Fundamentals of reaction engineering worked exam

FUNDAMENTALS OF REACTION ENGINEERING – WORKED EXAMPLES F R EE EE SST U D Y BBO OO OK S RAFAEL KANDIYOTI FREE STUDY BOOKS WWW.BOOKBOON.COM Rafael Kandiyoti Fundamentals of Reaction Engineering Worked Examples Download free books at BookBooN.com Fundamentals of Reaction Engineering – Worked Examples © 2009 Rafael Kandiyoti & Ventus Publishing ApS ISBN 978-87-7681-512-7 Download free books at BookBooN.com Fundamentals of Reaction Engineering – Worked Examples Contents Contents Chapter I: Homogeneous reactions – Isothermal reactors Chapter II: Homogeneous reactions – Non-isothermal reactors 13 Chapter III: Catalytic reactions – Isothermal reactors 22 Chapter IV: Catalytic reactions – Non-isothermal reactors 31 Please click the advert what‘s missing in this equation? You could be one of our future talents MAERSK INTERNATIONAL TECHNOLOGY & SCIENCE PROGRAMME Are you about to graduate as an engineer or geoscientist? Or have you already graduated? If so, there may be an exciting future for you with A.P Moller - Maersk www.maersk.com/mitas Download free books at BookBooN.com Fundamentals of Reaction Engineering – Worked Examples I Homogenepus reactions - Isothermal reactors Worked Examples - Chapter I Homogeneous reactions - Isothermal reactors Problem 1.1 Problem 1.1a: For a reaction AoB (rate expression: rA = kCA ) , taking place in an isothermal tubular reactor, starting with the mass balance equation and assuming plug flow, derive an expression for calculating the reactor volume in terms of the molar flow rate of reactant ‘A’ Solution to Problem 1.1a: Assuming plug flow, the mass balance over a differential volume element of a tubular reactor is written as: (MA nA )V Taking the limit as 'VR  - (MA nA)V+'V dn  A  rA dVR MA rA 'VR - ; rA dn  A ; VR dVR =  n Ae dn A rA ³ n A0 Problem 1.1b: If the reaction is carried out in a tubular reactor, where the total volumetric flow rate 0.4 m3 s-1, calculate the reactor volume and the residence time required to achieve a fractional conversion of 0.95 The rate constant k=0.6 s-1 Solution to Problem 1.1b: The molar flow rate of reactant “A” is given as, n A0 (  x A ) kn A Substituting, kC A vT nA Also nA C AvT The rate is then given as rA v  T k VR VR  n Ae ³ n A0 dn A nA vT k x Ae ³ n A0 dx A n A0 (  x A ) vT k x Ae ³ dx A (  xA ) vT ln(  x Ae ) # m3 k In plug flow the residence time is defined as W{ VR vT ê m3 ô 1 ằ ơô m s ẳằ > s@ ; W / 0.4 s Problem 1.2 The gas phase reaction Ao2S is to be carried out in an isothermal tubular reactor, according to the rate expression rA = k CA Pure A is fed to the reactor at a temperature of 500 K and a pressure of bar, at a rate of 1000 moles s-1 The rate constant k= 10 s-1 Assuming effects due to pressure drop through the reactor to be negligible, calculate the volume of reactor required for a fractional conversion of 0.85 Download free books at BookBooN.com Fundamentals of Reaction Engineering – Worked Examples Data: R, the gas constant P the total pressure nT0 total molar flow rate at the inlet k, reaction rate constant I Homogenepus reactions - Isothermal reactors : 8.314 J mol-1 K-1 = 0.08314 bar m3 kmol-1 K-1 : bar : 1000 moles s-1= kmol s-1 : 10 s-1 Solution to Problem 1.2 In Chapter I we derived the equation for calculating the volume of an isothermal tubular reactor, where the first order reaction, A o 2S, causes volume change upon reaction: ẵ ê RT (Eq 1.37) VR nT ®(  y A0 )ln ô ằ  y A0 x Ae ắ Pk  x Ae ẳ In this equation, for pure feed “A”, yA0 { nA0/nT0 = ( 0.08314 )( 500 ) VR ( )^2 ln( 6.67 )  0.85` ( )( 10 ) VR 6.1 m3 Let us see by how much the total volumetric flow rate changes between the inlet and exit of this reactor First let us review how the total molar flow rate changes with conversion: = nI0 nI nA0 – nA0 xA nA = nS0 + nA0 xA nS = = nT0 + nA0xA nT for the inert component for the reactant for the product nT is the total molar flow rate nT RT ( )( 0.08314 )( 500 ) / 20.8 m3 s-1 P vT ,exit nT ,exit RT / P; but nT,exit nT0  n A0 x A,exit vT and nT vT ,exit n A0 ; therefore (  0.85 )( 0.08314 )( 500 ) / 38.5 m3 s 1 vT 20.8 m3 s 1 ; vT ,exit 38.5 m3 s 1 The difference is far from negligible! In dealing with gas phase reactions, rates are often expressed in terms of partial pressures: rA= kpA, as we will see in the next example Download free books at BookBooN.com Fundamentals of Reaction Engineering – Worked Examples I Homogenepus reactions - Isothermal reactors Problem 1.3 The gas phase thermal decomposition of component A proceeds according to the chemical reaction k1 ZZZ X A YZZ Z BC k2 with the reaction rate rA defined as the rate of disappearance of A: rA k1C A  k2 CB CC The reaction will be carried out in an isothermal continuous stirred tank reactor (CSTR) at a total pressure of 1.5 bara and a temperature of 700 K The required conversion is 70 % Calculate the volume of the reactor necessary for a pure reactant feed rate of 4,000 kmol hr-1 Ideal gas behavior may be assumed Data k1 108 e10,000 / T s-1 (T in K) k2 108 e8,000 / T Gas constant, R = 0.08314 m3 kmol-1 s-1 (T in K) bar m3 kmol-1 K-1 www.job.oticon.dk Download free books at BookBooN.com Fundamentals of Reaction Engineering – Worked Examples I Homogenepus reactions - Isothermal reactors Solution to Problem 1.3 Using the reaction rate expression given above, the “design equation” (i.e isothermal mass balance) for the CSTR may be written as: VR n A0  n A k1C A  k2CB CC n A0 x AvT , where k2 k1n A  ( )nB nC vT ni vT Ci (Eq 1.1.A) We next write the mole balance for the reaction mixture: nA nA0  nA0 x A nB  nA x A nC  nA0 x A nT nA0  nA0 x A nA0 (1  x A ) This result is used in the “ideal gas” equation of state: vT nT RT PT n A0 RT (  xA ) PT Substituting into Eq 1.1.A derived above, we get: VR nA0 x A (nA0 RT )(1  x A ) k2 PT nA2 x A2 ẵ PT đk1nA0  k1nA0 xA  ¾ nA0 RT (1  x A ) ¿ ¯ Simplifying, § nA0 RT · x A (1  x A ) ă © PT ¹ ­k  k x  k2 PT xA ẵ đ 1 A ắ RT  x A ¿ ¯ kP To further simplify, we define a  T RT VR VR § nA0 RT · x A (1  x A ) ă 2 â PT k1  k1 x A  k1 x A  ax A  k1 x A (Eq 1.1.B) Next we calculate the values of the reaction rate constants k1 62.49 s 1 ; xA,exit = 0.70 ; T = 700 K k2 = 1088 m3s-1kmol-1; PT = 1.5 bar Substituting the data into Eq 1.1.B 4000 0.08314 700 (1000) 0.5 ¯°  0.85 ¿° 2.69 u 104 5.67 ; (k + k ) = 2.98u10-4 + 0.74u10-4 = 3.72u10-4 VR k VR = 4.1 m3 Download free books at BookBooN.com I Homogenepus reactions - Isothermal reactors Fundamentals of Reaction Engineering – Worked Examples Problem 1.5 A first order, liquid phase, irreversible chemical reaction is carried out in an isothermal continuous stirred tank reactor A o products The reaction takes place according to the rate expression rA = k CA In this expression, rA is defined as the rate of reaction of A (kmol m-3 s-1), k as the reaction rate constant (with units of s-1) and CA as the concentration of A (kmol m-3) Initially, the reactor may be considered at steady state At t = 0, a step increase takes place in the inlet molar flow rate, from nA0,ss to nA0 Assuming VR (the reactor volume) and vT (total volumetric flow rate) to remain constant with time, a Show that the unsteady state mass balance equation for this CSTR may be expressed as: n A0 dn A ê  ô  k ằ nA , dt ơW W ẳ Also write the appropriate initial condition for this problem The definitions of the symbols are given below b Solve the differential equation above [Part a], to derive an expression showing how the outlet molar flow rate of the reactant changes with time c How long would it take for the exit flow rate of the reactant A to achieve 80 % of the change between the original and new steady state values (i.e from nA,ss to nA) ? DEFINITIONS and DATA: nA0,ss steady state inlet molar flow rate of reactant A before the change : 0.1 kmol s-1 new value of the inlet molar flow rate of reactant A : 0.15 kmol s-1 nA0 reactor volume : m3 VR k reaction rate constant : 0.2 s-1 total volumetric flow rate : 0.1 m3 s-1 vT nA,ss nA t W rA CA steady state molar flow rate of reactant A exiting from the reactor, before the change, kmol s-1 the molar flow rate of reactant A exiting from the reactor at time t, kmol s-1 time, s average residence time, VR/vT , s rate of reaction of A , kmol m-3 s-1 concentration of A, kmol m-3 Download free books at BookBooN.com 10 I Homogenepus reactions - Isothermal reactors Fundamentals of Reaction Engineering – Worked Examples Solution to Problem 1.5 a CSTR mass balance with accumulation term: material balance on component A, over the volume VR for period ' t: ^n A0 ` (t )  n A (t )  r A (t )VR 't N A t ' t  N A t Bars indicate averages over the time interval 't, and NA(t) is the total number of moles of component A in the reactor at time t Also C A n A / vT N A / VR , and the average residence time, W VR / vT Taking the limit as 't o dt in the above equation, we get: dNA dt nA0  nA  rAVR Combining this with N A { VR C A VR (nA / vT ) leads to the equation: VR dnA nA0  nA  kC AVR vT dt Rearranging, nA0  nA  knAVR vT dnA , and, dt W dnA ª  ô  k ằ nA dt ơW ẳ nA W with initial condition: at t=0, ª1 nA=nA,ss º b We define E { «  k » and write the complementary and particular solutions for the first order ordinary ơW ẳ differential equation derived above nA,complementary = C1e -E t ; nA,particular = C2 (const) where the solution is the sum of the two solutions: nA (t) = nA,c (t)+ nA,p (t) We substitute the particular solution into differential equation to evaluate the constants: ê1 ôơW  k »¼ C2 nA0 W The most general solution can then be written as n A ; C2 C1e E t  nA0 EW n A0 EW To derive C1, we substitute the initial condition: nA=nA,ss at t=0, into the general solution: nA, ss C1  nA0 EW ; Ÿ C1 nA, ss  nA0 EW The solution of the differential equation can then be written as: nA c lim n A t of n A0 EW nA0 ẵ  E t nA0  đnA, ss  ¾e EW ¿ EW ¯ where EW =1+kW n A0 Ÿ The new steady state exit molar flow rate of A  kW Download free books at BookBooN.com 11 I Homogenepus reactions - Isothermal reactors Fundamentals of Reaction Engineering – Worked Examples The relationship linking t, nA,ss and nA is derived from the solution of the differential equation Đ Đ nA Ãẵ nA à ẵ  nA, ss ă nA, ss  ă EW W â W â EW t đln ắ đln ắ  kW ° §  kW ° § nA0 · ° nA · °  nA ¸ n  ăâ A E W áạ ăâ E W ¹ °¿ W 0.1 W 1 kW nA, final 20 s ; nA, ss 20  (0.2)(20) nA EW nA0  kW nA0, ss 1 kW 0.1  (20)(0.2) 0.02 kmol s 1 4s 0.15  (20)(0.2) 0.03 kmol s 1 For 80 % of the change to be accomplished: nA = nA,ss + 0.8 (nA,final - nA,ss) nA= 0.8 nA,final + 0.2 nA,ss = 0.8(0.03) + 0.2 (0.02) = 0.028 kmol s-1 ên ẵ W ê nA0  nA, ss »  ln « A0  nA » ắ t đln ô  kW E W ẳ ơEW ẳ t ^ln > 0.03  0.02@  ln > 0.03  0.028@` 6.44 s t 6.44 s Please click the advert Join the Accenture High Performance Business Forum © 2009 Accenture All rights reserved Always aiming for higher ground Just another day at the office for a Tiger On Thursday, April 23rd, Accenture invites top students to the High Performance Business Forum where you can learn how leading Danish companies are using the current economic downturn to gain competitive advantages You will meet two of Accenture’s global senior executives as they present new original research and illustrate how technology can help forward thinking companies cope with the downturn Visit student.accentureforum.dk to see the program and register Visit student.accentureforum.dk Download free books at BookBooN.com 12 II Homogeneous reactions - Non-isothermal reactors Fundamentals of Reaction Engineering – Worked Examples Worked Examples - Chapter II Homogeneous reactions – Non-isothermal reactors Problem 2.1 Problem 2.1a: Identify the terms in the energy balance equation for a continuous stirred tank reactor: T ¦ ni0 ³ C pi dT  VR ¦ H fi ( T )ri  Q T0 1st term: heat transfer by flow 2nd term: heat generation/absorption by the reaction(s) 3rd term: sensible heat exchange Steady state: no accumulation Solution to Problem 2.1a: Problem 2.1b: What form does the equation take for the single reaction: A o B + C Solution to Problem 2.1b: Hfi ri= HfA rA + HfB rB + HfC rC and rA = - rB = -rC = rA {HfA – HfB – HfC} = - rA'Hr Then the energy balance equation takes the form: T ¦ ni0 ³ C pi dT  VR ' H r rA  Q T0 Problem 2.1c: For a simple reversible-exothermic reaction A œ B, qualitatively sketch the locus of maximum reaction rates on an xA vs T diagram (together with the xA,eq line) Solution to Problem 2.1c: xA xA,eq Conversion xA,OPT Temperature T Download free books at BookBooN.com 13 II Homogeneous reactions - Non-isothermal reactors Fundamentals of Reaction Engineering – Worked Examples Problem 2.1d: The chemical reaction A o B is carried out in a continuous stirred tank reactor Calculate the adiabatic temperature rise for 100 % conversion The mole fraction of A in the input stream yA0 = (nA0/nT0) = The total heat capacity, C p = 80 J mol-1 K-1 The heat of reaction, ' H r  72 kJ mol-1 Solution to Problem 2.1d: From notes, the simplified energy balance for an adiabatic CSTR xA Solving for T  T0 C p ẵ đ ắ T  T0 °¯ y A0 ' H r °¿ 'H r ( T  T0 ) 72,000 80 Cp 900 K Problem 2.2 Problem 2.2a: Sketch the operating line on an xA vs T diagram (i) for an endothermic reaction carried out in an adiabatic CSTR; (ii) for an exothermic reaction carried out in an adiabatic CSTR Solution to Problem 2.2a: xA 'Hr > 'Hr < To T Problem 2.2b: Sketch the operating line on an xA vs T diagram for an exothermic reaction carried out in a CSTR, with cooling (Q < ): Solution to Problem 2.2b: xA Q