2.00) .
You are performing EDP-0.2, " Natural Circulation Cooldown", f ol l owi ng : a loss of offsite power, when you notice that pressurizer level is ' varying- by o large amount, inconsistent with the cooldown rate.
- 1. Which DNE statement best describes the cause f or the wide variations in pressurizer level in this situation?
- a. pressurizer level instruments are inaccurate because of the loss of of f si te power,
- b. -pressurizer l evel is swinging because of contraction of the RCS cool ant during the cooldown
- c. pressurizer' level is swinging because RCS pressure is varying arouno the accumulator injection point l d. pressurizer level is' swinging because of voiding in the reactor vessel head area .
- 2. Choose the-DNE statement that describes the action, if any, that would be taken to counteract the variations in pressurizer level,
- a. nothing can be done until offsite power is restored
- b. depressuri2e the RCS to f acilitate accumulator injection
- c. repressurize the RCS to collapse any voids
- d. increase the cooldown rate in order to increase subcooling
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6 4 Pcgo 2B' EMERGENCY AND ABNORM /4L PLANT EVOLUTIONS' [ ' 5.b L '32231' QUESTION 5.04 (1.50) . LER . BB-OO2 describes a situati on where electri c stri These keswere doors on ' fire doors were declared dzs3gned to fail open'on a loss.of power.- What were the two-inoperable in accordance with technical specifications. ~ f actors that were used as a basis f or declaring these doors inoperable? (***** CATEGDRY 5 CDNTINUED DN NEXT PAGE *****) __ ____m__.m_-___..m_._____ _ _ _ _ _ _ - _ _ _ _ _ _ _ _ . _ .
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Pcg2 29
. 5.- EMERGENCY AND ABNORMAL.' PLANT EVOLUTIONS I
.!E2Zd .
a. O ESTION 5.05 (1.00): , LER 88-010-01 describes a situation where 400 volt safeguards! busses 2BO4 cnd 2003 ' were; ti ed. together for maintenance, with the G01 diesel generator-rcd tagged'out of service.- Describe. what . technical- specification (s) was/were vi ol ated. (Specific Tech. Spec. number not required.) i O O (***** CATEGORY 5 CONT 1tIUED ON NEXT PAGE *****) i l
Pcgo 30 j 5i_iEDE$pEyCy_9yp_ggyp3D$(_gbgyJ,EygbUIJpy) l 3223F l i
- DUESTION 15.06' (1.00) ,
With Unit 2:at175% power, a saf ety injection and reactor- trip occurred.. JThe f ollowing statements summarize key indications that have been noted:
- c. _ One steam generator ' is depressurizing uncontrollably.
- b. The RCS is losing inventory at approximately 2OO 'gpm .
- c. Air ejector radiation alarms indicate a tube rupture.
- d. Core Exit Thermocouple are reading about 1300 degrees F.
Which statement must be given first priority for remedial ac ti on? O . (***** CATEGORY 5 CONTINUED ON NEXT PAGE *****) -_-x_-___________ _ _ _ _ _ _ _ _ _
\. y _ ;; . I x.
- Paga.31 E Di__EDEBgEygy,999,$gNggd@6,P69N1,Eyg6UllgNg.
.jg}3b DUE N .5.07' (1.00) -
[.
-Choose the NE correct, answer:,
Assume that rod trol is in' MANUAL'when mLdropped rod causes a runback. Pressurizer level wa :-
- o. decrease, because of drop in Tavg with the dropped rod.
decrease, because of a l ow setooint withLlower turbine power-b. T
- c. ' remain'the same, because of.Iow Tavg being offset by the runback.
- d. increase, because of heatup during t runback
.t cusa s.cn deMd O e e
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t.- Pcgo.32
- 15. EMERGENCY AND ABNORMAL' PLANT EVOLUTIONS h-32221-t l QUESTION 5.08' (1.00) ..
Following a transient in which PCV-430 (Pressurizer PORV) has stuck open, cnd MOV-516 (PORV block val ve) has f ailed to close, these readings are- = obtained in the control room: 1:0 ' A B Pcrameter l l SG Pressure (psig) 850 850 55 50
.SG NR Level. (%) 20 l AFW flow (gpm) llo Tevg-(deg F) <530 <530 RCS Pressure (psig) 850 890 TCWR (deg F) 526 525 Thermocouple (deg F) 725 725 off- off RCP's ,
Vessel Level (ft) 18 17.5 CTMT Pressure (psig) 4.0 Subcooling Meter (deg 1) >50 Superheat The best pump to start would be: (choose ONE)
- c. .a safety injection pump
- b. a second charging pump c.. a reactor coolant pump
- d. a residual heat removal pump S
e S 9 j (***** CATEGORY 5 CONTINUED ON NEXT PAGE *****) i
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.Pags:33 2:_ EDE6pEypy_3yp_ggyp3g3k_Ph$yI EypkUIJpy,S.
J3331 DUESTION- 5.09 (1.50) Assume that you are the SRO on Unit 2 f ollowing a reactor trip and saf ety , i injection because cf a steam generator tube rupture. While performing acti ons in EDP-3, you are informed by the RD that you are in the foldout page red-path f or heat sink based. on the f act that no feedwater i s being What actions do you take based on this cupplied to the . steam generators. rcport? I i l 1; 1 I (***** CATEGORY 5 CONTINUED DN NEXT PAGE *****) i
Pegg 34 L2i__EDES9ENCy,9Np_@pypBD36,fL$NI_gyg6Ullgyg 1223L ! l QUESTION 5.10 (1.50) . Unit 1 has suf f ered aSG
. Indicated total . loss wide of' alllevels
_ range AC from an-initial are 190" f or "A"condition of SG and ~ 195" l
' 200%' power..
can only f eed 'each SG at 75 gpm. for "B"'SG.- The turbine-driven AFW, tot ? Explain why you should or: should not initiate CSP-H.1, " Response.to Loss off 52condary Heat-Sink"7 e l l (***** CATEGDRY 5 CONTINUED ON NEXT PAGE *****)
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,' Pcge 35 5.- EMERGENCY AND-ABNORMAL PLANT EVOLUT1DNS' .
- ,, . .!3_331" j
7
'DUESTION. 5.11 (1.00) .
During'a fire in1the control room with of f-site AC available, reactor power.
]
hcs been ramped down.to.20% power and it becomes'necessary to evacuate the
. conteol ' room. Bef ore leaving . the control room, the CVCS system should'be: l' (choose DNE) aligned with one charging pump in manual at minimum-speed taking
- o. {
suction from the VCT . .
- b. aligned with one charging pump in manual at minimum. speed discharging .
to loop A cold leg-
- c. allgned with one charging pump in manual at minimum. speed ' taking ."
~
-suction from the RWST
- d. : aligned with two charging pumps'in manual at minimum speed taking-suction from the RWST' 1
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Paga 36 5:__EDEBgEggy_@9p_gpNg3D36_gb3yI,EygbyJJgNp 122'42 1
'DUESTION 5.12 (1.00)
Unit 1- i s ' shutting down and at 6*/. power = when intermediate range channel N36 fcils high. . Which ONE of the f ollowing. statements .best describes how this f ailure af f ects the reactor shutdown and subsequent operation of the Nucl ear Instrumentation system? a.. the reactor will trip on high IR flux, and source range NI's will re-energize when.N36 reaches the proper setpoint
- b. .the reactor will trip on high IR flux , and source range N1's will have to be manually re-energized
- c. the reactor will not trip, and source range NI's will re-energize when l N36 reaches the proper setpoint
- d. the reactor.will not trip, and source range NI's will have to be manually re-energized l 6 e
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Pcga 37
~2i__EDE6pEypI_Syp_SpypBd3b_fb3yI_EVphyIJpyp 12231 DUESTION 5.13 (1.50) ,
Unit 1. RCS is solid at 340 deg F and 350 psig, with.RCP A running. VCT pressure i s 25 psig. A control problem develops with letdown pressure control valve PCV-135 (in both auto and manual) and RCS pressure drops to 295 psig and then stabilizes. RCP A #1 seal leskoff flow dropped from 0.8 gpm to O.4 gpm, and #1 seal delta P dropped from 325 psig to 270 psig. Explain whether or not it is permissible to continue operating RCP A. Why i or why not? l l 1 (***** CATEGORY 5 CONTINUED DN NEXT PAGE *****) I l I
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y .i. Pcg:2 .- 38 . {g. 5. EMERGENCY AND ABNORMAL PLANT EVOLUTIONS I
. L3E1.
DUESTION 5.14 (2.00) . During a power ramp from 30% to 50% on Unit 1, - i t is discovered that water. 10 trickling-from the open pipe end of the containment cooler The valve will not service water cupply line' drain valve, SW210, at e rate of 0.1 GPM. (choose DNE) close any f urther. Continued power operation ist.
~ a. allowed, however the containment cooler must be declared inoperable.
~b.. not allowed because the containment cooler is inoperable.
- c. not allowed unless a pipe cap is installed to stop the leakage.
- d. .-all owed with no restri ctions.
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E .. y ,. p. P Pcgb 3$P. , 15.: .' EMERGENCY' AND' ABNDRMAL' PLANT EVOLUTIONS ' ( _37_,_ ) _3_ , g Lc n l ' ./
.DUESTION- 5.25, (2.00) .
Unit 2 is at 98% power, with one charging pump' operating in manual _at minimum speed.. The other charging pumps ARE available. You note the-f ollowing datas Parameter' 't=0 min t=2 min t=4 min 98 98 98-Rx Power ( */. ) - Tavg (deg F)~ 571- 571 571
- TCWR (deg F) 543 543 543 THWR-(deg F) 600 600 600 PZR Press (psig) 1985 1940 1900
- P2R Level (%) 46 42 38 SG "A" Press (psig) 700 700 700 SG "B" Press.(psig) 700 700 700 SF "A" (1b/hr) 3.3E6 3.3E6 3.3E6 SF "B" (1b/hr). 3.3E6 3.3E6 '3.3E6 EG Atmos'. A- closed closed closed SG Atmos. B. closed closed closed 2RE215 (uc/cc) 1.6E-5 1.OE+0 1.5E+0 Assume that 1% PZR level a 60'ga11ons Can'the charging system keep up with this transient?. Support your answer- q with appropriate calculations.
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e;. Pcg2 40 5a;_EDEBgEygy_99D_8pygBD9g,fkeyI_gyghyligyS. 12231i I-4 DUESTION 5.160 (1.50) . During a reactor :startup,, the f ollowing indications were observed on the nuclear: instruments:1
;N311 - 3 x 10E+4 and increasing,- positive SUR N32 - osci'11ating around 2 x 10E+3, oscillating SUR N35 .B.5 x 10E-31.and increasing, SUR' oscillating positive N36 - B.O x 10E-11 and. increasing, SUR oscillating positive N41 - O and steady N42 - O'and steady N43 - O and steady N44 - O and steady The DSS decided to suspend the reactor startup.
Fivo minuten after the startup is suspended, source range nuclear instrument channel N31 f ai l s l ow. Evaluate per Technical Specifications, and STATE what actions must be: token, BRIEFLY describing the reasoning behind these actions. - Assume that-rcpairsEwill take 72 hours to complete. 4 E . (***** CATEGDRY 5 CONTINUED ON NEXT PAGE *****)
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' Pago 41
[5.'
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EMERGENCY AND ABNDRMAL PLANT EVDLUTIONS ly '.l . (3. 3 */,
-- ) ;
L . t i; DUESTION 5.17 (1.50) Unit 1 :is starting up when the operator takes the f ollowing . readings: N-31 9x 10E3 cps ' N-32 9.1. x 10E3 cps ~ N-35~ 1 x 10E-11 amps' .l N-36 5x 10E-11. amps. o .- What is the'cause of the incorrect overlap in the above, readings? (0.5) .
- b. -BRIEFLY explain, per Technical Specifications, why the rtertup can or (1.0) cannot be continued.
0 0 0 4 # e
# e
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__ - = __ Peg 2 42 T h__EyEggEyCy_gyg_9pyggg(_ghgyI_gyg6911gyg 12231.
~
QUESTION lS.18' (1.00) A fire in the' Unit 1 side of the control room has necessitated evacuation. Prlor to this, both units'were at 100% equilibrium. Which DNE ofEthe if ol'iowing statements is ,true7
- c. - RCP's should be left running on Unit 2 and stopped on Unit.1.
- b. Each containment will be isolated via manual containment isolation.
- c. All technical specifications are to be adhered.to just Go they would be under other operating conditions.
._ d . Safety Injection is to be initiated on Unit 1, then unnecessary equipment will be locally secured.
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. pegy '43 Sh. EUEBEEbEY 6Np_$ pnp 3536_PkgNI,gyg([JIJggg-i .!HE 1
l
. QUESTION 5.19 (2.50)
Concerning a - f multed steam ' generator os List or ' describe 4 indications that a f ault is INSIDE containment. (3.0) i b.. Expl ain why' f eed flow is stopped to a-steam generator that is faulted (1.5) upstream of the MSIV and outside of containment. !I l l { 9 . (***** CATEGORY 5 CONTINUED ON NEXT PAGE *****)
Pags 44~ 51__EDEBgEyCy_9Np_ggNp393b_P69NI_EyphpJlpNS 122E . QUESTION 5.20- . (2.50) In ST-2 fcore cooling status tree), a transition to CSP-C.2 .is made at osveral poi nts. If no RCP's are running, thi s transi ti 7n .15 made' dependent on whether or not core exit thermoccuples are reading above or below 700 Thermocouple are not checked if an RCP is running. What.is the-degrees. basis for not using thermocouple' readings to determine the condition of.the otatus tree, if an RCP is running 7 i i l (***** CATEGDRY 5 CONTINUED ON NEXT PAGE *****) _______m.__m._ _ . _ _ . . _ _ _ _ _ _ _ _ _ _ _ '
1: Paga 45 5,,__EDEggEygL f ONp,$gyg3D$b,P69yI_Eygbu]Jgyg 122'61 , l QUESTIDN. 5.21 (3.00) Concerning the. Emergency Operating Procedures, BRIEFLY explain why each of-(1.O'each) the :f ollowing statements are F ALSE: o.- In general, a required task as stated in a procedural step MUST be completed prior to proceeding to the next step. l
- b. The procedures are applicable only at power and in. hot, standby.
- c. .The entry points into the procedures are through EDP-0, ECA O.0 and CSP-S.1.
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.Pcga 46 L l5. EMERGENCY AND ABNDRMAL PLANT EVOLUTIONS 13331 QUESTION 5.22 (1.00)
Choose the DNE correct answerr f Assume that the reactor has tripped, and one cuntrol rod has failed to insert into the core. How is the actual shutdown margin affected, as l
)
compared to the calculated. shutdown. margin?
- o. Actual shutdown margin is less than calculated shutdown margin, because
.of the positive reactivity associated with the . stuck rod.
- b. Actual shutdown margin is less than calculated shutdown margin, because MTC adds positive reactivity during the cooldown f ollowing the trip.
- c. Actual shutdown margin is the same as calculated shutdown margin, because it is assumed that a rod will stick on a trip.
- d. Actual shutdown margin is the same as calculated shutdown margin, because reactivity added by power def ect of f sets the stuck rod.
G. Actual shutdown margin is greater than calculated shutdown margin, because it is assumed that boration will compensate for a stuck rod. I
- f. Actual shutdown. margin is greater than calculated shutdown margin, because xenon peaking will offset-the stuck rod.
4 l
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. Pego 47,
;52 EMERGENCY'AND ABNORMAL PLANT' EVOLUTIONS.
12E31.
~DUESTION. 5.23 .(2.00)
- Choose the _ TWD entry conditions f or ' ECA-0.0, " Loss Of All AC' Power": ;
- c. a11 ' 480 VAC saf eguards busses de-energized ,
P b.: all 4160 VAC non-vital. busses de-energized c.. loss of all incoming transmission lines and turbine trip J. dual unit a trip with loss of all incoming transmission lines
- o. . response not obtained column from EDP-0,. step 3
- f. loss of 4160 VAC non-vital busses that f eed saf eguards busses e
9 i d 1 i i
. . l 1
a j l i (***** END OF CATEGDRY 5 *****) J
I . Pega 4B E6;; PLANT SYSTEMS'(307.) AND PLANT-WIDE GENERIC. RESPONg1916111ES_J1331 1 l I 3 GUESTION .6.01 (3.00) Assume that' RCS pressure is 2035 psig when the controlling channel' f or
. pressurizer pressure f ails low.
Limit your
- 1. . . Describe the changes that ' occur f or the f ollowing items.
. answer to INCREASE, DECREASE, REMAIN THE SAME, OPEN, SHUT, FULL ON,
(.5 ea) PARTIALLY ON, FULL OFF, ON, or OFF. L
- a. . Actual Pressurizer Pressure
- b. Spray Valves !
- c. . Proportional. Heaters
- d. Backup Heaters
- 2. Choose the ONE answer that describes the response of the OT delta (1.0)
T trin SETPOINTS.
~ a. The channel associated with the f ailed pressure instrument will decrease, and all others will not be affected. -
i
- b. All channels will decrease.
- c. All channels will be unaffected.
- d. The channel associated with the f ailed pressure instrument will decrease, and all others will increase. ,
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'l L-_-__-__-_______ __.
i
;Pcg2 49 U 6 *' PLANT' SYSTEMS-(30%) AND: PLANT-WIDE: GENERIC ,
. RESPONSIBILITIES (137)
.DUESTI'ON ca.02 .(1.00)
Which' DNE of -the f ollowing will cause the diHerential- rod worth to DECREASE 7 >< c.. Moderator temperatu're is' increased
'b4 Boron: concentration is decreased .,
.c,- . An adj acent burnabl e poison rod is depleted
.d. ' An adjacent rod is inserted to the same height l
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Pcge 50 6:__EL8NI_gy@lEDS_Jgg31_gND_E(@NI; WIDE _gENEBIC-BEgEDL451gl(111gS_J1}31 DUESTION 6,03 (1.00) Which DNE.of the f ollowing will best complete the statement?
"During plant cooldown, flow f rom the RHR system to the RCS is designed to
. be ...
- o. ... constant (around 3100 gpm) and this total flow is maintained by controlling the RHR heat exchanger flow bypass valve (RH-FCV-626)
- b. ... constant (around 1550 GPM).and this flow is maintained by controlling the RHR heat exchanger outlet valves (RH-HCV 624 & 625)
- c. ... varied (to a flow that meets the desired cooldown rate) and this flow is controlled by the RHR heat exchanger flow bypass valve (RH-FCV-626)
- d. ... varied (to a flow that meets the desired cooldown rate) 'and this flow is controlled by the RHR heat exchanger outlet valves (RH-HCV-624 L 626)
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g"l',6.. PLANT: SYSTEMS'(30%) ~ AND PLANT-WIDE GENERIC: ' d RESPONSIBILITIES (13%) [ .-
, DUESTION- 6.04 (2.00)-
' STATE the red path summary f or the: core. cooling status . tree as . listed. on
~the EDP-O' foldout page. Inc3ude setpoints and any coincidence. l
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Pega 52
- 6. PLANT SYSTEMS (30%) AND PLANT-WIDE GENERIC BEggDNglg16]IJEg,jl331
'QUE.. TION 6.05 (1.00)
Assume that RCS pressure i s stable at 2100 psig f ollowing i reactor trip cnd saf ety injection actuation. Which DNE of the following describes the ~ effect on subcooling margin when a saf ety injection pump is stopped? l
- 6. subcooling will decrease because RCS prer,sure will now be below the 1 -shutoff head for the 51 pump
- b. subcooling.will decrease.because break flow will now be greater than SI flow c~. subcooling will remain constant because pressure is above the shutoff head for the SI pump
- d. Subcooling will remain constant because charging pump flow will increase to maintain RCS pressure.
o (***** CATEGORY 6 CONTINUED DN NEXT PAGE *****) l l 1 i
- - - _ _ - _ _ _ _ _ _ - _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ . _ _ _ _ _ _ _ i
7 -_ - - -. .._ .. __. _ _ . _ _ _ J i. Pcga 53 l [ 64- ~ PLANT SYSTEMS'(30%)'AND' PLANT-WIDE GENERIC REpPON@lp!LJIlgS,J133[ 4 ! l L CiUESTION' 6.06 (1.50) LER BB-DO5' describes a situation where, during-calibration of an NIS p
' c h a n n e ) ,' a technical specification violation existed with respect to delta
.T . channel minimum degree of redundancy. What action must be taken during NIS chan.sel calibration' to ensure that the delta T channel minimum degree '
of redundancy is being met? s e e 9
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Pcge 54 64 ' PLANT SYSTEMS'(307.) AND PLANT-WIpg_ggNgg1C
.BggPQNpjplLJIJEg_JJg31 l-(.
DUESTION 6.07 (1.50) LER BB-DO2' describes a situation where a Unit 2 reactor trip occurred The irip was. caused by a during a shutdown af ter an-end-of-life coastdown. f culty source range instrument while the plant was in hot shutdown with Tovg at 540 degrees F. On the Why trip, both main feed regulating valves did the f eed regulating val ves go open automatically went full open. in thi s si tuati on, despite the f act that Tavg wasL less than 554 degrees F? Include any applicable setpoint(s) in your answer. O
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n Pag-# 55 6___E6eNI_SYSIEDS_1}Q31_9dD_E69MI;WlDE_@EGEgig BEggggg1pl61IlEg_J1}31
' QUESTION 6.08 (1.00)
Assume.that the reactor is at 60% power, when N42 begins to fail. Over the course of the entire shift, N42 drif ts up and eventually pegs high. All systems are in automatic, no operator action is taken, and turbine load remains constant. Which DNE of the f ollowing best describes the response of the rod control system to this failure?
- c. Rods will not move because Tavg and turbine power are remaining constant, with no power rate mismatch
- b. The rod control system will cause rods to withdraw, bringing Tavg up in i .accordance with programmed Tavg consistent with the N42 reading c, The rod drop circuitry will cause a turbine runback when the mismatch between N42 and the other power range channels reaches 2.5%, and-rods will step in to reduce Tavg. _l
- d. The rod control system will cause rods to insert to bring power down, since N42 is driving _the auctioneered nuclear power signal high.
l
)
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P&gs 56 16:__E66NI_SYSIEDg_j}p31_Syp_Ek991;yJpE_ggygBJC 5gsegygIg1611]Ep_jJg31-DUESTIDN 6.09 (1.00) Choose the ONE correct answer: Assume that the Unit 1 is operating at 95% when a rod drop occurs, ceasing on.NIS spike of -10% i n 1.5 Theseconds. All systems are in automatic and no operator action is taken. turbine will runback to 75% power because:
- a. 75% is the pre-determined turbine runback setpoint f or dropped rods i, b. tur!
- ne runbacks f or a dropped rod are-always 20%
c, the turbine runback will stop when Tavg reaches Tref
- d. the turbine load limiter is set at 75%
i i
- 1
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%__ELBUI_SYSIElig_lgg31,999_ELQUI;UlpE_gUEEIC Pcos 57 BEgEgy@lB16111Eg_ fig 31 'DUESTIDN
- 6.10 '(2.00)
Aosume that Unit 1 is at 100% power with rod control in MANUAL For when a dropped rod occurs, causing an NIS. spike of -5% in 1 second. each of the f ollowing parameters, would the final values be GREATER THAN, LESS THAN, or EQUAL TD the. final-values with rod control in AUTOMATIC? i
- a. Tavg
- b. Steam Flow.
- c. Pressurizer. Level d.- Generator Load (MWe) l 6
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Page 58
- 6. PLANT-SYSTEMS'(307.)-AND PLANT-WIDE GENERIC REpPQNpJp]LJIJEp_jl331-QUESTION 6.11 (2.00)
Assume that all systems are in automatic with the reactor at 100% when a turbine runback occuts. Power stabilizes at 70%.
- a. What overall effect will the runback have on the OT delta T setpoint? (0.5)
Limit your answer.to INCREASED, DECREASED, or REMAINED THE SAME.
- b. Identify the parameters that affect the OT delta T setpoint, and what effect that each will have on the DT delta T setpoint, following the completed runback.
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.Pago 59 6x__EkeNJ_gygIEDg_jpg31_BND,g63NJ;g]Dg_ggNEBJC SEggpN!!!J611]Eg_jl}31 GUESTION 6.12 (1.50)
Choo;4 the ONE correct answers Following a runback with al.1- systems in automatic, shutdown margin will:
- c. decrease, because rods are closer to the insertion limit
- b. remain the same, because boron concentration'has_not changed c, remain the same, because power defect is offset by contfoi rods
- o. increase, because of higher Tavg on the runback O
e (***** CATEGORY 6 CONTINUED DN NEXT PAGE *****) t-I.
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Pcgp-60. b.__PLgNI_@ySIEDg_13931_QU9_P(@NI;W19E_ GENE 61C BE@P9NSIB16111E@_J1331 DUESTION 6.13 (2.00) While reducing load, the " Power Range Channel Devi ati on" alarm comes in. Plant conditions prior to the alarm, and af ter receiving the alarm, are as fo)1ows: . Prior to Alarm After Alarm , Pcrameter N41 100% BS% l N42 99.5% B6.5% 200% B5% N43 100% B7% N44 215 steps 204 steps Rod G3 205 steps Rod C7 216 steps 215 steps 215 steps Rod G11 204 steps Rod K7 214 steps NTILT 15 1.001 0.989 NTILT 25 0.996 1.007 NTILT 35 - 1.001 0.989 NTILT 4S 1.001 1.013 What i s the cause of the " Power Range Deviation" alarm? ( 1. O) c.
- b. What actions should be taken to avoid violating Technical ( 1.0)
Specifications? (***** CATEGORY 6 CONTINUED ON NEXT PAGE *****) l _2.________._______ _1._ _ _ . _ _ _ _ . _ _ _ _ . _ . _ _
l f, Pags 61' L e 61_ _Eb eUI_5X SIE D S _ j }g'f1_89D,gh @yI;ylDg_ Q[Ng RJ C l SEEE9Np]p](JIJEp_jl}3]
' QUESTION 6.14 (1.50)
.When.a pressurizer pressure transmitter-is removed from service, several-bi st abl es - must' be ' tripped. Indicate whether or not each of the. f ollowing annunciators in the control room will be illuminated or dark when (.25 the ea) bistable tripping is complete.
- c. Pressurizer Low Pressure SI Channel Alert
- b. Pressurizer High Pressure Reactor Trip
- c. Auto Turbine Runback Overpower delta T
- d. Reactor ~ Coolant Overtemperature delta T Channel Alert
- o. Auto Turbine Runback Overtemperature delta T
- f. Pressurizer Low Pressure SI Trip 1
t
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1 , .s Pcgo'62 d f6 PLANT SYSTEMS'(30%)iAND' PLANT-WIDE GENERIC REpPQNgip1LIIJEg,jl}31 I i DUESTION -6.15 (1.00) : During. full power _ operation, the proportional. heaters are observed to be on all the time and the backup. heaters are cycling to maintain RCS pressure'at 1985-psig. The f ollowing indications are also notedt l .
- PRT conditions are normal "
- Charging flow is normal
- A and B. spray line temperatures differ by 25' degrees F
- P2R spray . valves. indicate' closed
--No radiation; alarms
~
Which'DNE of the. f ollowing 'is NOT a possible cause f or the .above readings?
- c. spray line leak .between the spray valve and the pressurizer
- b. spray va3veLhas excessive seat leakage
- c. spray bypass valve is throttled open t'oo f ar
- d. spray valve is stuck partially open
- o. spray valve controller malf unction 9
(***** CATEGDRY 6 CONTINUED ON NEXT PAGE *****) Ah _ _ _ _ _ . _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _
c- ). 61__E69NI,@y@][D@,j}Q31,$Np,E(gN1;yJDg,ggNg3}C Pcg'J 63! I' l ggSepNpJp]LJJJES_J1331 s DUESTION 6.16. ( 1 ~. 00) A large break LOCA occurs on Unit i during full power operation. At time t =0, an SI signal is generated; at time t=30. seconds, a containment spray cignal'is generated. The containment spray system components will operate in the 'f ollowing sequence: (choose DNE) spray pumps A and B start; spray discharge valves B60A, B,
- a. . t=30 sect C, D open t=2 min 30 sec: NaOH addition valves 836A and B open-
- b. t=30 sec: spray discharge valves B60A, B, C, D open t=40 see: spray pumps A and B start t=2 min 30 sect NaOH addition valves B36A and B open c, t=30 sec: spray pumps A and B start t=40 sec: spray discharge valves B60A, B, C, D open t=2 min 30 sec: NaOH addition valves 836A and B open
- d. t=0 sect spray, pumps A and B start t=30 sec: spray discharge valves 860A, B, C, D open t=2 min NaOH addition valves B36A and B open O
E 4 (***** CATEGORY 6 CONTINUED ON NEXT PAGE *****) l _ .I
Pcg2 64 .!
$1-_EbSNI_SygIggg_J39zq_eNp_fb@yI; Wipe _GgNERIC RESPONSIBILITIES
---------------- (13%)
l t
- DUESTION. 6.17 ( 1. 50 f Unit 2 is heating. up f ollowing a ref ueling outage when a Saf ety injection occurs. Given:the following data, STATE.why the Safety Injection occurred.
' Include any applicable 1 setpoints in your answer.
'l Time RCS Temp. PZR Press. RCS Boron Concentration '
j 1500 psig 2000 ppm
'1800 450 deg F.
2000 ppm 1900 475 deg F. 1600 psig j 2000 ppm-2000 500 deg F 1800 psig
.2100 500 deg F 1700 psig 2OOO' ppm n
k e e i e (***** CATEGORY 6 CONTINUED ON NEXT PAGE *****) .
Pcg2 65 6 __E6991_@ygJgDg_Jgp31_999_Eb@yI;ylpg_ggggg}C 555E99519161IlEp_jl}31 DUESTION 6.18 (1.50) A controller failure has caused ITCV-130 (component cooling toAssume no non-regenerati ve heat exchanger control val ve) to go closed. Isakage through 1TCV130, and the Letdown Gas Stripper is bypassed. Include any expected alarn and What effect would this have on CVCS? Setpoints not required.
-cutomatic actions that would occur.
G (***** CATEGORY 6 CONTINUED ON NEXT PAGE *****) e
Pcg2 66" 6:__E6991_5X5IEDS_1}gZ.1_99D_E(.991;MIDE_QENESIC - SEgEDWgig16111ES_113Z,1
~
1 l l DUESTION 6.19 (2.00) Ansume that Unit 2 is operating at 100% power, when the following events oCCurt 1433 "A" steam generator f eed reg . valve placed in manual 1442 rod control placed in manual f or replacement of e. blown moveable gripper fuse on a control bank D control rod - 1445 "A" charging pump is tagged 005. 1447 PC-486C- (first stage pressure) fails high RD attempts to move rods to match Tavg to Tref. A reactor trip 1449 occurs. Assume that all other systems / components'are in their. normal 100% power. lineups. D2 scribe the cause of the reactor trip, and the sequence of events that loads to the trip. Setpoints are not needed. 9 G . (***** CATEGORY 6 CONTINUED DN NEXT PAGE *****) I j
l I Pcge 67 6:__g(991_pfpIgyp,j}g31_ gyp _g(SyI-ylpg_ggyg31C 555E9021P161I159_l1231 1
.OUEST10N 6.20 (1.00)
Which DNE of the f ollowing is NOT true concerning Component Cooling Technical Specifications? l o. For single unit operation, the reactor shall not be made critical' i l unless the two component cooling pumps and the two component cooling heat exchangers assigned to that unit are operable,
- b. For single unit operation, during power operation, a single component cooling pump assigned-to that unit may be out of service for a maximum of 24 hours,
- c. For two-unit operation, both reactors shall not be made critical unless three component cooling pumps and three component cooling heat exchangers are operable.
- d. For two-unit operation, during power operation, two heat exchangers may be out of service for up to 48 hours.
4
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Pcga'6B
?
6.' PLANT SYSTEMS (307,)'AND PLANT-WIDE GENERIC-REpPpNpJp](JIJEp,JJ331
-QUESTION 6.21 (1.00)
Which DNE of the f ollowing is NOT true concerning the reactor makeup
.eystem?
- o. Automatic mode provides dilute boric acid solution, preset to. match boren concentra'cion in the RCS.
- b. Dilute mode permits addition of a preset quantity of water.at a preset flowrate to the VCT.
- c. Alternate Dilute mode is similar to Dilute mode except that water is supplied directly to the charging pump suction and to 'the VCT.
- d. Borate mode permits addition of a preset quantity of concentrated boric acid solution at a preset rate to the VCT.
l l (**w4* CATEGORY 6 CONTINUED ON NEXT PAGE *****)
61__f(gNIiSLSIgyp_13pM_3Np_g(3NI;MDE_gENEBJC Pcg;7 69 ]
-1 RESPONSIBILITIES
---------------------- ( 137. ) ;
J l 1
- i. i l 'DUESTION 6.22 (1.50) J Psr Ter.hni col - Speci f i cati on 15.3.14, " Fire Protection System", match the
~ f ollowing situations with the required action:
.(NOTE: items in;the ACTION column may be used more'than once)
SITUATION' ACTION 1.. penetration fire barrier
- a. continuous fire watch inoperabic
.2. DG sprinkler system inoperable b. periodic fire watch' inspection
- 3. fire detection instrument inoperable ,
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m,-_-__- _ _ -- _ _ _ __ q l POga 70 j
' 6.
PLANT'BYSTEMS (307)' AND PLANT-WIDE GENERIC ^ i RESPONSIBILITIES
---------------------- (13X) ;
1
- QUESTION 6.23 (2.00)
Choose the one term that best describes the process of determining an instrument's accuracy by visually comparing the indication to other (As'it
, independent ~ instrument channels measuring the same parameter. q 10 defined in Technical Specifications) I
, c. Instrument Check'
-b. Continuity Check
- c. Channel Functional Test
- d. Channel' Verification r
- o. Channel Calibration
.f. Channel Check e
e e 4 9 9 (***** CATEGORY 6 CONTINUED DN NEXT PAGE *****) 1
-r
Rn , 1Pcg2 71' [6.:. PLANT SYSTEMS: (30Z): AND PLANT-WIDE GENERIC: BESfgNS!!!L111ES_jl}31 35p
.DUESTION: 6.24" (2.00)
Choose the TWD correct answers: Under whatitwo circumstances is it permissible to violate a Technical Specification Limiting Condition of Operation (LCD)?
. o'. ?During an emergency to protectLpublic healthfand safety.
-b. During a General Area Emergency. .
- c. During a Site Area Emergency.
- d. When f ollowing the Emergency Operating Procedures.
- s. When f ollowing the . Abnormal Operating Procedures.
- f. When directed to by the Duty and Call Supervisor.
+
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L _ __ _ _ _ _ _ _ - - _ _ . . - _ - _ _ _ _ __ . . .. .- _ _
7- _ .- ,
'I E Paga 72- .I h__E69N]_SYSIEgg_j}p31,999_f($NI;ylpE_G,NgBJ.
1 BESE9Ng191(11]Eg_J1}31
'DUESTION 6.25- -(2.00)
- l. .
Par 10 CFR 55:
.a . " Controls" are apparatus and mechanisms, the manipulation of which directly affects the reactivity or power l evel of the reactor. Other than the rod control system, list three systems that are considered (.5 ea)
" control s" . .
- b. An unlicensed individual may manipulate.the controls under the direction and in the presence of a licensed operator or senior operator under two circumstances. STATE one of these two circumstances. (.5)
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m l Page 73
. 6___EbeUI_sySIEMg_fggzq_eyD_E6eMI: WIDE _@EGERIC RESgggglBJ.6111Eg_11@31 .t DUESTION 6.26 '(2.00)
A condition arises which ' requires entry into a high radiation area.. The operator entering the area will receive an estimated whole body dose of.40 mrem. . The personnel listed below, with their related personal information, are. recommended by the HP department to do the work. Each candidate is
-technically competent and physically capable of perf ormingPERMIT the task.DBTAINING Emergency ,litaits DO NOT apply and TIME CONSTRAINTS DO NDT:
AUTHORIZ ATION f or an ' exposure limit increase. Which candidate (s), if any, have acceptabl,? exposure margins to perform the task? . Indicate the reason (s) for rejecting a. candidate for the job,'if applicable. NOTE: Each exposure below (qtr, yr, life, etc.) includes the exposure-above it. Assume the current quarter is the f ourth calendar quarter. All exposures are in mrem. 2 3 4 Candidate '1 male malp f emal e female Sex 20 27 38 24
' Age 20 50 10 0 Today's exposure 250 Month exposure 270 450 40 1020 600 470 1030 Dtr. exposure 2010 Yr exposure 2200 2995 Illo 54730 5200 9770 Life exposure history form 4 3 months form 4 Remar *:s on file unavailable on file pregnant form 4 not on file pregnant G e
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46 . / PLANT SYSTEMS (30%) AND PLANT-WIDE-GENERIC 'P092 74-RESPONSIB1'_ITIES (13%) a i QUESTION 6.27 (2.00) _ Match the f ollowing definitions with the correct terms:- (NOTE: not all the TERMS will be used) DEFINITIONS
- 1. Any area, accessible to personnel, in which there exists radiation,
- originating in whole or in part within license. material, at such levels that a major portion of the body could receive in any one hour a dose in excess of 5 mrem, or in any 5 consecutive days a dose in excess of 100 meem. 1
- 2. Any area access to which is controlled by the licensee f or purposes of
. protection of individuals f rom exposure to radiation and radioactive meter i al s.
TERMS
- o. Radiation Area
- b. High Radiation Area
- c. Restricted Area
- d. Owner Controlled Area .
I i , l l l 1 1 1
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}
1 i E__ _ _ _ _ -- - -- - -
Wa .. Page 75
~
p ..6. ' PLANT SYSTEMS (3931_$Np_P(3NT-WJpg_ggNEBJC BEgEpNSJp]LJIJES_J1331 l JGUESTION 6.28 (1.00) Which DNE of the f ollowing is NOT a method used to control access to high radiation areas per 10 CFR 20?
- a. access controlled by a control device which shall cause the level of I
radiation to be reduced below tt.st at which individuals might receive a dose of 100 mr em i n 1 hour upon entry into the area .
- b. access controlled by a control-device which shall energize a conspicuous visible or audible alarm signal in such a manne- thata the individual entering the high radiation area and the licens ee. or supervisor of. the activi ty are made aware of the entry
- c. maintain the area locked except during periods when access to the area is requirec, with positive control over each individual entry.
- d. maintain the area locked with entrance and exit controlled by a key that is in the direct control of the Health Physics Supervisor or the seni or management offacial on shift I
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~1. l c l
. Pcgo i6 6 ~. PLANT SYSTEMS (307.).AND PLANT-WIDE GENERIC .j
' BEgPDyg191LIIJEg_J13Z1: i J i . J n i
. QUESTION 6.29 ( 1. 50 )'
- Metch the f ollowing TERMS with .the proper DEFINITIONS.
' TERMS DEFINITIONS isolated.with a red tag a. aquipment.not in's condition to j 1.
perform its normal function i o n 24- out of service tags installed and permission b. j 3. holdof f clearance granted to begin work (WE SCD & System Control) ( c. prevents operation that could result. in danger to personnel or. equipment 9 f 5 9 l (***** CATEGORY 6 f.ONTINUED DN NEXT PAGE *****) i e e
Page.77l
' 6 .- PLANT SYSTEMS (30%) AND PLANT-WIDE GENERIC BESPDySJgj6]IJES,jj}3[
/DOESTION 6.30 - ( 1. 00)
'Umit 1 is at - 100% power and Unit 2 is in a 'ref ueling outage. Fuel is baing loaded into Unit-2. Per technical specifications -
'c) What is the minimum control: room licensed personnel staffing - (0.5) requirement?
b)iWhat is the minimum local refueling licensed personnel-staffing (0.5)
-requirement?
4 9 9
?
a S e c. (***** END OF CATEGDRY 6 *****) (********** END OF EXAMINATION **********) L
- w. '< . i
'4.',REACTORTPRINCIPLES (7%)' THERMODYNAMICS. P692 78 ;
- (77) AND COMPONENTg_1107,1,JFUtJDAMENTALg_ EXAM) l
__ . 0, ,2. _ MASIER COPY ; (0.5 eal
- c. remain the same b, decrease {
- c. decrease
,d. increase
. ?
REFERENCE { Tschnical Specification 15.1.g,4
- 192OO2K114 , (KA's) .
i ANSWER 4.02 (1.00) d REFERENCE TRHB 32.4, pp 9 - 12 192OO3K106 ..(KA's) ANSWER 4.03 (1.00) O REFERENCE Westinghouse Reactor Core Control for Large PWRs, pp 2-11 to 2-12 192OO7K104 ..(KA's) ANSWER 4.04 (1.00) c . REFERENCE TRHB 32.5, pp 13 - 14 192OO4K111 ..(KA's) (***** CATEGORY 4 CONTINUED ON NEXT PAGE *****)
4, REACTOR PRINCIPLES (77) THERMODYNAMICS Pcga 79-JZ?l_099_C90PgyEUIg_figf1_f[Uyg@ DEMI 6Lg_E3901 ANSWER 4.05 (1.00) C b l REFERENCE TRHB 32.5, pp 13 - 14 192OO4K111 ..(KA's) ANSWER 4.06 (1.00) . NOTE: answer will depend on assumptions made by the candidate regtrding tha values of beta-effective, lambda and power defect. Acceptabl e ranges: { Power defett = 14 L g uiin. p uwm l'1.-2.0re*/9ofo# Lcmbda = 0.1 bsta eff = .0052 - .00'2 .CcLS tcu=26.06/SUR . NOTE: .25 credit for sol uti on rho =(beta eff)/(1+1ambdattau) NOTE:candidate .25 creditassumptions for solution based on power l evel arho / (p cm/*/. power ) NOTE: .5 credit for solution based on ; candidate assumptions (c1 ternate methods of calculation evaluated on a case basis) REFERENCE TRHB 32.4, pp 17 - 24 192OO8K113 192OOOK110 ..(KA's) ANSWER 4.07 (1.50) Eliminates effects of instantaneous variations in readings (Will accept: some indications of natural circulation depend on a trend ! in readings, not on the reading itself) l I
)
REFERENCE I ERG Executive Volume, Generic Issues, Natural Circulation , 193OOOK122 ..(KA's) ! l l l (***** CATEGORY 4 CONTINUED ON NEXT PAGE *****)
... .+. -
Idi-Pcge.:80. s4 THERMODYNAMICS
~~dREACTOR PRINCIPLES " (77.)
d?Zl_eND CQDPgNENT@_J1931_JEUNpSDgNT96@_El@M1 i
+
-ANSWER' '4.08 (I.00)- ,.
d
# v W ..
D REFERENCE-Y; x TRHB 39.3, pg.88
, W;stinghouse Thermo-Hydraulic Principles, pg.13-36 a 193OO9K105 ..(KA's) y, ..
. ANSWER 4.09 -(1.00)
- c
' REFERENCE
' Westinghouse Thermal Sciences, Mod D-5, 2. 3 ,
.193010K105 ..(KA's).
ANSWER 4.10 (2.00)
-(0.5 ea)
- a. decrease
- b. remain the-same
- c. increase'
- d. decrease REFERENCE TRHB 39.3, pp 82 - 83
' Westi nghouse Thermo-Hydraulic Princi,,il es, pp 13-22 to 13-24 193OOOK105 . 4KA's)
ANSWER 4.11 ei .50-; (0 5)
- 2. . . . . . . . . - t5: :::: bddd
- b. increase r- t : ;_sc hle{ej (.5 ea)
(***** CATEGORY 4 CONTINUED ON NEXT PAGE *****) i l L u_ _- :_ __ _ _ - __ -. _ _
~
-' -~ '~ -~~ - - - - - - - - - - - - - -
k 9 < j
]
o -: .
.Paga.01 '
A ^REACTDR PRINCIPLES (72) THERMODYNAMICS . dZZlteNP_C90E99ENIS_figfl,jEUypedEyl@b@,EXedl '! lj REFERENCE i bIP)-;TR'HB 39.2, pp 7'- 8,:21 i).y i Wactinghouse . Thermo-Hydraulic Principl es, pg 7-67 193OO5K103-
. .- ( KA' s ) -l
.' '{
d
<1-ANSWER... 4.12- (1.00). r L(654-557)=97' degrees-F (+/- 2)~
l
. REFERENCE ~ ,
.ctaam tables 193OO3K225 ...(KA's)
' ANSWER 4.13 .(1.00),
c b. REFERENCE TRHB'40.2, pp'52.- 58 193OO6K104 ..(KA's) ANSWER. 4.24- -(1.00) c' f.f , lN REFERENCE 4 TRHB 13.3, pg 29 L 19100BK104 ..(KA's) ANSWER 4.15 (2.00) b-
' REFERENCE LPO584, 2.4.13.d 191002K113 191002K114 ..(KA's)
(***** CATEGORY 4 CONTINUED ON NEXT PAGE '*****)
6: ,<.: > ! :- W.
, i PGgs-Q2;'
b 4
,;e a s- -y - a . , , _ .
k y. , , . y' l in 2. . RE ACTOR - PR I NC I PLES' - ( 7%) v THERMODYNAMIC CS 1
-(7Z) AND4 COMPONENTS'(10%)'(FUNDAMENTALS' EXAM)
~
j
.0'J I? ' :. ANSWER'- 4.16- (1.00) let el -REFERENCE -
TRHB'11.1, pg' 2 191001K101 ..(KA's) j? ANSWER- ~ 4.17 (1.00) .' g-C ,. REFERENCE
-OP-5D, sec E, prec.aution 2.3 191007K108 ..(KA's) 1
. ANSWER.. - 4.18 (1.00)
D L . REFERENCE TRHB'13.6, pp.5.- 6'
- 191002K109 ..(KA's)
ANSWER 4.19 (1.00) a-O
' REFERENCE TRHB 10.2, pg 31 191005K106 ..(KA's)
, , . .8 . E ' ANSWER 4.20 (1.00) ; e i
)
(***** CATEGDRY 4 CONTINUED ON NEXT PAGE *****) l J
.)
- - , - - . - - , - - , - - ,m-.,-,-_,_ - - _ _ , . , - - __ .' 0 ; REACTOR' PRINCIPLES-(7%) . THERMODYNAMICS Pag? 83 1Z3 ?._eU9_ C90E99ENIs_ fig 31_f gygpggggIeg S_Ef eg1 . REFERENCE-
- TRHB 10. 6, pg:33 191007K109E ..(KA's)
ANSWER 4.21. L ( 1. OO) hsad proportional to speed ' squared so: if head doubles, (N squared) doubles, or speed increases by a' factor of 1.414' '(.5) . . power proportior.a1'.to speed cubed so: . 1.414 cubed = 2.83 or power increases by a factor of.2.83 (.25). final' power =1nitial power
- increase = 2
- 2.83 = 5.66 KW (.25) (i,lkW)-
REFERENCE TRHB 40.2, pg 6 191005K104 ..(KA's) e (***** END OF CATEGORY 4 *****)
..: na- .i . i
~
Pcg7 84 25e EMERGENCY AND ABNORMAL PLANT EVOLUTIONS l}}l). ANSWER. 5.01 (3.00) Ecpen. charging pump suction from RWST (.5). Chut VCT outlet valve ('. 5 )
- ctart all avail able charging pumps (.5) open HCV-142 (charging flow control valve) for >= 70 gpm flow (.5) ctart boric acid transf er pumps (.5)-
cpen emergency boration valve (MOV-350) (.5) . REFERENCE' ADP-6E, pg 2 .. OOOO24G010 ..(KA's) ANSWER 5.02- (1.00) core deluge REFERENCE EDP-1, step'27 OOOO11K313 ..(KA's)
. ANSWER 5.03 .(2.00)
- 1. d
- 2. c (1.0 ea)
REFERENCE ERG Background Document, E S'- 0 . 2 , pg 45 g EOP-0.2, pg 11 l OOOO55K302 ..(KA's) , 1
. ANSWER 5.04 (1.50)
. fire near the door could cause loss of power (.75) and delta P generated by the fire could cause the door to open (.75)
{ k\ & & Y2. ce i k'. S c ts su & CeMI S W K Y h a $ " U'*) s$ 1 & N i I-(***** CATEGORY 5 CONTINUED ON NEXT PAGE *****)
- g. , ;...
- 5. EMERGENCY AND ABNORMAL PLANT EVDLUTIONS Paga 85
. dE2'$.)_
REFERENCE LER 88-002 OOOO67 GOO 4 ..(KA's) ANSWER 5.05 (1.00) hs l'ed 3hwe ws 'i ngekk G M k E5f *t p M Ane M d WM NM d tu :== ::r e cc-:! - :: i .2 = ; c- -b : : 2Sf* a .W: \c:h- : : p k.. p ykc, w k (.Q b pd Ml$ ( h , 2 (vi ol at ed TS 15. 3. 7. B.1. C AM. e) Oh,.4 kl% P.=fim((.3),M
.1) , Ritt L.\) ,4 Md 1 ip (.5)].
REFERENCE LER-88-010-01 TS 15.3.7.B.1.e OOOO56 GOO 3 ..(KA's) ANSWER 5.06 (1.00) d REFERENCE EDP status trees Point Beach Requal Guestion 031-03 000040G011 ..(KA's) A WER 5.07 ( 1 '. 00 3 e gesh 5.0 7 ocI6 d i REFERENCE TRHB 13. TRHB 1 .6 Poin Beach Req 1 Question 043-02 OOOO3A106 .(KA's) ANSWER 5.08 (1.00) e (***** CATEGORY 5 CONTINUED ON NEXT PAGE *****)
- 5. EMERGENCY AND ABNORMAL PLANT EVOLUTI']NG Pogs 86
(_ _3_3_%_) . REFERENCE CSP-C.1 Point Beach Requal Question'031-03-OO1 OOOO74A201 ..(KA's) ANSWER 5.09 (1.50) Check the status of all CSF status trees (.75), and go to the procedure with the highest priority. (.75) (alternate: have DTA run status trees - and exit to highest priority procedure)
.(Mkenk.RO M ard it tvs*M(* (.*75) Sb > O bei b N".8?
- b (*
" REFERENCE ERG Executive Volume, Generic Issues, Foldout Page Items, pg 6 OOOO38G012 ..(KA's)
ANSWER 5.10 (1.50) During a loss of'all AC status' trees are to be monitored for information onl y (.75) and CSP's are not to be performed (.75) REFERENCE ECA-O cautions Point Beach Requal Duestion 031-01 OOOO55G012 ..(KA's) ANSWER 5.11 (1.00) c l l REFERENCE TRHD 10.6 AOP-10A ) Point Beach Requal Question 055-02 l OOOO68A113 ..(KA's) l l ANSWER 5.12 (1.00) , b , i (***** CATEGORY 5 CONTINUED ON NEXT PAGE *****) I i 1 o
51__Eygg@@ygy_@yp_SByggygy_P66yl_gygkU119y@- Piga G7 12231
' REFERENCE TRHB 13.1 Logic Sheets.11 and 12 Point Beach.Requal Question 053-03 OOOO33A208 ..<KA's)
ANSWER. 5.13 (1.50) 01 seal delta P:is the limiting f actor (.5).and is greater than 200 psig (.5) and #1 seal flow is greater than .2 gpm (.5) (so continued operation 10 permi ssi bl e) REFERENCE 01-1
+
OP-1A l Point. Beach Requal Question 051-01 OOOO27A203 ..(KA's) ,
. ANSWER 5.14 ~(1.00) d
' REFERENCE Tech Spec 15.1.D, 15.3.6A Point Beach Requal Duestion 051-05 OOOO69A201 ..(KA's) i ANSWER 5.15 (2.00)
.Yas leak rate = 8% in 4 minutes = 2% = 120 gal / min (.75) charging system has 3 pumps at 60 gpm or 180 gpm capacity (.75)
REFERENCE ADP-3A Point Beach Requal Question 051-01-004 OOOO37A212 ..(KA's) l l (***** CATEGORY 5 CONTINUED ON NEXT PAGE *****) i i l:
l ,_ - ; , :. r
.s ~
5? " EMERGENCY ' AND ABNDRMAL PLANT EVDLUTIONS Pcgs BB
-d2221-
-ANSWER 5.16 (1.50)
Must-' maintain'(or: return to) hot shutdown (.75) becauseitwo SRN1's have Efeiled (.75) REFERENCE Tcchnical Specifications 15.3.5 000032G003 ..(KA's)
, ANSWER- 5.17 (1.50)
- c. ' N 7 35 -i s overcompensated (.5) .
.b. can continue since tech spect require only 'I operable 1R for startup (1.0)
REFERENCE Tsch Spec 15.3.5. 1TRHB 13.1 Point Deach Requal Question 055-02 OOOO33A211 ..(KA's) ANSWER. 5.18 (1.00) b
' REFERENCE ADP-10A' Point Beach Requal Question 055-02 OOOO67A213 ..(KA's)
(***** CATEGDRY 5 CONTINUED ON NEXT PAGE *****) a------_--_ _
v :
.c
' EMERGENCY AND'ABNDRMAL PLANT EVOLUTIDNSI Pog2 89 5
1333ZL ANSWER 5.19 (2.50) l
- c. containment pressure increase containment temperature. increase containment' sump level increase containment . relati ve humidity increase (.25 ea) b.- since the steam leak is unisolable (.5), the SG must' boil dry to stop the leak (.5). Continued f eed flow would result in continued steam generation (.5) ,
REFERENCE EDP-0 EOP-2-, Point Beach Requal Question 031-03 OOOO40K306 . -. (KA's) ANSWER 5.20 (1.50)
. (If an RCP i s operati ng, then).even under a highly voide'd RCS condition
(.75); the core exit thermocouple can be expected to indicate' saturated temperatures. (.75)
'R'EFERENCE ERG Background, Status Trees, F-0.2, pg 12 COOO74K311 ..(KA's)
ANSWER 5.21 (3.00) i
- s. - Simply starting the step is suf ficient.
b.. Some~ procedures are applicable in hot shutdown'and cold shutdown
- c. Entry is not allowed directly into CSP-S. I. (1.0 ea)
(eccept: entry into CSP-S.1 is through ST-1 or EOP-0) REFERENCE
' ERG Executive Vu tume, Users Guide, pp 5, 20-25 CSP-S.1 entry conditions ;
/
OOOO29GO12 OOOO29G011 ..(KA's) (***** CATEGORY 5 CONTINUED ON NEXT PAGE *****)
9, c EMERGENCY: ANQ,ABNQ@AL_% ANT _EVQLUTIQNg , Pcg2.90: 5.' ,
"12221 l
s: 1' < ANSWER 5.22 - ( 1. 00) C-
,., REFERENCE-LToch-Spec 15.3.'g.4 *
-000007G004 OOOOO7 GOO 3' ..(KA's) ,
ANSWER 5.23 (2.00) . c, e (1.O'ea) REFERENCE ECA-0.O' entry conditions Point- Beach Exam Bank Question 31-03-09 OOOO56G011 . .- ( K A ' s ) ,
-l I
(***** END DF CATEGORY 5 *****) 1 a
4: . <.. ,< ..
~ ' 162 ~ " PLANT SYSTEMS'-(307) AND PLANT-WIDE' GENERIC.. -
Pcg2 91 .r-RESPQ!LSig16111ES_11 E b t' y; . IANSWER" 6.' O l ' (3.00): 11 ? a. increase
;b. shut
.c.- full.on.
- d. 'on (.5 ea)
- 2. d . ( l'. O )
-R'EFERENCE
' TRHB 10.3, pp ~~ 12-14
- L.(TRHB '13. 3, . pp- 4-5 '
> 012OOOA205 ..(KA's) fANSWER '6.02 (1.00) d REFERENCE'
.TRHB 32.5,.pg 23 I
- OO1000K510 ..(KA's)
ANSWER. 6.03 (1.00) c
.. REFERENCE TRHB 10.7,'pg.3 OO5000K402 ..(KA's) .
ANSWER- '6. 04 ' (2.00) [ core exit thermocouple greater than 1200 degrees F (.5) L- DR core exit TC's greater than 700 degrees F (.5)
-AND vessel' level less than 29 ft (.5) with no RCP's running (.5)
I (***** CATEGORY 6 CONTINUED ON NEXT PAGE *****) l
)
q 4 i Poga 92 g L G___EbeNI_@y@ led @_f@g[1_QNp,,8(3N1;Wlpg_QgNgg1C BEEEQNg1B161I1Eg_11}}[ ] l s REFERENCE-
- ST-2 (core cooling)
EDP-6 foldout page 017000G015 ..(KA's) ANSWER h6.05 (1.00) f C l REFERENCE
- TRHB 10.8, pg 6 OO6050A102 ..(KA's)
' ANSWER 6.06 (1.50) place- associated delta T channels (.75) in a tripped condition. (.75)
REFERENCE LER 88-005 015000G005 015000A403 ..(KA's) ANSWER 6.07 (1.50) lowering ~ the l ow Tcvg range resistors are installed f or coastdown (.5),
-Tcvg.setpoint (.5) to 539 degrees F (.5)
REFERENCE LER 88-002 012OOOA101 ..(KA's) ANSWER 6.08 (1.00) 0 (***** CATEGDRY 6 CONTINUED ON NEXT PAGE *****)
- w. <
Pcgo 93
~6 ' PLANT SYSTEMS (30~/,)'AND PLANT-WIDE GENERIC-h
'- REgPONgip1LIIJEg_jl331 t. REFERENC8 fTRHB 13.8, pp.1-2
'OO1000K105 ...(KA's)
ANSWER 6.09 (1.00) 1 1
'b REFERENCE
-TRHB 13.11 Logic Sheet. 16 Point Beach Requal Question 052-03 045000K412 ..(KA's)
ANSWER 6.10 (2.00) (.5 ea)
- o. greater than
- b. greater than
- c. greater than
- d. equal to REFERENCE-TRHB 13.8 TRHB 13.9 Point Beach Requal Question 053-01 '
OO1000K104 2 ..(KA's)
' ANSWER. 6.11 (2.00)
- c. Increased'(.5)
- b. Tavg dropped which caused additional credit to the setpoint adding
(.5) flux
-Fidelta flux) changed due to large negative deltaW no d M $ O M s h ,di M penal ti es (.5) cab 4 FGAb4 %
pressurizer pressure will increase, causing additional credit (.5) (will accept: no change in pressurizer pressure) (***** CATEGORY 6 CONTINUED ON NEXT PAGE *****) - _ _ = . _ _ _ _ _ _ _
- i6. PLANTi SYSTEMS: (307.)'AND-PLANT-WIDE GENERIC .Paga 94
, BEgPON@lBlLillES_113Z1
' REFERENCE 16TRHB 13.3
~jr Poi nt Beach Requal Question 053-02 012OOOG010 ..(KA's)
- 7 l
l' ANSWER- 6.12. *(1.50) l_ 'C'
' REFERENCE
. Point Beach-Requal Question 055-01 OO1010A404 ..(KA's)
ANSWER 6.13 (2.00) e..' Rod G11 is misaligned (f ailed to drive in) ( 1. 0)[4M2 2*k N"f i* F'd *
- b. Rod G11. must be investigated to determine why it is misaligned ( . 5); and.
then realigned (.5) (alternate: f ollow actions of AOP-6B)
-REFERENCE fMM.CM YM P* N b6 . k rul% rods (.5))
Toch Specs 15.3.10.B.3, 15.3.10.D Point Beach Requal Question 039-03 OO1000K507 ..(KA's)
' ANSWER 6.14- (1.50)
-c. illuminated
- b. dark
.c. dark
.d. illuminated
- o. illuminated
- f. dark REFERENCE ARB, Logic Sheets 13, 15, 18 Point Beach Requal Question 053-02 ;
012000A404 ..(KA's) I 1 l (*****' CATEGORY 6 CONTINUED ON NEXT PAGE *****) _ - _ _ _ _ _ _ _ _ _ _ ~ - ._.
.6:__ELONI_SYEIEDS_J}9Z1_8Mp_C68MI;WJpg_ggyE819 .Phga;95 SESCgySIBJ611]ES_jl}Z.1 ]
'i i
q ANSWER' '6.15 (1.00) C REFERENCE TRHB 10.3
' Point Beach Requal Duestion- 055-02-002 010000G015 ..(KA's)
- ANSWER 6.16 (1.00) b-REFERENCE TRHB 10.12
~ Logics Sheet 8,9 Point Beach Requal Duestion 051-03 026000K404 ..(KA's)
ANSWER- 6.17 (1.50)
'when pressurizer pressure reached 1775 psig, SI auto-unblocked (.5) The cubsequent . pressure reduction to' less than 1735 psig (. 5 ) caused an SI on low pressurizer pressure. (.5)
REFERENCE OP-1A l Pcint Beach Requal Question 053-06 OO6000K405 ..(KA's) ANSWER 6.18 (1.50)
.lotdown temperature would increase. (.5) "non-regenerative heat exchanger lotdown outlet temperature high" alarm would actuate (.5) (exact title of clarm NOT required). TCV-145 would divert letdown flow around the damineralizers.(.5) 1-(***** CATEGDRY 6 CONTINUED DN NEXT PAGE *****)
i l
e F. .
.Pcga.96-16', PLANT-SYSTEMS (30%) AND PLANT-WIDE GENERIC- .
RESPONSIB161IJEp_J1Tfl. REFERENCE. TRHB.10.6 Figure 10.6.2-
- Point Beach-Requal Question 051-02 OO4010A401- ..(KA's)
~ ANSWER- 6.19 (2.00) t ' With the f aulty f use f or a bank D rod, this rod will drop when rod motion I
i n demanded : '(. 5) . - The NI mismatch will result in a (20%) turbine runback. (.5) With PC-486C failed, steam dumps will-not arm or open-(.5). With no cuto rod control'or steam dumps, Tavg will' rise to the 'OT delta T setpoint,
.ccusing the reactor to trip.(.5)
LNOTEs. other causes and scenarios will be evaluated on a case basis. REFERENCE TRHB213.9, pg.4 TRHB.13.1, pg 9 TRHB-13.3,-pp 4-5 OO1000K105 015000K103 OO1000A203 OO1000A101 ..(KA's) ANSWER 6.20 (1.00) c
. REFERENCE
.TRHB 10.9,-pp B - 9
'TGch Spec 15.3.3.C OOBOOOGOO5 ..(KA's)
ANSWER 6.21 (1.00) d REFERENCE TRHB 10.6, pp 30 - 32 OO4000K106 ..(KA's) I (***** CATEGORY 6 CONTINUED ON NEXT PAGE *****)
,7 . y .
F 61 PLANT SYSTEMg_J3p3)iANp, PLANT-WJpg_ggNgRJC Pcgs'97-4 REEPgNg}g}L))]Eg_JJ331 , 91 , LANSWER 6.22, . ( 1 '. 50 ) . i l
- 1. 3 a
- 2. 'b' 3.. b, (.5 ea)
REFERENCE ,
' Tcchnical Specification 15.3.14 194001K1.6 ..(KA's)
ANSWER 6.23- (1.00) f REFERENCE-Tcchnical Specification 15.1.f 194001A113' ..(KA*s)
-ANSWER- 6.24 (2.00) a, d -(1.0 ea)
I
~ REFERENCE EDP Format and Usage Text pp 12,13 10CFR50.54x 10CFR50.73 194001A116 .. . f K A ' s )
ANSWER 6.25 (2.00)-
- c. EHC system, reactor water makeup system, fees system (.5 ea)
(other answers evaluated on a case basis)
- b. as part of the training per, ram, or to load / unload f uel (.5 f or either) 1.
(***** CATEGORY 6 CONTINUED ON NEXT PAGE *****) m_ _ __ _ _ __ _ _ __ _
~ ~ - ~ ~ ~~ ~--- ~~
- 6. iP$ ANT' SYSTEMS ( 307. )' AND PLANT-W1DE GENERIC Page 98 REgPONgip16111E@_11331 REFERENCE-10 CFR 55.4
- 10 CFR 55.13
-194001A103 194001 A10'? 194001A111 ..(KA's)
ANSWER .6.26 (2.00) Candidate #1: Rejected (0.25). will exceed quarterl y admi.n limit (of 1050 mr/qte) (0.25). Cendidate #2: Rejected (0.25). will exceed monthly.atmin limit (of 450-mr/ month) (.25)
.Ccndidate #3: Rejected to.25). will exceed 500 mrem / pregnancy-(0.25).
Ccndidate #4: Rejected (0.25). will exceed quarterly admin limit (of 1050 mr/qtr) (.25) REFERENCE 10 CFR 20.101 HP 1.4, sec 5 194001K103 ..(KA's) c. ANSWER 6.27 (2.00)
'1. a
- 2. c (1.0 ea)
REFERENCE
.10 CFR 20.202 .
194001K103 ..(KA's) ANSWER- 6.28 (1.00) d REFERENCE i i 10 CFR 20.203 194001K103 ..(KA's) (***** CATEGDRY 6 CDNTINUED DN NTXT PAGE *****) ( 1
~ ..
n - w 66 ' PLANT : SYSTEMS (30Z) AND PLANT-WJQg_ggNg81C- Pogr 99 x"-BEEPQNgjBJLillgS;f}331 l ,- : s.
. ANSWER' 6.29- (1150) a y 1. ' c '
2 a fi, : ~3:~b (.5'ea) L
REFERENCE:
'PBNP 4.13, pp 4 - 5 194001K102- ..(KA's) r.'
ANSWER 6.30- .(1.00)
.c)sthree (licensed) operators (0.5) t.
b) A (l i censed ) SnD:or an SRD limited to f uel handling (either.for 0.5) REFERENCE TGehnical Specification 15.6.2 194001A103 ;.(KA's) 5 (***** END OF CATEGORY 6 *****) (********** END OF EXAMINATION **********) L-___-.---_:_--__-,___-_____-_____
TEST CROSS REFERENCE Page. I j; 99E9II99 _YO69E DESEggyCE_ ; l'
' 4.' 01 2.00 9000813 4.02 1.00 9000816 4.03 1.00 9000819 4.04 1.00 9000820 1 4.05 1.00 9000821 l 4.06 1.00 9000826 {
4.07 1.50 9000802 l 4.00 1.00 9000010 4.09 1.00 9000011 4.10 2.00 9000812 4.11 1.50 9000814 4.12 1.00 9000B18 4.13 1.00 9000825 4.14 1.00 9000803
- j 4.15 1.00 9000807 .
4.16 1.00 9000808 4.17 1.00 9000809 i 4.18 1.00 9000815 4.19 1 00 9000822. 4.20 1.00 9000827 4.21 1.00 9000828 24.00 5.01 3.00 9000817 5.02 1.00 9000832 5.03 2.00 9000833 5.04 1.50 9000837 5.05 1. CO 9000838 5.06 1.00 9000845 5.07 1.00 9000847 5.08 1.00 9000849 5.09 1.50 9000853 5.10 1.50 9000854 5.11 1.00 9000855 C 12 1.00 9000856 [ 5.13 1.50 9000857
- 5.14 1.00 9000858 5.15 2.00 9000859
. 5.16 1.50 9000861 l 5.17 1.50 9000862 5.18 1.00 9000863 5.19 2.50 9000864 3- 5.20 1.50 9000866 5.21 3.00 9000869 i
.5.22 1.00 9000870 5.23 2.00 9000871 35.00 6.01 3.00 9000829 6.02 1.00 9000830 6.03 1.00 9000831 6.04 2.00 9000834 f 6.05 1.00 9000835 i
TEST CROSS REFERENCE- Pcgo 2
' 99EEI.199 ._29k9g Rgggggggg, 6 06 1.50 9000836 6.07 1.50 9000839 6.08 1.00 9000840 6.09 1.00 9000841 6.10 2.00 9000842 6.11 2.00 9000843 6.12: 1.50 9000544 6.13 2.00 9000846 .
6.?A 1.50 9000848 6.15 1.00 9000850 6.16 1.00 9000851 6.17 1.50 9000852 6.18- 1.50 9000860 . 6.19 2.00 9000865
- 6,20 1.00 9000867 +
6.21 1.00 9000868 6.22 1.50 9000798 6.23 1.00 9000799 6.24 2.00 9000000 6.25 2.00 9000801 6.26 2.00 9000804 6.27 2.00 9000805 6.28 1.00 9000806 6.29 2.50 9000823 6.30 1.00 9000824 45.00 104.00 l i 1 1 _j}}