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P.O. Box 1686
Clemson, SC 29633
800-250-3196
www.ncees.org
ISBN 978-1-932613-47-6
Printed in the United States of America
November 2009
Certified Fiber Sourcing
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Morning Sample Questions
Morning Solutions
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Mechanical Afternoon Solutions
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Mechanical Afternoon Sample Questions
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Exam Specifications for the Afternoon Session
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about the exa1ninations-inclndingcu?cient exan1- specifications, exan1 policies,
éoi news and updates
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ES?? THE E}lA.M DEVELOPEF?
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required of candidates for licensure as professional engineers.
in the
These exai11inations n1easure n candidate's ability to de111onstrate n1inin1un1 cornpetency
each 1,r.cEES n1en1ber licensing board.
practice of enginee1ingand are administered by
1\JCEES follows the guidelines established in. the Standardsfor Educational and Psychological Testing
the A111erican Psychological P1.ssociation. These piocedures n1aximize the fairness and
1-.JCEES de?.1elops the exan1th1:?d:iüKllS
published by
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cu1Tent testing techniques.
specialists to guide the development of examinations using
the nation to
1\f CEES Jtelies on com1nittees cor11posed of professional engineers throughout
co111e fron1 diverse professional backgrou11ds
prepare the exar.ninations. These licensed engineers-who
the content expertise
including govemn1ent, industry, private consulting, and acaden1ia-supply
that is essential in developing exaruinations,
LICENSURE
professional engineer (P.E.) is to take and pass the
are
Fundamentals of Engineering (FE) examination. If you are a student or a recent college graduate, you
well advised to take this step while coursework is still fresh in your n1ind. After passing the exai11ination,
your state board will designate you an engineer intern (E.I.).
The first step
on
the path to licensure
as
a
To continue the licensure process, typically you must complete 4 years of progressive and verifiable
contact your licensing board to ensure
experience that is acceptable to your licensing board. You should
that you are on track to meet their requirements. If you are, you will be approved to take the Principles
and Practice of Enginee1ing (PE) examination. After you pass the PE examination, you may become
licensed as a professional engineer and use the distinguished P.E. designation.
Exam application procedures are available from the individual licensing boards in each state. Requirements
and fees vary among the boards.
DESCRIPTION OF EXAMINATIONS
The purpose of the FE examination is to determine if the examinee has an adequate understanding of basic
and discipline-specific subjects
science, mathematics, engineering science, engineering economics,
an engineeling bachelor degree program. The
normally covered in coursework taken in the last 2 years of
examination identifies those applicants who have demonstrated an acceptable level of competence in these
subjects.
is
The 8-hour FE examination is a no-choice examination in a multiple-choice fonnat The examination
administered in two 4-hour sessions. The morning session contains 120 questions, and the afternoon
are posed
session contains 60 questions. Each question has four answer options, Numerical questions
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Enits. The
exam
nonnally 1nternational Systen1 of Units (SI),
or
in U.S. Custo111aiy Syste:;-il (USCS)
on the exa1ninationº
specifications presented in this book give details of the
subjects covered
The rnorning session is com1non to
session is administered in the
exaininees, regardless of their engineering discipline. The aften1oon
follo\ving seven 1-i::1odules-Che1nical, Civil, Electrical, En_vironrnental,
Industfial, Mechanical, and Other Disciplines. h-1 general, if
your 1najor is a discipline other than
chemjcal, civil, electfica1, environrnental, industrial, or :tnecho.nical
engineering, you should choose the
Other Disciplines module in the aftenrnoon.
211
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the
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general content of the subject areasJ the level of
fonnat of the exaxn, the questions should be
helpful in
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difficulty of the
questions, an_d the
preparing for the exarnination. Solutions are
ine quesnons. ""! ne so I unon
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presenrerd rora
presenten n121-y nor ue -'-h e on y vvay ro so ve r_1e quesnon.
'l'h e
intent is to der11onstrate the typical effort
to solve each question. Exa:rn
required
questions, content, and
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SCORING
Both sessions of the FE exan1ination are worth the sai11e total
number of points, and questions are
weighted at one point for each n1oming question and tvvo points for each afternoon_
question. Within
each session, every question has
equal weight Points are not subtracted for incorrect
responses.
Therefore, it is to your advantage to answer every question. Y our final score on the
exainination is
deterrnined by summing the number of points obtained in each session.
One of the most critical considerations in
developing and administering examinations is establishing
scores that reflect a standard
of minimum competency. Before
passing
setting a minimum passing score
for a new examination or for the first examination after a
change in the specifications or standards,
NCEES conducts studies involving a representative
panel of engineers familiar with the examinee
population. This panel uses procedures widely recognized and accepted for
occupational licensing
purposes and develops a written standard of minimum
competency that clearly articulates what
knowledge is required of engineering interns. Panelists then take the
the
examination,
difficulty level of each question in the context of the minimum competency standard. evaluating
Finally, NCEES
reviews the panel's work and sets the passing score for the initial exam.
For subsequent exams, an
equating
method is used to set the passing score. The
passing (raw) score is never disclosed.
NCEES does not use a fixed-percentage pass rate. The
key issue is whether an individual candidate
meets the standard of minimal
co1npetence, not whether the candidate is better or worse than other
candidates. To avoid the confusion that might arise from fluctuations in the
passing score, exam results
are reported simply as pass or
fail. Some licensing jurisdictions may choose to report exam results of
failing candidates as a scaled score.
The legal authority for making licensure decisions rests with the individual
licensing boards and not
withNCEES.
l.?Yit?..1 POL?C?ES i.\ND PROCEDURES
A breach of an exc:n1iDation could lead to the ücensure of people who Zcre not cor_,_1petent to practice
at riskº Therefore, l\TCEES tak.es
engineering. This puts the health, safety, and welfare of the pu.blic
for exainple, restricting cell
111easures Decessm-y to protect the integrity of the exar11 process. This includes,
of paper, a:nd recording devices; controlling access into
phones, ce1iain calculators, peúcils, loose sheets
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could result in such rneasu?ces 2,5 disnnissaJ fron1 the exan1, cancellation of exc1111 results, and, in sorüe
a.n d
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cases,
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Be sure that you understand the
policies outlined in the 1,JCEES Candidate Jtgreen1ent, and read all
instructions fron1 your board or testing service before
expectations for exa1ninees are.
NCEES Candidate
exarr1
day
that you lmow exactly what the
so
i??greernent
The NCEES Candidate P1..gree1nente?cplains the policies, procedures, and conditions exarninees r11ust
to sign a statem,ent on their
agree to while taking an 1'TCEES exan1ination. Exan1inees are required
answer sheet before the exarnination starts to affinn that they have been provided the NCEES Candidate
Agreement, have read and understand the rnaterial, and agree to abide by the conditions cited. These
conditions apply to all l\f CEES exan1i11ationsº A cu1Tent 1'TCEES Candidate Agreement is available on the
NCEES Web site.
Special Accommodations
If you require special accomn1odations in the test-taking procedure, you should contact your state
licensing board office well in advance of the day of the examination so that appropriate aiTangements may
be determined. Only preapproved accommodations are allowed on exam day.
Exam Admission
Requirements
For exam
admission, examinees must present a current, signed, government-issued photographic
identification (such as a valid state driver's license or passpo1i). Student IDs are not acceptable. Examinees
must report to the exam site by the designated time. Examinees will not be admitted after the proctor
begins reading the exam instructions.
References
The FE examination is a closed-book examination. :However, since engineers rely heavily on reference
materials, you will be given a copy of the NCEES FE Supplied-Reference Handbook at the exam site.
The Handbook contains formulas and data that examinees cannot reasonably be expected to memorize.
The Handbook does not contain all the information required to answer every question on the examination. For example, basic theories, formulas, and definitions that examinees are expected to know have
not been included. To familiarize yourself with the content of the Handbook before the examination,
visit the NCEES Web site to purchase or print a copy of the Handbook. You will not be allowed to take
your copy of the Handbook into the examination; you must use the copy provided to you by the proctor in
..
the
exam
room.
Prohibited Items
A cu1Tent list of prohibited items is included in the NCEES Candidate Agreement on the NCEES Web
site. If a prohibited item is found in an examinee's possession after the exam begins, the item will be
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confiscated and sent to l\J CEES I-Ia-ving
and/or invalidation of your exa111 results,
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Exam ?rregu?arities
Fraud, deceit, dishonesty, and othef
examination
behavior 1n. connection w1
strictly prohibi:i=ed. Irregular behavioI i_ncludes but is not hn1ited.
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Copying or allo-,ving the copying of
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are
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exarn
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Disrupting other exan1inees
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Failing to
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Possessing, reproducing, or disclosing exarn questions, ansvvers, or other inf on11ation about the
exan1ination without authorization before, during, or after the exarn administration
,1,1
cease
work
before the proctor instructs you to do
on
so
the e:x.an1ination or put down the pencil whel1 i:in1e is called
Co1nmunicating with other exan1inees or with any outside
telephone, personal co1nputer, Internet, or any other means
during the exaínination by
source
Exam Results
Examinees are understandably eager to find out how they performed on the examination. To ensure that
the process is fair and equitable to examinees and to maintain the validity of the exam questions, NCEES
uses
a rigorous scoring process for each of the NCEES
multiple-choice examinations that takes
approximately 12 weeks to complete.
®
l1ll
'lf,
First, NCEES scans all answer sheets as they are received from the states. Answer sheets are
flagged for review when they are missing critical information, such as the candidate ID number.
The scoring process continues only when these issues are resolved.
Next, a psychometric analysis is performed on a sample population of answer sheets from each
multiple-choice examination to identify any questions with unusual statistics, which flag the
question for review.
Then, at least two subject-matter experts who are licensed engineers review the flagged items. In
addition, NCEES reviews all Examinee Comment Forms, and the subject-matter experts consider
the forms about specific exam questions. If the reviews confirm
question, credit may be given for more than one answer.
comments
®
s
on
an
error
in
a
When the analyses and reviews are completed, NCEES revises the answer keys as necessary. The
passing score and the final correct answers for each examination are then used to score all the answer
sheets. Scanners are calibrated before and during scoring. A percentage of the answer sheets are
hand-gradedand the results comparedto the machine score to ensure accuracy of results.
Finally, NCEES
releases the results to the state boards
results to examinees.
or
testing agencies, who in tum report the
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e;?c1rrlination: 120 ql:.estions \ú t?1e ii--hcv_r
The FE exari1ir1aüon ::s an 8-hour su:9plied-:¡_-eference
rnorning session and 60 questions ü1 the 4-hour zlternoon session.
The aften100E session is adrninistered in i:he following seve:;_-:1,_ 111oduJ.es-:henüca1,Civü, EiectácRl,
Environmental,Industrial, IVlechanical, and 'Üther Disciplines.
afternoon inodule they
Exaininees work all questions in the n1orning session and all questions in the
have chosen.
Topü:
Mathernatics
A. Analytic geornetry
B. Integral calculus
C. Matrix operations
D. Roots of equations
E. Vector analysis
F. Differential equations
G. Differential calculus
II.
7°/o
Engineering Probability and Statistics
A. Measures of central tendencies and dispersions ( e.g., 1nean, mode,
standard deviation)
B. Probability distributions ( e.g., discrete, continuous, normal, bino1nial)
C. Conditional probabilities
D. Estimation (e.g., point, confidence intervals) for a single mean
E. Regression and
F.
G.
III.
curve
fitting
Expected value (weighted average) in decision-1naking
Hypothesis testing
9o/o
Chemistry
A. Nomenclature
B. Oxidation and reduction
C. Periodic table
D. States of matter
E. Acids and bases
F. Equations (e.g.,
stoichiometry)
G. Equilibrium
H. Metals and nonmetals
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Topic
üf Te2<:
P_-1_rea
Computers
.A. Terminology ( e.g., rne:rn.sty types, CFU,
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Spreads"nlee·LLS (e
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prograrrnning ( e.g., assignxn.en.t sü:?ten.1e:;?i:i:s, loops and
branches, :function calls)
Ethics and Business; P'ractic,sG,
A_.
H.
C.
D.
E.
Code of ethics (professio112J and technical socieáes)
A.greernents and contracts
Ethical versus legal
Professional liability
Public protection issues (e.g., licensing boords)
Engineering Economics
J._.
Discounted cash flow ( e.g., equivalence, PVv, equivalent annual FVv,
rate of
return)
B. Cost ( e.g., incrernentall, average, sunk, estin1ating)
C. Analyses ( e.g., breakeven, benefit-cost)
D. Uncertainty ( e.g., expected value and risk)
VII.
Engineering Mechanics (Statics and Dyna1nics)
A. Statics
L Resultants off
2.
3.
4.
orce
systems
Concurrent force systems
Equilibrium of rigid bodies
Frames and trusses
Centroid of area
6. Area moments of inertia
7. Friction
5.
B. Dynamics
Linear motion
2. Angular motion
1.
3.
Mass moments of inertia
4.
Impulse and momentum
a.
particles
b.
5.
rigid bodies
Work, energy, and power
a. particles
b.
6.
applied to:
applied to:
as
rigid bodies
Friction
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D. D1eforr11ations (e.g., a;?ial, bending, torsion)
E. Combined stresses
F. Colur11ns
G. Indete1r11inant analysis
Plastic
I-I.
elastic def orn1ation
versus
IVIate:rial Properties.
Properties
l. Chemical
A_.
2.
3.
Electrical
Mechanical
4.
Physical
B. Con·osion mechanisms and control
C. Materials
1. Engineeredmaterials
2. Ferrous metals
3. N onf erro us metals
X.
XI.
Fluid Mechanics
A. Flow measurement
B. Fluid prope1iies
C. Fluid statics
D. Energy, impulse, and momentum
E. Pipe and other internal flow
7o/o
equations
Electricity and Magnetism
A. Charge, energy, current, voltage, power
B. Work done in moving a charge in an electric field
(relationship between voltage and work)
C. Force between charges
D. Cun·ent and voltage laws (Kirchhoff, Ohm)
E.
F.
Equivalent circuits (series, parallel)
I.
Basic complex algebra
Capacitance and inductance
G. Reactance and impedance, susceptance and admittance
H. AC circuits
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11
I
Enthalpy
Entropy
ORN? G SAMPLE EXAMINATION
HALF THE
¡I
CONTAI
S
QUESTIONS,
THE ACTUAL EXAMINATION,
UMBER
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cs-ncaviif"J' of the cu:cve
(]2)
Iocation_ of inflection points
( C)
nuJ211ber
(D)
area
Which
of the
-
2,..
of inflection points
under the
curve
on
the
curve
te uss?l to
o_,f·
tl1e
between certain bounds
curve
following
the
on
c2n
1s
the general solution to the diff
condition shown below?
dy +
5 y= O; y
(O)=
eren
ti al equation and boundary
1
dt
(A)
est
(B)
e
.
-5t
(C)
(D)
Copyright
201
O
by MCEES
13
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THE NEXT PAGE
+
e-4x
e
-2X
1
+
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A particle traveled in a straight line in such a way that its distance S fron1 a given
line after time twas S = 20? l. The rate of change of acceleration at time t = 2 is:
4®
-
(A)
(B)
(C)
(D)
72
144
192
208
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ii'_:
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point on that
+
=
+
+
1
+
?ls
-½-(
A/6
+
\
)
-
l
\
)
f6(
If
/1 denotes the derivative of
(A)
(B)
(C)
(D)
lim-
&
6.y?o
?y
lün
L3.y
6.y?O
ill
liin
lim
a
f(x
+
/tr)- f(x)
!tr
f(x
-
Ax)+ f(x)
?y
function of y= f(x), then/1 (x) is defined by:
u
",Nhat is the
x
=
(A)
(B)
(C)
(D)
8.,
of the :region in the first quadrant that is bounded by
area
y312, and the
y-axis?
2/5
3/S
2/3
1
Three lines
are
x+y
x-y
2x + y
TION
defined by the three equations:
=O
=O
=
1
The three lines form
a
triangle with vertices at:
(A)
(O, O),
G,! ),(1, -1)
(B)
(O, O),
(? ! ),H,
(C)
(1, 1), (1,
(D)
( 1, 1), (3,
,
-1), (2,
-
3), ( -2,
-1)
1)
-1)
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n1ean
A1 of a population fror11 a san1ple of size n drawn from the
You wish to estin1ate
population. For the sample, the rnean is x and the standard deviation is s. The probable accuracy
of the estin1ate improves with an increase in:
(A)
(B)
(C)
(D)
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The standard
(A)
?3
(B)
'\16
(C)
4
(D)
6
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The volun1-e (L)
(A_)
(B)
(C)
(D)
11.2
14.9
22.4
44.8
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and 1.00
pressure is 1nost nearly:
u::¡
A.tomic weights 1nay be taken as 75 fo:c axse::_·üc, í6 ior o??ygen, 8Ttd 12
car7os,n. i1-ccordi,_1g t?•
the equation above, t:b.e :reactirna
stzn1dard g[a:cn rnoie of iLs203 with carbon wül res?1:J,t
the
r
·
.
:i=on11at1on o?:f
(B)
-
(1"'\
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a
16"
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of J-As
gr"-11s
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or r0
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than
r
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If 60 1nL of 1\faOI-I solution neutralizes 40 1nL of 0.50 M
H2SO4, the concentration of the :NaOH
solution is 1nost nearly:
(C)
0.80 M
0.67 M
0.45 M
(D)
0.33 J\1
(A)
(B)
Copyright
.
greater arnount by weight
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by NCEES
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Consider the following equation:
The equation above is the formulation of the chemical
the following reactions?
(A)
C2 + D2 <-> A4 + B
(B)
4A + B
(C)
4C + 2D
(D)
A4 + B <? C2 + D2
p·,.,
'-·1•0yright 2(1'10 by NCEES
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equilibriumconstant equation for which of
2C + 2D
?>
2A + B
21
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20.
In
a
spreadsheet, the number in Cell A4 is set to 6. Then AS is
set to A4 + $A$4. This formula is
copied into Cells A6 and A 7. The number shown in Cell
A7
is most
(A)
12
(B)
(C)
(D)
24
nearly:
36
216
Copyright 201 O by NGEES
22
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-;¡
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0
0
-,
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u
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caJc11Jates the suxn:
This seg1nent
(A.)
S
=
1
+
ZT
(B)
S
=
1
+
ZT
(C)
S
+ 2
1
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l
=
+
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+
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2
z
z2
S=l+-+-+-+
2!
l!
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+ 3
ZT
+
ZT
1
2
1
(D)
+
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+
+
...
ZT +
...
+
3
+
3Z
+
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ZT
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+
)
(:NZ1
l\T
3
r
z3
...
3!
1-T
+
zN"'
-
"-
Nl)
in Colunms A-J, and Row 2 has
In a spreadsheet, Row 1 has the numbers 2, 4, 6, 8, 1 O,
, 20
columns. All other cells are zero except for Cell D3,
the numbers 1, 3, 5, 7, 9,
, 19 in the same
which contains the formula: Dl + D$1 *D2. This formula has been copied into cells D4 and D5.
The nu1nber that appears in cell D4 is most nearly:
...
...
(A)
(B)
64
(C)
519
(D)
4,216
20'10
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cornplete and objective analysis
eva_uatlon
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the professional backgrm-1:o_d_ of the defenda:nt
professional engineer originally licensed 30 years ago, you are asked to evaluate a
newly
developed computerized control system for a public transportation systern. You
rnay accept this
project if:
a
competent in the area of modem control systems
your professional engineering license has not lapsed
(A)
you
(B)
( C)
-
20"!
are
your original
area
of specialization was in transportation
systems
you have regularly attended meetings of
(D)
Cop'lright
L\
'-''""pí-,,..,,.d?1c;t
íJOssible oníy those questions posed by the attorneys
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provide an
->
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a
professional engineering society
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des?on
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y a0··e ccm.pet1"-1g
the design_ professional consortium fon_ned to con1pete fo:c the project. ·tcu.r
the lead gi"cruiJ
l J
_L,.-he n·,•,o:.ec;·
nwrerl
l
-L'·o enter +'ee J1ego·L'·1at-¡o-;1s wT
heen selec-'L·ect· as t-lJP f1i--·st
-'---ª
Cn·¡1so•r·t1·u-1n ly-s
?11r-;ng
the negotiations, the amount you have to cut fro:rn your fee to be avvarded the contract wiH
:,·equire dropping one of the consortiuiri. nrreinbers ,.vhose staff has special capa:bilities not
available fron1 the staff o:f the re'--T1Hiningconso:i.iitliTl 111e111bers. Csn your re1n2dning co:t1sortiuff1
,...n,1
cL_ _
_
-y,
,ffír
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ethically accept the contract?
contract to coordinate a project with other
hired
1 '0
be p··nvirlecl
'bv
l.L
í·l1at -"1t1s1..L
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-?m1s
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1J-·0£'ess1·onal
./
capa101ºlit-;es and cenrices
p-··ov1c1ting
1,To,
because
an
engineer n1ay not accept
e,
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a
1,Jl
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consultants.
-i es, if your ren1aining consortiurn rnernbers hire a fevv new lower-cost ern_ployees to do the
special work that ,vould have been provided by the consortiun1 n1en1ber thai: has been
(l>)
dropped.
(C)
1,To,
not if the
owner
is left with the ir.npression that the conso:i_iiun1 is still
fully qualified to
perfon11 all the required tasks.
'Y es, if in accepting an assignment to coordinate the project,
seal all the docun1ents for the entire work of the consortiun--i.
(D)
26..
a
single person will sign snd
student and have an on-site job interview with Company A. Just before you fly to the
intervievv, you get a call from Company B asking you to come for an on-site interview at their
offices in the same city. When you inform them of your interview with Company A, they suggest
you stop in after that. Co1npany A has already paid for your airfare and, at the conclusion of your
interview with the1n, issues you reimbursement forms for the balance of your trip expenses with
instructions to file for all your trip expenses. When you inform them of your added interview
stop at Company B, they tell you to go ahead and charge the entire cost of the trip to Company
A. You interview with Co1npany B, and at the conclusion, they give you travel reimbursement
forms with instructions to file for all your trip expenses. When you inform them of the
instructions of Company A, they tell you that the only expenses requiring receipts are airfare and
hotel rooms, so you should still file for all the other expenses with them even if Company A is
paying for it because s?dents always need a little spending money. What should you do?
You
are
a
(A)
Try to divide the expenses between both firms
(B)
Do
(C)
File for travel expenses with only
(D)
Tell all your classmates
as
as
best you
can.
both recruiting officers told you. It is their money and their travel policies.
20H, by MCEES
to
one
firm.
sign up to interview with these firms for the trips.
25
GO::)
OM TO THE NEXT PAGE
using hand tools. Tools vvill cost $1,000;i a:ud the JCCtmT :r[ac50. iis an alternative, an automated syste1n wiH cosi: $15,000 v1ith
01-c $0.t:;:O_ ~with an
anticipated annua1 volur.ne
5.000 units 2J1.6_
t
C,Q:St
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_...._
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(C)
2.8
3.6
15.0
(D)
never
(B)
2ft
the breakeven
point (years) is 1nos·t n.e2xly:
printer costs $900. Its salvage value after 5 years is $300. Annual n1aintenance is
$50. If the
interest rate is 8%, the equivalent unifom1 annual cost is n1ost
nearly:
.A.
(A=)
(B)
(C)
(D)
$224
$300
$327
$350
The need for a large-capacity water
supply system is forecast to occur 4 years from now. At that
the
time,
system required is estimated to cost $40,000. If an account earns 12%
per year
compounded annually, the amount that must be placed in the account at the end of each
year in
order to accumulate the necessary
is
purchase price most nearly:
29.,
(A)
(B)
(C)
(D)
Copyright
20·¡ O
$8,000
$8,370
$9,000
$10,000
by NCEES
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A
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project has the esóxü2tec:
n .--.
--,]
iL 1wt::i.1!. ,,'
(.
có:sh fiovvs shcvvn below.
I
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an
1ntetest rate
3
1
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Of
-¡
.,,
I
10
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,,.2
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per yes.r cor.t1pounaect
..
1
1
'lr,
a11:'.1ua.1.y,
-
tne annuo.I worth
('
o:c
.
U.e proJect
1
.
1s
e___
ne8..rly·
,
__
(A)
$L!-50
(B)
$361
(C)
$320
$226
(D)
,
0v,
.
.
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You 1nust choose between four pieces of comparable equipment based
below. All four pieces have a life of 8 years.
Parameter
First cost
Annual costs
Salvage value
on
the cash flows given
Equipment
A
B
C
D
$25,000
$8,000
$2,500
$35,000
$6,000
$3,500
$20,000
$9,000
$2,000
$40,000
$5,000
$4,000
The discount rate is 12%. Ignore taxes. The most preferable top two projects and the difference
between their present worth values are most nearly:
(A)
A and C, $170
(B)
Band D, $170
(C)
A and e, $234
(D)
Band D, $234
27
GO ON TO THE MEXT Pll.GE
'
m=
'
¢,
=
?=O
cylinder ·weighing 120
The
reaction (N) at
(A)
(B)
(C)
(D)
1\T
rests betvveen
Íl is rnost nearly:
frictionless
as
shovvn
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figtre
--
96
139
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150
/
200
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In the figure below, Block A weighs 50 1\l, Block B weighs 80 N, and Block C weighs 100 1\l.
The coefficient of friction at all surfaces is 0.30. The maximum force F (N) that can be applied to
Block B without disturbing equilibrium is most nearly:
(A)
(B)
(C)
(D)
CABLE
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5m
30.0
cm
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sn
(A)
belc,N
as
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cantilever bean1
statically indeterminate bean1
(C)
sin1ply supported beant
continuously loaded bean1
39.
The shear diagran1 for
a
particular bea1n is shown below. All lines in the diagrarn
are
straight.
The bending moment at each end of the beam is zero, and there are no concentrated couples
along the beam. The maximu1n 1nagnitude of the bending 1noment (kN°m) in the beam is most
nearly:
(A)
(B)
(C)
(D)
2 lTI
16
18
26
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6
4
SHEAR
(kN)
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of
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grain of
the grain parallel
(A)
(B)
(C)
(D)
used·
lS
a
process.
840 ld\J/m2, the
the neutral axis is 1nost nearly:
the
trans1nitted
shearing stress
a 200-inm-diarneter
500
1,200
1,320
1,500
The Euler formula for columns deals with:
(A)
relatively short colum_ns
(B)
shear stress
( C)
tensile stress
·
(D)
buckling
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L,l_,_
i.vv1.YSL.L3P1·'.z,-,·'-:.0.r
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-
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8.s:
(A)
v1orking
(B)
strain aging
( C)
grain growth
{D'\
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In general,
metal with high hardness will also have:
a
(A)
good formability
(B)
high ilnpact strength
(C)
high electrical conductivity
(D)
high yield strength
Glass is said to be
an
amorphous material. This
that it:
high melting point
(A)
has
a
(B)
is
supercooled vapor
( C)
has large cubic crystals
(D)
has
a
means
no
apparent crystal st111cture
,J
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fan ztilun1inu1n crünp connector
used
,vere
happ·en?
;:;_v·--?ec·L'L,..f':..._IJ.
.LLO
(A)
The copper ·wire Dnly
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(B)
(.ri\
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Both vviH
co::c-rode.
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47
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The rectangular homogeneous
gate shown below is 3.00 m high x LOO m wide and has a
frictionless hinge at the bottom. If the fluid
on the left side of the
gate has a densiíy of 1,600 kg/m3, the
n1agnitude of the force F (kN) required to
keep the gate closed is most nearly:
(A)
(B)
(C)
o
22
24
(D)
220
----------------------------------------------------------------------------------------------------------------------------------·-----------------------·------------------------
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FLUID-=========
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201
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by NCEES
34
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the
(
'
¡,.
\r-/
:J:Ywing stateE1ents is true of viscosity?
r
\
:·
-.-
.
x:e
ro\,
J
\
49.
a
,;
atwa.ys h as
a
__
is
t>
D
.e
•:i
Ine:ctrnJ L:) VlSCCH.lS 1orce.
o:i::
.
-,
('•('
iarge eriect
on t_1e va 1 ue
1
0
•
•
,..
•
I'
,
en t11e 1::1cü:::r1 xacccr.
usually 1ovv when turbulent forces predorninate.
horizontal jet of water (density= 1,000 kg/rn3) is deflected perpendicularly to the original jet
stream by a plate as shown below. The magnitude of force F (ld\J) required to hold the plate in
place is most nearly:
A?
(A)
(B)
(C)
(D)
JE\
4.5
9.0
FLOW
?
\
JET AREA= 0.01 m2
JET SPEED = 30 mis
45.0
90.0
_)
?
'
'
F
,--c::r:
'
.:_?PLATE
GO ON T() THE MEXT P,b,GE
I
I
I
I
I
Ill
11
\ji/hi-2h of the
(B\J
The
\_
folic\ving state:?nents 2tbo-it flov, through an insulated
tJ,·J stre8IIO_
-
The pitot tube shown below is placed at
1:91c
nrr:1-vi,1ty
_
u?-,('t11e
fln1·,r1
------~-
i1s
_
n
2 .v,
?.íld,-- -l?111e
e____
_
a
.l_
poini: where the velocity is 2.0 n1/s. The specific
lr"'")e-,·
?-ort:10-n·
ry,('-i-l.1e 1--1n1ano--,.-e-'.~er
Ltl l
-Vl
lJ
L_
,_1_
co·1l'-ai-,,,1s
-?
_
__,
__
..,__
_
on
the n1anon1eter is n1ost nearly:
(A_)
20.0
(B)
(C)
10.0
(D)
52.
(A)
(B)
(C)
(D)
0.-'1-1i"
º--·
--r11e
i..,
..__
·11·Pad1·-,-"G
1-b
h (n1)
----,1
0.40
0.20
If the complex power
(VAR)
::_s
a:nd dovv:o_stre2;r:n e:üth2ipies m:e eauaL
-
JL
ve
is most nearly:
1s
1,500 VA_ with
a
power factor of 0.866 lagging, the reactive power
O
750
1,300
1,500
GO ()M
TO TME NE:Cf Pl\GE
111
30
J
-Jso
T
is correct according to
300
143 Q
(B)
140
30Q
50Q
40Q
143 o
1400
(C)
300
(D)
400
son
300
resistor
25 Dis then connected
capacitor has been charged to a potential of 150 '\/,
across
the capacitor through a switch. Vlhen the switch is closed for ten tin1.e constants, the total
eneTgy (joules) djissipated by the resistor is 1110s nearly:
54?
.A...
10-1-,tF
i:
55.
rA,
\ c.)
,
l.º
(B)
Ll><lü-1
(C)
9.0
><
(D)
9.0
X 103
,/
o, A
0-7
·_1•
101
The com1ecting vvires and the battery in the circuit shown belovv have
cun·ent (amperes) through the 6-.Q resistor is rnost nearly:
(A)
(B)
negligible resistance. The
W'v
1/3
I
1/2
(C)
1
(D)
3/2
4.Q
6V
I
,,.
Joo
Copyright. 20·i O by MCEES
I
I
I
I
I
I
_I
111
11
/
I
I
II
I
I
111
I
_1
I
I
I
I
i
560
=
.,j
1·
?
is rnost nearly:
_-,
.L
o
-¡
?
57.
58.
+' l
insulated tank contains half liquid and half vapor by volu111e in equilibriurn. The release of
s111all quantity of the vapor without the addition of heat Vi.rill cause:
J.\.n
liquid in the tank
(P1.)
evaporation of
(B)
superheating of the vapor in the tank
( C)
a
(D)
an
some
rise in temperature
increase in enthalpy
The heat transfer during an adiabatic process is:
(A)
reversible
(B)
irreversible
(C)
dependent on temperature
(D)
zero
a
60.
,
l.7,
il
,¡?::t.
Cll?r•JINt?
_
(?:;;
!"-
l?,
j?
i.:1>1·\lvd-?
'/,"''7¡(;
Í1
.. ,o
.An isentropic process is one ·which:
occurs
(D)
is adiabatic and reversible
(C)
is reversible but not adiabatic
(B)
is adiabatic but not reversible
(A)
at constant pressure and
temperature
The universal
constant
is
8.
314
gas
kJ/
(kmol
K). The gas constan( [(kJ / (kg•K)] of
molecular weight of 44 is n1ost
nearly:
•
(A)
(B)
(C)
(D)
a gas
having a
0.19
0.38
0.55
5.3
I
I
11
Rr?? 1 f\J e;
s
L.IJTI
3
5
I
7
3
A
1
e
11
e
12
B
13
A
A
14
D
D
15
e
45
D
16
B
46
B
17
e
47
e
18
B
48
e
19
D
49
B
20
B
50
B
21
D
51
D
22
e
52
B
23
B
53
D
24
A
54
B
25
A
55
A
26
A
56
A
27
A
57
A
28
A
58
D
29
B
59
e
60
A
42
Í
The
0
under
area
TI-IE CORRECT
2.
is detennined
a curve
J1..NSWER IS?
The characteristic equation for
a
by integration, not diffe:te:atiation.
(D)
first-order linear hornogeneous differential equation is:
r+5=0
·wl11ch--~ªh"s
_,
"
-
1'
Ctl ··oo-•l a-·d,(--_J,
t:::
Refer to Differential Equations n1 the Mathernatics section of the
The fon11_ of the solution is then:
Ce
=
y
-at
vvhere
=
a
a
and
a=
5
_FE
Supplied-Reference _llandbook
for this problen1
C is determined from the boundary condition.
=
1
ce-S(O)
C=l
Th en,y
=
e
-5t
THE CORRECT
ANSWER IS: (B)
Refer to Differential Equations in the Mathematics section of the FE
Supplied-Reference Handbook.
The characteristic equation for a second-order linear homogeneous differential
equation is:
3.,
r2 + ar+
b
=
O
In this problem, D2 + 4D + 4
a=
4 and b
=
O, so:
4
=
In solving the characteristic equation, it is noted that there
Because
a2
=
4b, the solution for this
THE CORRECT
are
repeated real roots: r1
critically damped system is:
ANSWER IS: (D)
44
I
11
I
I
I
illi
I
11
I
I
II
I
I
I
I
1111
I
I
I
Ill
-
II
=
__
?_?
¡I
I_
I_
I
11111
I¡¡
II
I
II
=
r2
=
-2
tOFtNli\lG So
V
60t-
=
C'"
A _r.1_-u
-:i
?
S'
=
-
-,,
__
Finally, the
4r1
-
£.
'10-'-l
-12r'2
_,_
rate of change of acceleration is:
A'= sur= 120
Vvhen t
24t
-
2:
=
A'=
120
24(2)
-
120
=
-
48
=
THE C,ORRECT Af?SWEit IS:
5.
The
72
(A_)
product of vectors A and B is
cross
l
J
k
2
4
o
1
1
-1
To obtain
=
a
i( 4)
-
j(-2
-
-
O)+ k(2
4)
-
=
a
vector peí1Jendicularto
-4i +
2j
-
and B.
A_
2k
unit vector, divide by the magnitude.
Magnitude= ?(-4)2 +
-4i+2j-2k
-2i+
2?6
THE CORRECT
6.
UTI
22 +
(-2)2
=
J2A
=
2?6
j-k
?6
ANSWER
IS: (C)
Refer to Differential Calculus in the Mathematics section of the FE Supplied-Reference Handbook.
THE CORRECT
ANSWER IS: (C)
1
:
1111
I
1
\
1111
I
111
I¡
I
I
,...
-?
I
'e-'-,•·~e
1111
J_,/
-
.,,
CL
...:t
•
rr-
•
..
.-tL_l_ l
(ll·'-·1•e·•eclt'1n':
1_
CL-l
1
••
....
•
.
-
(
.,
"1--,·-·•~.
u tl 1 fr,_} -.w·:
·
_
1-..,11
1_,,..
l.!.J.
[,p_-'•-i.gLIJ?
\_?"
1L
1
y
1
l
1
=
=
dy
fdA fxdy fy312
·
·
o
o
THE CORRECT
=
y
5/2
5/2
l
2
5
o
Ai?SWER IS: (A)
8.
X
y=-x
y=x
y=-2x+I
From graph one, the intersection is at (O, O),
so
Also, the second intersection is at (1, ?1),
the vertices
THE CORRECT
so
ANSWER IS: (A)
46
Options (C) and (D)
are
are at (O, O),
incorrect
(! ?),(1, ?l),
,
9.
xdx
}10sin
10
=
o
=
1
Ü
[
:-cosxi(
COS
-
-
(
TC
-
COS Ü
JJ
=10[1+1]
=20
10.
Pkccuracy increases with increasing sample size.
THE CORRECT
ANSWER
IS: (B)
The key word is OR. What is the probability that the number is odd OR greater than 80? Refer
to Property 2 given under Laws of Probability in the Engineering Probability and Statistics
section of the FE Supplied-Reference Handbook.
P(A + B)
=
P(A)
+
P(B)
P(A,B)
-
Event A is removing a ball with
P(A)
=
50/100
=
0.5
Event Bis removing a ball with
P(B)
=
20/100
=
odd number,
an
a
number greater than 80.
0.2
Event A,B is removing a ball with
There are ten such balls.
P(A,B) I 0/100
=
Also
P(A,B)
=
P(A)
P(A + B)
x
=
odd number that is greater than 80.
0.1
P(B)
0.5 + 0.2
=
an
THE CORRECT ANSWER
-
=
0.5
x
0.2
(0.5
x
0.2)
=
=
0.1
0.6
IS: (C)
47
I
I
I
¡1
I
I
I
I
(
-3
7
=
a
../
3
·12·s,
1
...
12
3
8-15.5=7.5
1
7.5
3 stan d ar d
=
--
?
2.5
••
-i
oev1ations
Fron1 the Unit N on11al Distribution table in the
Engineering Probability and Statistics section of
the FE Supplied-ReferenceHandbook.
For
x
=
3,
R(x)
0.0013
=
THE CORRECT
ANSWER IS:
Pv=nRT
1
(l)(v)
v
=
=
(1)(0.08206)(546)
44.8 L
THE CORRECT
2 moles of As
15.,
x
75
THE CORRECT
ANSWER IS:
g/mole of As= 150 g of As
ANSWER IS: (C)
I
-
-
-
!!7!!!!!
I
I
I
I
_¡_
I-I1SOLi + 2 1\J::tOI-I -+ l"Ja2SOL1 + 2
-
'
'
0.5 lv1 fbS04
=
1.0 }\J H2S04
aO I-I
=
1. O
í. O IvI
J',J
40 rnL of LO }\J
·v
,JQ
/-...,,
I
x
17.
1
=
.L1..
fhSO4
aO I-I
=
60 rnL of
x
l\J
1--JaOI--:1
..Iv
6Qy
40/60
=
1'"4 l'-,T
H20
0.671\J
=
=
0.67 NI 1\JaOI-I
The n1olecular ·weight of 1?aOH is 40 g; therefore, 4 g/L of NaOI-I will fon11 1 L of 0.1 normal
NaOH solution. One liter of O. I nor111al HCI solution is required to neutralize the NaOH.
THE CORRECT .A.r?s-,:vER_ IS: (C)
Refer to the Che111istry section of the FE Supplied-Reference Ffandbook for the equilibriun1 constant
of
a
chemical reaction.
4A + B
2C
-?>
+
2D
THE CORRECT
Step
VAR
1
O
2
2
4
6
3
4
IS: (B)
ANSWER
EXIT LOOP
At the conclusion of the routine, V AR=
THE CORRECT ANSWER
6.
IS: (D)
49
I
I
I
I
I
/
4
6
o
5
+
2
+
()
b
+
7
Step
I
z
z
1\J
z
N
N
1
I
I
1\J
z
1
I
}'J
2
3
T
C'
K_
u
1
1
I
(NEXT I()
l+
-----------------------------------------------------------------
2
z2
1
+ z +
2
z2
2
(NEXT I()
-----------------------------------------------------------------
3
z3
1
(2)(3)
+ z +
z2
2+z3
(2)(3)
(NEXTK)
-----------------------------------------------------------------
4
z4
(2)(3)(4)
1
+
z
+
z2
2+ z3
(2)(3) + z3
(2)(3)( 4)
:.
The sequence
z
.
1s: S
=
I +
-
1!
+
z2
-
2!
+
z3
-
3!
+
z4
-
4!
+
...
+
zN
-
N!
Rorvs
Cohnnns
e
I
B
1
2
2
1
3
n
F,
6
o
o
10
5
7
9
64
k
4
:J
Q
-¡
_7
5
D3: Dl + D$1
?:?
D4: D2 + D$1
?:?
THE CORRECT
D2
=
8
D3
=
7 +
+ 8(7)
=
6L!,
=
519
AJ·{S'\iVER~ IS:
(C)
8(64)
Refer to the NCEES Rules of Professional Conduct, Section A.4., in the Ethics section of the
FE Supplied-Reference Handbook.
THE CORRECT
ANSWER IS: (B)'
Refer to the NCEES Rules of Professional Conduct, Section B. L, in the Ethics section of the
FE Supplied-Reference Handbook.
THE CORRECT ANSWER IS: (A)
Refer to the NCEES Rules of Professional Conduct, Section B.3. and Section C. I., in the Ethics
of the FE Supplied-Reference Handbook.
CORRECT ANSWER IS: (A)
to the NCEES Rules of Professional Conduct, Section B.5. and Section B.6., in the Ethics
of the FE Supplied-Reference Handbook.
CORRECT ANSWER IS: (A)
¡:·
.J
1
l
,..,$
·-·
-
...
I
$0.50 (5,000)
=
ORNING S<)LU
CH\!
$2,500
P:..nnual savings= $7,500
Additional invest1rLei1t
=
$2,500
-
$15,000
Payback= $14,000/$5,000
=
-
=
$5,000
$1,000
=
$14,000
2.8 years
TI-IE CORJtECT ltN"S"\IVER. IS: (A.)
Annual cost:
=
$900(A/P, 8%, 5)
=
$900(0.2505)
=
$225.45
=
$224.30
+
+
+ $50
$50
-
-
$300(AIF, 8%, 5)
$300(0.1705)
$50- $51.15
THE CORRECT ANSWER IS: (A)
29.
i= 12%
40,000
A
o
1
2
3
?
?
?
4
A
A= F (AIF, i, n)
=
THE CORRECT
40,000 (AIF, 12%, 4)
ANSWER
IS: (B)
=
$8,369 per year
..
·_-¡_,.:
..
30,
1,000
1,000
1,000
A
o
1
J
t
11
l
2
3
4
"if
-400
y
-1,l
P
/,
1201
,/0
-
=
-1,100 -400 (P/F, 12%,
=
-1,100 -400 (0.8929)
=
687.34
=
p·
?
=
,-o;
(AIP' LL I o'
--
¿_I\= \Jo
c:o7
•)
1) + 1,000
+ 1,000
,,
.
_)
Ll(O
,·
.
(PIF, 12%, 2)
(0.7972)
+ 1,000
+ 1,000
(0.7118)
(P/F, 12%, 3)
+ 1,000
+ 1,000
(PIF, 12%, 4)
(0.6355)
./
3?º2)
=
$226 per year
THE CORRECT
l.
-
00
,
J-1
:
ANSWER
IS: (D)
The easiest way to solve this problem is to look at the present worth of each alternative.
The present worth values are all given by
P
First Cost+ Annual Cost x (PIA, 12%, 8) Salvage Value x (PIF, 12%, 8)
First Cost+ Annual Cost>< 4.9676 Salvage Value x 0.4039
=
-
=
Then
-
P(A)
=
$63,731
P(B)
P(C)
P(D)
=
$63,392
=
$63,901
=
$63,222
The cash flows are all costs, so the most preferable two projects, those with the lowest present
worth costs, are B and D, and the difference between them is $170.
THE CORRECT
ANSWER
IS: (B)
53
I
I
11
J'1'onnal to the pla:t1e:
-
O: JV
I,F11
1ng cos
-
=
cp
O
\
fJi'r=
O:
•
.·.
-1ng
-
,l
sin
1ng sin?+ pJV
+ µ1ng cos
iy;
I
=
v;
O
/.,,,
µN ( ai impending slip)
5<
mg
JV
r,
=
u
?o
º
I
s111rn
'Y
=
tancb
=
I
?l
COS<p
tan?= 0.25
33.
THE CORRECT
ANB,VER IS: (D)
:í:Fy=0=-120+
-A.
A=
4
120N
5
150 N
B
5
4
3
A
THE CORRECT
34.
ANSWER IS: (C)
12
Ry= LFY
Rx
=LFx
=
(260)
13
+
I(300)
5
-
=-2_(260)
13
+
4
5
50
=
370
(300) =140
12
I
I
'(
I_
R
=
,
3961\T
THE CORRECT
-
-
4
-
-
-
-
-
50N
213
4
ANSWER IS: (D)
'-)j
__,
.
_
-
-
-
-
-
-
-
-
-
-
-
-
I so
'f
]\?
=
50 Ff
]YA
i
BLOCKB
?-
LFY
I so
JVB
'/¡/
!
=O= -50=
80 +
JVB
1301\T
BLOCKC
LF_v
¡100
?-
Ne
=O= -130-100
=
+ JVc
2301\T
µNC
j1
I Ne
Assume Blocks A and C re1nain stationary.
50
>
0.3(50)
F'
LFX =O= -F' + 0.3(50) + 0.3(130)
<Sl!---
F'
:>
JI
54 N
=
0.3(130)
130
Assume Blocks B and C
move.
,¡5°
-0.3(50)
F"
BLOCKB
LFX =O= -F11 + 0.3(50) + 0.3(230)
F"
BLOCKC
-?0.3(230)
fl
230
:.
F
THE
=
54
1\T
where A and C reinain stationary.
CORRECT
ANSWER
IS: (B)
55
=
84 N
f')
,...
;.. i
"'r-.?
_,... 1·1""J1
250
3b
.e,;_
Pu=
500 sin 30º
f
Mp= 250(0.30)
250
=
433
433(0,
-
1
O)
=
31 .7 N,,n1 ccvv
10.0 cm
30.0
-..J
1¡,,.,-:,/,..
1
Zero-fo-·ce
rnen1bers nsnally ncr-·L11·
v
v
-
,
_
·-
.L
----
-
--
-
--
,
e
a·c·
?1n·L'·s -v-v:l1e-·e
l
1
11_ l_
·-1e·'11bers
J:v_
_
cm
are c.___
.L
allg·¡1e-cl
__
_
•
--
e
?-s
V
f?Ilows,
..
.e
,
A
...
That is, joints where two 111embers are
along the sa111e line (OA and OC) and the third n1en1ber is
at son1e arbitrary angle. That n1e1nber
(OB) is a zero-force n1eraber because the forces in OA and
OC 1nust be equal and opposite,
For this specific proble1n,
we
in11nediately exarnine Joints Band E:
e
B:
E:
e
B
A
G
BG is
a
zero-force member
CE is
a
zero-force me1nber
Now, examine Joint G. Since BG is zero-force member, the
joint effectively looks like:
?
A
/e
?;}
G
F
and, therefore, CG is another zero-force member.
Finally, examine Joint
looks like:
C. Since both CG and CE are zero
D
B
F
and, therefore, CF
members.
THE CORRECT
force members, the joint
effectively
1s
another zero-force member. Thus, BG, CE, CG, CF
are
the zero-force
ANSWER IS: (A)
c6
.J
I
I
I
11
I
I
I¡
I
I
11
I
11
I
I
I
11
I
38.
By definition of
anu
;J
10
3.
9
•
rt
cantilever
a
_oa
1s
1
bean1'.)
.
?
•
d ea on,_y
1
·
at a
.
("
it is noír statically indeterrrii:n_ate, it is con1pletely supported0
spec1nc poínt.
6
lTI
()
10 kl\f
x=6m
12
.A.rea
1
}\.rea 2
=
=
=
13(2)
26 kN•111
2
@
G)
X
®
14
------
SHEAR DIAGRAlVl
6( 6)
=
18 kN
=
16
..
rn
2
A_rea 3
=
,4( 4)
l<l'-J·•n1
-16
-26
MOI\1ENT DIAGRAIVI
Maxünur.o. magnitude of the bending 1non1ent is 26 k}J "rn.
THE CORRECT
ANSvVER IS: (D)
(
40.
LF =PA
=
(1.4 106)
x
rr(O.S)
2?
4
\_
.F;-0d
=
275 kN
=
THE CORRECT
aA = 68
x
=
F:oc1
)
106 A
P=l.4MPa
ANSWER IS: (A)
C"'
,=--¡
,_.,
I
)
/
?m
........
,
Ji
T-
Tr
?
d
2
l.==-=
4
J
nd·
32
T=
1[ d3,,,
L
T[(O
\
16
T
42,
=
•
r¡)3 (8¿jl_"v''
\
L.,
,
V
/'-.
1
_
\,
V
n3)
Ir:,
v
1,319?N?1n
The Euler foíTD.ula is used for elastic
stability of relatively long colun111s, subjected to conceniTic
axial loads in co1npression.
THE CORRECT AJ?s,vER IS:
(D)
43.
The question statement is the definition of hot
vvorking.
THE CORRECT ANSWER IS:
(A)
44.
By definition, a metal with high hardness has
high tensile strength and a high yield strength.
a
THE CORRECT ANSWER IS:
(D)
45.
By definition, amorphous materials do
not have a crystal structure.
THE CORRECT ANSWER IS:
(D)
46.
Aluminum is anodic relative to copper and,
therefore, will corrode to protect the copper.
THE CORRECT ANSWER IS:
(B)
I
I
I
I
II
I
I
I
47
The
111ean
pressu:te of the fluid acting
.
?
ot pressure
,.,
.r
f
...
1s
=
J o:c
211.
-
•
1
e
the ne1g11t uon1
H11 J
Pg- (----\
=
2
1
i,
T,
-
fi=-=
Ff
3
=
3
-
º
and its point of application is 1.00
1::1(3)-Ji10)
"
l
t_1e
,
u \7 ºO
o 7J
6noro
1
the gate is evaluated at the n1ean height, an.d the center
,. ,
tne ?=11-nci 1s:
l
or
1orce
the
tota_
top; thus,
on
2
(3)
=
,..,
.
-
.
7 O, 61 O l\T
above the hinge. A n1on1ent balance about the hinge gives:
n1
o
70, 61 O
=
2,_
3-,.
f,
'Jc::37 -,J
L=I
'
3
TI-IE COR_RECT AJ?SvVER IS: (C)
48.
Refer to the Fluid Ivlechanics section of the FE Supplied-Reference Handbook.
where
dv
=
-
1:1
=
shear stress and
rate of shear defonnation
dy
Hence, µ is the ratio of shear stress to the rate of shear deformation.
THE CORRECT
ANSWER IS: (C)
49'o
()
=
A1
V1
=
(0.01ín2)(301n/s)
']
n
__,.:J lL
--•13/n
-
;::,
Q1º·"r-e
Llv
u
fl-,e
1..H
yp8-'-te¡;•
vv
,-,1.
j?ef ?s
1,
.
rPr?p·;1rF1·?11"1·
a.!_
_p'-'--'--•----'-'?
r,••·,1y,
vl_
a10-f-'i1e·c·1·eil
L
?-
JI.
.i.J,..,
_
·t'-?¡·1e
_
n1ust deflect the total hor:tzontal
f',·n,?.·1·r-,
-'º~-__,f';
n10:rnentun1 of the vvater.
F
pQV= (1,000
=
TI-IE CORRECT
50.
Flow through
kg/n13) (0.3 n13/s) (30
AJ\{S'WER_ IS?
111./s)
=
9,0001\J
=
9.0 kl,J
(B)
insulated valve closely follows
constant enthalpy.
an
a
throttling process
THE CORRECT ANS'WER IS: (B)
51.
pv2
=
z
:. h
gh(P-Pair)
pv
=
2g ( P
2
2
?
-
Pair )
THE CORRECT
52.
:!_
2g
?
( 2)2
( 2) (
9.
?
0.204
n1
8)
ANS"VER IS: (D)
S = apparent power
P
=
real power
Q
=
reactive power
s
=
cos
Q
=
P + JQ
e=
=
1s1
cos e +
i
ISi sin e
pf= o.866
(1,500
VA) sin[cos-10.866]
=
750 VAR
THE CORRECT ANSWER IS: (B)
60
111
11
I
.
.A"'
throttling process
is
ctt
I
-
T
30
40
5
30
Vc(t)
1
tJ
We (.,,
150
=
1
rr2
=-\._,Ve=-2
2
50
Ox/ 10-6
2
We= 0.113
initial stored energy.
J
·
Afcer ten tin1e constants,
AI'l'SWER IS:
THE CORRECT
Rr=
4Q + 3Q
Ry=
6
Q
?
IT
II
6Q
=
be dissipated.
energy
4Q + 2Q
=
6V
-
I
l A
=
6Q
6V
I
TI-IE CORRECT
-
_(1 i)2
(1 +
i)2
1
-
ANSWER IS: (A)
2i + i2
1
-
1
1
-
1
-
2i
=
-
1
+ 2i + ¡2
THE CORRECT
-i
=-=-1
+ 2i
ANSWER IS: (A)
l
57
As vapor escapes, the n1ass within the tank is reduced. \A/ith constant
volu:rne, the
volurne within the tank n1ust increase. This can happen only if liquid evaporates.
'1·
PruP
?\..
-.v~¡\
?.i--
:?1
1//////
1/
1/,
1/
1////(/
?
?
VAPOR
j
?
1/
1/
/
LIQUID
1/
?
?
1/
1///////////?
58?
By definition,
an
adiabatic process is
THE CORRECT
59..
a
process in which
heat is transferred.
no
ANSWER IS: (D)
An isentropic process is
one
equation:
for which the entropy remains constant. Entropy is defined
by the
T
ds=(ººJ
reversible
The entropy will be constant if 8Q O and the process is reversible. It is
theoretically possible
for a nonadiabatic, irreversible process to have a constant
entropy, but this is not one of the
responses. Option (D) describes a state, not a process.
=
THE CORRECT
60.
R
R
=
=
(MW)
8·314
0.1890
=
44
THE CORRECT
r
ANSWER IS: (C)
kJ
kg°K
ANSWER IS: (A)
I
I
I
11
I
II
I
I
I
¡I
11
I
I
I
111
?-
II
-á!f'
I
II
-
I
-
-
.
I
=
iiiiii-
!!•
E'rA
-?Á
;?
...
F{.
SPEC:IFIC./-tTI
THE AFTERNOON
NS
SESSI
IV!ECI-IANIC.PtL
RNGn,rEERH'FG
(60 questions in 8 topic aTreas)
Topic Area
lVIechanical Design and Analysis
A.
Stress analysis ( e.g., combined stresses,
torsion, non11éd, shear)
B. Failure theories ( e.g., static, dynamic,
buckling)
C. Failure analysis ( e.g., creep,
fatigue, fracture, buckling)
D. Def or=-mation and stiffness
E. Co111ponents (e.g.,
springs, pressure vessels, bea111s, piping, bearings,
colunms, power screvvs)
F. Power transn1ission (
e.g., belts, chains, clutches, gears, shafts, brakes,
axles)
G.
Joining (e.g., threaded fasteners, rivets, welds, adhesives)
!-L IVlanufacturability (
e.g., fits, tolerances, process capability)
II.
I.
Quality and reliability
J.
IV[echanical syste111s ( e.g., hydraulic,
pneu1natic, electro-hybrid)
Kinematics, Dynamics, and Vibrations
15%
A. I(ine1natics of 1nechanisrñs
B. Dynamics of rñechanisms
C.
Rigid body dynamics
D. Natural frequency and resonance
E. Balancing of rotating and
F.
III.
reciprocating equipment
Forced vibrations ( e.g., isolation, force
transmission, support rñotion)
Materials and Processing
A. Mechanical and thermal
properties ( e.g., stress/strain relationships,
ductility, endurance, conductivity, the1mal expansion)
10%
B. Manufacturing processes
( e.g., forming, machining, bending,
casting,
joining, heat treating)
C. Thermal processing (
e.g., phase transformations, equilibria)
D. Materials selection ( e.g.,
metals, composites, ceramics,
plastics, bio-materials)
E. Surface conditions ( e.g.,
corrosion, degradation, coatings, finishes)
F. Testing (e.g., tensile,
compression, hardness)
IV.
Measurements, Instrumentation, and Controls
A. Mathematical fundamentals
( e.g., Laplace transforms, differential
B.
C.
D.
E.
10%
equations)
System descriptions ( e.g., block diagrams, ladder logic,
transfer functions)
Sensors and signal
conditioning ( e.g., strain, pressure, flow, force,
velocity, displacement, temperature)
Data collection and
processing ( e.g., sampling theory, uncertainty,
digital/analog, data transmission rates)
Dynamic responses ( e.g., overshoot/time constant, poles and zeros,
stability)
64
I
I
I
II
I
I
I
,._
111
I
---;;,.-
I
I
I
11
11
I
II
111
I
I
I
111
I
I
I
11
1111!
I
111
I
111
1111
I
Thermodynamics and Energy Co:nve:tsioB
f'roeesse::::
Ideal and real gases
B. F?eversibility/ineversibility
C. The1n1odynan1icequilibriurn
A_.
D. Psychro111etrics
E. Perfo1n1ance of con1ponents
F. Cycles and processes (e.g., Otto, Diesel, Brayton, R_ankine)
G. Co1nbustion and combustion products
H. Energy storage
Cogeneration and regeneration/reheat
T
VI.
Fluid Mechanics and Fluid Iv1achinery
A. Fluid statics
B. Incompressible flow
C. Fluid transport systen1s (e.g., pipes, ducts, series/parallel operations)
D. Fluid n1achines: incon1pressible ( e.g., turbines, pumps,
15°/o
hydraulic motors)
E. Co1npressible flow
F. Fluid machines: con1pressible ( e.g., turbines, co1npressors, fans)
G. Operating characteristics ( e.g., fan laws, performance curves,
efficiencies, work/power equations)
VII.
H.
Lift/drag
I.
hnpulse/1non1entum
Heat Tran sf er
A. Conduction
B. Convection
C. Radiation
D. Composite walls and insulation
E.
Transient and periodic processes
F. Heat exchangers
G. Boiling and condensation heat transfer
VIII.
Refrigeration and HV AC
A. Cycles
B. Heating and cooling loads ( e.g., degree day data, sensible heat,
latent heat)
Psychrometric charts
D. Coefficient of perf onnance
E. Components( e.g., compressors, condensers, evaporators,
C.
expansion valve)
10°/4
I
1
I
I
¡
I
11
I
I
11
¡
1111
I
I
I
I
11
11111
._
1111
I
I
I
I
I
E:C;HA?JIC;t\t.
AF-rERNOON
SA
PLE QLJESl?I
CONTAINS 30 QUESTIONS,
HALF THE NUMBER ON THE ACTUAL EXAMINATION.
THE AFTERNOON SAMPLE EXAMINATION
67
I
I
I
I
11
111
-
I
I
I
111
¡I
11
I
I
I
I
¡I
"
•
.
•
.
?
-,
?
1
O
spnng constani: o-f 3 o. 5 2 5 J??T/111111 and a íree ?engttt
The force (l\J) required to cor11press the spring to a length of 125 1rir11 is n1ost nearly:
.
A hellcai con1press1on sp:nng has
1
(A)
1,500
(B)
(C)
(D)
2,500
4,800
6,500
1¡tions 2-3: l\_
pivoted lever
arn1
a
1
is in equilibriu111 under the force of
a
or
r-
"
A
1
'3
o nri:l-:t.
con1pression spring and
an
air
nder as shovvnbelow.
SPRING FORCE
?.
V
rcn 1.n71?o?o?c_co1
A
F
450mm
650 mn1
NOTTO
SCALE
AIR CYLINDER
FORCE
piston dian1eter of I 00 n1m, and the compression spring exerts a force
3,333 N. The pressure (kPa) in the air cylinder required to hold the lever ann in equilibrium is
The air
cylinder has
a
n1ost nearly:
150
270
294
305
(A)
(B)
(C)
(D)
If the piston diameter is reduced to 90 mm, the required release pressure will change by
of most nearly:
(A)
0.76
(B)
0.87
1.14
1.23
(C)
(D)
O
by NCEES
69
a
factor
GO OM TO THE NEXT PAGE
-
-
lVIATERJAL
E
=
210
V
=
Ü.24
a
=
=
8>,
10.5
x
E
--
6-,•~.J
-
1nm
610111111
7-¡
TIES:
t
103
/
x
/
I0-6/ºC
400 IvIPa
/
/
/
/
--
-?/
/
/
p.
/
I
l ' 000111rn
/
/
/
/
-?
/
/
/
/?
/
?/
I
VERTICAL-AXIS
PRESSURE VESSEL SECTION
u
:i
..
If the interna] pressure in the vertical axis of the cylindrical pressure vessel shov1n is 600 kPa, the
and B is rnost nearly:
hoop or tangential stress (Iv1Pa) in the vessel ~wall betvveer1 cross sections
P.}__
o
{A'j
\-
·1
_,
-
'
o i!,
-
36,8
(B)
'I
pc7·
147
(D)
5.
t::
J.,.J
(c\J
Assu111e the internal pressure is changed in the pressure vessel
stresses in the wall between cross sections A and B:
cr1
46.2 MPa
cr1
23 .1 I\1:Pa
CTr
Ü
The increase in length (111m) of the distance between
(A)
0.06
(B)
0.11
(C)
0.19
(D)
0.22
201
O
71
by NCEES
I
I
cross
so
that it produces the
following
sections A and B is most nearly:
GO ON TO THE NEXT P,l\GE
Questions 6-7:
object ,.vith
the object is
A11
below. Attached to
-,i0
whe-'"\
L_ ,:-he
l
obiect
1.0 a'-l Poi-1'_LL
J
_
_,
0
J..
,
of L50 kg 1noves without friction in a circular path as shovvn
a spring with a spring constant k of 400 N/rn. The
spring is undefon11ed
?Q
and the speed of tlie obiec·l e'lt Poi-·1-'LL O
--n/Q0,
J
< 10 ? 00 l?a 1r1ass 1n
--
_
_
?••
_,,_
_
k= 400 N/rn
125mm
\,
•.
,,,,\
NOTTO SCALE
6.
7.
The translational kinetic energy (J) of the object at Point Q is most
nearly:
3.00
(C)
6.00
(D)
29.40
The horizontal force (N) of the spring on the object at Point Q is most
nearly:
(A)
(B)
100
175
(C)
250
(D)
400
l
r-opvr·1gh.,,?0·1
/
,..,.:
-
1.50
(A)
(B)
·
E=-.-
_
(-'
,,cl
J_
b·J' N''E
·
,
"
..
...
,,...t,,,?
"?";,_,_-,,,
72
GO ON TO THE NEXT Pf.\.GE
,
,,
Quesuons
0
T1ne
_,_
s1nooth incline fo:t
5
1
2 -Kg
r11
.
,r
l
1
o_oc~z
s_1own
1n
-¡
.
t h_e f'1gure b e 1 ovv
until it clears the top of the rainp at
'
-,
1
a
r
acce1eracea rroTrt
1s
'
res{
-i
oy
r
rorce
speed of 8 nils.
?
ff
)?
?
#
/
/////////////////////////P-
8.
9.
The value of F
(A)
(B)
11.8
19.6
(C)
24.6
(D)
69.4
(J\T)
4.5m
Y
"--x
•
is most nearly:
The highest elevation h (m) above the x-axis the block will reach is most nearly:
(A)
(B)
(C)
(D)
1.2
3.3
5.7
7.8
,
Copyright 201 O by NCEES
73
GO ON TO THE NEXT PAGE
Constants for
Solute
Solvent
Carbon
Carbon
1 Oª
11.,
The
I
l\fickel
fee Iron
bee Iron
fee Iron
Copper
P.Juminun1
Copper
Carbon
Copper
hep Ti tani un1
7.41
X
1Q-l3
(B)
2.91
X
(C)
8.69
X
10-ll
10-ll
(D)
2.05
X
10
l,200ºC
l,500ºC
(C)
8,200ºC
8,500ºC
Copyí'igM 20"i O by NCEES
I
11
I
I
0.2
><
""2
2.
X ·¡o-11l
Q
34,000
29,300
67,000
30,200
47,100
43,500
1 O
X
10-4
10-4
.
at l ,000ºC is
(cal/niol)
·
1110s t
nearly:
-9
(A)
(B)
I
-1
0.15 X
0.2 X 10-4.
-L1
5.1 X 10
The temperature at which carbon has the
most nearly:
(D)
D0 (m2/s)
0.77
diffusivity (n12/s) of carbon in iron
(A)
Diffusivity
diffusivity in
same
74-
-I
fee iron
as
it has in hep titanium is
GOONTOTHENEXTPAGE
I
I
I
¡I
I
I
I
I
I
I
1
I
I
I
I
-1
1
I_
I
I
I
11
'-.__/
o
?
----
?
z
o
?
1--1
1-1
[I)
o
'"O
?
o
o
?
o
o
I=
o
00
O\
o
..¡::,,.
o
N
o
o
N
o
o
..¡::,,.
O\
o
o
00
(JQ
..¡::,,.
o
o
O\
o
o
.__]
o
o
--?
t,:)
+
Q
?-?0?
Ui
o
o
o
\O
N
\
?
o
00
.__]
I-'
00
N
00
00
r-1
+
t,:)
\
?
\
\\
+
?Q
\O
o
o
--------u
o
o
00
TEMPERATURE (ºC)
\
\
I-'
00
o
?
!?
\
o
o
I-'
V
1--'
w
+
R
+
+
~c::o
---.]
13,,
auton1atic controls block diagran1 is shown belo1N:
Ac-11
R(s)
C(s)
The single ele1nent relating the input to the output is best
represented by:
Copyright
(A_)
R(s)
C(s)
(B)
R(s)
C(s)
(C)
R(s)
C(s)
(D)
R(s)
C(s)
201
O
by NCEES
GO ON
TO THE MEJ'Cf PAGE
.•
•
,
-,,
i
-,.
1.::::--JL:,:
Quesnons
•
.
,
resrntance 1e1nperature detector
,
1?.
?
r?
\JtTV J
prov1 d es
.
.
-
,
res1Stance otrcput L1at
.
H
1
is
.
,
.
relRceo. ro
1
hu·
te·-n-}erature
J
lLl
·-
-
D
R-
_
•
1\.0 -Ll
_
T
I
1
Ot-err
rriy¡
101_,
-
Whe 1·e·
..
,
--
R
R esistance, Q
Ro
Ref eren
a
Coefficient, 0c-1
T
Ten1perature, ºC
T0
R ef eren
..
..
ce
ce
resistance, n
te1nperature, ºC
Consider an RTD with R0
14.
100 Q, a=
0.004 °c-1, and
The change in resistance (Q) of the RTD for
(A)
(B)
(C)
(D)
15"
=
a
T0
=
0ºC.
l0ºC change in terr1perature is n1ost nearly:
0.04
0.4
4.0
100.4
The RTD resistance (Q) at
(A)
(B)
2
(C)
100
(D)
200
a
temperature of 250ºC would be most nearly:
1
Copyright 20'1 O by NCEES
77
GO ON TO THE NEXT
P,11.GE
Questions 16-18: The pun1p shovvn in the figure is used to pump 50,000 kg of Tvvaterpet hour into a
boileL Pump suction conditions are 40ºC and 100 kPa. Pump discharge conditions are 40ºC and 14.0 MPa.
Boiler outlet conditions are 500ºC and 14.0 IvlPa. The boiler efficiency is 88%.
(D
T= 40ºC
P= l00kPa
í
?
I
I
.
Q)
--1
BOILER
? -1??iMPa
=
17
G) T
=
P
=
500ºC
14.0 MPa
88%
FLOW= 50,000 kg/h
16.,
(MPa)
(ºC)
14.0
14.0
14.0
336.75
336.75
14.0
500
40
h
Condition
(kJ/kg)
Compressed liquid
Saturated liquid
Saturated vapor
Superheated vapor
167.6
l,57Ll
2,637.6
3,322
150
190
240
300
The total heat transfer (MW) to the working fluid that
(A)
(B)
(C)
(D)
18..
T
If the pun1p efficiency is 80%, the pump power requirement
(kW) is n1ost nearly:
(A)
(B)
(C)
(D)
17.
p
occurs
in the boiler is most nearly:
10
15
29
44
If the coal used to fire the boiler has a heating value of 28,000
kJ/kg, the rate at which coal is
burned in the boiler (kg/s) is most nearly:
(A)
(B)
(C)
(D)
1.39
1.58
1.78
2.03
78
GO OM TO THE !\JEXT PAGE
Air is to be considered
Que§tio:n? 19-20:
,-,
I
._,,
p
1
-
.
as an
ideal gas vvith the
following prcperties:
o-? _'\-.
kJ1/tk
Y;\
\. -b
nv,
--
O
'-T'\
KJ'//'\kgwJ:-,-}
Cv
ílJ. 7
k
1.4
R
0.287 kJ/(kg•K)
"I
!CJ
1
s
One kilogram of air at 172 kPa and
344 kPa.
19.
00ºC is heated reversibly at constant vohnne until the pressure is
The specific volu1ne of the air (kJ/kg) at State
(A)
(B)
0.17
0.62
(C)
0.93
1.28
(D)
20.,
1
1
is
The change in entropy [kJ/(kg"l()] between States
(A)
(B)
1110s t
1
nearly:
and 2
(s2
-s1) is most nearly:
-0.498
O
(C)
0.498
(D)
0.693
70
,/
I
I
GO OM TO THE NEXT P,l\GE
I
1
11
Questions 21-23: In the figure below, the pipe is steel with an internal dian1eter of 100 DJ.In. Vlater is
pu1nped through the system; its velocit-y at Point C is 2.5 1n/s. The pressure at Point Ac. is atlnosphe:tic,
the gage pressure at Point Bis 125 kPa, and the gage pressure at Point C is 175 kPa. The
discharge at
Point D is to the atínosphere.
WELL ROTJl'--TDED,
í'
L,
'Viscosity, µ
K.inen1atic viscosity,
L _t)
De·1s-¡•-,__,
-
J
v
o
LO
x
LO
X
~
4
=
l. o
-•'1:-1 Ü
10-3 l\T?s/1n2
1
?t:=1
o-6 n12/s
1,000 kg/1113
I
J1
Lr?
I
17
6.o
e
111
c::J
D
@
U
Cy 'J_ J
T ,,,?P
\
E
0
®
(B)
lll-??
90º ELBOWS
,
1
)
(
CELBOW
=
o.9o
·11¡¡11
NOTTO SCALE
21.
The pu1npingrate
(C)
1.18
1.50
(D)
4.71
The pressure drop (Pa)
(A)
(B)
across
2,800
3,100
5,600
(D)
Ideally, the work that must
201
be
supplied to the pump (J/kg) is most nearly:
50
(A)
(B)
(C)
(D)
Copyright
each elbow is most nearly:
1,100
(C)
23.,
nearly:
1.02
(A)
(B)
22.
(1n3/min) is 1nost
125
175
50,000
O b?;r
NCEES
80
I
I
I
I
I
I
GO ON TO TME NEXT Pl,\GE
24
The centrifr1gal pump shov;n in the figure is to deliver 40 kg/s of v1ater
tained at O k Pa to a deaerating heater n1aintained at 200 kPa.
f:¡_-on1
H
condenser
111ain-·
1
suction, refeíTed to pun1p centerline
Friction head loss in suction line
Diaineter of suction line
Dian1eter of pump discharge line
ElevatioD head on discharge into heater, referred to pun1p centerline
Friction head loss in discharge line, including valves
Elevation head at pun1p discharge, referred to pun1p centerline
Elevation head
The total head
(111)
on
0.60
15
lll
Cí11
10 cm
20111
25 111
0.5 n1
at the discharge of the pun1p is n1ost nearly:
1.3
21
46
66
(A)
(B)
(C)
(D)
HEATER
?
P2
,,
A
10-cm DIA.?·
CONDENSER
P1= l0kPa
T1
+
Z1
=
'
5 m
t
7ºC
=
t
CENTRIFUGAL
PUMP
ttJS?TED
?
,
I
\
15-c,!n!A.I///
!f "''¡·)•'
£:::.1.?,I
r-f•
Convs•·1g',···'·
...
,""I
b·Y tllf'r.:C:?
l"':J"\.J?L."-c>
-
Z2
-J
,,
=
=
200 kPa
DENSITY=
958 kg/m3
20 m
?
,,
;__JI 0.5 m J
I
81
GO ON TO THE NEXT PAGE
.
.
O uestions
.
,,,.
P.'
l
2 5-L I :re1ate
to
.
-
•
('
0 A o "
o r--t
neat excnanger L1at ,vas d es1gne d to neat hqu1n vv-ater rrorn
1 J
Fo L to 1 ':::'.'V L
inside tubes using stemn condensing at 230ºC on the outer surface of the tubes. The
f:)llowing data
?
a
1
·¡
1
'
"
:l
-·
·
·¡
apply:
Data for compressed liquid vvater at l 70ºC:
S-·-erir'ic beaJ.,l
_IJ
__,_
-
-
Density
Dynamic viscosity
The1Tnal conductivity
Prandtl ntnnber
Tube rnaterial, copper
k
Tube I.D.
Tube O.D.
2.5
Cl11
3.8
Cll1
hi
6,000 W/(m2"I()
ho
12,000 W/(m2"l()
Outside fouling resistance
Inside fouling resistance
25.
For
26º
a
=
380 Vl/(n1·K.)
9 x 10-5 1112@1<.1vv
7
x
constant flow rate, the effect of
l 0-4
ni" I(J\"J./
fouling of the heat transfer surfaces
(A)
increase the temperature rise of the water
(B)
decrease the ten1perature rise of the water
( C)
increase heat exchanger effectiveness
(D)
make no change in heat exchanger effectiveness
is to:
The Reynolds number for the water flowing at 1.5 mis inside 3-m-long tubes with
is most nearly:
(A)
(B)
(C)
(D)
-
(A)
(B)
(C)
(D)
I.D. of 2.5
cm
53,000
106,000
212,000
424,000
The overall coefficient of heat transfer [W/(m2®K)] based upon inside surface
27.,
an
area is most
nearly:
1,000
1,100
4,250
4,500
GO ON TO
I
I
11
II
I
1111
TH¡:=,
NEXT P/:\GE
Question§ 28-30? P2- vapor-co1npression refrigeration cycíe using I-IFC-134a as the refrlgerant has the
pressure-enthalpy diagran1 shown belov1. The evaporator temperature is OºC, and the condenser
.
te1nperature
rn
4'Oº"r._,,.
p
3
4
LIOº"
r
L
OºC
h
28.
Assume the compression process is reversible and adiabatic. If the vapor entering the con1pressor
is saturated, the work done on the co1npressor (kJ/kg) is most nearly:
15
(A)
(B)
25
35
42
(C)
(D)
29"
The cooling producedby the evaporator (kJ/kg) is most nearly:
28
143
169
210
(A)
(B)
(C)
(D)
30..
The process 3-4 is:
(A)
constant entropy
(B)
constant
(C)
reversible
(D)
both constant entropy and enthalpy
Copy,·ight 20'i
O
enthalpy
9".1
0.J
by MCEES
I
lj
I
I
I
I
I
I
I
I
I
I
I
I
I
I
111111
I
11
I
I
I-
11
111
II
...
-------
--
111
I
I
I
1
?I
?I
CJ
CJ
u
?,-,,
-?
__¡¡_
.
L'
1
l ne 1orce
•
1
-
requirect to cnsplace
spring constant or rate. In this
free length
8
190 n1n1
65 1111n
-
-
con1ptessed
125 mn1
•
a.
sprnig
an
a111ount 8
('
rron1
.
?
-
'
rts íree 1engti1
.
?:s
--
-
"'
_fr'= kc,
-
1
vvt1ere
k
•
1
1s
tne
case:
length
The force required to deflect the spring this aruount is:
_F
=
k8
=
(38.525 l\T/n1í11)(65 n1n1)
THE CORRECT
2?
The lever
Drawing
A_f,lSVvER. IS:
A10
a
=
(650 1nrn)
(B)
( 450 111m) Fspring
-
450
O
?pring
"
II/Ill/I/I/I/////?
A
mm
3'
650mm
o
333 N
.///,;-;//
?
on
?iston
NOTTO
Fpiston
1C
d2
4
_
piston
1C
Fpiston
d2 piston
4
( 4) (2,307.5 N)
(100 mm)2
0.294 MPa
THE CORRECT
650mm
?
the piston will be the force divided by the piston
Fpiston
=
294 kPa
ANSWER IS: (C)
F
450mm
2,307.5 N
The pressure
=
=
3,333 N
650 mm
=
n
O:
450mm
Fpiston
=
Fpiston
pivot pin at
Fpiston:
=
_
}\T
will be in equilibriurn under the forces applied by the spnng and the piston.
free body diagrar.n vve have,
Solving for
p=
2,5Qij.
arn1
Sun11ning 1non1ents about the
L
=
area:
SCALE
.
·
,,,,
.:L
T·1· 1e
pressure
__
n-,-?essures
_\J-
t 1:1e systen1
.
r
rorce I area.
1s
diameter and p 2 be the pressure for
P1
and
P2
d2
7[-1-
=
,
ª2
n--
-----
100
_
2
p1
90
d2
·.
_
-'-
..
._,_,
1
111
pl
.LL·1,e
-L- e_,_L
_
-b-e-'
-
•
-,
1
ootn cases, tne :ratio oI
---··ec::s11re
,:-o-·
!
IJl u
1..
:e
"
ú
1nrn
1nm
-
90-n1n1 dian1eteL Then:
'\2
__
-1.-J2']
,
J
(D)
The fonnula for the hoop stress is:
IS (C)
ANSWER
THE CORRECT
The formula for the total
longitudinalstrain without
temperature rise is:
a
l
£axial
-v(crt +crr))= 210xl0
=_!_(cr¡
E
3
(23.1MPa-0.24(46.2MPa+0))=5.72xl0-6
-
This 1nust be conve1ied to displacementusing the following formula:
E
=
axial
8!
81,where l
is the
l
=
length of the section under consideration
£axial X l
=
5.72
=
0.0572
THE CORRECT
x
10-6
x
1,000
mm
mm
ANSWER
IS (A)
89
I
I
?
11
100-0-1·-,,
2
,A.i?"S'lvER IS:
TE.IE CORP?CT
4.
,
tl1e sarne
4
2
d1
_
"
1s
piston
4
p2
1
l?.
_,_
Fpiston
_
-
_
a
.
('
L1e rorce
1 ed
a11·amet0r so1ua
1
t-1'1p
--'--'
_
_
S 1nce
..
l
1
L l ·to
L
l
·1:01ve--sely
1J-·opo--?-L,_10·1a1
1
;.is
1.
_
111
u__,_
The kinetic energy T) when the object is at Q, is:
6º
T
7.
=
J1/2
;nv2
l/2 (1.5
=
1n/s )2 =
kg) (2
3 J
The horizontal force in the spring when the object is at Q is F
and 8 is the spring deflection. In this case:
o= length at Q
F
ko
=
=
+ U1 + f!Vl->2
T1
O
+
O
+ Fs
=
175
=
P
( 400 1,T/n1) (0.25 n1)
THE CORRECT
8.
length at
-
=
m111 +
100
2(125
1nnJ.)
-
175
=
ko where k is the spring constant
rnrn
250
=
n1n1
r,T
Al?SWER. IS: (A)
T2 + U2
=
1/2 mv2 + n1gh
SF= 1/2 (2) (8)2 + (2) (9.81) (3)
F=24.6N
THE COPJU:CT ANSWER IS: (C)
v,2
y
9•
2
=
Vyo
+ 2 as
r
2
Vy
=
=
O
j8
\
h
=
5.7
(3)"'2
-
I
)
5
-
2 ( 9. 81) ( h
-
4. 5)
m
THE CORRECT
ANSWER IS: (C)
90
I
I
11
I
I
11
11
I
I
I
I
I
I
¡I
I¡
I
I
111\1
=-
I
n
?r,
nr, -'.s
ll
·1C'c0
OU D' o Ck.!.
cl
UL.
-··-'tá. l' ·.ü ;s
000°r.__,,l ;l.!.J.:,..!.
-:i-r1d
r¡-•e .Co-·1 ·fer 1ºr'"'·1
T1r..,e
d1'_{'1··,ci1s1º0-n
l
v,
eql'"t?o-?
lVIaterials Science/Structure of Ivlatte:c section in the FR Supplied-Reference Iíandbook.
¡\
..
1)
.
..
l
-:-
_f---'d.,
?
.,_,
_
-
JL.
JL
):::'.'.,
V
---?
-
--
·-_
-,c-Ol11_·_._¡---,(1l
]011'
•l
---'
-
t111e
-0/RT
D· o e?
.,
-34,000
D
-··
-1
11.
0.2
=
e(1,273)(I.9S7)
x
(
10-4)
2.907
=
x
l 0-11
At what ten1perature does carbon have the san1e diffusivity in fee iron as in hep titaniur.n? The
diffusion equation is found in the IVIaterials Science/Structure of IVIatter section in the FE
Supplied-Reference J-Iandboolc
D=D
o
e-QIRT
Since D is the
sai11e
for fee iron and hep titaniun1
(Doe-QIRT).
tltanmm =(Doe-QIRT).
iron
.
-34,000
-43,500
x
(5.1
10-4) el.987xT
-21,892
T
e
0.0392e
=
-21?892
-4, 781
(0.2
x
10-4) el.987xT
-17,111
T
In 0_0392
=
=
_
(-l7;11
l)
-3.239
=
T
4,?Sl =1 476K=l
'
T=
3.239
THE CORRECT
'
203ºC
ANSWER
IS: (A)
q_,
J.
_/
I
I
I
1111
I
I
¡1
I
I
II
I
11¡
I
I
I
11
I
111
1¡1
11
111
1,100
LIQUID
900
¡?IL
800
20
,O.
8.8
1
780º____/
700
/
92
-
600
a+B
500
400
----·------?----------?
-
-
-?--1
o
20
40
60
80
Cu
Ag
80
60
40
20
o
COMPOSITION (w/o)
At 850ºC, ? + L phases
THE CORRECT
are
present
ANSWER IS: (B)
92
I
I
I
11
11
I
I
I
I
1
I¡
I
1
?
.-fhe solution requires
R(s)
a
step-by-step :;_-eduction of the systen1 loops.
--: -?-o
C(s)
I
1\Jext, co1nbine the forward blocks.
+
R(s)-»:o
;Y:\
£il.
Finally, reduce the
"outei-5' loop.
?
R(s)
ANSWER IS: (D)
Tl-IE CORRECT
14.
R=R0[l+a(T-T0)]
dR
11R
11T
=
dT
=
Raa11T
=
(100 Q) (0.004 ºC-1) (10 ºC)
=4.0Q
THE CORRECT ANSWER
15.
R
=
=
R0
[l
100
+
[l
a(T-
IS: (C)
To)]
+ 0.004
(250
-
O)]
=2000
THE CORRECT
A,NSWER IS: (D)
9').J
11
ii
I
---
I
I
I
Ill¡
I
C(s)
-
1?.u
u
=
v
1
1
o on
v
,
ffl 3¡-1
Kg
1
.
?
( p,
'1,-¡V
D
·
1i1\-2-11;
\
'
p
T]p
(
50,000 kg
I\. 3,600 s
J
001
(o,
'
,
-.-,
lÚ
?
3
/Ir
l"'-g
\
J
000
\
[r1¿1
t,
-:1
-
1
00)
1
eccl'?
T/-lll
2]
0.8
A
=
'Í
-1
2 Ll-1
.
.J
1
_(
TI-IE CORRECT
17s
T
YV
l.t.NS\VEFt IS: (C)
energy balance on the boiler gives:
.A?n
Q
l.l'.
=
rñ (h3
-
h2)
SO, OOO
=
3,600
'\
kg
[ (3,322 -167.6) kJ/kg] 43,8ll
=
s
J
kW= 44 MW
THE CORRECT ANSWER IS: (D)
18.
The mass rate of coal times its heating value divided by the boiler efficiency yields the heat added
to the water in the boiler.
(
.
m
m( h3
=
f
-hi)=,I
so, 000 kg
3,600
(HV)11b
s
'\
J
I[(3,322 -167.6) kJ/kg]
1.?78
=
(28,000 kJ/kg)(ü.88)
kg/s
THE CORRECT ANSWER IS: (C)
I
I
I
I
I
11
I
I
I
I
I
II
I
1¡
1
II
I
11-I
I
I
11
I¡
I
I
11
I
11
I
I
I
I
I
.,,_e
II.
11
I
_,
,.-;.
l
•
rrYI
1 _1e
??
'
í"
1
vo1un1e
specn1c
1s
f'
1
1dea1 gas
t11e
using
f'
•
•
•
·¡
rounn
equstlon or state.
pv=RT
RT'\
[0.287 kJ/(kg?K)][273 + 100]
v=
K?
172 kPa
p
Unit check
kJ
_¿
,....o
--13
11_
kl?
AJ"'?BVVEI-t IS:
THE CORRECT
I(
cg.h_
1
ni/kg
0.622
=
v
X
7
kg0I(.k1'T
(B)
The change in entropy is found fron1 the equations in the Then11odynan1ics
•
section of the FE
Supplied-Reference Handbook:
?s
=
cp
ln (T2IT1) -R ln (P2/P1)
a
constant volume process, -1
1i
=
__1.
-
=
therefore, P2IP1
?
=
=
=
cp
T2
T2IT1
ln (P2IP1) -R ln (P2/P1)
(cp
R) In (Pi/P1)
-
=
cv ln (P2IP1)
=
0.718 ln 2
0A98 kJ/(kg.K)
ANSWER IS: (C)
THE CORRECT
The cross-sectional
21.,
F
R
For
area
of the pipe is:
The flow rate is:
Q
Ac Ve= 0.007854 (2.5)(60)
=
THE CORRECT
11
I
I
I
=-
----
ANSWER IS: (B)
II
I
-
---
-=---
I
I
I
-
-
1.178 m3 /min
I
I
I
I
=
11
I
iii_
--
_¡¡¡¡¡¡
?---
..
-
1¡
I
II
I
11
11
I¡
\
I
I
11
I=
I
I
11
I
-
I
I
111
23.,
M'
F?
W·d1=-=
ea_
-
-
25
-¡
-
=
1
p
2,1,,.
175
-PB
1,000
p
The cross-sectional
A
4
2
TC
=-D
area
0.050 kJ/kg
_
-
4
4
( 0.10
50 J/kg
of the discharge pipe is:
TC
=
=
)2
=
0.007854
in
2
The velocity of flovv in the discharge pipe is:
m
40
pA
1,000 ( 0.007854)
V=-=
=5.093n1Js
The head at the discharge into the heater is:
P
1-Ih
v2
=-+-+Z2
pg
=
200, ooo
( 5.093 )2 +20=41.71n
2 ( 9.807 )
+
1,000 ( 9.807 )
2g
The head at the pump discharge is:
Hd= 41.7 -0.5
THE CORRECT
25.
+ 25
=
66.2
m
ANSWER IS: (D)
The effect of fouling on the heat-transfer surfaces is to reduce the heat-transfer rate by increasing
the surface resistance. The results on the water being heated would be to reduce the outlet
temperature of the water.
THE CORRECT
ANSWER IS: (B)
96
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Ill
-,
1
R
-'-eyno1cts
T-1ne
_,_
nun1
-b
e:t
found
1s
111
tbe Fluid lviechanics section of the FE Supplied-Reference
Handbook.
_
U.o
.l"?v
-
V_Dp
µ
V
=
1.5 nJ/s
D
=
2.5cn1
-
-:
D
.!L.
59
,/
/'-.
=
2.5x10-21n
10-4
l\f ºs
?
rn')
p
=
Re=
898 kg/111:i
( 1.5
nvs) (2.5 l 0-2
><
?
1.59
=
211,792
x
1 O
-4
n1)(898 kg/1113)
.
},T 0s/in
2
(rn/s )( 1n) (kg)
l
l\T
=
l kg.n11s2
(1\f 0S/1112)
(1n3)
Verify units
THE CORRECT
( din1ensionless)
ANSWER IS: (C)
97
The overall heat-transfer coeíficient
1-
tuoe
-
h_eat
.
.
exchanger equation
based
-
rr
the Heat
1n
inside surface
on
---
r·
JLTansrer
area
.
section
or
0
is found fror_c, the shell-and-0
F!l oupp
z·zea-_ R e1erence
the _1_LJ
--
1
J-lándbook.
D
_2___
ln
l
1
·-
=--+
UA
If
A
=
D-
R¡(·,l)
._
h;Ai
-
1
+
2nkL
A.
-·¡
A¡
-,
-1
/-1¡ 1n
l
_1
n
-
_
-
¡
-T
u
11._f(i)
Rf (i)
7
k
380
=
A-l
_
Aº
..
2rckl
rcD-L
l
Do
(2.5
1.667
u
X
10-2)ln
+ 7 X 10-4 +
(3.8
X
10-2)
(2.5
X
10-2)
X
10-4 + 7
10-4 + 0.138
X
X
+
2.5
2.5
+
X
3.8
2(380)
u
_!_ =
J
2.5xl0-2
3.8xl0-2
D-l
__
6,000
=
D-l
12,000
l
u
_!_
T
10-4
x
nD0L
1
I
6,000
=
ho
rn, º
\
I
h¡
hi
'\
(9 10-5) (3.8)(12, 000)
10-4 + 0.592x 10-4 + 0.548
X
10-4
9.945xl0-4
U= 1,005.5 W/(m2®I()
THE CORRECT
ANSWER IS: (A)
98
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Ill
¡I
11111
-11
II
,...
7
F'
:::,-
coo
Qevap
ling,
.
m
_
-
25
=
)
(}
11
kJ/kg
7
_
\
114 J
_
-
(
-
=
257)
kJ/kg
1
The process fron1 3 to 4 is a throttling process, which involves both a drop
temperature. is an irreversible process, generally considered constant enthalpy.
THE CORRECT
111
pressure and
ANSWER IS: (B)
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ii
FE Study i?Iaterial
by rqCEES
Published
FE Supplied-Reference I--Iandbook
FE San1ple Questions and Solutions and Internet-based Practice Tests for all 1nodules:
Che1nical
Civil
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Mechanical
Other Disciplines
For
1nore
inforn1ation about these and other Council publications and services,
visit us at www.ncees.org or contact our
Custo1ner Service Department at (800) 250-3196.
101
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lSBN 978- i -9326 ·13-4 7-6
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CJ
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