ASHRAE PRINCIPLES OF HEATING VENTILATING-2017 Principles of Heating Ventilating and Air Conditioning (Edition 8 90569).pdf

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1、Principles of HeatingVentilatingand Air Conditioning8th EditionRonald H. HowellPrinciples of HVAC, 8th Edition Solutions ManualSOLUTIONS MANUAL2007547819399Product Code: 90569 8/17ISBN: 978-1-939200-75-4 (paperback)ISBN: 978-1-393200-76-1 (PDF)1791 Tullie CircleAtlanta, GA 30329-2305404-636-8400 (wo

2、rldwide)www.ashrae.orgPHVAC SOLMAN_cover.indd 1 7/21/2017 12:23:50 PMPRINCIPLESOF HEATINGVENTILATINGANDAIR CONDITIONINGSOLUTIONS MANUALABOUT THE AUTHORSRonald H. Howell, PhD, PE, Fellow/Life Member ASHRAE, retired as professor and chair of mechanical engineeringat the University of South Florida and

3、 is also professor emeritus of the University of Missouri-Rolla. For 45 years hetaught courses in refrigeration, heating and air conditioning, thermal analysis, and related areas. He has been the prin-cipal or co-principal investigator of 12 ASHRAE-funded research projects. His industrial and consul

4、ting engineeringexperience ranges from ventilation and condensation problems to the development and implementation of a completeair curtain test program.Thefollowingauthorscontributedsignificantlytothetextbook Principles of Heating, Ventilating, and Air Conditioningand this Solutions Manual. They re

5、cently passed away and were not part of the 2017 revisions.William J. Coad, PE, Fellow ASHRAE, was ASHRAE president in 2001-2002. He was employed with McClure Engi-neering Associates, St. Louis, Mo., for 45 years. He was also president of Coad Engineering Enterprises. He served asa consultant to the

6、 Missouri state government and was a lecturer in mechanical engineering for 12 years and an affiliateprofessor in the graduate program for 17 years at Washington University, St. Louis. He was the author of Energy Engi-neering and Management for Building Systems (Van Nostrand Reinhold). Mr. Coad pass

7、ed away in August 2014.HarryJ.Sauer,Jr.,PhD,PE,FellowASHRAE,wasaprofessorofmechanicalandaerospaceengineeringattheUniver-sity of Missouri-Rolla. He taught courses in air conditioning, refrigeration, environmental quality analysis and control,and related areas. His research ranged from experimental bo

8、iling/condensing heat transfer and energy recovery equip-mentforHVACsystemstocomputersimulationsofbuildingenergyuseandactualmonitoringofresidentialenergyuse.He served as an advisor to the Missouri state government and has conducted energy auditor training programs for theUS Department of Energy. Dr.

9、 Sauer passed away in June 2008.PRINCIPLESOF HEATINGVENTILATINGANDAIR CONDITIONING8th EditionSOLUTIONS MANUALRonald H. HowellAtlantaISBN 978-1-939200-75-4 (paperback)ISBN 978-1-939200-76-1 (PDF) 1990, 1994, 1998, 2001, 2005, 2009, 2013, 2017 ASHRAE1791 Tullie Circle, N.E.Atlanta, GA 30329www.ashrae.

10、orgAll rights reserved.Printed in the United States of AmericaASHRAE is a registered trademark in the U.S. Patent and Trademark Office, owned by the American Society of Heating,Refrigerating and Air-Conditioning Engineers, Inc.ASHRAE has compiled this publication with care, but ASHRAE has not invest

11、igated, and ASHRAE expressly disclaimsany duty to investigate, any product, service, process, procedure, design, or the like that may be described herein. Theappearance of any technical data or editorial material in this publication does not constitute endorsement, warranty, orguaranty by ASHRAE of

12、any product, service, process, procedure, design, or the like. ASHRAE does not warrant that theinformation in the publication is free of errors, and ASHRAE does not necessarily agree with any statement or opinion inthis publication. The entire risk of the use of any information in this publication i

13、s assumed by the user.No part of this publication may be reproduced without permission in writing from ASHRAE, except by a reviewer whomay quote brief passages or reproduce illustrations in a review with appropriate credit, nor may any part of this publicationbe reproduced, stored in a retrieval sys

14、tem, or transmitted in any way or by any meanselectronic, photocopying, record-ing, or otherwithout permission in writing from ASHRAE. Requests for permission should be submitted atwww.ashrae.org/permissions.ASHRAE STAFF SPECIAL PUBLICATIONSMark S. Owen, Editor/Group Manager of Handbook and Special

15、PublicationsCindy Sheffield Michaels, Managing EditorJames Madison Walker, Managing Editor of StandardsLauren Ramsdell, Assistant EditorMary Bolton, Editorial AssistantMichshell Phillips, Editorial CoordinatorPUBLISHING SERVICESDavid Soltis, Group Manager of Publishing ServicesJayne Jackson, Publica

16、tion Traffic AdministratorPUBLISHERW. Stephen ComstockUpdates anderrata forthispublication willbeposted ontheASHRAE website at www.ashrae.org/publicationupdates.Notes to InstructorsThis manual contains solutions to most of the problems in the textbook Principles of Heating, Ventilating, and AirCondi

17、tioning, 8th edition, which is based on the 2017 ASHRAE HandbookFundamentals. Some of these problemsrequire the use of tables, figures, or equations in the 2017 Handbook that may not be found in Principles of Heating,Ventilating, and Air Conditioning.Thesolutionsinthismanualaregenerallypresentedinab

18、breviatedform,withsomeintermediatecomputationsomitted.Answers and solutions are included for the majority of the problems. The remaining problems are either those requiringdiscussion or those whose solutions depend on arbitrary assumptions or data selected by the instructor.Also, in some solutions,

19、the weather data from eight years earlier was used. This allows the user to assess the effect ofclimate change and/or annual weather variations.R.H. HowellCONTENTSSolutions toChapter 1 .3Chapter 2 .7Chapter 3 31Chapter 4 55Chapter 5 61Chapter 6 77Chapter 7 91Chapter 8 99Chapter 9 .111Chapter 10 133C

20、hapter 11 139Chapter 12 153Chapter 13 165Chapter 14 171Chapter 15 177Chapter 16 181Chapter 17 185Chapter 18 195Chapter 19 203Chapter 20 219Solutions toChapter 1BACKGROUNDChapter 1Background31.1 Estimate whether ice will form on a clear night whenambient air temperature is 45F (7.2C), if the water is

21、placed in a shallow pan in a sheltered location where theconvective heat transfer coefficient is 0.5 Btu/h ft2 F(2.8 W/m2 K).1.4 Estimate the size of cooling and heating equipmentthat is needed for a new bank building in middle Americathat is 14022012 ft high (42.7673.7 m high). Beconservative.1.5 E

22、stimate the size of heating and cooling equipmentthat will be needed for a residence in middle America thatis 28 78 8 ft high (8.5 23.8 2.4 m high).1.6 Estimate the initial cost of the complete HVACsystem (heating, cooling, and air moving) for an officebuilding, 4015010 ft high (12.245.73.1 m high).

23、Heat in by convection Heat out by radiation to space=(Assume space at 0R, Assume water is black body, 1=hA TairTwaterATwater4=0.5505 Tw0.1714Tw100-4= bytrialanderrorTw410R 50F will freeze=Floor area 14022030,800 ft2=Volume 14022012 370,000 ft3From Table 1.1: 250 ft2/ton and 3.0 Btu/hft3Cooling:30,80

24、0 ft2250 ft2/ton- 123 tons=Heating: 370,000 ft33.0 Btu/hft31,110,000 Btu/h=From Table 1.1: 700 ft2/ ton and 3.0 Btu/hft3Cooling:2878700- 3.12 tons or 3.12 12,000 37,400 Btu/h=Heating: 28788ft33.0 Btu/hft352,400 Btu/h=Cooling unit:40 150350 ft3ton- 1 7 t o n s=Heating unit: 40150103Btu/hft3180,000 Bt

25、u/h=Air movement: 17 tons 400 cfm/ton 6900 cfm or 1.2 cfm/ft240 150 7200 cfm=Costs: Cooling system ($1675/ton) 17 tons $28,500=Heating system ($2.92/cfm) 6900 7200cfm $20,000 21,000=Fans/ducting $7.846900 7200 cfm $54,100 $56,500=Total = $102,600 $106,0004Principles of HVAC, 8th EditionSolutions Man

26、ual1.7 Estimate the annual operating cost for the building inProblem 1.6 if it is all-electric.1.8 Open-ended design problem.From Table 1.2: 30.5 kWh/ft2yrEnergy 40 15030.5183,000 kWh=Cost $0.09 183,000$16,470Solutions toChapter 2THERMODYNAMICSAND PSYCHROMETRICSChapter 2Thermodynamics and Psychromet

27、rics72.6 Two pounds of air contained in a cylinder expandwithoutfrictionagainstapiston.Thepressureonthebackside of the piston is constant at 200 psia. The air initiallyoccupies a volume of 0.50 ft3. What is the work done bythe air in foot-poundsfif the expansion continues until thetemperature of the

28、 air reaches 100F?2.7 Determine the specific volume, enthalpy, andentropy of 1 kg of R-134a at a saturation temperature of5C and a quality of 14%.2.8 Saturated R-134a vapor at 42C is superheated atconstant pressure to a final temperature of 72C. What isthe pressure? What are the changes in specific

29、volume,enthalpy, entropy, and internal energy?PV mRT= V2mRT2P2-253.3560200144- 2.073 ft3= =WPvdPvd12Pv2v12001442.073 0.545,300 ftlbf= = = =From R-134a tables: 0.14 0.0830.8611310-+ 0.0123 m3/kg=h 0.14 395.80.86 192.9+ 221.3 kJ/kgs 0.14 1.73060.86 0.976+ 1.082 kJ/kgKFrom R-134a tables:1. 42C 315 K: P

30、 1.0721 MPa; vg0.0189 m3/kg hg420.44 kJ/kg=;=Sg1.7108 kJ/kgK= uhP 420.44 1.0721 1060.0189/1000=u 400.2 kJ/kg=2. 72C 345 K: P 1.0037 MPa;12- 42.0; h2453, s21.81=u2608.3=21142- 0.01890.0049m3/kg;= h2h1 453 420.44 32.6 kJ/kg=s2s1 1.81 1.71 0.10 kJ/kgK= u2u1 427.5 400.2 f 27.3 kJ/kg=8Principles of HVAC,

31、 8th EditionSolutions Manual2.9 A tank having a volume of 200 ft3contains saturatedvapor (steam) at a pressure of 20 psia. Attached to thistank is a line in which vapor at 100 psia, 400F flows.Steam from this line enters the vessel until the pressure is100 psia. If there is no heat transfer from the

32、 tank and theheat capacity of the tank is neglected, calculate the massof steam that enters the tank.2.10 Determine the heat required to vaporize 50 kg ofwater at a saturation temperature of 100C.2.11 The temperature of 150 kg of water is raised from15C to 85C by the addition of heat. How much heat

33、issupplied?2.12 Three cubic meters per second of water are cooledfrom 30C to 2C. Compute the rate of heat transfer inkilojoules per second (kilowatts).Qmhfg50 kg2256.28kJ/kg112 800 kJ= =Qmcpt 150 kg4.180kJkgK-85 15K 43 890 kJ= =QmcptV-cpt3m3/s0.001004 m3/kg-4.18 kJ/kgK30 2K=Q 350 000 kW=or Qmh30.001

34、004-125.72 8.39351 000 kW= =Steam line: P 100 psi=;t 400F=Tank: P120 psi,= sat vaporP2100 psi=120.09 ft3/lbm= m120020.09- 9.955 lbm=m2V22-=andmihim2u2m1u1= by trial and errorTry T2550F= u21195= m22005.9- 33.9 lbm=33.9 9.9551228?=33.911959.955108229,400 29,700 mi33.9 9.955 23.95 lbm=Chapter 2Thermody

35、namics and Psychrometrics92.13 Consider 10 lbmof air that is initially at 14.7 psia,100F. Heat is transferred to the air until the temperaturereaches 500F. Determine the change of internal energy,the change in enthalpy, the heat transfer, and the workdone fora. a constant-volume processb. a constant

36、-pressure process2.14 Thedischargeofapumpis10ftabovetheinlet.Waterentersatapressureof20psiaandleavesatapressureof200psia. The specific volume of the water is 0.016 ft3/lb. Ifthere is no heat transfer and no change in kinetic or internalenergy, what is the work per pound?2.15 The discharge of a pump

37、is 3 m above the inlet.Water enters at a pressure of 138 kPa and leaves at a pres-sure of 1380 kPa. The specific volume of the water is0.001 m3/kg. If there is no heat transfer and no change inkinetic or internal energy, what is the work per unit mass?a. Closed system, constant volume process, perfe

38、ct gasU mcvt 100.171500 100684 Btu= =H mcpt 100.240500 100960 Btu= =QW U= (closed system); w 0= since constant volumeQU 684 Btu=b. Closed system, constant pressure process, perfect gasU 684 Btu= H 960 Btu=;pc: Qmcpt H 960 Btu= =Q W U (closed system)= ; W Q u 960 684 276 Btu= =From 1st Law:wP11P22J-g

39、gc-z1z2J-+=0.01620 200144778-32.232.2-10778-+=0.533 0.013 0.546 Btu/lbm=mu1P11122- gz1+u2P22122- gz2+ QW+0=1380.00113800.00139.806 W0=W 30.66 J= Note 1 J (joule) Nm=10Principles of HVAC, 8th EditionSolutions Manual2.16 Air is compressed in a reversible, isothermal,steady-flow process from 15 psia, 1

40、00F to 100 psia.Calculate the work of compression per pound, the changeof entropy, and the heat transfer per pound of aircompressed.2.17 Liquid nitrogen at a temperature of 240F exists inacontainer,andboththeliquidandvaporphasesarepres-ent. The volume of the container is 3 ft3and the mass ofnitrogen

41、inthecontainerhasbeendeterminedas44.5lbm.What is the mass of liquid and the mass of vapor presentin the container?P115 psi=,t1100F= P2100 psi=,t2100F=1RT1P1-53.337315144- 9.21 ft3/lbm= =21P1P2-9.2115100-1.38 ft3/lbm= =P constant= pP11-=wPdP11d-12P1121-ln= =w 151449.211.389.21-ln 37,800 ftlbf/lbmS Cp

42、T2T1-ln RP2P1-ln 0.24373373-ln 53.310015-ln=0 101.2 101.2ftlbf/lbmR 0.13Btu/lbmR=V 3ft3=m 44.5 lbm=t 240F=220R=f0.02613 ft3/lbm=g0.0750 ft3/lbm= V/m 3 44.5 0.0674 ft3/lbm= =fgf+=xfgf-0.0674 0.026130.0750 0.02613- 0.8445= =mvmx 44.50.844537.58 lbmvapor= =mLm 1 x44.51 0.84456.92 lbmliquid= =Chapter 2T

43、hermodynamics and Psychrometrics112.18 A fan in an air-conditioning system is drawing1.25 hp at 1760 rpm. The capacity through the fan is0.85 m3/s of 24C air and the inlet and outlet ducts are0.31 m in diameter. What is the temperature rise of the airdue to this fan?2.19 Air is contained in a cylind

44、er. Initially, the cylindercontains 1.5 m3of air at 150 kPa, 20C. The air is thencompressedreversibly accordingto therelationship pvn=constant until the final pressure is 600 kPa, at which pointthe temperature is 120C. For this process determine:a. the polytropic exponent nb. the final volume of the

45、 airc. the work done on the air and the heat transfer2.20 Water at 20C is pumped from ground level to anelevated storage tank above ground level; the volume ofthetankis50m3.Initially,thetankcontainsairat100kPa,20C, and the tank is closed so that the air is compressedas the water enters the bottom of

46、 the tank. The pump isoperated until the tank is three-quarters full. The tempera-ture of the air and water remain constant at 20C. Deter-mine the work input to the pump.mh1v122-+h2V222-+ w0= AD24- 0.0755 m2=mh1h2w0= V1VA-0.850.0755- 11.26 m/s= =mCpt1t2w0=V2V1since small T and P=P1V1m1RT1=t1t21.250.7461.021.0