化工原理计算

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1、目录第一章流体流动与输送设备·····················································（2）第二章非均相物系分离·························································（26）第三章传热···································································（31）第四章蒸发···············································

2、····················（45）第五章吸收····································································（49）第六章蒸馏···································································（69）第七章干燥···································································（83）1PDF文件使用"pdfFact

3、oryPro"试用版本创建www.fineprint.com.cn第一章流体流动与输送设备1.燃烧重油所得的燃烧气，经分析知其中含CO28.5％，O27.5％，N276％，H2O8％（体积％），试求此混合气体在温度500℃、压力101.3kPa时的密度。解：混合气体平均摩尔质量-3M=SyM=0.085´44+0.075´32+0.76´28+0.08´18=28.86´10kg/molmii∴混合密度3-3pMm101.3´10´28.86´103r===0.455kg/mmRT8.31´(273+500)332．已知20℃下水和

4、乙醇的密度分别为998.2kg/m和789kg/m,试计算50％（质量％）乙醇水3溶液的密度。又知其实测值为935kg/m，计算相对误差。解：乙醇水溶液的混合密度1a1a20.50.5=+=+rrr998.2789m123r=881.36kg/mm相对误差：rm实-rmæ881.36ö´100%=ç1-÷´100%=5.74%rm实è935ø3．在大气压力为101.3kPa的地区，某真空蒸馏塔塔顶的真空表读数为85kPa。若在大气压力为90kPa的地区，仍使该塔塔顶在相同的绝压下操作，则此时真空表的读数应为多少？''解：p=p-p

5、=p-p绝a真a真''p=p-(p-p)=90-(101.3-85)=73.7kPa真aa真34．如附图所示，密闭容器中存有密度为900kg/m的液体。容器上方的压力表读数为42kPa，又在液面下装一压力表，表中心线在测压口以上0.55m，其读数为58kPa。试计算液面到下方测压口的距离。解：液面下测压口处压力p=p+rgDz=p+rgh题4附图013p1+rgh-p0p1-p0(58-42)´10Dz==+h=+0.55=2.36mrgrg900´9.812PDF文件使用"pdfFactoryPro"试用版本创建www.fin

6、eprint.com.cn5.如附图所示，敞口容器内盛有不互溶的油和水，油层和水层的厚度分别为700mm和600mm。33在容器底部开孔与玻璃管相连。已知油与水的密度分别为800kg/m和1000kg/m。（1）计算玻璃管内水柱的高度；（2）判断A与B、C与D点的压力是否相等。h1AB解：（1）容器底部压力h2p=p+rgh+rgh=p+rghCDa油1水2a水rh+rhr800题5附图油1水2油h==h+h=´0.7+0.6=1.16m12rr1000水水（2）p¹pp=pABCD6．水平管道中两点间连接一U形压差计，指示液为汞

7、。已知压差计的读数为30mm，试分别计算管内流体为（1）水；（2）压力为101.3kPa、温度为20℃的空气时压力差。解：（1）Dp=(r-r)Rg=(13600-1000)´0.03´9.81=3708.2Pa0（2）空气密度3-3'pM101.3´10´29´103r===1.206kg/mRT8.31´(273+20)''Dp=(r-r)Rg=(13600-1.206)´0.03´9.81=4002.1Pa0'∵空气密度较小，∴Dp»rRg07．用一复式U形压差计测量水流过管路中A、B两点的压力差。指示液为汞，两U形管之间充满

8、水，已知h1=1.2m，h2=0.4m，h4=1.4m，h3=0.25m，试计算A、B两点的压力差。解：图中1、2为等压面，即p=p12p=p+rghp=p+rgR1A123013p+rgh=p+rgR(1)344A1301又p=p