大作業(yè)二氧化硫轉(zhuǎn)化率最優(yōu)化(共10頁)

1、精選優(yōu)質(zhì)文檔-傾情為你奉上二氧化硫轉(zhuǎn)化率最優(yōu)化化研0801 馮鶴立 題目：SO21/2O2=SO3，四段絕熱反應(yīng)器，級(jí)間間接換熱。1.基礎(chǔ)數(shù)據(jù)：混合物恒壓熱容Cp 0.2549kcal/kg·KH =23135kcal/kmol床層空隙率y554kg/m3進(jìn)口SO2濃度8.0%，O2濃度9.0%，其余為氮?dú)?。處理?31kmolSO2/hr，要求最終轉(zhuǎn)化率98。2.動(dòng)力學(xué)方程：式中：3.基本要求：(1)在TX圖上，做出平衡線，至少4條等速率線；(2)以一維擬均相平推流模型為基礎(chǔ)，在催化劑用量最少的前提下，總的及各段的催化劑裝量；進(jìn)出口溫度、轉(zhuǎn)化率；并在T-X圖上標(biāo)出折線；4.討論：(

2、1)要求的最終轉(zhuǎn)化率從97變化到99對(duì)催化劑用量的影響；(2)如果有關(guān)系：YO2YSO221，SO2進(jìn)口濃度在79之間變化，對(duì)催化劑裝量的影響。計(jì)算過程：（1）由于反應(yīng)速率的表達(dá)式是溫度和轉(zhuǎn)化率的函數(shù)，所以在入口濃度一定的條件下，當(dāng)反應(yīng)速率等于零的時(shí)候，不同的溫度對(duì)應(yīng)不同的轉(zhuǎn)化率。根據(jù)這些數(shù)據(jù)就可以作出此反應(yīng)的平衡線。改變反應(yīng)速率的大小，分別可以得到不同反應(yīng)速率下的等速率線。平衡線和等速率線數(shù)據(jù)的計(jì)算程序如下：#include<math.h>#include<stdio.h>#define R 1.987#define h 0.0001double r(double x

3、, 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(10,18)*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)*ex

4、p(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/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);void main()double xso

5、2=0.08,x,t,r0=0,y; int i; for(i=0;i<5;i+)printf("R%d=%en",i+1,r0);for(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>0.001);printf("%5.2f %5.4fn",t,x);t=t+5;r0=r0+pow(10,-6);運(yùn)行上述程序后，獲得的數(shù)據(jù)見下表：溫度T/k反應(yīng)速率r×1060.0

6、1.0 2.0 3.0 4.0 693.150.9809 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

7、.8873 0.7828 0.6669 0.5500 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

8、0.7492 0.6794 783.150.9071 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

9、 818.150.8436 0.8077 0.7643 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

10、 0.7273 0.6912 0.6508 0.6070 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 根據(jù)上述數(shù)據(jù)制圖： 上圖以溫度為橫坐標(biāo)，轉(zhuǎn)化率為縱坐標(biāo)。最上面一條為平衡線，平衡線下的四條曲線為反應(yīng)速率分別為1e-6、2e-6、3e-6、4e-6(mol/gcat.sec)時(shí)的等速率線。（2）根據(jù)入口組成，設(shè)定

11、入口溫度，根據(jù)反應(yīng)對(duì)入口溫度所求偏導(dǎo)數(shù)在這一段內(nèi)對(duì)組成的積分為零，可以求得此段出口轉(zhuǎn)化率，即下一段的入口組成。由于段內(nèi)操作線的斜率為（），因此根據(jù)入口溫度、入口組成和出口組成計(jì)算可得出口溫度。要使催化劑用量最少，應(yīng)使，根據(jù)計(jì)算每段的最小催化劑用量。循環(huán)計(jì)算四段后可得最后的出口轉(zhuǎn)化率，若不能滿足要求（x98），則重新設(shè)定第一段入口溫度，再進(jìn)行計(jì)算直至滿足條件。上述計(jì)算過程的C語言程序如下：#include<stdio.h>#include<math.h>#define R 1.987#define h 0.0001double r(double x, double t,

12、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)*exp(-35992 / (R*t); K = 2.3*pow(10,-8)*exp(27200/(R*t); Kp=2.26203*pow(10,-5)*exp(11295.3 / t

13、); 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/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, doubl

14、e 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);double fun1(double x, double t, double xso2) double y; y=-dr(x,t,xso2)/(r(x,t,xso2)*r(x,t,xso

15、2); 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)+fun1(x2,t2,xso2)/20; x1=x2; while(sum<0);xout=x1-h/10;end: return(xout);double

16、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; x1=x2; while(x2<=xou);end: wcat=sum*131*1000/3600; return(wcat);void main()dou

17、ble xso2=0.08,xout,tout,x0=0.0001,t0,t00=719,t1,wcat,wsum;int i,j;loop1:wsum=0.0;x0=0.0001;t0=t00;printf("1 tin=%f xin=%en",t00,x0);for(i=0;i<=3;i+)xout=jifen(x0,t0,xso2);tout=t(t0,x0,xout);printf("%d tout=%f ",i+1,tout);printf("xout=%f n",xout);wcat=wjifen(x0,xout,t

18、0,xso2);printf("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;printf("%d tin= %f xin= %fn",i+2,t0,xout);printf("Wsum=%f n",wsum);t00=t00-0.1;printf("nn");if(x0<=0.

19、98) goto loop1;運(yùn)行上述程序的結(jié)果見下表：XTin/KXinTout/KXoutWcat/KGWsum/KG0.981717.800.00010873.0.667668349.91891.2723.600.66766778.0.9036611605.3715.920.90366729.0.9620421557.4693.190.96204697.0.9800850379.0.971719.000.00010873.0.662508010.81141.2725.870.66250780.0.8983910705.3718.650.89839732.0.9594018921.4696

20、.150.95940700.0.9785843503.本程序計(jì)算在一定SO2進(jìn)口含量（須滿足條件XSO2+XO2 = 0.21)的情況下使得出口轉(zhuǎn)化率達(dá)到0.98時(shí)所需要的催化計(jì)量。#include"math.h"#include"stdio.h"#define R 1.987#define h 0.0001double r(x,t,xso2)double x,t,xso2;double r,r1,r2,r3,keff,K,B,Kp,Pso2,Pso3,Po2; if(t>=693.15 && t<748.15) keff =

21、 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("7200/(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-xs

22、o2*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(x,t,xso2)double x,t,xso2;double y; y=(r(x,t+h,xso2)-r(x,t-h,xso2)/(2*h); return(y);double t(t0,x0,x)double t0,x0,x;double

23、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(x,t,xso2)double x,t,xso2; double y; y=-dr(x,t,xso2)/(r(x,t,xso2)*r(x,t,xso2); return(y);double jifen(x0,t0,xso2)double x0,t0,xso2; double sum=0.0,x1=x0,x2,t1,t2=693.15,xout; do t1=t(t0,x0,x1);

24、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(xin,xou,tin,xso2)double xin,xou,tin,xso2;double y,x1=xin,x2,t1,t2,sum=0.0,wcat;do t1=t(tin,xin,x1); x2=x1+h; t2=t(tin

25、,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);main() double xso2=0.08,xout,tout,x0=0.0001,t0,t00=717,t1,wsum=0.0,wcat; double xou5,tou5,tin5,xin5; int i,j;loop1: x0=0.0001; t0=t00; i=0; xini=

26、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; printf("i tin xin tout xout wcatn"); for(j=0;j<=3;j+) wcat=wjifen(xinj,xouj,tinj,xso2); printf("%d %5.5f %5.5f %5.5f %5.5f %5.5fn",j+1,tinj,xinj,touj,xouj,wcat); wsum=wsum+wcat; printf("wsum= %f (KG) %f (Ton)n",wsum,wsum/1000);XSO2Tin/KXinTout/KXoutWcat/KGW