化工原理精馏答案.docx

1、化工原理精馏答案五 蒸馏习题解答1解: (1)作x-y图及t-x(y)图,作图依据如下: xA=(p-pB0)/(pA0-pB0); yA=pA0xA/p 以t=90为例,xA=(760-208.4)/(1008-208.4)=0.6898 yA=10080.6898/760=0.9150 计算结果汇总: t80.0290100110120130131.8x10.68980.4483 0.2672 0.12870.01950 y10.91500.7875 0.61180.37770.07240 4.612x/(1+3.612x)10.91120.7894 0.6271 0.4052 0.084

2、00(2)用相对挥发度计算x-y值: y=x/1+(-1)x 式中=M=1/2(1+2) =pA0/pB0 1=760/144.8=5.249 ;2=3020/760=3.974M=1/2(1+2)=1/2(5.249+3.974)=4.612 y=4.612x/(1+3.612x) 由此计算x-y值亦列于计算表中,y-x图,t-x(y) 图如下: 1 题 附 图 2解: (1)求泡点: 在泡点下两组分的蒸汽分压之和等于总压P,即:pA+pB=pA0xA+xB0xB=p求泡点要用试差法,先设泡点为87 lgpA0=6.89740-1206.350/(87+220.237)=2.971 pA0=

3、102.971=935.41mmHg lgpB0=6.95334-1343.943/(87+219.337)=2.566 pB0=102.566=368.13mmHg 935.410.4+368.130.6=595600mmHg泡点为87,气相平衡组成为 y=pA/p=pA0xA/P=935.410.4/600=0.624 (2)求露点: 露点时,液滴中参与甲苯组成应符合下列关系: xA+xB=1或pA/pA0+pB/pB0=1式中 pA=0.4760=304mmHg; pB=0.6760=456mmHg求露点亦要用试差法,先设露点为103,则:lgpA0=6.8974-120.635/ (1

4、03+220.237)=3.165 pA0=1462.2mmHg lgpB0=6.95334-1343.943/(103+219.337)=2.784 pB0=608.14mmHg于是 : 304/1462.2+456/608.14=0.961再设露点为102,同时求得pA0=1380.4; pB0=588.84 304/1380.4+456/588.84=0.9951故露点为102,平衡液相组成为 xA=pA/pA0=304/1380.4=0.223解: (1)xA=(p总-pB0)/(pA0-pB0) 0.4=(p总-40)/(106.7-40) p总=66.7KPa yA=xApA0/p

5、=0.4106.7/66.7=0.64 (2)=pA0/pB0=106.7/40=2.674解: (1) yD= D =(y/x)A/(y/x)B =(yD /0.95)/(1-yD )/0.05)=2 yD =0.974 (2) L/VD = V=VD +L (V/VD )=1+(L/VD ) V0.96=VD 0.974+L0.95 (V/VD )0.96=0.974+(L/VD )0.95 (1+L/VD )0.96=0.974+(L/VD )0.95 (L/VD )=1.45解: 简单蒸馏计算: lnW1/W2= W2=(1-1/3)W1=2/3W1;y=0.46x+0.549,x1=

6、0.6,代入上式积分解得: 釜液组成:x2=0.498, 馏出液组成:WD xD =W1x1 -W2x2 (1/3W1)xD =W10.6-(2/3W1)0.498 xD =0.804 6解: FxF=Vy+Lx 0.4=0.5y+0.5x -(1) y=x/(1+(-1)x)=3x/(1+2x) -(2) (1),(2)联立求解,得y=0.528,x=0.272 回收率=(Vy)/(FxF )=0.50.528/0.4=66% 7.解: F=D+W FxF =DxD +WxW已知xF =0.24,xD =0.95,xW =0.03,解得: D/F=(xF -xW )/(xD -xW )=(0

7、.24-0.03)/(0.95-0.03)=0.228回收率 DxD /FxF =0.2280.95/0.24=90.4%残液量求取: W/D=F/D-1=1/0.228-1=3.38 W=3.38D=3.38(V-L)=3.38(850-670)=608.6kmol/h8解: (1) 求D及W,全凝量V F=D+W FxF =DxD +WxW xF =0.1,xD =0.95,xW =0.01(均为质量分率) F=100Kg/h,代入上两式解得: D=9.57Kg/h; W=90.43Kg/h由恒摩尔流得知: F(0.1/78+0.9/92)=V(0.95/78+0.05/92) 注意:如用

8、质量百分数表示组成,平均分子量Mm=1/(aA/MA+aB/MB) 解得 V=87Kg/h 由 于塔顶为全凝器,故上升蒸汽量V即为冷凝量, (2) 求回流比R V=D+L L=V-D=87-9.57=77.43Kg/h R=L/D=77.43/9.57=8.09(因为L与D的组成相同,故8.09亦即为摩尔比) (3) 操作线方程.因塔只有精馏段,故精馏段操作线方程为 yn+1 =Rxn /(R+1)+xD /(R+1)式中xD 应为摩尔分率 xD =( xD /MA)/xD /MA+(1-xD )/MB =(0.95/78)/(0.95/78+0.05/92)=0.961 yn+1=8.09x

9、n/9.09+0.961/9.09=0.89xn +0.106操作线方程为:yn+1 =0.89xn +0.1069解: y=R/(R+1)x+xD /(R+1) (1) R/(R+1)=0.75 R=0.75R+0.75 R=0.75/0.25=3 (2) xD /(R+1)=0.2075 xD /(3+1)=0.2079 xD =0.83 (3) q/(q-1)=-0.5 q=-0.5q+0.5 q=0.5/1.5=0.333 (4) 0.75x+0.2075=-0.5x+1.5xF 0.75xq+0.2075=-0.5xq +1.50.44 1.25xq =1.50.44-0.2075=

10、0.4425 xq =0.362 (5)0q1 原料为汽液混合物10解: (1) 求精馏段上升蒸汽量V和下降的液体量L,提馏段上升蒸汽量V和下降的液体量L.进料平均分子量: Mm=0.478+0.692=86.4 F=1000/86.4=11.6Kmol/h FxF =DxD +WxW F=D+W 11.60.4=D0.97+(11.6-D)0.02 D=4.64Kmol/h W=6.96Kmol/h R=L/D, L=3.74.64=17.17Kmol/h V=(R+1)D=4.74.64=21.8Kmol/h平均气化潜热r=308070.4+333200.6=32313.6KJ/Kmol从

11、手册中查得xF =0.4时泡点为95,则: q=r+cp(95-20)/r=(32313.6+159.275)/32313.6=1.37 L=L+qF=17.17+1.3711.6=33.1Kmol/h V=V-(1-q)F=21.8+0.3711.6=26.1Kmol/h (2) 求塔顶全凝器热负荷及每小时耗水量. Qc=Vr r=0.9730804+333200.03=30879.5KJ/Kmol Qc=21.830879.5=673172.7KJ/h 耗水量 Gc=673172.7/4.18(50-20)=5368.2Kg/h (3) 求再沸器热负荷及蒸汽耗量. 塔的热量衡算 QB+QF

12、 +QR=Qv+QW +QL QB=Qv+QW +QL-QF -QR 该式右边第一项是主要的,其它四项之总和通常只占很小比例,故通常有: QBQV=VIv Iv=(r+Cpt)=30879.5+159.28.2=43933.9KJ/Kmol QB=21.843933.9=957759.02KJ/h2.5KgF/cm2下蒸汽潜热r=522Kcal/Kg=5224.1818=39275.3KJ/Kmol 蒸汽需量为Gv Gv =QB/r=957759.02/39275.3=24.4Kmol/h =24.418=39.04Kg/h (4) 提馏段方程 y=Lx/(L-W)-WxW /(L-W)=1.

13、26x-0.00511解: 提馏段: ym+1=1.25xM-0.0187-(1) =LxM/V-WxW /V, L=L+qF=RD+F V=(R+1)D W=F-D, 精馏段: yn+1 =Rxn /(R+1)+xD /(R+1) =0.75xn +0.25xD -(2) q线:xF =0.50 -(3) 将(3)代入(1)得出: ym+1=1.250.5-0.0187=0.606,代入(2) 0.606=0.750.5+0.25xD , xD =0.92412解: (1) y1=xD =0.84, 0.84=0.45x1+0.55 x1=0.64, yW =30.64/(3+1)+0.84

14、/(3+1)=0.69, 0.69=0.45xW +0.55,xW =0.311, (2) D=100(0.4-0.311)/(0.84-0.311)=16.8(Kmol/h), W=100-16.8=83.2(Kmol/h)13解: (1) 求R,xD,xW精馏段操作线斜率为R/(R+1)=0.723 R=2.61提馏段方程 y=Lx/(L-W)-WxW/(L-W)=1.25x-0.0187精馏段操作线截距为 xD/(R+1)=0.263 xD =0.95提馏段操作线与对角线交点坐标为 y=x=xW xW =1.25 xW -0.0187 xW =0.0748 (2)饱和蒸汽进料时,求取进料

15、组成 将 y=0.723x+0.263 y=1.25x-0.0187 联立求解,得x=0.535,y=0.65因饱和蒸汽进料,q线为水平线,可得原料组成y=xF=0.6514解: (1) y1=xD =0.9,x1=0.9/(4-30.9)=0.692, (2) y2=10.692/(1+1)+0.9/2=0.796 (3) xD =xF =0.5, yD =0.5/2+0.9/2=0.715解: (1) FxF=Vyq+Lxq 0.45=(1/3)yq+(2/3)xq y q =2.5xq /(1+1.5xq) xq=0.375 yq=0.6 (2) Rmin=(xD-yq)/(yq-xq)

16、 =(0.95-0.6)/(0.6-0.375)=1.56 R=1.5Rmin=2.34 D=0.950.45/0.95=0.45 W=1-0.45=0.55 xW=(FxF-DxD)/W=(0.45-0.450.95)/0.55=0.041 L=RD=2.340.45=1.053; V=(R+1)D=1.503 L=L+qF=1.053+(2/3)1=1.72; V=V-(1-q)F=1.503-1/3=1.17 y=(L/V)x-WxW/V=1.72/1.17x-0.550.041/1.17 =1.47x-0.019316解: 精馏段操作线方程 yn+1 =3/4xn +0.24平衡线方程

17、 y=x/1+(-1)x=2.5x/(1+1.5x)提馏段操作线方程 y=1.256x-0.01278其计算结果如下: N0 x y 1 0.906 0.96 2 0.821 0.92 3 0.707 0.86 4 0.573 0.77 5 0.462 0.70 6 0.344 0.567 7 0.224 0.419 8 0.128 0.268 9 0.065 0.148 10 0.029 0.069由计算结果得知: 理论板为10块(包括釜), 加料板位置在第五块；17解: D/F=(xF -xW )/(xD -xW )=(0.52-xW )/(0.8-xW )=0.5 解得:xW =0.24

18、精馏段操作线方程: yn+1 =(R/(R+1)xn +xD /(R+1)=0.75xn +0.2 -(1)平衡线方程:y=x/(1+(-1)x)=3x/(1+2x) 或:x=y/(-(-1)y)=y/(3-2y) -(2)交替运用式(1),(2)逐板计算: xD =y1=0.8 .x1=0.571; y2=0.628,x2=0.360; y3=0.470,x3=0.228xW =0.24 共需NT=3块(包括釜). 18解:q=0,xD =0.9,xF =0.5,xW =0.1,R=5,精馏段操作线方程: yn+1=Rxn/(R+1)+xD/(R+1)=5xn/(5+1)+0.9/(5+1)

19、=0.833xn+0.15图解: 得理论板数为11块(不包括釜),包括釜为12块18题附图 19解: (1) F=D+W FxF =DxD +WxW D=F(xF -xW )/(xD -xW )=100(0.3-0.015)/(0.95-0.015) =30.48 Kmol/h=30.5 Kmol/hW=F-D=69.50 Kmol/h (2) NT及NF = xD =0.95、xW =0.015、q=1、R=1.5；xD /(R+1)=0.38作图得:NT =9-1=8(不含釜)进料位置: NF =6(3)L,V,yW 及xW-1 19题附图q=1,V=V=(R+1)D V=30.5(1.5

20、+1)=76.25Kmol/h L=L+qF=RD+F=1.530.5+100=145.8Kmol/h由图读得:yW =0.06, xW-1=0.0320解: (1) 原料为汽液混合物,成平衡的汽液相组成为x ,y平衡线方程 y=x/1+(-1)x=4.6x/(1+3.6x) - (1)q线方程 (q=2/(1+2)=2/3)则 y=q/(q-1)x-xF /(q-1)=-2x+1.35 - (2) 联解(1),(2)两式,经整理得: -2x+1.35=4.6x/(1+3.6x) 7.2x2 +1.740x-1.35=0 解知,x=0.329 y=0.693 (2) Rmin=(xD -ye)

21、/(ye-xe)=(0.95-0.693)/(0.693-0.329)=0.70621解: 因为饱和液体进料,q=1 ye=xe/1+(-1)xe=2.470.6/(1+1.470.6)=0.788 Rmin=(xD -ye)/(ye-xe)=(0.98-0.788)/(0.788-0.6)=1.02 R=1.5Rmin=1.53 Nmin=lg(xD /(1-xD )(1-xW )/xW )/lg =lg(0.98/0.02)(0. 95/0. 05)/lg2.47= 7.56 x=(R-Rmin)/(R+1)=(1.53-1.02)/(1.53+1)=0.202 Y=(N-Nmin)/(N

22、+1) Y=0.75(1-x0.567) (N-7.56)/(N+1)=0.75(1-0.2020.567) 解得N=14.5 取15块理论板(包括釜) 实际板数: N=(15-1)/0.7+1=21(包括釜)求加料板位置,先求最小精馏板数 (Nmin)精=lgxD /(1-xD )(1-xF )/xF /lg =lg0.98/0.020.4/0.6/lg2.47=3.85 N精/N=(Nmin)精/Nmin N精=N(Nmin)精/Nmin=14.53.85/7.56=7.4则精馏段实际板数为 7.4/0.7=10.6 取11块 故实际加料板位置为第12块板上.22解: (1) 由 y=x/

23、1+(-1)x=2.4x/(1+1.4x) 作y-x图由于精馏段有侧线产品抽出,故精馏段被分为上,下两段, 抽出侧线以上的操作线方程式: yn+1 =Rxn /(R+1)+xD /(R+1)=2/3xn +0.3 - (1)侧线下操作线方程推导如下: 以虚线范围作物料衡算 V=L+D1+D2Vys+1=Lxs+D1xD1+D2xD2 ;ys+1=Lxs/V +(D1xD1+D2xD2)/V=Lxs/(L+D1+D2)+(D1xD 1+D2xD2)/(L+D1+D2); L=L0-D2, 则:ys+1=(L0-D2)xs/(L0-D2+D1+D2)+(D1xD 1+D2xD 2)/(L0-D2+

24、D1+D2)=(R-D2/D1)xs/(R+1)+(xD1+D2xD2/D1)/(R+1)(R=L0/D1)将已知条件代入上式,得到: yS+1=0.5x+0.416 (2) 用图解法,求得理论塔板数为（5-1）块,见附图.22题附图23解: 根据所给平衡数据作x-y图.精馏段操作线 yn+1 =Rxn /(R+1)+xD /(R+1) =1.5xn /(1.5+1)+0.95/(1.5+1) =0.6xn +0.38q线方程与q线: 料液平均分子量:Mm=0.35+0.6518=22.9甲醇分子汽化潜热:r=252324.2=33868.8KJ/Kmol 水的分子汽化潜热:r=552184.

25、2=41731.2KL/Kmol 23题附图料液的平均分子汽化潜热: r=0.3533868.8+0.6541731.2=38979.4KL/Kmol 料液的平均分子比热 Cp=0.8822.94.2=84.6KL/Kmolq=r+Cp(ts-tF )/r=38979.4+84.6(78-20)/38979.4=1.13 q线斜率 q/(q-1)=1/13/0.13=8.7提馏段操作线方程与操作线：由于塔釜用直接蒸汽加热,故提馏段操作线过横轴上(xW ,0)一点,于是在x-y图上,作出三条线,用图解法所得理论板数为7.6块,可取8块(包括釜). 24解: 对全塔进行物料衡算: F1+F2=D+

26、W -(1) F1xF1+F2xF2=DxD +WxW 1000.6+2000.2=D0.8+W0.02 100=0.8D+0.02W -(2) 由式(1) W=F1+F2-D=100+200-D=300-D 代入式(2)得:D=120.5Kmol/h L=RD=2120.5=241kmol/h V=L+D=241+120.5=361.5Kmol/h 在两进料间和塔顶进行物料衡算,并设其间液汽流率为L,V,塔板序号为s. V+F1=D+L Vys+1+F1xF1=Lxs+DxD ys+1=(L/V)xs+(DxD -F1xF1)/V L=L+q1F1=241+1100=341Kmol/h V=

27、V=361.5 ys+1=(341/361.5)xs+(120.50.8-1000.6)/361.5 ys+1=0.943xs+0.125解: 对于给定的最大V,V=(R+1)D,回流比R愈小,塔顶产品量D愈大,但R 需满足产品的质量要求xD 0.98, 故此题的关键是求得回流比R.由题已知加料板为第14层,故精馏段实际板数为13层,精馏段板数为: 130.5=6.5取苯-甲苯溶液相对挥发度为=2.54 用捷算法求精馏段最小理论板数 (Nmin)精=ln0.98/0.02-0.5/0.5/ln2.54=4.175 y=N精馏段-(Nmin)精/(N精馏段+1)=(6.5-4.175)/(6.5

28、+1) =1.31 由y=0.75(1-x0.567) x=(1-Y/0.75)(1/0.567)=0.392=(R-Rmin)/(R+1) R=(0.392+Rmin)/(1-0.392) Rmin=(xD -ye)/(ye-xe) 对泡点进料xe=xF =0.5 ye=x/1+(-1)x =2.540.5/(1+1.540.5)=1.27/1.77=0.72 Rmin=(0.98-0.72)/(0.72-0.5)=0.26/0.22=1.18 R=(0.392+1.18)/(1-0.392)=1.572/0.608=2.59 D=V/(R+L)=2.5/(2.59+1)=0.696Kmol

29、/h 故最大馏出量为0.696Kmol/h26解:求n板效率: Emv =(yn -yn+1 )/(yn*-yn+1 ),因全回流操作,故有yn+1 =xn ,yn =xn-1与xn 成平衡的yn *=xn /1+(-1)xn =2.430.285/(1+1.430.285)=0.492于是: Emv=(xn-1 -xn )/(yn *-xn )=(0.43-0.285)/(0.492-0.285)=0.7求n+1板板效率: Emv=(yn+1 -yn+2)/(yn+1* -yn+2)=(xn-xn+)/(yn+1*-xn+1 ) yn+1 =2.430.173/(1+1.430.173)=0.337Emv=(0.2