第三章 理想流动均相反应器设计题解1、[间歇反应器与全混釜恒容一级]有一等温操作的间歇反应器进行某一级液相反应，13分钟后,反应物转化了70%.今拟将此反应转至全混流反应器,按达到相同的转化率应保持多大的空速解:㏑CA 0CA =kt, CA0CACA0- = ， C A = 间歇釜中∴㏑=－13k ， k= min-1在全混釜中τ=VR V0=CA0 XA k CA =0.70.30.0926⨯= min -1∴空速S=1τ=125.2=2、[平推流恒容一级]有一个活塞流管式反应器于555K,压力下进行A →P 气相反应,已知进料中含30%A(mol)，其余70%为惰性物料.加料流量为s.该反应的动力学方程为r A = mol/m 3·s,要求达到95%转化.试求⑴所需的空时 ⑵反应器容积解: τP =VR V0=1k ㏑CA 0CA =1k ㏑PA0PA =1k ㏑A0Ay y =1k ㏑11Ax -=10.27㏑110.95-= S∴V R =τP ·v 0=τP 00A A FC而C A0=0A P RT=30.30.082555⨯⨯=L=m 3V R =×36.3/19.8/mol s mol m =3.53m 33、[平推流变容过程一级]有一纯丙烷裂解反应方程式为C 3H 8→C 2H 4+CH 4.该反应在772℃等温条件下进行,其动力学方程式为-dP A /dt=kP A ,忽略存在的副反应,并已知k= 反应过程保持恒压.772℃和下的体积进料量为800L/h,求转化率为时所需的平推流反应器的体积.解: ∵εA =212-=∵k τP =－(1＋εA )㏑(1-ΧA )- εA ΧAfτP =－(1+㏑×∴τP =1.5ln 20.250.4-=V R =τP v 0=×800=1579L=1.579 m 34、[间歇釜变容一级]一级气相反应A →2R+S ,在等温等压间歇实验反应器中进行,原料中含75%A(mol),25%(mol)惰性气体,经8分钟后,其体积增加一倍.求此时达到了多大的转化率 速率常数多大 解: 膨胀因子 δA =3-11=2膨胀率 εA =y A0δA =×2=对应转化率X A 的反应体积 V=V 0(1＋εA ΧA ) 所以，ΧA =V V 0A1ε-=2-11.5=%K=1t ㏑11Ax -=18㏑110.667-= min -15、[全混流恒容二级反应]在全混流反应器中进行液相均相二级反应:A+B →C,在298K 下的动力学方程式为r A = mol/,该反应的进料速率为ν0 =0.018m 3/,B 的初始浓度相同,均为L,要求出口的转化率为90%,求需多大的全混釜 解:R 0V v =A0Af AC x r =A0Af A BC x kC C =A02Af A C x kC =A0220(1)AfA Af C x kCx -=τmτm =20.90.60.1(10.9)⨯-=150 min∴V R =v 0τm =0.018 m 3/min ×150min= m 36、[多釜串联液相二级]某一液相反应A+B →R+S,其速率常数k=(Kmol ·KS),初始浓度为m 3，在两个等体积的全混釜中进行反应,最终出口转化率.进料体积流量为0.278m 3/KS .求全混釜的总体积 解： τ1=10R V v =011A A C C r -=012201(1)A A A A C x kCx -τ2=20R V v =122A A A C C r -=021222()(1)A A A A A C xx kC x--∵ τ1=τ2 两釜相同所以， 121(1)A A x x -=2122(1)AA A xx x--, 而x A2 =整理有 2x A1= x A1)(1- x A1)2试差解得 x A1=所以，V R1=0012201(1)A A A A v C xkC x ⨯-=20.2780.75219.920.08(10.7521)⨯⨯⨯-= m 3总反应器体积 V R =2V R1=2×= m 37.【自催化反应优化】自催化反应 A+R →R+R ，速度方程为-r=kC A C R ，体系总浓度为C 0= C A +C R 。
若给你一个管式反应器和一个釜式反应器，为满足同一生产要求怎样联结设备费较少（5分）解：A+R → R+R -r A =kC A C R C 0 =C A + C R . 串联连接，管式反应器加釜式反应器 速度较快，同样转化时所用的体积较小。
只有当C Am = C Rm = (C A0 + C R0) = 0.5C 0 时速度最快。
∵AA A a r x x F --∇=010A A a r C C --=101τ AA a r C C --=212τ或 ⎰-=011A A AA C Cr dCτ ⎰-=122A A AA C Cr dCτA r -Am C A CAr 1-1+1 对 τ=AA a r C C --10+ ⎰-12A A AA C Cr dC对C A1求导时τ最小时V R 最少。
此时的x A 为全混釜出口最佳转化率。
8. An aqueous feed of A and B (400l/min,100mmolA/l 100mmolB/l) is to be converted to product in a play flow reactor ,the kinetics of the reaction is represented isA +B → R -r A = 200C A C B mol/l*minFind the volume of reactor needed for % conversion of A to product Answer :k τ= 1/C A – 1/C A0k τC A0 = x Af /1- x Af = = 999. K = 200l/mol *minτ= V R /V 0 = 1/k C A0 = 1/200* = V R /400 V R = 20L9. A gaseous feed of pure A (2mol/l 100mol/min) decomposes to give a variety of product in a plug flow reactor The kinetics of the conversion is represented byA → -r A = 10(min -1)Find the expected conversion in a 22-liter reactorAnswer: C A0 = 2 mol/l F A0 = 100 mol/min --r A = 10(min -1) C A τ= V R /V 0 =0A C A F R V = V R C A0 / F A0 = 22*2/100 =x A = 1 – e -10* = 1 –10、A liquid reactant stream (mol/l) passes through two mixed flowreactors in series The concentration of A in the exit of the first reactor is l find the concentration in the exit stream of the secondreactor .The reaction is second –order with respect to A and V 2/V 1 = 2Answer : -r A = Kc A 2 0=A ε 111A A r x V V-= 21202A A A A r x x F V--=∴210101a A A kC C x V V =∴2015.05.01-=V V k =2∴==)(20102V V k V V k 2*2 =2221A A A C C C -=2225.0A A C C -=44C A22 + C A2 – = 0 C A2 =4*21(-1 ±5.0*4*41+)C A2 = 81(3-1) = 1/4 = mol/l1. 在等温操作的间歇反应器中进行一级液相反应，13分钟反应物转化了80%，若把此反应移到活塞流反应器和全混流反应器中进行，达到同样转化所需的空时和空速（20分）解：㏑kt Ax =-11 k=124.08.011131111==--λλAfx t min -1 活塞流：τ= =-Afxk 111λ t = 13min 空速 ：s = 1/τ= 1/13 =全混流：τ= 8.018.0124.0111--=AfAfx x k =2. 自催化反应 A+R →R+R ，速度方程为-r=kC A C R ，体系总浓度为C 0= C A +C R 。
若给你一个管式反应器和一个釜式反应器，为满足同一生产要求怎样联结设备费较少（5分）解：A+R → R+R -r A =kC A C R C 0 =C A + C R . 串联连接，管式反应器加釜式反应器 速度较快，同样转化时所用的体积较小。
只有当C Am = C Rm = (C A0 + C R0) = 时速度最快。
∵AA A a r x x F --∇=01AA a r C C --=11τ AA a r CC --=212τ或 ⎰-=011A A A A C Cr dCτ ⎰-=122A A A A C Cr dCτA r -Am AAr 1-Am x Afx∴当总空时τ= τ1+τ1 最小时反应器体积最小。
对 τ= AA a rCC --10+ ⎰-12A A AA C C r dC对C A1求导时τ最小时V R 最少。
此时的x A 为全混釜出口最佳转化率。
3. An aqueous feed of A and B (400l/min,100mmolA/l 100mmolB/l) is to be converted to product in a play flow reactor ,the kinetics of the reaction is represented is A + B → R -r A = 200C A C B mol/l*minFind the volume of reactor needed for % conversion of A toproduct Answer :k τ= 1/C A – 1/C A0 k τC A0 = x Af /1- x Af = = 999. K = 200l/mol *minτ= V R /V 0 = 1/k C A0 = 1/200* = V R /400 V R = 20L4. A gaseous feed of pure A (2mol/l 100mol/min) decomposes to give a variety of product in a plug flow reactor The kinetics of the conversion is represented byA → -r A = 10(min -1)Find the expected conversion in a 22-liter reactor Answer: C A0 = 2 mol/l F A0 = 100 mol/min --r A = 10(min -1) C A τ= V R /V 0 =0A C A F R V = V R C A0 / F A0 = 22*2/100 =x A = 1 – e -10* = 1 –5. A liquid reactant stream (mol/l) passes through two mixed flow reactors in series The concentration of A in the exit of the first reactor is l find the concentration in the exit stream of the second reactor .The reaction is second –order with respect to A and V 2/V 1 = 2Answer : -r A = Kc A 2 0=A ε 111A A r x V V -= 21202A A A A r x x F V--=∴210101a A A kC C x VV= ∴2015.05.01-=V V k =2∴==)(20102V V k V V k 2*2 =2221A A A C C C -=2225.0A A C C -=44C A22 + C A2 – = 0C A2 = 4*21(-1 ±5.0*4*41+) C A2 = 81(3-1) = 1/4 = mol/l第三章均相反应器习题1、在等温操作的间歇反应器中进行一级液相反应，13分钟反应物转化了80%，若把此反应移到活塞流反应器和全混流反应器中进行，达到同样转化所需的空时和空速2、在平推流反应器进行液相均相反应:A+B →R(目的) r R = CA,Kmol/m 3·h CA 0=2 Kmol/m 32A →D r D = CA 2,Kmol/m 3·h 求⑴A 的转化率95%时所用的时间 ⑵此时R 的收率⑶若处理量为200L/min 时 V R =3在反应体积为1m 3的间歇操作釜式反应器中，环氧丙烷的甲醇溶液与水反应生成丙二醇 H 2COCHCH 3+H 2OH 2COHCHOHCH 3该反应对环氧丙烷为一级，反应温度下的速率常数为，原料液中环氧丙烷的浓度为m 3，环氧丙烷的最终转化率为90%。