Mechanical
Full Exam
PE
Engineering Pro Guides
HVAC & Refrigeration
80 exam difficulty level problems
Covers Mechanical PE HVAC & Refrigearation exam topics
Written in exam format
Also includes detailed solutions
Justin Kauwale, P.E.
SECTION 1
INTRODUCTION
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1.0 INTRODUCTION
One of the most important steps in an engineer's career is obtaining the professional
engineering (P.E.) license. It allows an individual to legally practice engineering in the state of
licensure. This credential can also help to obtain higher compensation and develop a credible
reputation. In order to obtain a P.E. license, the engineer must first meet the qualifications as
required by the state of licensure, including minimum experience, references and the passing of
the National Council of Examiners for Engineering and Surveying (NCEES) exam. Engineering
Pro Guides focuses on helping engineers pass the NCEES exam through the use of free
content on the website, http://www.engproguides.com and through the creation of books like this
sample exam and technical study guides.
This sample exam is intended to be a sample test on ONLY the key concepts and skills of the
HVAC & Refrigeration Mechanical P.E. Exam.
1.1 KEY CONCEPTS AND SKILLS
The key concepts and skills tested in this sample exam were first developed through an analysis
of the topics and information presented by NCEES. NCEES indicates on their website that the
P.E. Exam will cover an AM exam (4 hours) followed by the PM exam (4 hours). Within the
Mechanical Engineering field, there are three specialties to choose from for the depth exam:
HVAC & Refrigeration, Thermal & Fluids and Mechanical Systems & Materials.
This sample exam focuses on the HVAC and Refrigeration topic. NCEES indicates on their
website that the HVAC and Refrigeration exam will focus on the following topics:
(http://ncees.org/engineering/pe/):
1) Principles
a) Basic Engineering Practice - (4 questions)
i) Units and conversions
ii) Economic analysis
iii) Electrical concepts (e.g., power consumption, motor ratings, heat output, amperage)
b) Thermodynamics - (4 questions)
i) Cycles
ii) Properties
iii) Compression Processes
c) Psychrometrics - (8 questions)
i) Heating/cooling cycles, humidification/dehumidification, heating/cooling loads, sea
level and other elevations
d) Heat Transfer - (7 questions)
e) Fluid Mechanics - (4 questions)
f) Energy/Mass Balances (5 questions)
2) Applications
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a) Heating/Cooling Loads - (8 questions)
b) Equipment and Components - (18 questions)
i) Cooling towers and fluid coolers
ii) Boilers and furnaces
iii) Heat exchangers
iv) Condensers/evaporators
v) Pumps/compressors/fans
vi) Cooling/heating coils
vii) Control systems components
viii) Refrigerants
ix) Refrigeration components
c) Systems - (18 questions)
i) Air distribution
ii) fluid distribution
iii) refrigeration
iv) energy recovery
v) control concepts
d) Supportive Knowledges - (4 questions)
i) Codes and standards
ii) Air quality and ventilation
iii) Vibration control
iv) Acoustics, economic analysis, electrical concepts
Each of these topics were investigated and filtered by the test maker for concepts and skills that
meet the following criteria:
(1) First, the concept and skill must be commonly used in the HVAC & Refrigeration field of
Mechanical Engineering. For example, pump sizing, fan sizing, determining friction losses and
calculating net positive suction head are regular occurrences in the HVAC & Refrigeration field.
The breakdown of question topics is shown in the list above.
(2) Second, the skill and concept must be testable in roughly 6 minutes per problem. There are
(40) questions on the afternoon exam and you will be provided with 4 hours to complete the
test. This results in an average of 6 minutes per problem. This criterion limits the complexity of
the exam problems and the resulting solutions. For example, pressure drop calculations are
common in the HVAC & Refrigeration field, but the calculation is often very lengthy because of
the number of steps involved, especially if a unique fluid and flow condition is used. Thus,
common fluids like water/air and common pipe/duct materials are used.
(3) Third, the key concepts and skills must be used or be known by practicing HVAC &
Refrigeration Mechanical Engineers. This criterion is similar to the first criterion. However, this
criterion filters the concepts and skills further by limiting the field to material encountered and
used by practicing engineers. The HVAC & Refrigeration, Thermal & Fluids and Mechanical
Systems & Materials fields are vast and there are many different avenues an engineer can take.
Two diverging paths are those engineers involved in research and those who practice.
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Research engineers are pushing the boundaries of the field and are highly focused in their
specific area of the field. The Professional Engineering exam does not cover emerging
technologies or highly focused material.
(4) The P.E. Exam must test the principle or application of the skill and concept and not the
derivations or the background knowledge of the topic or concept. The exam also does not
cover background information on the NCEES topics. The P.E. Exam is meant to prove that the
test taker is minimally competent to practice in the Mechanical Engineering field. The exam is
less concerned with theory and more with the principle or application of the theory, skill or
concept. For example, the P.E. exam is less concerned with the theory of evaporation in a
cooling tower and more with the performance and selection of a cooling tower.
In summary, this book is intended to provide a sample of the necessary skills and concepts to
develop a minimally competent, practicing professional engineer in the Mechanical Engineering
field, capable of passing the P.E. exam. This book does this through the following means:
(1) Providing sample problems that can be completed in roughly 6 minutes per problem.
(2) Providing solutions to these problems that teach skills and concepts used by practicing
Mechanical Engineers.
1.2 UNITS
The primary units that are used in the P.E. Exam are United States Customary System Units
(USCS). As such, this guide focuses exclusively on the USCS. However, it is recommended
that the test taker have a conversion book, because certain areas of the P.E. Exam may use the
International System of Units (SI).
2.0 DISCLAIMER
In no event will Engineering Pro Guides be liable for any incidental, indirect, consequential,
punitive or special damages of any kind, or any other damages whatsoever, including, without
limitation, those resulting from loss of profit, loss of contracts, loss of reputation, goodwill, data,
information, income, anticipated savings or business relationships, whether or not Engineering
Pro Guides has been advised of the possibility of such damage, arising out of or in connection
with the use of this document or any referenced documents and/or websites.
This book was created on the basis of determining an independent interpretation of the
minimum required knowledge and skills of a professional engineer. In no way does this
document represent the National Council of Examiners for Engineers and Surveying views or
the views of any other professional engineering society.
3.0 HOW TO USE THIS SAMPLE EXAM
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This exam can be used in multiple ways, depending on where you are in your study process. If
you are at the beginning or middle, it can be used to test your competency, gain an
understanding and feel for the test format, and help to highlight target areas to study. If you are
at the end, it can be used to determine your preparedness for the real exam. Remember that
the questions are a sample of the many topics that may be tested and are limited to fit a full
exam length and therefore is not comprehensive of all concepts.
Because the exam is written to be similar to the difficulty and format of the NCEES exam, it is
recommended that the test be completed in one sitting and timed for four hours to simulate the
real exam. This will give you a better indication of your status of preparation for the exam. If
you are at the ending of your studying, it is recommended to couple this exam with the AM
section to simulate the full exam test day.
Review the exam day rules and replicate the environment for the real test as much as possible,
including the type of calculator you may use and the acceptable references. Keep a watch or
clock next to you to gauge your pace for 40 questions in 4 hours.
Based on the NCEES website, the following are general rules for exam day.
Allowed:
1. Snacks that are not disruptive to others
2. Watches and small clocks (highly recommended on test day, some test
facilities do not have a clock)
3. Religious head coverings
4. Two straight edges: e.g. ruler, scale, protractor, triangle
5. Approved references
6. Approved calculator (2 recommended for backup)
7. Eyeglasses
8. Non electronic magnifying glass
9. (Units conversion book is also recommended)
Prohibited:
1.
2.
3.
4.
5.
6.
7.
8.
Cell phones
Hats and hoods
Slide charts, wheel charts, drafting compasses
Weapons
Tobacco
Personal Chairs
Eyeglass/Magnifying glass cases
Scratch Paper (all writing must be done in the exam booklet)
For additional references on exam day policies, exam day processes, and items to bring
on your exam day, review the NCEES Examinee Guide:
http://ncees.org/exams/examinee-guide/
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Similar to the NCEES exam, the tested topics are presented in a random order. For best use of
your time, answer the questions that you know first and return to the questions that you are
unfamiliar with later. Once all the known questions are answered, go through the test again and
attempt to answer the remaining questions by level of difficulty. If time allots, review your
answers.
If you are stuck on a question, seek the following avenues.
1. Study Guide: It is important to understand your study guides and indices. During times
of uncertainty, these will likely lead you to your answers. Determine the key concept that
is being asked in the question and refer to your indices or pre-tabbed sections. The
answer will hopefully be found in the Mechanical Engineering Reference Manual
(MERM), the American Society of Heating, Refrigerating and Air-Conditioning Engineers
(ASHRAE) Handbooks or one of the other references, listed below.
2. Process of Elimination: There are only four possible choices for each question. Ask
yourself if there is an answer that does not make sense and eliminate it. Further narrow
down the answer that are derived from equations or concepts that you know are not right
and are instead meant to deceive the test taker. See if there are answers that are
similar or separated by something like a conversion error. This may be an indication that
the correct equation was used.
3. Educated Guess: Remember that there is no penalty for wrong answers. Hopefully with
the process of elimination you are able to narrow down as many answers as possible
and are able to create an educated guess.
4. Rules of Thumb: Rules of thumb can be used to not only speed up time, but to help lead
you in the right direction.
5. If the time is almost up and there are still unanswered questions remaining, determine
whether it makes sense to check for mistakes on the problems you do know how to
solve, or to tackle the unanswered problems.
Typical Exam Verbiage/Design:
1. Most Nearly: Due to rounding differences, the exam answers will not match yours
exactly and in fact may not closely resemble your answer. NCEES uses the term “most
nearly” to test your confidence in your solution. When the question prompts you with
“most nearly”, choose the answer that most closely matches yours, whether it be greater
than or lesser to your value.
2. Irrelevant Information: The exam is intended to test your overall understanding of
concepts. At times the question will include unnecessary information that is meant to
misdirect you.
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3. Deceiving Answers: NCEES wants to know that you are able to determine the
appropriate methods for the solutions. There are answers that were intentionally
produced from wrong equations to mislead the test taker. For example, you may forget
a 1/2 in the formula, KE = (1/2)MV2 and there would be two answers each off by a factor
of 1/2.
4. Do Not Overanalyze: The exam questions are meant to be completed in 6 minutes.
Therefore, they are intended to be written as straight forward as possible. Do not be
tempted to overanalyze the meaning of a question. This will only lead you down the
wrong path.
Review the Solutions:
Once the sample test is completed, grade your results. Measure your aptitude in speed,
concept comprehension, and overall score. If you score is above the 75% range then you are in
good shape. This 75% score is only applicable if you have prepared completely for the exam. If
you are just starting out, then please do not be worried about a low score. This is number is
also just a range; there is no finite score to determine passing the test. Instead, NCEES
calibrates the results against practicing professional engineers. See this
page http://ncees.org/exams/scoring-process/ for a better understanding of how NCEES grades
the scores.
Review the answers that you got wrong and use the solutions as a learning tool on how to
address these types of problems. Compare the types of questions you are missing with the
NCEES outline of topics and determine where you should focus your studying. Finally repeat as
many practice problems as you can to get a better grasp of the test and to continually improve
your score.
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4.0 RECOMMENDED REFERENCES
The following references are recommended to be reviewed prior to the exam and should be
used during the exam. When reviewing these references, make sure you first understand the
content. These references are intended for the afternoon depth exam and are used as
references by practicing Mechanical HVAC Engineers. Secondly, you should be very familiar
with the indices of these references and should be able to navigate the references to find
information quickly. This may require you to insert tabs into the references. Once you have
completed these two tasks then you should be ready to use these references during the exam.
(Tip: It is helpful to have the indices of your references printed separately to allow you to have
both the index and the reference material open at the same time, making for quicker searches.)
4.1 MECHANICAL ENGINEERING REFERENCE MANUAL (MUST HAVE)
By Michael Lindeburg PE
The Mechanical Engineering Reference Manual or MERM is the most popular and most
comprehensive manual designed for the Mechanical Professional Engineering exam. It is
recommended that the engineer be very familiar with the contents of this book and to bring this
book to the exam. This book and the engineering unit conversions book should allow you to
answer a good portion of the AM exam questions.
Amazon Link i: Mechanical Engineering Reference Manual for PE Exam, 13th Edition
4.2 ENGINEERING UNIT CONVERSIONS BOOK (MUST HAVE)
By Michael Lindeburg PE
Another book related to the MERM is the Engineering Unit Conversions book. This is also
another recommended book to use during the exam and while studying for the exam.
Amazon Link: Engineering Unit Conversions
4.3 HVAC EQUATIONS, DATA AND RULES OF THUMB (MUST HAVE)
By Arthur Bell, W. Larsen Angel
This reference book is a must have for practicing HVAC/R engineers. It is a concise reference
that can be used on the exam and in practice. This book will help you to answer any questions
missed by the MERM in the HVAC & Refrigeration section.
Amazon Link: HVAC Equations, Data and Rules of Thumb
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4.4 ASHRAE HANDBOOKS (MUST HAVE)
By ASHRAE
The American Society of Heating, Refrigerating and Air-Conditioning Engineers (ASHRAE) is
the guiding source for the HVAC engineer. The society publishes four handbooks that contain
the essential topics and knowledge for practicing engineer: HVAC Systems and Equipment,
HVAC Applications, Refrigeration, Fundamentals. Each of these handbooks is updated in a four
year rotation. The handbooks are comprehensive and detailed. It is not necessary to know the
extensive details of these books, but it is essential to understand their layout and know how to
navigate them. If there is a question that you are unsure of in the test, these books will likely
have the solution. But because there is so much information, it will be difficult to find the
necessary references without at least having a familiarity with these resources.
Amazon Link: 2016 ASHRAE Handbook: HVAC Systems and Equipment
Amazon Link: 2015 ASHRAE Handbook: HVAC Applications
Amazon Link: 2014 ASHRAE Handbook: Refrigeration
Amazon Link: 2013 ASHRAE Handbook: Fundamentals
4.5 ASHRAE CODES & STANDARDS
By ASHRAE
ASHRAE standards are another common tool used by practicing professional HVAC engineers.
The exam does not appear to be based on the latest version of the codes, since it is not
referenced as a resource by NCEES. However, it is recommended at a minimum that the
engineer in training be familiar with the information in each of the following codes:
Amazon Link: ASHRAE 15, Safety Standard for Refrigeration Systems
Amazon Link: ASHRAE 55, Thermal Environmental Conditions for Human Occupancy
Amazon Link: ASHRAE 62.1, Ventilation for Acceptable Indoor Air Quality
Amazon Link: ASHRAE 90.1, Energy Standard for Building, except Low-Rise Residential
Buildings
4.6 NFPA CODES
By NFPA
The National Fire Protection Agency provides codes and standards related to fire protection.
The only recommended NFPA codes are those relating to HVAC systems. These codes are
NFPA 90A and NFPA 90B. The test may also ask the name of the code required for a certain
application. Therefore, it is useful to also print the list of all the NFPA codes and standards.
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Amazon Link: NFPA 90A, Standard for the Installation of Air Conditioning and Ventilating
Systems
Amazon Link: 90B, Standard for the Installation of Warm Air Heating and Air Conditioning
Systems.
List of NFPA Codes and Standards
4.7 MECHANICAL PE: HVAC & REFRIGERATION TECHNICAL STUDY
GUIDE
By Justin Kauwale
This book is specifically written for the Mechanical PE – HVAC and Refrigeration exam. It is a
comprehensive study guide and that teaches the key concepts and skills needed for the test. It
is intended to direct your learning for the need to know materials and give you a sturdy
foundation in the HVAC and refrigeration principles and applications.
Mechanical PE: HVAC & Refrigeration Technical Study Guide
Additional free material is available at www.engproguides.com
4.8 COMMON PROPERTY TABLES AND CHARTS
General
Along with the compiled notes and indices that you bring to the exam, it is also a good idea to
bring the following tables and figures. Some charts are provided within the question of the
exam. A psychrometric chart is also provided in the test booklet, but it is useful to have these
tables and figures at your fingertips, both in SI and IP units. All these tables/figures are
available to print online.
•
•
•
•
Saturated/Superheated Water Table
Saturated/Superheated Refrigerant Table, R134a, R140a
Psychrometric Chart, Air
Mollier Chart, Pressure-Enthalpy Diagram: Water, Refrigerant R134a, R140a
i
Justin Kauwale is a participant in the Amazon Services LLC Associates Program, an affiliate advertising program designed to
provide a means for sites to earn advertising fees by advertising and linking to amazon.com
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SECTION 2
AM QUESTIONS
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QUESTION 1
PRINCIPLES BASIC ENGINEERING PRACTICE
Honey has a dynamic viscosity of 1000 poise, a specific heat capacity of 0.6 cal/g-oC, and a
density of 0.05 oz/mL. The kinematic viscosity of honey, in ft2/sec, is most nearly?
(A) 0.76
(B) 7.1
(C) 25
(D) 30
QUESTION 2
PRINCIPLES – BASIC ENGINEERING PRACTICE
A 3 hp, 1800 RPM motor operates on 208 volts, 3 phase, 60 hertz power. Assume a power
factor of 0.9 and a service factor of 1.15. If the motor is 85% efficient, how many amps must be
supplied to the motor?
(A) 7
(B) 8
(C) 14
(D) 16
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QUESTION 3
PRINCIPLES – BASIC ENGINEERING PRACTICE
A pump is required to pump 300 GPM at 100' of total dynamic head, if electricity costs $0.15 per
kWh and the pump operates 8 hours a day, 365 days a year, then what will be the total cost for
operating the pump? Pump efficiency = 0.65. Motor efficiency = 0.95. Specific gravity = 1.
(A) $2,620
(B) $3,140
(C) $3,670
(D) $4,030
QUESTION 4
PRINCIPLES BASIC ENGINEERING PRACTICE
A new machine is installed in order to increase productivity. This new machine costs $75,000
and has an ongoing operating and maintenance cost of $500 per month. The new machine will
save $2,000 per month and will have a salvage value of $10,000 after 10 years. If the interest
rate is 5%, then what is the annual cost/savings of the new machine?
(A) -$9,038
(B) +$8,243
(C) +$9,083
(D) +$12,038
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QUESTION 5
PRINCIPLES – THERMODYNAMICS
The following pressure-enthalpy diagram shows the vapor compression cycle for the refrigerant
R-134a. Which of the following statements is true?
(A) The compressor work occurs at Step 2 at constant enthalpy.
(B) The expansion device is used at Step 4 to conduct an isobaric process.
(C) At Step 2, the compressor conducts an isentropic process.
(D) At Step 4, the expansion device increases the pressure of the refrigerant at constant
enthalpy.
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QUESTION 6
PRINCIPLES – THERMODYNAMICS
A heat pump provides 70.0 MBTUH of cooling, 66.0 MBTUH of heating and uses 5.9 KW of
electricity. What is the COP of the heat pump during heating?
(A) 3.3
(B) 3.5
(C) 4.8
(D) 6.8
QUESTION 7
PRINCIPLES – THERMODYNAMICS
An R134a refrigeration cycle operates at pressures, 23.77 psia and 150 psia. If the refrigerant
is superheated to 10 F and refrigerant is subcooled to 100 F, then what is the COP? Assume
the compressor is 90% efficient.
(A) 1.0
(B) 2.0
(C) 3.0
(D) 4.0
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QUESTION 8
PRINCIPLES – THERMODYNAMICS
An R134a refrigerant cycle operates at the following evaporator and condenser temperatures, 5
F and 105 F. Assume a refrigerant flow rate of 180 pounds per hour. How many tons of cooling
is provided? Assume an ideal cycle.
(A) 0.86 tons
(B) 1.24 tons
(C) 1.83 tons
(D) 3.81 tons
QUESTION 9
PRINCIPLES – PSYCHROMETRICS
An air conditioning system provides air at 55 F DB/ 53 F WB, in order to maintain space
conditions at 75 F DB/50% relative humidity. What is the dew point of the space conditions?
(A) 47 F
(B) 50 F
(C) 55 F
(D) 63 F
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QUESTION 10
PRINCIPLES – PSYCHROMETRICS
A new electric heater is provided to heat 5,000 CFM of air at 60 F DB/ 55 F WB. If a 3 KW
heater is provided, then what is the resulting Humidity Ratio?
(A) 49 grains/lb
(B) 69 grains/lb
(C) .008 lb of water/lb of dry air
(D) .012 lb of water/lb of dry air
QUESTION 11
PRINCIPLES – PSYCHROMETRICS
How many lbs of water are in the air of a 10 ft x 50 ft x 10 ft room at 70oF DB air and 70%
relative humidity?
(A) 1.6
(B) 4.1
(C) 5.2
(D) 5.9
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QUESTION 12
PRINCIPLES – PSYCHROMETRICS
If you are designing an HVAC system at an elevation of 7,500 ft above sea level with outside air
conditions of 60 F DB and 90% relative humidity, then what is the density and dew point of the
outside air?
(A) 0.0631 lb/ft3 and 57.1 F dp
(B) 0.0631 lb/ft3 and 60.0 F dp
(C) 0.0574 lb/ft3 and 57.1 F dp
(D) 0.0759 lb/ft3 and 55.1 F dp
QUESTION 13
PRINCIPLES – PSYCHROMETRICS
A cooling tower is used to cool 10,000 lbs/hr of condenser water from 105 F to 80 F. If the air is
designed to enter at ambient air conditions of 85 F/50% relative humidty and leave at 95 F/80%
relative humidity, then what is the mass flow rate of air required?
(A) 1,950 lbs/hr
(B) 8,975 lbs/hr
(C) 10,000 lbs/hr
(D) 12,255 lbs/hr
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QUESTION 14
PRINCIPLES – PSYCHROMETRICS
An evaporative cooler is used to cool an airstream of 95 F/20% relative humidity. If the air
leaves the evaporative air cooler at 80 F, then what is the new relative humidity?
(A) 29%
(B) 48%
(C) 57%
(D) 69%
QUESTION 15
PRINCIPLES – PSYCHROMETRICS
45 F/80% relative humidity air is heated with a 8,100 Btu/hr electric heater. Then a humidifier is
used to increase the relative humidity to 50%. What is the amount of energy provided, if the
flow rate is 250 CFM. Assume sea level and a density of 0.075 lb/ft3.
(A) 7,000 Btu/hr
(B) 8,000 Btu/hr
(C) 10,000 Btu/hr
(D) 13,250 Btu/hr
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QUESTION 16
PRINCIPLES – PSYCHROMETRICS
Calculate the CFM at a cooling coil required to provide 12,000 Btu/hr of cooling to a space at
leaving coil conditions of 55 F DB/90% RH. Assume an elevation of 5,000 ft and the space is
kept at 75 F DB/50% RH. The cooling coil is also a 100% recirculation unit.
(A) 520 CFM
(B) 650 CFM
(C) 890 CFM
(D) 1060 CFM
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QUESTION 17
PRINCIPLES – HEAT TRANSFER
Determine the amount of energy required to cool 50 lbs of turkey from 70 F to 0 F, with the
following properties.
Property of Turkey
Specific Heat above Freezing
Enthalpy of Fusion (Freezing Point = 26 F)
Specific Heat below Freezing
(A) 7,200 Btu
(B) 9,600 Btu
Value
𝐵𝐵𝐵𝐵𝐵𝐵
𝑙𝑙𝑙𝑙 ∗ 𝐹𝐹
𝐵𝐵𝐵𝐵𝐵𝐵
105
𝑙𝑙𝑙𝑙
𝐵𝐵𝐵𝐵𝐵𝐵
0.35
𝑙𝑙𝑙𝑙 ∗ 𝐹𝐹
0.67
(C) 12,200 Btu
(D) 14,400 Btu
QUESTION 18
PRINCIPLES – HEAT TRANSFER
A new solar hot water heating system has two loops, (1) solar hot water loop and a (2) domestic
hot water loop. The loops are separated by a plate frame heat exchanger. If the solar hot water
fluid enters the heat exchanger at 160 F and leaves at 130 F and the domestic hot water enters
at 80 F and leaves at 120 F, then what is the LMTD?
(A) 30 F
(B) 31.3 F
(C) 33.7 F
(D) 35 F
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QUESTION 19
PRINCIPLES – HEAT TRANSFER
A feedwater heater is used to heat 50 lbm/min of water from 100 F to 120 F. Steam enters the
feedwater heater at 300 F. What is the mass flow rate of the steam, if the steam leaves at 200
F? Assume the feedwater heater is at a pressure of 14.7 psia.
(A) 47 lbm/hr
(B) 52 lbm/hr
(C) 60 lbm/hr
(D) 75 lbm/hr
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QUESTION 20
PRINCIPLES – HEAT TRANSFER
A client is considering replacing 1,200 ft2 of windows with a new high efficiency window. What
would be the reduction in conductive heat losses with the given below conditions?
(A) 1,500 Btu/hr
(B) 3,250 Btu/hr
(C) 8,900 Btu/hr
(D) 11,000 Btu/hr
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QUESTION 21
PRINCIPLES – HEAT TRANSFER
A chilled beam of 4” outer diameter and 100 feet in total length is located in a space with a dry
bulb temperature of 75 F. If the surface temperature of the chilled beam is 50 F, then what is
the radiative heat transfer to the space? Assume the emissivity of the chilled beam is 0.8.
(A) 1096 Btu/hr
(B) 658 Btu/hr
(C) 410 Btu/hr
(D) 2,000 Btu/hr
QUESTION 22
PRINCIPLES – HEAT TRANSFER
A motor is located in a mechanical room at 75 F DB. The motor operates at 10 BHP and is 95%
efficient. The remaining energy is released as heat to the space. The surface temperature of
the motor is 120 F. What is the overall heat transfer coefficient? The motor is approximated as
a box with dimensions, 2’ X 1’ X 1’. The motor is suspended, such that all surfaces are
exposed.
(A) 1.0 Btu/(hr-ft^2-Ԭ)
(B) 2.8 Btu/(hr-ft^2-Ԭ)
(C) 4.2 Btu/(hr-ft^2-Ԭ)
(D) 5.6 Btu/(hr-ft^2-Ԭ)
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QUESTION 23
PRINCIPLES – HEAT TRANSFER
A 100 ft long, 6” schedule 40 steel pipe carries chilled water at 45 F. The pipe has 2” of
insulation. The insulation k-value is 0.29 Btu-in/(hr-ft2-F). What is the surface temperature at
the outside of the insulation?
Assume the following:
Ambient temperature = 75 F, h_conv = 1.0 Btu/hr-ft2-F, h_rad = 1.0 Btu/hr-ft2-F, the heat transfer
throug the inner/outer pipe walls is negligible.
(A) 69 F
(B) 71 F
(C) 73 F
(D) 75 F
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QUESTION 24
PRINCIPLES – FLUID MECHANICS
500 GPM of water flows through a 6" schedule 40 pipe. If the pipe has a relative roughness
value of .0004 and a Reynolds number of 250,000, then what is the pressure drop through 100
ft of pipe?
(A) 1.7 ft of head
(B) 3.3 ft of head
(C) 4.2 ft of head
(D) 4.9 ft of head
QUESTION 25
PRINCIPLES – FLUID MECHANICS
50 F water flows through a 6" schedule 40 steel pipe at a volumetric flow rate of 200 gallons per
minute. What is the Reynolds number?
(A) 80,000
(B) 100,000
(C) 150,000
(D) 220,000
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QUESTION 26
PRINCIPLES – FLUID MECHANICS
A new 480 Volt, 3-phase motor is provided to serve the following pump with the design
conditions shown in the below figure. If the pump's impeller diameter is increased by a factor of
1.25 and the speed of the pump remains the same, then what will be the resulting flow rate?
(A) 150 GPM
(B) 200 GPM
(C) 250 GPM
(D) 313 GPM
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QUESTION 27
PRINCIPLES – FLUID MECHANICS
Rainwater at 60oF flows down a trench at velocity of 10 ft/s. The cross section of the trench is
illustrated below. What is the Reynolds number of the fluid?
3.5 ft
2 ft
6 ft
(A) 2.47 x 106
(B) 3.29 x 106
(C) 3.94 x 106
(D) 4.93 x 106
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QUESTION 28
PRINCIPLES – ENERGY/MASS BALANCE
A new heat exchanger uses steam to heat water. Steam (50 lb/min) enters the heat exchanger
at 20 psia, fully saturated. The condensate is not recovered. What mass flow rate of water can
be expected to provide a 40 degree change in temperature?
(A) 1,000 lb/min
(B) 1,200 lb/min
(C) 1,300 lb/min
(D) 1,400 lb/min
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QUESTION 29
PRINCIPLES – ENERGY/MASS BALANCE
50 GPM of water flows through a 2-1/2" pipe and then branches into (2) pipes shown in the
figure below. If the velocity in the 1-1/2" pipe is measured at 4 ft/sec, then what is the velocity
through the 1" pipe?
(A) 3.9 ft/sec
(B) 5.6 ft/sec
(C) 7.2 ft/sec
(D) 9.1 ft/sec
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QUESTION 30
PRINCIPLES – ENERGY/MASS BALANCE
A concentrated solar power system produces steam at 800oF, 400 psia and 1000 gpm. If the
steam expands through a turbine to 5 psia at 75% isentropic efficiency, and 85% mechanical
efficiency, how much power (KW) does the turbine generate? Assume a density of 0.0196
LBM/FT^3.
(A) 11.3
(B) 15.1
(C) 17.7
(D) 21.2
QUESTION 31
PRINCIPLES – ENERGY/MASS BALANCE
5,000 CFM of air at 78 F DB/55 % RH passes through a cooling coil with an apparatus dew
point of 53 F. The resulting discharge air temperature from the coil is 55 F DB/ 54 F WB.
What is the total amount of condensate produced?
(a) 0.05 GPM
(b) 0.12 GPM
(c) 0.37 GPM
(d) 0.65 GPM
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QUESTION 32
PRINCIPLES – ENERGY/MASS BALANCE
𝐶𝐶4 𝐻𝐻10 + 6.5(𝑂𝑂2 + 3.76𝑁𝑁2 ) → 4𝐶𝐶𝑂𝑂2 + 5𝐻𝐻2 𝑂𝑂 + 24.44𝑁𝑁2
What is the air to fuel ratio? Assume the equation is balanced.
(A) 10.1
(B) 12.9
(C) 14.8
(D) 15.4
QUESTION 33
APPLICATION – HEATING/COOLING LOADS
An air handler is used to cool 1,750 CFM of air at 78 F/60% RH to 55 F DB/54 F WB. What is
the sensible cooling provided by the air handler? The chilled water provided to the air handler
enters at 42 F and leaves at 52 F. Assume that the density of air is 0.075 lbs per cubic foot.
(A) 18,900 𝐵𝐵𝐵𝐵𝐵𝐵/ℎ
(B) 27,600 𝐵𝐵𝐵𝐵𝐵𝐵/ℎ
(C) 43,500 𝐵𝐵𝐵𝐵𝐵𝐵/ℎ
(D) 52,700 𝐵𝐵𝐵𝐵𝐵𝐵/ℎ
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QUESTION 34
APPLICATION – HEATING/COOLING LOADS
The net refrigeration effect of a refrigeration system is equal to 120 tons of cooling. The
compressor has a suction pressure of 40 PSIA and a discharge pressure of 200 PSIA. If the
compressor has a capacity equal to 100 kilowatts, then what is the COP of the refrigeration
system?
(A) 4.2
(B) 4.5
(C) 4.9
(D) 5.3
QUESTION 35
APPLICATION – HEATING/COOLING LOADS
A dedicated outside air unit cools 4,000 cfm of outside air to 55oF DB, 53oF WB. The outside
ambient condition is 87oF DB, 75oF WB. What is the total cooling load required by the air
handling unit? Assume that the density of air is 0.075 lbs per cubic foot.
(A) 576,000 𝐵𝐵𝐵𝐵𝐵𝐵/ℎ
(B) 297,000 𝐵𝐵𝐵𝐵𝐵𝐵/ℎ
(C) 138,000 𝐵𝐵𝐵𝐵𝐵𝐵/ℎ
(D) 59,840 𝐵𝐵𝐵𝐵𝐵𝐵/ℎ
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QUESTION 36
APPLICATION – HEATING/COOLING LOADS
A data center is maintained at 72oF, 50% RH. 8,000 cfm is supplied to the space, 520 cfm of
which is outside air. The outside air temperature is 30oF DB, 27oF WB. To maintain the space
humidity, steam is dispersed into the supply air stream of the duct. Calculate the flow rate of the
steam in lb/hr. Assume there is no latent load in the space and no latent cooling occurs.
Given: At 72oF, 50% RH, air density is 0.074
, humidity ratio is 58.6
(A) 0.23
(B) 13.8
(C) 186
(D) 211
QUESTION 37
APPLICATION – HEATING/COOLING LOADS
Space temperatures reach 65oF, 70% humidity during low load conditions. The total supply air
to the room is 160 cfm. How much reheat is required to maintain a maximum setpoint of 60%
relative humidity?
(A) 780
/
(B) 1730
/
(C) 6840
/
(D) 1720
/
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QUESTION 38
APPLICATION – HEATING/COOLING LOADS
Assume you have 2,000 CFM of outside air required and the design outside air conditions are
87 F DB/78 F WB. The building has a total volume of 200,000 ft3. The infiltration rate is found
to equal 2 air changes per hour. What is the total cooling load due to outside air and infiltration,
if the cooling coil cools the air to 55 F DB/53 F WB?
(A) 13 tons
(B) 29 tons
(C) 58 tons
(D) 63 tons
QUESTION 39
APPLICATION – HEATING/COOLING LOADS
A boiler is located in a mechanical room. The product data indicates that the boiler emits
75,000 Btu/hr of sensible heat to the space. There are also four 4’ long T5 28 watt light bulbs
located in the space. If the mechanical room is to be maintained at 85 F DB and the outside air
design condition is 78 F, then how much outside air must be supplied to the mechanical room?
Assume the same amount of air is exhausted from the mechanical room.
(A) 10,500 CFM
(B) 9,990 CFM
(C) 9,900 CFM
(D) 8,760 CFM
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QUESTION 40
APPLICATION – HEATING/COOLING LOADS
Calculate the building’s heat load given the following information. The building is to be
maintained at 69 F DB and the design outside air conditions are 5 F DB/75% RH. Assume no
infiltration and no outside air loads.
Wall area: 20,000 SF, Roof area: 2,500 SF, Window area: 2,000 SF
Wall: R-11; Roof: R-20; Window: U-Value = 0.75
(A) 296,000 Btu/hr
(B) 220,000 Btu/hr
(C) 184,000 Btu/hr
(D) 124,000 Btu/hr
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SECTION 3
PM QUESTIONS
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QUESTION 41
APPLICATIONS – EQUIPMENT & COMPONENTS
An air handler supplies 5,000 CFM at a temperature of 55 °F. The air handler was designed for
1,000 CFM of outside air at 87 °F DB and 78 °F WB. The remaining return air from the space is
at 77 °F DB and 55% relative humidity. What are the entering conditions of the air into the coil,
in DB and WB?
(A) 79 ℉ 𝐷𝐷𝐷𝐷, 68.3 ℉ 𝑊𝑊𝑊𝑊
(B) 79 ℉ 𝐷𝐷𝐷𝐷, 59.6 ℉ 𝑊𝑊𝑊𝑊
(C) 85 ℉ 𝐷𝐷𝐷𝐷, 74.3 ℉ 𝑊𝑊𝑊𝑊
(D) 85 ℉ 𝐷𝐷𝐷𝐷, 75.7 ℉ 𝑊𝑊𝑊𝑊
QUESTION 42
APPLICATIONS – EQUIPMENT & COMPONENTS
A classroom of 25 people has the following heat gains:
People: 250 Btu/h per person (Sensible); 200 Btu/h per person (Latent)
Lighting: 4,000 Btu/h; Computers: 8,000 Btu/h ; Walls, Roofs, Windows: 22,000 Btu/h
Ventilation: 7,500 Btu/h (Sensible); 7,500 Btu/h (Latent)
The air handler serving the classroom has a supply air temperature of 55 °F and the space is to
be maintained at 75 °F DB and 50% Relative humidity. What CFM is required?
(A) 2,210 𝐶𝐶𝐶𝐶𝐶𝐶
(B) 2,675 𝐶𝐶𝐶𝐶𝐶𝐶
(C) 2,790 𝐶𝐶𝐶𝐶𝐶𝐶
(D) 3,865 𝐶𝐶𝐶𝐶𝐶𝐶
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QUESTION 43
APPLICATIONS – EQUIPMENT & COMPONENTS
A dedicated outside air unit is used to pre-cool 2,500 CFM of outside air at 88 °F DB and 85%
relative humidity to 60 °F DB, 58 °F WB. What is the quantity of water removed by the coil in
GPM?
(A) 0.29 𝐺𝐺𝐺𝐺𝐺𝐺
(B) 0.33 𝐺𝐺𝐺𝐺𝐺𝐺
(C) 2.9 𝐺𝐺𝐺𝐺𝐺𝐺
(D) 2.4 𝐺𝐺𝐺𝐺𝐺𝐺
QUESTION 44
APPLICATIONS – EQUIPMENT & COMPONENTS
An air handling unit cools 2,000 CFM of outside air (90 F DB/80% RH) and 6,000 CFM of return
air (77 °F DB, 50% RH) to 52 °F DB, 51 °F WB. If chilled water enters the coil at 44 °F and
leaves at 56 °F, what is the required GPM?
(A) 42 𝐺𝐺𝐺𝐺𝐺𝐺
(B) 80 𝐺𝐺𝐺𝐺𝑀𝑀
(C) 105 𝐺𝐺𝐺𝐺𝐺𝐺
(D) 126 𝐺𝐺𝐺𝐺𝐺𝐺
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QUESTION 45
APPLICATIONS – EQUIPMENT & COMPONENTS
A cooling coil has a surface temperature of 55 °F. If 5,000 CFM of air enters the coil at 80 °F
DB, 65% RH, then what is the bypass factor that produces exiting air at 60 °F DB? Assume that
there are no minor heat gains or losses across the coil and that the coils temperature is
maintained at 55 °F.
(A) 0.03
(B) 0.10
(C) 0.15
(D) 0.20
QUESTION 46
APPLICATIONS – EQUIPMENT & COMPONENTS
A chilled water pump is used to pump 200 GPM of chilled water through a water cooled chiller,
with a 10 °F temperature difference. The condenser water circuit of the chiller operates on a 15
°F temperature difference and the compressor produces 250,000 Btu/hr. What GPM is required
by the condenser water pump? Assume no minor heat gains or losses in the chilled/condenser
water circuits and negligible bypass factors.
(A) 106 GPM
(B) 133 GPM
(C) 167 GPM
(D) 200 GPM
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QUESTION 47
Which of the following is not a type of fan?
(A) Airfoil
(B) Backward inclined
(C) Vertical Inline
(D) Propeller
QUESTION 48
An evaporative air cooler is used to cool air at 90 F, 30% relative humidity. Water is supplied to
the evaporative air cooler at 70 F. What is the dry bulb temperature of the air leaving the
evaporative cooler, if the evaporative cooler is 80% effective?
(A) 67.2 ℉ 𝐷𝐷𝐷𝐷
(B) 70.0 ℉ 𝐷𝐷𝐷𝐷
(C) 71.8 ℉ 𝐷𝐷𝐷𝐷
(D) 75.0 ℉ 𝐷𝐷𝐷𝐷
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QUESTION 49
APPLICATIONS – EQUIPMENT & COMPONENTS
An R-134A chiller has a suction pressure of 80 PSIA and discharge pressure of 200 PSIA. The
refrigerant undergoes 15 °F of superheat and 0 °F of sub-cooling. What is the COP Of the
chiller? Assume a refrigerant flow of 25 lb/min.
(A) 4.9
(B) 5.7
(C) 7.7
(D) 10.2
QUESTION 50
APPLICATIONS – EQUIPMENT & COMPONENTS
An R-410A chiller has a suction pressure of 100 PSIA with 20 °F super-heat and a discharge
pressure of 200 PSIA. Assume 15 °F of sub-cooling. What is the required refrigerant flow rate
in (lb/min) in order to produce 20 tons of cooling?
(A) 1,470 lbs/hr
(B) 1,870 lbs/hr
(C) 2,570 lbs/hr
(D) 2,890 lbs/hr
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QUESTION 51
APPLICATIONS – EQUIPMENT & COMPONENTS
An R-134A chiller has a suction pressure of 100 PSIA and discharge pressure of 200 PSIA.
The refrigerant undergoes 15 °F of superheat and 0 °F of sub-cooling. What is the quality of the
refrigerant entering the evaporator? Assume a refrigerant flow of 25 lb/min.
(A) 0.15
(B) 0.22
(C) 0.32
(D) 0.42
QUESTION 52
APPLICATIONS – EQUIPMENT & COMPONENTS
An R-410A chiller has a suction pressure of 150 PSIA with 20 °F super-heat and a discharge
pressure of 400 PSIA. Assume 15 °F of sub-cooling and a refrigerant flow rate of 20 lb/min.
What is the condenser leaving temperature of the refrigerant?
(A) 90 ℉
(B) 99 ℉
(C) 114 ℉
(D) 129 ℉
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QUESTION 53
APPLICATIONS – EQUIPMENT & COMPONENTS
20 lb/hr of 15 PSIA steam is delivered to a heating coil. 2,000 CFM of air enters the coil at 60
°F DB and 90% relative humidity. What is the exiting dry bulb temperature of the air, assume no
super heat or sub-cooling. Bypass factor and minor heat gains/losses are negligible.
(A) 69℉
(B) 73℉
(C) 75℉
(D) 79℉
QUESTION 54
A pump is sized for 250 GPM at 100’ TDH. What is the total flow produced by two of these
pumps in series?
(A) 100 GPM
(B) 200 GPM
(C) 250 GPM
(D) 500 GPM
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QUESTION 55
APPLICATIONS – EQUIPMENT & COMPONENTS
A cooling tower has 150 GPM of water at entering and leaving temperatures of 100 F and 85 F.
If the outside air conditions are 82 F DB/75% relative humidity. What is the effectiveness of the
cooling tower?
(A) 62%
(B) 68%
(C) 74%
(D) 79%
QUESTION 56
APPLICATIONS – EQUIPMENT & COMPONENTS
A new air handler with a VFD at 60 HZ provides 2,000 CFM of conditioned air to multiple
classrooms. 500 CFM of the 2,000 CFM is outside air that is constantly provided to maintain
acceptable indoor air quality. If during low load conditions the VFD is ramped down to 45 HZ,
then what is the percentage of outside air compared to the total air supplied?
(A) 25%
(B) 33%
(C) 44%
(D) 67%
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QUESTION 57
APPLICATIONS – EQUIPMENT & COMPONENTS
An existing chiller is served by a chilled water pump (150 GPM, 75 TDH), 65% efficient pump,
90% efficient motor. A recent study was conducted and it was found that the pump was
oversized and should be replaced with a new higher efficiency pump at 150 GPM, 50 TDH, 80%
efficient pump and 95% premium efficiency motor. If the pump runs 8 hours a day, 5 days a
week, 52 weeks a year, how many kWh per year will be saved by switching to the new pump?
(A) 3,640 kWh
(B) 3,990 kWh
(C) 4,150 kWh
(D) 7,250 kWh
QUESTION 58
APPLICATIONS – EQUIPMENT & COMPONENTS
A 10 BHP supply fan is provided with both the motor and the fan in the air conditioned space.
The motor efficiency is 95%. What is the total heat gain from the fan and motor to the space?
(A) 22,680 𝐵𝐵𝐵𝐵𝐵𝐵ℎ
(B) 24,230 𝐵𝐵𝐵𝐵𝐵𝐵ℎ
(C) 25,460 𝐵𝐵𝐵𝐵𝐵𝐵ℎ
(D) 26,800 𝐵𝐵𝐵𝐵𝐵𝐵ℎ
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QUESTION 59
APPLICATIONS – SYSTEMS & COMPONENTS
A 100 GPM condenser water pump is supplied with water by a cooling tower basin that is 10 ft
above the centerline of the pump. The suction line of the pump consists of 40 ft of 3" Schedule
40 steel pipe and 3 elbows. Condenser water pump serves a cooling tower with entering and
leaving conditions of 95 F and 85 F. What is the net positive suction head available at the
condenser water pump?
(A) 7
(B) 34
(C) 41
(D) 48
QUESTION 60
APPLICATIONS – SYSTEMS & COMPONENTS
A north facing wall is 20’ long by 10’ high. There are (2) 2’ X 4’ windows with 1/8” clear glass.
The wall consists of 8” Concrete (2.1 (h*ft^2* )/Btu), 2” Insulation (8.2 (h*ft^2* )/Btu) and 5/8”
Gypsum (1.5 (h*ft^2* )/Btu). The CLTD of the wall and window is found to equal 40 F. The
windows have a SC of 0.6, U-factor of 0.95 Btu/(h*ft^2* ) and a SCL of 80. What is the total
heat transferred through the wall and windows, Btu/h? Do not subtract the area of the window
from the wall.
(A) 2,050
(B) 2,300
(C) 2,750
(D) 3,330
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QUESTION 61
APPLICATIONS – SYSTEMS & COMPONENTS
A pump is required to supply 200 GPM of water to (2) outlets at a pier. The first outlet is 500
feet from the pump, the second outlet along the same path of travel is 500 feet from the first
outlet. The first length of pipe is 6" STD Steel, the second length of pipe is 4" STD Steel pipe.
What is the total pressure drop in FT of Head? Assume all minor losses are negligible.
(A) 12 feet
(B) 16 feet
(C) 18 feet
(D) 21 feet
QUESTION 62
APPLICATIONS – SYSTEMS & COMPONENTS
An air side economizer should operate when which of the following is absolutely true?
(A) When the dry-bulb of the outside air is less than the re-circulated air.
(B) When the humidity of the outside air is less than the re-circulated air.
(C) When the enthalpy of the outside air is less than the re-circulated air.
(D) When the humidity ratio of the outside air is less than the re-circulated air.
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QUESTION 63
APPLICATIONS – SYSTEMS & COMPONENTS
A humidifier evaporates 0.011 GPM of water into an incoming air stream. Entering air
conditions are 85 F DB, 70% relative humidity. What is the required flow rate (CFM) of the fan?
Assume the air leaving the spray humidifier is at 85 F DB, 90% relative humidity.
(A) 230 CFM
(B) 400 CFM
(C) 510 CFM
(D) 640 CFM
QUESTION 64
APPLICATIONS – SYSTEMS & COMPONENTS
A new diffuser is placed in the center of a 40’ X 40’ room. For the given CFM, the diffuser has
values of T150 of 5’, T1o0 of 10’ and T50 of 15’. At what distance from the nearest wall will the
velocity of the air be 100 feet per minute?
(A) 5’
(B) 10’
(C) 15’
(D) 20’
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QUESTION 65
APPLICATIONS – SYSTEMS & COMPONENTS
An existing 500 ton chiller with a COP of 4.5, runs for an equivalent of 3 full load hours a day,
365 days a year. The chiller is being replaced with a new 500 ton chiller with an efficiency of
0.6 kW/ton for a total cost of $750,000. If the interest rate is 4% and the lifetime of the new
chiller is 25 years, then what is the annual value of replacing the existing chiller with a new
chiller? Assume no annual maintenance costs; include annual electricity cost savings with unit
cost of $0.25 per kWh.
(A) $23,090
(B) $24,910
(C) $36,000
(D) $48,000
QUESTION 66
APPLICATIONS – SYSTEMS & COMPONENTS
A refrigeration unit is required to store 1,000 lbs of salmon (Heat Capacity Above Freezing: 0.88
Btu/lbm*F, Heat Capacity Below Freezing: 0.51 Btu/lbm*F, Latent heat of Fusion: 110 Btu/lbm,
Initial Freezing Point: 28 F). If the salmon arrives to the unit at 70 F and must be cooled to 10 F
in 2 hours, then what is the required size of the air conditioning system, in Btu/hr.
(A) 18,230
(B) 37,210
(C) 78,070
𝐵𝐵𝐵𝐵𝐵𝐵
ℎ
𝐵𝐵𝐵𝐵𝐵𝐵
ℎ
𝐵𝐵𝐵𝐵𝐵𝐵
(D) 110,000
ℎ
𝐵𝐵𝐵𝐵𝐵𝐵
ℎ
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QUESTION 67
APPLICATIONS – SYSTEMS & COMPONENTS
A new cooling coil provides sensible cooling of 250,000 Btu/hr. The entering air conditions into
the coil are 80 F DB. Leaving conditions from the coil at 55 F DB. If the coil is at an elevation of
5,000 FT, then what is the air flow rate in CFM? Assume negligible bypass factor and
miscellaneous heat gains/losses.
Density = 0.062 lb/ft^3; Heat Capacity = 0.24 Btu/lb*F
(A) 5,125
(B) 6,065
(C) 9,565
(D) 11,200
QUESTION 68
A 100% outside air handler serving a theater supplies 10,000 CFM of OAIR at 55 F DB/54 F WB
to maintain space conditions at 75 F DB and 50% Relative Humidity. Outside air conditions are
at 85 F DB and 80% Relative Humidity. How many tons of cooling can be saved if a total
enthalpy wheel is provided with 75% effectiveness? Assume negligible bypass factor and no
minor heat gains/losses.
(A) 25 tons
(B) 43 tons
(C) 56 tons
(D) 60 tons
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QUESTION 69
APPLICATIONS – SYSTEMS & COMPONENTS
A sensible heat recovery device is used to preheat entering outdoor air 3,500 CFM, (40 F, 60%
RH) with 4,000 CFM exhaust air (77 F, 55% RH). The sensible effectiveness of the device is
60%. What is the leaving supply air dry bulb temperature? Assume zero leakage.
(A) 49.3 ℉
(B) 62.2 ℉
(C) 67.8 ℉
(D) 74.2 ℉
QUESTION 70
APPLICATIONS – SYSTEMS & COMPONENTS
A new pump is required to pump 350 GPM from a well up to a holding tank that is 200 FT above
the pump. Assume a total friction loss of 20 FT of head from piping and fittings. The pump is
required to pump 8 hours a day, 7 days a week, 52 weeks a year. The pump is 85% efficient
and the motor is 95% efficient. Electricity costs are $0.25 per kWh. What are the yearly cost
savings if a new location for the holding tank is found at an elevation of 100 FT above the
pump? Assume the same efficiencies and friction losses between the two scenarios.
(A) $1,720 per year
(B) $2,090 per year
(C) $4,170 per year
(D) $5,970 per year
Mechanical PE - HVAC and Refrigeration Full Exam
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QUESTION 71
A 22" X 10" galvanized steel duct is used to convey 2,000 CFM of industrial exhaust. What is
the pressure drop in units of IN. WG per 100 ft.
Density = 0.075 lb/ft^3; roughness factor = 0.0003 ft.
(A) 0.07 in. WG per 100 ft
(B) 0.10 in. WG per 100 ft
(C) 0.14 in. WG per 100 ft
(D) 0.19 in. WG per 100 ft
QUESTION 72
APPLICATIONS – SYSTEMS & COMPONENTS
A new steam boiler provides 100 lb/hr of steam at 30 PSIA, 0 degrees super heat to various hot
water heaters. If the hot water heaters are designed to provide a 40 degree delta to incoming
water at 80 F, then what is the total GPM of hot water that the boiler can support?
(A) 4.73 GPM
(B) 10.2 GPM
(C) 15.7 GPM
(D) 21.9 GPM
Mechanical PE - HVAC and Refrigeration Full Exam
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QUESTION 73
APPLICATIONS – SYSTEMS & COMPONENTS
A new building with dimensions of 200' (L) X 150' (W) X 10' (H) is classified as having average
construction tightness, which relates to 0.3 air changes per hour of infiltration. If outside air is at
88 F DB/80% RH and the indoor design conditions are 75 F DB/50% RH, then what is the total
cooling load in tons added by infiltrated air? Use standard air conditions.
(A) 89,560
𝐵𝐵𝐵𝐵𝐵𝐵
(B) 104,600
ℎ
(C) 124,400
(D) 159,400
𝐵𝐵𝐵𝐵𝐵𝐵
ℎ
𝐵𝐵𝐵𝐵𝐵𝐵
ℎ
𝐵𝐵𝐵𝐵𝐵𝐵
ℎ
QUESTION 74
APPLICATIONS – SYSTEMS & COMPONENTS
Which of the following filters provides greater than 99% arrestance?
(A) MERV 1
(B) MERV 7
(C) MERV 13
(D) MERV 18
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QUESTION 75
APPLICATIONS – SYSTEMS & COMPONENTS
A new dedicated outside air handling unit is used to cool 5,000 CFM of OAIR (88 F DB, 70%
RH) to 55 F DB/54 F WB. The outside air handling unit is equipped with a wrap around heat
pipe. The heat pipe pre-cools the outside air by 10 degrees F. What is the total reduction in
tons of cooling by installing a heat pipe? Assume density = 0.075 lb/ft^3, elevation = sea level.
(A) 4.5 𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇
(B) 6.2 𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇
(C) 7.9 𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇
(D) 10 𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇
QUESTION 76
APPLICATIONS – SYSTEMS & COMPONENTS
A new counter-current heat exchanger is designed for incoming cold water at 43 F and leaving
water at 52 F. If the entering/exiting hot water of 60 F and 50 F, then what will be the LMTD?
(A) 7.5℉
(B) 8.5℉
(C) 10.0℉
(D) 11.5℉
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QUESTION 77
APPLICATIONS – SUPPORT KNOWLEDGE
What is the maximum allowable flame spread index of gypsum board air ducts, in accordance
with NFPA 90A?
(A) 0
(B) 25
(C) 50
(D) 100
QUESTION 78
APPLICATIONS – SUPPORT KNOWLEDGE
A new classroom is designed for 25 people and has a total area of 500 square feet. What is the
required CFM of ventilation?
(A) 60 𝐶𝐶𝐶𝐶𝐶𝐶
(B 250 𝐶𝐶𝐶𝐶𝐶𝐶
(C) 310 𝐶𝐶𝐶𝐶𝐶𝐶
(D) 520 𝐶𝐶𝐶𝐶𝐶𝐶
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QUESTION 79
APPLICATIONS – SUPPORT KNOWLEDGE
The unit "clo" is used to describe the thermal insulation provided by which of the below?
(A) Wall insulation
(B) Roof insulation
(C) Garment insulation
(D) Heating equipment insulation
QUESTION 80
APPLICATIONS – SUPPORT KNOWLEDGE
Which of the following codes are least likely to be required to be checked with regards to the
installation of a commercial gas furnace?
(A) NFPA 54
(B) ASHRAE 90.1
(C) NFPA 70
(D) International Fuel Gas Code
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SECTION 4
AM SOLUTIONS
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SOLUTION 1
PRINCIPLES BASIC ENGINEERING PRACTICE
Honey has a dynamic viscosity of 1000 poise, a specific heat capacity of 0.6 cal/g-oC, and a
density of 0.05 oz/mL. The kinematic viscosity of honey, in ft2/sec, is most nearly?
Kinematic viscosity is defined as:
The specific heat capacity is irrelevant.
ν=
𝜇𝜇
𝜌𝜌
The dynamic (or absolute) viscosity is irrelevant:
The density is:
lbf
1 Pa ∙ s 0.020885 ft 2
𝑙𝑙𝑙𝑙𝑙𝑙 ∙ 𝑠𝑠
𝜇𝜇 = (1000 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝) ∗
∗
= 2.09
10 poise
𝑓𝑓𝑡𝑡 2
𝑃𝑃𝑃𝑃
𝜌𝜌 = 0.05
𝑜𝑜𝑜𝑜 0.0625 𝑙𝑙𝑙𝑙𝑙𝑙 28317 𝑚𝑚𝑚𝑚
𝑙𝑙𝑙𝑙𝑙𝑙
∗
∗
= 88.4 3
3
𝑚𝑚𝑚𝑚
1 𝑜𝑜𝑜𝑜
1 𝑓𝑓𝑡𝑡
𝑓𝑓𝑡𝑡
Therefore the kinematic viscosity is:
𝑓𝑓𝑓𝑓
2
𝑙𝑙𝑙𝑙𝑙𝑙 ∗ 2
𝑙𝑙𝑙𝑙𝑙𝑙 ∙ 𝑠𝑠
𝑓𝑓𝑡𝑡 3
𝑠𝑠 = 0.76 𝑓𝑓𝑡𝑡
ν = 2.09
∗
∗
32.2
𝑓𝑓𝑡𝑡 2
88.4 𝑙𝑙𝑙𝑙𝑙𝑙
𝑙𝑙𝑙𝑙𝑙𝑙
𝑠𝑠
The correct answer is A
(A) 0.76
Mechanical PE AM Session Sample Exam 1 Solutions -1
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SOLUTION 2
PRINCIPLES – BASIC ENGINEERING PRACTICE
A 3 hp, 1800 RPM motor operates on 208 volts, 3 phase, 60 hertz power. Assume a power
factor of 0.9 and a service factor of 1.15. If the motor is 85% efficient, how many amps must be
supplied to the motor?
The service factor and RPM are irrelevant to the problem. Service factor is the overload
capacity that is built into the motor. Since the motor has a service factor of 1.15, the motor can
handle 115% of the rated motor capacity. The motor speed is given in units of revolutions per
minute (RPM).
The power supplied to the motor is the apparent power (S), which consists of real (P) and
reactive (Q) power. The useful power used to turn the motor is given as P. Q is a result of a
portion of the current producing a magnetic field and does not contribute to useful power.
Therefore, the total current supplied to the motor must be found from the apparent power.
𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆 𝑆𝑆 =
The real power delivered to the motor is:
The apparent power is:
𝑃𝑃 =
𝑃𝑃
𝐵𝐵𝐵𝐵𝐵𝐵
𝑎𝑎𝑎𝑎𝑎𝑎 𝑃𝑃 =
,
𝑃𝑃𝑃𝑃
𝜂𝜂
(3 𝐻𝐻𝐻𝐻)
𝑊𝑊
∗ 746
= 2,633 𝑊𝑊
𝐻𝐻𝐻𝐻
0.85
𝑆𝑆 =
2,633 𝑊𝑊
= 2,925 𝑊𝑊
0.9
For 3-phase power, 𝑆𝑆 = 𝐼𝐼 ∗ 𝑉𝑉 ∗ √3. Therefore, the current supplied to the motor is:
Correct answer is B.
𝐼𝐼 =
𝑆𝑆
𝑉𝑉 ∗ √3
=
2,925 𝑊𝑊
208 𝑉𝑉 ∗ √3
= 8 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎
(B) 𝟖𝟖 𝑨𝑨
Mechanical PE AM Session Sample Exam 1 Solutions -2
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SOLUTION 3
PRINCIPLES – BASIC ENGINEERING PRACTICE
A pump is required to pump 300 GPM at 100' of total dynamic head, if electricity costs $0.15 per
kWh and the pump operates 8 hours a day, 365 days a year, then what will be the total cost for
operating the pump? Pump efficiency = 0.65. Motor efficiency = 0.95. Specific gravity = 1.
First, calculate the total mechanical power required to move 300 GPM at 100' TDH.
𝑃𝑃𝑚𝑚𝑚𝑚𝑚𝑚ℎ 𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤,𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝[𝐻𝐻𝐻𝐻] =
ℎ𝑓𝑓𝑓𝑓 ∗ 𝑄𝑄𝑔𝑔𝑔𝑔𝑔𝑔 ∗ (𝑆𝑆𝑆𝑆)
3956
𝑃𝑃𝑚𝑚𝑚𝑚𝑚𝑚ℎ 𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤,𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝[𝐻𝐻𝐻𝐻] =
100 ∗ 300 ∗ (1)
3956
𝑃𝑃𝑚𝑚𝑚𝑚𝑚𝑚ℎ 𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤,𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝[𝐻𝐻𝐻𝐻] = 7.58 𝐻𝐻𝐻𝐻
Next, remember that additional electricity will be required to run the pump and motor due to the
inefficiencies in the equipment
𝑃𝑃𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒 [𝐻𝐻𝐻𝐻] =
𝑃𝑃𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒 [𝐻𝐻𝐻𝐻] =
Now convert HP to KW.
7.58 𝐻𝐻𝐻𝐻
0.65 ∗ 0.95
7.58 𝐻𝐻𝐻𝐻
0.65 ∗ 0.95
𝑃𝑃𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒 [𝐻𝐻𝐻𝐻] = 12.3 𝐻𝐻𝐻𝐻
𝑃𝑃𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒 [𝐾𝐾𝐾𝐾] = 9.2 𝐾𝐾𝐾𝐾
Find kWh by multiplying kW by run time (hours per year)
𝑃𝑃𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑡𝑡𝑦𝑦 [𝐾𝐾𝐾𝐾𝐾𝐾] = 9.2 ∗ 8 ∗ 365
𝑃𝑃𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒 [𝐾𝐾𝐾𝐾𝐾𝐾] = 26,864 𝑘𝑘𝑘𝑘ℎ
Finally, use the utility rate to determine the cost of running the pump.
Correct answer is D.
$0.15
� = $4,030
𝑘𝑘𝑘𝑘ℎ
𝑌𝑌𝑌𝑌𝑌𝑌𝑌𝑌𝑌𝑌𝑌𝑌 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 = 26,864 𝑘𝑘𝑘𝑘ℎ ∗ �
(D) $𝟒𝟒, 𝟎𝟎𝟎𝟎𝟎𝟎
Mechanical PE AM Session Sample Exam 1 Solutions -3
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SOLUTION 4
PRINCIPLES BASIC ENGINEERING PRACTICE
A new machine is installed in order to increase productivity. This new machine costs $75,000
and has an ongoing operating and maintenance cost of $500 per month. The new machine will
save $1,000 per month and will have a salvage value of $7,500 after 10 years. If the interest
rate is 5%, then what is the annual cost of the new machine?
This problem involves finding the total annual cost. First, convert all your terms to an annual
value.
Initial Cost [Negative value = money lost at the beginning of the lifetime]
𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹 𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛 𝑡𝑡𝑡𝑡 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦 𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒 𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡, 5%
𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣 (𝑃𝑃)𝑡𝑡𝑡𝑡 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣 (𝐴𝐴)
𝐴𝐴
𝐴𝐴 = 𝑃𝑃 ∗ ( , 5%, 10)
𝑃𝑃
𝐴𝐴 = −$75,000 ∗ 0.12950
𝐴𝐴𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖 = −$9,712.50
Salvage Value [Positive value = money gained at the end of the lifetime]
𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹 𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣 (𝐹𝐹)𝑡𝑡𝑡𝑡 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣 (𝐴𝐴)
𝐴𝐴
𝐴𝐴 = 𝐹𝐹 ∗ ( , 5%, 10)
𝐹𝐹
𝐴𝐴 = $10,000 ∗ 0.07950
𝐴𝐴𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 = $795
Operating & Maintenance Cost [Negative value = Money lost]
𝐴𝐴𝑂𝑂&𝑀𝑀 =
−$500
∗ 12 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚ℎ𝑠𝑠 = −$6,000 𝑝𝑝𝑝𝑝𝑝𝑝 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦
𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚ℎ
Savings [Positive value = Money gained]
𝐴𝐴𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 =
Finally, sum up all annual values.
$2,000
∗ 12 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚ℎ𝑠𝑠 = $24,000 𝑝𝑝𝑝𝑝𝑝𝑝 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦
𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚ℎ
𝐴𝐴𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 = 𝐴𝐴𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖 + 𝐴𝐴𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 + 𝐴𝐴𝑂𝑂&𝑀𝑀 + 𝐴𝐴𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠
𝐴𝐴𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 = −$9,712.50 + $795 + −$6,000 + $24,000
𝐴𝐴𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 = $9,082.50
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Correct answer is C.
(C)
$ ,
Mechanical PE AM Session Sample Exam 1 Solutions -5
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SOLUTION 5
PRINCIPLES – THERMODYNAMICS
The following pressure-enthalpy diagram shows the vapor compression cycle for the refrigerant
R-134a. Which of the following statements is true?
A is incorrect, the Compressor work does occur at Step 2, however the enthalpy is shown to
change. The compressor work occurs at constant entropy, not constant enthalpy.
B is incorrect, the expansion device is used at Step 4, however the process is isobaric or
constant pressure.
C is correct, the compressor work does occur at Step 2 and the process is isentropic
since it follows the constant entropy line.
D is incorrect; the expansion device decreases the pressure.
Mechanical PE AM Session Sample Exam 1 Solutions -6
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(A) The compressor work occurs at Step 2 at constant enthalpy.
(B) The expansion device is used at Step 4 to conduct an isobaric process.
(C) At Step 2, the compressor conducts an isentropic process.
(D) At Step 4, the expansion device increases the pressure of the refrigerant at constant
enthalpy.
Mechanical PE AM Session Sample Exam 1 Solutions -7
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SOLUTION 6
PRINCIPLES – THERMODYNAMICS
A heat pump provides 70.0 MBTUH of cooling, 66.0 MBTUH of heating and uses 5.9 KW of
electricity. What is the COP of the heat pump during heating?
The coefficient of performance for a heat pump during heating is the heat energy delivered by
the heat pump divided by the work done by the compressor.
𝐶𝐶𝐶𝐶𝐶𝐶 =
𝐶𝐶𝐶𝐶𝐶𝐶 =
The correct answer is A.
𝑄𝑄𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻
𝑊𝑊𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑟𝑟𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒
66.0 𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀ℎ
= 3.3
𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀ℎ
�
5.9 𝑘𝑘𝑘𝑘 ∗ �3.412
𝑘𝑘𝑘𝑘
(A) 3.3
(B) 3.5
(C) 4.8
(D) 6.8
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SOLUTION 7
PRINCIPLES – THERMODYNAMICS
An R134a refrigeration cycle operates at pressures, 23.77 psia and 150 psia. If the refrigerant
is superheated to 10 F and refrigerant is subcooled to 100 F, then what is the COP? Assume
the compressor is 90% efficient.
The first step in completing this problem is to find the conditions on the R134a, pressureenthalpy diagram.
Point D is found first as the intersection of the horizontal pressure line 150 psia and the subcooled liquid temperature of 100 F.
Point A is found by a vertical downward line from point D, which is the constant enthalpy line,
and the intersection of the suction pressure 23.8 psia line.
Point B is found by the intersection of the constant pressure line 23.8 psia and the superheated
temperature line 10 F.
The next step is to calculate the net refrigeration.
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𝐵𝐵𝐵𝐵𝐵𝐵
� = 𝑚𝑚̇ ∗ (ℎ1 − ℎ4 )
𝑄𝑄 �
ℎ𝑟𝑟
ℎ4 = ℎ𝑓𝑓,𝑠𝑠𝑠𝑠𝑠𝑠 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐− 150 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝/100
𝐹𝐹
= 45.1
𝐵𝐵𝐵𝐵𝐵𝐵
𝑙𝑙𝑙𝑙𝑙𝑙
ℎ1 = ℎ𝑏𝑏 = ℎ𝑔𝑔,𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 ℎ𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒−23.8 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝/ 10𝐹𝐹 = 105
𝑈𝑈𝑈𝑈𝑈𝑈 𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒 𝑡𝑡𝑡𝑡 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑓𝑓𝑓𝑓𝑓𝑓 𝑛𝑛𝑛𝑛𝑛𝑛 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒 → 𝑄𝑄 �
𝐵𝐵𝐵𝐵𝐵𝐵
𝑙𝑙𝑙𝑙𝑙𝑙
𝐵𝐵𝐵𝐵𝐵𝐵
� = 𝑚𝑚̇ ∗ (ℎ1 − ℎ4 )
ℎ𝑟𝑟
𝑁𝑁𝑁𝑁𝑥𝑥𝑥𝑥 𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 𝐶𝐶, 𝑖𝑖𝑖𝑖 𝑡𝑡ℎ𝑒𝑒 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑤𝑤𝑤𝑤𝑤𝑤 100% 𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒
𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒 𝑖𝑖𝑖𝑖 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 → 𝑠𝑠𝑏𝑏 = 𝑠𝑠𝑐𝑐
𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒 𝑖𝑖𝑖𝑖 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 → 0.227
ℎ𝑐𝑐 = ℎ2,
𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖
𝐵𝐵𝐵𝐵𝐵𝐵
= 𝑠𝑠𝑐𝑐
𝑙𝑙𝑙𝑙𝑙𝑙 °𝑅𝑅
= 122.5 𝐵𝐵𝐵𝐵𝐵𝐵/𝑙𝑙𝑙𝑙𝑙𝑙
𝑠𝑠𝑠𝑠𝑠𝑠𝑐𝑐𝑒𝑒 𝑡𝑡ℎ𝑒𝑒 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑖𝑖𝑖𝑖 90% 𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒 𝑡𝑡ℎ𝑒𝑒𝑒𝑒 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤 𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤 𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛 𝑡𝑡𝑡𝑡 𝑏𝑏𝑏𝑏 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑,
𝑠𝑠𝑠𝑠 𝑡𝑡ℎ𝑒𝑒 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 ℎ2 𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣 𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤 𝑏𝑏𝑏𝑏 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙
90% =
90% =
ℎ2,
ℎ2,
122.5 − 101
ℎ2, 𝑎𝑎𝑎𝑎𝑡𝑡𝑢𝑢𝑢𝑢𝑢𝑢 − 101
𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎
𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎
𝐶𝐶𝐶𝐶𝐶𝐶 =
𝐶𝐶𝐶𝐶𝐶𝐶 =
ℎ2,𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖 − ℎ1
ℎ2, 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 − ℎ1
= 124.9
= 124.9
𝐵𝐵𝐵𝐵𝐵𝐵
𝑙𝑙𝑙𝑙𝑙𝑙
𝐵𝐵𝐵𝐵𝐵𝐵
𝑙𝑙𝑙𝑙𝑙𝑙
𝑛𝑛𝑛𝑛𝑛𝑛 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟
𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤
ℎ1 − ℎ4
105 − 45.1
=
= 3.01
ℎ2,𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 − ℎ1 124.9 − 105
The correct answer is C, 3.0
(C) 3.0
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SOLUTION 8
PRINCIPLES – THERMODYNAMICS
An R134a refrigerant cycle operates at the following evaporator and condenser temperatures, 5
F and 105 F. Assume a refrigerant flow rate of 180 pounds per hour. How many tons of cooling
is provided? Assume an ideal cycle.
The first step in completing this problem is to find the conditions on the R134a, pressureenthalpy diagram.
Point D is found first as the intersection of the saturated liquid curve and the discharge
temperature of 105 F.
Point A is found by a vertical downward line from point D, which is the constant enthalpy line,
and the intersection of the suction temperature 5 F line.
Point B is found by the intersection of the suction temperature 5 F line and the saturated vapor
line.
Point C is found as the intersection of the constant entropy line from Point B and the discharge
temperature horizontal line of 105.
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The next step is to calculate the net refrigeration.
𝐵𝐵𝐵𝐵𝐵𝐵
� = 𝑚𝑚̇ ∗ (ℎ1 − ℎ4 )
𝑄𝑄 �
ℎ𝑟𝑟
ℎ4 = ℎ𝑓𝑓,105 𝐹𝐹 = 46.9
ℎ1 = ℎ𝑏𝑏 = ℎ𝑔𝑔,5 𝐹𝐹
𝐵𝐵𝐵𝐵𝐵𝐵
𝑙𝑙𝑙𝑙
ℎ1 = 103.9 𝐵𝐵𝐵𝐵𝐵𝐵/𝑙𝑙𝑙𝑙𝑙𝑙
𝑈𝑈𝑈𝑈𝑈𝑈 𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒 𝑡𝑡𝑡𝑡 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑓𝑓𝑓𝑓𝑓𝑓 𝑄𝑄 → 𝑄𝑄 �
𝑄𝑄 �
𝐵𝐵𝐵𝐵𝐵𝐵
� = 𝑚𝑚̇ ∗ (ℎ1 − ℎ4 )
ℎ𝑟𝑟
𝐵𝐵𝐵𝐵𝐵𝐵
� = 180 ∗ (103.9 − 46.9)
ℎ𝑟𝑟
𝐵𝐵𝐵𝐵𝐵𝐵
𝐵𝐵𝐵𝐵𝐵𝐵
� = 10,260
𝑄𝑄 �
ℎ𝑟𝑟
ℎ𝑟𝑟
𝐵𝐵𝐵𝐵𝐵𝐵
ℎ𝑟𝑟 = 0.86 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 𝑜𝑜𝑜𝑜 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑔𝑔
𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 𝑡𝑡𝑡𝑡 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 → 𝑄𝑄 =
𝐵𝐵𝐵𝐵𝐵𝐵
12,000
ℎ𝑟𝑟
𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇
10,260
The correct answer is A, 0.86 tons.
(A) 0.86 tons
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SOLUTION 9
PRINCIPLES – PSYCHROMETRICS
An air conditioning system provides air at 55 F DB/ 53 F WB, in order to maintain space
conditions at 75 F DB/50% relative humidity. What is the dew point of the space conditions?
Use your psychrometric chart to find the dew point of the space conditions.
The correct answer is C, 55 F.
(A) 47 F
(B) 50 F
(C) 55 F
(D) 63 F
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SOLUTION 10
PRINCIPLES – PSYCHROMETRICS
A new electric heater is provided to heat 5,000 CFM of air at 60 F DB/ 55 F WB. If a 3 KW
heater is provided, then what is the resulting Humidity Ratio?
For this equation, use your psychrometric chart and find the point identified by 60 F DB/55 F
WB. The psychrometric chart below shows the point as red dot. After the air passes the heater,
the air will be sensibly heated, which will not change the moisture content. Thus the humidity
ratio will remain the same.
The value shown is nearly .008 lb of water/lb of dry air.
The correct answer is C.
(C) .008 lb of water/lb of dry air
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SOLUTION 11
PRINCIPLES – PSYCHROMETRICS
How many lbs of water are in the air of a 10 ft x 50 ft x 10 ft room at 70oF DB air and 70%
relative humidity?
For this problem, you should have your psychrometric chart.
From the psychrometric chart, the humidity ratio and density values at 70oF DB, 70% relative
humidity are as follows.
ℎ𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 = 𝜔𝜔 = 0.011
𝑙𝑙𝑙𝑙 𝑜𝑜𝑜𝑜 𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤
𝑙𝑙𝑙𝑙 𝑜𝑜𝑜𝑜 𝑎𝑎𝑎𝑎𝑎𝑎
𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 𝑜𝑜𝑜𝑜 𝑎𝑎𝑎𝑎𝑎𝑎 = 𝜌𝜌 = 0.074
𝑙𝑙𝑙𝑙
𝑓𝑓𝑡𝑡 3
Use the density, humidity ratio and the total volume of air to find the water content in the air.
𝑚𝑚𝑚𝑚𝑚𝑚𝑠𝑠𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤 = 𝜔𝜔 ∗ 𝜌𝜌 ∗ 𝑉𝑉 = 0.011
𝑙𝑙𝑙𝑙 𝑜𝑜𝑜𝑜 𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤
𝑙𝑙𝑙𝑙 𝑜𝑜𝑜𝑜 𝑎𝑎𝑎𝑎𝑎𝑎
∗ 0.074
∗ (10𝑓𝑓𝑓𝑓 ∗ 50𝑓𝑓𝑓𝑓 ∗ 10𝑓𝑓𝑓𝑓)
𝑙𝑙𝑙𝑙 𝑜𝑜𝑜𝑜 𝑎𝑎𝑎𝑎𝑎𝑎
𝑓𝑓𝑡𝑡 3
𝑚𝑚𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤 = 4.07 𝑙𝑙𝑙𝑙 𝑜𝑜𝑜𝑜 𝑤𝑤𝑎𝑎𝑡𝑡𝑡𝑡𝑡𝑡
Correct answer is B.
(A) 1.6 lbs
(B) 4.1lbs
(C) 5.2 lbs
(D) 5.9 lbs
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SOLUTION 12
PRINCIPLES – PSYCHROMETRICS
If you are designing an HVAC system at an elevation of 7,500 ft above sea level with outside air
conditions of 60 F DB and 90% relative humidity, then what is the density and dew point of the
outside air?
In order to complete this problem, you need psychrometric charts for different elevations. For
the exam you should have the necessary psychrometric charts to complete this problem. You
can get Psychrometric Charts from ASHRAE at the following website or at the following free
websites.
http://www.techstreet.com/ashrae/standards/complete-set-of-psychrometric-charts-ip?gateway_code=ashrae&product_id=1724246
http://www.coolerado.com/pdfs/Psychrmtrcs/5000Psychrmtrc11x17.pdf
http://www.coolerado.com/pdfs/Psychrmtrcs/7500Psychrmtrc11x17.pdf
Based on Psychrometric Chart 5 (elevation 7,500 ft), the dew point is 57.1 F dp and the density
is 0.0574 lb/ft3. The dew point is found by navigating to 60 F dry bulb and 90% relative humidity
and then drawing a horizontal line to the saturation curve (left).
Correct answer is C.
(A) 0.0631 lb/ft3 and 57.1 F dp
(B) 0.0631 lb/ft3 and 60.0 F dp
(C) 0.0574 lb/ft3 and 57.1 F dp
(D) 0.0759 lb/ft3 and 55.1 F dp
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SOLUTION 13
PRINCIPLES – PSYCHROMETRICS
A cooling tower is used to cool 10,000 lbs/hr of condenser water from 105 F to 80 F. If the air is
designed to enter at ambient air conditions of 85 F/50% relative humidty and leave at 95 F/80%
relative humidity, then what is the mass flow rate of air required?
The energy from the condenser water is transferred to the air.
∗
∗
,
,
,
10,000
,
∗
,
,
∗ 1.0
∗ 105
80
∗ 55.0
34.6
12,255
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,
Correct answer is D.
(D) 12,255 lbs/hr
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SOLUTION 14
PRINCIPLES – PSYCHROMETRICS
An evaporative cooler is used to cool an airstream of 95 F/20% relative humidity. If the air
leaves the evaporative air cooler at 80 F, then what is the new relative humidity?
In an evaporative air cooler, the dry bulb temperature is lowered by the removal of sensible heat
and the replacement of the sensible heat with the addition of latent heat/moisture. It is assumed
that the air enters and leaves at constant enthalpy, thus you must follow the 95 F/20% point
along the constant enthalpy line until it hits the 80 F dry bulb line.
The correct answer is B, 48% relative humidity.
(B) 48%
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SOLUTION 15
PRINCIPLES – PSYCHROMETRICS
45 F/80% relative humidity air is heated with a 8,100 Btu/hr electric heater. Then a humidifier is
used to increase the relative humidity to 50%. What is the amount of energy provided, if the
flow rate is 250 CFM. Assume sea level and a density of 0.075 lb/ft3.
First calculate the exiting dry bulb temperature of the heater.
1.08 ∗
∗ ∆ → 8,100
/
45
1.08 ∗ 250 ∗
75
Now find the enthalpy of the final condition, 75 F DB/50% RH and the beginning condition, 45
F/80% RH.
28.1
→
4.5 ∗ 250 ∗ 28.1
4.5 ∗
16.2
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∗∆
13,250
The correct answer is most nearly, (D), 13,250 Btu/hr.
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SOLUTION 16
PRINCIPLES – PSYCHROMETRICS
Calculate the CFM at a cooling coil required to provide 12,000 Btu/hr of cooling to a space at
leaving coil conditions of 55 F DB/90% RH. Assume an elevation of 5,000 ft and the space is
kept at 75 F DB/50% RH. The cooling coil is also a 100% recirculation unit.
First, you need to find the enthalpy and the entering and leaving conditions of the coil, using the
5,000 ft elevation psychrometric chart.
𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃ℎ𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 𝑐𝑐ℎ𝑎𝑎𝑎𝑎𝑎𝑎, 55 𝐹𝐹𝐹𝐹𝐹𝐹& 90% 𝑅𝑅𝑅𝑅 → ℎ𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐,
75 𝐹𝐹𝐷𝐷𝐷𝐷& 50% 𝑅𝑅𝑅𝑅 → ℎ𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐,
𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒
𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙
= 30.23
Next use the total heat equation, but with a non-standard density
75 𝐹𝐹𝐹𝐹𝐹𝐹& 50% 𝑅𝑅𝑅𝑅 → 𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷 = .06135
55 𝐹𝐹𝐹𝐹𝐹𝐹& 90% 𝑅𝑅𝑅𝑅 → 𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷 = .06377
𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝑖𝑖𝑡𝑡𝑡𝑡 =
𝑄𝑄 = 𝐶𝐶𝐶𝐶𝐶𝐶 ∗
The correct answer is A, 520 CFM.
𝑙𝑙𝑙𝑙
𝑓𝑓𝑡𝑡 3
𝑙𝑙𝑙𝑙
𝑓𝑓𝑡𝑡 3
0.06135 + .06377
= 0.0625
2
𝑙𝑙𝑙𝑙𝑙𝑙
1 ℎ𝑜𝑜𝑜𝑜𝑜𝑜
∗ (𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷 3 )(ℎ𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐,
𝑓𝑓𝑡𝑡
60 𝑚𝑚𝑚𝑚𝑚𝑚
12,000 = 𝐶𝐶𝐶𝐶𝐶𝐶 ∗
𝐵𝐵𝐵𝐵𝐵𝐵
𝐿𝐿𝐿𝐿
𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒
= 24.04
− ℎ𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐,
𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙
60 𝑚𝑚𝑚𝑚𝑚𝑚 0.0625 𝑙𝑙𝑙𝑙𝑙𝑙
∗(
)(30.23 − 24.04)
1 ℎ𝑜𝑜𝑜𝑜𝑜𝑜
𝑓𝑓𝑡𝑡 3
𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 = 517 𝐶𝐶𝐶𝐶𝐶𝐶
(A) 520 CFM
Mechanical PE AM Session Sample Exam 1 Solutions -22
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𝐵𝐵𝐵𝐵𝐵𝐵
𝑙𝑙𝑙𝑙
SOLUTION 17
PRINCIPLES – HEAT TRANSFER
Determine the amount of energy required to cool 50 lbs of turkey from 70 F to 0 F, with the
following properties.
Property of Turkey
Specific Heat above Freezing
Value
𝐵𝐵𝐵𝐵𝐵𝐵
𝑙𝑙𝑙𝑙 ∗ 𝐹𝐹
𝐵𝐵𝐵𝐵𝐵𝐵
105
𝑙𝑙𝑙𝑙
𝐵𝐵𝐵𝐵𝐵𝐵
0.35
𝑙𝑙𝑙𝑙 ∗ 𝐹𝐹
0.67
Enthalpy of Fusion (Freezing Point = 26 F)
Specific Heat below Freezing
Use the following equations to calculate the total amount of energy to cool turkey from 70 F to 0
F.
𝑄𝑄70 𝑡𝑡𝑡𝑡 26 = 50 𝑙𝑙𝑙𝑙𝑠𝑠 ∗ 0.67
𝐵𝐵𝐵𝐵𝐵𝐵
∗ (70℉ − 26℉)
𝑙𝑙𝑙𝑙 ∗ 𝐹𝐹
𝑄𝑄70 𝑡𝑡𝑡𝑡 26 = 1,474 𝐵𝐵𝐵𝐵𝐵𝐵
𝑄𝑄𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 = 50 𝑙𝑙𝑙𝑙𝑙𝑙 ∗ 105
𝐵𝐵𝐵𝐵𝐵𝐵
𝑙𝑙𝑙𝑙
𝑄𝑄𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 = 5,250 𝐵𝐵𝐵𝐵𝐵𝐵
𝑄𝑄26 𝑡𝑡𝑡𝑡 0 = 50 𝑙𝑙𝑙𝑙𝑙𝑙 ∗ 0.35
𝐵𝐵𝐵𝐵𝐵𝐵
∗ (26℉ − 0℉)
𝑙𝑙𝑙𝑙 ∗ 𝐹𝐹
𝑄𝑄26 𝑡𝑡𝑡𝑡 0 = 455 𝐵𝐵𝐵𝐵𝐵𝐵
Correct answer is A.
𝑄𝑄𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 = 7,179 𝐵𝐵𝐵𝐵𝐵𝐵
(A) 7,200 Btu
(B) 9,600 Btu
(C) 12,200 Btu
(D) 14,400 Btu
Mechanical PE AM Session Sample Exam 1 Solutions -23
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SOLUTION 18
PRINCIPLES – HEAT TRANSFER
A new solar hot water heating system has two loops, (1) solar hot water loop and a (2) domestic
hot water loop. The loops are separated by a plate frame heat exchanger. If the solar hot water
fluid enters the heat exchanger at 160 F and leaves at 130 F and the domestic hot water enters
at 80 F and leaves at 120 F, then what is the LMTD? Assume a parallel flow heat exchanger.
𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿 =
∆𝑇𝑇𝑎𝑎 − ∆𝑇𝑇𝑏𝑏
∆𝑇𝑇
ln( 𝑎𝑎 )
∆𝑇𝑇𝑏𝑏
∆𝑇𝑇𝑎𝑎 = 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 𝑎𝑎𝑎𝑎 𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒 = 160 − 80 = 80 𝐹𝐹
∆𝑇𝑇𝑏𝑏 = 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 𝑎𝑎𝑎𝑎 𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒 = 130 − 120 = 10 𝐹𝐹
𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿 =
The correct answer is C.
80 − 10
= 33.7 𝐹𝐹
80
ln( )
10
(A) 30 F
(B) 31.3 F
(C) 33.7 F
(D) 35 F
Mechanical PE AM Session Sample Exam 1 Solutions -24
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SOLUTION 19
PRINCIPLES – HEAT TRANSFER
A feedwater heater is used to heat 50 lbm/min of water from 100 F to 120 F. Steam enters the
feedwater heater at 300 F. What is the mass flow rate of the steam, if the steam leaves at 200
F? Assume the feedwater heater is at a pressure of 14.7 psia.
The first step is to calculate the gain in energy by the water.
𝑄𝑄 �
𝐵𝐵𝐵𝐵𝐵𝐵
� = 𝑚𝑚̇ ∗ 𝑐𝑐𝑝𝑝 (𝑇𝑇𝑓𝑓 − 𝑇𝑇𝑖𝑖 )
𝑄𝑄 �
ℎ𝑟𝑟
𝑙𝑙𝑙𝑙𝑙𝑙
𝑚𝑚𝑚𝑚𝑚𝑚
𝐵𝐵𝐵𝐵𝐵𝐵
𝐵𝐵𝐵𝐵𝐵𝐵
� = 50
∗ 60
∗ 1.0
∗ (120 − 100 𝐹𝐹)
𝑚𝑚𝑚𝑚𝑚𝑚
ℎ𝑟𝑟
𝑙𝑙𝑙𝑙𝑙𝑙 ℉
ℎ𝑟𝑟
𝑄𝑄 = 60,000
𝐵𝐵𝐵𝐵𝐵𝐵
ℎ𝑟𝑟
The second step is to find the enthalpy of the steam entering and leaving the feedwater heater.
Use your steam tables and you will find that the 300 F steam entering is in the superheated
region.
ℎ𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠,
ℎ𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠,
𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒
= 1,190
𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑔𝑔
= 168
𝐵𝐵𝐵𝐵𝐵𝐵
𝑙𝑙𝑙𝑙𝑙𝑙
𝐵𝐵𝐵𝐵𝐵𝐵
𝑙𝑙𝑙𝑙𝑙𝑙
Finally, the heat gain by the water is equal to the lost by the steam.
𝑄𝑄 = 60,000
𝐵𝐵𝐵𝐵𝐵𝐵
= 𝑚𝑚̇ ∗ (ℎ𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠,𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒 − ℎ𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠,𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 )
ℎ𝑟𝑟
𝑄𝑄 = 60,000
The correct answer is C, 60 lbm/hr
𝐵𝐵𝐵𝐵𝐵𝐵
= 𝑚𝑚̇ ∗ (1,190 − 168)
ℎ𝑟𝑟
𝑚𝑚̇ = 58.7
𝑙𝑙𝑙𝑙𝑙𝑙
ℎ𝑟𝑟
(C) 60 lbm/hr
Mechanical PE AM Session Sample Exam 1 Solutions -25
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SOLUTION 20
PRINCIPLES – HEAT TRANSFER
A client is considering replacing 1,200 ft2 of windows with a new high efficiency window. What
would be the reduction in conductive heat losses with the given below conditions?
You need to find the overall heat transfer coefficient, U. But it is easier to find the overall heat
transfer coefficient by first converting all the heat transfer coefficients to R-values.
1
1
1
1.5
1
1.5
1
2.0
1
2.0
2∗
2∗
1/4"/ 12"/1
0.05
2∗
1/4"/ 12"/1
0.03
∗
;
0.8
2.8
1"/ 12"/1
0.8
3.4
Next find conductive heat gains
∗
1/
1
∗ 1,200 ∗ 95
2.8
75
8,571
/
1
∗ 1,200 ∗ 95
3.4
75
7,152
/
The difference is 1,419 Btu/hr and the correct answer is most nearly, (A)
Mechanical PE AM Session Sample Exam 1 Solutions -26
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SOLUTION 21
PRINCIPLES – HEAT TRANSFER
A chilled beam of 4” outer diameter and 100 feet in total length is located in a space with a dry
bulb temperature of 75 F. If the surface temperature of the chilled beam is 50 F, then what is
the radiative heat transfer to the space? Assume the emissivity of the chilled beam is 0.8.
In order to calculate the radiative heat gain, you should use the following equation.
∗
°
∗ ∗
1.714
∗
50
1.714
460
10
∗
10
°
∗ 4/12 ∗ 100
510 ° ;
∗ 105 ∗ 0.8 ∗ 510
75
535
105
460
535 ° ;
2,054
/
The correct answer is most nearly, (D) 2,000 Btu/hr
Mechanical PE AM Session Sample Exam 1 Solutions -27
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SOLUTION 22
PRINCIPLES – HEAT TRANSFER
A motor is located in a mechanical room at 75 F DB. The motor operates at 10 BHP and is 95%
efficient. The remaining energy is released as heat to the space. The surface temperature of
the motor is 120 F. What is the overall heat transfer coefficient? The motor is approximated as
a box with dimensions, 2’ X 1’ X 1’. The motor is suspended, such that all surfaces are
exposed.
First find the amount of heat that is emitted to the space.
𝑄𝑄ℎ𝑒𝑒𝑒𝑒𝑒𝑒 = 10 𝐻𝐻𝐻𝐻 ∗
𝐵𝐵𝐵𝐵𝐵𝐵
ℎ𝑟𝑟 ∗ 0.05 = 1,273 𝐵𝐵𝐵𝐵𝐵𝐵/ℎ𝑟𝑟
1 𝐻𝐻𝐻𝐻
2,545
Next the amount of heat emitted to the space is also found with the following equation
𝑄𝑄ℎ𝑒𝑒𝑒𝑒𝑒𝑒 = 1,273
1,273
𝐵𝐵𝐵𝐵𝐵𝐵
= ℎ𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜 ∗ 𝐴𝐴𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 ∗ (𝑇𝑇𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 − 𝑇𝑇𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 )
ℎ𝑟𝑟
𝐵𝐵𝐵𝐵𝐵𝐵
= ℎ𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜 ∗ (2 ∗ 1 ∗ 2 + 2 ∗ 1 ∗ 1 + 2 ∗ 2 ∗ 1) ∗ (120 − 75)
ℎ𝑟𝑟
ℎ𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜 = 2.8 𝐵𝐵𝐵𝐵𝐵𝐵/(ℎ𝑟𝑟 − 𝑓𝑓𝑡𝑡 2 − ℉)
The correct answer is (B) 2.8 Btu/(hr-ft^2-℉).
Mechanical PE AM Session Sample Exam 1 Solutions -28
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SOLUTION 23
PRINCIPLES – HEAT TRANSFER
A 100 ft long, 6” schedule 40 steel pipe carries chilled water at 45 F. The pipe has 2” of
insulation. The insulation k-value is 0.29 Btu-in/(hr-ft2-F). What is the surface temperature at
the outside of the insulation?
Assume the following:
Ambient temperature = 75 F, h_conv = 1.0 Btu/hr-ft2-F, h_rad = 1.0 Btu/hr-ft2-F, the heat transfer
throug the inner/outer pipe walls is negligible.
In order to find the surface temperature of the pipe, you must equate the heat transfer through
the insulation to the heat transfer from the surface of the pipe to convection and radiation.
,
∗
∗
∗
0.29
2
100
∗
1.0
0.145 ∗ 157.1 ∗
∗6
1.0
45
6.525
0.145 ∗
2.145 ∗
∗
0.145
1
12
157.1
2.0
2.0 ∗ 157.1 ∗ 75
150
2.0 ∗
156.525
73.0
The correct answer is most nearly, (C) 73.0 F.
Mechanical PE AM Session Sample Exam 1 Solutions -29
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SOLUTION 24
PRINCIPLES – FLUID MECHANICS
500 GPM of water flows through a 6" schedule 40 pipe. If the pipe has a relative roughness
value of .0004 and a Reynolds number of 250,000, then what is the pressure drop through 100
ft of pipe?
Use the Darcy Weisbach equation to find the pressure drop.
ℎ=
𝑓𝑓𝑓𝑓𝑣𝑣 2
[𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷 𝑊𝑊𝑊𝑊𝑊𝑊𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠ℎ 𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸]
2𝐷𝐷𝐷𝐷
𝑓𝑓𝑓𝑓
𝑤𝑤ℎ𝑒𝑒𝑒𝑒𝑒𝑒 ℎ = 𝑓𝑓𝑓𝑓 𝑜𝑜𝑜𝑜 ℎ𝑒𝑒𝑒𝑒𝑒𝑒; 𝑓𝑓 = 𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷 𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓; 𝑣𝑣 = 𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣 � �,
𝑠𝑠𝑠𝑠𝑠𝑠
𝐷𝐷 = 𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 [𝑓𝑓𝑓𝑓], 𝑔𝑔 = 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 [32.2
𝑓𝑓𝑓𝑓
]
𝑠𝑠𝑠𝑠𝑠𝑠 2
First, you need to use the Moody Diagram to find the friction factor. [Use your MERM]
𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 = 0.004 𝑎𝑎𝑎𝑎𝑎𝑎 𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑑𝑑 ′ 𝑠𝑠 𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁 = 250,000
𝑓𝑓 = .018
𝑓𝑓𝑡𝑡 3
𝑓𝑓𝑓𝑓
1
𝑠𝑠𝑠𝑠𝑠𝑠
� = 5.55
𝑉𝑉 = 500 𝑔𝑔𝑔𝑔𝑔𝑔 ∗ �
�∗�
2
𝑠𝑠
448.83 𝑔𝑔𝑔𝑔𝑔𝑔
0.20063 𝑓𝑓𝑡𝑡
1
Mechanical PE AM Session Sample Exam 1 Solutions -30
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Use the Darcy Weisbach equation
ℎ=
The correct answer is A.
. 018 ∗ 100 𝑓𝑓𝑓𝑓 ∗ 𝑣𝑣 2
𝑓𝑓𝑓𝑓
2 ∗ 0.506 𝑓𝑓𝑓𝑓 ∗ 32.2
𝑠𝑠𝑠𝑠𝑠𝑠 2
ℎ = 1.7 𝑓𝑓𝑓𝑓 𝑜𝑜𝑜𝑜 ℎ𝑒𝑒𝑒𝑒𝑒𝑒
(A) 1.7 ft of head
(B) 3.3 ft of head
(C) 4.2 ft of head
(D) 4.9 ft of head
Mechanical PE AM Session Sample Exam 1 Solutions -31
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SOLUTION 25
PRINCIPLES – FLUID MECHANICS
Water at 50 °F flows through a 6" schedule 40 steel pipe at a volumetric flow rate of 200 gallons
per minute. What is the Reynolds number?
First, find the kinematic viscosity of water at 50 F [refer to your Mechanical Engineering
Reference Manual]
𝑣𝑣 = 1.410 𝑥𝑥 10−5 𝑓𝑓𝑡𝑡 2 /𝑠𝑠𝑠𝑠𝑠𝑠
Next find the velocity of water through the 6" schedule 40 steel pipe.
𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼 𝐴𝐴𝐴𝐴𝐴𝐴𝑎𝑎6" 𝑆𝑆𝑆𝑆ℎ 40 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 = 0.2006 𝑓𝑓𝑡𝑡 2
𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼 𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷6" 𝑆𝑆𝑆𝑆ℎ 40 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 = 0.5054 𝑓𝑓𝑓𝑓
[𝑓𝑓𝑓𝑓𝑓𝑓 𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 𝑎𝑎𝑎𝑎𝑎𝑎 𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 𝑜𝑜𝑜𝑜 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝,
𝑟𝑟𝑟𝑟𝑟𝑟𝑒𝑒𝑒𝑒 𝑡𝑡𝑡𝑡 𝑀𝑀𝑀𝑀𝑀𝑀ℎ𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸 𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅 𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀]
𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 200
200
𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔
𝑓𝑓𝑡𝑡 3
𝑡𝑡𝑡𝑡
𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚
𝑠𝑠𝑠𝑠𝑠𝑠
𝑓𝑓𝑡𝑡 3
𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔
1
� = 0.4456
∗�
𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚
𝑠𝑠𝑠𝑠𝑠𝑠
448.83
𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉 = 0.4456
Finally, find Reynolds number.
𝑓𝑓𝑡𝑡 3
÷ 0.2006 𝑓𝑓𝑡𝑡 2
𝑠𝑠𝑠𝑠𝑠𝑠
𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉 = 2.22 𝑓𝑓𝑓𝑓/𝑠𝑠𝑠𝑠𝑠𝑠
𝑅𝑅𝑅𝑅 =
𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣 ∗ 𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑
𝑣𝑣
𝑓𝑓𝑓𝑓
∗ 0.5054𝑓𝑓𝑓𝑓
𝑠𝑠𝑠𝑠𝑠𝑠
𝑅𝑅𝑅𝑅 =
−5
1.41 𝑥𝑥 10 𝑓𝑓𝑡𝑡 2 / sec
Correct answer is A.
2.22
𝑅𝑅𝑅𝑅 = 79,573
(A) 𝟖𝟖𝟖𝟖, 𝟎𝟎𝟎𝟎𝟎𝟎
(B) 100,000
(C) 150,000
(D) 220,000
Mechanical PE AM Session Sample Exam 1 Solutions -32
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SOLUTION 26
PRINCIPLES – FLUID MECHANICS
A new 480 Volt, 3-phase motor is provided to serve the following pump with the design
conditions shown in the below figure. If the pump's impeller diameter is increased by a factor of
1.25 and the speed of the pump remains the same, then what will be the resulting flow rate?
First, find the current design point at the intersection of the system curve and the pump curve.
200 𝐺𝐺𝐺𝐺𝐺𝐺 𝑎𝑎𝑎𝑎 80 𝑓𝑓𝑓𝑓 𝑜𝑜𝑜𝑜 ℎ𝑒𝑒𝑒𝑒𝑒𝑒
If the impeller diameter is increased and the speed remains the same, then the affinity laws can
be used.
𝐷𝐷𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖 𝐹𝐹𝐹𝐹𝐹𝐹𝑤𝑤𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖
=
𝐷𝐷𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓
𝐹𝐹𝐹𝐹𝐹𝐹𝑤𝑤𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓
200
𝐷𝐷𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖
=
1.25 ∗ 𝐷𝐷𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖 𝐹𝐹𝐹𝐹𝐹𝐹𝑤𝑤𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓
𝐹𝐹𝐹𝐹𝐹𝐹𝑤𝑤𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 = 250 𝐺𝐺𝐺𝐺𝐺𝐺
Mechanical PE AM Session Sample Exam 1 Solutions -33
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Correct answer is C.
(C) 250 GPM
Mechanical PE AM Session Sample Exam 1 Solutions -34
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SOLUTION 27
PRINCIPLES – FLUID MECHANICS
Rainwater at 60oF flows down a trench at velocity of 10 ft/s. The cross section of the trench is
illustrated below. What is the Reynolds number of the fluid?
3.5 ft
2 ft
6 ft
Reynolds number is found by the following equation:
𝑅𝑅𝑅𝑅 =
𝜌𝜌𝜌𝜌𝐷𝐷ℎ 𝑣𝑣𝑣𝑣ℎ
=
𝜇𝜇
𝜈𝜈
where Dh is the equivalent hydraulic diameter of the trench.
For fluid partially flowing through a rectangular channel, the hydraulic diameter is:
𝐷𝐷ℎ =
4ℎ𝑤𝑤
4 ∗ 2𝑓𝑓𝑓𝑓 ∗ 6𝑓𝑓𝑓𝑓
=
= 4.8𝑓𝑓𝑓𝑓
𝑤𝑤 + 2ℎ 6𝑓𝑓𝑓𝑓 + 2 ∗ 2𝑓𝑓𝑓𝑓
where h is the height of the fluid in the trench and w is the width of the trench.
The kinematic viscosity, 𝜈𝜈, of water at 60oF is 1.217x10-5 ft2/s.
Solve for the Reynolds number.
𝑓𝑓𝑓𝑓
� ∗ 4.8𝑓𝑓𝑓𝑓
𝑠𝑠
6
𝑅𝑅𝑅𝑅 =
2 = 3.94 × 10
𝑓𝑓𝑡𝑡
1.217 × 10−5
𝑠𝑠
�10
The correct answer is C.
(A) 2.47 x 106
(B) 3.29 x 106
(C) 3.94 x 106
(D) 4.93 x 106
Mechanical PE AM Session Sample Exam 1 Solutions -35
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SOLUTION 28
PRINCIPLES – ENERGY/MASS BALANCE
A new heat exchanger uses steam to heat water. Steam (50 lb/min) enters the heat exchanger
at 20 psia, fully saturated. The condensate is not recovered. What mass flow rate of water can
be expected to provide a 40 degree change in temperature?
This problem is an energy balance type problem. The energy from the steam is transferred to
water.
Steam Available Energy
𝑄𝑄 = 𝑚𝑚̇ ∗ ℎ𝑓𝑓𝑓𝑓
From the steam tables, find the latent heat of vaporization of 20 psia steam.
ℎ𝑓𝑓𝑓𝑓 = 960.05
Energy Balance Equation
50
𝐵𝐵𝐵𝐵𝐵𝐵
𝑙𝑙𝑙𝑙
∗ 960.05
= 𝑚𝑚̇ ∗ 𝑐𝑐𝑝𝑝 ∗ ∆𝑇𝑇
𝑙𝑙𝑙𝑙
𝑚𝑚𝑚𝑚𝑚𝑚
50
𝑙𝑙𝑙𝑙
𝐵𝐵𝐵𝐵𝐵𝐵
∗ 960.05
= 𝑚𝑚̇ ∗ 1.0 ∗ 40
𝑚𝑚𝑚𝑚𝑚𝑚
𝑙𝑙𝑙𝑙
𝑐𝑐𝑝𝑝 = 1.0
The correct answer is B.
𝐵𝐵𝐵𝐵𝐵𝐵
𝑙𝑙𝑙𝑙
𝐵𝐵𝐵𝐵𝐵𝐵
𝑓𝑓𝑓𝑓𝑓𝑓 𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤, 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 𝑡𝑡𝑡𝑡 𝑀𝑀𝑀𝑀𝑅𝑅𝑅𝑅
𝑙𝑙𝑙𝑙𝑙𝑙 ∗ 𝐹𝐹
𝑚𝑚̇ = 1200 𝑙𝑙𝑙𝑙/𝑚𝑚𝑚𝑚𝑚𝑚
(A) 1,000 lb/min
(B) 1,200 lb/min
(C) 1,300 lb/min
(D) 1,400 lb/min
Mechanical PE AM Session Sample Exam 1 Solutions -37
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SOLUTION 29
PRINCIPLES – ENERGY/MASS BALANCE
50 GPM of water flows through a 2-1/2" pipe and then branches into (2) pipes shown in the
figure below. If the velocity in the 1-1/2" pipe is measured at 4 ft/sec, then what is the velocity
through the 1" pipe?
This is a flow rate balance type problem.
50
𝑄𝑄2.5" = 𝑄𝑄1.5" + 𝑄𝑄1"
𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 1 𝑚𝑚𝑚𝑚𝑚𝑚
1 𝑓𝑓𝑡𝑡 3
𝑓𝑓𝑓𝑓
𝑠𝑠𝑠𝑠𝑠𝑠
𝑓𝑓𝑓𝑓
𝑠𝑠𝑠𝑠𝑠𝑠
∗
∗
=4
∗ 60
∗ 0.01417 𝑓𝑓𝑡𝑡 2 + 𝑋𝑋
∗ 60
∗ 0.00597 𝑓𝑓𝑡𝑡 2
𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 60 sec 7.48 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔
𝑠𝑠𝑠𝑠𝑠𝑠
𝑚𝑚𝑚𝑚𝑚𝑚
𝑠𝑠𝑠𝑠𝑠𝑠
𝑚𝑚𝑚𝑚𝑚𝑚
50
𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔
1 𝑓𝑓𝑡𝑡 3
𝑓𝑓𝑓𝑓
𝑠𝑠𝑠𝑠𝑠𝑠
𝑓𝑓𝑓𝑓
𝑠𝑠𝑠𝑠𝑠𝑠
∗
=4
∗ 60
∗ 0.01417 𝑓𝑓𝑡𝑡 2 + 𝑋𝑋
∗ 60
∗ 0.00597 𝑓𝑓𝑡𝑡 2
𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 7.48 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔
𝑠𝑠𝑠𝑠𝑠𝑠
𝑚𝑚𝑚𝑚𝑚𝑚
𝑠𝑠𝑠𝑠𝑠𝑠
𝑚𝑚𝑚𝑚𝑚𝑚
𝑿𝑿 = 𝟗𝟗. 𝟏𝟏𝟏𝟏
𝒇𝒇𝒇𝒇
𝒔𝒔𝒔𝒔𝒔𝒔
The correct answer is most nearly D, 9.1 ft per second.
(A) 3.9 ft/sec
(B) 5.6 ft/sec
(C) 7.2 ft/sec
(D) 9.1 ft/sec
Mechanical PE AM Session Sample Exam 1 Solutions -38
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SOLUTION 30
PRINCIPLES – ENERGY/MASS BALANCE
A concentrated solar power system produces steam at 800oF, 400 psia and 1000 gpm. If the
steam expands through a turbine to 5 psia at 75% isentropic efficiency, and 85% mechanical
efficiency, how much power (KW) does the turbine generate?
The power generated in the turbine is defined as
𝑃𝑃𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 = 𝜂𝜂𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒 ∗ 𝜂𝜂𝑚𝑚𝑚𝑚𝑚𝑚ℎ𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 ∗ 𝑚𝑚̇𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖 ∗ (ℎ𝑖𝑖𝑖𝑖 – ℎ𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒 )
Refer to your steam tables for the density of steam at 800oF
𝑙𝑙𝑙𝑙𝑙𝑙
)
𝑓𝑓𝑡𝑡 3
𝜌𝜌 = 0.0196 (
So, the ideal mass flow rate is:
𝑚𝑚̇𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖 = 1000 𝑔𝑔𝑔𝑔𝑔𝑔 ∗
𝑓𝑓𝑡𝑡 3
𝑙𝑙𝑙𝑙𝑙𝑙
𝑚𝑚𝑚𝑚𝑚𝑚
𝑙𝑙𝑙𝑙𝑙𝑙
∗ 0.0196 3 ∗ 60
= 157.2
𝑓𝑓𝑡𝑡
ℎ𝑟𝑟
ℎ𝑟𝑟
7.48 𝑔𝑔𝑔𝑔𝑔𝑔
From the superheated steam tables at 800oF and 400 psia:
ℎ𝑖𝑖𝑖𝑖 = 1416.9
𝐵𝐵𝐵𝐵𝐵𝐵
𝐵𝐵𝐵𝐵𝐵𝐵
𝑎𝑎𝑎𝑎𝑎𝑎 𝑠𝑠𝑖𝑖𝑖𝑖 = 1.6849
𝑙𝑙𝑙𝑙𝑙𝑙
𝑙𝑙𝑙𝑙𝑙𝑙 ∗ °𝑅𝑅
Since the expansion of an ideal turbine is isentropic, the leaving entropy of the steam is equal to
the entering entropy.
𝑠𝑠𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒 = 𝑠𝑠𝑖𝑖𝑖𝑖 = 1.6849
𝐵𝐵𝐵𝐵𝐵𝐵
𝑙𝑙𝑙𝑙𝑙𝑙 ∗ °𝑅𝑅
The quality, x, of the steam leaving the turbine may be found with this entropy. So, from the
saturated steam tables at 5 psia,
𝑠𝑠𝑓𝑓 = 0.2349
𝐵𝐵𝐵𝐵𝐵𝐵
𝐵𝐵𝐵𝐵𝐵𝐵
𝑎𝑎𝑎𝑎𝑎𝑎 𝑠𝑠𝑓𝑓𝑓𝑓 = 1.6089
𝑙𝑙𝑙𝑙𝑙𝑙 ∗ °𝑅𝑅
𝑙𝑙𝑙𝑙𝑙𝑙 ∗ °𝑅𝑅
ℎ𝑓𝑓 = 130.2
𝐵𝐵𝐵𝐵𝐵𝐵
𝐵𝐵𝐵𝐵𝐵𝐵
𝑎𝑎𝑎𝑎𝑎𝑎 ℎ𝑓𝑓𝑓𝑓 = 1000.5
𝑙𝑙𝑙𝑙𝑙𝑙
𝑙𝑙𝑙𝑙𝑙𝑙
𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆 𝑠𝑠𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒 = 𝑠𝑠𝑓𝑓 + 𝑥𝑥 ∗ 𝑠𝑠𝑓𝑓𝑓𝑓 , 𝑡𝑡ℎ𝑒𝑒 𝑞𝑞𝑞𝑞𝑞𝑞𝑞𝑞𝑞𝑞𝑞𝑞𝑞𝑞 𝑥𝑥 𝑖𝑖𝑖𝑖:
𝑥𝑥 =
𝑠𝑠𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒 – 𝑠𝑠𝑓𝑓
1.6849 – 0.2349
=
= 0.90
1.6089
𝑠𝑠𝑓𝑓𝑓𝑓
Then, the quality can be used to find the leaving enthalpy.
ℎ𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒 = ℎ𝑓𝑓 + 𝑥𝑥 ∗ ℎ𝑓𝑓𝑓𝑓
Mechanical PE AM Session Sample Exam 1 Solutions -39
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130.2
𝐵𝐵𝐵𝐵𝐵𝐵
𝐵𝐵𝐵𝐵𝐵𝐵
𝐵𝐵𝐵𝐵𝐵𝐵
� = 1031.9
+ 0.90 ∗ �1000.5
𝑙𝑙𝑙𝑙𝑙𝑙
𝑙𝑙𝑙𝑙𝑙𝑙
𝑙𝑙𝑙𝑙𝑙𝑙
Finally, the turbine power can be solved:
𝑃𝑃𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 = 0.75 ∗ 0.85 ∗ 157.2
Now convert to MW.
Correct answer is A.
𝑙𝑙𝑙𝑙𝑙𝑙
𝐵𝐵𝐵𝐵𝐵𝐵
𝐵𝐵𝐵𝐵𝐵𝐵
� = 38,583𝐵𝐵𝑡𝑡𝑢𝑢ℎ
∗ �1416.9
– 1031.9
ℎ𝑟𝑟
𝑙𝑙𝑙𝑙𝑙𝑙
𝑙𝑙𝑙𝑙𝑙𝑙
= 38,583 𝐵𝐵𝐵𝐵𝐵𝐵ℎ ∗
1 𝐾𝐾𝐾𝐾
= 11.3 𝐾𝐾𝐾𝐾
3,412 𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀ℎ
(A) 11.3
(B) 15.1
(C) 17.7
(D) 21.2
Mechanical PE AM Session Sample Exam 1 Solutions -40
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SOLUTION 31
PRINCIPLES – ENERGY/MASS BALANCE
In order to determine the amount of condensate produced, the change in humidity ratio must
first be determined.
∆𝑊𝑊𝐿𝐿𝐿𝐿 = 𝑊𝑊𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖 − 𝑊𝑊𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 = .0113 − .0087 [
∆𝑊𝑊𝐿𝐿𝐿𝐿 = .0026
. 0026
𝑙𝑙𝑙𝑙𝑙𝑙 𝑜𝑜𝑜𝑜 𝐻𝐻20
]
𝑙𝑙𝑙𝑙𝑙𝑙 𝑜𝑜𝑜𝑜 𝑑𝑑𝑑𝑑𝑑𝑑 𝑎𝑎𝑎𝑎𝑎𝑎
𝑙𝑙𝑙𝑙𝑙𝑙 𝑜𝑜𝑜𝑜 𝐻𝐻20
𝑙𝑙𝑙𝑙𝑙𝑙 𝑜𝑜𝑜𝑜 𝑑𝑑𝑟𝑟𝑟𝑟 𝑎𝑎𝑎𝑎𝑎𝑎
𝑙𝑙𝑙𝑙𝑙𝑙 𝑜𝑜𝑜𝑜 𝑑𝑑𝑑𝑑𝑑𝑑 𝑎𝑎𝑎𝑎𝑎𝑎
𝑙𝑙𝑙𝑙𝑙𝑙 𝑜𝑜𝑜𝑜 𝐻𝐻20
𝑙𝑙𝑙𝑙𝑙𝑙 𝑜𝑜𝑜𝑜 𝐻𝐻20
∗ 5,000 𝐶𝐶𝐶𝐶𝑀𝑀 𝑜𝑜𝑜𝑜 𝑑𝑑𝑑𝑑𝑑𝑑 𝑎𝑎𝑎𝑎𝑎𝑎 ∗ 0.075
= 0.98
3
𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚
𝑓𝑓𝑓𝑓
𝑙𝑙𝑙𝑙𝑙𝑙 𝑜𝑜𝑜𝑜 𝑑𝑑𝑑𝑑𝑑𝑑 𝑎𝑎𝑎𝑎𝑎𝑎
0.98
𝑓𝑓𝑓𝑓 3
7.48 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔
𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔
𝑙𝑙𝑙𝑙𝑙𝑙 𝑜𝑜𝑜𝑜 𝐻𝐻20
∗
∗
= 0.12
3
1 𝑓𝑓𝑓𝑓
𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚
𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚
62.4 𝑙𝑙𝑙𝑙𝑙𝑙 𝑜𝑜𝑜𝑜 𝐻𝐻20
The correct answer is (B) 0.12 GPM.
Mechanical PE AM Session Sample Exam 1 Solutions -41
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SOLUTION 32
PRINCIPLES – ENERGY/MASS BALANCE
𝐶𝐶4 𝐻𝐻10 + 6.5(𝑂𝑂2 + 3.76𝑁𝑁2 ) → 4𝐶𝐶𝑂𝑂2 + 5𝐻𝐻2 𝑂𝑂 + 24.44𝑁𝑁2
What is the air to fuel ratio? Assume the equation is balanced.
𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹 𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 𝑡𝑡ℎ𝑒𝑒 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 𝑜𝑜𝑜𝑜 𝑎𝑎𝑎𝑎𝑎𝑎, 6.5(𝑂𝑂2 + 3.76𝑁𝑁2 )
6.5 ∗ (16 ∗ 2 + 3.76 ∗ 14 ∗ 2) = 892
𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆 𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 𝑡𝑡ℎ𝑒𝑒 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 𝑜𝑜𝑜𝑜 𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓, 𝐶𝐶4 𝐻𝐻10
4 ∗ 12 + 10 ∗ 1 = 58
𝐴𝐴𝐴𝐴𝐴𝐴 𝑡𝑡𝑡𝑡 𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 =
𝑚𝑚𝑎𝑎𝑎𝑎𝑎𝑎
892
=
= 15.4
58
𝑚𝑚𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓
The answer is most nearly, (D), 15.4.
Mechanical PE AM Session Sample Exam 1 Solutions -42
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SOLUTION 33
APPLICATION – HEATING/COOLING LOADS
An air handler is used to cool 1,750 CFM of air at 78 F/60% RH to 55 F DB/54 F WB. What is
the sensible cooling provided by the air handler? The chilled water provided to the air handler
enters at 42 F and leaves at 52 F. Assume that the density of air is 0.075 lbs per cubic foot.
The sensible heat equation is as shown below:
𝐵𝐵𝐵𝐵𝐵𝐵
] = 𝑚𝑚̇𝑐𝑐𝑝𝑝 ∆𝑇𝑇
ℎ
𝑄𝑄𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 [
𝑤𝑤ℎ𝑒𝑒𝑒𝑒𝑒𝑒 𝑐𝑐𝑝𝑝 = 0.240
𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 𝐹𝐹𝐹𝐹
𝑚𝑚̇= (
𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀
)*(
60 𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀
1 𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻
𝐵𝐵𝐵𝐵𝐵𝐵
𝑙𝑙𝑙𝑙𝑙𝑙−℉
)*(
;
.075 𝑙𝑙𝑙𝑙𝑙𝑙
)
𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 𝐹𝐹𝐹𝐹.
𝐵𝐵𝐵𝐵𝐵𝐵
] = 𝟏𝟏. 𝟎𝟎𝟎𝟎 ∗ ∆𝑻𝑻 ∗ 𝑪𝑪𝑪𝑪𝑪𝑪
ℎ
𝑸𝑸𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔 [
Now, plug in the CFM and the change in dry bulb temperature.
𝐵𝐵𝐵𝐵𝐵𝐵
� = 1.08 ∗ (78 − 55) ∗ 1,750
𝑄𝑄𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 �
ℎ
𝑄𝑄𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 �
𝐵𝐵𝐵𝐵𝐵𝐵
� = 43,470 𝐵𝐵𝐵𝐵𝐵𝐵/ℎ
ℎ
Correct answer is C.
(A) 18,900 𝐵𝐵𝐵𝐵𝐵𝐵/ℎ
(B) 27,600 𝐵𝐵𝐵𝐵𝐵𝐵/ℎ
(C) 𝟒𝟒𝟒𝟒, 𝟓𝟓𝟓𝟓𝟓𝟓 𝑩𝑩𝑩𝑩𝑩𝑩/𝒉𝒉
(D) 52,700 𝐵𝐵𝐵𝐵𝐵𝐵/ℎ
Mechanical PE AM Session Sample Exam 1 Solutions -43
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SOLUTION 34
APPLICATION – HEATING/COOLING LOADS
The net refrigeration effect of a refrigeration system is equal to120 tons of cooling. The
compressor has a suction pressure of 40 PSIA and a discharge pressure of 200 PSIA. If the
compressor has a capacity equal to 100 kilowatts, then what is the COP of the refrigeration
system?
COP stands for coefficient of performance, which is the ratio of the “work out” to the “work in”.
In this case the “work out” of the refrigeration system is the net refrigeration effect and the “work
in” is the compressor work.
𝐶𝐶𝐶𝐶𝐶𝐶 =
𝑊𝑊𝑜𝑜𝑜𝑜𝑜𝑜 𝑁𝑁𝑁𝑁𝑁𝑁 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒
=
𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 𝑤𝑤𝑜𝑜𝑜𝑜𝑜𝑜
𝑊𝑊𝑖𝑖𝑖𝑖
In order to complete this equation, convert to like terms:
100 𝐾𝐾𝐾𝐾 ∗
1 𝑡𝑡𝑡𝑡𝑡𝑡 𝑜𝑜𝑜𝑜 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟
= 28.4 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 𝑜𝑜𝑜𝑜 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟
3.517 𝐾𝐾𝐾𝐾
𝐶𝐶𝐶𝐶𝐶𝐶 =
120 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡
= 4.2
28.4 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡
Correct answer is A.
(A) 𝟒𝟒.2
(B) 4.5
(C) 4.9
(D) 5.3
Mechanical PE AM Session Sample Exam 1 Solutions -44
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SOLUTION 35
APPLICATION – HEATING/COOLING LOADS
A dedicated outside air unit cools 4,000 cfm of outside air to 55oF DB, 53oF WB. The outside
ambient condition is 87oF DB, 75oF WB. What is the total cooling load required by the air
handling unit? Assume that the density of air is 0.075 lbs per cubic foot.
The total heat equation for air with a density of 0.075 lb/ft3 is as shown below:
𝐵𝐵𝐵𝐵𝐵𝐵
lbm ∗ min
] = 4.5
∗ (∆h) ∗ CFM
ℎ
ft 3 ∗ h
𝑄𝑄𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 [
Use the psychometric chart to find the enthalpies at the entering and leaving conditions.
At 87oF DB, 75oF WB, the air entering the dedicated outside air unit has an enthalpy of:
ℎ𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒 = 38.5 𝐵𝐵𝐵𝐵𝐵𝐵/𝑙𝑙𝑙𝑙
At 55oF DB, 53oF WB, the air entering the dedicated outside air unit has an enthalpy of:
ℎ𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 = 22.0 𝐵𝐵𝐵𝐵𝐵𝐵/𝑙𝑙𝑙𝑙
Therefore, the total cooling load is:
𝐵𝐵𝐵𝐵𝐵𝐵
lbm ∗ min
Btu
ft 3
] = 4.5
∗
(38.5
−
22.0)
∗
4.000
ℎ
ft 3 ∗ h
lb
min
𝑄𝑄𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 [
𝐵𝐵𝐵𝐵𝐵𝐵
� = 297,000 𝐵𝐵𝐵𝐵𝐵𝐵/ℎ
𝑄𝑄𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 �
ℎ
Correct answer is B.
(A) 576,000 𝐵𝐵𝐵𝐵𝐵𝐵/ℎ
(B) 𝟐𝟐𝟐𝟐𝟐𝟐, 𝟎𝟎𝟎𝟎𝟎𝟎 𝑩𝑩𝑩𝑩𝑩𝑩/𝒉𝒉
(C) 138,000 𝐵𝐵𝐵𝐵𝐵𝐵/ℎ
(D) 59,840 𝐵𝐵𝐵𝐵𝐵𝐵/ℎ
Mechanical PE AM Session Sample Exam 1 Solutions -45
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SOLUTION 36
APPLICATION – HEATING/COOLING LOADS
A data center is maintained at 72oF, 50% RH. 8,000 cfm is supplied to the space, 520 cfm of
which is outside air. The outside air temperature is 30oF DB, 27oF WB. To maintain the space
humidity, steam is dispersed into the supply air stream of the duct. Calculate the flow rate of the
steam in lb/hr. Assume there is no latent load in the space and no latent cooling occurs.
Given: At 72oF, 50% RH, air density is 0.074
𝑙𝑙𝑙𝑙
𝑓𝑓𝑓𝑓 3
, humidity ratio is 58.6
𝑔𝑔𝑔𝑔𝑤𝑤𝑤𝑤𝑤𝑤
𝑙𝑙𝑙𝑙𝑑𝑑𝑑𝑑𝑑𝑑
Since there is no latent load in the space, the steam is the only source of moisture. No latent
cooling occurs, meaning moisture is only required to be made up at the outside air. The supply
air flow rate (8,000 cfm) is unnecessary information.
To find the total addition of steam required, find the difference between the humidity ratios at the
final design space conditions (𝜔𝜔𝑓𝑓 ) and the outside air conditions (𝜔𝜔𝑜𝑜𝑜𝑜 ).
∆𝜔𝜔 = 𝜔𝜔𝑓𝑓 − 𝜔𝜔𝑜𝑜𝑜𝑜
At 30oF DB, 27oF WB, the outside air humidity ratio is 𝜔𝜔𝑜𝑜𝑜𝑜 = 16.9
The final design humidity ratio is given as 𝜔𝜔𝑓𝑓 = 58.6
Therefore,
∆𝜔𝜔 = 58.6
𝑔𝑔𝑔𝑔𝑤𝑤𝑤𝑤𝑤𝑤
𝑙𝑙𝑙𝑙𝑑𝑑𝑑𝑑𝑑𝑑
.
𝑔𝑔𝑔𝑔𝑤𝑤𝑤𝑤𝑤𝑤
𝑙𝑙𝑙𝑙𝑑𝑑𝑑𝑑𝑑𝑑
.
𝑔𝑔𝑔𝑔𝑤𝑤𝑤𝑤𝑤𝑤
𝑔𝑔𝑔𝑔𝑤𝑤𝑤𝑤𝑤𝑤
𝑔𝑔𝑔𝑔𝑤𝑤𝑤𝑤𝑤𝑤
− 16.9
= 41.7
.
𝑙𝑙𝑙𝑙𝑑𝑑𝑑𝑑𝑑𝑑
𝑙𝑙𝑙𝑙𝑑𝑑𝑑𝑑𝑑𝑑
𝑙𝑙𝑙𝑙𝑑𝑑𝑑𝑑𝑑𝑑
Calculate the steam flow rate based on the outside air flow rate:
𝑚𝑚̇𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 �
𝑚𝑚̇𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 �
𝑙𝑙𝑙𝑙
𝑔𝑔𝑔𝑔𝑤𝑤𝑤𝑤𝑤𝑤
𝑙𝑙𝑙𝑙𝑤𝑤𝑤𝑤𝑤𝑤
𝑙𝑙𝑙𝑙
60𝑚𝑚𝑚𝑚𝑚𝑚
� = 𝐶𝐶𝐶𝐶𝐶𝐶 ∗ 𝜌𝜌 � 3 � ∗ ∆𝜔𝜔 �
�∗�
�
� ∗ �0.000143
ℎ𝑟𝑟
𝑙𝑙𝑙𝑙𝑑𝑑𝑑𝑑𝑑𝑑
𝑔𝑔𝑔𝑔𝑤𝑤𝑤𝑤𝑤𝑤
𝑓𝑓𝑓𝑓
ℎ𝑟𝑟
𝑙𝑙𝑙𝑙
520 𝑓𝑓𝑡𝑡 3 0.074 𝑙𝑙𝑙𝑙
41.7 𝑔𝑔𝑔𝑔𝑤𝑤𝑤𝑤𝑤𝑤
𝑙𝑙𝑙𝑙𝑤𝑤𝑤𝑤𝑤𝑤
60𝑚𝑚𝑚𝑚𝑚𝑚
�=
�∗�
�∗�
�
∗�
� ∗ �0.000143
3
ℎ𝑟𝑟
𝑚𝑚𝑚𝑚𝑚𝑚
𝑙𝑙𝑙𝑙𝑑𝑑𝑑𝑑𝑑𝑑
𝑔𝑔𝑔𝑔𝑤𝑤𝑤𝑤𝑤𝑤
ℎ𝑟𝑟
𝑓𝑓𝑓𝑓
Correct answer is B.
(A) 0.23
𝑚𝑚̇𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 = 13.8
𝑙𝑙𝑙𝑙
ℎ𝑟𝑟
𝑙𝑙𝑙𝑙
ℎ𝑟𝑟
(B) 𝟏𝟏𝟏𝟏. 𝟖𝟖
𝒍𝒍𝒍𝒍
𝒉𝒉𝒉𝒉
Mechanical PE AM Session Sample Exam 1 Solutions -46
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(C) 186
(D) 211
𝑙𝑙𝑙𝑙
ℎ𝑟𝑟
𝑙𝑙𝑙𝑙
ℎ𝑟𝑟
Mechanical PE AM Session Sample Exam 1 Solutions -47
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SOLUTION 37
APPLICATION – HEATING/COOLING LOADS
Space temperatures reach 65oF, 70% humidity during low load conditions. The total supply air
to the room is 160 cfm. How much reheat is required to maintain a maximum setpoint of 60%
relative humidity?
To maintain 60% humidity in the space, sensible heating is provided to the supply air. The
sensible heating equation is:
1.08 ∗ ∆ ∗
Use the psychometric chart to find the initial conditions of 65oF, 70% RH. Move right
(horizontal) along the chart for sensible heating until 60% humidity is reached. At 60% humidity,
the dry bulb is 69.5oF.
Use the equation above to calculate the sensible heating.
1.08 ∗ 69.5
65
∗ 160
778
Correct answer is A.
(A)
/
(B) 1730
/
(C) 6840
/
(D) 1720
/
Mechanical PE AM Session Sample Exam 1 Solutions -48
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SOLUTION 38
APPLICATION – HEATING/COOLING LOADS
Assume you have 2,000 CFM of outside air required and the design outside air conditions are
87 F DB/78 F WB. The building has a total volume of 200,000 ft3. The infiltration rate is found
to equal 2 air changes per hour. What is the total cooling load due to outside air and infiltration,
if the cooling coil cools the air to 55 F DB/53 F WB?
First calculate the infiltration CFM
𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 → 200,000 𝑓𝑓𝑡𝑡 3 ∗ �2
1 ℎ𝑜𝑜𝑜𝑜𝑜𝑜
𝑐𝑐ℎ𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎
�∗
= 6,666.7 𝐶𝐶𝐶𝐶𝐶𝐶
60 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚
ℎ𝑜𝑜𝑜𝑜𝑜𝑜
Next calculate the change in enthalpy from the outside air conditions to the cooling coil
conditions. Use your psychrometric chart to find the enthalpy of the two conditions.
ℎ𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 = 22.0 𝐵𝐵𝐵𝐵𝐵𝐵/𝑙𝑙𝑙𝑙
ℎ𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜 = 41.5 𝐵𝐵𝐵𝐵𝐵𝐵/𝑙𝑙𝑙𝑙
𝑄𝑄 = 4.5 ∗ 𝐶𝐶𝐶𝐶𝐶𝐶 ∗ (∆ℎ) → = 4.5 ∗ (6,666.7 + 2000) ∗ (41.4 − 22)
𝑄𝑄 = 756,603
The correct answer is (D) 63 tons.
𝐵𝐵𝐵𝐵𝐵𝐵
1 𝑡𝑡𝑡𝑡𝑡𝑡
∗
= 63.1 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡
ℎ𝑟𝑟 12,000 𝐵𝐵𝐵𝐵𝐵𝐵
ℎ𝑟𝑟
Mechanical PE AM Session Sample Exam 1 Solutions -49
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SOLUTION 39
APPLICATION – HEATING/COOLING LOADS
A boiler is located in a mechanical room. The product data indicates that the boiler emits
75,000 Btu/hr of sensible heat to the space. There are also four 4’ long T5 28 watt light bulbs
located in the space. If the mechanical room is to be maintained at 85 F DB and the outside air
design condition is 78 F, then how much outside air must be supplied to the mechanical room?
Assume the same amount of air is exhausted from the mechanical room.
This is a simple sensible heat equation, but first you need to convert the light wattage to Btu/hr.
𝐵𝐵𝐵𝐵𝐵𝐵
ℎ𝑟𝑟 = 382 𝐵𝐵𝐵𝐵𝐵𝐵/ℎ𝑟𝑟
4 ∗ (28 𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤 ∗
1 𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤
3.412
Then use the sensible heat equation to find the volumetric flow rate
(75,000
𝑄𝑄 = 1.08 ∗ 𝑉𝑉̇ ∗ ∆𝑇𝑇
𝐵𝐵𝐵𝐵𝐵𝐵
𝐵𝐵𝐵𝐵𝐵𝐵
+ 382
) = 1.08 ∗ 𝑉𝑉̇ ∗ (85 − 78)
ℎ𝑟𝑟
ℎ𝑟𝑟
𝑉𝑉̇ = 9,971 𝐶𝐶𝐶𝐶𝐶𝐶
The correct answer is most nearly (B) 9,980 CFM
Mechanical PE AM Session Sample Exam 1 Solutions -50
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SOLUTION 40
APPLICATION – HEATING/COOLING LOADS
Calculate the building’s heat load given the following information. The building is to be
maintained at 69 F DB and the design outside air conditions are 5 F DB/75% RH. Assume no
infiltration and no outside air loads.
Wall area: 20,000 SF, Roof area: 2,500 SF, Window area: 2,000 SF
Wall: R-11; Roof: R-20; Window: U-Value = 0.75 Btu/(hr-ft2-F)
In heating the conduction load is the primary load in a building, which is governed by the below
equation.
∗
∗ ∆
1
∗ 20,000 ∗ 69
11
5
116,363
1
∗ 2,500 ∗ 69
20
5
8,000
0.75 ∗ 2,000 ∗ 69
5
96,000
/
/
/
220,363
The correct answer is (B) 220,000 Btu/hr.
Mechanical PE AM Session Sample Exam 1 Solutions -51
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SECTION 5
PM SOLUTIONS
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SOLUTION 41
An air handler supplies 5,000 CFM at a temperature of 55 F. The air handler was designed for
1,000 CFM of outside air at 87 F DB and 78 F WB. The remaining return air from the space is
at 77 F DB and 55% relative humidity. What are the entering conditions of the air into the coil,
in DB and WB?
This problem involves finding the mixed air condition of two airstreams. Remember that only
the dry bulb temperature, humidity ratio and enthalpy are linearly related.
First, find the mixed dry bulb temperature, using the lever rule.
𝐶𝐶𝐶𝐶𝑀𝑀𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 ∗ 𝑇𝑇𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 = 𝐶𝐶𝐶𝐶𝑀𝑀𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 ∗ 𝑇𝑇𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 + 𝐶𝐶𝐶𝐶𝑀𝑀𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜 ∗ 𝑇𝑇𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜 [𝐷𝐷𝐷𝐷]
5,000 ∗ 𝑇𝑇𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 = 1,000 ∗ 87℉ + 4,000 ∗ 77℉ [𝐷𝐷𝐷𝐷]
𝑇𝑇𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 = 79℉ [𝐷𝐷𝐷𝐷]
Next we are going to use the same equation, but with enthalpies.
From the psychrometric chart, ℎ𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 = 30.45
𝐵𝐵𝐵𝐵𝐵𝐵
𝑙𝑙𝑙𝑙
; ℎ𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜 = 41.47
𝐵𝐵𝐵𝐵𝐵𝐵
𝑙𝑙𝑙𝑙
𝐶𝐶𝐶𝐶𝑀𝑀𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 ∗ ℎ𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 = 𝐶𝐶𝐶𝐶𝑀𝑀𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 ∗ ℎ𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 + 𝐶𝐶𝐶𝐶𝑀𝑀𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜 ∗ ℎ𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜
5,000 ∗ 𝑇𝑇𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 = 1,000 ∗ 41.47 + 4,000 ∗ 30.45
ℎ𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 = 32.65
The mixed air condition is 79℉ [𝐷𝐷𝐷𝐷], ℎ𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 = 32.65
𝐵𝐵𝐵𝐵𝐵𝐵
𝑙𝑙𝑙𝑙
𝐵𝐵𝐵𝐵𝐵𝐵
𝑙𝑙𝑙𝑙
Finally, use the psychrometric chart to find the Wet Bulb condition.
Correct answer is A.
(A) 𝟕𝟕𝟕𝟕 ℉ 𝑫𝑫𝑫𝑫, 𝟔𝟔𝟔𝟔. 𝟑𝟑 ℉ 𝑾𝑾𝑾𝑾
(B) 79 ℉ 𝐷𝐷𝐷𝐷, 59.6 ℉ 𝑊𝑊𝑊𝑊
(C) 85 ℉ 𝐷𝐷𝐷𝐷, 74.3 ℉ 𝑊𝑊𝑊𝑊
(D) 85 ℉ 𝐷𝐷𝐷𝐷, 75.7 ℉ 𝑊𝑊𝑊𝑊
Mechanical PE - HVAC and Refrigeration Full Exam
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SOLUTION 42
The air handler serving the classroom has a supply air temperature of 55 F and the space is to
be maintained at 75 F DB and 50% Relative humidity. What CFM is required?
First calculate the total loads.
Category
People
Lighting
Computers
External
Ventilation
Total
Sensible (Btu/h)
6,250
4,000
8,000
22,000
7,500
47,750
Total (Btu/h)
11,250
4,000
8,000
22,000
15,000
60,250
𝑄𝑄𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 = 𝑚𝑚̇ ∗ 𝑐𝑐𝑝𝑝 ∗ ∆𝑇𝑇
𝑄𝑄𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 = 1.08 ∗ 𝐶𝐶𝐶𝐶𝐶𝐶 ∗ ∆𝑇𝑇
47,750
Correct answer is A.
𝐵𝐵𝐵𝐵𝐵𝐵
= 1.08 ∗ 𝐶𝐶𝐶𝐶𝐶𝐶 ∗ (75 − 55℉)
ℎ𝑟𝑟
𝐶𝐶𝐶𝐶𝐶𝐶 = 2,210
(A) 𝟐𝟐, 𝟐𝟐𝟐𝟐𝟐𝟐 𝑪𝑪𝑪𝑪𝑪𝑪
(B) 2,675 𝐶𝐶𝐶𝐶𝐶𝐶
(C) 2,790 𝐶𝐶𝐶𝐶𝐶𝐶
(D) 3,865 𝐶𝐶𝐶𝐶𝐶𝐶
Mechanical PE - HVAC and Refrigeration Full Exam
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SOLUTION 43
A dedicated outside air unit is used to pre-cool 2,500 CFM of outside air at 88 F DB and 85%
relative humidity to 60 F DB, 58 F WB. What is the quantity of water removed by the coil in
GPM?
First find the humidity ratios of the entering and leaving air, through the use of the psychrometric
chart.
Entering Air: 88 DB, 85% RH: 172.46
Leaving Air: 60 DB, 58 WB: 68.89
𝑔𝑔𝑔𝑔
𝑙𝑙𝑙𝑙
𝑔𝑔𝑔𝑔
𝑙𝑙𝑙𝑙
𝑜𝑜𝑜𝑜 0.0246
𝑜𝑜𝑜𝑜 0.0098
Next, find the amount of water removed.
𝑙𝑙𝑙𝑙𝑙𝑙 𝑜𝑜𝑜𝑜 𝐻𝐻20
𝑙𝑙𝑙𝑙𝑙𝑙 𝑜𝑜𝑜𝑜 𝑑𝑑𝑑𝑑𝑑𝑑 𝑎𝑎𝑎𝑎𝑎𝑎
𝑙𝑙𝑙𝑙𝑙𝑙 𝑜𝑜𝑜𝑜 𝐻𝐻20
𝑙𝑙𝑙𝑙𝑙𝑙 𝑜𝑜𝑜𝑜 𝑑𝑑𝑑𝑑𝑑𝑑 𝑎𝑎𝑎𝑎𝑎𝑎
∆𝑊𝑊𝐿𝐿𝐿𝐿 = 𝑊𝑊𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖 − 𝑊𝑊𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 = .0246 − .0098 [
∆𝑊𝑊𝐿𝐿𝐿𝐿 = .0148
. 0148
𝑙𝑙𝑙𝑙𝑙𝑙 𝑜𝑜𝑜𝑜 𝐻𝐻20
]
𝑙𝑙𝑙𝑙𝑙𝑙 𝑜𝑜𝑜𝑜 𝑑𝑑𝑑𝑑𝑑𝑑 𝑎𝑎𝑎𝑎𝑎𝑎
𝑙𝑙𝑙𝑙𝑙𝑙 𝑜𝑜𝑜𝑜 𝐻𝐻20
𝑙𝑙𝑙𝑙𝑙𝑙 𝑜𝑜𝑜𝑜 𝑑𝑑𝑑𝑑𝑑𝑑 𝑎𝑎𝑎𝑎𝑎𝑎
. 075 𝑙𝑙𝑙𝑙𝑙𝑙 𝑜𝑜𝑜𝑜 𝑑𝑑𝑑𝑑𝑑𝑑 𝑎𝑎𝑎𝑎𝑎𝑎
𝑙𝑙𝑙𝑙𝑙𝑙 𝑜𝑜𝑜𝑜 𝐻𝐻20
𝑙𝑙𝑙𝑙𝑙𝑙 𝑜𝑜𝑜𝑜 𝐻𝐻20
∗ 2,500 𝐶𝐶𝐶𝐶𝑀𝑀 𝑜𝑜𝑜𝑜 𝑑𝑑𝑑𝑑𝑑𝑑 𝑎𝑎𝑎𝑎𝑎𝑎 ∗
= 28
3
𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚
1 𝑓𝑓𝑓𝑓
𝑙𝑙𝑙𝑙𝑙𝑙 𝑜𝑜𝑜𝑜 𝑑𝑑𝑑𝑑𝑑𝑑 𝑎𝑎𝑎𝑎𝑎𝑎
2.8
𝑓𝑓𝑓𝑓 3
7.48 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑛𝑛𝑛𝑛
𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔
𝑙𝑙𝑙𝑙𝑙𝑙 𝑜𝑜𝑜𝑜 𝐻𝐻20
∗
∗
= 0.33
3
1 𝑓𝑓𝑓𝑓
𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚
𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚
62.4 𝑙𝑙𝑙𝑙𝑙𝑙 𝑜𝑜𝑜𝑜 𝐻𝐻20
Correct answer is B.
(A) 0.29 𝐺𝐺𝐺𝐺𝐺𝐺
(B) 𝟎𝟎. 𝟑𝟑𝟑𝟑 𝑮𝑮𝑮𝑮𝑮𝑮
(C) 2.9 𝐺𝐺𝐺𝐺𝐺𝐺
(D) 2.4 𝐺𝐺𝐺𝐺𝐺𝐺
Mechanical PE - HVAC and Refrigeration Full Exam
PM Session Solutions -3
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SOLUTION 44
An air handling unit cools 2,000 CFM of outside air (90 F DB/80% RH) and 6,000 CFM of return
air (77 F DB, 50% RH) to 52 F DB, 51 F WB. If chilled water enters the coil at 44 F and leaves
at 56 F, what is the required GPM?
The first part of this solution involves finding the conditions of the air entering the coil, through
the mixed air equation.
𝐶𝐶𝐶𝐶𝑀𝑀𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 ∗ 𝑇𝑇𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 = 𝐶𝐶𝐶𝐶𝑀𝑀𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 ∗ 𝑇𝑇𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 + 𝐶𝐶𝐶𝐶𝑀𝑀𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜 ∗ 𝑇𝑇𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜 [𝐷𝐷𝐷𝐷]
8,000 ∗ 𝑇𝑇𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 = 2,000 ∗ 90℉ + 6,000 ∗ 77℉ [𝐷𝐷𝐷𝐷]
𝑇𝑇𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 = 80.25℉ [𝐷𝐷𝐷𝐷]
Next if we connect the return air and the outside air points on the psychrometric chart, then we
know that the mixed air must lie on this line and the intersection of the mixed dry bulb
temperature of 80.25 F.
From the psychrometric chart we find that ℎ𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 = 34.21
𝐵𝐵𝐵𝐵𝐵𝐵
𝑙𝑙𝑙𝑙
.
Next we find the leaving enthalpy from the psychrometric chart ℎ𝑙𝑙𝑙𝑙𝑙𝑙 = 20.85
𝐵𝐵𝐵𝐵𝐵𝐵
𝑙𝑙𝑙𝑙
.
Next find the heat transferred by the coil using the entering and leaving enthalpies of the air.
𝑄𝑄 = 4.5 ∗ 8000 𝐶𝐶𝐶𝐶𝐶𝐶 ∗ (34.21 − 20.85) = 480,960
𝐵𝐵𝐵𝐵𝐵𝐵
ℎ
Finally, assume all the heat transferred to the coil is from the chilled water.
480,960
Correct Answer is B
𝐵𝐵𝐵𝐵𝐵𝐵
= 500 ∗ 𝐺𝐺𝐺𝐺𝐺𝐺 ∗ (56 − 44 ℉)
ℎ𝑟𝑟
𝐺𝐺𝐺𝐺𝐺𝐺 = 80.2
(A) 42 𝐺𝐺𝐺𝐺𝐺𝐺
(B) 𝟖𝟖𝟖𝟖 𝑮𝑮𝑮𝑮𝑮𝑮
(C) 105 𝐺𝐺𝐺𝐺𝐺𝐺
(D) 126 𝐺𝐺𝐺𝐺𝐺𝐺
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SOLUTION 45
A cooling coil has a surface temperature of 55 °F. If 5,000 CFM of air enters the coil at 80 °F
DB, 65% RH, then what is the bypass factor that produces exiting air at 60 °F DB? Assume that
there are no minor heat gains or losses across the coil and that the coils temperature is
maintained at 55 °F.
In this question, (1-x) % of the air entering the coil leaves at 55 °F DB. While (x) % of the air is
bypassed the coil and leaves at the same entering condition of 80 °F DB.
Use the mixed air equation to find the amount of CFM that will produce a 60 °F leaving mixed air
temperature.
𝐶𝐶𝐶𝐶𝑀𝑀𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 ∗ 𝑇𝑇𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 = 𝐶𝐶𝐶𝐶𝑀𝑀𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏 ∗ 𝑇𝑇𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏 + 𝐶𝐶𝐶𝐶𝑀𝑀𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 ∗ 𝑇𝑇𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 [𝐷𝐷𝐷𝐷]
5,000 ∗ 60 = 5,000 ∗ (𝑥𝑥) ∗ 80℉ + 5,000 ∗ (1 − 𝑥𝑥) ∗ 55℉ [𝐷𝐷𝐷𝐷]
Cancel the 5,000 CFM
60 = 80𝑥𝑥 + 55 − 25𝑥𝑥
5 = 25𝑥𝑥
Bypass Factor = 0.20
𝑥𝑥 = 0.2
Correct answer is D
(A) 0.03
(B) 0.10
(C) 0.15
(D) 0.20
Mechanical PE - HVAC and Refrigeration Full Exam
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SOLUTION 46
A chilled water pump is used to pump 200 GPM of chilled water through a water cooled chiller,
with a 10 F temperature difference. The condenser water circuit of the chiller operates on a 15
F temperature difference and the compressor produces 250,000 Btu/hr. What GPM is required
by the condenser water pump? Assume no minor heat gains or losses in the chilled/condenser
water circuits and negligible bypass factors.
The condenser needs to be sized to release the heat from the chilled water circuit and the
compressor.
First find the total heat released by the chilled water.
𝑄𝑄
𝐵𝐵𝐵𝐵𝐵𝐵
= 500 ∗ 𝐺𝐺𝐺𝐺𝐺𝐺 ∗ (∆𝑇𝑇 ℉)
ℎ𝑟𝑟
𝑄𝑄
𝐵𝐵𝐵𝐵𝐵𝐵
= 500 ∗ 200 ∗ (10 ℉)
ℎ𝑟𝑟
𝑄𝑄𝑐𝑐ℎ𝑤𝑤 = 1,000,000
Find the total heat into the condenser water.
𝐵𝐵𝐵𝐵𝐵𝐵
ℎ𝑟𝑟
𝑄𝑄𝑐𝑐𝑐𝑐𝑐𝑐 = 𝑄𝑄𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 + 𝑄𝑄𝑐𝑐ℎ𝑤𝑤
𝑄𝑄𝑐𝑐𝑐𝑐𝑐𝑐 = 250,000 + 100,000 = 1,250,000
Next find the GPM of the condenser water.
1,250,000
Correct answer is C.
𝐵𝐵𝐵𝐵𝐵𝐵
ℎ𝑟𝑟
𝐵𝐵𝐵𝐵𝐵𝐵
= 500 ∗ 𝐺𝐺𝐺𝐺𝐺𝐺 ∗ (15 ℉)
ℎ𝑟𝑟
𝐺𝐺𝐺𝐺𝐺𝐺 = 167
(A) 106 GPM
(B) 133 GPM
(C) 167 GPM
(D) 200 GPM
Mechanical PE - HVAC and Refrigeration Full Exam
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SOLUTION 47
Which of the following is not a type of fan?
Airfoil, backward inclined and propeller are all types of fans. Vertical inline is a type of pump.
Correct answer is C.
(A) Airfoil
(B) Backward inclined
(C) Vertical Inline
(D) Propeller
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SOLUTION 48
An evaporative air cooler is used to cool air at 90 F, 30% relative humidity. Water is supplied to
the evaporative air cooler at 70 F. What is the enthalpy of the air leaving the evaporative cooler,
if the evaporative cooler is 80% effective?
This question involves the use of the effectiveness equation of an evaporative cooler.
𝜀𝜀 =
𝜀𝜀 =
𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑
𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑
90 − 𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿 𝐴𝐴𝐴𝐴𝐴𝐴 𝑇𝑇𝑇𝑇𝑇𝑇𝑝𝑝 𝐷𝐷𝐷𝐷
90 − 𝑊𝑊𝑊𝑊𝑊𝑊 𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵 𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇
From the psychrometric chart find the wet bulb of the air.
0.8 =
90 − 𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿 𝐴𝐴𝐴𝐴𝐴𝐴 𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 𝐷𝐷𝐷𝐷
90 − 67.2
𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿 𝐴𝐴𝐴𝐴𝐴𝐴 𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 = 71.8 ℉ 𝐷𝐷𝐷𝐷
Correct answer is C.
(A) 67.2 ℉ 𝐷𝐷𝐷𝐷
(B) 70.0 ℉ 𝐷𝐷𝐷𝐷
(C) 𝟕𝟕𝟕𝟕. 𝟖𝟖 ℉ 𝑫𝑫𝑫𝑫
(D) 75.0 ℉ 𝐷𝐷𝐷𝐷
Mechanical PE - HVAC and Refrigeration Full Exam
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SOLUTION 49
An R-134A chiller has a suction pressure of 80 PSIA and discharge pressure of 200 PSIA. The
refrigerant undergoes 15 F of superheat and 0 F of sub-cooling. What is the COP Of the
chiller? Assume a refrigerant flow of 25 lb/min.
This problem involves the use of your Pressure Enthalpy diagram which can be found in the
ASHRAE Fundamentals book.
Calculating the COP of the chiller you must find the net refrigeration effect and the compressor
work.
𝐶𝐶𝐶𝐶𝐶𝐶 =
𝑚𝑚̇ ∗ (ℎ𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒,𝑙𝑙𝑙𝑙𝑙𝑙 − ℎ𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒,𝑒𝑒𝑒𝑒𝑒𝑒 )
𝑛𝑛𝑛𝑛𝑛𝑛 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒
=
𝑐𝑐𝑐𝑐𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤
𝑚𝑚̇ ∗ (ℎ𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐,𝑙𝑙𝑙𝑙𝑙𝑙 − ℎ𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐,𝑒𝑒𝑒𝑒𝑒𝑒 )
Find the enthalpy of the refrigerant entering the evaporator, which is equal to the enthalpy of the
refrigerant leaving the condenser. This point is located at the intersection of (1) the discharge
pressure of 200 PSIA and (2) the saturated liquid curve [0 degree sub-cooling].
ℎ𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒,𝑒𝑒𝑒𝑒𝑒𝑒 = ℎ𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐,𝑙𝑙𝑙𝑙𝑙𝑙 = 54.3
𝐵𝐵𝐵𝐵𝐵𝐵
𝑙𝑙𝑙𝑙
The enthalpy of the refrigerant leaving the evaporator is found at the intersection of (1) the
suction pressure 80 PSIA and (2) 15 F super-heating [15 F past the saturated vapor line].
ℎ𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒,𝑙𝑙𝑙𝑙𝑙𝑙 = 116
𝐵𝐵𝐵𝐵𝐵𝐵
𝑙𝑙𝑙𝑙
The compressor work is found through the below equation:
𝑚𝑚̇ ∗ (ℎ𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐,𝑙𝑙𝑙𝑙𝑙𝑙 − ℎ𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐,𝑒𝑒𝑒𝑒𝑒𝑒 )
The enthalpy of the refrigerant entering the compressor is equal to the refrigerant leaving the
evaporator.
ℎ𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐,𝑒𝑒𝑒𝑒𝑒𝑒 = ℎ𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒,𝑙𝑙𝑙𝑙𝑙𝑙 = 116
𝐵𝐵𝐵𝐵𝐵𝐵
𝑙𝑙𝑙𝑙
The enthalpy of the refrigerant leaving the compressor is found by following the constant
entropy line from the point leaving the evaporator up to the intersection of the discharge
pressure line.
ℎ𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐,𝑙𝑙𝑙𝑙𝑙𝑙 = 124
𝐵𝐵𝐵𝐵𝐵𝐵
𝑙𝑙𝑙𝑙
Mechanical PE - HVAC and Refrigeration Full Exam
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Now plug in all values to the COP equation.
𝐶𝐶𝐶𝐶𝐶𝐶 =
𝐶𝐶𝐶𝐶𝐶𝐶 =
𝑛𝑛𝑛𝑛𝑛𝑛 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒 𝑚𝑚̇ ∗ (116 − 54.3)
=
𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤
𝑚𝑚̇ ∗ (124 − 116)
𝑛𝑛𝑛𝑛𝑛𝑛 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒 𝑚𝑚̇ ∗ (116 − 54.3)
=
𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤
𝑚𝑚̇ ∗ (124 − 116)
Correct answer is C, 7.7
(A) 4.9
(B) 5.7
(C) 7.7
(D) 10.2
Mechanical PE - HVAC and Refrigeration Full Exam
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SOLUTION 50
An R-410A chiller has a suction pressure of 100 PSIA with 20 F super-heat and a discharge
pressure of 200 PSIA. Assume 15 F of sub-cooling. What is the required refrigerant flow rate in
(lb/min) in order to produce 20 tons of cooling?
This problem involves finding the net refrigeration effect through the following equation.
𝑁𝑁𝑁𝑁𝑁𝑁 𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅 𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸 = 𝑚𝑚̇ ∗ (ℎ𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒,𝑙𝑙𝑙𝑙𝑙𝑙 − ℎ𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒,𝑒𝑒𝑒𝑒𝑒𝑒 )
Use the ASHRAE fundamentals book and navigate to the Thermo-physical Properties of
Refrigerants in order to find the Pressure-Enthalpy diagram of R-410A.
Find the enthalpy of the refrigerant entering the evaporator, which is equal to the enthalpy of the
refrigerant leaving the condenser. This point is located at the intersection of (1) the discharge
pressure of 200 PSIA and (2) the 15 degree sub-cooling [15 F towards the liquid region past the
saturated liquid line].
ℎ𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒,𝑒𝑒𝑒𝑒𝑒𝑒 = ℎ𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐,𝑙𝑙𝑙𝑙𝑙𝑙 = 31.7
𝐵𝐵𝐵𝐵𝐵𝐵
𝑙𝑙𝑙𝑙
The enthalpy of the refrigerant leaving the evaporator is found at the intersection of (1) the
suction pressure 100 PSIA and (2) 15 F super-heating [15 F past the saturated vapor line].
ℎ𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒,𝑙𝑙𝑙𝑙𝑙𝑙 = 125
𝐵𝐵𝐵𝐵𝐵𝐵
𝑙𝑙𝑙𝑙
Next plug the numbers into the net refrigeration equation.
20 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 ∗
12,000 𝐵𝐵𝐵𝐵𝐵𝐵ℎ
= 𝑚𝑚̇ ∗ (125 − 31.7)
1 𝑡𝑡𝑡𝑡𝑡𝑡
2,572
𝑙𝑙𝑙𝑙𝑙𝑙
= 𝑚𝑚̇
ℎ𝑟𝑟
Correct answer is C.
(A) 1,470 lbs/hr
(B) 1,870 lbs/hr
(C) 2,570 lbs/hr
(D) 2,890 lbs/hr
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SOLUTION 51
An R-134A chiller has a suction pressure of 100 PSIA and discharge pressure of 200 PSIA.
The refrigerant undergoes 15 F of superheat and 0 F of sub-cooling. What is the quality of the
refrigerant entering the evaporator? Assume a refrigerant flow of 25 lb/min.
This question involves finding the properties of refrigerant prior to entering the evaporator.
Use the ASHRAE fundamentals book and navigate to the Thermo-physical Properties of
Refrigerants in order to find the Pressure-Enthalpy diagram of R-134A.
Find the location of the refrigerant entering the evaporator, which is the point located at the
intersection of (1) the suction pressure of 100 PSIA and (2) the constant enthalpy line from the
point where the refrigerant leaves the condenser. The point at which the refrigerant leaves the
condenser is found at the intersection of the discharge pressure line and the saturated liquid
curve, since there is no sub-cooling.
Condenser Leaving Point: 200 PSIA, x = 0% quality (saturated liquid, no sub-cooling),
ℎ𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐,𝑙𝑙𝑙𝑙𝑙𝑙 = 54.3 𝐵𝐵𝐵𝐵𝐵𝐵/𝑙𝑙𝑙𝑙
Evaporator Entering Point: 100 PSIA,
ℎ𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐,𝑙𝑙𝑙𝑙𝑙𝑙 = ℎ𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒,𝑒𝑒𝑒𝑒𝑒𝑒 = 54.3 𝐵𝐵𝐵𝐵𝐵𝐵/𝑙𝑙𝑙𝑙
From the P-H Diagram, find the intersection and read the quality:
Correct answer is B.
𝑥𝑥𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒,𝑒𝑒𝑒𝑒𝑒𝑒 = 0.216
(A) 0.15
(B) 0.22
(C) 0.32
(D) 0.42
Mechanical PE - HVAC and Refrigeration Full Exam
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SOLUTION 52
An R-410A chiller has a suction pressure of 150 PSIA with 20 °F super-heat and a discharge
pressure of 400 PSIA. Assume 15 °F of sub-cooling and a refrigerant flow rate of 20 lb/min.
What is the condenser leaving temperature of the refrigerant?
This question involves finding the properties of the refrigerant as it leaves the condenser.
Use the ASHRAE Fundamentals book and navigate to the Thermo-Physical properties of
refrigerants in order to find the Pressure-Enthalpy diagram of R-410A.
The point at which the refrigerant leaves the condenser is at the intersection of (1) the discharge
pressure of 200 PSIA and (2) the 15 degree sub-cooling
Intersection of discharge pressure of 400 PSIA and saturated liquid line is at a temperature of
114 ℉.
But the condenser provides 15 more degrees of sub-cooling, thus the temperature leaving the
condenser is 99 ℉
Correct answer is B.
(A) 90 ℉
(B) 𝟗𝟗𝟗𝟗 ℉
(C) 114 ℉
(D) 129 ℉
Mechanical PE - HVAC and Refrigeration Full Exam
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SOLUTION 53
20 lb/hr of 15 PSIA steam is delivered to a heating coil. 2,000 CFM of air enters the coil at 60 F
DB and 90% relative humidity. What is the exiting dry bulb temperature of the air, assume no
super heat or sub-cooling. Bypass factor and minor heat gains/losses are negligible.
This question is an energy balance equation. The heat lost by condensing the steam (latent
heat) is gained by the air passing through the coil.
𝑚𝑚̇𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 ∗ ℎ𝑓𝑓𝑓𝑓 = 1.08 ∗ 𝐶𝐶𝐶𝐶𝐶𝐶 ∗ (∆𝑇𝑇)
First, go to your steam tables as a function of pressure, since Pressure is given and find the
enthalpy of evaporation.
ℎ𝑓𝑓𝑓𝑓 = 969.5
𝐵𝐵𝐵𝐵𝐵𝐵
𝑙𝑙𝑙𝑙
Then plug in the enthalpy of evaporation and the other variables into the energy balance
equation.
20
𝑙𝑙𝑙𝑙
𝐵𝐵𝐵𝐵𝐵𝐵
∗ 969.5
= 1.08 ∗ 2,000 ∗ (∆𝑇𝑇)
ℎ𝑟𝑟
𝑙𝑙𝑙𝑙
∆𝑇𝑇 = 9.0 ℉
The final dry bulb temperature of the air leaving the coil will be 60 + 9.0 = 69℉
Correct answer is A.
(A) 𝟔𝟔𝟔𝟔℉
(B) 73℉
(C) 75℉
(D) 79℉
Mechanical PE - HVAC and Refrigeration Full Exam
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SOLUTION 54
A pump is sized for 250 GPM at 100’ TDH. What is the total flow produced by two of these
pumps in series?
If pumps are arranged in SERIES then the pressures at each point along their pump curve are
added. Thus the point 250 GPM, 100' TDH will now be located at 250 GPM, 200' TDH.
If pumps are arranged in PARALLEL then the flows at each point along their pump curve are
added. Thus the point 250 GPM, 100' TDH will now be located at 500 GPM, 100' TDH.
Correct answer is C.
(A) 100 GPM
(B) 200 GPM
(C) 250 GPM
(D) 500 GPM
Mechanical PE - HVAC and Refrigeration Full Exam
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SOLUTION 55
A cooling tower has 150 GPM of water at entering and leaving temperatures of 100 F and 85 F.
If the outside air conditions are 82 F DB/75% relative humidity. What is the effectiveness of the
cooling tower, in %?
Cooling tower effectiveness is governed by the following equation:
𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸 [%] =
𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 [℉]
𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 ∆ [℉]
=
𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 + 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎ℎ [℉]
𝑚𝑚𝑚𝑚𝑚𝑚 ∆ [℉]
The range is found through the following equation:
𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅 = 𝑇𝑇𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤,𝑖𝑖𝑖𝑖 [℉] − 𝑇𝑇𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤,𝑜𝑜𝑜𝑜𝑜𝑜 [℉]
𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅 = 100 − 85 [℉] = 15℉
The approach is found through the following equation:
𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴ℎ = 𝑇𝑇𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤,𝑜𝑜𝑜𝑜𝑜𝑜 [℉] − 𝑇𝑇𝑎𝑎𝑎𝑎𝑎𝑎 𝑖𝑖𝑖𝑖,𝑊𝑊𝑊𝑊 [℉]
Find the wet bulb temperature through the psychrometric chart:
𝑇𝑇𝑎𝑎𝑎𝑎𝑎𝑎 𝑖𝑖𝑖𝑖,𝑊𝑊𝑊𝑊 = 75.7℉
𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴ℎ = 85 − 76 [℉] = 9.3 ℉
Correct answer is A.
𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸 [%] =
15
= 61.7%
15 + 9.3 [℉]
(A) 62%
(B) 68%
(C) 74%
(D) 79%
Mechanical PE - HVAC and Refrigeration Full Exam
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SOLUTION 56
A new air handler with a VFD at 60 HZ provides 2,000 CFM of conditioned air to multiple
classrooms. 500 CFM of the 2,000 CFM is outside air that is constantly provided to maintain
acceptable indoor air quality. If during low load conditions the VFD is ramped down to 45 HZ,
then what is the percentage of outside air compared to the total air supplied?
When the VFD is ramped down the total supply air will be reduced, but the amount of outside air
will remain constant, thereby increasing the percentage of outside air in the total supply air.
This question uses the fan laws. From the fan laws, Flow (CFM) increases linearly with Speed
which is related to the Frequency of the VFD.
𝐶𝐶𝐶𝐶𝑀𝑀1 𝑁𝑁1
=
, 𝑤𝑤ℎ𝑒𝑒𝑒𝑒𝑒𝑒 𝑁𝑁 𝑖𝑖𝑖𝑖 𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 (𝐻𝐻𝐻𝐻)
𝐶𝐶𝐶𝐶𝑀𝑀2 𝑁𝑁2
2,000 60
=
𝐶𝐶𝐶𝐶𝑀𝑀2 45
𝐶𝐶𝐶𝐶𝑀𝑀2 = 1,500
Then find the percentage of outside air in the new total supplied air:
%𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜 =
500
= 33%
1,500
Correct answer is B.
(A) 25%
(B) 33%
(C) 44%
(D) 67%
Mechanical PE - HVAC and Refrigeration Full Exam
PM Session Solutions -17
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SOLUTION 57
An existing chiller is served by a chilled water pump (150 GPM, 75 TDH), 65% efficient pump,
90% efficient motor. A recent study was conducted and it was found that the pump was
oversized and should be replaced with a new higher efficiency pump at 150 GPM, 50 TDH, 80%
efficient pump and 95% premium efficiency motor. If the pump runs 8 hours a day, 5 days a
week, 52 weeks a year, how many kWh per year will be saved by switching to the new pump?
First find the Pump horsepower, specific gravity (SG) of water is equal to 1.0.
𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 𝐻𝐻𝑃𝑃𝑜𝑜𝑜𝑜𝑜𝑜 =
𝐺𝐺𝐺𝐺𝐺𝐺 ∗ 𝑇𝑇𝑇𝑇𝑇𝑇 ∗ 𝑆𝑆𝑆𝑆
3956
𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 𝐻𝐻𝑃𝑃𝑜𝑜𝑜𝑜𝑜𝑜 =
150 ∗ 75 ∗ 1.0
3956
𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 𝐻𝐻𝑃𝑃𝑜𝑜𝑜𝑜𝑜𝑜 = 2.84 𝐻𝐻𝐻𝐻
Then find the total electricity used by dividing the Pump Horsepower by the efficiency of the
pump and the motor and convert to KW.
𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸 𝑈𝑈𝑈𝑈𝑈𝑈 = 2.84 𝐻𝐻𝐻𝐻 ∗ �
0.746 𝐾𝐾𝐾𝐾
1
1
�∗�
�∗
= 3.62 𝐾𝐾𝐾𝐾
1 𝐻𝐻𝐻𝐻
0.90
0.65
Multiply by the number of hours the pump is used
𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸 𝑈𝑈𝑈𝑈𝑈𝑈𝑈𝑈𝑈𝑈 = 3.62 𝐾𝐾𝐾𝐾 ∗
8 ℎ𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜 5 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 52 𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤
∗
∗
= 7,530 𝑘𝑘𝑘𝑘ℎ
𝑑𝑑𝑑𝑑𝑑𝑑
𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤
𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦
Second find the Pump horsepower of the new Pump, specific gravity (SG) of water is equal to
1.0.
𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 𝐻𝐻𝑃𝑃𝑛𝑛𝑛𝑛𝑛𝑛 =
𝐺𝐺𝐺𝐺𝐺𝐺 ∗ 𝑇𝑇𝑇𝑇𝑇𝑇 ∗ 𝑆𝑆𝑆𝑆
3956
𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 𝐻𝐻𝑃𝑃𝑛𝑛𝑛𝑛𝑛𝑛 =
150 ∗ 50 ∗ 1.0
3956
𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 𝐻𝐻𝑃𝑃𝑛𝑛𝑛𝑛𝑛𝑛 = 1.90 𝐻𝐻𝐻𝐻
Then find the total electricity used by dividing the Pump Horsepower by the efficiency of the
pump and the motor and convert to KW.
𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸 𝑈𝑈𝑈𝑈𝑈𝑈 = 1.90 𝐻𝐻𝐻𝐻 ∗ �
0.746 𝐾𝐾𝐾𝐾
1
1
�∗�
�∗
= 1.87 𝐾𝐾𝐾𝐾
1 𝐻𝐻𝐻𝐻
0.95
0.80
Multiply by the number of hours the pump is used
𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸 𝑈𝑈𝑈𝑈𝑈𝑈𝑈𝑈𝑈𝑈 = 1.87 𝐾𝐾𝐾𝐾 ∗
8 ℎ𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜 5 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 52 𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤
∗
∗
= 3,890 𝑘𝑘𝑘𝑘ℎ
𝑑𝑑𝑑𝑑𝑦𝑦
𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤
𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦
Mechanical PE - HVAC and Refrigeration Full Exam
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The total electricity savings is found below:
7,530 𝑘𝑘𝑘𝑘ℎ − 3,890 𝑘𝑘𝑘𝑘ℎ = 3,640 𝑘𝑘𝑘𝑘ℎ
Correct answer is A.
(A) 3,640 kWh
(B) 3,990 kWh
(C) 4,150 kWh
(D) 7,250 kWh
Mechanical PE - HVAC and Refrigeration Full Exam
PM Session Solutions -19
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SOLUTION 58
A 10 BHP supply fan is provided with both the motor and the fan in the air conditioned space.
The motor efficiency is 95%. What is the total heat gain from the fan and motor to the space?
The total heat gain from the fan and motor into the space is found by determining the total
electrical input. Which is found by the below equation:
𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸 𝑈𝑈𝑈𝑈𝑈𝑈𝑈𝑈𝑈𝑈 = 10 𝐵𝐵𝐵𝐵𝐵𝐵 ∗ (
1
)
0.95
𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸 𝑈𝑈𝑈𝑈𝑈𝑈𝑈𝑈𝑈𝑈 = 10.53 𝐻𝐻𝐻𝐻
The total heat gain to the space is equal to the total electricity usage of the motor.
10 HP of heat is from the fan [since the fan is in the space]
It is important to note that 10 BHP value has already included the efficiency of the fan.
0.53 HP of heat is from the motor due to its inefficiencies [since the motor is in the space]
Next convert HP to Btu/h.
10.53 𝐻𝐻𝐻𝐻 ∗
𝐵𝐵𝐵𝐵𝐵𝐵
ℎ = 26,804 𝐵𝐵𝐵𝐵𝐵𝐵ℎ
1 𝐻𝐻𝐻𝐻
2,546
Correct answer is D.
(A) 22,680 𝐵𝐵𝐵𝐵𝐵𝐵ℎ
(B) 24,230 𝐵𝐵𝐵𝐵𝐵𝐵ℎ
(C) 25,460 𝐵𝐵𝐵𝐵𝐵𝐵ℎ
(D) 𝟐𝟐𝟐𝟐, 𝟖𝟖𝟖𝟖𝟖𝟖 𝑩𝑩𝑩𝑩𝑩𝑩𝑩𝑩
Mechanical PE - HVAC and Refrigeration Full Exam
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SOLUTION 59
A 100 GPM condenser water pump is supplied with water by a cooling tower basin that is 10 ft
above the centerline of the pump. The suction line of the pump consists of 40 ft of 3" Schedule
40 steel pipe and 3 90 degree ells. Condenser water pump serves a cooling tower with entering
and leaving conditions of 95 F and 85 F. What is the net positive suction head available at the
condenser water pump?
NPSHA is found through the following equation:
𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁 = 𝑃𝑃𝑎𝑎𝑎𝑎𝑎𝑎 ±𝑃𝑃𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒 − 𝑃𝑃𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 − 𝑉𝑉𝑃𝑃𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤
𝑃𝑃𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒 = +10′ (𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏 𝑡𝑡ℎ𝑒𝑒 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏 𝑖𝑖𝑖𝑖 10′ 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 𝑡𝑡ℎ𝑒𝑒 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑜𝑜𝑜𝑜 𝑡𝑡ℎ𝑒𝑒 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝)
The pipe friction is found through the ASHRAE Fundamentals tables for Pipe Sizing.
First find the total equivalent length which is found by adding the total length of pipes and the
total equivalent length of the elbows.
Velocity through the pipe is
100
1 𝑓𝑓𝑡𝑡 3
𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 1 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚
1
� = 4.34 𝑓𝑓𝑓𝑓/𝑠𝑠𝑠𝑠𝑠𝑠
∗
∗
∗�
𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 60 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 7.48 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 . 05134 𝑓𝑓𝑡𝑡 2
According to ASHRAE Fundamentals, the total equivalent length for Schedule 40 Regular 90
Degree Elbow, 3" pipe with velocity of 4.34 ft/sec is equal to 8.4 ft per elbow
𝑇𝑇𝑇𝑇𝑇𝑇 = 40′ + (3 𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒) ∗ 8.4′ = 65.2′
Second check ASHRAE Fundamentals or the MERM for the pressure drop per 100 ft, for 3"
Schedule 40 and a 100 GPM.
𝑃𝑃𝑃𝑃 = 2.6 𝑓𝑓𝑓𝑓 𝑜𝑜𝑜𝑜 ℎ𝑒𝑒𝑒𝑒𝑒𝑒 𝑝𝑝𝑝𝑝𝑝𝑝 100′
Multiply the pressure drop factor per length of pipe by the total equivalent length of pipe:
𝑃𝑃𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 = 2.6
𝑓𝑓𝑓𝑓 𝑜𝑜𝑜𝑜 ℎ𝑒𝑒𝑒𝑒𝑒𝑒
∗ 65.2′ = 1.7 𝑓𝑓𝑓𝑓 𝑜𝑜𝑜𝑜 ℎ𝑒𝑒𝑒𝑒𝑒𝑒
100′
Mechanical PE - HVAC and Refrigeration Full Exam
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Find the vapor pressure of the water as a function of temperature use the average temperature
of 90 F.
𝑃𝑃𝑣𝑣𝑣𝑣 = 1.6 𝑓𝑓𝑓𝑓 𝑜𝑜𝑜𝑜 ℎ𝑒𝑒𝑒𝑒𝑒𝑒
Finally, the first term for atmospheric pressure must be added, since the basin is open to the
atmosphere and is subject to this pressure.
𝑃𝑃𝑎𝑎𝑎𝑎𝑎𝑎 = 14.7 𝑃𝑃𝑃𝑃𝑃𝑃 ∗
2.31 𝑃𝑃𝑃𝑃𝑃𝑃
= 33.9 𝑓𝑓𝑓𝑓 𝑜𝑜𝑜𝑜 ℎ𝑒𝑒𝑒𝑒𝑒𝑒
𝑓𝑓𝑓𝑓 𝑜𝑜𝑜𝑜 ℎ𝑒𝑒𝑒𝑒𝑒𝑒
Plug in all the variables into the below equation.
𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁 = 𝑃𝑃𝑎𝑎𝑎𝑎𝑎𝑎 ±𝑃𝑃𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒 − 𝑃𝑃𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 − 𝑉𝑉𝑃𝑃𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤
𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁 = 33.9′ + 10′ − 1.7 − 1.6
Correct answer is C.
𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁 = 40.6 𝑓𝑓𝑓𝑓 𝑜𝑜𝑜𝑜 ℎ𝑒𝑒𝑒𝑒𝑒𝑒
(A) 7 𝑓𝑓𝑓𝑓 𝑜𝑜𝑜𝑜 ℎ𝑒𝑒𝑒𝑒𝑒𝑒
(B) 34 𝑓𝑓𝑓𝑓 𝑜𝑜𝑜𝑜 ℎ𝑒𝑒𝑒𝑒𝑒𝑒
(C) 𝟒𝟒𝟒𝟒 𝒇𝒇𝒇𝒇 𝒐𝒐𝒐𝒐 𝒉𝒉𝒉𝒉𝒉𝒉𝒉𝒉
(D) 48 𝑓𝑓𝑓𝑓 𝑜𝑜𝑜𝑜 ℎ𝑒𝑒𝑒𝑒𝑒𝑒
Mechanical PE - HVAC and Refrigeration Full Exam
PM Session Solutions -22
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SOLUTION 60
A north facing wall is 20’ long by 10’ high. There are (2) 2’ X 4’ windows with 1/8” clear glass.
The wall consists of 8” Concrete (2.1
(1.5
ℎ∗𝑓𝑓𝑡𝑡 2 ∗℉
𝐵𝐵𝐵𝐵𝐵𝐵
ℎ∗𝑓𝑓𝑡𝑡 2 ∗℉
𝐵𝐵𝐵𝐵𝐵𝐵
), 2” Insulation (8.2
ℎ∗𝑓𝑓𝑡𝑡 2 ∗℉
𝐵𝐵𝐵𝐵𝐵𝐵
) and 5/8” Gypsum
). The CLTD of the wall and window is found to equal 40 F. The windows have a
SC of 0.6, U-factor of 0.95
𝐵𝐵𝐵𝐵𝐵𝐵
ℎ∗𝑓𝑓𝑡𝑡 2 ∗℉
and a SCL of 80. What is the total heat transferred through
the wall and windows, Btu/h? Do not subtract the area of the window from the wall.
First calculate the total equivalent heat transfer value for the entire wall assembly:
𝑅𝑅𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 = 𝑅𝑅𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 + 𝑅𝑅𝑖𝑖𝑖𝑖𝑖𝑖 + 𝑅𝑅𝑔𝑔𝑔𝑔𝑔𝑔
Luckily all the values given in the problem are converted to the units of R-values.
𝑅𝑅𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡
Convert to R-Value to U-factor.
ℎ ∗ 𝑓𝑓𝑡𝑡 2 ∗ ℉
= 2.1 + 8.2 + 1.5 = 11.8
𝐵𝐵𝐵𝐵𝐵𝐵
𝑈𝑈𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 =
1
𝐵𝐵𝐵𝐵𝐵𝐵
= .0848
11.8
ℎ ∗ 𝑓𝑓𝑡𝑡 2 ∗ ℉
Second calculate the total heat through the wall:
𝑄𝑄𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤 = 𝑈𝑈 ∗ 𝐴𝐴 ∗ 𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶
𝑄𝑄𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤 = .0848
𝐵𝐵𝐵𝐵𝐵𝐵
∗ [20′ 𝑋𝑋 10′] ∗ 40
ℎ ∗ 𝑓𝑓𝑡𝑡 2 ∗ ℉
𝑄𝑄𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤 = 678.4
Next calculate the conduction through the window
𝑄𝑄𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤,𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 = .95
𝐵𝐵𝐵𝐵𝐵𝐵
ℎ
𝐵𝐵𝐵𝐵𝐵𝐵
∗ [2 ∗ (4 ∗ 2)] ∗ 40℉
ℎ ∗ 𝑓𝑓𝑡𝑡 2 ∗ ℉
𝑄𝑄𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤,𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 = 608
𝐵𝐵𝐵𝐵𝐵𝐵
ℎ
Mechanical PE - HVAC and Refrigeration Full Exam
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Next calculate the solar radiation through the window
𝑄𝑄𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤,𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 = 𝑆𝑆𝑆𝑆 ∗ 80 ∗ 𝐴𝐴
𝑄𝑄𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤,𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 = 0.6 ∗ 80 ∗ [2 ∗ (4 ∗ 2)]
Sum up all heat gains.
𝑄𝑄𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤,𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 = 768
𝐵𝐵𝐵𝐵𝐵𝐵
ℎ
𝑄𝑄𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 = 𝑄𝑄𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤,𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 + 𝑄𝑄𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤,𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 + 𝑄𝑄𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤
𝑄𝑄𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 = 768
𝐵𝐵𝐵𝐵𝐵𝐵
𝐵𝐵𝐵𝐵𝐵𝐵
𝐵𝐵𝐵𝐵𝐵𝐵
+ 608
+ 678.4
ℎ
ℎ
ℎ
𝑄𝑄𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 = 2,054.4
𝐵𝐵𝐵𝐵𝐵𝐵
ℎ
Correct answer is A.
(A) 𝟐𝟐, 𝟎𝟎𝟎𝟎𝟎𝟎
(B) 2,300
(C) 2,750
(D) 3,330
𝑩𝑩𝑩𝑩𝑩𝑩
𝒉𝒉
𝐵𝐵𝐵𝐵𝐵𝐵
ℎ
𝐵𝐵𝐵𝐵𝐵𝐵
ℎ
𝐵𝐵𝐵𝐵𝐵𝐵
ℎ
Mechanical PE - HVAC and Refrigeration Full Exam
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SOLUTION 61
A pump is required to supply 200 GPM of water to (2) outlets at a pier. The first outlet is 500
feet from the pump, the second outlet along the same path of travel is 500 feet from the first
outlet. The first length of pipe is 6" STD Steel, the second length of pipe is 4" STD Steel pipe.
What is the total pressure drop in FT of Head? Assume all minor losses are negligible.
The quickest way to solve this problem is to use your resources, specifically ASHRAE
Fundamentals.
Navigate to the Pipe Sizing chapter and find the graph that plots Head Loss as a function of flow
rate and pipe size. Use the graph for standard Schedule 40 pipe.
Since each outlet supplies 200 GPM, the total flow through the 6” pipe segment is 200 GPM +
200 GPM = 400 GPM.
Find 400 GPM and 6" pipe, Pressure loss ≈ 1.1
Multiply by total length of 6" pipe 1.1
𝑓𝑓𝑓𝑓 𝑜𝑜𝑜𝑜 ℎ𝑒𝑒𝑒𝑒𝑒𝑒
1𝑜𝑜𝑜𝑜 𝑓𝑓𝑓𝑓
𝑓𝑓𝑓𝑓 𝑜𝑜𝑜𝑜 ℎ𝑒𝑒𝑒𝑒𝑒𝑒
1𝑜𝑜𝑜𝑜 𝑓𝑓𝑓𝑓
1𝑜𝑜𝑜𝑜 𝑓𝑓𝑓𝑓
∗ 500 𝑓𝑓𝑓𝑓 = 5.5 𝑓𝑓𝑓𝑓 𝑜𝑜𝑜𝑜 ℎ𝑒𝑒𝑒𝑒𝑒𝑒
Find 200 GPM and 4" pipe, Pressure loss ≈ 2.4
Multiply by total length of 4" pipe 2.4
𝑓𝑓𝑓𝑓 𝑜𝑜𝑜𝑜 ℎ𝑒𝑒𝑒𝑒𝑑𝑑
𝑓𝑓𝑓𝑓 𝑜𝑜𝑜𝑜 ℎ𝑒𝑒𝑒𝑒𝑒𝑒
1𝑜𝑜𝑜𝑜 𝑓𝑓𝑓𝑓
∗ 500 𝑓𝑓𝑓𝑓 = 12.0 𝑓𝑓𝑓𝑓 𝑜𝑜𝑜𝑜 ℎ𝑒𝑒𝑒𝑒𝑒𝑒
Total pressure drop = 5.5 + 12.0 = 17.5 𝑓𝑓𝑓𝑓 𝑜𝑜𝑜𝑜 ℎ𝑒𝑒𝑒𝑒𝑒𝑒
Correct answer is C
(A) 12 feet
(B) 16 feet
(C) 18 feet
(D) 21 feet
Mechanical PE - HVAC and Refrigeration Full Exam
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SOLUTION 62
An air side economizer should operate when which of the following is absolutely true?
The primary purpose of the air side economizer is to save energy. An air-side economizer lets
in outside air to the air conditioning equipment when it is more economical to cool the outside air
rather than the re-circulated air. This occurs when the enthalpy of the outside air is less than
the re-circulated air.
Economizers are often used in cool & dry climates.
Correct answer is C.
(A) When the dry-bulb of the outside air is less than the re-circulated air.
(B) When the humidity of the outside air is less than the re-circulated air.
(C) When the enthalpy of the outside air is less than the re-circulated air.
(D) When the humidity ratio of the outside air is less than the re-circulated air.
Mechanical PE - HVAC and Refrigeration Full Exam
PM Session Solutions -26
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SOLUTION 63
A humidifier evaporates 0.011 GPM of water into an incoming air stream. Entering air
conditions are 85 F DB, 70% relative humidity. What is the required flow rate (CFM) of the fan?
Assume the air leaving the spray humidifier is at 85 F DB, 90% relative humidity.
In this question, 0.011 GPM is evaporated into the air, thereby raising the moisture content of
the air. Since the entering and exiting conditions of the air are known, the total CFM can be
found through the following equation:
∆𝑊𝑊𝐿𝐿𝐿𝐿
𝑙𝑙𝑙𝑙𝑙𝑙 𝑜𝑜𝑜𝑜 𝐻𝐻20
. 075 𝑙𝑙𝑙𝑙𝑙𝑙 𝑜𝑜𝑜𝑜 𝑑𝑑𝑑𝑑𝑑𝑑 𝑎𝑎𝑎𝑎𝑎𝑎
𝑙𝑙𝑙𝑙𝑙𝑙 𝑜𝑜𝑜𝑜 𝐻𝐻20
∗ 𝑋𝑋 𝐶𝐶𝐶𝐶𝑀𝑀 𝑜𝑜𝑜𝑜 𝑑𝑑𝑑𝑑𝑑𝑑 𝑎𝑎𝑎𝑎𝑎𝑎 ∗
= 𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸 𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅
𝑙𝑙𝑙𝑙𝑙𝑙 𝑜𝑜𝑜𝑜 𝑑𝑑𝑑𝑑𝑑𝑑 𝑎𝑎𝑎𝑎𝑎𝑎
𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚
1 𝑓𝑓𝑡𝑡 3
𝑤𝑤ℎ𝑒𝑒𝑒𝑒𝑒𝑒 ∆𝑊𝑊𝐿𝐿𝐿𝐿 = 𝑐𝑐ℎ𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 𝑖𝑖𝑖𝑖 ℎ𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 (𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒 − 𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒)
Find the humidity ratio of the entering and exiting conditions from the Psychrometric chart.
𝑙𝑙𝑙𝑙𝑙𝑙 𝑜𝑜𝑜𝑜 𝐻𝐻20
]
𝑙𝑙𝑙𝑙𝑙𝑙 𝑜𝑜𝑜𝑜 𝑑𝑑𝑑𝑑𝑑𝑑 𝑎𝑎𝑎𝑎𝑎𝑎
∆𝑊𝑊𝐿𝐿𝐿𝐿 = 𝑊𝑊𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒 − 𝑊𝑊𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒 = .0237 − .0183 [
∆𝑊𝑊𝐿𝐿𝐿𝐿 = .0054
Next convert GPM to lbs per minute.
0.011
𝑙𝑙𝑙𝑙𝑙𝑙 𝑜𝑜𝑜𝑜 𝐻𝐻20
𝑙𝑙𝑙𝑙𝑙𝑙 𝑜𝑜𝑜𝑜 𝑑𝑑𝑑𝑑𝑑𝑑 𝑎𝑎𝑎𝑎𝑎𝑎
1 𝑓𝑓𝑡𝑡 3
62.4 𝑙𝑙𝑙𝑙𝑙𝑙 𝑜𝑜𝑜𝑜 𝐻𝐻20
𝑙𝑙𝑙𝑙𝑙𝑙 𝑜𝑜𝑜𝑜 𝐻𝐻20
𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔
∗
∗
= 0.092
3
𝑓𝑓𝑡𝑡
𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚
𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 7.48 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔
Plug in the values into the first equation.
. 0054
𝑙𝑙𝑙𝑙𝑙𝑙 𝑜𝑜𝑜𝑜 𝐻𝐻20
. 075 𝑙𝑙𝑙𝑙𝑙𝑙 𝑜𝑜𝑜𝑜 𝑑𝑑𝑑𝑑𝑑𝑑 𝑎𝑎𝑎𝑎𝑎𝑎
𝑙𝑙𝑙𝑙𝑙𝑙 𝑜𝑜𝑜𝑜 𝐻𝐻20
∗ 𝑋𝑋 𝐶𝐶𝐶𝐶𝐶𝐶 𝑜𝑜𝑜𝑜 𝑑𝑑𝑑𝑑𝑑𝑑 𝑎𝑎𝑎𝑎𝑎𝑎 ∗
= 0.092
3
𝑙𝑙𝑙𝑙𝑙𝑙 𝑜𝑜𝑜𝑜 𝑑𝑑𝑑𝑑𝑑𝑑 𝑎𝑎𝑎𝑎𝑎𝑎
1 𝑓𝑓𝑓𝑓
𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚
Correct answer is A.
𝑋𝑋 = 226 𝐶𝐶𝐶𝐶𝐶𝐶
(A) 230 CFM
(B) 400 CFM
(C) 510 CFM
(D) 640 CFM
Mechanical PE - HVAC and Refrigeration Full Exam
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SOLUTION 64
A new diffuser is placed in the center of a 40’ X 40’ room. For the given CFM, the diffuser has
values of T150 of 5’, T100 of 10’ and T50 of 15’. At what distance from the nearest wall will the
velocity of the air be 100 feet per minute?
Performance data for diffusers includes a table describing the air velocities as a function of CFM
and the distance from the diffuser. The term “ T150 of 5’ ” indicates that at 5’ from the diffuser the
air velocity is equal to 150 feet per minute. The term “ T100 of 10’ ” indicates that at 10’ from the
diffuser the air velocity is equal to 100 feet per minute. This point is also 10’ from the nearest
wall because the diffuser is placed in the center of a 40’ X 40’ room.
Correct answer is B.
(A) 5’
(B) 10’
(C) 15’
(D) 20’
Mechanical PE - HVAC and Refrigeration Full Exam
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SOLUTION 65
An existing 500 ton chiller with a COP of 4.5, runs for an equivalent of 3 full load hours a day,
365 days a year. The chiller is being replaced with a new 500 ton chiller with an efficiency of
0.6 kW/ton for a total cost of $750,000. If the interest rate is 4% and the lifetime of the new
chiller is 25 years, then what is the annual value of replacing the existing chiller with a new
chiller? Assume no annual maintenance costs; include annual electricity cost savings with unit
cost of $0.25 per kWh.
This question involves Engineering Economics and calculating the Present Value of a design
alternative.
First calculate the yearly amount of energy savings between the old and proposed new chiller.
Convert COP to KW/Ton
4.5 =
0.782
12
(𝑋𝑋) ∗ (3.412)
𝑥𝑥 = 0.782
𝑘𝑘𝑘𝑘
𝑡𝑡𝑡𝑡𝑡𝑡
𝑘𝑘𝑘𝑘
ℎ𝑟𝑟𝑟𝑟
𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑
∗ 500 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 ∗ 3
∗ 365
= 428,145 𝑘𝑘𝑘𝑘ℎ
𝑡𝑡𝑡𝑡𝑡𝑡
𝑑𝑑𝑑𝑑𝑑𝑑
𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦
Calculate yearly energy costs of the old chiller.
428,145 𝑘𝑘𝑘𝑘ℎ ∗
$0.25
= $107,036
𝑘𝑘𝑘𝑘ℎ
Calculate yearly energy costs of the new chiller.
0.6
𝑘𝑘𝑘𝑘
ℎ𝑟𝑟𝑟𝑟
𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 $0.25
∗ 500 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 ∗ 3
∗ 365
∗
= $82,125
𝑡𝑡𝑡𝑡𝑡𝑡
𝑑𝑑𝑑𝑑𝑑𝑑
𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦 𝑘𝑘𝑘𝑘ℎ
The yearly energy savings of installing the new chiller is approximately:
𝑌𝑌𝑌𝑌𝑌𝑌𝑌𝑌𝑌𝑌𝑌𝑌 𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆 = $107,036 − $82,125 = $24,911
The annual owning costs of the new chiller initial cost is found by navigating to
𝐴𝐴
𝑌𝑌𝑌𝑌𝑌𝑌𝑌𝑌𝑌𝑌𝑌𝑌 𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂 𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 = $750,000 ∗ � , 𝑖𝑖 = 4%, 𝑛𝑛 = 25 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦� ;
𝑃𝑃
Mechanical PE - HVAC and Refrigeration Full Exam
PM Session Solutions -29
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𝑌𝑌𝑌𝑌𝑌𝑌𝑌𝑌𝑌𝑌𝑌𝑌 𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂 𝐶𝐶𝐶𝐶𝑠𝑠𝑠𝑠𝑠𝑠 = $750,000 ∗ .0640 = $48,000
The annual value is equal to the summation of the yearly savings and yearly owning costs.
𝐴𝐴𝐴𝐴 = $48,000 − $24,91 = $23,089
Correct answer is A.
(A) $23,090
(B) $24,910
(C) $36,000
(D)$48,000
Mechanical PE - HVAC and Refrigeration Full Exam
PM Session Solutions -30
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SOLUTION 66
A refrigeration unit is required to store 1,000 lbs of salmon (Heat Capacity Above Freezing: 0.88
Btu/lb*°F, Heat Capacity Below Freezing: 0.51 Btu/lb*°F, Latent heat of Fusion: 110 Btu/lb,
Initial Freezing Point: 28 F). If the salmon arrives to the unit at 70 F and must be cooled to 10 F
in 2 hours, then what is the required size of the air conditioning system, in Btu/hr.
It is important to note that these values of specific heat and latent heat of fusion for salmon and
many other types of food can be found in ASHRAE Refrigeration.
Calculate the total amount of cooling to bring the salmon from 70 F to Freezing.
𝑄𝑄 = 1,000 𝑙𝑙𝑙𝑙𝑙𝑙 ∗ 0.88 ∗ (70 − 28 ℉)
𝑄𝑄 = 36,960 𝐵𝐵𝐵𝐵𝐵𝐵
Calculate the total amount of cooling to freeze the salmon.
𝑄𝑄 = 1,000 𝑙𝑙𝑙𝑙𝑙𝑙 ∗ 110
𝑄𝑄 = 110,000 𝐵𝐵𝐵𝐵𝐵𝐵
Calculate the total amount of cooling to bring the salmon from freezing to 10 F.
𝑄𝑄 = 1,000 𝑙𝑙𝑙𝑙𝑙𝑙 ∗ 0.51 ∗ (28 − 10)
𝑄𝑄 = 9,180 𝐵𝐵𝐵𝐵𝐵𝐵
Sum up the total cooling and divide by 2 hours in order to calculate the required size of the air
conditioning system.
𝑄𝑄 =
𝐵𝐵𝐵𝐵𝐵𝐵
36,960 + 110,000 + 9,180 𝐵𝐵𝐵𝐵𝐵𝐵
= 78,070
ℎ
2 𝐻𝐻𝐻𝐻
Correct answer is C.
(A) 18,230
(B) 37,210
𝐵𝐵𝐵𝐵𝐵𝐵
ℎ
𝐵𝐵𝐵𝐵𝐵𝐵
(C) 𝟕𝟕𝟕𝟕, 𝟎𝟎𝟎𝟎𝟎𝟎
ℎ
(D) 110,000
𝑩𝑩𝑩𝑩𝑩𝑩
𝒉𝒉
𝐵𝐵𝐵𝐵𝐵𝐵
ℎ
Mechanical PE - HVAC and Refrigeration Full Exam
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SOLUTION 67
A new cooling coil provides sensible cooling of 250,000 Btu/hr. The entering air conditions into
the coil are 80 F DB. Leaving conditions from the coil at 55 F DB. If the coil is at an elevation of
5,000 FT, then what is the air flow rate in CFM? Assume negligible bypass factor and
miscellaneous heat gains/losses.
Density = 0.062 lb/ft^3; Heat Capacity = 0.24 Btu/lb*F
This problem is an energy balance uses the original sensible heat equation.
𝑄𝑄𝑎𝑎𝑎𝑎𝑎𝑎 = 𝑄𝑄𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐
250,000
𝑄𝑄𝑎𝑎𝑎𝑎𝑎𝑎 = 𝑚𝑚̇ ∗ 𝑐𝑐𝑝𝑝 ∗ ∆𝑇𝑇
𝑓𝑓𝑡𝑡 3 60 𝑚𝑚𝑚𝑚𝑚𝑚
𝑙𝑙𝑙𝑙
𝐵𝐵𝐵𝐵𝐵𝐵
𝐵𝐵𝐵𝐵𝐵𝐵
= 𝑄𝑄𝑎𝑎𝑎𝑎𝑎𝑎 = 𝑥𝑥
∗
∗ 0.062 3 ∗ 0.24
(80 − 55 𝐹𝐹)
hr
𝑓𝑓𝑡𝑡
𝑙𝑙𝑙𝑙𝑙𝑙 ∗ 𝐹𝐹
ℎ𝑟𝑟
min
𝑥𝑥 = 11,200
𝑓𝑓𝑡𝑡 3
min
Correct answer is D.
(A) 5,125
(B) 6,065
(C) 9,565
(D) 11,200
Mechanical PE - HVAC and Refrigeration Full Exam
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SOLUTION 68
A 100% outside air handler serving a theater supplies 10,000 CFM of OAIR at 55 F DB/54 F WB
to maintain space conditions at 75 F DB and 50% Relative Humidity. Outside air conditions are
at 85 F DB and 80% Relative Humidity. How many tons of cooling can be saved if a total
enthalpy wheel is provided with 75% effectiveness? Assume negligible bypass factor and no
minor heat gains/losses.
First calculate the total cooling provided without an enthalpy wheel. Find the enthalpies of the
air entering and leaving the coil, from the psychrometric chart.
ℎ𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐,𝑖𝑖𝑖𝑖 = 43.42
ℎ𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐,𝑜𝑜𝑜𝑜𝑜𝑜 = 22.6
Calculate the total cooling load:
𝐵𝐵𝐵𝐵𝐵𝐵
; 85 ℉, 80% 𝑅𝑅𝑅𝑅
𝑙𝑙𝑙𝑙
𝐵𝐵𝐵𝐵𝐵𝐵
; 55 ℉ 𝐷𝐷𝐷𝐷, 54 ℉ 𝑊𝑊𝑊𝑊
𝑙𝑙𝑙𝑙
𝑄𝑄 = 4.5 ∗ 𝐶𝐶𝐶𝐶𝐶𝐶 ∗ (ℎ𝑖𝑖𝑖𝑖 − ℎ𝑜𝑜𝑜𝑜𝑜𝑜 )
𝑄𝑄 = 4.5 ∗ 10,000 ∗ (43.42 − 22.6) = 936,900 𝐵𝐵𝐵𝐵𝐵𝐵ℎ
Next calculate the enthalpy exiting the enthalpy wheel.
ℎ𝑒𝑒𝑒𝑒𝑒𝑒 − ℎ𝑙𝑙𝑙𝑙𝑙𝑙,𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎
= 0.75
ℎ𝑒𝑒𝑒𝑒𝑒𝑒 − ℎ𝑙𝑙𝑙𝑙𝑙𝑙,𝑚𝑚𝑚𝑚𝑚𝑚
Where the maximum leaving enthalpy is that of the air leaving the space at 75 F/50% RH.
43.42 − ℎ𝑙𝑙𝑙𝑙𝑙𝑙,𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎
= 0.75
43.42 − 28.14
ℎ𝑙𝑙𝑙𝑙𝑙𝑙 = 31.96
𝐵𝐵𝑡𝑡𝑢𝑢
𝑙𝑙𝑙𝑙
Plug in the new coil entering enthalpy into the previous equation and find the new total cooling.
𝑄𝑄 = 4.5 ∗ 10,000 ∗ (31.96 − 22.6) = 421,200 𝐵𝐵𝐵𝐵𝐵𝐵ℎ
Mechanical PE - HVAC and Refrigeration Full Exam
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Next calculate the total difference and convert to tons of cooling.
Correct answer is B.
936,900 𝐵𝐵𝐵𝐵𝐵𝐵ℎ − 421,200 𝐵𝐵𝐵𝐵𝐵𝐵ℎ
= 43 𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 𝑜𝑜𝑜𝑜 𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶
12,000
(A) 25 tons
(B) 43 tons
(C) 56 tons
(D) 60 tons
Mechanical PE - HVAC and Refrigeration Full Exam
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SOLUTION 69
A sensible heat recovery device is used to preheat entering outdoor air 3,500 CFM, (40 F, 60%
RH) with 4,000 CFM exhaust air (77 F, 55% RH). The sensible effectiveness of the device is
60%. What is the leaving supply air dry bulb temperature? Assume zero leakage.
Use the air-to-air heat exchanger effectiveness equation.
𝜀𝜀 =
𝐶𝐶𝐶𝐶𝑀𝑀𝑂𝑂𝑂𝑂 ∗ (𝐷𝐷𝐵𝐵𝑆𝑆𝑆𝑆 − 𝐷𝐷𝐵𝐵𝑂𝑂𝑂𝑂 )
𝐶𝐶𝐶𝐶𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀 ∗ (𝐷𝐷𝐵𝐵𝑅𝑅𝑅𝑅 − 𝐷𝐷𝐵𝐵𝑂𝑂𝑂𝑂 )
0.60 =
3,500 ∗ (𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆 𝐷𝐷𝐷𝐷 − 40)
3,500 ∗ (77 − 40)
𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆 𝐷𝐷𝐷𝐷 = 62.2 ℉
Correct answer is B.
(A) 49.3 ℉
(B) 𝟔𝟔𝟔𝟔. 𝟐𝟐 ℉
(C) 67.8 ℉
(D) 74.2 ℉
Mechanical PE - HVAC and Refrigeration Full Exam
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SOLUTION 70
A new pump is required to pump 350 GPM from a well up to a holding tank that is 200 FT above
the pump. Assume a total friction loss of 20 FT of head from piping and fittings. The pump is
required to pump 8 hours a day, 7 days a week, 52 weeks a year. The pump is 85% efficient
and the motor is 95% efficient. Electricity costs are $0.25 per kWh. What are the yearly cost
savings if a new location for the holding tank is found at an elevation of 100 FT above the
pump? Assume the same efficiencies and friction losses between the two scenarios.
In this question you must first calculate the pump horsepower, in order to determine the
electricity usage.
Pump 1: Total Dynamic Head = 200′ + 20′ = 220′
Use the pump horsepower equation
𝑃𝑃ℎ𝑝𝑝 =
𝐺𝐺𝐺𝐺𝐺𝐺 ∗ 𝑇𝑇𝑇𝑇𝑇𝑇 ∗ 𝑆𝑆𝑆𝑆
; 𝑆𝑆𝑆𝑆 = 1.0 [𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤]
3960 ∗ 𝜀𝜀𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝
𝑃𝑃ℎ𝑝𝑝,1 =
350 ∗ 220 ∗ 1.0
3960 ∗ 0.85
𝑃𝑃ℎ𝑝𝑝,1 = 22.9 𝐻𝐻𝐻𝐻
Next calculate energy used by the motor which powers the pump, by dividing by the efficiency of
the motor.
𝑃𝑃𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚,1 =
22.9 𝐻𝐻𝐻𝐻
= 24.1 𝐻𝐻𝐻𝐻 𝑜𝑜𝑜𝑜 18 𝐾𝐾𝐾𝐾; 1 𝐻𝐻𝐻𝐻 = 0.746 𝐾𝐾𝐾𝐾
0.95
18 𝐾𝐾𝐾𝐾 ∗
8 ℎ𝑟𝑟𝑟𝑟 7 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 52 𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤
∗
∗
= 52,416 𝑘𝑘𝑘𝑘ℎ;
𝑑𝑑𝑑𝑑𝑑𝑑
𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤
1 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦
Total Yearly cost for the 1st scenario is: 52,416 𝑘𝑘𝑘𝑘ℎ ∗
$0.25
𝑘𝑘𝑘𝑘ℎ
= $13,104 𝑝𝑝𝑝𝑝𝑝𝑝 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦
Repeat the process with the new TDH of = 100′ + 20′ = 120′
𝑃𝑃𝑘𝑘𝑘𝑘,1 =
9.8 𝐾𝐾𝐾𝐾 ∗
𝑃𝑃ℎ𝑝𝑝 =
350 ∗ 120 ∗ 1.0
3960 ∗ 0.85
0.746 𝑘𝑘𝑘𝑘
350 ∗ 120 ∗ 1.0
1
�∗
∗�
1 𝐻𝐻𝐻𝐻
3960 ∗ 0.85
0.95
𝑃𝑃𝑘𝑘𝑘𝑘,2 = 9.8 𝐾𝐾𝐾𝐾
8 ℎ𝑟𝑟𝑟𝑟 7 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 52 𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤 $0.25
∗
∗
∗
= $7,134 𝑝𝑝𝑝𝑝𝑝𝑝 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦
𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤
1 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦
𝑘𝑘𝑘𝑘ℎ
𝑑𝑑𝑑𝑑𝑑𝑑
Total Yearly cost for the 2nd scenario is: $7,134 𝑝𝑝𝑝𝑝𝑝𝑝 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦
Mechanical PE - HVAC and Refrigeration Full Exam
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The savings of scenario 2 over scenario 1 is approximately:
Correct answer is D
$13,104 − $7,134 = $5,970 𝑝𝑝𝑝𝑝𝑝𝑝 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦
(A) $1,720 per year
(B) $2,090 per year
(C) $4,170 per year
(D) $5,970 per year
Mechanical PE - HVAC and Refrigeration Full Exam
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SOLUTION 71
A 22" X 10" galvanized steel duct is used to convey 2,000 CFM of industrial exhaust. What is
the pressure drop in units of IN. WG per 100 ft.
Density = 0.075 lb/ft^3; roughness factor = 0.0003 ft.
In order to calculate the pressure drop through a duct, use ASHRAE Fundamentals and
navigate to the equivalent rectangular duct table.
Find 22" X 10": The equivalent circular duct is 16" D.
Next go to the Friction Chart for Round Duct (Density = 0.075 lb/ft^3; roughness factor = 0.0003
ft.)
Navigate to 16" D and 2,000 CFM and read the friction loss.
Friction loss = 0.19 in. wg per 100 ft.
Correct answer is D.
(A) 0.07 in. WG per 100 ft
(B) 0.10 in. WG per 100 ft
(C) 0.14 in. WG per 100 ft
(D) 0.19 in. WG per 100 ft
Mechanical PE - HVAC and Refrigeration Full Exam
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SOLUTION 72
A new steam boiler provides 100 lb/hr of steam at 30 PSIA, 0 degrees super heat to various hot
water heaters. If the hot water heaters are designed to provide a 40 degree delta to incoming
water at 80 F, then what is the total GPM of hot water that the boiler can support?
Create an energy balance equation between the steam and the hot water.
𝑄𝑄𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 = 𝑚𝑚̇ ∗ ℎ𝑓𝑓𝑓𝑓
Find ℎ𝑓𝑓𝑓𝑓 in the MERM, Steam Tables as a function of pressure, Navigate to 30 PSIA. Read the
enthalpy of evaporation.
𝑄𝑄𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 = 100
𝑙𝑙𝑙𝑙
𝐵𝐵𝐵𝐵𝐵𝐵
𝐵𝐵𝐵𝐵𝐵𝐵
∗ 945.2
= 94,520
ℎ𝑟𝑟
𝑙𝑙𝑙𝑙
ℎ
𝑄𝑄ℎ𝑜𝑜𝑜𝑜 𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤 = 𝑚𝑚̇ ∗ 𝑐𝑐𝑝𝑝 ∗ ∆𝑇𝑇
𝑄𝑄ℎ𝑜𝑜𝑜𝑜 𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤 = 500 ∗ 𝐺𝐺𝐺𝐺𝐺𝐺 ∗ ∆𝑇𝑇
𝑄𝑄𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 = 94,520
𝐵𝐵𝐵𝐵𝐵𝐵
= 𝑄𝑄ℎ𝑜𝑜𝑜𝑜 𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤 = 500 ∗ 𝑥𝑥 𝐺𝐺𝐺𝐺𝐺𝐺 ∗ 40 ℉
ℎ
𝑥𝑥 = 4.73 𝐺𝐺𝐺𝐺𝐺𝐺
Correct answer is A.
(A) 4.73 GPM
(B) 10.2 GPM
(C) 15.7 GPM
(D) 21.9 GPM
Mechanical PE - HVAC and Refrigeration Full Exam
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SOLUTION 73
A new building with dimensions of 200' (L) X 150' (W) X 10' (H) is classified as having average
construction tightness, which relates to 0.3 air changes per hour of infiltration. If outside air is at
88 F DB/80% RH and the indoor design conditions are 75 F DB/50% RH, then what is the total
cooling load in tons added by infiltrated air?
First find the total amount of infiltrated air by first finding the total air volume.
200′ ∗ 150′ ∗ 10′ = 300,000 𝑓𝑓𝑡𝑡 3
300,000 𝑓𝑓𝑡𝑡 3 ∗ 0.3
𝑎𝑎𝑎𝑎𝑎𝑎 𝑐𝑐ℎ𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎
1 ℎ𝑜𝑜𝑜𝑜𝑜𝑜
∗
= 1,500 𝐶𝐶𝐶𝐶𝐶𝐶
ℎ𝑜𝑜𝑜𝑜𝑜𝑜
60 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚
Find the enthalpies of the indoor and outside air in order to determine the total cooling load in
tons.
𝑄𝑄 = 4.5 ∗ 𝐶𝐶𝐶𝐶𝐶𝐶 ∗ ∆ℎ
𝑄𝑄 = 4.5 ∗ 1,500 ∗ (46.57 − 28.14 )
𝑄𝑄 = 124,403
𝐵𝐵𝐵𝐵𝐵𝐵
ℎ
Correct answer is C.
(A) 89,560
𝐵𝐵𝐵𝐵𝐵𝐵
(B) 104,600
ℎ
𝐵𝐵𝐵𝐵𝑢𝑢
(C) 𝟏𝟏𝟏𝟏𝟏𝟏, 𝟒𝟒𝟒𝟒𝟒𝟒
(D) 159,400
ℎ
𝑩𝑩𝑩𝑩𝑩𝑩
𝒉𝒉
𝐵𝐵𝐵𝐵𝐵𝐵
ℎ
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SOLUTION 74
Which of the following filters provides greater than 99% arrestance?
Refer to ASHRAE Fundamentals, Filters section to find that the MERV 18 filter provides 99%
arrestance.
Correct answer is D.
(A) MERV 1
(B) MERV 7
(C) MERV 13
(D) MERV 18
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SOLUTION 75
A new dedicated outside air handling unit is used to cool 5,000 CFM of OAIR (88 F DB, 70%
RH) to 55 F DB/54 F WB. The outside air handling unit is equipped with a wrap around heat
pipe. The heat pipe pre-cools the outside air by 10 degrees F. What is the total reduction in
tons of cooling by installing a heat pipe? Assume density = 0.075 lb/ft^3, elevation = sea level.
𝑄𝑄 = 1.08 ∗ 𝐶𝐶𝐶𝐶𝐶𝐶 ∗ ∆𝑇𝑇
𝑄𝑄 = 1.08 ∗ 5,000 ∗ (10 ℉)
𝑄𝑄 = 54,000
𝐵𝐵𝐵𝐵𝑢𝑢
= 4.5 𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 𝑜𝑜𝑜𝑜 𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶
ℎ
Correct answer is A.
(A) 𝟒𝟒. 𝟓𝟓 𝑻𝑻𝑻𝑻𝑻𝑻𝑻𝑻
(B) 6.2 𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇
(C) 7.9 𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇
(D) 10 𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇
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SOLUTION 76
A new counter-current heat exchanger is designed for incoming cold water at 43 F and leaving
water at 52 F. If the entering/exiting hot water of 60 F and 50 F, then what will be the LMTD?
Use the Log Mean Temperature Difference Equation
𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿 =
∆𝑇𝑇𝐴𝐴 − ∆𝑇𝑇𝐵𝐵
∆𝑇𝑇
ln � 𝐴𝐴 �
∆𝑇𝑇𝐵𝐵
∆𝑇𝑇𝐵𝐵 = 50 − 43 = 7
∆𝑇𝑇𝐴𝐴 = 60 − 52 = 8
𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿 =
8−7
= 7.5℉
8
ln � �
7
Correct answer is A.
(A) 𝟕𝟕. 𝟓𝟓℉
(B) 8.5℉
(C) 10.0℉
(D) 11.5℉
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SOLUTION 77
What is the maximum allowable flame spread index of gypsum board air ducts, in accordance
with NFPA 90A?
According to NFPA 90A, Standard for the Installation of Air Conditioning and Ventilation
Systems, gypsum board air ducts must have a flame spread index of 25 or below.
It is also important to note that NFPA 90A also requires a smoke developed index of 50 or
below.
Correct answer is B.
(A) 0
(B) 25
(C) 50
(D) 100
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SOLUTION 78
A new classroom is designed for 25 people and has a total area of 500 square feet. What is the
required CFM of ventilation?
This problem involves the use of ASHRAE 62.1. Refer to ASHRAE 62.1 and find the ventilation
rates required for a classroom. ASHRAE 62.1 requires 10 CFM per person and 0.12 CFM per
square feet.
𝑥𝑥 = 25 ∗ 10 + 500 ∗ 0.12 = 310 𝐶𝐶𝐶𝐶𝐶𝐶
Correct answer is C.
(A) 60 𝐶𝐶𝐶𝐶𝐶𝐶
(B 250 𝐶𝐶𝐶𝐶𝐶𝐶
(C) 𝟑𝟑𝟑𝟑𝟑𝟑 𝑪𝑪𝑪𝑪𝑪𝑪
(D) 520 𝐶𝐶𝐶𝐶𝑀𝑀
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SOLUTION 79
The unit "clo" is used to describe the thermal insulation provided by which of the below?
This question involves knowledge of ASHRAE 55, Thermal Environmental Conditions for
Human Comfort. It is used to describe the insulation provided by clothing and garments.
Walls, Roofs and Equipment insulation are typically described by R-Values, U-Factors or kfactors.
Correct answer is C.
(A) Wall insulation
(B) Roof insulation
(C) Garment insulation
(D) Heating equipment insulation
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SOLUTION 80
Which of the following codes are least likely to be required to be checked with regards to the
installation of a commercial gas furnace?
A commercial gas furnace installation will need to conform to the requirements of NFPA 54, the
National Fuel Gas Code and the International Fuel Gas Code, depending on the jurisdiction.
ASHRAE 90.1 determines the energy efficiency requirements for commercial equipment
including commercial gas furnaces, depending on the jurisdiction.
The code that is least likely to be consulted is NFPA 70, the National Electric Code. This code
is primarily about the installation of electrical equipment.
Correct answer is C.
(A) NFPA 54
(B) ASHRAE 90.1
(C) NFPA 70
(D) International Fuel Gas Code
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SECTION 6
CONCLUSION
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5.0 CONCLUSION
If you have any questions on this sample exam or any other Engineering Pro Guides product,
then please contact:
Justin Kauwale at contact@engproguides.com
Hi. My name is Justin Kauwale, the creator of Engineering Pro Guides. I will be happy to
answer any questions you may have about the PE exam. Good luck on your studying! I hope
you pass the exam and I wish you the best in your career. Thank you for your purchase!
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SECTION 7
DIAGNOSTICS
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Mechanical PE Exam – HVAC & Refrigeration
AM Session -Sample Exam Diagnostics
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Major Category
Principles – Basic Engineering Practice
Principles – Basic Engineering Practice
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Principles – Thermodynamics
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Principles – Heat Transfer
Principles – Heat Transfer
Principles – Heat Transfer
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Principles – Energy/Mass Balances
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Application – Heating/Cooling Loads
Application – Heating/Cooling Loads
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Application – Heating/Cooling Loads
Application – Heating/Cooling Loads
Application – Heating/Cooling Loads
Application – Heating/Cooling Loads
Correct?
Mechanical PE Exam – HVAC & Refrigeration
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Applications – Equipment & Components
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Correct?