ML18009A820
| ML18009A820 | |
| Person / Time | |
|---|---|
| Site: | Harris  |
| Issue date: | 01/24/1991 |
| From: | Richey R CAROLINA POWER & LIGHT CO. |
| To: | Casto C NRC OFFICE OF INSPECTION & ENFORCEMENT (IE REGION II) |
| Shared Package | |
| ML18009A818 | List: |
| References | |
| 91NTS061, 91NTS61, NUDOCS 9103050100 | |
| Download: ML18009A820 (40) | |
Text
ENCLOSURE 3
IIL Carolina Power & Light Company Shearon Harris Energy & Environmental Center Route t ~ Box 327 New Hill,North Carolina 27562 January 24, 1991 File: NTS-3501 Serial:
91NTS061
,Mr. Chuck Casto US NRC-Region II 101 Marietta Street, N.W.
Atlanta, Georgia 30323
Subject:
SRO NRC Written Exam Comments
Dear Mr. Casto:
On January 22, 1991, Shearon Harris Nuclear Power Plant received NRC SRO examinations.
The examination comments are submitted by CP&L. Copies of reference material are included where indicated.
Should you need any-explanations or additional reference material, please do not hesitate to contact the SHNPP Manager - Training, Mr. A. W. Powell, at (919) 362-3219.
R, B. Richey Vice President - Harris Nuclear Project ATB/ljw Enclosures c; Mr. J. E, Tedrow (NRC-SHNPP)
@103050100 5000400
@10220 DR gDOCV' pDR
- JANUARY22, 1991, NRC EXAM R
EXAM ENERAL MMEN 1.
We would like to take this opportunity to express our appreciation to NRC Region II for the initial review of the written exam.
We especially appreciate the effort by the NRC representative to open the NRC office on the weekend to continue the review of CPAL's comments.
During the written exam review conducted in the NRC Region II office, numerous discrepancies affecting approximately 70% ofthe exam questions were identified and corrected prior to administration of the exam.
2.
The walk-through and simulator exams were reviewed on site before they were administered to the candidates.
A number of changes recommended to the JPM questions were incorporated in the walk-through portion of the exam.
3.
Written exam comments:
a.
Several questions require memorization that would not normally be expected of an SRO candidate.
These questions, along with comments, can be located on Attachment I. These comments are provided for information only.
b.
Technical comments are on Attachment II. It was our intent to identify and correct technical inaccuracies on the preview. However, due to the extent of changes made and limited time available for review, six questions stillremain technically incorrect.
Page 2 of 10
lt
ATFACHMENYI MEMORIZATIONCOMMENTS
'6 47 56 Question required memorization of steps contained within PLP-500. Allof the titles listed are indeed notified by this procedure; The question asks which one is notified immediately.
This was commented on during exam preview.
Question requires the operator to have the main steam isolation signal logic diagram memorized in detail.
Question requires the operator to have pressure switch setpoints committed to memory.
This was commented on during exam preview.
29, 64, and 75 Questions require the operator to have Tech Spec LCO text as well as applicable action statements committed to memory.
79 This question requires memorization of Tech Spec Interpretation 89-004.
Interpretations are not normally committed to memoiy.
Page 3 of 10
ATTACHMENTII DETAILEDCOMMENTS ON SPECIFIC QUESTIONS Page 4 of 10
TI N4 QUESTION: 040 (1.00)
What operator action is required to regain use of pressurizer heater group "C" following a loss of pressurizer level (level less than 174)
- a. With level less than 124 cycle the heater breaker at MCC.
- b. With level less than 174 cycle the heater control switch.
- c. With level less than 124 cycle the heater control switch.
- d. With level less than 17% cycle the heater breaker at MCC..
ANSWER:
REFERENCE:
PZRLC-LP-3.0, L.O. 1.1.6, page 15 CPPLMMRNT Page 15, Section 2.5.3 of PZRLC-LP-3.0 states that pressurizer level must be greater than 17 percent to allow htr Group "C'o be reenergized manually.
Since all choices state level is less than 17 percent, no answer is correct (see attached).
RE MMENDATI N Delete question because no answer is correct.
Page 5 of 10
IJ
Co PZRLC-LP-3.0 KEY AIDS 2.
Hagan type controller a..
Receives input from l.
Auctioneered high Tavg (program level) 2.
Reference pressurizer level (actual level)
B.
Pressurizer Level Controller Selector 1
CHAN 460/461 g
CHAN 459/460 g
CHAN 459/461 2.
Normally selected to CHAN 459/460 Interlocks A.
Heaters l.
Interlocked with PZR'level (decreasing) a.
De-energize when 1 of 2 selected channels fails below 17%
(may be overridden at local control b.
station)
Backup heaters automatically re-energize when 2 of 2 channels greater than 17% if control switches in ON or AUTO c.
Variable control heaters must be re-energized manually when level re uirements satisfied 2.
Interlocked with PZR level (increasing) a.
Backup heaters energize at
+5%
above.program level Heatup of insurge water
- ensured, 15 of 20
R E
TI N4 QUESTION:
046 (1 ~ 00)
Given the following:
- The plant is operating normally at 504 power.
- All systems are in automatic.
WHICH one of the following correctly describes the effect on the listed parameters resulting from a sudden closure of the "B" loop main steam line isolation valve7
- a. Total steam flow remains the same, loops "A" and "C" Tavg decreases, loop "B" Tavg increases.
- b. Total steam flow and loops "A" and "C" Tavg remain the
- same, loop "B" Tavg increases.
- c. Total steam flow and loops "A" and "C" Tavg increases, loop "B" Tavg decreases.
- d. Total steam flow remains the same, loops "A" and "C" Tavg increases, loop "B" Tavg decreases.
ANSWER'EFERENCE:
T&AA-LP-2.25, pages 19 and 20, LO. 1.1.3
~PL MMENT TAAA-LP-2.25,pages 18-20, calculate the expected response of an MSIVclosure assuming rod control was initiallyin manual.
The. lesson plan states that Loop "A"and "C'avg will decrease vice remain the same.
Since rod control is given in the initial conditions as being in automatic, the absolute value of Tavg willdecrease; however, the trend is still the same.
The actual answer is therefore "a" vice "b."
RE MMENDATI N Change answer key to "a" vice "b."
Page 6 of 10
T6AA-LP-2.25 KEY AIDS 2.
Feed flow increases attempting to recover level 2.
Overpressure reactor trip is initiated a.
Causes a turbine trip 1.
Steam flow drops to nearly zero 2.
hT across SGs is nearly zero b.
Steam pressure increases even more 1.
Follows Tavg after trip 2.
Increases to PORV setpoint (-
1106 psig) c.
PORVs relieve steam at a rate'atching reactor heat input 1.
From decay heat 2.
From 'RCPs d.
Tavg will stabilize around Tsat for 1106 psig (- 5594F) 1.
Pressurizer level follows avg 2.
Stabilizes at the program level for new Tavg 3.
Pressure will eventually return to normal Closure of Hain Steam Isolation Valve A.
Given:
the reactor is operating at 50%
power with turbine control in automatic and rod control in manual when a main steam isolation valve shuts.
What would happen to the-following plant parameters'P 1.
Turbine power 2.
Reactor power 3.
Tav (all loops)
TAA-TP-160 TAA-TP-161 18 of 23
(a T&AA-LP-2.25 KEY AIDS 4.
SG pressures 5.
Core Tavg, dT 6.
TH, Tc.(all loops)
B.
Assumptions l.
No plant trip 2.
No SG safety or relief valve I
actuation 3.
No operator action C.
Parameters
~I~z fA.~
7 Sfpl F~l'.
3.
4 ~
Turbine power Remains at 50%
(EHC in auto.)
Reactor power a;
Remains constant at 50%
b.
Determined by turbine power Tavg of affected loop
<SG of affected -
0%
b.
aT of affected
- 04F c.
Tc goes to TH
~
Tavg of affected
= TH Tavg of unaffected loops a.
50% power removed by 2
SGs b.
Larger primary to secondary hT across the unaffected SG QSG
= UA(Tavg Tstm) avg stm) after 3/2(Tavg Tstm) before avg Tstm 48'F at 100% or 244F at 50%
(Tavg Tstm) after
= 3/2(24 F) 360F c.
Loop hT increases by 3/2 for the unaffected loops.
New loop hT = (1.5)(31.8)
= 47.7'F Tc Tavg H
hT 557 572.9 588.8 31.8 Tstm
= 548.9 At 100%
Tc 557 T
= 588.8 TH = 620+6 hT = 63.6 Tstm
= 540.8 At 50%
19 of 23
T&AA-LP-2.25 KEY AIDS:
2Tav (unaffected)
+ Tav (aff) d Tavg ( core )
Tavg (core) must be constant since power constant, no rod
- motion, no change in boron T
T 2( H+
H 47 7)'72 9oF 2
H 3
5 ~
6.
TH = 588.8'F (did not change)
Tavg
( unaf fected )
= 588 ~ 8 - ~ 5 ( 47 ~ 7 )
564.95oF (or 565oF)
SG pressure of affected SG Tavg
= Tstm av
'tm c.
Pstm'at 588.8'F --1418.7 psia, SG pressure of unaffected SGs avg " Tstm Tstm = 565 F 36 F
529oF 7 ~
8 ~
9.
10.
Pstm at 528'F
= 878 psia TH in all loops remains at 588.8'F T< in affected loop goes to TH TC in unaffected loops TH is 588.8oF Tavg is 564.95'F (or 565'F)
C -
5 Note that if rods were in auto g / 7, g 4/l7IALL)
YES control, they would step in to reduce.
auctioneered high Tavg from 588.8 F
to 572.9'F
~Qi~ig
)p, uo f lv+v4'J8-20 of 23
I t
E M
TI N QUESTION:
059 (1. 00)
Technical Specifications state that 1All shutdown and control bank rods shall be OPERABLE and within + or -.12 steps of their group step counter demand position.'HICH one of the following will cause entry into the associated Technical Specification action statement?
(Assume only one rod in control bank D is effected and all other bank D rods move in response to a 3 degree Tavg Tref mismatch.)
a.
The CRDM liftcoil energizes and the movable gripper de-energizes before the stationary gripper energizes during rod insertion.
b.
The CRDM liftcoil and the movable gripper coil remain de-energized while the stationary gripper functions normally during rod withdrawal.
c.
The CRDM liftcoil and the movable gripper coil remain energized while the stationary gripper functions normally during rod withdrawal.
- d. The CRDM liftcoil and the movable gripper energize at the same time while the stationary gripper functions normally during rod insertion.
ANSWER:
REFERENCE:
RODCS-LP-3.0, pages 16 through 21, LO. 1.1,6 RODCS-LP-3.0, pages 16 through 20, explain the sequence of events necessary to move control rods.
Answer "b" is a correct answer as it will result in an eventual misalignment due to simultaneous deenergization ofboth the movable and stationary gripper. Answer "a" is also correct as it too results in deenergizing both the movable and stationary grippers simultaneously.
RE ME ATI N Accept either answer a, or b as correct.
Page 7 of 10
RODCS-LP-3.0 KEY AIDS b.
Stationary gripper latches engaged 2.
Reactor trip a.
All coil current interrupted byl.
Choice (manual)
RPS 2 ~
3.
Electrical malfunction F.
CRDM operation 1.
Hold a.
Lower coil (stationary gripper coil) energized at lower current b.
c ~
d.
All gripper latches disengaged under spring force Drive rod drops Rods inserted into core 3.
Rod repositioning a.
Withdrawal l.
Initial conditions Hold Stationary gripper coil energized Low current (4.4 amps)
Movable gripper coil deenergized Lift coil CRDM Model 2 ~
deenergized Stationary gripper coil current increases to eight amps lifting rod 1/16" 16 of 54 p
7cl.
RODCS-I P-3.0 KEY AIDS NOTE:
CRDM model does not lift rod this 1/16" with stationary grippers because of design limitations.
3 ~
4
~
Movable gripper coil energized Movable gripper latches engage drive rod groove 1/16th inch below ridge Allows movable grippers to freely engage without trying to simultaneously lift rod NOTE:
CRDM model lifts rod 1/16" as movable grippers engage as evidenced by chipped paint on rod.
(This is not the way the real device functions)l This step occurs almost simultaneously with the stationary gripper coil current increases.
5.
Stationary gripper coil current decreases to zero Weight transferred to movable gripper as stationary coil disengages during current decreases-Rod lowered 1/16th inch by this action NOTE:
CRDM model does not demonstrate this action because of design limitations 17 of 54
RODCS-LP-3.0 KEY AEDS 6.
7 ~
Lift coil energizes by high current (40 amps)
Movable gripper assembly rises 5/8th inch Drive rod raised 5/8th inch Stationary gripper coil reenergized at high current (8.0 amps)
Engages drive rod Raises drive rod 1/16th inch NOTE:
CRDM model does not demonstrate this action because of design limitations Load transfers to stationary grippers No load on movable grippers 8.
Lift coil current profiled (reduced) to low
=current (16 amps) to reduce coil heating Drive rod still held by stationary 9.'rippers Movable gripper deenergizes Latches disengage and swing out freely NOTE:
CRDM model drops 1/16". inch as movable latches disengage.
(This is not the way the real device functions).
18 of 54 p, 7c
~f
RODCS-LP-3.0 KEY AIDS 10.
Lift coil deenergizes-movable gripper assembly lovers 5/8th inch ll.
Stationary gripper coil current reduced to 4.4 amps Rod drops 1/16" NOTE:
CRDM model will not demonstrate the 1/16" inch drop because of design limitations b.
12.
Rod raised 5/8 inch in 780 milliseconds Rod insertion NOTE:
CRDM model exhibits similar design limitations as in the vithdrav'equence 1.
Hold position Stationary gripper coil at low current Movable gripper coil deenergized Lift coil 2 ~
3.
deenergized Stationary gripper coil current increases to 8.0 amps Rod lifted 1/16" Lift coil energized High current (40 amps)
Movable gripper assembly raised 5/8th inch 19 of 54
RODCS-LP-3.0 KEY AIDS 4.
Movable gripper coil energized.
Movable gripper 5.
6.
7 ~
8.
engages Movable gripper latches 1/16th inch below ridge Stationary gripper coil current decreases to zero Weight transferred to movable gripper as stationary coil current decreases and stationary gripper disengages Rod lowered 1/16th inch by this action Lift coil deenergized Movable gripper assembly lowers rod 5/8th inch Stationary gripper coil.
reenergized at high current (8 amps)
Stationary grippers engage drive rod Stationary gripper latches raise drive rod 1/16th inch Weight transferred to stationary gripper Movable gripper coil deenergizes Latches freely disengage 20 of 54
RODCS=LP-3.0 KEY AIDS 3 ~
4 ~
Logic error: zero current to'both stationary and movable gripper coils simultaneously (could cause rod drop)
Multiplexing error:
more than one group in a power cabinet attempts to move at the same time 5.
Loose card b.
Detection of errors 1.
Feedback from sampling resistors compared to demand current 2.
Logic Coincidence Error must be sensed on all rods within group (4
out of 4)
Allows movement of single rods if Lift Coil Disconnect C ~
Switches open Energizes both stationary and movable gripper coils simultaneously 1.
Low current 2.
Locks rods d.
"Rod Control Urgent Failure" annunciator l.
ALB-13 2.
Power cabinet e.
Inhibit signal prevents all ro4 motion l.
All rods in that cabinet "Automatic" mode "Manual" mode 41 of 54
'I
QUESTZON:
070 (l. 00)
WHZCH one of.the following describes the parameters the ERFZS computer uses to derive the RCS subcooling margin during a loss of all AC?
a.
RCS narrow range pressure to calculate T-sat, highest one of the core exit thermocouples or loop wide range T-hot.
b.
RCS narrow range pressure to calculate T-sat, highest of the average core exit thermocouples -in the hottest quadrant or loop wide range T-hot.
c.
RCS wide range pressure to calculate T-sat, highest one of the core exit thermocouples or loop wide range T-hot.
d.
RCS wide range pressure to calculate T-sat, highest of the average core exit thermocouples in. the hottest: quadrant or loop wide range T-hot.
ANSWER:
REFERENCE:
RCTEMP-LP-3.0, page 7, LO. 1.1.11 P L MMENT Although this lesson plan implies "d" as the correct answer, a more recently developed lesson plan (COMP-LP-3.2, Section 2,3) explains the correct way subcooling is determined.
Lesson plan COMP-LP-3.2 was not provided with the exam reference materials.
These are RO level lesson plans and only selected ones are reviewed during the SRO training program.
CP&Lwillupdate Lesson Plan RCTEMP-LP-3.0.
RE MENDATI N Delete the question because no answer is correct.
Page 8 of 10
COMP-LP-3.2 KEY AIDS 2.
When started, previous eight minutes of data plotted from a historical buffer'.
Accessed by using paging keys on the second level display 4.
Contain SPDS status boxes at bottom of each display 2 '
INSTRUMENTATION AND SIGNAL USAGE
)(; f ( p"i /'.Q5 L
A.
Top Level Display 1.
Subcooling a.
Average of five hottest incore thermocouples b.
Pressurizer pressure
> 1700 psig and PT 402, PT 403 when ( 1700 psig B.
Subcriticality 1.
Power Range
>5%
a.
Takes 2 of 4 channels b.
Current value display averages those power ranges that are
>5%
2.
Intermediate Range Startup Rate a.
Takes average of 5 consecutive readings over 10 seconds b.
Compared to next block of 5
reading to calculate SUR c.
Startup rates calculated for both channels then combined in 1 of 2
logic 3.
Source Range Energized a.
Looks at P-6 setpoint b.
1 of 2 logic 4.
Source range startup rate calculated same way as intermediate range COMP-TP-58.0 COMP-TP-46.0 COMP-TP-59.0 6 of 10
RCTEMP-LP-3.0 KEY AIDS C.
Loop Temperature Difference (aT) 1.
Generated from narrow range Tc and Th RTD's by the following calculation RCTEMP-TP-3.0 RCTEMP-TP-S.O zT ='h Tc 2.
Used to generate inputs for control and protection instrumentation a.
Protective functions derived from temperature elements described in 2.3.B.2.a b.
Control functions derived from temperature elements described in 2.3.B.2.b C ~
4T display 1.
Two separate aT's for each loop
- 2.,Indication for both channels in all three loops on MCB 3.
hT provides inputs to RPS and control systems D.
RCS Subcooling 1.
Calculated by one of two 100%
redundant ERFIS computers 2.
ERFIS calculate saturation temperature (TSAT) corresponding to
,the RCS wide range pressure 3.
ERFIS screens all 51 core exits thermocouples and calculates the average core exit temperature in the hottest core quadrant 4.
The core exit temprature is auctioneered high against the RCS loop wide range TH RTD's 5.
The auctioneered TH is compared to 7 of 15
I-r
RCTEMP-LP-3.0 KEY AIDS 6.
the calculated TSAT to derive the subcooling margin Calculations are updated at a minimum of every 30 seconds 2.4 SYSTEM ARRANGEMENT MAJOR COMPONENTS A.
Protection Circuits 1.
Computed Tavg and aT output signals from each loop provide inputs to reactor protection a.
Three loop channels independent of each other b.
Coincidence is two-out-of-three (2/3) 2.
Each loop inputs to four protection circuits (Tavg) a.
Overtemperature aT (OTaT) set'point development b.
Overpower hT (OPaT) --setpoint development c.
Low Tavg (feedwater isolation with reactor trip) d.
Low-Low Tavg (steam dump block P-12) 3.
4T inputs for two protection circuits a.
OTdT provides reactor power input b.
OPdT provides reactor power input RCTEMP-TP-2.0 RCTEMP-TP-3.0 B.
Control Circuits
.RCTEMP-TP-4.0 l.
Operation of Tavg in control circuitry RCTEMP-TP-S.O a.
Three Tavg signals from control Tavg signals are auctioneered 8 of 15
~
~
R XAM QUESTZON:
08 1
( 1 ~ 00)
Below PATH-l, entry point C, the path is designed to initiate recovery from WHZCH one of the following events?
a.
Steam Generator tube rupture.
P
- b. Stuck open PORV and associated block valve.
- c. Faulted Steam Generator.
- d. Loss of coolant flow.
ANSWER:
REFERENCE:
EOP-LP-3.1, page 8, L.O. 1.A.1
~PL MMBNT EOP-LP-3.1, pages 8 and 9, state that this point'of PATH-1 is for both a loss of reactors pry coolant.
EPP-14 also directs a transition to this point. Therefore, both answer "b" and "c" are correct.
R MMENDATI N Accept both answers "b" and "c" as correct.
Page 9 of 10
EOP-LP-3.1 OUTLINE KEYAIDS A2 Ifthere are no intact SGs available, the actions of EPP-015 can be performed out of sequence to enhance plant control Q3 What are recent changes to PATH-1, revision 1A?
A3 1.
Added check forRHR in shutdown cooling mode 2.
Moved step to reset SI to early in PATH-1 I.
3.
Modify FRP-H.1 transition to check SG level C.
PATH-1 (from entry point C)
- 1. Use associated WOG guideline to facilitate discussion a.
Purpose-provides actions to start recovery from loss of reactor or second coolant b;
Reference events (1) Reactor Coolant leak (< 3/8 inches)
(a)
Within capacity of charging (b)
Normal plant shutdown (2) Small break (3/8'inch to 1 inch)
(a)
RCS willdepressurize (b)
Reaches equilibrium pressure when break Qow equals injection Qow (c)
Containment indicates break but pressure probably remains below 3 psig (d)
RCS subcooling increases throughout transient due to cold injection water (e)
RCS repressurization possible if
- 1) Injection flow > > break Qow or
- 2) Heat sink not maintained (reach satu-ration)
(3) Medium break LOCA (1 inch to 1 foot )
(a)
RCS pressure drops rapidly to saturation (b)
Injection Qow < break Qow initially WOG E-1 Page 8 of 17 Rev. 1
I
EOP-LP-3.1 OUTLINE KEYAIDS (c)
(d)
RCS drains until break Qow switches &om liquid to steam
- 1) Draining of SG tubes reduces heat removal
- 2) Core uncovery possible for cold-leg break When break Qow becomes steam
- 1) RCS pressure decreases
- 2) Injection Qow increases
- 3) Heat removal increases (4) Large break LOCA (> 1 foot ) has four stages (a)
(b)
(c)
(d)
Blowdown of RCS to CNMT pressure Re611 when cooling water'lls vessel lower plenum ReQood as downcomer annulus Qlls pro-
. viding a static head for pushing w'ater into the core Long term recirc when ECCS suction is swapped to the CNMT sumps (5) Loss of secondary coolant (a)
-RCS pressure decreases with cooldown (b)
(c)
When faulted SG drys out and AFW is isolated, decay heat begins to restore RCS temperature and pressure Since there is no break Qow, injection increases RCS pressure and level (d) "If injection is not terminated, the PRZ becomes water solid and PORVs lif't c.
Major action categories (1) Monitor plant equipment for optimal mode of operation (a)
RCP seal injection (b)
SG levels EOP-TP-158.01 Page 9 of 17
'7b Rev. 1
EOPHP3 FAULTED STEAM GENERATOR ISOLATION Instructions 6.
Check CST Level - GREATER THAN aors Res onse Not Obtained Switch to alternate AFM water supply.
M8r%8*~&
CAUTION A maximum of 1600 PSID between the RCS and SG enhances SG tube integrity.
7.
Check Faulted SG(s):
a.
Pressure - STABLE a.
WHEN 'pressure has stabilized, THEN dump steam from intact SGs. to stabilize RCS temperature.
b.
Dump steam from intact SGs to stabilize RCS temperature.
GO TO Step 8.
8.
Check Secondary Radiation:
a.
Periodically sample SGs for activity.
b.
Secondary radiation-NORMAL b.
GO TO PATH-2, entry point J.
9.
GO TO PATH-1, Entry Point C.
- END-I)
Ig plCGC(g(.J r0 j'K "'C+p ]p f'pg-l JAP%/ j'<7 C pg/" li.-
Q'j~r.'OP-EPP-014 Rev.
3 FINAL PAGE Page 5 of 5
QUESTION:
090 (1.00)
The reactor has tripped from a 100 day run at 1004 power.
The
<>C<>
Steam Generator PORV is not functioning.
WHICH one of the following describes the plant response to this event shortly after the trip7 (Assume no operator action is taken.)
a.
RCS T-Cold would be at the saturation temperature of SG safeties setpoint and RCS delta-T would be about 2-3 degrees F.
b.
RCS T-Cold would be at the saturation temperature of SG safeties setpoint and RCS delta-T would be about 30 degrees F.
c.
RCS T-Cold would be at the no-load value of T-Avg and RCS delta-T would be about 2-3 degrees F.
d.
RCS T-Cold would be at the no<<load value of T-Avg and RCS delta-T would be about 30 degrees F.
ANSWER:
REFERENCE:
T&AA-LP-2.17, page 12, LO. 1.1.2 CPPLMMHNT The listed reference deals with an analysis of an SGTR event vice a Rx trip without a significant problem. In CP&L's recommended rewrite of the question during the preview visit, a loss of AC power was also included in the question.
This would result. in a loss of all RCPs, both "A"and "B" SG PORVs, and condenser steam dumps. This scenario results in the answer key response of "b" as correct.
However, since no loss of all AC-has been stated, answer "b" is not correct due to the functioning of all RCPs and steam dumps.
RE MEND TI Change the correct answer to "c."
Page 10 of 10
,t
ENCLOSURE 4
NRC Resolution of Facility Comments SRO Examination Question 040 NRC Resolution:
This question was provided by the facility during the pre-exam review.
Reference material (PZRLC-LP-3.0, section 2.5.3c),
states that "variable control heaters must be re-energized manually when level requirements satisfied" and section 2.5.3a states that the interlock that de-energizes the heaters when 1 of 2
selected channels falls below 17 percent may be
'overridden at the local control station.
"Manual" operation may be construed as "local" station operation or "manual" operation at the control boards.
In either case, it is not clear in the question stem or destructors.
The facility's recommendation is accepted.
The question was deleted.
Question 046 NRC Resolution:
Both destructors "a"
and "b"
may be evaluated as correct depending on the instantaneous event time the parameters are evaluated.
Shearon Harris'utomatic rod control system uses auctioneered high Tavg.
At the instant of "B" loop MSIV
- closure, one of the 3 Tavg instruments is providing input to automatic rod control.
At this point "A" and "C" loop Tavg are essentially the same until the automatic rod control dead band is overcome by the increasing "B" loop Tavg and rods commence driving in.
Once rod motion commences, the Tavg in all loops begin to decrease until'he Tavg-Tref and nuclear power inputs to automatic'od control is nulled within the summing amplifier.
Since the evaluation period was not made clear in the stem of the question the facility's recommendation is accepted. 'he answer key was changed from llbll to lla11 Question 059 NRC Resolution:
The facility's comments states that both distractors "a" and I
I'or control rod OPERABILITY.
That is in fact true.
Distractor "a" will cause the rod to "ratchet" inward, while distractor "b" will result in a dropped rod.
The question
Enclosure 4
stem clearly states that the response is to be evaluated for only a three degree Tavg-Tref mismatch. 'hat implies a 'finite period.
Since-the "ratchet" effect is unpredictable and may or may not result in a deviation of greater than 12 steps, the facility recommendation is accepted.
The answer key was changed to accept either "a" or'"b".
/
Question 070 NRC Resolution:
The reference material provided to the examiner for examination development supported the question and its correct response.
During the post exam review the facility presented updated and revised reference material that resulted in making all distractors incorrect.
Therefore, the facility's recommendation is accepted.
The question was deleted.
Question 081 NRC Resolution:
The question and distractors as originally written provided only one correct response.
During the facility's pre-exam review, distractor "c" was changed making both distractors "b" and "c" correct.
The facility recommendation was not accepted.
The answer key remains as is.
,Question 090 NRC Resolution:
The original question stem is quoted as follows:
"The reactor tripped and main condenser vacuum was lost due to a loss of all AC power,.a Steam generator Tube Rupture (SGTR) has also occurred.
WhiCh one of the following describes the plant response to this event?,"
During the pre-exam review the stem content of the question was changed such that it yielded a complete subject change.
No subsequent changes were made to the distractors including failure to define the correct'esponse.
This question stem change deleted all reference to the loss of AC power and the SGTR making the correct response "c".
Therefore, the facility's recommendation is accepted.
The answer key was changed from "b" to "c".
r ENCLOSURE 5
SIMULATOR FACILITY REPORT Facility Licensee:
Carolina Power and Light Company Facility Docket No.:
50-400 Operating Tests Administered on:
January 23-25, 1991 This form is to be used only to report observations.
These observations do not constitute audit or inspection findings
'nd are
- not, without
= further verification and
- review, indicative of non-compliance with 10 CFR 55.45(b).
These observations do not affect NRC certification or approval of the simulation facility other than to provide information which may be used in future evaluations.
No licensee action is required in response to these observations.
During the conduct of the simulator portion of the operating
- tests, no simulator fidelity items were identified.