二氧化硫轉(zhuǎn)化率最優(yōu)化剖析

1、 二氧化硫轉(zhuǎn)化率最優(yōu)化題目：SO2+1/2O2=SO3,四段絕熱反應(yīng)器，級間間接換熱。基礎(chǔ)數(shù)據(jù)：混合物恒壓熱容Cp=0.2549kcaVkgK-AH=23135kcabzkinol床層空隙率Py=554kg/m3進(jìn)口SO2濃度8.0%,Ch濃度9.0%,其余為氮?dú)?。處理?31kinolSO2/lu,要求最終轉(zhuǎn)化率98%。動力學(xué)方程：?KPsoJPQ-Cgcw.sec(JB+(B_1比+JK弦他J式中:(-76062)、RT)(一35992)、RT)o何“(-73555B=48148exp|T丿27200、keff=7.6915x10lsexpkeff=1.5128xl07exp(420-47

2、5C)(475-600C)K=23xl0-sexpRT丿/?=1.987(1190s2Kp=2.26203xl0-5exp*基本要求:在T-X圖上，做出平衡線，至少4條等速率線；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；討論：要求的最終轉(zhuǎn)化率從97%變化到99%對催化劑用量的影響；如果有關(guān)系：YO2+YSO2=21%,SO?進(jìn)口濃度在79%之間變化，對催化劑裝量的影響。計算過程：(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)速率等于零的時候，不同的溫度對應(yīng)不同的轉(zhuǎn)化率。根據(jù)這些

3、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)的計算程序如下：#iiiclude#iiiclude#defineR1.987#defineh0.0001doubler(doublex、doublet,doublexso2)doubler,ilj2,r3.B,keff,K.Kp,Pso2,Pso3,Po2;if(t=693.15&t748.15)keff=7.6915*pow(10,18)*exp(-76062/(R*t);if=748.15&t=873.15)keff=1.5128*pow(10J)*exp(-35992/(R*t)

4、;K=2.3*pow(10r8)*exp(27200/(R*t);Kp=2.26203*pow(105)*exp(l1295.3/t);Pso2=(xso2-xso2*x)/(l-xso2*x/2);Pso3=xso2*x/(l-xso2*x/2);Po2=(0.17xso2xso2*x/2)/(l-xso2*x/2);rl=Po2*Pso2/Pso3;r2=Pso3/(Pso2*sqit(Po2)*Kp);B=48148*exp(-7355.5/t);r3=sqrt(B+(B-1)*(1-x)/x)+sqrt(K*(1-x)/x);r=keff*K*rl*(1-r2*r2)/(r3*r3);

5、retum(r);voidmain()doublexso2=00&x，tj0=0.y;inti;for(i=0;i5;i+)prmtf(nR%d=%eiiM,i+lj-O);for(t=693.15;t=l)break:y=fabs(pow(10,5)*r(x,t.xso2)-pow(10,5)*10);while(v0.001);pimtf(H%5.2f%54fm工x);t=t+5;r0=r(Kpow(l03e-6、4e-6（moVgcat.sec）時的等速率線。（2）根據(jù)入口組成，設(shè)定入口溫度，根據(jù)反應(yīng)對入口溫度所求偏導(dǎo)數(shù)在這一段內(nèi)對組成的積分為零，可以求得此段出口轉(zhuǎn)化率，即下一段的入口組

6、成。由于段內(nèi)操作線的斜率為1/幾（A=）,因此根據(jù)入口溫度、入口組成和出口組ax成計算可得出口溫度。要使催化劑用量最少，應(yīng)使/心胡“嚴(yán)E（X*,7；井）,根據(jù)catdx計算每段的最小催化劑用量。循環(huán)計算四段后可得最后的出口轉(zhuǎn)化率，若不能滿足要求(x=98%),則重新設(shè)定第一段入口溫度，再進(jìn)行計算直至滿足條件。上述計算過程的c語言程序如下：#iiiclude#iiiclude#defineR1.987#defineh0.0001doubler(doublex、doublet,doublexso2)doubler,ilj2,r3.keffK.B,Kp,Pso2Tso3.Po2;if(t=693.1

7、5&t=748.15&t=873.15)keff=1.5128*pow(10J)*exp(-35992/(R*t);K=2.3*pow(10r8)*exp(27200/(R*t);Kp=2.26203*pow(10873.15)xout=xl;gotoend;sum=sum+h*(fiin1(x1,t1,xso2)+fuii1(x2,t2,xso2)/20;xl=x2;while(sum=873.15)gotoend;sum=sum+(l/r(xl,tl,xso2)+l/r(x2,2xso2)*lv1000;xl=x2;while(x2=xou);end:wcat=sum*131*1000/3

8、600;retuin(wcat);voidmain()doublexso2=0.08.xout5tout0=0.0001j0,t00=719,t1,wcat,wsum;intiJ;loopl:wsum=0.0:x0=0.0001;tO=tOO;printf(M1tm=%fxm=%en,tOO.xO);fbr(i=O;ih);x0=xout;tO=tl;tiii=%fxiii=%fiT，i+2,tO,xout);printf(MWsum=%fnM,wsuni);tOO=tOO-O.l;if(x0=693.15&t=748.15&t=873.15)keff=1.5128*pow(10J)*exp(

9、-35992/(R*t);K=2.3*pow(10r8)*exp(n7200/(R*t);Kp=2.26203*pow(10873.15)xout=xl;gotoend;sum=sum+h*(fiinl(xl,t1,xso2)+funl(x2,t2xso2)/20;xl=x2;while(sum=873.15)gotoend;sum=sum+(l/r(x1,t1,xso2)+l/r(x2,t2、xso2)*lv1000;xl=x2;while(x2h);x0=xout;tO=tl;i+;xiiii=xout;tmi=t0;if(i=3)gotoloop2;tOO=tOO-O.l;if(x0=0

10、.98)gotoloopl;printf(MitinxmtoutxoutwcatiiH);for(j=0j=3j+)wcat=jifen(xiiij,xouIjJmIj,xso2);%5.5f%55f%5.5f%5.5f%5.5farj+ljiiij.xmjjouj,xouj,wcat);wsum=wsum+wcat;printf(Mwsum=%f(KG)%f(Ton)iiHAvsumvsunV1000);XSO2Tiii/KXinTout/KXoutWcavKGWsum/KG0.0701717.000.00010873.149360.671104648.3185126480.67022738

11、.420.67110790.777740.896094167.403023732.770.89609747.912530.961165352.485124708.450.96116712.883840.9801712312.463580.0751716.500.00010873.149690.673254962.0218929195.769532737.200.67325789.466980.897854568.286863761.310.89785746.056920.961226040.785224707.280.96122711.652650.9800113624.675560.0801

12、715.900.00010873.147760.675825334.8236833107.72562735.660.67582787.852520.900105079.878493729.480.90010743.773140.961526999.832924705.640.96152709.942830.9800115693.194060.0851715.400.00010873.148090.677975743.4568438125.310072733.870.67797786.071820.902295720.004983727.380.90299741.249600.961898184.484144703.790.96189708.016040.97