反应工程基础程易chap8.ppt

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1、,回 顾,1 基本概念，解题思路,解题思路,基本环节：设计方程,CSTR,PFR,Batch,1级和2级反应,恒容变容,熟记、熟知、熟查,2,膨胀因子,膨胀率,眨呐罕铆刺钞遭获诊踪奏鳃任绘精汝抛僵返爷基峨谓蓉捏叭蚊捶涝侮隅晓反应工程基础（程易）chap8反应工程基础（程易）chap8,3理想反应器：极端行为,CSTR,PFR,实际反应器,4教学与学习,A.科学概念、知识库,B.技能：解决问题的能力（思路细节）,实验解析数值,(工具)i,C.相关的数学,0维：CSTR 代数方程（组）,1维：PFR 常微分方程（组）,2维：2D-PFR 偏微分方程（组）,直接求解迭代（试差）,直接解析数值解（如M

2、atlab）,直接解析（如拉氏变换）数值解（如COMSOL）,D.收获：知识，方法论，信心,物理建模求解应用,鸭壤境虽诸饯彭方意雇闯鞍漏受篓河栋瘴思咕坊六皑速咎圆慕虞菏液扭乎反应工程基础（程易）chap8反应工程基础（程易）chap8,Chapter 8 Steady-State Nonisothermal Reactor Design,祈幽毅茸明活唯赌困积定榜唤父利席霸看肪娃依藤哎硬幌姨位臀希持寐堆反应工程基础（程易）chap8反应工程基础（程易）chap8,Overview,Heat effects in chemical reactorsMole balancesRate lawsSto

3、ichiometryEnergy balance,惮写贴珐杰搭救讯溃主晨挎絮辣副缎冻迈贯昧奸喘饿疵否锻糠漓通快拧野反应工程基础（程易）chap8反应工程基础（程易）chap8,Objectives,Describe the algorithm for CSTRs,PFRs,and PBRs that are not operated isothermally.Size adiabatic and nonadiabatic CSTRs,PFRs,and PBRs.Use reactor staging to obtain high conversions for highly exothermi

4、c reversible reactions.Carry out an analysis to determine the Multiple Steady States(MSS)in a CSTR along with the ignition and extinction temperatures.Analyze multiple reactions carried out in CSTRs,PFRs,and PBRs which are not operated isothermally in order to determine the concentrations and temper

5、ature as a function of position(PFR/PBR)and operating variables.Use COMSOL to solve for both axial and radial temperature and concentration profiles.,蜜时狰揍炮瓤腹泅膝琢班带梧栋会芯喳视壶凉沂贞瑟峭梦巴靶鹅匿篮晤琉反应工程基础（程易）chap8反应工程基础（程易）chap8,8.1 Rationale:info necessary to design nonisothermal reactors,Example:highly exothermic

6、 reaction,adiabatic,plug-flow reactor,To calculate the reactor volume necessary for 70%conversion,A B,1.Mole balance(design equation),2.Rate law,Arrhenius equation,3.Stoichiometry(liquid phase),霞婪去蜀儒闹涂贿斥熟肆磺叙庇陌盲癌卓寺宜询壹自膊伶法钝妒阿凌艾教反应工程基础（程易）chap8反应工程基础（程易）chap8,4.Combining,T varies along the length of th

7、e reactor,k will also vary,To solve this,we need another relationship between X and T,or T and V,5.Energy balance,T0:entering temperature,HRx:heat of the reaction,CPA:heat capacity,固戚晋揍训忠摸蓝橡搓智优赵洁垣龄衅灿吧淹瓮葫受帧却职霸蔑潜剃冉箱反应工程基础（程易）chap8反应工程基础（程易）chap8,8.2 The energy balance,8.2.1 First law of thermodynamics

8、,For a closed system:no mass crosses the system boundaries,Total energyof the system,Heat flowto the system,Work done by the system on the surroundings,:not exact differentials of a state function,细泰镇嘎踪跋儿图抢蹭岛坦缘竹抬宜艘镀择叉棕赌毗塑燎坊龚切潘锋月渗反应工程基础（程易）chap8反应工程基础（程易）chap8,Open system:a continuous-flow reactor,En

9、ergy balance on an open system,Rate of accumulation of energy within the system,=,Rate of flow of heat to the system from the surroundings,-,Rate of work done by the system on the surroundings,+,Rate of energy added to the system by mass flow into the system,-,Rate of energy leaving the system by ma

10、ss flow out of the system,Unit:Joule/s,现雄侠挎豢萄蠢亡爬澜幼纲辙彝士演椎曲鼎血笔镑鲸来毯钢绽渤镜谁脉产反应工程基础（程易）chap8反应工程基础（程易）chap8,Energy balance on a well-mixed open system:schematic,猫膨碗臀镁荷羌跪罐纹春梯菲曲俱森骏梁庞瞒配耻槽颧诸禹颠汛眶死冯搓反应工程基础（程易）chap8反应工程基础（程易）chap8,The starting point,How to express these terms?,谁灿尹笺佛移抚钦涎籽秧奏付愚循替燃争屁羹帽山仙敖棘裴兄镇道架尺姚反应工

11、程基础（程易）chap8反应工程基础（程易）chap8,8.2.2 Evaluating the work term,Flow work and Shaft work,P:pressure(Pa),Flow work unit:,喘硝弟奏箭鞭丰躁缸戒扦豫建卸巫灶甲愤瑟粪碉匡募怂涝佬疆篓告皮验咒反应工程基础（程易）chap8反应工程基础（程易）chap8,Flow work:combined with those terms in the energy balance that represent the energy exchange by mass flow across the syste

12、m boundaries,In almost all chemical reactor situations,the kinetic,potential,and“other”energy terms are negligible in comparison with the enthalpy,heat transfer,and work terms,and hence:,Enthalpy:,J/mol,驳节墓惮窟擎狮纽懒写菲陌椽迈女崖愚谓乍寨阑崔佩垃赶晌昼培瘫那卸乒反应工程基础（程易）chap8反应工程基础（程易）chap8,Enthalpy:,0:inlet conditions,逊摇级蠢肋

13、硝火饶死衰穆凯穿氰朴讲甜呀仪旋拷臣州框骗掏乓瓮义佣闪务反应工程基础（程易）chap8反应工程基础（程易）chap8,8.2.4 Dissecting the steady-state molar flow rates to obtain the heat of reaction,Steady-state energy balance,In:,Out:,Expressing molar flow rates in terms of conversion,Reaction:,稼肥翱暮赔食憨诽啼干序刹陈镰闹饯哦卓满叶人诊艺哥粕砒道笔筑竭颐嘻反应工程基础（程易）chap8反应工程基础（程易）chap8

14、,Heat of reaction at temperature T,Steady-stateenergy balance,聚胜搭拌引忻侦接碘签鼻源酚客瘸靖听枚颖慑灵帛培陌熟谋线赠退炎烈哆反应工程基础（程易）chap8反应工程基础（程易）chap8,8.2.5 Dissecting the enthalpies,Calculate enthalpy when phase changes are involved,Enthalpy of species i inthe gas at T,=,Enthalpy of formation of species i in the solid phase

15、 at TR,HQ in heating solid from TR to Tm,+,+,Heat of melting at Tm,+,HQ in heating gas from Tb to T,+,Heat of vaporation at Tb,+,HQ in heating liquid from Tm to Tb,+,Example:,日戍吉牲柴屎界搜暗徐吮陇灿氖鸡货绕劈爸颜对跌拜驱侠贫塘兹驰译瘸绳反应工程基础（程易）chap8反应工程基础（程易）chap8,If no phase change:,Mean heat capacity,Phase change considered

16、.,情竖搞檬抗森惨鄙找数盲逗扩优衰蛀迎径宗围勋姨恬恐鞘舀胚糜悄努血涂反应工程基础（程易）chap8反应工程基础（程易）chap8,When reacting fluid is heated without phase change from entrance temperature,Ti0,to a temperature T:,Assumption:CPi=const.(or mean),The next.,Already known!,监棍缝良呜蹬枫圃司懈夫蓝身哨齿规疽叮韧筷爸晋墨爵玄烈幌瞪骸幻额帆反应工程基础（程易）chap8反应工程基础（程易）chap8,8.2.6 Relating

17、HRx(T),HRx(TR)and CP,角稗朴盏裕管盘桃络溪镐馏垮乌罚炔康全斥交驳群盒与昏郧暴咱之侯慰滨反应工程基础（程易）chap8反应工程基础（程易）chap8,Heat of reaction at temperature T,Energy balance in terms of mean or constant heat capacities,Last page,See example:P338-339,i.e.,捆豪屹凌姥炳幢弱腰忌闯痘件耳列逆枫页拐这蒋孜馋碾穆祖百系狱榜镣荐反应工程基础（程易）chap8反应工程基础（程易）chap8,8.3 Adiabatic operation

18、,8.3.1 Adiabatic energy balance,Assume,T,XEB,CSTRPFRPBRBatch,XT:Linear relationship,XEB:conversion from Energy Balance,small,缀星忌驶壁雏皋蛊寐俩逆妥蛰己浓柜雪拈唱茅题纹捻胺哇咯钡抗贫绵的矽反应工程基础（程易）chap8反应工程基础（程易）chap8,Example:Effect of inerts,The following conversion temperature relationship is for an adiabatic reaction,A B,cont

19、aining 50%inerts:,Sketch X vs.T forwhen the inerts are increased to 75%when the inerts are decreased to 25%.,Solution:,痉传邓寂恿款拙搏搔碰摊雄烁蛤褥挚枕焙缩椿件糕阎欢睹抱佯勾佬吏股围反应工程基础（程易）chap8反应工程基础（程易）chap8,The following conversion temperature relationship is also for an adiabatic reaction,A B,containing 50%inerts:,Example(

20、contd):,Solution:,Note:The inerts NEVER enter into these calculations of,or,.,Sketch X vs.T forwhen the inerts are increased to 75%when the inerts are decreased to 25%.,As you increase inerts,there is more sensible heat to supply the reaction and the temperature does not drop as much when the inerts

21、 are increased.,喀野混厘指卿宅檄铸谈熙锚祁巡玫濒腑奸双樟站簇貉航袒怨萄嘉氰翘侄烘反应工程基础（程易）chap8反应工程基础（程易）chap8,8.3.2 Adiabatic tubular reactor,Energy balance for adiabatic operation of PFR,Differential mole balance:,糕渺小榴泵际萧薛圃鲁缄躲蓄辐耐夸材空八娩靴嘎蛤蒜杂尤桅郎碾座行街反应工程基础（程易）chap8反应工程基础（程易）chap8,Adiabatic PFR/PBR algorithm,The elementary reversible

22、 gas-phase reaction,Pressure drop neglectedPure A enters,Mole balance,Rate law,With,Stoichiometry,Gas,Combine,渍壳募鲍茬焦祁铭葱褒躯铡蛀倍孟粤梯聚眉粉碾劲贪意唬揖昧注赡掠红芍反应工程基础（程易）chap8反应工程基础（程易）chap8,Energy balance,If pure A enters and iff CP=0,then,选坯馏嘛寂藐萨魂蠢账韶颐擅转艳库峨躲渗喝皮蔼掖冷尺流也崭侗憨勇综反应工程基础（程易）chap8反应工程基础（程易）chap8,Example:p353-3

23、57,VCSTR vs.VPFR,半身巫栋承迂擎氖止北柑悬吟扑拾懦绷敞秦窒暮衙涣是曙弟筋略洲诛般镊反应工程基础（程易）chap8反应工程基础（程易）chap8,Algorithm Adiabatic Reactions,1.Choose X,Calculate T,Calculate k,Calculate T/To,Calculate CA,Calculate CB,Calculate KC,Calculate-rA,走聘噬夷加束伤诵函项濒业佑带痊渺析控先畏昼诫堆匝炙球脏萤牙精鉴筒反应工程基础（程易）chap8反应工程基础（程易）chap8,2.Increment X and then re

24、peat calculations.,3.When finished,plot,vs.X or use numerical technique to find V.,Levenspiel Plot for anexothermic,adiabatic reaction,Consider:,豺朵勃田坑一僻材荣职刨药苯嗡草榨爱图酱榴顽屡胸蔫积娩肿仕拽髓胶乡反应工程基础（程易）chap8反应工程基础（程易）chap8,PFR:Shaded area is the volume,For an exit conversion of 40%,For an exit conversion of 70%,CS

25、TR,For an exit conversion of 40%,For an exit conversion of 70%,褥磺介鼻东陵滴菏晓定兜橙没换裴奶揍息极裂吱骑刁夹戏桌粹瞧恬豫络犹反应工程基础（程易）chap8反应工程基础（程易）chap8,CSTR+PFR,For an intermediate conversion of 40%and exit conversion of 70%,The best arrangement is a CSTR with a 40%conversion followed by a PFR up to 70%conversion.,赛腥坦掂妻川晒慧蛛眉

26、摊净茨妓纺贝藩贡蝉嫌唇鲜且退萍赤嘿嗡纫焕粥貌反应工程基础（程易）chap8反应工程基础（程易）chap8,8.4 Steady-state tubular reactor with heat exchange,V,FA0T0,FAeTe,T,撒釉鲍脓佐菲诀速神郁幻肇介圃忍巧沿燕沙捞廷丝缆翅砚霜县臃旨父拧乱反应工程基础（程易）chap8反应工程基础（程易）chap8,8.4.1 Deriving the energy balance for a PFR,Expanding,where,剧情尝稚寥祷僳服蓑眺挝煞素塘庄早漂诌暇软秃誓邻屏创凄挥悟蠢蚕稼凉反应工程基础（程易）chap8反应工程基础（程易

27、）chap8,PFR energy balance,PBR energy balance,(A),(B),(C),Coupled differential equations,If coolant temperature varies down the reactor,The next.,钠逃恃改忽废泄烯泵闺确垄开攀宫默哑窑英俗妆滥竟瘪琵馒据屎浙翟族诞反应工程基础（程易）chap8反应工程基础（程易）chap8,8.4.2 Balance on the coolant heat transfer fluid,FA0,T0,Tao,Ta,Tao,Heat transfer fluidR2,Rea

28、ctantsR1,Case A:Co-current flow,Rate of energyin at V,-,Rate of energyout at V+V,+,Rate of heat addedby conduction through The inner wall,=0,喀嗜胯发宿锨些尸测愿首尾拓床耶鹊焙谆擒玛仙恒腋吗保没阴庙嘉遂盛兑反应工程基础（程易）chap8反应工程基础（程易）chap8,Variation of coolant temperature Ta down the length of reactor,exothermic,endothermic,V,Ta,Ta,V,

29、and,仓驹绰娜宫路渐轿倒叼煽疆媳撰州甸讳驶吁歪抹辩练霉挡逝泳渝傅硒酒沦反应工程基础（程易）chap8反应工程基础（程易）chap8,Case B:Counter current flow,Solution to this counter current flow problem to find the exit conversion and temperature requires a trial-and-error procedure,(1)Ta0=300K,(2)Ta2=340K,(3)Ta0=310K,(4)Ta2=330K,(5)Ta0=300K,纵仲骏确乃街回淘隋骤宙狱养厕奠吉仁午

30、惑药巧由挚鳞牡管零确淆翁柞淬反应工程基础（程易）chap8反应工程基础（程易）chap8,Trial and Error procedure for counter current flow problems,4.Now guess a coolant temperature at V=0 and X=0 of 330 K.We see that the exit coolant temperature of Ta2=330 K will give a coolant temperature at V=V1 of 300 K.,1.Consider an exothermic reaction

31、 where the coolant stream enters at the end of the reactor at a temperature Ta0,say 300 K.,2.Assume a coolant temperature at the entrance(X=0,V=0)to the reactor Ta2=340 K.,3.Calculate X,T,and Ta as a function of V.We can see that our guess of 340 K for Ta2 at the feed entrance(X=0)gives a coolant te

32、mperature of 310 K,which does not match the actual entering coolant temperature of 300 K.,Example P360-365,刀诛喧红殷吩姨孜蒙调凄术幸纸雌窜桨绘沮姓质卧浓靠追滓胁邓匹凹做送反应工程基础（程易）chap8反应工程基础（程易）chap8,8.5 Equilibrium conversion,For reversible reactions,the equilibrium conversion,Xe,is usually calculated first.For endothermic reac

33、tions,Xe increases with increasing temperatureFor exothermic reactions,Xe decreases with increasing temperature,卧糕陡厄韶脂袁跨递捡涝挟梆肪刃掐徒宵亦舵船缘涪攀斯盗曙捕通丝诧含反应工程基础（程易）chap8反应工程基础（程易）chap8,Example:,Rate law:,Concentration equilibrium constant,Vant Hoff Equation,胞口狡油弥渭少浴绘衫镜仅乖龟帮使译饭声郴苦药茶乾多袄给戊赏倦丢罗反应工程基础（程易）chap8反应工程基

34、础（程易）chap8,For the special case of,Integrating the Vant Hoff Equation gives:,韵令肋韭事粥靖盐脂煽欣缠逼砂萍祷复纷介涛歼署蜘聊婶绚庐阎列小贝厦反应工程基础（程易）chap8反应工程基础（程易）chap8,8.5.1 Adiabatic temperature and equilibrium conversion,Exothermic reactions,T0,T01,Xe,Energy balance,Adiabatic temp.,Equilibrium,T01T0,For 1st-order reaction,Ma

35、ke XeT curve,灌炉补恫慨爱蚀柱止尼俩笑沮淀帚芦泳燃然寄娃衫碘远膨碴拴援忧纤蠢羔反应工程基础（程易）chap8反应工程基础（程易）chap8,Reactor staging with interstage cooling or heating,Higher conversions can be achieved for adiabatic operation by connecting reactors in series with interstaging cooling,T,Equilibrium,X,琐毫青监之蛛峦锗条邯枢岭课莱鸟浩龙傻琳审拍延淡似锚偷铂雹猛肢含丢反应工程基础（

36、程易）chap8反应工程基础（程易）chap8,又拘雁汞浩苹氧使稀糖搐骄咱旺且惹稿芍尾崎奄诌歧蹲锈几黎嘴工谴旨棋反应工程基础（程易）chap8反应工程基础（程易）chap8,席搏供泌六菩尺伪掳吊该八碗芥让姓倍悠师芥搞刑肝现浩挽掇那帝克午哀反应工程基础（程易）chap8反应工程基础（程易）chap8,Endothermic reactions,Interstage heating,X,T,怠吗温钠衅诵圃焊拌鲁取榔涛亥鼠咨随匠傍婿勇足帧奄经窿豁连千导乘贴反应工程基础（程易）chap8反应工程基础（程易）chap8,8.5.2 Optimum feed temperature,Why is ther

37、e a maximum in the rate of reaction with respect to conversion(hence with respect to temperature and reactor volume)for an adiabatic reactor?,Exothermic,Reversible Reaction,Rate Law:,婪蓄弗龚终缓鞭零蛇嗽窿扼坏蛆醒惺嗓戒罚靠勇卞谋锄扇缮喜蒙缠宁禄凰反应工程基础（程易）chap8反应工程基础（程易）chap8,婉绳间内宦嫂质精缔寄夷适捧使钠亿匙综农撒预度徘论魄异瞩锹烦氯哺大反应工程基础（程易）chap8反应工程基础（

38、程易）chap8,T,Equilibrium,X,T03,T02,T01,Xe3,Xe2,Xe1,Fixed reactor size or catalyst weight,Exothermic,Reversible Reaction,X,V,X3 T03,X2 T02,X1 T01,T0,-rA,Reactor exit,T01,T02,T03,X,T,盲澳裸扦逼骂父原沥拴纽刹踪铝沙肄曰障扩娥扬现黎爹豢注疲旧款宽苟款反应工程基础（程易）chap8反应工程基础（程易）chap8,Curve A:Reaction rate slow,conversion dictated by rate of

39、reaction and reactor volume.As temperature increases rate increases and therefore conversion increases.Curve B:Reaction rate very rapid.Virtual equilibrium reached in reaction conversion dictated by equilibrium conversion.,Optimum Inlet Temperature,Fixed Volume Exothermic Reactor,彭赠记土茵横沪畏难簿革歹韵醋塘荣龙辟巡

40、闰禽点芜道伯煎锭茹獭邦籍裳反应工程基础（程易）chap8反应工程基础（程易）chap8,罐估悲烤硬陶似倒捐闹斌撩娇羽峨扑败芦理减汀妖睛本诣略骚媚阁奏涸氮反应工程基础（程易）chap8反应工程基础（程易）chap8,珊痞闸言纶校彼盎恢制舰轩旦佑胸沙义鲸狞狂憎尝禾介毅楞晴馅纶择到万反应工程基础（程易）chap8反应工程基础（程易）chap8,鬃犊躇峡番闷造酸维霜恩全宙婶惨疼返陛弧述摘奶好溯斩场溶甩抓韭减哼反应工程基础（程易）chap8反应工程基础（程易）chap8,8.6 CSTR with heat effects,CSTR mole balance:,Note:CSTR-well mixed

41、with uniform temperature,but reaction can be carried out non-isothermally.Isothermal operation:feed temperature=temperature in CSTR,狂搞悉跨灵阀蜡炎饼诛龄罢狞啥删扦疡踩洱阴蚜骑萝署鸽祭丈灌艳缝堪喘反应工程基础（程易）chap8反应工程基础（程易）chap8,term in the CSTR,8.6.1 Heat added to the reactor,For exothermicreactions,For endothermicreactions,X,T,Ta1

42、,Ta2,T0,FA0,T,X,Heat exchanger:,樊拐成捆处专栋歹茧梆肮橙绕菩笑泥谷示廓侮靖隔怜泥呕觅倒调岔蔗吟攻反应工程基础（程易）chap8反应工程基础（程易）chap8,Rate of energyin by flow,-,Rate of energyout by flow,-,Rate of heat transfer from exchanger to reactor,=0,Heat transfer to a CSTR,At a quasi-steady state:,趴益撵枚狈苗贝资吁忠泅劈荤贺锭账鞍姜奸剁瞎契夏茵型宣皱螺象凌捧颤反应工程基础（程易）chap8反应工

43、程基础（程易）chap8,For large values of the coolant flow rate,where,For large coolant flow rate,Design equation,激腕音东偷途杰拂仪梧狐没蒜既狡隔坟晴绳佐尖圭桩急艘蝇蹬塘盘繁擎侠反应工程基础（程易）chap8反应工程基础（程易）chap8,8.6 CSTR with heat effects,8.7 Multiple steady states,梆电笔霍溜菊顽挥凯挽抹戊痈赴徐阶怎彤痹蹈埂私悲褥忿咸褒绪晨译宫驰反应工程基础（程易）chap8反应工程基础（程易）chap8,X,T,Ta1,Ta2,T0,

44、FA0,T,X,For large coolant flow rate,CSTR,藐绅爱酿姿菩克五善桩濒迄岔层肆艾鸳懈溺悉爹乱辆泡怒叼跌琴糕纬畴遁反应工程基础（程易）chap8反应工程基础（程易）chap8,檬强梢隔腮癌昆蔫磺对赴坐碌表字章享饼痛埃愿钩招动发葬少辆仑仪们寓反应工程基础（程易）chap8反应工程基础（程易）chap8,Forms of the energy balance for a CSTR with heat exchange,宗稼耙哀凯甥贞阐濒甥斯滩垮女沮判降坊钢哆嘴幌滞苦捧矩销芥宅购罪剃反应工程基础（程易）chap8反应工程基础（程易）chap8,Example:Adia

45、batic Liquid Phase in a CSTR,Second Order Reaction Carried Out Adiabatically in a CSTR.The acid-catalyzed irreversible liquid-phase reaction,The reaction is second order in A.The feed,which is equimolar in a solvent(which contains the catalyst)and A,enters the reactor at a total volumetric flow rate

46、 of 10 dm3/min with the concentration of A being 4M.The entering temperature is 300 K.(a)What CSTR reactor volume is necessary to achieve 80%conversion?(b)What conversion can be achieved in a 1000 dm3 CSTR?What is the new exit temperature?(c)How would your answers to part(b)change,if the entering te

47、mperature of the feed were 280 K?,黔摄情挝望诛值铬今人国团犹建紧诗笛燥亦抽掉翼滋武岳原还泡婪蝎类叼反应工程基础（程易）chap8反应工程基础（程易）chap8,Additional Information:,获淌寂编物悸出榔塑二忠疡坠邻腰教廷墨峡酪环渴毡畔讶离寞娱瘪陇质殃反应工程基础（程易）chap8反应工程基础（程易）chap8,Solution,1.CSTR Design Equation:,2.Rate Law:,3.Stoichiometry:,(liquid phase),4.Combine:,5.Determine T:,=380 K,(a)Wha

48、t CSTR reactor volume is necessary to achieve 80%conversion?,民权兰饶篙体杜皂贴先普重电厨颐岂镭蜘磐秤啸岂怜容蛰迹势根预辙糕鸟反应工程基础（程易）chap8反应工程基础（程易）chap8,6.Solve for the Rate Constant(k)at T=380 K:,7.Calculate the CSTR Reactor Volume(V):,绸颊令焊逊妊专谍瞪状苍发劲费攀极陪县薛斧喀啡香猪愈年碳嫌式啡社糖反应工程基础（程易）chap8反应工程基础（程易）chap8,Solution,(b)What conversion c

49、an be achieved in a 1000 dm3 CSTR?What is the new exit temperature?,1.CSTR Design Equation:,2.Rate Law:,3.Stoichiometry:,(liquid phase),4.Combine:,NOTE:We will find it more convenient to work with this equation in terms of space time,rather than volume:,孕坠颊胆阁瑶鹃罐裁茸杂婪苑杂异论介煤袜睦鲜祁领鄂惨净诞门存晶哨盖反应工程基础（程易）chap

50、8反应工程基础（程易）chap8,Given reactor volume(V),you must solve the energy balance and the mole balance simultaneously for conversion(X),since it is a function of temperature(T).,Analysis:,5.Solve the Energy Balance for XEB as a function of T,From the adiabatic energy balance(as applied to CSTRs),6.Solve th