化工原理杨祖荣著高等教育课后答案

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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）1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％，O

3、27.5％，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yM.008544.007532.07628.0081828.8610kg/molmii∴混合密度33pMm1013.1028.86103.0455kg/mmRT.831(273500)332．已知20℃下水和乙醇的密度分别为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1a1a25

4、.05.09982.789m123881.36kg/mm相对误差：m实m881.36100%1100%.574%m实9353．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pp绝a真a真''pp(pp)90(1013.85)737.kPa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

5、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gzpgh题4附图013p1ghp0p1p0(5842)10zh.055.236mgg900.98125.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

6、ghghpghCDa油1水2a水hh800题5附图油1水2油hhh7.06.0.116m121000水水（2）ppppABCD6．水平管道中两点间连接一U形压差计，指示液为汞。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）p()Rg(136001000).003.98137082.Pa0（2）空气密度33'pM1013.1029103.1206kg/mRT.831(27320)''p()R

7、g(13600.1206).003.98140021.Pa0'∵空气密度较小，∴pRg07．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ghppgR1A123013pghpgR(1)344A1301又pppghgR56B3021256pppg(hh)34542题7附图pghgRg(hh)(2)B30242将

8、(2)代入(1)中:pghpghgR