Chemical Reactor Design Youn-Woo Lee School of Chemical and Biological Engineering Seoul National University 155-741, 599 Gwanangro, Gwanak-gu, Seoul, Korea  ywlee@snu.ac.kr  http://sfpl.snu.ac.kr

Multiple Reactions Chapter 8 Chemical Reaction Engineering Seoul National University

개요 개요. 화학반응기에서 오직 한 가지 원하는 반응만이 일어나는 것은 좀처럼 없는 일이다. 주로 복합반응이 일어나며, 반응 중의 일부는 원하는 반응이지만, 일부는 원하지 않는 반응이다. 화학공장의 경제적 성공을 위한 중요한 요소들 중의 하나는 원하는 주반응과 함께 일어나는 원하지 않는 부반응을 최소화시키는 것이다. 이 장에서는 복합반응에서 반응기의 선정과 일반 몰수지, 순반응속도, 상대반응속도에 대하여 논의한다. 첫째, 복합반응의 네 가지 기본 반응을 설명한다:   직렬반응    병렬반응    독립반응    복합반응 다음으로 선택도를 정의하고, 반응조건과 반응기들을 적절히 선택함으로써 원치 않는 부가반응을 최소화하는 방법들에 대해 논의한다. 그리고 복합반응이 포함된 반응공학 문제를 풀기 위해서 어떻게 CRE 알고리즘을 수정하는 지를 보여 준다. 마지막으로 이 알고리즘들이 실제 반응들에 어떻게 적용되는지 이해하기 위해 많은 예제들이 주어진다. Seoul National University

Introduction  Seldom is the reaction of interest the only one that occurs in a chemical reactor. Typically, multiple reactions will occur, some desired and some undesired. One of the key factors in the economic success of a chemical plant is minimization of undesired side reactions that occur along with the desired reaction.  In this chapter, we discuss reactor selection and general mole balances for multiple reactions. First, we describe the four basic types of multiple reactions: series, parallel, independent, and complex.  Next, we define the selectivity parameter and discuss how it can be used to minimize unwanted side reactions by proper choice of operating conditions and reactor selection. Seoul National University

Objectives  Define different types of selectivity and yield  Choose a reaction system that would maximize the selectivity of the desired product given the rate laws for all the reactions occurring in the system.  Size reactors to maximize the selectivity and to determine the species concentrations in a batch reactor, semi-batch reactor, CSTR, PFR, and PBR, systems. Seoul National University

8.1 Definition of Multiple Reaction  Parallel rxns (competing rxns) B A C  Series rxns (consecutive rxns) A B C  Complex rxns (Parallel + Series rxns) A + B C + D A + C E  Independent rxns A B + C D E + F k1 k2 Seoul National University

8.1.1 Examples of Multiple Reaction  Parallel rxns (Oxidation of ethylene to ethylene oxide) 2CO2 + 2H2O CH2=CH2 +O2 CH2-CH2  Series rxns (rxn of EO with NH3) O O CH2-CH2 + NH3 HOCH2CH2NH2 EO EO (HOCH2CH2)2NH (HOCH2CH2)3N  Complex rxns (formation of butadiene from ethanol) C2H5OH C2H4 + H2O C2H5OH CH3CHO + H2 C2H4 + CH3CHO C4H6 + H2O  Independent rxns (The cracking of crude oil to form gasoline) C15H32 C12H26 + C3H6 C8H18 C6H14 + C2H4 k1 k2 Monoethanolamine Diethanolamine Triethanolamine Seoul National University

8.1.2 Desired and Undesired Reaction The economic incentive Maximize the formation of D Minimize the formation of U D (Desired Product) A U (Undesired Product) kD kU Competing or side rxn Total cost Reaction cost Separation cost cost Low High M A D A, U reaction separation Rxn-separation system producing both D & U Efficiency of a reactor system Seoul National University

8.1.2 Instantaneous Selectivity, SDU D (Desired Product) A U (Undesired Product) kD kU Selectivity tells us how one product is favored over another when we have multiple reactions. The rate laws are 순간선택도는 D의생성속도와 U의 생성속도와의 비율이다. Rate of formation of D = Rate of formation of U Seoul National University

Overall Selectivity, SDU ~ For Flow Reactor SD/U = ~ FU FD = Exit molar flow rate of undesired product Exit molar flow rate of desired product 총괄선택도는 D의 유출속도와 U의 유출속도의 비율이다. For Batch Reactor SD/U = ~ NU ND = No. of moles of undesired product at the end of rxn time No. of moles of desired product at the end of rxn time Seoul National University

Comparison between SD/U and SD/U for a CSTR ~ ~ Mission: Develop a relationship between SD/U and SD/U in a CSTR SD/U = rU rD Solution D (Desired Product) A U (Undesired Product) kD SD/U = ~ FU FD kU Mole balance of D and U yields FD=rDV and FU=rUV, respectively rDV rD FD ~ SD/U SD/U = = = = FU rUV rU SD/U = SD/U ~ Seoul National University

8.1.3 Yields rD YD = -rA ND YD = NAo-NA FD YD = FAo-FA ). Instantaneous Yield (Basis: Reaction Rate) rD YD = -rA Overall Yield (Basis: Mole or Molar Flow Rate) ND Mole of desired product formed at the end of reaction ~ YD = For a batch system: = NAo-NA Number of moles of key reactant consumed FD ~ YD = For a flow system: FAo-FA 선택도와 마찬가지로 CSTR에서의 순간 수율과 총괄 수율은 동일하다 (i.e., ) Seoul National University

Note Different definitions for selectivity and yield Check carefully to ascertain the definition intended by the author From an economic standpoint, overall selectivities and yields are important in determining profits The instantaneous selectivities give insights in choosing reactors and reaction schemes that will help maximize the profit Seoul National University

8.2 복합반응의 풀이를 위한 알고리즘 복합반응의 알고리즘은 병렬반응, 직렬반응, 복합반응, 그리고 독립반응에 적용될 수 있다. 소프트웨어 패키지들 (ODE풀이자)을 사용할 경우에 전화율을 사용하는 것보다 몰 Nj 또는 몰유량 Fj를 사용하는 것이 문제를 풀기가 더 쉽다. 액상계에서는 몰수지식에서 농도 Cj 가 대개 선호되는 변수이다. 지금까지 공부해온 여러 형태의 반응기에 대한 몰수지식을 표8-1에 나타내었다. Seoul National University

Algorithm for solution to complex reactions Steps in in Analyzing Mutiple Reaction 1. Number each reaction. 2. Mole balance for each species. 3. Determine rij 4. Relate to rate of reaction of each species to the species for which the rate law is given. 5. Determine the net rate of formation of each species 6. Express rate laws as a function of Cj when eX<<1. 7. Express rate laws as a function of mole when eX>>1. 8. Combine all the above and solve the resulting set of equations 몰수지: 속도식: 화학양론: 결합: CRE 알고리즘의 주된 변화는 알고리즘에서의 속도식 단계가 지금은 아래의 3개의 하위단계들을 포함하는 속도들 단계로 바뀐 것이다. (1) 속도식들 (2) 알짜속도들 (3) 상대속도들

Table 8-2 CRE 알고리즘의 수정 1. 모든 하나하나의 반응에 개별적으로 숫자를 부여 2. 모든 하나하나의 성분에 대한 몰수지식 3. 모든 하나하나의 반응의 속도법칙, 예: 첨자 “i” 는 반응 번호, 첨자“, j ” 는 성분을 나타낸다. 4. 각 성분의 반응 알짜 속도, 예, j N개의 반응에 대하여, A 항의 알짜생성속도는 다음과 같다: 5. 모든 하나하나의 반응에 대한 상대속도들 주어진 반응에서의 i : 표6-2의 알고리즘의 남은 단계는 변하지 않는다. Table 8-2 CRE 알고리즘의 수정 Seoul National University

We want to maximize SD/U. 8.3 Parallel Reactions D (Desired Product) A U (Undesired Product) kD kU The rate laws are (8-6) (8-7) (8-8) We want to maximize SD/U. (8-9) = Rate of formation of U Rate of formation of D (8-10) a1 and a2 are positive reaction orders Seoul National University

8.3.2 Maximizing the Desired Product for One Reactant D (Desired Product) A U (Undesired Product) kD kU  Case 1: a1 > a2 , a = a1- a2 (8-11) Maximize SD/U - keeping the concentration of reactant A as high as possible during the rxn - in gas phase rxn, we should run it w/o inerts and at high pressures to keep CA high - in liquid phase rxn, the use of diluents should be keep to a minimum - use a batch or plug-flow reactor Seoul National University

8.3.2 Maximizing the Desired Product for One Reactant D (Desired Product) A U (Undesired Product) kD kU  Case 2: a2 > a1 , b = a2- a1 (8-12) Maximize SD/U - keeping the concentration of reactant A as low as possible during the rxn - in gas phase rxn, we should run it with inerts and at low pressures to keep CA low - in liquid phase rxn, the use of diluents should be keep to a maximum - use a CSTR or recycle reactor (product stream act as a diluent) Seoul National University

Maximizing SDU for one reactant Whether the reaction should be run at high or low T?  Case 3: ED > EU kD (rD) increases more rapidly with increasing temp. than does the kU. keeping the temperature as high as possible to maximize SDU.  Case 4: EU > ED - keeping the temperature as low as possible to maximize SDU - not so low that the desired rxn does not proceed to any significant extent. Seoul National University

Ex 8-1: Maximizing SB/XY for the Trambouze Reaction Reactant A decomposes by three simultaneous reactions to form three products, B (desired), X (undesired), and Y (undesired). These liquid phase reactions, along with the appropriate rate laws, are called the Trambouze reactions. X (0차) A B (1차) Y (2차) k1 k2 k3 How and under what conditions (e.g., reactor types, temperature, concentrations) should the reaction be carried out to maximize the selectivity of B ?  The specific reaction rates are given @ 300K  E1=10,000, E2=15,000, and E3=20,000 kcal/mole  CA0=0.4M, v0=2.0 dm3/s Seoul National University

Maximizing SB/XY for the Trambouze Reaction CA SB/(X+Y) 0.025 0.357143 0.05 0.625 0.075 0.775862 0.1 0.833333 0.125 0.15 0.803571 0.175 0.76087 0.2 0.714286 0.225 0.668317 0.25 0.275 0.585106 0.3 0.54878 0.325 0.515873 0.35 0.486111 0.375 0.459184 0.4 0.434783 0.425 0.412621 0.45 0.392442 0.475 0.374016 0.5 SB/(X+Y)=(0.0015*CA)/(0.0001+0.008*CA2) 0.84 SB/(X+Y) 0.112 CA (mol/dm3) Seoul National University

Maximizing SB/XY for the Trambouze Reaction As we see, the selectivity reaches a maximum at a concentration C*A. Because the concentration changes down the length of a PFR, we cannot operate at this maximum. Consequently, we will use a CSTR and design it to operate at this maximum. To find the maximum C*A, we differentiate SB/(X+Y) w.r.t. CA, set the derivative to zero, and solve for C*A. That is, Seoul National University

Maximizing SB/XY for the Trambouze Reaction We now calculate this CSTR volume and conversion. The net rate of formation of A from reactions (1), (2), and (3) is CSTR volume for this liquid-phase reaction, CSTR volume for maximum selectivity Seoul National University

Maximizing SB/XY for the Trambouze Reaction Maximum the selectivity w.r.t. temperature Case 1: Run @ high temperature (w/o side rxn) Case 2: Run @ low temperature (with insured conversion) For the activation energies given above So the selectivity for this combination of activation energies is independent of temperature ! Seoul National University

Maximizing SB/XY for the Trambouze Reaction What is the conversion of A in the CSTR? If greater than 72% conversion of A is required, then the CSTR operated with a reactor concentration of 0.112 mol/dm3 should be followed by a PFR because the concentration and selectivity will decrease continuously from C*A as we move down the PFR to an exit concentration CAf. Hence the system Would be the highest selectivity while forming more the desired product B, beyond what was formed at C*A in a CSTR. How can we increase the conversion and still have a high selectivity SB/(X+Y)? Seoul National University

Maximizing SB/XY for the Trambouze Reaction Optimum CSTR followed by a PFR. The exit concentration of X, Y, and B can be found from the CSTR mole balances Let’s check to make sure the sum of all the species in solution equals the initial concentration CA0 = 0.4. CA+CX+CB+CY=0.112+0.0783+0.132+0.0786=0.4 //// QED Seoul National University

Maximizing SB/XY for the Trambouze Reaction The reason we want to use a PFR after we reach the maximum selectivity, SB/XY, is that the PFR will continue to gradually reduce CA. Thus, more B will be formed than if another CSTR were to follow. If 90% conversion were required, then the exit concentration would be CAf=(1-0.9)(0.4 mol/dm3)=0.04 mol/dm3. The PFR mole balances for this liquid-phase reaction (v=v0) are Combining mole balances with their respective rate laws yields t=0, the entering concentrations to the PFR are the exit concentrations from CSTR. Seoul National University

Maximizing SB/XY for the Trambouze Reaction The conversion can be calculated We will use Polymath to plot the exit concentrations as a function of t and then determine the volume (V=v0t) for 90% conversion (CA=0.04 mol/dm3) and then find CX, CB, and CY at this volume. This volume turns out to be approximately 600 dm3. At the exit of the PFR, CA=0.037, CX=0.11, CB=0.16, and CY=0.09 all in mol/dm3. One now has to make a decision as to whether adding the PFR to increase the conversion of A from 0.72 to 0.9 and the molar flow rate of B from 0.26 to 0.32 mol/s is worth not only the added cost of the PFR, but also the decrease in selectivity. This reaction was carried out isothermally; nonisothermal multiple reactions will be discussed in Chapter 8. Seoul National University

Maximizing SB/XY for the Trambouze Reaction FA0=0.8 mol/s FB=0.26 mol/s X*=0.72 SB/XY=0.84 FB=0.32 mol/s X=0.9 SB/XY=0.8 V=600 dm3 CA* V=1564 dm3 ~ 분석: 이제 PFR을 덫붙여 A의 전화율을 0.72에서 0.9로 증가시키고 B의 몰유량을 0.26에서 0.32mol/s로 증가시키는 것이 PFR의 추가 설치비용 뿐만 아니라 선택도가 0.84에서 0.8로 감소하는 만큼의 가치가 있는지 결정해야 한다. 이 예제에서 트람보제(Trambouze) 반응을 통해 CSTR에서 성분 B에 대한 선택도를 최적화하는 방법을 보여 주었다. 여기서 최적 출구 조건 (CA=0.112 mol/dm3), 전화율 (X=0.72), 선택도 (SB/XY=0.84)를 찾았다. 이 때의 반응기 부피는 V = 1564 dm3 이었다. 만약 전화율을 90%로 증가시키고 싶다면, CSTR 뒤에 PFR을 설치할 수 있고 이때 선택도는 감소하는 것을 알 수 있다. SB/XY X* X PFR CSTR CAf C*A CA0 Seoul National University

PFR concentration & selectivity profiles for the Trambouze Reaction Seoul National University

PFR concentration & selectivity profiles for the Trambouze Reaction Seoul National University

PFR concentration & selectivity profiles for the Trambouze Reaction Seoul National University

PFR concentration & selectivity profiles for the Trambouze Reaction Seoul National University

PFR concentration & selectivity profiles for the Trambouze Reaction Seoul National University

8.3.3 Reactor Selection and Operating Conditions D (Desired Product) A + B U (Unwanted Product) The rate laws are Rate selectivity parameter (= Instantaneous selectivity) is to maximized by choosing reactor schemes. Two simultaneous reactions with two reactants k1 k2 Rate selectivity parameter Seoul National University

Reactor Selection Criteria  Selectivity  Yield  Temperature control  Safety  Cost Seoul National University

Different reactors and schemes for maximizing the desired product Figure 8-2 Different reactors and schemes for maximizing the desired product A B B A A B A B A B (a) CSTR (b) tubular reactor (c ) batch (d) semi-batch 1 (e) semi-batch 2 A B A B (f) Tubular reactor with side streams (g) Tubular reactor with side streams B A (h) Series of small CSTRs Seoul National University

Different reactors and schemes for maximizing the desired product Figure 8-2 Different reactors and schemes for maximizing the desired product The two reactors with recycle shown in (i) and (j) can be used for highly exothermic reactions. Here recycle stream is cooled and returned to the reactor to dilute and cool the inlet stream thereby avoiding hot spots and run-away reactions. The PFR with recycle is used for gas-phase reactions, and the CSTR is used for liquid-phase reactions. A B A B (i) Tubular reactor with recycle (j) CSTR with recycle Seoul National University

Different reactors and schemes for maximizing the desired product Figure 8-2 Different reactors and schemes for maximizing the desired product The last two reactors, (k) and (l), are used for thermodynamically limited reactions where the equilibrium lies far to the left (reactant side) A + B C + D And one of the products must be removed (e,g., C) for the reaction to continue to completion. The membrane reactor (k) is used for thermodynamically limited gas-phase reactions, while reactive distillation (l) is used for liquid-phase reaction when one of the products has a high volatility (e.g., C) than the other species in the reactor. ← A+B C D A B D C (k) Membrane reactor (l) Reactive distillation Seoul National University

Example 8-2: Minimizing unwanted products for two reactants For the parallel reactions, consider all possible combinations of reaction orders and select the reaction scheme that will maximize SD/U. D (Desired Product) A + B U (Undesired Product) Case I : a1 > a2, b1 > b2, a = a1-a2 > 0, b = b1-b2 > 0 the rate selectivity parameter k1 k2 To maximize the SDU, maintain the concentration of both A and B as high as possible  a tubular reactor (Figure 8-2 (b))  a batch reactor (Figure 8-2 (c))  high pressures (if gas phase), and reduce inerts Seoul National University

Example 8-2: Minimizing unwanted products for two reactants for the parallel reactions D (Desired Product) A + B U (Undesired Product) Case II : a1 > a2, b1 < b2, a = a1-a2 > 0, b = b2-b1 > 0 the rate selectivity parameter k1 k2 To maximize the SDU, maintain CA high and CB low.  a semibatch reactor in which B is fed slowly into A. (Figure 8-2(d))  a tubular reactor with side stream of B continually (Figure 8-2(f))  a series of small CSTRs with A fed only to the first reactor and small B fed to each reactor. (Figure 8-2(h)) Seoul National University

Example 8-2: Minimizing unwanted products for two reactants for the parallel reaction D (Desired Product) A + B U (Undesired Product) Case III : a1 < a2, b1 < b2, a = a2-a1 > 0, b = b2-b1 > 0 the rate selectivity parameter k1 k2 To maximize the SDU, maintain the concentration of both A and B as low as possible  a CSTR (Figure 8-2(a))  a tubular reactor in which there is a large recycle ratio (Figure 8-2(i))  a feed diluted with inert material  low pressures (if gas phase) Seoul National University

Example 8-2: Minimizing unwanted products for two reactants for the parallel reaction D (Desired Product) A + B U (Undesired Product) Case IV : a1 < a2, b1 > b2, a = a2-a1 > 0, b = b1-b2 > 0 the rate selectivity parameter k1 k2 To maximize the SDU, maintain the concentration of both A and B as high as possible  a semibatch reactor in which A is slowly fed to a large amount of B (Figure 8-2(e))  a membrane reactor or tubular reactor with side stream of A (Figure 8-2(g))  a series of small CSTRs with fresh A fed to each reactor Seoul National University

8.4 Maximizing the desired product in series reaction k1 k2 A B C  In parallel rxns, maximize the desired product  by adjusting the reaction conditions (e.g. CA, CB)  by choosing the proper reactor  In series rxns, maximize the desired product  by adjusting the space-time for a flow reactor  by choosing real-time for a batch reactor Seoul National University

Maximizing the desired product in series reaction k1 k2 A B C Desired Product  If the first reaction is slow and second reaction is fast, it will be extremely difficult to produce species B.  If the first reaction (formation of B) is fast and the reaction to form C is slow, a large yield of B can be achieved.  However, if the reaction is allowed to proceed for a long time in a batch reactor or if the tubular flow reactor is too long, the desired product B will be converted to C.  In no other type reaction is exactness in the calculation of the time needed to carry out the reaction more important than in series reactions. Seoul National University

Example 8-3: Maximizing the yield of the intermediate product 다음 단일 액상 직렬 반응이 회분식 반응기에서 수행된다. k1 k2 A → B → C 반응은 반응온도까지 빠르게 가열되고 유지되다가 급랭시킨다. (a) A, B, C의 농도를 시간의 함수로 그리고 분석하여라. (b) B의 농도가 최대가 될 때 반응을 급랭시키 위한 시간을 계산하여라. (c) 급랭하는 시간에 총괄선택도와 수율을 어떻게 되는가? 추가 정보 CA0 = 2M, k1 = 0.5h-1, k2=0.2h-1 Seoul National University

풀이 부분 (a) 반응들의 번호 정하기: 앞에 기술한 직렬반응은 다음 두 개의 반응들로 나타낼 수 있다. k1 반응 (1) : A → B -r1A = k1CA k2 반응 (2) : B → C -r2B = k2CB 1. 몰수지 2A. A에 대한 몰수지: a. V = V0 일 때의 농도 항으로 된 몰수지 (E8-3.1) b. 반응 1에 대한 속도법칙 : 반응은 단일반응이다. (E8-3.2) c. 몰수지와 속도법칙을 결합하고 (E8-3.3) 초기조건 t=0에서 CA=CA0를 가지고 적분하면 (E8-3.4) CA에 대해 풀면 (E8-3.5) Seoul National University

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Reaction paths for different ks in series reaction A B C k1 k2 For k1/k2>1, a Large quantity of B Can be obtained B For k1/k2<1, a Little quantity of B Can be obtained 1st rxn is slow 2nd rxn is fast A C Long rxn time in batch or long tubular reactor  B will be converted to C Seoul National University

8.5 복합반응 복합 반응계는 상호 작용하는 직렬반응과 병렬반응이 결합되어있다. 이 복합반응에 대한 알고리즘의 개요는 몰유량과 농도로 몰수지식을 쓰는 제 6장에 주어진 것과 매우 비슷하다 (그림 6-1). 각각의 반응에 번호를 매긴 후에 그림 6-1과 같이 각각의 성분에 대해 몰수지를 작성한다. 이 두 가지 알고리즘의 가장 큰 차이는 반응 속도법칙 단계이다. 표 8-2와 같이 반응물의 농도 항으로 각각 성분에 대한 알짜반응속도를 찾는 3 단계 (3, 4, 5)를 가지고 있다. 예로써, PFR, CSTR, 반회분식 반응기에서 수행되는 다음의 복합 반응을 공부하자. A + 2B → C 2A + 3C → D Seoul National University

예제 8-5 PBR에서의 복합 기상 반응 아래의 복합기상반응은 단일반응의 속도법칙을 따르며, PBR에서 등온에서 진행된다.   아래의 복합기상반응은 단일반응의 속도법칙을 따르며, PBR에서 등온에서 진행된다. (1) A + 2B → C (2) 2A + 3C → D 원료는 A와 B가 같은 몰수이고 FA0 = 10 mol/min 이며 부피유량은 100 dm3/min이다. 촉매 질량은 1000kg, 압령강하는 a= 0.0019 kg-1, 그리고 총 유입 농도는 CTO = 0.2 mol/dm3이다. (a) FA, FB, FC, FD, y, 를 촉매질량 W에 대한 함수로 그래프를 그리고 분석하여라.

풀이 PBR 기상 몰수지 속도 알짜속도

속도법칙 상대속도 반응1: 반응2: 선택도 W = 0, FD = 0일 때 SC/D는 무한대이다. 그러므로 ODE 풀이자가 충돌이 일어나는 것을 방지하기 위해 W = 0으로부터 매우 작은 수, W=0.0001 kg 사이에서 SC/D = 0으로 설정한다.

화학양론 등온 T=T0

매개변수 이 식들을 폴리매스의 ODE 풀이자에 입력하면, 표 E8-5.1와 그림E8-5.1, 그림E8-5.2의 결과를 얻을 수 있다. 분석: 그림 E8-5.2로부터 선택도가 바로 입구 근처 (W= 60kg) 최대값을 갖고 그 이후로 급격히 감소하는 것을 볼 수 있다. 그러나 원하는 생성물인 C가 최대 유량을 갖는 촉매 질량이 200kg까지도 A의 90%는 소모되지 않는다. 반응 (1)의 활성화 에너지가 반응 (2)의 활성화 에너지보다 크다면, 온도를 높여 C의 몰유량과 선택도를 높일 수 있다. 그러나 그것도 소용이 없다면, 원하는 생성물의 선택도와 몰유량 중 어떤 것이 더 중요한지 선택해야 한다. 전자의 경우 PBR 촉매 무게는 60kg이 된다. 후자의 경우 PBR 촉매 무게는 200kg이다.

Relationship for Multiple Rxns Occuring in a PFR For equations for species j and reaction i that are to be combined when we have q reactions and n species Mole Balance: Rate laws: Stoichiometry: (gas-phase) (liquid-phase)

Multiple reactions Reaction 1: 4NH3 + 6NO  5N2 + 6H2O Reaction 2: 2NO  N2 + O2 Reaction 3: N2 + 2O2  2NO2 The rates for rxns 1, 2 and 3 are given in terms of species NO, N2, and O2, respectively. Consequently, to relate each reacting species in each rxn to its rate law more clearly, we divide each rxn through by the stoichiometric coefficient of the species for which the rate law is given. 1: NO + 2/3NH3  5/6N2 + H2O 2: 2NO  N2 + O2 3: O2 + 1/2N2  NO2 The corresponding rate laws are related by:

Reaction 1: The rate law w.r.t. NO is The relative rates are Then the rate of NH3, N2, and H2O Reaction 2: Reaction 3:

Net rate of formation NO: Net rate of formation N2: Net rate of formation O2:

Multiple reactions in a PFR For gas-phase rxns, concentration of species j is DT=DP=0 For gas-phase rxns, concentration of species j is (1) Mole balance on NO:

(2) Mole balance on NH3: (3) Mole balance on H2O:

(4) Mole balance on N2: (5) Mole balance on O2: (6) Mole balance on NO2:

Hydrodealkylation of Mesitylene in a PFR 1,3,5-trimethylbenzene CH3 + H2 + CH4 k1 k2 The hydrodealkylation of mesitylene is to be carried out isothermally at 1500 R and 35 atm in a packed-bed reactor in which the feed is 66.7 mol% hydrogen and 33.3 mol% mesitylene. The volumetric feed rate is 476 ft3/h and the reactor volume (i.e. V=W/rb) is 238 ft3. M=mesitylene, X=xylene, T=toluene, Me=methane, H=hydrogen The bulk density of the catalyst has been included in the specific reaction rate (i.e., k1=k1’ rb) Plot the concentrations of hydrogen, mesitylene, and xylene as a function of space-time. Calculate the space-time where the production of xylene is a maximum (i.e., topt) .

Reaction 1: M + H  X + Me Reaction 2: X + H  T + Me Mole balances: Hydrogen Mesitylene Xylene Toluene Methane 2. Rate laws:

Relative Reaction rates: Reaction 1: -r1H = -r1M = r1X = r1Me Reaction 2: -r2H = -r2X = r2T = r2Me 3. Stoichiometry: (no volume change with reaction, v=v0) Flow rates: Combining and substituting in terms of the space-time ( ) If we know CM, CH, and CX, then CMe and CT can be calculated from the reaction stoichiometry. Consequently, we need only to solve the following three equations: 3개 연립방정식

Ci t (hr) topt Parameter evaluation: We now solve these three equations simultaneously using POLYMATH. 2.0 1.5 1.0 0.5 0.0 t (hr) Ci topt CH CM CX H2 0.0 0.1 0.2 0.3 0.4 0.5

Multiple reactions in a CSTR For a CSTR, a coupled set of algebraic eqns analogous to PFR differential eqns must be solved. Rearranging yields where After writing a mole balance on each species in the reaction set, we substitute for concentrations in the respective rate laws. If there is no volume change with reaction, we use concentrations, Cj, as variables. If the reactions are gas-phase and there is volume change, we use molar flow rates, Fj as variables. q reactions in gas-phase with N different species to be solved

Hydrodealkylation of Mesitylene in a CSTR Calculate the conversion of hydrogen and mesitylene along with the exiting concentrations of mesitylene, hydrogen, and xylene in a CSTR. 1. Mole Balances: 3. Stoichiometry: 2. Rate laws: 4. Combining and letting yields:

Ex 6-7: Hydrodealkylation of Mesitylene in a CSTR We put these equations in a form such that they can be readily solved using POLYMATH. For t=0.5, the exiting concentrations are CH=0.0089, CM=0.0029 and CX=0.0033. The overall conversion is 0.020 0.015 0.010 0.005 0.000 Ci tCSTR 0.0 1.0 2.0 CH CX CM

Hydrodealkylation of Mesitylene in a CSTR The moles of hydrogen consumed in reaction 1 are equal to the moles of mesitylene consumed. Therefore, the conversion of hydrogen in reaction 1 is The conversion of hydrogen in reaction 2 is The yield of xylene from mesitylene based on molar flow rates exiting the CSTR for t=0.5 is The overall selectivity of xylene relative to toluene is

Selectivity and overall selectivity in a CSTR For a CSTR the instantaneous selectivity and the overall selectivity are the same. For example, the instantaneous selectivity of xylene w.r.t. toluene is The instantaneous selectivity The overall selectivity Mole balances of xylene and toluene yields Substituting in the instantaneous selectivity equation for rX and rT For all ideal CSTR

The attainable Region A B C 2A D k1 k2 k3 PFR CSTR

Phase plane plot of CB as a function of CA 0.00014 0.00012 0.00010 0.00008 0.00006 0.00004 0.00002 0.00000 A B C 2A D k1 k2 k3 CSTR+PFR PFR CB (kmol/m3) For maximum CB CA>0.38, CSTR with by-pass CA=0.38, single CSTR CA<0.38, CSTR+PFR CSTR 0.0 0.2 0.4 0.6 0.8 1.0 CA (kmol/m3)

8.6 Membrane Reactors to Improve Selectivity in Multiple Reactions In addition to using membrane reactors to remove a reaction product in order to shift the equilibrium toward completion, we can use membrane reactors to increase selectivity in multiple reactions. This increase can be achieved by injecting one of the reactants along the length of the reactor. It is particularly effective in partial oxidation of hydrocarbons, chlorination, ethoxylation, hydrogenation, nitration, and sulfunation reactions to name a few.

8.6 Membrane Reactors to Improve Selectivity in Multiple Reactions O C In the top two reactions, the desired product is the intermediate. However, because there is oxygen present, the reactants and intermediates can be completely oxidized to form undesired products CO2 and water.

8.6 Membrane Reactors to Improve Selectivity in Multiple Reactions The desired product in the bottom reaction is xylene. By keeping one of the reactants at a low concentration, we can enhance selectivity. By feeding a reactant through the sides of a membrane reactor, we can keep its concentration low.

Example 8-8 Membrane Reactors to Improve Selectivity in Multiple Reactions The reactions above take place in the gas phase. The overall selectivity, , are to be compared for a membrane reactor (MR) and a conventional PFR. First, we use the instantaneous selectivity to determine which species should be fed through the membrane.

Example 8-8 Membrane Reactors to Improve Selectivity in Multiple Reactions We see that to maximize SD/U we need to keep the concentration of A high and the concentration of B low; therefore, we feed B through the membrane. The molar flow rate of A entering the reactor is 4 mol/s and that of B entering through the membrane is 4 mol/s as shown in Figure E8.8.1. For the PFR, B enters along with A. The reactor volume is 50 dm3 and the entering total concentration is 0.8 mol/dm3. Plot the molar flow rates and overall selectivity, , as a function of reactor volume for both the MR and PFR.

Figure E8-8.1 B A CB0

PFR MR Solution: Mole Balances for both the PFR and MR Species A: Species B: Darcy’s law Species D: Species U:

Net Rates and Rate Laws Species A: Species B: Species D: Species U:

Transport Law (Darcy’s law) Pressure (kPa) at shell side Pressure (kPa) at tube side Volumetric flow rate (m3/s) through the membrane membrane surface area (m2) At=aVt where a=m2/m3 K: membrane permeability (m/s·kPa) The flow rate through the membrane can be controlled by pressure drop across the membrane surface area per unit volume of reactor. The total molar flow rate of B through the sides of the reactor is

Stoichiometry: isothermal, no DP For both the PFR and MR for no pressure drop down the length of the reactor and isothermal operation, the concentrations are Species A: Species B: Species D: Species U:

Combine The Polymath Program will combine the mole balance, net rates, and stoichiometric equations to solve for the molar flow rate and selectivity profiles for both the conventional PFR and the MR and also the selectivity profile. A note of caution on calculating the overall selectivity. We have to fool Polymath because at the entrance of the reactor FU=0. Polymath will look at and will not run because it will say you are dividing by zero. Therefore, we need to add a very small number to the denominator, say 0.0001; that is

Conventional PFR Example 8-8 Membrane Reactors to Improve Selectivity in Multiple Reactions 10 20 30 40 50 4.0 3.0 2.0 1.0 0.0 Conventional PFR FA=FB FD Fi (mol/s) FU V (dm3)

Membrane Reactor Example 8-8 Membrane Reactors to Improve Selectivity in Multiple Reactions 10 20 30 40 50 4.0 3.0 2.0 1.0 0.0 Membrane Reactor FA Fi (mol/s) FB FD FU V (dm3)

PFR vs MR Example 8-8 Membrane Reactors to Improve Selectivity in Multiple Reactions 10 20 30 40 50 4.0 3.0 2.0 1.0 0.0 PFR vs MR FA FA=FB Fi (mol/s) FB V (dm3)

PFR vs MR Example 8-8 Membrane Reactors to Improve Selectivity in Multiple Reactions 10 20 30 40 50 4.0 3.0 2.0 1.0 0.0 PFR vs MR Fi (mol/s) FD FU FD FU V (dm3)

PFR vs MR Example 8-8 Membrane Reactors to Improve Selectivity in Multiple Reactions 10 20 30 40 50 80 60 PFR vs MR SD/U 14 0.65 V (dm3)