1、化學(xué)工程基礎(chǔ)(武漢大學(xué)第二版)1、在間歇反應(yīng)器中進(jìn)行等溫一級(jí)反應(yīng)A→RS已知反應(yīng)進(jìn)行30S時(shí)反應(yīng)物A的轉(zhuǎn)化率為90%,求轉(zhuǎn)化率達(dá)到99%時(shí)還需要多少時(shí)間?計(jì)算結(jié)果說明了什么?2、在間歇反應(yīng)器中進(jìn)行等溫液相反應(yīng)AB→R已知(rA)=kCACBmolL1min1k=0.3Lmol1min1CA0=CB0=0.2molL1試計(jì)算反應(yīng)物A的轉(zhuǎn)化率分別達(dá)到90%和99%時(shí)所需要的時(shí)間,并做比較。3、在間歇反應(yīng)器中用乙酸和丁醇生產(chǎn)乙酸丁酯,其反應(yīng)式
2、為:3493492CHCOOHCHOHCHCOOCHHOABCD?反應(yīng)在100℃等溫進(jìn)行,進(jìn)料物質(zhì)的量比為A:B=1:4.97,并以少量硫酸為催化劑。由于丁醇過量,其動(dòng)力學(xué)方程為(rA)=kC2A,式中k為1.74102m3kmol1min1。已知反應(yīng)物密度ρ為750kgm3(反應(yīng)前后基本不變),若每天生產(chǎn)乙酸丁酯2450kg(不考慮分離過程損失),每批物料的非生產(chǎn)時(shí)間取0.5h,求乙酸轉(zhuǎn)化率為55%時(shí)所需間歇反應(yīng)器的體積(裝料系數(shù)φ=
3、0.75)。??3433492A0CHCOOHCHOHCHCOOCHHOF0.0326min.kmol????解:2450由化學(xué)反應(yīng)方程可知:=2460116105533A0011.753601744.97750F0.03260.0186min1.753AOAOCkmolmvmC???????????????222020()10.5540.07min11.74101.75310.55AAAAAArAkCkCxxtkCx?????????
4、?????????3030.018640.07301.301.301.730.75RRRTVvttmVVm???????????答:略。4、在間歇反應(yīng)器中進(jìn)行二級(jí)液相反應(yīng)AB→RS兩種反應(yīng)物的起始濃度均為1kmolm3反應(yīng)10min后的轉(zhuǎn)化率為80%。如果將該反應(yīng)改在全混流反應(yīng)器中進(jìn)行,求達(dá)到相同轉(zhuǎn)化率所需要的反應(yīng)時(shí)間。????AA0Ax解:間歇反應(yīng)器:t=kC1x??A0A0A00A00850011F7.1424620.80.04C=
5、=0.02L()2F7.14357L5.95minC0.025.950.876.5(1)0.31110.8CSTRARAkmolhkmolvhLvxVLkx??????????????解:()環(huán)氧乙烷等體積體積流量:12121211201111102122(2)40(1)(1)5.954067.6%(1)0.311(1)4089.5%(1)CSTRCSTRCSTRCSTRCSTRAAARRRAAAARAAAAARAAxxxVVVkxkx
6、vxxVxkxxvxxVxkx?????????????????????串聯(lián):有:或:()又00AA32=40=80.7%(1)CSTRRAvvxVxkx???1()并聯(lián):兩個(gè)反應(yīng)器的體積流量各為0ARPA15.951(4)PFR=80%V=lnln30.810.31110.8vxLkx????反應(yīng)器:答:略。7、若將3題的酯化反應(yīng)改在活塞流反應(yīng)器中進(jìn)行,請(qǐng)利用3題的已知條件和計(jì)算結(jié)果計(jì)算:(1)轉(zhuǎn)化率達(dá)55%所需要的反應(yīng)器有效體積;(
7、2)反應(yīng)進(jìn)行80min時(shí)的殘余乙酸濃度。3RP03020V=vt=0.018640.07=0.75m1.753(2)0.510111.7410801.753AAACCkmolmkC?????????????解:(1)達(dá)到相同轉(zhuǎn)化率所需的反應(yīng)時(shí)間相同:答:略。8、在活塞流反應(yīng)器中進(jìn)行乙醛分解反應(yīng):CH3CHO→CH4CO在791K,1.013105Pa,進(jìn)料為純乙醛的條件下,該反應(yīng)的速率方程為(rA)=kC2Ak=0.33m3kmol1。