二氧化硫转化率最优化剖析

1、二氧化硫转化率最优化题目：S02+ 1/2O2=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 = 0.2549kcal/kg K- H =23135kcal/kmol床层空隙率p尸554kg/m3进口 SO2浓度8.0%, O2浓度9.0%，其余为氮气处理量131kmolSO2/hr，要求最终转化率98%，KE。？巳。3 1Imol/gcat.secl2. 动力学方程：RSO2 二 keff 巳2 2.B B-1 甌 . KFS2 险式中:keff =7.6915 1018expkeff =1.5128 107 expJ 76062 )、RT 丿J35992) RT丿

2、420 - 475oC475-600oCB =48148expK =2.3 10$ exp-7355.5)iT丿S7200、I RT丿p fSo2 FO22*11295.3、il T丿KP =2.26203 10 expR= 1.9873.基本要求:(1) 在T X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) 要求的最终转化率从97%变化到99%对催化剂用量的影响；(2) 如果有关系：YO2+ YSO2= 21%，SQ进口浓度在7 9%之间变化，对催化 剂装

3、量的影响。计算过程：(1)由于反应速率的表达式是温度和转化率的函数，所以在入口浓度一定 的条件下，当反应速率等于零的时候，不同的温度对应不同的转化率。 根据这些数据就可以作出此反应的平衡线。改变反应速率的大小，分别可以得到不同反应速率下的等速率线。平衡线和等速率线数据的计算程序如下：#in clude#in clude#defi ne R 1.987#defi ne h 0.0001double(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

4、 & 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/Pso3;r2 =Pso3/(Pso2*sqrt(Po2)*Kp);B =48148*exp(-7355.5/t

5、);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 mai n()double xso2=0.08,x,t,r0=0,y;int i;for(i=0;i5;i+)prin tf(R%d=%en,i+1,r0);for(t=693.15;t=1) break;y=fabs(pow(10,5)*r(x,t,xso2)-pow(10,5)*r0);while(y0.001);prin tf(%5.2f%5.4fn,t,x);t=t+5;r0=r0+pow(10,-6);运行

6、上述程序后，获得的数据见下表：温度T/k反应速率=r 1060.01.02.03.04.0693.150.9809698.150.9797703.150.97820.0117708.150.97620.2719713.150.97400.4750718.150.97150.62050.1844723.150.96860.72140.38630.0784728.150.96550.79010.53730.29030.0638733.150.96200.83650.64730.45360.2689738.150.95820.86740.72650.57610.4285743.150.95410.8

7、8730.78280.66690.5500748.150.94960.89950.82240.73360.6416753.150.94470.89640.82370.73980.6529758.150.93940.89270.82390.74440.6617763.150.93380.88850.82310.74740.6685768.150.92780.88370.82140.74910.6736773.150.92130.87840.81880.74970.6772778.150.91440.87250.81540.74920.6794783.150.90710.86620.81130.7

8、4770.6804788.150.89940.85930.80650.74530.6803793.150.89120.85190.80100.74210.6792798.150.88260.84400.79480.73800.6772803.150.87350.83570.78810.73320.6743808.150.86400.82680.78070.72770.6706813.150.85400.81750.77270.72150.6661818.150.84360.80770.76430.71460.6609823.150.83270.79750.75520.70710.6550828

9、.150.82150.78680.74570.69900.6484833.150.80970.77570.73570.69040.6412838.150.79760.76420.72520.68120.6335843.150.78510.75230.71420.67160.6252848.150.77220.74000.70290.66140.6163853.150.75900.72730.69120.65080.6070858.150.74540.71430.67910.63980.5972863.150.73150.70100.66660.62850.5870868.150.71740.6

10、8740.65390.61670.5764873.150.70300.67360.64080.60470.5654根据上述数据制图:1.00.90.80.70.60.5 X0.40.30.20.10.0-R=0c R=1e-6* R=2e-6R=3e-6:R=4e-6700720740760780800820840860880上图以温度为横坐标，转化率为纵坐标。最上面一条为平衡线，平衡线下的 四条曲线为反应速率分别为1e-6、2e-6、3e-6、4e-6（mol/gcat.sec时的等速率线。（2）根据入口组成，设定入口温度，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

11、段出口转化率，即下一段的入口组成。由 于段内操作线的斜率为1/& （九=dT），因此根据入口温度、入口组成和出口组 dx成计算可得出口温度。要使催化剂用量最少，应使ri（Xi,out，Tj,out）=g（Xi卄,T卄），dxr(x,T)计算每段的最小催化剂用量。循环计算四段后可得最后XoutWCat r 根据厂二-FA0Xin的出口转化率，若不能满足要求(x= 98%),则重新设定第一段入口温度，再进 行计算直至满足条件。上述计算过程的C语言程序如下：#in clude#in clude#define R 1.987#defi ne h 0.0001double r(double x, dou

12、ble t, double xso2)double r,r1,r2,r3,keff,K,B,Kp,Pso2,Pso3,Po2;if(t=693.15 & t=748.15 & t873.15)xout = x1;goto end;sum=sum+h*(fu n1(x1,t1,xso2)+fu n1(x2,t2,xso2)/20; x仁 x2;while(sum=873.15) goto end ;sum=sum+(1/r(x1,t1,xso2)+1/r(x2,t2,xso2)*h/1000;x仁 x2;while(x2=xou);en d: wcat=sum*131*1000/3600;ret

13、urn(wcat);void mai n()double xso2=0.08,xout,tout,x0=0.0001,t0,t00=719,t1,wcat,wsum; int i,j;loop1:wsum=0.0;x0=0.0001;tO=tOO;prin tf(1tin=%fxin=%en,t00,x0);for(i=0;ih); x0=xout;t0=t1;prin tf(%d tin= %f xi n= %fn ,i+2,t0,xout);prin tf(Wsum=%fn ,wsum);too=too-o.i；prin tf(nn);if(x0=693.15 & t=748.15 & t

14、873.15)xout = x1;goto end;sum=sum+h*(fu n1(x1,t1,xso2)+fu n1(x2,t2,xso2)/20; x仁 x2;while(sum=873.15) goto end ;sum=sum+(1/r(x1,t1,xso2)+1/r(x2,t2,xso2)*h/1000;x仁 x2;while(x2h);xO=xout;t0=t1;i+;xi ni=xout;tin i=tO;if(i=3) goto loop2;tOO=tOO-O.1;if(x0=0.98) goto loop1;prin tf(i tinxintoutxoutwcat n);f

15、or(j=0;j=3;j+)wcat=wj ifen(xin j,xouj,t in j,xso2);prin tf(%d %5.5f%5.5f%5.5f%5.5f%5.5fn,j+1,ti nj,xi nj,touj,xouj,wcat);wsum=wsum+wcat;prin tf(wsum= %f (KG) %f (To n)n,wsum,wsum/1000);XsO2Tin/KXinTout/KXoutWcat/KGWsum/KG0.0701717.000.00010873.149360.671104648.3185126480.67022738.420.67110790.777740

16、.896094167.403023732.770.89609747.912530.961165352.485124708.450.96116712.883840.9801712312.463580.0751716.500.00010873.149690.673254962.0218929195.769532737.200.67325789.466980.897854568.286863761.310.89785746.056920.961226040.785224707.280.96122711.652650.9800113624.675560.0801715.900.00010873.147760.675825334.8236833107.72562735.660.67582787.852520.900105079.878493729.480.90010743.773140.961526999.832924705.640.96152709.942830.9800115693.194060.0851715.400.00010873.148090.677975743.4568438125.310072733.870.67797786.071820.90229572