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1、化学工程基础(武汉大学)第二章课后答案(FundamentalsofChemicalEngineering(WuhanUniversity)secondchapters)Thesecondchapterisfluidflowandtransportation1,anengineeringatmosphericpressureis9.81104Pa,anengineeringatmosphericpressureequivalenttohowmanymillimetersofmercury?Howmuchisit?Ricewatercolumn?Aliqu
2、idequivalentto850kgm3liquid.Howmanymetersofliquidcolumn?Solution:known:PHg=13600kgm3PH2O=1000kgM3P=850materialP=9.81104PakgM3Request:hHg,h,H,2O,HmaterialForP=g;PHghHgPH2OghH=2O=PGH9.81x104P==0.735mHg=735mmHgSo:hHg=PHgg136009.819.81x104P==10.0mH2OH,H,2O=PH2Og1000x9.819.81x104P==11
3、.8MliquidcolumnH=materialG850x9.81p2,aphysicalatmosphericpressureis760mmHg,andaskhowmuchPaitisequivalentto?Howmanymetersofwatercolumn?Solution:forP=PHgGPH2OghhHg=H2OPHghHg13600x0.760==10.33mH2OSo:H,H,2O=...PH2O1000P=hHgHgGP=13600x0.760x9.81=1.013105PaA0.80kgm3gasdensityisshownint
4、hefigureofthefilledgasholder,holderandopeningUtubemanometer,3,IndicatormercuryreadingRis0.4m,theopeningofthebranchpipemercurysurfaceisfilledwithaheightofR'0.01Mwater.ZuoceshuiTheverticaldistancebetweenthesilversurfaceandthecenterlineofthesidebarish=0.76m,andtheabsolutepressureatt
5、hecentersectionofthepressureopeningismeasured.(localatmosphericpressure)FiveStrongfor1.0133x10Pa)Solution:lettheatmosphericpressurePa,pressuremeasuringholeincentersectionoftheabsolutepressureisPA,waterdensityH2O=1000kgM3Hg=13600kg,mercurydensitym3,mandnfacetothereferencesurfaceca
6、lculation:(1)Pm=PA+H/ggasPn=Pa+G+2OR'pHRPHgg(2)Forsimultaneous(1)(2)expressionsandarranged:,Pm=PnPA=Pa+G+2OR'pHRPHgghpg(3)=1.0133x105+0.01x1000x9.81+0.4x13600x9.810.76x0.8x9.81=154788.5Pa=1.55x105PaTheppHgandP7、ing(3)inH2OgR'.Handpgtwogastype(3),willbesimplifiedas:PA=Pa+R=Hgg=1.0133x105+0.4136009.81=154696.4=1.55105Pa4.MeasurethepressureinthecontainerwithU-tubemanometer(suchasdrawings).InthecaseofFig.1,theabsolutepressureintheinstrumentisnotapparentHowmanymillimetersofmercuryarethey?InF
8、igure2,howmanymillimetersofmercuryarethe