谭天恩 化工原理题解(1~6)

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1、谭天恩化工原理题解(1~6)本文由z461014237贡献doc文档可能在WAP端浏览体验不佳。建议您优先选择TXT，或下载源文件到本机查看。化工原理题解1-1解：ρ=pMRTp=450mmHg，M=29kg/kmol,R=8.314kJ/(kmol·K),T=233Kρ=450×29×101.3=0.898kg/m38.314×233×7601-2解：真空度=大气压-绝对压力即：760-738=655-真空度真空度=633mmHg1-3解：P=P水+P油=ρ水gh1+ρ油gh2=1000×0.52×9.8+3.46×9.8×916=36300Pa（表压）=

2、271mmHg（表压）=3.69mH2O（表压）1-4解：P下=P上+ρgh（58?42）103=1000×9.8(?Z?0.55)×?Z=2.18m1-5解752mmHg=100200Pa=100.2kPa绝压=大气压-真空度=100.2-83.4=16.8kPa表压=-真空度=-83.4kPaP=ρghH=?P83.4==8.51mρg1000×9.8即：液柱高度为8.51m。1-6解Pbottom=Ptop+ρgh，ρ=Pbottom+PtoppM，p=RT2Ptop=Pbottom2000RT?Mgh2000RT+Mgh2000×8.314×293?

3、9.807×102×4×292000×8.314×293+9.807×102×4×29Ptop=101.3×103×=96.5869×103Pa=724.64mmHgPTop又解pMRTpMdP=?P=P2?P=?gdh1RTPtdP102×4M=?∫∫PbP01000RTgdh408MglnPPb=Pt1000RTP408Mglnb=Pt1000RTP2=P?ρgdh,ρ=1Pt=Pbe408Mg1000RT?408×29×9.8071000×8.314×293P2dhP1hPBottom=101.3×103×e=96587.80Pa=96.5878kPa

4、A12345678BCDPtop=PbottomEXP(-MgH/1000RT)Ptop=PbottomEXP(-MgH/1000RT)Ptop96.5878kPa=Ptop96.5878kPa=Pbottom101.3Pbottom101.329MM29408HH408293TT293层高层高44层数层数102102A1234567BCDPtop=PbottomEXP(-MgH/1000RT)Ptop=PbottomEXP(-MgH/1000RT)Ptop96.58780107kPa96.58780107kPaPtopPbottom101.3Pbottom

5、101.329MM29408HH408293TT293层高层高44层数102层数102==1-7解P=PA+ρH2Ogh11P=P2+ρHggR11AH2OR1h12B4h2R23HgP3=P2+ρH2Og(h2-h1+R1)P3=PB+ρH2Og(h2-R2)+ρHggR2所以1PA=P2+ρHggR1?ρH2Ogh1PB=P2+ρH2Og(h2-h1+R1)?ρH2Og(h2-R2)?ρHggR2PAB=PA?PB=(P2+ρHggR1?ρH2Ogh1)?(P2+ρH2Og(h2-h1+R1)?ρH2Og(h2-R2)?ρHggR2)=ρHggR1?ρH

6、2OgR1?ρH2OgR2+ρHggR2=ρHg?ρH2O）gR1+ρHg?ρH2O）gR2（（?PAB=(ρHg?ρH2O)g(R1+R2)=(13600?1000)×9.807×(0.9+0.95)=228601.17Pa=1715.1mmHg1-8解h1h2P1=p1+ρcg(h1+R)P1=p2+ρcgh2+ρAgR?P=p1?p2=ρcgh2+ρAgR-ρcg(h1+R)真实压差11RP=ρcgh2+ρAgR-ρcg(h1+R)=910×9.807×h2+1000×9.807×0.320-910×9.807×h1?910×9.807×0.320=9

7、10×9.807×0.004+1000×9.807×0.320?910×9.807×0.320=318.13908Pa估计压差P=ρA?ρc）gR（=1000?910）9.807×0.320（×=282.4416Pa误差%=318.13908-282.4416=0.112207151664612318.139081-9解qm=1000γqv=1000×1.83×150×10?3/60=4.575kg/su小==qvA150×10?3/603.14×[(57?2×3.5)×10?3]24=1.27388535031847m/sG小=ρu=1000×1.83×1

8、.27388535031847=2331.20960