【《烧碱制作中蒸发阶段物料衡算计算》1100字】.docx
烧碱制作中蒸发阶段物料衡算计算目录烧球制作中蒸发阶段物料衡算计算1.1 蒸发工或物MWWw1.÷w2÷w3=w1.2 效的.m.a.121第三效衡算1.2.2第二效衡算1.2.3第一效衡%2蒸发衡尊2.1 2.1.1 估算乐力、淑度S5SS662.1.2 溶质浓度对溶液温度造成的损失2.1.3 液柱静压对T造成的损失2.1.4 各管路之间由助力导致的温差损失2.1.5 各效沸点22蒸发J2.3根据工艺要求NaOH出蒸发器时的质量分数为50%,碱液进入蒸发器的质量分数为31.5%。NaC1.fNaOH>40pni:NaC1.OJ./NaOH>20ppm；Ee2O3,NaOH>0.3ppn1.1蒸发工段物料衡算根据任务书已知要求生产100%NaoH为82kVa则每小时的产后为：,82×IO682kta=.（）.=9761.9048kg,h盐损失量为1%，NaoH进蒸发涔了量为：9761.9084-99%=9860.5135kg碱液进入蒸发器的fit为：G9860.5135Fo=-=,=31808.1081k1hXU.51U各组分质量流量：NaCI：9860.5135×40×1Q6=03944kg.zhNaC1.Ou9860.5135×2O×10=0.1972kghFCQ3：9860.5135×03×1(T6=0002958kg'hH,O=31808.1081-9860.5135-0.3944-0.1972-0.002958=21947.0000kg,h底液出蒸发器的量为：x9761.9084=1.9523.81.68kg'hUJ各组分质量流域：NaCh9761.9084×40×106三0.3905kgNaCIOu9761.9084×20×1"=.1952kg'hFe2C)3:9761.9084×0.3×IO6=O.2929kg,hH,0=19523.8168-9761.9084-0.3905-0.1952-0.002929=9761.3198kgh总蒸发水量：W=Fw-FWj1.=21947.00-9761.3198=12185.6802kg'h估算水量：水量设为：Wi：W2:W3=1.2:I.I:Iw1+w2+wji=wW1=443i.1.564kgh解得：W2=4061.8934kghW3=3692.6304kgh表2.IO物料平衡表进蒸发系统物料出蒸发系统物料组分质员流盘（kgh）组分侦员:流里kgh）NaOH9860.5135NaOH9761.9084NaCI0.3944NaCI0.3905NaC1.Oi0.1972NaCIOs0.1952Fc>Oj0.002958FeO0.002929H2O21947.0(XX)HiO9761.3198总计31808.1081总计19523.81681.2各效的物料衡算各效损失相同得：(9860.5135-9761.9084)+3=32.8684kg出三效NaOH的流域为：Fnioh=9860.5135-32.8684=9647.6451kg'h出二效NaOH的流量为：Fnj0h=9647.6451-32.8684=9614.7767kg'h出效NaoH的流量为：FNaoH=9614.7767-32.8684=9581.9083kg1.h1.2.1第三效衡算出三效浓度为：X.F3.9647.64513(F0-W,)31808.1081-3692.6304,各成分质量流量：NaC1.:9647.6451×40×106=0.3859kg.'hNaCIO:9647.6451×20×IO6=O.1930kghFeQ：9647.6451×0.3×106=0.002894kg,'hH：O：21947.0000-3692.6304=18524.3696kg'h总流量：18524.3696+9647.6451+0.3859-0.1930+0.002894=27902.5965kg,1h表2.11第三效的物料组分进蒸发系统物料质量流员(kg,1h)组分出蒸发系统物料质量流量(kgh)NaOH9860.5135NaOH9647.6451NaCI0.3944NaCI0.3859NaC1.OA0.1972NaCIOs0.1930Fc>0j0.002958FciOj0.002894Hg21947.00H；O182543696总计31808.1081总计27902.59651.2.2第二效衡算出二效浓度为：,F29614.7767“2=(FI-W、)=27902.5965-4061.8934=°，4°33各成分质量流量：Nad：9614.7767×40×106=0.3846kgNaC1.OA:9614.7767×20×10«=0.1923kghFciOx：9614.7767×0.3×i06=0.002884kg'hHiO：18254.3696-4061.8934=14192.4762kg,h总流量：14192.4762+9614.7767+0.3846-0.192340.00288423807.8331kg,h表2.12第二效的物料进蒸发系统物料出蒸发系统物料组分质量流量(kgh)组分质量潦量(kgh)NaOH9647.6451NaOH9614.7767NaCI0.3859NaCI0.3846NaC1.O;0.1930NaC1.O;0.1923FeXh().(X)2894Fe2O30.002884H:O18254.3696Mo14192.4762总计27902.5965总计23087.83311.2.3第一效衡算出一效浓度为：F1.9581.9083X1.=（F2-W1）=23087.8331-4431.1564=015,36各成分质量流量：NaC1.:9581.9083×40×106=03833kg.zhNaC1.Oi:9581.9083×20×1"=.1916kg,'hFeQ:9581.9083×0.3×106=0.002875kgH2O:14192.4762-4431.1564=9761.3198kgh总流量：9761.3198+9581.9083+0.3833+0.1916+0.002875=19343.8059kg,h我2.”第效的物料进蒸发系统物料出蒸发系统物料组分质量流量kg/1h)组分质量流量(kgh)NaOH9614.7767NaOH9581.9083NaCI0.3846NaCI0.3833NaC1.O50.1923NaC1.O50.1916FeO0.002884FeiO?0.002875Hg14192.4762HiO9761.3198总计23087.8331总计19343.80592蒸发衡算2.1温差损失2.1.1 估算压力、温度估算压力和温度；蒸汽压力P=0.69MPa（绝）和尾效压力Pko.0139MPU（绝）等分配。即：（P1.-P0（0.69-0.0139）Ag21._1._!r=：=O.2253MPaP'=P1.-P=0.69-0.2253=0.4647MPaI>2=P-2AP=O.69-20.2253=0.2394MpaP；=Pr3AP=0.69-3x0.2253=0.01.41.MPa杳找资料可得：P'=0.4647MPa（绝）T'=1.48.919r1.'=211.9.8552kJ.'kgP；=0.2394MPa（绝）T2=I25.526r,'=2356.6924kJkgP；=0.01.41MPa（绝）Tj'=51.87=2373.65kJkg2.1.2 溶质浓度对溶液温度造成的损失根据杜比公式=ktu,+m-twk=1.+O.I42xm=150.75x2-2.71x解得：（,1=42.2299卜2=3061931.1=19.3422总T损失：i=42.2299+30.6193+19.3422=92.19142.1.3 液柱静压对T造成的损失由经验可得：1"+2",+3"=I.52.1.4 各管路之间由助力导致的温差损失有经验可得：1.t»»（+2+3=12.1.5 各效沸点一效：»»»»»t1=T1+1+1.+1.=148.919+42.2299+1.5+1=193.6489二效：（，*»t2=T22÷2+2=125.526-30.6193+1.5+1=158.6453:效:t3=T3+3+j+j=51.87+19.3422+1.5+1=73.71222.2 蒸发水碱液入口温度：60C由公式：j+Diri+(FCpo-1WnCpw)(ti.,-t0)JWi=查资料可得：Cpo=3.61.92kJ(kg),Cpw=4.2751kJ(kg)系数：,=0.98-0.7×(0.5136-0.4033)=0.9028,=0.98-0.7×(0.4033-03431)=0.9379j=0.98-0.7×(0.3431-0.310)=0.9568代入公式：一效水量：WU=02X)"27I.NMj<kt<1.92-(1.93.ZM1.SXW3)21193552二效水量:1.20,94WnK552p¾IOSI>361.92-427wu)(1.47V7I22)2356.6924三效水量:,v-.0.9S6dVi223S66924(31.1CI8"3.612427S1."W1.rWi2)A(73.71.224)总水量：W1.I+W1.2+Wu=W解得:D=9693.2790kg'hWu=6835.2874kghWu=2865.2418kg4Wu=2485.1509kgh2.3 总温差和分配A1.=T-TK-E-"=1.64.7-51.9-92.1914-4.5-3=13.1086传热量：效：QI=DOR«=9693.2790x2071.5=20079627.4485W二效:Q2=D1.iRi=6835.2874×2I20.2028=I4492I95.4843W三效:Q=DuR2=2865.2418×2188.2032=6269731.2755W传热系数：K1.-K,=K,48W)W.1m2k由于公式：乙KA1K2A2K3Aj因为：A=A,=Aq所以：13.1086=20079627.448514492195.48436269731.27551,'+I，I×4X004X004800)A解得:A=559.4539m't=7.4774t2=5.3967t3=2.316i从各效蒸发器中流出的溶液的海点和二次蒸汽的温度：一效溶液沸点：t=-t1=164.7-7.4774=157.2226oC一效二次蒸汽温度：T=t1.-,-1''=161.0074-42.2299-1-1.5=116.2755二效溶液沸点：t2=7-t2=116.2755-5.3967=110.8788oC二效二次蒸汽温度：T2=t2-2'-2-2"=110.8788-30.6193-1-1.5=77.7595三效溶液沸点：t3=T2-Ab=77.7595-2.3161=75.4434通过查资料可得:P1=0.1513MPa验算条件：P1.i'-P；pPu=0.04038MPaO.15I3-O.46470.4647卜OW=OeP1.；?«P20.04038-0.23940.2394=O.83I3<Z2=O.OIWiJ-w；6835.2874-4431.15644431.15640.5426<Zr0.01W1.2-W;W212865.2418-3396.70493396.7049=0.1565<=0.01.f=O.327(XZj=O.OI24X5.1509-3692.63043692.6304不符合条件，重新计算.
