반응공학1 Youn-Woo Lee School of Chemical and Biological Engineering

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1 반응공학1 Youn-Woo Lee School of Chemical and Biological Engineering
Seoul National University , 599 Gwanangro, Gwanak-gu, Seoul, Korea  

2 Seoul National University
Chapter 5 Isothermal Reactor Design: Conversion Reaction Engineering I 등온반응기설계:전화율 Seoul National University

3 개 요 제1장과 제2장에서는 반응기에 대한 몰수지식과 이 몰수지식을 이용하여 반응기의 크기를 예측하는 것에 대해 논의하였다. 제3장에서는 반응식들에 대해서, 제4장에서는 반응의 화학양론에 대해서 알아보았다. 제5장과 6장에서는 앞에서 배운 네 장의 모든 내용을 가져다가 반응과 반응기를 서로 결합시켜, 여러 가지 형태의 반응기들을 설계할 수 있는 논리적 체계를 세우고자 한다. 이러한 논리적 체계를 이용함으로써, 독자들은 수많은 수식과 더불어 수반되는 여러 가지 제약 및 조건 (즉 전체 몰수의 변화가 있는지의 여부 등과 같은)들을 암기에 의하기 보다는, 논리적 추론에 의하여 반응공학문제를 풀 수 있게 된다.

4 이 장에서는 등온 반응기의 설계를 공부하기 위해 제2장의 표 S2-1 에 나타난 전화율로 표현된 몰수지식을 이용하기로 한다
이 장에서는 등온 반응기의 설계를 공부하기 위해 제2장의 표 S2-1 에 나타난 전화율로 표현된 몰수지식을 이용하기로 한다. 전화율은 회분식 반응기, CSTR 및 PFR에서 발생하는 단일 반응의 진행 정도를 측정하기 위해 선호되는 변수이다. 어떤 전화율에 도달하기 위한 회분식 반응기의 반응 시간과 흐름 반응기의 부피를 모두 계산 할 수 있다. 전화율을 변수로 하는 등온 반응기의 핵심 원리를 설명하기 위해 4 가지의 반응과 4 가지 다른 반응기를 선택하였다. (1) 에틸렌글리콜을 생산하는 액상반응에서 비반응속도상수 k를 구하기 위해 실험적 회분 반응기를 이용한다. (2) 회분 반응기의 실험으로부터 구한 k를 이용하여 에틸렌글리콜을 생산하기 위한 공업용 CSTR을 설계한다. (3) 에탄의 기상 열분해반응에 의해 에틸렌을 생산하는 PFR을 설계한다. (4) 에틸렌의 부분산화반응에 의해 에틸렌옥사이드를 생산하고 압력강하를 수반하는 충전층 반응기 (PBR)를 설계한다. 이 반응기를 모두 합치면 연간 20억 파운드의 에틸렌글리콜을 생산하는 화학공장을 설계하게 된다는 것을 알 수 있을 것이다.

5 Seoul National University
Objectives • Describe the algorithm that allows the reader to solve chemical reaction engineering problems through logic rather than memorization. • Size batch reactors, semibatch reactors, CSTRs, PFRs, and PBRs for isothermal operation given the rate law and feed conditions. • Studying a liquid-phase batch reactor to determine the specific reaction rate constant needed for the design of a CSTR. • Design of a tubular reactor for a gas-phase pyrolysis reaction. • Account for the effects of pressure drop on conversion in packed bed tubular reactors and in packed bed spherical reactors. Seoul National University

6        반응기를 설계할 수 있는 논리적 체계  와 를 결합하여 시작 끝 일반몰수지식 설계방정식 계산 네
회분식: CSTR: 플러그흐름: 설계방정식  계산 반응시간 반응기부피 촉매무게  네  –rA = f(X) 가 주어져 있는가 아니오 반응속도식  화학양론  결합  와 를 결합하여 -rA = f(X)를 얻는다  액상: 기상:

7 Follow the Reaction Design Algorithm Follow the Yellow Brick Road

8 Isothermal Reaction Design Algorithm

9 Seoul National University
Fig. 5-1 Isothermal Reaction Design Algorithm for Conversion Seoul National University

10 Algorithm for isothermal reactor design
1. Mole balance and design equation 2. Rate law 3. Stoichiometry 4. Combine 5. Evaluate We can solve the equations in the combine step either A. Graphically (Chapter 2) B. Numerical (Appendix A4) C. Analytical (Appendix A1) D. Software packages (polymath) Seoul National University

11 Seoul National University
French Menu Analogy Seoul National University

12 Algorithm for isothermal reactors
French Menu Analogy Fig. 5-2 Seoul National University

13 Scale-up of Liquid-Phase Batch Reactor
to the Design of a CSTR Pilot plant Operation Full-scale Production Laboratory Experiment PAST Full-scale Production Microplant (Lab-bench-scale unit) High cost of a pilot-plant leads to jump pilot plant operation FUTURE To make this jump successfully requires a through understanding of the chemical kinetics and transport limitations.

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5.2 Batch Operation For constant volume batch reactor, the mole balance can be written in terms of concentration Generally, when analyzing laboratory experiments, it is best to process the data in terms of the measured variable. Because concentration is the measured variable for most liquid-phase reactions, the general mole balance equation applied to reactions in which there is no volume change becomes This is the form we will use in analyzing reaction rate data in Chap 7. Seoul National University

15 Reaction time in Batch Operation
Algorithm for isothermal reactor design Mole balance & Design equation 2. Rate law 3. Stoichiometry 4. Combination 2nd order Isothermal Liquid-phase Batch reaction 5. Analytical Evaluation Seoul National University

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Batch Reaction Times Table 5-1 Mole balance First-order Second-order Rate law Stoichiometry (V=V0) Combine Integration Seoul National University

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Batch Reaction Times Seoul National University

18 Batch Reaction Times 1st-order k (s-1) 2nd-order kCA0 (s-1)
Table 5-2 The order of magnitude of time to achieve 90% conversion For first- and second-order irreversible batch reactions 1st-order k (s-1) 2nd-order kCA0 (s-1) Reaction time tR 10-4 10-3 Hours 10-2 10-1 Minutes 1 10 Seconds 1,000 10,000 Milliseconds Seoul National University

19 tt = tf + te + tR + tc Reaction time in Batch Operation
Typical cycle times for a batch polymerization process tt = tf + te + tR + tc Activity Time (h) 1. Charge feed to the reactor and agitate, tf 2. Heat to reaction temperature, te 3. Carry out reaction, tR 4. Empty and clean reactor, tc Varies Total cycle time excluding reaction Batch polymerization reaction times may vary between 5 and 60 hours. Decreasing the reaction time with a 60-h reaction is a critical problem. As the reaction time is reduced (e.g. 2.5 h for a 2nd-order reaction with X=0.9 and kCA0=10-3 s-1), it becomes important to use large lines and pumps to achieve rapid transfer and to utilize efficient sequencing to minimize the cycle time. Seoul National University

20 Design a Reactor to Produce of ethylene glycol
Design a CSTR to produce 200 million pounds of ethylene glycol per year by hydrolyzing ethylene oxide. However, before the design can be carried out, it is necessary to perform and analyze a batch reactor experiment to determine the specific reaction rate constant (kA). Since the reaction will be carried out isothermally, kA will need to be determined only at the reaction temperature of the CSTR. At high temperature there is a significant by-product formation, while at temperature below 40oC the reaction does not proceed at a significant rate; consequently, a temperature of 55oC has been chosen. Since the water is usually present in excess, its concentration may be considered constant during the course of the reaction. The reaction is first-order in ethylene oxide. O CH2-OH CH2-CH2 + H2O CH2-OH H2SO4 A B C Catalyst Seoul National University

21 Example 5-1 Determining k from Batch Data
In the lab experiment, 500mL of a 2 M solution (2 kmol/m3) of EO in water was mixed with 500mL of water containing 0.9 wt % sulfuric acid catalyst. At T=55oC, the CEG was recorded with time. Determine the specific reaction rate at 55oC. Time Concentration of EG (min) (kmol/m3) EO + H2O → EG A + B → C Seoul National University

22 Problem Solving Algorithm
Example 4-1 Determining k from Batch Data Problem statement. Determine the kA Sketch Identify C1. Relevant theories Rate law: Mole balance: C2. Variables Dependent: concentrations, Ci Independent: time, t C3. Knowns and unknowns Knowns: CEG = f(time) Unknowns: 1. CEO = f(time) 2. kA 3. Reactor volume C4. Inputs and outputs: reactant fed all at once a batch reactor C5. Missing information: None Assumptions and approximations: Assumptions 1. Well mixed 2. All reactants enter at the same time 3. No side reactions 4. Negligible filling and emptying time 5. Isothermal operation Approximations 1. Water in excess (CH2O~constant) CB~ CBO E. Specification. The problem is neither overspecified nor underspecified. F. Related material. This problem uses the mole balances developed in Chap. 1 for a batch reactor and the stoichiometry and rate laws developed in Chap. 3. G. Use an Algorithm.(figs 5-1 & 5-2) A, B, C batch Seoul National University

23 Problem Solving Algorithm
Example 4-1 Determining k from Batch Data 1. MOLE BALANCE Batch reactor that is well-mixed Since water is present in such excess, the concentration of water at any time t is virtually the same as the initial concentration and the rate law is independent of the concentration of H2O. (CB~CB0) 2. RATE LAW 3. STOICHIOMETRY A B C Species symbol Initial Change Remaining Concentration CH2CH2O A NA0 - NA0X NA=NA0(1-X) CA=CA0(1-X) H2O B ΘBNA0 - NA0X NB=NA0(ΘB-X) CB=CA0(ΘB-X) CB~ CA0 ΘB = CB0 (CH2OH) C 0 NA0X NC =NA0X CC=CA0X NT0 NT =NT0 - NA0X Seoul National University

24 Problem Solving Algorithm
Example 5-1 Determining k from Batch Data , 4. COMBINING Mole balance Rate law 5. EVALUATE For isothermal operation, k is constant: Seoul National University

25 Problem Solving Algorithm
Example 5-1 Determining k from Batch Data The concentration of EG at any time t can be obtained from the reaction stoichiometry A + B C Seoul National University

26 Example 5-1 Determining k from Batch Data
Rearranging and taking the logarithm of both side yields We see that a plot ln[(CA0-CC)/CA0] as a function of t will be a straight line with a slope –k. The rate law can now be used in the design of an industrial CSTR. Note that this rate law was obtained from the lab-scale batch reactor (1000 mL). 0.6 t CC (min) (kmol/m3) 0.06 1.55 8.95 Seoul National University

27 Example 5-1 Determining k from Batch Data
The rate law can now be used in the design of an industrial CSTR. Note that this rate law was obtained from the lab-scale batch reactor (1000 mL). 분석: 본 예제에서는 시간 t에 대한 함수로 화학종 C의 농도, CC를 계산하기 위해 CRE 알고리즘을 사용하였다. (몰수지식 → 속도법칙 → 화학양론 → 결합) 그리고 1차 반응의 반응식에 대한 규명과 비반응속도상수 k를 결정하기 위해 t에 대한 CC의 회분식 실험 자료를 이용하였다.

28 5.3 Design of CSTR Design Equation for a CSTR Mole balance (5-6) the space time (5-7) For a 1st-order irreversible reaction, the rate law is Rate law Combine Rearranging CSTR relationship between space time and conversion for a 1st-order liquid-phase rxn (5-8) Seoul National University

29 A single CSTR We could also combine Eq (4-12) and (5-8) to find the exit concentration of A, CA: exit concentration of A (5-9) Seoul National University

30 Reaction Damköhler number
The Damkohler is a dimensionless number that can give us a quick estimate of the degree of conversion that can be achieved in continuous-flow reactor. For 1st-order irreversible reaction For 2nd-order irreversible reaction RULE OF THUMB Da  0.1 will usually give less than 10% conversion. Da  10.0 will usually give greater than 90% conversion.

31 5.3.1.2 A Second-Order Reaction in a CSTR
For a 2nd-order liquid-phase reaction being carried out in a CSTR, the combination of the rate law and the design equation yields We solve the above eq. for X: (5-10) For const density v=v0, FA0X=v0(CA0-CA) (5-12) The minus sign must be chosen in the quadratic equation because X cannot be greater than 1. Using our definition of conversion, we have (5-11) Seoul National University

32 A Second-Order Reaction in a CSTR
60 6 0.67 0.88 At high conversion, a 10-fold increase in Da will increase the conversion only to 88% due to lowest value of reactant concentration in CSTR. Seoul National University

33 Seoul National University
CSTRs in Series CA1, X1 CA2, X2 CA0 v0 -rA1, V1 -rA2, V2 For 1st-order irreversible reaction with no volume change (v=v0) is carried out in two CSTRs placed in series. The effluent concentration of A from reactor 1 is From a mole balance on reactor 2, Seoul National University

34 Seoul National University
CSTRs in Series Solving for CA2, the concentration exiting the second reactor, we get (5-13) If instead of two CSTRs in series we had n equal-sized CSTRs connected in series (t1 = t2 = … = tn = ti = (Vi/v0)) operating at the same temperature (k1 = k2 = … = kn = k), the concentration leaving the last reactor would be (5-14) The conversion and the rate of disappearance of A for these n tank reactors in series would be (5-15) (5-16) Seoul National University

35 Conversion as a function of reactors in series
for different Damköhler numbers for a first-order reaction tk=0.1 tk=0.5 tk=1 Da  1, 90% conversion is achieved in two or three reactors; thus the cost of adding subsequent reactors might not be justified Da ~0.1, the conversion continues to increase significantly with each reactor added Seoul National University

36 CSTRs in Parallel A balance on any reactor i, gives the individual
reactor volume -rA1, V1 -rAi, Vi -rAn, Vn FA0 FA0 i 1 i n The volume of each individual reactor, Vi, is related to the total volume, V, of all the reactors, and similar relationship exists for the total molar flow rate

37 CSTRs in Parallel Substituting these values into Eq (5-10) yields
-rA1, V1 -rAi, Vi -rAn, Vn FA0 FA0 i 1 i n The conversion achieved in any one of the reactors in parallel is identical to what would be achieved if the reactant were fed in one stream to one large reactor of volume V!

38 Example 5-2: Producing 200,000,000 lb/yr in a CSTR
~91 ton/yr CA0=1 vA0=vB0 It is desired to produce 200 x 106 pounds per year of EG. The reactor is to operated isothermally. A 1 lb mol/ft3 solution of ethylene oxide (EO) in water is fed to the reactor together with an equal volumetric solution of water containing 0.9 wt% of the catalyst H2SO4. The specific reaction rate constant is min-1 as determined in Ex 5-1. (a) If 80% conversion is to be achieved, determine the necessary CSTR volume. (b) If two 800-gal reactors were arranged in parallel, what is the corresponding conversion? (c) If two 800-gal reactors were arranged in series, Seoul National University

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M.W. of EG=62 M.W. of EO=58 Seoul National University

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v0 = vA0 + vB0 =1 5’ 10’ 3m ~2 gps 1.5m 5 gal~19.8L 1 gal ~3.78 L Seoul National University

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(5-15): Seoul National University

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46 Producing 200,000,000 lb/yr of EG in a CSTR
one CSTR two equal-sized CSTRs in parallel two equal-sized CSTRs in series 1480gal X=0.8 800gal X=0.81 800gal X1=0.68 X2=0.90 Conversion in the series arrangement is greater than in parallel for CSTRs. The two equal-sized CSTRs in series will give a higher conversion than two CSTRs in parallel of the same size when the reaction order is greater than zero. Seoul National University

47 two equal-sized CSTRs in parallel two equal-sized CSTRs in series
분석: 단일 CSTR, 2개의 직렬연결 CSTR, 2개의 병렬연결 CSTR에서 수행된 등온 1차-비가역-액상 반응에 대해서 CRE 알고리즘을 적용하였다. 그리고 각 경우에 대한 방정식을 대수적으로 풀었다. 병렬로 연결된 2개의 CSTR로 몰유량이 동등하게 배분되어 주입되는 경우, 전체 전화율은 단일 CSTR과 같았다. 직렬로 연결된 2개의 CSTR의 경우에는 전체 전화율이 단일 CSTR보다 컸다. 0보다 큰 반응차수의 속도법칙을 따르는 등온 반응의 경우 항상 이러한 결과를 갖는다. one CSTR two equal-sized CSTRs in parallel two equal-sized CSTRs in series 1480gal X=0.8 800gal X=0.81 800gal X1=0.68 X2=0.90 Seoul National University

48 PEL (Permissible Exposure Limit)

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5.4 Tubular Reactors 2nd-order liquid-phase rxn (A Products) Turbulent, No dispersion - No radial gradients in T, u, or C PLUG-FLOW REACTOR PFR mole balance Rate law: Stoichiometry for liq. phase rxn T & P = constant Combination The differential form of PFR design equation must be used when there is a DP or heat exchange between PFR & the surrounds. In the absence of DP or heat exchange, the integral form of the PFR design equation is used. Da2 is the Damkohler number for a second-order reaction Seoul National University

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5.4 Tubular Reactors 2nd-order gas-phase rxn (A Products) Turbulent, No dispersion - No radial gradients in T, u, or C PLUG-FLOW REACTOR PFR mole balance Rate law: Stoichiometry for gas phase rxn T & P = constant The differential form of PFR design equation must be used when there is a DP or heat exchange between PFR & the surrounds. In the absence of DP or heat exchange, the integral form of the PFR design equation is used. Combination Seoul National University

53 5.3 Tubular Reactors CA0 is not function of X; k=constant (isothermal)
Integration yields (see Appendix A.1 Eq. page1009) Reactor length will be Cross sectional area Seoul National University

54 Conversion as a function of distance down the reactor
A  0.5B (e=-0.5) A  B (e=0) A  2B (e=1) A  3B (e=2) Seoul National University

55 The reaction that has a decrease in the total number of moles
will have the highest conversion for a fixed reactor length. the reactant spends more time the reactant spends less time The volumetric flow rate decreases with increasing conversion, and the reactant spends more time in the reactor than reactants that produce no net change in the total number of moles. Seoul National University

56 Change in Gas-Phase Volumetric Flow Rate Down the Reactor
v=vo(1+X) (5-17) =1 : (A→2B) v = vo(1+X) = 0 : (A→B) v = vo =0.5 : (2A→B) v = vo(1-0.5X) Complete conversion When there is a decrease in the number of moles in the gase phase, the volumetric flow rate decreases with increasing conversion, and the reactant spends more time in the reactor than reactants that produce no net change in the total number of moles. Seoul National University

57 Example 5-3: Determination of a PFR Volume
Determine the PFR volume necessary to produce 300 million pounds of ethylene a year from cracking a feed stream of pure ethane. The reaction is irreversible and follows an elementary rate law. We want to achieve 80% conversion of ethane, operating the reactor isothermally at 1100K at a pressure of 6 atm. C2H6  C2H4 + H2 A  B + C FB = 300x106 lb/year = lb-mol/sec FB = FAoX FAo = FB/X = 0.340/0.8 = lb-mol/sec Seoul National University

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.2) Ind. Eng. Chem. Process Des. Dev., 14, 218 (1975) Ind. Eng. Chem., 59(5), 70 (1967) Seoul National University

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C2H6  C2H4 + H2 Seoul National University

64 5.5 Pressure Drop in Reactors
 In liquid-phase reaction - the concentration of reactants is insignificantly affected by even relatively large change in the total pressure - ignore the effect of pressure drop on the rate of reaction when sizing liquid-phase chemical reactors - that is, pressure drop is ignored for liquid-phase kinetics calculations  In gas-phase reaction - the concentration of the reacting species is proportional to total pressure - the effects of pressure drop on the reaction system are a key factor in the success or failure of the reactor operation - that is, pressure drop may be very important for gas-phase reactions (Micro-reactors packed with solid catalyst) Seoul National University

65 Pressure drop and the rate law
기상반응에서는 반응 성분의 농도가 반응압력에 비례하므로 압력강하에 대한 고려가 필수적이다. • for an ideal gas, • For isothermal operation - determine the ratio P/P0 as a function of V or W - combine the concentration, rate law, and design equation - the differential form of the mole balance (design equation) must be used Seoul National University

66 Pressure drop and the rate law
• For example, - the second order isomerization reaction in a packed-bed reactor 2A  B + C -the mole balance (differential form) - rate law - stoichiometry for gas-phase reactions The differential form of PFR design equation must be used when there is a DP Seoul National University

67 Pressure drop and the rate law
• Then, the rate law - the larger the pressure drop from frictional losses, the smaller the reaction rate • Combining with the mole balance (assuming isothermal operation: T=T0) • Dividing by FA0(=v0CA0) (5-20) Seoul National University

68 Pressure drop and the rate law
The right-hand side is a function of only conversion and pressure -Another equation is needed to determine the conversion as a function of catalyst weight: that is, we need to relate the pressure drop to the catalyst weight (5-21) We need Seoul National University

69 Flow through a packed bed
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70 Flow through a packed bed
We need • The majority of gas-phase reactions are catalyzed by passing the reactant through a packed of catalyst particles • Ergun equation: to calculate pressure drop in a packed porous bed 에르군 식 (5-22) r laminar turbulent G=ru=superficial mass velocity [kg/m2s]; u=superficial velocity [m/s]; Dp=diameter of particle in the bed [m]; f=porosity=volume of void/total bed volume; 1- f =volume of solid/total bed volume, gc=1.0 [m/s2]; m=viscosity [kg/ms] • The gas density (r) is the only parameter that varies with pressure on the right-hand side. We are now going to calculate the pressure drop through the bed. Seoul National University

71 Flow through a packed bed
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72 Flow through a packed bed
We need • Equation of continuity - The reactor is operated at steady state, the mass flow rate at any point is equal to the entering mass flow rate • Gas-phase volumetric flow rate • Then, Seoul National University

73 Pressure drop in a packed bed reactor
• then, Ergun equation • Simplifying (5-24) (5-25) We need Ac • The catalyst weight, z Volume of solid Density of solid catalyst (5-26) Seoul National University

74 Pressure drop in a packed bed reactor
• Simplifying (5-27) (5-29) (5-30) Seoul National University

75 Pressure drop in a packed bed reactor
(5-30) ε < 0, the pressure drop (DP) will be less than ε = 0 ε > 0, the pressure drop (DP) will be greater than ε = 0 • For isothermal operation and (5-31) • The two expressions are coupled ordinary differential equations. We can solve them simultaneously using an ODE solver such as Polymath. • For the isothermal operation and ε = 0, we can obtain an analytical solution. • Polymath will combine the mole balance, rate law and stoichiometry Seoul National University

76 Pressure drop in a packed bed reactor
1 For the isothermal operation and ε = 0, (4-30) Analytical Solution If ε = 0 or ε X ≪ 1, we can obtain an analytical solution to Eq. (4-30) for isothermal operation (i.e., T=T0). For isothermal operation with ε = 0, Eq. (4-30) becomes Isothermal with ε = 0 Rearranging gives us Taking P/P0 inside the derivative (4-32) Seoul National University

77 Pressure Drop in a Packed Bed Reactor Integrating with P=P0 @ W=0
ε = 0 (5-33) T = T0 Seoul National University

78 Pressure drop in a packed bed reactor
If ε = 0 or ε X ≪ 1, we can obtain an analytical solution to Eq. (4-30) for isothermal operation (i.e., T=T0). For isothermal operation with ε = 0, Eq. (4-30) becomes Pressure ratio only for ε = 0 or ε X ≪ 1 (5-33) Pressure as a function of reactor length, z (5-34) Seoul National University

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5.5 Pressure Drop in Pipes Pressure drop for gases along the length of the pipe w/o packing Integrating with P=P0 at L=0, and assuming that f = constant Rearranging, we get Example 4-4: 1½” schedule 40 x1000-ft L (ap=0.018), DP<10% However, for high volumetric flow rates through microreactors, the DP may be significant. Example 4-4 Seoul National University

80 5.5.4 Analytical Solution for Reaction with Pressure Drop
For gas phase reactions, as the pressure drop increases, the concentration decreases, resulting in a decreased rate of reaction, hence a lower conversion when compared to a reactor without a pressure drop. 5-8 Seoul National University

81 Reaction with Pressure Drop Separating variable and Integrating
Conversion as a function of catalyst weight 2nd-order isothermal reaction A B Combining Mole balance: Separating variable and Integrating Rate law: (5-19) Stoichiometry: Gas-phase isothermal with e=0 (5-37) (5-33) Seoul National University

82 Reaction with Pressure Drop
Conversion as a function of catalyst weight Conversion for 2nd-order isothermal reaction in PFR with DP (5-38) Catalyst weight for 2nd-order isothermal reaction in PFR with DP (5-39) Seoul National University

83 The Optimum Catalyst Particle Diameter Why not pack the catalyst
Pressure drop dominant Internal diffusion inside catalyst dominant k ~ 1/Dp Conversion X Dp, opt Particle Diameter, Dp Why not pack the catalyst into a large diameter to reduce DP? Problems with large diameter tubes: Bypassing or Channeling Little efficient of heat transfer rate Seoul National University

84 Example 4-6 Production of ethylene glycol
Calculate the catalyst weight necessary to achieve 60% conversion when EO is to be made by the vapor phase catalytic oxidation of ethylene with air. 3 C2H4 4 Air W=? X=0.6 C2H4+ ½ O2  C2H4O Ag mol/s X=0.6 5 260oC, 10atm

85 Example 4-6 Production of ethylene glycol
2 H2, C2H4 C2H6 separator C2H6  C2H4 + H2 1 V=81 ft3, X=0.8 402 million lbC2H6 /yr 3 C2H4 4 Air W=45,440 lb, X=0.6 C2H4+ ½ O2  C2H4O Ag 5 O2, C2H4, N2, C2H4O 6 separator 260oC, 10bar H2O C2H4O absorber C2H4O(aq) 7 8 H2O, 0.9wt% H2SO4 200 million lb EG/yr 9 V=197 ft3, X=0.8 CH2OH C2H4O + H2O  Cat. Seoul National University

86 Example 4-6 Calculating X in a Reactor with Pressure Drop
Seoul National University

87 Example 4-6 Calculating X in a Reactor with Pressure Drop
Seoul National University

88 Example 4-6 Calculating X in a Reactor with Pressure Drop
Seoul National University

89 Example 4-6 Calculating X in a Reactor with Pressure Drop
Seoul National University

90 Example 4-6 Calculating X in a Reactor with Pressure Drop
Seoul National University

91 Negligible Pressure Drop in Pipes
Pressure drop along the length of the pipe Integrating with P=P0 at L=0, and assuming that f = constant Rearranging, we get Example 4-4: 1½” sch. 40 x1000-ft L (ap=0.018), DP<10% Seoul National University

92 5.6 Synthesizing a Chemical Plant
화학공장의 통합 설계 이 장의 예제문제에서 사용된 여러 가지 반응, 반응기 및 반응물과 생성물의 몰흐름에 대해 주의 깊게 연구하면 연간 4억 2백만 파운드의 에탄을 원료로 하여 2억 파운드의 에틸렌글리콜을 생산하는 화학공장을 창제할 수 있게 배치할 수 있다.  The profit from a chemical plant will be the difference between income from sales and the cost to produce the chemical Profit = value of products – cost of reactants – operating costs – separation costs  The operating cost: energy, labor, overhead, and depreciation of equipment Seoul National University

93 Production of ethylene glycol
2 H2, C2H4 C2H6 separator C2H6  C2H4 + H2 1 V=81 ft3, X=0.8 402 million lbC2H6 /yr 3 C2H4 4 Air W=45,440 lb, X=0.6 C2H4+ ½ O2  C2H4O Ag 5 O2, C2H4, N2, C2H4O 6 separator H2O C2H4O absorber C2H4O(aq) 7 8 H2O, 0.9wt% H2SO4 200 million lb EG/yr 9 V=197 ft3, X=0.8 CH2OH C2H4O + H2O  Cat. Seoul National University

94 Production of ethylene glycol
lb mol s M.W. bp(oC) 1 2 3 4 5 6 7 8 9 $/lb C2H6 0.425 0.040 C2H4 H2 O2 N2 EO EG 0.102 0.380 H2SO4 - 0.043 H2O Total 0.425 0.102 Seoul National University

95 Synthesizing a Chemical Plant
 Ethylene glycol = $0.69/lb (2x108 lb/yr) Ethane = $0.17/lb (4x108 lb/yr) Sulfuric acid = $0.15/lb (2.26x106 lb/yr) Operating cost = $1.2x107/yr  Profit = $138MM – $68MM - $0.34MM - $12MM = $57.7 MM  How the profit will be affected by conversion, separation, recycle stream, and operating costs? 2006년의 에탄, 황산 및 에틸렌글리콜의 가격은 파운드당 각각 $0.17, $0.15 그리고 $0.69이다. 최근 시세에 대해서는 을 참조하여라. Seoul National University

96 Seoul National University
Fig. 5-1 Isothermal Reaction Design Algorithm for Conversion Seoul National University

97 Seoul National University
마무리 이 장은 등온반응기에 대한 화학반응공학의 핵심적인 내용을 제공하고 있다. 이 장을 완전히 배운 뒤 독자들은 이 장에서 논의된 모든 반응기들, 즉 batch, CSTR, PFR, PBF, 멤브레인 반응기 및 반회분반응기에 대하여 다음과 같은 알고리즘 블록 쌓기를 적용할 수 있다. Evaluation Combine Stoichiometry Rate Law Mole Balance 독자들은 압력강하를 계산할 수 있으며, PBR의 촉매입자크기와 같은 시스템 변수들이 전화율에 미치는 영향을 기술할 수 있고, 촉매입자크기가 변할 때 최적의 전화율이 존재하는 이유를 설명할 수 있게 된다. 독자들은 전화율을 사용하여 화학반응공학문제들을 풀 수 있게 된다. 끝으로, 이 장을 완성한 후 독자들은 캘리포니아 기술사 시험문제들을 약 30분 안에 풀 수 있으며[P5-11B에서 P5-15B까지 참조] 오작동을 하는 반응기를 진단하고 문제점을 해결할 수 있게 된다[P5-8B 참조]. Seoul National University

98 Why not pack the catalyst
Closure After completing this chapter, you should be able to account for pressure drop and describe the effect of system variables such as particle size on the conversion and explain why there is an optimum in the conversion when catalyst particle size is varies. Pressure drop dominant Internal diffusion inside catalyst dominant k ~ 1/Dp Conversion X Dp, opt Particle Diameter, Dp Why not pack the catalyst into a large diameter to reduce DP? Problems with large diameter tubes: Bypassing or Channeling Smaller heat transfer area