二氧化硫转化率最优化王晓冰

1、二氧化硫转化率最优化化研1313 王晓冰 2013200186SO21/2O2=SO3，四段绝热反应器，级间间接换热。1.基础数据：混合物恒压热容Cp 0.2549 kcal/kgKH =23135 kcal/kmol催化剂堆密度b554 kg/m3进口SO2浓度8.0%，O2浓度9.0%，其余为氮气处理量131 kmol SO2/hr，要求最终转化率98 2.动力学方程：式中：3.基本要求：(1)在TX图上，做出平衡线，至少4条等速率线；(2)以一维拟均相平推流模型为基础，在催化剂用量最少的前提下，总的及各段的催化剂装量；进出口温度、转化率；并在T-X图上标出折线；4.讨论：(1)要求的最终

2、转化率从98变化到99对催化剂用量的影响；(2)YO2YSO221，SO2进口浓度在79之间变化，对催化剂装量的影响。计算过程：编写c+程序，如下：#include<math.h>#include<stdio.h>#define R 1.987#define h 0.00001double r(double x, double t, double xso2)double r,r1,r2,r3, B,keff,K, Kp,Pso2,Pso3,Po2; if(t>=693.15 && t<748.15)keff = 7.6915*pow(10,1

3、8)*exp(-76062 / (R*t); if(t>=748.15 && t<=873.15) keff = 1.5128*pow(10,7)*exp(-35992 / (R*t); K = 2.3*pow(10,-8)*exp(27200/(R*t); Kp=2.26203*pow(10,-5)*exp(11295.3 / t); Pso2=(xso2-xso2*x) / (1-xso2*x/2); Pso3=xso2*x / (1-xso2*x/2); Po2=(0.17-xso2-xso2*x/2) / (1-xso2*x/2); r1 =Po2*Pso2

4、/Pso3; r2 =Pso3/(Pso2*sqrt(Po2)*Kp); B =48148*exp(-7355.5/t); r3 =sqrt(B+(B-1)*(1-x)/x)+sqrt(K*(1-x)/x); r =keff*K*r1*(1-r2*r2)/(r3*r3); return(r);int main()double xso2=0.08,x,t,r0=0,y; int i; FILE *fp;fp=fopen("data.txt","w"); for(i=0;i<5;i+)fprintf(fp,"R=%en",r0);f

5、or(t=693.15;t<=873.15;)x=0.01;dox=x+h;if(x>=1) break;y=fabs(pow(10,5)*r(x,t,xso2)-pow(10,5)*r0);while(y>0.0001);fprintf(fp,"%5.2f %5.4fn",t,x);t=t+5;r0=r0+pow(10,-6);fclose(fp);return(0);运行上述程序后，获得的数据见下表：温度T/k反应速率0.0 1.0 ×10-62.0×10-6 3.0 ×10-64.0 ×10-6693.150.

6、9809 698.150.9797 703.150.9782 0.0117 708.150.9762 0.2719 713.150.9740 0.4750 718.150.9715 0.6205 0.1844 723.150.9686 0.7214 0.3863 0.0784 728.150.9655 0.7901 0.5373 0.2903 0.0638 733.150.9620 0.8365 0.6473 0.4536 0.2689 738.150.9582 0.8674 0.7265 0.5761 0.4285 743.150.9541 0.8873 0.7828 0.6669 0.55

7、00 748.150.9496 0.8995 0.8224 0.7336 0.6416 753.150.9447 0.8964 0.8237 0.7398 0.6529 758.150.9394 0.8927 0.8239 0.7444 0.6617 763.150.9338 0.8885 0.8231 0.7474 0.6685 768.150.9278 0.8837 0.8214 0.7491 0.6736 773.150.9213 0.8784 0.8188 0.7497 0.6772 778.150.9144 0.8725 0.8154 0.7492 0.6794 783.150.90

8、71 0.8662 0.8113 0.7477 0.6804 788.150.8994 0.8593 0.8065 0.7453 0.6803 793.150.8912 0.8519 0.8010 0.7421 0.6792 798.150.8826 0.8440 0.7948 0.7380 0.6772 803.150.8735 0.8357 0.7881 0.7332 0.6743 808.150.8640 0.8268 0.7807 0.7277 0.6706 813.150.8540 0.8175 0.7727 0.7215 0.6661 818.150.8436 0.8077 0.7

9、643 0.7146 0.6609 823.150.8327 0.7975 0.7552 0.7071 0.6550 828.150.8215 0.7868 0.7457 0.6990 0.6484 833.150.8097 0.7757 0.7357 0.6904 0.6412 838.150.7976 0.7642 0.7252 0.6812 0.6335 843.150.7851 0.7523 0.7142 0.6716 0.6252 848.150.7722 0.7400 0.7029 0.6614 0.6163 853.150.7590 0.7273 0.6912 0.6508 0.

10、6070 858.150.7454 0.7143 0.6791 0.6398 0.5972 863.150.7315 0.7010 0.6666 0.6285 0.5870 868.150.7174 0.6874 0.6539 0.6167 0.5764 873.150.7030 0.6736 0.6408 0.6047 0.5654 根据上述数据制图： 上图以温度为横坐标，转化率为纵坐标。5个速率，0（对比，即反应平衡时）、1×10-6、2×10-6、3×10-6、4×10-6，作T-X的等速率线。因为是一维拟均相平推流，则；根据入口组成，设定入口温度

11、，根据反应对入口温度所求偏导数在这一段内对组成的积分为零，可以求得此段出口转化率，即下一段的入口组成。催化剂用量最少时应使得；由于段内操作线的斜率为，因此根据入口温度、入口组成和出口组成计算可得出口温度。循环计算四段后可得最后的出口转化率，若不能满足要求（x98），则重新设定第一段入口温度，再进行计算直至满足条件。编写c+程序，如下：#include<stdio.h>#include<math.h>#define R 1.987#define h 0.0001double r(double x, double t, double xso2)double r,r1,r2,

12、r3,keff,K,B,Kp,Pso2,Pso3,Po2; if(t>=693.15 && t<748.15) keff = 7.6915*pow(10,18)*exp(-76062 / (R*t); if(t>=748.15 && t<=873.15) keff = 1.5128*pow(10,7)*exp(-35992 / (R*t); K = 2.3*pow(10,-8)*exp(27200/(R*t); Kp=2.26203*pow(10,-5)*exp(11295.3 / t); Pso2=(xso2-xso2*x) / (1-

13、xso2*x/2); Pso3=xso2*x / (1-xso2*x/2);Po2=(0.17-xso2-xso2*x/2) / (1-xso2*x/2); r1 =Po2*Pso2/Pso3; r2 =Pso3/(Pso2*sqrt(Po2)*Kp); B =48148*exp(-7355.5/t); r3 =sqrt(B+(B-1)*(1-x)/x)+sqrt(K*(1-x)/x); r =keff*K*r1*(1-r2*r2)/(r3*r3); return(r);double dr(double x, double t, double xso2)double y; y=(r(x,t+h

14、,xso2)-r(x,t-h,xso2)/(2*h); return(y);double t(double t0, double x0, double x)double y,lamda,H=-23135,Cp=254.9,rou=0.500,c=1.282; lamda=-H*c/(rou*Cp); y=t0+lamda*(x-x0); return(y);double fun1(double x, double t, double xso2) double y; y=-dr(x,t,xso2)/(r(x,t,xso2)*r(x,t,xso2); return(y);double jifen(

15、double x0, double t0, double xso2) double sum=0.0,x1=x0,x2,t1,t2=693.15,xout; do t1=t(t0,x0,x1); x2=x1+h/10; t2=t(t0,x0,x2); if(t2>873.15) xout = x1; goto end; sum=sum+h*(fun1(x1,t1,xso2)+fun1(x2,t2,xso2)/20; x1=x2; while(sum<0);xout=x1-h/10;end: return(xout);double wjifen(double xin, double x

16、ou, double tin, double xso2)double y,x1=xin,x2,t1,t2,sum=0.0,wcat;do t1=t(tin,xin,x1); x2=x1+h; t2=t(tin,xin,x2); if(t2>=873.15) goto end ; sum=sum+(1/r(x1,t1,xso2)+1/r(x2,t2,xso2)*h/1000; x1=x2; while(x2<=xou);end: wcat=sum*131*1000/3600; return(wcat);int main()double xso2=0.08,xout,tout,x0=0

17、.0001,t0,t00=719,t1,wcat,wsum;int i,j;loop1:wsum=0.0;x0=0.0001;t0=t00;FILE*fp;fp=fopen("data2.txt","w");fprintf(fp,"1 tin=%f xin=%en",t00,x0);for(i=0;i<=3;i+)xout=jifen(x0,t0,xso2);tout=t(t0,x0,xout);fprintf(fp,"%d tout=%f ",i+1,tout);fprintf(fp,"xout=

18、%f n",xout);wcat=wjifen(x0,xout,t0,xso2);fprintf(fp,"Wcat=%f ",wcat);wsum=wsum+wcat;t1= 693.15;dot1=t1+0.01;while(fabs(pow(10,5)*r(xout,t1,xso2)-pow(10,5)*r(xout,tout,xso2)>h);x0=xout;t0=t1;fprintf(fp,"%d tin= %f xin= %fn",i+2,t0,xout);fprintf(fp,"Wsum=%f n",wsu

19、m);t00=t00-0.1;fprintf(fp,"nn");if(x0<=0.98) goto loop1;return(0);所得数据如下：第一节反应器： tin=717.800000 xin=1.000000e-004 tout=873.148833 xout=0.667660 Wcat=8349.667114（kg） 第二节反应器： tin= 723.600000 xin= 0.667660 tout=778.519894 xout=0.903660 Wcat=11605.588273（kg） 第三节反应器： tin= 715.920000 xin= 0.9

20、03660 tout=729.505692 xout=0.962040 Wcat=21557.219801（kg） 第四节反应器： tin= 693.190000 xin= 0.962040 tout=697.388114 xout=0.980080 Wcat=50379.460102（kg） 总催化剂：Wsum=91891.935289（kg） 作折线图如下：讨论：（1） 将第二个程序改为99%，因为在方程给出的最低温度693.15K时转化率为98.38%，无法达到99%，所以得不出99%转化率时的催化剂用量。（2） 程序如下（举例，YSO2=7%）：#include"math.h

21、"#include"stdio.h"#define R 1.987#define h 0.0001double r(double x,double t,double xso2)double r,r1,r2,r3,keff,K,B,Kp,Pso2,Pso3,Po2; if(t>=693.15 && t<748.15) keff = 7.6915*pow(10,18)*exp(-76062 / (R*t); if(t>=748.15 && t<=873.15) keff = 1.5128*pow(10,7)*ex

22、p(-35992 / (R*t); K = 2.3*pow(10,-8)*exp(27200/(R*t); Kp=2.26203*pow(10,-5)*exp(11295.3 / t); Pso2=(xso2-xso2*x) / (1-xso2*x/2); Pso3=xso2*x / (1-xso2*x/2); Po2=(0.21-xso2-xso2*x/2) / (1-xso2*x/2); r1 =Po2*Pso2/Pso3; r2 =Pso3/(Pso2*sqrt(Po2)*Kp); B =48148*exp(-7355.5/t); r3 =sqrt(B+(B-1)*(1-x)/x)+sq

23、rt(K*(1-x)/x); r =keff*K*r1*(1-r2*r2)/(r3*r3); return(r);double dr(double x,double t,double xso2)double y; y=(r(x,t+h,xso2)-r(x,t-h,xso2)/(2*h); return(y);double t(double t0,double x0,double x)double y,lamda,H=-23135,Cp=254.9,rou=0.500,c=1.282; lamda=-H*c/(rou*Cp); y=t0+lamda*(x-x0); return(y);doubl

24、e fun1(double x,double t,double xso2) double y; y=-dr(x,t,xso2)/(r(x,t,xso2)*r(x,t,xso2); return(y);double jifen(double x0,double t0,double xso2) double sum=0.0,x1=x0,x2,t1,t2=693.15,xout; do t1=t(t0,x0,x1); x2=x1+h/10; t2=t(t0,x0,x2); if(t2>873.15) xout = x1; goto end; sum=sum+h*(fun1(x1,t1,xso2

25、)+fun1(x2,t2,xso2)/20; x1=x2; while(sum<0);xout=x1-h/10;end: return(xout);double wjifen(double xin,double xou,double tin,double xso2)double y,x1=xin,x2,t1,t2,sum=0.0,wcat;do t1=t(tin,xin,x1); x2=x1+h; t2=t(tin,xin,x2); if(t2>=873.15) goto end ; sum=sum+(1/r(x1,t1,xso2)+1/r(x2,t2,xso2)*h/1000;

26、x1=x2; while(x2<=xou);end: wcat=sum*131*1000/3600; return(wcat);int main() double xso2=0.07,xout,tout,x0=0.0001,t0,t00=717,t1,wsum=0.0,wcat; double xou5,tou5,tin5,xin5; int i,j; FILE*fp; fp=fopen("data4.txt","w");loop1: x0=0.0001; t0=t00; i=0; xini=x0; tini=t00;loop2: xout=jifen(x0,t0,xso2); tout=t(t0,x0,xout); xoui=xout; toui=tout;t1= 693.15;do t1=t1+0.01; while(fabs(pow(10,5)*r(xout,t1,xso2)-pow(10,5)*r(xout,tout,xso2)>h);x0=xout; t0=t1;i+; xini=xout; tini=t0; if(i<=3) goto loop2;t00=t00-0.1;if(x0<=0.98) goto loop1; fprintf(fp,"