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(1)Fundamentals of Reaction Engineering - Examples Rafael Kandiyoti. Download free books at.

(2) Rafael Kandiyoti. Fundamentals of Reaction Engineering Worked Examples. 2 Download free eBooks at bookboon.com.

(3) Fundamentals of Reaction Engineering – Worked Examples © 2009 Rafael Kandiyoti & Ventus Publishing ApS ISBN 978-87-7681-512-7. 3 Download free eBooks at bookboon.com.

(4) Fundamentals of Reaction Engineering – Worked Examples. Contents. Contents Chapter I:. Homogeneous reactions – Isothermal reactors. 5. Chapter II:. Homogeneous reactions – Non-isothermal reactors. 13. Chapter III:. Catalytic reactions – Isothermal reactors. 22. Chapter IV:. Catalytic reactions – Non-isothermal reactors. 31. www.sylvania.com. We do not reinvent the wheel we reinvent light. Fascinating lighting offers an infinite spectrum of possibilities: Innovative technologies and new markets provide both opportunities and challenges. An environment in which your expertise is in high demand. Enjoy the supportive working atmosphere within our global group and benefit from international career paths. Implement sustainable ideas in close cooperation with other specialists and contribute to influencing our future. Come and join us in reinventing light every day.. Light is OSRAM. 4 Download free eBooks at bookboon.com. Click on the ad to read more.

(5) I. Homogenepus reactions - Isothermal reactors. Fundamentals of Reaction Engineering – Worked Examples. 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 0. - (MA nA)V+'V. dn A rA dVR. MA rA 'VR. -. 0 ; rA. dn A ; VR dVR. = 0. . 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 ( 1 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. 0. n A0 dx A n A0 ( 1 x A ). vT k. x Ae. ³. 0. dx A ( 1 xA ). vT ln( 1 x Ae ) # 2 m3 k. In plug flow the residence time is defined as. W{. VR vT. ª m3 º « 3 1 » ¬« m s ¼». > s@ ;. W. 2 / 0.4. 5 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 2 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.. 5 Download free eBooks at bookboon.com.

(6) I. Homogenepus reactions - Isothermal reactors. 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. : 8.314 J mol-1 K-1 = 0.08314 bar m3 kmol-1 K-1 : 2 bar : 1000 moles s-1= 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: ½° ª 1 º RT ° (Eq. 1.37) VR nT 0 ®( 1 y A0 )ln « » y A0 x Ae ¾ Pk °¯ °¿ ¬ 1 x Ae ¼ In this equation, for pure feed “A”, yA0 { nA0/nT0 = 1. ( 0.08314 )( 500 ) VR ( 1 )^2 ln( 6.67 ) 0.85` ( 2 )( 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 + 2 nA0 xA nS = ----------------------------------= nT0 + nA0xA nT. for the inert component for the reactant for the product nT is the total molar flow rate. nT 0 RT ( 1 )( 0.08314 )( 500 ) / 2 20.8 m3 s-1 P vT ,exit nT ,exit RT / P; but nT,exit nT0 n A0 x A,exit vT 0. and nT 0 vT ,exit. n A0 ; therefore. ( 1 0.85 )( 0.08314 )( 500 ) / 2 38.5 m3 s 1. vT 0. 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.. 6 Download free eBooks at bookboon.com.

(7) I. Homogenepus reactions - Isothermal reactors. Fundamentals of Reaction Engineering – Worked Examples. 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. m3 kmol-1 s-1 (T in K) bar m3 kmol-1 K-1. Gas constant, R = 0.08314. 360° thinking. .. 360° thinking. .. 360° thinking. .. Discover the truth at www.deloitte.ca/careers. © Deloitte & Touche LLP and affiliated entities.. Discover the truth at www.deloitte.ca/careers. © Deloitte & Touche LLP and affiliated entities.. 7. © Deloitte & Touche LLP and affiliated entities.. Discover the truth at www.deloitte.ca/careers Click on the ad to read more Download free eBooks at bookboon.com. © Deloitte & Touche LLP and affiliated entities.. D.

(8) I. Homogenepus reactions - Isothermal reactors. Fundamentals of Reaction Engineering – Worked Examples. 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 0 x A. nC. nA0 x A. nT. nA0 nA0 x A. nA0 (1 xA ) .. This result is used in the “ideal gas” equation of state:. vT. nT RT PT. n A0 RT ( 1 xA ) . PT. Substituting into Eq. 1.1.A derived above, we get:. VR. nA0 x A (nA0 RT )(1 x A ) . k2 PT nA2 0 x A2 ½ PT ®k1nA0 k1nA0 xA ¾ nA0 RT (1 x A ) ¿ ¯. Simplifying, § nA0 RT · x A (1 x A ) . ¨ ¸ 2 © PT ¹ k k x k2 PT xA ½ ® 1 1 A ¾ RT 1 x A ¿ ¯ kP To further simplify, we define a 2 T . RT VR. VR. § nA0 RT · x A (1 x A ) 2 ¨ ¸ 2 2 © PT ¹ 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 2 4000

(9) 0.08314

(10) 700

(11) ª 0.7 1 0.7

(12). VR « 2 2 3600

(13) 1.5

(14) «¬ 62.49 k1 x A a 0.7

(15) VR. º »;a »¼. 1088

(16) 1.5

(17) 0.08314

(18) 700

(19). 43.11

(20) 0.11

(21) # 4.74 m3. 8 Download free eBooks at bookboon.com. 28.

(22) I. Homogenepus reactions - Isothermal reactors. Fundamentals of Reaction Engineering – Worked Examples. Problem 1.4 The liquid phase reaction for the formation of compound C k1 A B o C. with reaction rate rA1. k1C AC B , is accompanied by the undesirable side reaction k2 A B o D. with reaction rate rA2. k2 C ACB . The combined feed of 1000 kmol per hour is equimolar in A and B. The two. components are preheated separately to the reactor temperature, before being fed in. The pressure drop across the reactor may be neglected. The volumetric flow rate may be assumed remain constant at 22 m3 hr-1. The desired conversion of A is 85 %. However, no more than 20 % of the amount of “A” reacted may be lost through the undesirable side reaction. Find the volume of the smallest isothermally operated tubular reactor, which can satisfy these conditions. Data k1 = 2.09u1012 e-13,500/T k2 = 1017 e-18,000/T. m3 s-1 kmol-1 m3 s-1 kmol-1. ; ;. (T in K) (T in K). Solution to Problem 1.4 For equal reaction orders: (n1 / n2 ) (k1 / k2 ) , where n1 is moles reacted through Reaction 1, and n2 is moles reacted through Reaction 2. Not allowing more than 20 % loss through formation of by-product “D” implies: (n1 / n2 ) (k1 / k2 ) t 4 . Since 'E2 > 'E1, the rate of Reaction 2 increases faster with temperature than the rate of Reaction 1. So the maximum temperature for the condition ^ n1 / n2 t 4` will be at (n1 / n2 ) (k1 / k2 ) 4 . This criterion provides the equation for the maximum temperature:. 2.09 u 10 12 4500 / T e 10 17. 4 ;. 4 u 1017 ½ . Solving, T ln ® 12 ¾ ¯ 2.09 u 10 ¿. 4500 T. 4500 12.16. 370 K .. The temperature provides the last piece of information to write the integral for calculating the volume. 0.85. n dx. A ; ³0 k1 A0 k2

(23) C ACB. VR. 0.85. ³0. VR. vT2 n A0 dx A kn 2A0 1 x A

(24). 2. define k k1 k2. vT2. 0.85. ° ½° 22

(25) 1 VR 1¾ ® k 3600

(26) (1000) 0.5

(27) ¯° 1 0.85

(28) ¿° 2.69 u 104 5.67

(29) ; (k + k ) = 2.98u10-4 + 0.74u10-4 = 3.72u10-4 2. VR. dx. A ³ 0 kn A0 1 x A

(30) 2. k. 1. 2. VR = 4.1 m3. 9 Download free eBooks at bookboon.com.

(31) 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 ª 1 º « k » nA , dt ¬W W ¼ 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 : 2 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. 10 Download free eBooks at bookboon.com.

(32) 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 1 º « k » nA dt ¬W ¼. nA 0. 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 ¬W ¼ 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:. ª1 º «¬W k »¼ C2. nA0. W. The most general solution can then be written as n A. nA0. ; C2. C1e E. t. . 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. ;. . 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 ½ E t nA0 ®nA, ss ¾e EW ¿ EW ¯ where EW =1+kW.. n A0 The new steady state exit molar flow rate of A 1 kW. 11 Download free eBooks at bookboon.com.

(33) 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 § § nA 0 ·½ nA 0 · ½ nA, ss ¸ ° ° ¨ nA, ss ° ¨ ¸° EW ¹° W ° © W ° © EW ¹° t ®ln ¾ ®ln ¾ 1 kW ° § 1 kW ° § nA0 · ° nA 0 · ° nA ¸ n °¯ ¨© A E W ¸¹ °¿ °¯ ¨© E W ¹ °¿. W. 2 0.1. W 1 kW. nA, final. 20 s. ;. nA, ss. 20 1 (0.2)(20) nA 0. EW. nA0 1 kW. nA0, ss 1 kW. 0.1 1 (20)(0.2). 0.02 kmol s 1. 4s. 0.15 1 (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 º ªn º °½ W ° ª nA0 nA, ss » ln « A0 nA » ¾ t ®ln « 1 kW ¯° ¬ E W ¼ ¬EW ¼ ¿° t. 4 ^ln > 0.03 0.02@ ln > 0.03 0.028@`. 6.44 s. t. 6.44 s. We will turn your CV into an opportunity of a lifetime. Do you like cars? Would you like to be a part of a successful brand? We will appreciate and reward both your enthusiasm and talent. Send us your CV. You will be surprised where it can take you.. 12 Download free eBooks at bookboon.com. Send us your CV on www.employerforlife.com. Click on the ad to read more.

(34) 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 0 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:. 6 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 0 T0. Problem 2.1c: For a simple reversible-exothermic reaction A 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. 13 Download free eBooks at bookboon.com. T.

(35) 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) = 1. 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

(36). ° C p ½° ® ¾ T T0

(37) . °¯ 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 > 0. 'Hr < 0. 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 < 0 ): Solution to Problem 2.2b: xA. Q<0. (heat removal). B. A. Q = 0 (adiabatic). T. 14 Download free eBooks at bookboon.com.

(38) II. Homogeneous reactions - Non-isothermal reactors. Fundamentals of Reaction Engineering – Worked Examples. Problem 2.3 The gas phase reaction k1. oB A m k2. with reaction rate expression rA k1 p A k2 pB will be carried out in a continuous stirred tank reactor. The feed is pure A. The required conversion of xA is 0.62. Calculate a) the minimum reactor size in which the required conversion can be achieved, and, b) the amount of heat to be supplied or removed for steady state operation.. Ideal gas behaviour may be assumed. DATA: Volumetric flow rate of feed vT0 : 10 m3 sec-1 Average heat capacities of A and B : 42 kJ kmol-1 K-1 : 573 K Feed temperature, T0 Pressure of feed line and reactor : 10 bar Gas constant, R : 0.08314(bar m3K-1kmol-1) = 8.314 kJ K-1 kmol-1 Heat of reaction, 'Hr : - 41570 kJ kmol-1 3 kmol m-3 s-1 bar-1 k1 = 10 exp{-41570 / RT} 6 k2 = 10 exp{-83140 / RT} kmol m-3 s-1 bar-1 Activation energies are given in kJ kmol-1. I joined MITAS because I wanted real responsibili� I joined MITAS because I wanted real responsibili�. Real work International Internationa al opportunities �ree wo work or placements. �e Graduate Programme for Engineers and Geoscientists. Maersk.com/Mitas www.discovermitas.com. � for Engin. M. Month 16 I was a construction M supervisor ina cons I was the North Sea supe advising and the N he helping foremen advi ssolve problems Real work he helping International Internationa al opportunities �ree wo work or placements ssolve p. 15 Download free eBooks at bookboon.com. Click on the ad to read more.

(39) II. Homogeneous reactions - Non-isothermal reactors. Fundamentals of Reaction Engineering – Worked Examples. Solution to Problem 2.3 The minimum reactor size for achieving a given conversion implies operating at the temperature where the reaction rate is a maximum. The reaction rate expression is written as:. PT 1 x A

(40) and pB. k1 p A k2 pB , where p A. rA. k1 PT 1 x A

(41) k2 PT x A. rA. PT x A .. PT ^k1 k1 k2

(42) x A `. § dr · The temperature for the maximum reaction rate can be found from ¨ A ¸ © dT ¹ x A. § drA · ¨ ¸ © dT ¹ x A. PT. xA dk2 dT. where and. k20. ' E2 RT. 2. dk1 dk1 dk2 ½ ® ¾ PT x A dT ¯ dT dT ¿. 0. 0. 1 dk2 / dT

(43) 1 dk1 / dT

(44) e ' E2. RT. ;. dk1 dT. k10. ' E1 RT. 2. e ' E1. RT. ' E2 ' E1

(45) ½ k20 ' E2 exp ® ¾ k10 ' E1 RT ¯ ¿ 83140 41570

(46) ½ 106 § 83200 · § 5000 · exp ® ¾ 2000 exp ¨ ¸ 3 ¨ 41600 ¸ 8.314 T ¹ © T ¹ 10 © ¯ ¿ 1 xA . 1 2000 e 5000 / T dk2 dT dk1 dT. dk2 dT dk1 dT. Solving for the temperature:. 5000. T. § 1 · 1 ½ ln ®¨¨ 1 ¸¸ ¾ ¹ 2000 ¿ ¯© x A. 618 K. The reactor volume is obtained from the mass balance equation. VR n A0. PT vT 0 RT. n A0 x A PT ^k1 k1 k2

(47) x A `. 10

(48) 10

(49) 0.08314

(50) 573

(51). 2.099kmol / sec. At 618 K, k1 = 0.306 ; k2 = 0.094. Combining, we find:. VR = 2.24 m3 b) The heat balance:. Q. ' H r n A0 x A nT 0 C p T T0

(52). Q. 54098 3967. 50131 # 50000 kJ / sec. The amount of heat to be removed is. Q # 50000 kJ / sec. 16 Download free eBooks at bookboon.com.

(53) II. Homogeneous reactions - Non-isothermal reactors. Fundamentals of Reaction Engineering – Worked Examples. Problem 2.4 A continuous stirred tank reactor is used for carrying out the liquid phase reaction: A + B o products. The reaction rate expression is rA = k(T) CA CB kmol h-1 m-3. The total molar flow rate at the inlet is 3 kmol h1 . The inlet flow is equimolar in A and B. The total volumetric flow rate (vT) is: 1 m3 h-1. Calculate the operating temperature required for a fractional conversion of 0.7. Also calculate how much heat must be added to the reactor to maintain the system at steady state, if the feed is introduced at the reactor operating temperature. The vessel is designed to withstand the vapour pressure of the reaction mixture. No reaction takes place in the vapour phase. Density changes due to (i) the chemical reaction and (ii) changes in temperature may be ignored. Data: Reactor volume, VR Heat of reaction, 'Hr Reaction rate constant, k(T),. : 3 m3 : 35,000 kJ (kmol A reacted)-1 : 1015 e-13000/T m3 kmol-1 h-1 (T in K). Solution to Problem 2.4. The material balance equation for a CSTR may be written as:. VR. n A0 n A rA. n A0 x AvT2. k( T )n 2A0 ( 1 x A )2. where ni ni0 vT xA nA. = molar flow rate of component i = initial molar flow rate of component i = total volumetric flow rate = fractional conversion of A = CAvT. Solving for k(T), we get:. k( T ). x AvT2. ( 0.7 )( 1 )2 2. VR n A0 ( 1 x A ). 1015 exp{ 13000 / T }. 1.73 m3 kmol 1 h 1. 2. ( 3 )( 1.5 )( 0.3 ). 13000 T 382.3 K. 1.73 ; . T. 1.73 ½ ln ® 15 ¾ 34.0 ¯ 10 ¿. Heat balance. nT 0 C p ( T0 T ) ' H R x An A0 Q 0 for 'T Q. 0, heat must be added. ' H R x An A0 35000( 0.7 )( 1.5 ) 36750 kJ h 1 Q. 36750 kJ h 1. 17 Download free eBooks at bookboon.com.

(54) II. Homogeneous reactions - Non-isothermal reactors. Fundamentals of Reaction Engineering – Worked Examples. Problem 2.5 The gas phase reaction A o B + C is carried out in an adiabatic continuous stirred tank reactor. The reaction rate expression is given by the equation: rA = k(T) pA, where k(T) is the reaction rate constant and pA the partial pressure of the reactant A. The feed stream is pre-heated to 600 K; it contains an equimolar amount of reactant A and inert component D and. The total pressure is 1 bar; it may be assumed constant throughout the system. Calculate the steady state reactor operating temperature and the conversion of the reactant. (Note: There are two equally valid solutions to this problem. One of the two solutions will suffice as an adequate answer.) Ideal gas behaviour may be assumed. You may also assume total heat capacities of the feed and product streams to be equal. Data:. Reactor volume, VR Inlet total molar flow rate, nT0 Reaction rate constant, k(T) Heat of reaction, 'Hr Heat capacity of the feed stream, C p. : 10 m3 : 10 kmol s-1 : 1.0 x 1020 e-30,000/T kmol bar-1 m-3 s-1 : 48 000 kJ (kmol A reacted)-1 : 200 kJ kmol-1 K-1. 18 Download free eBooks at bookboon.com. (T in K). Click on the ad to read more.

(55) II. Homogeneous reactions - Non-isothermal reactors. Fundamentals of Reaction Engineering – Worked Examples. Solution to Problem 2.5 The mass and energy balances must be solved simultaneously. The energy balance equation:. nT 0 C p ( T0 T ) nA0 x A ' H R where xA : vT : VR: ni : nT0 :. 0 ; nT 0 2nA0. fractional conversion total volumetric flow rate Reactor volume molar flow rate of component i total inlet molar flow rate. From the energy balance equation. nT 0 C P T ( 2 )( 200 ) ( T0 T ) ( T 600 ) ; xA 5 n A0 ' H r 48000 120 n A0 x A The mass balance equation: VR . Next we need to calculate nT as a function of xA. k pA xA. = nA0 – nA0 xA nA nI = nI0 = nA0xA nB nC = nA0xA ___________________ = nT0 + nA0xA nT. pA. m inert component. nA ptotal nT. n A0 ( 1 x A ) ptotal ; then nT 0 n A0 x A. ½ nT 0 nA0 x A ½ xA 1 ® ¾® ¾ ¯VR k (T ) ptotal ¿ ¯ 1 x A ¿. Solve 1. ½ 10 5 x A ½ xA ® 21 30000 / T ¾ ® ¾ with ) ¿ ¯ 1 xA ¿ ¯10 (e. o. VR. nA0 x A nT 0 nA0 x A ; k (T ) nA0 (1 x A ) ptotal. ½ 10 5 x A ½ xA 1 ® ¾ 20 30000 / T ¾ ® ) ¿ ¯ 1 xA ¿ ¯10(1)(1)(10 )(e. xA. T 5 with 120. 640 0.333 3.884 4.390u10-21 1.326. 645 0.375 4.453 6.313u10-21 1.129. Two possible solutions T xA numerator exp(-30000/T) RHS. 635 0.292 3.346 3.035u10-21 1.557. ANS: T = ~ 647 K ; xA = ~0.39. 19 Download free eBooks at bookboon.com. 650 0.417 5.039 9.029u10-21 0.957. 655 0.458 5.629 1.284u10-20 0.809.

(56) II. Homogeneous reactions - Non-isothermal reactors. Fundamentals of Reaction Engineering – Worked Examples. Alternative solution: T 601 xA 0.00833 numerator 0.0836 exp(-30000/T) 2.096u10-22 RHS 0.402. 602 0.0167 0.168 2.277u10-22 0.750. 603 0.025 0.281 2.437u10-22 1.183. 604 0.0333 0.339 2.686u10-22 1.306. 605 0.0417 0.426 2.916u10-22 1.524. ANS: T ~ 602.5 ; xA= ~ 0.021. Problem 2.6 The rate of the first order irreversible gas-phase reaction, A o B is given by rA = k pA. The reaction will be carried out in a tubular adiabatic reactor. The reaction is exothermic. Calculate the reactor volume, VR (m3), required for a fractional conversion of “A”: xA = 0.9. Plug flow and ideal gas behaviour may be assumed. Definitions & data: Inlet mole fractions: yA0 = 0.20; yI0 (inerts) = 0.8 T0 nT0 PT k 'Hr. Feed temperature inlet total molar flow rate total pressure reaction rate constant Heat of Reaction. :400 K :1 kmol s-1 :1 bar :1.8 x 104 exp{- 5000/T} kmol s-1 m-3 bar-1 (T in K) :- 50,000 kJ (kmol A reacted)-1. Cp. Heat capacity of the reaction mixture. : 40 kJ kmol-1 K-1. Q rA pA nA0. Heat exchange with surroundings Reaction rate of component A, kmol m-3 s-1 partial pressure of A, bar molar flow rate of A, kmol s-1. Note 1. The operating line for a tubular reactor assumed to operate in plug flow is given by:. dT dx A. n A0 nT 0. ° Q ( ' H r ® Cp °¯ rA C p. ) ½° ¾ ¿°. Note 2. The Trapezoidal Rule for integration is given by x. ³ f ( x )dx |. 0. f ½ f ( ' x ) ® 0 f1 f 2 ... f n1 n ¾ 2¿ ¯2. where, the interval of integration is divided into n equal segments and 'x= x/n . For the purposes of this exercise, use n = 3. Solution to Problem 2.6 The mass balance equation for a plug flow reactor is given by: n A ,exit. VR. . ³. n A0. dnA rA. 20 Download free eBooks at bookboon.com.

(57) II. Homogeneous reactions - Non-isothermal reactors. Fundamentals of Reaction Engineering – Worked Examples. where rA kp A k( 1 x A ) p A0 , VR is the reactor volume (m3), nA the molar flow rate of “A” (kmol m-3 s-1) and pA0 the inlet partial pressure of “A” (bar). x A ,exit. VR. ³. 0. VR. x A ,exit. n A0 dx A. ³. k( 1 x A ) p A0. n A0 p A0 k0. x A,exit. ³. 0. 0. n A0 dx A [ k0 exp( 5000 / T )]( 1 x A ) p A0. dx A. ;. ( 1 x A ) exp( 5000 / T ) 0.9. VR. ( 5.55 u 10 5 ) ³ 0. x A,exit. 0.9. dx A ( 1 x A ) exp( 5000 / T ). (Eq. 2.6.A). To express T in terms of xA, we use the operating line for an adiabatic tubular reactor, assumed to operate in plug flow [i.e Q=0]:. dT dx A. nA0 ( ' H r ) nT 0 C p. ( 0.2 )( 50 000 ) ( 1 )( 40 ). 250 K .. Integrating: T = T0 + 250 xA . This equation can then be substituted for T in Equation Eq. 2.6.A.. VR. ( 5.55 u 10. 5. 0.9. )³ 0. dx A ( 1 x A ) exp[ 5000 /( T0 250x A )]. We use the Trapezoidal Rule for the integration, with. f ( xA ). 1. ;. ( 1 x A ) exp[ 5000 /( 400 250x A )] [f(xA)]-1 3.73 x 10-6 1.88 x 10-5 4.51 x 10-5 3.35 x 10-5. xA 0 0.3 0.6 0.9. 'xA= xA/3 = (0.9/3) = 0.3 f(xA)/2 13.4 x 104 1.49 x104. f3 ½ f0 ® f1 f 2 ¾ 2¿ ¯2. f(xA) 5.33 x 104 2.22 x 104. 2.24 x 105. 0.9. VR. f ½ f ( 5.55 u 10 5 ) ³ f ( x )dx | ( 5.55 u 10 5 ) ( ' x ) ® 0 f1 f 2 3 ¾ 2¿ ¯2 0 VR. (5.55 u105 ) (0.3) (2.24 u105 ) # 3.73 m3. 21 Download free eBooks at bookboon.com.

(58) III. Catalytic reactions - Isothermal reactors. Fundamentals of Reaction Engineering – Worked Examples. Worked Examples - Chapter III Catalytic reactions – Isothermal reactors Problem 3.1 The first order irreversible gas phase reaction. AoB + C. is to be carried out in an isothermal tubular fixed bed catalytic reactor operating at atmospheric pressure. The intrinsic reaction rate expression is rA k C A . Pure “A” will be fed to the reactor at a rate of 0.13 kmol s-1. a. Compare the relative magnitudes of the rates of external mass transport and chemical reaction. b. Calculate the mass of catalyst required for a conversion of 80 %. For purposes of calculating km, an average Reynolds number of 500 may be assumed over the length of the reactor. Plug flow and ideal gas behaviour may be assumed. The pressure drop along the length of the reactor may be neglected. Data: Reaction rate constant, k: 0.042 Average gas density, Ug: Average gas viscosity, μg : Average diffusivity of “A”, DA :. m3 (kg-cat s)-1 1.0 kg m-3 -5 kg m-1 s-1 5.0 x 10 5.0 x 10-5 m2 s-1. External surface area of catalyst pellets, am:. 0.667. Bed void fraction, HB: Mass flow rate, G : Effectiveness factor, K : Reactor temperature, T :. 0.5 8.3 0.12 423. jD. 0.46. HB. Re. 0.4. where Re. m2 kg-1 kg m-2 s-1 K. d p G / P B and jD. km U B P B ½ ® ¾ G ¯ Ub DAB ¿. where km is defined by the equation: rA,p = km am (Cb - Cs). R (gas constant) = 8.314 kJ kmol-1 K-1 = 0.08314 bar m3 K-1 kmol-1 Solution to Problem 3.1 From the course notes rp Solving for Cs , Cs. km am ( Cb Cs ) K kCs .. km am Cb and rp K k km am. 1 1½ ® ¾ ¯ km am K k ¿. 1. 22 Download free eBooks at bookboon.com. Cb .. 2/ 3.

(59) III. Catalytic reactions - Isothermal reactors. Fundamentals of Reaction Engineering – Worked Examples. 1 1 and , given Re constant at a value of 500. km am Kk 0.458 ½ 0.4 0.077 ® ¾ u ( 500 ) ¯ 0.5 ¿. (a) Compare. jD Sc. P. 5 u 10 5. U DA. 1 u 5 u 10 5. 1.. ( 0.077 ) u ( 8.3 )( 1 ) 0.64 m s 1 1 U 1 1 1 1 2.34 ; 198.4 K k ( 0.12 ) u ( 0.042 ) km am ( 0.64 ) u ( 0.667 ) ( Sc )2 / 3. Comparing magnitudes, chemical reaction turns out to be the controlling (slower) step.. no.1 en. nine years in a row. ed. jD G. Sw. km. STUDY AT A TOP RANKED INTERNATIONAL BUSINESS SCHOOL Reach your full potential at the Stockholm School of Economics, in one of the most innovative cities in the world. The School is ranked by the Financial Times as the number one business school in the Nordic and Baltic countries.. Stockholm. Visit us at www.hhs.se. 23 Download free eBooks at bookboon.com. Click on the ad to read more.

(60) III. Catalytic reactions - Isothermal reactors. Fundamentals of Reaction Engineering – Worked Examples. 1 1 ½ (b) Let us define the overall rate constant as . { ® ¾ ¯ km am K k ¿ rp . Cb , and since n A C AvT , we can write rp. 1. . Then,. . ( n A / vT ) .. (Eq. A). Also ( 2.34 198.4 )1. .. 4.98 u 10 1. (Eq. B). Since the reaction “A o B + C” in the gas phase leads to a molar change upon reaction, we next need to express vT in terms of the conversion. We start with the change in molar flow rate.. nA. n A0 n A0 x A. nB. n A0 x A. nC. n A0 x A. nT n A0 ( 1 x A ) Here, nT is the total volumetric flow rate expressed as a function of the fractional conversion of “A”. Using the ideal gas law: nT RT n A0 RT vT ( 1 xA ) . PT PT Substituting in Eq. (A), we get. . n A0 ( 1 x A ). rp. n A0 RT ( 1 xA ) PT. . PT ( 1 x A ). RT ( 1 x A ). To calculate the mass of catalyst required: n A,exit. . W. ³. n A0 x A,exit. ³. 0. ( 1 xA ) dx A ( 1 xA ). W. . W. . x A,exit. ³. 0. dn A rA. n A0 RT . PT. x A,exit. ³. 0. ( 1 xA ) dx A ( 1 xA ). 1 ½ 1 1 ® ¾ dx A ( 1 xA ) ¿ ¯( 1 xA ). 2 ln( 1 x A,exit ) x A,exit. n A0 RT ^2 ln( 1 x A,exit ) x A,exit ` . PT. ( 0.13 )( 0.08314 )( 423 )^2 ln( 0.2 ) 0.8` [ 4.98 u 10 3 ]( 1 ) Weight of catalyst: 2220 kg. 24 Download free eBooks at bookboon.com. # 2220 kg. (Eq. C).

(61) III. Catalytic reactions - Isothermal reactors. Fundamentals of Reaction Engineering – Worked Examples. Problem 3.2 A tubular reactor packed with spherical pellets will be used for carrying out the catalytic reaction: 2A oB + C . Reactant “A” is available as 20 (mole) % A in an inert gas and will be fed to the reactor at the rate of 0.02 kmol s-1. The intrinsic reaction rate given by. rA,V. kV C A2 kmol m3 s 1 ,. where CA denotes the concentration of A. The intended conversion is 75 %. (a) The reactor is to be operated isothermally at 600 K and 1 bar pressure. Estimate whether intraparticle diffusion resistances affect the overall reaction rate. For the purposes of this part of the calculation CA,s # CA,b may be assumed. (b) Estimate whether bulk to catalyst surface mass transport resistances affect the overall reaction rate. Due to dilution by the inert gas, bulk properties of the gas mixture may be considered as approximately constant. The pressure drop over the length of the reactor and deviations from the ideal gas law may be neglected. Data Catalyst pellet density, U p :. 2200. kg m-3. Effective diffusivity of A within the pellet, DA,eff:. 5 x 10-3 1 x 10-6. m m2 s-1. Molecular diffusivity of A in the gas mixture, DA:. 1 x 10-4. Reaction rate constant, kv:. 3.46 x 104. m2 s-1 m3 kmol-1 s-1. Mass flow rate, G: Density of the gas mixture, U g :. 10 1. kg m-2 s-1 kg m-3. Viscosity of the gas mixture, μ: Void fraction of the bed, H b:. 1.5 x 10-5 0.42. kg m-1 s-1. Catalyst pellet diameter, dp. For integral reaction rate orders the effectiveness factor may be expressed as 1/ 2. Sx 2 V p k [ C A,s ] n. K. C A,s ½ ° ° n D k [ C ] dC ® ³ A,eff ¾ °¯ 0 ¿°. .. The dimensionless mass flux jD is given by the correlation. jD. ( 0.46 / H ) (Re)0.4 , where Re. d pG / P. where jD is related to km, the bulk stream to particle surface mass transfer coefficient, by. 25 Download free eBooks at bookboon.com.

(62) III. Catalytic reactions - Isothermal reactors. Fundamentals of Reaction Engineering – Worked Examples. ° P ½° ( km U g / G ) ® ¾ ¯° U g DA ¿°. jD. 2/ 3. .. R (gas constant) = 8.314 kJ kmol-1 K-1 = 0.08314 bar m3 K-1 kmol-1 Solution to Problem 3.2 If intraparticle diffusion resistances are significant, K # 1 / ) where ) is usually >5. Now let us calculate ) by assuming that K # 1 / ) . From the main text (Chapter 6), we have: C ½ V p k [ C A,s ] n ° A,s ° n ® ³ DA,eff k [ C ] dC ¾ Sx 2 °¯ 0 °¿. ). ). V p kV [ C A,s ] n ° k [ C A,s ] n1 ½° D ® A,eff ¾ Sx n1 2 °¯ °¿. 1 / 2. 1 / 2. V p ° 2Deff 1 ½° ® n1 ¾ S x °¯ n 1 kV C A,s °¿. (Eq. 6.74) 1 / 2. For n = 2, we get 1/ 2. ½° d p ° 3 k C ® V A,s ¾ 6 °¯ 2Deff °¿. ). /2 1.9 u 10 2 C 1A,s. (Eq. A). Neglecting external transport resistances, at the reactor inlet:. CA. p A,partial. nA vT. RT. 0.2( 1 ) ( 0.08314 ) ( 600 ). 4 u 10 3 kmol m 3. At the reactor exit:. CA. ( 0.05 )( 1 ) 1 u 10 3 kmol m 3 ( 0.08314 )( 600 ). Substituting these values into Eq. A above, we get. ) | 12 at the reactor inlet and ) | 6 at the reactor exit. So we must conclude that diffusion resistances are not negligible within this reactor. Part (b): At steady state, the rate of reaction within a particle must be equal to the net rate of diffusion of the reactant.. net flux. From jD. 5 u 10 3 ( 10 ). 3333 1.5 u 10 5 ( 0.46 / H ) (Re)0.4 we can calculate jD 0.043 . Meanwhile jD G 1 km 2/ 3 Ug ° P ½° ® ¾ ¯° U g DA ¿°. Let us calculate km am .. Re. d pG. km am ( Cb Cs ). P. ° P °½ ® ¾ °¯ U g DA °¿. 2/ 3. ° 1.5 u 10 5 °½ ® 4 ¾ °¯ 1 u 1 u 10 °¿. 2/ 3. 26 Download free eBooks at bookboon.com. 0.28.

(63) III. Catalytic reactions - Isothermal reactors. Fundamentals of Reaction Engineering – Worked Examples. ( 0.043 )( 10 ) 1 1.54 and 1 0.28 4S rp2 3 6 am 3 4 3 S rp U p rp U p 5 u 10 u 2200 3 km am 0.84 m3 kg 1 s 1. km. 0.545 m 2 kg 1. The catalyst bed density is calculated from U B ( 1 H ) U p ( 0.58 ) u ( 2200 ) 1276 kg m 3 Then we use the concept that, at steady state, net diffusion of reactant to the catalyst pellet equals the amount of reactant that has been converted. 2 U B km am ( C Ab C As ) K k C As. Define a constant D . At the inlet D. kmol /( m2 s ) U B km am ( 1276 ) u ( 0.84 ) 0.372 kmol m 3 Kk § 1 · 4 ¨ ¸ ( 3.46 u 10 ) © 12 ¹. The equation to solve now takes the form: 2 D ( C Ab C As ) C As. (Eq. B). and at the inlet C Ab 4 u 10 3 kmol m 3 . Eq. B can be solved as a quadratic equation or by trial and error.. C As # 3.95 u 10 3 kmol m 3 Comparing C Ab and C As , we can see the difference is small. Therefore, in this case, the bulk-stream to pellet surface mass transport resistance appears negligible at the reactor inlet.. 27 Download free eBooks at bookboon.com. Click on the ad to read more.

(64) III. Catalytic reactions - Isothermal reactors. Fundamentals of Reaction Engineering – Worked Examples. Problem 3.3 Differential reactors are used to simulate an infinitesimally thin (“slice”) section of, say, a tubular reactor. In the laboratory, this is done to measure reaction rates and other parameters of catalytic reactions, in the context of a small amount of conversion.. The first order irreversible reaction A o B C will be carried out over 0.002 kg packed catalyst in a differential test reactor. The intrinsic rate expression for the reaction is. kmol ( kg cat u s )1. rA 7 u 1015 exp ^18,000 / T ` C A. where the temperature T is given in K and CA, the concentration of the reactant “A”, is expressed in kmol m-3. Assuming isothermal operation at 500 K and given the data listed below, (a) Estimate whether intraparticle diffusion resistances would be expected to affect the overall rate. (b) Estimate whether bulk to surface diffusion resistances would be expected to affect the overall rate. (c) Calculate the conversion through the reactor. Data: Catalyst pellet diameter, dp:. 0.001. m. Catalyst pellet density Up : Effective diffusivity of A within the pellet, DA,eff :. 1500. kg m-3. 1 x 10-6. m2 s-1. Molecular diffusivity of A in the bulk stream, DA :. 1 x 10-4. Mass flow rate, G: Viscosity of the gas mixture, μ: Density of the gas mixture, Ug: Void fraction of the bed, H: Total volumetric flow rate, vT :. 0.1 1 x 10-6 1 kg m-3 0.42 0.005. m2 s-1 kg m-2 s-1 kg m-1 s-1. m3 s-1. The dimensionless mass flux jD is given by the correlation. jD. ( 0.46 / H ) (Re)0.4 , where Re. d pG / P. where jD is related to km, the bulk stream to particle surface mass transfer coefficient, by. jD. ° P ½° ( km U g / G ) ® ¾ ¯° U g DA ¿°. 2/ 3. .. The Thiele modulus for spherical catalyst pellets is given by:. · ° k U p ¸® ¹ °¯ DA,eff. 1/ 2. °½ )s ¾ ¿° where Rs is the pellet radius, k the reaction rate constant; Up and Deff have been defined above. § Rs ¨ © 3. 28 Download free eBooks at bookboon.com. (Eq. A).

(65) III. Catalytic reactions - Isothermal reactors. Fundamentals of Reaction Engineering – Worked Examples. Solution to Problem 3.2 Part (a). Rs. 0.0005 m .. At 500 K, the reaction rate constant k. 0.0005 3. )s. ( 1.62 )( 1500 ) 1 u 10 6. 1.62. Substituting into Eq. A, we can calculate ) s .. 8.225. ) s is large. We conclude that intraparticle diffusion resistances are significant and would be expected to affect the overall rate. Part (b):. 4S rp2. am. 3 rp U p. 4 3 S rp U p 3. 3 5 u 10. 4. u 1500. 4.0 m 2 kg 1. To calculate km , we first need the Reynolds number. Re. jD. ( 0.001 )( 0.1 ) 1 u 10 6. 100 . Using this value we can get. 0.174. Then. km. jD G ° P ½° ® ¾ U g ¯° U g DA ¿°. 2 / 3. 0.375. m s 1. The basic equation to use for comparing external and intraparticle diffusion effects is:. rp. 1 1 1 K k km am. km am. C Ab . We now compare. 1 1 and . Kk km am. 1 0.66 km am 1 1.62 / 8.225 0.197; 5.1 Kk. ( 4 )( 0.375 ) 1.5 m3 s 1 ;. Kk # k / )s. External diffusion resistance less significant! Part (c): For a “differential reactor” the mass balance equation dW. 'W # . 'n A rp. , where 'n A. . n A0 ' x A . We also have rp. dn A can be written as: rp. 1 C Ab and 1 1 K k km am. 29 Download free eBooks at bookboon.com.

(66) III. Catalytic reactions - Isothermal reactors. Fundamentals of Reaction Engineering – Worked Examples. CA. n A / vT . Thus we can write C Ab. n A0 ( 1 x A ) and rp vT. In the case of the differential reactor, x A # ' x A . Thus. 1 1 1 K k km am. u. n A0 ( 1 x A ) . vT. 1 1 ½ ® ¾ ¯K k km am ¿ 0.005 ^5.1 0.667` 14.42 0.002. 1 x A vT # 'W xA. 'W. 1 1 xA n A0 ' x A 1 ½ u® ¾ vT ; then xA n A0 ( 1 x A ) ¯K k km am ¿. xA. 1 0.066 15.41. xA. 0.066. 30 Download free eBooks at bookboon.com. Click on the ad to read more.

(67) IV. Catalytic reactions - No isothermal reactors. Fundamentals of Reaction Engineering – Worked Examples. Worked Examples - Chapter IV Catalytic reactions – Non-isothermal reactors Problem 4.1 The exothermic catalytic gas phase cracking reaction A o light products is carried out with a feed of 10 % “A” in an inert diluent, “B”. The intrinsic rate for the reaction is given by:. rA. kmol s 1 ( kg cat )1. 125 exp ^5000 / T ` p A. where pA denotes the partial pressure (bar) of “A” and T denotes the local temperature in K. The catalyst pellets (dia. 0.004 m) may be considered isothermal and the value of the effectiveness factor is estimated as 0.08 throughout the reactor. We select a point within the reactor where, G, the superficial mass velocity of the gas stream is 0.8 kg m-2 s-1, the bulk temperature, Tb, is 500 K, and the partial pressure of “A” in the bulk, pA,b, is 0.05 bar. Calculate: (i) (ii) (iii). the global rate of reaction, the external concentration gradient (expressed in terms of the partial pressures), and, the temperature gradient between the bulk gas stream and external catalyst pellet surfaces,. at the selected point in the reactor. Bulk gas properties may be taken as those of the diluent. The pressure drop along the reactor axis may be neglected. Data: Effectiveness factor, K : Total reactor pressure, pT :. 0.08 1 bar. Void fraction of the packed bed, HB : Surface area of catalyst pellets, am: Heat of reaction, 'Hr : Molecular diffusivity of A in B, DAB :. (see statement of problem). 0.5 0.4 m2 kg-1 20 000 kJ (kmol A reacted)-1 2 u 10-4 m2s-1. Density of “B”, U B :. 0.1 kg m-3. Viscosity of “B”, P B : Heat capacity of “B”, Cp,B :. 1.0 u 10-5 kg m-1 s-1 1.0 kJ kg-1 K-1. Thermal conductivity of “B”, OB :. jD. jD. 1.5 u 10-5 kJ m-1 s-1 K-1. ( 0.7 ) jH. 0.46. HB. km U B P B ½ ® ¾ G ¯ Ub DAB ¿. Re 0.4 ; Re. 2/ 3. ;. jH. d p G / PB. C p PB ½ ® ¾ C p G ¯ OB ¿ h. 31 Download free eBooks at bookboon.com. 2/ 3. ..

(68) IV. Catalytic reactions - No isothermal reactors. Fundamentals of Reaction Engineering – Worked Examples. In these equations, km, the bulk to surface mass transfer coefficient is defined in terms of the equation: rA,p = (kmam/RTb) ( pb - ps ), and, h is the surface to bulk heat transfer coefficient, defined by h am (Ts - T b) = ( 'Hr) rA,p . Finally, R (gas constant) = 8.314 kJ kmol-1K-1= 0.08314 bar m3 K-1 kmol-1 Solution to Problem 4.1. ( 0.004 ) u ( 0.8 ). Re. d p G / PB. jD. 9.16 u 10 2. 320 ; substituting this value into the correlations, we get. 1 u 10 5 and jH 0.131 .. Next, we calculate the Prandtl and Schmidt numbers. Pr. C pB P B. 1 u 1 u 10 5. OB. 1.5 u 10 5. PB. 0.67 and Sc. U B DAB. 1 u 10 5. 0.5 .. 0.1 u 2 u 10 4. On the other hand. ham ( Ts Tb ) rA,p u ' H r , leads to. h. jH C p G (Pr)2 / 3. ( Ts Tb ). rA,p ( ' H r )(Pr)2 / 3. ( rA,p ) ( 20,000 ) ( 0.67 )2 / 3. jH C p G am. ( 0.131 )( 1 )( 0.8 )( 0.4 ). recalling that Tb. 3.65 u 10 5 rA,p. 500 K , Ts. 500 3.65 u 10 5 rA,p .. (Eq. A). Similarly. km am ( pb ps )A RTb. rA,p and km. jD G. 1. U B Sc 2 / 3. Then. ( pb ps )A. leads to ( pb ps )A. rA,p U B ( Sc 2 / 3 ) R Tb. rA,p ( 0.1 ) ( 0.5 )2 / 3 ( 0.08314 )( 500 ). am jD G. ( 0.4 )( 9.16 u 10 2 )( 0.8 ). 89.1 u rA,p , where p A,b. 0.05 leads to p A,s. The intrinsic reaction rate was given as rA. 0.05 89.1 u rA,p. 125 exp ^5000 / T ` p A. 32 Download free eBooks at bookboon.com. (Eq. B).

(69) Fundamentals of Reaction Engineering – Worked Examples. Since rA,p. IV. Catalytic reactions - No isothermal reactors. K rA,s K rA ( at surface conditions ) , and the intrinsic reaction rate expression was. given in the problem statement as rA can be written as. rA,p K rA,s. ( 125 ) exp ^5000 / T ` p A , the global reaction rate expression. ( 0.08 ) u ( 125 ) exp ^5000 / Ts ` p A,s .. Substituting expressions for Ts from Eq. (A) and p A,s from Eq. (B), we get rA,p. K rA,s. ^. `. ( 0.08 ) u ( 125 ) u ª¬0.05 89.1 u rA,p º¼ u exp 5000 /( 500 3.65 u 10 5 rA,p ). p. p. p A,s. Ts. This equation can be solved by a simple trial and error procedure. rA,p 2.6 u 10 5 kmol ( kg cat u s )1 Then Ts. 500 3.65 u 10 5 [ 2.6 u 10 5 ] # 509.5. 'T ps. 0.05 89.1 rA,p. 9.5 K. Not negligible!. 4.77 u 10 2. ' p 0.05 0.0477 0.0023 bar o ~ 5 % of total pressure. Probably negligible…just.. Excellent Economics and Business programmes at:. “The perfect start of a successful, international career.” CLICK HERE. to discover why both socially and academically the University of Groningen is one of the best places for a student to be. www.rug.nl/feb/education 33 Download free eBooks at bookboon.com. Click on the ad to read more.

(70) Fundamentals of Reaction Engineering – Worked Examples. IV. Catalytic reactions - No isothermal reactors. Problem 4.2 The exothermic gas phase reaction. A+BoC is carried out over a porous catalyst, in a non-isothermal/non-adiabatic fixed bed catalytic reactor. The feed stream is equimolar in A and B. (a) Using a one-dimensional, pseudo-homogeneous model framework, derive steady-state mass and energy balance equations and state boundary conditions to describe the behaviour of the reactor. Define your terms carefully. (b) Calculate the maximum temperature that the exit gas stream would attain if the reactor were to be operated adiabatically. Judging by this result, what change would you recommend in reactor operating conditions. (c) Discuss briefly for the present case, advantages and disadvantages of using a one-dimensional reactor model over a two-dimensional model.. Data: Heat of reaction, 'Hr : Average heat capacities. CpA # CpB : CpC :. 300 000. kJ (kmol A reacted)-1. 30. kJ kmol-1 K-1 kJ kmol-1 K-1. 60. Solution to Problem 4.1 Part A: Mass balance: The mass balance over an infinitesimally thin (“slice”) element of the catalyst packed reactor (dW) gives: dn A rA dW , where n A is defined by the equation n A n A0 ( 1 x A ) and dn A n A0 dx A . In these equations W denotes the mass of catalyst and nA the molar flow rate of “A”. nT 0 C p is the total heat capacity of the inlet stream. It is assumed not to change with conversion; so we shall use it as the total heat capacity of the reaction mixture. From dn A. rA dW we can then write the full mass balance equation as:. Energy balance:. ª ¦ ni C pi º dT ¬ ¼. where Q is the external heat transfer term and. dx A dW. rA . n A0. (Eq. A). dW ( Q rA ' Hr ). ¦ ni C pi #. nT 0 C p as explained in Chapter 3 of the main. text. The energy balance equation may then be written as:. dT dW. Q rA ' H r nT 0 C p. The initial condition is given as: At W = 0, T = T0 = Twall .. 34 Download free eBooks at bookboon.com. (Eq. B).

(71) Fundamentals of Reaction Engineering – Worked Examples. IV. Catalytic reactions - No isothermal reactors. There are many ways of formulating Q, the heat loss term. One common way would be: Q A h ( T Twall ) , where A denotes the contact surface area and h the heat transfer coefficient. Part B: Combining Eqs. A and B, we can write. n A0 rA. dT dx A For adiabatic operation, Q = 0 .. dT dx A. . ° Q r ' H ½° A r . ® ¾ ¯° nT 0 C p ¿°. 'H r [ nT 0 C p ]. n A0. Integrating. 'T. . ' H r n A0 C p nT 0. ( 0.5 )( 3 u 10 5 ) 30. 5000 K. Unacceptably high! Improve cooling and/or reduce reactant concentrations and/or dilute the catalyst. Part C: The main difficulty arises from ignoring the radial temperature distribution. In dealing with highly exothermic reactions, when the maximum temperature along the central axis turns out to be much larger than the radially averaged temperature, runaway may occur under conditions where the 1-dimensional model would predict “safe” operation. The disadvantage of 2-dimensional operation is the requirement for accurate additional data for radial heat (and mass) transport: Orad ,effective , Drad ,effective .. Problem 4.3 The exothermic reaction. A+BoC. was carried out using two different catalyst particle sizes: 8 mm diameter catalyst pellets and 50 μ (average) diameter catalyst particles. Both sets of experiments were carried out under conditions where external temperature and concentration gradients were negligible. Global rates of reaction, measured as a function of temperature, using approximately similar external surface concentrations of A and B for all experiments, are given in the table below. Assuming the 50 μ particles to be isothermal under all operating conditions, (a) Calculate from this data, the energy of activation most likely to approximate the true value for the chemical reaction step at the catalytic sites. (b) For the 8 mm catalyst pellets, calculate the effectiveness factor at each temperature. State your assumptions clearly. c. Briefly explain observed changes in the values of effectiveness factors as a function of the experimental temperature. What do values above unity imply?. 35 Download free eBooks at bookboon.com.

(72) Fundamentals of Reaction Engineering – Worked Examples. IV. Catalytic reactions - No isothermal reactors. Table A. Experimental global reaction rates as a function of catalyst particle size and temperature.. Catalyst Particles Rate, kmol s-1. Temp (K) 425.5 458.7 495.1 549.5 637.0. 8 mm pellets rA,p (kg cat)-1. 9.2 x 10-8 2.5 x 10-7 6.95 x 10-7 1.85 x 10-6 4.12 x 10-6. 5.12 x 10-7 5.95 x 10-7 7.18 x 10-7 8.58 x 10-7 9.39 x 10-7. Solution to Problem 4.3 Part (a): “Series 2” Arrhenius plot for 50 Pm particles (Figure A) shows the line bending above 495 K. The slope of the straight component of the line gives the correct ('Ea/R) value for chemical control, i.e. for K=1. Using. values of the rate between 425 and 495 K, we can calculate the slope of the straight part of the line. ln r1 ln r2 1.42 1.62 ° °½ Slope ® 3 ¾ 1 1 °¯ ( 2.02 2.35 ) u 10 °¿ T2 T2. ' Ea R. ' Ea. 6060 K ; the gas cons tan t R. 8.314 kJ /( kmol u K ). 5.04 kJ kmol 1. In the past four years we have drilled. 89,000 km That’s more than twice around the world.. Who are we?. We are the world’s largest oilfield services company1. Working globally—often in remote and challenging locations— we invent, design, engineer, and apply technology to help our customers find and produce oil and gas safely.. Who are we looking for?. Every year, we need thousands of graduates to begin dynamic careers in the following domains: n Engineering, Research and Operations n Geoscience and Petrotechnical n Commercial and Business. What will you be?. careers.slb.com Based on Fortune 500 ranking 2011. Copyright © 2015 Schlumberger. All rights reserved.. 1. 36 Download free eBooks at bookboon.com. Click on the ad to read more.

(73) Fundamentals of Reaction Engineering – Worked Examples. IV. Catalytic reactions - No isothermal reactors. ln r vs. 1000 x (1/T) 0 0. 0.5. 1. 1.5. 2. 2.5. ln r. -5 Series1. -10. Series2. -15 -20 1000 x (1/T) [T in K]. Figure A. Arrhenius plot based on data for 50 Pm particles in Table A. Part (b): The first three rates in rows 1-3 of column-2 in Table A and Table B are the same. The data corresponds to the part of the Series 2 curve in Figure A before it begins to bend. Since the line is straight up to and including 495 K, we can safely conclude we have chemical control and K = 1. Above that temperature, the line bends. If there had been no diffusive limitations the line would have continued as a straight line. That is why the extrapolated line still corresponds to K = 1. We can read off (in an enlarged diagram) that the extrapolated rate for 549.5 K is 2.26u10-6 and for 637 K, the rate corresponding to K = 1 is 1.07u10-5. We then use Eq. 6.76 (Chapter 6) from the main text.. (robs )1 (Eq. 6.76) Ș1 /Ș2 (robs )2 where (robs )1 is the second column in Table B and (robs )2 third column in Table B. With K1 taken as unity, we can calculate the effectiveness factors for the 8 mm pellets at different temperatures. Table B. Experimental global reaction rates as a function of catalyst particle size and temperature.. Rate, kmol s-1 Temp (K) 425.5 458.7 495.1 549.5 637.0. rA,p (kg cat)-1 8 mm pellets (data from Table A). 50 Pm catalyst particles (K1=1) 9.2 x 10-8 2.5 x 10-7 6.95 x 10-7 2.26 x 10-6 1.07 x 10-6. 5.12 x 10-7 5.95 x 10-7 7.18 x 10-7 8.58 x 10-7 9.39 x 10-7. 37 Download free eBooks at bookboon.com. K Effectiveness factor for 8 mm pellets. 5.56 2.38 1.03 0.38 0.228.

(74) Fundamentals of Reaction Engineering – Worked Examples. IV. Catalytic reactions - No isothermal reactors. Part (c): For exothermic reactions K=1 means that the temperature at the centre of the pellet is greater than the surface temperature Ts. That in turn implies that the rate at the centre of the pellet is greater than the rate at the. surface. In this example, the effect was more visible at relatively low surface temperatures. This explains the change in the nature of the value of effectiveness factor with increasing temperature.. Problem 4.4 Differential reactors are used to simulate an infinitesimally thin (“slice”) section of, say, a tubular reactor. In the laboratory, this is done to measure reaction rates and other parameters of catalytic reactions, in the context of a small amount of conversion.. The intrinsic rate for the endothermic catalytic gas phase cracking reaction “A o products”, is given by the first order reaction rate expression. rA. 4,000 ^exp( 10,000 / T )` p A. kmol /( kg cat u s )1 ,. where pA is the partial pressure (bar) of A, and T denotes the local temperature in K. The reaction is carried out in a differential reactor, where reactant “A” is supplied, mixed 1:1 with an inert diluent. The superficial mass velocity of the gas stream flowing over the catalyst particles, G, is 1 kg m-2 s-1, and the temperature of the bulk stream is 750 K. The overall conversion in the reactor has been determined experimentally as 3.6 %. The catalyst pellets have a diameter of 0.004 m and may be considered isothermal. Calculate the global rate of reaction, the magnitude of the effectiveness factor, and the external concentration and external temperature gradients. The bulk gas properties can be taken as those of the diluent, B. The pressure drop through the reactor may be neglected. Data: Molecular diffusivity of A in B, DAB: Density of diluent B, UB: Viscosity of B, μB: Heat capacity of B, Cp,B: Thermal conductivity of B, OB: Reactor pressure, pT: Void fraction of the packed bed, H: Surface area of the catalyst pellets, am: Heat of reaction, 'HR: Pellet effective diffusivity, Deff: Pellet density, Uc:. jD. jD. ( 0.7 ) jH. 3 x 10-4 0.1 1.5 x 10-5 1.0 2.5 x 10-5 1 0.48 0.4 25,000 1 x 10-6 2,000. 0.46. HB. km U B P B ½ ® ¾ G ¯ Ub DAB ¿. m2 s-1 kg m-3 kg m-1 s-1 kJ kg-1 K-1 kJ m-1 s-1 K-1 bar m2 kg-1 kJ (kmol A)-1 m2 s-1 kg m-3. Re 0.4 ; Re. 2/ 3. ;. jH. d p G / PB. C p PB ½ ® ¾ C p G ¯ OB ¿ h. 2/ 3. .. In these equations, km, the bulk to surface mass transfer coefficient is defined in terms of the equation: rA,p = (kmam/RTb) ( pA,b – pA,s ),. 38 Download free eBooks at bookboon.com.

(75) Fundamentals of Reaction Engineering – Worked Examples. IV. Catalytic reactions - No isothermal reactors. and, h is the surface to bulk heat transfer coefficient, defined by h am (Ts - T b) = ( 'Hr) rA,p . Finally, R (gas constant) = 8.314 kJ kmol-1K-1= 0.08314 bar m3 K-1 kmol-1 Solution to Problem 4.4. h am ( Ts Tb ) rA,p ( ' H r ) , where h ( Ts Tb ) Pr. Re. C pB P B. OB. 1 u 1.5 u 10 5. d pG. 2.5 u 10 ( 0.004 )( 1 ). PB. 1.5 u 10 5. 0.6;. 5. ( Ts Tb ). (Pr)2 / 3. (Eq. A). jH C pB G am 0.71. ( rA,p ) u ( 25000 ) u ( 0.71 ) ( 0.147 )( 1 )( 1 )( 0.4 ) Ts. (Pr)2 / 3. rA,p ( ' H r )(Pr)2 / 3. 267 ; from equations in the data section jD. Substituting in Eq. A, we get. jH C p G. 750 3.02 u 10 5 rA,p. 0.103 and jH. 0.147 .. 3.02 u 10 5 rA,p. (Eq. B). American online LIGS University is currently enrolling in the Interactive Online BBA, MBA, MSc, DBA and PhD programs:. ▶▶ enroll by September 30th, 2014 and ▶▶ save up to 16% on the tuition! ▶▶ pay in 10 installments / 2 years ▶▶ Interactive Online education ▶▶ visit www.ligsuniversity.com to find out more!. Note: LIGS University is not accredited by any nationally recognized accrediting agency listed by the US Secretary of Education. More info here.. 39 Download free eBooks at bookboon.com. Click on the ad to read more.

(76) Fundamentals of Reaction Engineering – Worked Examples. IV. Catalytic reactions - No isothermal reactors. km am ( p A,b p A,s ) ; we can write RTb. Using the mass transfer equation: rA,p. (Eq. C). 2 / 3. km km. jD G P B ½ jD G ( Sc )2 / 3 ® ¾ U B ¯ Ub DAB ¿ UB º ( 0.103 )( 1 ) ª 1.5 u 10 5 1.59 « 4 » 0.1 ¬« ( 0.1 ) u ( 3 u 10 ) ¼». Substituting this result in Eq. C, we get:. p A,b p A,s. rA,p RTb. ( 0.08314 )(750 )rA,p. km am. ( 1.59 )( 0.4 ). 98.04 rA,p. Reactor pressure: 1 bar. Conversion is 3.6%. Exit stream composition assumed equal to reactor composition. This basically means that we treat the “differential reactor” as a mini-CSTR. The feed is 50 % “A”. Therefore:. pb. ( 0.5 )( 1 0.036 ) 0.482 bar . Thus: ps 0.482 98.04 rA,p rA,p K rA,s and K. 1 1 1 ½ ® ¾ ; rearranging ) s ¯ tanh 3) s 3) s ¿ 1/ 2. ° k RTs Uc ½° )s ® ¾ °¯ Deff °¿ ) s in the above equation is form of the Thiele modulus appropriate for the given reaction rate Rs 3. expression.. )s. 0.002 3. 1/ 2. ° ª 10000 º ( 0.08314 )( 2000 ) ½° u u T ®4,000 exp « ¾ » s 1 u 10 6 ¬ Ts ¼ ¯° ¿°. )s. 1/ 2. 544 ^Ts exp( 10000 / Ts )`. (Eq. D). At Ts 750 K , ) s # 14 ; therefore we can safely consider K # 1 / ) s . Going back to the global reaction rate expression: rA,p 4000 K u exp ^10,000 / Ts ` u p A,s (Eq. E). rA,p. ^. `. 4000 K u exp 10,000 /(750 3.02 u 10 5 rA,p u ( 0.482 98.04 rA,p ). (Eq. F). We need to substitute for K in Eq. E. As already mentioned, K # 1 / ) s ; ) s is given by Eq. C, and Ts by. Eq. B. Then: 1/ 2. ª º °½ 10,000 ° 5 »¾ ) s 544 ®(750 3.02 u 10 rA,p ) u exp « «¬ (750 3.02 u 10 5 rA,p ) »¼ ° °¯ ¿ The easier way to solve for rA,p is by trial-and-error between K # 1 / ) s , Eq. G and Eq. F.. 40 Download free eBooks at bookboon.com. (Eq. G).

(77) Fundamentals of Reaction Engineering – Worked Examples. IV. Catalytic reactions - No isothermal reactors. Close enough answers:. rA,p # 1.15 u 10 4. K 0.075 Ts. 'T p A,s. 715.3 K 34.7 K 0.47 bar. p A,b p A,s. 0.011 bar. .. 41 Download free eBooks at bookboon.com. Click on the ad to read more.

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