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目录第一章流体流动与输送机械·····················································（2）第二章非均相物系分离·························································（38）第三章传热···································································（44）第四章蒸发···································································（70）第五章气体吸收·······························································（74）第六章蒸馏···································································（96）第七章固体干燥·······························································（120）11.某烟道气的组成为CO213％，N276％，H2O11％（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Mm=ΣyiMi=(0.13×44+0.76×28+0.11×18)×10?3=28.98×10?3kg/mol∴混合密度ρm=pMm101.3×103×28.98×10?3==0.457kg/m38.31×(273+500)RT2．已知20℃时苯和甲苯的密度分别为879kg/m3和867kg/m3,试计算含苯40％及甲苯60％（质量％）的混合液密度。解：1ρm=ρ1a1+ρ2a2=0.40.6+879867混合液密度ρm=871.8kg/m33．某地区大气压力为101.3kPa，一操作中的吸收塔塔内表压为130kPa。若在大气压力为75kPa的高原地区操作该吸收塔，且保持塔内绝压相同，则此时表压应为多少？解：p绝=pa+p表=pa＋p表（∴p表=(pa＋p真)－pa=101.3＋130)?75=156.3kPa4．如附图所示，密闭容器中存有密度为900kg/m3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=p0+ρgΔz=p1+ρghΔz=p1+ρgh?p0p?p0(58?42)×103=1+h=+0.55=2.36mρgρg900×9.81∴5.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。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/m3和1000kg/m3。（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。解：（1）容器底部压力h1h2C题5附图ABDp=pa+ρ油gh1+ρ水gh2=pa+ρ水ghh=ρ∴油h1+ρ水h2ρ水=ρ油ρ水h1+h2=800×0.7+0.6=1.16m1000（2）pA≠pBpC=pD6．为测得某容器内的压力，采用如图所示的U形压力计，指示液为水银。已知该液体密度为900kg/m3，h=0.8m，R=0.45m。试计算容器中液面上方的表压。解：如图，1-2为等压面。212p1=p+ρghp2=pa+ρ0gR题6附图p+ρgh=pa+ρ0gR则容器内表压：p?pa=ρ0gR?ρgh=13600×0.45×9.81?900×0.8×9.81=53.0kPa7．如附图所示，水在管道中流动。为测得A-A′、B-B′截面的压力差，在管路上方安装一U形压差计，指示液为水银。已知压差计的读数R＝180mm，试计算A-A′、B-B′截面的压力差。已知水与水银的密度分别为1000kg/m3和13600kg/m3。解：图中，1-1′面与2-2′面间为静止、连续的同种流体，且处于同一水平面，因此为等压面，即题7附图p1=p1'，p2=p2'2又p1'=pA?ρgmp1=p2+ρ0gR=p2'+ρ0gR=pB?ρg(m+R)+ρ0gR所以pA?ρgm=pB?ρg(m+R)+ρ0gR整理得pA?pB=(ρ0?ρ)gR由此可见，U形压差计所测压差的大小只与被测流体及指示液的密度、读数R有关，而与U形压差计放置的位置无关。代入数据pA?pB=(13600?1000)×9.81×0.18=22249Pa8．用U形压差计测量某气体流经水平管道两截面的压力差，指示液为水，密度为1000kg/m3，读数R为12mm。为了提高测量精度，改为双液体U管压差计，指示液A为含40％乙醇的水溶液，密度为920kg/m3，指示液C为煤油，密度为850k...