Text

Exam Rept 50-440/OL-85-01 of Exams Administered During Wks of 850326 & 0420.Exam results:15 Reactor Operator & Senior Reactor Operator Candidates Passed Exam & 11 Failed
	ML20129A283
Person / Time
Site:	Perry
Issue date:	05/30/1985
From:	Cliff W, Dimmock L, Hill D, Lang T, Mcmillen J, Sly G; NRC OFFICE OF INSPECTION & ENFORCEMENT (IE REGION III)
To:	;
Shared Package
ML20129A268	List: ML20129A261; ML20129A283;
References
	50-440-OL-85-01, 50-440-OL-85-1, NUDOCS 8506040519
	Download: ML20129A283 (69)
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U.S. NUCLEAR REGULATORY COMMISSION REGION III ReportNo(s). 50-440/0L-85-01 Docket No(s). 50-440 LicenseNo(s).CPPR-148 Licensee: Cleveland Electric Illuminating Company-Post Office Box 5000 Cleveland, OH 44101 Facility Name: Perry Nuclear Power Station

' Examination Administered At: Perry Nuclear Power Station Examination Conducted: The Weeks of March 26 and April 2, 1985 Examiner (s):

L g-z9*S-0 k, %

VcMillen h/Jd// h 0=:)

L. Dimmock J/.7d/ra-h t}) b%

V. S1y f//d//$

Date '

p. Hill f/70//$

Date '

I S /So}ks ll.C11ff Approved By:

.M 1

ief 6/34/f[

dperating License Section Dite /

Examination Summary Examination administered the weeks of March 26 and May 20, 1985 (Report No(s). 50-440/0L-85-01)

The written examination was administered to eight Reactor Operators and eighteen Senior Reactor Operators.

Four candidates did not take simulator examinations since they passed them previously.

Results: Of the eighteen Senior and eight Reactor Operator examinations given, 15 passed the examination.

Nogo

!!$h40 0

PDR L_

e, REPORT DETAILS 1.

Examiners T. Lang, J. McMillen, L. Dimmock, G. Sly, W. Cliff, D. Hill 2.

Examination Review Meeting The examinations were reviewed by the following personnel with all changes incorporated into the exam:

S.-Garchow, M. Morrow, T. Silakoski, F. Kearny, M. Haskins 3.

Exit Meeting T. Lang and L. Dimmock held a short exit meeting and told Mr. Silakoski who clearly passed the oral and simulator portions of the exam.

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NUCLEAR REGULATORY COMMISSION v

REACTOR OPERATOR LICENSE EXAMINATION FACILITY:

PERRY 1 i

REACTOR TYPE:

BWR-GE6 DATE ADMINISTERED: 85/03/26 EXAMINER:

LANG,T.

APPLICANT:

INSTRUCTIONS TO APPLICANT:

Uso separate paper for the answers.

Write answers on one side only.

Staple question sheet on top of the answer sheets.

Points for each gyestion are indicated in parentheses after the question. The passins grade requires at least 70% in each category and a final grade of at lomst 80%.

Enamination papers will be picked up six (6) hours after the e>: amination star ts.

% OF

. CATEGORY

% OF APPLICANT'S CATEGORY VALUE T01AL SCORE VALUE CATEGORY

!I I

1.

PRINCIPLES OF NUCLEAR POWER PLANT OPERATION, THERMODYNAMICS, HEAT TRANSFER AND FLUID FLOW

'4 50 I I.__. __24 75

________ 2.

PLANT DESIGN INCLUDING SAFETY

_I__

L AND EMERGENCY SYSTEMS

- f0I____I'_4.7'"

________ 3.

INSTRUMENTS AND CONTROLS 25.00

'S 25

_I.I._

________ 4.

PROCEDURES - NORMAL, ABNORMAL, EMERGENCY AND RADIOLOGICAL

/

CONTROL 99.00 100.00 TOTALS FINAL GRADE _________________%

All ucrk done on this examination is my own. I have neither

!Sivon nor received aid.

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m 1 '.. PRINCIPLES OF NUCLEAR POWER PLANT OPERATION, PAGE 2

~~~~ UER 66YUdk5657~UEST TEEU5 FEE ~d D FLUID FL6U

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T QUESTION 1.01 (3.00)

Briefly explain or define the following terms:

a. Thermal Neutron.

(1.0) f b.

Intrinsic Neutron Source.

(1 0)

c. Reactivity (if equation is used in your answer, explain the

%s equation.)

(1.0)

QUESTION 1.02 (2.00)

More control rods must be withdrawn to go critical when the reactor is b[gJustifyyouranswerwithrespecttocontrolrodworth.)

hot than when it is cold. Is the above statement TRUE or FALSE? (Note:

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QUESTION 1 03 (1.00) g Ag

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a. Define ' Condensate Depression'.

(0.5 k b. Is it necessary for plants to operate with

condensate depression'?

Explain your answer.

(0.5)

\\c5 f 5# f ho m mt DUgJTION 1.04 (2.00)

%*O gg gpyl, A centrifugal pump is operating at 3600 RPM with a pump head of 160 FT.

punp speed is then reduced so that pump head is 100 FT. What is the new puap speed? Show all work.

QUESTION 1.05 (3.50)

o. List the three reactivity coefficients in a DWR at power and give approximate values for each.

(1.5) b.

What effect (increase, decrease, or no effect) do each of the three coefficients have on total core reactivity following a saftey/ relief valve failing open? Driefly explain why the dominant coefficient effects reactivity in the manner you indicate.

(2.0)

GUESTION 1.06 (1 00)

D0 fine or explain the following terms

a. Convective heat transfer. 3 J

% b>nt $(

Q ",rg (0.5)

[

b. Critical power Ratio (0.5)

II

'1.

PRINCIPLES OF NUCLEAR POWER PLANT OPERATION, PAGE 3

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t DUESTION 1.07 (1.50)

NPSH at 100% power or at 10% power?

kg(Dscsarecirculationpumphavemore Explain your' answer. Include the sources of NPSH in your answer.

QULSTION 1.08 (3.00)

In your ctor you pr uce and remove Xenon during' normal operation.

a.How is Xenon produced and removed in your core?

(1.0) i s Xenon f ound. % Cuds "W b * ( 1. 0 )

b.Where physically in your core

.,a wb 94 T'#

b c.Does Shutdown Margin Change when Xenon concentration changes?

Explain your answer.

(1.0)

QUESTION 1.09 (3.00)

Th2 following statement relate to rod worth. For each of the followin3 otote whether it is TRUE or FALSE and explain the reason for your answer.

a. Control Rod Worth is greater at the center of'the core than at anywhere else t ik or m :. Sr gd cue \\g y (1.0) b.As moderator temperature increases rod worth increases.

(1.0) tg l

c.As voids increase rod worth increases.

.(1 0)

.DU STION 1 10 (2.00)

[Jf${tetafor b

U235 is 0.0065. Why is Dett ef f ective appro:timately 0.007? (1.0) b.How does the above affect reactor period?

(1 0)

DUESTION 1 11 (3 00)

Rosetor power is being increased on a 50 second period.

a.How long does it take to increase power from 2kw to imw?

(1.0) b.What reactivity is associated with the 50 second period?

(1.0) l)

c.What is the Kef f durir.g the power increase?

(1.0) i l

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2.

PLANT DESIGN INCLUDING SAFETY AND EMERGENCY SYSTEMS PAGE 4

QUESTION 2.01-(2.00)

Following-the automatic start of.a Diesel Generator six (6) conditions cost be satisfied in order for the Diesel Output breaker to close. What cro four (4) of the six (6) conditions?

(0 5 each)

DUESTION 2.02 (2.00)

For each of the RCIC system component failures listed below, state weather or'not RCIC will auto inject into the reactor vessel.

4 -If it will inject, provide one potential adverse effect or consequence.

--If it will not inject, breifly explain why.

Assume no operator actions, and the component is in the failed or oissligned condition at the time RCIC receives an auto initiation signal.

c. The turbine exhaust valve (F068) is misaligned closed, and the RCIC system receives an auto initiation.

(1.0) b.

The minimum' flow valve fails to auto open (stays shut) when the system conditions require it to be open.

(1.0)

GUESTION 2 03 (2 00)

In re3ards to the CRD system:

a.How does the on-line flow control valve respond following a scram?

(1.0) b.Dreifly explain the operational consequences of the scram inlet valve sticking shut on a scram. Consider the following two situations and the effect(s) on a single CRD HCU mechanism.

(1.0 1.At 200 psis. Reactor Pressure. L 2.At 800 psis. Reactor Pressure. 2 Ota Od*I %

MID IM OUESTION 2.04 (2.00) a.What is a Stator Cooling Run-Back? Be specific in your answer include all appropriate set points. teP we, D p p.

15'/o nrS M */,, AL D (1 0)

J y gn b.What signal (s) will initiate a Stator Cooling Run-Back?

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PLANT DESIGN INCLUDING SAFETY AND EMERGENCY SYSTEMS PAGE 5

QUESTIONL 2.05' (3.00)

Fct the following components in the 0ff-Gas system signify if they are BEFORE or AFTER the DESICANT DRYER in the Off-Gas flow path.

O. Water Seperator.

b.0ff-Gas Condenser.

c.After Filter.

d. Cooler ~ Condenser.

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o. Gas Cooler.

f. Moisture Seperator.

QUESTION 2.06 (2.00)

Acccrding to your lesson plan on extraction steam, there are two reasons Gr Purposes of the Positive Assist Non-Return check valve. What are these two purposes?

{ qpw%

1, QUE. TION 2.07 (2.50) g What are five indications you could check to verify Standby Liquid Control (0.5 each)

.q'Anitiation?

QUESTION (3.00) 3 g

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Thore are five instruments ranges used at your plant for level.

o.What are these five ranges and to what are they referenced?

(2.5) b.Why are most of the reference les piping run outside the drywell into the containment?

(0.5)

' QUESTION 2 09 (2.50) a.

Why does the Control Rod Drive Hydraulic (CRDH) system g

supply water to the Reactor Recirculation system?

(0.75) b.

Why does the Nuclear Closed Cooling Water system supply water to the CRDH system? Sun CO - Ao CLR illD 3C'BU (0.75)

Sbm NS Seds S. 6t@

b4

M7.

c.

Why is a continuous bypass flow maintained from the CRDH Sg drive water pump. discharge to the condensate storage tank?

(1.0)

y; b'

26 PLANT DESIGN INCLUDING SAFETY AND EMERGENCY SYSTEMS PAGE 6

' QUESTION 2 101 (1 50) a.

Following WHAT MAJOR ACCIDENT would the Emergency Closed

-),

Cooling (ECC) System be used to supply the Fuel Pool Cooling f/

and Cleanup System heat exchangers?

WHY?

(1.0) k)

b.

Which ECC loop-can be operated from the Remote Shutdown Panel?'

(0 5)

GUESTION 2.11 (2.00)

In regards to the Perry Fire Protection Systemt a.

What two methods are used to maintain the WATER Fire Protec-tion: system static pressure greater than 80 psis?

(1 0)

/ b.

How'do each of following CO2 Fire Protection system valves

,g fail upcn loss of electrical power?

1) Master-Control Valves (0 5)

2) Selector Valves (0.5)

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INSTRUMENTS AND CONTROLS jy,(%,

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DUESTION 3.01 (3.00)

I In regards to the 125VDC system:

/D t'

d c.What are the two methods of charging the batteries, and when would each method be used?

(2.0) b.It is estiniated that it will take 24 hours2.777778e-4 days 0.00667 hours 3.968254e-5 weeks 9.132e-6 months to recharge the batteries following a capacity test. It is sus 3ested that the chargins time can be reduced to 12 hours1.388889e-4 days 0.00333 hours 1.984127e-5 weeks 4.566e-6 months if both chargers are placed in parallel.

Would you permit this operation, explain your answer.

(1 0)

GUESTION 3.02 (2.00)

Th. SRV's have two modes of operation. One of which is the Pneumatic Actuating Mode.

O.What are threq ways in which the pneumstic modp of aperatkon can be c% pad.i MV p M k b hg At%*r d (1.5) initiated?

b. Pneumatic actuating is one of the two methods of SRV actuation. What is the other mode?

(0.5) 7 g

DUESTION 3.03 (4.00)

Roserding the RPS system;

c. Indicate weather the solenoids associated with the following valves are enersi:ed or de-enersi:ed. Assume a SCRAM signal is present.

1. Pilot Scram Valves.

(0.5) 2.Back Up Scram Valves.

(0.5)

3. Scram Discharse Vent and Drain Valves.

(0.5) b.Within the RPS trip system the pilot scram valves solenoids are devided into 4 groups (8 total). What indication is avalable to the operator that power is available and each group of solenoids is enersized? 9puAl 7kgo (0.5) c.What alarms and/or trips are associated with the Scram Dischase water level? Set points required for full credit.

(1.5) d.Specifiely, where is (are) the sensor (s) located for the variable

'W' in the APRM Scram Set Point formula.66W +50.

(0.5)

Flow Et W s. in 3 bec.

qh OUESTION 3.04 (2.00)

Liot the trip functions and tr ip actions with normal set points that are asccciated with the intermediate ran3e monitors.(IRM's). Four required for full credit.

i

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"3.

INSTRUMENTS AND CONTROLS PAGE 8

x DUESTION 3.05~

(4.00)-

1Tha Recirculation system (recirculation pumps, jet pumps, and FCV ) needs prctection from two-conditions which can prodyce cavitation.

s.What are these two conditions? qe4 yos. na48 Sdpoinh,

F/Mr%gLno.fo(20) e 21.6%

b.How is protection against cavitation.4ccomplished if the two conditions in part (a.) were to occur?

(2.0)

'0UESTION 3.06 (1.50)

In regards to the Feedwater Control System:

a.Why is " Level Programins' necessary?

(0.75) b.Is ' Level Prostamins' continuously done through out a power increase

.to fvil power? Explain.

(0.75)

DUESTION 3.07 (2.50)

Answer the followins in regards to the Turbine Generator and Control Syctemt

a. Explain the operation of the Intercept Valves on a Turbine Overspeed.

(1 0)

% b.What is the purpose of the Man. Combined Flow Limiter?

(1 0) c.The following speeds can be selected for the turbine, 100,800,1500, and 1800 RPM. In which speed (s) is the wobbulator circuit in effect?

riot p teco stm t> pDs. > /voo W 4 MO i> 1500 %.

(0'5)

DUESTION 3.08 (2 00)

Yev over hear an operator candidate te111n3 a second operator candidate of his superior performance in the NRC simulator' exam.

'They save me e loss of seal water to the cire. pumps and then failed the automatic cire. pump trip on me. It was easy to tell because I had to manually trip the cite. pump. Also, latter on I re-started if even with the low flow seal water alarm up, and you shouldn't be able to do that!'

new A Tp tm w.kr ha insteeD o( h M.

What if anythins did you find he did correctly or incorrectly? Two cnswers required for full credit. False assumptions he made will count as answers as well as correct or incorrect actions.

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3.

PAGE

.9

__'__' INSTRUMENTS AND CONTROLS'

-DUESTION 3.09

. (2.00)

\\

If the'following alarm were to annunciate xxxxxxxxxxxxxxxxxxx x

ISO' PH ASE E:US x

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x HYOROGEN x

x TROUBLE x

-l_

xxxxxxxxxxxxxxxxxxx a.What'would be your immediate-actions if it were not an instrument

- failure?

(1.5) b.What would be the cause of the alarm annunciating? Assume-that it was not an instrument failure.

(0.5)

-00ESTION-3.10 (1.50)

_. What.are the three temperature interlocks associated with the following e ola t ni?

uxxxxxuxxxuxxxxxxxx Q

'x Recire Pun.p B x

x Temp x

x Interlock x

xxxxxxxxxxxxxxxxxxx (1.5)

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' PROCEDURES - NORMAL, ABNORMAL, EMERGENCY AND PAGE 10 REDf6L66I6dL 66UTR6L

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QUESTION 4.01 (1.50)

In regards to the Fire Protection System:

a. What is the difference between a deluge and a wet-pipe system? (0.75)

b. How does a preaction system function?

(0.75)

V QUESTION 4.02 (2.00)

While withdrawins control rods for startup you inadvertantly acheive a custained short period. What action would you take if the sustained period wat e:

S

'a.

20 seconds.-

(1.0) c

((

b.

4 seconds (1.0) s QUESTION 4.03 (2.50)

'In regards to a Loss Of Feedwater Heating:

a.The immediate action for a Loss of Feedwater Heating is to:

' Reduce recireviation flow such that thermal power is at least 20%

.below the level prior to the reduction in feedwater heating.'

If you complete the above action would you e::pect a scram? (Assume you were at full power prior to the lose of feedwater heating) If your answer is yes than what caused the scram and what could you have done to prevent it?

(2.0) b.Why is there a limit on the meSawatt output of the senerator when you lose feedwater heating?

(0.5) b, TION 4.04 (2.00)

In regard to Uncoupled control rods; e.Assumins the Rod Pattern Controller will permit it, how can re-coupling be performed?

(0.5) b.How do you verify that re-coupling was successful?

(1.5)

GUEETION 4.05 (2.00)

Dofine Hot Standby and Hot Shutdown.

(2.0)

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PROCEDURES - NORMAL, ABNORMAL, EMERGENCY AND PAGE 11 RA656E6656AL"66 TR6[~~~~~~~~~~~~~~~~~~~~~~~~

~~~~

QUESTION 4.06 (2.00)

C}$ At 8% Reactor power, and prior to transfering the mode switch to 'Run' 6 tha operator is required to verify six (6) conditions exist. What are D fcur (4) of these conditions?

(N

' QUESTION 4.07 (1.50)

In accordance with the startup procedure a normal reactor shutdown should ccamence if the operator notes that he does not have a ' proper SRM/IRM qyvarlap.'

s a.What is-SRM/IRM overlap?

(0.5) b.What is required in order for it to be ' proper *?

(1.0)

QUESTION 4.08 (2.00)

Du.ing a Reactor Cold Startup the operator is cautioned to use extreme caution whenever the continuous withdrawal mode on rod movement is vorformed. What are the restrictions for use of ' Continuous Withdrawal'?

es GUESTION 4.09 (3.00)

CD If a control room evacuation were required according to ONI-C61 ' Evacuation of the Control Room' the reactor operator must perform six (6) things be

( for he leaves. What are these six (6) things?

DUESTION 4.10 (2.00)

, prior to' securing a safety system once initiated the operator must ensure hiaself that certain conditions are met. What are these conditions?

OUESTION 4.11 (2.00) c)Your reactor is in cold shutdown with all rods full in. Maintenance has f((just'finishedworkingonMSLlow pressure interlocks. They ask you to go into 'Run' to verify correct operation of the interlock. Assuming there is no other work in progress, what Tech. Spee. restrictions apply to the Ocde switch change?

5

'4 PROCEDURES - NORMAL,. ABNORMAL, EMERGENCY AND PAGE 12

~~~~ d656E66E65[~66UiR6E""~~~~~~~~~~~~~~~~~~~~~~

R QUESTION ~4.12 (2.50)

According to your Turbine and/or Generator Trip procedure your inmediate cetions require you to PERFORM one action and verify that five (5) others occur.

a.What is the first action the operator is required to PERFORM?'

(0.5)

(h b.The operator is required to verify five automatic actions occur; one of which is to verify that the Control, Main Stop, and Intermediate g

valves shut. What are the other four (4) actions?

(2.0)

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15 EcunT1ons/ DATA SHEET t

F = F, e / T M = 1/(1-k)

Ici - 3.7 x 10105q N(t) = No e-AT op = - 1 x 10 5 3x[ y

'8*

(L +L ) (erod)2 f s K

(4 avg)

= - 1 x 10 3 LK/: voids n - v/(1 + d) ay K

P = I $ v/(3.7 x 1010) og = - 4.5 x 10 6 LK/~*F K

t= (g p)/1p up = -4.5 x 10 4 5K/% power t = 1/p + (g p)/ Ap K

t= 1/(p-8) 1(c) = Io e-AC 4

v=Vg + xys.

TV2 = in(2)/ A E = xhg + (1 x) by Cp = (CP ase) (K8) (E )

b A

S = xSg + (1-x) Sg Q = 1:Cp At 1 in = 2.54 em L oVI op = f D 2ge 1 gal. - 3.785 liters = 8.33 lb.

f = 64/F.e 1 kg = 2.005 lb p = k(eff) -1 N = OAo/A K(efi) 17.58 vatts = 1 BT;;/ min

{

1 CR1 1-K(eff)2 1 psi = 6.895 Pa 1 psi - 2.036 -

E. (0 OC)

I H O (G 4C)

M CR2 1-K(eff)1 1 psi = 27.68

li =.0071 Q = !!ah li = 2 x 10 5 sec Q = UAAT l

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1, PRINCIPLES OF NUCLEAR POWER PLANT OPERATION, PAGE 13

~~~~ UER566Y 555C5,~555T TE5 5FER"506~EL656 FL6U ~ T ANSWERS '- PERRY 1 -85/03/26-LANG,T. ANSWER 1 01 (3.00)

c. A neutron whose kinitic energy is comparable to the moleevlar kinitic energy of the surrounding medium.

(1.0)

b. Intrinsic neutron sources are sources of neutrons in a reactor which do not originate in the fission process. Some of-these sources provide neutrons that allow the reactor to be started up for the first time.

(1.0) c. Reactivity simply relates the state of the reactor with respect to criticality and can be thought of as a measure of the deviation from criticality. (1.0) REFERENCE GP eral Theory 1 A?4SWER 1 02 (2.00) An important plant parameter which influences rod worth is the moderator temperature. As moderator temperature increases, control rod worth also increases because leakage out of the fuel bundle has increased. This increase results in more interactions with control rods, and more neutrons are lost. This explains why more control rods must be. withdrawn to go critical when the reactor is hot then when it is cold-- the control rods are inserting more negitive-reactivity. REFERENCE General Theory ANSWER 1.03 (1.00) 8. The subcooling of condensate i.e., the condensed liquid existing in the hotwell'usually is about 8 degrees F. below the saturation temperature. (0.5) b. Without ' Condensate Depression

the condensate pumps would cavitate due to the water at the eye of the pump being at saturation temperature.

(0.5) i. l l ?

T ..'1..

PRINCIPLES OF NUCLEAR POWER PLANT OPERATION, PAGE-14

--- isEER55?sssics-REsi isissFEs As5 FEGi5 FE5s ANSWERS - -PERRY 1 -85/03/26-LANG,T. -ANSWER 1.04 (2.00) If Flow is proportional to RPM kf' .Hasd is. proportional to RPM squared. Power is proportonal to RPM cubed. 160.FT. HEAD 100 FT. HEAD = . 3e2 2 (3600 RPH-) (X) 2-2 160 X 100(3600) = 2 '100(3600) 2 _----------- = x 160 __________________ 2 -1100(3600) l-------- =X .=2846 RPM \\160 REFERENCE GGneral Theory ANSWER 1.05 (3.50) 0., 1. moderator temperature coefficient alpha-T=-1 x10 -4 per degree change in temperature. 2. moderator void coefficient alpha V=1 x 10 -3 per'% change in voids. 3. fuel temperature coefficient alpha D =1 x 10 -5 per degree change in' fuel temperature.

b. Alpha T increases core delta K/K.

Alpha V decreases core delta K/K. Dominant effect--- relief valve opening results in decreasing.'eactor pressure which increased voids and decreases moderator density resulting in more neutrons leaking out of the core and reducing power. . REFERENCE G2neral Theory L

=,' 1. ' PRINCIPLES OF NUCLEAR POWER PLANT OPERATION, PAGE 15 ~~~~iE5R566 55fC57~AEST iR5U5fER"506~fL656"fL6U ~ -ANSWERS -- PERRY 1 -85/03/26-LANG,T. ANSWER 1.06-(1.00)

c. Convection is the process of transmitting heat'from a heaJed surface or area to a fluid by circulation or mixing of the fluid (Convection takes place only in fluids)dE noi r-cq,c!

(0.5) b. Critical power ratio is the ratio of critical bundle power / actual bundle power where critical bundle power is the bundle power where the onset of transition boiling occurs. (0.5) REFERENCE General Theory AN'dWER 1.07 (1.50) More NPSH at 100%. Recite pump NPSH at power >20% is primarily dependent.on F.W. Flow Subcooling. There is substantially more F.W. Flow at 100% than st'10%. In addition there is proportionately more cool F.W. Flow than Hot R= turn Flow from separators and dryers at the higher power. Subcooling at 100% is approximately 20 degrees F. REFERENCE Fluid Theeory md 4ER 1.08 (3.00)

a. produced = direct from fission and decay of iodine 135.

aghVp h. (3 removed = decay and from neutron absorption. (1.0) .c 8 b Xenon is located in the gas plenum and in the fuel.- f b Q" '(1.0) c.Yes. SDM is based on a. clean / cold core, meaning Xenon free and cold. If_the core.is not clean then SDM will change. (1.0) REFERENCE ' Introduction to Nuclear Reactor Operations, page 8-5

\\ j 9 ~ 1. PRINCIPLES OF NUCLEAR POWER PLANT OPERATION, PAGE 16 - __________________________________AND FLUID FLOW THERMODYNAMICS, HEAT TRANSFER ANSWERS'-- PERRY 1 -85/03/26-LANG,T. ANSWER 1.09 (3.00)

s. False.-It is possible for s thern.a1 f lu:: peak to e::ist anywhere radially within the core which would increase the worth of the rod within that fiv:: peak even if the rod SpVm<near the edge of the core.

b.True. As nioder a tor temperature increases moderator density decreases. Neutrons tr avel e greater distance in thermally diffusing and are more likely to come in contact-with and be absorbed by a control rod.

e. False. With En incr ease in void content,the rieutr ori slowing down length increases to the point where a neutron nisy not be thernialized where it reaches a control rod. Since control rods are thermal absorbers fewer neutrons are captured by the control rod and hence rod. worth decr eases.

REFERENCE .] ,toduction to Nuclear Ope r ati oris Psse 7-11 s ANSWER 1.10 (2.00) r*Dcts effectivt is gretter thar beta because delayed neutrons are bo r ri at lower erier gy than pr on.pt rieu t r o n s so a.Isrser percentage of the delsyed rieutr ont will b e c o r, c thern.a1 rieut r ons and esusc fission. To account for this 1stger p e r c e rit a g e et the thern.s1 rie u t r o n level than at the fast rieut r on level we use the ef f ective delayed neutr on f rsetiori r ather than delayed neutr ori f r setion in our calculations. 1.The dif f er erier pr oduces r l oriser per iod f or

s. g i v e ri Eddition of rcactivity which cari be seers in the rezetor period equatior T=(E-p)/Lp With beta effective used i res t ea d of beta, the tern. ( E:e f f p ) is lar ser giving a l orige r per iod.

PE TERE NCE 3 r.ti oduct ion to Reettor Operations

'1.- ' PRINCIPLES'0F NUCLEAR POWER PLANT OPERATION, PAGE 17- '--- isEEs55isisi55-sEii iEissFEE As5 FEUi5 FE5s ANSWERS -- PERRY 1 -85/03/26-LANG,T. ' ANSWER 1.11 (3.00) t o.P=Poe---- 2 t= Tin (P/Po)- k (7/g g1 3 o g g (1.0) M* = 501 n ( EG44-) - % 3o s j =?45eece 2 A) N { 30 3T ((, g) ~- 9 3/O. 7 h. b.T=(B-p)/Lp =B/(1+LT) =.0075/(1+(.1)(50)) =.0013 (1.0) c.Keff=1/(1 p) =1/(1.0013) =1.0013 (1.0) REFERENCE G2neral Reactor Theory 5 se --t+ w + ,-,e--m,y-.w_,-r-w-ww-m, y

-2.- PLANT DESIGN INCLUDING SAFETY AND EMERGENCY SYSTEMS PAGE 18 ANSWERS ---PERRY 1 -85/03/26-LANG,T. ' ANSWER 2.01 (2.00)

1. Engine speed greater than 425 RPM.

2.Up to rated voltage. 3.Prefered and alternate. breakers open.

4. Low bus volta 3e+

5. Breaker racked in.

6.No bus lockout. (0.5 each) REFERENCE Perry Lesson Plan, Standby Diesel Generator and Auxilary, page 8. ANSWER 2.02. (2.00) a.iWill not inject. (0.25) Steam isolation valve (F045) is interlocked closed if the exhaust valve (F068) is closed ().75).

b. Will inject. (0.25) Possible damage to pump from overheating or at low flows.'

(0.75)'. . REFERENCE Perry Lesson Plan (E51) RCIC ANSWER 2.03 (2.00) s.The flow control will see a high flow and the FCV will close. (1.0) b.1. Rod will not scr am. C-oWI-c tt-t Su4 vit% M k O l k SCfo # (0.5)

2. Rod will scram.(0.25)but scram time wflT"FP longer.(ore 5)

(0.5) REFERENCE Perry Lesson Plan,C11 Control Rods. fO $b(2.00) a t. 'I R 2.04 '? a.A Stator Cpdling Run-Back automaticly reduc generator output to'less than 30,297 amps. within 2 min. or/and 9871 amps. within 3.5 min. (1.0) b.Either 3 pslI'3 Stator Cooling inlet pressure o 203 egrees F. (1.0) Stator et e. AO[/d Y-REFERENCE n43 Stator Cooling Water ~

'2. PLANT DESIGN INCLUDING SAFETY AND EMERGENCY SYSTEMS PAGE 19 ANSWERS -- PERRY 1 -85/03/26-LANG,T. ANSWER 2.05 (3.00)

0. Water Seperator-before.

b.0ff-Gas Condenser-before. c.After Filter-after.

d. Cooler Condenser-before.

o. Gas Cooler-after.

Lf.Hoisture Serperator-before. (0.5 each) REFERENCE N64 Off Gas ANSWER 2.06 (2.00) followin3aturbinetripd(by

1. Prevent overspeedin3 the main turbine lines.)sfromreturningto heater preventing steam trapped in the feedwater g4pl A{ ;/.

(1.0) the turbine through the extraction steam

2. Prevent water from entering the main turbine through the s< traction

team lines in the event the feedwater heaters become flooded.

(1.0) REFERENCE 'd36 Extraction Steam ANSWER 2.07 (2.50) 1.~ Squib continuity lamp of explosive valve F004A (F004B) extinguishe.s indicating that the squibs have received a firing permissive. 2.RWCU system outboard (inboard) isolation valve indicate closed.

3. Pump Starter Energized indicator of the selected pump is illuminated.

4. Tank Shutoff valve F001A (F001B) has opened.

5. Pump discharge pressure increases to approximately 25 psis, above reactor pressure.

6.Storase tank level dropping.

7. Reactor power dropping.

REFERENCE SLC, C41

~~2.-_ PLANT DESIGN INCLUDING SAFETY AND EMERGENCY SYSTEMS PAGE 20 .' ANSWERS -- PERRY 1 -85/03/26-LANG,T. iA CLo k k COM.fo b - ANSWER 2.08 (3.00) IA / Tkp gj %d M M/ [c

o. Wide Range-Inst. Zero Narrow Range-Inst. Zero

@hY Upset Range-Inst. Zero Shutdown Range-Inst. Zero Fuel Zone Range-TAF (0.5 each) b.Most of the Reference les piping is run out of the Drywell into the containment to minimize density changes because of changing drywell ' temperatures.~ (0.5) ' REFERENCE Reactor Vessel,B21 ANSWER 2.09 (2.50) a. To provide seal purse for the recire pumps. (.75) c o o l e r s. (,6f ak** + NM (.75) b. To cool the CRD pump lube oil c. To prevent immediate pump damage in the event the containment outboard isolation valve (F083) is closed. (1.0) (Valve number not required for full credit.) REFERENCE SDN, C11, pg. 28-30 EDH-192 ANSWER 2.10 (1.50) a. LOCA [0.53. Nuclear Closed Cooling is isolated from the Fuel Pool Cooling.and Cleanup system E0.53. (1.0) b.' Loop A (0.5) REFERENCE SDM, P42, pg. 10 and 17 EDH-195 e

e

2i ' PLANT' DESIGN INCLUDING' SAFETY AND EMERGENCY SYSTEMS PAGE 21

ANSWERS -- PERRY-1

.-85/03/26-LANG,T. ~ ANSWER -2.11 (2.00) o. A pressure maintenance tank [0.53 and's jockey pump [0.53 (1.0) b. 1. OPEN (0.5) 2.- CLOSED (0.5) REFERENCE SDM, P54(WTR) ps.2, P54(CO2) ps. 29 I, f e w

.PAGE 22 .____' INSTRUMENTS AND CONTROLS .3, ANSWERS - PERRY 1 -85/03/26-LANG,T. Q$$b grifLk 4 ANS,WER .3.01 (3.00) .o.The two methods of charging are Floa and Equalize. In Float the batteries receive a trikle charg ontinuously during normal operation. b.No. Battery chargers can be operate in parallel operation for only a chort time._ Parallel operation deve ops e::cessive circulating currents between chargers which could damage e rectifiers. (1.0) REFERENCE R42 D.C. Systems,page 19 and 20. I., B kb cf (-o f & 'f ' I g 4g A of we 1 M' ec Ct$ of on ANSWER 3.02 (2.00) V a.1. Manually by positioning control itches in the control room. (0.5) 2. Automaticly, on recept of two independent high reactor pressure (0.5) vessel signals.

3. Automatylg on recept of an ADS signal, b.'Self Actuating Mode. Sa( g nu k (0.5)

RLFERENCE D21/N11/C85, Main Steam, page 7 and 8. ANSWER 3.03 (4.00) c.1.De-energized,2. energized,3.de-energized. (1.5) b.Each group has power available lights on. panel.' (0,5) o' Rod block =24 sal.- T4.Sr.'7, ' Scram =48 gal. LO 5)' P

d. Flow restrictors in each reciculation loop.

M O W e k etp t (E5 ) ~ REFERENCE RPS and Control Rod Drive, C/71 and C/51. i 4 i i l l t

I

3.

. INSTRUMENTS AND CONTROLS PAGE 23 ANSWERS -- PERRY 1 -85/03/26-LANG,T. ANSWER 3.04 (2.00) IRM Downsc' ale-5/125 of scale = rod block. (0.5) IRM High Flux-108/125 of scale = rod block. (0.5) IRM Inop.-low high voltage -module unplugged -mode switch not in operate (0.5) IRM High-High Flux-120/125 of scale-scram (0 5) REFERENCE Nuclear-Instrumentation. ANSWER 3.05 (4.00) 0 1. Low Feedwater Flow (3.43 E6 lbm/hr) combined with FCV position less than 26.'4 % open, (43.4% Recic Flow),and (1.0)

2. Steam Dome / pump suction line temperature differential low (8 degrees delta T).

(1.0) b.1.The low flow interlock monitors the operation of the Feedwater systeci and will initiate at a predetermined low flow setpoint (22% of rated feedwater flow') if the flow. control valve is less than 26.4% open. Will initiate a fast to slow downshift of tne reciculation pumps. (1.0) 2.If the temperature between the Rx steam dome and each respective pumps suction temperature becomes excessively low (8 degrees) the recie. pump will downshift to low speed. (1.0) REFERENCE B33 Rx Reciculation System. ANSWER 3.06 (1.50) -c.At high steam flows a lower actual level in the vessel is required to caintain a constant carryover.- (0.75) b.No. Biasing does not. occur until >45% steam flow. (0.75) REFERENCE .Fecdwater Control,N27.

l 3. INSTRUMENTS AND. CONTROLS PAGE 24 ANSWERS -- PERRY 1 -85/03/26-LANG,T. a I ANSWER 3.07 (2.50) o.For: turbine overspeed conditions from 100 to 105% the control valves will throttle closed. However, due to the large quantity of steam in the Moisture Seperator. and crcssatound piping speed could increase even though the C.V. are_ full closed. The intercept valves will throttle closed from 105 to 107% overspeed-to prevent further steam inlet to the L.P. turbine. (1.0) .b. Limits the sum of turbine and bypass steam to a preset limit such that on a Pressure Regulator failure blowdown would be limited. -(1.0) c.1500 n.d 1200-(0.5) REFERENCE .T.L. and N32/C85 Steam B/P and Pressure Control. c,7c ugojgtr ANSWER 3.08 (2.00) Correctly--Manually tripped the reci_culation pumps. Incorrectly---1. Thought that there was a low seal water trip. 2.Thovsh that you couldn't start up a circulatiori pump with low flow alarm up. / W g r.,g h/ r [r.,3

3. Started circulatio& g wat pump with-dow seal condition present.

ny two for full credit. REFERENCE Recirculation Lession Plan i

"3. INSTRUMENTS AND CONTROLS PAGE 25 ANSWERS -- PERRY 1 -85/03/26-LANG,T. ANSWER 3.09 (2.00) e.Immediate actions a's. listed are:

1. Determine whether the alarm is due to high hydrogen or analyzer

'malfuction. 2.If hydrogen concentration is above 40% of the lower flamability (0.5) limit, perform the following actions:

a. Trip the Generator.

(0.5)

b. Conduct a Fast Reactor Shutdown if necessary.

(0.5) NOTE:0NLY PART TWO OF THE ANSWER IS GRADED BECAUSE THE QUESTION SPECIFIES THAT THE ALARM IS NOT DUE TO AND INSTRUMENT FAILURE. B.The cause of the alarm to annunciate would be the following: 1 25% of the lower flamability limit for hydrosen is reached. (0.5)

2. Combustible Gas Vapor Analyzer inoperable.

NOTE: ONLY PART ONE OF THE ANSWER WAS GRADED BECAUSE THE QUESTION SPECIFIES THAT THE ALARM IS NOT DUE TO AND INSTRUMENT FAILURE. RE'ERENCE OM6,ARI-R13-1 ANSWER 3.10 (1.50)

0. Temperature difference between reactor vessel bottom drain and reactor steam dome temperature is greater than 100 degrees F.

(0.5)

b. Temperature difference between Recirculation Pump

_ction temperature end Reactor steam dome temperature is greater than 70 degrees F. (0.5)

c. Temperature difference between the two loop suction ines is greater than 50 degrees F.

(0.5) REFERENCE ARI-833-35 70 k 7.5

4. PROCEDURES - NORMAL, ABNORMAL, EMERGENCY AND PAGE 26 --- EA5i5t5EiEAt E5sTR6t

~~~---------

ANSWERS -- PERRY 1 -85/03/26-L'ANG,T. ANSWER 4.01 _ (1.50)

c. Wet' pipe sprinkler systems are actuated by melting fuseable links in.the eprinkler head (pipe is filled with water up to the sprinkler head).

Deluge systems are actuated by rate of rise-temperature detectors which cpen an injection valve-water is not up to the sprinkler head, only up 1 to the injection valve. (0.75) b.A preaction system operates in conjuction with a standard deluge system. However, the sprinkler heads are of the type used in the wet pipe system. Actuation of the detectors will open the deluge valve but injection will not occur until the fusable link in the sprinkler head melts. (0.75) REFERENCE Emergency Plan ANSWER 4.02 (2.00) a.If a sustained period of less than 30 seconds is indicated, take prompt cetion to correct this condition by inserting cotrol rods until a reactor period of greater than 30 seconds is indicated._ Contact the Unit-Supervisor prior to re-withdrawing the control rods. (1.0) b.1.If a reactor period less than 5 seconds occurs, take corrective actions per the ARI for window B-1 on P680-6A (which states to insert control rods until a period of greater than 30 seconds is acheived) (0.5)

2. Notify the NRC within one hour.

(0.5) REFERENCE IOI-1,ARI for window B-1,T.S. ANSWER 4.03 (2.50) dM e.There would be a scram ---- the cold water addition which would cause flux levels to increase until you get a High High Flux trip on the APRMs. The action which the operator could take to prevent the scram would be to insert control rods. (2.0) b.To preven undue loading and overstressins of any turbine part. (0.5) REFERENCE. ONI-N36,page 3 and 4 4

' 4. -PROCEDURES - NORMAL, ABNORMAL, EMERGENCY AND PAGE 27 R 656L6656d6~66UTUUL '~~~~~~~~~~~~~~~~~~~~~~~ ~~~~ ANSWERS -- PERRY 1 -85/03/26-LANG,T. ANSWER 4.04 (2.00) 0.If permitted by the Rod Pctten Contr11er, attempt to re-couple the control rod by insertin3 it two notches. Monitor drive water pressure during insertion. (0.5) b.1. Select and fully withdraw the control rod and drive mechanism to be checked, using the continuous out mode.

2. Verify that the rod position display for the selected CRD provides o changins readout in increasin3 increments until

'48' is observed. If flux-level permit observe changes in LPRM readings to verify that the rod follows the drive.

3. Depress the rod movement control ' withdraw' push button, and also the

' Cont. Withdraw

push button and observe that the rod pattern display remains unchanged. Thus verifying that the control rod did not so to the

. overtravel position. (1.5) REFERENCE ONI-C11-2 pase 2 ANSWER 4.05 (2.00) CONDITION MODE SWITCH POSITION RX. COOLANT TEMP. Hot Standby Startup/ Hot Standby Any temperature. Hot Shutdown Shutdown Above 200' des.F. REFERENCE 10I-5 page 1 l 4

G..

PROCEDURES - NORMAL, ABNORMAL, EMERGENCY AND PAGE 28 ~~~~k bEUL55iEAt E5sTR5t ANSWERS -- PERRY 1 -85/03/26-LANG,T. ANSWER 4.06 (2.00)

1. Main steam pressure is greater than 850 psig., and the MSL Isol. Logic Main Steam Line Pressure Low annunciator has cleared.

2. Condenser Vacuum is greater than 26 inches Hg. vac., and the MSL Isol.

Logic Main-Condenser Vacuum Low annunciator has cleared. 3.The APRM Downscale annunciators have cleared. 4.The reactor coolant has been verified to be within the chemistry limits p co.f r Operational Condition 1 per tech. s 5.86.IRM/APRM OEVRLAP per tech. specs. and' all APRMs are indicating between 5% and 12% all IRMs are indicating on scale. aAny four.for full credit. at 0.5 each. REFERENCE IDI lepage 18 ANSWER 4.07 .(1.50) a-tverlap between the SRM/IRM means that-the flux level on the IRhs should sincrease prior to increasing flux higher than the SRM can read. Flux level should be indicated on both the SRMs and the IRMs at the same time. (0.5) b.In order to be proper at least three IRMs in each trip system should be indicating an increase in flux level when the SRMs are reading 4-6E4 with the SRMs fully inserted. (1.0) REFERENCE Startup Procedure ANSWER 4.08 (2.00) It is recommended that from the point of reactor criticality until the first bypass valve is partially opened, the ' Continuous Withdrawal' mode not be used between positions 12 and 24. (2.0) REFERENCE Cold Startup Procedure OM4A IOI-1,PAGE 10 [ v

-*4. PROCEDURES - NORMAL,. ABNORMAL, EMERGENCY AND PAGE 29 ~~~~kkbEbLUbEbkl bbNTkbl ~ ~~~~~~~~~~~~~~~~~~~~~~~~ ANSWERS -- PERRY 1 -85/03/24-LANG,T. ANSWER 4.09 (3.00)

1. Arm and depress div.

1,2,3, and 4 Manual Scram push buttons and verify cil control rods inserted.

2. Place.the Reactor System Mode Switch in the " Shutdown

position.

3. Trip the turbine generator.

4. Verify neutron monitors indicate a decreasing power lever.

5. Verify station loads have autoniaticly transfered to the Startup Trans-former on tripping the turbine generator.

6. Place the div.3 diesel generator " Diesel Control Transfer

switch to local.

(0.5) each, six required for full credit. REFERENCE ONI-C61, Evacuation of the control room. ANSWER 4.10 (2.00) Thr initiation of an emergency system shall not be assumed to be -inadvertant. Prior to securing an emergency system, the operator shall verify, by multiple indications, that a valid initiation signal does not .oxist or that the system operation.is no longer required to maintain core cooling.. 96 l's o r 4 cc,H pocQ er,t T G t< c)(kvM _ MO h y REFERENCE Safety System Operation. ANSWER 4.11 (2.00) The reactor mode switch may be placed in the Run or Startup/ Hot Standby position to test the switch interlock functions provided that the_ control ' rods.are verified to remain fully inserted by a second licensed operator or other technically qualified member of the unit technical staff. REFERENCE Tech. Specs. Table 1.2 ' Operational Conditions' ' ANSWER 4.12 (2.50)

a. Trip the Main Turbine b.1. Generator brks open.

2. Normal Supply brks trip.

3.Startup Supply brks close.

4. Generator Field brks trip.

(0.5) each - - ~

4.

-PROCEDURES' NORMAL, ABNORMAle EMERGENCY AND PAGE 30 RADIOLOGICAL. CONTROL a ANSWERS -- PERRY 1 -85/03/26-LANGeT. i REFERENCE ONI-N32 E 4 e '5P k

se ( U. S. NUCLEAR REGULATORY COHHISSION SENIOR REACTOR OPERATOR LICENSE EXAMINATION FACILITY: Perry Unit 1 REACTOR TYPE: BWR DATE ADMINISTERED: 3/26/85 EXAMINER: T. Lang APPLICANT: ' INSTRUCTIONS TO APPLICANT: Uat separate paper for the answers. Write answers on one side only.. Staple question chsst on. top of the answer sheets. Points for each question'are indicated in parentheses after the question. The passing grade requires at least 70% in each estigory and a final grade of at least 80%. % of Category % Of Applicant's Category Value Total Score Value -Category 26 26 5. Theory of Nuclear Power Plant Operation, Fluids, and Thermodynamics 25 25 6. Plant Systems Design, Control, and Instrumentation 25 25 7. Procedures - Normal, Abnormal, Emergency, and Radiological Control 24 24 8. Administrative Procedures, Conditions, and Limitations 100 100 TOTALS Final Grade All work done on this exam is my own, I have neither given nor received aid. Applicant's Signature

5. Theory of Nuclear Power Operations, Fluids, and Thermodynamics (26.0) 5.1 You increase core power by pulling control rods ajr und the center fuel bundle. Assuming that recirculation.,K'd is kept - constant would the-flow through the center bundle increase, decrease, or stay the same? Explain your answer. (2.0) 5.2 Your latest computer printout of MFLPD and MAPRAT shows the following values for Regions 1 to 3. Region 1 2 3 MFLPD 1.10 0.95 1.05 MAPRAT 1.02 0.92 1.00 a. Which, if any, of these values are bevond their thermal limits? (1.0) b. Why are each of the above limits imposed? (What do they protectaoainst?) (1.5) c. Compared to E0L, would values for MAPRAT at BOL be laraer Whyptw/cs ace ~.THs5 mM@A. is wr AacMM cbutte NC. (1.5) (orsmaller? enamoc. rafme e,ee,amt 5.3 a. List the three (3) reactivity coefficients in a BWR at power and 'give approximate values for. each. (1.5) b. What effect (increase, decrease, or no effect) do each of the three coefficients have on total core reactivity following a safety / relief valve failina open? Briefly explain why the dominant coefficient effects reactivity in the manner you indicate. (2.0) 5.4 What effect does an increase in feedwater flow have on recirculation pump NPSH? Explain why. '(1.0) 5.5 The following statement relates to rod worth. For each of the following state whether it is TRUE or FALSE and explain the reason for your answer. a. Control Rod Worth is areater at the center of the core than at anywhere else. (1.0) b. As moderator temperature increases rod worth increases. -(1.0)

c..As voids increase rod worth increases.

(1.0)

l' 2 5.6 The Perry-1 reactor is taken to criticality from a cold condition and then placed on an 80 second positive period. a. From control room nuclear instrumentation, how can the operator tell when the heating reange has been reached? -(Rod position and recirculation are held constant.) ~ (1.0) Y b. In which of the following intervals was the heatina range entered? Explain the reason for your answer. (Show all work) (1.5) Interval 1 - reactor power increased by a factor of 6 in 143.3 seconds. Interval 2 . reactor power increased by a factor of 3 in 99.0 seconds. -Interval 3 - reactor power increased by a factor of 5 in 128.8 seconds. (N5te: The intervals may 'not be in sequence.) 5.7 With regard to Reactivity Coefficients: Which. reactivity coefficient is the most dominant under the following conditions: (2.0).

1) During rod drop' accident at 15% power

2) MSIV' closure at-100% power

3) Pulling rods at 1% power '

4)' Feedwater contro1Ter~ fails high at 100% power (i.e., water level initially drops)? 5.8 a. Beta for U235 is 0.0065. Why is Beta effective approximately (1.0) 0,.007?' ./ b. How does 'the above affect reactor period? (1.0) 5.9 ' Perry-1 reactor is operating a.t 85% power when one' recirculation pump trips. Indicate how the.followina parameters would initially change (increase or decrease) and briefly explain why the change occurs: a. Reactor power (1.0) b. Reactor water level (1.0) c. Feedwater flow (1.0) .o --'w- - - -~- ,-ve+-- _.,,.,7

3 5.10 In your reactor you produce and remove Xenon'during normal operation. a. How is Xenon produced and removed in your core? (1.0) b. Where physically in your core is Xenon found? (1.0) c.- Does shutdown margin change when' Xenon concentration changes? Explain your answer. (1.0) - End of Catenory 5 - i + e o-- ,-v,

T-6. Plant Systems Design, Control, and Instrumentatf5n (25.0) 6.1-There are five (5) instrument ranges used at your plant for level: a. What are these five (5) ranges and to what are they referenced? - (2.5)' ~ b. Why is the piping for most. reference legs run outside the drywell into the_ containment? (0.5)- 6.2. For the.following components in the Off-Gass System signify if I they are before or after the Desicant Dryer in the Off-Gas flow } path. (3.0) a. Water Separator 2 b. Off-Gas-Condenser

c. 'After Filter id.

Cooler Condenser i e. Gas Cooler f. Moisture' Separator 6.3 With regard to the Standby Diesel Generators: a. List the five (5)' prerequisites'that must exist before an ' auto-start will be initiated..(Do not include the auto-start signals.) (1.5) { b. What is the purpose of the 150 psig interlock in the. starting air system? (0.5)

c. ~With regard to'the Fuel Oil System:

-1) What causes the Fuel -Oil Booster Pump to automatically start? (0.5). i'

2) On a fuel pump overspeed failure, a pressure switch initiates an alarm and affects another Fuel Oil System component. What component ~is affected and in what way?

. (1.0) A o e

2 6.4 With respect to the Automatic Depressurization System (ADS): a. What Levels (1, 2, 3, etc.) provide input to ADS? (1,0) b. Under what normal condition (s) will pushing the four ADS manual initiation pushbuttons not result in ADS initiation? What is the reason for this? (1.0) c. What are the two (2) Seal-In functions contained in each ADS-logic channel? (1.0) 6.5 In regards to the CRD System: a. How does the on-line control valve respond following a scram? (1.0) ~ b. Briefly explain the operational consequences of the scram inlet valve sticking shut on a scram. Consider the followina two situations and the effect(s) on a single CRD HCU mechanism. (1.0) 1. At 200 psig reactor pressure. 2. At 800 psig reactor pressure. 6.6 With regard to the Remote Reactor Shutdown Panel: List the five (5) systems or system modes (portions of a. systems) that are controlled from the Remote Reactor Shutdown (1.5) Panel. b. In what position should the " NORM-EMERG" switches be left when panel is not in operation? (0.5) 6.7 With regard to the Rod Control and Information System (RCIS): The operator accidentally selected a rod out of secuence and a. received a rod block. He depressed the P.03 SELECT CLEAR pushbutton on the Operator Control Module and then attempted to select the correct rod but could not. Why? (1.0)- b. What indication (s) would the operator see when a ganged rod is: 1. more than-one (1) notch out of alignment while driving? (0.5) 2. more than two (2) notches out of alignment while stationary (0.5) or moving? A control rod drive (CRD) is bypassed by the " Drive Bypass" c. function on the Rod Gang Drive System. Under what normal (not scram) condition (s) can the CRD move while in this bypassed (0.5) condition? d. When withdrawing Group 5, at what notch is it banked and what is its notch withdrawal limit? (1.0)

3 6.8 a. Following.KHAT MAJOR ACCIDENT would the Emergency Closed Cooling (ECC) System be used to supply the Fuel Pool Cooling and Cleanup: System heat exchangers? WHY? (1.0) b. Which ECC loop can tue operated from the Remote Shutdown Panel?. (0.5) -6.9

a. 'The reactor recirculation pumps require protection from two (2) conditions.which can produce cavitation.

Explain each condition. Include in your answer any automatic action or interlock associated with each condition. (2.0) b. In'regards to the recirc pumps, what action will occur when an ATWS signal is received? Be specific. (1.5) - End of Category 6 -

7. Procedures - Normal, Abnormal, Emeraency,-and Radiolooical Control (25.0) 7.1 Accordino to procedures for Emeroency RPV Depressurization: a. What is~ the primary system used to cause depressurization? (0.5) b. If the system in (a) is partially or totally unavailable, what action is to be taken? (1.0) 7.2

a. -Give four (4) of the nine (9) automatic actions that you would expect to occur on a loss of service water.

(7.0) b. What are two (2) of the three (3) immediate operator + actions that should be perfonned? (2.0) 7.3 The procedures for the Operational ALARA Prooram describe three (3) simple conceots that station personnel should utilize to minimize their own exposure and that of their fellow workers. What are these concepts? (1.5) 7.4 Accordina to procedures for Gross Fuel Claddino Failure (0NI-J11-1): a. If the "OG POST-TREAT PRCS RAD MON A/B RAD" Hi Hi Hi alarm clears, the operator is to reset the 0FF-GAS POST-TREATMENT trip module. What is the reason for this action? (1.0) b. On e the OFF-GAS POST-TREATMENT trip is reset the positions of the seven valves listed below are to be verified. What should be the new positions of each of these valves? (2.5) ~ 1. Absorber train bypass valve (N64-F045) 2. Bypass line blockvalve (N64-F062) 3. Bypass-line air seal valve (N64-F061) 4. Inlet-absorber train A (N64-F051A) 5. Inlet-absorber train B (N64-F051B Abso~ber train-A D012 A bypass valve 6. r 7. Absorber train-B Do12 B bypass valve 7.5 Accordino to the orocedures for Reactor Shutdown (101 1, Section 4.2, Power Decrease, and 10I-4 Shutdown), at-what oercent power (approximate) should the followino actions (2.0) be performed? a. Transfer normal station electrical loads from the Auxiliary Transformer to the Startup Transformer. b. Secure one of the runnino feedwater pumns and one feedwater booster pump. c. Verify operability of RPCS. d. Transfer recirculation pump to slow speed. e. Unload and shut down main turbine.

i 2 I 7.6 While operatino at 65% power, the only operating pressure regulator fails open, such that the control valves.open (the alternate regulator had earlier failed and was beino worked on by I&C personnel). What immediate action should the operator take to control reactor pressure? (Assume a slow failure) (2.0) 7.7 According to the Personnel Radiation Protection Requirements: a. What is the administrative whole body dose limit (s) for an individual 20 years or older, who has a completed lifetime occupational exposure history record? (0.5) b. Based on 10CFR20, what is the maximum allowable accumulated whole body exposure for a 38 year old person? (0.5) -c. When tapino protective coveralls, when wearing two pairs of coveralls or when a wet suity is worn over one pair of coveralls, only the outer pair normally need tapino. (TRUE or FALSE) (0.5) d. What are the whole body dose limits for life-savina actions? (0.5) e. What are the whole body dose limits when immediate action is required to prevent serious in.iury? (0.5) 7.8 While operatino at 100% power, the olant sutfers a complete loss of instrument air. How will valve operations be affected (open, close, or fail as is) for the following valves: (1.5) a. Feedwater pump recirculation flow control valves b. NCC surge tank make-up valves c. MSIVs d. Temperature control valves for TBCC heat exchangers e. Scram valves f. Fitw control valve A(B) of CRD system g. SJAEs A(B) suction valve on Main Condenser 7.9 ONI-N36 " Loss of Feedwater Heatinn" states that the immediate action would be to reduce recirculation flow. The first subsequent action is to insert CRDs. a. Of the two actions.above which is more likely to prevent a scram? Explain why. (1.0) b. Why is each action oerformed? (1.0) c. If a scram were to occur, what would be the cause? (0.5) N 4 [ ~A f f y

m M. - 3 7.10' According to.I0I-1 four (d) conditions must be verified prior to A _ placing the mode switches in "Run." What are these four (4) conditions?- (2.0) .End of Category 7 - A w t i e t ---w-+r---t +a w -w-s--v -,e -.w ww r,,e w - ~ww w w ---em.,m_-y. w,-w., .e ,,.,,e w. m.--,e,--m,,,, --,a-+,n.,,s,m-, -s en-,.,. - -, ,m-r,,-,,w,.,,--

1 8. Administrative Procedures, Conditions,-and Limitations (24.0) 8.1 During a unit startup, you discover that a control rod is immovable. What action must be taken if it failed to move because of excessive friction? Be specific. (2.0) 8.2 During plant' cold shutdown, the maintenance supervisor informs you that on routine checking he found the Division 1125 volt battery discharged, the reason unknown: Do Tech Specs. require action (Yes or No)? Explain. (2.0) 8.3 Which of the following occurrences require I hour reports to the NRC: (2.0) a. Inadvertent HPCS' initiation b. Reactor heatup rate of 150'F in a one-hour period c. Site boundary dose > 50 MR/hr whole body

d.. Stuck open main steam relief valve 8.4 With regard to the Fire Brigade:

a. What is the minimum number of personnel required? (0.5) b. Who are specifically excluded from the Fire Brigade? (1.0) 8.5 According to Tech. Specs., Secondary Containment Integrity: a. What parameter (s) is monitored to demonstrate secondary containment integrity and what should its value(s) be? (1.0) b. What five (5) conditions must be met for Secondary { Containment to exist? (2.5) 8.6 With regard to Equipment Tagging, Bypasses, Lifted, Leads and Jumpers: a. The presence of a Danger tag on a component allows repositioning of the component but only with permission of-the designated Job Supervisor. (True or False) (n.5) b. Under what condition (s) may the independent verification requirements for return of safety equipment to standby readiness be waived? (1.0) c. What does a pink half-dot on an annunciator window signify? (1.0) d. Who performs the technical evaluation required for all proposed temporary bypasses, lifted leads, and jumpers? (0,5)

2 s 8.7 With regard to certain shift personnel and.their functions: a. There need only be one (1) licensed operator "at the controls" during Operational Condition 4. (TrueorFalse)? (0.5) b. During new fuel handling operations, in the Fuel Handling Building, a licensed Senior Reactor Operator must be supervising. (True or False)? (0.5) c. During what Operational Conditions shall_the Shift Technical Advisor be on shift? (0.5) d. Who approves surveillance test results to determine if they meet Tech. Spec. requirements? (0.5) 8.8.With regard to Accident Monitoring Instrumentation, according to Tech. Specs., under what operational conditions (use number only) are the following-instruments to be operable? (2.0) 1. Primary Containment and Drywell Hydrogen Concentration Analyzer and Monitor s 2. Drywell Air Temperature 3. Offgas Ventilation Exhaust Monitor 4. Safety / Relief Valve Position Indicators 8.9 Prior to securing a safety system once initiated the operator must ensure himself that certain conditions are met. What are these conditions? (2.0)

~ 3 8.10 -With regard to the Emergency Plannina Instruction: a. In which of the four (4) emergency classes would you place the following: (2.5) 1. ransport of a contaminated individual to an offsite hospital 2. Tornado striking facility 3. Control room evacuation with shutdown controlled at Remote Shutdown Panel 4. A 7_ gpm' increase in unidentified leakage within the last 4 hrs ~ 15. An ATWS b. Concerning the_0perations Support Center (OSC) 1. Who is in charge of the OSC? (0.5) 2._ Where 'is it located? (0.5) 3. Under what condition (s) is it activated? (0.5) - End of Exam - t

15 EQUATIONS / DATA SHEET t P = Pa. e / t M = 1/(1-k) 5q N(c) = No e-AT 1Ci = 3.7 x 1010 an = - l' x 10 5 ag/ y a = (1.g+L ) (cred)2 r K ($ avg) = - 1 x 10 3 LK/* voids n - v/(1 + d) ay K P = I e v/(3.7 x 1010) an = - 4.5 ' x 10 . AK/ *F K t= (8 p)/Ap ap. = -4~.5 x 10 4 5K/% power t = T/p + (s-p)/ Ap K k-T= 1/(p-S) 1(c) = Io e-At - Tl/2 - in(2)/ A E = xhg + (1 x) hg Cp = (CP ase) (Ks) (Kg) b S = xSg + (1-x) Sg Q = 11Cp At 1 in.= 2.54 en Ap=f I oV2 D 2ge- - 1 gal. = 3.785 liters = 8.33 lb. f = 64/Re 1 kg = 2.205 lb p = k(eff) -1 N = 94/A K(eff) 17.58 vatts = 1 BTU / min 1 CR1 1-K(eff)2 1 psi = 6.895 Pa t ~ ' ' E (0 OC) 1 psi = 2.036 - H M CR2 1-K(eff)1 1 psi = 27.68 - E 0 (G.4C) 8 =.0071 Q = !!ah I = 2 x 10 5 ,,e Q = UAAT

16 m Table 1. Saturated Sicam: Temperature Table AMrim ^ ~ lpetific Delune 'l arnairy I atIo0,~ ' ~ ' ~ te.no lie p.4 s et bl Let bt L.: "..t torp f ane so in Lileved hap Vapor bwd E.40 V etw tu;m o (v.'s bcar 74*r i O u, s 's 's ht " 's h le b e e 1r 8 32 8 ' 0 0PnS9 60:50?2 3 ml 7 3104 7 - 0 0171 10t% 1075 5 O Woo 2 1473 7 :481 12 8 ' Ht 0 0 % 01 00M?l 30%I 9 3061 9 1 956 lots a 1074 4 0 0041 7 1761 7 :s J 3a I 34 0 0 10395 0 0160?O 2019 0 7419 0 8*0 10712 10772 0 0061 2 16%i 2 1711 38 1 38 8 0 !!24 9 00:6019 7el41 2634 7 6 0i t 1072 1 107:1 0 0122 2 154l 2 165) 38 8 48 9 O l?t41 00:F.r 9 724s 8 2aM 8 s 0'1 19r10 W10 0 0152 7143? 7Mu als 47 8 c il;43 e d Mn. 9 ??tJ a ??/14 IC 63 lr44 8 Ifl% 9 0 0?02 f ilM ? t'77 ar t se l 0 1s197 n 0 3419

II? 8 2182 1

!?.41 1064 7 loni t O e? !? 2 1217 f la*9 sa l 44 8 0 15)i4 OA6TO IF%F 1%%F la na F In6 7 6 tral 6 89?i? 2 ll!I 21191 al t all 0 16514 00W!! 18J0 0 1830 0 15 054 1066 4 1C815 0 03?1 2 10 % 24327 48 8 58 1 0 17796 C ONC?) 1764 8 1704 I It M4 10 6 ) last a 00'%I 2 090: 7 178? Se t 12 8 O th6) 0 0:f 0's Ma* ? I'41 2 !0 0$ F 15.42 1 *= 17 00 10 2 0'44 2 1197 li t $4 8 0 20625 00M26 1487 4 18474 27 0's 10634 196 1 0 021's 7 0'n ?ine la s 16 I 0 17183 0 016014 1383 6 0436 24 M9 I(151 9 4cn6 0 0 04;8 2 OS1J 21070 sie le t 0 23843 00160Ji I?*2 ; 1792 2 24 060 1060 0 1084 9 0 0516 7 0a91 2 i018 58 1 88 8 0 25611 001833 17076 I?076 2n rm0 toS9 7 Ins: 7 0 can 7 S'9i ?CN6 68 8 82 0 0 27494 0 0!6014 II?9 7 11?9 ? J0M9 IMES less 4 G e%13 2 0791 2 n84% 61 8 to 8 0 29s97 0 016031 10% $ 10 % 5 3' 038 los7 4 141S 4 0612 2 0117 2 r,e;s 6a I 64 8 0 31426 0016n43 949 0 gal l 34 0g in% ) 1090 4 00C0 2 0c9a 70 41 66 8 68 8 0 33889 0 016046 916 5 926 5 % 054 1055 2 10'l 2 0 0'08 1 9996 2 070a 61 8 ft 0 0 36?92 00;&M0 I'83 868 4 18 es? losa O IM71 0 0785 1 **cr3 2 ess'r re a F2 8 0 35144 0 0160s4 8i43 814 3 as ia t

nt?9 10950 O nts) 8 9sia
rg 11 8 74 0 0 aIMO O 0ln059 7Aa i 164 I a ' 1a6 lesl 8 101: 8 0 ca;i i treg irg?p pa a 74 1 0 4aa?0 0 016M3 FI P s 117 4 44 043 1051 1 10% F 0 C4 194is 1 04 :7 re 8 ill 0 47s61 0 016067 6738 673 9 a6 040 10a9 5 lon 6 0 08 B 195IO 2 0s:S re s 83 8 05483 0 016017 633 3 653 3 es ij 7 loss a 1044 4 0 0832 1 94 's 7 m*1 81 8 81 1 O S4093 003378 Stil ti) $

50 613 134 ' ) 1%F3 00%9 89114 26")1 32 8 88 8 0 $7 707 0 016041 MO) %03 51 t79 Its! Ibt? 0A 1 9?47 7 0748 64 8 86 8 0 6;$10 0 016947 ??F S 577 5 S4 P?4 1i450 1994 3 0 1083 19!S1 7031 It s 88 8 O MMI 0 01609) 4%8 4%8 M 012 10439 1099 9 9 8079 1 9060 2 0:31 88 8 94 9 0 69113 0040eg att 1 448 i Ss 0:3 194? F 1100 8 0 1115 1 ti?O 2 0*E' 9: 8 97 1 0 14113 0019 0% del 3 441 3 go a: t 104:6 1101 6 r t ti? l '88; 2 0t.ll til le t 0 7M62 0 056111 816 3 ail) 8 ' 010 leaO S lla? S 0;;st 1 8*12 193 1 94 g 96 8 0 84117 O C16)li 197 8 391 9 6a 0% 10 m 3 Il01 ) O I??8 4 47C4 19'?8 te l l 98 8 0 893 % 0 016823 370 9 370 9 66 003 1038 2 Ilos 2 0!!60 1861F i 9876 H8 198 0 0 94174 0 0:4110 h04 IVI 8 6? Tl1 10371 lid l 0171% I IUfl 1 9:15 198 8 182 0 1 00189 0 0:6137 1111 331 1 61 49% 103%

1105 9 0 1331 l esas 1 97'S Isr 0 164 0 6 06965 O c;6;a4 3:31 313 1 18 992 1038 8 1136 8 OIM6 193$4 i tF?S 16s t 196 8 8 347 0 0:4151

?% B6 2S6 i8 13 91 1033 6 1107 6 01:07 I s??) 1967S IIt s Isl e 1 2030 0G16 58 280 18 28330 75 98 1012 S 1108 5 01437 1 8.88 1 9676 -les s lit s 4 7190 0 0161 5 7sS 3r psS j9 7 F eq icil s 1109 3 01472 1 1:05 i9577 li8 s ille I MPS 0 f16171 Mi 3,F 751 38 19 94 10)0 2 1110 7 01907 1 8':71 .9578 lit t 184 0 1 4?99 fle16184 /le y

ja pp gg e7 1079 1 1118 0 O IS41 t illa 9440 1%8 186 0 ISill 0 0 6188 2;S 84 275 05 819' 1071 9 1818 9 01577 1 78 %

.l %38 6 1 88 1833 Its e l'89 1 6009 4 0.61 % 214 70 214 21 85 97 1074 4 1817 7 0 1611 67774 1 78 I i 6977 0 014704 ?03 FS 20376 8 P 97 10?% 11:16 0 ms4 i ts93 1 9311 . Ist 8 172 8 1 7791 0 066213 197 94 19796 81 16 1074 5 18:4 4 O l64J I F613 1 9?13 I?f t j 17a 8 1 8501 0 016??! 18373 18324 91 % 1023 3 1115 3 C I FIS 1 7S33 89747 le4 9 176 8 I 9959 0 064??9 174 08 17409 9396 1022 7 til61 01149 81st) 1 9202 1 88 f 118 0 2 1064 0 046238 16545 1654F 93 % 1021 0 1117 0 01783 1 7374 1 9157 178 8 l 138 0 2 7730 0 0i6247 15737 15733 97 % 1019 8 tilf t 01017 1 7295 191:t 131 8 e 121 8 2 3445 0 016754 889 64 149 66 99 95 1018 7 1818 6 0 1851 3 7217 19054 132 9 IHB 2447 O cia 265 142a0 18141 101 95 10175 1119 5 01184 1 71s0 89re IH s 138 0 2 604F 0 016714 13555 1355F 103 9% loi6 e 18 03 0 14.8 17053 1 8143 134 8 134 0 2 7a 31 0 016784 11909 179 81 105 9% 10lS 2 1111 1 01951 1 6986 1 8937 138 8 d I la8 8 2 8892 0 016713 172 *8 I?)00 10F ei 1014 0 11?? 0 O l'85 18910 l etts la8 3 a tel t 3 Cdil 00:6303 18178 187 ?? 101 94 10:2 9 II?? I 0 70;8 16584 I Iss; isg g 184 0 3 1997 0 0163i2 t il 74 all 76 lil 45 1011 7 1873 6 175 16'S9 l 9910 144 8 184 8 33'S3 0 0163?2 106 58 106 i9 ll3 9) 1011 5 1174 5 C ;'84 1661s 4 8 rs9 las t j 148 8 3 $381 0 016332 10168 iOI 70 !!S 95 1009 3 1875 3 0ill7 1 6610 8 8727 144 8 1 110 0 3 7Isa 0 016143 170% 9707 187 95 1008 2 It?61 02:50 .6S?6 1896 158 0 153 8 3 906% 00:4193 9?66 9760 11* *S 1C07 0 1876 9 0 7181 6463 1 8646 lif e I Ila t 4 t0?$ 00le M 3 P450 8952 I?l15 1005 8 1827 7 S ?til 6390 18606 194 5 194 8 4 3068 0 016374 84 56 8457 123 95 1004 6 1I28 6 n;;al. 6318 105.66 198 9 lis t 4 Sill 0 086384 8001 80 83 175 % B003 4 11:1 a 0 7?81 44145 88%?6 154 0 f 681 4 7414 0 616195 1777 7773 177 % IfIn' ? 1980 7 G ? til 1 6174 18487 188 0 187 0 4 17/2 O niksdb 73'*0 lie? !?9 96 let:5 0 ll'8 0 0. 64 % 16103 10444 187 0 164 8 % ?!?4 0f:5417 10 10 70 1? t il *6 tria 1111 0 0 7317 I f o t2 1 8409 16a t 166 1 5 4423 0 0164?8 6167 AffR 13397 944 6 1831 6 0:409 I S*6l 1 8371 164 0 les t $ 171J 0 016410 6478 f 4 83 13597 917 e 1833 8 0 244l 1 5492 18JJ3 168 1 1700 % e9'6 90t* ni 67 64 6706 13797 494 7 1114 7 07473 1 5477 1 8795 S FI S 171 0 67H6 O elf 48.1 $941 5945 liill w%0 1160 C ho) 151%3 1 8734 177 8 I ra g 6 S656 O ili64 '4 % 1) %4 97 lal 18 991 8 IIM8 O h37 1 % 44 1 8171 174 8 178 0 6 8090 0 0i&JA4 S459 S4 61 14399 99/ 6 ll16 4 07%$ j $&l6 1 8184 176 0 110 0 71640 0 016494 52 35 5136 las tl 991 s 18374 0 2600 15548 18147 ITI S e 0 89 segwg tog 4g g.e egeoggpg gg e e

n 17 on Table 1. Saturated steam: Temperature Table-Continued ~ dre,si ~" 5pecific volume l im...r, t rireni, Al o le*O th per $. t %t Sal 54t 5.el ut fra p Fahr Le la 1.epen d f=4p var <ir l*n 4 f e sp V.ipar I.g 1 f.e V erw I< l y e, v,, og he h og h g 4 g 9 f 180 0 7 9110 0 010%I0 %8 71 %0 72 Itt no

90 7 1114 7 0 7431 154'1 18f:1 181 0 187 4 70's0 O nM/?

4R it? 19181 140 01 win siis fa GA? 1 %1s t t w '% 197 0 184 0 2 711 0 0165 4 s= ?!? an let isfor 9.i0 li i n g o pgu 14 ta I sm es4 8 tH e 8 %69 0 0l&N47 44 sa) 41400 144:1? 9*4 % IHJS 0 ?r?g t g?ig lert nes t 160 0 4 447 0 016%S9 87 611 81 634 1% tt i 98% ) l1411 0 7 F% l i?ll IISS 184 8 190 0 91t3 O n16%F2 40 948 4C 951 l%eos out I: t/ l 02787 191st 1 7384 v13 8 192 8 9 747 O G !6'al) 39 317 39 M4 160 01 Sa? 8 1842 9 0 7t!8 I M92 I NO 997 0 194 0 10 168 00'6%8 37408 37 814 181 0% 9a: 4 11412 076:t 1 $Cli l ine ita s IMO 10 6C5 0 014411 36 344 36 364 164 06 esia 1844 4 0 7479 1 4352 17e31 196 8 IMO 110$8 0 0.6624 34 954 34 970 166 08 919 1 lies t 0 2910 1 8888 1 7?90 198 8 200 0 II S?6 0 016A 37 33677 31639 isa01 97!9 li t4 0 0 7410 l as's 1 77*8 ilt a 758 8 !? S!? 0 0lM64 3t 63) 311SI I!? i n 914 4 11 14 0]Pi 1 4611 l ir?+a 73s s 280 8 19564 0 0lM31 20 8t? 28 0's 114 ta g?;a 11:10 Ol%t l a% 7: t ?A>2 ret t lif t 14 696 00:4719 ?9 787 26 T99 18117 9:03 li'A S 0 1171 1 4447 I Fu 8 Ft7 8 216 0 15 901 0 016147 248 3 2s 894 184 20 Mit 14S2 0 0 3181 1 4323 I ?)CS 216 0 fre t 17184 0 014'15 23 Ill 23 :44 leg ?1 a'5 7 1153 4 0 3?41 l a'il 17447 278 0 224 0 it SM 0016A% 21 429 21 54) 19?27 9' ' 6 Ilta ? 3 33H l aani 18!41

2t l

!!B 8 20 Ct1 - 00:683a 700% 20 073 IS4 J1 94a 0 11 % 3 33339 1 358i 1 r3;o lig e 232 0 21 %F 0 016a*4 18 '01 18 718 200 3% '574 IIS7 8 01417 I JIJ2 I ?:43 232 0 234 8 23 216 OG36835 47454 17418

04 s0 954 0 llS9 2 0 3414 1 3725 1 7201 235 0 288 0 24 M8 0 016976 16 104 16 321 204 8%

9'? I 1160 6 0 !)13 1 36?1 1 ?!a? 289 8 Isa e 26 826 0 015158 IS 74 3 15 ?61 ill 41 111 % iM70 0 Mtt i Jasa i :sas tie: la8 8 29 196 0 016'40 14 ?64 14 /81 214 V Sua lis) 4 0 M41 13)'8 1 ?::4 780 0 212 0 30 88J 0 0170?? 13 3$1 13 375 ??C 62 941 ll(a 7 03702 1 3766 169'1 21:0 236 0 33 091 001705$ 12 $20 12 SJ8 22469 941 8 110 1 0 3163 1 3154 1 6917 234 8 fu t 35 4?? 0 0170A9 ll 7tl 11 76' 774 16 914 4 1147 4 0 1919 1 304) 154'? 261 0 fla t 37014 00lfi?) Il t'% 1904 23'03 ? '% e ilsa 7 0 3e ?% 17933 l ar 4 764 0 2H I 40 500 0 0178S7 10 354 10 115 2h 11 t'il 18 7n 0 0 3117 1 78'1 161n 160 s !!! t 41749 0 01719) e riq g r%g 7:n w

in t 1111 3 O tie r I ppig I s rs?

f r7 s 278 0 46 447 0 017?l8 9 3*? 918J 28108 9/ / % ll?? S 0 3.e41 11%07 l e+5 3 fil a l 283 5 49 130 0 nt ??s4 s 6'r 0ua 24117 e a e, i! Fi 0 0 401s 12'Al I6*91 't10 i tee t $7 414 0 01730 0 1294 t ies) 7'.61 971 7 1814 0 O tist I?rn if'te

ss 0 a

284 8 $$ 19% 0 01714 16* 1t issot 7%; 4 ein a l i te. 7 c a '04 Ilog l a t.4 iss 8 4 217 4 59 350 0 01730 7 ?J01 7247% 261 5 Sis t 16774 0443 1 2186 46441 211 0 2M O 63 084 0 01741 68nt 6 84J3 265 4 91J O lif8 6 04317 12082 16a30 196 0 I a 3eg 67 001 0 0! ?al 64483 6 86'l 259 1 9;C O 1179 7 O s372 I ll'1 16158 2:18 3ss a 11 119 0 01749 6 03SS 61130 ??)4 9e ' 0 1186 9 C as?6 Ile'F 44231 3ii s 3 00 7% 413 0 04153 S F6% S 1823 214 0 9480 lip? ? J a s79 81776 15'% 301 0 lif t 79 933 0 01757 $ 4')66 S a re? 282 1 901 0 !! Al l 0 4)33 114 6 ItN9 312 8 all t 84 688 0 01761 $ 1673 51849 286 1 0979 116a 4 0 4586 1 1576 14162 3ils 370 8 59 4a1 0 01766 4 8951 4 9134 296 4 Sit t lia% 7 04ue l l e'? 1 61:6 !?t 0 324 0 ta l?6 0 7,. !!O 4 6418 46%% 214 6 lit 6 1196 2 0 4*.i? I l 378 l 'C il Ita l 320 $ 100 ;45 0 01ile asus a 4?;5 259 7 688 $ 18872 0 4 f44 1 !?lo 15't 5 3:e a 332 8 105 907 0 0i779 4188 e let6 3c? S pe) 3 Ilss 2 O s ?v8 lin) I tss; 317 s 134 0 188 820 0 04183 3 9681 3 98S9 307 1 842 4 1189 1 0 18'O l itA6 15)H 136 8 348 0 187997 0 017s7 3 7699 1 7978 31: 3 a:g e l130 4 csy? It93 lan? 3:30 348 0 824 430 0 01792 3 583a 36013 315 5 p74 % 4151 0 04M4 1794 161.:n lia l 340 8 131 143 0 01797 340 8 3 4?iA 3 37 877 2 1111 1 CW6 10?a1 ; *,V 9 340 3 352 9 118 138 0 01801 3 7473 3 2603 323 9 666 9 til2 7 0570 10?CS 15 9J 352 0 .j 356 8 185 424 0 01506 30863 3 80a4 320 8 865 5 1193 6 0 5113 1 06:4 1 5:21 3H O t 368 9 153 0:0 0 016: n 293*? 29%?3 3 t? ) t I 1144 4 0 51'l 109F l "I 363 8 3p t l'iO 903 00.9.6 2942 2 0:54 t '6 5 0'46 1195 2 05?!? I f 474 15+ ! ? 362 8 368 0 169153 0 0. t?! 7 t491 26073 34:9 89 1 1169 9 OW3 10H2 I ?'M 360 0 I SS'.a 3it t 177640 0 01026 2 5451 2 % 33 385 0 4'll 1126 7 0$lia inNn 172 0 j all e 186 517 0 0.8J4 24??9 2:46 2 3:13 424 1 lasia 0 S?65 I Glat 465 3 3is t 4 l 300 0 19') '?9 0 61036 ? !!?0 ?)MI 33 3 % Als 4 1896 0 04:4 1037 1 5473 20:3 3M t 20$ 244 0 01047 2 2170 2 21M 3%79 8"4 1.$* 7 0 5466 04%4 14:2 3ta l 348 8 41$ 2?0 0 01847 2 11'4 2 1311 J8 ' ? I'1 ' 1:91 3 0%4 0 9876 IM2 3:30 317 4 ??S S16 0 014%) 2 0!M 2 0169 Ju ) 11 :54 li ti t O'%' 0 9?P6 IS"? 317 0 3MI 2M 493 0 0:054 4 9711 14477 3?06 B.'11 !?tC 4 0 %I7 096% I t313 196 8 9 490 8 23 F ?% 0 0io64 i M id I pr *l 3 Ft 1 859 l't 0 0597 0 4'e ? I t'?4 a:10 448 0 iS8 7?S 0 0.B 'O l iutt f le?? 3'94 8 '/ 0 IAls 0 918 c "t o I t 'i4 a:i 8 4H I 213 4n0 0 06875 16777 l FO%s 3e t e a:42 INi9 05'*6 Mi?9 1 61 % a:8 e 470 282 3)4 0 0iOEI 18152 16640 3844 514 ? I?o? 4 0 $4 4 0 9611 1 5147 4'28 415 0 295 647 0 01847 ! $4e 3 8 % 53 391 1 810 7 6202 8 0 586e 09?il lit:5 4i68 l 7 3*s t es2 17031 fl **l) 0 916,% l 'sn*? .s;s t e3 470 0 )c9 740 09:514 l aele 1 e11,4 03r i ?i s iii mi .o i?m em 47 0 3?? ni Own i iu i +a 0 I?3? T 9tNo O P110 l 'aca a:0 8 470 0 3J6 861 0 0.M B 3S41 137 A? 4"91 .e 41 F*19 I?sa 0 0 6093 08M3 IJ*6 a;t e 432 8 3$100 Oc 963 1 r;64 1 3?i19 438 0 366 03 0 019i9 4 24467 I ?s406 414 6 Fell 1708 1 0 5112 0 8016 1 4978 ers t of. 38: s. O m. s i is,6i i ri... .,,0 ,... I ?,.. e cli 0 779 i 44,0 .3. ess a 397 % 0011) i lot 4 I lka% 4 14 7Ai n !?q 6 0 $?s0 O svtl 1 44*3 444 3 1 488 9 444 01 O nli4a s tr?t; 1 87137 4;80 Pt4 7 I?ot ? 0 E7'9 08%7 l is t g sea s I 812 0 41114 0 S1947 B C%fba 1971ll a *? S ll? t 17tha 8 0608 0 84;i 14;:s eg2 0 434 8 448 73 0 01954 10iill 101412 a370 76 t 8 170e 8 C else 0 0i85 i s tal asa s I e

18 .6 g Table 1. Saturated Steam: TempeFature Table-Continued - ~ ~ ~ An' Prm Spe6# 'c volume f atheio,- tet t0f

femo, t h oce sal be

$ 4s ',4 t v

f. :

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O 638 % 1198 5 0 15?5 06311 3817 117 8 lut 1097 55 0 02894 0 319e6 0 40160 5572 632 0 1184 1 0 7515 0 6227 : 3197 184 8 $60 0 1133 10 0 07737 O M501 0 74714 $6 ? 4 6?) 3 Ill' 7 0 ?4?% 08;32 ! ?'5? 181 8 564 0 1810 13 0 02??l 0 15099 0 3'170 56! 6 688 6 lite t 076's 0 6041 1 37's $6a 8 564 0 1707 12 0 022?% 0 33141 01*9?5 51'9 618 $ llN S 0 17?S 0$M3 I ?' '1 951 8 $72 8 1746 76 0017a9 0 32479 0 34618 $74 ) 604 $ alR? F 0 7775 0 $859 4 36 ?4 172 8 576 8 1789 14 0 0??be 0 31167 0 33424 58J 7 527 2 1180 9 0 78?5 OS'M 1 3517 Sf68 8 Mtl 4376 37 0 07779 0 29917 0 37716 5898 SA19 1879 0 0 78 ?6 0 5873 139C 588 8 Sea t 1.1677 0 07295 0 78M3 0 11048 9 16 914 l176 9 0 7127 80 1 3507 led 8 0 5'ag let t 1410 0 0 0?111 0 77608 0 79919 8401 U47 1874 4 0 7978 0 %4 4 3444 les s 197 9 1853 ) c o/l/R 0 76499 02:427 605 7 sta 8 111? 6 0 4610 0 5 t90 1 34 ?0 197 8 i HS8 1497 8 0 02J45 0 25425 0 21170 6:14 lis t 1810 2 0 A082 0 $79J 13Jrl 184 8 l 688 8 15487 0 0?)*4 0 74 'n4 e ?0747 4171 von 118.7 7 01 14 0 %196 131 e) s27 3 las t IM97 0 47147 0 ?J i'4 15 7 6?? e 5472 11'9 1 0 8lg7 O s:n7 I t;n.s 464 9 O M.795 588 1 16173 0 074n? 0 ??194 07 E? 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I 0 8'la 044 4 l iest 61s t n all t 19410 0 075J9 0 17044 0 89'83 645 9 4764 114? ? 0 0578 O s i64 8 7sia 812 e 636 0 2007 8 007%4 0 16??6 0 18797 6124 est 7 1118 1 0 8628 0 42$1 3 : sit a3s e 644 0 20999 0 07394 015477 0 18071 6191 4*46 Ill! ? 0 8646 04:14 1 74?! 640 0 I ka8 2114 3 0 0767$ 0 14444 0 11769 68% 9 4418 IIMO 08:46 0 4*li 1 7764 644 0 { las t 21784 0 0/657 013876 0 16%)4 6ve s ti l lila c 0 8806 0 ?t? l 1 7'19 640 0 612 8 2719 2 0 02693 0 13174 0 16816 109 0 418 7 1841 1 0 8364 O U67 1 7634 siz e 454 8 2301 7 0 02124 0 12387 0 15115 7974 8057 lil31 0 89)I 0 3637 12%7 til t 864 4 2365 7 0 07768 0 11663 0 14411 714 9 79? 1 11070 0 8993 03M? 1 7499 64t s $44 0 2431 1 0 0:811 013947 013M7 772 9 377 7 1100 6 0 9064 0 31's 1 7875 464 8 Sell 24991 0 02858 0 10??9 013047 7)l S 3621 10* 5 0 9117 0 17i0 1 147 lie s 672 0 25'6 6 0 07911 0 09%is 0 12474 1402 345 7 1085 9 0 9212 03054 I U66 6'20 $18 0 7636 8 0 02970 0 08/99 0 11769 7497 3 05 1017 6 0918' 0 :417 1 2119 ' lit t lef t ??o84 0 03037 0 G808 01 117 7%4 ) 1:01 10a s S 0 * *6S 0 7'20 I ?e86 611 1 684 9 718? l 0 01:14 00049 010161 ff,# 1 ?*i ? 40' 0 144? 0 7%)? I I:ss saa g 648 0 ?%74 0 037 % O n*SM 0 0'1779 77a 0 7%82 10470 0 9535 0 733' i ll?? las t 891 8 2914 ) 0 03363 00V91 0 09110 190 % till 109 6 0 *s 14 0 ?l10 1 1144 497 0 H60 30134 0 03455 0 04916 0 00)71 804 4 2178 10112 0 9189 01641 ! !591 454 8 790 8 3094 3 0 01647 0 038$7 0 0'519 A*7 4 I?? 7 9'52 0 9901 0:490 1 1110 7tt e 192 0 313) $ 0 014?4 001171 0 06917 AMO 1447 979 7 1 0006 O l?ag IBM; 7s28 794 0 3177 2 O ndios 0 07 1? 0 0610n asa i 807 4 9%i 1 0149 0 08 % 1 8046 Fea t i 781 0 3199 3 0 04477 0 0:304 0 05110 8 10 61 4 914 4 1 0179 0 0s77 l os% 710 9n's, 0 781 47* 3204 2 0 05078 0 0nuuo 0 05078 00 ton o l u612 0 0000 1 n617 7814 t*

C7stical tem 0erstm C

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r / 19 Table 2: Satu7aled Steam: Pressure Table C Specific volume intner07

inimp, A0s Press temp Set tat

$at sat S et be AM Pms Lb/$0 la f4hr bquad Iv4D V400r DQuid (v40 Very leQwd feJD VJg2r it 2 19 p i eg v,, o hp h,g h Sg 8,, S 3 g g g 0 90063 31 018 0 014027 .13074 33c? 4 0 mn) 10 m it,7) 5 0 cono 7 i8F? 7 I4 T' 4 HILS all $9 37) O nism; I? ps g g/ 5 5 271A7 10#0 I 1f47 4 0 0' !? 7 04 '% 7 e4 7 O li alt 79 586 0 0!sJ111 641 % 641 % el a?) 10116 tr es.1 no t?% $4:4 in ? F0 8 18 18 108 74 0 0;6114 13151 33340 4913 trul lins a O t t?' 11tb f.'el Il la 16? ?4 0 016401 73 515 73 937 170 10 1G9 9 till t 0 i'49 494 l ass) se 18 8 19123 0 0665S? 18804 Jg a?0 lei tg qs; I liaf 3 0l M V:3

n '9

'8 I 03'1/1 14441 37'4* 14 696 tella 212 00 0 046719 26 787 76 7$4 (2047 970 3 li%0 5 13 8 28303 0 0161'6 26214 76790 lal /l 'i59 7 1150 9 0 1137 1 4445 31%i? 11 8 26 9 l'7 46 C el4814 70 0?0 ?ine7 19677 .ir 01 1:ss t a us g 3*(?

en
8 l

38 3 750 14 o us toa9 13 1 44 1114*4 218 9 9a% 2 1144 1 0 M97 I lit) t*3% 38 8 at 8 2472S 00171$1 10 arts to49% 734 1 9414 1861 8 0 Jill 4 2844 Ia'n a3 e le t 20102 0 017274 84*67 8 5180 7% 2 971 9 18141 0 4612 1 24?4 16%4 $4 8 1 68 8 292 73 0 :187343 78%2 7 8736 26? ? 914 4 ll17 4 0 4??) 12387 I A440 68 8 fo g 30193 0 017492 6 1879 6 10%0 77? 1 161 8 1:40 6 0 44i1 IIPS 1014 Ft 1 04 8 312 04 0 017573 5494 % s711 ?#2 t

o09 11:3 1 O a534

! !675 1 6?04 08 0 se s 3?0 28 00:1659 4 8779 4 8153 793 7 814 6 138% 3 0 a443 11410 14ill He ige l 37782 0 817140 4 4113 e s t:c

as s sat 6 114' ;

e t ti I t?na twr tag e lig a 3}s it 0 01787 4 01 % a 1:44 3m g 041 1 lia4 9 0 44:4 11:t% 19M tit t 17s 0 38377 0 011111 3 7697 J i?FS Ji'4 8'F S 11 7)4 0 4511 10s's J l ' a rt tin g 138 0 34733 0 017'6 3844 3 as44 319 0 sit e I:91 7 0:nq l e#i I gni t 130 s 140 1 - 35304 0 018'3 370 0 3?D0 3/s O P'4 3 1151 0 0 'ri 4 0'a s I YS2 148 8 Ile 8 35883 001N9 2 *** 8 3a09 330 6 84 4 1I94 1 0 M 41 1 0*54 1 * * $5 in I IH I M3 SS 0 0'815 7 81 % 7 83.4 336 I 8%90 till I 0 $?n6 1 0435 1 % 41 ens 3 tra g M8 47 O C.578 2 6656 16134 38i ? AS4 8 tt% 0 04:'4. 10122 1 N; till 184 8 37308 00it? F 2 Si29 2 5312 3 t9 7 4W F 1;*69 05'8 10?iS a tta) les s 190 0 377 53 0 01833 2 38a7 2 40JO 330 9 846 7 lis7 4 0 U44 10li3 1 4444 sta 8 180 0 381 80 0 01819 7 76p1 ? 28'3 3% 5 a4? e llen 1 0 ga se i rf14

s454 2:C 8 till 189 9i 00:2:4 7 i6171 7 18?i7 3% 9 8158 4 t99 0 C M 'O C U?)

85411 t'll 278 0 189 " 00:850 7 06719 2 f%;1 34 7 8K4 I t *' 6 0*S41 O M14 15:7a 1:38 238 0 39370 0 014*S 4 97**1 19m6 3 83 Sit e 1*00 1 4 5984 0 F48 1%H4 218 8 Fa8 8 39F J9 0 c1450 8 9F09 1 91759 l'i 3 578 4 I?o0 6 0 % 34 CMH I 1;'t 2a3 0 718 1 4M 97 00184 1874S2 18J317 376 1 8 50 !?Si l 0 56 ?9 O Mit 15?4 218 1 763 8 404 de 0 0:470

1*544 77418 314 1 8?i 6 110l i O t*??

08Y9 15:!S 2ns t 378 3 40180 00ia?5 6137 7FI) 393 4 8.83 I?ti t 0 $7ea o s433 1 3.31 3rg e 2s48 41107 0 3,an0 ,63t69 6 % 49 3471 a;)I I?c? ) 0 ss'14 O se ) 09101 151'A 288 8 798 0 414 25 0 0188) ,51591 59482 J90 6 8! / 0 110/ 6 0 91'73 1 Sili its 8 3es t 41F l5 0 018e9 I S?lta 15a714 3*40 3"9 n 1967 9 O spp? 09'?3 1 %165 lit t j nog all 3 0 0tti? 810F47 1 37 % 4 a09 8 rw ; t?rd o 0 60%9 C a909 44%4 lit t ste g 444 60 0 01934 1 14161 1 16095 824 7 Jac a lloa 6 0 6187 0 8630 4 4841 assa S 454 8 4% ?8 0 01 % 4 1 011?: t ell'9 sa'? $ 4 11 1st s t ot a nguO cal.s ,47 a e:L e 188 0 46701 0 0i91) 0 9t'8 7 G ert; 4l,4 7 e r :'O 08144 t s..a g gg 558 g 47648 0 019#4 0 8?? q) 0 84117 as0 9 74 t 3 I?O4 ) c ail; tt :1% 1 4*17 to:8 644 8 48620 0 0;063 0 ?J.'l 0 76175 471 7 117 0 I?PJ F O i. *? ) 0 F138 14 WI Ere 658 8 4*a 89 08:7c17 0 68411 0 FG443 dell Fy 'i 9 1;0? s 0 68/4 07%' 1 43s; lif t

80 8

$03 08 0 0/0$0 0 63Sc5 0 6%% 491 6 Flo ? 1201 8 0 69'8 0 7371 1 437,5 7c4 8 F18 8 110 84 00?059 G e,8s a3 0 6t449 SM 9

- - e i:00 7 0 70??

0 7?PJ 14*1? F58 0 038 8 S1821 0 070A? 0 $4L's 0 %e16 $rA 3 sees t r i,4 0 1141 C M*l 141'. ) fna lit t 579 14 0 01:3% 0 51127 O l s '07 %I9 a 671 ) liu e O Mi 0 6 a,9 14'.4 Iss e lef t $119% 0 4?l?) G a M54 O W91 S/57 641 7 1864 r P '1

15 7U iM?

9:s s 958 8 S38 I4 0 07141 08%188 C a t?0S Sit 7 Mai lial 1 ei'4 Det; I - 1 ts4 3 1888 9 $49%4 Un'l% G i?a *4 041%4

41 %

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U14 t'as t tile t

% I87 0 0?714 0 35459 0 19373 56J S 4?? / Ita? O 0 ?447 OWs 1 11'9 l'it a 1280 0 % 719 0 07232 0 34013 0 36145 571 9 61' O 1184 8 0714 0 W64 I ?'91 titt e l 1751 8 5F719 0 07750 0 3?106 0 345 % S 18 8 ani g

1a? 6 07?*0 0 $f 50 l *4 M I?$s t 1388 0 577 a2 0 0'?t9 0 30112 0 1'991 88i6 5114 llO2 0 *s t) 0 5;33 t U1 ruta i

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s # 4 il t? g saw a u;;

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7-l i Answers to Perry-1 SR0 Exam 5. Theory of Nuclear' Power Plant Operations, Fluids and Thermodynamics (26.0) 5.1 As fuel temperature is increased (due to control rod pull) _more voiding is created. Therefore more back pressure, which would mean less flow. However, because of core orificing back pressure is less significant and flow through bundle is almost the same. (2.0) Ref: Thermodynamics 5.2 a. MFLPD Region 3 ; RE6 ion 1. (0.5) MAPRAT Region 1 (0.5) b. MFLPD -~ Maintains 41% cladding strain, fuel failure. (0.75) MAPRAT - Maintains 42200"F followino LOCA, c: cay heat removal. (0.75) WIE 9

c. Pima 11e8(%0.5); MAPLHGR limit is larger at BOL (0.5) since, with exposure, local peaking factor gets smaller as heat transfer is reduced (0.5).

(1.5) Ref: Perry-NPP Thermal Sciences, pg. 10.3. t Moderatortegperaturecoefficientalpha 5.3 a. 1. T = -1 x 10- per degree change in temperature. 2. Moderator void coefficient alpha V =--I x 10-3 perichange in voids. 3. Fuel temperature coefficient alpha D =-1 x 10-5 per degree change in fuel temperature. (1.5) b. Alpha T increases core delta K/K - Alpha V decreases core delta K/K Dominant effect - relief valve opening results in decreasing reactor pressure which increased voids and decreases moderator density resulting in more neutrons leaking out the core and reducing power. (2.0) Ref: General Theory O

Answers Section 5 2 5.4 Increase (0.5); increased subcooling (0.5) (1.0) Ref: Morris, T. C. ; Thermo/HT/ Fluid flow (3/83), pg. 7-96 5.5 a. False. It is possible for a thermal flux peak to exist anywhere radially within the core which would increase 1he worth of the rod within that flux peak even if the rod SpVm) near the edge (1.0) of the core. b. True. As moderator temperature increases moderator density decreases. Neutrons travel a greater distance in thermally diffusing and are more likely to come in contact with and be absorbed by a control rod. (1.0) c. False. With an increase in void content, the neutron slowina down length increases to the point where a neutron may not be thermalized when it reaches a control rod. Since control'rois are thermal absorbers fewer neutrons are captured by the control rod and hence rod worth decreases. (1.0) Ref: Introduction to Nuclear Operations, pg 7-11 5.6

a. - Operator can notice that period has become longer and that power change on IRMs, SRMs is' leveling off and turning around.

(1.0)

b.. (From P = Poe /T -> T =

In huo ) Interval 2 (0.5): t the period has lengthened from 80 seconds. The other intervals have 80 second periods (1.0). (1.5) Ref: General control room indications; Perry, Intro to NR Operations, pg 4.18-4.19. 5.7.(0.5 for each) (2.0)

1) Doppler coefficient g "',1 s k b e y <*( ) k A y.

2) Void coefficient

3) Moderator coefficient -

O '~n g, g.- - u.

4) Void coefficient Ref: Standard Reactor Theory

Answers Section.5 3 5.8

a.. Beta effective is greater than beta because delayed neutrons are born at lower energy than prompt neutrons so a larger percentage of the delayed neutrons will become thermal neutrons and cause fission. To account for this larger percentage at the thermal neutron level than at the fast neutron level we use the effective -delayed neutron fraction rather than delayed neutron fraction in our calculations.

(1.0) b. The difference produces a longer period for a given addition of reactivity which can be seen in the reactor period equation T=(B-p)/Lp. With beta effective used instead of beta, the term (Beff-p) is-larger giving a longer period. (1.0) Ref: Introduction to Reactor Operations 5.9 a.' Decrease (0.34) due to increased void content in the core (0.33) as flow decreases (0.33). (1.0) b. Increase (0.34) due to increased voiding in the core (0.33) and recirc pump no longer taking suction on the annulus (0.33). (1.0) c. Decrease (0.34) due to steam flow decrease (0.33) and level increase (0.33). (1.0) Ref: Standard BWR Transient Analysis 5.10 a. Produced = direct from fission and decay of iodine 135. (1.0) Removed = decay and from neutron absorption. b. Xenon is located in the gas plenum and in the fuel. (1.0) c. Yes. SDM will change as Xenon changes. (1.0) Ref: Introduction to Nuclear Reactor Operations 4 a i

Answers to Perry-1 SR0 Exam 6. Plant Systems Design, Control, and Instrumentation (25.0) 6.1

a. -Wide Range - Inst Zero Narrow Range - Inst Zero I?6 ID {d F veszT "y :t Range - Inst Zero DS

' T (" N Shutdown Range - Inst Zero ros. Level Zone Range - TAF (2.5) b. Most of the reference leg piping is run out of the drywell into the containment to minimize density changes because of changing drywell temperatures. (0.5) Ref: Reactor Vessel B21 6.2 a. Before b. Before c. After d. Before e. After. f. Before .(3.0) Ref: N64 Off-Gas 6.3 a. (0.3 pt. each) (1.5) 1.- Normal /Inop. switch in normal .2. Local / Control Room switch _in_ CONTROL ROOM position 3. Air bank pressure abou1(250Jpsig~ 5**4 84 JS r** 4. Control room 0FF/AUT0/ START switch in AUTO or-M e d 5. Barring device retaining pin installed

6. pc oven.Vo reie

7. m c.w w e <

'e c '< cs f. l b. To allow manual restart after a failure to start (0.5) c.

1) Automatic diesel start signals (0. 5 ).

2) DC fuel oil booster pump (0.5); prevents (1.0) automatic start (0.5); ca m s bi a<.t Dtt e Ref: Perry, System Description Manual, R43, pg. 8; R44, pg Sa; R45, pg. 5, respectively.

I Answers Section 6 2 6.4 a. Level 1 (0.5) and Level 3 (0.5) (1.0) b. If one RHR or the LPCS pump is not operating (0.5); ensures ECCS' capability to deliver water to cool the core before depressurization (0.5) (1.0) g c.[HIdrywellpressure(0.5) ! ADS Logic (0.5) (1.0) M .Ny $1 + Ref: Perry System Description Manual, B21C, pg.13,11, and g,, g Ipcet 10 respectively. 6.5 a. The flow control will see a high flow and the FCV will close. (1.0), b. 1. Rodwillbtscram. (0.5) 2. Rod will scram (0.25) but scram time will be longer (.25) (0.5) Ref: Perry Lesson Plan C11 Control Rods 6.6 a. (0.30each) (1.5)

1. Reactor pressure relief

2. RCIC flow

3. RHR flow

--r maus of RM. LPer; 5,ivfbo~ou 'cu w ev-) %.'. % urou L -

4. ESW loop A g

5. ECC loop A b.

NORM (0.5) Ref: Perry, Systen Description Manual, C61, pg. 2 and 24, respectively. Saerz ~ 6.7 a. Depressing ROD select clear inhibits further rod footion i d ' 14 (1.0) unless re-depressed. .c, t b. 1. "NEXT" on Rod Display moduie Gang Position display, oA ($.5)' Pov-. aw 's msa u suiv " 2. " Misaligned" on Rod Display Module Gang Position display (0.5)- / c. If " gang" mode is selected (0.5) Banked at notch 12 (0,5); 1 notch limit below notch 12M*"& or($ d. m s au n u mar Mus zo % k s : h r w a,,rar M e. s a uc m m n aus2c% rum.) Ref: Perry, Systems Description Manual - C11(RCIS), 7c% pg. 30, 26, 43, and 9 respectively.

Answers Section 6 3 ~'-~~ 6.8 a. LOCA (0.5). Nuclear Closed Cooling is isolated from the Fuel Pool Cooling and Cleanup system (0.5). Powee. Onwa (,6*ow) (1.0) Cost. :sset Mc hPs oc sw :n.siw ou,% wu b. Loop A (0.5) Ref: Perry, Systems Description Manual, P 44, pg. 5, 8, and 6, respectively. 6.9 a. 1. Low Total Feedwater Flow - If during fast speed pump operation feedwater flow is excessively low, water temperature in the downcomer region will increase. This condition could result in cavitation in the flow control valve. This interlock monitors the operation of the F.W. system and will initiate a fast to slow speed transfer if the flow control valve is less than 26.4% open. 2. Steam Dome / Pump Suction Line Temperature Differential. If hav>v g the differential temp. between the[mL B and/or Cland #l p.u each respective pump suction becomes excessively low ~ 8'F," / p see,. cavitation can occur. The recirc pumps are therefore down @r M shifted to slow speed. (2.0) b. High Reactor Vessel pressure and low reactor water level are both ATWS signals. The recirc pumps receive the recirc pump trip sianals from the Redundant Reactivity Control System. Low reactor vessel water level will cause the system to trip the LFMGs input and output breakers, and will trip d p ecirc pump fast speed breakers. High reactor pressure will trip the fast speed breakers and will t initiate an auto shift to slow speed. The slow speed pumps will trip 25 seconds later unless 'the APRMs are downscale. (1.5) --a

Answers to Perry-1 SRO Exam 7. Procedures - Norwal, Abnormal, Emergency, and Radiological Control (25.0) 7.1 a. ADS (0.5) b. Open other SRV's (until 8 are open) (1.0: Ref: Perry PEI-1, pg. 31, 32 respectively 7.2 a. (any'4: 0.5each) (2.0) . Turbine runback will occur '(as ' temp. increases to 81 C); turbine trip could occur if runback is too slow. Service an'd instrument air compressor will trip. RWCU pumps will trip. 5 sen.a RWCU outboard jaet. valves shut (when non-recen. Hx outlet T reaches 140 F). Isolated ph ase bus coolina fan will trip; standby will start. Offgas glycol coolers refrig unit compressors will trip. . - Offgas vault refrig, unit compressor will trip. Service water traveling screens will automatically start. Service water strainers backwash cycle will start. b. (any 2: 1.0each) (2.0) If the SW HDR PRESS is less than 37 psig, start the standby Service water Pump per S0I-P41. If NCC HX OUT TEMP cannot be restored to less than 95 F, place the third NCC heat exchanger on service per S01-P43. If TBCC HX OUTLET TEMP cannot be restored to'less than 95 F, place the second TBCC heat exchanger on service per S01-P44. Ref: Perry ONI-P41, pg.1, 2, 2, respectively.

k Answers Section 7 2 7.3 Time: to minimize time in radiation areas. (1.5) Distance: maintain as much distance as possible between worker and rad. source. Shielding: shield the source or the worker. Ref: PAP-0118, pg. 2. 7.4 a. To prevent further loss of Main Condenser vacuum (1.0) b. (0.357each) ~(2.5) 1. open 2. open 3. close 4. close 5. close 6. close 7. close Ref: Perry ONI-J11-1, pg. 3, 3. 7.5-(0.4 each) (2.0) a. 10% 15% b. 40% 35% c. 20% 15% d. 35% e. 5% Ref: Perry 101-4, pg. 14; 10I-3, pg. 8 and 9; 101-4, pa.4; 101-3, pg. 9; and 10I-4, pg. 15. 7.6 Reduce the MAX COMBINED Flow Limit setpoint until steam flow is compatible with reactor power. Maintain turbine inlet pressure constant at approximately 920 psig. (2.0) Ref: Perry ONI-CBS-2 mrem / (0.5) 5(N-18) quarter = 5 (38-18) = 5 x 20 = 100 rem (0.5) 7.7 a. b. c. False (0.5 d. 100 rems (0.5 e. 25 rems (0.5) Ref: Perry, PAP-0208, pg 7, 9, 16, 13; PR-2, pg. 4; and PAP-0208, pg. 13, respectively

~ o. ~ Answers Section 7-3 - 7.8 (0.5 each) (3.5) a. Open b. Closed c. Closed - d. As is e.- ' Closed f. Closed g. Closed Ref: Perry ONI-P52 i 7.9 a. Inserting a CRD will be more likely to prevent a scram because it will brina you farther away from the flow biased scram setpoint. (1.0) b. You decrease recirc flow to decrease power so the effects of a -scram would be less, and by decreasino power you decrease steam flow, thereby decreasing feed flow. CRD insertion is performed to increase the margin between the flow biased scram setpoint and actual power, and to prevent exceeding any local thermal power. (1.0) l-c. High Flux - APRM High g M Fw g,,w grem uge (0.5) 7.10 1. Main Steam line pressure greater than 850 psig. - W / 2. Condenser. Vacuum is greater than 26 inches Hg. 3. APRM downscale lights are cleared. ' coolant has been verified to be.within chemistry limits. (2.0) 4. Reactor //be s(L 6 /Wh* rgamp Ref:. 101-1 9 s ) i --,,-ws-------,~.-r-,-~,e, t,.,.~, e+ ,w.- - -.--,.vw---, .y -,.w...w--+,.w-y

w.

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Answers to Perry-1 SR0 Exam 8. Administrative Procedures, Conditions, and Limitations (24.0) 8.1 1. Verify that the inoperable control rod, if withdrawn, is separated, from all other inoperable control rods by at least two control (.5 cells in all directions. 2. Disarm the associated directional control valves either: - hydrolically by closino the drive water and exhaust wated { d

1) Electricallyly b;;;;i; c *he DS :::Wr r~d, or i

isolation valves. 3. Comply with SDM Tech. Specs. (.O (2.0) Ref: Tech. Specs. 8.2 Yes, (0,5); with less than Div. I and/or Div. 2 required battery operable, suspend core alterations, handling of irradiated fuel in sec. containment and operations with potential of draining vessel (1.5) Ref: Perry T/S (2.0) 8.3 All of the 4 occurrences (since each requires a declaration of an emergency event and this is a category of reportable events). (2.0) .Ref: Perry EPI-A1; EPI-B1 8.4 - a. 5 (0.5) b. The Shift Supervisor; the STA nor the 2 other members of the minimum shift crel.necessary for safe shutdown of the unit and any personnel required for other essential functions during a fire emergency. (1.0) i Ref: Perry Tech. Spec. pg 6-1 1

Answers Section 8 2 8.5 a. pressure within containment 0.40 inches vacuum H O '(1,0) 2 b.- 1. All penetrations terminating in the annulus and required to be closed during accident conditions are either: a) capable of being closed by an operable automatic isolation valve; or b) closed by at least one manual valve, fluid flange or deactivated automatic valve. 2. The primary containment equipment hatch is closed and sealed and the shield blocks are installed. 3. The door in the access to the annulus is closed. 4. The saaling mechanism associated with each shield penetration is operable. 5. The pressure within secondary containnent is less than (2.5) or equal to T.S. requirements. (.5each) Ref: Perry T.S. pg 3/4 6-37 and 1-6 respectively 8.6 a. False (0.5) b. If radiation exposure of 25 millirems per lineup or if the lineup requires entry into a Hiah Radiation Zone. (1.0) c. One or more logic inputs to an annunciator are defeated but the annunciator is otherwise operable. (1.0) d. The STA (0.5)- Ref: Perry PAP-1401, pg 5; PAP-0205, pg 6; PAP-1402, pg 3, 8, 10, and 4, respectively

4 Answers Section 8 3 8.7 a. True (0.5) b. False (0.5) c. 1, 2, and 3 (0.5) d. Shift Supervisor (0.5) Ref: Perry, PAP-0110, pg 5 and 3 (for a and c); PAP-1301, pg 4 for (b); PAP-1105, pg 4 for (d) 8.8 1. 1, 2 (0.5) 2. 1, 2 (0.5) 3. 1, 2, 3 (0.5) 4. 1, 2 (0.5) Ref: Perry, Tech. Specs,. pg.3/4 3/77 8.9 The initiation of an emergency system shall not be assumed to be inadvertant,. Prior to securing an emergency system, the operator shall verify, by multiple indications, that a valid initiation signal does not exist or that the system operation is no longer required to maintain core cooling. (2.0) Ref: Safety System Operation f 8.10 a. (0.5each) (2.5) 1. Unusual Event 2. Alert 3. Alert 4. Unusual Event 5. Site Area Emergency Ref: Perry EPI: EPI-A1, pg 7, 49, 7, 9 (with T.S. 3.43.2, pg 8), and 6 respectively b. 1. Maintenance Coordinator oc m.ar m e g,u g,r,4_ (0,5) 2. On the 599' and 574' elevation of Control Complex adjacent to the Rod Controlled area control point and the Health Physics / Chem. areas. (0.5) 3. At an Alert or at discretion of Emer91ncy Director. (0.5) Ref: Perry EPI: EPI-A7, pg.2, I and 3 respectively i ,}}

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