full course file (including all tutorial/hw solutions)

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

mailto:[email protected]

http://home.iitk.ac.in/~samkhan

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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

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Course File: ESO201A Prof. Sameer Khandekar

INDIAN INSTITUTE OF TECHNOLOGY KANPUR ESO 201A: Thermodynamics Instructor: Sameer Khandekar

Timings: Lectures: MWF 10.00-10.50 AM in L16 Tutorials: Th 10.00- 10.50 AM (10 class rooms, as indicated below)

Faculty coordinates:

# Name Dep.

E-mail Tel Office Section Classroom

Instructor

Sameer Khandekar ME samkhan 7038 SL-109 - L16

Tutors 1 Dr. Jayant Singh CHE jayantks 6141 NL2-202 1 TB-203

2 Dr. Nitin Kaistha CHE nkaistha 7513 FB-457 2 TB-204

3 Dr. Indranil Saha Dalal CHE indrasd 6072 FB-467 3 TB-205

4 Dr. Shikha Prasad ME shikhap 6973 FB-438 4 TB-206

5 Dr. Prabhat Munshi ME pmunshi 7243 FB-402 5 TB-207

6 Dr. Subrata Sarkar ME subra 7942 FB-353 6 TB-208

7 Dr. Jishnu Bhattacharya ME jishnu 7684 NL-302 7 TB-209

8 Mr. Srihari Dinesh AE jsdinesh 7727 KANZI House 8 TB-210

9 Mr. Pradeep K. S. AE pradeeps 6843 NWTF - 302 9 TB-211

10 Mr. Azmira Nageshwar Rao AE azmiran 7727 KANZI House 10 TB-212

Grading Policy:

Item Points Remarks

Mid-semester Examination 25 points Closed book/notes

Two announced quizzes 25 points (12.5 points each) Closed book/notes

Final Examination 40 points Closed book/notes

Tutorial marks 10 points Attendance + Tutor input

Total 100 points

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: 978-0-07107254-0, SI Units, of the above book (available in the Shopping Centre at approx. Rs. 700/-).

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Monday Tuesday Wednesday Thursday Friday Saturday Sunday

July 27 28 29 30 (T1) 31 - -

- - - - - 1 2

3 4 5 6 (T2) 7 8 9

10 11 12 13 (T3) 14 15 16

17 18 19 20 (T4) 21 22 23

24 25 26 27 (T5-QUIZ#1) 28 29 (Friday plan) 30

31 - - - - - -

- 1 2 3 (T6) 4 5 6

7 8 9 10 (T7) 11 12 13

14 15 16 (MIDSEM) 17 18 19 20 MIDSEM

21 22 23 24 (T8) 25 26 27

28 29 30 - - - -

- - - 1 (T9) 2 3 (EXTRA CLASS) 4

5 6 7 8 (T10) 9 10 11

12 13 14 15 (T11-QUIZ#2) 16 17 18

19 20 21 22 23 24 25 RECESS

26 27 28 29 (T12) 30 31

2 3 4 5 (T13) 6 7 8

9 10 11 12 (T14) 13 14 15

16 17 18 (ENDSEM) 19 20 21 22 ENDSEM

23 24 25 26 27 28 29 ENDSEM

30

Institute holiday

Tutorial = 12 + 2 Quizes

Lecture = 39 + 2 Extra Classes

Examination

August

September

October

November

ESO201A: Thermodynamics

Instructor: Dr. Sameer Khandekar

2015-2016 (First Semester)

Semester Plan

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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, non-

mechanical 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.

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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 – Webb-

Ruben 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

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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

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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.

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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

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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)

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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 14

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 21

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 32

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

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Tutorial #1

ESO 201A

Thermodynamics

Instructor Prof. Sameer Khandekar

Contact

Office: SL-109, Tel: 7038 E-mail: [email protected]






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 one-

seventh 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.

Blood h

Analysis For a given gage pressure, the relation ghP ρ= can be expressed for mercury and blood as bloodblood ghP ρ= and mercurymercury ghP ρ= . Setting these two relations equal to each other we get

mercurymercurybloodblood ghghP ρρ ==

Solving for blood height and substituting gives

m 1.55=== m) 12.0(kg/m 1050kg/m 600,13

3

3

mercuryblood

mercuryblood hh

ρ

ρ

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 h

P

33 kg/m 1030)kg/m 0(1.03)(100SG2

==×= OHρρ

The pressure exerted on a diver at 30 m below the free surface of the sea is the absolute pressure at that location:

kPa 404=⎟⎟⎠

⎞⎜⎜⎝

⎛+=

+=

223

atm

N/m 1000kPa 1m) )(30m/s )(9.807kg/m (1030kPa) (101

ghPP ρ

preparation. If you are a student using this Manual, you are using it without permission.

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course

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

D =12 m

m = 170 kg

N 296,10m/skg 1N 1

)m )(904.8m/s)(9.81kg/m (1.16

m 8.904/3m) π(64/3r4π

2323

balloonair

333

=⎟⎟⎠

⎞⎜⎜⎝

⎛

⋅=

====

V

V

gFB

balloon

ρ

The total mass is

kg 9.3198529.149

kg 9.149)m (904.8kg/m7

1.16

peopleHetotal

33HeHe

=×+=+=

=⎟⎠⎞

⎜⎝⎛==

mmm

m Vρ

The total weight is

N 3138m/skg 1N 1

)m/s kg)(9.81 (319.92

2total =⎟

⎟⎠

⎞⎜⎜⎝

⎛

⋅== gmW

Thus the net force acting on the balloon is

N 71573138296,10net =−=−= WFF B

Then the acceleration becomes

2m/s 22.4=⎟⎟⎠

⎞⎜⎜⎝

⎛ ⋅==

N 1m/skg1

kg 319.9N 7157 2

total

net

mF

a



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.

P

Patm

W = mg

Analysis Drawing the free body diagram of the piston and balancing the vertical forces yield

⎟⎟⎠

⎞⎜⎜⎝

⎛ ⋅×−=

−=−=

−

kPa1skg/m 1000

)m10kPa)(25 100(180)m/s (9.81)(

)(2

242atm

atm

m

APPmgAPPAW

It yields 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.

P

Patm

W = mg

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

kg 0.0408=

⎟⎟⎠

⎞⎜⎜⎝

⎛ ⋅×==

=

−

kPa 1skg/m 1000

m/s 9.81)m10kPa)(4 (100 2

2

26gage

gage

gAP

m

APW

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.

Water

Patm= 99 kPa h

Analysis The pressure at the bottom of the tube can be expressed as

tubeatm )( hgPP ρ+=

Solving for h,

m 2.14=

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛ ⋅−=

−=

kPa 1N/m 1000

N 1m/skg 1

)m/s )(9.81kg/m (1000kPa 99)(120 22

23

atm

gPP

hρ



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 suppliedConversion efficiency

=

Substituting,

Natural gas heater: Btu/103.21$Btu 1025

ft 10.55

$0.012/ftenergy useful ofcost Unit 633

−×=⎟⎟⎠

⎞⎜⎜⎝

⎛=

Heating by oil heater: Btu/101.15$Btu 138,700

gal 10.55

$1.15/galenergy useful ofcost Unit 6−×=⎟⎟⎠

⎞⎜⎜⎝

⎛=

Electric heater: Btu/104.27$Btu 3412

kWh 10.90

)$0.084/kWhenergy useful ofcost Unit 6−×=⎟

⎠⎞

⎜⎝⎛=

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 suppliedConversion efficiency

=

Substituting,

Natural gas heater: kJ/105.13$kJ 105,500

therm10.87

m$1.24/therenergy useful ofcost Unit 6−×=⎟⎟⎠

⎞⎜⎜⎝

⎛=

Heating oil heater: kJ/104.10$kJ 138,500

gal 10.87

$1.25/galenergy useful ofcost Unit 6−×=⎟⎟⎠

⎞⎜⎜⎝

⎛=

Electric heater: kJ/100.25$kJ 3600

kWh 11.0

$0.09/kWh)energy useful ofcost Unit 6−×=⎟

⎠⎞

⎜⎝⎛=

Therefore, the system with the lowest energy cost for heating the house is the heating oil heater.

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.

1-9

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

NV

v =

and the molar specific volume of one of the smaller systems is

NN

/ VVv ==

2/2

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.



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,

R 762

F302=+=+°=

°=+=+°=°460302460)F()R(

32)150)(8.1(32)C(8.1)F(TT

TT

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



1-13

1-39E The flash point temperature of engine oil given in °F unit is to be converted to K and R units.


K 457

R 823

===

=+=+°=

8.1823

1.8)R()K(

460363460)F()R(TT

TT

1-40E The temperature of ambient air given in °C unit is to be converted to °F, K and R units.


R 419.67K 233.15

C40

=+−==+−=

°−=+−=°−=

67.4594015.27340

32)8.1)(40(C40

TTT

1-41E The change in water temperature given in °F unit is to be converted to °C, K and R units.


R 10K 5.6C5.6

=°=∆==∆

°==∆

F108.1/108.1/10

TTT

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

C18.3°=−

=−°

=°8.13265

8.132)F()C( TT

C23.9°=−

=−°

=°8.13275

8.132)F()C( TT

A temperature change of 10°F in various units are

K 5.6

C5.6

R 10

=°∆=∆

°==°∆

=°∆

=°∆=∆

)C()K(8.1

101.8

)F()C(

)F()R(

TT

TT

TT



Tutorial #2

ESO 201A

Thermodynamics


Contact

Office: SL‐109, Tel: 7038 E‐mail: [email protected]




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 air-conditioning 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.


R 762

F302=+=+°=

°=+=+°=°460302460)F()R(

32)150)(8.1(32)C(8.1)F(TT

TT

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



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 for a given mass flow rate: 2/2Vm&

Vj

Nozzle

Shaft

kJ/kg 8.1/sm 1000

kJ/kg 12

)m/s 60(2 22

22

mech =⎟⎠⎞

⎜⎝⎛===

Vkee

kW 216=⎟

⎠⎞

⎜⎝⎛=

==

kJ/s 1kW 1kJ/kg) kg/s)(1.8 120(

mechmechmax emEW &&&

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.


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

gal/year 6.428miles/gal 35

miles/year 15,000gallonper Miles

yearper driven MilesnConsumptio Fuel Annual High ===

gal/year 4.405miles/gal 37

miles/year 15,000gallonper Miles

yearper driven MilesnConsumptio Fuel Annual Low ===

Then the fuel and money saved per year become

gal/year 23.2gal/year 405.4gal/year 6.428

nConsumptio Fuel AnnualnConsumptio Fuel AnnualSavings Fuel LowHigh

=−=

−=

$51.0/year=== $2.20/gal)gal/year)( 2.23()fuel ofcost Unit )(savings Fuel(savingsCost

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

kJ/kg 0.45=⎟⎠

⎞⎜⎝

⎛−=

−=∆

22

2221

22

/sm 1000kJ/kg 1

2)m/s 0()m/s 30(

2ke

VV


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

kJ/kg 0.98=⎟⎠

⎞⎜⎝

⎛−=−=∆22

212

/sm 1000kJ/kg 1m )0100)(m/s 8.9()(pe zzg

2-46E A water pump increases water pressure. The power input is to be determined.

Analysis The power input is determined from

50 psia

hp 12.6=

⎟⎠

⎞⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛

⋅−=

−=

Btu/s 0.7068hp 1

ftpsia 5.404Btu 1psia)1050)(/sft 2.1(

)(

33

12 PPW V&&

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 Q Q Qcooling lights people heat gain= + +

15,000 kJ/h Qcool·

Room

40 people 10 bulbs

where

&

&

&

Q

Q

Q

lights

people

heat gain

10 100 W 1 kW

40 360 kJ / h 4 kW

15,000 kJ / h 4.17 kW

= × =

= × =

= =

Substituting,

& .Qcooling 9.17 kW= + + =1 4 4 17

Thus the number of air-conditioning units required is

units 2⎯→⎯= 1.83kW/unit 5

kW 9.17


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

4342143421&&

energies etc. otential, p kinetic,internal, in change of Rate

system

massand work, heat, by transferenergy netof Rate

/ dtdEEE outin =− → inEdtdE &=/room

- Lights - TV - Refrig - Iron

Electricity

ROOM since no energy is leaving the room in any form, and thus . Also, 0=outE&

W 1410

W 1000200110100ironrefrigTVlightsin

=+++=

+++= EEEEE &&&&&

Substituting, the rate of increase in the energy content of the room becomes

W1410== inroom / EdtdE &

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

0/

energies etc. potential, kinetic, internal,in change of Rate

(steady) 0system

mass and work,heat,by nsferenergy tranet of Rate

==−444 3444 2143421

&& dtdEEE outin → outin EE && =

2ke

2out

airoutairin sh,V

mmW &&& ==

where

kg/s 62.10/s)m )(9kg/m 18.1( 33air === V&& ρm

Substituting, the minimum power input required is determined to be

W340==⎟⎠

⎞⎜⎝

⎛== J/s 340/sm 1

J/kg 12

m/s) 8(kg/s) .6210(2 22

22out

airin sh,V

mW &&

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.


Tutorial #3

ESO 201A

Thermodynamics


Contact





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.


[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 CO2 (carbondioxide), which is the principal gas causing the greenhouse effect and thus global climate change. On average, 0.59 kg of CO2 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 CO2 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

kJ/kg 491.0/sm 1000

kJ/kg 1m) 50)(m/s (9.8122

2

outmech,inmech,

=

⎟⎠

⎞⎜⎝

⎛=

==− ghPeeρ

Then the rate at which mechanical energy of fluid supplied to the turbine and the overall efficiency become

kW 2455kJ/kg) 491kg/s)(0. 5000()(|| inmech,inmech,fluidmech, ==−=∆ eemE &&

0.760==∆

==kW 2455kW 1862

|| fluidmech,

outelect,gen-turbineoverall E

W&

&ηη

(b) Knowing the overall and generator efficiencies, the mechanical efficiency of the turbine is determined from

0.800===→=95.076.0

generator

gen-turbineturbinegeneratorturbinegen-turbine η

ηηηηη

(c) The shaft power output is determined from the definition of mechanical efficiency,

kW 1960≈==∆= kW 1964kW) 2455)(800.0(|| fluidmech,turbineoutshaft, EW && η

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.


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. Wind

turbine

80 m 7 m/s

WindProperties 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 for a given mass flow rate:

2/2Vm&

kJ/kg 0245.0/sm 1000

kJ/kg 12

)m/s 7(2 22

22

mech =⎟⎠

⎞⎜⎝

⎛===Vkee

kg/s 982,434m) (80m/s) 7)(kg/m 25.1(

4

23

2

====ππ

ρρD

VVAm&

kW 1078==== kJ/kg) 45kg/s)(0.02 982,43(mechmechmax emEW &&&

The actual electric power generation is determined by multiplying the power generation potential by the efficiency,

kW 323=== kW) 1078)(30.0(max turbinewindelect WW && η

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.


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. 2Analysis (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 1

Storagetank

Pump 20 m

kg/s70/s)m 070.0)( kg/m1000( 33 === V&& ρm

kJ/kg196.0/sm 1000

kJ/kg1m) 20)(m/s (9.81 22

222 =⎟

⎠⎞

⎜⎝⎛== gzpe

Then the rate of increase of the mechanical energy of water becomes

kW 13.7kJ/kg) 6kg/s)(0.19 70()0()( 22inmech,outmech,fluidmech, ===−=−=∆ pempemeemE &&&&

The overall efficiency of the combined pump-motor unit is determined from its definition,

67.2%or 0.672 kW20.4 kW7.13

inelect,

fluidmech,motor-pump ==

∆=

W

E&

&η

(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:

PPP

meemE ∆=−

=−=∆ V&&&&ρ

12inmech,outmech,fluidmech, )(

Solving for ∆P and substituting,

kPa 196=⎟⎟⎠

⎞⎜⎜⎝

⎛ ⋅=

∆=∆

kJ1m kPa1

/sm 0.070 kJ/s13.7 3

3fluidmech,

V&

&EP

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.


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.

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 for a given mass flow rate:

2/2Vm&

kJ/kg032.0/sm 1000

kJ/kg12

)m/s 8(2 22

22

mech =⎟⎠⎞

⎜⎝⎛===

Vkee

kg/s540,784

m) (100m/s) 8)( kg/m25.1(4

23

2

====ππ

ρρD

VVAm&

100 m

Wind turbine

8 m/s

Wind

kW2513 kJ/kg)032 kg/s)(0.540,78(mechmechmax ==== emEW &&&

The actual electric power generation is determined from

kW 804.2=== kW)2513)(32.0(maxnewind turbielect WW && η

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.

PUMP

Pump inlet

6 hp

∆P = 1.2 psiAssumptions 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

hp 4.71 Btu/s .333ftpsi 5.404

Btu 1psi) /s)(1.2ft 15( )(

)(])()[()(

33

12

1212inmech,outmech,fluidmech,

==⎟⎟⎠

⎞⎜⎜⎝

⎛

⋅=−=

−=−=−=∆

PP

PPmPPmeemE

V

vvv

&

&&&&

since 1 hp = 0.7068 Btu/s, , and there is no change in kinetic and potential energies of the fluid. Then the mechanical efficiency of the pump becomes

vVV /&&& == ρm

78.6%or 0.786hp 6

hp 71.4

shaft pump,

fluidmech,pump ==

∆=

WE&

&η

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.


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 . 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

2/2mech VPvghe ++=

⎟⎟⎠

⎞⎜⎜⎝

⎛ −+−=⎟

⎟⎠

⎞⎜⎜⎝

⎛−−+=−=∆

2)(

2)(

2)()(

21

22

12

21

1

22

2inmech,outmech,fluidmech,VV

PPV

PvV

PvmeemE ρV&&&&

since , and there is no change in the potential energy of the fluid. Also,

vVV /&&& == ρm

m/s 9.194/m) (0.08

/sm 1.04/ 2

3

211

1 ====ππDA

V VV &&

m/s 84.84/m) (0.12

/sm 1.04/ 2

3

222

2 ====ππDA

V VV &&

Substituting, the useful pumping power is determined to be

kW3.26

m/s kN1 kW1

m/s kg1000 kN1

2m/s) (19.9)m/s 84.8( ) kg/m860( kN/m400/s)m (0.1 2

22323

fluidmech,upump,

=

⎟⎠⎞

⎜⎝⎛

⋅⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛

⋅−

+=

∆= EW &&

PUMP

Pump inlet

1

2

Motor

35 kW

Then the shaft power and the mechanical efficiency of the pump become

kW 5.31kW) 35)(90.0(electricmotorshaftpump, === WW && η

83.6%==== 836.0 kW31.5 kW3.26

shaft pump,

upump,pump W

W&

&η

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.


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

m 42.88

s 60min 1L/min) 15(

km/h 3.6m/s 1km/h) 250(

tip =⎟⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛

==ππn

VD

&

222

m 61404

m) 42.88(4

===ππDA

Then the average velocity of air through the wind turbine becomes

m/s 5.23===)m )(6140 kg/m31.1(

kg/s000,4223A

mVρ&

(b) The kinetic energy of the air flowing through the turbine is

kW3.574m/s) 3 kg/s)(5.2000,42(21V

21EK 22 === m&&

Then the conversion efficiency of the turbine becomes

31.3%0.313 ==== kW3.574

kW180EK &

&Wη

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.


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

ton/year CO 123,000 2=×=

==

kg/year CO 1023.1

kg/kWh) (0.59household)kWh/year (700household) (300,000kWh)per CO ofAmount consumed)(y electricit ofAmount (produced CO ofAmount

28

22

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

ton/year CO 231,000 2=×=

==

kg/year CO 1031.2

kg/kWh) old)(1.1kWh/househ (700household) (300,000kWh)per CO ofAmount consumed)(y electricit ofAmount (produced CO ofAmount

28

22

Therefore, the refrigerators in this city are responsible for the production of 231,000 tons of CO2.


Tutorial #4

ESO 201A

Thermodynamics


Contact





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]


[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


e initial temperature and the final pressure are to be determined.

nalysis This is a constant volume process. The specific volume is

3-32 A rigid container that is filled with water is cooled. Th

A

/kgm 150.0 3=kg 1

m 150.0 3

1 = mvv

initial stat rheated va e temperature is determined

A (Table /kgm

a13

1

1 C⎭⎬⎫

==

TPv

is a const e cooling process (v = V /m = stant). Theixture and thus the pressure is the saturation pressure at

e final temperature:

4)-A (Table /kgm 150.0

C40C40 @sat 23

12

2 kPa 7.385==⎭⎬⎫

==°=

°PPT

vv

3-33 initial pressure are to be determined.

Analysis

2 ==V

The e is supe por. Th to be

6)-150.0MP 2

395°=

This ant volum con final state is saturated mth

A rigid container that is filled with R-134a is heated. The final temperature and

This is a constant volume process. The specific volume is

/kgm 1348.0kg m 348 3=

e pressure is th

ined by

13)-A lekPa 200 ⎫=P

10m.1 3

21 ===V

vv

The initial state is determined to be a mixture, and thus th e saturation pressure at the given temperature

11)-A (Table C40- @sat 1 kPa 51.25== °PP

The final state is superheated vapor and the temperature is determinterpolation to be

(Tab /kgm 1348.0 23

2

2 C66.3°=⎭⎬=

Tv

R-134a -40°C 10 kg

1.348 m3

P

v

2

1

v

2

1

Q H2O Pa

kg 2 M1

150 L

P


3-26


3-57 The properties of compressed liquid water at a specified state are to be determined using the compressed liquid tables,

error) (4.53% kJ/kg .02335C80@

C80@

3

=≅ °

°

f

f

hh

liquid table (Table A-7),

kJ/kg .50330=⎬

and also sing 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,

by u

T = 80°C ⇒ error) (1.35%kJ/kg 97.334

error) (0.90% /kgm 0.001029C80@

=≅=≅ °f

uuvv

From compressed

=kJ/kg 0.9035

C08=⎭°

hT

/kgm 0.00102MPa 20

3=⎫= u

id approximation are listed above in parentheses.

Pv

The percent errors involved in the saturated liqu


3-28


3-59 Superheated steam in a piston-cylinder device is cooled at constant pressure until half of the mass condenses. The final

C179.88°== MPa sat@1TT (Table A-5)

the final state is specified to be x2 = 0.5. The specific tial and the final states are

/kgm 0.25799 3 (Table A-6)

/kg)001127.019436.0(5.0

3−×

3m 20.128−=− /kgm0.25799)5977 3

tank is cooled until the vapor starts condensing. The initial pressure in the tank is to be etermined.

is a constant volume process (v = V /m = constant), and the ume is equal to the final specific volume that is

/kgm 79270.0 3= (Table A-4)

s condensing at 150°C. Then from

MPa 0.30=⎭⎬⎫

1/kgP

tempe 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 liquid-

rature

vapor mixture, and thus the final temperature must be the saturation temperature at the final pressure,

H2O 300°C 1 MPa

T

v

2

1

H2O T1= 250°C

P1 = ?

T

v

2

1 25

15

°C

(c) The quality at olumes at the iniv

C300

MPa 1

1

1 =⎭⎬⎫

=voT

1.0=P

MPa 1.022

2 +=⎬⎫=

f xP

vvv001127.0

5.02 +=⎭= fgx

m .097750=

Thus,

=−= kg)(0.0 (0.8)(∆ 12 vvV m

3-60 The water in a rigidd

Analysis Thisinitial specific vol

C124@21 == °gvvv

since the vapor startTable A-6,

=1 025T °C= 3

1 m 0.79270v


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.


0.001157 m3/kg and u = 850.46 kJ/kg (Table A-4).

e volume of the cylinder at the initial state is

0011.0kg)( 4.1(== vV m

Properties The saturated liquid properties of water at 200°C are: vf =

Analysis (a) The cylinder initially contains saturated liquid water. Thf

33 m 001619.0/kg)m 57 = 11

The volume at the final state is

3m 0.006476== )001619.0(4V

(b) The final state properties are

kg/m 004626.0kg 1.4

m 0.006476 32 ===

mV

v

kg/m 004626.02

232 =

°=

⎪⎪⎬⎫

== P

TkPa 21,367

C371.3v

3

kJ/kg 5.220112

2 =⎭ux (Table A-4 or A-5 or EES)

kJ 1892

(c) The total internal energy change is determined from

==−=∆ kJ/kg 850.46)-kg)(2201.5 4.1()( 12 uumU

ror involved in using the enthalpy of water by the incompressible liquid approximation is to be determined.

7E)-A B 51.376 F400

psia 1500=

⎭⎬⎫

°==

hTP

4E)-A (TableBtu/lbm 04.75

he error involved is

3-62E The er

Analysis The state of water is compressed liquid. From the steam tables,

(Table tu/lbm

Based upon the incompressible liquid approximation,

3 F400

psia 1500F400 @ =≅

⎭⎬⎫

°==

°fhhTP

T

0.39%=×−

= 10051.376

04.37551.376ErrorPercent

which is quite acceptable in most engineering calculations.

Water 1.4 kg, 200°C

sat. liq. Q


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.


-100 0 100 200 300 400 500101

102

103

104

105 R134a

70°C

Pa]

40°C

P [k

10°C

-10°C

h [kJ/kg]

-30°C

0.2

0

.3

0.5

0.8

1

1.2

kJ/

kg-K

3-68 xture is heated until it reaches the critical state. The mass of the quid 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 l specific volume of water,

he total mass is

A rigid vessel that contains a saturated liquid-vapor mili

will be equal to the final specific volume, which is equal to the critica

/kgm 0.003106 321 === crvvv (last row of Table A-4)

T

kg .6096/kgm 0.003106

m 0.33

3===

vVm

At 150°C, vf = 0.001091 m3/kg and vg = 0.39248 m3/kg (Table A-4). Then the quality of water at the initial state is

0.0051490.0010910.392480.0010910.0031061

1 =−−

=−

=fg

fxv

vv

Then the mass of the liquid phase and its volume at the initial state are determined from

3m 0.105

kg 96.10

===

=−=−=

/kg)m 91kg)(0.0010 (96.10

96.60)0.005149)((1)1(3

1

fff

tf

m

mxm

vV

T

v

cp

vcr

H2O 150°C


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

A-1).

ideal gas, the volume of the second tank and e mass of air in the first tank are determined to be

and th l 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

e fina

/kg.K (Table

Analysis Let's call the first and the second tanks A and B. Treating air as an th

kg 5.846K) K)(298/kgmkPa (0.287

)m kPa)(1.0 (500

kPa 200

K) K)(308/kgmkPa kg)(0.287 (5

3

3

1

3

1

11

=⋅⋅⎠⎝

=⋅⋅

=⎟⎟⎠

⎞⎜⎜⎝

=

A

BPRTm 3m2.21

kg 10.8465.0m 3.211 3

==

BA

he al eq pressure b

⎛BV

1 =⎟⎟⎞

⎜⎜⎛

=A RTPm V

Thus,

5.846

2.21.0+=+=

+=+= BA

mmmVVV

T n the fin uilibrium ecomes

kPa284.1 m 3.21

K) K)(293/kgmkPa kg)(0.287 0.8463

3=

⋅⋅=

VmR

3-7 lastic ntains air at a specified state. The volume is doubled at the same pressure. The initial volume and the final temperature are to be determined.

ecified conditions, air behaves as an ideal gas.

nalysis According to the ideal gas equation of state,

(1T2 =2P

9E An e tank co

Assumptions At sp

A

F590R 1050

ft 404.9 3

°==⎯→⎯+

=⎯→⎯=

=

+⋅⋅=

=

22

1

2

1

2

3

R 460)(652

R 460)R)(65/lbmolftpsia 73lbmol)(10. 3.2(psia) (32

TT

TT

TnRP u

V

V

V

V

V

Air V = 1 m3

T = 25°C P = 500 kPa

Air m = 5 kg T = 35°C

P = 200 kPa

×

A B


3-39

Compressibility Factor


it is from re the gas deviates from

factor Z at the same reduced temperature and pressure.

-86C Reduced pressure is the pressure normalized with respect to the critical pressure; and reduced temperature is the mperature normalized with respect to the critical temperature.

cr 22.06 MPa

3-84C It represent the deviation from ideal gas behavior. The further away 1, the moideal gas behavior.

3-85C All gases have the same compressibility

3te

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, T = 647.1 K, P =cr

Analysis (a) From the ideal gas equation of state,

error) 7.0%(/kgm3K) K)(623.15/kgmkPa (0.4615 3RT

(b) From

0.01917===kPa 15,000P

v⋅⋅ 6

the compressibility chart (Fig. A-15),

65.0K

0.453MPa 22.06

MPa 10

=

⎪⎪⎭

⎪⎪⎬

⎫

===

===

Z

TT

PPP

crR

R

error) (8.5%

(c) From

1.04K 647.1

673Tcr

Thus,

/kgm 0.01246 3=== /kg)m 1917(0.65)(0.0 3idealvv Z

the superheated steam table (Table A-6),

} /kgm 0.01148 3=°== v C035

MPa 15TP

H2O 15 MPa 350°C


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

1555 1 2

88.16 1 V (m3)

P (kPa)

/kgm 10200.0

)001157.012721.0(80.0001157.0

/kgm 001157.0

kPa 9.1554

3

2

3C200 @ f1

C200 @sat 21

=

−+=

+=

==

===

°

°

fgf x

PPP

vvv

vv

The definition of specific volume gives

33

33

1

212 m 16.88

/kgm 0.001157/kgm 0.10200

)m (1 ===v

vVV

The work done during the process is determined from

kJ 101.355 5×=⎟⎠

⎞⎜⎝

⎛

⋅−=−== ∫ 3

312

2

1 out,

mkPa 1kJ 1)m1kPa)(88.16 (1554.9)( VVV PdPWb

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). P

2

1

T = 12°C

Analysis The boundary work is determined from its definition to be

kJ 272−=

⋅=

=== ∫

kPa 600kPa 150lnK) K)(285kJ/kg kg)(0.287 (2.4

lnln2

1 2

1

1

211ou, P

PmRTPdPW tb VV

VV

V

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.


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.

PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.

. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course

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 . The boundary work done during this process is to be determined. 2−= VaP

Assumptions The process is quasi-equilibrium. P

2

1

P = aV--2

V (m3)

Analysis The boundary work done during this process is determined from

kJ 53.3−=

⎟⎟⎠

⎞⎜⎜⎝

⎛

⋅⎟⎟⎠

⎞⎜⎜⎝

⎛−⋅−=

⎟⎟⎠

⎞⎜⎜⎝

⎛−−=⎟

⎠

⎞⎜⎝

⎛== ∫∫

3336

12

2

1 2

2

1 out,

mkPa 1kJ 1

m 0.31

m 0.11)mkPa (8

11VV

VV

V adadPWb

0.3 0.1

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.


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:

psia 20

)ft )(7psia/ft (5psia 15 3311

−=+=

+=

bb

baP V P (psia)

1

2 P = aV + b

V (ft3)

100

At state 2: 15

3

2

23

22

ft 24

psia) 20()psia/ft (5psia 100

=

−+=

+=

V

V

V baP

7

and,

Btu 181=

⎟⎟⎠

⎞⎜⎜⎝

⎛

⋅−

+=−

+==

33

1221

out,ftpsia 5.4039

Btu 17)ft(24

2psia15)(100

)(2

Area VVPP

Wb

Discussion The positive sign indicates that work is done by the system (work output).



Tutorial #5

(QUIZ #1)

ESO 201A

Thermodynamics







Tutorial #6

ESO 201A

Thermodynamics


Contact







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.


[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 well-insulated 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


)(V

0)=PE=KE (since )(

12

12in,

energies etc. potential, kinetic, internal,in Change

system

mass and work,heat,by nsferenergy traNet

outin

uumtIQuumUW

EEE

e

−=∆

=−=∆=

∆=−4342143421

The properties of water are (Tables A-4 through A-6)

( )[ ]( )

kJ/kg 2569.7vaporsat.

/kgm 0.29065

kJ/kg 980.032052.30.25466.97

/kgm 0.290650.0010531.15940.250.001053

kJ/kg 3.2052,97.466/kgm 1.1594,001053.0

25.0kPa 150

/kgm 0.29065@2

312

11

311

3

1

1

3 ==⎪⎭

⎪⎬⎫==

=×+=+=

=−×+=+=

====

⎭⎬⎫

==

g

fgf

fgf

fgf

gf

uu

uxuu

x

uuxP

vv

vvv

vv

We

H2O V = const.

T

1

2

v

Substituting,

min 60.2==∆

⎟⎟⎠

⎞⎜⎜⎝

⎛−=∆

s 33613kJ/s 1

VA 1000.03)kJ/kg980kg)(2569.7 (2)A 8)(V 110(

t

t


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).


&Analysis (a) There is only one inlet and one exit, and thus & &m m m1 2= = . We take diffuser 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

outin


(steady) 0system


outin 0

EE

EEE

&&

44 344 21&

43421&&

=

=∆=−

AIR 1 2

20

0)peW (since /2)V+()2/(2

12

212

222

211

VVhh

QhmVhm

−+−=

≅∆≅≅=+ &&&&

,

or,

(

) ( )kJ/kg 426.98

/sm 1000kJ/kg 1

2m/s 230

22

222=⎟

⎟⎠

⎞⎜⎜⎝

⎛−

rom Table A-17,

b) The specific volume of air at the diffuser exit is

m/s 30kJ/kg 400.98

212

12 −=−

−=VV

hh2

F

T2 = 425.6 K

(

( )( )( ) /kgm 1.221

kPa 100K 425.6K/kgmkPa 0.287 3

3

2

22 =

⋅⋅==

PRT

v

From conservation of mass,

2m 0.0678===⎯→⎯=m/s 30

)/kgm 1.221)(kg/s 36006000(1 3

2

2222

2 Vm

AVAmv

v

&&


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.


&

P1 = 300 kT

Pa 200°C 1 =

V1 = 45 m/s A1 = 110 cm2

P2 = 100 kPa V2 = 180 m/s

AIR

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 . Using the ideal gas relation, the specific volume and

the mass flow rate of air are determined to be & &m m m1 2= =

/kgm 0.4525kPa 300

)K 473)(K/kgmkPa 0.287( 33

1

11 =

⋅⋅==

PRT

v

kg/s 1.094=== )m/s 45)(m 0.0110(/kgm 0.4525 311

1VAm

v&

b

11 2

stem, which is a control volume since mass crosses the boundary. The energy balance for this stem can be expressed in the rate form as

fRoutin 0

EE

EEE

&&

44 344 21&

43421&&

=

=∆=−

( ) We take nozzle as the systeady-flow sy

outin


(steady) 0system

mass and work,heat,by nsferenergy tranet o ate

( )2

02

0

0)peW (since /2)+()2/(2

12

212,

21

22

12

222

211

VVTTc

VVhh

QVhmVhm

avep−

+−=⎯→⎯−

+−=

≅∆≅≅=+ &&&&

Substituting,

⎟⎟⎠

⎞⎜⎜⎝

⎛−+−⋅=

22

22

2/sm 1000

kJ/kg 12

)m/s 45()m/s 180()C200)(KkJ/kg 1.02(0 oT

It yields

T2 = 185.2°C

(c) The specific volume of air at the nozzle exit is

/kgm 1.315kPa 100

)K 273185.2)(K/kgmkPa 0.287( 33

2

22 =

+⋅⋅==

PRT

v

( )m/s 180/kgm 1.315

1kg/s .094112322

2AVAm =⎯→⎯=

v& → A2 = 0.00799 m2 = 79.9 cm2


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

kJ/kg 3375.1C500

MPa 101

1

1 =⎭⎬⎫

°==

hTP


kPa 01=×

⎫=P H2OkJ/kg

90.0 222

+=+=⎭⎬= fgf hxhh

x

Analysis There is only one inlet and one exit, and thus & & &m m m1 2

2344.72392.10.90.811912

= = . We take the turbine as the sy , which is a control volumfo ad

stem e since mass crosses the boundary. The energy balance r this ste y-flow system can be expressed in the rate form as

0E44 344 21

& =∆=

Substituting, the required mass flow rate of the steam is determined to be

2


(steady) 0system


outin EE && =

outin EE43421&& −

& & & &

& & ( )

mh W mh Q ke pe

W m h h

1 2

2 1

= + ≅ ≅ ≅

= − −out

out

(since 0)∆ ∆

kg/s 4.852=⎯→⎯−−= mm && kJ/kg )3375.12344.7(kJ/s 5000


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 mmm &&& == 21 . 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


woheat,by

outin


(steady) 0system

mass and rk,nsferenergy tranet of Rate

outin 0

EE

EEE

&&

444 344 21&

43421&&

=

=∆=−

p −=−=

≅∆≅∆=+

&&&

&&

120 kPa 20°C 10 L/s

1 MPa 300°C

Compressor W&

)()(

0)peke (since

1212in

21in

TTcmhhmW

hmhm

Thus,

kJ/kg 285.0=−⋅=−= 0)K2K)(300kJ/kg (1.018)( 12in TTcw p

(b) The specific volume of air at the inlet and the mass flow rate are

/kgm 7008.0K) 273K)(20/kgmkPa 287.0( 33

11 =

+⋅⋅==

RTv

kPa 1201P

kg/s 0.01427/kgm 0.7008/sm 010.0

3

3

1

1 ===v

V&&m

Then the power input is determined from the energy balance equation to be

kW 4.068=−⋅=−= 0)K2K)(300kJ/kg 8kg/s)(1.01 (0.01427)( 12in TTcmW p&&


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 mmm &&& == 21 . 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

outin


(steady) 0system


outin 0

EE

EEE

&&

444 344 21&

43421&&

=

=∆=−

⎟⎟⎠

⎞⎜⎜⎝


⎛⎞⎛ − 22

21 VV −

+−=⎟⎟⎠

⎜⎜⎝

+−=

+⎟⎟⎠

⎞⎜⎜⎝

⎛+=⎟

⎟⎠

⎞⎜⎜⎝

⎛+

2)(

2

222

22

12121out

out

22

2

21

1

VVTTcmhhmW

WV

hmV

hm

p&&&

&&&

The specific volume of air at the inlet and the mass flow rate are

/kgm 1707.0kPa 1300

K) 273K)(500/kgmkPa 287.0( 33

1

11 ==

Pv =

+⋅⋅RT

kg/s 46.88===/kgm 0.1707m/s) )(40m 2.0(

3

2

1

11

v

VAm&

Similarly at the outlet,

/kgm 148.1kPa 100

K) 273K)(127/kgmkPa 287.0( 33

2

2 =+⋅⋅

==P

RTv 2

m/s 82.53m 1

/kg)m 8kg/s)(1.14 88.46(2

3

2

22 ===

Am

Vv&

(b) Substituting into the energy balance equation gives

kW 18,300=⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠

⎞⎜⎝

⎛−+−⋅=

⎟⎟⎠

⎞⎜⎜⎝

⎛ −+−=

22

22

22

21

21out

/sm 1000kJ/kg 1

2m/s) 82.53(m/s) 40(

)K127K)(500kJ/kg 048.1(kg/s) 88.46(

2)(

VVTTcmW p&&

100 kPa 127°C

1.3 MPa500°C 40 m/s

Turbine



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.

5 MPa 600°C 20 kg/s Properties From the steam tables (Tables A-5 and A-6)

STEAM 20 kg/s

kPa 10

kJ/kg 1.27481

MPa 0.5

kJ/kg 3666.9C060

33

22

2

11

+=⎫=

=⎭⎬⎫

==

=⎭⎬°=

fgf xhhhP

hxP

hT

II

0.1 MPa 2 kg/s

sat. vap.

10 kPa x=0.85

I

kJ/kg 1.2225)1.2392)(85.0(81.19185.0

MPa 5

2 =+=⎭⎬=

⎫=

x

P

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 steady-flow system can be expressed in the rate form as

mass and work,heat,b

hhhmW

Whmhmhm

EE

−−=

++=

=

=

&&

&&&&

&&

3

1

)9.01.0(

0

3211out

out332211

outin


(steady) 0system

nsferenergy tranet of Rateoutin EEE ∆=−

444444 21

y

&

Substituting, the power output of the turbine is

43421&&

kW 27,790=×−×−=

−−=kJ/kg )1.22259.01.27481.0.9kg/s)(3666 20(

)9.01.0( 3211out hhhmW &&

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 mm &&m == 21 . We take the throttling valve as the system, which icontrol volume since mass crosses the boundary. The energy balance for this steady-flow system can be expressed in the rate

s a

form as

WQ & .

2 =⎭⎬⎫

°==

hTP

0

21

21

outin

(steady) 0systemoutin

hhhmhm

EE

EEE

===

=∆=−

&&

&&

&&&

100 kPa 120°C

Throttling valve Steam 2 MPa

Since ≅& 0∆peke ≅≅∆=

The enthalpy of steam at the exit is (Table A-6),

kJ/kg 1.2716C120kPa 100

22

at the inlet is (Table A-5) The quality of the steam

0.957=−

=−

=⎭⎬⎫

===

8.188947.9081.2716

kJ/kg 1.2716kPa 2000 2

121

1

fg

f

hhh

xhh

P


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

N2130 kPa 180°C

3

11 m 2586.0

kPa) (130K) 27380kJ/kg.K)(1 2968.0(kg) 25.0(

=+

==P

mRTV

3

22 m 4202.0

kPa 80K) 27380kJ/kg.K)(1 2968.0(kg) 25.0(

=+

==P

mRTV

kJ 16.3=⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛= 3

33

1

211

m 2586.0m 4202.0ln)m 6kPa)(0.258 (130ln

V

VVPWb

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).

Air 2 MPa 350°C

Analysis For the isothermal expansion process:

3

11 m 01341.0

kPa) (2000K) 27350kJ/kg.K)(3 287.0(kg) 15.0(

=+

==P

mRTV

3

22 m 05364.0

kPa) (500K) 27350kJ/kg.K)(3 287.0(kg) 15.0(

=+

==P

mRTV

kJ 37.18=⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=− 3

33

1

21121, m 01341.0

m 05364.0ln)m 41kPa)(0.013 (2000lnVV

VPWb

For the polytropic compression process:

33

2.13

2.133322 m 01690.0kPa) (2000)m 64kPa)(0.053 (500 =⎯→⎯=⎯→⎯= VVVV nn PP

kJ -34.86=−

−=

−−

=− 2.11)m 64kPa)(0.053 (500)m 90kPa)(0.016 (2000

1

332233

32, nPP

WbVV

For the constant pressure compression process:

kJ -6.97=−=−=−3

31313, 0.01690)m41kPa)(0.013 (2000)( VVPWb

The net work for the cycle is the sum of the works for each process

kJ -4.65=−+−+=++= −−− )97.6()86.34(18.3713,32,21,net bbb WWWW



Tutorial #7

ESO 201A

Thermodynamics


Contact







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


[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

20 L Evacuated

)0 (since initialout2systemoutin ===→∆=− mmmmmmm i

Energy balance:

C


)0 (since initialout22in ≅=+ hmQ ii

energies etc. potential,kinetic, internal,in hange

systemoutin

≅≅==

∆=−

pekeEEWum

EEE4342143421

ing the two balances:

Combin

( )ihumQ −= 22in

where


kJ/kg 214.07K 300

22 =

⎯⎯⎯⎯ →⎯==u

TTi

Substitutin

kJ/kg .19300

kg 0.02323)K 300)(K/kgmkPa 0.287(

17-A Table

32

22

=

=⋅⋅

==

hRT

m

i

g,

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.

)m 0.020)(kPa 100( 3P V


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)

kJ/kg 02.250K 350kJ/kg 49.210K 295kJ/kg 17.295K 295

22

11=⎯→⎯==⎯→⎯==⎯→⎯=

uTuThT ii

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:


mmmmmm iPi = 600 kPa Ti = 22°C

V1 = 2 m3

P1 = 100 kPa T1 = 22°C

Q·

12systemoutin −=→∆=−

Energy balance:

)0 (since 1122in


system


outin

≅≅≅−=+

∆=−

pekeWumumhmQ

EEE

ii

4342143421

The initial and the final masses in the tank are

kg

11.946)K 350)(K/kgmkPa 0.287( 3

22 =

⋅⋅==

RTm )m 2)(kPa 600(

kg 2.362)K 295)(K/kgm

)m 2)(kPa 100(

32

3

3=

⋅⋅

P

P

V

V

hen from e mass balance,

(b) The heat transfer during this process is determined from

Discussion The negative sign for heat transfer indicates that the assumed direction is wrong. Therefore, we reversed the direction.

kPa 0.287(1

11 ==

RTm

T th

m m mi = − = − =2 1 11 946 2 362. . 9.584 kg

( )( ) ( )( ) ( )( )kJ 339=→−=

−+−=−+−=

out

1122in

kJ 339kJ/kg 210.49kg 2.362kJ/kg 250.02kg 11.946kJ/kg 295.17kg 9.584

Q

umumhmQ ii


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)

STEAM 2 MPa

Q

kJ/kg 3468.3,kJ/kg 3116.9/kgm 0.17568

C500MPa 2

kJ/kg 3024.2,kJ/kg 2773.2/kgm 0.12551

C300MPa 2

22

32

2

2

11

31

1

1

===

⎭⎬⎫

°==

===

⎭⎬⎫

°==

huTP

huTP

v

v

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:


mmmmmm e 21systemoutin −=→∆=−

Energy balance:

)0 (since 1122in


system


outin

≅≅≅−=−

∆=−

pekeWumumhmQ

EEE

ee

4342143421

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,

kJ/kg 3246.222

21 =≅hekJ/kg 3468.33024.2

=++ hh


kg1.138/kgm 0.17568 3

22 v

m 0.2

kg 1.594/kgm 0.12551

m 0.2

32

3

3

1

11

===

===

V

v

V

m

m

hen from e mass and energy balance relations,

)

T th

kg 0.456138.1594.121 =−=−= mmme

( )( ) ( )( ) ( )(kJ 606.8=

−+=−+=

kJ/kg 2773.2kg 1.594kJ/kg 3116.9kg 1.138kJ/kg 3246.2kg 0.4561122 umumhmQ eein


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.


kPa 002

⎭

=

=×+=

+=⎫=

hxhhP

Analysis (a) The cylinder contains saturated vapor at the final state at a pressure of 200 kPa, thus the final temperature in be

T Tsat @ 200 kPa = 120.2°C

, 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

(P = 200 kPa) m1 = 10 kg

H2O Pi = 0.5 MPa Ti = 350°C

Properties The properties of steam are (Tables A-4 through A-6)

6.0 111 ⎭

⎬= fgfx1

kJ/kg .618252201.60.6504.71

kJ/kg 3168.1C350

MPa 0.5

kJ/kg 2706.3 vaporsat. kPa 200@2

=⎬⎫

°==

⎭⎬

ii

g

hTP

kPa 0022 =⎫=

hhP

i

the cylinder must

2 =

(b) We take the cylinder as the system

12systemoutin mmmmmm i −=→∆=− Mass balance:

Energy balance:

≅≅−+= pekeQumumWhm ii

43421

,

since the boundary work and ∆U combine into ∆H for constant pressure expansion and compression processes. Solving for m2 and substituting,

)0 (since 1122outb,


system


outin ∆=− EEE43421

≅

Combining the two relations gives

( ) 112212outb,0 umumhmmW i −+−−=

or

( ) 1122120 hmhmhmm i −+−−=

( )( ) ( ) kg 9.072kg 10

kJ/kg 2706.33168.1kJ/kg .618253168.1

12

12 =

−−

=−−

= mhhhhm

i

i

Thus,

mi = m2 - m1 = 29.07 - 10 = 19.07 kg


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.


Mass balance:


mmmmm

mmm

e

e

−=−=

∆=−

Energy balance:

hmumumQumu

EEE

+−=−

∆=−

1122in

1122


system


outin 4342143421

he initial state properties of R-134a in the t3=

e

v

quid state, and found the exiting enthalpy

The final specific volume in the container is

A-C line

Liquid R-134a 5 kg 24°C

21

12

systemoutin

m−

ee mhmQ =−in

ee

Combining the two balances:

ehmmumumQ )( 211122in −+−=

T ank are

kJ/kg 98.84kJ/kg 44.84

0C24

1

11

==

⎭⎬⎫

=°=

hu

xT

(Table A-11) /kgm 0008261.0

Note that we assumed that the refrigerant leaving the tank is at saturated liaccordingly. The volume of the tank is

3311 m 004131.0)/kgm 0008261.0)(kg 5( === vV m

/kgm 0.01652kg 25.0

m 004131.0 33

2

2 ===mV

v

The final state is now fixed. The properties at this state are (Table A-11)

kJ/kg 73.164)kJ/kg 65.158)(5061.0(kJ/kg 44.84

0008261.0031834.00008261.001652.0

/kgm 01652.0

C24

22

22

32

2

=+=+=

=−−

=−

=

⎪⎭

⎪⎬⎫

=

°=

fgf

fg

f

uxuu

xT 0.5061v

vv

v

Substituting into the energy balance equation,

kJ 22.64=

+−=−+−=

)kJ/kg 98.84)(kg 75.4()kJ/kg 44.84)(kg 5()kJ/kg 73.164)(kg 25.0()( 211122in ehmmumumQ


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)

kJ/kg 852.26liquid sat.

C200

kJ/kg 850.46/kgm 0.001157

liquid sat.C200

C200@

C200@1

3C200@11

==⎪⎭

⎪⎬⎫=

====

⎭⎬⎫°=

o

o

o

o

fee

f

f

hhT

uuT vv



mmmmmm eMass balance: in 21systemout −=→∆=−

Energy balance: H2O Sat. liquid T = 200°C V = 0.3 m3 Q

)0 (since 1122in


system


outin

≅≅≅−+=

∆=−

pekeWumumhmQ

EEE

ee

4342143421


( ) kg 129.7kg 259.4

kg 259.4/kgm0.0011571

11 ==m

v m 0.3

21

121

2

3

3

===

=

mm

V

=−=−= mmm

Now we determine the final internal energy,

Then from the mass balance,

kg 129.77.1294.25921e

( )( ) kJ/kg 866.467.1743009171.046.850009171.0

C200

009171.0001157.012721.0001157.0002313.0

/kgm 0.002313kg 129.7

m 0.3

222

2

22

33

22

=+=+=⎭⎬⎫

=°=

=−−

=−

=

===

fgf

fg

f

uxuuxT

x

m

v

vv

Vv

Then the heat transfer during this process is determined from the energy balance by substitution to be

( )( ) ( )( ) ( )( )

kJ 2308=−+= kJ/kg 850.46kg 259.4kJ/kg 866.46kg 129.7kJ/kg 852.26kg 129.7Q


5-105

5-133 A rigid tank initially contains saturated R-134a liquid-vapor mixture. The tank is connected to a supply line, and R-134a 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)

( )

kJ/kg 335.06


100 ⎭⎬°=i CT

MPa 1.0

kJ/kg 244.48/kgm 0.02936

vaporsat.kPa 007

26.1670.7.57077.0

kPa 700@2

3kPa 700@22

111

⎫=

====

⎭⎬⎫=

×+=+=⎭⎬=

i

g

g

fgf

P

uuP

uxuux

vv

kJ/kg 187.65/kgm 0.030630.00080200.043420.70.0008020C14 3

11

=

==−×+=+=⎫°=

i

fgf

h

xT vvv

the tank as the system, which is a control volume nce mass crosses the boundary. Noting that the microscopic energies f flowing and nonflowing fluids are represented by enthalpy h and

xpressed as

ass balance:

1

1 MPa 100°C

0.4 m3

R-134a

R-134a

Analysis We takesiointernal energy u, respectively, the mass and energy balances for this uniform-flow system can be e

12systemoutin mmmmmm i −=→∆=− M

Energy balance:

Change


)0 (since 1122in

energies etc. potential,kinetic, internal,in

systemoutin

≅≅−=+

∆=−

pekeWumumhmQ

EEE

ii

4342143421

perature is the saturation mperature at this pressure,

(b) The initial and the final masses in the tank are

≅

(a) The tank contains saturated vapor at the final state at 800 kPa, and thus the final temte

C26.7°== kPa 700 @sat 2 TT

kg .6213/kgm 0.02936

m 0.4

kg .0613/kgm 0.03063

m 0.4

3

3

22

3

3

11

===

===

vV

vV

m

m

Then from the mass balance

(c) The heat transfer during this process is determined from the energy balance to be

)

kg 0.5653=−=−= 06.1362.1312 mmmi

( )( ) ( )( ) ( )(kJ 691=

−+−=−+−=

kJ/kg 187.65kg 13.06kJ/kg 244.48kg 13.62kJ/kg 335.06kg 0.5653 1122in umumhmQ ii


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:

)0 (since initialout2systemoutin ===→∆=− mmmmmmm i P0 T0

V Evacuated

Energy balance:

)0peke (since initialout22in


system


outin

≅≅==≅=+

∆=−

EEWumhmQ

EEE

ii

4342143421

Combining the two balances:

)()( 2222in ipi TcTcmhumQ −=−= v

but

Ti = T2 = T0

and

v = R.

Substituting,

cp - c

( ) VV

v 000

00202in PRT

RTPRTmTccmQ p −=−=−=−=

Therefor

Qout = P0V (Heat is lost from the tank)

e,



Tutorial #8

ESO 201A

Thermodynamics


Contact







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]


[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

sink

HE

Fuel

22 L/h

Engine

55 kW

kg/h .617)L/h 22)(kg/L 0.8()( fuelfuel === V&& ρm

The rate of heat supply to the car is

kW 215.1kJ/h ,400774)kJ/kg 44,000)(kg/h .617(

coalHV,coal

===

= qmQH &&

Then the thermal efficiency of the car becomes

25.6%==== 0.256kW 215.1

kW 55outnet,th

HQ

W&

&η

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

kWh 103.646 12×=

×−×

=

−=

+==

kWh 10878.134.0

kWh 10878.1

1212

coalth

coalout

coalout

coal

in

coalth

WW

Q

WQW

QW

η

η

sink

HE

Coal

Furnace

1.878×1012 kWh

ηth = 34% outQ&



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

MW 5.93732.0MW 300

th

outnet,in ===

ηW

Q&

&

MJ 101.8s) 360024(MJ/s) 5.937( 7inin ×=×=∆= tQQ &

The amount and rate of coal consumed during this period are

kg/s 48.33s 360024kg 10893.2

MJ/kg 28MJ 101.8

6coal

coal

7

HV

incoal

=××

=∆

=

×=×

==

tmm

qQm

&

kg 102.893 6

(b) Noting that the air-fuel ratio is 12, the rate of air flowing through the furnace is

kg/s 401.8=== kg/s) 48.33(fuel) air/kg kg 12(AF)( coalair mm &&



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.

Reservoir

Reservoir

HP

HQ&

5 hp

COP = 2.4

Assumptions The heat pump operates steadily.

Analysis Applying the definition of the heat pump coefficient of performance to this heat pump gives

Btu/h 30,530=⎟⎟⎠

⎞⎜⎜⎝

⎛==

hp 1Btu/h 2544.5)hp 5)(4.2(COP innet,HPWQH

&&

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.

Reservoir

Reservoir

AC innet,W&HQ&

LQ&

Analysis Applying the first law to the air conditioner gives

kW 5.025.2innet, =−=−= LH QQW &&&

Applying the definition of the coefficient of performance,

4===kW 0.5kW 2.0COP

innet,R W

QL&

&

6-41 The power input and the COP of a refrigerator are given. The cooling effect of the refrigerator is to be determined.

Reservoir

Reservoir

R innet,W&HQ& COP=1.4

LQ&

Assumptions The refrigerator operates steadily.

Analysis Rearranging the definition of the refrigerator coefficient of performance and applying the result to this refrigerator gives

kW 4.2=== kW) 3)(4.1(COP innet,RWQL&&



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.


−12°C

30°C

R

HQ&

inW&

4800 kJ/h

3300 kJ/h LQ&

Analysis The power input is determined from

kW 0.417=⎟⎠⎞

⎜⎝⎛=

=−=−=

kJ/h 3600kW 1kJ/h) 1500(

kJ/h 150033004800in LH QQW &&&

The COP is

2.2===kJ/h 1500kJ/h 3300COP

inWQL&

&

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.


cool space

Kitchen air

R LQ&

COP=1.2Analysis (a) Using the definition of the coefficient of performance, the power input to the refrigerator is determined to be

kW 0.83==== kJ/min 051.2kJ/min 60

COPRinnet,

LQW&

&

(b) The heat transfer rate to the kitchen air is determined from the energy balance,

kJ/min 110=+=+= 5060innet,WQQ LH&&&

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

Reservoir

HP

HQ&

2 hp

LQ&


Analysis Applying the definition of the heat pump coefficient of performance to this heat pump gives

2.97=⎟⎠

⎞⎜⎝

⎛==Btu/h 2544.5

hp 1hp 2Btu/h 15,090COP

innet,HP W

QH&

&



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.

Reservoir

Reservoir

R innet,W&

HQ& COP=1.6

LQ&


Analysis Rearranging the definition of the refrigerator coefficient of performance and applying the result to this refrigerator gives

kW 4.34=⎟⎠⎞

⎜⎝⎛==

s 3600h 1

1.60kJ/h 25,000

COPRinnet,

LQW

&&

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

( ) ( )( )( ) kJ 2520C820CkJ/kg 4.2kg 105swatermelon =−°⋅×=∆= oTmcQL

cool space

Kitchen air

R 450 WCOP = 2.5

The rate at which this refrigerator removes heat is

( )( ) ( )( ) kW 1.125kW 0.452.5COP innet,R === WQL&&

That is, this refrigerator can remove 1.125 kJ of heat per second. Thus the time required to remove 2520 kJ of heat is

min 37.3s 2240 ====∆kJ/s 1.125kJ 2520

L

L

QQ

t&

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.



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.


Outside

House

HQ&

HP COP = 2.5

60,000kJ/hAnalysis 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,

& , ,QH = − =60 000 4 000 56 000 kJ h, /

Using the definition of COP, the power input to the heat pump is determined to be

kW6.22 kJ/h 3600

kW 12.5

kJ/h 56,000COPHP

innet, =⎟⎟⎠

⎞⎜⎜⎝

⎛== HQW

&&

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.

&QL

Ice Machine

ice 25°F

water 55°F

Outdoors

R

COP = 2.4 Analysis The cooling load of this ice machine is

( )( ) Btu/h 4732Btu/lbm 169lbm/h 28 === LL qmQ &&

Using the definition of the coefficient of performance, the power input to the ice machine system is determined to be

hp 0.775=⎟⎟⎠

⎞⎜⎜⎝

⎛==

Btu/h 2545hp 1

2.4Btu/h 4732

COPRinnet,

LQW&

&

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,

Outside

AC

Computer room

7000 Btu/h

Btu/h 13,853= W 40607003360

W700 W)100(7)people of No.(

kW 3.36=kW)(0.4) (8.4=factor) (Usagepower) Rated(

peoplecomputerstotal

personpeople

computers

=+=+=

=×=×=

×=

QQQ

QQ

Q

&&&

&&

&

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

rsconditione Air2≈=

==

98.1Btu/h 7000Btu/h 853,13

A/C ofcapacity Coolingload Coolingrsconditioneair of No.



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.


Properties The enthalpies of R-134a at the condenser inlet and exit are

kJ/kg 47.95

0kPa 800

kJ/kg 22.271C35kPa 800

22

2

11

1

=⎭⎬⎫

==

=⎭⎬⎫

°==

hxP

hTP

Analysis (a) An energy balance on the condenser gives the heat rejected in the condenser

kW 164.3kJ/kg )47.9522.271(kg/s) 018.0()( 21 =−=−= hhmQH &&

QH800 kPa x=0

Win

Condenser

Evaporator

Compressor

Expansion valve

QL

800 kPa 35°C

The COP of the heat pump is

2.64===kW 2.1

kW 164.3COPinW

QH&

&

(b) The rate of heat absorbed from the outside air

kW 1.96=−=−= 2.1164.3inWQQ HL&&&

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

100 kPa -26°C

Condenser

Evaporator

Compressor

Expansion valve

100 kPa x=0.2

Win

QL

Properties The properties of R-134a at the evaporator inlet and exit states are (Tables A-11 through A-13)

kJ/kg 74.234C26

kPa 100

kJ/kg 71.602.0

kPa 100

22

2

11

1

=⎭⎬⎫

°−==

=⎭⎬⎫

==

hTP

hxP

Analysis (a) The refrigeration load is

kW 72.0kW) 600.0)(2.1(COP)( in === WQL&&

The mass flow rate of the refrigerant is determined from

kg/s 0.00414=−

=−

=kJ/kg )71.6074.234(

kW 72.0

12 hhQ

m LR

&&

(b) The rate of heat rejected from the refrigerator is

kW 1.32=+=+= 60.072.0inWQQ LH&&&


6-33

6-94 The validity of a claim by an inventor related to the operation of a heat pump is to be evaluated.


Analysis Applying the definition of the heat pump coefficient of performance,

273 K

293 K

HP

HQ&

LQ&75

2.67kW 75kW 200COP

innet,HP ===

WQH&

&

The maximum COP of a heat pump operating between the same temperature limits is

7.14K) K)/(293 273(1

1/1

1COP maxHP, =−

=−

=HL TT

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.

TL

26°C

HP

HQ&

LQ&

4.25 kW


Analysis The temperature of the low-temperature reservoir is

K 264.6=⎯→⎯−

=⎯→⎯−

= LLLH

H TTTT

TCOP

K )299(K 2997.8maxHP,

The heating load is

kW 37.0=⎯→⎯=⎯→⎯= HH

in

H QQ

WQ

COP &&

&

&

kW 25.47.8maxHP,

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.


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

-8°C

25°C

R

300 kJ/min

( ) ( ) ( ) 03.81K 2738/K 27325

11/

1revR, =

−+−+=

−=

LH TTCOP

The power input to this refrigerator is determined from the definition of the coefficient of performance of a refrigerator,

kW0.623 kJ/min 37.368.03kJ/min 300

maxR,minin,net, ====

COPQW L&

&



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.


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

( ) ( ) ( ) 14.75K 27322/K 27321

1/1

1COP revHP, =++−

=−

=HL TT

5 kW

House 22°C

HP

110,000 kJ/h

The required power input to this reversible heat pump is determined from the definition of the coefficient of performance to be

kW 2.07=⎟⎟⎠

⎞⎜⎜⎝

⎛==

s 3600h 1

14.75kJ/h 110,000

COPHPminin,net,

HQW&

&

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.


450 R

540 R

R Wnet,in.

15,000 Btu/h

Analysis The COP of this reversible refrigerator is

5R 045R 540

R 450COP maxR, =−

=−

=LH

L

TTT

Using this result in the coefficient of performance expression yields

kW 0.879=⎟⎠⎞

⎜⎝⎛==

Btu/h 14.3412kW 1

5Btu/h 000,15

COP maxR,innet,

LQW

&&

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.


TL

300 K

R

2000 kW

HQ&

LQ&200 kW

Analysis Applying the definition of the refrigerator coefficient of performance,

kW 1800=−=−= 2002000innet,WQQ HL&&&

Applying the definition of the heat pump coefficient of performance,

9kW 200kW 1800COP

innet,R ===

WQL&

&

The temperature of the heat source is determined from

C3°−==⎯→⎯−

=⎯→⎯−

= K 270 300

9 COP maxR, LL

L

LH

L TT

TTT

T



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

0.744K 1173K 30011Cth,maxth, =−=−==

H

L

TTηη

27°C

900°C

HE

800 kJ/min

R

-5°C

Then the maximum power output of this heat engine is determined from the definition of thermal efficiency to be

( )( ) kJ/min 595.2kJ/min 8000.744thoutnet, === HQW && η

which is also the power input to the refrigerator, . innet,W&

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

( ) ( ) ( ) 8.371K 2735/K 27327

11/

1COP revR, =−+−+

=−

=LH TT

Then the rate of heat removal from the refrigerated space becomes

( )( ) ( )( ) kJ/min4982 kJ/min 595.28.37COP innet,revR,R, === WQL&&

(b) The total rate of heat rejection to the ambient air is the sum of the heat rejected by the heat engine ( ) and the heat

discarded by the refrigerator ( ), HE,LQ&

R,HQ&

kJ/min 5577.2595.24982

kJ/min 204.8595.2800

innet,R,R,

outnet,HE,HE,

=+=+=

=−=−=

WQQ

WQQ

LH

HL

&&&

&&&

and

kJ/min5782 2.55778.204R,HE,ambient =+=+= HL QQQ &&&



Tutorial #9

ESO 201A

Thermodynamics


Contact







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.


[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 (cp = 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 (cp = 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

outin

outin


(steady) 0system


outin 0

QW

EE

EEE

&&

&&

44 344 21&

43421&&

=

=

=∆=−

P2


ut ==WQ &

AIR T = const.

Q ·

30 kW

Therefore,

o& kW 30in

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 P1

kW/K 0.101−=−=−=∆K 298

kW 30

sys

airout,air T

QS

&&

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.

600 K

1200 K

100 kJ

Analysis The entropy change of the source and sink is given by

kJ/K 0.0833=+=+=∆K 600K 1200L

L

H

H

TTS − kJ 100kJ 100QQ

ince the entropy of everything involved in this process has creased, this transfer of heat is possible.

Sin


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

kJ/K 08333.0K 1200

kJ 100===∆

HH T

QS

TL

TH

Q =100 kJ and the entropy change of the low-temperature reservoir is

kJ/K 1667.0


K 600LL T

kJ 100−=

−==∆

Q

he total entropy change of everything involved with this system is en

S

T th

kJ/K 0.0833−=−=∆+∆=∆ 1667.008333.0total LH SSS

which violates the increase in entropy principle since the total entropy change is negative.

e of two reservoirs is to pump satisfies the increase in entropy principle.

efficient of erformance expression, first Law, and thermodynamic temperature scale gives

7-27 A reversible heat pump with specified reservoir temperatures is considered. The entropy changbe calculated and it is to be determined if this heat


Analysis Since the heat pump is completely reversible, the combination of the co

7°C

netW&

LQ&

24°C

HP

300 kWp

47.1711COP revHP, === )K 297/()K 280(1/1 −− HL TT

he power required to drive this heat pump, according to the coefficient of ance, is then

Tperform

kW 17.17kW 300=== HQ

W&

& 17.47COP revHP,

in

17.17kW 300innet, =−=−= WQQ HL&&&

The rate at which the entropy of the high temperature reservoir changes, according to the definition of the entropy, is

net,

According to the first law, the rate at which heat is removed from the low-temperature energy reservoir is

kW kW 8.282

kW/K 1.01===∆K 297

kW 300Q&

H

HH T

S&

and that of the low-temperature reservoir is

kW/K 1.01−===∆K 280L

L TS − kW 17.17LQ&&

The net rate of entropy change of everything in this system is

as it must be since the heat pump is completely reversible.

kW/K 0=−=∆+∆=∆ 01.101.1total LH SSS &&&


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

95°F

Carnot heat engine

SINK95°F

Heat

∆S QT

=

is equal to the heat ansferred to the sink, the heat transfer become

outfluid,

Noting that heat transferred from the working fluidtr

( )( ) 388Btu/R 0.7R 555fluidfluidfluid Btu 388.5. Btu 5−=−=∆= STQ =→ Q

) The entropy change of the sink is determined from (b


Btu/R0.7 Btu 388.5insink, ===∆Q

SR 555sink

sink T

(c) Thus the total entropy change of the process is

0=+−=∆+∆=∆= totalgen SS 7.07.0sinkfluid SS

his is expected since all processes of the Carnot cycle are reversible processes, and no entropy is generated during a versible process.

rant

othermal, internally reversible process.

Analysis Noting that both the refrigerant and the cooled space undergo reversible isothermal processes, the entropy change

Tre

7-29 R-134a enters an evaporator as a saturated liquid-vapor at a specified pressure. Heat is transferred to the refrigefrom 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 is

for them can be determined from

∆S QT

=

pr sure of the refrigerant is maintained constant. Therefore, the temperature of the refrigerant also remains

(Table A-12)

Then,

(a) The esconstant at the saturation value,

K 257.4C15.6kPa @160sat =°−== TT

R-134a 160 kPa

-5°C

180 kJ

kJ/K 0.699===∆K 257.4trefrigeranT

kJ 180int,refrigerantrefrigeran

QS

) Similarly,

(b

kJ/K 0.672−=−=−=K 268kJ 180

space

outspace,space T

QS∆

(c) The total entropy change of the process is

kJ/K 0.027=−=∆+∆=∆= 672.0699.0spacetrefrigerantotalgen SSSS


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

/lbmft 25.1lbm 2

ft 5.2 33

11 ===

mV

v


which is between vf and vg for 300 psia. The initial quality and the entropy are then (Table A-5E)

RBtu/lbm 3334.1)RBtu/lbm 92289.0)(8075.0(RBtu/lbm 58818.0

8075.0/lbmft )01890.05435.1(

1−

==fg

xv

/lbmft )01890.025.1(

11

3

31

⋅=⋅+⋅=+=

=−−

fgf

f

sxss

vv

⋅=⎭⎬⎫

==°=

sPP

T

ence, the change in the total entropy is

)( 12 ssmS

7-32 An insulated rigid tank contains a saturated liquid-vapor mixture of water at a specified pressure. An electric heater ide 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)

H2O 300 psia

2 lbm 2.5 ft3

The final state is superheated vapor and P

v

2 1 6E)-A (TableRBtu/lbm 5706.1

psia 300F005

212

2

H

−=∆

Btu/R 0.4744=⋅−= RBtu/lbm )3334.15706.1)(lbm 2(

ins

( )( )( )( )

KkJ/kg 6.7298 vaporsat.

KkJ/kg 2.88107894.525.04337.1/kgm 0.290650.0010531.15940.25

We

H2O 5 kg

150 kPa

0.001053kPa 150 111 +=+=⎬⎫= xP fgf vvv

2.0

212

11

3

1

⋅=⎭⎬⎫=

⋅=+=+==−

⎭=

s

sxssx fgf

vv

Then the entropy change of the steam becomes

5

( ) kJ/K 19.2=⋅−=−=∆ KkJ/kg )2.88106.7298)(kg 5(12 ssmS


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


5)-A (TableKkJ/kg 8902.5

kJ/kg 6.2784

1 kPa 6000

1

1

1

1 ⋅=

=

⎭⎬⎫

==

sh

xP

For the maximum velocity at the exit, the entropy will be constant during the process. The exit state enthalpy is (Table A-6)

kJ/kg 5.24924.19858533.033.7983058.4

K kJ/kg 8902.5 kPa 1200

2

2

12

2

=×+=+=

8533.02159.28902.52 =−

=−

= fssx

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

⎭⎬⎫

⋅===

fgf

fgxhhh

sss

P

6000 kPa

1

2

1200 kPa

s

T

⎟⎟⎠

⎞⎛ − 22 VV⎜⎜⎝

=−

≅∆≅≅⎟⎟⎠

⎞⎜⎜⎝

⎛+=⎟

⎟⎠

⎞⎜⎜⎝

⎛+

=

=∆=−

2

0)pe (since 22

0

1221

22

2

21

1

outin


(steady) 0system


outin

hh

QWV

hmV

hm

EE

EEE

&&&&

&&

444 3444 21&

43421&&

Solving for the exit velocity and substituting,

[ ]m/s 764.3=

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛−+=−+=

⎟⎟⎠

⎞⎜⎜⎝

⎛ −=−

5.022

25.021

212

21

22

21

kJ/kg 1/sm 1000kJ/kg )5.24926.2784(2m/s) 0()(2

2

hhVV

VVhh


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 well-insulated 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

U

EEE

∆=

∆=−

0energies etc. potential,

kinetic, internal,in Change

system


outin 4342143421


160 L

Copper 75 kg

WATER

or,

U 0waterCu =∆+∆ U

0)]([)]([ water12Cu12 =−+− TTmcTTmc

where

kg 159.5)m 0.160)(kg/m 997( 33water === Vρm

Using specific heat values for copper and liquid water at room temperature and substituting,

0C5)1(C)kJ/kg kg)(4.18 (159.5C0)11(C)kJ/kg kg)(0.386 (75 22 =°−°⋅+°−°⋅ TT

T = 19.0 C = 292 K °2

The entropy generated during this process is determined from

( )( )

( )( ) kJ/K 20.9K 288K 292.0

ln KkJ/kg 4.18kg 159.5ln

kJ/K .857K 383K 292.0

ln KkJ/kg 0.386kg 75ln

1

2avgwater

1

2avgcopper

=⎟⎟⎠

⎞⎜⎜⎝

⎛⋅=⎟⎟

⎠

⎞⎜⎜⎝

⎛=∆

−=⎟⎟⎠

⎞⎜⎜⎝

⎛⋅=⎟⎟

⎠

⎞⎜⎜⎝

⎛=∆

TT

mcS

TT

mcS

Thus,

kJ/K 1.35=+−=∆+∆=∆ 20.985.7watercoppertotal SSS


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

( )( )

( )( ) kJ/K 1.571K 353K 288lnKkJ/kg 0.386kg 20ln

kJ/K 4.579K 353K 288lnKkJ/kg 0.45kg 50ln

1

2avgcopper

1

2avgiron

−=⎟⎟⎠

⎞⎜⎜⎝

⎛⋅=⎟⎟

⎠

⎞⎜⎜⎝

⎛=∆

−=⎟⎟⎠

⎞⎜⎜⎝

⎛⋅=⎟⎟

⎠

⎞⎜⎜⎝

⎛=∆

TTmcS

TTmcS

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

Lake 15°C

Copper 20 kg 80°C

Iron 50 kg 80°C

work,heat,by enNet

copperironout


system

mass and nsferergy tra

outin

UUUQ

EEE

∆+∆=∆=−

∆=−4342143421

uting,

or,

copper21iron21out )]([)]([ TTmcTTmcQ −+−=

Substit

( )( )( ) ( )( )( )kJ 1964

K288353KkJ/kg 0.386kg 20K288353KkJ/kg 0.45kg 50out

=−⋅+−⋅=Q

Thus,

kJ/K 6.820K 288kJ 1964

lake

inlake,lake ===∆

TQ

S

Then the total entropy change for this process is

kJ/K 0.670=+−−=∆+∆+∆=∆ 820.6571.1579.4lakecopperirontotal SSSS



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

)(0

12

12

12


system


outin

TTuu

uumU

EEE

==

−=∆=

∆=−4342143421

since u = u(T) for an ideal gas. Then the entropy change of the gas becomes

( )(8.314kmol 5

lnlnln111

avg,

=

=⎟⎠

⎜⎝

+=∆VVv uu NRR

TcNS

) ( )

kJ/K28.81

2ln KkJ/kmol

220

2

=

⋅

⎟⎞

⎜⎛ VVT

This also represents the total entropy change since the tank does not contain anything else, and there are no interactions with the surroundings.

IDEAL GAS

5 kmol40°C



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).

Air 5 kg

327°C 100 kPa

Analysis (a) The entropy change of air is determined from

kJ/K 2.488−=++

=

=∆

K


273)(327K 273)(27kJ/kg.K)ln kg)(0.718 (5

ln 2Tmc

ut

−=−= TTmcQ v

The entropy change of the surroundings is

S1

air Tv

1 T

s

2

1 2

air

surr

327ºC

27ºC

(b) An energy balance on the system gives

)2727kJ/kg.K)(3 kg)(0.718 5()( 21o

kJ 1077=

kJ/K 3.59K 300kJ 1077

surr

outsurr ===∆

TQs

The entropy change of universe due to this process is

kJ/K 1.10=+−=∆+∆=∆= 59.3488.2surrairtotalgen SSSS


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)

200 1

0.002

600 2

v (m3/kg)

∫=2

112 dPw v

When combined with the ideal gas equation of state


PRT

v

The work expression reduces to

=

kJ/kg 1.32−=

⎟⎠

⎞⎜⎝

⎛⋅

−=

−=−=−== ∫∫

33

1

222

1

22

1

2

112

mkPa 1kJ 1

kPa 200kPa 600/kg)lnm kPa)(0.002 (600

lnlnPP

PPP

RTP

dPRTdPw vv

he negative sign indicates that work is done on the system in the amount of 1.32 kJ/kg.

s to be pumped by a 25-kW pump at a specified rate. The highest pressure the water can be pumped to

ial energy changes are negligible. 3 The

pressure the liquid can have at the pump exit can be determined from the reversible steady-flow work fo liquid,

Thus,

T

7-111 Liquid water iis to be determined.

Assumptions 1 Liquid water is an incompressible substance. 2 Kinetic and potentprocess 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 highestrelation r a

( )

⎟⎟⎠

⎞⎜⎜⎝

⎛

⋅−=

−=⎟⎠⎞

⎜⎝⎛ ∆+∆+= ∫

323

1212

1

00in

mkPa 1kJ 1Pak )100)(/kgm 0.001)(kg/s 5(kJ/s 25 P

PPmpekedPmW vv &&&

It yields

PUMP

P2

100 kPa

25 kW

kPa 5100=2P


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


0

12in

outin

energies etc. potential,

(steady) 0systemoutin

TTcmQ

EE

EEE

p −=

=

=∆=−

&&

&&

&&&

sfer to the cold water in this heat exchanger becom

watercold °−°°=

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

)(

0)peke (since 21in hmhmQ ≅∆≅∆=+ &&&

kinetic, internal,in change of Ratemass and work,heat,by

nsferenergy tranet of Rate44 344 2143421

Then the rate of heat tran es

Cold water10°C

0.95 kg/s

Hot water

85°C1.6 kg/s

70°C

([ inoutin −= TTcmQ p&& kW 238.3=C)10CC)(70kJ/kg. kg/s)(4.18 95.0()]

C.549C)kJ/kg. kg/s)(4.19 6.1(

kW 3.238 C85)]([ inouthot wateroutin °=°

−°=−=⎯→⎯−=p

p cmQTTTTcmQ&

&&&

(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:

generation

entropy of Ratemass andheat by

ansferentropy trnet of Rate

ssmssmS

Ssmsmsmsm

QSsmsmsmsm

−+−=

=+−−+

==+−−+

&&&

&&&&&

&&&&&

44 344 21{(steady) 0

systemgenoutin SSSS ∆=+−

)()(

0

)0 (since 0

34hot12coldgen

gen4hot2cold3hot1cold

gen43223311

entropy of change of Rate

&

Noting that both fluid streams are liquids (incompressible substances), the rate of entropy generation is determined to be

&43421&&

K/kW 0.06263=

+=

+=

273+85273+49.5kJ/kg.K)ln kg/s)(4.19 6.1(

273+10273+70kJ/kg.K)ln kg/s)(4.18 95.0(

lnln3

4hot

1

2coldgen T

Tcm

TT

cmS pp &&&


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

nal 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

Analysis ( in tank A undergoes a reversible, adiabatic process, and thus s2 = s1. From the steam tables (Tables A-4 through A-6),

ank A:

fi

and thus kinetic and potential energy changes are negligible. 5 There are no work interactions.

a) The steam

T

( )( )( )( )( )( )

( ) ( )( )( )( ) kJ/kg 7.1704kJ/kg 2024.65928.050.504

/kgm 0.52552001061.08858.05928.0001061.0

5928.059680.5

5302.18479.4

mixture sat.

kPa 002

KkJ/kg .847941191.56.07765.1kJ/kg 6.17739.19486.022.604

/kgm 0.27788001084.046242.06.0001084.0

6.0kPa 400

,2,2

3,2,2

,2,2

kPa 200@sat,2

12

1

1,1

1,1

31,1

1

1

=+=+==−+=+=

=−

=−

=

°==

⎪⎭

⎪⎬⎫

==

⋅=+=+==+=+=

=−+=+=

⎭⎬⎫

==

fgAfA

fgAfA

fg

fAA

A

fgfA

fgfA

fgfA

uxuux

sss

x

TT

ssP

sxssuxuu

x

xP

vvv

vvv

C120.2

Tank B:

The initial and the final masses in tank A are

and

A steam

V = 0.3 m3

P = 400 kPa x = 0.6

B steam

m = 2 kg T = 250°C

P = 200 kPa

×

300 kJ

KkJ/kg 7.7100kJ/kg 2731.4

/kgm 1.1989

C250kPa 200

,1

,1

3,1

1

1

⋅===

⎭⎬⎫

°==

B

B

B

su

TP

v

kg 0.5709/kgm 0.52552

m 0.3

kg .0801/kgm 0.27788

m 0.3

3

3

,2,2

3

3

,1,1

===

===

A

AA

A

AA

m

m

v

V

v

V

Thus, 1.080 - 0.5709 = 0.5091 kg of mass flows into tank B. Then,

kg .50925091.025091.0,1,2 =+=+= BB mm

The final specific volume of steam in tank B is determined from

( ) ( )( )

/kgm 9558.0kg 2.509

/kgm 1.1989kg 2 33

,2

11

,2,2 ====

B

B

B

BB m

mm

vVv

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

Substituting,

( ) ( )BA

BA

umumumumQWUUUQ

EEE

11221122out

out


system


outin

0)=PE=KE (since )()(−+−=−

=∆+∆=∆=−

∆=−4342143421

( )( ) ( )( ){ } ( ) ( )( ){ } kJ/kg 2433.3

4.27312509.26.1773080.17.17045709.0300

,2

,2

=−+−=−

B

B

uu


7-168


⎭

T C116.1

mined by applying the entropy balance on an extended system temperature of the extended system is the

Thus,

/kgm .95580 3⎪⎫=v

KkJ/kg .91566kJ/kg 2433.3 ,2

,2

,2

,2

⋅=⎪⎬

= B

B

B

B

su

°=

(b) The total entropy generation during this process is deterthat includes both tanks and their immediate surroundings so that the boundarytemperature of the surroundings at all times. It gives

{

BAgensurrb,

out

entropyin Change

system

generationEntropy

gen

mass andheat by ansferentropy trNet

outin

SSSTQ

SSSS

∆+∆=+−

∆=+−4342143421

Rearranging and substituting, the total entropy generated during this process is determined to be

( ) ( )

( )( ) ( )( ){ } ( )( ) ( )( ){ } kJ/K 0.498=

+−+−=

+−+−=+∆+∆=

K 290kJ 300

surrb,surrb,

7100.729156.6509.28479.4080.18479.45709.0

out11221122

outgen T

Qsmsmsmsm

TQ

SSS BABA


Tutorial #10

ESO 201A

Thermodynamics









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


[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 T0 = 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 (cp = 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 (cp = 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 T0 = 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,

298 K

1500 K

HE

kW 33.4=kJ/s) 3600/000,150)(8013.0(

Exergy

8013.0K 1500K 29811

inrevth,outrev,outmax,

0revth,maxth,

=

===

=−=−==

QWW

TT

H&&& η

ηη &Wrev

8-20 A heat engine receives heat from a source at a specified temperature at a specified rate, and rejects the waste 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,

kW 283.6 =kJ/s) 400)(7091.0(

7091.0K 1100K 32011

inrevth,outrev,

revth,maxth,

==

=−=−==

QW

TT

H

L

&& η

ηη

320 K

1100 K

HE

400 kJ/s

120 kW(b) The irreversibility rate is the difference between the reversible power and the actual power output:

kW 163.6=−=−= 1206.283outu,outrev, WWI &&&

(c) The second law efficiency is determined from its definition,

42.3%==== 423.0kW 6.283

kW 120

outrev,

outu,II W

Wη



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.

530 R

TH

HE η th = 36% η II = 60%

Analysis From the definition of the second law efficiency,

Thus,

R 1325=R)/0.40 530()1/(1

60.060.036.0

revth,revth,

II

threvth,

revth,

thII

=−=⎯→⎯−=

===⎯→⎯=

ηη

ηη

ηηη

η

LHH

L TTTT

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

kW 13.89=kJ/h 000,50outin === HQQW &&&

W ·

50,000 kJ/h

4 °C

House 25 °C

The COP of a reversible heat pump operating between the specified temperature limits is

20.1415.298/15.2771

1/1

1COP revHP, =−

=−

=HL TT

Thus,

and

kW 12.91

kW 0.978

=−=−=

===

978.089.13

20.14kW 89.13

COP

inrev,inu,

revHP,inrev,

WWI

QW H

&&&

&&



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

Steam 1 kg

800 kPa 180°C

4)-A (Table KkJ/kg 3672.0

/kgm 001003.0kJ/kg 83.104

kPa 100 C25


/kgm 24720.0kJ/kg 7.2594

C180

kPa 800

C25@0

3C25@0

C25@0

0

0

3

⋅=≅=≅

=≅

⎭⎬⎫

=°=

⋅===

⎭⎬⎫

°==

°

°

°

f

f

f

ss

uu

PT

s

u

TP

vv

v

The exergy of steam is

[ ]

kJ 622.7=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⋅−−

⎟⎠

⎞⎜⎝

⎛

⋅−+−

=

−−−+−=Φ

KkJ/kg)3672.0K)(6.7155 298(mkPa 1

kJ 1/kgm)001003.020kPa)(0.247 100(kJ/kg)83.1047.2594(kg) 1(

)()(

33

00000 ssTPuum vv

For R-134a;


/kgm 0008286.0kJ/kg 85.85

kPa 100 C25


/kgm 044554.0kJ/kg 99.386

C180

kPa 800

C25@0

3C25@0

C25@0

0

0

3

⋅=≅=≅

=≅

⎭⎬⎫

=°=

⋅===

⎭⎬⎫

°==

°

°

°

f

f

f

ss

uu

PT

s

u

TP

vv

v

R-134a

1 kg 800 kPa 180°C

[ ]

kJ 5.02=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⋅−−

⎟⎠

⎞⎜⎝

⎛

⋅−+−

=

−−−+−=Φ

KkJ/kg)32432.0K)(1.3327 298(mkPa 1

kJ 1/kgm)0008286.0554kPa)(0.044 100(kJ/kg)85.8599.386(kg) 1(

)()(

33

00000 ssTPuum vv

The steam can therefore has more work potential than the R-134a.



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)

Vacuum

4 kg 200 kPa

80°C WATER

KkJ/kg 0756.1=

kJ/kg 334.97=kg/m 001029.0=

C80kPa 200

C80@1

C80@1

3C80@1

1

1

⋅≅≅≅

⎭⎬⎫

°==

°

°

°

f

f

f

ssuu

TP

vv

Noting that , kg/m 002058.0001029.022 312 =×== vv

KkJ/kg 0278.16430.60002584.00261.1

kJ/kg 14.3188.21580002584.058.317

0002584.0001026.09933.3

001026.0002058.0

kg/m 002058.0

kPa 40

22

22

22

32

2

⋅=×+=+==×+=+=

=−−

=−

=

⎪⎭

⎪⎬⎫

=

=

fgf

fgf

fg

f

sxssuxuu

xP v

vv

v

Taking the direction of heat transfer to be to the tank, the energy balance on this closed system becomes

)( 12in


system


outin

uumUQ

EEE

−=∆=

∆=−4342143421

or

kJ 30.67 kJ .3067=kg334.97)kJ/kg)(318.14 4( outin =→−−= QQ

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,

{

surr

out12gen

12systemgenoutb,

out

entropyin Change

system

generationEntropy

gen


outin

)(

)(

TQ

ssmS

ssmSSTQ

SSSS

+−=

−=∆=+−

∆=+−4342143421

Substituting,

kJ 10.3=

⎥⎦⎤

⎢⎣⎡ ⋅−=

⎟⎟⎠

⎞⎜⎜⎝

⎛+−==

K 298kJ 67.30+K)kJ/kg0756.1kg)(1.0278 (4K) 298(

)(surr

out120gen0destroyed T

QssmTSTX



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

kg 00234.0K) K)(298kg/mkPa 287.0(

)m kPa)(0.002 100(3

3

1

11 =⋅⋅

==RTP

mV

AIR V1 = 2 L P1 = 100 kPa T1 = 25°C

Also,

L 0.473=L) 2(K) kPa)(298 600(K) kPa)(423 100(

112

212

1

11

2

22 ==⎯→⎯= VVVV

TPTP

TP

TP

and

KkJ/kg 1608.0kPa 100kPa 600ln K)kJ/kg 287.0(

K 298K 423ln K)kJ/kg 009.1(

lnln0

2

0

2avg,02

⋅−=

⋅−⋅=

−=−PP

RTT

css p

Thus, the exergy of air at the final state is

[ ][ ]

kJ 0.171=⋅+

⋅⋅=

−+−−−=Φ=

kPa]kJ/m[m0.002)-473kPa)(0.000 100(

K)kJ/kg K)(-0.1608 (298-298)K-K)(423kJ/kg (0.722kg) 00234.0(

)()()(

33

02002002avg,22 VVv PssTTTcmX

(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,

kJ 0.171=−

−=

∆=−−

0171.0=12inrev,

exergyin Change

system

ndestructioExergy

e)(reversibl 0destroyed

mass and work,heat,by nsferexergy traNet outin

XXW

XXXX43421444 3444 2143421

(c) The second-law efficiency of this process is

14.3%===kJ 2.1

kJ 171.0

inu,

inrev,II W

Wη



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 E E

U m u uu u T T

in out− =

= = −= → =

Net energy transferby heat, work, and mass

system

Change in internal, kinetic, potential, etc. energies

1 24 34 124 34∆

∆0 2 1

2 1 2 1

( ) Vacuum Argon

300 kPa 70°C

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,

{

)( 12systemgen

entropyin Change

system

generationEntropy

gen


outin

ssmSS

SSSS

−=∆=

∆=+−4342143421

where

kJ/K 0.433=(2)lnK)kJ/kg kg)(0.2081 3(

lnlnln)(1

2

1

2

1

02

avg,12system

⋅=

=⎟⎟⎠

⎞⎜⎜⎝

⎛+=−=∆

VV

VV

v mRRT

TcmssmS

Substituting,

kJ 129=kJ/K) K)(0.433 298()( 120gen0destroyed =−== ssmTSTX



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 well-insulated 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

waterironinpw,


system


outin

UUUW

EEE

∆+∆=∆=

∆=−4342143421

Wpw

Water

Iron 85°C

100 L 20°C

water12iron12inpw, )]([)]([ TTmcTTmcW −+−=

where

kJ 240)s 6020)(kJ/s 2.0(

kg 7.99)m 1.0)(kg/m 997(

inpw,pw

33water

=×=∆=

===

tWW

m&

Vρ

Substituting,

kg 52.0=

°−°⋅+°−°⋅

iron

iron C)20C)(24kJ/kg kg)(4.18 7.99(C)85C)(24kJ/kg 45.0(=kJ 240m

m

(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,

{

waterironsystemgen

entropyin Change

system

generationEntropy

gen


outin

SSSS

SSSS

∆+∆=∆=

∆=+−4342143421

where

kJ/K 651.5

K 293K 297lnK)kJ/kg kg)(4.18 7.99(ln

kJ/K 371.4K 358K 297lnK)kJ/kg kg)(0.45 0.52(ln

1

2avgwater

1

2avgiron

=⎟⎠⎞

⎜⎝⎛⋅=⎟⎟

⎠

⎞⎜⎜⎝

⎛=∆

−=⎟⎠⎞

⎜⎝⎛⋅=⎟⎟

⎠

⎞⎜⎜⎝

⎛=∆

TT

mcS

TT

mcS

Substituting,

kJ 375.0=kJ/K )651.5371.4K)( 293(gen0destroyed +−== STX



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

copperironout


system


outin

UUUQ

EEE

∆+∆=∆=−

∆=−4342143421

Copper

Iron

Lake 15°C

Iron 85°C

or,

copperout TTmcTTmcQ )]([)]([ 21iron21 −+−=

Substituting,

( )( )( ) ( )( )( )

kJ 1964K288353KkJ/kg 0.386kg 20K288353KkJ/kg 0.45kg 50out

=−⋅+−⋅=Q

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,

{

lake

outcopperirongen

copperironsystemgenoutb,

out

entropyin Change

system

generationEntropy

gen


outin

TQ

SSS

SSSSTQ

SSSS

+∆+∆=

∆+∆=∆=+−

∆=+−4342143421

where

( )( )

( )( ) kJ/K 1.571K 353K 288


kJ/K 4.579K 353K 288


1

2avgcopper

1

2avgiron

−=⎟⎟⎠

⎞⎜⎜⎝

⎛⋅=⎟⎟

⎠

⎞⎜⎜⎝

⎛=∆

−=⎟⎟⎠

⎞⎜⎜⎝

⎛⋅=⎟⎟

⎠

⎞⎜⎜⎝

⎛=∆

TT

mcS

TT

mcS

Substituting,

kJ 196=kJ/KK 288kJ 1964

571.1579.4K) 293(gen0destroyed ⎟⎟⎠

⎞⎜⎜⎝

⎛+−−== STX



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)

KkJ/kg 4244.0

kJ/kg 77.117 0

kPa 1200

1

1

1

1

⋅==

⎭⎬⎫

==

sh

xP

KkJ/kg 4562.0 kJ/kg 77.117kPa 200

2

2

12

2

⋅=°−=

⎭⎬⎫

===

sT

hhP C10.1

(b) Noting that the throttling valve is adiabatic, the entropy generation is determined from

KkJ/kg 0.03176 ⋅=⋅−=−= KkJ/kg)4244.0(0.456212gen sss

Then the irreversibility (i.e., exergy destruction) of the process becomes

kJ/kg 9.464=⋅== K)kJ/kg K)(0.03176 (298gen0dest sTex



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 1500 kPa

300°C

Helium KkJ/kg 2295.2kPa 1500kPa 100lnK)kJ/kg (2.0769

K 573K 298lnK)kJ/kg 1926.5(

lnln1

2

1

212

⋅=

⋅−⋅=

−=−PP

RTT

css p

The maximum (reversible) work is the exergy difference between the inlet and exit states

kJ/kg 2092=⋅−−−⋅=

−−−=

−−−=

K)kJ/kg 2295.2K)( 298(25)KK)(300kJ/kg 1926.5(

)()(

)(

21021

21021outrev,

ssTTTc

ssThhw

p100 kPa

25°C

There is only one inlet and one exit, and thus mmm &&& == 21 . 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

out21out

out21out

2outout1

outin


(steady) 0system


outin

)()(

0

qhhwQhhmW

hmQWhm

EE

EEE

−−=−−=

++=

=

=∆=−

&&&

&&&&

&&

44 344 21&

43421&&

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,

⎟⎟⎠

⎞⎜⎜⎝

⎛−−−−−=

⎟⎟⎠

⎞⎜⎜⎝

⎛−−−=

+⎟⎟⎠

⎞⎜⎜⎝

⎛−+=

=

=∆=−−

TT

qssThh

TT

qw

mTT

QWm

XX

XXXX

0out21021

0out21outrev,

20

outoutrev,1

outin

exergy of change of Rate

(steady) 0system

ndestructio exergy of Rate


mass and work,heat,by nsferexergy tranet of Rate

outin

1)()(

1)(

1

0

ψψ

ψψ &&&&

&&

444 3444 21&

444 3444 21&

43421&&

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.



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)

KkJ/kg 1693.7

kJ/kg 8.3658C600

MPa 6

1

1

1

1

⋅==

⎭⎬⎫

°==

sh

TP

KkJ/kg 6953.7

kJ/kg 4.2682C100

kPa 50

2

2

2

2

⋅==

⎭⎬⎫

°==

sh

TP

Analysis (b) There is only one inlet and one exit, and thus & &m m m1 2 &= = . 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

outin


(steady) 0system


outin 0

EE

EEE

&&

444 344 21&

43421&&

=

=∆=− 80 m/s 6 MPa 600°C


50 kPa 100°C

140 m/s

STEAM

⎥⎥⎦

⎤

⎢⎢⎣

⎡ −+−=

++=+

2

)2/()2/(2

22

121out

222out

211

VVhhmW

VhmWVhm

&&

&&&

5 MW

Substituting,

kg/s 5.156s/m 1000

kJ/kg 12

m/s) 140(m/s) 80(4.26828.3658kJ/s 5000

22

22

=

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠

⎞⎜⎝

⎛−+−=

m

m

&

&

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,

]∆pe∆ke)()[()(

0

02102121rev,out

2rev,out1

outin

exergy of change of Rate

(steady) 0system

ndestructio exergy of Rate


mass and work,heat,by nsferexergy tranet of Rate

outin

−−−−−=−=

+=

=

=∆=−−

ssThhmmW

mWm

XX

XXXX

&&&

&&&

&&

444 3444 21&

444 3444 21&

43421&&

ψψ

ψψ

Substituting,

kW 5808=⋅−−=

−−=

KkJ/kg )7.6953K)(7.1693 kg/s)(298 156.5(kW 5000)( 210outoutrev, ssTmWW &&&

(b) The second-law efficiency of a turbine is the ratio of the actual work output to the reversible work,

86.1%===MW 808.5

MW 5

outrev,

outII W

W&

&η

Discussion Note that 13.9% percent of the work potential of the steam is wasted as it flows through the turbine during this process.


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)

KkJ/kg 5432.6

kJ/kg 3.3178C400

MPa 6

1

1

1

1

⋅==

⎭⎬⎫

°==

sh

TP Steam

2

1

KkJ/kg 0225.7MPa 2

212

2 ⋅=⎭⎬⎫

==

shh

P

Analysis The decrease in exergy is of the steam is the difference between the inlet and exit flow exergies,

kJ/kg 143 =⋅−=

−=−−∆−∆−∆−=−=KkJ/kg)5432.6K)(7.0225 (298

)()]([exergyin Decrease 12021021000

ssTssTpekehψψ

Discussion Note that 143 kJ/kg of work potential is wasted during this throttling process.



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,

KkJ/kg .07561kJ/kg .02335

C08kPa 200

KkJ/kg 7.5081kJ/kg .42870

C200kPa 200

KkJ/kg 0.22447kJ/kg .9862

C15kPa 200

C80@3

C08@3

3

3

2

2

2

2

C15@1

C15@1

1

1

⋅=≅=≅

⎭⎬⎫

°==

⋅==

⎭⎬⎫

°==

⋅=≅=≅

⎭⎬⎫

°==

°

°

f

f

f

f

sshh

TP

sh

TP

sshh

TP

o

o

2

MIXING CHAMBER

200 kPa

15°C4 kg/s

200°C

1

600 kJ/min

80°C 3

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

Mass balance: 321(steady) 0

systemoutin 0 mmmmmm &&&&&& =+⎯→⎯=∆=−

Energy balance:

33out221

outin


(steady) 0system


outin 0

hmQhmhm

EE

EEE

&&&&

&&

444 344 21&

43421&&

+=+

=

=∆=−

Combining the two relations gives ( ) ( ) ( 3223113212211out hhmhhmhmmhmhm )−+−=+−+= &&&&&&&Q

Solving for and substituting, the mass flow rate of the superheated steam is determined to be &m2

( ) ( )( )

( ) kg/s 0.429=−

−−=

−−−

=kJ/kg02.3352870.4

kJ/kg02.33562.98kg/s 4kJ/s) (600/60

32

311out2 hh

hhmQm

&&&

Also, kg/s .42940.4294213 =+=+= mmm &&&

(b) The exergy destroyed during a process can be determined from an exergy balance or directly from its definition where the entropy generation Sgen0destroyed STX = gen is determined from an entropy balance on an extended system that includes the mixing chamber and its immediate surroundings. It gives

{

0

out221133gengen

surrb,

out332211


0system

generation entropy of Rate

gen

mass andheat by ansferentropy trnet of Rate

outin

0

0

TQ

smsmsmSSTQ

smsmsm

SSSS

&&&&&&

&&&&

43421&&

43421&&

+−−=→=+−−+

=∆=+−

Substituting, the exergy destruction is determined to be

kW 202

=

+×−×−×=

⎟⎟⎠

⎞⎜⎜⎝

⎛+−−==

kW/K)298/1022447.045081.7429.0.07561K)(4.429 298(,

1122330gen0destroyedsurrb

out

TQ

smsmsmTSTX&

&&&&&



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


)(

0)peke (since

0

12in

21in

outin


(steady) 0system


outin

TTcmQ

hmhmQ

EE

EEE

p −=

≅∆≅∆=+

=

=∆=−

&&

&&&

&&

444 344 21&

43421&&

Then the rate of heat transfer to the cold water in this heat exchanger becomes

kW5940=C)20CC)(70kJ/kg. kg/s)(4.18 5.4()]([ waterinout .°−°°=−= TTcmQ p&&

Oil 170°C 10 kg/s

(12 tube passes)

Water 20°C

4.5 kg/s

70°C

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

C129.1°=°

−°=−=→−=C)kJ/kg. kg/s)(2.3 10(

kW 5.940C170 )]([ inoutoiloutinp

p cmQ

TTTTcmQ&

&&&

(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:

{

)()(

0

)0 (since 0

34oil12watergen

gen4oil2water3oil1water

gen43223311


(steady) 0system

generation entropy of Rate

gen

mass andheat by ansferentropy trnet of Rate

outin

ssmssmS

Ssmsmsmsm

QSsmsmsmsm

SSSS

−+−=

=+−−+

==+−−+

∆=+−

&&&

&&&&&

&&&&&

44 344 21&&

43421&&

Noting that both fluid streams are liquids (incompressible substances), the rate of entropy generation is determined to be

kW/K 0.736273+170273+129.1ln kJ/kg.K) kg/s)(2.3 10(

273+20273+70ln kJ/kg.K) kg/s)(4.18 5.4(

lnln3

4oil

1

2watergen

=+=

+=TT

cmTT

cmS pp &&&

The exergy destroyed during a process can be determined from an exergy balance or directly from its definition , gen0destroyed STX =

kW 219=== )kW/K K)(0.736 298(gen0destroyed STX &&


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)

Mixture45°C

Sat. vap. 0.23 kg/s

Mixing chamber

Water 15°C 4.6 kg/s

kJ/kg.K 63862.0kJ/kg 44.188

0C45

kJ/kg.K 22447.0kJ/kg 98.62

0C15

3

3

1

3

01

01

1

1

==

⎭⎬⎫

=°=

====

⎭⎬⎫

=°=

sh

xT

sshh

xT

An energy balance on the chamber gives

kJ/kg 5.2697kJ/kg) 44.188)(kg/s 23.06.4(kg/s) 23.0(kJ/kg) 98.62(kg/s) 6.4(

)(

2

2

321332211

=+=+

+==+

hh

hmmhmhmhm &&&&&

The remaining properties of the saturated steam are

kJ/kg.K 1907.71

kJ/kg 5.2697

2

2

2

2

=°=

⎭⎬⎫

==

sT

xh C114.3

(b) The specific exergy of each stream is

01 =ψ

kJ/kg 28.628)kJ/kg.K22447.0K)(7.1907 273(15kJ/kg)98.625.2697()( 020022

=−+−−=−−−= ssThhψ

kJ/kg 18.6)kJ/kg.K22447.0K)(0.63862 273(15kJ/kg)98.6244.188()( 030033

=−+−−=−−−= ssThhψ

The exergy destruction is determined from an exergy balance on the chamber to be

kW 114.7=+−+=

+−+=kJ/kg) 18.6)(kg/s 23.06.4(kJ/kg) 28kg/s)(628. 23.0(0

)( 3212211dest ψψψ mmmmX &&&&&

(c) The second-law efficiency for this mixing process may be determined from

0.207=+

+=

++

=kJ/kg) 28kg/s)(628. 23.0(0kJ/kg) 18.6)(kg/s 23.06.4()(

2211

321II ψψ

ψηmm

mm&&

&&



Tutorial #11

(QUIZ #2)

ESO 201A

Thermodynamics







Tutorial #12

ESO 201A

Thermodynamics


Contact







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.


[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 (non-combustibles) 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

2222283 NOHCO)N76.3O(HC pzyx ++⎯→⎯++

Balancing the carbon in this reaction gives C3H8

Air

100% theoretical

Combustion chamber

CO2, H2O, N2 y = 3

and the hydrogen balance gives

482 =⎯→⎯= zz

The oxygen balance produces

52/432/22 =+=+=⎯→⎯+= zyxzyx


balance of the nitrogen in this reaction gives

=×== x

balanced form, the reaction is

N8.8

The mass fraction of carbon dioxide is determined from

A

276.32 ⎯→⎯=× ppx 8.18576.376.3

In

222283 1OH4CO3N8.18O5HC ++⎯→⎯++ 2

0.181==kg 4.730

++=

++==

kg 132kg/kmol) 28)(kmol 8.18(kg/kmol) 18)(kmol 4(kg/kmol) 44)(kmol 3(

kg/kmol) 44)(kmol 3(

mfN2N2H2OH2OCO2CO2

CO2CO2

products

CO2CO2 MNMNMN

MNm

m

) The mo ctions of water vapor are

(b le and mass fra

0.155==++

=++

==kmol 8.25

kmol 4kmol 8.18kmol 4kmol 3

kmol 4

N2H2OCO2

H2O

products

H2OH2O NNN

NN

Ny

0.0986==

++=

++==

kg 4.730kg 72

kg/kmol) 28)(kmol 8.18(kg/kmol) 18)(kmol 4(kg/kmol) 44)(kmol 3(kg/kmol) 18)(kmol 4(

mfN2N2H2OH2OCO2CO2

H2OH2O

products

H2OH2O MNMNMN

MNm

m


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

[ ] 2th2th2222th63 N3.76)(1.5O0.5O3H3CO3.76NO1.5HC aaa ×+++⎯→⎯++

where ath is the stoichiometric coefficient for air. It is determined from


.Products C3H6

50% excess air O2 balance: 15 3 15 0 5 4 5. . .a a ath th th= + + ⎯→⎯ =

Substituting,

C H O N CO H O O N3 6 2 2 2 2 2 26 75 3 76 3 3 2 25 25 38+ + ⎯ →⎯ + + +. . . .

The air-fuel ratio is determined by taking the ratio of the mass of the air to the mass of the fuel,

( )( )

( )( ) ( )( ) fuel air/kg kg 22.2=×

==kg/kmol 29kmol 4.766.75

AF airm

mixture is the saturation temperature of the water vapor in the product gases orrespond g to its partial pressure. That is,

+ kg/kmol 2kmol 3kg/kmol 12kmol 3fuelm

(b) The dew-point temperature of a gas-vaporc in

( ) kPa .3679kPa 105 k mol33.63

kmol 3prod

prod=⎟⎟

⎠

⎞⎜⎜⎝

⎛=⎟

⎟⎠

⎞⎜⎜⎝

⎛P

NN v

v

hus,

=P

T

C44.5°== kPa .3679@satdp TT


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

[ ] 22222188 3.76)N(12.5OH9CO83.76NO5.12HC ×++⎯→⎯++

The combustion equation with 100% excess air is


94O 5.12O9H8CO3.76NO25HC +++⎯→⎯++ [ ] 222222188 N

The heat transfer for this combustion process is determined from the energy balance systemoutin applied on the combustion chamber with W = 0. It reduces to

EEE ∆=−

( ) ( )∑ ∑ RfRPfP hhhNhhhNQout −+−−+=− oooo

ssuming the air and the combustion products to be ideal gases, we have h = h(T). From the tables,

A

Substance

ofh

kJ/kmol K298h

kJ/kmol K530h

kJ/kmol 18 (g) 8,450 C8H -20 --- ---

O2 0 8682 15,708 N2 0 8669 15,469 H2O (g) CO2 -393,520 9364 19,029

ubstitutin ,

-241,820 9904 17,889

S g

( )( ) ( )( ) ( )( )( )( ) ( )( )

188

out

HC kmolkJ 880,239,4 /−=00450,20818669469,15094

8682708,1505.129904889,17820,241964029,19520,3938−−−−−++

−++−+−+−+−=−Q

93

188out HC kmolkJ 880,239,4 /=Q or

Then the heat transfer per kg of fuel is

188HC kJ/kg 37,200===kg/kmol 114

fuel kJ/kmol 880,239,4

fuel

outout M

QQ

Products

257°C

C8H18

25°C

100% excess air

25°C

Qout

Combustion chamber

P = 1 atm


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


6 1.5 7.5= + =

stem

1000 K

Q

C6H6+ Air 25C, 1 atm

( ) ( ) 2th2222th66 N3.76O3H6CO3.76NOgHC aa ++⎯→⎯++

where ath is the stoichiometric coefficient and is determined from the O2 balance,

tha

Then the actual combustion equation with 30% excess air becomes

( ) ( ) 22222266 36.66N2.49OO3H0.48CO5.52CO3.76NO9.75gHC ++++⎯→⎯++

The heat transfer for this constant volume combustion process is determined from the energy balance E Ein out syE − = ∆ It reduces to applied on the combustion chamber with W = 0.

( ) ( )∑ ∑ −−+−−−+=−RP

Since both the reactants and the products behave as ideal gases,

fRfP PhhhNPhhhNQ vv ooooout

all the internal energy and enthalpies depend on temperature

vP terms in this equation can be replaced by Ronly, and the uT.

It yields

( ) ( )∑ ∑ −−−−+=−RP

°

ufRufP TRhNTRhhhNQ ooK 298K 1000out

reactants are at the standard reference temperature of 25 C. From the tables, since the

Substance hfo

kJ/kmol

h298 K

kJ/kmol

h1000 K

kJ/kmol

H6 (g) 82,930 C6 --- ---

O2 0 8682 31,389

N2 0 8669 30,129

H2O (g) -241,820 9904 35,882

CO -

CO

110,530

-393,520 9

8669

364

30,355

42,769 2

Thus,

or

( )( )( )( )( )( )( )( )( )( )( )( ) ( )( )( )

kJ 4332,200,298314.876.475.9298314.8930,821

1000314.88669129,30066.361000314.88682389,31049.2

1000314.89904882,35820,24131000314.88669355,30530,11048.0

1000314.89364769,42520,39352.5out

−=×−−×−−

×−−++×−−++

×−−+−+×−−+−+×−−+−=−Q

kJ 2,200,433=outQ


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

[ ] 2th2th2222th104 N)76.30.2(O0.1OH5CO43.76NO0.2HC aaa ×+++⎯→⎯++

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: 5.60.15.240.2 ththth =⎯→⎯++= aaa


N88.48O5.6OH5CO4N6 +++⎯→⎯

The mole fractions of the products are

N

Substituting, [ ]2104 7.3O13HC ++ 22222

kmol 38.6488.485.654 =+++=m

0.7592

0.1010

0.0777

0.0621===kmol 4CO2

CO2N

y

===

===

kmol 64.38kmol 48.88kmol 64.38

kmol 64.38kmol 5

kmol 64.38

N2N2

O2

H2OH2O

m

m

m

m

NN

y

N

NN

y

N

The mass of carbon dioxide in the products per unit mass of fuel burned is

ProductsC4H10

Air

100% excess

===kmol 6.5O2N

y

1042 HC /kgCO kg 3.034=××

=kg )58(1kg )44(4

C4H10

CO2

mm


fuel air/kg kg 30.94=×

==)kmolkg/ 58)(kmol 1(

kg/kmol) 29)(kmol 4.7613(AFfuel

air

mm


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

Products

60% CH440% C2H6O

Air

100% theoretical kmol 8696.0

kg/kmol 46kg 40mf

kmol 75.3kg/kmol 16CH4

CH4 M

kg 60mf

C2H6O

C2H6OC2H6O

CH4

===

===

MN

N

ole fraction of methane and ethanol in the fuel mixture are

The m

1882.0


kmol )8696.075.3(C2H6OCH4 ++ NN

The combustion

kmol 0.8696

8118.0kmol 3.75

C2H6O

CH4

==

===

Ny

Nx

equation in this case can be written as

N OH CO 3.76NO FDBa ++⎯→⎯++

where ath c coefficient for air. The coefficient ath and other coefficients are to be determined from the mass bal

2

kmol )8696.075.3(C2H6OCH4 ++ NN

=

[ ] 22222th624 OHC CH yx +

is the stoichiometriances

Carbon balance: + Byx =

Hydrogen balance: Dyx 264 =+

DBya +=+ 22 th Oxygen balance:

Fa =th76.3 Nitrogen balance:

Substituting x and y values into the equations and solving, we find the coefficients as

1882.08118.0

th

==

==

DB

Then, we write the balanced reaction equation as

188.2188.1

228.8188.2 == Fa

yx

[ ] 22222624 N 228.8OH 188.2CO 188.13.76NO 188.2OHC 1882.0 CH 8118.0 ++⎯→⎯+++


fuel air/kg kg 94.13kg/kmol)1616122(kmol) 1882.0(kg/kmol)1412(kmol) 8118.0(

kg/kmol) 29)(kmol 76.4188.2(AFfuel

air

=+×+×+×+

×==

mm

Then, the required flow rate of air becomes

kg/s 139.4=== kg/s) 10)(94.13(AF fuelair mm &&


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

( ) 2th2222th4 N3.76O2HCO3.76NOCH aa ++⎯→⎯++


=+=a Products

25°C

CH4

25°C

Air

100% theoretical

Q

Combustion chamber

P = 1 atm

where ath is determined from the O2 balance,

th 211

Substituting,

( ) 222224 5.64NO2HCO3.76NO2CH ++⎯→⎯++

The heat transfer for this combustion process is determined from the energy balance E Ein out systemE − = ∆ applied on the combustion chamber with W = 0. It reduces to

( ) ( ) ∑ ∑∑ ∑ −=−+−−+=− ooooooRfRPfPRP

since both the rea the tables,

fRfP hNhNhhhNhhhNQ ,,out

ctants and the products are at 25°C and both the air and the combustion gases can be treated as ideal gases.

stance

From

Sub

h fo

kJ/kmol CH4 850 -74,O2 0 N2 0 H2O (l) CO2 -393,520

Thus,

or

-285,830

( )( ) ( )( ) ( )( )

4CH kmol kJ / 890,330=outQ

If combustion is achiev

/,−=−−−−+−+−=− 4out CH kmol kJ 33089000850,7410830,2852520,3931Q

ed with 100% excess air, the answer would still be the same since it would enter and leave at 25°C, and absorb no energy.


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

kmol 0247.0kg/kmol 32

kg 0.79


kg 0.72


kg 41.11

kmol 465.3kg/kmol 2

kg 6.93


kg 39.25

S

SS

N2

N2N2

O2

O2O2

H2

H2H2

C

CC

===

===

===

===

===

Mm

N

Mm

N

Mm

N

Mm

N

Mm

N

39.25% C 6.93% H2

41.11% O20.72% N20.79% S

11.20% ash (by mass)

Air 40% excess

Products 127°C

Coal

Combustion chamber The mole number of the mixture and the mole fractions are

kmol 071.80247.00257.0285.1465.3271.3 =++++=mN

0.00306kmol 8.071kmol 0.0247

0.00319kmol 8.071kmol 0.0257

0.1592kmol 8.071kmol 1.285

0.4293kmol 8.071kmol 3.465

4052.0kmol 8.071kmol 3.271

SS

N2N2

O2O2

H2H2

CC

===

===

===

===

===

m

m

m

m

m

NN

y

NN

y

NN

y

NN

y

NN

y

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

2th22th22

22th222

N76.34.1SO00306.0O4.0OH4293.0CO4052.0 ⎯

)N76.3O(4.1S00306.0N00319.0O1592.00.4293H0.4052C

×++++⎯→

++++++

aa

a

ccording to the O2 mass balance,

92 ththth =⎯→⎯++×+=+ aaa

Substituting,

N441.2SO00306.0O1855.0OH4293.0CO

)N76.

++++

A

15.0 4637.000306.04.04293.05.04052.04.1

22222

22222

4052.0

3O(6492.0S00306.0N00319.0O1592.00.4293H0.4052C

⎯→⎯

++++++

The heat transfer for this combustion process is determined from the energy balance outin EEE ∆=− system applied on the ombustion chamber with W = 0. It reduces to

c

( ) ( )∑ ∑ −+−+=RfRPfP hhhNhhhNQ oooo

out −−



15-46


Assumin he com oducts l gases, = h(T). From the tables,

g the air and t bustion pr to be idea we have h

Substance

ofh

kJ/kmol

K 298h

ol kJ/km

K 400h

kJ/kmol

O2 0 8682 11,711

N2 0 8669 ,640

H2O (g) -241,820 9904 13,356

SO2 -297,100 - -

The enthalpy change of sulfur dioxide between the standard temperature and the product temperature using constant specific heat assumption is

11

CO2 -393,520 9364 13,372

kJ/kmol 425325)KK)(127kJ/kmol 7.41(SO2 =−⋅=∆=∆ Tch p

Substituting into the energy balance relation,

( )( ) ( )( )( )( ) ( )( ) ( )( )

188HC kmolkJ 244,25304253100,29700306.08669640,110441.28682711,1101855.0

9904356,13820,2414293.09364372,

/−=−+−+−++−++

−+−+−

r

out 13520,3934052.0 +−=−Q

fuel kmolkJ 244,253out /=Q o

Then the heat transfer per kg of fuel is

coal kJ/kg 23,020=

=

×+×+×+×+×==

kg/kmol 11.00fuel kJ/kmol 244,253

kg/kmol )3200306.02800319.0321592.024293.012(0.4052fuel kJ/kmol 244,253

fuel

outout M

QQ


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 systemoutin EEE ∆=− applied on the combustion chamber with Q = W = 0 reduces to

( ) ( ) ( ) oooooooRfRPfPRfRPfP hNhhhNhhhNhhhN ,∑ ∑∑ ∑ =−+⎯→⎯−+=−+

since all the reactants are at the standard reference temperature of 25°C. Then, for the stoichiometric oxygen

222 +⎯→⎯

From the tables,

hus,

HC 22 + OH 1CO 2O5.2

Products

TP

C2H2

25°C

100% theoretical O2

25°C

Combustion chamber


T

( ) ( ) ( ) 00730,226)1(9904820,241)1(9364520( ,393)2 H2OCO2 ++=−+−+−+− hh

It yields kJ 220,284,112 H2OCO2 =+ hh

The adiabatic flame temperature is obtained from a trial and error solution. A first guess is obtained by dividing the right-hand 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.

Table A-2b, the highest available value of specific heat is cp = 1.234 kJ/kg·K or CO2 at 1000 K. The specific heat of water vapor is c = 1.8723 kJ/kg·K (Table A-2a). Using these specific heat values, In f

p

( ) ( ) ( ) 00730,226)1(820,241)1(520,393)2( ++=∆+−+∆+− Tc Tc pp

here . The specific heats on a molar base are w C)25( af °−=∆ TT

KkJ/kmol .733kg/kmol) K)(18kJ/kg 8723.1(

KkJ/kmol 54.3kg/kmol) K)(44kJ/kg 234.1(

H2O,

CO, 2

⋅=⋅==

⋅=⋅==

Mcc

Mc

p

p

c p

p

Substituting,

( ) ( )

K 8824KkJ/kmol )7.333.542(

kJ/kmol 590,255,1590,255,17.33)3.542(

730,2267.33820,241)1(3.54520,393)2(

=⋅+×

=∆

=∆+∆×=∆+−+∆+−

T

TTTT

Then the adiabatic flame temperature is estimated as

e

Substanc

ofh

kJ/kmol

K298h

kJ/kmol

2 (g) 226,730 C2H ---

O2

2 (g)

-393,520 9364

0 8682

N2 0 8669

H O -241,820 9904

CO2

C8849°=+=+∆= 25882425af TT


Tutorial #13

ESO 201A

Thermodynamics









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, P1=95 kPa, T1 = 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]


[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

P

4

1

3

2

K 4.537K)(8) 288( 13.111

1

2

112 ===⎟⎟

⎠

⎞⎜⎜⎝

⎛= −−

−n

n

rTTTv

v

And the temperature at the end of the expansion is

K 4.7898

K) 1473(34

34 =⎟⎠

⎜⎝

=⎟⎠

⎜⎝

=⎟⎟⎠

⎜⎜⎝

=r

TTTv

11 13.1113 ⎞⎛⎞⎛⎞⎛ −−− nn

v

for the polytropic compression gives The integral of the work expression


kJ/kg 6.238)18(13.1

11 2

21 −⎥⎦⎢⎣⎟⎠

⎜⎝−− n

wv

K) K)(288kJ/kg 287.0( 13.11

11 =−⋅

=⎥⎤

⎢⎡

−⎟⎞

⎜⎛

= −−n

RT v

imilarly, the work produced during the expansion is S

kJ/kg 0.654181

13.1K) K)(1473kJ/kg 287.0(

11

13.11 ⎤⎡ −n

4

3343 =

⎥⎥⎦

⎤

⎢⎢⎣

⎡−⎟

⎠⎞

⎜⎝⎛

−⋅

−=⎥⎥

⎦⎢⎢

⎣−⎟⎟

⎠

⎞⎜⎜⎝

⎛−

−=−

− nRT

wv

v

pplication of the first law to each of the four processes gives

A

1 kJ/kg 53.59K)2884.537)(KkJ/kg 718.0(kJ/kg 6.238)( 12212 =−⋅−=−−= − TTcwq v −

1473)(KkJ/kg 718.0()( 2332 ⋅=−=− TTcq v kJ/kg 8.671K)4.537 =−

kJ/kg 2.163K)4.7891473)(KkJ/kg 718.0(kJ/kg 0.654)( 434343 =−⋅−=−−= −− TTcwq v

kJ/kg 0.360K)2884.789)(KkJ/kg 718.0()( 1414 =−⋅=−=− TTcq v

The head added and rejected from the cycle are

The thermal efficiency of this cycle is then

kJ/kg 419.5kJ/kg 835.0

=+=+=

=+=+=

−−

−−

0.36053.592.1638.671

1421out

4332in

qqqqqq

0.498=−=−=0.8355.41911

in

outth q

qη


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.


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

P

4

1

3

2Qin Qout

( )( )

( ) ( ) kPa 2338kPa 100K


308K 757.9

9.5

K 757.99.5K

11

2

2

12

1

11

2

22

0.41

=⎟⎟⎠

⎞⎜⎜⎝

⎛==⎯→⎯=

=⎞⎛

−

PTT

PT

PT

P

k

v

vvv

v

308

2

112 =⎟⎟

⎠⎜⎜⎝

= TTv

Process 3-4: isentropic expansion.

( )( ) K 1969==⎟⎟⎠

⎞⎜⎜⎝

⎛=

−0.4

1

3

443 9.5K 800

k

TTv

v

Process 2-3: v = constant heat addition.

( ) kPa 6072=⎟⎟⎠

⎞⎜⎜⎝

⎛==⎯→⎯= kPa 2338

K 757.9K 1969

22

323 TTT

3223 PT

PPP vv

(b)

3

( )( )( )( )

kg10788.6K 308K/kgmkPa 0.287

m 0.0006kPa 100 43

3

1

11 −×=⋅⋅

==RTP

mV

( ) ( ) ( )( )( ) kJ 0.590=−⋅×=−=−= − K757.91969KkJ/kg 0.718kg106.788 42323in TTmcuumQ v

(c) Process 4-1: v = constant heat rejection.

( ) ( )( )( ) kJ 0.240K308800KkJ/kg 0.718kg106.788)( 41414out =−⋅×−=−=−= −TTmcuumQ v

kJ 0.350240.0590.0outinnet =−=−= QQW

59.4%===kJ 0.590kJ 0.350

in

outnet,th Q

Wη

( )( )kPa 652=⎟

⎟⎠

⎞⎜⎜⎝

⎛ ⋅

−=

−=

−=

==

kJmkPa

1/9.51m 0.0006kJ 0.350

)/11(MEP

3

31

outnet,

21

outnet,

max2min

rWW

r

VVV

VVV (d)


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.


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

P

4

1

2 3 qin

qout

( )( ) K 971.120K 2932

112 =⎟⎟

⎠⎜⎜⎝

= TTV

0.41

=⎞⎛

−kV

rocess 2-3: P = constant heat addition.

P

2.265K971.1K2200

2

3

2

3

2

22

3

33 =⎯→⎯=TT V

==TTPP VVV

rocess 3- isentropic expansion.

P 4:

( )

( ) ( )( )( ) ( )( )

63.5%===

=−=−=

=−⋅=−=−=

=−⋅=−=−=

=⎟⎠⎞

⎜⎝⎛=⎟

⎠⎞

⎜⎝⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

−−−

kJ/kg 1235kJ/kg 784.4

kJ/kg 784.46.4501235

kJ/kg 450.6K293920.6KkJ/kg 0.718

kJ/kg 1235K971.12200KkJ/kg 1.005

K 920.620

2.265K 2200265.2265.2

in

outnet,th

outinoutnet,

1414out

2323in

0.41

3

1

4

23

1

4

334

qw

qqw

TTcuuq

TTchhq

rTTTT

p

kkk

η

v

V

V

V

V

(b) ( )( )

( ) ( )( )kPa933

kJmkPa

1/201/kgm 0.885kJ/kg 784.4

/11MEP

/kgm 0.885kPa 95

K 293K/kgmkPa 0.287

3

31

outnet,

21

outnet,

max2min

max3

3

1

11

=⎟⎟⎠

⎞⎜⎜⎝

⎛ ⋅

−=

−=

−=

==

==⋅⋅

==

rww

r

PRT

vvv

vvv

vv



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.

s

T

3

4

1

5qin1200 K

300 K

86

7

10

9

2

Properties The properties of air are given in Table A-17.

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,

( )( )

( )

( ) ( )( ) ( ) kJ/kg 62.86636.94679.127722

kJ/kg 2.142219.30026.41122

kJ/kg 946.3633.7923831

5

⎞⎛P

238kJ/kg 77.7912

K 2001

kJ/kg 411.26158.4386.13

386.1kJ/kg 300.19

K 300

65outT,

12inC,

865

6

755

421

2

11

56

12

1

=−=−=

=−=−=

==⎯→⎯=⎟⎠

⎜⎝

==

===

⎯→⎯=

==⎯→⎯===

==

⎯→⎯=

hhw

hhw

hhPP

P

Phh

T

hhPPP

P

Ph

T

rr

r

rr

r

Thus,

33.5%===kJ/kg 662.86kJ/kg 222.14

outT,

inC,bw w

wr

( ) ( ) ( ) ( )

36.8%===

=−=−=

=−+−=−+−=

kJ/kg 1197.96kJ/kg 440.72

kJ/kg 440.72222.1486.662

kJ/kg 1197.9636.94679.127726.41179.1277

in

netth

inC,outT,net

6745in

qw

www

hhhhq

η

(b) When a regenerator is used, rbw remains the same. The thermal efficiency in this case becomes

( ) ( )( )

55.3%===

=−=−=

=−=−=

kJ/kg 796.63kJ/kg 440.72

kJ/kg 796.6333.40196.1197

kJ/kg 33.40126.41136.94675.0

in

netth

regenoldin,in

48regen

qw

qqq

hhq

η

ε



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),

( )

( )( )

kJ/kg 198.8706.781.191

kJ/kg 7.06mkPa 1

kJ 1kPa 107,000/kgm 0.00101

/kgm 00101.0

kJ/kg .81191

in,12

33

121in,

3kPa 10 @1

kPa 10 @1

=+=+=

=

⎟⎟⎠

⎞⎜⎜⎝

⎛

⋅−=

−=

==

==

p

p

f

f

whh

PPw

hh

v

vv


( )( ) kJ/kg 3.62151.23928201.081.191

8201.04996.7

6492.08000.6kPa 10

KkJ/kg 8000.6kJ/kg 411.43

C500MPa 7

44

44

34

4

3

3

3

3

=+=+=

=−

=−

=⎭⎬⎫

==

⋅==

⎭⎬⎫

°==

fgf

fg

f

hxhh

sss

xss

P

sh

TP

qin

qout

10 kPa 1

3

2

4

7 MPa

T

s

Thus,

kJ/kg 7.12508.19615.3212kJ/kg 8.196181.1916.2153kJ/kg 5.321287.1984.3411

outinnet

14out

23in

=−=−=

=−=−=

=−=−=

qqwhhqhhq

and

38.9%===kJ/kg 3212.5kJ/kg 1250.7

in

netth q

wη

(b) skg 036 /.kJ/kg 1250.7kJ/s 45,000

net

net ===wWm&

&

(c) The rate of heat rejection to the cooling water and its temperature rise are

( )( )

( )( ) C8.4°=°⋅

==∆

===

CkJ/kg 4.18kg/s 2000kJ/s 70,586

)(

kJ/s ,58670kJ/kg 1961.8kg/s 35.98

watercooling

outwatercooling

outout

cmQ

T

qmQ

&

&

&&


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.


Analysis From the steam tables (Tables A-4, A-5, and A-6),

/kgm 001026.0kJ/kg 03.314

C753.63.813.6

kPa 503

C75 @1

C75 @ 1

kPa 50 @sat 1

1

==

=≅

⎭⎬⎫

°=−=−==

°

°

f

fhhTT

Pvv

kJ/kg 10.6 mkPa 1kJ 1 kPa)506000)(/kgm 001026.0(

)(

33

121inp,

=⎟⎠

⎞⎜⎝

⎛

⋅−=

−= PPw v


kJ/kg 13.32010.603.314inp,12 =+=+= whh

kJ/kg 4.2336)7.2304)(8660.0(54.340

8660.05019.6

0912.17219.6

kPa 50

KkJ/kg 7219.6kJ/kg 9.3302

C450kPa 6000

44

44

34

4

3

3

3

3

=+=+=

=−

=−

=

⎭⎬⎫

==

⋅==

⎭⎬⎫

°==

fgsfs

fg

fs

hxhhs

ssx

ssP

sh

TP

qin

qout

50 kPa 1

3

2

4

6 MPa

4s

s

T

kJ/kg 4.2394)4.23369.3302)(94.0(9.3302)( 4s3T3443

43T =−−=−−=⎯→⎯

−−

= hhhhhhhh

sηη

Thus,

kW 18,050

kW 122

kW 59,660

=−=−=

===

=−=−=

=−=−=

122170,18

kJ/kg) kg/s)(6.10 20(

kW 18,170kJ/kg)4.2394.9kg/s)(3302 20()(

kJ/kg)13.320.9kg/s)(3302 20()(

inP,outT,net

inP,inP,

43outT,

23in

WWW

wmW

hhmW

hhmQ

&&&

&&

&&

&&

and

0.3025===660,59050,18

in

netth Q

W&

&η


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.


T


Analysis From the steam tables (Tables A-4, A-5, and A-6),

kJ/kg 87.19806.781.191kJ/kg 06.7 mkPa 1

kJ 1 kPa)107000)(/kgm 001010.0()(

/kgm 001010.0

kJ/kg 81.191

inp,12

33

121inp,

3kPa 10 @1

kPa 10 @1

=+=+==

⎟⎠

⎞⎜⎝

⎛⋅

−=

−=

==

==

whh

PPw

hh

f

f

v

vv

C373.3°==

⎭⎬⎫

==

⋅=+=+==+=+=

⎭⎬⎫

==

3

3

43

3

44

44

4

4

kJ/kg 5.3085

kPa 7000

KkJ/kg 3385.6)6160.4)(93.0(0457.2kJ/kg 0.2625)5.2047)(93.0(87.720

93.0

kPa 800

Th

ssP

sxsshxhh

xP

fgf

fgf

1

5

2

6

3

4

800 kPa

10 kPa

7 MPa

s

C416.2°==

⎭⎬⎫

==

⋅=+=+==+=+=

⎭⎬⎫

==

5

5

65

5

66

66

6

6

kJ/kg 0.3302

kPa 800

KkJ/kg 6239.7)4996.7)(93.0(6492.0kJ/kg 4.2416)1.2392)(93.0(81.191

90.0kPa 10

Th

ssP

sxsshxhh

xP

fgf

fgf

Thus,

kJ/kg 6.222481.1914.2416

kJ/kg 6.35630.26250.330287.1985.3085)()(

16out

4523in

=−=−==−+−=−+−=

hhqhhhhq

and

37.6%==−=−= 3757.06.35636.222411

in

outth q

qη


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.


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),

)throttling( kJ/kg 32.107

kJ/kg 32.107 liquid sat.

MPa 1

kJ/kg 29.275 MPa 1

KkJ/kg 92927.0kJ/kg 77.252

vapor sat.

C4

34

MPa 1 @ 33

212

2

C4 @ 1

C4 @ 11

=≅

==⎭⎬⎫=

=⎭⎬⎫

==

⋅====

⎭⎬⎫°=

°

°

hh

hhP

hss

P

sshhT

f

g

g


The mass flow rate of the refrigerant is

kg/s 750.2kJ/kg 107.32)(252.77

kJ/s 400)(41

41 =−

=−

=⎯→⎯−=hh

QmhhmQ L

L

&&&&

QH

QL

4°C 1

23

4

1 MPa

·

Win·

·4s

s

T

The power requirement is

kW 61.93=−=−= kJ/kg 252.77)29kg/s)(275. 750.2()( 12in hhmW &&

The COP of the refrigerator is determined from its definition,

6.46===kW 61.93

kW 400COPin

R WQL&

&


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.


Analysis (a) From the refrigerant tables (Tables A-12 and A-13),

( )

kJ/kg 68.234MPa 10173.0

vapor sat.C26

throttlingkJ/kg 83.87

kJ/kg 83.87C26MPa 75.0

kJ/kg 07.284MPa 8.0

/kgm 19841.0KkJ/kg 97207.0

kJ/kg 50.239

C20kPa 100

5

55

34

C 26 @ 33

3

212

2

31

1

1

1

1

==

⎭⎬⎫°−=

=≅

=≅⎭⎬⎫

°==

=⎭⎬⎫

==

=⋅=

=

⎭⎬⎫

°−==

°

hPT

hh

hhTP

hss

P

sh

TP

f

ss

v

T

QH

QL

0.10 MPa

1

2s

3

4

0.75 MPa

s

0.8 MPa

·

·

2

Win·

0.10 M-20°C

-26°C

Pa

Then the mass flow rate of the refrigerant and the power input becomes

( ) ( ) ( )[ ] ( ) kW2.40 78.0/kJ/kg 239.50284.07kg/s 0.0420/

kg/s 0.0420/kgm 0.19841/sm 0.5/60

12in

3

3

1

1

=−=−=

===

Cs hhmW

m

η&&

&&

v

V

(b) The rate of heat removal from the refrigerated space is

( ) ( )( ) kW 6.17=−=−= kJ/kg .8387234.68kg/s 0.042045 hhmQL &&

(c) The pressure drop and the heat gain in the line between the evaporator and the compressor are

and

( ) ( )( ) kW0.203

1.73

kJ/kg 234.68239.50kg/s 0.0420

10073.101

51gain

15

=−=−=

=−=−=∆

hhmQ

PPP

&&



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,

.

3

45

6

QH

Compressor

12

QL

Turbine

Heat Exch.

Heat Exch.

Regenerator

.

( )

( )( ) K 4.4325K 2.273 4.1/4.0k/1k

1

212 ==⎟⎟

⎠

⎞⎜⎜⎝

⎛=

−

PP

TT s

K 5.472

2.2732.2734.43280.0 2

2

12

12

12

12

=⎯→⎯−

−=

−−

=−−

=

TT

TTTT

hhhh ss

Cη

The temperature at state 4 can be determined by solving the following two equations simultaneously:

( ) 4.1/4.0

4

k/1k

4

545 5

1⎟⎠⎞

⎜⎝⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

−

TPP

TT s

ssT TT

Thhhh

54

4

54

54 2.19385.0

−−

=→−−

=η

Using EES, we obtain T4 = 281.3 K.

s

1

3

56

4

2sQH·

QRefrig·

Qrege

5

2TAn energy balance on the regenerator may be written as

or, ( ) ( )

K 3.2463.2812.3082.2734316

61436143

=+−=+−=

−=−⎯→⎯−=−

TTTT

TTTTTTcmTTcm pp && 35°C

0°C

The effectiveness of the regenerator is

0.434=−−

=−−

=−−

=3.2462.3083.2812.308

63

43

63

43regen TT

TThhhh

ε -80°C

(b) The refrigeration load is

kW 21.36=−=−= K)2.19346.3kJ/kg.K)(2 5kg/s)(1.00 4.0()( 56 TTcmQ pL &&

(c) The turbine and compressor powers and the COP of the cycle are

kW 13.80K)2.27372.5kJ/kg.K)(4 5kg/s)(1.00 4.0()( 12inC, =−=−= TTcmW p&&

kW 43.35kJ/kg)2.19381.3kJ/kg.K)(2 5kg/s)(1.00 4.0()( 54outT, =−=−= TTcmW p&&

0.478=−

=−

==43.3513.80

36.21COPoutT,inC,innet, WW

QW

Q LL&&

&

&

&



11-65

(d) The simple gas refrigeration cycle analysis is as follows: ( )

( ) K 6.19451K 2.3081 4.1/4.0k/1k

34 =⎟⎠⎞

⎜⎝⎛=⎟

⎠⎞

⎜⎝⎛=

−

rTT s

1

2QH

3

4s QRefrig·

2

4 1

·T

K 6.2116.1942.308

2.30885.0 4

4

43

43 =⎯→⎯−

−=⎯→⎯

−−

= TT

TTTT

sTη

35°C 0°C

kW 24.74=−=

−=

kJ/kg)6.21173.2kJ/kg.K)(2 5kg/s)(1.00 4.0(

)( 41 TTcmQ pL &&

s

[ ]kW 32.41

kJ/kg)6.211(308.2)2.273(472.5kJ/kg.K) 5kg/s)(1.00 4.0(

)()( 4312innet,

=−−−=

−−−= TTcmTTcmW pp &&&

0.599===32.4174.24COP

innet,WQL&

&



Tutorial #14

ESO 201A

Thermodynamics









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.


[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,

( ) ( )

K/kgm 101.602K/kgm 101.621

C0)3(70/kgm0.029966)(0.036373

kPa00)5(900KkJ/kg1.0309)(0.9660

C3070kPa500900

3434

3?

kPa700

C30C70?

C50

kPa 500kPa 900

kPa700

?

C50

?

⋅×−≅⋅×−

°−−

−≅−

⋅−

⎟⎟⎠

⎞⎜⎜⎝

⎛

°−

−−≅⎟

⎟⎠

⎞⎜⎜⎝

⎛

−

−

⎟⎠⎞

⎜⎝⎛∆∆

−≅⎟⎠⎞

⎜⎝⎛∆∆

⎟⎠⎞

⎜⎝⎛−=⎟

⎠⎞

⎜⎝⎛

−−

=

°°

°=

=°=

PT

PT

PT

ss

TPs

TPs

vv

v

v∂∂

∂∂

since kJ ≡ kPa·m³, and K ≡ °C for temperature differences. Thus the last Maxwell relation is satisfied.



12-12

12-19 It is to be proven that v

⎟⎠⎞

⎜⎝⎛∂∂

−=⎟

⎠⎞

⎜⎝⎛∂∂

TP

kk

TP

s 1

Analysis Using the definition of cv ,

vvv

v ⎟⎠⎞

⎜⎝⎛∂∂

⎟⎠⎞

⎜⎝⎛∂∂

=⎟⎠⎞

⎜⎝⎛∂∂

=TP

PsT

TsTc

Substituting the first Maxwell relation sTP

s⎟⎠⎞

⎜⎝⎛∂∂

−=⎟⎠⎞

⎜⎝⎛∂∂ v

v

,

v

vv

⎟⎠⎞

⎜⎝⎛∂∂

⎟⎠⎞

⎜⎝⎛∂∂

−=TP

TTc

s

Using the definition of cp,

PPP

p TsT

TsTc ⎟

⎠⎞

⎜⎝⎛∂∂

⎟⎠⎞

⎜⎝⎛∂∂

=⎟⎠⎞

⎜⎝⎛∂∂

=v

v

Substituting the second Maxwell relation sP T

Ps⎟⎠⎞

⎜⎝⎛∂∂

=⎟⎠⎞

⎜⎝⎛∂∂v

,

Ps

p TTPTc ⎟

⎠⎞

⎜⎝⎛∂∂

⎟⎠⎞

⎜⎝⎛∂∂

=v

From Eq. 12-46,

TP

pP

TTcc ⎟

⎠⎞

⎜⎝⎛∂∂

⎟⎠⎞

⎜⎝⎛∂∂

−=−v

vv

2

Also,

vcc

ck

k

p

p

−=

−1

Then,

TPs

TP

Ps

PT

TP

PT

TTP

kk

⎟⎠⎞

⎜⎝⎛∂∂

⎟⎠⎞

⎜⎝⎛∂∂

⎟⎠⎞

⎜⎝⎛∂∂

−=

⎟⎠⎞

⎜⎝⎛∂∂

⎟⎠⎞

⎜⎝⎛∂∂

⎟⎠⎞

⎜⎝⎛∂∂

⎟⎠⎞

⎜⎝⎛∂∂

−=−

vv

vv

v

21

Substituting this into the original equation in the problem statement produces

v

vv

⎟⎠⎞

⎜⎝⎛∂∂

⎟⎠⎞

⎜⎝⎛∂∂

⎟⎠⎞

⎜⎝⎛∂∂

⎟⎠⎞

⎜⎝⎛∂∂

−=⎟⎠⎞

⎜⎝⎛∂∂

TP

PT

TP

TP

TPss

But, according to the cyclic relation, the last three terms are equal to −1. Then,

ss T

PTP

⎟⎠⎞

⎜⎝⎛∂∂

=⎟⎠⎞

⎜⎝⎛∂∂



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,

kJ/kg 2159.9=

⎟⎟⎠

⎞⎜⎜⎝

⎛°−

−+=

⎟⎟⎠

⎞⎜⎜⎝

⎛

−−

−=

⎟⎠⎞

⎜⎝⎛∆∆

−≅

⎟⎠⎞

⎜⎝⎛=

C130.58)(136.27kPa 50

/kg)m 0.001073K)(0.60582 273.15(133.52

kPa75)2(325)(

)(

3

kPa @275satkPa 253@satkPa 003@kPa 300@sat

kPa 300 sat,kPa 300@

sat

TTT

TPT

dTdPTh

fg

fg

fgfg

vv

vv

v

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,

kgkJ 82206 /.K 10

kPa.18)169(232.23/kg)m 0.001060K)(0.89133 273.15(120

C115C125)(

)(

3

C115@satC125@satC120@

C120sat,C120@

sat

=

⎟⎟⎠

⎞⎜⎜⎝

⎛ −−+=

⎟⎟⎠

⎞⎜⎜⎝

⎛

°−°

−−=

⎟⎠⎞

⎜⎝⎛∆∆

−≅

⎟⎠⎞

⎜⎝⎛=

°°°

°°

PPT

TPT

dTdPTh

fg

fg

fgfg

vv

vv

v

Also, Kkg/kJ 6131.5 ⋅=+

==K273.15)(120

kJ/kg 2206.8T

hs fg

fg

The tabulated values at 120°C are hfg = 2202.1 kJ/kg and sfg = 5.6013 kJ/kg·K.



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,

( ) ( )

R/lbmft 101.635R/lbmft 101.639

F00)7(900/lbmft.6507)1(1.9777

psia0)35(450RBtu/lbm1.7009)(1.6706

F007900psia035450

3333

3?

psia400

F700F900?

F800

psia 350psia 450

psia400

?

F800

?

⋅×−≅⋅×−

°−−

−≅−

⋅−

⎟⎟⎠

⎞⎜⎜⎝

⎛

°−

−−≅⎟

⎟⎠

⎞⎜⎜⎝

⎛

−

−

⎟⎠⎞

⎜⎝⎛∆∆

−≅⎟⎠⎞

⎜⎝⎛∆∆

⎟⎠⎞

⎜⎝⎛∂∂

−=⎟⎠⎞

⎜⎝⎛∂∂

−−

=

°°

°=

=°=

PT

PT

PT

ss

TPs

TPs

vv

v

v

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 bP

RT+=v . Then,

PR

T P=⎟

⎠⎞

⎜⎝⎛∂∂v

From the fourth Maxwell relation,

PR

−=⎟⎠⎞

⎜⎝⎛∂∂

−=⎟⎠⎞

⎜⎝⎛∂∂

PT TPs v

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 2vv

ab

RTP +−

= . Then,

b

RTP

−=⎟

⎠⎞

⎜⎝⎛∂∂

vv

From the third Maxwell relation,

b

R−

=⎟⎠⎞

⎜⎝⎛∂∂

=⎟⎠⎞

⎜⎝⎛∂∂

vv vTPs

T



12-35

12-48 An expression for the isothermal compressibility of a substance whose equation of state is RTaP =− )(v is to be derived.

Analysis The definition for the isothermal compressibility is

TP⎟⎠⎞

⎜⎝⎛∂∂

−=αv

v1

Solving the equation of state for specific volume,

aP

RT+=v

The specific volume derivative is then

2P

RTP T

−=⎟⎠⎞

⎜⎝⎛∂∂v

Substituting these into the isothermal compressibility equation gives

)(2 aPRTP

RTaPRT

PPRT

+=⎟

⎠⎞

⎜⎝⎛

+=α

12-49 An expression for the isothermal compressibility of a substance whose equation of state is 2/1)( Tba

bRTP

+−

−=

vvv

is to be derived.

Analysis The definition for the isothermal compressibility is

TP⎟⎠⎞

⎜⎝⎛∂∂

−=αv

v1

The derivative is

222/12 )(2

)( bb

Ta

bRTP

T +

++

−−=⎟

⎠⎞

⎜⎝⎛∂∂

vv

v

vv

Substituting,

22/12222/12 )(2

)(

1

)(2

)(

111

bb

Ta

bRT

bb

Ta

bRTP T

+

++

−−

−=

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

+

++

−−

−=⎟⎠⎞

⎜⎝⎛∂∂

−=

v

v

v

v

vv

v

vv

vv

α



12-44

12-65 The equation of state of a gas is given by 2T

bPP

RT−=v . An equation for the Joule-Thomson coefficient inversion

line using this equation is to be derived.

Analysis From Eq. 12-52 of the text,

⎥⎦

⎤⎢⎣

⎡−⎟

⎠⎞

⎜⎝⎛∂∂

µ= v

v

Pp T

Tc 1

When µ = 0 as it does on the inversion line, this equation becomes

vv

=⎟⎠⎞

⎜⎝⎛∂∂

PTT

Using the equation of state to evaluate the partial derivative,

32

TbP

PR

T P+=⎟

⎠⎞

⎜⎝⎛∂∂v

Substituting this result into the previous expression produces

032 222 =⎯→⎯−=+TbP

TbP

PRT

TbP

PRT

The condition along the inversion line is then

0=P

12-66 It is to be demonstrated that the Joule-Thomson coefficient is given by ( )Pp T

TcT

⎟⎠⎞

⎜⎝⎛

∂∂

=/2 vµ .

Analysis From Eq. 12-52 of the text,

⎥⎦

⎤⎢⎣

⎡−⎟

⎠⎞

⎜⎝⎛∂∂

µ= v

v

Pp T

Tc 1

Expanding the partial derivative of v/T produces

21/

TTTTT

PP

vvv−⎟

⎠⎞

⎜⎝⎛∂∂

=⎟⎠⎞

⎜⎝⎛

∂∂

When this is multiplied by T2, the right-hand side becomes the same as the bracketed quantity above. Then,

( )Pp T

TcT

⎟⎠⎞

⎜⎝⎛

∂∂

=/2 vµ



ESO 201A

Thermodynamics

End of file


Department of Mechanical Engineering Indian Institute of Technology Kanpur

Kanpur (UP) 208016 India






full course file (including all tutorial/hw solutions)

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