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作业习题解答第六章除尘装置6.1解:计算气流水平速度smAQv/1087.257.414.92.120。设粒子处于Stokes区域,取sPa51082.1。按《大气污染控制工程》P162(6-4)mmgLHvdp2.17102.1719.1281.91021.157.41087.21082.1181863250min即为能被100%捕集的最小雾滴直径。6.2解:按层流考虑,根据《大气污染控制工程》P163(6-5)2.229.64801812122121nnnn,因此需要设置23层。6.3解:sPahmkg51086.1)./(067.0mmmgLHvdp10084104.8781.9105.2123.01086.118185350min,符合层流区假设。6.4解:设空气温度为298K,首先进行坎宁汉修正:smMRTv/6.4661097.28142.3298314.8883,mv85106.66.466185.1499.01082.1499.0,21.063.0106.622Kn264.1]4.0257.1[21.0121.010.1eC。故smgCdupps/1058.11852525.060/1061.3202.05.01058.1)1(35QnLWusi。用同样方法计算可得0.83m粒子的分级效率为0.864。因此总效率695.0)864.0525.0(5.0i6.5解:按《AirPollutionControlEngineering》公式)]9(exp[12ipcWDNV。令=50%,N=5,Vc=15m/s,p=2.9×103kg/m3,W=0.76m,sPa5102,代入上式得dc=11.78m。利用《大气污染控制工程》P170(6-18)22)/(1)/(cpicpiidddd计算各粒径粉尘分级效率,由此得总效率%3.55iig6.6解:根据《大气污染控制工程》P144(5-53)iiigPg32/(P=0.1)计算分级效率,结果如下表所示:粉尘间隔/m0~55~1010~1515~2020~2525~3030~3535~4040~4545质量频率/%捕集g30.51.41.92.12.12.02.02.02.084.0出口g276.012.94.52.11.50.70.50.40.31.1%/i5.5949.4179.1790.0092.6596.2697.3097.8398.3699.85据此可作出分级效率曲线。由上表可见,5~10m去除效率为49.41。因此在工程误差允许范围内,dc=7.5m。6.7解:据《大气污染控制工程》P169(6-13)Pavp144015293.19.92121221。6.8解:根据《AirPollutionControlEngineering》P258公式)]9(exp[12ipcWDNV。因)/(1000322mkgDDppppa单位取单位,故pD2=10002paD;由题意,当smVc/20%,50。取sPa51082.1,N=10,代入上式)]1082.191000)100.1(2010(exp[1%50526iW,解得Wi=5.5mm。根据一般旋风除尘器的尺寸要求,D0=4Wi=2.2cm;H=2Wi=1.1cm。气体流量Q=A.V=H.W.Vc=1.21×10-3m3/s6.9解:按《大气污染控制工程》P170(6-18)22222225)5/(1)5/()/(1)/(pipipipicpicpiidddddddd;0221025pipipipiiqddddqdd。dg=20m,25.1,])32.020ln(exp[79.1])ln2ln(exp[ln2122pipiggpigpidddddq代入上式,利用Matlab积分可得%3.9610piiqdd。6.10解:驱进速度按《大气污染控制工程》P187(6-33)smdqEwpp/176.01011081.1310100103.0365315。2885.123.0mdLA,Q=0.075m3/s,代入P188(6-34)%8.98)176.0075.0885.1exp(1)exp(1iiwQA。6.11解:1)Q’=2/3=0.667m3/s,S=3.662=13.4m2,%3.99)122.02/667.04.13exp(1i。2)5.13/15.0maxvv,查图6-27得Fv=1.75故%8.9875.1%)3.991(1)1(1Fvi。6.12解:1)由题意77.0)9.0exp(15.0kkdp=3.5m,%2.93)5.377.0exp(11dp=8.0m,%8.99)0.877.0exp(12dp=13.0m,%100)0.1377.0exp(13故%98%6.9832.01%8.992.0%2.932.02)301%6.982i,则i2=0.42g/m30.1g/m3。不满足环保规定和使用者需要。6.13解:1)由《大气污染控制工程》P183(6-31)电场荷电为CEdqp16526120201004.3104.3)105(1085.85.35.1323扩散荷电按P184(6-32)计算,与电场荷电相比很小,可忽略。因此饱和电荷值3.04×10-16C。2)电场荷电为CEdqp19526120201086.4104.3)102.0(1085.85.35.1323扩散荷电与电场荷电相比很小,可忽略,故粉尘荷电量4.86×10-19C。3)取sPa5105.2dp=5m时,smdqEwpp/088.0105105.23104.31004.3365516;dp=0.2m时,smdqEwpp/1051.3102.0105.23104.31086.43365519。6.14解:查图得集气板面积约1000m3.(1000m3/min)-1。根据)exp(1iiwQA,0.995=1-exp(-wi)解得wi=5.30m/min。6.15解:%95)exp(1wQA,故05.0)exp(wQA,0025.0)2exp(wQA因此%75.990025.01)2exp(1'wQA。6.16解:设3种粒子的分级效率分别为1、2、3,则6586.095.03)1()1()1(3710321keeekkk因此%9.991,%0.992,%1.863。6.17解:1)粉尘粒径dp=10m当液滴直径为50m时,R=0.2;碰撞数3.36618)(2CDpppIDuudN,14.19IN。由给出计算公式可得%3.50同理可得液滴直径为100m、500m时捕集效率为42.6%、10.1%。2)dp=50m用同样方法计算可得颗粒在直径为50m、100m、500m的液滴上捕集效率分别为0、10.2%、25.0%。6.18解:按《大气污染控制工程》P211(6-53)OcmHQQvpglT23223235.961036.1)1083(1003.1)(1003.1由(6-55)233.02229]101.6exp[pdgpCgliepfdCP粒径小于0.1m所占质量百分比太小,可忽略;粒径大于20.0m,除尘效率约为1;因此%0152.01000.81000.121000.161000.1310078.010021.02222225.1733.05.1233.05.733.0333.075.033.03.033.0eeeeeeP故%48.981P。6.19解:OcmHQQvpglT23232316631012)11600(1003.1)(1003.1坎宁汉修正143.12.1172.01172.01pCdC0])10845.1(166322.02.1143.1789.11101.6exp[]101.6exp[242292229gpCglipfdCP6.20解:设气液比1L/m3,dp=1.2m,3/8.1cmgp,f=0.25。在1atm与510.K下查得sPag51099.2。由OcmHvQQvpglT2323234.152100.11003.1)(1003.1可解得v=121.6m/s。故喉管面积2058.06.1211.7mS,DT=272mm。取喉管长度300mm,通气管直径D1=544mm。241,62,则mmmctgDDLT64.064022111,mctgDDLT13.322222(取D2=600mm)。6.21解:由《AirPollutionControlEngineering》P3009.48式tDzcDM24。t通过P293Figure9.18读取。取33/102mkgp,雨滴Db=2mm,处于牛顿区,利用《大气污染控制工程》P150(5-83)smv/0.7]205.1/81.9)205.1100.1(100.2[74.12/133。因此,912.01021082.1180.7)103(10218352632bppsDvdN。从Figure9.18读出t=0.11(Cylinder)。故M=g0083.011.080300)102(423。而液滴本身gDM331019.461'。故质量增加了1.98×10-4%。6.22解:由《AirPollutionControlEngineering》公式tAQDCCLD5.1ln0。代入已知数据httAA3.12105.21.01025.11.0ln33,即需持续半天左右的时间。6.23解:%5.99%10015.90458.015.9设破裂2个布袋后气体流量分配不变,近似求得出口浓度如下:300/0761.06002)1(600598'mgCCC。因此%2.99%10015.90761.015.9。6.24解:设恒定速度v1,则401vKxffg,40011vKxvKxppgffg。若在400Pa压降下继续,则40022212vKxvKxvKxppgppgffg4008.70360400400360360401222121122121vvQvvvQvQvvvv40015.25.1694008.70303608.70304002222222dtdQQdtdQdtdQQdtdQ解此微分方程得Q2=90.1m3。6.25解:当T=300K时,sPa51086.1,v=1.8m/min=0.03m/s。SxMp,1210100102.143MMSMxppKMbp/0
本文标题:第六章除尘装置
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