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COURSE FILE
ESO 201A Thermodynamics
1st Semester 2015-2016
Instructor
Prof. Sameer Khandekar
Department of Mechanical Engineering
Indian Institute of Technology Kanpur
Kanpur (UP) 208016 India
Contact
Office: SL-109, Tel: 7038
E-mail: [email protected]
URL: home.iitk.ac.in/~samkhan
ESO201A: Thermodynamics
Instructor: Dr. Sameer Khandekar
2015-2016 (First Semester)
COURSE OUTLINE

Chapter 1, Chapter 2
Definitions and concepts: System and CV, Macroscopic and microscopic view points;
Property, Thermodynamics Static and Equilibrium, Energy, Work interaction and various
modes of work, Heat: Zeroth Law of Thermodynamics, Temperature Scale.

Chapter 3
Properties of Pure Substances, Phase, Simple compressible substance, Mathematical,
Tabular and Graphical representation of data; Ideal gas Van der Waals Equation of state;
Compressibility chart; Thermodynamic Diagrams including Mollier diagram and Steam
Tables.

Quiz #1
======================================================================
Chapter 4, Chapter 5
First law of thermodynamics and its applications to non-flow processes, Applications of
first law of thermodynamics to flow processes; Steady flow and Transient flow processes.

MID SEMESTER EXAMINATION
======================================================================
Chapter 6, Chapter 7, Chapter 8
Second Law of Thermodynamics and its Applications, Availability.

Quiz #2
======================================================================
Chapter 15
Applications of first law of thermodynamics to chemically reacting systems.

Chapter 9, Chapter 10, Chapter 11
Gas power cycles, Vapor power cycles, Refrigeration cycles.

Chapter 12
Thermodynamic potentials, Maxwell relations; Thermodynamic relations.
END SEMESTER EXAMINATION
Course File: ESO201A
Prof. Sameer Khandekar
INDIAN INSTITUTE OF TECHNOLOGY KANPUR
ESO 201A: Thermodynamics
Instructor: Sameer Khandekar
Timings:
Tutorials:
Lectures: MWF 10.00-10.50 AM in L16
Th 10.00- 10.50 AM (10 class rooms, as indicated below)
Faculty coordinates:
#
Name
Dep
.
E-mail
Tel
Office
Section
Classroom
7038
SL-109
-
L16
6141
7513
6072
6973
7243
7942
7684
7727
6843
7727
NL2-202
FB-457
FB-467
FB-438
FB-402
FB-353
NL-302
KANZI House
NWTF - 302
KANZI House
1
2
3
4
5
6
7
8
9
10
TB-203
TB-204
TB-205
TB-206
TB-207
TB-208
TB-209
TB-210
TB-211
TB-212
Instructor
Sameer Khandekar
ME
samkhan
Tutors
1
2
3
4
5
6
7
8
9
10
Dr. Jayant Singh
Dr. Nitin Kaistha
Dr. Indranil Saha Dalal
Dr. Shikha Prasad
Dr. Prabhat Munshi
Dr. Subrata Sarkar
Dr. Jishnu Bhattacharya
Mr. Srihari Dinesh
Mr. Pradeep K. S.
Mr. Azmira Nageshwar Rao
CHE
CHE
CHE
ME
ME
ME
ME
AE
AE
AE
jayantks
nkaistha
indrasd
shikhap
pmunshi
subra
jishnu
jsdinesh
pradeeps
azmiran
Grading Policy:
Item
Mid-semester Examination
Two announced quizzes
Final Examination
Tutorial marks
Points
25 points
25 points (12.5 points each)
40 points
10 points
Total
100 points
Remarks
Closed book/notes
Closed book/notes
Closed book/notes
Attendance + Tutor input
Attendance: is mandatory, as per institute rules, both in the lectures and in the tutorials.
Re-grading: You can return the Examination/Quiz copies on the day the answer-scripts are distributed
(in the tutorials) for re-grading, after writing your comments on the top of the answer-scripts. If you
are absent (except official leaves, if granted) on that day, you lose the right of appeal.
Make-up Exams: Only one make-up examination will be given, only to the students with regular
attendance in Tutorials. Students missing mid-semester examination, or only one central quiz, can
take this upon production of a valid medical/other relevant certificate/official reason. The entire
material covered in the course will be included, irrespective of which test is missed. This make-up will
be at the same time as the final make-up exam (date of which will be decided by the DOAA). Thus,
there will be only ONE make-up test for the entire course, after the final examination. Use of unfair
means of any kind during the course, exams, quizzes, etc., including use of proxy-attendance or taking
help during examinations, copying, etc.) would result in severe punishment, as per institute rules.
I-cards: Carrying I-cards is mandatory in all examinations/quizzes.
Text Book: (Sufficient copies in the Reserved Books Section, Central Library)
●
●
Y. A. Cengel and M. A. Boles, Thermodynamics: An Engineering Approach,
6th Edition, SI Units, Tata McGraw Hill (2008 Indian Edition) in already the library (Reading
Section). Procurement of 7th edition is in progress (Order placed in July 2015).
You may also purchase 7th Edition, 2011 (Mc. Graw Hill Education (India), ISBN: 978-0-071072540, SI Units, of the above book (available in the Shopping Centre at approx. Rs. 700/-).
Course File: ESO201A
Prof. Sameer Khandekar
ESO201A: Thermodynamics
Instructor: Dr. Sameer Khandekar
2015-2016 (First Semester)
Semester Plan
Monday
27
3
10
August
17
24
31
7
September 14
21
28
5
October 12
19
26
2
9
November 16
23
30
July
Tuesday
28
4
11
18
25
1
8
15
22
29
6
13
20
27
3
10
17
24
Institute holiday
Tutorial
Lecture
Examination
Course File: ESO201A
Prof. Sameer Khandekar
Wednesday
29
5
12
19
26
2
9
16 (MIDSEM)
23
30
7
14
21
28
4
11
18 (ENDSEM)
25
=
=
Thursday
30 (T1)
6 (T2)
13 (T3)
20 (T4)
27 (T5-QUIZ#1)
3 (T6)
10 (T7)
17
24 (T8)
1 (T9)
8 (T10)
15 (T11-QUIZ#2)
22
29 (T12)
5 (T13)
12 (T14)
19
26
Friday
31
7
14
21
28
4
11
18
25
2
9
16
23
30
6
13
20
27
12 + 2 Quizes
39 + 2 Extra Classes
Saturday
1
8
15
22
29 (Friday plan)
5
12
19
26
3 (EXTRA CLASS)
10
17
24
31
7
14
21
28
Sunday
2
9
16
23
30
6
13
20
27
4
11
18
25
8
15
22
29
MIDSEM
RECESS
ENDSEM
ENDSEM
ESO201A: Thermodynamics
Instructor: Sameer Khandekar
First Semester: 2015 – 2016
Course File
Lecture #1:
Introduction to Thermodynamics, Importance, Definitions – Continuum, System:
closed, open and isolated, Boundary: real and imaginary, Control mass and
control volume, Properties of a system, State of a system, Equilibrium:
Mechanical, Thermal, Chemical, Phase equilibrium, Quasi-equilibrium process,
Continuum Thermodynamics, Simple-compressible and complex systems, The State
Postulate
Lecture #2
Overview of 2-stroke, 4-stroke and Wankel engine (To highlight unsteady and
non-uniform processes), Intensive and Extensive properties, Concept of
temperature from Zeroth law of thermodynamics, Linear dependence of temperature
on pressure for constant volume gas thermometers, temperature as a manifestation
of kinetic energy, absolute thermodynamic scale of temperature - Kelvin scale,
Other scales, Overview of temperature, measuring devices:-liquid crystal sheet,
RTD, infrared thermography, thermocouples, mercury thermometers, Pressure
measurement - absolute and gauge pressure, piezoelectric transducer, manometer,
Pascal’s law, hydrostatic law, Barometer.
Lecture #3
Classification of Energy-mechanical, electrical, chemical, thermal, nuclear,
magnetic etc., Microscopic and Macroscopic forms, Internal energy, Macroscopic
- kinetic, gravitational potential, Flow work, Mechanical forms of energy,
Microscopic - translational, rotational and vibrational energies of atoms and
molecules, electronic translational, rotational and spin, Chemical bond energy,
Nuclear energy, Interactions in Open and Closed Systems, Latent and specific
heat, manifestation of heat energy into either temperature or in other
microscopic forms of energy.
Lecture #4
Heat and Work interactions, Different processes-Adiabatic, isothermal,
isobaric, isochoric, Sign convention for heat and work, Similarities between
heat and work, Point and Path functions, Mechanical form of work-displacement
work, shaft work, surface tension work, spring work, stretching work, nonmechanical forms of work-electrical work, magnetic work, generalized expression
for work done.
Lecture #5
Overview of Energy conversion systems – Examples (on PPT) jet engine, marine
engine, roots blower, engine turbocharger, Pelton wheel etc., Introduction to
First law of thermodynamics, Concept of Total energy and its change, Work done
in an adiabatic process, Work done and energy change in a cyclic process,
Thermodynamic efficiency of a sub-system and total efficiency of a system,
Highlighting the connection between Energy and Environment.
Course File: ESO201A
Prof. Sameer Khandekar
Lecture #6
Definition of pure substance, phases: solids liquid and gases, principal phase
and sub-phases, Demonstration of mechanical boiling, Introduction to phase
diagrams, T-v diagram, Saturation pressure and saturation temperature, Sensible
heating, Latent heat of vaporization, Compressed or sub-cooled liquid, Saturated
liquid, Saturated and superheated vapor, Critical temperature, pressure and
volume, Maintaining isothermal conditions for a system.
Lecture #7
(On PPT) Phase diagrams and tables, PV, TV and PT diagrams, P-V-T surfaces (for
substances which expand and contract on freezing, respectively), Reading phase
diagrams, latent heat of fusion and evaporation, triple point, concept of
regelation, allotropic forms of solid phases, Boyles law, Charles law for gas
phases, Introduction to Property Tables, Enthalpy as a combination property.
Lecture #8
Introduction to supercritical fluids, Phase diagram of CO2, H2O, Properties of
wet mixture, dryness fraction or quality, locating a mixture on 2-phase diagram,
Formulation of mixture properties, iso-quality lines on 2-phase diagrams,
quality as a thermodynamic property inside the 2-phase dome, reading property
data from superheated vapor, 2-phase region and compressed liquid table,
characteristics of superheated vapor and compressed liquids, 2 example problems
Lecture #9
Ideal Gas law and its validity, reasons of deviation from ideal gas behavior,
van der Waals correction for pressure and volume, Estimation of constants ‘a’
and ‘b’ using critical point data, Beattie-Bridgeman equation, Benedict – WebbRuben Equation, Virial equation, (On PPT) - solution of van der Waals equation
of state, Comparison of percentage errors for different equations of state, Use
of Ideal gas equation with compressibility chart.
Lecture #10
Limitations of van der Waals equation, Introduction to metastable states,
reduced pressure and temperature, generalized compressibility chart, Principle
of corresponding states, 3 example problems on the use of charts and tables
Lecture #11
First law applied to a closed system, Moving boundary work, boundary work done
in different quasi-equilibrium processes -isobaric, isochoric, isothermal and
polytropic processes, example problem involving shaft work, friction work and
spring work, (On PPT) Pistons in different IC engine configurations.
Lecture #12
Introduction to heat capacity, Expression for heat addition at constant pressure
and constant volume, Calculation of ‘u’, ‘h’, ‘Cp’, ‘Cv’ for an ideal gas,
Relation between Cp and Cv for an ideal gas, heat capacity for solids and
liquids.
Lecture #13
Specific Heat Ratio for mono-atomic, di-atomic and polyatomic gases, adiabatic
process for an ideal gas for a simple compressible system, Comparison of
compression work done in adiabatic and isothermal processes, Slope of adiabatic
and isothermal processes on P_V diagram, Constant internal energy process, (On
Course File: ESO201A
Prof. Sameer Khandekar
PPT) Different types of compressors-air cooled, water cooled, 2-stage etc., one
example problem.
Lecture #14
First law applied to an open system, comparison of control mass and control
volume approach, conservation of mass principle applied to an open system,
steady and unsteady system, Incompressible system, Concept of flow work.
Lecture #15
Derivation of energy equation for control volume, Examples of typical
engineering systems, Turbines, Compressors, Heat Exchangers, Pipe flow, pumps.
Lecture #16
Completion of discussion on steady flow processes, Mixing, Isenthalpic process,
Joule-Thompson effect (only introduce), Discussion of Quiz #1, two example
problems of Chapter 4.
Lecture #17
Unsteady Control Volume Processes, Charging and discharging of gas cylinders,
need for second law of thermodynamics, discussion on quality of energy, thermal
efficiency and level of perfection of thermodynamic systems, concept of thermal
source and sink, block diagram of a typical thermal power plant to illustrate
second law.
Lecture #18
Need of heat rejection in a cyclic process, Clausius and Kelvin-Planck,
statements of second law of thermodynamics, Efficiency analysis of a thermal
power plant and a refrigeration cycle, coefficient of performance for
refrigerator and heat pump cycle, equivalence of Kelvin-Planck and Clausius
statements.
Lecture #19
Reversible and irreversible processes, internal, external and totally
reversible processes and their examples, internally reversible iso-thermal
process, heat transfer processes, The Carnot’s heat engine cycle.
Lecture #20
Carnot cycle implications, The Carnot principles, Thermodynamic temperature
scale, Thermal efficiency of a reversible engine as a function of high and low
temperature reservoirs and its functional form, Kelvin scale of temperature or
absolute temperature scale, proof of Q_add/Q_rej = T_H/T_L
Lecture #21
Clausius inequality its validity, Proof and implications, the increase of
entropy principle.
Lecture #22
Entropy change of a control mass during an irreversible process, entropy
generation for reversible and irreversible processes, Entropy change for an
isolated system, Discussion on equilibrium states w.r.t. increase of entropy
principle
MID TERM COMPLETE
Course File: ESO201A
Prof. Sameer Khandekar
Lecture #23
Property diagrams involving entropy, TdS relations, T-S diagram, H-S diagram,
Representation of Carnot’s cycle on T-S, P-V diagram. Entropy change for solid
and liquid phases, Entropy change for an Ideal Gas, Isentropic processes for an
Ideal Gas.
Lecture #24
Entropy change in rate form, entropy balance for control mass and control
volume, generalized equation for entropy change in a CV, entropy change and
steady flow energy equation for reversible adiabatic process and reversible
isothermal process, entropy change for an adiabatic nozzle, Bernoulli's equation
derivation.
Lecture #25
Isentropic efficiency of steady flow systems - gas turbine, steam turbine,
nozzle, detailed analysis of a compressor, comparison of reversible adiabatic
and reversible isothermal processes to get an ideal benchmarking process for a
compressor, isentropic efficiency of a compressor, use of intercooler in
intermediate stages of compression to achieve nearly isothermal process.
Lecture #26
Condition for minimum work associated with compression with intercooling,
reversible adiabatic and reversible isothermal efficiencies of compressor,
Introduction to the concept of exergy or work potential, definitions, dead
state, forms of exergy, exergy of kinetic and potential energy.
Lecture #27
Exergy, reversible work, useful work, irreversibility (destruction of
exergy), second law efficiency, Exergy calculation for a closed system
(control mass), example problem, difference between first law and second law
efficiency.
Lecture #28
Exergy due to flowing mass, ways of increasing exergy of a system (heat, work
and mass), exergy balance for a closed system, decrease of exergy principle,
exergy of an isolated system for a completely reversible and irreversible
system, exergy balance for a closed system.
Lecture #29
Exergy balance equations in rate form (Similarities/differences of the
functional form with energy conservation and entropy balance equation),
Exergy analysis for a control volume and for a steady flow system, exergy
destruction expression for an isentropic turbine, adiabatic nozzle and
compressor, example problem - calculation of rate of entropy generation and
rate of exergy destruction associated with heat transfer, system and extended
system, second law efficiency (exergy based) of turbines and compressors.
Lecture #30
Thermodynamic analysis of combustion, Conventional and non-conventional
fuels, complete and incomplete combustion, stoichiometric analysis,
air/fuel ratio, relative air/fuel ratio, equivalence ratio, constant
volume and constant pressure combustion, heat of reaction at constant
volume and constant pressure.
Course File: ESO201A
Prof. Sameer Khandekar
Lecture #31
Concept of higher and lower heating/calorific value, reference state for
energy calculations, internal energy/enthalpy vs. temperature graph for
reactants and products, Adiabatic flame temperature of fuel, concept of
formation reaction, heat of formation, standard heat of formation, the net
enthalpy of a substance relative to a standard state.
Lecture #32
Introduction to thermodynamic power cycles, classification, gas power cycles,
heat addition and rejection at constant temperature and pressure,
characteristics of fuel (gasoline and diesel), operation of a typical four
stroke engine, Limitations of Carnot cycle for real-time engineering systems,
Lecture #33
Analysis of Otto and diesel cycles and derivation of thermal efficiencies,
compression ratio and cut-off ratio, concept of relative efficiency, analysis
of dual cycle (as homework), Real cycles and some brief differences, Solved
problem.
Lecture #34
Vapor Power cycles, Carnot cycle and its limitations, Modified Carnot cycle
or Rankine cycle, Detailed analysis of components of Rankine cycle, heat
input, work output and thermal efficiency of a Rankine cycle, ways to improve
thermal efficiency and network output from a Rankine cycle and their
limitations-by increasing boiler pressure or lowering of condenser
pressure, Reheating or turbine staging, super critical boilers.
Lecture #35
Actual Rankine cycle, isentropic pump and turbine efficiencies, pressure
drops in a typical power plant, ways to improve thermal efficiency, reheat
and regeneration processes as a means of improving thermal efficiency of a
plant, detailed analysis of regeneration, numerical problem on steam power
plant with reheat.
Lecture #36
Discussion on second law of thermodynamics and entropy, Refrigerator and
heat pump, the reversed Carnot cycle as refrigeration cycle and its
limitations, the reversed Rankine cycle and its limitations, the ideal
and actual vapor compression refrigeration cycles, discussion on
refrigerants to be used in the refrigeration cycles - ammonia, sulfur
di-oxide, Freon 12, R134a, criteria for selecting refrigerants and their
limitations.
Lecture #37
(Course reaction Survey completed in first 10 minutes)
Introduction to Brayton Cycle, Pressure ration, Analysis and thermal
efficiency, Gas refrigeration cycle, Reversed Brayton cycle and its COP,
example problem on reversed Brayton cycle, Discussion on working fluids and
tonnage rating of refrigerators.
Lecture #38
Thermodynamic Potentials - Internal energy, Enthalpy, Helmholtz free energy,
Gibbs free energy, expressions for U, H, F, G in terms of measurable
quantities, derivation of Maxwell equations using properties of partial
Course File: ESO201A
Prof. Sameer Khandekar
derivatives and point functions, extraction of primary definition of
temperature, pressure and volume from thermodynamic potentials, derivation of
Clapeyron and Clausius-Clapeyron equations, slope of P-T diagram.
Lecture #39
Generalized relation for change of internal energy, enthalpy and entropy,
recovering the ideal gas change relations from the generalized relations,
Isothermal compressibility and Volumetric expansivity, Difference between Cp
and Cv, Discussion on use of Maxewell equations for generating property
tables.
Lecture #40
Joule-Thomson Coefficient, Inversion temperature, Implications for Ideal gas
and real gases, Isentropic compressibility, Ratio of Specific heats, Speed of
sound, Example problem.
(End of Course)
Course File: ESO201A
Prof. Sameer Khandekar
ESO 201A
Thermodynamics
Instructor: Prof. Sameer Khandekar
SL-109, ME, IITK Tel: 7038, [email protected]
Tutorial Problem Set
Tutorial
Date
Problems
Additional Home Work
Classes
completed
Tutorial #1
30/07/2015
1-75, 1-109, 1-114, 2-116
1-19C, 1-36, 1-42E
02
Tutorial #2
06/08/2015
1-35C, 2-42C, 2-47, 2-51
2-14, 2-40C, 2-50
05
Tutorial #3
13/08/2015
2-68, 2-71, 2-78
2-69, 2-72, 2-76, 2-85
08
Tutorial #4
20/08/2015
3-57, 3-59, 3-68, 3-87
3-32, 3-60, 3-61, 3-78, 4-13, 4-21
11
Tutorial #5
27/08/2015
QUIZ #1
Tutorial #6
03/09/2015
4-32, 5-29, 5-51, 5-53
4-23, 4-24, 5-30, 5-57, 5-59, 5-67
17
Tutorial #7
10/09/2015
5-121, 5-130, 5-133, 5-201
5-120, 5-122, 5-123, 5-125
20
16/09/2015
MID SEMESTER EXAMINATION
Tutorial #8
24/09/2015
6-28, 6-56, 6-101, 6-107
6-22, 6-40, 6-41, 6-42, 6-46, 6-51, 6-95, 6-96
23
Tutorial #9
01/10/2015
7-27, 7-29, 7-57, 7-64
7-25, 7-32, 7-69, 7-89, 7-100, 7-110, 7-111, 7-144, 7-212
25
Tutorial #10
08/10/2015
8-23, 8-31, 8-40, 8-45
8-19, 8-34, 8-43, 8-46, 8-52, 8-53, 8-61, 8-62, 8-86, 8-88, 8-92
27
Tutorial #11
15/10/2015
QUIZ #2
Tutorial #12
29/10/2015
15-14, 15-26, 15-58, 15-71
15-20, 15-28, 15-60, 15-62, 15-79
35
Tutorial #13
05/11/2015
9-35, 9-129, 10-23, 11-21
9-39, 9-57, 10-21, 10-39, 11-13, 11-80
38
Tutorial #14
12/11/2015
12-13, 12-19, 12-23
12-16, 12-49, 12-66
40
Course File: ESO201A
Prof. Sameer Khandekar
14
21
32
Tutorial #1
ESO 201A
Thermodynamics
Instructor
Prof. Sameer Khandekar
Contact
Office: SL-109, Tel: 7038
E-mail: [email protected]
URL: home.iitk.ac.in/~samkhan
ESO 201A Thermodynamics
Instructor: Sameer Khandekar
Tutorial 1
[1.75] The maximum blood pressure in the upper arm of a healthy person is about 120
mm Hg. If a vertical tube open to the atmosphere is connected to the vein in the arm of
the person, determine how high the blood will rise in the tube. Take the density of the
blood to be 1050 kg/m3.
[1-109] Balloons are often filled with helium gas because it weighs only about oneseventh of what air weighs under identical conditions. The buoyancy force, which can be
expressed as Fh = row_air*g*Volume_/balloon, will push the balloon upward. If the
balloon has a diameter of 12 m and carries two people, 85 kg each, determine the
acceleration of the balloon when it is first released. Assume the density of air is p = 1.16
kg/m3, and neglect the weight of the ropes and the cage.
[1-114] A pressure cooker cooks a lot faster than an ordinary pan by maintaining a
higher pressure and temperature inside. The lid of a pressure cooker is well sealed, and
steam can escape only through an opening in the middle of the lid. A separate metal
piece, the petcock, sits on top of this opening and prevents steam from escaping until
the pressure force overcomes the weight of the petcock. The periodic escape of the
steam in
this manner prevents any potentially dangerous pressure buildup and keeps the
pressure inside at a constant value. Determine the mass of the petcock of a pressure
cooker whose operation pressure is 100 kPa gage and has an opening cross-sectional
area of 4 mm2. Assume an atmospheric pressure of 101 kPa, and draw the free-body
diagram of the petcock.
[2-116] A river flowing steadily at a rate of 175 m3/s is considered for hydroelectric
power generation. It is determined that a dam can be built to collect water and release
it from an elevation difference of 80 m to generate power. Determine how much power
can be generated from this river water after the dam is filled.
Additional Homework Problems (Tutorial 1)
[1-19C] You have been asked to do a metabolism (energy) analysis of a person. How
would you define the system for this purpose? What type of system is this?
[1-36] The deep body temperature of a healthy person is37°C. What is it in kelvins?
[1-42E] Humans are most comfortable when the temperature is between 65°F and 75°F.
Express these temperature limits in °C. Convert the size of this temperature range (10°F)
to K, °C, and R. Is there any difference in the size of this range as measured in relative or
absolute units?
1-28
1-75 A vertical tube open to the atmosphere is connected to the vein in the arm of a person. The height that the blood will
rise in the tube is to be determined.
Assumptions 1 The density of blood is constant. 2 The gage pressure of blood is 120 mmHg.
Properties The density of blood is given to be ρ = 1050 kg/m3.
Analysis For a given gage pressure, the relation P = ρgh can be expressed
for mercury and blood as P = ρ blood ghblood and P = ρ mercury ghmercury .
Blood
h
Setting these two relations equal to each other we get
P = ρ blood ghblood = ρ mercury ghmercury
Solving for blood height and substituting gives
hblood =
ρ mercury
ρ blood
hmercury =
13,600 kg/m 3
1050 kg/m 3
(0.12 m) = 1.55 m
Discussion Note that the blood can rise about one and a half meters in a tube connected to the vein. This explains why IV
tubes must be placed high to force a fluid into the vein of a patient.
1-76 A diver is moving at a specified depth from the water surface. The pressure exerted on the surface of the diver by
water is to be determined.
Assumptions The variation of the density of water with depth is negligible.
Properties The specific gravity of seawater is given to be SG = 1.03. We take the density of water to be 1000 kg/m3.
Analysis The density of the seawater is obtained by multiplying
its specific gravity by the density of water which is taken to be
1000 kg/m3:
Patm
Sea
ρ = SG × ρ H 2 O = (1.03)(100 0 kg/m 3 ) = 1030 kg/m 3
h
The pressure exerted on a diver at 30 m below the free surface
of the sea is the absolute pressure at that location:
P = Patm + ρgh
⎛ 1 kPa
= (101 kPa) + (1030 kg/m 3 )(9.807 m/s 2 )(30 m)⎜
⎜ 1000 N/m 2
⎝
= 404 kPa
P
⎞
⎟
⎟
⎠
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
preparation. If you are a student using this Manual, you are using it without permission.
1-46
1-109 A helium balloon tied to the ground carries 2 people. The acceleration of the balloon when it is first released is to be
determined.
Assumptions The weight of the cage and the ropes of the balloon is negligible.
Properties The density of air is given to be ρ = 1.16 kg/m3. The density of helium gas is 1/7th of this.
Analysis The buoyancy force acting on the balloon is
V balloon = 4π r 3 /3 = 4 π(6 m) 3 /3 = 904.8 m 3
FB = ρ air gV balloon
⎛
1N
= (1.16 kg/m )(9.81m/s )(904.8 m )⎜
⎜ 1 kg ⋅ m/s 2
⎝
3
2
3
⎞
⎟ = 10,296 N
⎟
⎠
D =12 m
The total mass is
⎛ 1.16
⎞
m He = ρ HeV = ⎜
kg/m 3 ⎟(904.8 m 3 ) = 149.9 kg
⎝ 7
⎠
m total = m He + m people = 149.9 + 2 × 85 = 319.9 kg
The total weight is
⎛ 1N
W = m total g = (319.9 kg)(9.81 m/s 2 )⎜
⎜ 1 kg ⋅ m/s 2
⎝
⎞
⎟ = 3138 N
⎟
⎠
m = 170 kg
Thus the net force acting on the balloon is
Fnet = FB − W = 10,296 − 3138 = 7157 N
Then the acceleration becomes
a=
Fnet
7157 N ⎛⎜ 1kg ⋅ m/s 2
=
m total 319.9 kg ⎜⎝ 1 N
⎞
⎟ = 22.4 m/s 2
⎟
⎠
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1-49
1-113 The pressure of a gas contained in a vertical piston-cylinder device is measured to be 180 kPa. The mass of the piston
is to be determined.
Assumptions There is no friction between the piston and the cylinder.
Patm
Analysis Drawing the free body diagram of the piston and balancing the
vertical forces yield
W = PA − Patm A
mg = ( P − Patm ) A
⎛ 1000 kg/m ⋅ s 2
( m)(9.81 m/s 2 ) = (180 − 100 kPa)(25 × 10 − 4 m 2 )⎜
⎜
1kPa
⎝
It yields
P
⎞
⎟
⎟
⎠
W = mg
m = 20.4 kg
1-114 The gage pressure in a pressure cooker is maintained constant at 100 kPa by a petcock. The mass of the petcock is to
be determined.
Assumptions There is no blockage of the pressure release valve.
Analysis Atmospheric pressure is acting on all surfaces of the petcock, which
balances itself out. Therefore, it can be disregarded in calculations if we use the
gage pressure as the cooker pressure. A force balance on the petcock (ΣFy = 0)
yields
W = Pgage A
(100 kPa)(4 × 10 − 6 m 2 ) ⎛⎜ 1000 kg/m ⋅ s 2
⎜
g
1 kPa
9.81 m/s 2
⎝
= 0.0408 kg
m=
Pgage A
=
Patm
P
W = mg
⎞
⎟
⎟
⎠
1-115 A glass tube open to the atmosphere is attached to a water pipe, and the pressure at the bottom of the tube is
measured. It is to be determined how high the water will rise in the tube.
Properties The density of water is given to be ρ = 1000 kg/m3.
Analysis The pressure at the bottom of the tube can be expressed as
P = Patm + ( ρ g h) tube
Solving for h,
h=
Patm= 99 kPa
P − Patm
ρg
⎛ 1 kg ⋅ m/s 2
⎜
=
1N
(1000 kg/m 3 )(9.81 m/s 2 ) ⎜⎝
= 2.14 m
(120 − 99) kPa
h
Water
⎞⎛ 1000 N/m 2
⎟⎜
⎟⎜ 1 kPa
⎠⎝
⎞
⎟
⎟
⎠
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2-57
2-116E The energy contents, unit costs, and typical conversion efficiencies of various energy sources for use in water
heaters are given. The lowest cost energy source is to be determined.
Assumptions The differences in installation costs of different water heaters are not considered.
Properties The energy contents, unit costs, and typical conversion efficiencies of different systems are given in the
problem statement.
Analysis The unit cost of each Btu of useful energy supplied to the water heater by each system can be determined from
Unit cost of useful energy =
Unit cost of energy supplied
Conversion efficiency
Substituting,
⎛ 1 ft 3 ⎞
−6
⎜
⎟
⎜ 1025 Btu ⎟ = $21.3 × 10 / Btu
⎝
⎠
Natural gas heater:
Unit cost of useful energy =
$0.012/ft 3
0.55
Heating by oil heater:
Unit cost of useful energy =
$1.15/gal ⎛
1 gal
⎞
⎟ = $15.1× 10 − 6 / Btu
⎜
0.55 ⎜⎝ 138,700 Btu ⎟⎠
Electric heater:
Unit cost of useful energy =
$0.084/kWh) ⎛ 1 kWh ⎞
−6
⎜
⎟ = $27.4 × 10 / Btu
0.90
⎝ 3412 Btu ⎠
Therefore, the lowest cost energy source for hot water heaters in this case is oil.
2-117 A home owner is considering three different heating systems for heating his house. The system with the lowest
energy cost is to be determined.
Assumptions The differences in installation costs of different heating systems are not considered.
Properties The energy contents, unit costs, and typical conversion efficiencies of different systems are given in the
problem statement.
Analysis The unit cost of each Btu of useful energy supplied to the house by each system can be determined from
Unit cost of useful energy =
Unit cost of energy supplied
Conversion efficiency
Substituting,
Natural gas heater:
Unit cost of useful energy =
$1.24/therm ⎛ 1 therm ⎞
−6
⎜⎜ 105,500 kJ ⎟⎟ = $13.5 × 10 / kJ
0.87
⎠
⎝
Heating oil heater:
Unit cost of useful energy =
$1.25/gal ⎛ 1 gal ⎞
⎟ = $10.4 × 10 − 6 / kJ
⎜
0.87 ⎜⎝ 138,500 kJ ⎟⎠
Electric heater:
Unit cost of useful energy =
$0.09/kWh) ⎛ 1 kWh ⎞
−6
⎜
⎟ = $25.0 × 10 / kJ
1.0
⎝ 3600 kJ ⎠
Therefore, the system with the lowest energy cost for heating the house is the heating oil heater.
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1-9
Systems, Properties, State, and Processes
1-19C This system is a region of space or open system in that mass such as air and food can cross its control boundary.
The system can also interact with the surroundings by exchanging heat and work across its control boundary. By tracking
these interactions, we can determine the energy conversion characteristics of this system.
1-20C The system is taken as the air contained in the piston-cylinder device. This system is a closed or fixed mass system
since no mass enters or leaves it.
1-21C Any portion of the atmosphere which contains the ozone layer will work as an open system to study this problem.
Once a portion of the atmosphere is selected, we must solve the practical problem of determining the interactions that occur
at the control surfaces which surround the system's control volume.
1-22C Intensive properties do not depend on the size (extent) of the system but extensive properties do.
1-23C If we were to divide the system into smaller portions, the weight of each portion would also be smaller. Hence, the
weight is an extensive property.
1-24C If we were to divide this system in half, both the volume and the number of moles contained in each half would be
one-half that of the original system. The molar specific volume of the original system is
v =
V
N
and the molar specific volume of one of the smaller systems is
v =
V/ 2 V
=
N /2 N
which is the same as that of the original system. The molar specific volume is then an intensive property.
1-25C For a system to be in thermodynamic equilibrium, the temperature has to be the same throughout but the pressure
does not. However, there should be no unbalanced pressure forces present. The increasing pressure with depth in a fluid,
for example, should be balanced by increasing weight.
1-26C A process during which a system remains almost in equilibrium at all times is called a quasi-equilibrium process.
Many engineering processes can be approximated as being quasi-equilibrium. The work output of a device is maximum
and the work input to a device is minimum when quasi-equilibrium processes are used instead of nonquasi-equilibrium
processes.
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1-12
Temperature
1-33C The zeroth law of thermodynamics states that two bodies are in thermal equilibrium if both have the same
temperature reading, even if they are not in contact.
1-34C They are Celsius (°C) and kelvin (K) in the SI, and fahrenheit (°F) and rankine (R) in the English system.
1-35C Probably, but not necessarily. The operation of these two thermometers is based on the thermal expansion of a
fluid. If the thermal expansion coefficients of both fluids vary linearly with temperature, then both fluids will expand at the
same rate with temperature, and both thermometers will always give identical readings. Otherwise, the two readings may
deviate.
1-36 A temperature is given in °C. It is to be expressed in K.
Analysis The Kelvin scale is related to Celsius scale by
T(K] = T(°C) + 273
Thus,
T(K] = 37°C + 273 = 310 K
1-37E The temperature of air given in °C unit is to be converted to °F and R unit.
Analysis Using the conversion relations between the various temperature scales,
T (°F) = 1.8T (°C) + 32 = (1.8)(150) + 32 = 302°F
T (R ) = T (°F) + 460 = 302 + 460 = 762 R
1-38 A temperature change is given in °C. It is to be expressed in K.
Analysis This problem deals with temperature changes, which are identical in Kelvin and Celsius scales. Thus,
∆T(K] = ∆T(°C) = 45 K
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1-13
1-39E The flash point temperature of engine oil given in °F unit is to be converted to K and R units.
Analysis Using the conversion relations between the various temperature scales,
T (R ) = T (°F) + 460 = 363 + 460 = 823 R
T (K ) =
T (R ) 823
=
= 457 K
1.8
1.8
1-40E The temperature of ambient air given in °C unit is to be converted to °F, K and R units.
Analysis Using the conversion relations between the various temperature scales,
T = −40°C = ( −40)(1.8) + 32 = −40°C
T = −40 + 273.15 = 233.15 K
T = −40 + 459.67 = 419.67 R
1-41E The change in water temperature given in °F unit is to be converted to °C, K and R units.
Analysis Using the conversion relations between the various temperature scales,
∆T = 10 / 1.8 = 5.6°C
∆T = 10 / 1.8 = 5.6 K
∆T = 10°F = 10 R
1-42E A temperature range given in °F unit is to be converted to °C unit and the temperature difference in °F is to be
expressed in K, °C, and R.
Analysis The lower and upper limits of comfort range in °C are
T (°C) =
T (°F) − 32 65 − 32
=
= 18.3 °C
1. 8
1. 8
T (°C) =
T (°F) − 32 75 − 32
=
= 23.9 °C
1. 8
1. 8
A temperature change of 10°F in various units are
∆T (R ) = ∆T (°F) = 10 R
∆T (°F) 10
∆T (°C) =
=
= 5.6°C
1.8
1 .8
∆T (K ) = ∆T (°C) = 5.6 K
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Tutorial #2 ESO 201A Thermodynamics Instructor Prof. Sameer Khandekar Contact Office: SL‐109, Tel: 7038 E‐mail: [email protected] URL: home.iitk.ac.in/~samkhan ESO 201A Thermodynamics
Instructor: Sameer Khandekar
Tutorial 2
[1-35C] Consider an alcohol and a mercury thermometer that read exactly 0°C at the ice point
and 100°C at the steam point. The distance between the two points is divided into 100 equal
parts in both thermometers. Do you think these thermometers will give exactly the same
reading at a temperature of, say, 60°C? Explain.
[2-42C] On a hot summer day, a student turns his fan on when he leaves his room in the
morning. When he returns inthe evening, will the room be warmer or cooler than
theneighboring rooms? Why? Assume all the doors and windowsare kept closed.
[2-47] A classroom that normally contains 40 people is tobe air-conditioned with window airconditioning units of 5-kW cooling capacity. A person at rest may be assumed todissipate
heat at a rate of about 360 kJ/h. There are 10 lightbulbsin the room, each with a rating of
100 W. The rate ofheat transfer to the classroom through the walls and thewindows is
estimated to be 15,000 kJ/h. If the room air is tobe maintained at a constant
temperature of 21°C, determinethe number of window air-conditioning units required.
[2-51] A fan is to accelerate quiescent air to a velocity of 8 m/s at a rate of 9 m3/s.
Determine the minimum power thatmust be supplied to the fan. Take the density of air to be
1.18 kg/m3.
Additional Homework Problems
[2-14]A water jet that leaves a nozzle at 60 m/s at a flowrate of 120 kg/s is to be used to
generate power by strikingthe buckets located on the perimeter of a wheel. Determinethe
power generation potential of this water jet.
[2-40C] For a cycle, is the net work necessarily zero? Forwhat kind of systems will this be the
case?
[2-50]Consider a room that is initially at the outdoor temperatureof 20°C. The room contains a
100-W lightbulb, a110-W TV set, a 200-W refrigerator, and a 1000-W iron.Assuming no heat
transfer through the walls, determine therate of increase of the energy content of the room
when all ofthese electric devices are on.
1-12
Temperature
1-33C The zeroth law of thermodynamics states that two bodies are in thermal equilibrium if both have the same
temperature reading, even if they are not in contact.
1-34C They are Celsius (°C) and kelvin (K) in the SI, and fahrenheit (°F) and rankine (R) in the English system.
1-35C Probably, but not necessarily. The operation of these two thermometers is based on the thermal expansion of a
fluid. If the thermal expansion coefficients of both fluids vary linearly with temperature, then both fluids will expand at the
same rate with temperature, and both thermometers will always give identical readings. Otherwise, the two readings may
deviate.
1-36 A temperature is given in °C. It is to be expressed in K.
Analysis The Kelvin scale is related to Celsius scale by
T(K] = T(°C) + 273
Thus,
T(K] = 37°C + 273 = 310 K
1-37E The temperature of air given in °C unit is to be converted to °F and R unit.
Analysis Using the conversion relations between the various temperature scales,
T (°F) = 1.8T (°C) + 32 = (1.8)(150) + 32 = 302°F
T (R ) = T (°F) + 460 = 302 + 460 = 762 R
1-38 A temperature change is given in °C. It is to be expressed in K.
Analysis This problem deals with temperature changes, which are identical in Kelvin and Celsius scales. Thus,
∆T(K] = ∆T(°C) = 45 K
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2-5
2-14 A water jet strikes the buckets located on the perimeter of a wheel at a specified velocity and flow rate. The power
generation potential of this system is to be determined.
Assumptions Water jet flows steadily at the specified speed and flow rate.
Analysis Kinetic energy is the only form of harvestable mechanical
energy the water jet possesses, and it can be converted to work entirely.
Therefore, the power potential of the water jet is its kinetic energy,
which is V2/2 per unit mass, and m& V 2 / 2 for a given mass flow rate:
emech = ke =
V 2 (60 m/s)2 ⎛ 1 kJ/kg ⎞
=
= 1.8 kJ/kg
⎜
2 2⎟
2
2
⎝ 1000 m /s ⎠
W&max = E& mech = m& emech
⎛ 1 kW ⎞
= (120 kg/s)(1.8 kJ/kg)⎜
⎟ = 216 kW
⎝ 1 kJ/s ⎠
Shaft
Nozzle
Vj
Therefore, 216 kW of power can be generated by this water jet at the
stated conditions.
Discussion An actual hydroelectric turbine (such as the Pelton wheel) can convert over 90% of this potential to actual
electric power.
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2-16
The First Law of Thermodynamics
2-40C No. This is the case for adiabatic systems only.
2-41C Energy can be transferred to or from a control volume as heat, various forms of work, and by mass transport.
2-42C Warmer. Because energy is added to the room air in the form of electrical work.
2-43E The high rolling resistance tires of a car are replaced by low rolling resistance ones. For a specified unit fuel cost, the
money saved by switching to low resistance tires is to be determined.
Assumptions 1The low rolling resistance tires deliver 2 mpg over all velocities. 2 The car is driven 15,000 miles per year.
Analysis The annual amount of fuel consumed by this car on high- and low-rolling resistance tires are
Annual Fuel Consumption High =
Miles driven per year 15,000 miles/year
=
= 428.6 gal/year
Miles per gallon
35 miles/gal
Annual Fuel Consumption Low =
Miles driven per year 15,000 miles/year
=
= 405.4 gal/year
Miles per gallon
37 miles/gal
Then the fuel and money saved per year become
Fuel Savings = Annual Fuel Consumptio n High − Annual Fuel Consumptio n Low
= 428.6 gal/year − 405.4 gal/year = 23.2 gal/year
Cost savings = (Fuel savings)( Unit cost of fuel) = (23.2 gal/year)($2.20/gal) = $51.0/year
Discussion A typical tire lasts about 3 years, and thus the low rolling resistance tires have the potential to save about $150
to the car owner over the life of the tires, which is comparable to the installation cost of the tires.
2-44 The specific energy change of a system which is accelerated is to be determined.
Analysis Since the only property that changes for this system is the velocity, only the kinetic energy will change. The
change in the specific energy is
∆ke =
V 22 − V12 (30 m/s) 2 − (0 m/s) 2 ⎛ 1 kJ/kg ⎞
=
⎜
⎟ = 0.45 kJ/kg
2
2
⎝ 1000 m 2 /s 2 ⎠
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2-17
2-45 The specific energy change of a system which is raised is to be determined.
Analysis Since the only property that changes for this system is the elevation, only the potential energy will change. The
change in the specific energy is then
⎛ 1 kJ/kg ⎞
∆pe = g ( z 2 − z1 ) = (9.8 m/s 2 )(100 − 0) m⎜
⎟ = 0.98 kJ/kg
⎝ 1000 m 2 /s 2 ⎠
2-46E A water pump increases water pressure. The power input is to be determined.
Analysis The power input is determined from
W& = V& ( P2 − P1 )
50 psia
⎛
1 Btu
= (1.2 ft 3 /s)(50 − 10)psia ⎜
⎜ 5.404 psia ⋅ ft 3
⎝
= 12.6 hp
⎞⎛
1 hp
⎞
⎟⎜
⎟⎝ 0.7068 Btu/s ⎟⎠
⎠
Water
10 psia
The water temperature at the inlet does not have any significant effect on the required power.
2-47 A classroom is to be air-conditioned using window air-conditioning units. The cooling load is due to people, lights,
and heat transfer through the walls and the windows. The number of 5-kW window air conditioning units required is to be
determined.
Assumptions There are no heat dissipating equipment (such as computers, TVs, or ranges) in the room.
Analysis The total cooling load of the room is determined from
Q& cooling = Q& lights + Q& people + Q& heat gain
where
Q& lights = 10 × 100 W = 1 kW
Q& people = 40 × 360 kJ / h = 4 kW
Q& heat gain = 15,000 kJ / h = 4.17 kW
Room
15,000 kJ/h
40 people
10 bulbs
·
Qcool
Substituting,
Q& cooling = 1 + 4 + 4.17 = 9.17 kW
Thus the number of air-conditioning units required is
9.17 kW
5 kW/unit
= 1.83 ⎯
⎯→ 2 units
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2-19
2-50 A room contains a light bulb, a TV set, a refrigerator, and an iron. The rate of increase of the energy content of the
room when all of these electric devices are on is to be determined.
Assumptions 1 The room is well sealed, and heat loss from the room is negligible. 2 All the appliances are kept on.
Analysis Taking the room as the system, the rate form of the energy balance can be written as
E& − E&
1in424out
3
Rate of net energy transfer
by heat, work, and mass
=
→
dE system / dt
14243
dE room / dt = E& in
Rate of change in internal, kinetic,
potential, etc. energies
ROOM
since no energy is leaving the room in any form, and thus E& out = 0 . Also,
E& in = E& lights + E& TV + E& refrig + E& iron
= 100 + 110 + 200 + 1000 W
Electricity
= 1410 W
- Lights
- TV
- Refrig
- Iron
Substituting, the rate of increase in the energy content of the room becomes
dE room / dt = E& in = 1410 W
Discussion Note that some appliances such as refrigerators and irons operate intermittently, switching on and off as
controlled by a thermostat. Therefore, the rate of energy transfer to the room, in general, will be less.
2-51 A fan is to accelerate quiescent air to a specified velocity at a specified flow rate. The minimum power that must be
supplied to the fan is to be determined.
Assumptions The fan operates steadily.
Properties The density of air is given to be ρ = 1.18 kg/m3.
Analysis A fan transmits the mechanical energy of the shaft (shaft power) to mechanical energy of air (kinetic energy). For
a control volume that encloses the fan, the energy balance can be written as
E& − E&
1in424out
3
Rate of net energy transfer
by heat, work, and mass
= dE system / dt ©0 (steady) = 0
144424443
→
E& in = E& out
Rate of change in internal, kinetic,
potential, etc. energies
V2
W& sh, in = m& air ke out = m& air out
2
where
m& air = ρV& = (1.18 kg/m 3 )(9 m 3 /s) = 10.62 kg/s
Substituting, the minimum power input required is determined to be
V2
(8 m/s) 2 ⎛ 1 J/kg ⎞
W& sh, in = m& air out = (10.62 kg/s)
⎜
⎟ = 340 J/s = 340 W
2
2
⎝ 1 m 2 /s 2 ⎠
Discussion The conservation of energy principle requires the energy to be conserved as it is converted from one form to
another, and it does not allow any energy to be created or destroyed during a process. In reality, the power required will be
considerably higher because of the losses associated with the conversion of mechanical shaft energy to kinetic energy of air.
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Instructor: Sameer Khandekar
Tutorial 3
[2-68] The water in a large lake is to be used to generateelectricity by the installation of
a hydraulic turbine-generatorat a location where the depth of the water is 50 m (Fig. 2-62).
Water is to be supplied at a rate of 5000 kg/s. If theelectric power generated is measured to
be 1862 kW and the generator efficiency is 95 percent, determine (a) the
overallefficiency of the turbine—generator, (b) the mechanical efficiency of the turbine,
and (c) the shaft power supplied by theturbine to the generator.
[2-71] Water is pumped from a lake to a storage tank 20 m above at a rate of 70 L/s
while consuming 20.4 kWof electric power. Disregarding any frictional losses in the pipes
and any changes in kinetic energy, determine(a) the overall efficiency of the pump-motor
unit and (b) the pressure difference between the inlet and the exitof the pump.
[2-78] A wind turbine is rotating at 15 rpm under steadywinds flowing through the turbine at
a rate of 42,000 kg/s.The tip velocity of the turbine blade is measured to be 250 km/h. If 180
kW power is produced by the turbine, determine(a) the average velocity of the air and
(b) the conversionefficiency of the turbine. Take the density of air to be 1.31 kg/m3.
Additional Homework Problems (Tutorial 3)
[2-69]At a certain location, wind is blowing steadily at7 m/s. Determine the mechanical energy
of air per unit massand the power generation potential of a wind turbine with 80-m-diameter
blades at that location. Also determine theactual electric power generation assuming an overall
efficiencyof 30 percent. Take the air density to be 1.25 kg/m3.
[2-72]Large wind turbines with blade span diameters ofover 100 m are available for electric
power generation. Considera wind turbine with a blade span diameter of 100 m installed at a
site subjected to steady winds at 8 m/s. Takingthe overall efficiency of the wind turbine to be
32 percent and the air density to be 1.25 kg/m3, determine the electric powergenerated by this
wind turbine. Also, assuming steady winds of 8 m/s during a 24-hour period, determine the
amount ofelectric energy and the revenue generated per day for a unitprice of $0.06/kWh for
electricity.
[2-76]An oil pump is drawing 35 kW of electric powerwhile pumping oil with ρ= 860 kg/m3 at a
rate of 0.1 m3/s.The inlet and outlet diameters of the pipe are 8 cm and 12cm, respectively. If
the pressure rise of oil in the pump ismeasured to be 400 kPa and the motor efficiency is 90
percent,determine the mechanical efficiency of the pump.
[2-85]When a hydrocarbon fuel is burned, almost all of thecarbon in the fuel burns completely
to form CO 2 (carbondioxide), which is the principal gas causing the greenhouse effect and thus
global climate change. On average, 0.59 kg of CO 2 is produced for each kWh of electricity
generated from a power plant that burns natural gas. A typical new householdrefrigerator uses
about 700 kWh of electricity per year. Determine the amount of CO 2 production that is due to
therefrigerators in a city with 300,000 households.
2-30
2-68 A hydraulic turbine-generator is to generate electricity from the water of a lake. The overall efficiency, the turbine
efficiency, and the shaft power are to be determined.
Assumptions 1 The elevation of the lake and that of the discharge
site remains constant. 2 Irreversible losses in the pipes are
negligible.
Properties The density of water can be taken to be ρ = 1000
kg/m3. The gravitational acceleration is g = 9.81 m/s2.
Analysis (a) We take the bottom of the lake as the reference level
for convenience. Then kinetic and potential energies of water are
zero, and the mechanical energy of water consists of pressure
energy only which is
e mech,in − e mech,out =
P
ρ
= gh
⎛ 1 kJ/kg ⎞
= (9.81 m/s 2 )(50 m)⎜
⎟
⎝ 1000 m 2 /s 2 ⎠
= 0.491 kJ/kg
Then the rate at which mechanical energy of fluid supplied to the turbine and the overall efficiency become
| ∆E& mech,fluid |= m& (e mech,in − e mech,in ) = (5000 kg/s)(0.491 kJ/kg) = 2455 kW
η overall = η turbine-gen =
W& elect,out
1862 kW
=
= 0.760
&
| ∆E mech,fluid | 2455 kW
(b) Knowing the overall and generator efficiencies, the mechanical efficiency of the turbine is determined from
η turbine-gen = η turbineη generator → η turbine =
η turbine-gen
η generator
=
0.76
= 0.800
0.95
(c) The shaft power output is determined from the definition of mechanical efficiency,
W& shaft,out = η turbine | ∆E& mech,fluid |= (0.800)(2455 kW) = 1964 kW ≈ 1960 kW
Therefore, the lake supplies 2455 kW of mechanical energy to the turbine, which converts 1964 kW of it to shaft work that
drives the generator, which generates 1862 kW of electric power.
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2-31
2-69 Wind is blowing steadily at a certain velocity. The mechanical energy of air per unit mass, the power generation
potential, and the actual electric power generation are to be determined.
Assumptions 1 The wind is blowing steadily at a constant
uniform velocity. 2 The efficiency of the wind turbine is
independent of the wind speed.
Properties The density of air is given to be ρ = 1.25 kg/m3.
Analysis Kinetic energy is the only form of mechanical energy
the wind possesses, and it can be converted to work entirely.
Therefore, the power potential of the wind is its kinetic energy,
which is V2/2 per unit mass, and m& V 2 / 2 for a given mass
flow rate:
e mech = ke =
Wind
Wind
turbine
7 m/s
80 m
V 2 (7 m/s) 2 ⎛ 1 kJ/kg ⎞
=
⎜
⎟ = 0.0245 kJ/kg
2
2
⎝ 1000 m 2 /s 2 ⎠
m& = ρVA = ρV
πD 2
4
= (1.25 kg/m 3 )(7 m/s)
π (80 m) 2
4
= 43,982 kg/s
W& max = E& mech = m& emech = (43,982 kg/s)(0.0245 kJ/kg) = 1078 kW
The actual electric power generation is determined by multiplying the power generation potential by the efficiency,
W& elect = η wind turbineW& max = (0.30)(1078 kW) = 323 kW
Therefore, 323 kW of actual power can be generated by this wind turbine at the stated conditions.
Discussion The power generation of a wind turbine is proportional to the cube of the wind velocity, and thus the power
generation will change strongly with the wind conditions.
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2-33
2-71 Water is pumped from a lake to a storage tank at a specified rate. The overall efficiency of the pump-motor unit and
the pressure difference between the inlet and the exit of the pump are to be determined.
Assumptions 1 The elevations of the tank and the lake remain constant. 2 Frictional losses in the pipes are negligible. 3 The
changes in kinetic energy are negligible. 4 The elevation difference across the pump is negligible.
Properties We take the density of water to be ρ = 1000 kg/m3.
Analysis (a) We take the free surface of the lake to be point 1 and the
free surfaces of the storage tank to be point 2. We also take the lake
surface as the reference level (z1 = 0), and thus the potential energy at
points 1 and 2 are pe1 = 0 and pe2 = gz2. The flow energy at both points
is zero since both 1 and 2 are open to the atmosphere (P1 = P2 = Patm).
Further, the kinetic energy at both points is zero (ke1 = ke2 = 0) since the
water at both locations is essentially stationary. The mass flow rate of
water and its potential energy at point 2 are
2
Storage
tank
20 m
Pump
1
m& = ρV& = (1000 kg/m 3 )(0.070 m 3/s) = 70 kg/s
⎛ 1 kJ/kg ⎞
pe 2 = gz 2 = (9.81 m/s 2 )(20 m)⎜
⎟ = 0.196 kJ/kg
⎝ 1000 m 2 /s 2 ⎠
Then the rate of increase of the mechanical energy of water becomes
∆E& mech,fluid = m& (e mech,out − e mech,in ) = m& ( pe 2 − 0) = m& pe 2 = (70 kg/s)(0.196 kJ/kg) = 13.7 kW
The overall efficiency of the combined pump-motor unit is determined from its definition,
η pump-motor =
∆E& mech,fluid 13.7 kW
=
= 0.672 or 67.2%
20.4 kW
W& elect,in
(b) Now we consider the pump. The change in the mechanical energy of water as it flows through the pump consists of the
change in the flow energy only since the elevation difference across the pump and the change in the kinetic energy are
negligible. Also, this change must be equal to the useful mechanical energy supplied by the pump, which is 13.7 kW:
∆E& mech,fluid = m& (e mech,out − e mech,in ) = m&
P2 − P1
ρ
= V&∆P
Solving for ∆P and substituting,
∆P =
∆E& mech,fluid
13.7 kJ/s ⎛ 1 kPa ⋅ m 3
⎜
=
0.070 m 3 /s ⎜⎝ 1 kJ
V&
⎞
⎟ = 196 kPa
⎟
⎠
Therefore, the pump must boost the pressure of water by 196 kPa in order to raise its elevation by 20 m.
Discussion Note that only two-thirds of the electric energy consumed by the pump-motor is converted to the mechanical
energy of water; the remaining one-third is wasted because of the inefficiencies of the pump and the motor.
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2-34
2-72 A large wind turbine is installed at a location where the wind is blowing steadily at a certain velocity. The electric
power generation, the daily electricity production, and the monetary value of this electricity are to be determined.
Assumptions 1 The wind is blowing steadily at a constant
uniform velocity. 2 The efficiency of the wind turbine is
independent of the wind speed.
Properties The density of air is given to be ρ = 1.25 kg/m3.
Wind
Analysis Kinetic energy is the only form of mechanical
energy the wind possesses, and it can be converted to work
entirely. Therefore, the power potential of the wind is its
kinetic energy, which is V2/2 per unit mass, and m& V 2 / 2 for
a given mass flow rate:
e mech = ke =
Wind
turbine
8 m/s
100 m
V 2 (8 m/s ) 2 ⎛ 1 kJ/kg ⎞
=
⎜
⎟ = 0.032 kJ/kg
2
2
⎝ 1000 m 2 /s 2 ⎠
m& = ρVA = ρV
πD 2
4
= (1.25 kg/m 3 )(8 m/s)
π (100 m) 2
4
= 78,540 kg/s
W& max = E& mech = m& e mech = (78,540 kg/s)(0.032 kJ/kg) = 2513 kW
The actual electric power generation is determined from
W& elect = η wind turbineW& max = (0.32)(2513 kW) = 804.2 kW
Then the amount of electricity generated per day and its monetary value become
Amount of electricity = (Wind power)(Operating hours)=(804.2 kW)(24 h) =19,300 kWh
Revenues = (Amount of electricity)(Unit price) = (19,300 kWh)($0.06/kWh) = $1158 (per day)
Discussion Note that a single wind turbine can generate several thousand dollars worth of electricity every day at a
reasonable cost, which explains the overwhelming popularity of wind turbines in recent years.
2-73E A water pump raises the pressure of water by a specified amount at a specified flow rate while consuming a known
amount of electric power. The mechanical efficiency of the pump is to be determined.
∆P = 1.2 psi
Assumptions 1 The pump operates steadily. 2 The changes in velocity and elevation
across the pump are negligible. 3 Water is incompressible.
Analysis To determine the mechanical efficiency of the pump, we need to know the
increase in the mechanical energy of the fluid as it flows through the pump, which is
6 hp
PUMP
∆E& mech,fluid = m& (emech,out − e mech,in ) = m& [( Pv ) 2 − ( Pv )1 ] = m& ( P2 − P1 )v
⎛
1 Btu
= V& ( P2 − P1 ) = (15 ft 3 /s)(1.2 psi)⎜
⎜ 5.404 psi ⋅ ft 3
⎝
⎞
⎟ = 3.33 Btu/s = 4.71 hp
⎟
⎠
Pump
inlet
since 1 hp = 0.7068 Btu/s, m& = ρV& = V& / v , and there is no change in kinetic and potential energies of the fluid. Then the
mechanical efficiency of the pump becomes
η pump =
∆E& mech,fluid 4.71 hp
=
= 0.786 or 78.6%
6 hp
W& pump, shaft
Discussion The overall efficiency of this pump will be lower than 83.8% because of the inefficiency of the electric motor
that drives the pump.
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2-36
2-76 A pump is pumping oil at a specified rate. The pressure rise of oil in the pump is measured, and the motor efficiency is
specified. The mechanical efficiency of the pump is to be determined.
Assumptions 1 The flow is steady and incompressible. 2 The elevation difference across the pump is negligible.
Properties The density of oil is given to be ρ = 860 kg/m3.
Analysis Then the total mechanical energy of a fluid is the sum of the potential, flow, and kinetic energies, and is expressed
per unit mass as emech = gh + Pv + V 2 / 2 . To determine the mechanical efficiency of the pump, we need to know the
increase in the mechanical energy of the fluid as it flows through the pump, which is
⎛
V2
V2
∆E& mech,fluid = m& (e mech,out − e mech,in ) = m& ⎜ ( Pv ) 2 + 2 − ( Pv ) 1 − 1
⎜
2
2
⎝
⎞ &⎛
V 2 − V12
⎟ = V ⎜ ( P2 − P1 ) + ρ 2
⎟
⎜
2
⎠
⎝
since m& = ρV& = V& / v , and there is no change in the potential
energy of the fluid. Also,
V1 =
V2 =
V&
A1
V&
A2
V&
=
πD12
=
/4
V&
πD 22 / 4
=
0.1 m 3 /s
π (0.08 m) 2 / 4
=
3
0.1 m /s
π (0.12 m) 2 / 4
2
⎞
⎟
⎟
⎠
35 kW
= 19.9 m/s
PUMP
= 8.84 m/s
Motor
Pump
inlet
Substituting, the useful pumping power is determined to be
1
W& pump,u = ∆E& mech,fluid
⎛
(8.84 m/s) 2 − (19.9 m/s) 2
= (0.1 m 3 /s)⎜ 400 kN/m 2 + (860 kg/m 3 )
⎜
2
⎝
= 26.3 kW
⎛
1 kN
⎜
⎜ 1000 kg ⋅ m/s 2
⎝
⎞ ⎞⎛ 1 kW ⎞
⎟ ⎟⎜
⎟ ⎟⎝ 1 kN ⋅ m/s ⎟⎠
⎠⎠
Then the shaft power and the mechanical efficiency of the pump become
W& pump, shaft = η motor W& electric = (0.90)(35 kW) = 31.5 kW
η pump =
W& pump, u
W&
pump, shaft
=
26.3 kW
= 0.836 = 83.6%
31.5 kW
Discussion The overall efficiency of this pump/motor unit is the product of the mechanical and motor efficiencies, which is
0.9×0.836 = 0.75.
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2-38
2-78 A wind turbine produces 180 kW of power. The average velocity of the air and the conversion efficiency of the
turbine are to be determined.
Assumptions The wind turbine operates steadily.
Properties The density of air is given to be 1.31 kg/m3.
Analysis (a) The blade diameter and the blade span area are
Vtip
D=
=
πn&
A=
πD 2
4
=
⎛ 1 m/s ⎞
(250 km/h)⎜
⎟
⎝ 3.6 km/h ⎠ = 88.42 m
⎛ 1 min ⎞
π (15 L/min)⎜
⎟
⎝ 60 s ⎠
π (88.42 m) 2
4
= 6140 m 2
Then the average velocity of air through the wind turbine becomes
V=
42,000 kg/s
m&
=
= 5.23 m/s
ρA (1.31 kg/m 3 )(6140 m 2 )
(b) The kinetic energy of the air flowing through the turbine is
KE& =
1
1
m& V 2 = (42,000 kg/s)(5.23 m/s) 2 = 574.3 kW
2
2
Then the conversion efficiency of the turbine becomes
η=
W&
180 kW
=
= 0.313 = 31.3%
&
KE 574.3 kW
Discussion Note that about one-third of the kinetic energy of the wind is converted to power by the wind turbine, which is
typical of actual turbines.
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2-40
2-84E A person trades in his Ford Taurus for a Ford Explorer. The extra amount of CO2 emitted by the Explorer within 5
years is to be determined.
Assumptions The Explorer is assumed to use 940 gallons of gasoline a year compared to 715 gallons for Taurus.
Analysis The extra amount of gasoline the Explorer will use within 5 years is
Extra Gasoline
= (Extra per year)(No. of years)
= (940 – 715 gal/yr)(5 yr)
= 1125 gal
Extra CO2 produced
= (Extra gallons of gasoline used)(CO2 emission per gallon)
= (1125 gal)(19.7 lbm/gal)
= 22,163 lbm CO2
Discussion Note that the car we choose to drive has a significant effect on the amount of greenhouse gases produced.
2-85 A power plant that burns natural gas produces 0.59 kg of carbon dioxide (CO2) per kWh. The amount of CO2
production that is due to the refrigerators in a city is to be determined.
Assumptions The city uses electricity produced by a natural gas power plant.
Properties 0.59 kg of CO2 is produced per kWh of electricity generated (given).
Analysis Noting that there are 300,000 households in the city and each household consumes 700 kWh of electricity for
refrigeration, the total amount of CO2 produced is
Amount of CO 2 produced = (Amount of electricity consumed)(Amount of CO 2 per kWh)
= (300,000 household)(700 kWh/year household)(0.59 kg/kWh)
= 1.23 × 10 8 CO 2 kg/year
= 123,000 CO 2 ton/year
Therefore, the refrigerators in this city are responsible for the production of 123,000 tons of CO2.
2-86 A power plant that burns coal, produces 1.1 kg of carbon dioxide (CO2) per kWh. The amount of CO2 production that
is due to the refrigerators in a city is to be determined.
Assumptions The city uses electricity produced by a coal power plant.
Properties 1.1 kg of CO2 is produced per kWh of electricity generated (given).
Analysis Noting that there are 300,000 households in the city and each
household consumes 700 kWh of electricity for refrigeration, the total amount
of CO2 produced is
Amount of CO 2 produced = (Amount of electricity consumed)(Amount of CO 2 per kWh)
= (300,000 household)(700 kWh/household)(1.1 kg/kWh)
= 2.31 × 10 8 CO 2 kg/year
= 231,000 CO 2 ton/year
Therefore, the refrigerators in this city are responsible for the production of 231,000 tons of CO2.
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Tutorial #4 ESO 201A Thermodynamics Instructor Prof. Sameer Khandekar Contact Office: SL‐109, Tel: 7038 E‐mail: [email protected] URL: home.iitk.ac.in/~samkhan ESO 201A Thermodynamics
Instructor: Sameer Khandekar
Tutorial 4
[3-57] Determine the specific volume, internal energy, and enthalpy of compressed liquid
water at 80°C and 20 M pausing the saturated liquid approximation. Compare these values
to the ones obtained from the compressed liquid tables.
[3-59] A piston-cylinder device contains 0.8 kg of steam at300°C and 1 MPa. Steam is cooled
at constant pressure until one-half of the mass condenses (a) Show the process on a T-v
diagram. (b) Find the final temperature. (c) Determine the volume change.
[3-68] A 0.3-m3 rigid vessel initially contains saturated liquid-vapor mixture of water at 150°
C. The water is now heated until it reaches the critical state. Determine the mass of the
liquid water and the volume occupied by the liquid at the initial state.
[3-87] Determine the specific volume of superheated water vapor at 15 MPa and 350°C,
using (a) the ideal-gas equation,(b) the generalized compressibility chart, and (c) the steam
tables. Also determine the error involved in the first two cases. [Answers: (a) 0 .01917 m3/
kg, 67.0 percent, (b) 0 .01246 m3/kg, 8.5 percent, (c) 0 .01148 m3/kg]
Additional Homework Problems (Tutorial 4)
[3-32] One kilogram of water fills a 150-L rigid container at an initial pressure of 2 MPa. The
container is then cooled to40°C. Determine the initial temperature and the final pressure of the
water.
[3-60] A rigid tank contains water vapor at 250°C and an unknown pressure. When the tank is
cooled to 124°C, the vapor starts condensing. Estimate the initial pressure in the tank. [Answer:
0 .30 MPa]
[3-61] 3-61 A piston-cylinder device initially contains 1.4-kg saturated liquid water at 200°C.
Now heat is transferred to the water until the volume quadruples and the cylinder contains
saturated vapor only. Determine (a) the volume of the tank,(b) the final temperature and
pressure, and (c) the internal energy change of the water.
[3-78] A 1-m3 tank containing air at 25°C and 500 kPa is connected through a valve to another
tank containing 5 kg of air at 35°C and 200 kPa. Now the valve is opened, and the entire system
is allowed to reach thermal equilibrium with the surroundings, which are at 20°C. Determine
the volume of the second tank and the final equilibrium pressure of air.
[4-13] 1-m3 of saturated liquid water at 200°C is expanded isothermally in a closed system until
its quality is 80 percent. Determine the total work produced by this expansion, in kJ.
[4-21] Carbon dioxide contained in a piston-cylinder device is compressed from 0.3 to 0.1 m3.
During the process, the pressure and volume are related by P = aV-2, where a = 8 kPa-m6.
Calculate the work done on the carbon dioxide during this process.
3-11
3-32 A rigid container that is filled with water is cooled. The initial temperature and the final pressure are to be determined.
Analysis This is a constant volume process. The specific volume is
H2O
2 MPa
1 kg
150 L
0.150 m 3
= 0.150 m 3 /kg
v1 = v 2 = =
m
1 kg
V
Q
The initial state is superheated vapor. The temperature is determined to be
P1 = 2 MPa
⎫
⎬ T1 = 395°C
v 1 = 0.150 m 3 /kg ⎭
(Table A - 6)
P
This is a constant volume cooling process (v = V /m = constant). The final
state is saturated mixture and thus the pressure is the saturation pressure at
the final temperature:
1
2
T2 = 40°C
⎫
⎬ P = Psat @ 40°C = 7.385 kPa (Table A - 4)
v 2 = v 1 = 0.150 m /kg ⎭ 2
v
3
3-33 A rigid container that is filled with R-134a is heated. The final temperature and initial pressure are to be determined.
Analysis This is a constant volume process. The specific volume is
v1 = v 2 =
V
m
=
R-134a
-40°C
10 kg
1.348 m3
1.348 m 3
= 0.1348 m 3 /kg
10 kg
The initial state is determined to be a mixture, and thus the pressure is the
saturation pressure at the given temperature
P1 = Psat @ -40°C = 51.25 kPa (Table A - 11)
The final state is superheated vapor and the temperature is determined by
interpolation to be
P2 = 200 kPa
⎫
⎬ T2 = 66.3°C (Table A - 13)
v 2 = 0.1348 m 3 /kg ⎭
P
2
1
v
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3-26
3-57 The properties of compressed liquid water at a specified state are to be determined using the compressed liquid tables,
and also by using the saturated liquid approximation, and the results are to be compared.
Analysis Compressed liquid can be approximated as saturated liquid at the given temperature. Then from Table A-4,
T = 80°C ⇒
v ≅ v f @ 80°C = 0.001029 m 3 /kg (0.90% error)
u ≅ u f @ 80°C = 334.97 kJ/kg
h ≅ h f @ 80°C = 335.02 kJ/kg
(1.35% error)
(4.53% error)
From compressed liquid table (Table A-7),
v = 0.00102 m 3 /kg
P = 20 MPa ⎫
u = 330.50 kJ/kg
T = 80°C ⎭⎬
h = 350.90 kJ/kg
The percent errors involved in the saturated liquid approximation are listed above in parentheses.
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3-28
3-59 Superheated steam in a piston-cylinder device is cooled at constant pressure until half of the mass condenses. The final
temperature and the volume change are to be determined, and the process should be shown on a T-v diagram.
Analysis (b) At the final state the cylinder contains saturated liquidvapor mixture, and thus the final temperature must be the saturation
temperature at the final pressure,
T = Tsat@1 MPa = 179.88°C
(Table A-5)
H 2O
300°C
1 MPa
(c) The quality at the final state is specified to be x2 = 0.5. The specific
volumes at the initial and the final states are
P1 = 1.0 MPa
T1 = 300 o C
⎫
3
⎬ v 1 = 0.25799 m /kg
⎭
P2 = 1.0 MPa
x2 = 0.5
⎫
⎬ v 2 = v f + x2v fg
⎭ = 0.001127 + 0.5 × (0.19436 − 0.001127 )
= 0.09775 m3/kg
(Table A-6)
T
1
2
Thus,
v
∆V = m(v 2 − v 1 ) = (0.8 kg)(0.09775 − 0.25799)m 3 /kg = −0.1282 m 3
3-60 The water in a rigid tank is cooled until the vapor starts condensing. The initial pressure in the tank is to be
determined.
Analysis This is a constant volume process (v = V /m = constant), and the
initial specific volume is equal to the final specific volume that is
25
3
v 1 = v 2 = v g @124°C = 0.79270 m /kg (Table A-4)
since the vapor starts condensing at 150°C. Then from
Table A-6,
T1 = 250°C
⎫
⎬ P = 0.30 MPa
3
v 1 = 0.79270 m /kg ⎭ 1
T °C
1
H2O
T1= 250°C
P1 = ?
15
2
v
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3-29
3-61 Heat is supplied to a piston-cylinder device that contains water at a specified state. The volume of the tank, the final
temperature and pressure, and the internal energy change of water are to be determined.
Properties The saturated liquid properties of water at 200°C are: vf = 0.001157 m3/kg and uf = 850.46 kJ/kg (Table A-4).
Analysis (a) The cylinder initially contains saturated liquid water. The volume of the cylinder at the initial state is
V1 = mv 1 = (1.4 kg)(0.001157 m 3 /kg) = 0.001619 m 3
The volume at the final state is
V = 4(0.001619) = 0.006476 m 3
Water
1.4 kg, 200°C
sat. liq.
(b) The final state properties are
v2 =
V
m
=
0.006476 m3
= 0.004626 m3 / kg
1.4 kg
v 2 = 0.004626 m3 / kg ⎫⎪
x2 = 1
Q
T2 = 371.3 °C
⎬ P2 = 21,367 kPa
⎭⎪ u = 2201.5 kJ/kg
2
(Table A-4 or A-5 or EES)
(c) The total internal energy change is determined from
∆U = m(u 2 − u1 ) = (1.4 kg)(2201.5 - 850.46) kJ/kg = 1892 kJ
3-62E The error involved in using the enthalpy of water by the incompressible liquid approximation is to be determined.
Analysis The state of water is compressed liquid. From the steam tables,
P = 1500 psia ⎫
⎬ h = 376.51 Btu/lbm (Table A - 7E)
T = 400°F ⎭
Based upon the incompressible liquid approximation,
P = 1500 psia ⎫
⎬ h ≅ h f @ 400° F = 375.04 Btu/lbm (Table A - 4E)
T = 400°F ⎭
The error involved is
Percent Error =
376.51 − 375.04
× 100 = 0.39%
376.51
which is quite acceptable in most engineering calculations.
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3-32
3-67
The Pessure-Enthalpy diagram of R-134a showing some constant-temperature and constant-entropy lines are
obtained using Property Plot feature of EES.
P [kPa]
1.2
kJ
/k g
1
0.8
0.5
0.3
0.2
104
-K
R134a
105
70°C
103
40°C
10°C
-10°C
102
-30°C
101
-100
0
100
200
300
400
500
h [kJ/kg]
3-68 A rigid vessel that contains a saturated liquid-vapor mixture is heated until it reaches the critical state. The mass of the
liquid water and the volume occupied by the liquid at the initial state are to be determined.
Analysis This is a constant volume process (v = V /m = constant) to the critical state, and thus the initial specific volume
will be equal to the final specific volume, which is equal to the critical specific volume of water,
v 1 = v 2 = v cr = 0.003106 m 3 /kg
(last row of Table A-4)
The total mass is
0.3 m 3
V
m= =
= 96.60 kg
v 0.003106 m 3 /kg
At 150°C, vf = 0.001091 m3/kg and vg = 0.39248 m3/kg (Table A4). Then the quality of water at the initial state is
x1 =
v1 −v f
v fg
=
T
cp
H2O
150°C
0.003106 − 0.001091
= 0.005149
0.39248 − 0.001091
vcr
v
Then the mass of the liquid phase and its volume at the initial state are determined from
m f = (1 − x1 )mt = (1 − 0.005149)(96.60) = 96.10 kg
V f = m f v f = (96.10 kg)(0.001091 m3/kg) = 0.105 m 3
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3-36
3-78 Two rigid tanks connected by a valve to each other contain air at specified conditions. The volume of the second tank
and the final equilibrium pressure when the valve is opened are to be determined.
Assumptions At specified conditions, air behaves as an ideal gas.
Properties The gas constant of air is R = 0.287 kPa.m3/kg.K (Table A-1).
Analysis Let's call the first and the second tanks A and B. Treating air as an ideal gas, the volume of the second tank and
the mass of air in the first tank are determined to be
⎛ m1RT1 ⎞
(5 kg)(0.287 kPa ⋅ m 3/kg ⋅ K)(308 K)
⎟ =
= 2.21 m 3
⎟
P
200
kPa
⎝ 1 ⎠B
V B = ⎜⎜
⎛ PV ⎞
(500 kPa)(1.0 m3 )
m A = ⎜⎜ 1 ⎟⎟ =
= 5.846 kg
3
⎝ RT1 ⎠ A (0.287 kPa ⋅ m /kg ⋅ K)(298 K)
Thus,
A
B
Air
V = 1 m3
T = 25°C
P = 500 kPa
×
Air
m = 5 kg
T = 35°C
P = 200 kPa
V = V A + V B = 1.0 + 2.21 = 3.21 m 3
m = m A + mB = 5.846 + 5.0 = 10.846 kg
Then the final equilibrium pressure becomes
P2 =
mRT2
V
=
(10.846 kg)(0.287 kPa ⋅ m3 /kg ⋅ K)(293 K)
3.21 m3
= 284.1 kPa
3-79E An elastic tank contains air at a specified state. The volume is doubled at the same pressure. The initial volume and
the final temperature are to be determined.
Assumptions At specified conditions, air behaves as an ideal gas.
Analysis According to the ideal gas equation of state,
PV = nRu T
(32 psia)V = (2.3 lbmol)(10.73 psia ⋅ ft 3 /lbmol ⋅ R)(65 + 460) R
V = 404.9 ft 3
V 2 T2
T2
=
⎯
⎯→ 2 =
⎯
⎯→ T2 = 1050 R = 590°F
V 1 T1
(65 + 460) R
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3-39
Compressibility Factor
3-84C It represent the deviation from ideal gas behavior. The further away it is from 1, the more the gas deviates from
ideal gas behavior.
3-85C All gases have the same compressibility factor Z at the same reduced temperature and pressure.
3-86C Reduced pressure is the pressure normalized with respect to the critical pressure; and reduced temperature is the
temperature normalized with respect to the critical temperature.
3-87 The specific volume of steam is to be determined using the ideal gas relation, the compressibility chart, and the steam
tables. The errors involved in the first two approaches are also to be determined.
Properties The gas constant, the critical pressure, and the critical temperature of water are, from Table A-1,
R = 0.4615 kPa·m3/kg·K,
Tcr = 647.1 K,
Pcr = 22.06 MPa
Analysis (a) From the ideal gas equation of state,
v=
RT (0.4615 kPa ⋅ m 3 /kg ⋅ K)(623.15 K)
=
= 0.01917 m 3 /kg (67.0% error)
P
15,000 kPa
(b) From the compressibility chart (Fig. A-15),
10 MPa
P
⎫
=
= 0.453 ⎪
Pcr 22.06 MPa
⎪
⎬ Z = 0.65
673 K
T
⎪
TR =
=
= 1.04
⎪⎭
Tcr 647.1 K
PR =
H2O
15 MPa
350°C
Thus,
v = Zv ideal = (0.65)(0.01917 m 3 /kg) = 0.01246 m 3 /kg (8.5% error)
(c) From the superheated steam table (Table A-6),
P = 15 MPa
T = 350°C
} v = 0.01148 m /kg
3
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preparation. If you are a student using this Manual, you are using it without permission.
4-9
4-13 Water is expanded isothermally in a closed system. The work produced is to be determined.
Assumptions The process is quasi-equilibrium.
Analysis From water table
P1 = P2 = Psat @ 200°C = 1554.9 kPa
v 1 = v f @ 200°C = 0.001157 m 3 /kg
v 2 = v f + xv fg
= 0.001157 + 0.80(0.12721 − 0.001157)
P
(kPa)
1
1555
2
= 0.10200 m 3 /kg
The definition of specific volume gives
V 2 = V1
1
88.16
V (m3)
3
v2
0.10200 m /kg
= (1 m 3 )
= 88.16 m 3
v1
0.001157 m 3 /kg
The work done during the process is determined from
Wb,out =
∫
2
1
⎛ 1 kJ ⎞
P dV = P(V 2 −V1 ) = (1554.9 kPa)(88.16 − 1)m 3 ⎜
⎟ = 1.355 × 10 5 kJ
3
1
kPa
m
⋅
⎝
⎠
4-14 Air in a cylinder is compressed at constant temperature until its pressure rises to a specified value. The boundary
work done during this process is to be determined.
Assumptions 1 The process is quasi-equilibrium. 2 Air is an ideal gas.
Properties The gas constant of air is R = 0.287 kJ/kg.K (Table A-1).
Analysis The boundary work is determined from its definition to be
Wb, out =
∫
2
1
P dV = P1V1 ln
P
2
V2
P
= mRT ln 1
V1
P2
150 kPa
= (2.4 kg)(0.287 kJ/kg ⋅ K)(285 K)ln
600 kPa
T = 12°C
1
V
= −272 kJ
Discussion The negative sign indicates that work is done on the system (work input).
4-15 Several sets of pressure and volume data are taken as a gas expands. The boundary work done during this
process is to be determined using the experimental data.
Assumptions The process is quasi-equilibrium.
Analysis Plotting the given data on a P-V diagram on a graph paper and evaluating the area under the process curve, the
work done is determined to be 0.25 kJ.
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4-15
4-21 CO2 gas in a cylinder is compressed until the volume drops to a specified value. The pressure changes during the
process with volume as P = aV −2 . The boundary work done during this process is to be determined.
Assumptions The process is quasi-equilibrium.
P
Analysis The boundary work done during this process is determined from
Wb,out =
2
∫ PdV = ∫
1
2
⎛ 1
a ⎞
1 ⎞
− ⎟⎟
⎜ 2 ⎟dV = −a⎜⎜
⎝V ⎠
⎝ V 2 V1 ⎠
2⎛
1
⎛ 1
1
= −(8 kPa ⋅ m 6 )⎜
−
⎜ 0.1 m 3 0.3 m 3
⎝
= −53.3 kJ
⎞⎛ 1 kJ
⎟⎜
⎟⎜ 1 kPa ⋅ m 3
⎠⎝
P = aV--2
⎞
⎟
⎟
⎠
1
0.3
0.1
V
(m3)
Discussion The negative sign indicates that work is done on the system (work input).
4-22E A gas in a cylinder is heated and is allowed to expand to a specified pressure in a process during which the pressure
changes linearly with volume. The boundary work done during this process is to be determined.
Assumptions The process is quasi-equilibrium.
Analysis (a) The pressure of the gas changes linearly with volume, and thus the process curve on a P-V diagram will be a
straight line. The boundary work during this process is simply the area under the process curve, which is a trapezoidal.
Thus,
At state 1:
P1 = aV 1 + b
3
3
15 psia = (5 psia/ft )(7 ft ) + b
b = −20 psia
P
(psia)
100
At state 2:
P2 = aV 2 + b
P = aV + b
2
15
1
100 psia = (5 psia/ft 3 )V 2 + (−20 psia)
V 2 = 24 ft 3
V
7
(ft3)
and,
Wb,out = Area =
⎛
P1 + P2
(100 + 15)psia
1 Btu
(V 2 −V1 ) =
(24 − 7)ft 3 ⎜
3
⎜
2
2
⎝ 5.4039 psia ⋅ ft
⎞
⎟
⎟
⎠
= 181 Btu
Discussion The positive sign indicates that work is done by the system (work output).
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
preparation. If you are a student using this Manual, you are using it without permission.
Tutorial #5
(QUIZ #1)
ESO 201A
Thermodynamics
Instructor
Prof. Sameer Khandekar
Contact
Office: SL-109, Tel: 7038
E-mail: [email protected]
URL: home.iitk.ac.in/~samkhan
Tutorial #6
ESO 201A
Thermodynamics
Instructor
Prof. Sameer Khandekar
Contact
Office: SL-109, Tel: 7038
E-mail: [email protected]
URL: home.iitk.ac.in/~samkhan
ESO 201A Thermodynamics
Instructor: Sameer Khandekar
Tutorial 6
[4-32] A well-insulated rigid tank contains 2 kg of a saturated liquid-vapor mixture of water at
150 kPa. Initially, three-quarters of the mass is in the liquid phase. An electric resistor placed in
the tank is connected to a 110-V source and a current of 8 A flows through the resistor when
the switch is turned on. Determine how long it will take to vaporize all the liquid in the tank.
Also, show the process on a T-v diagram with respect to saturation lines.
[5-29] Air at 80 kPa and 127°C enters an adiabatic diffuser steadily at a rate of 6000 kg/h
and leaves at 100 kPa. The velocity of the air stream is decreased from 230 to 30 m/s as it
passes through the diffuser. Find (a) the exit temperature of the air and (b) the exit area
of the diffuser.
[5-51] Steam enters an adiabatic turbine at 10 MPa and500°C and leaves at 10 kPa with a
quality of 90 percent. Neglecting the changes in kinetic and potential energies, determine
the mass flow rate required for a power output of 5 MW.
[5-53] An adiabatic air compressor compresses 10 L/s of air at 120 kPa and 20°C to 1000 kPa
and 300°C. Determine (a) the work required by the compressor, in kJ/kg, and (b) the power
required to drive the air compressor, in kW.
Additional Homework Problems (Tutorial 6)
[4-23] A piston-cylinder device initially contains 0.25 kg of nitrogen gas at 130 kPa and 180°C.
The nitrogen is now expanded isothermally to a pressure of 80 kPa. Determine the boundary
work done during this process.
[4-24] A piston-cylinder device contains 0.15 kg of air initially at 2 MPa and 350°C. The air is first
expanded isothermally to 500 kPa, then compressed polytropically with a polytrophic exponent
of 1.2 to the initial pressure, and finally compressed at the constant pressure to the initial state.
Determine the boundary work for each process and the net work of the cycle.
[5-30] Air enters an adiabatic nozzle steadily at 300 kPa,200°C, and 45 m/s and leaves at 100
kPa and 180 m/s. The inlet area of the nozzle is 110 cm2. Determine (a) the mass flow rate
through the nozzle, (b) the exit temperature of their, and (c) the exit area of the nozzle.
[Answers: (a) 1.09 kg/s, (b) 185°C, (c) 79.9 cm2]
[5-57] An adiabatic gas turbine expands air at 1300 kPa and 500°C to 100 kPa and 127°C.
Air enters the turbine through a 0.2-m2 opening with an average velocity of 40m/s, and
exhausts through a 1-m2 opening. Determine (a) the mass flow rate of air through the turbine
and (b) the power produced by the turbine. [Answers: (a) 46 .9 kg/s, (b) 18.3 MW]
[5-59] Steam enters a steady-flow turbine with a mass flow rate of 20 kg/s at 600°C, 5 MPa,
and a negligible velocity. The steam expands in the turbine to a saturated vapor at 500 kPa
where 10 percent of the steam is removed for some other use. The remainder of the steam
continues to expand to the turbine exit where the pressure is 10 kPa and quality is 85
percent. If the turbine is adiabatic, determine the rate of work done by the steam
during this process. [Answer: 27,790 kW]
[5-67] Saturated liquid-vapor mixture of water, called wet steam, in a steam line at 2000 kPa is
throttled to 100 kPa and120°C. What is the quality in the steam line? [Answer: 0.957]
[5-76] A hot-water stream at 80°C enters a mixing chamber with a mass flow rate of 0.5 kg/
s where it is mixed with a stream of cold water at 20°C. If it is desired that the mixture leave
the chamber at 42°C, determine the mass flow rate of the cold-water stream. Assume all
the streams are at a pressure of 250 kPa. [Answer: 0.865 kg/s]
4-21
4-32 An insulated rigid tank is initially filled with a saturated liquid-vapor mixture of water. An electric heater in the tank is
turned on, and the entire liquid in the tank is vaporized. The length of time the heater was kept on is to be determined, and
the process is to be shown on a P-v diagram.
Assumptions 1 The tank is stationary and thus the kinetic and potential energy changes are zero. 2 The device is wellinsulated and thus heat transfer is negligible. 3 The energy stored in the resistance wires, and the heat transferred to the tank
itself is negligible.
Analysis We take the contents of the tank as the system. This is a closed system since no mass enters or leaves. Noting that
the volume of the system is constant and thus there is no boundary work, the energy balance for this stationary closed
system can be expressed as
E − Eout
1in424
3
Net energy transfer
by heat, work, and mass
=
∆Esystem
1
424
3
H2O
Change in internal, kinetic,
potential, etc. energies
We,in = ∆U = m(u2 − u1 )
V = const.
(since Q = KE = PE = 0)
VI∆t = m(u2 − u1 )
The properties of water are (Tables A-4 through A-6)
P1 = 150 kPa ⎫ v f = 0.001053, v g = 1.1594 m 3 /kg
⎬
x1 = 0.25
⎭ u f = 466.97, u fg = 2052.3 kJ/kg
We
T
v 1 = v f + x1v fg = 0.001053 + [0.25 × (1.1594 − 0.001053)] = 0.29065 m 3 /kg
2
v 2 = v 1 = 0.29065 m 3 /kg ⎫⎪
1
u1 = u f + x1u fg = 466.97 + (0.25 × 2052.3) = 980.03 kJ/kg
sat.vapor
⎬ u 2 = u g @0.29065 m3 /kg = 2569.7 kJ/kg
⎪⎭
v
Substituting,
⎛ 1000 VA ⎞
⎟⎟
(110 V)(8 A)∆t = (2 kg)(2569.7 − 980.03)kJ/kg⎜⎜
⎝ 1 kJ/s ⎠
∆t = 33613 s = 60.2 min
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5-14
5-29 Air is decelerated in a diffuser from 230 m/s to 30 m/s. The exit temperature of air and the exit area of the diffuser are
to be determined.
Assumptions 1 This is a steady-flow process since there is no change with time. 2 Air is an ideal gas with variable specific
heats. 3 Potential energy changes are negligible. 4 The device is adiabatic and thus heat transfer is negligible. 5 There are
no work interactions.
Properties The gas constant of air is 0.287 kPa.m3/kg.K (Table A-1). The enthalpy of air at the inlet temperature of 400 K
is h1 = 400.98 kJ/kg (Table A-17).
&1 = m
&2 = m
& . We take diffuser as the system, which is a control
Analysis (a) There is only one inlet and one exit, and thus m
volume since mass crosses the boundary. The energy balance for this steady-flow system can be expressed in the rate form
as
E& − E& out
1in424
3
=
Rate of net energy transfer
by heat, work, and mass
∆E& systemÊ0 (steady)
1442443
=0
Rate of change in internal, kinetic,
potential, etc. energies
1
E& in = E& out
AIR
2
& ≅ ∆pe ≅ 0)
m& (h1 + V12 / 2) = m& (h2 + V22 /2) (since Q& ≅ W
0 = h2 − h1 +
V 22 − V12
2
,
or,
h2 = h1 −
V 22 − V12
(30 m/s)2 − (230 m/s)2
= 400.98 kJ/kg −
2
2
⎛ 1 kJ/kg
⎜
⎜ 1000 m 2 /s 2
⎝
⎞
⎟ = 426.98 kJ/kg
⎟
⎠
From Table A-17,
T2 = 425.6 K
(b) The specific volume of air at the diffuser exit is
v2 =
(
)
RT2
0.287 kPa ⋅ m 3 /kg ⋅ K (425.6 K )
=
= 1.221 m 3 /kg
(100 kPa )
P2
From conservation of mass,
m& =
1
v2
A2V 2 ⎯
⎯→ A2 =
m& v 2 (6000 3600 kg/s)(1.221 m 3 /kg)
=
= 0.0678 m 2
V2
30 m/s
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5-15
5-30 Air is accelerated in a nozzle from 45 m/s to 180 m/s. The mass flow rate, the exit temperature, and the exit area of the
nozzle are to be determined.
Assumptions 1 This is a steady-flow process since there is no change with time. 2 Air is an ideal gas with constant specific
heats. 3 Potential energy changes are negligible. 4 The device is adiabatic and thus heat transfer is negligible. 5 There are
no work interactions.
Properties The gas constant of air is 0.287 kPa.m3/kg.K (Table A-1).
The specific heat of air at the anticipated average temperature of 450
K is cp = 1.02 kJ/kg.°C (Table A-2).
Analysis (a) There is only one inlet and one exit, and thus
&1 = m
&2 = m
& . Using the ideal gas relation, the specific volume and
m
the mass flow rate of air are determined to be
AIR
P2 = 100 kPa
V2 = 180 m/s
RT1 (0.287 kPa ⋅ m 3 /kg ⋅ K)(473 K )
=
= 0.4525 m 3 /kg
P1
300 kPa
v1 =
m& =
P1 = 300 kPa
T1 = 200°C
V1 = 45 m/s
A1 = 110 cm2
1
v1
A1V1 =
1
0.4525 m 3 /kg
(0.0110 m 2 )(45 m/s) = 1.094 kg/s
(b) We take nozzle as the system, which is a control volume since mass crosses the boundary. The energy balance for this
steady-flow system can be expressed in the rate form as
E& − E& out
1in424
3
=
Rate of net energy transfer
by heat, work, and mass
∆E& systemÊ0 (steady)
1442443
=0
Rate of change in internal, kinetic,
potential, etc. energies
E& in = E& out
& ≅ ∆pe ≅ 0)
m& (h1 + V12 / 2) = m& (h2 + V 22 /2) (since Q& ≅ W
0 = h2 − h1 +
V 22 − V12
V 2 − V12
⎯
⎯→ 0 = c p , ave (T2 − T1 ) + 2
2
2
Substituting,
0 = (1.02 kJ/kg ⋅ K )(T2 − 200 o C) +
(180 m/s) 2 − (45 m/s) 2
2
⎛ 1 kJ/kg
⎜
⎜ 1000 m 2 /s 2
⎝
⎞
⎟
⎟
⎠
It yields
T2 = 185.2°C
(c) The specific volume of air at the nozzle exit is
v2 =
m& =
RT2 (0.287 kPa ⋅ m 3 /kg ⋅ K)(185.2 + 273 K )
=
= 1.315 m 3 /kg
P2
100 kPa
1
v2
A2V2 ⎯
⎯→ 1.094 kg/s =
1
3
1.315 m /kg
A2 (180 m/s ) → A2 = 0.00799 m2 = 79.9 cm2
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preparation. If you are a student using this Manual, you are using it without permission.
5-33
5-51 Steam expands in a turbine. The mass flow rate of steam for a power output of 5 MW is to be determined.
Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potential energy changes are
negligible. 3 The device is adiabatic and thus heat transfer is negligible.
Properties From the steam tables (Tables A-4 through 6)
1
P1 = 10 MPa ⎫
⎬ h1 = 3375.1 kJ/kg
T1 = 500°C ⎭
P2 = 10 kPa ⎫
⎬ h2 = h f + x2 h fg = 191.81 + 0.90 × 2392.1 = 2344.7 kJ/kg
x2 = 0.90 ⎭
&1 = m
&2 = m
& . We take the turbine as
Analysis There is only one inlet and one exit, and thus m
the system, which is a control volume since mass crosses the boundary. The energy balance
for this steady-flow system can be expressed in the rate form as
E& − E& out
1in424
3
=
Rate of net energy transfer
by heat, work, and mass
∆E& systemÊ0 (steady)
1442443
H2O
2
=0
Rate of change in internal, kinetic,
potential, etc. energies
E& in = E& out
& 1 = W& out + mh
& 2 (since Q& ≅ ∆ke ≅ ∆pe ≅ 0)
mh
W& out = − m& (h2 − h1 )
Substituting, the required mass flow rate of the steam is determined to be
5000 kJ/s = −m& (2344.7 − 3375.1) kJ/kg ⎯
⎯→ m& = 4.852 kg/s
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preparation. If you are a student using this Manual, you are using it without permission.
5-35
5-53 Air is compressed at a rate of 10 L/s by a compressor. The work required per unit mass and the power required are to
be determined.
Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potential energy changes are
negligible. 3 Air is an ideal gas with constant specific heats.
Properties The constant pressure specific heat of air at the average temperature of (20+300)/2=160°C=433 K is cp = 1.018
kJ/kg·K (Table A-2b). The gas constant of air is R = 0.287 kPa⋅m3/kg⋅K (Table A-1).
Analysis (a) There is only one inlet and one exit, and thus m& 1 = m& 2 = m& . We take the compressor as the system, which is a
control volume since mass crosses the boundary. The energy balance for this steady-flow system can be expressed in the
rate form as
E& − E&
1in424out
3
=
Rate of net energy transfer
by heat, work, and mass
∆E& system Ê0 (steady)
1442444
3
=0
Rate of change in internal, kinetic,
potential, etc. energies
E& in = E& out
W& in + m& h1 = m& h2 (since ∆ke ≅ ∆pe ≅ 0)
W& in = m& (h2 − h1 ) = m& c p (T2 − T1 )
1 MPa
300°C
Compressor
Thus,
win = c p (T2 − T1 ) = (1.018 kJ/kg ⋅ K)(300 − 20)K = 285.0 kJ/kg
120 kPa
20°C
10 L/s
(b) The specific volume of air at the inlet and the mass flow rate are
v1 =
m& =
RT1 (0.287 kPa ⋅ m 3 /kg ⋅ K)(20 + 273 K)
=
= 0.7008 m 3 /kg
P1
120 kPa
V&1
0.010 m 3 /s
=
= 0.01427 kg/s
v 1 0.7008 m 3 /kg
Then the power input is determined from the energy balance equation to be
W& in = m& c p (T2 − T1 ) = (0.01427 kg/s)(1.01 8 kJ/kg ⋅ K)(300 − 20)K = 4.068 kW
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
preparation. If you are a student using this Manual, you are using it without permission.
5-39
5-57 Air is expanded in an adiabatic turbine. The mass flow rate of the air and the power produced are to be determined.
Assumptions 1 This is a steady-flow process since there is no change with time. 2 The turbine is well-insulated, and thus
there is no heat transfer. 3 Air is an ideal gas with constant specific heats.
Properties The constant pressure specific heat of air at the average temperature of (500+127)/2=314°C=587 K is cp = 1.048
kJ/kg·K (Table A-2b). The gas constant of air is R = 0.287 kPa⋅m3/kg⋅K (Table A-1).
Analysis (a) There is only one inlet and one exit, and thus m& 1 = m& 2 = m& . We take the turbine as the system, which is a
control volume since mass crosses the boundary. The energy balance for this steady-flow system can be expressed in the
rate form as
E& − E&
1in424out
3
=
Rate of net energy transfer
by heat, work, and mass
∆E& system Ê0 (steady)
1442444
3
=0
1.3 MPa
500°C
40 m/s
Rate of change in internal, kinetic,
potential, etc. energies
E& in = E& out
2 ⎞
⎞
⎛
⎟ = m& ⎜ h2 + V 2 ⎟ + W& out
⎜
⎟
2 ⎟⎠
⎠
⎝
⎛
V 2 − V 22
W& out = m& ⎜ h1 − h2 + 1
⎜
2
⎝
⎛
V2
m& ⎜ h1 + 1
⎜
2
⎝
Turbine
2
2
⎞
⎛
⎟ = m& ⎜ c p (T1 − T2 ) + V1 − V 2
⎟
⎜
2
⎠
⎝
⎞
⎟
⎟
⎠
100 kPa
127°C
The specific volume of air at the inlet and the mass flow rate are
v1 =
m& =
RT1 (0.287 kPa ⋅ m 3 /kg ⋅ K)(500 + 273 K)
=
= 0.1707 m 3 /kg
P1
1300 kPa
A1V1
v1
=
(0.2 m 2 )(40 m/s)
0.1707 m 3 /kg
= 46.88 kg/s
Similarly at the outlet,
v2 =
RT2 (0.287 kPa ⋅ m 3 /kg ⋅ K)(127 + 273 K)
=
= 1.148 m 3 /kg
P2
100 kPa
V2 =
m& v 2 (46.88 kg/s)(1.148 m 3 /kg)
=
= 53.82 m/s
A2
1m2
(b) Substituting into the energy balance equation gives
⎛
V 2 − V 22
W& out = m& ⎜ c p (T1 − T2 ) + 1
⎜
2
⎝
⎞
⎟
⎟
⎠
⎡
(40 m/s) 2 − (53.82 m/s) 2 ⎛ 1 kJ/kg
= (46.88 kg/s) ⎢(1.048 kJ/kg ⋅ K)(500 − 127)K +
⎜
2
⎝ 1000 m 2 /s 2
⎣⎢
⎞⎤
⎟⎥
⎠⎦⎥
= 18,300 kW
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
preparation. If you are a student using this Manual, you are using it without permission.
5-41
5-59 Steam expands in a two-stage adiabatic turbine from a specified state to another state. Some steam is extracted at the
end of the first stage. The power output of the turbine is to be determined.
Assumptions 1 This is a steady-flow process since there is no change
with time. 2 Kinetic and potential energy changes are negligible. 3 The
turbine is adiabatic and thus heat transfer is negligible.
Properties From the steam tables (Tables A-5 and A-6)
P1 = 5 MPa ⎫
⎬ h1 = 3666.9 kJ/kg
T1 = 600°C ⎭
P2 = 0.5 MPa ⎫
⎬ h2 = 2748.1 kJ/kg
x2 = 1
⎭
5 MPa
600°C
20 kg/s
STEAM
20 kg/s
I
P3 = 10 kPa ⎫ h3 = h f + xh fg
⎬
x 2 = 0.85 ⎭ = 191.81 + (0.85)(2392.1) = 2225.1 kJ/kg
Analysis We take the entire turbine, including the connection part
between the two stages, as the system, which is a control volume since
mass crosses the boundary. Noting that one fluid stream enters the
turbine and two fluid streams leave, the energy balance for this steadyflow system can be expressed in the rate form as
E& − E&
1in424out
3
=
Rate of net energy transfer
by heat, work, and mass
∆E& system ©0 (steady)
3
144
42444
II
10 kPa
x=0.85
0.1 MPa
2 kg/s
sat. vap.
=0
Rate of change in internal, kinetic,
potential, etc. energies
E& in = E& out
m& 1 h1 = m& 2 h2 + m& 3 h3 + W& out
W& out = m& 1 ( h1 − 0.1h2 − 0.9h3 )
Substituting, the power output of the turbine is
W& out = m& 1 (h1 − 0.1h2 − 0.9h3 )
= (20 kg/s)(3666.9 − 0.1 × 2748.1 − 0.9 × 2225.1) kJ/kg
= 27,790 kW
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
preparation. If you are a student using this Manual, you are using it without permission.
5-45
5-67 Steam is throttled from a specified pressure to a specified state. The quality at the inlet is to be determined.
Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potential energy changes are
negligible. 3 Heat transfer to or from the fluid is negligible. 4 There are no work interactions involved.
Analysis There is only one inlet and one exit, and thus m& 1 = m& 2 = m& . We take the throttling valve as the system, which is a
control volume since mass crosses the boundary. The energy balance for this steady-flow system can be expressed in the
rate form as
E& in − E& out = ∆E& system Ê0 (steady) = 0
E& in = E& out
m& h1 = m& h2
h1 = h2
Since
Throttling valve
Steam
2 MPa
100 kPa
120°C
Q& ≅ W& = ∆ke ≅ ∆pe ≅ 0 .
The enthalpy of steam at the exit is (Table A-6),
P2 = 100 kPa ⎫
⎬ h2 = 2716.1 kJ/kg
T2 = 120°C ⎭
The quality of the steam at the inlet is (Table A-5)
h2 − h f
⎫
2716.1 − 908.47
=
= 0.957
⎬ x1 =
h1 = h2 = 2716.1 kJ/kg ⎭
1889.8
h fg
P1 = 2000 kPa
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
preparation. If you are a student using this Manual, you are using it without permission.
4-16
4-23 A piston-cylinder device contains nitrogen gas at a specified state. The boundary work is to be determined for the
isothermal expansion of nitrogen.
Properties The properties of nitrogen are R = 0.2968 kJ/kg.K , k = 1.4 (Table A-2a).
Analysis We first determine initial and final volumes from ideal gas relation, and find the boundary work using the relation
for isothermal expansion of an ideal gas
V1 =
mRT (0.25 kg)(0.2968 kJ/kg.K)(180 + 273 K)
=
= 0.2586 m 3
P1
(130 kPa)
V2 =
mRT (0.25 kg)(0.2968 kJ/kg.K)(180 + 273 K)
=
= 0.4202 m 3
P2
80 kPa
⎛V
Wb = P1V1 ln⎜⎜ 2
⎝ V1
⎛ 0.4202 m 3 ⎞
⎞
⎟ = 16.3 kJ
⎟⎟ = (130 kPa)(0.2586 m 3 ) ln⎜
⎜ 0.2586 m 3 ⎟
⎠
⎠
⎝
N2
130 kPa
180°C
4-24 A piston-cylinder device contains air gas at a specified state. The air undergoes a cycle with three processes. The
boundary work for each process and the net work of the cycle are to be determined.
Properties The properties of air are R = 0.287 kJ/kg.K , k = 1.4 (Table A-2a).
Analysis For the isothermal expansion process:
V1 =
V2 =
mRT (0.15 kg)(0.287 kJ/kg.K)(350 + 273 K)
=
= 0.01341 m 3
P1
(2000 kPa)
mRT (0.15 kg)(0.287 kJ/kg.K)(350 + 273 K)
=
= 0.05364 m 3
P2
(500 kPa)
Air
2 MPa
350°C
⎛ 0.05364 m3 ⎞
⎛V ⎞
⎟ = 37.18 kJ
Wb,1− 2 = P1V1 ln⎜⎜ 2 ⎟⎟ = (2000 kPa)(0.01341 m3 ) ln⎜
⎜ 0.01341 m3 ⎟
⎝ V1 ⎠
⎠
⎝
For the polytropic compression process:
P2V 2n = P3V 3n ⎯
⎯→(500 kPa)(0.05364 m 3 )1.2 = (2000 kPa)V 31.2 ⎯
⎯→V 3 = 0.01690 m 3
Wb , 2 − 3 =
P3V 3 − P2V 2 (2000 kPa)(0.01690 m 3 ) − (500 kPa)(0.05364 m 3 )
=
= -34.86 kJ
1− n
1 − 1.2
For the constant pressure compression process:
Wb,3−1 = P3 (V1 −V 3 ) = (2000 kPa)(0.01341 − 0.01690)m3 = -6.97 kJ
The net work for the cycle is the sum of the works for each process
Wnet = Wb,1− 2 + Wb, 2 −3 + Wb,3−1 = 37.18 + (−34.86) + (−6.97) = -4.65 kJ
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
preparation. If you are a student using this Manual, you are using it without permission.
Tutorial #7
ESO 201A
Thermodynamics
Instructor
Prof. Sameer Khandekar
Contact
Office: SL-109, Tel: 7038
E-mail: [email protected]
URL: home.iitk.ac.in/~samkhan
ESO 201A Thermodynamics
Instructor: Sameer Khandekar
Tutorial 7
[5-121] A 2 m3 rigid tank initially contains air at 100 kPa and 22°C. The tank is connected to a
supply line through a valve. Air is flowing in the supply line at 600 kPa and 22°C. The valve is
opened, and air is allowed to enter the tank until the pressure in the tank reaches the line
pressure, at which point the valve is closed. A thermometer placed in the tank indicates that
the air temperature at the final state is 77°C. Determine (a) the mass of air that has entered the
tank and (b) the amount of heat transfer. [Answers: (a) 9.58 kg, (b) 339 kJ]
[5-130] A 0.3-m3 rigid tank is filled with saturated liquid water at 200°C. A valve at the bottom
of the tank is opened, and liquid is withdrawn from the tank. Heat is transferred to the water
such that the temperature in the tank remains constant. Determine the amount of heat that
must be transferred by the time one-half of the total mass has been withdrawn.
[5-133] A 0.4-m3 rigid tank initially contains refrigerant-134a at 14°C. At this state, 70 percent
of the mass is in the vapor phase, and the rest is in the liquid phase. The tank is connected
by a valve to a supply line where refrigerant at 1 MPa and100°C flows steadily. Now the
valve is opened slightly, and the refrigerant is allowed to enter the tank. When the pressure
in the tank reaches 700 kPa, the entire refrigerant in the tank exists in the vapor phase only.
At this point the valve is closed. Determine (a) the final temperature in the tank, (b) the mass
of refrigerant that has entered the tank, and (c) the heat transfer between the system and the
surroundings.
[5-201] Consider an evacuated rigid bottle of volume V that is surrounded by the atmosphere
at pressure P0 and temperatureT0. A valve at the neck of the bottle is now opened and
the atmospheric air is allowed to flow into the bottle. The air trapped in the bottle
eventually reaches thermal equilibrium with the atmosphere as a result of heat transfer
through the wall of the bottle. The valve remains open during the process so that the
trapped air also reaches mechanical equilibrium with the atmosphere. Determine the net heat
transfer through the wall of the bottle during this filling process in terms of the
properties of the system and the surrounding atmosphere
Additional Homework Problems (Tutorial 7)
[5-120] Consider a 20-L evacuated rigid bottle that is surrounded by the atmosphere at 100
kPa and 27°C. A valve at the neck of the bottle is now opened and the atmospheric air is
allowed to flow into the bottle. The air trapped in the bottle eventually reaches thermal
equilibrium with the atmosphere as a result of heat transfer through the wall of the bottle.
The valve remains open during the process so that the trapped air also reaches mechanical
equilibrium with the atmosphere. Determine the net heat transfer through the wall of the
bottle during this filling process. [Answer = 2.0 kJ]
[5-122] A 0.2-m3 rigid tank equipped with a pressure regulator contains steam at 2 MPa
and 300°C. The steam in the tank is now heated. The regulator keeps the steam pressure
constant by letting out some steam, but the temperature inside rises. Determine the
amount of heat transferred when the steam temperature reaches 500°C
[5-123] An insulated, vertical piston-cylinder device initially contains 10 kg of water, 6 kg
of which is in the vapor phase. The mass of the piston is such that it maintains a constant
pressure of 200 kPa inside the cylinder. Now steam at 0.5 MPa and350°C is allowed to enter
the cylinder from a supply line until all the liquid in the cylinder has vaporized. Determine
(a) the final temperature in the cylinder and (b) the mass of the steam that has entered.
[Answers: (a) 120.2°C, (b) 19.07 kg]
[5-125] An air-conditioning system is to be filled from a rigid container that initially contains 5
kg of liquid R-134a at 24°C.The valve connecting this container to the air-conditioning system is
now opened until the mass in the container is 0.25 kg, at which time the valve is closed. During
this time, only liquid R-134a flows from the container. Presuming that the process is isothermal
while the valve is open, determine the final quality of the R-134a in the container and the total
heat transfer. [Answers: 0.506, 22.6 kJ]
5-92
5-120 An evacuated bottle is surrounded by atmospheric air. A valve is opened, and air is allowed to fill the bottle. The
amount of heat transfer through the wall of the bottle when thermal and mechanical equilibrium is established is to be
determined.
Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process, but it
can be analyzed as a uniform-flow process since the state of fluid at the inlet remains constant. 2 Air is an ideal gas with
variable specific heats. 3 Kinetic and potential energies are negligible. 4 There are no work interactions involved. 5 The
direction of heat transfer is to the air in the bottle (will be verified).
Properties The gas constant of air is 0.287 kPa.m3/kg.K (Table A-1).
Analysis We take the bottle as the system, which is a control volume since mass crosses the boundary. Noting that the
microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u, respectively,
the mass and energy balances for this uniform-flow system can be expressed as
Mass balance:
100 kPa
27°C
min − mout = ∆msystem → mi = m2 (since mout = minitial = 0)
Energy balance:
E − Eout
1in
424
3
Net energy transfer
by heat, work, and mass
=
∆Esystem
1
424
3
Change in internal, kinetic,
potential, etc. energies
20 L
Evacuated
Qin + mi hi = m2u2 (since W ≅ Eout = Einitial = ke ≅ pe ≅ 0)
Combining the two balances:
Qin = m2 (u2 − hi )
where
m2 =
P2V
(100 kPa )(0.020 m 3 )
=
= 0.02323 kg
RT2 (0.287 kPa ⋅ m 3 /kg ⋅ K )(300 K )
-17
Ti = T2 = 300 K ⎯Table
⎯ ⎯A⎯
⎯→
hi = 300.19 kJ/kg
u 2 = 214.07 kJ/kg
Substituting,
Qin = (0.02323 kg)(214.07 − 300.19) kJ/kg = − 2.0 kJ
or
Qout = 2.0 kJ
Discussion The negative sign for heat transfer indicates that the assumed direction is wrong. Therefore, we reverse the
direction.
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
preparation. If you are a student using this Manual, you are using it without permission.
5-93
5-121 A rigid tank initially contains air at atmospheric conditions. The tank is connected to a supply line, and air is allowed
to enter the tank until mechanical equilibrium is established. The mass of air that entered and the amount of heat transfer are
to be determined.
Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process, but it
can be analyzed as a uniform-flow process since the state of fluid at the inlet remains constant. 2 Air is an ideal gas with
variable specific heats. 3 Kinetic and potential energies are negligible. 4 There are no work interactions involved. 5 The
direction of heat transfer is to the tank (will be verified).
Properties The gas constant of air is 0.287 kPa.m3/kg.K (Table A-1). The properties of air are (Table A-17)
Ti = 295 K ⎯
⎯→ hi = 295.17 kJ/kg
T1 = 295 K ⎯
⎯→ u1 = 210.49 kJ/kg
T2 = 350 K ⎯
⎯→ u 2 = 250.02 kJ/kg
Analysis (a) We take the tank as the system, which is a control volume since mass crosses the boundary. Noting that the
microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u, respectively,
the mass and energy balances for this uniform-flow system can be expressed as
Mass balance:
min − mout = ∆msystem → mi = m2 − m1
Pi = 600 kPa
Ti = 22°C
Energy balance:
E − Eout
1in424
3
=
Net energy transfer
by heat, work, and mass
∆Esystem
1
424
3
Change in internal, kinetic,
potential, etc. energies
Qin + mi hi = m2u2 − m1u1 (since W ≅ ke ≅ pe ≅ 0)
V1 = 2 m3
·
Q
P1 = 100 kPa
T1 = 22°C
The initial and the final masses in the tank are
m1 =
P1V
(100 kPa )(2 m3 )
=
= 2.362 kg
RT1 (0.287 kPa ⋅ m3/kg ⋅ K )(295 K )
m2 =
P2V
(600 kPa )(2 m3 )
=
= 11.946 kg
RT2 (0.287 kPa ⋅ m3/kg ⋅ K )(350 K )
Then from the mass balance,
mi = m2 − m1 = 11.946 − 2.362 = 9.584 kg
(b) The heat transfer during this process is determined from
Qin = −mi hi + m2u2 − m1u1
= −(9.584 kg )(295.17 kJ/kg ) + (11.946 kg )(250.02 kJ/kg ) − (2.362 kg )(210.49 kJ/kg )
= −339 kJ → Qout = 339 kJ
Discussion The negative sign for heat transfer indicates that the assumed direction is wrong. Therefore, we reversed the
direction.
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
preparation. If you are a student using this Manual, you are using it without permission.
5-94
5-122 A rigid tank initially contains superheated steam. A valve at the top of the tank is opened, and vapor is allowed to
escape at constant pressure until the temperature rises to 500°C. The amount of heat transfer is to be determined.
Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process, but it
can be analyzed as a uniform-flow process by using constant average properties for the steam leaving the tank. 2 Kinetic
and potential energies are negligible. 3 There are no work interactions involved. 4 The direction of heat transfer is to the
tank (will be verified).
Properties The properties of water are (Tables A-4 through A-6)
P1 = 2 MPa ⎫ v 1 = 0.12551 m 3 /kg
⎬
T1 = 300°C ⎭ u1 = 2773.2 kJ/kg, h1 = 3024.2 kJ/kg
P2 = 2 MPa ⎫ v 2 = 0.17568 m 3 /kg
⎬
T2 = 500°C ⎭ u 2 = 3116.9 kJ/kg, h2 = 3468.3 kJ/kg
STEAM
2 MPa
Q
Analysis We take the tank as the system, which is a control volume since mass crosses the boundary. Noting that the
microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u, respectively,
the mass and energy balances for this uniform-flow system can be expressed as
Mass balance:
min − mout = ∆msystem → me = m1 − m2
Energy balance:
E − Eout
1in424
3
Net energy transfer
by heat, work, and mass
=
∆Esystem
1
424
3
Change in internal, kinetic,
potential, etc. energies
Qin − me he = m2u2 − m1u1 (since W ≅ ke ≅ pe ≅ 0)
The state and thus the enthalpy of the steam leaving the tank is changing during this process. But for simplicity, we assume
constant properties for the exiting steam at the average values. Thus,
he ≅
h1 + h2 3024.2 + 3468.3 kJ/kg
=
= 3246.2 kJ/kg
2
2
The initial and the final masses in the tank are
m1 =
V1
0.2 m 3
=
= 1.594 kg
v 1 0.12551 m 3 /kg
m2 =
V2
0.2 m 3
=
= 1.138kg
v 2 0.17568 m 3 /kg
Then from the mass and energy balance relations,
me = m1 − m 2 = 1.594 − 1.138 = 0.456 kg
Qin = me he + m2u2 − m1u1
= (0.456 kg )(3246.2 kJ/kg ) + (1.138 kg )(3116.9 kJ/kg ) − (1.594 kg )(2773.2 kJ/kg )
= 606.8 kJ
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
preparation. If you are a student using this Manual, you are using it without permission.
5-95
5-123 A cylinder initially contains saturated liquid-vapor mixture of water. The cylinder is connected to a supply line, and
the steam is allowed to enter the cylinder until all the liquid is vaporized. The final temperature in the cylinder and the mass
of the steam that entered are to be determined.
Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process, but it
can be analyzed as a uniform-flow process since the state of fluid at the inlet remains constant. 2 The expansion process is
quasi-equilibrium. 3 Kinetic and potential energies are negligible. 3 There are no work interactions involved other than
boundary work. 4 The device is insulated and thus heat transfer is negligible.
Properties The properties of steam are (Tables A-4 through A-6)
P1 = 200 kPa ⎫
⎬ h1 = h f + x1 h fg
x1 = 0.6
⎭
= 504.71 + 0.6 × 2201.6 = 1825.6 kJ/kg
P2 = 200 kPa ⎫
⎬ h 2 = h g @ 200 kPa = 2706.3 kJ/kg
⎭
Pi = 0.5 MPa ⎫
⎬ hi = 3168.1 kJ/kg
Ti = 350°C ⎭
sat. vapor
(P = 200 kPa)
m1 = 10 kg
H2O
Pi = 0.5 MPa
Ti = 350°C
Analysis (a) The cylinder contains saturated vapor at the final state at a pressure of 200 kPa, thus the final temperature in
the cylinder must be
T2 = Tsat @ 200 kPa = 120.2°C
(b) We take the cylinder as the system, which is a control volume since mass crosses the boundary. Noting that the
microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u, respectively,
the mass and energy balances for this uniform-flow system can be expressed as
min − mout = ∆msystem → mi = m2 − m1
Mass balance:
Energy balance:
E − Eout
1in
424
3
Net energy transfer
by heat, work, and mass
=
∆Esystem
1
424
3
Change in internal, kinetic,
potential, etc. energies
mi hi = Wb,out + m2u2 − m1u1 (since Q ≅ ke ≅ pe ≅ 0)
Combining the two relations gives
0 = Wb, out − (m2 − m1 )hi + m2u 2 − m1u1
or,
0 = −(m2 − m1 )hi + m2h2 − m1h1
since the boundary work and ∆U combine into ∆H for constant pressure expansion and compression processes. Solving for
m2 and substituting,
m2 =
(3168.1 − 1825.6) kJ/kg (10 kg ) = 29.07 kg
hi − h1
m1 =
(3168.1 − 2706.3) kJ/kg
hi − h2
Thus,
mi = m2 - m1 = 29.07 - 10 = 19.07 kg
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
preparation. If you are a student using this Manual, you are using it without permission.
5-97
5-125 R-134a from a tank is discharged to an air-conditioning line in an isothermal process. The final quality of the R-134a
in the tank and the total heat transfer are to be determined.
Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process, but it
can be analyzed as a uniform-flow process since the state of fluid at the exit remains constant. 2 Kinetic and potential
energies are negligible. 3 There are no work interactions involved.
Analysis We take the tank as the system, which is a control volume since mass crosses the boundary. Noting that the
microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u, respectively,
the mass and energy balances for this uniform-flow system can be expressed as
Mass balance:
min − mout = ∆msystem
A-C line
− m e = m 2 − m1
m e = m1 − m 2
Energy balance:
E −E
1in424out
3
Net energy transfer
by heat, work, and mass
=
Liquid R-134a
5 kg
24°C
∆E system
1
424
3
Change in internal, kinetic,
potential, etc. energies
Qin − m e he = m 2 u 2 − m1u1
Qin = m 2 u 2 − m1u1 + m e he
Combining the two balances:
Qin = m 2 u 2 − m1u1 + (m1 − m 2 )he
The initial state properties of R-134a in the tank are
3
T1 = 24°C ⎫ v 1 = 0.0008261 m /kg
⎬ u1 = 84.44 kJ/kg
x=0
⎭ h = 84.98 kJ/kg
e
(Table A-11)
Note that we assumed that the refrigerant leaving the tank is at saturated liquid state, and found the exiting enthalpy
accordingly. The volume of the tank is
V = m1v 1 = (5 kg)(0.0008261 m 3 /kg) = 0.004131 m 3
The final specific volume in the container is
v2 =
V
m2
=
0.004131 m 3
= 0.01652 m 3 /kg
0.25 kg
The final state is now fixed. The properties at this state are (Table A-11)
v −v f
0.01652 − 0.0008261
⎫⎪ x 2 = 2
=
= 0.5061
0
.031834 − 0.0008261
v
⎬
fg
v 2 = 0.01652 m 3 /kg ⎪⎭
u 2 = u f + x 2 u fg = 84.44 kJ/kg + (0.5061)(158.65 kJ/kg) = 164.73 kJ/kg
T2 = 24°C
Substituting into the energy balance equation,
Qin = m 2 u 2 − m1u1 + ( m1 − m 2 ) he
= (0.25 kg )(164.73 kJ/kg ) − (5 kg )(84.44 kJ/kg ) + (4.75 kg )(84.98 kJ/kg )
= 22.64 kJ
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
preparation. If you are a student using this Manual, you are using it without permission.
5-102
5-130 A rigid tank initially contains saturated liquid water. A valve at the bottom of the tank is opened, and half of the mass
in liquid form is withdrawn from the tank. The temperature in the tank is maintained constant. The amount of heat transfer
is to be determined.
Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process, but it
can be analyzed as a uniform-flow process since the state of fluid leaving the device remains constant. 2 Kinetic and
potential energies are negligible. 3 There are no work interactions involved. 4 The direction of heat transfer is to the tank
(will be verified).
Properties The properties of water are (Tables A-4 through A-6)
3
T1 = 200°C ⎫ v1 = v f @ 200 o C = 0.001157 m /kg
⎬
sat. liquid ⎭ u1 = u f @ 200o C = 850.46 kJ/kg
Te = 200o C ⎫⎪
⎬ he = h f @ 200 o C = 852.26 kJ/kg
sat. liquid ⎪⎭
Analysis We take the tank as the system, which is a control volume since mass crosses the boundary. Noting that the
microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u, respectively,
the mass and energy balances for this uniform-flow system can be expressed as
min − mout = ∆msystem → me = m1 − m2
Mass balance:
Energy balance:
E − Eout
1in
424
3
=
Net energy transfer
by heat, work, and mass
∆Esystem
1
424
3
Change in internal, kinetic,
potential, etc. energies
H2O
Sat. liquid
T = 200°C
V = 0.3 m3
Q
Qin = me he + m2u2 − m1u1 (since W ≅ ke ≅ pe ≅ 0)
The initial and the final masses in the tank are
m1 =
V1
0.3 m 3
=
= 259.4 kg
v 1 0.001157 m 3 /kg
m 2 = 12 m1 =
1
2
(259.4 kg ) = 129.7 kg
Then from the mass balance,
me = m1 − m 2 = 259.4 − 129.7 = 129.7 kg
Now we determine the final internal energy,
v2 =
x2 =
V
m2
=
0.3 m 3
= 0.002313 m 3 /kg
129.7 kg
v 2 −v f
v fg
=
0.002313 − 0.001157
= 0.009171
0.12721 − 0.001157
T2 = 200°C
⎫
⎬ u 2 = u f + x 2 u fg = 850.46 + (0.009171)(1743.7 ) = 866.46 kJ/kg
x 2 = 0.009171 ⎭
Then the heat transfer during this process is determined from the energy balance by substitution to be
Q = (129.7 kg )(852.26 kJ/kg ) + (129.7 kg )(866.46 kJ/kg ) − (259.4 kg )(850.46 kJ/kg )
= 2308 kJ
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
preparation. If you are a student using this Manual, you are using it without permission.
5-105
5-133 A rigid tank initially contains saturated R-134a liquid-vapor mixture. The tank is connected to a supply line, and R134a is allowed to enter the tank. The final temperature in the tank, the mass of R-134a that entered, and the heat transfer
are to be determined.
Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process, but it
can be analyzed as a uniform-flow process since the state of fluid at the inlet remains constant. 2 Kinetic and potential
energies are negligible. 3 There are no work interactions involved. 4 The direction of heat transfer is to the tank (will be
verified).
Properties The properties of refrigerant are (Tables A-11 through A-13)
T1 = 14°C ⎫ v 1 = v f + x1v fg = 0.0008020 + 0.7 × (0.04342 − 0.0008020 ) = 0.03063 m 3 /kg
⎬
x1 = 0.7
u1 = u f + x1u fg = 70.57 + 0.7 × 167.26 = 187.65 kJ/kg
⎭
P2 = 700 kPa ⎫ v 2 = v g @700 kPa = 0.02936 m 3 /kg
⎬
sat. vapor
⎭ u 2 = u g @700 kPa = 244.48 kJ/kg
Pi = 1.0 MPa ⎫
⎬ hi = 335.06 kJ/kg
⎭
Ti = 100°C
R-134a
Analysis We take the tank as the system, which is a control volume
since mass crosses the boundary. Noting that the microscopic energies
of flowing and nonflowing fluids are represented by enthalpy h and
internal energy u, respectively, the mass and energy balances for this
uniform-flow system can be expressed as
1 MPa
100°C
0.4 m3
R-134a
min − mout = ∆msystem → mi = m2 − m1
Mass balance:
Energy balance:
E − Eout
1in424
3
=
Net energy transfer
by heat, work, and mass
∆Esystem
1
424
3
Change in internal, kinetic,
potential, etc. energies
Qin + mi hi = m2u2 − m1u1 (since W ≅ ke ≅ pe ≅ 0)
(a) The tank contains saturated vapor at the final state at 800 kPa, and thus the final temperature is the saturation
temperature at this pressure,
T2 = Tsat @ 700 kPa = 26.7°C
(b) The initial and the final masses in the tank are
m1 =
0.4 m 3
V
=
= 13.06 kg
v 1 0.03063 m 3 /kg
m2 =
0.4 m 3
V
=
= 13.62 kg
v 2 0.02936 m 3 /kg
Then from the mass balance
mi = m 2 − m1 = 13.62 − 13.06 = 0.5653 kg
(c) The heat transfer during this process is determined from the energy balance to be
Qin = − mi hi + m 2 u 2 − m1u1
= −(0.5653 kg )(335.06 kJ/kg ) + (13.62 kg )(244.48 kJ/kg )− (13.06 kg )(187.65 kJ/kg )
= 691 kJ
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
preparation. If you are a student using this Manual, you are using it without permission.
5-169
5-201 An evacuated bottle is surrounded by atmospheric air. A valve is opened, and air is allowed to fill the bottle. The
amount of heat transfer through the wall of the bottle when thermal and mechanical equilibrium is established is to be
determined.
Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process, but it
can be analyzed as a uniform-flow process since the state of fluid at the inlet remains constant. 2 Air is an ideal gas. 3
Kinetic and potential energies are negligible. 4 There are no work interactions involved. 5 The direction of heat transfer is
to the air in the bottle (will be verified).
Analysis We take the bottle as the system. It is a control volume since mass crosses the boundary. Noting that the
microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u, respectively,
the mass and energy balances for this uniform-flow system can be expressed as
Mass balance:
min − mout = ∆msystem → mi = m2 (since mout = minitial = 0)
P0
T0
Energy balance:
E − Eout
1in
424
3
=
Net energy transfer
by heat, work, and mass
∆Esystem
1
424
3
Change in internal, kinetic,
potential, etc. energies
Qin + mi hi = m2u2 (since W ≅ Eout = Einitial = ke ≅ pe ≅ 0)
V
Evacuated
Combining the two balances:
Qin = m2 (u2 − hi ) = m2 (cv T2 − c pTi )
but
Ti = T2 = T0
and
cp - cv = R.
Substituting,
(
)
Qin = m2 cv − c p T0 = −m2 RT0 = −
P0V
RT0 = − P0V
RT0
Therefore,
Qout = P0V
(Heat is lost from the tank)
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
preparation. If you are a student using this Manual, you are using it without permission.
Tutorial #8
ESO 201A
Thermodynamics
Instructor
Prof. Sameer Khandekar
Contact
Office: SL-109, Tel: 7038
E-mail: [email protected]
URL: home.iitk.ac.in/~samkhan
ESO 201A Thermodynamics
Instructor: Sameer Khandekar
Tutorial 8
[6-28] A coal-burning steam power plant produces a net power of 300 MW with an overall
thermal efficiency of 32 percent. The actual gravimetric air-fuel ratio in the furnace is calculated
to be 12 kg air/kg fuel. The heating value of the coal is 28,000 kJ/kg. Determine (a) the amount
of coal consumed during a 24-hour period and (b) the rate of air flowing through the furnace.
Answers: (a) 2.89 x 106 kg,(b) 402 kg/s
[6-56] Refrigerant-134a enters the condenser of a residential heat pump at 800 kPa and 35°C
at a rate of 0.018 kg/s and leaves at 800 kPa as a saturated liquid. If the compressor
consumes1.2 kW of power, determine (a) the COP of the heat pump and (b) the rate of heat
absorption from the outside air.
[6-101] A refrigerator operating on the reversed Carnot cycle has a measured work input of 200
kW and heat rejection of 2000 kW to a heat reservoir at 27°C. Determine the cooling load
supplied to the refrigerator, in kW, and the temperature of the heat source, in °C. [Answers:
1800 kW, -3 °C]
[6-107] A Carnot heat engine receives heat from a reservoir at900°C at a rate of 800 kJ/min and
rejects the waste heat to the ambient air at 27°C. The entire work output of the heat engine
is used to drive a refrigerator that removes heat from the refrigerated space at-5°C and
transfers it to the same ambient air at 27°C. Determine (a) the maximum rate of heat
removal from the refrigerated space and (b) the total rate of heat rejection to the ambient air.
[Answers: (a) 4982 kJ/min, (b) 5782 kJ/min]
Additional Homework Problems (Tutorial 8)
[6-22] An automobile engine consumes fuel at a rate of 22 L/h and delivers 55 kW of power to
the wheels. If the fuel has a heating value of 44,000 kJ/kg and a density of 0.8 g/cm3, determine
the efficiency of this engine.
[6-40] An air conditioner produces a 2-kW cooling effect while rejecting 2.5 kW of heat. What is
its COP?
[6-41] A refrigerator used to cool a computer requires 3 kW of electrical power and has a COP
of 1.4. Calculate the cooling effect of this refrigerator, in kW.
[6-42] A food department is kept at — 12°C by a refrigerator in an environment at 30°C. The
total heat gain to the food department is estimated to be 3300 kJ/h and the heat rejection in
the condenser is 4800 kJ/h. Determine the power input to the compressor, in kW and the COP
of the refrigerator.
[6-46] A household refrigerator that has a power input of 450 W and a COP of 2.5 is to cool five
large watermelons, 10kg each, to 8°C. If the watermelons are initially at 20°C, determine how
long it will take for the refrigerator to cool them. The watermelons can be treated as water
whose specific heat is 4.2 kJ/kg °C. Is your answer realistic or optimistic? Explain.
[6-51] A heat pump is used to maintain a house at a constant temperature of 23°C. The house is
losing heat to the outside air through the walls and the windows at a rate of 60,000 kJ/h while
the energy generated within the house from people, lights, and appliances amounts to 4000
kJ/h. For a COP of 2.5, determine the required power input to the heat pump.
[6-95] A heat pump operates on a Carnot heat pump cycle with a COP of 8.7. It keeps a space at
26°C by consuming 4.25 kW of power. Determine the temperature of the reservoir from which
the heat is absorbed and the heating load provided by the heat pump.
[6-96] A refrigerator is to remove heat from the cooled space at a rate of 300 kJ/min to
maintain its temperature at -8°C. If the air surrounding the refrigerator is at 25°C, determine
the minimum power input required for this refrigerator.
6-6
6-22 The power output and fuel consumption rate of a car engine are given. The thermal efficiency of the engine is to be
determined.
Assumptions The car operates steadily.
Properties The heating value of the fuel is given to be 44,000 kJ/kg.
Analysis The mass consumption rate of the fuel is
m& fuel = ( ρV& ) fuel = (0.8 kg/L)(22 L/h) = 17.6 kg/h
Fuel
Engine
The rate of heat supply to the car is
22 L/h
Q& H = m& coal q HV,coal
55 kW
HE
= (17.6 kg/h )(44,000 kJ/kg)
= 774,400 kJ/h = 215.1 kW
sink
Then the thermal efficiency of the car becomes
η th =
W& net,out
55 kW
=
= 0.256 = 25.6%
&
215.1 kW
QH
6-23 The United States produces about 51 percent of its electricity from coal at a conversion efficiency of about 34 percent.
The amount of heat rejected by the coal-fired power plants per year is to be determined.
Analysis Noting that the conversion efficiency is 34%, the amount of heat rejected by the coal plants per year is
η th =
Qout =
Wcoal
Wcoal
=
Qin
Qout + Wcoal
Wcoal
η th
Furnace
− Wcoal
1.878 × 1012 kWh
− 1.878 × 1012 kWh
0.34
= 3.646 × 10 12 kWh
=
Coal
Q& out
HE
1.878×1012 kWh
ηth = 34%
sink
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
preparation. If you are a student using this Manual, you are using it without permission.
6-11
6-28 A coal-burning power plant produces 300 MW of power. The amount of coal consumed during a one-day period and
the rate of air flowing through the furnace are to be determined.
Assumptions 1 The power plant operates steadily. 2 The kinetic and potential energy changes are zero.
Properties The heating value of the coal is given to be 28,000 kJ/kg.
Analysis (a) The rate and the amount of heat inputs to the power plant are
W& net,out 300 MW
Q& in =
=
= 937.5 MW
0.32
η th
Qin = Q& in ∆t = (937.5 MJ/s)(24 × 3600 s) = 8.1 × 10 7 MJ
The amount and rate of coal consumed during this period are
mcoal =
Qin 8.1 × 107 MJ
=
= 2.893 × 106 kg
qHV
28 MJ/kg
m& coal =
mcoal 2.893 × 10 6 kg
= 33.48 kg/s
=
24 × 3600 s
∆t
(b) Noting that the air-fuel ratio is 12, the rate of air flowing through the furnace is
m& air = (AF)m& coal = (12 kg air/kg fuel)(33.48 kg/s) = 401.8 kg/s
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
preparation. If you are a student using this Manual, you are using it without permission.
6-13
6-39E The COP and the power input of a residential heat pump are given. The rate of heating effect is to be determined.
Assumptions The heat pump operates steadily.
Reservoir
Analysis Applying the definition of the heat pump coefficient of performance
to this heat pump gives
Q& H
HP
⎛ 2544.5 Btu/h ⎞
⎟⎟ = 30,530 Btu/h
Q& H = COPHPW& net,in = (2.4)(5 hp)⎜⎜
1 hp
⎝
⎠
5 hp
COP = 2.4
Reservoir
6-40 The cooling effect and the rate of heat rejection of an air conditioner are given. The COP is to be determined.
Assumptions The air conditioner operates steadily.
Analysis Applying the first law to the air conditioner gives
W& net,in = Q& H − Q& L = 2.5 − 2 = 0.5 kW
Applying the definition of the coefficient of performance,
COPR =
Q& L
W&
net,in
=
2.0 kW
=4
0.5 kW
Reservoir
Q& H
W& net,in
AC
Q& L
Reservoir
6-41 The power input and the COP of a refrigerator are given. The cooling effect of the refrigerator is to be determined.
Assumptions The refrigerator operates steadily.
Reservoir
Analysis Rearranging the definition of the refrigerator coefficient of
performance and applying the result to this refrigerator gives
Q& H
Q& L = COPR W& net,in = (1.4)(3 kW) = 4.2 kW
COP=1.4
R
W& net,in
Q& L
Reservoir
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
preparation. If you are a student using this Manual, you are using it without permission.
6-14
6-42 A refrigerator is used to keep a food department at a specified temperature. The heat gain to the food department and
the heat rejection in the condenser are given. The power input and the COP are to be determined.
Assumptions The refrigerator operates steadily.
Analysis The power input is determined from
W& in = Q& H − Q& L
= 4800 − 3300 = 1500 kJ/h
⎛ 1 kW ⎞
= (1500 kJ/h)⎜
⎟ = 0.417 kW
⎝ 3600 kJ/h ⎠
The COP is
COP =
30°C
Q& H
4800 kJ/h
W& in
R
Q& L
3300 kJ/h
−12°C
Q& L 3300 kJ/h
=
= 2.2
W& in 1500 kJ/h
6-43 The COP and the refrigeration rate of a refrigerator are given. The power consumption and the rate of heat rejection
are to be determined.
Assumptions The refrigerator operates steadily.
Kitchen air
Analysis (a) Using the definition of the coefficient of performance, the power
input to the refrigerator is determined to be
W&net,in =
COP=1.2
R
Q& L
60 kJ/min
=
= 50 kJ/min = 0.83 kW
COPR
1.2
Q& L
cool space
(b) The heat transfer rate to the kitchen air is determined from the energy balance,
Q& H = Q& L + W&net,in = 60 + 50 = 110 kJ/min
6-44E The heat absorption, the heat rejection, and the power input of a commercial heat pump are given. The COP of the
heat pump is to be determined.
Reservoir
Assumptions The heat pump operates steadily.
Analysis Applying the definition of the heat pump coefficient of
performance to this heat pump gives
COPHP
Q& H
=
W&
net, in
1 hp
15,090 Btu/h ⎛
⎞
=
⎜
⎟ = 2.97
2 hp
2544.5
Btu/h
⎝
⎠
Q& H
HP
2 hp
Q& L
Reservoir
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
preparation. If you are a student using this Manual, you are using it without permission.
6-15
6-45 The cooling effect and the COP of a refrigerator are given. The power input to the refrigerator is to be determined.
Assumptions The refrigerator operates steadily.
Reservoir
Analysis Rearranging the definition of the refrigerator coefficient of
performance and applying the result to this refrigerator gives
Q& H
Q& L
25,000 kJ/h ⎛ 1 h ⎞
W& net,in =
=
⎜
⎟ = 4.34 kW
COPR
1.60
⎝ 3600 s ⎠
COP=1.6
W& net,in
R
Q& L
Reservoir
6-46 The COP and the power consumption of a refrigerator are given. The time it will take to cool 5 watermelons is to be
determined.
Assumptions 1 The refrigerator operates steadily. 2 The heat gain of the refrigerator through its walls, door, etc. is
negligible. 3 The watermelons are the only items in the refrigerator to be cooled.
Properties The specific heat of watermelons is given to be c = 4.2 kJ/kg.°C.
Analysis The total amount of heat that needs to be removed from the watermelons is
QL = (mc∆T )watermelons = 5 × (10 kg )(4.2 kJ/kg ⋅ °C)(20 − 8)o C = 2520 kJ
Kitchen air
The rate at which this refrigerator removes heat is
(
)
Q& L = (COPR ) W& net,in = (2.5 )(0.45 kW ) = 1.125 kW
That is, this refrigerator can remove 1.125 kJ of heat per second. Thus the time
required to remove 2520 kJ of heat is
R
450 W
COP = 2.5
cool space
Q
2520 kJ
∆t = L =
= 2240 s = 37.3 min
&
Q L 1.125 kJ/s
This answer is optimistic since the refrigerated space will gain some heat during this process from the surrounding air,
which will increase the work load. Thus, in reality, it will take longer to cool the watermelons.
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6-19
6-51 The rate of heat loss, the rate of internal heat gain, and the COP of a heat pump are given. The power input to the heat
pump is to be determined.
Assumptions The heat pump operates steadily.
Analysis The heating load of this heat pump system is the difference between the heat
lost to the outdoors and the heat generated in the house from the people, lights, and
appliances,
60,000
kJ/h
House
Q& H
Q& H = 60,000 − 4,000 = 56,000 kJ / h
HP
COP = 2.5
Using the definition of COP, the power input to the heat pump is determined to be
W&net,in =
Outside
Q& H
56,000 kJ/h ⎛ 1 kW ⎞
⎜⎜
⎟⎟ = 6.22 kW
=
COPHP
2.5
⎝ 3600 kJ/h ⎠
6-52E The COP and the refrigeration rate of an ice machine are given. The power consumption is to be determined.
Assumptions The ice machine operates steadily.
Outdoors
Analysis The cooling load of this ice machine is
COP = 2.4
Q& L = m& q L = (28 lbm/h)(169 Btu/lbm) = 4732 Btu/h
R
Using the definition of the coefficient of performance, the power input
to the ice machine system is determined to be
W&net,in =
⎞
Q& L
4732 Btu/h ⎛
1 hp
⎜⎜
⎟⎟ = 0.775 hp
=
COPR
2.4
⎝ 2545 Btu/h ⎠
&
Q
L
water
55°F
Ice
Machine
ice
25°F
6-53E An office that is being cooled adequately by a 12,000 Btu/h window air-conditioner is converted to a computer
room. The number of additional air-conditioners that need to be installed is to be determined.
Assumptions 1 The computers are operated by 7 adult men. 2 The computers consume 40 percent of their rated power at
any given time.
Properties The average rate of heat generation from a person seated in a room/office is 100 W (given).
Analysis The amount of heat dissipated by the computers is equal to the
amount of electrical energy they consume. Therefore,
Q& computers = (Rated power) × (Usage factor) = (8.4 kW)(0.4) = 3.36 kW
Outside
Q& people = ( No. of people) × Q& person = 7 × (100 W) = 700 W
Q& total = Q& computers + Q& people = 3360 + 700 = 4060 W = 13,853 Btu/h
since 1 W = 3.412 Btu/h. Then noting that each available air conditioner provides
7000 Btu/h cooling, the number of air-conditioners needed becomes
Cooling load
13,853 Btu/h
No. of air conditioners =
=
Cooling capacity of A/C 7000 Btu/h
= 1.98 ≈ 2 Air conditioners
AC
7000 Btu/h
Computer
room
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
preparation. If you are a student using this Manual, you are using it without permission.
6-21
6-56 Refrigerant-134a flows through the condenser of a residential heat pump unit. For a given compressor power
consumption the COP of the heat pump and the rate of heat absorbed from the outside air are to be determined.
Assumptions 1 The heat pump operates steadily. 2 The
kinetic and potential energy changes are zero.
QH
800 kPa
x=0
Properties The enthalpies of R-134a at the condenser
inlet and exit are
Condenser
P1 = 800 kPa ⎫
⎬h1 = 271.22 kJ/kg
T1 = 35°C ⎭
P2 = 800 kPa ⎫
⎬h2 = 95.47 kJ/kg
x2 = 0
⎭
Expansion
valve
800 kPa
35°C
Win
Compressor
Evaporator
Analysis (a) An energy balance on the condenser gives
the heat rejected in the condenser
QL
Q& H = m& (h1 − h2 ) = (0.018 kg/s)(271.22 − 95.47) kJ/kg = 3.164 kW
The COP of the heat pump is
COP =
Q& H 3.164 kW
=
= 2.64
1.2 kW
W&in
(b) The rate of heat absorbed from the outside air
Q& L = Q& H − W& in = 3.164 − 1.2 = 1.96 kW
6-57 A commercial refrigerator with R-134a as the working fluid is considered. The evaporator inlet and exit states are
specified. The mass flow rate of the refrigerant and the rate of heat rejected are to be determined.
Assumptions 1 The refrigerator operates steadily. 2 The kinetic and
potential energy changes are zero.
QH
Properties The properties of R-134a at the evaporator
inlet and exit states are (Tables A-11 through A-13)
Condenser
P1 = 100 kPa ⎫
⎬h1 = 60.71 kJ/kg
x1 = 0.2
⎭
P2 = 100 kPa ⎫
⎬h2 = 234.74 kJ/kg
T2 = −26°C ⎭
Analysis (a) The refrigeration load is
Q& L = (COP)W& in = (1.2)(0.600 kW) = 0.72 kW
Expansion
valve
Win
Compressor
Evaporator
100 kPa
x=0.2
QL
100 kPa
-26°C
The mass flow rate of the refrigerant is determined from
m& R =
Q& L
0.72 kW
=
= 0.00414 kg/s
h2 − h1 (234.74 − 60.71) kJ/kg
(b) The rate of heat rejected from the refrigerator is
Q& H = Q& L + W& in = 0.72 + 0.60 = 1.32 kW
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
preparation. If you are a student using this Manual, you are using it without permission.
6-33
6-94 The validity of a claim by an inventor related to the operation of a heat pump is to be evaluated.
Assumptions The heat pump operates steadily.
Analysis Applying the definition of the heat pump coefficient of performance,
COPHP
Q& H
=
W&
net,in
200 kW
=
= 2.67
75 kW
The maximum COP of a heat pump operating between the same temperature
limits is
COPHP, max =
1
1 − TL / TH
=
1
= 14.7
1 − (273 K)/(293 K)
293 K
Q& H
HP
75
Q& L
273 K
Since the actual COP is less than the maximum COP, the claim is valid.
6-95 The power input and the COP of a Carnot heat pump are given. The temperature of the low-temperature reservoir and
the heating load are to be determined.
Assumptions The heat pump operates steadily.
Analysis The temperature of the low-temperature reservoir is
COPHP, max =
TH
299 K
⎯
⎯→ 8.7 =
⎯
⎯→ TL = 264.6 K
TH − TL
(299 − TL ) K
26°C
Q& H
HP
4.25 kW
Q& L
The heating load is
COPHP, max =
Q& H
Q& H
⎯
⎯→ 8.7 =
⎯
⎯→ Q& H = 37.0 kW
4.25 kW
W& in
TL
6-96 The refrigerated space and the environment temperatures for a refrigerator and the rate of heat removal from the
refrigerated space are given. The minimum power input required is to be determined.
Assumptions The refrigerator operates steadily.
Analysis The power input to a refrigerator will be a minimum when the refrigerator operates in a reversible manner. The
coefficient of performance of a reversible refrigerator depends on the temperature limits in the cycle only, and is
determined from
COPR, rev =
1
1
=
= 8.03
(TH / TL ) − 1 (25 + 273 K )/ (− 8 + 273 K ) − 1
The power input to this refrigerator is determined from the definition of
the coefficient of performance of a refrigerator,
W&net,in, min
Q& L
300 kJ/min
=
=
= 37.36 kJ/min = 0.623 kW
COPR, max
8.03
25°C
R
300 kJ/min
-8°C
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
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6-35
6-99 A heat pump maintains a house at a specified temperature. The rate of heat loss of the house and the power
consumption of the heat pump are given. It is to be determined if this heat pump can do the job.
Assumptions The heat pump operates steadily.
Analysis The power input to a heat pump will be a minimum when the heat pump operates in a reversible manner. The
coefficient of performance of a reversible heat pump depends on the temperature limits in the cycle only, and is determined
from
COPHP, rev =
1
1
=
= 14.75
1 − (TL / TH ) 1 − (2 + 273 K )/ (22 + 273 K )
The required power input to this reversible heat pump is determined from the
definition of the coefficient of performance to be
W&net,in, min =
Q& H
110,000 kJ/h ⎛ 1 h ⎞
⎜⎜
⎟⎟ = 2.07 kW
=
COPHP
14.75
⎝ 3600 s ⎠
110,000 kJ/h
House
22°C
5 kW
HP
This heat pump is powerful enough since 5 kW > 2.07 kW.
6-100E The power required by a reversible refrigerator with specified reservoir temperatures is to be determined.
Assumptions The refrigerator operates steadily.
Analysis The COP of this reversible refrigerator is
COPR, max =
TL
450 R
=
=5
T H − T L 540 R − 450 R
Using this result in the coefficient of performance expression yields
W& net,in =
540 R
Q& L
15,000 Btu/h ⎛
1 kW
⎞
=
⎜
⎟ = 0.879 kW
COPR, max
5
3412
.
14
Btu/h
⎝
⎠
.
W
net,in
R
15,000 Btu/h
450 R
6-101 The power input and heat rejection of a reversed Carnot cycle are given. The cooling load and the source
temperature are to be determined.
Assumptions The refrigerator operates steadily.
Analysis Applying the definition of the refrigerator coefficient of performance,
Q& L = Q& H − W& net,in = 2000 − 200 = 1800 kW
300 K
Q& H
R
Applying the definition of the heat pump coefficient of performance,
Q& L
COPR =
W&
net,in
1800 kW
=
=9
200 kW
2000 kW
Q& L
200 kW
TL
The temperature of the heat source is determined from
COPR, max =
TL
TL
⎯
⎯→ 9 =
⎯
⎯→ T L = 270 K = −3°C
TH − TL
300 − T L
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6-39
6-107 A Carnot heat engine is used to drive a Carnot refrigerator. The maximum rate of heat removal from the refrigerated
space and the total rate of heat rejection to the ambient air are to be determined.
Assumptions The heat engine and the refrigerator operate steadily.
Analysis (a) The highest thermal efficiency a heat engine operating between two specified temperature limits can have is
the Carnot efficiency, which is determined from
η th, max = η th,C = 1 −
300 K
TL
= 1−
= 0.744
1173 K
TH
900°C
-5°C
800 kJ/min
Then the maximum power output of this heat engine is
determined from the definition of thermal efficiency to be
HE
W& net, out = η th Q& H = (0.744 )(800 kJ/min ) = 595.2 kJ/min
R
27°C
which is also the power input to the refrigerator, W& net,in .
The rate of heat removal from the refrigerated space will be a maximum if a Carnot refrigerator is used. The COP of the
Carnot refrigerator is
COPR, rev =
1
=
1
(TH / TL ) − 1 (27 + 273 K )/ (− 5 + 273 K ) − 1
= 8.37
Then the rate of heat removal from the refrigerated space becomes
(
)(
)
Q& L , R = COPR, rev W& net,in = (8.37 )(595.2 kJ/min ) = 4982 kJ/min
(b) The total rate of heat rejection to the ambient air is the sum of the heat rejected by the heat engine ( Q& L , HE ) and the heat
discarded by the refrigerator ( Q& H , R ),
Q& L, HE = Q& H , HE − W&net, out = 800 − 595.2 = 204.8 kJ/min
Q& H , R = Q& L , R + W&net,in = 4982 + 595.2 = 5577.2 kJ/min
and
Q& ambient = Q& L , HE + Q& H , R = 204.8 + 5577 .2 = 5782 kJ/min
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
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Tutorial #9
ESO 201A
Thermodynamics
Instructor
Prof. Sameer Khandekar
Contact
Office: SL-109, Tel: 7038
E-mail: [email protected]
URL: home.iitk.ac.in/~samkhan
ESO 201A Thermodynamics
Instructor: Sameer Khandekar
Tutorial 9
[7-27] A completely reversible heat pump produces heat at arate of 300 kW to warm a house
maintained at 24°C. Theexterior air, which is at 7°C, serves as the source. Calculate the rate of
entropy change of the two reservoirs and determineif this heat pump satisfies the second law
according tothe increase of entropy principle.
[7-29] Refrigerant-134a enters the coils of the evaporator ofa refrigeration system as
a saturated liquid-vapor mixture at a pressure of 160 kPa. The refrigerant absorbs 180 kJ
of heatfrom the cooled space, which is maintained at -5°C, andleaves as saturated vapor at
the same pressure. Determine (a) the entropy change of the refrigerant, (b) the entropy
change of the cooled space, and (c) the total entropy change for this process.
[7-57] Steam enters a steady-flow adiabatic nozzle with alow inlet velocity as a saturated vapor
at 6 MPa and expands to 1.2 MPa.(a) Under the conditions that the exit velocity is to be the
maximum possible value, sketch the T-s diagram with respect to the saturation lines for this
process.(b) Determine the maximum exit velocity of the steam, in m/s.
[7-64] A 75-kg copper block initially at 110°C is droppedinto an insulated tank that contains 160
L of water at 15°C.Determine the final equilibrium temperature and the total entropy change
for this process.
Additional Homework Problems (Tutorial 9)
[7-25] Heat in the amount of 100 kJ is transferred directlyfrom a hot reservoir at 1200 K to a
cold reservoir at 600 K.Calculate the entropy change of the two reservoirs and determine if the
increase of entropy principle is satisfied.
[7-32] A well-insulated rigid tank contains 5 kg of a saturatedliquid-vapor mixture of water at
150 kPa. Initially, three-quartersof the mass is in the liquid phase. An electric resistance heater
placed in the tank is now turned on and kept on until allthe liquid in the tank is vaporized.
Determine the entropy change of the steam during this process. [Answer: 19.2 kJ/K]
[7-69] A 50-kg iron block and a 20-kg copper block, bothinitially at 80°C, are dropped into a
large lake at 15°C. Thermalequilibrium is established after a while as a result of heat transfer
between the blocks and the lake water. Determine thetotal entropy change for this process.
[7-89] An insulated rigid tank is divided into two equal partsby a partition. Initially, one part
contains 5 k mol of an ideal gas at 250 kPa and 40°C, and the other side is evacuated. The
partition is now removed, and the gas fills the entire tank.Determine the total entropy change
during this process. [Answer:28.81 kJ/K]
[7-100] A constant-volume tank contains 5 kg of air at 100 kPa and 327°C. The air is cooled to
the surroundings temperature of27°C. Assume constant specific heats at 300 K. (a) Determine
the entropy change of the air in the tank during the process, inkJ/K, (b) determine the net
entropy change of the universe due to this process, in kJ/K, and (c) sketch the processes for the
airin the tank and the surroundings on a single T-s diagram. Be sure to label the initial and final
states for both processes.
[7-110] Calculate the work produced, in kJ/kg, for thereversible isothermal, steady-flow
process 1-3 shown in Fig. P7-110 when the working fluid is an ideal gas.
[7-111] Liquid water enters a 25-kW pump at 100-kPa pressureat a rate of 5 kg/s. Determine
the highest pressure theliquid water can have at the exit of the pump. Neglectthe kinetic and
potential energy changes of water, and takethe specific volume of water to be 0.001 m3/kg.
[Answer: 5100 kPa]
[7-144] Cold water (c p = 4.18 kJ/kg-°C) leading to a showerenters a well-insulated, thin-walled,
double-pipe, counterflowheat exchanger at 10°C at a rate of 0.95 kg/s and is heated to 70°C by
hot water (c p = 4.19 kJ/kg-°C) that entersat 85°C at a rate of 1.6 kg/s. Determine (a) the rate of
heat transfer and (b) the rate of entropy generation in the heatexchanger.
[7-212] Two rigid tanks are connected by a valve. Tank Ais insulated and contains 0.3 m3 of
steam at 400 kPa and60 percent quality. Tank B is uninsulated and contains 2 kg of steam at
200 kPa and 250°C. The valve is now opened, andsteam flows from tank A to tank B until the
pressure in tank A drops to 200 kPa. During this process 300 kJ of heat istransferred from tank B
to the surroundings at 17°C. Assuming the steam remaining inside tank A to have undergone
areversible adiabatic process, determine (a) the final temperature in each tank and (b) the
entropy generated during this process. [Answers: (a) 120.2°C, 116.1°C, (b) 0.498 kJ/K]
7-5
7-24 Air is compressed steadily by a compressor. The air temperature is maintained constant by heat rejection to the
surroundings. The rate of entropy change of air is to be determined.
Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potential energy changes are
negligible. 3 Air is an ideal gas. 4 The process involves no internal irreversibilities such as friction, and thus it is an
isothermal, internally reversible process.
Properties Noting that h = h(T) for ideal gases, we have h1 = h2 since T1 = T2 = 25°C.
Analysis We take the compressor as the system. Noting that the enthalpy of air remains constant, the energy balance for
this steady-flow system can be expressed in the rate form as
E& − E& out
1in424
3
=
Rate of net energy transfer
by heat, work, and mass
∆E& systemÊ0 (steady)
1442443
=0
Rate of change in internal, kinetic,
potential, etc. energies
E& in = E& out
W&in = Q& out
P2
·
Q
AIR
T = const.
30 kW
Therefore,
Q& out = W& in = 30 kW
Noting that the process is assumed to be an isothermal and internally
reversible process, the rate of entropy change of air is determined to be
∆S& air = −
Q& out,air
Tsys
=−
P1
30 kW
= −0.101 kW/K
298 K
7-25 Heat is transferred directly from an energy-source reservoir to an energy-sink. The entropy change of the two
reservoirs is to be calculated and it is to be determined if the increase of entropy principle is satisfied.
Assumptions The reservoirs operate steadily.
Analysis The entropy change of the source and sink is given by
∆S =
Q H Q L − 100 kJ 100 kJ
+
=
+
= 0.0833 kJ/K
1200 K 600 K
TH
TL
Since the entropy of everything involved in this process has
increased, this transfer of heat is possible.
1200 K
100 kJ
600 K
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7-6
7-26 It is assumed that heat is transferred from a cold reservoir to the hot reservoir contrary to the Clausius statement of the
second law. It is to be proven that this violates the increase in entropy principle.
Assumptions The reservoirs operate steadily.
Analysis According to the definition of the entropy, the entropy change of the
high-temperature reservoir shown below is
∆S H =
Q
100 kJ
=
= 0.08333 kJ/K
T H 1200 K
TH
Q =100 kJ
and the entropy change of the low-temperature reservoir is
∆S L =
Q −100 kJ
=
= −0.1667 kJ/K
TL
600 K
TL
The total entropy change of everything involved with this system is then
∆S total = ∆S H + ∆S L = 0.08333 − 0.1667 = −0.0833 kJ/K
which violates the increase in entropy principle since the total entropy change is negative.
7-27 A reversible heat pump with specified reservoir temperatures is considered. The entropy change of two reservoirs is to
be calculated and it is to be determined if this heat pump satisfies the increase in entropy principle.
Assumptions The heat pump operates steadily.
Analysis Since the heat pump is completely reversible, the combination of the coefficient of
performance expression, first Law, and thermodynamic temperature scale gives
COPHP,rev =
1
1 − TL / TH
=
1
= 17.47
1 − (280 K ) /(297 K )
The power required to drive this heat pump, according to the coefficient of
performance, is then
W& net,in
24°C
300 kW
HP
W& net
Q& L
7°C
Q& H
300 kW
=
=
= 17.17 kW
COPHP, rev
17.47
According to the first law, the rate at which heat is removed from the low-temperature energy reservoir is
Q& L = Q& H − W& net,in = 300 kW − 17.17 kW = 282.8 kW
The rate at which the entropy of the high temperature reservoir changes, according to the definition of the entropy, is
∆S& H =
Q& H 300 kW
=
= 1.01 kW/K
TH
297 K
and that of the low-temperature reservoir is
∆S& L =
Q& L − 17.17 kW
=
= −1.01 kW/K
280 K
TL
The net rate of entropy change of everything in this system is
∆S& total = ∆S& H + ∆S& L = 1.01 − 1.01 = 0 kW/K
as it must be since the heat pump is completely reversible.
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7-7
7-28E Heat is transferred isothermally from the working fluid of a Carnot engine to a heat sink. The entropy change of the
working fluid is given. The amount of heat transfer, the entropy change of the sink, and the total entropy change during the
process are to be determined.
Analysis (a) This is a reversible isothermal process, and the entropy change
during such a process is given by
∆S =
Heat
SINK
95°F
Q
T
95°F
Carnot heat engine
Noting that heat transferred from the working fluid is equal to the heat
transferred to the sink, the heat transfer become
Qfluid = Tfluid ∆S fluid = (555 R )(−0.7 Btu/R ) = −388.5 Btu → Qfluid, out = 388.5 Btu
(b) The entropy change of the sink is determined from
∆Ssink =
Qsink, in
Tsink
=
388.5 Btu
= 0.7 Btu/R
555 R
(c) Thus the total entropy change of the process is
Sgen = ∆S total = ∆Sfluid + ∆Ssink = −0.7 + 0.7 = 0
This is expected since all processes of the Carnot cycle are reversible processes, and no entropy is generated during a
reversible process.
7-29 R-134a enters an evaporator as a saturated liquid-vapor at a specified pressure. Heat is transferred to the refrigerant
from the cooled space, and the liquid is vaporized. The entropy change of the refrigerant, the entropy change of the cooled
space, and the total entropy change for this process are to be determined.
Assumptions 1 Both the refrigerant and the cooled space involve no internal irreversibilities such as friction. 2 Any
temperature change occurs within the wall of the tube, and thus both the refrigerant and the cooled space remain isothermal
during this process. Thus it is an isothermal, internally reversible process.
Analysis Noting that both the refrigerant and the cooled space undergo reversible isothermal processes, the entropy change
for them can be determined from
∆S =
Q
T
(a) The pressure of the refrigerant is maintained constant. Therefore, the temperature of the refrigerant also remains
constant at the saturation value,
T = Tsat @160 kPa = −15.6°C = 257.4 K
Then,
∆Srefrigerant =
Qrefrigerant,in
Trefrigerant
=
(Table A-12)
180 kJ
= 0.699 kJ/K
257.4 K
(b) Similarly,
∆S space = −
Qspace,out
Tspace
180 kJ
=−
= −0.672 kJ/K
268 K
R-134a
160 kPa
180 kJ
-5°C
(c) The total entropy change of the process is
S gen = ∆S total = ∆S refrigeran t + ∆S space = 0.699 − 0.672 = 0.027 kJ/K
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7-8
Entropy Changes of Pure Substances
7-30C Yes, because an internally reversible, adiabatic process involves no irreversibilities or heat transfer.
7-31E A piston-cylinder device that is filled with water is heated. The total entropy change is to be determined.
Analysis The initial specific volume is
v1 =
V1
m
=
2.5 ft 3
= 1.25 ft 3 /lbm
2 lbm
H2O
300 psia
2 lbm
2.5 ft3
which is between vf and vg for 300 psia. The initial quality and the entropy are then
(Table A-5E)
x1 =
v1 −v f
v fg
=
(1.25 − 0.01890) ft 3 /lbm
(1.5435 − 0.01890) ft 3 /lbm
= 0.8075
s1 = s f + x1 s fg = 0.58818 Btu/lbm ⋅ R + (0.8075)(0.92289 Btu/lbm ⋅ R ) = 1.3334 Btu/lbm ⋅ R
The final state is superheated vapor and
P
T2 = 500°F
⎫
⎬ s 2 = 1.5706 Btu/lbm ⋅ R (Table A - 6E)
P2 = P1 = 300 psia ⎭
1
2
Hence, the change in the total entropy is
∆S = m( s 2 − s1 )
v
= (2 lbm)(1.5706 − 1.3334) Btu/lbm ⋅ R
= 0.4744 Btu/R
7-32 An insulated rigid tank contains a saturated liquid-vapor mixture of water at a specified pressure. An electric heater
inside is turned on and kept on until all the liquid vaporized. The entropy change of the water during this process is to be
determined.
Analysis From the steam tables (Tables A-4 through A-6)
P1 = 150 kPa ⎫ v 1 = v f + x1v fg = 0.001053 + (0.25)(1.1594 − 0.001053) = 0.29065 m 3 /kg
⎬
x1 = 0.25
s1 = s f + x1 s fg = 1.4337 + (0.25)(5.7894 ) = 2.8810 kJ/kg ⋅ K
⎭
v 2 = v1
⎫
⎬ s 2 = 6.7298 kJ/kg ⋅ K
sat. vapor ⎭
H2O
5 kg
150 kPa
We
Then the entropy change of the steam becomes
∆S = m(s 2 − s1 ) = (5 kg )(6.7298 − 2.8810) kJ/kg ⋅ K = 19.2 kJ/K
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7-26
7-57 Steam enters a nozzle at a specified state and leaves at a specified pressure. The process is to be sketched on the T-s
diagram and the maximum outlet velocity is to be determined.
Analysis (b) The inlet state properties are
6000 kPa
T
P1 = 6000 kPa ⎫ h1 = 2784.6 kJ/kg
(Table A - 5)
⎬
x1 = 1
⎭ s1 = 5.8902 kJ/kg ⋅ K
1200 kPa
1
For the maximum velocity at the exit, the entropy will be constant
during the process. The exit state enthalpy is (Table A-6)
2
s2 − s f
5.8902 − 2.2159
=
= 0.8533
⎫
s fg
4.3058
⎬
s 2 = s1 = 5.8902 kJ/kg ⋅ K ⎭ h2 = h f + xh fg = 798.33 + 0.8533 × 1985.4 = 2492.5 kJ/kg
x2 =
P2 = 1200 kPa
s
We take the nozzle as the system, which is a control volume since mass crosses the boundary. Noting that one fluid stream
enters and leaves the nozzle, the energy balance for this steady-flow system can be expressed in the rate form as
E& − E&
1in424out
3
∆E& system ©0 (steady)
144
42444
3
=
Rate of net energy transfer
by heat, work, and mass
=0
Rate of change in internal, kinetic,
potential, etc. energies
E& in = E& out
⎛
⎛
V2 ⎞
V2 ⎞
m& ⎜ h1 + 1 ⎟ = m& ⎜ h2 + 2 ⎟
⎜
⎜
2 ⎟⎠
2 ⎟⎠
⎝
⎝
⎛ V 2 − V12
h1 − h2 = ⎜ 2
⎜
2
⎝
(since W& ≅ Q& ≅ ∆pe ≅ 0)
⎞
⎟
⎟
⎠
Solving for the exit velocity and substituting,
⎛ V 2 − V12
h1 − h2 = ⎜ 2
⎜
2
⎝
V2
= [V
2
1
+ 2(h1
⎞
⎟
⎟
⎠
− h )]
0.5
2
⎡
⎛ 1000 m 2 /s 2
= ⎢(0 m/s) 2 + 2(2784.6 − 2492.5) kJ/kg⎜
⎜ 1 kJ/kg
⎢⎣
⎝
⎞⎤
⎟⎥
⎟
⎠⎥⎦
0.5
= 764.3 m/s
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preparation. If you are a student using this Manual, you are using it without permission.
7-32
Entropy Change of Incompressible Substances
7-63C No, because entropy is not a conserved property.
7-64 A hot copper block is dropped into water in an insulated tank. The final equilibrium temperature of the tank and the
total entropy change are to be determined.
Assumptions 1 Both the water and the copper block are incompressible substances with constant specific heats at room
temperature. 2 The system is stationary and thus the kinetic and potential energies are negligible. 3 The tank is wellinsulated and thus there is no heat transfer.
Properties The density and specific heat of water at 25°C are ρ = 997 kg/m3 and cp = 4.18 kJ/kg.°C. The specific heat of
copper at 27°C is cp = 0.386 kJ/kg.°C (Table A-3).
Analysis We take the entire contents of the tank, water + copper block, as the system. This is a closed system since no mass
crosses the system boundary during the process. The energy balance for this system can be expressed as
E − Eout
1in
424
3
=
Net energy transfer
by heat, work, and mass
∆Esystem
1
424
3
Change in internal, kinetic,
potential, etc. energies
0 = ∆U
WATER
or,
Copper
75 kg
∆U Cu + ∆U water = 0
[mc(T2 − T1 )]Cu + [mc(T2 − T1 )]water = 0
160 L
where
m water = ρV = (997 kg/m 3 )(0.160 m 3 ) = 159.5 kg
Using specific heat values for copper and liquid water at room temperature and substituting,
(75 kg)(0.386 kJ/kg ⋅ °C)(T2 − 110)°C + (159.5 kg)(4.18 kJ/kg ⋅ °C)(T2 − 15)°C = 0
T2 = 19.0°C = 292 K
The entropy generated during this process is determined from
⎛T ⎞
⎛ 292.0 K ⎞
⎟⎟ = −7.85 kJ/K
∆S copper = mcavg ln⎜⎜ 2 ⎟⎟ = (75 kg )(0.386 kJ/kg ⋅ K ) ln⎜⎜
⎝ 383 K ⎠
⎝ T1 ⎠
⎛T ⎞
⎛ 292.0 K ⎞
⎟⎟ = 9.20 kJ/K
∆S water = mc avg ln⎜⎜ 2 ⎟⎟ = (159.5 kg )(4.18 kJ/kg ⋅ K ) ln⎜⎜
⎝ 288 K ⎠
⎝ T1 ⎠
Thus,
∆S total = ∆S copper + ∆S water = −7.85 + 9.20 = 1.35 kJ/K
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7-37
7-69 An iron block and a copper block are dropped into a large lake. The total amount of entropy change when both blocks
cool to the lake temperature is to be determined.
Assumptions 1 The water, the iron block and the copper block are incompressible substances with constant specific heats at
room temperature. 2 Kinetic and potential energies are negligible.
Properties The specific heats of iron and copper at room temperature are ciron = 0.45 kJ/kg.°C and ccopper = 0.386 kJ/kg.°C
(Table A-3).
Analysis The thermal-energy capacity of the lake is very large, and thus the temperatures of both the iron and the copper
blocks will drop to the lake temperature (15°C) when the thermal equilibrium is established. Then the entropy changes of
the blocks become
⎛T ⎞
⎛ 288 K ⎞
⎟⎟ = −4.579 kJ/K
∆Siron = mcavg ln⎜⎜ 2 ⎟⎟ = (50 kg )(0.45 kJ/kg ⋅ K )ln⎜⎜
T
⎝ 353 K ⎠
⎝ 1⎠
⎛T ⎞
⎛ 288 K ⎞
⎟⎟ = −1.571 kJ/K
∆Scopper = mcavg ln⎜⎜ 2 ⎟⎟ = (20 kg )(0.386 kJ/kg ⋅ K )ln⎜⎜
⎝ 353 K ⎠
⎝ T1 ⎠
We take both the iron and the copper blocks, as the system. This is a
closed system since no mass crosses the system boundary during the
process. The energy balance for this system can be expressed as
E − Eout
1in
424
3
=
Net energy transfer
by heat, work, and mass
Iron
50 kg
80°C
∆Esystem
1
424
3
Change in internal, kinetic,
potential, etc. energies
− Qout = ∆U = ∆U iron + ∆U copper
Lake
15°C
Copper
20 kg
80°C
or,
Qout = [mc(T1 − T2 )]iron + [mc(T1 − T2 )]copper
Substituting,
Qout = (50 kg )(0.45 kJ/kg ⋅ K )(353 − 288)K + (20 kg )(0.386 kJ/kg ⋅ K )(353 − 288)K
= 1964 kJ
Thus,
∆Slake =
Qlake,in
Tlake
=
1964 kJ
= 6.820 kJ/K
288 K
Then the total entropy change for this process is
∆S total = ∆Siron + ∆Scopper + ∆Slake = −4.579 − 1.571 + 6.820 = 0.670 kJ/K
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preparation. If you are a student using this Manual, you are using it without permission.
7-48
7-89 One side of a partitioned insulated rigid tank contains an ideal gas at a specified temperature and pressure while the
other side is evacuated. The partition is removed, and the gas fills the entire tank. The total entropy change during this
process is to be determined.
Assumptions The gas in the tank is given to be an ideal gas, and thus ideal gas relations apply.
Analysis Taking the entire rigid tank as the system, the energy balance can be expressed as
E − Eout
1in
424
3
=
Net energy transfer
by heat, work, and mass
∆Esystem
1
424
3
Change in internal, kinetic,
potential, etc. energies
0 = ∆U = m(u2 − u1 )
u2 = u1
IDEAL
GAS
5 kmol
40°C
T2 = T1
since u = u(T) for an ideal gas. Then the entropy change of the gas becomes
⎛
T ©0
V ⎞
V
∆S = N ⎜ cv ,avg ln 2 + Ru ln 2 ⎟ = NRu ln 2
⎜
⎟
V1 ⎠
V1
T1
⎝
= (5 kmol)(8.314 kJ/kmol ⋅ K ) ln (2)
= 28.81 kJ/K
This also represents the total entropy change since the tank does not contain anything else, and there are no interactions
with the surroundings.
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preparation. If you are a student using this Manual, you are using it without permission.
7-60
7-100 Air contained in a constant-volume tank s cooled to ambient temperature. The entropy changes of the air and the
universe due to this process are to be determined and the process is to be sketched on a T-s diagram.
Assumptions 1 Air is an ideal gas with constant specific heats.
Properties The specific heat of air at room temperature is cv = 0.718 kJ/kg.K
(Table A-2a).
Analysis (a) The entropy change of air is determined from
∆Sair = mcv ln
Air
5 kg
327°C
100 kPa
T2
T1
= (5 kg)(0.718 kJ/kg.K)ln
(27 + 273) K
(327 + 273) K
= −2.488 kJ/K
(b) An energy balance on the system gives
Q out = mcv (T1 − T2 )
= (5 kg)(0.718 kJ/kg.K)(327 − 27)
= 1077 kJ
T
327ºC
air
2
27ºC
The entropy change of the surroundings is
∆ssurr =
1
Qout 1077 kJ
=
= 3.59 kJ/K
Tsurr
300 K
1
surr
2
s
The entropy change of universe due to this process is
Sgen = ∆S total = ∆Sair + ∆Ssurr = −2.488 + 3.59 = 1.10 kJ/K
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7-66
7-110 The reversible work produced during the process shown in the figure is to be determined.
Assumptions The process is reversible.
Analysis The reversible work relation is
P
(kPa)
2
∫
w12 = vdP
600
1
2
When combined with the ideal gas equation of state
v=
200
RT
P
2
2
∫
1
∫
1
v (m3/kg)
0.002
The work expression reduces to
w12 = vdP = − RT
1
P
P
dP
= − RT ln 2 = − P2v 2 ln 2
P
P1
P1
= −(600 kPa)(0.002 m 3 /kg)ln
600 kPa ⎛ 1 kJ ⎞
⎜
⎟
200 kPa ⎝ 1 kPa ⋅ m 3 ⎠
= −1.32 kJ/kg
The negative sign indicates that work is done on the system in the amount of 1.32 kJ/kg.
7-111 Liquid water is to be pumped by a 25-kW pump at a specified rate. The highest pressure the water can be pumped to
is to be determined.
Assumptions 1 Liquid water is an incompressible substance. 2 Kinetic and potential energy changes are negligible. 3 The
process is assumed to be reversible since we will determine the limiting case.
Properties The specific volume of liquid water is given to be v1 = 0.001 m3/kg.
Analysis The highest pressure the liquid can have at the pump exit can be determined from the reversible steady-flow work
relation for a liquid,
Thus,
⎛
W&in = m& ⎜
⎝
∫
2
1
⎞
vdP + ∆ke©0 + ∆pe©0 ⎟ = m& v1 (P2 − P1 )
P2
⎠
⎛ 1 kJ ⎞
⎟
25 kJ/s = (5 kg/s)(0.001 m3/kg )( P2 − 100) k Pa ⎜⎜
3⎟
⎝ 1 kPa ⋅ m ⎠
25 kW
PUMP
It yields
P2 = 5100 kPa
100 kPa
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
preparation. If you are a student using this Manual, you are using it without permission.
7-100
7-144 Cold water is heated by hot water in a heat exchanger. The rate of heat transfer and the rate of entropy generation
within the heat exchanger are to be determined.
Assumptions 1 Steady operating conditions exist. 2 The heat exchanger is well-insulated so that heat loss to the
surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid. 3 Changes
in the kinetic and potential energies of fluid streams are negligible. 4 Fluid properties are constant.
Properties The specific heats of cold and hot water are given to be 4.18 and 4.19 kJ/kg.°C, respectively.
Analysis We take the cold water tubes as the system, which is a control volume.
The energy balance for this steady-flow system can be expressed in the rate form as
E& − E& out
1in
424
3
=
Rate of net energy transfer
by heat, work, and mass
∆E& systemÊ0 (steady)
1442443
Cold water
10°C
0.95 kg/s
=0
Rate of change in internal, kinetic,
potential, etc. energies
Hot water
E& in = E& out
Q& in + m& h1 = m& h2 (since ∆ke ≅ ∆pe ≅ 0)
Q& in = m& c p (T2 − T1 )
85°C
1.6 kg/s
70°C
Then the rate of heat transfer to the cold water in this heat exchanger becomes
Q& in = [ m& c p (Tout − Tin )] cold water = (0.95 kg/s)(4.18 kJ/kg. °C)(70°C − 10°C) = 238.3 kW
Noting that heat gain by the cold water is equal to the heat loss by the hot water, the outlet temperature of the hot water is
determined to be
Q&
238.3 kW
⎯→ Tout = Tin −
= 85°C −
= 49.5°C
Q& = [m& c p (Tin − Tout )]hot water ⎯
(1.6 kg/s)(4.19 kJ/kg.°C)
m& c p
(b) The rate of entropy generation within the heat exchanger is determined by applying the rate form of the entropy balance
on the entire heat exchanger:
S&in − S&out
1424
3
Rate of net entropy transfer
by heat and mass
+
S&gen
{
Rate of entropy
generation
= ∆S&system©0 (steady)
1442443
Rate of change
of entropy
m& 1s1 + m& 3s3 − m& 2 s2 − m& 3s4 + S&gen = 0 (since Q = 0)
m& cold s1 + m& hot s3 − m& cold s2 − m& hot s4 + S&gen = 0
S&gen = m& cold ( s2 − s1 ) + m& hot ( s4 − s3 )
Noting that both fluid streams are liquids (incompressible substances), the rate of entropy generation is determined to be
T
T
S& gen = m& cold c p ln 2 + m& hot c p ln 4
T3
T1
= (0.95 kg/s)(4.18 kJ/kg.K)ln
49.5 + 273
70 + 273
+ (1.6 kg/s)(4.19 kJ/kg.K)ln
85 + 273
10 + 273
= 0.06263 kW/K
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preparation. If you are a student using this Manual, you are using it without permission.
7-167
7-212 Two rigid tanks that contain water at different states are connected by a valve. The valve is opened and steam flows
from tank A to tank B until the pressure in tank A drops to a specified value. Tank B loses heat to the surroundings. The
final temperature in each tank and the entropy generated during this process are to be determined.
Assumptions 1 Tank A is insulated, and thus heat transfer is negligible. 2 The water that remains in tank A undergoes a
reversible adiabatic process. 3 The thermal energy stored in the tanks themselves is negligible. 4 The system is stationary
and thus kinetic and potential energy changes are negligible. 5 There are no work interactions.
Analysis (a) The steam in tank A undergoes a reversible, adiabatic process, and thus s2 = s1. From the steam tables (Tables
A-4 through A-6),
Tank A:
v 1, A = v f + x1v fg = 0.001084 + (0.6 )(0.46242 − 0.001084 ) = 0.27788 m 3 /kg
P1 = 400 kPa ⎫
⎬u1, A = u f + x1u fg = 604.22 + (0.6 )(1948.9) = 1773.6 kJ/kg
x1 = 0.6
⎭ s = s + x s = 1.7765 + (0.6 )(5.1191) = 4.8479 kJ/kg ⋅ K
1, A
f
1 fg
T2, A = Tsat @ 200 kPa = 120.2 °C
s 2, A − s f
4.8479 − 1.5302
P1 = 200 kPa ⎫ x
=
= 0.5928
⎪ 2, A =
s
5.59680
s 2 = s1
fg
⎬
(sat. mixture) ⎪⎭ v 2, A = v f + x 2, Av fg = 0.001061 + (0.5928)(0.8858 − 0.001061) = 0.52552 m 3 /kg
u 2, A = u f + x 2, A u fg = 504.50 + (0.5928)(2024.6 kJ/kg ) = 1704.7 kJ/kg
Tank B:
300 kJ
v 1, B = 1.1989 m 3 /kg
P1 = 200 kPa ⎫
⎬ u1, B = 2731.4 kJ/kg
T1 = 250°C ⎭
s1, B = 7.7100 kJ/kg ⋅ K
A
steam
V = 0.3 m3
P = 400 kPa
x = 0.6
The initial and the final masses in tank A are
m1, A =
VA
0.3 m 3
=
= 1.080 kg
v 1, A 0.27788 m 3 /kg
m 2, A =
VA
0.3 m 3
=
= 0.5709 kg
v 2, A 0.52552 m 3 /kg
and
×
B
steam
m = 2 kg
T = 250°C
P = 200 kPa
Thus, 1.080 - 0.5709 = 0.5091 kg of mass flows into tank B. Then,
m 2, B = m1, B + 0.5091 = 2 + 0.5091 = 2.509 kg
The final specific volume of steam in tank B is determined from
v 2, B =
VB
m 2, B
=
(m1v 1 )B
m 2, B
=
(2 kg )(1.1989 m 3 /kg ) = 0.9558 m 3 /kg
2.509 kg
We take the entire contents of both tanks as the system, which is a closed system. The energy balance for this stationary
closed system can be expressed as
E −E
1in424out
3
Net energy transfer
by heat, work, and mass
=
∆E system
1
424
3
Change in internal, kinetic,
potential, etc. energies
− Qout = ∆U = (∆U ) A + (∆U ) B
(since W = KE = PE = 0)
− Qout = (m 2 u 2 − m1u1 ) A + (m 2 u 2 − m1u1 ) B
Substituting,
{
}
− 300 = {(0.5709 )(1704.7 ) − (1.080 )(1773.6 )} + (2.509 )u 2, B − (2 )(2731.4 )
u 2, B = 2433.3 kJ/kg
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7-168
Thus,
T2, B = 116.1 °C
v 2, B = 0.9558 m 3 /kg ⎫⎪
⎬
s
u 2, B = 2433.3 kJ/kg ⎭⎪ 2, B = 6.9156 kJ/kg ⋅ K
(b) The total entropy generation during this process is determined by applying the entropy balance on an extended system
that includes both tanks and their immediate surroundings so that the boundary temperature of the extended system is the
temperature of the surroundings at all times. It gives
S − S out
1in424
3
Net entropy transfer
by heat and mass
−
+ S gen = ∆S system
{
1
424
3
Entropy
generation
Change
in entropy
Qout
+ S gen = ∆S A + ∆S B
Tb,surr
Rearranging and substituting, the total entropy generated during this process is determined to be
S gen = ∆S A + ∆S B +
Qout
Q
= (m 2 s 2 − m1 s1 ) A + (m 2 s 2 − m1 s1 ) B + out
Tb,surr
Tb,surr
= {(0.5709)(4.8479) − (1.080 )(4.8479)} + {(2.509)(6.9156) − (2 )(7.7100)} +
300 kJ
290 K
= 0.498 kJ/K
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
preparation. If you are a student using this Manual, you are using it without permission.
Tutorial #10
ESO 201A
Thermodynamics
Instructor
Prof. Sameer Khandekar
Contact
Office: SL-109, Tel: 7038
E-mail: [email protected]
URL: home.iitk.ac.in/~samkhan
ESO 201A Thermodynamics
Instructor: Sameer Khandekar
Tutorial 10
[8-23] A house that is losing heat at a rate of 50,000 kJ/hwhen the outside temperature drops
to 4°C is to be heated byelectric resistance heaters. If the house is to be maintained at 25°C at
all times, determine the reversible work input for thisprocess and the irreversibility. [Answers:
0.978 kW, 12.91 kW]
[8-31] Which has the capability to produce the most work in a closed system — 1 kg of steam
at 800 kPa and 180°C or 1 kgof R-134a at 800 kPa and 180°C? Take T0 = 25°C and P0 = 100
kPa. [Answers: 623 kJ, 5.0 kJ]
[8-40] A piston-cylinder device initially contains 2 L of air at100 kPa and 25°C. Air is now
compressed to a final state of600 kPa and 150°C. The useful work input is 1.2 kJ. Assuming the
surroundings are at 100 kPa and 25°C, determine (a) the energy of the air at the initial and the
final states, (b) the minimum work that must be supplied to accomplish this compression
process, and (c) the second-law efficiency of this process. Answers: (a) 0, 0.171 kJ, (b) 0.171 kJ,
(c) 14.3 percent
[8-45] An iron block of unknown mass at 85°C is dropped into an insulated tank that contains
100 L of water at 20°C.At the same time, a paddle wheel driven by a 200-W motor is activated
to stir the water. It is observed that thermal equilibrium is established after 20 min with a final
temperature of24°C. Assuming the surroundings to be at 20°C, determine (a) the mass of the
iron block and (b) the energy destroyed during this process. Answers: (a) 52.0 kg, (b) 375 kJ
Additional Homework Problems (Tutorial 10)
[8-19] Consider a thermal energy reservoir at 1500 K that can supply heat at a rate of
150,000 kJ/h. Determine the energy of this supplied energy, assuming an environmental
temperature of 25°C.
[8-34] A rigid tank is divided into two equal parts by a partition. One part of the tank contains 4
kg of compressed liquid water at 200 kPa and 80°C and the other side is evacuated. Now the
partition is removed, and the water expands to fill the entire tank. If the final pressure in the
tank is 40 kPa, determine the energy destroyed during this process. Assume the surroundings to
be at 25°C and 100 kPa.
[8-43] An insulated rigid tank is divided into two equal parts by a partition. Initially, one part
contains 3 kg of argon gas at300 kPa and 70°C, and the other side is evacuated. The partition is
now removed, and the gas fills the entire tank. Assuming the surroundings to be at 25 °C,
determine the energy destroyed during this process. [Answer: 129 kJ]
[8-46] A 50-kg iron block and a 20-kg copper block, both initially at 80°C, are dropped into a
large lake at 15°C. Thermal equilibrium is established after a while as a result of heat transfer
between the blocks and the lake water. Assuming the surroundings to be at 20°C, determine
the amount of work that could have been produced if the entire process were executed in a
reversible manner.
[8-52] Refrigerant-134a enters an expansion valve at 1200 k Paas a saturated liquid and leaves
at 200 kPa. Determine (a) the temperature of R-134a at the outlet of the expansion valve and
(b) the entropy generation and the energy destruction during this process. Take T 0 = 25°C.
[8-53] Helium is expanded in a turbine from 1500 kPa and300°C to 100 kPa and 25 °C.
Determine the maximum work this turbine can produce, in kJ/kg. Does the maximum work
require an adiabatic turbine?
[8-61] Steam enters an adiabatic turbine at 6 MPa, 600°C, and 80m/s and leaves at 50 kPa,
100°C, and 140 m/s. If the power output of the turbine is 5 MW, determine (a) the
reversible power output and (b) the second-law efficiency of the turbine. Assume the
surroundings to be at 25°C. [Answers:(a) 5.81 MW. (b) 86.1 percent]
[8-62] Steam is throttled from 6 MPa and 400°C to a pressure of 2 MPa. Determine the
decrease in energy of the steam during this process. Assume the surroundings to be at 25°
C. [Answer: 143 kJ/kg]
[8-86] Liquid water at 200 kPa and 15°C is heated in a chamber by mixing it with superheated
steam at 200 kPa and200°C. Liquid water enters the mixing chamber at a rate of 4 kg/s, and the
chamber is estimated to lose heat to the surrounding air at 25°C at a rate of 600 kJ/min. If the
mixture leaves the mixing chamber at 200 kPa and 80°C, determine(a) the mass flow rate of the
superheated steam and (b) the wasted work potential during this mixing process.
[8-88] A well-insulated shell-and-tube heat exchanger is used to heat water (c p = 4.18 kJ/kg-°C)
in the tubes from 20 to 70°C at a rate of 4.5 kg/s. Heat is supplied by hot oil (c p = 2.30 kJ/kg-°C)
that enters the shell side at 170°C at a rate of 10 kg/s. Disregarding any heat loss from the heat
exchanger, determine (a) the exit temperature of oil and (b) the rate of energy destruction in
the heat exchanger. Take T 0 = 25 °C.
[8-92] Liquid water at 15°C is heated in a chamber by mixing it with saturated steam. Liquid
water enters the chamber at the steam pressure at a rate of 4.6 kg/s and the saturated steam
enters at a rate of 0.23 kg/s. The mixture leaves the mixing chamber as a liquid at 45°C. If the
surroundings are at 15°C, determine (a) the temperature of saturated steam entering the
chamber, (b) the energy destruction during this mixing process, and (c) the second-law
efficiency of the mixing chamber. [Answers: (a) 114.3°C, (b) 114.7 kW, (c) 0.207]
8-6
8-19 A heat reservoir at a specified temperature can supply heat at a specified rate. The exergy of this heat supplied is to be
determined.
Analysis The exergy of the supplied heat, in the rate form, is the amount
of power that would be produced by a reversible heat engine,
T0
298 K
= 1−
= 0.8013
1500 K
TH
= W& rev,out = η th, rev Q& in
η th, max = η th, rev = 1 −
Exergy = W& max,out
= (0.8013)(150,000 / 3600 kJ/s)
1500 K
&
W
rev
HE
298 K
= 33.4 kW
A heat engine receives heat from a source at a specified temperature at a specified rate, and rejects the waste
8-20
heat to a sink. For a given power output, the reversible power, the rate of irreversibility, and the 2nd law efficiency are to be
determined.
Analysis (a) The reversible power is the power produced by a reversible heat engine operating between the specified
temperature limits,
η th,max = η th,rev = 1 −
W& rev,out
TL
320 K
=1−
= 0.7091
TH
1100 K
1100 K
= η th,rev Q& in = (0.7091)(400 kJ/s) = 283.6 kW
(b) The irreversibility rate is the difference between the reversible
power and the actual power output:
400 kJ/s
HE
120 kW
I& = W& rev,out − W& u,out = 283.6 − 120 = 163.6 kW
(c) The second law efficiency is determined from its definition,
η II =
Wu,out
Wrev,out
=
320 K
120 kW
= 0.423 = 42.3%
283.6 kW
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
preparation. If you are a student using this Manual, you are using it without permission.
8-9
8-22E The thermal efficiency and the second-law efficiency of a heat engine are given. The source temperature is to be
determined.
TH
Analysis From the definition of the second law efficiency,
η II =
η th
η
0.36
⎯
⎯→ η th, rev = th =
= 0.60
η II 0.60
η th, rev
HE
Thus,
η th, rev = 1 −
TL
⎯
⎯→ T H = T L /(1 − η th, rev ) = (530 R)/0.40 = 1325 R
TH
η th = 36%
η II = 60%
530 R
8-23 A house is maintained at a specified temperature by electric resistance heaters. The reversible work for this heating
process and irreversibility are to be determined.
Analysis The reversible work is the minimum work required to accomplish this process, and the irreversibility is the
difference between the reversible work and the actual electrical work consumed. The actual power input is
W& in = Q& out = Q& H = 50,000 kJ/h = 13.89 kW
50,000 kJ/h
The COP of a reversible heat pump operating between the
specified temperature limits is
COPHP,rev =
1
1
=
= 14.20
1 − TL / TH 1 − 277.15 / 298.15
House
25 °C
4 °C
Thus,
W& rev,in =
Q& H
13.89 kW
=
= 0.978 kW
COPHP, rev
14.20
and
·
W
I& = W& u,in − W& rev,in = 13.89 − 0.978 = 12.91 kW
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
preparation. If you are a student using this Manual, you are using it without permission.
8-16
8-31 Steam and R-134a at the same states are considered. The fluid with the higher exergy content is to be identified.
Assumptions Kinetic and potential energy changes are negligible.
Analysis The properties of water at the given state and at the dead state are
u = 2594.7 kJ/kg
P = 800 kPa ⎫
3
⎬ v = 0.24720 m /kg
T = 180°C ⎭
s = 6.7155 kJ/kg ⋅ K
Steam
1 kg
800 kPa
180°C
(Table A - 6)
u 0 ≅ u f @ 25°C = 104.83 kJ/kg
⎫
3
⎬ v 0 ≅ v f @ 25°C = 0.001003 m /kg
P0 = 100 kPa ⎭
s 0 ≅ s f @ 25°C = 0.3672 kJ/kg ⋅ K
T0 = 25°C
(Table A - 4)
The exergy of steam is
Φ = m[u − u 0 + P0 (v − v 0 ) − T0 ( s − s 0 )]
⎡
⎛ 1 kJ
3
⎢(2594.7 − 104.83)kJ/kg + (100 kPa)(0.24720 − 0.001003)m /kg⎜
= (1 kg) ⎢
⎝ 1 kPa ⋅ m 3
⎢⎣− (298 K)(6.7155 − 0.3672)kJ/kg ⋅ K
⎞⎤
⎟⎥
⎠⎥
⎥⎦
= 622.7 kJ
For R-134a;
u = 386.99 kJ/kg
P = 800 kPa ⎫
3
⎬ v = 0.044554 m /kg
T = 180°C ⎭
s = 1.3327 kJ/kg ⋅ K
(Table A - 13)
u 0 ≅ u f @ 25°C = 85.85 kJ/kg
⎫
3
⎬ v 0 ≅ v f @ 25°C = 0.0008286 m /kg
P0 = 100 kPa ⎭
s 0 ≅ s f @ 25°C = 0.32432 kJ/kg ⋅ K
T0 = 25°C
(Table A - 11)
R-134a
1 kg
800 kPa
180°C
Φ = m[u − u 0 + P0 (v − v 0 ) − T0 ( s − s 0 )]
⎡
⎛ 1 kJ
3
⎢(386.99 − 85.85)kJ/kg + (100 kPa)(0.044554 − 0.0008286)m /kg⎜
= (1 kg) ⎢
⎝ 1 kPa ⋅ m 3
⎢⎣− (298 K)(1.3327 − 0.32432)kJ/kg ⋅ K
⎞⎤
⎟⎥
⎠⎥
⎥⎦
= 5.02 kJ
The steam can therefore has more work potential than the R-134a.
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
preparation. If you are a student using this Manual, you are using it without permission.
8-19
8-34 A rigid tank is divided into two equal parts by a partition. One part is filled with compressed liquid while the other
side is evacuated. The partition is removed and water expands into the entire tank. The exergy destroyed during this process
is to be determined.
Assumptions Kinetic and potential energies are negligible.
Analysis The properties of the water are (Tables A-4 through A-6)
v 1 ≅ v f @ 80°C = 0.001029 m 3 / kg
P1 = 200 kPa ⎫
⎬ u1 ≅ u f @ 80°C = 334.97 kJ/kg
T1 = 80°C ⎭
s1 ≅ s f @ 80°C = 1.0756 kJ/kg ⋅ K
4 kg
200 kPa
80°C
WATER
Vacuum
Noting that v 2 = 2v 1 = 2 × 0.001029 = 0.002058 m 3 / kg ,
v2 −v f
0.002058 − 0.001026
= 0.0002584
v fg
3.9933 − 0.001026
P2 = 40 kPa
⎫⎪
⎬ u 2 = u f + x 2 u fg = 317.58 + 0.0002584 × 2158.8 = 318.14 kJ/kg
v 2 = 0.002058 m 3 / kg ⎪⎭
s 2 = s f + x 2 s fg = 1.0261 + 0.0002584 × 6.6430 = 1.0278 kJ/kg ⋅ K
x2 =
=
Taking the direction of heat transfer to be to the tank, the energy balance on this closed system becomes
E −E
1in424out
3
=
Net energy transfer
by heat, work, and mass
∆E system
1
424
3
Change in internal, kinetic,
potential, etc. energies
Qin = ∆U = m(u 2 − u1 )
or
Qin = (4 kg)(318.14 − 334.97)kJ/kg = −67.30 kJ → Qout = 67.30 kJ
The irreversibility can be determined from its definition Xdestroyed = T0Sgen where the entropy generation is determined from
an entropy balance on an extended system that includes the tank and its immediate surroundings so that the boundary
temperature of the extended system is the temperature of the surroundings at all times,
S in − S out
1424
3
Net entropy transfer
by heat and mass
−
+ S gen = ∆S system
{
1
424
3
Entropy
generation
Change
in entropy
Qout
+ S gen = ∆S system = m( s 2 − s1 )
Tb,out
S gen = m( s 2 − s1 ) +
Qout
Tsurr
Substituting,
⎛
Q
X destroyed = T0 S gen = T0 ⎜⎜ m( s 2 − s1 ) + out
Tsurr
⎝
⎞
⎟
⎟
⎠
67.30 kJ ⎤
⎡
= (298 K) ⎢(4 kg)(1.0278 − 1.0756)kJ/kg ⋅ K +
298 K ⎥⎦
⎣
= 10.3 kJ
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
preparation. If you are a student using this Manual, you are using it without permission.
8-25
8-40 A cylinder initially contains air at atmospheric conditions. Air is compressed to a specified state and the useful work
input is measured. The exergy of the air at the initial and final states, and the minimum work input to accomplish this
compression process, and the second-law efficiency are to be determined
Assumptions 1 Air is an ideal gas with constant specific heats. 2 The kinetic and potential energies are negligible.
Properties The gas constant of air is R = 0.287 kPa.m3/kg.K (Table A-1). The specific heats of air at the average
temperature of (298+423)/2=360 K are cp = 1.009 kJ/kg·K and cv = 0.722 kJ/kg·K (Table A-2).
Analysis (a) We realize that X1 = Φ1 = 0 since air initially is at the dead state. The mass of air is
P1V1
(100 kPa)(0.002 m 3 )
=
= 0.00234 kg
RT1 (0.287 kPa ⋅ m 3 / kg ⋅ K)(298 K)
m=
Also,
P2V 2 P1V1
PT
(100 kPa)(423 K)
=
⎯
⎯→ V 2 = 1 2 V 1 =
(2 L) = 0.473 L
T2
T1
P2 T1
(600 kPa)(298 K)
AIR
V1 = 2 L
and
s 2 − s 0 = c p ,avg
P1 = 100 kPa
T1 = 25°C
P
T
ln 2 − R ln 2
P0
T0
= (1.009 kJ/kg ⋅ K) ln
423 K
600 kPa
− (0.287 kJ/kg ⋅ K) ln
298 K
100 kPa
= −0.1608 kJ/kg ⋅ K
Thus, the exergy of air at the final state is
[
]
X 2 = Φ 2 = m cv ,avg (T2 − T0 ) − T0 ( s 2 − s 0 ) + P0 (V 2 −V 0 )
= (0.00234 kg)[(0.722 kJ/kg ⋅ K)(423 - 298)K - (298 K)(-0.1608 kJ/kg ⋅ K)]
+ (100 kPa)(0.000473 - 0.002) m 3 [kJ/m 3 ⋅ kPa]
= 0.171 kJ
(b) The minimum work input is the reversible work input, which can be determined from the exergy balance by setting the
exergy destruction equal to zero,
X − X out
1in
4243
Net exergy transfer
by heat, work,and mass
− X destroyed 0 (reversible) = ∆X system
144424443 1
424
3
Exergy
destruction
Change
in exergy
Wrev,in = X 2 − X1
= 0.171− 0 = 0.171kJ
(c) The second-law efficiency of this process is
η II =
Wrev,in
Wu,in
=
0.171 kJ
= 14.3%
1.2 kJ
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
preparation. If you are a student using this Manual, you are using it without permission.
8-28
8-43 One side of a partitioned insulated rigid tank contains argon gas at a specified temperature and pressure while the
other side is evacuated. The partition is removed, and the gas fills the entire tank. The exergy destroyed during this process
is to be determined.
Assumptions Argon is an ideal gas with constant specific heats, and thus ideal gas relations apply.
Properties The gas constant of argon is R = 0.2081 kJ/kg.K (Table A-1).
Analysis Taking the entire rigid tank as the system, the energy balance can be expressed as
E − Eout
1in
424
3
=
Net energy transfer
by heat, work, and mass
∆Esystem
1
424
3
Change in internal, kinetic,
potential, etc. energies
0 = ∆U = m(u2 − u1 )
u2 = u1
→ T2 = T1
Argon
300 kPa
70°C
Vacuum
since u = u(T) for an ideal gas.
The exergy destruction (or irreversibility) associated with this process can be determined from its definition
Xdestroyed = T0Sgen where the entropy generation is determined from an entropy balance on the entire tank, which is an
insulated closed system,
S − S out
1in424
3
Net entropy transfer
by heat and mass
+ S gen = ∆S system
{
1
424
3
Entropy
generation
Change
in entropy
S gen = ∆S system = m( s 2 − s1 )
where
⎛
T 0
V ⎞
V
∆Ssystem = m( s2 − s1 ) = m⎜ cv , avg ln 2 + R ln 2 ⎟ = mR ln 2
⎜
⎟
T
V
V1
1
1 ⎠
⎝
= (3 kg)(0.2081 kJ/kg ⋅ K) ln (2) = 0.433 kJ/K
Substituting,
X destroyed = T0 S gen = mT0 ( s 2 − s1 ) = (298 K)(0.433 kJ/K) = 129 kJ
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
preparation. If you are a student using this Manual, you are using it without permission.
8-30
8-45 A hot iron block is dropped into water in an insulated tank that is stirred by a paddle-wheel. The mass of the iron
block and the exergy destroyed during this process are to be determined. √
Assumptions 1 Both the water and the iron block are incompressible substances with constant specific heats at room
temperature. 2 The system is stationary and thus the kinetic and potential energies are negligible. 3 The tank is wellinsulated and thus there is no heat transfer.
Properties The density and specific heat of water at 25°C are ρ = 997 kg/m3 and cp = 4.18 kJ/kg.°F. The specific heat of
iron at room temperature (the only value available in the tables) is cp = 0.45 kJ/kg.°C (Table A-3).
Analysis We take the entire contents of the tank, water + iron block, as the system, which is a closed system. The energy
balance for this system can be expressed as
E −E
1in424out
3
=
Net energy transfer
by heat, work, and mass
∆E system
1
424
3
Change in internal, kinetic,
potential, etc. energies
W pw,in = ∆U = ∆U iron + ∆U water
100 L
20°C
Iron
85°C
Wpw,in = [mc(T2 − T1 )]iron + [mc(T2 − T1 )] water
where
m water = ρV = (997 kg/m 3 )(0.1 m 3 ) = 99.7 kg
Wpw = W& pw,in ∆t = (0.2 kJ/s)(20 × 60 s) = 240 kJ
Wpw
Water
Substituting,
240 kJ = m iron (0.45 kJ/kg ⋅ °C)(24 − 85)°C + (99.7 kg)(4.18 kJ/kg ⋅ °C)(24 − 20)°C
m iron = 52.0 kg
(b) The exergy destruction (or irreversibility) can be determined from its definition Xdestroyed = T0Sgen where the entropy
generation is determined from an entropy balance on the system, which is an insulated closed system,
S in − S out
1424
3
Net entropy transfer
by heat and mass
+ S gen = ∆S system
{
1
424
3
Entropy
generation
Change
in entropy
S gen = ∆S system = ∆S iron + ∆S water
where
⎛T ⎞
⎛ 297 K ⎞
∆S iron = mc avg ln⎜⎜ 2 ⎟⎟ = (52.0 kg)(0.45 kJ/kg ⋅ K) ln⎜
⎟ = −4.371 kJ/K
⎝ 358 K ⎠
⎝ T1 ⎠
⎛T ⎞
⎛ 297 K ⎞
∆S water = mc avg ln⎜⎜ 2 ⎟⎟ = (99.7 kg)(4.18 kJ/kg ⋅ K) ln⎜
⎟ = 5.651 kJ/K
T
⎝ 293 K ⎠
⎝ 1⎠
Substituting,
X destroyed = T0 S gen = (293 K)(−4.371 + 5.651) kJ/K = 375.0 kJ
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
preparation. If you are a student using this Manual, you are using it without permission.
8-31
8-46 An iron block and a copper block are dropped into a large lake where they cool to lake temperature. The amount of
work that could have been produced is to be determined.
Assumptions 1 The iron and copper blocks and water are incompressible substances with constant specific heats at room
temperature. 2 Kinetic and potential energies are negligible.
Properties The specific heats of iron and copper at room temperature are cp, iron = 0.45 kJ/kg.°C and cp,copper = 0.386
kJ/kg.°C (Table A-3).
Analysis The thermal-energy capacity of the lake is very large, and thus the temperatures of both the iron and the copper
blocks will drop to the lake temperature (15°C) when the thermal equilibrium is established.
We take both the iron and the copper blocks as the system, which is a closed system. The energy balance for this
system can be expressed as
E −E
1in424out
3
=
Net energy transfer
by heat, work, and mass
∆E system
1
424
3
Lake
15°C
Change in internal, kinetic,
potential, etc. energies
− Qout = ∆U = ∆U iron + ∆U copper
or,
Iron
85°C
Copper
Iron
Qout = [mc(T1 − T2 )]iron + [mc(T1 − T2 )]copper
Substituting,
Qout = (50 kg )(0.45 kJ/kg ⋅ K )(353 − 288)K + (20 kg )(0.386 kJ/kg ⋅ K )(353 − 288)K
= 1964 kJ
The work that could have been produced is equal to the wasted work potential. It is equivalent to the exergy destruction (or
irreversibility), and it can be determined from its definition Xdestroyed = T0Sgen . The entropy generation is determined from an
entropy balance on an extended system that includes the blocks and the water in their immediate surroundings so that the
boundary temperature of the extended system is the temperature of the lake water at all times,
S in − S out
1424
3
Net entropy transfer
by heat and mass
−
+ S gen = ∆S system
{
1
424
3
Entropy
generation
Change
in entropy
Qout
+ S gen = ∆S system = ∆S iron + ∆S copper
Tb,out
S gen = ∆S iron + ∆S copper +
Qout
Tlake
where
⎛T ⎞
⎛ 288 K ⎞
⎟⎟ = −4.579 kJ/K
∆S iron = mc avg ln⎜⎜ 2 ⎟⎟ = (50 kg )(0.45 kJ/kg ⋅ K ) ln⎜⎜
T
⎝ 353 K ⎠
⎝ 1⎠
⎛T ⎞
⎛ 288 K ⎞
⎟⎟ = −1.571 kJ/K
∆S copper = mc avg ln⎜⎜ 2 ⎟⎟ = (20 kg )(0.386 kJ/kg ⋅ K ) ln⎜⎜
T
⎝ 353 K ⎠
⎝ 1⎠
Substituting,
⎛
1964 kJ ⎞
⎟kJ/K = 196 kJ
X destroyed = T0 S gen = (293 K)⎜⎜ − 4.579 − 1.571 +
288 K ⎟⎠
⎝
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preparation. If you are a student using this Manual, you are using it without permission.
8-37
Exergy Analysis of Control Volumes
8-52 R-134a is is throttled from a specified state to a specified pressure. The temperature of R-134a at the outlet of the
expansion valve, the entropy generation, and the exergy destruction are to be determined.
Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potential energy changes are
negligible. 3 Heat transfer is negligible.
(a) The properties of refrigerant at the inlet and exit states of the throttling valve are (from R134a tables)
P1 = 1200 kPa ⎫ h1 = 117.77 kJ/kg
⎬
x1 = 0
⎭ s1 = 0.4244 kJ/kg ⋅ K
P2 = 200 kPa
⎫ T2 = −10.1°C
⎬
h2 = h1 = 117.77 kJ/kg ⎭ s 2 = 0.4562 kJ/kg ⋅ K
(b) Noting that the throttling valve is adiabatic, the entropy generation is determined from
s gen = s 2 − s1 = (0.4562 − 0.4244 )kJ/kg ⋅ K = 0.03176 kJ/kg ⋅ K
Then the irreversibility (i.e., exergy destruction) of the process becomes
exdest = T0 s gen = (298 K)(0.03176 kJ/kg ⋅ K) = 9.464 kJ/kg
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preparation. If you are a student using this Manual, you are using it without permission.
8-38
8-53 Heium expands in an adiabatic turbine from a specified inlet state to a specified exit state. The maximum work output
is to be determined.
Assumptions 1 This is a steady-flow process since there is no change with time. 2 The device is adiabatic and thus heat
transfer is negligible. 3 Helium is an ideal gas. 4 Kinetic and potential energy changes are negligible.
Properties The properties of helium are cp = 5.1926 kJ/kg.K and R = 2.0769 kJ/kg.K (Table A-1).
Analysis The entropy change of helium is
s 2 − s1 = c p ln
T2
P
− R ln 2
T1
P1
= (5.1926 kJ/kg ⋅ K) ln
= 2.2295 kJ/kg ⋅ K
1500 kPa
300°C
298 K
100 kPa
− (2.0769 kJ/kg ⋅ K) ln
573 K
1500 kPa
Helium
The maximum (reversible) work is the exergy difference between the inlet and exit states
wrev,out = h1 − h2 − T0 ( s1 − s 2 )
100 kPa
25°C
= c p (T1 − T2 ) − T0 ( s1 − s 2 )
= (5.1926 kJ/kg ⋅ K)(300 − 25)K − (298 K)(−2.2295 kJ/kg ⋅ K)
= 2092 kJ/kg
There is only one inlet and one exit, and thus m& 1 = m& 2 = m& . We take the turbine as the system, which is a control volume
since mass crosses the boundary. The energy balance for this steady-flow system can be expressed in the rate form as
E& − E&
1in424out
3
=
Rate of net energy transfer
by heat, work, and mass
∆E& system 0 (steady)
1442443
=0
Rate of change in internal, kinetic,
potential, etc. energies
E& in = E& out
m& h1 = W& out + Q& out + m& h2
W& = m& (h − h ) − Q&
out
1
2
out
wout = (h1 − h2 ) − q out
Inspection of this result reveals that any rejection of heat will decrease the work that will be produced by the turbine since
inlet and exit states (i.e., enthalpies) are fixed.
If there is heat loss from the turbine, the maximum work output is determined from the rate form of the exergy
balance applied on the turbine and setting the exergy destruction term equal to zero,
X& − X&
1in424out
3
Rate of net exergy transfer
by heat, work,and mass
− X& destroyed 0 (reversible) = ∆X& system 0 (steady) = 0
144424443 144
42444
3
Rate of exergy
destruction
Rate of change
of exergy
X& in = X& out
⎛ T ⎞
m& ψ 1 = W& rev,out + Q& out ⎜⎜1 − 0 ⎟⎟ + m& ψ 2
⎝ T ⎠
⎛ T ⎞
wrev,out = (ψ 1 −ψ 2 ) − q out ⎜⎜1 − 0 ⎟⎟
⎝ T ⎠
⎛ T
= (h1 − h2 ) − T0 (s1 − s 2 ) − q out ⎜⎜1 − 0
⎝ T
⎞
⎟⎟
⎠
Inspection of this result reveals that any rejection of heat will decrease the maximum work that could be produced by the
turbine. Therefore, for the maximum work, the turbine must be adiabatic.
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
preparation. If you are a student using this Manual, you are using it without permission.
8-49
8-61 Steam expands in a turbine from a specified state to another specified state. The actual power output of the turbine is
given. The reversible power output and the second-law efficiency are to be determined.
Assumptions 1 This is a steady-flow process since there is no change with time. 2 Potential energy change is negligible. 3
The temperature of the surroundings is given to be 25°C.
Properties From the steam tables (Tables A-4 through A-6)
P1 = 6 MPa ⎫ h1 = 3658.8 kJ/kg
⎬
T1 = 600°C ⎭ s1 = 7.1693 kJ/kg ⋅ K
P2 = 50 kPa ⎫ h2 = 2682.4 kJ/kg
⎬
T2 = 100°C ⎭ s 2 = 7.6953 kJ/kg ⋅ K
&1 = m
&2 = m
& . We take the turbine as the system, which is a
Analysis (b) There is only one inlet and one exit, and thus m
control volume since mass crosses the boundary. The energy balance for this steady-flow system can be expressed in the
rate form as
E& − E&
1in424out
3
=
Rate of net energy transfer
by heat, work, and mass
∆E& system 0 (steady)
1442444
3
=0
80 m/s
6 MPa
600°C
Rate of change in internal, kinetic,
potential, etc. energies
E& in = E& out
m& (h1 + V12 / 2) = W& out + m& (h2 + V 22 / 2)
W& out
STEAM
⎡
V 2 − V 22 ⎤
= m& ⎢h1 − h2 + 1
⎥
2
⎢⎣
⎥⎦
5 MW
Substituting,
⎛
(80 m/s) 2 − (140 m/s) 2
5000 kJ/s = m& ⎜ 3658.8 − 2682.4 +
⎜
2
⎝
m& = 5.156 kg/s
⎛ 1 kJ/kg
⎜
⎝ 1000 m 2 / s 2
⎞ ⎞⎟
⎟⎟
⎠⎠
50 kPa
100°C
140 m/s
The reversible (or maximum) power output is determined from the rate form of the exergy balance applied on the turbine
and setting the exergy destruction term equal to zero,
X& − X&
1in424out
3
Rate of net exergy transfer
by heat, work,and mass
− X& destroyed 0 (reversible) = ∆X& system 0 (steady) = 0
144424443 144
42444
3
Rate of exergy
destruction
Rate of change
of exergy
X& in = X& out
& ψ = W&
m
1
rev,out
&ψ2
+m
& (ψ 1 −ψ 2 ) = m
& [(h1 − h2 ) − T0 ( s1 − s2 ) − ∆ke − ∆pe 0 ]
W& rev,out = m
Substituting,
W& rev, out = W& out − m& T0 ( s1 − s2 )
= 5000 kW − (5.156 kg/s)(298 K)(7.1693 − 7.6953) kJ/kg ⋅ K = 5808 kW
(b) The second-law efficiency of a turbine is the ratio of the actual work output to the reversible work,
W& out
η II = &
W
rev, out
=
5 MW
= 86.1%
5.808 MW
Discussion Note that 13.9% percent of the work potential of the steam is wasted as it flows through the turbine during this
process.
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
preparation. If you are a student using this Manual, you are using it without permission.
8-50
8-62 Steam is throttled from a specified state to a specified pressure. The decrease in the exergy of the steam during this
throttling process is to be determined.
Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potential energy changes are
negligible. 3 The temperature of the surroundings is given to be 25°C. 4 Heat transfer is negligible.
Properties The properties of steam before and after throttling are (Tables A-4 through A-6)
P1 = 6 MPa ⎫ h1 = 3178.3 kJ/kg
⎬
T1 = 400°C ⎭ s1 = 6.5432 kJ/kg ⋅ K
Steam
1
P2 = 2 MPa ⎫
⎬ s 2 = 7.0225 kJ/kg ⋅ K
h2 = h1
⎭
Analysis The decrease in exergy is of the steam is the difference between the inlet and
exit flow exergies,
Decrease in exergy = ψ 1 − ψ 2 = −[∆h
0
− ∆ke
0
− ∆pe
0
2
− T0 ( s1 − s 2 )] = T0 ( s 2 − s1 )
= (298 K)(7.0225 − 6.5432)kJ/kg ⋅ K
= 143 kJ/kg
Discussion Note that 143 kJ/kg of work potential is wasted during this throttling process.
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
preparation. If you are a student using this Manual, you are using it without permission.
8-77
8-86 Liquid water is heated in a chamber by mixing it with superheated steam. For a specified mixing temperature, the mass
flow rate of the steam and the rate of exergy destruction are to be determined.
Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potential energy changes are
negligible. 3 There are no work interactions.
Properties Noting that T < Tsat @ 200 kPa = 120.23°C, the cold water and the exit mixture streams exist as a compressed liquid,
which can be approximated as a saturated liquid at the given temperature. From Tables A-4 through A-6,
P1 = 200 kPa ⎫ h1 ≅ h f @15o C = 62.98 kJ/kg
⎬s ≅ s
= 0.22447 kJ/kg ⋅ K
T1 = 15°C
f @15o C
⎭ 1
P2 = 200 kPa ⎫ h2 = 2870.4 kJ/kg
⎬
⎭ s 2 = 7.5081 kJ/kg ⋅ K
T2 = 200°C
1
P3 = 200 kPa ⎫ h3 ≅ h f @ 80°C = 335.02 kJ/kg
⎬
T3 = 80°C
⎭ s 3 ≅ s f @80°C = 1.0756 kJ/kg ⋅ K
Analysis (a) We take the mixing chamber as the system, which is a
control volume. The mass and energy balances for this steady-flow
system can be expressed in the rate form as
m& in − m& out = ∆m& system
Mass balance:
600 kJ/min
0 (steady)
15°C
4 kg/s
MIXING
CHAMBER
200 kPa
80°C
3
2 200°C
=0 ⎯
⎯→ m& 1 + m& 2 = m& 3
Energy balance:
E& − E&
1in424out
3
Rate of net energy transfer
by heat, work, and mass
∆E& system 0 (steady)
1442444
3
=
=0
Rate of change in internal, kinetic,
potential, etc. energies
E& in = E& out
m& h1 + m& 2 h2 = Q& out + m& 3 h3
Combining the two relations gives
Q& out = m& 1 h1 + m& 2 h2 − (m& 1 + m& 2 )h3 = m& 1 (h1 − h3 ) + m& 2 (h2 − h3 )
& 2 and substituting, the mass flow rate of the superheated steam is determined to be
Solving for m
m& 2 =
Also,
Q& out − m& 1 (h1 − h3 ) (600/60 kJ/s) − (4 kg/s )(62.98 − 335.02)kJ/kg
=
= 0.429 kg/s
(2870.4 − 335.02)kJ/kg
h2 − h3
m& 3 = m& 1 + m& 2 = 4 + 0.429 = 4.429 kg/s
(b) The exergy destroyed during a process can be determined from an exergy balance or directly from its
definition X destroyed = T0 S gen where the entropy generation Sgen is determined from an entropy balance on an extended
system that includes the mixing chamber and its immediate surroundings. It gives
S& − S& out
1in424
3
Rate of net entropy transfer
by heat and mass
m& 1 s1 + m& 2 s 2 − m& 3 s 3 −
+
S& gen
{
Rate of entropy
generation
= ∆S& system 0 = 0
14243
Rate of change
of entropy
Q&
Q& out
+ S& gen = 0 → S& gen = m& 3 s 3 − m& 1 s1 − m& 2 s 2 + out
T0
Tb,surr
Substituting, the exergy destruction is determined to be
⎛
⎞
Q&
X& destroyed = T0 S& gen = T0 ⎜ m& 3 s 3 − m& 2 s 2 − m& 1 s1 + out ⎟
⎜
Tb, surr ⎟⎠
⎝
= (298 K)(4.429 × 1.0756 − 0.429 × 7.5081 − 4 × 0.22447 + 10 / 298)kW/K
= 202 kW
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
preparation. If you are a student using this Manual, you are using it without permission.
8-79
8-88 Water is heated by hot oil in a heat exchanger. The outlet temperature of the oil and the rate of exergy destruction
within the heat exchanger are to be determined.
Assumptions 1 Steady operating conditions exist. 2 The heat exchanger is well-insulated so that heat loss to the
surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid. 3 Changes
in the kinetic and potential energies of fluid streams are negligible. 4 Fluid properties are constant.
Properties The specific heats of water and oil are given to be 4.18 and 2.3 kJ/kg.°C, respectively.
Analysis We take the cold water tubes as the system, which is a control volume.
The energy balance for this steady-flow system can be expressed in the rate form as
E& − E&
1in424out
3
=
Rate of net energy transfer
by heat, work, and mass
∆E& system 0 (steady)
1442444
3
=0
Rate of change in internal, kinetic,
potential, etc. energies
Oil
170°C
10 kg/s
70°C
E& in = E& out
Q& in + m& h1 = m& h2 (since ∆ke ≅ ∆pe ≅ 0)
Q& in = m& c p (T2 − T1 )
Water
20°C
4.5 kg/s
(12 tube passes)
Then the rate of heat transfer to the cold water in this heat exchanger becomes
Q& = [ m& c p (Tout − Tin )] water = ( 4.5 kg/s)(4.18 kJ/kg. °C)(70°C − 20°C) = 940.5 kW
Noting that heat gain by the water is equal to the heat loss by the oil, the outlet temperature of the hot water is determined
from
Q&
940.5 kW
= 170°C −
= 129.1°C
Q& = [m& c p (Tin − Tout )] oil → Tout = Tin −
(10 kg/s)(2.3 kJ/kg.°C)
m& c p
(b) The rate of entropy generation within the heat exchanger is determined by applying the rate form of the entropy balance
on the entire heat exchanger:
S& in − S& out
1424
3
Rate of net entropy transfer
by heat and mass
+
S& gen
{
Rate of entropy
generation
= ∆S& system 0 (steady)
1442443
Rate of change
of entropy
m& 1 s1 + m& 3 s 3 − m& 2 s 2 − m& 3 s 4 + S& gen = 0 (since Q = 0)
m& water s1 + m& oil s 3 − m& water s 2 − m& oil s 4 + S& gen = 0
S& gen = m& water ( s 2 − s1 ) + m& oil ( s 4 − s 3 )
Noting that both fluid streams are liquids (incompressible substances), the rate of entropy generation is determined to be
T
T
S& gen = m& water c p ln 2 + m& oil c p ln 4
T1
T3
= (4.5 kg/s)(4.18 kJ/kg.K) ln
70 + 273
129.1 + 273
+ (10 kg/s)(2.3 kJ/kg.K) ln
= 0.736 kW/K
20 + 273
170 + 273
The exergy destroyed during a process can be determined from an exergy balance or directly from its
definition X destroyed = T0 S gen ,
X& destroyed = T0 S& gen = ( 298 K)(0.736 kW/K ) = 219 kW
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
preparation. If you are a student using this Manual, you are using it without permission.
8-83
8-92 Water is heated in a chamber by mixing it with saturated steam. The temperature of the steam entering the chamber,
the exergy destruction, and the second-law efficiency are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. 3 Heat loss from
the chamber is negligible.
Analysis (a) The properties of water are (Tables A-4 through A-6)
T1 = 15°C⎫ h1 = h0 = 62.98 kJ/kg
⎬
x1 = 0 ⎭ s1 = s 0 = 0.22447 kJ/kg.K
T3 = 45°C⎫ h3 = 188.44 kJ/kg
⎬
x1 = 0
⎭ s3 = 0.63862 kJ/kg.K
Water
15°C
4.6 kg/s
Mixing
chamber
Mixture
45°C
Sat. vap.
0.23 kg/s
An energy balance on the chamber gives
m& 1h1 + m& 2h2 = m& 3h3 = (m& 1 + m& 2 )h3
(4.6 kg/s)(62.98 kJ/kg) + (0.23 kg/s)h2 = (4.6 + 0.23 kg/s)(188.44 kJ/kg)
h2 = 2697.5 kJ/kg
The remaining properties of the saturated steam are
h2 = 2697.5 kJ/kg ⎫ T2 = 114.3 °C
⎬
x2 = 1
⎭ s 2 = 7.1907 kJ/kg.K
(b) The specific exergy of each stream is
ψ1 = 0
ψ 2 = h2 − h0 − T0 ( s2 − s0 )
= (2697.5 − 62.98)kJ/kg − (15 + 273 K)(7.1907 − 0.22447)kJ/kg.K = 628.28 kJ/kg
ψ 3 = h3 − h0 − T0 ( s3 − s0 )
= (188.44 − 62.98)kJ/kg − (15 + 273 K)(0.63862 − 0.22447)kJ/kg.K = 6.18 kJ/kg
The exergy destruction is determined from an exergy balance on the chamber to be
X& dest = m& 1ψ 1 + m& 2ψ 2 − (m& 1 + m& 2 )ψ 3
= 0 + (0.23 kg/s)(628.28 kJ/kg) − (4.6 + 0.23 kg/s)(6.18 kJ/kg)
= 114.7 kW
(c) The second-law efficiency for this mixing process may be determined from
ηII =
(m& 1 + m& 2 )ψ 3
(4.6 + 0.23 kg/s)(6.18 kJ/kg)
=
= 0.207
&
&
m1ψ 1 + m2ψ 2 0 + (0.23 kg/s)(628.28 kJ/kg)
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
preparation. If you are a student using this Manual, you are using it without permission.
Tutorial #11
(QUIZ #2)
ESO 201A
Thermodynamics
Instructor
Prof. Sameer Khandekar
Contact
Office: SL-109, Tel: 7038
E-mail: [email protected]
URL: home.iitk.ac.in/~samkhan
Tutorial #12
ESO 201A
Thermodynamics
Instructor
Prof. Sameer Khandekar
Contact
Office: SL-109, Tel: 7038
E-mail: [email protected]
URL: home.iitk.ac.in/~samkhan
ESO 201A Thermodynamics
Instructor: Sameer Khandekar
Tutorial 12
[15-14] Propane fuel (C3H8) is burned in the presence of air. Assuming that the combustion is
theoretical— that is, only nitrogen (N2), water vapor (H20), and carbon dioxide (C02) are present
in the products—determine (a) the mass fraction of carbon dioxide and (b) the mole and mass
fractions of the water vapor in the products.
[15-26] Propylene (C3H6) is burned with 50 percent excess air during a combustion process.
Assuming complete combustion and a total pressure of 105 kPa, determine (a) the air-fuel ratio
and (h) the temperature at which the water vapor in the products will start condensing.
[15-58] n-Octane gas (C8H18) is burned with 100 percent excess air in a constant pressure burner.
The air and fuel enter this burner steadily at standard conditions and the products of combustion
leave at 257°C. Calculate the heat transfer, in kJ/kg fuel, during this combustion.
[15-71] A constant-volume tank contains a mixture of 1 kmol of benzene (C6H6) gas and 30
percent excess air at 25°C and 1 atm. The contents of the tank are now ignited, and all the
hydrogen in the fuel burns to H20 but only 92 percent of the carbon burns to CO2, the remaining
8 percent forming CO. If the final temperature in the tank is 1000 K, determine the heat transfer
from the combustion chamber during this process.
Additional Homework Problems (Tutorial 12)
[15-20] n-Butane fuel (C4H10) is burned with a 100 percent excess air. Determine the mole
fractions of each of the products. Also, calculate the mass of carbon dioxide in the products per
unit mass of the fuel and the air-fuel ratio.
[15-28] A fuel mixture of 60 percent by mass methane (CH4) and 40 percent by mass ethanol
(C2H6OH), is burned completely with theoretical air. If the total flow rate of the fuel is 10 kg/s,
determine the required flow rate of air.
[15-60] Methane (CH4) is burned completely with the stoichiometric amount of air during a
steady-flow combustion process. If both the reactants and the products are maintained at 25°C
and 1 atm and the water in the products exists in the liquid form, determine the heat transfer
from the combustion chamber during this process. What would your answer be if combustion
were achieved with 100 percent excess air?
[15-62] A coal from Texas which has an ultimate analysis (by mass) as 39.25 percent C, 6.93
percent H2, 41.11 percent O2, 0.72 percent N2, 0.79 percent S, and 11.20 percent ash (noncombustibles) is burned steadily with 40 percent excess air in a power plant boiler. The coal and
air enter this boiler at standard conditions and the products of combustion in the smokestack are
at 127°C. Calculate the heat transfer, in kJ/kg fuel, in this boiler. Include the effect of the sulfur
in the energy analysis by noting that sulfur dioxide has an enthalpy of formation of —297,100
kJ/kmol and an average specific heat at constant pressure of Cp = 41.7 kJ/kmol-K.
[15-79] Estimate the adiabatic flame temperature of an acetylene (C2H2) cutting torch, in °C,
which uses a stoichiometric amount of pure oxygen.
15-5
15-14 Propane is burned with theoretical amount of air. The mass fraction of carbon dioxide and the mole and mass
fractions of the water vapor in the products are to be determined.
Properties The molar masses of C3H8, O2, N2, CO2, and H2O are 44, 32, 28, 44, and 18 kg/kmol, respectively (Table A-1).
Analysis (a) The reaction in terms of undetermined coefficients is
C 3 H 8 + x(O 2 + 3.76 N 2 ) ⎯
⎯→ yCO 2 + zH 2 O + pN 2
Balancing the carbon in this reaction gives
C3H8
y=3
Air
and the hydrogen balance gives
2z = 8 ⎯
⎯→ z = 4
Combustion
chamber
CO2, H2O, N2
100% theoretical
The oxygen balance produces
2x = 2 y + z ⎯
⎯→ x = y + z / 2 = 3 + 4 / 2 = 5
A balance of the nitrogen in this reaction gives
2 × 3.76 x = 2 p ⎯
⎯→ p = 3.76 x = 3.76 × 5 = 18.8
In balanced form, the reaction is
C 3 H 8 + 5O 2 + 18.8 N 2 ⎯
⎯→ 3CO 2 + 4H 2 O + 18.8N 2
The mass fraction of carbon dioxide is determined from
mf CO2 =
m CO2
N CO2 M CO2
=
m products N CO2 M CO2 + N H2O M H2O + N N2 M N2
(3 kmol)(44 kg/kmol)
(3 kmol)(44 kg/kmol) + (4 kmol)(18 kg/kmol) + (18.8 kmol)(28 kg/kmol)
132 kg
=
= 0.181
730.4 kg
=
(b) The mole and mass fractions of water vapor are
y H2O =
N H2O
N H2O
4 kmol
4 kmol
=
=
=
= 0.155
N products N CO2 + N H2O + N N2 3 kmol + 4 kmol + 18.8 kmol 25.8 kmol
mf H2O =
m H2O
N H2O M H2O
=
m products N CO2 M CO2 + N H2O M H2O + N N2 M N2
(4 kmol)(18 kg/kmol)
(3 kmol)(44 kg/kmol) + (4 kmol)(18 kg/kmol) + (18.8 kmol)(28 kg/kmol)
72 kg
=
= 0.0986
730.4 kg
=
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
preparation. If you are a student using this Manual, you are using it without permission.
15-15
15-26 Propylene is burned with 50 percent excess air during a combustion process. The AF ratio and the temperature at
which the water vapor in the products will start condensing are to be determined.
Assumptions 1 Combustion is complete. 2 The combustion products contain CO2, H2O, O2, and N2 only. 3 Combustion
gases are ideal gases.
Properties The molar masses of C, H2, and air are 12 kg/kmol, 2 kg/kmol, and 29 kg/kmol, respectively (Table A-1).
Analysis (a) The combustion equation in this case can be written as
C 3 H 6 + 1.5a th [O 2 + 3.76N 2 ] ⎯
⎯→ 3CO 2 + 3H 2 O + 0.5a th O 2 + (1.5 × 3.76)a th N 2
where ath is the stoichiometric coefficient for air. It is determined from
O2 balance:
15
. a th = 3 + 15
. + 0.5a th
⎯
⎯→
C3H6
a th = 4.5
Products
50% excess air
Substituting,
C 3 H 6 + 6.75 O 2 + 3.76N 2
⎯⎯→ 3CO 2 + 3H 2 O + 2.25O 2 + 25.38N 2
The air-fuel ratio is determined by taking the ratio of the mass of the air to the mass of the fuel,
AF =
m air
(6.75 × 4.76 kmol)(29 kg/kmol)
=
= 22.2 kg air/kg fuel
m fuel (3 kmol)(12 kg/kmol) + (3 kmol)(2 kg/kmol)
(b) The dew-point temperature of a gas-vapor mixture is the saturation temperature of the water vapor in the product gases
corresponding to its partial pressure. That is,
⎛ Nv
Pv = ⎜
⎜ N prod
⎝
⎞
⎟ Pprod = ⎛⎜ 3 kmol ⎞⎟(105 kPa ) = 9.367 kPa
⎜ 33.63 kmol ⎟
⎟
⎝
⎠
⎠
Thus,
Tdp = [email protected] kPa = 44.5°C
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
preparation. If you are a student using this Manual, you are using it without permission.
15-41
15-58 n-Octane is burned with 100 percent excess air. The heat transfer per kilogram of fuel burned for a product
temperature of 257°C is to be determined.
Assumptions 1 Steady operating conditions exist. 2 Air and combustion gases are ideal gases. 3 Kinetic and potential
energies are negligible. 4 Combustion is complete. 5 The fuel is in vapor phase.
Properties The molar masses of propane and air are 44 kg/kmol and 29 kg/kmol, respectively (Table A-1).
Analysis The combustion reaction for stoichiometric air is
C 8 H 18 + 12.5[O 2 + 3.76N 2 ] ⎯
⎯→8CO 2 + 9H 2 O + (12.5 × 3.76)N 2
Qout
The combustion equation with 100% excess air is
C 8 H 18 + 25[O 2 + 3.76N 2 ] ⎯
⎯→ 8CO 2 + 9H 2 O + 12.5 O 2 + 94 N 2
The heat transfer for this combustion process is determined from
the energy balance E in − E out = ∆E system applied on the
combustion chamber with W = 0. It reduces to
− Qout =
∑ (
N P h fo
+h −h
o
) −∑ (
P
N R h fo
+h −h
o
C8H18
25°C
100% excess air
Combustion
chamber
P = 1 atm
Products
257°C
25°C
)
R
Assuming the air and the combustion products to be ideal gases, we have h = h(T). From the tables,
Substance
C8H18 (g)
O2
N2
H2O (g)
CO2
h fo
h298K
h530K
kJ/kmol
-208,450
0
0
-241,820
-393,520
kJ/kmol
--8682
8669
9904
9364
kJ/kmol
--15,708
15,469
17,889
19,029
Substituting,
− Qout = (8)(− 393,520 + 19,029 − 9364 ) + (9 )(− 241,820 + 17,889 − 9904 ) + (12.5)(0 + 15,708 − 8682 )
+ (94 )(0 + 15,469 − 8669 ) − (1)(− 208,450 ) − 0 − 0
= −4,239,880 kJ/kmol C 8 H 18
or
Qout = 4,239,880 kJ/kmol C 8 H 18
Then the heat transfer per kg of fuel is
Qout =
Qout
4,239,880 kJ/kmol fuel
=
= 37,200 kJ/kg C 8 H18
114 kg/kmol
M fuel
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
preparation. If you are a student using this Manual, you are using it without permission.
15-56
15-71 A mixture of benzene gas and 30 percent excess air contained in a constant-volume tank is ignited. The heat transfer
from the combustion chamber is to be determined.
Assumptions 1 Both the reactants and products are ideal gases. 2 Combustion is complete.
Analysis The theoretical combustion equation of C6H6 with
stoichiometric amount of air is
Q
C6 H 6 (g ) + ath (O 2 + 3.76N 2 ) ⎯
⎯→ 6CO 2 + 3H 2O + 3.76ath N 2
C6H6+ Air
25C, 1 atm
where ath is the stoichiometric coefficient and is determined from the O2 balance,
a th = 6 + 1.5 = 7.5
1000 K
Then the actual combustion equation with 30% excess air becomes
C6 H 6 (g ) + 9.75(O 2 + 3.76N 2 ) ⎯
⎯→ 5.52CO 2 + 0.48CO + 3H 2O + 2.49O 2 + 36.66N 2
The heat transfer for this constant volume combustion process is determined from the energy balance E in − E out = ∆E system
applied on the combustion chamber with W = 0. It reduces to
− Qout =
∑ N (h
P
o
f
+ h − h o − Pv
) − ∑ N (h
R
P
o
f
+ h − h o − Pv
)
R
Since both the reactants and the products behave as ideal gases, all the internal energy and enthalpies depend on temperature
only, and the Pv terms in this equation can be replaced by RuT.
It yields
− Q out =
∑ N (h
P
o
f
+ h1000 K − h298 K − Ru T
) − ∑ N (h
P
R
o
f
− Ru T
)
R
since the reactants are at the standard reference temperature of 25°C. From the tables,
hfo
h 298 K
h1000 K
kJ/kmol
kJ/kmol
kJ/kmol
C6H6 (g)
82,930
---
---
O2
0
8682
31,389
N2
0
8669
30,129
H2O (g)
-241,820
9904
35,882
CO
-110,530
8669
30,355
CO2
-393,520
9364
42,769
Substance
Thus,
−Qout = (5.52 )(−393,520 + 42,769 − 9364 − 8.314 × 1000 )
+ (0.48)(− 110,530 + 30,355 − 8669 − 8.314 × 1000 )
+ (3)(− 241,820 + 35,882 − 9904 − 8.314 × 1000 )
+ (2.49 )(0 + 31,389 − 8682 − 8.314 × 1000 )
+ (36.66 )(0 + 30,129 − 8669 − 8.314 × 1000 )
− (1)(82,930 − 8.314 × 298) − (9.75)(4.76 )(− 8.314 × 298)
= −2,200,433 kJ
or
Qout = 2,200,433 kJ
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
preparation. If you are a student using this Manual, you are using it without permission.
15-10
15-20 n-Butane is burned with 100 percent excess air. The mole fractions of each of the products, the mass of carbon
dioxide in the products per unit mass of the fuel, and the air-fuel ratio are to be determined.
Assumptions 1 Combustion is complete. 2 The combustion products contain CO2, H2O, O2, and N2 only.
Properties The molar masses of C, H2, O2, and air are 12 kg/kmol, 2 kg/kmol, 32 kg/kmol, and 29 kg/kmol, respectively
(Table A-1).
Analysis The combustion equation in this case can be written as
C 4 H 10 + 2.0a th [O 2 + 3.76N 2 ] ⎯
⎯→ 4CO 2 + 5H 2 O + 1.0a th O 2 + (2.0 × 3.76)a th N 2
where ath is the stoichiometric coefficient for air. We have automatically accounted for the 100% excess air by using the
factor 2.0ath instead of ath for air. The stoichiometric amount of oxygen (athO2) will be used to oxidize the fuel, and the
remaining excess amount (1.0athO2) will appear in the products as free oxygen. The coefficient ath is determined from the O2
balance,
O2 balance:
2.0a th = 4 + 2.5 + 1.0a th ⎯
⎯→ a th = 6.5
Substituting,
C 4 H 10 + 13[O 2 + 3.76 N 2 ] ⎯
⎯→ 4CO 2 + 5H 2 O + 6.5O 2 + 48.88 N 2
The mole fractions of the products are
C4H10
N m = 4 + 5 + 6.5 + 48.88 = 64.38 kmol
y CO2
N
4 kmol
= CO2 =
= 0.0621
Nm
64.38 kmol
y H2O
N
5 kmol
= H2O =
= 0.0777
Nm
64.38 kmol
y O2 =
N O2
6.5 kmol
=
= 0.1010
N m 64.38 kmol
y N2 =
N N2 48.88 kmol
=
= 0.7592
64.38 kmol
Nm
Products
Air
100% excess
The mass of carbon dioxide in the products per unit mass of fuel burned is
mCO2
(4 × 44) kg
=
= 3.034 kg CO 2 /kg C 4 H10
m C4H10 (1 × 58) kg
The air-fuel ratio is determined by taking the ratio of the mass of the air to the mass of the fuel,
AF =
mair
(13 × 4.76 kmol)(29 kg/kmol)
=
= 30.94 kg air/kg fuel
mfuel
(1 kmol)(58 kg/ kmol)
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
preparation. If you are a student using this Manual, you are using it without permission.
15-17
15-28 A fuel mixture of 60% by mass methane, CH4, and 40% by mass ethanol, C2H6O, is burned completely with
theoretical air. The required flow rate of air is to be determined.
Assumptions 1 Combustion is complete. 2 The combustion products contain CO2, H2O, and N2 only.
Properties The molar masses of C, H2, O2 and air are 12 kg/kmol, 2 kg/kmol, 32 kg/kmol, and 29 kg/kmol, respectively
(Table A-1).
Analysis For 100 kg of fuel mixture, the mole numbers are
N CH4 =
N C2H6O
mf CH4
60 kg
=
= 3.75 kmol
M CH4 16 kg/kmol
mf C2H6O
40 kg
=
=
= 0.8696 kmol
M C2H6O 46 kg/kmol
60% CH4
40% C2H6O
Air
Products
100% theoretical
The mole fraction of methane and ethanol in the fuel mixture are
x=
N CH4
3.75 kmol
=
= 0.8118
N CH4 + N C2H6O (3.75 + 0.8696) kmol
y=
N C2H6O
0.8696 kmol
=
= 0.1882
N CH4 + N C2H6O (3.75 + 0.8696) kmol
The combustion equation in this case can be written as
x CH 4 + y C 2 H 6 O + a th [O 2 + 3.76N 2 ] ⎯
⎯→ B CO 2 + D H 2 O + F N 2
where ath is the stoichiometric coefficient for air. The coefficient ath and other coefficients are to be determined from the
mass balances
Carbon balance:
x + 2y = B
Hydrogen balance:
4 x + 6 y = 2D
Oxygen balance:
2ath + y = 2 B + D
Nitrogen balance:
3.76a th = F
Substituting x and y values into the equations and solving, we find the coefficients as
x = 0.8118
y = 0.1882
a th = 2.188
B = 1.188
D = 2.188
F = 8.228
Then, we write the balanced reaction equation as
0.8118 CH 4 + 0.1882 C 2 H 6 O + 2.188 [O 2 + 3.76N 2 ] ⎯
⎯→1.188 CO 2 + 2.188 H 2 O + 8.228 N 2
The air-fuel ratio is determined by taking the ratio of the mass of the air to the mass of the fuel,
AF =
mair
(2.188 × 4.76 kmol)(29 kg/kmol)
=
mfuel (0.8118 kmol)(12 + 4 × 1)kg/kmol + (0.1882 kmol)(2 × 12 + 6 × 1 + 16)kg/kmol
= 13.94 kg air/kg fuel
Then, the required flow rate of air becomes
m& air = AFm& fuel = (13.94)(10 kg/s) = 139.4 kg/s
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
preparation. If you are a student using this Manual, you are using it without permission.
15-43
15-60 Methane is burned completely during a steady-flow combustion process. The heat transfer from the combustion
chamber is to be determined for two cases.
Assumptions 1 Steady operating conditions exist. 2 Air and combustion gases are ideal gases. 3 Kinetic and potential
energies are negligible. 4 Combustion is complete.
Analysis The fuel is burned completely with the stoichiometric amount of air, and thus the products will contain only H2O,
CO2 and N2, but no free O2. Considering 1 kmol of fuel, the theoretical combustion equation can be written as
CH 4 + a th (O 2 + 3.76N 2 )⎯
⎯→CO 2 + 2H 2O + 3.76a th N 2
Q
CH4
where ath is determined from the O2 balance,
25°C
a th = 1 + 1 = 2
Substituting,
Air
CH 4 + 2(O 2 + 3.76N 2 )⎯
⎯→CO 2 + 2H 2O + 5.64N 2
100% theoretical
Combustion
chamber
P = 1 atm
Products
25°C
The heat transfer for this combustion process is determined from the energy balance Ein − E out = ∆E system applied on the
combustion chamber with W = 0. It reduces to
− Qout =
∑ N (h
P
o
f
+ h − ho
) − ∑ N (h
R
P
o
f
+ h − ho
) =∑N
R
o
Ph f ,P
−
∑N
o
Rh f ,R
since both the reactants and the products are at 25°C and both the air and the combustion gases can be treated as ideal gases.
From the tables,
h fo
Substance
CH4
O2
N2
H2O (l)
CO2
kJ/kmol
-74,850
0
0
-285,830
-393,520
Thus,
−Qout = (1)(−393,520 ) + (2)(−285,830 ) + 0 − (1)(−74,850 ) − 0 − 0 = −890,330 kJ / kmol CH 4
or
Qout = 890,330 kJ / kmol CH 4
If combustion is achieved with 100% excess air, the answer would still be the same since it would enter and leave at 25°C,
and absorb no energy.
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preparation. If you are a student using this Manual, you are using it without permission.
15-45
15-62 A certain coal is burned steadily with 40% excess air. The heat transfer for a given product temperature is to be
determined.
Assumptions 1 Combustion is complete. 2 The combustion products contain CO2, CO, H2O, SO2, and N2. 3 Combustion
gases are ideal gases.
Properties The molar masses of C, H2, N2, O2, S, and air are 12, 2, 28, 32, 32, and 29 kg/kmol, respectively (Table A-1).
Analysis We consider 100 kg of coal for simplicity. Noting that the mass percentages in this case correspond to the masses
of the constituents, the mole numbers of the constituent of the coal are determined to be
NC =
mC
39.25 kg
=
= 3.271 kmol
M C 12 kg/kmol
N H2 =
m H2
6.93 kg
=
= 3.465 kmol
M H2 2 kg/kmol
N O2 =
mO2
41.11 kg
=
= 1.285 kmol
M O2 32 kg/kmol
N N2 =
m N2
0.72 kg
=
= 0.0257 kmol
M N2 28 kg/kmol
NS =
mS
0.79 kg
=
= 0.0247 kmol
M S 32 kg/kmol
39.25% C
6.93% H2
41.11% O2
0.72% N2
0.79% S
11.20% ash
(by mass)
Coal
The mole number of the mixture and the mole fractions are
N m = 3.271 + 3.465 + 1.285 + 0.0257 + 0.0247 = 8.071 kmol
yC =
N C 3.271 kmol
=
= 0.4052
N m 8.071 kmol
y H2 =
N H2 3.465 kmol
=
= 0.4293
Nm
8.071 kmol
y O2 =
N O2 1.285 kmol
=
= 0.1592
N m 8.071 kmol
y N2 =
N N2 0.0257 kmol
=
= 0.00319
Nm
8.071 kmol
yS =
N S 0.0247 kmol
=
= 0.00306
Nm
8.071 kmol
Air
40% excess
Combustion
chamber
Products
127°C
Ash consists of the non-combustible matter in coal. Therefore, the mass of ash content that enters the combustion chamber
is equal to the mass content that leaves. Disregarding this non-reacting component for simplicity, the combustion equation
may be written as
0.4052C + 0.4293H 2 + 0.1592O 2 + 0.00319 N 2 + 0.00306S + 1.4a th (O 2 + 3.76 N 2 )
⎯
⎯→ 0.4052CO 2 + 0.4293H 2 O + 0.4a th O 2 + 0.00306SO 2 + 1.4a th × 3.76 N 2
According to the O2 mass balance,
0.1592 + 1.4a th = 0.4052 + 0.5 × 0.4293 + 0.4a th + 0.00306 ⎯
⎯→ a th = 0.4637
Substituting,
0.4052C + 0.4293H 2 + 0.1592O 2 + 0.00319 N 2 + 0.00306S + 0.6492(O 2 + 3.76 N 2 )
⎯
⎯→ 0.4052CO 2 + 0.4293H 2 O + 0.1855O 2 + 0.00306SO 2 + 2.441N 2
The heat transfer for this combustion process is determined from the energy balance E in − E out = ∆E system applied on the
combustion chamber with W = 0. It reduces to
− Q out =
∑ N (h
P
o
f
+h −ho
) − ∑ N (h
P
R
o
f
+h −ho
)
R
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15-46
Assuming the air and the combustion products to be ideal gases, we have h = h(T). From the tables,
h fo
h298 K
h 400 K
kJ/kmol
kJ/kmol
kJ/kmol
O2
0
8682
11,711
N2
0
8669
11,640
H2O (g)
-241,820
9904
13,356
CO2
-393,520
9364
13,372
SO2
-297,100
-
-
Substance
The enthalpy change of sulfur dioxide between the standard temperature and the product temperature using constant specific
heat assumption is
∆hSO2 = c p ∆T = ( 41.7 kJ/kmol ⋅ K)(127 − 25)K = 4253 kJ/kmol
Substituting into the energy balance relation,
− Qout = (0.4052)(− 393,520 + 13,372 − 9364) + (0.4293)(− 241,820 + 13,356 − 9904)
+ (0.1855)(0 + 11,711 − 8682) + (2.441)(0 + 11,640 − 8669) + (0.00306)(− 297,100 + 4253) − 0
= −253,244 kJ/kmol C 8 H 18
or
Qout = 253,244 kJ/kmol fuel
Then the heat transfer per kg of fuel is
Qout =
Qout
253,244 kJ/kmol fuel
=
M fuel (0.4052 × 12 + 0.4293 × 2 + 0.1592 × 32 + 0.00319 × 28 + 0.00306 × 32) kg/kmol
253,244 kJ/kmol fuel
11.00 kg/kmol
= 23,020 kJ/kg coal
=
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
preparation. If you are a student using this Manual, you are using it without permission.
15-64
15-79 Acetylene is burned with stoichiometric amount of oxygen. The adiabatic flame temperature is to be determined.
Assumptions 1 Steady operating conditions exist. 2 Air and combustion gases are ideal gases. 3 Kinetic and potential
energies are negligible. 4 There are no work interactions. 5 The combustion chamber is adiabatic.
Analysis Under steady-flow conditions the energy balance E in − E out = ∆E system applied on the combustion chamber with
Q = W = 0 reduces to
∑ N (h
P
o
f
+h −ho
) = ∑ N (h
R
P
o
f
+h −ho
)
R
⎯
⎯→
∑ N (h
P
o
f
+h −ho
) =∑N
P
o
R h f ,R
since all the reactants are at the standard reference temperature of 25°C. Then, for the stoichiometric oxygen
C 2 H 2 + 2.5O 2 ⎯
⎯→ 2 CO 2 + 1 H 2 O
From the tables,
C2H2
h fo
h 298K
kJ/kmol
kJ/kmol
C2H2 (g)
226,730
---
O2
0
8682
N2
0
8669
H2O (g)
-241,820
9904
CO2
-393,520
9364
Substance
25°C
100% theoretical O2
25°C
Combustion
chamber
Products
TP
Thus,
(
)
(
)
(2) − 393,520 + hCO2 − 9364 + (1) − 241,820 + hH2O − 9904 = (1)(226,730) + 0 + 0
2hCO2 + 1hH2O = 1,284,220 kJ
It yields
The adiabatic flame temperature is obtained from a trial and error solution. A first guess is obtained by dividing the righthand side of the equation by the total number of moles, which yields 2,284,220/(2+1) = 428,074 kJ/kmol. The ideal gas
tables do not list enthalpy values this high. Therefore, we cannot use the tables to estimate the adiabatic flame temperature.
In Table A-2b, the highest available value of specific heat is cp = 1.234 kJ/kg·K for CO2 at 1000 K. The specific heat of
water vapor is cp = 1.8723 kJ/kg·K (Table A-2a). Using these specific heat values,
(
)
(
)
(2) − 393,520 + c p ∆T + (1) − 241,820 + c p ∆T = (1)(226,730) + 0 + 0
where ∆T = (Taf − 25)°C . The specific heats on a molar base are
c p ,CO2 = c p M = (1.234 kJ/kg ⋅ K)(44 kg/kmol) = 54.3 kJ/kmol ⋅ K
c p ,H2O = c p M = (1.8723 kJ/kg ⋅ K)(18 kg/kmol) = 33.7 kJ/kmol ⋅ K
Substituting,
(2)(− 393,520 + 54.3∆T ) + (1)(− 241,820 + 33.7∆T ) = 226,730
(2 × 54.3)∆T + 33.7∆T = 1,255,590
1,255,590 kJ/kmol
∆T =
= 8824 K
(2 × 54.3 + 33.7) kJ/kmol ⋅ K
Then the adiabatic flame temperature is estimated as
Taf = ∆T + 25 = 8824 + 25 = 8849 °C
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
preparation. If you are a student using this Manual, you are using it without permission.
Tutorial #13
ESO 201A
Thermodynamics
Instructor
Prof. Sameer Khandekar
Contact
Office: SL-109, Tel: 7038
E-mail: [email protected]
URL: home.iitk.ac.in/~samkhan
ESO 201A Thermodynamics
Instructor: Sameer Khandekar
Tutorial 13
[9-35] Someone has suggested that the air-standard Otto cycleis more accurate if the two
isentropic processes are replacedwith polytropic processes with a polytropic exponent n = 1.3.
Consider such a cycle when the compression ratio is 8, P 1 =95 kPa, T 1 = 15°C, and the maximum
cycle temperature is 1200°C. Determine the heat transferred to and rejected fromthis cycle, as
well as the cycle’s thermal efficiency. Use constantspecific heats at room temperature.
[Answers: 835 kJ/kg, 420 kJ/kg, 49.8 percent]
[9-129] Consider an ideal gas-turbine cycle with two stagesof compression and two stages
of expansion. The pressureratio across each stage of the compressor and turbine is 3.
The air enters each stage of the compressor at 300 K andeach stage of the turbine at
1200 K. Determine the back work ratio and the thermal efficiency of the cycle,
assuming (a) noregenerator is used and (b) a regenerator with 75 percent effectiveness is
used. Use variable specific heats.
[10-23] A simple Rankin cycle uses water as the workingfluid. The boiler operates at 6000 kPa
and the condenser at50 kPa. At the entrance to the turbine, the temperature is 450°C. The
isentropic efficiency of the turbine is 94 percent, pressure and pump losses are negligible, and
the water leaving the condenser is sub cooled by 6.3°C. The boiler is sizedfor a mass flow rate of
20 kg/s. Determine the rate at which heat is added in the boiler, the power required to
operatethe pumps, the net power produced by the cycle, and the thermal efficiency. [Answers:
59,660 kW, 122 kW, 18,050 kW, 30.3 percent.]
[11-21] Refrigerant-134a enters the compressor of a refrigerator at 100 kPa and -20°C at a
rate of 0.5 m3/min and leave sat 0.8 MPa. The isentropic efficiency of the compressor is 78
percent. The refrigerant enters the throttling valve at0.75 MPa and 26°C and leaves the
evaporator as saturated vapor at -26°C. Show the cycle on a T-s diagram withrespect to
saturation lines, and determine (a) the power input to the compressor, (b) the rate
of heat removal from therefrigerated space, and (c) the pressure drop and rate of
heatgain in the line between the evaporator and the compressor. [Answers: (a) 2.40 kW, (b)
6.17 kW, (c) 1.73 kPa, 0.203 kW]
Additional Homework Problems (Tutorial 13)
[9-39] The compression ratio of an air-standard Otto cycle is 9.5. Prior to the
isentropic compression process, the air is at100 kPa, 35°C, and 600 cm3. The temperature
at the end of the isentropic expansion process is 800 K. Using specificheat values at
room temperature; determine (a) the highest temperature and pressure in the cycle;
(b) the amount of heattransferred in, in kJ; (c) thermal efficiency; and (d) mean
effective pressure. [Answers: (a) 1969 K, 6072 kPa, (b) 0.59 kJ,(c) 59.4 percent, (d) 652 kPa]
[9-57] An ideal diesel engine has a compression ratio of 20and uses air as the working fluid. The
state of air at thebeginning of the compression process is 95 kPa and 20°C. Ifthe maximum
temperature in the cycle is not to exceed 2200 K, determine (a) the thermal efficiency and (b)
the mean effective pressure. Assume constant specific heats for air at roomtemperature.
[Answers: (a) 63.5 percent, (b) 933 kPa]
[10-21] Consider a steam power plant that operates on a simple ideal Rankin cycle and has a
net power output of 45MW. Steam enters the turbine at 7 MPa and 500°C and is cooled in the
condenser at a pressure of 10 kPa by running cooling water from a lake through the tubes of
the condenserat a rate of 2000 kg/s. Show the cycle on a T-s diagram with respect to
saturation lines, determine (a) thermal efficiencyof the cycle, (b) mass flow rate of the steam,
and (c) temperature rise of cooling water. [Answers: (a) 38.9 percent, (b) 36 kg/s, (c) 8.4°C]
[10-39] Consider a steam power plant that operates on the ideal reheat Rankin cycle. The plant
maintains the boiler at7000 kPa, the reheat section at 800 kPa, and the condenser at 10 kPa.
The mixture quality at the exit of both turbinesis 93 percent. Determine the temperature at the
inlet of each turbine and the cycle’s thermal efficiency. [Answers: 373°C,416°C, 37.6 percent]
[11-13] An ideal vapor-compression refrigeration cycle thatuses refrigerant-134a as its
working fluid maintains a condenser at 1000 kPa and the evaporator at 4°C. Determine this
system’s COP and the amount of power required to service a 400 kW cooling load. [Answers:
6.46, 61.9 kW]
[11-80] A gas refrigeration system using air as the workingfluid has a pressure ratio of 5. Air
enters the compressor at0°C. The high-pressure air is cooled to 35°C by rejecting heat to the
surroundings. The refrigerant leaves the turbine at -80°C and then it absorbs heat from
the refrigerated spacebefore entering the regenerator. The mass flow rate of air is 0.4
kg/s. Assuming isentropic efficiencies of 80 percent forthe compressor and 85 percent for the
turbine and using constantspecific heats at room temperature, determine (a)
theeffectiveness of the regenerator, (b) the rate of heat removal from the refrigerated
space, and (c) the COP of the cycle.Also, determine (d) the refrigeration load and the COP
if this system operated on the simple gas refrigeration cycle. Usethe same compressor inlet
temperature as given, the same turbine inlet temperature as calculated, and the same
compressorand turbine efficiencies. Answers: (a) 0.434, (b) 21.4 kW,(c) 0.478, (d) 24.7 kW,
0.599
9-21
9-35 The two isentropic processes in an Otto cycle are replaced with polytropic processes. The heat added to and rejected
from this cycle, and the cycle’s thermal efficiency are to be determined.
Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is
an ideal gas with constant specific heats.
Properties The properties of air at room temperature are R = 0.287 kPa·m3/kg·K, cp = 1.005 kJ/kg·K, cv = 0.718 kJ/kg·K,
and k = 1.4 (Table A-2a).
Analysis The temperature at the end of the compression is
⎛v
T2 = T1 ⎜⎜ 1
⎝v2
⎞
⎟⎟
⎠
P
n −1
= T1 r
n −1
1.3−1
= (288 K)(8)
= 537.4 K
4
And the temperature at the end of the expansion is
⎛v
T4 = T3 ⎜⎜ 3
⎝v4
⎞
⎟⎟
⎠
3
n −1
⎛1⎞
= T3 ⎜ ⎟
⎝r⎠
n −1
⎛1⎞
= (1473 K)⎜ ⎟
⎝8⎠
2
1.3−1
= 789.4 K
1
v
The integral of the work expression for the polytropic compression gives
w1− 2 =
RT1 ⎡⎛ v 1
⎢⎜
n − 1 ⎢⎜⎝ v 2
⎣
⎞
⎟⎟
⎠
n −1
⎤ (0.287 kJ/kg ⋅ K)(288 K)
− 1⎥ =
(81.3−1 − 1) = 238.6 kJ/kg
1
.
3
1
−
⎥
⎦
Similarly, the work produced during the expansion is
w3 − 4 = −
RT3 ⎡⎛ v 3
⎢⎜
n − 1 ⎢⎜⎝ v 4
⎣
⎞
⎟⎟
⎠
n −1
1.3−1
⎤
⎤
(0.287 kJ/kg ⋅ K)(1473 K) ⎡⎛ 1 ⎞
− 1⎥ = −
− 1⎥ = 654.0 kJ/kg
⎢⎜ ⎟
1.3 − 1
⎥
⎣⎢⎝ 8 ⎠
⎦⎥
⎦
Application of the first law to each of the four processes gives
q1− 2 = w1− 2 − cv (T2 − T1 ) = 238.6 kJ/kg − (0.718 kJ/kg ⋅ K )(537.4 − 288)K = 59.53 kJ/kg
q 2−3 = cv (T3 − T2 ) = (0.718 kJ/kg ⋅ K )(1473 − 537.4)K = 671.8 kJ/kg
q 3− 4 = w3− 4 − cv (T3 − T4 ) = 654.0 kJ/kg − (0.718 kJ/kg ⋅ K )(1473 − 789.4)K = 163.2 kJ/kg
q 4−1 = cv (T4 − T1 ) = (0.718 kJ/kg ⋅ K )(789.4 − 288)K = 360.0 kJ/kg
The head added and rejected from the cycle are
q in = q 2 −3 + q 3− 4 = 671.8 + 163.2 = 835.0 kJ/kg
q out = q1− 2 + q 4 −1 = 59.53 + 360.0 = 419.5 kJ/kg
The thermal efficiency of this cycle is then
η th = 1 −
q out
419.5
= 1−
= 0.498
835.0
q in
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
preparation. If you are a student using this Manual, you are using it without permission.
9-25
9-39 An ideal Otto cycle with air as the working fluid has a compression ratio of 9.5. The highest pressure and temperature
in the cycle, the amount of heat transferred, the thermal efficiency, and the mean effective pressure are to be determined.
Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is
an ideal gas with constant specific heats.
Properties The properties of air at room temperature are cp = 1.005 kJ/kg·K, cv = 0.718 kJ/kg·K, R = 0.287 kJ/kg·K, and
k = 1.4 (Table A-2).
Analysis (a) Process 1-2: isentropic compression.
⎛v
T2 = T1 ⎜⎜ 1
⎝v 2
k −1
⎞
⎟⎟
⎠
P
= (308 K )(9.5)0.4 = 757.9 K
3
⎛ 757.9 K ⎞
v T
P2v 2 P1v 1
⎟⎟(100 kPa ) = 2338 kPa
=
⎯
⎯→ P2 = 1 2 P1 = (9.5)⎜⎜
v 2 T1
T2
T1
⎝ 308 K ⎠
Qin
2
Process 3-4: isentropic expansion.
⎛v
T3 = T4 ⎜⎜ 4
⎝v3
⎞
⎟
⎟
⎠
k −1
Qout 4
1
v
= (800 K )(9.5)0.4 = 1969 K
Process 2-3: v = constant heat addition.
⎛ 1969 K ⎞
P3v 3 P2v 2
T
⎟⎟(2338 kPa ) = 6072 kPa
=
⎯
⎯→ P3 = 3 P2 = ⎜⎜
T3
T2
T2
⎝ 757.9 K ⎠
( b)
m=
(
)
P1V1
(100 kPa ) 0.0006 m 3
=
= 6.788 × 10 − 4 kg
3
RT1
0.287 kPa ⋅ m /kg ⋅ K (308 K )
(
)
(
)
Qin = m(u 3 − u 2 ) = mcv (T3 − T2 ) = 6.788 ×10 −4 kg (0.718 kJ/kg ⋅ K )(1969 − 757.9)K = 0.590 kJ
(c) Process 4-1: v = constant heat rejection.
(
)
Qout = m(u 4 − u1 ) = mcv (T4 − T1 ) = − 6.788 ×10 −4 kg (0.718 kJ/kg ⋅ K )(800 − 308)K = 0.240 kJ
Wnet = Qin − Qout = 0.590 − 0.240 = 0.350 kJ
η th =
(d)
W net,out
Qin
V min = V 2 =
MEP =
=
0.350 kJ
= 59.4%
0.590 kJ
V max
r
W net,out
V 1 −V 2
=
W net,out
V1 (1 − 1 / r )
=
⎛ kPa ⋅ m 3
⎜
0.0006 m 3 (1 − 1/9.5) ⎜⎝ kJ
(
0.350 kJ
)
⎞
⎟ = 652 kPa
⎟
⎠
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
preparation. If you are a student using this Manual, you are using it without permission.
9-38
9-57 An ideal diesel engine with air as the working fluid has a compression ratio of 20. The thermal efficiency and the
mean effective pressure are to be determined.
Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is
an ideal gas with constant specific heats.
Properties The properties of air at room temperature are cp = 1.005 kJ/kg·K, cv = 0.718 kJ/kg·K, R = 0.287 kJ/kg·K, and
k = 1.4 (Table A-2).
Analysis (a) Process 1-2: isentropic compression.
⎛V
T2 = T1 ⎜⎜ 1
⎝V 2
⎞
⎟⎟
⎠
k −1
= (293 K )(20)
P
2
0.4
qin
3
= 971.1 K
4
Process 2-3: P = constant heat addition.
qout
V
P3V 3 P2V 2
T
2200K
=
⎯
⎯→ 3 = 3 =
= 2.265
V 2 T2 971.1K
T3
T2
1
v
Process 3-4: isentropic expansion.
⎛V
T4 = T3 ⎜⎜ 3
⎝V 4
⎞
⎟⎟
⎠
k −1
⎛ 2.265V 2
= T3 ⎜⎜
⎝ V4
⎞
⎟⎟
⎠
k −1
⎛ 2.265 ⎞
= T3 ⎜
⎟
⎝ r ⎠
k −1
⎛ 2.265 ⎞
= (2200 K )⎜
⎟
⎝ 20 ⎠
0.4
= 920.6 K
q in = h3 − h2 = c p (T3 − T2 ) = (1.005 kJ/kg ⋅ K )(2200 − 971.1)K = 1235 kJ/kg
q out = u 4 − u1 = cv (T4 − T1 ) = (0.718 kJ/kg ⋅ K )(920.6 − 293)K = 450.6 kJ/kg
wnet,out = q in − q out = 1235 − 450.6 = 784.4 kJ/kg
η th =
(b)
v1 =
wnet,out
q in
=
784.4 kJ/kg
= 63.5%
1235 kJ/kg
(
)
RT1
0.287 kPa ⋅ m 3 /kg ⋅ K (293 K )
=
= 0.885 m 3 /kg = v max
P1
95 kPa
v min = v 2 =
MEP =
v max
r
wnet,out
v1 −v 2
=
wnet,out
v 1 (1 − 1 / r )
=
⎛ kPa ⋅ m 3
⎜
0.885 m 3 /kg (1 − 1/20 ) ⎜⎝ kJ
(
784.4 kJ/kg
)
⎞
⎟ = 933 kPa
⎟
⎠
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
preparation. If you are a student using this Manual, you are using it without permission.
9-96
9-129 An ideal gas-turbine cycle with two stages of compression and two stages of expansion is considered. The back work
ratio and the thermal efficiency of the cycle are to be determined for the cases of with and without a regenerator.
Assumptions 1 The air standard assumptions are applicable. 2 Air is an ideal
gas with variable specific heats. 3 Kinetic and potential energy changes are
negligible.
Properties The properties of air are given in Table A-17.
T
1200 K
qin
9
Analysis (a) The work inputs to each stage of compressor are identical, so are
the work outputs of each stage of the turbine since this is an ideal cycle. Then,
⎯→
T1 = 300 K ⎯
h1 = 300.19 kJ/kg
Pr1 = 1.386
300 K
P
⎯→ h2 = h4 = 411.26 kJ/kg
Pr 2 = 2 Pr1 = (3)(1.386) = 4.158 ⎯
P1
T5 = 1200 K ⎯
⎯→
4
2
3
1
5
7
6
8
10
s
h5 = h7 = 1277.79 kJ/kg
Pr5 = 238
P6
⎛1⎞
Pr5 = ⎜ ⎟(238) = 79.33 ⎯
⎯→ h6 = h8 = 946.36 kJ/kg
P5
⎝ 3⎠
= 2(h2 − h1 ) = 2(411.26 − 300.19 ) = 222.14 kJ/kg
Pr6 =
wC,in
wT,out = 2(h5 − h6 ) = 2(1277.79 − 946.36) = 662.86 kJ/kg
Thus,
rbw =
wC,in
wT,out
=
222.14 kJ/kg
= 33.5%
662.86 kJ/kg
q in = (h5 − h4 ) + (h7 − h6 ) = (1277.79 − 411.26 ) + (1277.79 − 946.36 ) = 1197.96 kJ/kg
w net = wT,out − wC,in = 662.86 − 222.14 = 440.72 kJ/kg
η th =
wnet
440.72 kJ/kg
=
= 36.8%
1197.96 kJ/kg
q in
(b) When a regenerator is used, rbw remains the same. The thermal efficiency in this case becomes
q regen = ε (h8 − h4 ) = (0.75)(946.36 − 411.26 ) = 401.33 kJ/kg
q in = q in,old − q regen = 1197.96 − 401.33 = 796.63 kJ/kg
η th =
wnet 440.72 kJ/kg
=
= 55.3%
796.63 kJ/kg
q in
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10-13
10-21 A steam power plant operates on a simple ideal Rankine cycle between the specified pressure limits. The thermal
efficiency of the cycle, the mass flow rate of the steam, and the temperature rise of the cooling water are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.
Analysis (a) From the steam tables (Tables A-4, A-5, and A-6),
h1 = h f @ 10 kPa = 191.81 kJ/kg
T
v 1 = v f @ 10 kPa = 0.00101 m 3 /kg
w p ,in = v 1 (P2 − P1 )
(
)
⎛ 1 kJ
= 0.00101 m /kg (7,000 − 10 kPa )⎜
⎜ 1 kPa ⋅ m 3
⎝
= 7.06 kJ/kg
3
⎞
⎟
⎟
⎠
h2 = h1 + w p ,in = 191.81 + 7.06 = 198.87 kJ/kg
P3 = 7 MPa ⎫ h3 = 3411.4 kJ/kg
⎬
T3 = 500°C ⎭ s 3 = 6.8000 kJ/kg ⋅ K
3
2
7 MPa
qin
10 kPa
1
qout
4
s
s 4 − s f 6.8000 − 0.6492
P4 = 10 kPa ⎫
=
= 0.8201
⎬ x4 =
s 4 = s3
s fg
7.4996
⎭
h4 = h f + x 4 h fg = 191.81 + (0.8201)(2392.1) = 2153.6 kJ/kg
Thus,
q in = h3 − h2 = 3411.4 − 198.87 = 3212.5 kJ/kg
q out = h4 − h1 = 2153.6 − 191.81 = 1961.8 kJ/kg
wnet = q in − q out = 3212.5 − 1961.8 = 1250.7 kJ/kg
and
η th =
(b)
m& =
wnet 1250.7 kJ/kg
=
= 38.9%
q in
3212.5 kJ/kg
W&net
45,000 kJ/s
=
= 36.0 kg/s
wnet 1250.7 kJ/kg
(c) The rate of heat rejection to the cooling water and its temperature rise are
Q& out = m& q out = (35.98 kg/s )(1961.8 kJ/kg ) = 70,586 kJ/s
Q& out
70,586 kJ/s
∆Tcooling water =
=
= 8.4°C
(m& c) cooling water (2000 kg/s )(4.18 kJ/kg ⋅ °C )
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10-15
10-23 A simple Rankine cycle with water as the working fluid operates between the specified pressure limits. The rate of
heat addition in the boiler, the power input to the pumps, the net power, and the thermal efficiency of the cycle are to be
determined.
Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.
Analysis From the steam tables (Tables A-4, A-5, and A-6),
P1 = 50 kPa
T1 = Tsat @ 50 kPa
⎫ h1 ≅ h f @ 75°C = 314.03 kJ/kg
⎬
− 6.3 = 81.3 − 6.3 = 75°C ⎭ v 1 = v f @ 75°C = 0.001026 m 3 /kg
wp,in = v 1 ( P2 − P1 )
⎛ 1 kJ ⎞
= (0.001026 m 3 /kg )(6000 − 50)kPa ⎜
⎟
1 kPa ⋅ m 3 ⎠
⎝
= 6.10 kJ/kg
h2 = h1 + wp,in = 314.03 + 6.10 = 320.13 kJ/kg
T
6 MPa
2
P3 = 6000 kPa ⎫ h3 = 3302.9 kJ/kg
⎬
1
T3 = 450°C
⎭ s 3 = 6.7219 kJ/kg ⋅ K
s4 − s f
6.7219 − 1.0912
P4 = 50 kPa ⎫ x 4 s =
=
= 0.8660
s fg
6.5019
⎬
s 4 = s3
⎭ h = h + x h = 340.54 + (0.8660)(2304.7) = 2336.4 kJ/kg
4s
f
4 s fg
ηT =
3
qin
50 kPa
qout 4s 4
s
h3 − h 4
⎯
⎯→ h4 = h3 − η T (h3 − h4s ) = 3302.9 − (0.94)(3302.9 − 2336.4) = 2394.4 kJ/kg
h3 − h 4 s
Thus,
Q& in = m& (h3 − h2 ) = (20 kg/s)(3302.9 − 320.13)kJ/kg = 59,660 kW
W& T,out = m& (h3 − h4 ) = (20 kg/s)(3302.9 − 2394.4)kJ/kg = 18,170 kW
W& P,in = m& wP,in = (20 kg/s)(6.10 kJ/kg) = 122 kW
W& = W&
− W&
= 18,170 − 122 = 18,050 kW
net
T, out
P,in
and
η th =
W& net 18,050
=
= 0.3025
59,660
Q& in
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10-31
10-39 An ideal reheat Rankine with water as the working fluid is considered. The temperatures at the inlet of both turbines,
and the thermal efficiency of the cycle are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Kinetic and
potential energy changes are negligible.
Analysis From the steam tables (Tables A-4, A-5, and A-6),
T
h1 = h f @ 10 kPa = 191.81 kJ/kg
3
v 1 = v f @ 10 kPa = 0.001010 m /kg
wp,in = v 1 ( P2 − P1 )
⎛ 1 kJ ⎞
= (0.001010 m 3 /kg )(7000 − 10)kPa ⎜
⎟
⎝ 1 kPa ⋅ m 3 ⎠
= 7.06 kJ/kg
h2 = h1 + wp,in = 191.81 + 7.06 = 198.87 kJ/kg
5
7 MPa
3
2
800 kPa
10 kPa
1
P4 = 800 kPa ⎫ h4 = h f + x 4 h fg = 720.87 + (0.93)(2047.5) = 2625.0 kJ/kg
⎬
x 4 = 0.93
⎭ s 4 = s f + x 4 s fg = 2.0457 + (0.93)(4.6160) = 6.3385 kJ/kg ⋅ K
P3 = 7000 kPa ⎫ h3 = 3085.5 kJ/kg
⎬
s3 = s 4
⎭ T3 = 373.3°C
4
6
s
P6 = 10 kPa ⎫ h6 = h f + x 6 h fg = 191.81 + (0.93)(2392.1) = 2416.4 kJ/kg
⎬
x 6 = 0.90 ⎭ s 6 = s f + x 6 s fg = 0.6492 + (0.93)(7.4996) = 7.6239 kJ/kg ⋅ K
P5 = 800 kPa ⎫ h5 = 3302.0 kJ/kg
⎬
s5 = s6
⎭ T5 = 416.2°C
Thus,
q in = (h3 − h2 ) + (h5 − h4 ) = 3085.5 − 198.87 + 3302.0 − 2625.0 = 3563.6 kJ/kg
q out = h6 − h1 = 2416.4 − 191.81 = 2224.6 kJ/kg
and
η th = 1 −
q out
2224.6
=1−
= 0.3757 = 37.6%
3563.6
q in
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11-6
11-13 An ideal vapor-compression refrigeration cycle with refrigerant-134a as the working fluid is considered. The COP
and the power requirement are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.
Analysis In an ideal vapor-compression refrigeration cycle, the compression process is isentropic, the refrigerant enters the
compressor as a saturated vapor at the evaporator pressure, and leaves the condenser as saturated liquid at the condenser
pressure. From the refrigerant tables (Tables A-11, A-12, and A-13),
T1 = 4°C ⎫ h1 = h g @ 4°C = 252.77 kJ/kg
⎬
sat. vapor ⎭ s1 = s g @ 4°C = 0.92927 kJ/kg ⋅ K
T
P2 = 1 MPa ⎫
⎬ h2 = 275.29 kJ/kg
s 2 = s1
⎭
P3 = 1 MPa ⎫
⎬ h = hf
sat. liquid ⎭ 3
@ 1 MPa
·
QH
·
Win
= 107.32 kJ/kg
h4 ≅ h3 = 107.32 kJ/kg ( throttling )
The mass flow rate of the refrigerant is
Q& L = m& (h1 − h4 ) ⎯
⎯→ m& =
2
3 1 MPa
Q& L
400 kJ/s
=
= 2.750 kg/s
h1 − h4 (252.77 − 107.32) kJ/kg
4°C
4s
4
·
QL
1
s
The power requirement is
W& in = m& (h2 − h1 ) = (2.750 kg/s)(275.29 − 252.77) kJ/kg = 61.93 kW
The COP of the refrigerator is determined from its definition,
COPR =
Q& L
400 kW
=
= 6.46
&
Win 61.93 kW
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11-14
11-21 A refrigerator with refrigerant-134a as the working fluid is considered. The power input to the compressor, the rate
of heat removal from the refrigerated space, and the pressure drop and the rate of heat gain in the line between the
evaporator and the compressor are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.
Analysis (a) From the refrigerant tables (Tables A-12 and A-13),
h1 = 239.50 kJ/kg
P1 = 100 kPa ⎫
s = 0.97207 kJ/kg ⋅ K
T1 = −20°C ⎬⎭ 1
v 1 = 0.19841 m 3 /kg
T
P2 = 0.8 MPa ⎫
⎬ h2 s = 284.07 kJ/kg
s 2 s = s1
⎭
P3 = 0.75 MPa ⎫
⎬ h3 ≅ h f
T3 = 26°C
⎭
@ 26 °C
2s
·
QH
0.8 MPa
·
Win
3
= 87.83 kJ/kg
h4 ≅ h3 = 87.83 kJ/kg (throttling )
T5 = −26°C
sat. vapor
2
0.75
MPa
0.10 MPa
4
⎫ P5 = 0.10173 MPa
⎬ h = 234.68 kJ/kg
⎭ 5
·
QL
1
0.10 MPa
-20°C
-26°C
s
Then the mass flow rate of the refrigerant and the power input becomes
m& =
V&1
0.5/60 m 3 /s
=
= 0.0420 kg/s
v 1 0.19841 m 3 /kg
W& in = m& (h2 s − h1 ) / η C = (0.0420 kg/s )[(284.07 − 239.50 ) kJ/kg ] / (0.78) = 2.40 kW
(b) The rate of heat removal from the refrigerated space is
Q& L = m& (h5 − h4 ) = (0.0420 kg/s)(234.68 − 87.83) kJ/kg = 6.17 kW
(c) The pressure drop and the heat gain in the line between the evaporator and the compressor are
∆P = P5 − P1 = 101.73 − 100 = 1.73
and
Q& gain = m& (h1 − h5 ) = (0.0420 kg/s )(239.50 − 234.68) kJ/kg = 0.203 kW
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
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11-64
11-80 A regenerative gas refrigeration cycle using air as the working fluid is considered. The effectiveness of the
regenerator, the rate of heat removal from the refrigerated space, the COP of the cycle, and the refrigeration load and the
COP if this system operated on the simple gas refrigeration cycle are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal
gas with constant specific heats.
Properties The properties of air at room temperature are cp = 1.005 kJ/kg·K and k = 1.4 (Table A-2).
Analysis (a) From the isentropic relations,
⎛P
T2 s = T1 ⎜⎜ 2
⎝ P1
ηC =
0.80 =
⎞
⎟⎟
⎠
(k −1) / k
.
QL
= (273.2 K )(5)0.4 / 1.4 = 432.4 K
Heat
Exch.
h2 s − h1 T2 s − T1
=
h2 − h1
T2 − T1
432.4 − 273.2
⎯
⎯→ T2 = 472.5 K
T2 − 273.2
T5 s
ηT =
⎞
⎟⎟
⎠
(k −1) / k
⎛1⎞
= T4 ⎜ ⎟
⎝5⎠
Regenerator
Heat
Exch.
3
5
The temperature at state 4 can be determined by solving
the following two equations simultaneously:
⎛P
= T4 ⎜⎜ 5
⎝ P4
6
.
QH
4
0.4 / 1.4
Turbine
Compressor
h4 − h5
T − 193.2
→ 0.85 = 4
h 4 − h5 s
T4 − T5 s
Using EES, we obtain T4 = 281.3 K.
T
·
QH
An energy balance on the regenerator may be written as
m& c p (T3 − T4 ) = m& c p (T1 − T6 ) ⎯
⎯→ T3 − T4 = T1 − T6
or,
T6 = T1 − T3 + T4 = 273.2 − 308.2 + 281.3 = 246.3 K
2
2s
3
35°C
0°C
Qrege
1
4
The effectiveness of the regenerator is
ε regen =
1
2
h3 − h4 T3 − T4 308.2 − 281.3
=
=
= 0.434
h3 − h6 T3 − T6 308.2 − 246.3
-80°C
5
·6
5 QRefrig
s
(b) The refrigeration load is
Q& L = m& c p (T6 − T5 ) = (0.4 kg/s)(1.00 5 kJ/kg.K)(2 46.3 − 193.2) K = 21.36 kW
(c) The turbine and compressor powers and the COP of the cycle are
W& C,in = m& c p (T2 − T1 ) = (0.4 kg/s)(1.00 5 kJ/kg.K)(4 72.5 − 273.2) K = 80.13 kW
W& T, out = m& c p (T4 − T5 ) = (0.4 kg/s)(1.00 5 kJ/kg.K)(2 81.3 − 193.2)kJ/kg = 35.43 kW
COP =
Q& L
W&
net,in
=
Q& L
W& C,in − W& T, out
=
21.36
= 0.478
80.13 − 35.43
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11-65
(d) The simple gas refrigeration cycle analysis is as follows:
⎛1⎞
T4 s = T3 ⎜ ⎟
⎝r⎠
ηT =
(k −1) / k
⎛1⎞
= (308.2 K )⎜ ⎟
⎝5⎠
0.4 / 1.4
= 194.6 K
T3 − T4
308.2 − T4
⎯
⎯→ 0.85 =
⎯
⎯→ T4 = 211.6 K
T3 − T4 s
308.2 − 194.6
Q& L = m& c p (T1 − T4 )
2
T
35°C
0°C
= (0.4 kg/s)(1.005 kJ/kg.K)(273.2 − 211.6)kJ/kg
= 24.74 kW
·
QH
2
3
1
·
4 QRefrig
4s
s
W& net,in = m& c p (T2 − T1 ) − m& c p (T3 − T4 )
= (0.4 kg/s)(1.005 kJ/kg.K)[(472.5 − 273.2) − (308.2 − 211.6)kJ/kg ]
= 41.32 kW
COP =
Q& L
W&
net,in
=
24.74
= 0.599
41.32
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
preparation. If you are a student using this Manual, you are using it without permission.
Tutorial #14
ESO 201A
Thermodynamics
Instructor
Prof. Sameer Khandekar
Contact
Office: SL-109, Tel: 7038
E-mail: [email protected]
URL: home.iitk.ac.in/~samkhan
ESO 201A Thermodynamics
Instructor: Sameer Khandekar
Tutorial 14
[12-13] Verify the validity of the last Maxwell relation(Eq. 12-19) for refrigerant-134a at 50°C
and 0.7 MPa.
[12-19] Prove that:
[12-23]Using the Clapeyron equation, estimate the enthalpy of vaporization of steam at 300
kPa, and compare it to the tabulated value.
Additional Homework Problems (Tutorial 14)
[12-16] Using the Maxwell relations, determine a relation for (ds/dP)T for a gas whose equation
of state is P(v-b) = RT. [Answer: R/P]
[12-49] Derive an expression for the isothermal compressibility of a substance whose equation
of state is
where a and b are empirical constants.
[12-66] Demonstrate that the Joule-Thomson coefficient is given by
12-8
The Maxwell Relations
12-13 The validity of the last Maxwell relation for refrigerant-134a at a specified state is to be verified.
Analysis We do not have exact analytical property relations for refrigerant-134a, and thus we need to replace the
differential quantities in the last Maxwell relation with the corresponding finite quantities. Using property values from the
tables about the specified state,
⎛ ∂s ⎞ ? ⎛ ∂v ⎞
⎜ ⎟ =− ⎜
⎟
⎝ ∂P ⎠ T
⎝ ∂T ⎠ P
?
⎛ ∆s ⎞
⎛ ∆v ⎞
≅− ⎜
⎜
⎟
⎟
⎝ ∆P ⎠ T =50°C
⎝ ∆T ⎠ P =700kPa
⎛ s 900 kPa − s 500 kPa
⎜
⎜ (900 − 500 )kPa
⎝
? ⎛v
⎞
70°C − v 30°C
⎟
≅
−⎜⎜
⎟
⎠ T =50°C ⎝ (70 − 30 )°C
⎞
⎟
⎟
⎠ P =700kPa
(0.9660 − 1.0309)kJ/kg ⋅ K ? (0.036373 − 0.029966)m 3 /kg
≅−
(900 − 500)kPa
(70 − 30)°C
− 1.621 × 10 − 4 m 3 /kg ⋅ K ≅ −1.602 × 10 − 4 m 3 /kg ⋅ K
since kJ ≡ kPa·m³, and K ≡ °C for temperature differences. Thus the last Maxwell relation is satisfied.
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12-12
k ⎛ ∂P ⎞
⎛ ∂P ⎞
12-19 It is to be proven that ⎜
⎟ =
⎜
⎟
∂
T
k
− 1 ⎝ ∂T ⎠v
⎝
⎠s
Analysis Using the definition of cv ,
⎛ ∂s ⎞
⎛ ∂s ⎞ ⎛ ∂P ⎞
cv = T ⎜
⎟ = T⎜ ⎟ ⎜
⎟
T
∂
⎝
⎠v
⎝ ∂P ⎠ v ⎝ ∂T ⎠v
⎛ ∂s ⎞
⎛ ∂v ⎞
Substituting the first Maxwell relation ⎜ ⎟ = −⎜
⎟ ,
⎝ ∂P ⎠ v
⎝ ∂T ⎠ s
⎛ ∂v ⎞ ⎛ ∂P ⎞
cv = −T ⎜
⎟ ⎜
⎟
⎝ ∂T ⎠ s ⎝ ∂T ⎠ v
Using the definition of cp,
⎛ ∂s ⎞
⎛ ∂s ⎞ ⎛ ∂v ⎞
c p = T⎜
⎟ = T⎜ ⎟ ⎜
⎟
⎝ ∂T ⎠ P
⎝ ∂v ⎠ P ⎝ ∂T ⎠ P
⎛ ∂s ⎞
⎛ ∂P ⎞
Substituting the second Maxwell relation ⎜
⎟ =⎜
⎟ ,
⎝ ∂v ⎠ P ⎝ ∂T ⎠ s
⎛ ∂P ⎞ ⎛ ∂v ⎞
c p = T⎜
⎟ ⎜
⎟
⎝ ∂T ⎠ s ⎝ ∂T ⎠ P
From Eq. 12-46,
2
⎛ ∂v ⎞ ⎛ ∂P ⎞
c p − cv = −T ⎜
⎟ ⎜
⎟
⎝ ∂T ⎠ P ⎝ ∂v ⎠ T
Also,
cp
k
=
k −1 c p − cv
Then,
⎛ ∂P ⎞ ⎛ ∂v ⎞
⎟
⎜
⎟ ⎜
k
⎝ ∂T ⎠ s ⎝ ∂T ⎠ P
⎛ ∂P ⎞ ⎛ ∂T ⎞ ⎛ ∂v ⎞
=−
= −⎜
⎟ ⎜
⎟ ⎜
⎟
2
k −1
⎝ ∂T ⎠ s ⎝ ∂v ⎠ P ⎝ ∂P ⎠ T
⎛ ∂v ⎞ ⎛ ∂P ⎞
⎜
⎟ ⎜
⎟
⎝ ∂T ⎠ P ⎝ ∂v ⎠ T
Substituting this into the original equation in the problem statement produces
⎛ ∂P ⎞
⎛ ∂P ⎞ ⎛ ∂T ⎞ ⎛ ∂v ⎞ ⎛ ∂P ⎞
⎜
⎟ = −⎜
⎟ ⎜
⎟ ⎜ ⎟ ⎜
⎟
⎝ ∂T ⎠ s
⎝ ∂T ⎠ s ⎝ ∂v ⎠ P ⎝ ∂P ⎠ T ⎝ ∂T ⎠ v
But, according to the cyclic relation, the last three terms are equal to −1. Then,
⎛ ∂P ⎞
⎛ ∂P ⎞
⎜
⎟ =⎜
⎟
⎝ ∂T ⎠ s ⎝ ∂T ⎠ s
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12-14
The Clapeyron Equation
12-21C It enables us to determine the enthalpy of vaporization from hfg at a given temperature from the P, v, T data alone.
12-22C It is assumed that vfg ≅ vg ≅ RT/P, and hfg ≅ constant for small temperature intervals.
12-23 Using the Clapeyron equation, the enthalpy of vaporization of steam at a specified pressure is to be estimated and to
be compared to the tabulated data.
Analysis From the Clapeyron equation,
⎛ dP ⎞
h fg = Tv fg ⎜
⎟
⎝ dT ⎠ sat
⎛ ∆P ⎞
≅ T (v g − v f ) @300 kPa ⎜
⎟
⎝ ∆T ⎠ sat, 300 kPa
⎛
⎞
(325 − 275)kPa
⎟
= Tsat @300 kPa (v g − v f ) @300 kPa ⎜
⎜ Tsat @325 kPa − Tsat @275 kPa ⎟
⎝
⎠
⎞
⎛
50 kPa
3
⎟⎟
= (133.52 + 273.15 K)(0.60582 − 0.001073 m /kg)⎜⎜
⎝ (136.27 − 130.58)°C ⎠
= 2159.9 kJ/kg
The tabulated value of hfg at 300 kPa is 2163.5 kJ/kg.
12-24 The hfg and sfg of steam at a specified temperature are to be calculated using the Clapeyron equation and to be
compared to the tabulated data.
Analysis From the Clapeyron equation,
⎛ dP ⎞
h fg = Tv fg ⎜
⎟
⎝ dT ⎠ sat
⎛ ∆P ⎞
≅ T (v g − v f ) @120°C ⎜
⎟
⎝ ∆T ⎠ sat,120°C
⎛ Psat @125°C − Psat @115°C
= T (v g − v f ) @120°C ⎜⎜
125°C − 115°C
⎝
⎞
⎟
⎟
⎠
⎛ (232.23 − 169.18)kPa ⎞
⎟⎟
= (120 + 273.15 K)(0.89133 − 0.001060 m 3 /kg)⎜⎜
10 K
⎝
⎠
= 2206.8 kJ/kg
Also,
s fg =
h fg
T
=
2206.8 kJ/kg
= 5.6131 kJ/kg ⋅ K
(120 + 273.15)K
The tabulated values at 120°C are hfg = 2202.1 kJ/kg and sfg = 5.6013 kJ/kg·K.
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
preparation. If you are a student using this Manual, you are using it without permission.
12-10
12-15E The validity of the last Maxwell relation for steam at a specified state is to be verified.
Analysis We do not have exact analytical property relations for steam, and thus we need to replace the differential
quantities in the last Maxwell relation with the corresponding finite quantities. Using property values from the tables about
the specified state,
⎛ ∂s ⎞ ? ⎛ ∂v ⎞
⎜
⎟ =− ⎜
⎟
⎝ ∂P ⎠ T
⎝ ∂T ⎠ P
?
⎛ ∆s ⎞
⎛ ∆v ⎞
≅−⎜
⎜
⎟
⎟
P
∆
⎝
⎠ T =800° F
⎝ ∆T ⎠ P = 400 psia
⎛ s 450 psia − s 350 psia
⎜
⎜ (450 − 350)psia
⎝
? ⎛ v 900° F − v 700° F
⎞
⎟
≅
−⎜⎜
⎟
⎠ T =800° F
⎝ (900 − 700 )°F
⎞
⎟
⎟
⎠ P = 400 psia
(1.6706 − 1.7009)Btu/lbm ⋅ R ? (1.9777 − 1.6507)ft 3 /lbm
≅−
(450 − 350)psia
(900 − 700)°F
− 1.639 × 10 −3 ft 3 /lbm ⋅ R ≅ −1.635 × 10 −3 ft 3 /lbm ⋅ R
since 1 Btu ≡ 5.4039 psia·ft3, and R ≡ °F for temperature differences. Thus the fourth Maxwell relation is satisfied.
12-16 Using the Maxwell relations, a relation for (∂s/∂P)T for a gas whose equation of state is P(v-b) = RT is to be
obtained.
Analysis This equation of state can be expressed as v =
RT
+ b . Then,
P
R
⎛ ∂v ⎞
⎜
⎟ =
⎝ ∂T ⎠ P P
From the fourth Maxwell relation,
R
⎛ ∂s ⎞
⎛ ∂v ⎞
⎜ ⎟ = −⎜
⎟ =−
P
⎝ ∂P ⎠ T
⎝ ∂T ⎠ P
12-17 Using the Maxwell relations, a relation for (∂s/∂v)T for a gas whose equation of state is (P-a/v2)(v-b) = RT is to be
obtained.
Analysis This equation of state can be expressed as P =
RT
a
+ 2 . Then,
v −b v
R
⎛ ∂P ⎞
⎜
⎟ =
⎝ ∂T ⎠v v − b
From the third Maxwell relation,
R
⎛ ∂s ⎞
⎛ ∂P ⎞
⎜ ⎟ =⎜
⎟ =
⎝ ∂v ⎠ T ⎝ ∂T ⎠ v v − b
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
preparation. If you are a student using this Manual, you are using it without permission.
12-35
12-48 An expression for the isothermal compressibility of a substance whose equation of state is P (v − a ) = RT is to be
derived.
Analysis The definition for the isothermal compressibility is
α=−
1 ⎛ ∂v ⎞
⎜ ⎟
v ⎝ ∂P ⎠ T
Solving the equation of state for specific volume,
v=
RT
+a
P
The specific volume derivative is then
RT
⎛ ∂v ⎞
⎜
⎟ =− 2
P
⎝ ∂P ⎠ T
Substituting these into the isothermal compressibility equation gives
α=
RT ⎛
P
RT
⎞
⎜
⎟=
2 RT + aP
P
(
RT
+ aP)
P ⎝
⎠
12-49 An expression for the isothermal compressibility of a substance whose equation of state is P =
RT
a
−
v − b v (v + b)T 1 / 2
is to be derived.
Analysis The definition for the isothermal compressibility is
α=−
1 ⎛ ∂v ⎞
⎜ ⎟
v ⎝ ∂P ⎠ T
The derivative is
RT
a
2v + b
⎛ ∂P ⎞
+ 1/ 2 2
⎜ ⎟ =−
2
(v − b)
T
v (v + b) 2
⎝ ∂v ⎠ T
Substituting,
⎛
⎜
1 ⎛ ∂v ⎞
1⎜
1
α =− ⎜ ⎟ =− ⎜
2v + b
RT
a
v ⎝ ∂P ⎠ T
v
+ 1/ 2 2
⎜⎜ −
2
(
v
−
b
)
T
v
(v + b) 2
⎝
⎞
⎟
1
⎟
⎟=−
RTv
a 2v + b
−
+ 1/ 2
⎟⎟
2
−
(
v
b
)
T
(v + b) 2
⎠
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
preparation. If you are a student using this Manual, you are using it without permission.
12-44
12-65 The equation of state of a gas is given by v =
RT bP
−
. An equation for the Joule-Thomson coefficient inversion
P T2
line using this equation is to be derived.
Analysis From Eq. 12-52 of the text,
cp =
⎤
1 ⎡ ⎛ ∂v ⎞
⎟ −v ⎥
⎢T ⎜
µ ⎣ ⎝ ∂T ⎠ P
⎦
When µ = 0 as it does on the inversion line, this equation becomes
⎛ ∂v ⎞
T⎜
⎟ =v
⎝ ∂T ⎠ P
Using the equation of state to evaluate the partial derivative,
R
bP
⎛ ∂v ⎞
⎜
⎟ = +2 3
∂
T
P
T
⎝
⎠P
Substituting this result into the previous expression produces
RT
bP RT bP
bP
+2 2 =
− 2 ⎯
⎯→ 3 2 = 0
P
P
T
T
T
The condition along the inversion line is then
P=0
12-66 It is to be demonstrated that the Joule-Thomson coefficient is given by µ =
T2
cp
⎛ ∂ (v / T ) ⎞
⎟ .
⎜
⎝ ∂T ⎠ P
Analysis From Eq. 12-52 of the text,
cp =
1
µ
⎡ ⎛ ∂v ⎞
⎤
⎟ −v ⎥
⎢T ⎜
⎣ ⎝ ∂T ⎠ P
⎦
Expanding the partial derivative of v/T produces
1
⎛ ∂v / T ⎞
⎜
⎟ =
⎝ ∂T ⎠ P T
v
⎛ ∂v ⎞
⎜
⎟ − 2
⎝ ∂T ⎠ P T
When this is multiplied by T2, the right-hand side becomes the same as the bracketed quantity above. Then,
µ=
T2
cp
⎛ ∂ (v / T ) ⎞
⎜
⎟
⎝ ∂T ⎠ P
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
preparation. If you are a student using this Manual, you are using it without permission.
ESO 201A
Thermodynamics
End of file
Instructor
Prof. Sameer Khandekar
Department of Mechanical Engineering
Indian Institute of Technology Kanpur
Kanpur (UP) 208016 India
Contact
Office: SL-109, Tel: 7038
E-mail: [email protected]
URL: home.iitk.ac.in/~samkhan