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sample questions + solutionselectrical and computer: POWER

PEPEprinciples and practiceof engineering

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Copyright © 2011 by NCEES. All rights reserved. P.O Box 1686 Clemson, SC 29633 800-250-3196 www.ncees.org ISBN: 978-1-932613-56-8 Printed in the United States of America November 2010

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iii

CONTENTS

Introduction to NCEES Exams ................................................................ 1

About the publisher

Updates on exam content and procedures

Exam-day schedule

Admission to the exam site

Candidate Agreement

Scoring and reporting

Exam Specifications .......................................................................... 2

Sample Questions and Solutions

Morning Sample Questions .................................................................. 5

Afternoon Sample Questions ............................................................. 23

Morning Solutions .............................................................................. 41

Afternoon Solutions ............................................................................ 53

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1

About the publisher NCEES is a national nonprofit organization dedicated to advancing professional licensure for engineers and surveyors. It develops, administers, and scores the examinations used for engineering and surveying licensure in the United States. It also facilitates professional mobility and promotes uniformity of the U.S. licensure processes through services for its member licensing boards and licensees. These services include the records program, study materials, credentials evaluations, exam administration, and more. NCEES is composed of member licensing boards from throughout the United States and its territories. Updates on exam content and procedures NCEES.org is our home on the Web. Visit us there for updates on everything exam-related, including specifications, exam-day policies, scoring, and corrections to published study materials. This is also where you will register for the exam and find the additional steps you should follow in your state to be approved for the exam. Exam-day schedule Be sure to arrive at the exam site on time. Late-arriving examinees will not be allowed into the exam room once the proctor has begun to read the exam script. The report time for the exam will be printed on your Exam Authorization. Normally, you will be given 1 hour between morning and afternoon sessions. Admission to the exam site To be admitted to the exam, you must bring two items: (1) your Exam Authorization and (2) a current, signed, government-issued photo identification. Candidate Agreement The NCEES Candidate Agreement describes references and other personal materials allowed in the exam room. It also describes other exam policies designed to protect exam content. This document will be printed in your exam book. It is also available for download at NCEES.org. If you register to take an NCEES exam, you are strongly encouraged to review this document prior to exam day, as you will be asked to affirm that you abide by the policies and procedures it describes. Scoring and reporting NCEES typically releases exam results to its member licensing boards 8–10 weeks after the exam. Depending on your state, you will be notified of your exam result online through your My NCEES account or via postal mail from your state licensing board. Detailed information on the scoring process can be found at NCEES.org.

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5

POWER MORNING

SAMPLE QUESTIONS

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POWER MORNING SAMPLE QUESTIONS

Copyright 2011 by NCEES 6 GO ON TO THE NEXT PAGE

101. A SCADA system configuration requires maximum values (i.e., the actual value of current when the transducer is at full scale). The transducer has a full-scale ac current of 5 A. Using a current transformer with a turns ratio of 400:5, the maximum current value (amperes) is most nearly: (A) 1,200 (B) 800 (C) 400 (D) 300

102. A differential relay circuit connected to protect a delta-wye transformer using current trans-

formers (CTs) must be balanced with properly connected auxiliary CTs to allow for the 30° phase shift and the turns ratio. If this is not accomplished, which of the following is most correct? (A) The circuit could cause the relay to be picked up under non-fault conditions.

(B) The ANSI standard for designating terminals H1 and X1 on wye-delta transformers would not be followed.

(C) A positive-sequence voltage drop from Bushing H1 to the neutral would lead the positive-sequence voltage drop from Bushing X1 to the neutral by more than 30°.

(D) The delta winding must be located on the high-voltage side of the transformer. 103. An industrial plant is served by a 12,470-V delta/480-V grounded-wye transformer. The high-

voltage side of the transformer is served from a 12,470-V wye-connected distribution line that has a grounded neutral. The best protection of the transformer against lightning strikes on the 12,470-V line will result from surge arrester connections of:

(A) phase-to-ground on the 12,470-V side of the transformer

(B) phase-to-phase on the 12,470-V side of the transformer

(C) phase-to-ground on the 480-V side of the transformer

(D) phase-to-neutral-to-ground on the 480-V side of the transformer

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POWER MORNING SAMPLE QUESTIONS

Copyright 2011 by NCEES 17 GO ON TO THE NEXT PAGE

MAIN DISTRIBUTION

PANEL

PANEL

A

LOAD

F1

128. Which of the following conditions would most likely result in a saturated magnetic flux within the indicated equipment?

(A) A 50-Hz power transformer operated at 60 Hz and at rated MVA

(B) A 3-phase induction motor driven by an inverter at the rated voltage of the motor and half its rated frequency

(C) A power transformer, operating at rated MVA, connected delta to a long transmission line that is exposed to a heavy geomagnetic solar storm

(D) A current transformer with its primary carrying rated current and its secondary winding shorted

129. Answer this question in accordance with the 2008 edition of the National Electrical Code®

(NEC®). Assume the voltage at the main distribution panel is 480 V, 3-phase. Feeder F1 consists of three 500-kcmil THWN copper phase conductors and a neutral in a steel conduit that runs a distance of 250 feet. Panel A serves a constant, balanced load of 400 A at 0.80 lagging pf. The voltage (V) at Panel A is most nearly:

(A) 475 (B) 470 (C) 465 (D) 460

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23

POWER AFTERNOON

SAMPLE QUESTIONS

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POWER AFTERNOON SAMPLE QUESTIONS

Copyright 2011 by NCEES 24 GO ON TO THE NEXT PAGE

V = 7,200 Vrms

jXs

IF

X

RF

501. A single-phase transformer serves a 7.2-kW resistive load located 200 ft from the transformer. Each conductor from the transformer to the load has an impedance of (0.78 + j0.052) per 1,000 ft. If the voltage at the load terminals is 240 V, the voltage magnitude (V) at the transformer secondary is most nearly:

(A) 265 (B) 250 (C) 245 (D) 230

502. A 3-phase, 460-V, 25-hp induction motor draws 34 A at 0.75 lagging power factor from a 480-V

source. The reactive power (kvar) required to correct the power factor to 0.90 lagging is most nearly: (A) –2.8 (B) –4.9 (C) –8.4 (D) –14.6

503. The 3-phase load at a bus is 8,000 kW at 0.80 lagging power factor. The total reactive power

(kvar) supplied by a 3-phase capacitor bank that will increase the power factor to 0.95 lagging is most nearly:

(A) 6,000 (B) 3,400 (C) 2,900 (D) 2,600

504. The diagram below represents the Thevenin equivalent of a single-phase distribution system. A

fault occurs between point X and ground. RF represents the fault resistance. The current IF is 3,600 A when RF is 0 . If RF is changed to 1.0 , the current IF (amperes) is most nearly:

(A) 2,000 (B) 2,400 (C) 3,200 (D) 4,600

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POWER AFTERNOON SAMPLE QUESTIONS

Copyright 2011 by NCEES 39

G1 LOAD

VTT1

230-kV BUS

G1: 3 PHASE, 834 MVA, 22 kV, Xd = 23%, 0.85 LAGGING pf'

T1: 3 PHASE, 933 MVA, 22/230-kV, Z = 15%, DELTA/GRD-WYE

540. A generating unit is shown in the figure. The 3-phase, short-circuit MVA contribution of G1 for a 230-kV bus fault is most nearly:

(A) 1,995 (B) 2,290 (C) 3,625 (D) 6,220

This completes the afternoon session. Solutions begin on page 53.

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41

POWER MORNING SOLUTIONS

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POWER MORNING SOLUTIONS

43

101. Iprimary = 5 A (400 turns/5 turns) = 400 A THE CORRECT ANSWER IS: (C)

102. The circuit would need to be balanced out to account for the 30 phase shift between wye and

delta windings and, also, for the turns ratio of the transformer. Otherwise, there would be a non-zero current through the relay restraint windings during non-fault conditions. Answer (A). The idea that CTs must be physically wired to maintain some arbitrary terminal nomenclature seems nonsensical; thus, eliminate (B). The ANSI standard is, indeed, to label the high-side terminal H1 and the low-side terminal X1 such that voltage H1-neutral leads X1-neutral by 30, and the phase shift will be 30, not more than 30. But no matter how the CTs are connected, it will not affect the phasing of the transformer windings themselves; thus, eliminate (C).

There is no reason a delta winding must be on either the low or high side of a transformer; thus, eliminate (D).

THE CORRECT ANSWER IS: (A)

103. Even though the high side of the transformer is connected delta (ungrounded), a strike to a 12-kV

phase conductor will impose a high voltage to ground, which is best dissipated by surge arresters connected between the phases and ground. Answer (A). Phase-to-phase arrester location will not be as sensitive to ground currents caused by lightning strikes as a phase-to-ground location. This eliminates (B). The problem statement refers to protection against lightning strikes on the 12-kV line. Arresters on the 480-V side would not protect the transformer against such an event. This eliminates (C) and (D).

THE CORRECT ANSWER IS: (A)

104. 10-hp load is 50 A 1.25 = 62.5 A [NEC® Table 430.248, NEC® Article 430.22 (A)]

Resistance heating load = A2.4V240

W000,1

Total load = 62.5 + 4.2 = 66.7 A

Select #4 copper (NEC® Table 310.16) THE CORRECT ANSWER IS: (C)

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53

POWER AFTERNOON SOLUTIONS

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POWER AFTERNOON SOLUTIONS

55

501. ft) 200(ft 000,1

052.078.0ft) 200(

ft 000,1

052.078.0

jIV

jIV lineloadlinesec

V 4.249

V 143.04.249

)052.078.0(5

1)030)(2(0240

secV

j

THE CORRECT ANSWER IS: (B)

502. 1 13 cos (pf ) 3(480)(34) cos (0.75) 28.3 41.4 kVAmotor l lS V I

Qnew = P[tan(cos–1)]

= 21.2[tan(cos–1 0.9)] = 10.3 1[sin(cos 0.75)][ 3(480)(34)] 18.7oldQ

Qnew = Qold + Qadd

Qadd = Qnew – Qold = 10.3 – 18.7

= –8.4 kvar

THE CORRECT ANSWER IS: (C)

503. kW 000,8loadP

2 2

2 2

pf 0.80

8,000 kW 10,000 kVA0.80

10,000 8,000 6,000 kvar

8,000 kW 8, 421 kVA0.95

8, 421 8,000 2,630 kvar

6,000 2,630 3,370 kvar

load

load

new

new

cap load new

S

Q

S

Q

Q Q Q

THE CORRECT ANSWER IS: (B)

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PE Sample Questions and Solutions Published by NCEES

Chemical

Civil: Construction

Civil: Geotechnical

Civil: Structural

Civil: Transportation

Civil: Water Resources and Environmental

Electrical and Computer: Computer Engineering

Electrical and Computer: Electrical and Electronics

Environmental

Mechanical: HVAC and Refrigeration

Mechanical: Mechanical Systems and Materials

Mechanical: Thermal and Fluids Systems

Structural

For more information about these and other NCEES publications and services,

visit NCEES.org or call Customer Service at (800) 250-3196.

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