分离过程典型例题

1、典型例题-计算题1 .第一脱甲烷塔的进料组成及操作条件下各组分的相平衡常数如下表所示,要求甲烷的蒸出率为98%,乙烯的回收率为96%,试分别按清楚分割和不清楚分割方法计算储出液和釜液的组成,并比拟两种计算结果.序号1234567组分H2CH4C2H4C2H6C3H6C3H8C4XiF/摩尔33.85.833.225.70.500.200.70Ki/1.70.280.180.0330.0220.004解：取100Kmol进料为基准,选取甲烷为轻关键组分,乙烯为重关键组分.按清楚分割的概念,组分1为轻组分,全部从塔顶储出液在采出,组分4、5、6、7为重组分,全部从塔釜液中采出.由于甲烷的蒸出率为9

2、8%D25.80.985.684Kmol;B25.85.6840.116Kmol乙烯的回收率为96%B333.20.9631.872Kmol.D333.231.8721.328KmolDD1D2D333.85.6841.32840.812KmolBFD10040.81259.188Kmol或B7Bi0.11631.872：25.70.500.300.7059.188Kmoli2Di33.80.828；X2DD25.6840.139X1DD40.812D40.812X3DD31.3280.033D40.812B20.1160.002B331.8720.538X2BX3BB59.188；B59.1

3、88B425.700.434；B50.500.008X4BB59.188X5BB59.188B60.300.005；B70.700.012X6BB59.188X7BB59.188将计算结果列表如下:组分ZiXiDXiBDiBiH20.3380.828/33.8CH40.580.1390.0025.6840.116C2H40.3320.0330.5381.32831.872C2H60.257/0.434/25.70C3H60.005/0.008/0.50C3H80.003/0.005/0.30C40.007/0.012/0.701.0001.0001.00040.81259.188不清口分割k料

4、衡算由iH请计算各组分的相对挥发度,结果列表序号1234567组分H2CH4C2H4C2H6C3H6C3H8C4XiF/摩尔33.85.833.225.70.500.200.70Ki/1.70.280.180.0330.0220.004ai3/6.0711.00.6430.1180.0790.014D2B35.68431.8721lglgNmD3B21.3280.1163.92lg23lg6.071由于氢气的相对挥发度很大,该组分全部从塔顶储出液中采出.由曳iNmD和FiDiBi分别计算组分4、5、6、7的分布.BiB3对组分4:-D4:mS0.6433.921.3280.0074；而25.7

5、D4B4B4B331.872由此计算得D40.189;B425.511对组分5：Di5mD-0.1183.92jj28_0.00001;而0.5D5B5B5B331.872由此计算得D50.0；B50.50将计算结果列表如下：组分ZiXiDXiBDiBiH20.3380.824/33.8CH40.580.1390.0025.6840.116C2H40.3320.0320.5401.32831.872C2H60.2570.0050.4320.18925.511C3H60.005/0.008/0.50C3H80.003/0.005/0.30C40.007/0.012/0.701.0001.0001

6、.00041.00158.9992 .要求在常压下别离环己烷a.沸点80.8C和苯b.沸点80.2C,它们的包沸组成为苯0.502摩尔分数,共沸点77.4C,现以丙酮为恒沸剂进行恒沸精储,丙酮与环己烷形成恒沸物,共沸组成为0.60环己烷摩尔分数,假设希望得到几乎纯洁的苯,试计算：所需恒沸剂量.塔顶、塔釜储出物量各为多少.以100kmol/h进料计解：以100kmol/h进料为基准,设丙酮包沸剂的用量为Skg,恰好与料液中的环己烷组成恒沸物,进料量和塔顶包沸物的量和组成分别为F,D,xFi,xDioD对环己烷作物料平衡DxdiFxfi;Fxfi1000.498XD10.6083kmol恒沸剂S的

7、用量为SDXD3830.40332kmol由于塔釜希望得到几乎纯洁的苯,WW2FXF2100.50250.2kmol3 .甲醇a.-丙酮b.在55.7C时形成恒沸物,其恒沸组成为Xi0.198,水和苯均可作为萃取剂进行萃取精储以别离甲醇和丙酮,试通过计算确定水c.与苯d.的选择度,并据理说明哪种萃取剂更佳及塔顶储出液各为何种物质？：XS0.8mol；A120.2798；A210.2634；A130.3794；A310.2211；A230.9709；A320.5576；A140.8923；A410.7494；A240.2021；A420.1533解：以水作萃取剂A12A21)0.27980.26

8、340.2716A1S2(A1SAs1)0.37940.22110.3002A2S2(A2SAS2)0.97090.55760.7643lgS12XsAsA2sA12(12x1)0.80.30020.76430.2716(120.198)0.5025S120.3144以苯作萃取剂,A1s12(A1SAS1)0.89230.74940.82085A2S(A2sAS2)0.20210.15330.17725lgS12XsA1sA2sA12(12X1)0&0.820850.177250.2716(120,198)0.3836S122.4191以苯作为萃取剂进行萃取精微别离甲醇和丙酮更佳,而水

9、作为萃取剂比不参加萃取剂时的效果更差.由于甲醇的沸点64.7C高于丙酮55.7C,那么塔顶储出物为丙酮,塔釜储出物为甲醇和苯.4 .乙酸甲酯a,和甲醇b,混合物在45c时为恒沸物,今以水为溶剂进行萃取精储,其组成为X10.7,xs0.8;A120.447;A210.411;A131.3;A310.82;A230.36;A320.22.试求其选择度,并说明塔顶储出何物？解:A12一(A122A21)0.4470.4110.429Ais12(A1SAsi)1.30.821.062A2slgS121一(A2s2xsA1sAs2)0.360.220.292A12(12x1)0.81.060.290.4

10、29(120.7)0.7533S125.6661由于甲醇的沸点64.7C低于乙酸甲酯C,那么塔顶储出物为甲醇酮,塔釜储出物为乙酸甲酯和水.5.某裂解气组成如下表所小.组分H2CH4C2H4C2H6C3H6i-C4H10Ey.0.1320.37180.30200.0970.0840.01321.000现拟以i-C4H10微分作吸收剂,从裂解气中回收99%的乙烯,原料气的处理量为100kmol/h,塔的操作压力为4.052Mpa,塔的平均温度按-14C计,求:为完成此吸收任务所需最小液气比.操作液气比假设取为最小液气比的1.5倍,试确定为完成吸收任务所需理论板数.各个组分的吸收分率和出塔尾气的量和

11、组成.塔顶应参加的吸收剂量.解：选乙烯为关键组分,查得在4.052Mpa和-14C下各组分的相平衡常数列于下表.a最小液气比的计算在最小液气比下N,人关关0.99,%：关K关0.720.990.7128Vminb.理论板数的计算1.50.71281.0692操作液气比LV15LVmin关键组分的吸收因子为上31.485VK关0.72AlgNL1理论板数lgA,0.991.485lg0.99118.868lg1.485C.各个组分的吸收分率和出塔尾气的量和组成A由.N1A关K关LAiAiiiN1KiVKi和A1以及v1,i1iVN1,i；V1vi,i；yi,iv先进行计算结果见表组分H2CH4C

12、2H4C2H6C3H6i-C4H10EyN1,i0.1320.37180.30200.0970.0840.01321.000Ki/3.150.720.500.1450.056n1,i,kmol/h13.237.1830.29.78.41.32100.0A/0.33941.4852.13847.37419.09300.33940.990.9951.0001.00013.224.5610.3020.04850038.1115yu0.34640.64450.00790.0013d.吸收剂量V10038.111569.056kmol/hL0(L061.8885)Lo30.9442塔内气体的平均流率为：

13、L塔内液体的平均流率为：由LV1.0692,得Lo42.89kmol/h6 .在一精微塔中别离苯B,甲苯T,二甲苯X和异丙苯C四元混合物.进料量200mol/h,进料组成zB=0.2,zT=0.1,zX=0.4mol.塔顶采用全凝器,饱和液体回流.相对挥发度数据为：BT=2.25,TT=1.0,XT=0.33,ct=0.21.规定异丙苯在釜液中的回收率为99.8%,甲苯在储出液中的回收率为99.5%.求最少理论板数和全回流操作下的组分分配解：根据题意顶甲苯为轻关键组分,异丙苯为重关键组分,从相对挥发度的大小可以看出,二甲苯为中间组分,在作物料衡算时,初定它在储出液和釜液中的分配比,并通过计算修

14、正物料衡算如下储出液di40600.99520.00240DW=98.14釜液W-600.00518800.998W组分进料B40T60X2080C一200解得D=101.86xTD=0.5861xTW=0.0031xCD=0.00157xCW=0.8135NmlgC-DWXTXClg工CT,0.58610.8135lg0.00160.0031lg21=7.35生生LNmWxWccTdxWX20解得dx=1.05mol;WX=18.95mol组分进料diWB4040一T60600.995600.005X201.0518.95C80800.002800.998100.9199.09组分xdixw

15、iB0.39640T0.59160.0030X0.01040.1912C0.001580.8057lgJD坛W,0.59160.8051Nxcxt0.00160.0030=7.37Nmglg10.21CT2&xxTNmWx%cTdxWx20组分进料储出液釜液数量Mol%数量mol%数量mol%苯40204039.643一甲苯603059.759.1670.33.02二甲苯20101.041.0318.9619.13异丙苯80400.160.15979.8480.55200100100.9100.0099.1100.007 .某原料气组成如下：组分CH4C2H6C3H8i-C4H10n-

16、C4H10i-C5H12n-C5H12n-C6H14y0(摩尔分率)0.7650.0450.0350.0250.0450.0150.0250.045先拟用不挥发的烂类液体为吸收剂在板式塔吸收塔中进行吸收,平均吸收温度为38C,压力为1.013Mpa,如果要求将i-C4H10回收90%.试求：为完成此吸收任务所需的最小液气比.操作液气比为组小液气比的1.1倍时,为完成此吸收任务所需理论板数.各组分的吸收分率和离塔尾气的组成.求塔底的吸收液量解：由题意知,i-C4H10为关键组分由P=1.013Mpa,t平=38C查得K关=0.56(P-T-K图)在最小液气比下N=0°,A关=中关=0.

17、9(LZ)minK关隈V=0.560.9=0.504.LV1.1LVmin=1.10.504=0.55440.55440.560.99Alog()N1所以理论板数为10gAlog0.990.910.9log0.9919.48组分进料量相平衡常数KiAii被吸收量塔顶尾气数量组成CH476.517.40.0320.0322.44874.050.923C2H64.53.750.1480.1480.6683.8340.048C3H83.51.30.4260.4261.4912.0090.025i-C4H102.50.560.990.902.2500.2500.003n-C4H104.50.41.38

18、60.994.4550.0450.0006AiN11计算结果如下:AiN1LVKi它组分吸收率公式i-C5H121.50.183.081.001.5000.00.0n-C5H122.50.1443.851.002.5000.00.0n-C6H144.50.0569.91.004.5000.00.0合计100.0一一19.81080.190以CH4为例:LAi=VKi0.554417.40.0329481_0.03210.99ccc9.481=0.03210.32V1(CH4)=(1-i)VN+1=(1-0.032)76.5=74.05V1(CH4)y1(CH4)-V1塔内气体平均流率:74.0

19、50.92376.510080.190v290.10Kmol/h塔内液体平均流率:Lo(L.19.81)L=2Lo9.905由%=0.5544;Lo=40.05Kmol/h8 .乙醇-苯-水系统在101.3kPa,64.86C形成恒沸物,其组成为22.8mol%乙醇a.,53.9mol%苯b.和23.3mol%水c.,利用恒沸点气相平衡组成与液相组成相等这样有利条件,计算在64.86C等温别离该包沸混合液成为三个纯液体产物所需的最小功.解：在等温等压条件下,将其别离成纯组分时所需最小功为Wmin,TRTnFxFilnFixFi设为理想溶液,Fi1,假设nF1kmolWmin,TRTXFilnX

20、Fi8.31(64.38273.15)0.2281n0.2280.539ln0.5390.2331n0.2332836kJ/kmol9 .含乙烯32.4%的乙烯-乙烷混合物于2Mpa压力下进行蒸储,塔顶为纯乙烯,温度为239K,塔釜为纯乙烷、温度为260K,正常操作下,塔釜参加热量为8800kJ/kg乙烯,试计算别离净耗功为多少？解：设环境温度T0298K,nF1kmol,乙烯的量为10.324268.424kg_1111W争QTo(一一)8.4248800298()7465.6kJ/kmolTlTh23926010 .计算题9的热力学效率解:Wmin,TnFRTyFilnyFi8.31298

21、0.3241n0.3240.6761n0.6761560kJ/kmolWmin1560mn_0.209W净7465.611.试求环境温度下,下表所示的气体别离过程的最小功组分C2C3n-C4n-C5n-C6E进料kmol/h30200370350501000产品1kmol/h301924226产品1kmol/h836635050774解：求出各物流的组成组分C2C3n-C4n-C5n-C6E进料kmol/h30200370350501000XiF0.030.200.370.350.051.00产品1kmol/h301924226Xi10.1330.8500.0171.00产品1kmol/h83

22、6635050774Xi20.0100.4730.4520.0651.00Wmin,tRTnk(yi,klnyi,k)nj(yi,jlnyi,j)出进8.314298226(0.1331n0.1330.85ln0.850.0171n0.017)774(0.01ln0.010.473ln0.4730.452ln0.4520.0651n0.065)1000(0.031n0.030.21n0.20.371n0.370.351n0.350.051n0.05)1.188106kJ/kmol12 .将含丙稀80%摩尔和丙烷20%的料液在常压下别离为含内稀99.6面口5%的两股产品,试求别离最小功.设料液和

23、产品均处于环境温度298K,料液可以当作理想溶液.解：设料液量为1kmo1,计算两股产品的量：nQ10.996nQ20.0510.8nQ1nQ21联列解得：nQ10.7928kmol,nQ20.2072kmol别离最小功由式6-15计算：wmin=0.7928(0.9961n0.996+0.0041n0.004)+0.2072(0.051n0.05+0.951n0.95)-(0.81n0.8+0.21n0.2)RT.=0.4386RT0=1087.4KJ/Kmo1料液13 .一轻姓混合物进行精储塔的计算,求得最小回流比Rm=1.45,最小理论板数Nm=8.6.求回流化为R=1.25Rm,全塔平

24、均板效率为n=0.60,所需最小塔板板数.R=1.251.45=1.8125=L8125-M5=013氏+118125+1由此图查,得:"+10,58由于,Nm=8.6所以,N=19.43实际板数为：19.43/0.6=32.4=33块14 .某厂裂解气组成(摩尔)如下：13.2%氢、37.18%甲烷、30.2乙烯、9.7%乙烷、8.4%丙烯、1.32异丁烷、所用的吸收剂中不含所吸收组分.要求乙烯的回收率到达99%.该吸收塔处理的气体量为100kmo1/h,操作条件下各组分的相平衡常数如下:丙烷0.520.15甲烷异丁烷3.10.058乙烯0.72操作液气化为最小液气比的1.5倍试求

25、：(1)最小液气化；(2) 所需理论板数；(3) 各组份的吸收因子、吸收率(4) 塔顶尾气的组成及数量；(5) 塔顶应参加的吸收剂量解：(1)最小液气化=mA=0.72+0.P?=0,712S(2)所需理论板数N=15*07123=1口692%=15(/同A=(-)=1,0692x=1435旷K0.72N二1=8,86%(3)各组份的吸收因子、吸收率吸收因子,以甲烷为例：工十七十M.941Q口十沼Q61.0692护=(9)-rr-=1.0692x-=0.3-149中阮vK甲烷3.1吸收率,以甲烷为例:中甲烷:节篇一由烷_*T-SM4炉m_0344中邮413449=0.341塔顶尾气的组成及数量(见表)W+1A.W+1(J)1y1H13.200013