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时间:2020-04-05
《化工原理1_7章习题答案解析.docx》由会员上传分享,免费在线阅读,更多相关内容在教育资源-天天文库。
1、完美格式整理版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·························································(32)第三章传热···································································(42)第四章蒸发·····················································
2、··············(69)第五章气体吸收·······························································(73)第六章蒸馏···································································(95)第七章固体干燥·······························································(119)学习好帮手完美格式整理版第一章流体流动与输送机械1.某烟道气的组成为CO213
3、%,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MmyiMi(0.13440.76280.1118)10328.98103kg/mol∴混合密度ρpMm101.310328.981030.457kg/m3mRT8.31(273500)2.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:1a1a20.40.6879867m12混合液密度ρ871.8kg/m3m3.某地区大气压力为101.3kPa,一操作中的吸收
4、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pp表p'+p表'aap表'(pa+p真)-pa'(101.3+130)75156.3kPa4.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题4附图pp0gzp1ghp1ghp0p1p0h(5842)103zgg0.552.36m9009.81学习好帮手完美格式整理版
5、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。h1AB已知油与水的密度分别为800kg/m3和1000kg/m3。h2CD(1)计算玻璃管内水柱的高度;题5附图(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ppa油gh1水gh2pa水ghh油h1水h2油h28000.70.61.16mh11000水水(2)pApBpCpD6.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
6、方的表压。解:如图,1-2为等压面。12p1pghp2pa0gR题6附图pghpa0gR则容器内表压:ppa0gRgh136000.459.819000.89.8153.0kPa7.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p1p1',p2p2'又p1'pAgm学习好帮
7、手完美格式整理版p1p20gRp2'0gRpBg(mR)0gR所以pAgmpBg(mR)0gR整理得pApB(0)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pApB(136001000)9.810.1822249Pa8.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850kg/m3。问读数可以放大多少倍?此时读
8、数为多少?解:用U形压差计测量时,因被测流体为气体,
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