Chapter 3 PROPERTIES OF PURE SUBSTANCES
Solutions Manual for Thermodynamics An Engineering Approach 7th Edition by CengelFull Download: y-cengel/3-1Solutions Manual forThermodynamics: An Engineering ApproachSeventh EditionYunus A. Cengel, Michael A. BolesMcGraw-Hill, 2011Chapter 3PROPERTIES OF PURE SUBSTANCESPROPRIETARY AND CONFIDENTIALThis Manual is the proprietary property of The McGraw-Hill Companies, Inc. (“McGraw-Hill”) andprotected by copyright and other state and federal laws. By opening and using this Manual the useragrees to the following restrictions, and if the recipient does not agree to these restrictions, the Manualshould be promptly returned unopened to McGraw-Hill: This Manual is being provided only toauthorized professors and instructors for use in preparing for the classes using the affiliatedtextbook. No other use or distribution of this Manual is permitted. This Manual may not be soldand may not be distributed to or used by any student or other third party. No part of thisManual may be reproduced, displayed or distributed in any form or by any means, electronic orotherwise, without the prior written permission of McGraw-Hill.PROPRIETARY MATERIAL. 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for coursepreparation. If you are a student using this Manual, you are using it without permission.Full download all chapters instantly please go to Solutions Manual, Test Bank site: TestBankLive.com
3-2Pure Substances, Phase Change Processes, Property Diagrams3-1C A liquid that is about to vaporize is saturated liquid; otherwise it is compressed liquid.3-2C A vapor that is about to condense is saturated vapor; otherwise it is superheated vapor.3-3C No.3-4C The temperature will also increase since the boiling or saturation temperature of a pure substance depends onpressure.3-5C Because one cannot be varied while holding the other constant. In other words, when one changes, so does the otherone.3-6C At critical point the saturated liquid and the saturated vapor states are identical. At triple point the three phases of apure substance coexist in equilibrium.3-7C Yes.3-8C Case (c) when the pan is covered with a heavy lid. Because the heavier the lid, the greater the pressure in the pan, andthus the greater the cooking temperature.3-9C At supercritical pressures, there is no distinct phase change process. The liquid uniformly and gradually expands intoa vapor. At subcritical pressures, there is always a distinct surface between the phases.PROPRIETARY MATERIAL. 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for coursepreparation. If you are a student using this Manual, you are using it without permission.
3-3Property Tables3-10C A perfectly fitting pot and its lid often stick after cooking as a result of the vacuum created inside as the temperatureand thus the corresponding saturation pressure inside the pan drops. An easy way of removing the lid is to reheat the food.When the temperature rises to boiling level, the pressure rises to atmospheric value and thus the lid will come right off.3-11C The molar mass of gasoline (C8H18) is 114 kg/kmol, which is much larger than the molar mass of air that is 29kg/kmol. Therefore, the gasoline vapor will settle down instead of rising even if it is at a much higher temperature than thesurrounding air. As a result, the warm mixture of air and gasoline on top of an open gasoline will most likely settle downinstead of rising in a cooler environment3-12C Yes. Otherwise we can create energy by alternately vaporizing and condensing a substance.3-13C No. Because in the thermodynamic analysis we deal with the changes in properties; and the changes areindependent of the selected reference state.3-14C The term hfg represents the amount of energy needed to vaporize a unit mass of saturated liquid at a specifiedtemperature or pressure. It can be determined from hfg hg - hf .3-15C Yes. It decreases with increasing pressure and becomes zero at the critical pressure.3-16C Yes; the higher the temperature the lower the hfg value.3-17C Quality is the fraction of vapor in a saturated liquid-vapor mixture. It has no meaning in the superheated vaporregion.3-18C Completely vaporizing 1 kg of saturated liquid at 1 atm pressure since the higher the pressure, the lower the hfg .3-19C No. Quality is a mass ratio, and it is not identical to the volume ratio.PROPRIETARY MATERIAL. 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for coursepreparation. If you are a student using this Manual, you are using it without permission.
3-43-20C The compressed liquid can be approximated as a saturated liquid at the given temperature. Thus v T , P v f @ T .3-21C Ice can be made by evacuating the air in a water tank. During evacuation, vapor is also thrown out, and thus thevapor pressure in the tank drops, causing a difference between the vapor pressures at the water surface and in the tank. Thispressure difference is the driving force of vaporization, and forces the liquid to evaporate. But the liquid must absorb theheat of vaporization before it can vaporize, and it absorbs it from the liquid and the air in the neighborhood, causing thetemperature in the tank to drop. The process continues until water starts freezing. The process can be made more efficientby insulating the tank well so that the entire heat of vaporization comes essentially from the water.3-22Complete the following table for H2 O:T, CP, kPav, m3 / kgPhase description5012.357.72Saturated mixture143.64000.4624Saturated vapor2505000.4744Superheated vapor1103500.001051Compressed liquidPROPRIETARY MATERIAL. 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for coursepreparation. If you are a student using this Manual, you are using it without permission.
3-53-23Problem 3-22 is reconsidered. The missing properties of water are to be determined using EES, and thesolution is to be repeated for refrigerant-134a, refrigerant-22, and ammonia.Analysis The problem is solved using EES, and the solution is given below."Given"T[1] 50 [C]v[1] 7.72 [m 3/kg]P[2] 400 [kPa]x[2] 1T[3] 250 [C]P[3] 500 [kPa]T[4] 110 [C]P[4] 350 [kPa]"Analysis"Fluid 'steam iapws' "Change the Fluid to R134a, R22 and Ammonia and solve"P[1] pressure(Fluid , T T[1], v v[1])x[1] quality(Fluid , T T[1], v v[1])T[2] temperature(Fluid , P P[2], x x[2])v[2] volume(Fluid , P P[2], x x[2])v[3] volume(Fluid , P P[3], T T[3])x[3] quality(Fluid , P P[3], T T[3])v[4] volume(Fluid , P P[4], T T[4])x[4] quality(Fluid , P P[4], T T[4])"x 100 for superheated vapor and x -100 for compressed liquid"SOLUTION for waterT [C]P [kPa]xv 00500.001000.4744110.00350.00-1000.001051SOLUTION for R-134aT [C]P [kPa]xv 00--110.00350.001000.08666PROPRIETARY MATERIAL. 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for coursepreparation. If you are a student using this Manual, you are using it without permission.
3-6SOLUTION for R-22T [C]P [kPa]Xv 0.001000.09959110.00350.001000.103SOLUTION for AmmoniaT [C]P [kPa]Xv 0.001000.5076110.00350.001000.5269Steam700600500T [C]4003008600 kPa2600 kPa200500 kPa10045 kPa00.01.02.03.04.05.06.07.08.09.010.0s [kJ/kg-K]Steam700600500T [C]4003008600 kPa2600 kPa200500 kPa100010-445 kPa10-310-210-1100101102103v [m 3/kg]PROPRIETARY MATERIAL. 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for coursepreparation. If you are a student using this Manual, you are using it without permission.
3-7Steam105104250 CP [kPa]103170 C102110 C75 C10110010-310-210-11001011023v [m /kg]Steam105104250 CP [kPa]103170 C102110 C75 C101100050010001500200025003000h [kJ/kg]Steam40008600 kPah [kJ/kg]35002600 kPa3000500 kPa250045 0.0s [kJ/kg-K]PROPRIETARY MATERIAL. 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for coursepreparation. If you are a student using this Manual, you are using it without permission.
3-83-24E Complete the following table for H2 O:T, FP, psiau, Btu / lbmPhase description30067.03782Saturated mixture267.2240236.02Saturated liquid5001201174.4Superheated vapor400400373.84Compressed liquidProblem 3-24E is reconsidered. The missing properties of water are to be determined using EES, and the3-25Esolution is to be repeated for refrigerant-134a, refrigerant-22, and ammonia.Analysis The problem is solved using EES, and the solution is given below."Given"T[1] 300 [F]u[1] 782 [Btu/lbm]P[2] 40 [psia]x[2] 0T[3] 500 [F]P[3] 120 [psia]T[4] 400 [F]P[4] 420 [psia]"Analysis"Fluid 'steam iapws'P[1] pressure(Fluid , T T[1], u u[1])x[1] quality(Fluid , T T[1], u u[1])T[2] temperature(Fluid , P P[2], x x[2])u[2] intenergy(Fluid , P P[2], x x[2])u[3] intenergy(Fluid , P P[3], T T[3])x[3] quality(Fluid , P P[3], T T[3])u[4] intenergy(Fluid , P P[4], T T[4])x[4] quality(Fluid , P P[4], T T[4])"x 100 for superheated vapor and x -100 for compressed liquid"Solution for steamT, ºFP, psiaxu, 00400-100373.8PROPRIETARY MATERIAL. 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for coursepreparation. If you are a student using this Manual, you are using it without permission.
3-93-26 Complete the following table for H2 O:T, CP, kPah, kJ / kgxPhase description120.212002045.80.7Saturated mixture140361.5318000.565Saturated mixture177.66950752.740.0Saturated liquid80500335.37---Compressed liquid350.08003162.2---Superheated vapor3-27 Complete the following table for Refrigerant-134a:T, CP, kPav, m3 / kgPhase description-123200.000750Compressed liquid30770.640.0065Saturated mixture18.735500.03741Saturated vapor606000.04139Superheated vapor3-28 Complete the following table for water:*P, kPaT, oCv, m3/kgh, kJ/kg200120.20.88582706.3Condition description and quality, ifapplicablex 1, Saturated vapor270.31301959.3x 0.650, Two-phase mixture201.84001.53583277.0Superheated vapor800300.001004*125.74*Compressed liquid450147.90--Insufficient informationApproximated as saturated liquid at the given temperature of 30oCPROPRIETARY MATERIAL. 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for coursepreparation. If you are a student using this Manual, you are using it without permission.
3-103-29E Complete the following table for Refrigerant-134a:T, FP, psiah, Btu / lbmxPhase description65.8980780.566Saturated mixture1529.75969.920.6Saturated mixture107015.35---Compressed liquid160180129.46---Superheated vapor110161.16117.231.0Saturated vapor3-30 A piston-cylinder device contains R-134a at a specified state. Heat is transferred to R-134a. The final pressure, thevolume change of the cylinder, and the enthalpy change are to be determined.Analysis (a) The final pressure is equal to the initial pressure, which is determined fromP2 P1 Patm mpg2πD /4 88 kPa (12 kg)(9.81 m/s 2 ) 1 kN2 π (0.25 m) /4 1000 kg.m/s 2 90.4 kPa (b) The specific volume and enthalpy of R-134a at the initial state of 90.4 kPa and -10 C and at the final state of 90.4 kPaand 15 C are (from EES)v1 0.2302 m3/kg3v 2 0.2544 m /kgh1 247.76 kJ/kgh2 268.16 kJ/kgThe initial and the final volumes and the volume change areV1 mv 1 (0.85 kg)(0.2302 m 3 /kg) 0.1957 m 3V 2 mv 2 (0.85 kg)(0.2544 m 3 /kg) 0.2162 m 3R-134a0.85 kg-10 CQ V 0.2162 0.1957 0.0205 m 3(c) The total enthalpy change is determined from H m(h2 h1 ) (0.85 kg)(268.16 247.76) kJ/kg 17.4 kJ/kg3-31E The temperature of R-134a at a specified state is to be determined.Analysis Since the specified specific volume is higher than vg for 120 psia, this is a superheated vapor state. From R-134atables,P 120 psia T 140 F (Table A - 13E)v 0.4619 ft /lbm 3PROPRIETARY MATERIAL. 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for coursepreparation. If you are a student using this Manual, you are using it without permission.
3-113-32 A rigid container that is filled with water is cooled. The initial temperature and the final pressure are to be determined.Analysis This is a constant volume process. The specific volume isH2O2 MPa1 kg150 L0.150 m 3 0.150 m 3 /kgv1 v 2 m1 kgVQThe initial state is superheated vapor. The temperature is determined to beP1 2 MPa T1 395 Cv 1 0.150 m 3 /kg (Table A - 6)PThis is a constant volume cooling process (v V /m constant). The finalstate is saturated mixture and thus the pressure is the saturation pressure atthe final temperature:12T2 40 C P Psat @ 40 C 7.385 kPa (Table A - 4)v 2 v 1 0.150 m 3 /kg 2v3-33 A rigid container that is filled with R-134a is heated. The final temperature and initial pressure are to be determined.Analysis This is a constant volume process. The specific volume isv1 v 2 Vm R-134a-40 C10 kg1.348 m31.348 m 3 0.1348 m 3 /kg10 kgThe initial state is determined to be a mixture, and thus the pressure is thesaturation pressure at the given temperatureP1 Psat @ -40 C 51.25 kPa (Table A - 11)The final state is superheated vapor and the temperature is determined byinterpolation to beP2 200 kPa T2 66.3 C (Table A - 13)v 2 0.1348 m 3 /kg P21vPROPRIETARY MATERIAL. 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for coursepreparation. If you are a student using this Manual, you are using it without permission.
3-123-34E Left chamber of a partitioned system contains water at a specified state while the right chamber is evacuated. Thepartition is now ruptured and heat is transferred to the water. The pressure and the total internal energy at the final state areto be determined.Analysis The final specific volume is3 ft 3v2 1.5 ft 3 /lbmm 2 lbmV2At this specific volume and the final temperature, the state is a saturatedmixture, and the pressure is the saturation pressureWater500 psia2 lbm1.5 ft3Evacuated1.5 ft3P2 Psat @ 300 F 67.03 psia (Table A - 4E)The quality and internal energy at the final state arex2 v 2 v f(1.5 0.01745) ft 3 /lbm 0.2299(6.4663 0.01745) ft 3 /lbmu 2 u f x 2 u fg 269.51 (0.2299)(830.25) 460.38 Btu/lbmv fg The total internal energy is thenU 2 mu 2 (2 lbm)(460.38 Btu/lbm) 920.8 Btu3-35 The enthalpy of R-134a at a specified state is to be determined.Analysis The specific volume isv V9 m3 0.03 m 3 /kgm 300 kgInspection of Table A-11 indicates that this is a mixture of liquid and vapor. Using the properties at 10 C line, the qualityand the enthalpy are determined to bex v v fv fg (0.03 0.0007930) m 3 /kg(0.049403 0.0007930) m 3 /kg 0.6008h h f xh fg 65.43 (0.6008)(190.73) 180.02 kJ/kg3-36 The specific volume of R-134a at a specified state is to be determined.Analysis Since the given temperature is higher than the saturation temperature for 200 kPa, this is a superheated vapor state.The specific volume is thenP 200 kPa 3 v 0.11646 m /kg (Table A - 13) T 25 CPROPRIETARY MATERIAL. 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for coursepreparation. If you are a student using this Manual, you are using it without permission.
3-133-37E A spring-loaded piston-cylinder device is filled with R-134a. The water now undergoes a process until its volumeincreases by 40%. The final temperature and the enthalpy are to be determined.Analysis From Table A-11E, the initial specific volume isv 1 v f x1v fg 0.01143 (0.80)(4.4300 0.01143) 3.5463 ft 3 /lbmand the initial volume will beV1 mv 1 (0.2 lbm)(3.5463 ft 3 /lbm) 0.7093 ft 3PWith a 40% increase in the volume, the final volume will be3V 2 1.4V1 1.4(0.7093 ft ) 0.9930 ft231The distance that the piston moves between the initial and final conditions is x v4(0.9930 0.7093)ft 3 V V 0.3612 ftA p πD 2 / 4π (1 ft) 2As a result of the compression of the spring, the pressure difference between the initial and final states is P 4(37 lbf/in)(0.3612 12 in) F k xk x 1.42 lbf/in 2 1.42 psiaApA p πD 2 / 4π (12 in) 2The initial pressure isP1 Psat @ -30 F 9.87 psia (Table A - 11E)The final pressure is thenP2 P1 P 9.87 1.42 11.29 psiaand the final specific volume isv2 V2m 0.9930 ft 3 4.965 ft 3 /lbm0.2 lbmAt this final state, the temperature and enthalpy areP2 11.29 psia T1 81.5 F v 2 4.965 ft /lbm h1 119.9 Btu/lbm3(from EES)Note that it is very difficult to get the temperature and enthalpy from Table A-13E accurately.PROPRIETARY MATERIAL. 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for coursepreparation. If you are a student using this Manual, you are using it without permission.
3-143-38E A piston-cylinder device that is filled with water is cooled. The final pressure and volume of the water are to bedetermined.Analysis The initial specific volume isv1 V1m 2.4264 ft 3 2.4264 ft 3 /lbm1 lbmH2O600 F1 lbm2.4264 ft3This is a constant-pressure process. The initial state is determined to be superheatedvapor and thus the pressure is determined to beT1 600 F P P2 250 psia (Table A - 6E)3v 1 2.4264 ft /lbm 1PThe saturation temperature at 250 psia is 400.1 F. Since the final temperatureis less than this temperature, the final state is compressed liquid. Using theincompressible liquid approximation,12v 2 v f @ 200 F 0.01663 ft 3 /lbm (Table A - 4E)vThe final volume is thenV 2 mv 2 (1 lbm)(0.01663 ft 3 /lbm) 0.01663 ft 33-39 A piston-cylinder device that is filled with R-134a is heated. The final volume is to be determined.Analysis This is a constant pressure process. The initial specific volume isv1 V 1.595 m 3 0.1595 m 3 /kg10 kgmR-134a-26.4 C10 kg1.595 m3The initial state is determined to be a mixture, and thus the pressure is thesaturation pressure at the given temperatureP1 Psat @ -26.4 C 100 kPa (Table A - 12)The final state is superheated vapor and the specific volume isP2 100 kPa 3 v 2 0.30138 m /kg (Table A - 13)T2 100 C P12The final volume is thenV 2 mv 2 (10 kg)(0.30138 m 3 /kg) 3.0138 m 3vPROPRIETARY MATERIAL. 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for coursepreparation. If you are a student using this Manual, you are using it without permission.
3-153-40E The total internal energy and enthalpy of water in a container are to be determined.Analysis The specific volume isv Vm 2 ft 3 2 ft 3 /lbm1 lbmWater100 psia2 ft3At this specific volume and the given pressure, the state is a saturated mixture.The quality, internal energy, and enthalpy at this state are (Table A-5E)x v v f (2 0.01774) ft 3 /lbm 0.4490(4.4327 0.01774) ft 3 /lbmu u f xu fg 298.19 (0.4490)(807.29) 660.7 Btu/lbmh h f xh fg 298.51 (0.4490)(888.99) 697.7 Btu/lbmv fgThe total internal energy and enthalpy are thenU mu (1 lbm)(660.7 Btu/lbm) 660.7 BtuH mh (1 lbm)(697.7 Btu/lbm) 697.7 Btu3-41 The volume of a container that contains water at a specified state is to be determined.Analysis The specific volume is determined from steam tables by interpolation to beP 100 kPa 3 v 2.4062 m /kg (Table A - 6)T 250 C The volume of the container is thenV mv (3 kg)(2.4062 m 3 /kg) 7.22 m 3Water3 kg100 kPa250 CPROPRIETARY MATERIAL. 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
Thermodynamics: An Engineering Approach Seventh Edition Yunus A. Cengel, Michael A. Boles McGraw-Hill, 2011 Chapter 3 PROPERTIES OF PURE SUBSTANCES PROPRIETARY AND CONFIDENTIAL This Manual is the proprietary property of The McGraw-Hill Companies, Inc. (“McGraw-Hill”) and protected by copyright and other state and federal laws.
Part One: Heir of Ash Chapter 1 Chapter 2 Chapter 3 Chapter 4 Chapter 5 Chapter 6 Chapter 7 Chapter 8 Chapter 9 Chapter 10 Chapter 11 Chapter 12 Chapter 13 Chapter 14 Chapter 15 Chapter 16 Chapter 17 Chapter 18 Chapter 19 Chapter 20 Chapter 21 Chapter 22 Chapter 23 Chapter 24 Chapter 25 Chapter 26 Chapter 27 Chapter 28 Chapter 29 Chapter 30 .
TO KILL A MOCKINGBIRD. Contents Dedication Epigraph Part One Chapter 1 Chapter 2 Chapter 3 Chapter 4 Chapter 5 Chapter 6 Chapter 7 Chapter 8 Chapter 9 Chapter 10 Chapter 11 Part Two Chapter 12 Chapter 13 Chapter 14 Chapter 15 Chapter 16 Chapter 17 Chapter 18. Chapter 19 Chapter 20 Chapter 21 Chapter 22 Chapter 23 Chapter 24 Chapter 25 Chapter 26
DEDICATION PART ONE Chapter 1 Chapter 2 Chapter 3 Chapter 4 Chapter 5 Chapter 6 Chapter 7 Chapter 8 Chapter 9 Chapter 10 Chapter 11 PART TWO Chapter 12 Chapter 13 Chapter 14 Chapter 15 Chapter 16 Chapter 17 Chapter 18 Chapter 19 Chapter 20 Chapter 21 Chapter 22 Chapter 23 .
About the husband’s secret. Dedication Epigraph Pandora Monday Chapter One Chapter Two Chapter Three Chapter Four Chapter Five Tuesday Chapter Six Chapter Seven. Chapter Eight Chapter Nine Chapter Ten Chapter Eleven Chapter Twelve Chapter Thirteen Chapter Fourteen Chapter Fifteen Chapter Sixteen Chapter Seventeen Chapter Eighteen
18.4 35 18.5 35 I Solutions to Applying the Concepts Questions II Answers to End-of-chapter Conceptual Questions Chapter 1 37 Chapter 2 38 Chapter 3 39 Chapter 4 40 Chapter 5 43 Chapter 6 45 Chapter 7 46 Chapter 8 47 Chapter 9 50 Chapter 10 52 Chapter 11 55 Chapter 12 56 Chapter 13 57 Chapter 14 61 Chapter 15 62 Chapter 16 63 Chapter 17 65 .
HUNTER. Special thanks to Kate Cary. Contents Cover Title Page Prologue Chapter 1 Chapter 2 Chapter 3 Chapter 4 Chapter 5 Chapter 6 Chapter 7 Chapter 8 Chapter 9 Chapter 10 Chapter 11 Chapter 12 Chapter 13 Chapter 14 Chapter 15 Chapter 16 Chapter 17 Chapter
Chapter 3 Chapter 4 Chapter 5 Chapter 6 Chapter 7 Chapter 8 Chapter 9 Chapter 10 Chapter 11 Chapter 12 Chapter 13 Chapter 14 Chapter 15 Chapter 16 Chapter 17 Chapter 18 Chapter 19 Chapter 20 . Within was a room as familiar to her as her home back in Oparium. A large desk was situated i
Methyl Ethyl Ketone pure (liquid) 3 Methyl Isobutyl Ketone pure (liquid) 3 Methyl Methacrylate pure (liquid) 2 Methyl Propyl Ketone pure (liquid) 3 Methyl Salicylate pure (liquid) 3 Methylene Chloride pure (liquid) 3 Nitric Acid 1-10% 1 10-20% 2 20% 3
