第三章 理想流动均相反应器设计题解

1、. . . 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:=kt, =0.7 ， CA=0.3CA0间歇釜中0.3=13k， k=0.0926 min-1 在全混釜中=25.2 min-1空速S=0.0397min-12、平推流恒容一级有一个活塞流管式反应器于555K,0.3MPa压力下进行AP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6.3mol/s.该反应的动力学方程为rA=0.27CA mol/m3s,要求达到95%转化.试求所需的空时? 反应器容积?解: P =11.1 SVR =Pv0=P而CA0=0.0198mol/L=19.8mol/m3VR=11.1s=3.53m33、平推流变容过程一级有一纯丙烷裂解反应方程式为C3H8C2H4+CH4.该反应在772等温条件下进行,其动力学方程式为-dPA/dt=kPA,忽略存在的副反应,并已知k=0.4h-1 反应过程保持恒压0.1MPa

2、. 772和0.1MPa下的体积进料量为800L/h,求转化率为0.5时所需的平推流反应器的体积.解: A=0.5kP=(1A)(1-A)- AAf0.4P=(1+0.5)(1-0.5)-0.50.5P=1.974hVR=Pv0=1.974800=1579L=1.579 m34、间歇釜变容一级一级气相反应A2R+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 A=2膨胀率 A=yA0A=0.752=1.5对应转化率XA的反应体积 V=V0(1AA)所以，A=66.7%K=0.0735 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+BC,在298K下的动力学方程式为rA=0.6CACB mol/(L.min),该反应的进料速率为0 =0.018m3/min.A,B 的初始浓度相同,均为0.1mol/L,要求出口的转化率为90%,求需多大的全混釜?解: =mm =150 minVR=v0m=0.018 m3/min150min=2.7

3、 m36、多釜串联液相二级某一液相反应A+BR+S,其速率常数k=9.92m3/(KmolKS),初始浓度为0.08Kmol/m3 ，在两个等体积的全混釜中进行反应,最终出口转化率0.875.进料体积流量为0.278m3/KS .求全混釜的总体积?解： 1= 2= 1=2 两釜相同所以， =, 而xA2 =0.875整理有 (1-0.875)2 xA1=(0.875- xA1)(1- xA1)2试差解得 xA1=0.7251所以，VR1=4.16 m3总反应器体积 VR=2VR1=24.16=8.33 m37.【自催化反应优化】自催化反应 A+RR+R，速度方程为-r=kCACR，体系总浓度为C0= CA+CR 。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 -rA =kCACR C0 =CA + CR. 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CAm = CRm = 0.5(CA0 + CR0) = 0.5C0 时速度最快。 或 当总空时= 1+1 最小时反应器体积最小。 对 = + 对

4、CA1求导时最小时VR最少。 此时的xA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 is A + B R -rA = 200CACB mol/l*minFind the volume of reactor needed for 99.9% conversion of A to product Answer : k= 1/CA 1/CA0 kCA0 = xAf /1- xAf = 0.999/1-0.999 = 999. K = 200l/mol *min = VR/V0 = 1/k CA0 = 1/200*0.1 = VR/400 VR = 20L9. A gaseous feed of pure A (2mol/l 100mol/min) decomposes to give a var

5、iety of product in a plug flow reactor The kinetics of the conversion is represented by A 2.5B -rA = 10(min-1) Find the expected conversion in a 22-liter reactor Answer: CA0 = 2 mol/l FA0 = 100 mol/min -rA = 10(min-1) CA= VR/V0 = = VR CA0 / FA0 = 22*2/100 = 0.44minx A = 1 e-10*0.44 = 1 e-4.4 10、 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 0.5mol/l find the concentration in the exit stream of the seco

6、nd reactor .The reaction is second order with respect to A and V2/V1 = 2Answer : -rA = KcA2 =22*2 =44CA22 + CA2 0.5 = 0CA2 = (-1 )CA2 = (3-1) = 1/4 = 0.25 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： k=min-1 活塞流：= t = 13min 空速 ：s = 1/= 1/13 = 0.077min-1 全混流：= = 32.3min2. 自催化反应 A+RR+R，速度方程为-r=kCACR，体系总浓度为C0= CA+CR 。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 -rA =kCACR C0 =CA + CR. 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CAm = CRm = 0.5(CA0 + C

7、R0) = 0.5C0 时速度最快。 或 当总空时= 1+1 最小时反应器体积最小。 对 = + 对CA1求导时最小时VR最少。 此时的xA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 -rA = 200CACB mol/l*minFind the volume of reactor needed for 99.9% conversion of A to product Answer : k= 1/CA 1/CA0 kCA0 = xAf /1- xAf = 0.999/1-0.999 = 999. K = 200l/mol *min = VR/V0 = 1/k CA0 = 1/200*0.1 = VR/400 VR = 20L4. A gaseous feed of p

8、ure A (2mol/l 100mol/min) decomposes to give a variety of product in a plug flow reactor The kinetics of the conversion is represented by A 2.5B -rA = 10(min-1) Find the expected conversion in a 22-liter reactor Answer: CA0 = 2 mol/l FA0 = 100 mol/min -rA = 10(min-1) CA= VR/V0 = = VR CA0 / FA0 = 22*2/100 = 0.44minx A = 1 e-10*0.44 = 1 e-4.4 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 0.5mol/l find the concentration in the exit stream of the second reactor .The reaction is second order with respect to A and V2/V1 = 2Answer : -rA = KcA2

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