96全热回收风冷热泵机组在星级酒店的应用探讨

2643631671-6612201204-363-05TK115ATotalHeatRecoveryandAir-cooledHeatPumpUnitsAppliedResearchinStarHotelHuaGuoliangLiuDaopingZhangXiaoliLiGuang(InstituteofRefrigerationTechnology,UniversityofShanghaiforScienceandTechnology,Shanghai,200093)AbstractCondensingheatrecoverytechnologyappliedinHeatpumpsystem.Useofairconditioningsystemofcondensingheattohotwaterpreparationcanbothreduceenergyconsumptionandavoidtosetaaloneheatsourcesofdrawbacks.Firstlythepaperintroducedtheworkingprincipleandadvantagesofthetotalheatrecoveryair-cooledheatpumpunitsanddiscussedtheheatpumpheatrecoverytechnologywithanengineeringexample.Secondlyheatpumpheatrecoveryschemecomparedwiththeconventionalschemefromenergyconservation,theinitialinvestmentandannualrunningcosts.Finally,dotheanalysisfromdynamicrecoveryperiod,thestaticpaybackperiod.Theuseofheatpumpheatrecoveryunitsinhotelbuildingisfeasibleintechnicalandeconomicaspectsandenergysavingeffectobviously.Keywordstotalheatrecovery;aircooledheatpump;hotel;energysaving;[e-mail][e-mail]2011-09-1665%[1]241.1[2][phone]RefrigerationandAirConditioningVol.26No.4Aug.2012.363367364201211Fig.1Air-cooledheatpumpheatrecoveryunitsinsummeroperationmode1+CBAAB22Fig.2Air-cooledheatpumpheatrecoveryunitsinwinteroperationmode2ABCAC451.25065[3]R417a[4][5][6]2.13.614265264365[7]3800kW2200kW90t/d30XQ[8],1000kW1000kW1000kW33Fig.3Heatrecoverysystemschematicdiagram2.22.2.1190T550.78/5944Fig.4Tapwatertemperatureoftendencychartinshanghaiarea11Table1Domestichotwaterenergyconsumptioninothermonths3.53.43.22.72.32.72.92.82.72.52.11.82.12.3S=C×M(55T)×N/3600/COP1SC4.1868kJ/(kg·)M90TNCOPT20.716.32W1=Q1×N1/COP1×g1=3800×3000/3.7×0.6/1000=184.92Q1kWN1COP1g1184.9×0.78=144.23W2=Q2×N2/COP2×g2=2200×2400/3.6×0.7/10000=102.73Q2kWN2COP2g2102.7×0.78=80.12.2.2+11233.9/m34113.6/m335564kJ/Nm336620120.85V1=4.1868××55×/35564/22Table2Annualdomestichotwaterenergyconsumption1.91.81.71.40.81.10.971.01.011.21.41.67.47.026.65.042.883.963.493.63.964.325.045.071658.382COP3=4.5W3=Q1×N1/COP3×g14152118.63V2=(kW)×3600×//×=2200×3600×2400/35564/0.85×0.7=44.023.8/m3167.233Table3Initialinvestmentandoperatingcostsaboutthetwoschemes600475240.6344.18652.31N=SCNC/(NRSR5NSCNCNRSRN=1.22NNRSR=(SCNC)(1+i)n/[(1+i)n1]/i6N=lg[(NRSR)/(NRSR)i(SCNC)]/lg(1+i)i6.5%N=3.12643671103.63.11.2234[1],.[J].,2008,(2):2729.[2],.[J].,2008,(1):5658.[3],,,.[J].,2006,27(4):3740.[4],.R417aR22[J].,2003,24(4):14.[5].[J].,2008,22(6):7376.[6],.[J].,2008,34():6063.[7]GB50019-2003,[S].:,2004.[8].30XQ[M].2012.