化工原理作业解析(精品)

单击此处编辑母版标题样式,单击此处编辑母版文本样式,第二级,第三级,第四级,第五级,*,过程工程原理第,5,章作业解析,5-9.,拟设计一常压填料吸收塔,用清水处理,3000m,/h,含,5%NH,3,的空气,要求,NH,3,的回收率为,99%,取塔底空塔气速,1.1m/s,实际用水量为最小用水量的,1.5,倍,.,已知塔内操作温度为,25,平衡关系为,y=1.3x,气相体积传质系数,Ky,a,为,270kmol/m,h,试求,:(a),用水量和出塔溶液的浓度,;(b),填料层高度,;(c),若入塔水中已含氨,0.1%,所需填料层高度可随意增加,能否达到,99%,的回收率,?,解,:,(a):,塔底,塔顶的气,液相组成,:,y,1,=5%=0.05,y,2,=y,1,(1-y)=5%,1%,=0.005,x,2,=0,由物料守恒可得,:L(x,1,-x,2,)=G(y,1,-y,2,),则,:L/G=(y,1,-y,2,)/(x,1,-x,2,),当,x,1,=,x,1,*,时,用水量最小,.,即,:,(L/,G),min,=(y,1,-y,2,)/(x,1,*-x,2,)=(y,1,-y,2,)/,(y,1,/m)-0,=(0.05-0.0005)/(0.05/3)=1.29,L,min,=1.29G,又,G=(3000,3600m,/s,)/(30003600m,)=1.1m/,m,s,= (,1.1m/,m,s)/(22.4L/mol)=0.049kmol/m,s,用水量,L=1.5L,min,=1.51.290.049=0.095kmol/m,s,此时,(L/G)=(y,1,-y,2,)/x,1,所以出塔溶液的浓度,:,x,1,=(0.05-0.0005)/(0.095/0.049)=0.0256,(b):,塔顶,:,y,2,=y,2,-y,2,*=y,2,-mx,2,=y,2,=0.0005,塔底,: y,1,=y,1,-y,1,*=y,1,-mx,1,=0.05-1.30.0256=0.0167,所以,: ,y,m,=(y,1,- y,2,)/,(,y,1,/y,2,) =(0.0167-0.0005)/,(0.0137/0.0005) =0.0046 h,0,=G/Ky,a,(y,1,-y,2,)/,y,m,=0.049/(270/3600),(0.05-0.0005)/0.0046m =7.03m,(c): L/G=,0.05/95%-0.05(1-98%)/95%/0.0256=2.02,s=1.3/2.021,操作线在塔顶位置与平衡线相交。,X,2,=0.1%,y,2,min,=1.30.1%=0.0013,故,y,max,=,(0.05/0.95)-0.0013/(0.05/0.95)=97.5%99%,故不能达到。,5-10.,试求例,5-6,中的填料塔，其每平方米截面积共可回收多少丙酮,(,以,kg/h,计,)?,若将原设计的填料层高度减少,1/3,回收量会减少多少,?,解,:,丙酮的回收量,:,M,=98%y,1,G=98%,3%0.02kmol/m,s =5.88,10,kmol/m,s=122.8kg/,m,h,填料层的减少即减少了传质单元数,h,0,=H,OG,N,OG, h,0,=H,OG,N,OG,N,OG,/N,OG,=,(1-s),(y,1,-mx,2,/y,2,-mx,2,)+s,/,(1- s),(y,1,-mx,2,/y,2,-mx,2,)+s,=2/3,又,s=,m/(l/v,)=0.714,(1-s),(y,1,/y,2,)+s,= 2/3,(1-s),(y,1,/y,2,)+s,解得,:y,2,=0.0016,则回收量减少,=0.02(0.0016- 0.0006) =0.00002,kmol/m,s =4.18kg/,m,h,5-11.,若例,5-6,填料塔出塔水中的丙酮为,80%,饱和,其于数据不变,求所需的水量及填料层高度,?,解,:,此时塔底,塔顶的气液相组成为,: y,1,=0.03, y,2,=y,1,(1-y)=0.03(1-0.98)=0.0006 x,2,=0, x,1,=80%x*,1,=80%y,1,/m=80%0.03/1.75=0.0137,则塔底,:,y,1,=y,1,-y,1,*=y,1,-mx,1,=0.03-1.75,0.0137=0.006,塔顶,:,y,2,=y,2,-y,2,*=y,2,-mx,2,=y,2,=0.0006,所以,: ,y,m,=(y,1,- y,2,)/,(,y,1,/ y,2,) =(0.006-0.0006)/,(0.006/0.0006)=0.0023 h,0,=,G,/,Ky,a,(y,1,-y,2,)/,y,m,=(0.02,/0.016)(0.03-0.0006)/0.0023m=15.98m,由物料守恒有：,G(y,1,-y,2,)=L(x,1,-x,2,) L=G(y,1,-y,2,)/(x,1,-x,2,)=0.02(0.03-0.0006)/(0.0137-0) =0.0429kmol/m,s,5-12.,在填料塔内用稀硫酸吸收空气中的氨,当溶液中存在游离酸时,氨的平衡分压为零,.,下列三种情况下的操作条件基本相同,试求所需填料高度的比例,:(a),混合气含氨,1%,要求吸收率为,90%;(b),混合气含氨,1%,要求吸收率,99%;(c),混合气含氨,5%,要求吸收率为,99%;(,对上述低浓度气体,=(,y,b,-y,a,)/y,b,),。,解,:,(a): y1=1%=0.01;y2=(1-,)y1=10%1%=0.001 (b): y1=1%=0.01;y2=(1-,)y1=1%1%=0.0001 (c): y1=5%=0.05;y2=(1-,)y1=1%5%=0.0005 ,当溶液中存在游离酸时,P*,NH,3,=0(P*,A,=EX,A,),则此时氨在稀硫酸中的摩尔分数,X,A,=0,吸收过程中,溶液中始终存在游离,H,2,SO,4,所以有,x,1,=x,2,=0,y,1,=y,1,-y*,1,=y,1,-mx,1,=y,1,;,y,2,=y,2,-y*,2,=y,2,-mx,2,=y,2,N,OG,=(y,1,-y,2,)/,y,m,=(y,1,-y,2,)(y,1,/y,2,)/(y,1,-y,2,)=(y,1,/y,2,),ha:hb:hc,=(G/Ky,a,N,OG(a),):(G/Ky,a,N,OG(b),):(G/,Ky,a,N,OG(c,),=(0.01/0.001):(0.01/0.0001): (0.05/0.0005) = 10: 100: 100 =1:2:2,即上述三种情况下,所需填料高度之比为,:1:2:2,。,