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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：In CCA0 =kt,CA0A0CA =0.7 , Ca=0.3Ca0间歇釜中.In-10.3= 13k,k=0.0926 min在全混釜中TVRCA0 XA0.7-1=V0 = k CA = 0.3 0.0926 =252 讪空速 5=2 = =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 P气相反应，已知进料中含30%A(mol
2、)，其余70%惰性物料.加料流量为6.3mol/s.该反应的动力学方程为j=0.27Camol/m3 s,要求达到95%专化.试求所需的空时？反应器容积？解:VR 1 lnT p = V0=k 山番=1 Inn rXA = 027 In 1 0.95 =11.1 S Vr =T P-V0= t PCA0CA0而 Ca0= R= 0 舄红=0.0198mol/L=19.8mol/mVr=11.1sX6.3mol /s19.8mol /m3=3.53m33、平推流变容过程一级有一纯丙烷裂解反应方程式为Gn QH+CH.该反应在 772 C等温条件下进行，其动力学方程式为-dPMdt=kP a,忽略
3、存在的副反应 ，并已知 k=0.4h-1反应过程保持恒压 0.1MPa. 772C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：Tg A= 221 =0.5 kT F=- (1 +e a) In (1- X A)- a X Af0.4 t p= (1+0.5) In (1-0.5)-0.5 X 0.51.51 n2 0.250.4=1.974hVr= T PV0=1.974X 800=1579L=1.579 m3 4、间歇釜变容一级一级气相反应 22R+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
4、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 S a=2膨胀率 A=y A0 S A=0.75 X 2=1.5对应转化率 沦的反应体积 V=V 0(1 + s A X A)所以，V 1X A=w =黄=66.7%A5K=? ln 一=8 h 7 =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+B t C,在298K下的动力学方程式为a=0.6CaCb mol/(L.min),该反应的进料速率为v0 =0.018m /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？Vr
5、 _ CA0XAf _ CA0XAf _ CA0XAf _CA0 XAf_解:=厂=22 = T mVrAkCACBkCA2kCA02 (1 xAf )2T m =0.6 0.：(9 0.9)2 =150 min Vr=vo t n=0.018 m3/mi nx 150min=2.7 m3 6、多釜串联液相二级33某一液相反应 A+4R+S，其速率常数k=9.92m/(Kmol KS),初始浓度为0.08Kmol/m , 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:T 沁-CA0 CA1 -CA0XA1V)T 2=VR
6、2 =C-= CA0(XA2 XA1)V0rA2kCA02 (1 XA2)2所以,整理有 试差解得T 1= T 2两釜相同xA2 xA 1(1 XA1)2(1 XA2)2 2(1-0.875)xa1=(0.875- xX a1=0.7251而 XA2 =0.8752A1)(1- XA1)所以，Vr=S冷二遊霊需山=416 m3总反应器体积 Vr=2Vr1=2 x 4.16=8.33 m37. 【自催化反应优化】自催化反应 A+RtR+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+
7、R t 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 = C Rm =0.5(Ca0 + Cr0)=O.5Co时速度最快。XA1XA0Ca0 C A1Ca1 C A2-Fa0rA1rA2rA或1CA0 dCACA1 dCArACA12CA2rA此时的XA为全混釜出口最佳转化率8. An aqueous feed of A and B (4001/mi n,100mmolA/l100mmolB/l) is to beconv erted to product in a play flow reactor ,
8、the kin etics of the react ion is represe nted isA + B R-rA = 200CaCb mol/l*mi nFind the volume of reactor n eeded for 99.9% conv ersi on of A to productAn swer :kT = 1/Ca -1/Ca0k T CA0 = xAf /1- xAf = 0.999/1-0.999 = 999.K = 200l/mol *mi nT = VR/V0 = 1/k Ca0 = 1/200*0.1 = Vr/400Vr = 20L9. A gaseous
9、 feed of pure A (2mol/l 100mol/m in) decomposes to give a variety of product in a plug flow reactor The kinetics of the conversion is represented byA 2.5B-rA = 10(mi n-1)Find the expected conversion in a 22-liter reactor-1Answer: Ca0 = 2 mol/l Fa0 = 100 mol/min -rA = 10(min ) Cat = Vr/V0 = 噩二 Vr Ca0
10、 / Fa0 = 22*2/100 = 0.44minCA0A -10*0 44“-4 4x a = 1 _e 0.44 = 1 e10. 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 theconcen trati on in the exit stream of the sec ond reactor .The reacti on
11、 is sec ond-order with respect to A and V/V1 = 2An swer : -rA = Kca2A 0V1 字 翼斗1A10A1A0A2 V1XA1C A0.kV11 0.5-V00.52=2二 v7 2( vT)CC0 . 5C A 22*2 = ACT = C A2 2=424CA2 + CA2 -0.5=01Ca2 = 2*4 (-1 14*4*0.5)1Ca2 = 8 (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
12、转化所需的空时和空速？ （20分）解：In JAkt k-1 1 xAf13 1 0.8 0.124min活塞流:T - 1 1 XAft - 13min空速:s - 1/t - 1/13 - 0.077min-1全混流:1XAf1.8T k 1 XAf0.124 1 0.82. 自催化反应 A+R f R+R,速度方程为-r二kCACR，体系总浓度为C。二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 f R+R-rA =kCACR Co =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
13、当 CAm = CRm = 0.5（Ca0 + Cro） = 0.5Co 时速度最快。 XA1 XA0Ca0 CA1Ca1 CA2 Fa0rA1rA2U1A当总空时T = T 1+T 1最小时反应器体积最小。c对 T = C2占+詈 对Cai求导时T最小时Vr最少ACA2 A此时的Xa为全混釜出口最佳转化率。3. An aqueous feed of A and B (4001/mi n,100mmolA/l100mmolB/l)is to be converted to product in a play flow reactor ,the kinetics of the reacti on
14、 is represe nted isA + B t R-a = 200CaCb mol/l*minFind the volume of reactor n eeded for 99.9% conversion of A to productAn swer :k t = 1/Ca -1/Caok t Cao = XAf /1- XAf = 0.999/1-0.999 = 999.K = 2001/mol *mint = Vr/Vo = 1/k Cao = 1/200*0.1 = Vr/400Vr = 20L4. A gaseous feed of pure A (2mol/l 100mol/m in) decomposes to give a variety of product in a plug flow reactor The kin etics of the conversion is represe nted byA f 2.5B-rA = 10(min-1)Find the expected conversion in a 22-liter reactorAn swer: Cao = 2 mol/lFao = 100 mol/min Ta =
